mas3111: partial differential equations with...

MAS3111 PartialDifferential Equations

with Applications

Andrew W Baggaleyandrewbaggaleynclacuk

httpabagwikidotcom

Licensed under the Creative Commons Attribution-NonCommercial 30 Unported License (theldquoLicenserdquo) You may not use this file except in compliance with the License You may ob-tain a copy of the License at httpcreativecommonsorglicensesby-nc30 Unlessrequired by applicable law or agreed to in writing software distributed under the License isdistributed on an ldquoAS ISrdquo BASIS WITHOUT WARRANTIES OR CONDITIONS OF ANY KINDeither express or implied See the License for the specific language governing permissions andlimitations under the License

Contents

1 Motivation amp Notation 7

11 Partial Derivatives 7

12 Notation for Derivatives 8121 Ordinary derivatives 8122 Partial derivatives 8123 vectors 8

2 Ordinary Differential Equation Refresher 9

21 Introduction 9211 2nd order linear ODEs 9212 Solutions of the Cauchy-Euler ODE 11

3 Introduction to PDEs 13

31 General remarks 13311 Examples 14

32 Prototypical second order linear PDEs 15321 The diffusion equation 15322 The Laplace equation 16323 The wave equation 16

33 PDEs vs ODEs 16331 Geometrical Interpretation of solutions 17

34 Solution methods 18341 Solution properties 18

35 Trivial Partial Differential Equations 19351 Integration wrt different variables 19352 No derivatives wrt one the variables of u = u(xy) 20

353 Equations which are solvable for ux or uy (not involving u) 20354 Special Tricks 21

4 1st-order Linear PDEs 23

41 The truncated PDE 23411 Finding a particular solution 25

42 Solution to strictly-linear first-order PDEs by change of variables 26421 examples 28

43 Characteristic curves 31

44 Linear waves 32

5 Quasilinear PDEs and nonlinear waves 37

51 Solution to first-order quasilinear PDEs by Lagrangersquos method of charac-teristics 37

511 Tangent and normal to a surface 37512 The method of characteristics 38

52 The Cauchy Problem 43

53 Nonlinear waves 44531 Shocks 45

54 Traffic Flow 46541 The Traffic Flow Equation 46542 The quadratic model 47

6 1st order nonlinear PDEs 49

61 Introduction 49

62 Charpitrsquos equations 49

63 Boundary data 51

64 Examples 51

65 Sand Piles 53

66 Derivation of the Eikonal equation from the Wave Equation 55

7 Classification of 2nd-order PDEs 57

71 Coordinate transformations and classification 57

72 Characteristics and their properties 58

73 Properties of characteristics 59

74 Canonical forms 60741 Examples 61

8 Separation of variables 63

81 Cartesian coordinates 63

82 Polar coordinates 65

83 Laplacersquos equation in 3D Cartesians 66

84 Spherical geometry and Legendre polynomials 68841 Legendre polynomials 69

85 Legendrersquos associated equation 70

Partial DerivativesNotation for Derivatives

Ordinary derivativesPartial derivativesvectors

1 Motivation amp Notation

Just what are differential equations Why are they important Wersquoll leave the definition tolater on in these notes however a sense of their importance can be drawn from their ability tomathematically describe or model real-life situations The equations come from the diversedisciplines of demography ecology chemical kinetics architecture physics mechanical en-gineering quantum mechanics electrical engineering civil engineering meteorology and arelatively new science called chaos theory The same differential equation may be important toseveral disciplines although for different reasons For example demographers ecologists andmathematical biologists would immediately recognize

dNdt

= rN

This equation is used to predict populations of certain kinds of organisms reproducing underideal conditions In contrast physicists chemists and nuclear engineers would be more inclinedto regard the equation as a mathematical model of radioactive decay Many economists and math-ematically minded investors would recognize this differential equation but in a totally differentcontext it also models future balances of investments earning interest at rates compoundedcontinuously

This however is an example of an ordinary differential equation which you will have metin earlier courses Here the unknown function N(t) is a function of only one variable But welive in a three-dimensional world and so many important physical phenomenon can only beunderstood through the study of partial differential equations Here the unknown function is afunction of more than one variable

In this course we will discuss various analytic methods to solve partial differential equationsBy the end of the course we will be able to solve a number of important equations with a vastrange of real world applications

11 Partial DerivativesThe differential (or differential form) of a function f of n independent variable (x1x2 xn) isa linear combination of the basis form (dx1dx2 dxn)

d f =n

sumi=1

part fpartxi

dxi =part fpartx1

dx+part fpartx2

dx2 + +part fpartxn

dxn

8 Chapter 1 Motivation amp Notation

where the partial derivatives are defined by

part fpartxi

= limhrarr0

f (x1x2 xi +h xn)minus f (x1x2 xi xn)

h

The usual differentiation identities apply to the partial differentiations (sum product quotientchain rules etc)

12 Notation for DerivativesWe will usually stick to the standard notation denoting derivatives by

121 Ordinary derivatives

yprime = y =dydt

yprimeprime︸︷︷︸prime notation

= y︸︷︷︸dot notation

=d2ydt2︸︷︷︸

full notation

where y = y(t)

122 Partial derivativesLet f (xy) be a function of x and y Then

fx equivpart fpartx

fy equivpart fparty

fxx equivpart 2 fpartx2 fyy equiv

part 2 fparty2 fxy equiv

part 2 fpartxparty

123 vectorsWe shall also use interchangeably the notations

~uequiv uequiv u

for vectors

Introduction2nd order linear ODEsSolutions of the Cauchy-Euler ODE

2 Ordinary Differential Equation Refresher

21 IntroductionIn previous modules you will have met various techniques to solve both 1st and 2nd order ordinarydifferential equations (ODEs)

211 2nd order linear ODEsDefinition 211 mdash 2nd order linear ODEs The general 2nd order linear differential equationmay be written as

L[y] = p(x)d2ydx2 +q(x)

dydx

+ r(x)y = f (x) (21)

If f (x) = 0 the equation is said to be homogenous if f (x) 6= 0 the equation is inhomogenous orforced The homogeneous equation L[y] = 0 has two non-trivial linearly independent solutionsy1(x) and y2(x) its general solution is called the complementary function

yCF = Ay1 +By2 (22)

here A and B are arbitrary constants For the forced equation ( f (x)) describes the forcing on thesystem it is usual to seek a particular integral solution yPI(x) which is just any single solution ofit Then the general solution of Eq (21) is

y = yCF + yPI (23)

Finding particular solutions can often involve some inspired guesswork eg substituting asuitable guessed form for yPI with some free parameters which are then constrained by the ODEHowever more systematic methods have been developed unfortunately these often add to thecomplexity of the problem See the no free lunch theorem

In this course we shall be primarily interested in two common forms of the 2nd order linearODE constant coefficients and Cauchy-Euler form

Definition 212 mdash 2nd order constant coeffient ODEs A 2nd order constant coeffient ODE

10 Chapter 2 Ordinary Differential Equation Refresher

(homogenous) takes the form

ad2ydx2 +b

dydx

+ cy = 0 (24)

Definition 213 mdash 2nd order constant coeffient ODEs A 2nd order Cauchy-Euler ODE(homogenous) takes the form

ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)

In both cases a b and c are (for the sake of simplicity real) constantsSolutions to these special forms and others can be found by taking an educated guess (an

ansatz) for the form of the solution based on the eigenfunction of the operator which underpinsthe equation For example the operator which effectively defines the 2nd order constant coefficientODE is the operator

L =dn

dxn (26)

Recall in analogy to the matrix eigenvalue problem Ax = λx we consider the eigenvalue problem

L[y] = microy (27)

where y = y(x) and micro is a constant It is straightforward to show that for the operator defined inEq (211) the eigenfunction y(x) is given by

y = emx (28)

where m is a constant For example consider if

L =ddx

(29)

then the eigenfunction is defined by

dydx

= microyminusrarr dyy

= microdxminusrarr y = emicrox (210)

Note we take the simplest form for the eigenfunction and so ignore the constant of integrationhere One can readily check from substituting y = e

radicmicrox into

d2ydx2 = microy (211)

that this is the eigenfunction of the differential operator L = d2dx2 and so on for higher orderderivatives of y

Hence to solve equations of the form of (24) one first looks for a solution in the form y = emxSubstituting this into Eq (24) yields

am2emx +bmemx + cemx = 0 (212)

which reduces to the quadratic equation

am2 +bm+ c = 0 (213)

21 Introduction 11

known as the auxiliary equation This has solution(s)

m =minusbplusmn

radicb2minus4ac

2a (214)

Thus one has three cases with which to deal depending on whether the quadratic has two realroots two complex roots or a repeated (double) root The nature of the solution is dependent onthe form the roots of the auxiliary equation take as can be seen in the table below

Roots General solution of homogeneous equationm1m2 isin Rm1 6= m2 yCF = Aem1x +Bem2x

m1m2 isin Rm1 = m2 yCF = (A+Bx)em1x

m1m2 = r+ is isin C yCF = erx(Acos(sx)+Bsin(sx))with rs isin R

212 Solutions of the Cauchy-Euler ODEConsider rewriting Eq (25) by dividing through by a

x2yprimeprime+αxyprime+βy = 0 where αβ =const (215)

Note x = 0 is a regular singular point The equation is built from applying linear combinationsof the differential operator

L[y] = xn dnydxn (216)

The eigenfunction of this operator can readily be shown to be a a simple power function iey = xr where r is a constant If we assume the solution is of the form y = xr then yprime = rxrminus1yprimeprime = r(rminus1)xrminus2 Substituting into the ODE we find

[r(rminus1)+αr+β ]xr = 0 (217)

ie r2 +(αminus1)r+β = 0 This is called the indicial equation that determines r It is quadraticand so there are 3 cases for the solutions

r12 =minus(αminus1)plusmn

radic(αminus1)2minus4β

2(218)

1 2 distinct real roots ndash straightforward solution2 1 repeated real root ndash problematic case as only one root is found immediately3 2 complex conjugate roots

Exercise 21

2x2yprimeprime+3xyprimeminus y = 0 (219)

Let y = xr then

2r(rminus1)+3rminus1 = 0hArr 2r2 + rminus1 = 0

(2rminus1)(r+1) = 0hArr r1 =minus1r2 = 12 ndash two real roots

General solution


y = Axminus1 +Bradic

x AB arbitrary constants

Exercise 22

x2yprimeprime+5xyprime+4y = 0 (220)

First we assume y=xr hence

r(rminus1)+5r+4 = 0lArrrArr r2 +4r+4 = (r+2)2 = 0 (221)

r1 = r2 =minus2 ndash repeated root (222)

Hence we have found only one solution y = axminus2 This is problematic as we need a secondlinearly independent solution We can find it by differentiating the original ODE with respectto r First we write ODE in operator form L[y] = 0 We have already shown that if y = xr

then

L[xr] = (r+2)2xr = 0 for r =minus2 (223)

Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part

part r(r+2)2xr = 2(r+2)xr +(r+2)2 part

part rxr

= (r+2)xr(2+(r+2) logx) = 0 if r =minus2

Comparing both resultsL[xr logx] = 0

Hencey2 = xr logx is a second solution

General solution

y = axr +bxr logx = xr(a+b logx)

Before we move on to discuss Partial Differential Equations we are reminded of one importantproperty of linear equations

R The principal of superposition - A linear equation has the useful property that if u1 andu2 both satisfy the equation then so does αu1 +βu2 for any αβ isinR This is often used inconstructing solutions to linear equations This is not true for nonlinear equations whichhelps to make this sort of equations more interesting but much more diffcult to deal with

General remarksExamples

Prototypical second order linear PDEsThe diffusion equationThe Laplace equationThe wave equation

PDEs vs ODEsGeometrical Interpretation of solutions

Solution methodsSolution properties

Trivial Partial Differential EquationsIntegration wrt different variablesNo derivatives wrt one the variables ofu = u(xy)Equations which are solvable for ux or uy(not involving u)Special Tricks

3 Introduction to PDEs

31 General remarks

Recall the definition of a Partial Differential Equation

Definition 311 A Partial Differential Equation is an equation for one (or several) unknownfunction of several independent variables involving its derivatives of various orders anddegrees

F(

xyupartupartx

partuparty

part 2u

partxpartypart 2upartx2

)= 0 (31)

where F is a given function of the independent variables x y and of the unknown functionu(xy)

R The order of a PDE is the order of the partial derivative(s) of highest order that appear inthe equation

R The degree of a PDE is the highest power of the highest order derivative occurring in theequation

A PDE is linear if it is of first degree in the unknown function and its derivatives

Definition 312 mdash 1st order linear PDE

P(xy)ux +Q(xy)uy +R(xy)u = S(xy) (32)

Definition 313 mdash 2nd order linear PDE

A(xy)uxx +2B(xy)uxy +C(xy)uyy +D(xy)ux +E(xy)uy +F(xy)u = G(xy) (33)

A PDE is quasilinear if it is linear in the highest order derivatives which appear in the equa-tion

14 Chapter 3 Introduction to PDEs

Definition 314 mdash 1st order quasilinear PDE

P(xyu)ux +Q(xyu)uy = R(xyu) (34)

Definition 315 mdash 2nd order quasilinear PDE

A(xyuuxuy)uxx +2B(xyuuxuy)uxy +C(xyuuxuy)uyy = D(xyuuxuy) (35)

A PDE is semi-linear if it is quasilinear and the coefficients of the highest order derivatives arefunctions of the independent variables only

Definition 316 mdash 1st order semi-linear PDE

P(xy)ux +Q(xy)uy = R(xyu) (36)

Definition 317 mdash 2nd order semi-linear PDE

A(xy)uxx +2B(xy)uxy +C(xy)uyy = D(xyuuxuy) (37)

PDEs which are neither linear nor quasilinear are said to be nonlinear In this course we shallassume the independent variables are real

311 Examples

xpartupartx

+ ypartuparty

= sinxy 1st order linear

partupart t

+upartupartx

= 0 1st order quasilinear

(partupartx

)2

+u3(

partuparty

)4

+partupart z

= u 1st order nonlinear

partupart t

+upartupartx

= νpart 2upartx2 (Burgers eq) 2nd order semi-linear

(x+3)partupartx

+ xy2 partuparty

= u3 1st order semi-linear

xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2

= t +1 2nd order semi-linear

part 2upart t2 = c2 part 2u

partx2 (Wave equation) 2nd order linear

part 2upartx2 +

part 2uparty2 = 0 (Laplace equation) 2nd order linear

partupart t

= νpart 2upartx2 (Heat equation) 2nd order linear

32 Prototypical second order linear PDEs 15

Partial Differential Equations (as well as Ordinary Differential Equations) arise most naturally inthe process of mathematical modelling of natural phenomena This is the process of describingmathematically a physical phenomenon of interest The process involves ldquoidealizationrdquo of thephenomenon ie making simplifying assumptions designed to capture the essential features ofthe phenomenon but to leave out the less significant ones

Examples of Partial Differential Equations arise in but are not limited to Physics ChemistryBiology Economics Engineering and many others Indeed most physical theories can besummarized in terms of Partial Differential Equations egClassical Mechanics Lagrange-Euler equations (of the Lagrangian formulation) Hamilton-

Jacobi equations (of the Hamiltonian formulation)Fluid Mechanics Navier-Stokes equationElectrodynamics Maxwellrsquos equationsGeneral Relativity Einsteinrsquos field equationsQuantum Mechanics Schroumldingerrsquos equationOne can then say that much of Physics is devoted to the formulation of appropriate PartialDifferential Equations and to the attempts to find their solutions in various cases of interest

As a branch of science becomes better understood it becomes more formalized and math-ematical in form Thus most of the other sciences and branches of engineering follow in thefootsteps of Physics and formulate their fundamental theories in terms of Partial DifferentialEquations

32 Prototypical second order linear PDEs321 The diffusion equation

The Partial Differential Equation

partupart t

= Dnabla2u (38)

is called the diffusion equation Here u(~x t) = u(xyz t) is function in four real variables~x = (xyz)T is the position vector of a point in space with Cartesian co-ordinates (xyz) t isthe time variable and D is called the coefficient of diffusion which is a tensor in general

Definition 321 The linear differential operator nabla2 is called the Laplace operator (akaLaplacian or nabla squared or del squared) and in coordinate-independent form is defined by

nabla2 = nabla middotnabla = divgradequiv ∆

In 3D Cartesian coordinates this takes the explicit form

nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2

with obvious reductions in the 2D and 1D cases

The diffusion equation describes the non-uniform distribution and the evolution of somequantity For examplebull temperature In this context the diffusion equation is called the heat equation u represents

the temperature and D represents the so called the thermal diffusivity of the material inquestionbull concentration of a chemical componentbull magnetic field Now u = ~B the magnetic induction vector and D is the electric resistivity

of the medium


The diffusion equation is the prototypical example of a parabolic equation to be discussed laterin the course

322 The Laplace equationThe equation

nabla2u = 0 (39)

is called the Laplace equation This is a special case of the diffusion equation for an equilibriumprocess ie part 2u

part t2 = 0 The Laplace equation is the prototypical example of an elliptic equation tobe discussed later in the course

R The solutions of the Laplace equations are called harmonic functions and represent thepotentials of irrotational and solenoidal vector fields

Definition 322 A vector field ~w is called irrotational if nablatimes~w = 0 A vector field ~w issolenoidal if nabla middot~w = 0

If ~w is irrotational it can be represented as a gradient of a scalar potential ie ~w = nablau becausenablatimes~w = nablatimesnablau = 0 If ~w is solenoidal then nabla middot~w = 0 = nabla middotnablau = nabla2u = 0 so the Laplaceequation follows

For these reasons the Laplace equation describesbull incompressible inviscid fluid flow were ~w is the fluid velocitybull gravitational theory where ~w = ~F is the gravity force and minusu is the gravity potential in

free spacebull electrostatic theory where ~w= ~E is the electric field in free space andminusu is the electrostatic

potential

323 The wave equationThe equation

nabla2u =minus 1

c2part 2

part t2 u (310)

where c is the wave speed is called the wave equation The wave equation (perhaps unsur-prisingly) describes a variety of waves The wave equation is the prototypical example of anhyperbolic equation to be discussed later in the course

33 PDEs vs ODEsThe main difference between Ordinary Differential Equations and Partial Differential Equa-tions is the number of independent variables on which the unknown function (solution) dependsie the domain where the solutions are defined (sought) In particular from the appropriatedefinitions we note that the solutions ofbull Ordinary Differential Equationsare defined in R1bull Partial Differential Equationsare defined in R2 (or in general Rn)

Example 31 Solve the trivial equation

ux = 0

by treating it as (a) Ordinary Differential Equation and (b) Partial Differential Equation

33 PDEs vs ODEs 17

(a) Suppose u is defined in R1 ie u = u(x) Then ux = 0 is an Ordinary DifferentialEquationwith solution

u =C

for an arbitrary constant C(b) Suppose u is defined in R2 ie u = u(xy) Then ux = 0 is a Partial Differential

Equationwith solution

u = f (y)

where f (middot) is an arbitrary functionThe two solutions are obviously very different

331 Geometrical Interpretation of solutionsDefinition 331 A solution if it exists written in the form

u = f (x) or u = g(xy) is called an explicit solution and

w(xy) = 0 or v(xyu) = 0 is called an implicit solutionto an Ordinary Differential Equation or a Partial Differential Equation respectively

From the explicit expressions it is clear that geometricallybull the solutions to Ordinary Differential Equationsrepresent curves in R2 whilebull the solutions to Partial Differential Equationsrepresent surfaces (hypersurfaces) in

R3 (Rn)

Example 32 Find the general solution of the Partial Differential Equationuxy = 0Integrate the equation uxy = 0 once wrt y to get

ux = g(x)

Integrate a second time wrt x to get the general solution

u =int

g(x)dx+ f (y) = w(x)+ f (y)

where f w are arbitrary functions Note that the general solution defines a surface in R3

IMPORTANT Always remember to include appropriate constants (ODE) or functions ofintegration (PDE) where necessary

Exercise 31 Solve uxx = f (y) where f is a given functionIntegrate the equation uxx = f (y) once wrt x to get

ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)

where gw are arbitrary functions Note that the general solution defines a surface in R3

Note we can split u into two components

u(xy) =12

f (y)x2︸︷︷︸particular integral

+ g(y)x+w(y)︸︷︷︸complementary function


Just as in the ODE case the particular integral is the part of the solution generated by thepresence of the inhomogeneous term and the complementary function is the part of the solutioncorresponding to the homogeneous equation

R As the solutions to Partial Differential Equations define surfaces the theory of PartialDifferential Equations has an important relationship to geometry

Exercise 32 Find a Partial Differential Equation which has solutions all surfaces of revolu-tion

1 Surfaces of Revolutionbull Consider some curve z = w(x)bull Rotate the curve around the z-axis to obtain a surface of revolutionbull Cut the surface by a plane by taking z = constant to form a circle x2 + y2 = r2

Thus the equation to the surface of revolution is z = u(x2 + y2)2 Find a Partial Differential Equationwith solution z = u(x2 + y2) By taking partial

derivatives we get the equations

ux = uprime(x2 + y2)2x

uy = uprime(x2 + y2)2y

which gives rise to the equation

yuxminus xuy = 0 (311)

So a Partial Differential Equationcan serve as a definition of a surface of revolution

34 Solution methodsPartial Differential Equations are incredibly difficult to solve so much so more often that notit is impossible to solve a Partial Differential Equation In the absence of an explicit analyticalexpression for the solutions of a given Partial Differential Equation in question the goal of theldquoadvancedrdquo mathematical analysis is to establish certain important properties of the PDE and itssolutions

341 Solution propertiesWhen analytical solutions of a Partial Differential Equation cannot be found it is important toobtain as much information as possible about the following propertiesbull Existence - can one prove that solutions exist even if one cannot find thembull Non-existence - can one prove that a solution does not existbull Uniquenessbull Continuous dependence on parameters and or initial and boundary conditionsbull Equilibrium states and their stabilitybull RegularitySingularity ie can one prove smoothness (ie continuity and differentiability)

of the solutionsIt is perhaps best to motivate the investigation of these properties by first considering illustrativeexamples from ODEs

1dudt

= u u(0) = 1

35 Trivial Partial Differential Equations 19

The solution u = et exisits for 0le t lt infin2

dudt

= u2 u(0) = 1

The solution u = 1(1minus t) exisits for 0le t lt 13

dudt

=radic

u u(0) = 0

has two solutions u = 0 and u = t24 hence non-uniquenessIf we turn back to PDEs the extension is natural

Example 33 Solve the PDE

part

part tuminus∆u = 2

radicu

for x isin R and t gt 0 and the initial condition u(0x) = 0We can quickly check that

u(tx) = 0 is a solution

and u(tx) = t2 is also a solution

Hence the solution to this PDE is not unique

Definition 341 mdash Well-posedness We say that a PDE with boundary (or intial) conditionsis well-posed if solution exists (globally) is unique and depends continuously on the auxillarydata If any of these properties (ie existence uniqueness and stability) is not satisfied theproblem is said to be ill-posed It is typical that problems involving linear equations (orsystems of equations) are well-posed but this may not be always the case for nonlinearsystems

35 Trivial Partial Differential EquationsSome Partial Differential Equations are immediately solvable by direct integration OtherPartial Differential Equations can be easily reduced to Ordinary Differential Equations eitherimmediately or after an appropriate change of variables The resulting Ordinary DifferentialEquations can then be solved by standard techniques We demonstrate some cases with examples

351 Integration wrt different variables Example 34 Find the general solution to the Partial Differential Equation

uxy = 0

Integrating this with respect to y keeping x constant we get

ux = w(x)

where w(x) is an arbitrary function Integrating again this time with respect to x and keeping yconstant we have

u =int

w(x)dx+w2(y) = w1(x)+w2(y)

where w1(x)w2(y) are arbitrary functions


R The general solution of an n-th order Partial Differential Equation contains n arbitraryfunctions For instance the general solution ofbull a first order Partial Differential Equation contains one arbitrary functionbull a second order Partial Differential Equation contains two arbitrary function

This is similar to the case of Ordinary Differential Equations where the general solutionof an n-th order Ordinary Differential Equation contains n arbitrary constants

352 No derivatives wrt one the variables of u = u(xy)In this case the it can immediately be observed that the Partial Differential Equation is effectivelyequivalent to an Ordinary Differential Equation and can be solved by standard methods

Example 35 Find the general solution to the Partial Differential Equations

(a) uxx +u = 0 (b) uyy +u = 0

(a) This is effectively an ODE wrt x

uprimeprime+u = 0

with the general solutionu(xy) = A(y)sinx+B(y)cosx

where A(y)B(y) are arbitrary functions(b) Similarly but wrt y so the general solution is

u(xy) =C(x)siny+D(x)cosy

where C(y)D(y) are arbitrary functions

353 Equations which are solvable for ux or uy (not involving u) Example 36 Find the general solution to the Partial Differential Equation

uxy +ux + f (xy) = 0

where f (xy) = x+ y+1Let

p = ux

then the PDE becomes

py + p+ f (xy) = 0

This is a first order Partial Differential Equation for p = p(xy) where x is treated as a constantand can be solved by an integrating factor methodThe integrating factor is

micro = ey

and in the particular case when f (xy) = x+ y+1 we have

part

party(ey p) =minus(x+ y+1)ey =minus(x+1)eyminus yey

pey =minusint(x+1)eydyminus

intyeydy︸︷︷︸

by parts

=minus(x+1)eyminus yey + ey +C(x)

So ux equiv pequivminus(x+1)minus y+1+C(x)eminusy

=minus(x+ y)+C(x)eminusy


To find u(xy) we integrate the last expression with respect to x

u =minusint(x+ y)dx+ eminusy

intC(x)dx

=minusx2

2minus yx+D(x)eminusy +E(y)

where D(x) =int

C(x)dx and E(x) are arbitrary functions

354 Special TricksA variety of other cases are possible for instance

Example 37 Find the general solution of the Partial Differential Equation

uuxyminusuxuy = 0

We can rearrange this to get

uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx

Integrating with respect to x

lnuy = lnu+ a(y)︸︷︷︸lnb(y)

= lnu+ ln(b(y))

rArr uy = ub(y)

This is now a separable ODE

1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int

b(y)dy+ e(x) = lnD(y)+ lnE(x)

rArr u = E(x)D(y)

where E(x)D(y) are arbitrary functions

The truncated PDEFinding a particular solution

Solution to strictly-linear first-order PDEs bychange of variables

examplesCharacteristic curvesLinear waves

4 1st-order Linear PDEs

Recall our earlier definitionDefinition 401 mdash strictly-linear first order Partial Differential Equationin two variables

a(xy)ux +b(xy)uy + c(xy)u+d(xy) = 0 (41)

where a b c and d are given functions of x and y

41 The truncated PDEIn the method of solution by change of variables we will first need to solve the so called truncatedPDE We consider this here

Definition 411 mdash the truncated PDE The Partial Differential Equation

a(xy)ux +b(xy)uy = 0 (42)

is called the truncated PDE associated with the strictly linear first-order PDE (41)

Let v(xy) be any one possible solution of (42) then the general solution is given by

u = w(v(xy)

)

Proof Taking the partial derivatives of u(xy) we get that

ux = wvvx uy = wvvy

Substituting these into equation (42)

wv(avx +bvy) = 0

which is satisfied since v(xy) is already one possible solution

24 Chapter 4 1st-order Linear PDEs

Definition 412 Let v(xy) be any one possible solution of (42) then the curve given by theequation

c = v(xy)

where c is an arbitrary constant is called a characteristic curve or simply a characteristic ofthe truncated PDE (42)

R The characteristics are curves wholly contained in the solution surface of the PartialDifferential Equation

Clearly if characteristics of the truncated PDE are known we can find the general solution Thefollowing Lemma states how a characteristic can be found The characteristics c = v(xy) of(42) satisfy the so-called characteristic Ordinary Differential Equation

dydx

=b(xy)a(xy)

Proof Select x as the independent parameter along the curve

c = v(xy(x))

and differentiate both sides of c = v(xy) wrt x

vx + vyyx = 0

to find thatyx =minus

vx

vy

Use the truncated PDE (42) to express

minusvx

vy=

b(xy)a(xy)

Substitute to find the characteristic ODE

dydx

=b(xy)a(xy)

R Recall that the solution of an ODE such as the characteristic ODE can always be writtenin implicit form c = v(xy)

Example 41 Find the general solution of the PDE


In this case a = y b =minusx So the characteristic ODE is

dydx

=minusxy

41 The truncated PDE 25

This is a separable equation that we can integrate immediately to find

12

y2 =minus12

x2 + c1

This solution can be easily put in implicit form

c = x2 + y2

and by Lemma 41 is the characteristic c = v(xy) while

v(xy) = x2 + y2

is one possible solution of the PDENow by Lemma 41 the general solution is

u = w(x2 + y2)

where w is an arbitrary function in x and y

411 Finding a particular solutionTo find a particular solution means to determine w of Lemma 41 To do this one auxiliarycondition (aka boundary condition) must be given

R Typically the auxiliary condition is given as a requirement that the solution surface containsa particular specified curve The curve is usually specified in parametric form

x = x(s) y = y(s) u = u(s) (44)

This requirement fixes w when substituted into u = w(v(xy)

)

Exercise 41 Find the particular solution of the PDE


containing the curves specified by

(a) x = sy = su = s (b) x = 1y = su = s gt 1

Note that this is the same PDE as in Example 41 So the general solution is

u = w(x2 + y2)

(a) Substitute x = s y = s and z = s we have

s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2


and we have found w Then the particular solution surface is

z =

radicx2 + y2

2

(b) Substitute x = 1 y = s and z = s gt 1 we have

s = w(1+ s2)

Letting r = 1+ s2 s =radic

rminus1 so w(r) =radic

rminus1 So the general solution is

z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1

which is a hyperboloid


uxminusuy = 0

and then the particular solution containing the curve

x = sy = 0 and u = s2

Identifya = 1 b =minus1

Characteristic ODE isdydx

=minus1

Its solution isy =minusx+ c

Rearrange to get the characteristic curve

c = x+ y

The general solution then isu = w(x+ y)

To find the particular solution substitute x = s y = 0 u = s2

w(s) = s2

which immediately defines the function w So the particular solution is

u = (x+ y)2

42 Solution to strictly-linear first-order PDEs by change of variablesThe basic idea is that we wish to find a transformation to a new pair of independent variablessay ξ η which will transform PDE (41) into a PDE with one of the partial derivatives absentThen we can treat it as an ODE The specific transformation we need to make is given by thefollowing

42 Solution to strictly-linear first-order PDEs by change of variables 27

Theorem 421 The first-order strictly-linear PDE (41) can be transformed into an OrdinaryDifferential Equation by a change-of-variables transformation

η = η(xy) ξ = ξ (xy)

whereη(xy) = v(xy)

is any possible solution of the truncated PDE (42)

Proof The ldquooldrdquo independent variables are expressed in terms of the ldquonewrdquo ones by the inversetransformation

x = x(η ξ ) y = y(η ξ )

Then the unknown function is transformed by

u(xy) = u(x(η ξ )y(η ξ )

)= u(η ξ )

The derivatives are transformed by

ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ

partx= uηηx +uξ ξx (46)

uy =partuparty

=partupartη

partη

party+

partupartξ

partξ

party= uηηy +uξ ξy

which may be written in matrix form as[ux

uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]

Substituting all into PDE (41) it is finally transformed into

(aηx +bηy)uη +(aξx +bξy)uξ + cu+d = 0

This equation will become an Ordinary Differential Equationif we require that the coefficientsin front of the derivatives vanish As the coefficients have the same form up to notation thisrequirement can be written as

avx +bvy = 0

But this is now exactly the truncated equation associated with equation (41)We can conclude that if one of the equations in the change-of-variable transformation is

chosen to beη = v(xy)

where v(xy) is any solution to the truncated PDE equation (41) will reduce to an ODE

Definition 421 mdash Jacobian The matrix

J =

[ηx ξx

ηy xiy

]is called Jacobian matrix of the transformation

R The other equation in the change-of-variable transformation can be chosen arbitrarily aslong as the transformation is non-singular Non-singularity is checked by the conditionthat the Jacobian determinant

J =

∣∣∣∣ηx ξxηy ξy

∣∣∣∣ 6= 0


421 examples Example 43 Find the particular solution of the PDE

uxminusuy +u+ xminus y+2 = 0

containing the curve x = s y = 0 and u = s

Step 1 ndash Form and solve the associated truncated PDEavx +bvy = 0


Form the characteristic ODEdydx

=ba=minus1

Solvec = x+ y

The general solution is

v = w(x+ y)

Step 2 ndash Select and perform a coordinate transformationSelect the simplest particular solution of the truncated equation

v = x+ y

as one of the needed co-ordinate transformations We select the simplest transformation w(x) = xas we donrsquot want to complicate life So let

η = x+ y

Choose the second transformation arbitrarily Eg we can take

ξ = xminus y

as both being simple enough and ldquosymmetricrdquo to the first transformation Then the inversetransformations are

x =12(s+ t)

y =12(ηminusξ )

Note that since ηx = 1 ηy = 1 ξx = 1 and ξy =minus1 the Jacobian is

J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0

So the chosen coordinate transformation is acceptable as it is non-singular Now we have thederivative transformations

ux = uη +uξ uy = uη minusuξ

Substituting all into the PDE we obtain

2uξ +u+(ξ +2) = 0

which is lacking one of the derivatives as intended and so we can solve it as an ODE


Step 3 ndash Solve the ODE

uξ +12

u =minus12

ξ minus1

This is a first-order linear ODE solvable by finding an integrating factor

micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )

=minus2e12 ξ minusξ e

12 ξ +2e

12 ξ +C(η)

rArr u =minusξ +C(η)eminus12 ξ

Converting to the original variables xy

u(xy) =minus(xminus y)+C(x+ y)eminus12 (xminusy)

Step 4 ndash Find the particular solutionRequire that the general solution contains the given curve x = s y = 0 and u = s

s =minuss+C(s)eminus12 s

Rearrange to find that the particular function C(s)

C(s) = 2se12 s

Then the particular solution is given by

u(xy) =minus(xminus y)+2(x+ y)eminus12 (x+yminus(xminusy))

= yminus x+2(x+ y)ey

Exercise 42 Find the general solution of

xux + yuyminusu = 0


x = coss y = sins and u = 1

We have a = x b = y and c =minusu which gives the truncated Partial Differential Equation

xzx + yzy = 0

The characterising ODE is given by

dydx

=ba=

yxrArr lny = lnx+ lnC

rArr yxequiv elnC

rArr v(xy) =yx=C


So the general solution of the truncated Partial Differential Equationz = w(yx)Now we change the variables again by choosing the simplest solution of the truncated

Partial Differential Equation for the first change and then choosing an arbitrary non-sigularchange of variable for the second

η =yx ξ = xy

Calculating |J| gives

det[J] =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x

∣∣∣∣=minusyxminus y

x

=minus2yx6= 0

so this is non-singular and so we can make a change of variablesThen

ux = uηηx +uξ ξx

uy = uηηy +uξ ξy

Then the Partial Differential Equation transforms into

minusyx

uη + xyuξ +yx

uη + xyuξ minusu = 0

2xyuξ minusu = 0 a 1st order separable ODE Then

2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)

This gives the general solution

u = c(η)radic

ξ = c(y

x

)radicxy

Now we have the general solution and so it remains to find the particular solution givenby x = coss y = sins and u = 1 Substituting these conditions into the general solution gives

1 = c(tans)radic

cosssins

Setting r = tans we get

sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1


So the particular solution to the Partial Differential Equationwith the given conditions is

u =

radic(xy+

yx

)xy =

radicx2 + y2

Example 44 Find the general solution of the linear first order equation

x2ux + yuy + xyu = 1

We have a = x2 b = y and c = xy which gives the truncated Partial Differential Equation

x2zx + yzy = 0


dydx

=ba=

yx2 rArr lny =minus1

x+C

rArrC = lny+1x for y gt 0 x 6= 0

Hence we change the variables (choosing perhaps the simplest arbitrary non-sigular changeof variable for the second)

η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0

∣∣∣∣=minusηy

ηy =1y6= 0

Then

ux = uηηx +uξ ξx = uξ minus1x2 uη

uy = uηηy +uξ ξy =1y

uη

Given we can write ξ = x y = eηminus1ξ the PDE transforms into

uξ +1ξ

eηminus1ξ u =1ξ

which can be solved using the integrating factor method

43 Characteristic curvesWe now investigate the importance of the characteristics let us consider the homogenous first-order PDE

a(xy)ux +b(xy)uy = c(xyu) (47)


(note here the form is slightly different from Eq (41)) The characteristics are defined by theODE

dydx

=b(xy)a(xy)

(48)

which represent a one parameter family of curves whose tangent at each point is in the diretionof the vector e = (ab) Note that

aux +buy = (ab) middot (uxuy) = e middotnablau

ie the derivative of u in the direction of the vector e If we represent the characteristic curvesparametrically such that x = x(τ) y = y(τ) where τ is the parametric variable along the curvethen

dxdτ

= a(xy)dydτ

= b(xy)

Then the variation of u with respect to x along the characteristic curves is

dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty

Using the PDE (Eq (47)) we immediately see

dudx

=c(xy)a(xy)

In terms of curvilinear coordinates τ the variation of u along the curves is

dudτ

=dudx

dxdτ

= c(xy)

Hence a solution to the PDE can be found by considering the system of equations given by

dxdτ

= adydτ

= bdudτ

= c (49)

Note in this context these equations are called the Monge equations in honour of the Frenchmathematician Gaspard Monge We shall see in the next chapter that these extend to encompass1st-order quasilinear PDEs as well For now we shall use them to investigate linear waves

44 Linear wavesLet us consider the first order linear wave equation

partupart t

+ cpartupartx

= 0 (410)

Given that we have spent the bulk of the chapter focusing on a change of variable approach wecould apply this technique to find

η = xminus ct ξ = x+ ct

works well and the PDE reduces to

partu(ξ η)

partξ= 0rarr u(x t) = F(xminus ct)

44 Linear waves 33

Figure 41 (left) A surface plot of a particular solution to the linear wave equation given byu(x t) = exp

minus(xminus ct)2

for c = 05 (right) A plot of the characteristics in the xt-plane

However we could also have used the Monge equations

dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt

= crarr xminus ct = const = x0 u = const = u0

In the next chapter we shall prove that the general solution to the PDE is given by

G(uxminus ct) = 0lArrrArr u = F(xminus ct)

However this could also be seen for this example by letting x0 = s which defines the choice ofcharacteristic and as the initial form for u ie u0 only depends on s we have u = F(s) equivalentto saying u(x t = 0) = F(x) Note whatever reasoning is applied we have the characteristicsdefined as a one parameter family of straight lines

x = s+ ct or t =1c(xminus s)

which have gradient 1c and pass through (s0) as shown in Fig 41 If we are given u(x0) =eminusx2

then the particular solution to the 1st-order linear wave equation is

u(x t) = expminus(xminus ct)2 (412)

Figure 41 shows a plot of the characteristics and the particular solution a 3D figure can be foundin Fig 43 We now consider a modified form of Eq 410 which is still a linear PDE (1st-order)

partupart t

+ cxpartupartx

= 0 (413)

subject to the same initial condition u(x0) = eminusx2 The Monge equations are given by

dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)


The first equation givest = τ + t0

we are free to choose t0 = 0 as the Monge equations are invariant under the transformationτ rarr τ + const Hence the second equation gives

lnx = ct + lnx0 rarr x = x0ect

as in the above example we let x0 be represented by the parametric variable s which defines thechoice of characteristic finally u = u0(s) Hence the characteristics are (as above) defined as aone parameter family of lines

x = sect or t = ln(x

s

)Note in this example the characteristics are not straight lines as the wave speed is not constantbut varies with x Now the general solution can be written down as

u(x t) = F(xeminusct)

and the particular solution (for u(x0) = eminusx2) as

u(x t) = expminus(xeminusct)2 (415)

Figure 42 shows a plot of the characteristics and the particular solution a 3D figure can be foundin Fig 44 Both these problems involved homogenous problems ie there was no forcing term(c(xy) = 0) in Eq (47) and we have already seen

dudτ

= c(xy)

Hence for homogenous problems we can make an important statement

R For homogenous (unforced) problems the value of u is fixed along characteristics

This closes our discussion of 1st-order linear PDEs we now move on to consider how to solvequasilinear problems and how the nonlinearity effects the nature of the solution

44 Linear waves 35

Figure 42 (left) A surface plot of a particular solution to Eq (413) give by expminus(xeminusct)2


Figure 43 A surface plot of a particular solution to the linear wave equation given by u(x t) =expminus(xminus ct)2

for c = 05

var ocgs=hostgetOCGs(hostpageNum)for(var i=0iltocgslengthi++)if(ocgs[i]name==MediaPlayButton0)ocgs[i]state=false


Figure 44 A surface plot of a particular solution to Eq (413) given by expminus(xeminusct)2

for

c = 05


Solution to first-order quasilinear PDEs byLagrangersquos method of characteristics

Tangent and normal to a surfaceThe method of characteristics

The Cauchy ProblemNonlinear waves

ShocksTraffic Flow

The Traffic Flow EquationThe quadratic model

5 Quasilinear PDEs and nonlinear waves

We being by recalling our earlier definition

Definition 501 mdash quasilinear first order Partial Differential Equationin two variables

a(xyu)ux +b(xyu)uy = c(xyu) (51)

where a b c are given functions of x y and u which are possibly non-linear

51 Solution to first-order quasilinear PDEs by Lagrangersquos method of charac-teristics

The first-order quasilinear PDE in 2 variables has a convenient geometrical interpretation whichis very useful in finding its solutions

511 Tangent and normal to a surfaceDefinition 511 A surface in R3 can be defined in several equivalent ways includingbull Explicit form z = z(xy)bull Implicit form F

(xyz(xy)

)= 0

bull Parametric form (a) as set of equations

x = x(s t)y = y(s t)z = z(s t)

(b) as aposition-vector equation ~R(s t) =

(x(s t)y(s t)z(s t)

)

Definition 512 A curve in R3 can be defined in several equivalent ways includingbull Parametric form (a) as set of equations

x = x(τ)y = y(τ)z = z(τ)

(b) as a

position-vector equation~r(τ) =(x(τ)y(τ)z(τ)

)

bull As an intersection of two surfaces

F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0

Definition 513 The vector

~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ

)is called the tangent vector to the curve~r(τ)

38 Chapter 5 Quasilinear PDEs and nonlinear waves

Theorem 511 If P = (xyz)) is a ldquoregularrdquo point on a surface F(xyz

)= 0 then the set of

tangents to all possible curves wholly contained in the surface and passing through P forms asingle plane called the tangent plane of the surface through the point

Proof Omitted

Definition 514 A vector perpendicular to the tangent plane is called normal

R Recall that in Cartesian coordinates nabla is the vector differential operator given bynabla = (partxpartypartz)

Theorem 512 (The equation of a normal) The normal vector to the surface F(xyz) = 0is given by nablaF

Proof Let F(xyz) = 0 be a surface in R3 Let~r =(x(τ)y(τ)z(τ)

)be a curve which is wholly

contained in this surface Then the following equation holds

F(x(τ)y(τ)z(τ)) = 0

Differentiate with respect to s

partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely

~T middotnablaF = 0 (52)

where ~T =(

partxpartτ party

partτ part z

partτ

)is the tangent to the curve~r(τ) Since~r(τ) is an arbitrary curve it

follows that the vector nablaF is normal to the surface

512 The method of characteristics

R Recall that a solution to a PDE in two variables xy can be written in the implicit formG(xyu(xy)

)= 0 and represents a surface called the solution (integral) surface

R Recall that any curve c = v(xyu) wholly contained in the solution surface of a PDE iscalled a characteristic curve of the PDE

Theorem 513 The general solution of the first-order quasilinear PDE (51) is given by

v2(xyu) = w(v1(xyu)

) (53)

where w(middot) is an arbitrary function and c1 = v1(xyu) and c2 = v2(xyu) are any solutions ofthe ldquocharacteristic ODEsrdquo

dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)

51 Solution to first-order quasilinear PDEs by Lagrangersquos method ofcharacteristics 39Proof Let G(xyu(xy)) = 0 be an solution to the first-order quasilinear PDE (51) It can bewritten in implicit form as

G(xyu(xy)) = 0

By implicit differentiation wrt x and wrt y

ux =minusGz

Gx uy =minus

Gz

Gy

Substitute in the first-order quasilinear PDE (51) to obtain

aGx +bGy + cGz = 0

or equivalently as a scalar product of two vectors

~A middotnablaG = 0 (55)

where

~A = (abc)

nablaG = (GxGyGu)

By Lemma 512 we know that nablaG is normal to the solution surface Then by comparison ofthe last equation with the equation for the normal (52) we conclude that ~A is tangent to thesolution surface and in particular to any characteristic curve~r(τ) =

(x(τ)y(τ)z(τ)

)contained

in it Thus the equations defining a characteristic curve are

(abc) =(

dxdτ

dydτ

dzdτ

)

They can be rewritten asdxa

= dτdyb

= dτdzc

= dτ

These are the Monge equations we met in the previous chapter The LHS of any two of them arethus equal and can be combined into the sought after ldquocharacteristic ODESrdquo

dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)

Next let the surfaces given by

c1 = v1(xyu) c2 = v2(xyu)

where c1c2 are arbitrary constants be any solutions of the above characteristic ODEs Theintersection of these two surfaces specifies a characteristic curve of the first-order quasilinearPDE (51) For intersection we require

c1 = c2

However because both c1 and c2 are arbitrary we can impose the more general requirement

c2 = w(c1)

Hence the general solution to (51) written in implicit form is

v2(xyu) = w(v1(xyu)

)

where w(middot) is an arbitrary function


R Finally any particular solution can be found by fixing w(middot) from the requirement that theintegral surface must contain some given curve

x = x(s) y = y(s) u = u(s)

Quite when this auxiliary information provides us with a well-posed problem will be discussedlater in the chapter

Example 51 mdash Trivial PDE Find the general solution to the PDE

ux = 0

Identify~A = (100)

Characteristic ODEs aredydx

=ba= 0

dudx

=ca= 0

with implicit solution in the form of surfaces

y = c1 u = c2

The intersection of these two surfaces is in general

c2 = w(c1)

and so the general solution to the equation is

u(xy) = w(y)

where w is an arbitrary function

Example 52 mdash Strictly Linear Partial Differential Equation Find the particular solution ofthe Partial Differential Equation

yuxminus xuy = 2xyu

that contains the line x = y = u = τ gt 0Identify

~A = (yminusx2xyu)

Characteristic ODEs are

dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2

v2(xyu) = ueminusx2= c2

The characteristic line is the intersection of the two surfaces

c2 = w(c1)

51 Solution to first-order quasilinear PDEs by Lagrangersquos method ofcharacteristics 41so the general solution is

ueminusx2= w(x2 + y2)

where w(middot) is an arbitrary functionTo find the particular solution we substitute the given conditions

x = y = u = s

into the general solution to get

seminuss2= w(2s2)

Select a change of variable

r = 2s2 s =radic

r2

converting the last equation into

w(r) =radic

r2

eminusr2

so that w(r) can be immediately recognized Hence the particular solution is

u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2

Example 53 mdash Quasi-linear Partial Differential Equation Find the general solution to

(x+u)ux +(y+u)uy = 0

Identify~A = (x+uy+u0)

Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions

ln(y+u) = ln(x+u)+ lnc1 lnv2(xyu) = u = c2

y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1

The intersection of these two surfaces is the characteristic curve

c2 = w(c1)

so the general solution is

u = w(

y+ux+u

)where w is an arbitrary function


R The first-order quasi-linear PDE is the most general PDE considered so far and many ofthe other types we have discussed are particular cases of a first-order quasi-linear PDEIt follows that the Method of Characteristics can be used for all such cases in additionto other methods that might be available In particular note that a strictly-linear PartialDifferential Equationcan be solved by

(a) the Lagrange Method of Characteristics(a) the change-of-variables method of the previous section

Example 54 mdash A system of two PDEs Find the general solution to the system of two PDEs


xux + yuy = u (57)

In an earlier example we found that the general solution of 56 is

u = w(x2 + y2)

or written in an alternative way

u = w(v) v = x2 + y2

Substituting into 57 withux = wv2x uy = 2ywv

gives a first-order separable ODE2(x2 + y2)wv = w

Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v

So the general solution satisfying both equations is

u = cradic

x2 + y2

where c is an arbitrary constant

R Note that the second PDE served as an effective additional condition to fix the arbitraryfunction w(middot) of the first solution However this condition is in the form of a ODE anotherarbitrary constant arose


2yux +uuy = 2yu

and the particular solution containing the curve

x = cos2 s y = sins and u = 1


Identify~A = (2yu2yu)


dydx

=u2y

dudx

= u

with solutions

2ydy = c2exdx lnu = x+ lnc2

y2 = c2ex + c1 u = c2ex

v1(xy) = c2exminus y2 v2(xyu) = ueminusx = c2

= uminus y2 = c1

The general solution is given by c2 = w(c1) so

ueminusx = w(uminus y2)

where w is an arbitrary functionTo find the particular solution we substitute the given conditions for xyu to get

eminuscos2 s = w(1minus sin2 s) = w(cos2 s)

Setting r = cos2(s) we find the function

w(r) = eminusr

Then the particular solution is

u = exeminus(uminusy2) = ex+y2minusu

52 The Cauchy ProblemSo far we have been happy to solve the PDE and then apply the boundary data However anatural question to ask is can we always apply the boundary data Or more formally what arethe requirements on the boundary data for the problem to be well posed

The term Cauchy data refers to the boundary data that when applied to a PDE in principledetermine the solution at least locally For the first-order quasilinear PDE Cauchy data is theprescription of u on some curve Γ in the (xy)-plane that is we set u = u0(s) when x = x0(s)and y = y0(s) where s parametrises Γ The combination of the PDE and Cauchy data is calledthe Cauchy problem For the moment we assume that x0 y0 and u0 are smooth functions of s(although there are interesting cases where this is not true eg where Γ has corners) and thatthere are no values of s for which xprime0(s) = yprime0(s) = 0 (this ensures that s is a sensible parameterfor Γ) We have seen that the method of characteristics outlined above usually allows a solutionsurface to be constructed in a neighbourhood of Γ However the procedure fails if the initialcurve Γ is at any point tangent to (ab)T (where we refer to Eq (51))

This is best understood by the means of an exampleConsider the linear PDE

ux +uy = 0


hence ~A = (abc) = (110) and so characteristic ODEs are

dydx

= 1dudx

= 0

and we find the general solutionu = w(xminus y)

Now consider three different sets of initial data1 u = s2x = sy = 0lArrrArr s2 = w(s)

u = (xminus y)2

2 u = sx = sy = slArrrArr s2 = w(0) This is impossible hence there is no solution Notex = sy = s gives x = y which is a characteristics projection Γ is tangent to (ab)T

3 u = 0x = sy = slArrrArr 0 = w(0) here w can be essentially any function subject tow(0) = 0 This is the non-generic case in which it just happens that the initial data and thecharacteristic equation agree and the solution is consequently non-unique

This example illustrates three possibilities for the problem1 If Γ is not tangent to a characteristic projection then there should be a unique solution at

least locally2 If Γ is at any point tangent to a characteristic projection then there is in general no solution3 There is however an exceptional case in which the data for u specified on Γ agree with

the ODE satisfied by u along characteristic projections If this happens then there is anon-unique solution

53 Nonlinear wavesAt the end of the previous chapter we considered linear waves homogenous problems of theform

ut + c(x t)ux = 0

Armed with the method of characteristics we now focus on the more interesting case where

ut + c(x tu)ux = 0

this is an example of a nonlinear wave problem In particular we shall focus on the problem

ut + c(u)ux = 0 u(x0) = f (x) (58)

We define the Monge equations as

dtdτ

= 1dxdτ

= c(u)dudτ

= 0

As we know for a homogeneous problem u is constant along the characteristics and so thesystem of equations can be readily integrated to give

x = tc(u)+ s

and henceu = f (xminus tc(u))

an implicit equation which defines u(x t) Such a form means that the wave speed depends onthe initial height of the wave and can lead to interesting consequences

53 Nonlinear waves 45

Figure 51 (left) A surface plot of the solution to the nonlinear wave problem whose solution isgiven in Eq (59) Note the formation of a shock at t = 1

531 ShocksNow lets assume we are give

c(u) = 1+u

and an initial form for the wave as

u(x0) = f (x) =

1 for xle 01minus x for 0 lt x lt 10 for xge 1

or in parametric form

u(x0) = f (s) =

1 for sle 01minus s for 0 lt s lt 10 for sge 1

Then solution is then given by

u(x t) =

1 for xle 2t1+ tminus x

1minus tfor 2t lt x lt 1+ t

0 for xge 1+ t

(59)

We can see that the solution blows up at t = 1 This is where the characteristics of the equationcross one another

R When characteristics cross the solution breaks down

This is typically indicative of shock waves where the linear ramp becomes vertical infinitegradient implies derivatives donrsquot exist Figure 51 which shows the formation of the shock forthe above problem In the context of a wave equation this is called wave breaking The onset ofwave breaking is characterised by the solution having a vertical tangent (ie ux becomes infinite)at some point The time t = tb at which this first occurs is called the breaking time Taking thex-derivative of the solution u(x t) = u0(s) and the equation for the characteristics (x = F(s)t+ s)where F(s) = c(u0(s))) we get

ux = uprime0(s)sx

and1 = F prime(s)sxt + sx


Eliminating sx gives

ux =uprime0(s)

1+ tF prime(s)

Hence the breaking time istb = min

sG(s)

where G(s) =minus1F prime(s) Generally we only accept tb as a genuine value if it is non-negative andso F prime(s)lt 0 Hence F(s) is decreasing ie the gradient of characteristics is increasing Thiscorresponds to the fact that breaking occurs where characteristics converge If no non-negativevalue exists the wave does not break The value of s = sb which leads to this minimum alsodetermines the characteristic on which breaking first occurs and hence gives the location wherebreaking first occurs

xb = F(sb)tb + sb

Exercise 51 Find the breaking time and location for the initial value problem

ut +uux = 0 u(x0) = eminusx2

The characteristics arex = eminuss2

t + s

Hence F(s) = eminuss2 Now F prime(s) =minus2seminuss2

hence G(s) = es22s To find the turning points

of this function consider

Gprime(s) =(

1minus 12s2

)es2 lArrrArr s =plusmn 1radic

2

Sketching the function G shows that it must have a minimum for s gt 0 and a maximum fors lt 0 Hence the breaking occurs on the characteristic with s = sb = 1

radic2 The breaking

time is given by

tb = G(sb) =

radice2

Finally the breaking location is found at

xb = eminuss2btb + sb =

radic2

54 Traffic Flow

A classical theory of traffic flow based on 1D quasilinear waves was developed in Manchester in1955 by Sir James Lighthill (founder of the IMA) and G B Whitham

541 The Traffic Flow EquationLet x be distance along a road (not necessarily straight) Traffic density ρ(x t) on a road isdefined as the number of cars (or other vehicles) per unit distance at the point x and time t Thenthe number of cars at time t in the region a lt x lt b isint b

aρ(x t)dx

54 Traffic Flow 47

Figure 52 Total gridlock

ρ is really a subtle kind of average Conservation Law No cars can be created or destroyedand so can use the conservation law

partρ

part t=minuspartφ

partx

where φ(x t) is the flux In this context the flux is the rate at which cars are crossing the fixedpoint x ie it is (density of cars) times (speed of cars)

φ = ρu

where u(x t) is the traffic speed Hence we have

0 = ρt +(ρu)x = ρt +ρxu+ρux

We now need another relation which links ρ and u to close the model It is logical to proposeu(x t) = u(ρ(x t)) a cars speed is not dependent on where it is on the road or what time it isonly on the density of the traffic This gives

φ = ρu = ρu(ρ) = f (ρ)

We are now left withpartρ

part t+

part

partxf (ρ) = ρt + f primeρx = 0

a quasilinear PDE

542 The quadratic model

Two assumptions1 When ρ = 0 u = umax the maximum speed a car can travel at2 If ρ = ρc u = 0 where ρc=1spacing between cars in a traffic jam

Consider the simplest model in which u is a linear function of ρ

u = umax

(1minus ρ

ρc

)Now f (ρ) = ρu = umaxρ(1minusρρc) which gives the quadratic traffic model problem

ρt + c(ρcminus2ρ)ρx = 0 c =umax

ρc (510)


Now we see the wave speed is given by 1(slope of the characteristics) which is c(ρcminus2ρ) thiscan be positive or negative depending upon the density of traffic The traffic speed is c(ρcminusρ)hence the wave speed is less than the traffic speed

c(ρcminus2ρ)lt c(ρcminusρ)

This means that changes in density travel more slowly than cars So when you drive you gofaster than the changes in density thatrsquos why you have to slow down to avoid thickening oftraffic This is the other way round from water waves where as you float with the wave breakingwaves come up behind you The reason is the nonlinear term ρρx in Eq 510 has a minus signwhere the nonlinear term in for a an equation modelling a water wave

ut +uux = 0

has a plus sign

IntroductionCharpitrsquos equationsBoundary dataExamplesSand PilesDerivation of the Eikonal equation from theWave Equation

6 1st order nonlinear PDEs

61 IntroductionNow we consider general first-order nonlinear scalar PDEs ie Eqns that are not necessarilyquasilinear The general form of such an equation is

F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)

Note for a quasilinear PDE F is a linear function of p and q

F(pquxy) = a(xyu)p+b(xyu)qminus c(xyu) (64)

62 Charpitrsquos equationsA starting point for finding a solution to Eq (61) is to consider taking the derivative of Eq (61)with respect to both x and y to give

partFpart p

part ppartx

+partFpartq

partqpartx

=minuspartFpartxminus p

partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty

=minuspartFpartyminusq

partFpartu

(66)

Which making use of Eq (63) reduce to

partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)

50 Chapter 6 1st order nonlinear PDEs

and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)

So if we define characteristics or rays as curves x(τ) y(τ) satisfying

dxdτ

=partFpart p

dydτ

=partFpartq

(69)

then along these curves

dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)

We therefore have a system of four ODEs for x y p and q along the rays Recall though that ingeneral F depends on u also so to close the system we also need an ODE for u along the raysnamely

dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)

In summary we have the following system of ODEs for x y p q and u known as Charpitrsquosequations

dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)

It easy to verify that these reduce to the usual characteristic equations

dxdτ

= adydτ

= bdudτ

= c (614)

For the quasilinear form However we are not finished with just Charpitrsquos equations we mustalso consider how to incorporate boundaryinitial data

63 Boundary data 51

63 Boundary data

As for quasilinear equations Cauchy data specifies u along some curve Γ in the (xy)-plane

x = x0(s) y = y0(s) u = u0(s) (615)

We also require initial conditions for p and q p = p0(s) q = q0(s) which are obtained bydifferentiating u0 with respect to s and using the PDE Eq (61)

du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)

We shall now demonstrate how to use Charpitrsquos method to solve nonlinear 1st-order PDEs usingsome examples

64 Examples

Example 61 Find the solution to the nonlinear PDE

uxuy = u (617)

Given the solution to the PDE satisfies

u = s2 on x = s y = s+1 (618)

We first make use of the initial data to satisfy 616 we require

2s = p0 +q0 p0q0 = s2 (619)

Now the PDE can be written as

F = pqminusu = 0 (620)

Charpitrsquos equations Eq (627) give

dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)

These can be solved to find parametric forms which satisfy Eq (625)

p = seτ q = seτ u = s2e2τ x = seτ y = seτ +1 (622)

We can express u = u(xy) in a number of different ways u = x2 u = (yminus1)2 or u = x(yminus1)However it is easy to check that the only possible solution to the PDE is

u = x(yminus1)

which indeed satisfies both the original PDE and the Cauchy data


Example 62 Find the solution to the following PDE

u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)

These can be solved firstly we find

p = α q =minusα2 (628)

hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)

Finally we solve for u to give

u = α2τ +αs (631)

Now we eliminate the parametric variables s and τ finding

τ = y s = xminus2αy (632)

From this and Eq (631) we can write down the solution as

u = α2y+α(xminus2αy) = α(xminusαy) (633)

which satisfies both the original PDE and the Cauchy data

65 Sand Piles 53

N F

mg

Figure 61 A schematic for modelling sugar piled on a spoon

65 Sand PilesSand piles are common in nature the physics involved has important applications to industryparticularly pharmaceuticals Letrsquos imagine a very simple situation we take a spoon and poursugar onto it until we can pour no more A very simple modelling approach is to assume thatthe sugar particles are in a limiting equilibrium Hence the frictional force on it F is as largeas it can be (otherwise we could pile more sugar on to the spoon) Furthermore the frictionalforce is proportional to the normal reaction N (see Fig 61) thus F = microN where micro is thecoefficient of friction (how rough is the surface of the spoon) Resolving horizontally we haveN sinθ = F cosθ where θ is as shown hence tanθ = micro The height of the sandpile is given byu = u(xy) and we know cosθ = (001) middotn where

n =(uxuyminus1)radic

u2x +u2

y +1 (634)

is the unit normal to the surface From this it is straightforward to show that(partupartx

)2

+

(partuparty

)2

= micro2 (635)

a famous equation known as the Eikonal equation typically found when considering thepropagation of (eg electromagnetic) waves

Exercise 61 mdash Sugar on a spoon Consider sugar piled up on a spoon such that its heightis given by u(xy) At criticality (just before the sugar would start to slide off the spoon) thesugar makes a constant angle of repose with the horizontal We have seen that we can modelthe pile using

|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)


We can renormalise the equation to give(partupartx

)2

+

(partuparty

)2

= 1 (637)

the Eikonal equation a nonlinear PDE of the form

F(xyu pq) = (p2 +q2minus1)2 = 0

note the factor 05 is purely for convenience Given this form of F Charpitrsquos equations are

dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1

Note p and q are constant along rays and hence given by their boundary values

p = p0(s) q = q0(s)

We integrate the remaining ODEs to give

x = x0(s)+ p0(s)τ y = y0(s)+q0(s)τ u = u0(s)+ τ

At the spoons edge the height of the sugar pile must be 0 hence

dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1

This can readily be solved to give

p0 =plusmnyprime0radic

(xprime0)2 +(yprime0)2 q0 =

plusmnxprime0radic(xprime0)2 +(yprime0)2

where the primes denote differentiation with respect to s Note the vector (p0q0) is the unitnormal to the boundary (the edge of the spoon) Hence the rays are straight lines perpendicularto the spoons edge and u(xy) is the distance of the point (xy) from the edge

Also note that there are two possible solutions corresponding to the plusmn in the expressionsfor p0q0 The correct solution is chosen by ensuring that the rays propagate into the regionof interest not out of it Hence here we choose (p0q0) to be the inward pointing normalOtherwise the solution corresponds to the sandpile outside of a hole

Now we assume that the spoon is elliptical and so we can write

x0(s) = acos(s) y0(s) = bsin(s) 0le s lt 2π

for some constants a and b The solution is given parametrically by

x= acos(s)minus bτ cos(s)radica2 sin2(s)+b2 cos2(s)

y= bsin(s)minus aτ sin(s)radica2 sin2(s)+b2 cos2(s)

u= τ

(638)

The solution surface (along with the corresponding rays) are plotted in Fig 62 Notice thereis a ridge across which p and q are discontinuous along the x-axis between x =(a2b2)aand x =+(a2b2)a such ridges are common in granular materials and arise naturally whenwe model such systems as PDEs see Fig 63


Figure 62 (left) A surface plot of the solution to the nonlinear modelling of sugar on a spoonwhose solution is given in Eq (638) with a = 15 b = 1 (right) The corresponding rays for theproblem straight lines which propagate into the centre of the spoon Here there is a ridge acrosswhich p and q are discontinuous

Figure 63 Sand dunes in Mesquite Spring (northernmost part of Death Valley USA)

66 Derivation of the Eikonal equation from the Wave Equation

In the previous section we used the Eikonal equation to model sand piles However the equationis most commonly found in the field of geometric optics Here we consider how the Eikonalequation is derived from the wave equation The derivation is classic and can be found in manypopular textbooks

We begin by stating the wave equation in 2D

φtt = c2(φxx +φyy)


We assume φ = eminusiωtψ(xy) Substituting this into the wave equation leaves

ψxx +ψyy + k2ψ = 0

where k = ωc The equation can be non-dimensionlised by setting xprime = xL yprime = yL Droppingthe primes we have

ψxx +ψyy +κ2ψ = 0

where κ = L2k We letψ = A(xy)eiκu(xy)

where A is the wave amplitude and u is the phase We compute

ψx = iκuxAeiκu +Axeiκu

andψxx =minusκ

2u2xAeiκu + iκuxxAeiκu +2iκuxAxeiκu +Axxeiκu

Substituting this and the corresponding term for ψyy into the equation for ψ gives

minusκ2A(u2

x +u2y)+ iκ[(uxx +uyy)A+2nablau middotnablaA]+ (Axx +Ayy)+κ

2A = 0

Assuming high spatial frequency (κ 1) the two largest terms (proportional to κ2) balance toleave the eikonal equation

u2x +u2

y = 1

A Special solution is u =minusx A = 1 so ψ = eiκx and

φ = eminusi(κx+ct)

a wave propagating to the left see Fig 64

Figure 64 Plane waves of the form φ = eminusiκ(kxminusct) with k = 1 c =minus1 (left) k = 5 c =minus10(left)

Coordinate transformations and classifica-tionCharacteristics and their propertiesProperties of characteristicsCanonical forms

Examples

7 Classification of 2nd-order PDEs

Definition 701 The equation

a(middot)uxx +2b(middot)uxy + c(middot)uyy +F(middot) = 0 ()

is a general second order Partial Differential Equation Furthermore the equation isa) quasi-linear is abcF are functions of xyuuxuy

b) strictly linear is abcF are functions of xy and if F = e(xy)ux + f (xy)uy +g(xy)u+h(xy)

The part

a(middot)uxx +2b(middot)uxy + c(middot)uyy

is called the principal part of ()

R The mathematical properties of () and its solutions are largely determined by its principalpart and not by F

71 Coordinate transformations and classificationIdea Find a coordinate transformation which simplifies the principal part of ()Consider the change of variables

xyminusrarr ξ (xy) η(xy)

The transformation must be non-singular ie

J

(ξ η

xy

)=

∣∣∣∣ ξx ηx

ξy ηy

∣∣∣∣ 6= 0infin

Then derivatives transform as

ux = uξ ξx +uηηx

uxx = (uξ ξ ξx +uξ ηηx)ξx +uξ ξxx +(uηξ ξx +uηηηx)ηx +uηηxx

58 Chapter 7 Classification of 2nd-order PDEs

and so on for uyuyyuxy etcSubstituting into Eqn () it transforms to

αuξ ξ +2βuξ η + γuηη +Φ(middotmiddot) = 0 (dagger)

whereΦ(ξ η uξ uη u) = F(xyuxuyu)+

and

α = aξ2x +2bξxξy + cξ

2y

β = aξxηx +b(ξxηy +ξyηx)+ cξyηy

γ = aη2x +2bηxηy + cη

2y

We seek conditions under which (dagger) reduces to

2βuξ η +Φ = 0

ie we need α = γ = 0 hence

a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0

These are two identical quadratic equations of the form

ap2 +2bp+ c = 0

They are called characteristic equations and have 2 1 or 0 real solutions depending on sgn(b2minusac) Equation () is called

case I hyperbolic if b2minusac gt 0case I parabolic if b2minusac = 0case I elliptic if b2minusac lt 0

R The type of Partial Differential Equationis invariant under coordinate transformationsUsing direct manipulation it is easy to show that

αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)

72 Characteristics and their propertiesDefinition 721 The solutions of the characteristic equations are called characteristic curves

The characteristics equations can be solved to give

ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a


These expressions are simply 1st order ODEs masking as PDEs In general their solutions willhave the implicit form of curves in the xy-plabe

ξ (xy) =C1 η(xy) =C2

On any such curve the derivativedξ

dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy

Similarly for η(xy) =C2 This gives a recipe for finding the characteristic curves in the xy-plane

dydx

=bplusmnradic

b2minusaca

solve these equations and put the solutions in the implicit form


73 Properties of characteristics1) The characteristics define coordinate transformations which transform the general secondorder PDE to a particular simple canonical form2) The characteristics are exceptional curves in the sense that knowledge of the values uuxuy

along the curves does not uniquely determine the values of uxxuyyuxy along the curves (ieessential physical discontinuities propagate along characteristics)This can be seen in the construction of the characteristics however we can also give a moreformal proof

Proof Let ψ = (x(s)y(s)) be a parametric curve Suppose uuxuy are specified along ψ as

u = F(s) ux = G(s) uy = H(s)

Thendux

ds= uxxxs +uxyys = Gs

duy

ds= uyxxs +uyyys = Hs

in addition PDE () holdsauxx +2buxy + cuyy =minusF

These 3 equations form a linear system for uxxuyxuyya 2b cxs ys 00 xs ys

uxx

uxy

uyy

=

minusFHs

Gs

This system has a unique solution unless the determinant of the matrix is zero ie

a(

dydx

)2

minus2b(

dydx

)+ c = 0

this is the characteristic equation of ()


74 Canonical formsCase I Hyperbolic equation b2minusac gt 0The 2 real solutions of the characteristic equation define 2 characteristic curves through everypoint

dydx

=bminusradic

b2minusaca

minusrarr ξ (xy) =C1 = const

dydx

=b+radic

b2minusaca

minusrarr η(xy) =C2 = const

Equation () reduces to the canonical form

uξ η +1

2βΦ = 0 (first form)

this can be further transformed to

uξ ξ minusuηη +1α

Φ = 0 (second form)

Prototype Wave equationCase II Parabolic equation b2minusac = 0The one real solution of the characteristic equation defines only one characteristic curve throughevery point

dydx

=baminusrarr ν(xy) =C = const

Since b2minus ac = β 2minusαγ = 0 and only one of α and γ can be made zero (say α 6= 0 γ = 0)then β = 0 So equation (dagger) takes the canonical form

uξ ξ +1α

Φ = 0

where the coordinate ξ = ξ (xy) is arbitrary C2 function as long as

J

(ξ η

xy

)6= 0

Prototype Diffusion equationCase III Elliptic equation b2minusac lt 0No real characteristics The characteristic equations are complex

dydx

=b+ i

radic|b2minusac|a

which will have a solution of the form

z(xy) = ξ (xy)+ iη(xy) = const

for real ξ η Direct manipulations then shows

0 = az2x +2vzxzy + cz2

y = (αminus γ)+2iβ

So α = γ and β = 0 (If we choose ξ η to take the form above z = ξ + iη) and the canonicalequation becomes

uξ ξ +uηη +1α

Φ = 0

Prototype Laplace equation

74 Canonical forms 61

741 ExamplesClassify the following PDEsbull uxx +2uxy +uyy = uxminus xuybull uxx +2uxy +5uyy = 3uxminus yuy

bull uxx + x2uyy = yuy

and find their canonical formsa = 1b = 1c = 1 b2minusac = 0minusrarr parabolic

Characteristic equation

dydx

=ba= 1 =rArr y = x+ cminusrarr ξ (xy) = yminus x =C

Choose as a coordinate transformation

ξ = yminus xlarrminus from the characteristic equation

η = ylarrminus arbitrary as long as non-singular

Important to check that this transformation is non-singular∣∣∣∣J (ξ η

xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0

Thenux = uξ ξx +uηηx =minusuξ uy = uξ ξy +uηηy = uξ +uη

uxx = uξ ξ uxy =minusuξ ξ minusuηη uyy = uξ ξ +2uξ η +uηη

The equation becomesuηη =minusuξ minus (ηminusξ )(uξ +uη)

which is the canonical form(ii) a = 1b = 1c = 5 b2minusac =minus4 lt 0larrminus ellipticCharacteristic equation

dydx

=1plusmnradicminus4

1= 1plusmn2i

y = (1plusmn2i)x+ crArr (yminus x)plusmn i(2x) =C

Choose as characteristic coordsξ = yminus x

η = 2x

This transformation is non-singular∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0

Thenux =minusuξ +2uη uy = uξ uxx = uξ ξ minus4uξ η +4uηη

uxy =minusuξ ξ +2uξ η uyy = uξ ξ

Equation transforms into canonical form

uξ ξ +uηη = 3(minusuξ +2uη)minus (ξ +η2)uξ


(iii) a = 1b = 0c = x2 b2minusac =minusx2 le 0if x 6= 0 ndashellipticif x = 0 ndashparabolicCharacteristic equation

dydx

=0plusmnradicminusx2

1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C

Characteristic coordinates

ξ = y η = x22larrminus non-singular

ux = xuη uxx = x2uηη +uη = 2ηuηη +2uη

uy = uξ uyy = uξ η

Equation takes the canonical form

uξ ξ +uηη =1

2η(ξ uξ minus2uη)

Cartesian coordinatesPolar coordinatesLaplacersquos equation in 3D CartesiansSpherical geometry and Legendre polyno-mials

Legendre polynomialsLegendrersquos associated equation

8 Separation of variables

81 Cartesian coordinatesThe basic idea is to replace a single Partial Differential Equationin n independent variablesx1x2 xn by n Ordinary Differential Equationby writing

u(x1x2 xn) = u1(x1)u2(x2) un(xn)

and then substitute in the Partial Differential Equation

Example 81 The one dimensional wave equation

uxx =1c2 utt 0 lt x lt ` t ge 0

bcs u(0 t) = 0u(` t) = 0 t ge 0

ics u(x0) =U(x)ut(x0) =V (x)0le xle `

Assume solution can be separated

u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ

and henceX primeprimeminusλX = 0 (i)

T primeprimeminusλc2T = 0 (ii)

At this stage we donrsquot know if λ gt 0 or lt 0 Consider first (i) with λ gt 0 The general solutionof (i) is then

X = Aeradic

λx +Beminusradic

λx

64 Chapter 8 Separation of variables

Boundary conditions require X(0) = X(`) = 0 ie

A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0

the solution of which is A = B = 0 similarly if λ = 0 Hence we must have λ lt 0 and we set

λ =minusp2

so that (i) and (ii) becomeX primeprime+ p2X = 0 (iii)

T primeprimeprime+ p2c2T = 0 (iv)

which have the general solutions

X = Acos(px)+Bsin(px)

T = Acos(pct)+Bsin(pct)

Boundary conditions X(0) = X(`) = 0 give

A = 0 Bsin(pl) = 0

Clearly B 6= 0 otherwise the solution is trivial hence

pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`

)satisfies the equation and bcs for each n Write the partial solution un as

un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`

)since the equation is linear we can add up theses for n = 12 infin to get (superposition)

u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`

)which satisfies the equation and the boundary conditions The constants Cn and Dn are to befound from the initial conditions as follows

u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)

ndash each of these is a Fourier sine series the coefficients of CnDn are given by

Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime

Note that u(x t) may also be written

u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ

`(xminus ct)minus cos

nπ

`(x+ ct)


Example 82 Apply the method of separation of variables to the heat conduction (diffusion)equation ut = kuxx (k gt 0 constant)Set

u(x t) = X(x)T (t)

which gives XT prime = kX primeprimeT and hence

1k

T prime

T=

X primeprime

X= const =minusω

2

where ω gt 0 hence we have X primeprime+ω2X = 0 which as above has trigonometric solutions Thisleaves T prime =minuskω2T so that

T (t) = Aexp(minuskω

2t)

where A is an arbitrary constant the x-dependence is oscillatory but the t-dependence is adecaying exponential

Example 83 The wave equation in 2D

utt = c2nabla

2u = c2(uxx +uyy)

assume a solution of the form u(xy t) = X(x)Y (y)T (t) Plugging this into the PDE gives

XY T primeprime = c2(X primeprimeY T +XY primeprimeT )

T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y

HenceT primeprime+(cω)2T = 0

andX primeprime

X=minusY primeprime

Yminusω

2

So we can sayX primeprime

X=minusΩ

2 X primeprime+Ω2X = 0

andY primeprime

Y= ω

2minusΩ2 Y primeprime+(Ω2minusω

2)Y = 0

If we have appropriate boundary conditions these will yield oscillating (trigonometric) solutionsin t x and y This solution would be relevant for the vibrations of a rectangular membrane

82 Polar coordinates Example 84 The wave equation in 2D (cylindrical polar coordinates)

utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2

)Assume u(rθ t) = R(r)Θ(θ)T (t) For bounded solutions as trarr infin

T primeprime

T=minusω

2c2


which givesRprimeprime

R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω

2r2 =minusΘprimeprime

Θ= Ω

2

The second relation givesΘprimeprime

Θ=minusΩ

2

and trigonometric solutions which we would expect as Θ(θ) is periodic with period 2π Finally

r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0

r2Rprimeprime+ rRprime+(ω2r2minusΩ2)R = 0

An equation we have met before Besselrsquos equation So the solutions of this equation containBesselrsquos functions Hence Besselrsquos functions are crucial for understanding the vibrations on thesurface of a drum for example

83 Laplacersquos equation in 3D Cartesians

nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0

Setφ(xyz) = X(x)Y (y)Z(z)

ThenX primeprimeY Z +XY primeprimeZ +XY Zprimeprime = 0

Divide by XY ZX primeprime

X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime

Zwhere the lhs is independent of z and the rhs is a function of z onlyHence

X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ

2 =minusY primeprime

Ywhere the lhs is independent of y and the rhs is a function of y and so we can write

X primeprime

Xminus γ


Y= const = β

2 (say)


ThenY primeprime+βY = 0

andX primeprimeminus (β 2 + γ

2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0

We have transformed a three dimensional PDE into 3 ODEs

R Choice of exactly how to separate depends on the geometry of the problem applying thebcs is usually the most difficult part

Here we have

Zprimeprime+ γ2Z = 0 Y primeprime+β

2Y = 0 X primeprime+αX = 0

with α2 +β 2 + γ2 = 0 Suppose the bcs are

φ = 0 for z = 0c y = 0b x = 0

φ = f (yz) on x = a

ThenX(0) = 0 X(a) = f (yz) Y (0) = Y (b) = Z(0) = Z(c) = 0

For Y and Z these are satisfied by

Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )

so α2 lt 0 set λ 2 =minusα2 ThenX primeprimeminusλ

2X = 0

which has solutionX =C sinhλx+Dcoshλx

X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx

satisfies the PDE and bcs (expect on x = a) with λ 2 = λ 2mn = β 2

m + γ2n and by superposition

φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx

It remains to satisfy the bc φ(ayz) = f (yz)infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)

which is a double Fourier series

R If f = 0 then φ = 0 if nabla2φ = 0 in D isin Rn and φ = φ0 on the boundary of a simplyconnected region D then φ = φ0 in D


84 Spherical geometry and Legendre polynomialsThe Laplacian in spherical polar coordinates (rθ φ) where x = r sinθ cosφ y = r sinθ sinφ

and z = r cosθ is given by

∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2

Recall that r isin [0infin) θ isin [0π] and φ isin [02π) In particular Laplacersquos equation ∆u = 0 is

urr +2r

ur +1

r2 sinθ(sinθ uθ )θ +

1r2 sin2

θuφφ = 0

Separable solutions u = R(r)Θ(θ)Φ(φ) satisfy

Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ

(sinθ Θprime)prime

Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0

After multiplying by r2 sin2θ we see that

Φprimeprime

Φ=minusm2

This gives

Φ = Acos(mφ)+Bcos(mφ)

and since the Φ(φ +2π) = Φ(φ) for all φ is it is clear that m isin Z+ Multiplying by r2 we nowhave

r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)

and we get the equations for R and Θ

d2Rdr2 +

2r

dRdrminus λ

r2 R = 0 (an Euler equation) (81)

and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)

In spherical geometry it is the θ -dependence that needs to be studied most carefully We willsee that the r-dependence is easily obtained later We rewrite (82) in terms of new independentvariable micro = cosθ so that

ddθ

=minussinθd

dmicro

and we get Legendrersquos associated equation

ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)

where micro isin [minus11] Since 0 le θ le π and minus1 le micro le 1 the change of variables micro = cosθ isa bijection and hence invertible The special case in which m = 0 corresponding to axially

84 Spherical geometry and Legendre polynomials 69

symmetry (ie φ -independent) solutions of Laplacersquos equation is called Legendrersquos equationFrom the form

d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ

1minusmicro2 minusm2

(1minusmicro2)2

)Θ = 0

we see that is has regular singular points at micro =plusmn1 It may be shown that unless λ is of the formn(n+1) for integer nge m there is no solution that is non-singular at both micro = cosθ = 1 andmicro = cosθ =minus1 In spherical polars these correspond to θ = 0 (the positive z-axis) and θ = π

(the negative z-axis) and so usually in applications we are only interested in the case λ = n(n+1)for integers nge mge 0 In this case the R equation (81) is the Cauchy-Euler equation

d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)

for which the solutions are R = ra where a(aminus 1)+ 2aminus n(n+ 1) = (a+ n+ 1)(aminus n) = 0Hence the general solution is R =Crn +Drminusnminus1

841 Legendre polynomialsThe polynomial

Pn(x) =1

2nndn

dxn [(x2minus1)n] (85)

is called the Legendre polynomial of degree n and is know as Rodriguesrsquo formula It is thenth derivative of a polynomial of degree 2n and hence has degree n The first few Legendrepolynomials are

P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2

Theorem 841 For all non-negative integers n(i) Pn(z) satisfies Legendrersquos equation

(1minus x2)yprimeprimeminus2xyprime+n(n+1)y = 0

(ii) Pn(1) = 1(iii) Pn(minusx) = (minus1)nPn(x)

Proof (i) Let w = (x2minus1)n Then wprime = 2nx(x2minus1)nminus1 and so (1minus x2)wprime+2nxw = 0 Differ-entiating n+1 times using Leibnizrsquos rule we get

(1minus x2)w(n+2)+

(n+1

n

)(1minus x2)primew(n+1)+

(n+1nminus1

)(1minus x2)primeprimew(n)

+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0

and after simplifying

(1minus x2)w(n+2)minus2xw(n+1)+n(n+1)w(n) = 0

This shows that w(n) is a solution of Legendrersquos equation Since this ODE is linear and Pn(x) prop

w(n) this proves (i)


(ii) Using the Leibniz rule again

2nnPn(x) =dn

dxn [(x+1)n(xminus1)n]

= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1

)n(z+1)nminus1 dnminus1

dxnminus1 [(xminus1)n]+ middot middot middot

= (x+1)nn+O(xminus1)

Now setting x = 1 we get 2nnPn(1) = 2nn and so (ii) is proved(iii) We have

Pn(minusx) =1

2nndn

d(minusx)n [((minusx)2minus1)n] = (minus1)n 12nn

dn

dxn [(x2minus1)n] = (minus1)nPn(x)

as required

The second solution of Legendrersquos equation Qn(x) may be found using for example theWronskian method but has logarithmic singularities at x = plusmn1 and is not relevant to mostapplications The general solution of Legendrersquos equation is y = aPn(x)+bQn(x)

85 Legendrersquos associated equation

We now return to the most general equation giving the θ -dependence in separable solutions ofLaplacersquos equation in spherical polars Legendrersquos associated equation

(1minus x2)yprimeprimeminus2xyprime+(

n(n+1)minus m2

1minus x2

)y = 0 (86)

where mn isin Z+ Legendrersquos equation is the special case m = 0

Theorem 851 If u is a solution of Legendrersquos equation then

y = (1minus x2)m2 dmudxm (87)

is a solution of Legendrersquos associated equation (86)

Proof We have that

(1minus x2)uprimeprimeminus2xuprime+n(n+1)u = 0

and so by the Leibniz rule the mth derivative is

(1minusx2)u(m+2)+m(minus2x)u(m+1)+m(mminus1)

2(minus2)u(m)minus2(xum+1+mu(m))+n(n+1)u(m) = 0

that is

(1minus x2)u(m+2)minus2(m+1)xu(m+1)+(n(n+1)minusm(m+1)

)u(m) = 0 (lowast)


On the other hand for y = (1minus x2)m2 u(m) as given in (87)

(1minus x2)yprimeprimeminus2zyprime =ddx

((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx

((1minus x2)m2+1 u(m+1)minusmx(1minus x2)m2 u(m)

)= (1minus x2)m2+1 u(m+2)minus (m+2)x(1minus x2)m2 u(m+1)

minusmx(1minus x2)m2 u(m+1)minusm((1minus x2)m2minusmx2(1minus x2)m2minus1)u(m)

= (1minus x2)m2((1minus x2)u(m+2)minus2(m+1)xu(m+1)minusm

(1minus mx2

1minus x2

)u(m)

)and using (lowast) this becomes

= (1minus x2)m2u(m)

(minusn(n+1)+m(m+1)minusm+

m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y

This verifies that y as given by (87) satisfies (86)

The Legendre polynomials Pn(x) and the singular solutions Qn(x) give solutions of Legendrersquosassociated equation for nge mge 0

Pmn (x) = (1minus x2)m2 dmPn(x)

dxm Qmn (x) = (1minus x2)m2 dmQn(x)

dxm

associated Legendre functions of the first and second kind respectively Notice in particular thatPm

n (x) = 0 for n lt m Functions of the second kind are singular for x =plusmn1 and are not relevantto most applications

Motivation amp Notation

Partial Derivatives

Notation for Derivatives

Ordinary derivatives

Partial derivatives

vectors

Ordinary Differential Equation Refresher

Introduction

2nd order linear ODEs

Solutions of the Cauchy-Euler ODE

Introduction to PDEs

General remarks

Examples

Prototypical second order linear PDEs

The diffusion equation

The Laplace equation

The wave equation

PDEs vs ODEs

Geometrical Interpretation of solutions

Solution methods

Solution properties

Trivial Partial Differential Equations

Integration wrt different variables

No derivatives wrt one the variables of u = u(xy)

Equations which are solvable for ux or uy (not involving u)

Special Tricks

1st-order Linear PDEs

The truncated PDE

Finding a particular solution

Solution to strictly-linear first-order PDEs by change of variables

examples

Characteristic curves

Linear waves

Quasilinear PDEs and nonlinear waves

Solution to first-order quasilinear PDEs by Lagranges method of characteristics

Tangent and normal to a surface

The method of characteristics

The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow

The Traffic Flow Equation

The quadratic model

1st order nonlinear PDEs

Introduction

Charpits equations

Boundary data

Examples

Sand Piles

Derivation of the Eikonal equation from the Wave Equation

Classification of 2nd-order PDEs

Coordinate transformations and classification

Characteristics and their properties

Properties of characteristics

Canonical forms

Examples

Separation of variables

Cartesian coordinates

Polar coordinates

Laplaces equation in 3D Cartesians

Spherical geometry and Legendre polynomials

Legendre polynomials

Legendres associated equation

fdrm0

fdrm1

httpabagwikidotcom

Licensed under the Creative Commons Attribution-NonCommercial 30 Unported License (theldquoLicenserdquo) You may not use this file except in compliance with the License You may ob-tain a copy of the License at httpcreativecommonsorglicensesby-nc30 Unlessrequired by applicable law or agreed to in writing software distributed under the License isdistributed on an ldquoAS ISrdquo BASIS WITHOUT WARRANTIES OR CONDITIONS OF ANY KINDeither express or implied See the License for the specific language governing permissions andlimitations under the License

Contents

















44 Linear waves 32








61 Introduction 49


63 Boundary data 51

64 Examples 51

65 Sand Piles 53

















dNdt

= rN





d f =n

sumi=1

part fpartxi

dxi =part fpartx1

dx+part fpartx2

dx2 + +part fpartxn

dxn



part fpartxi

= limhrarr0


h




yprime = y =dydt



=d2ydt2︸︷︷︸

full notation

where y = y(t)


fx equivpart fpartx

fy equivpart fparty



part 2 fpartxparty


~uequiv uequiv u

for vectors






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

Contents

















44 Linear waves 32








61 Introduction 49


63 Boundary data 51

64 Examples 51

65 Sand Piles 53

















dNdt

= rN





d f =n

sumi=1

part fpartxi

dxi =part fpartx1

dx+part fpartx2

dx2 + +part fpartxn

dxn



part fpartxi

= limhrarr0


h




yprime = y =dydt



=d2ydt2︸︷︷︸

full notation

where y = y(t)


fx equivpart fpartx

fy equivpart fparty



part 2 fpartxparty


~uequiv uequiv u

for vectors






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1






44 Linear waves 32








61 Introduction 49


63 Boundary data 51

64 Examples 51

65 Sand Piles 53

















dNdt

= rN





d f =n

sumi=1

part fpartxi

dxi =part fpartx1

dx+part fpartx2

dx2 + +part fpartxn

dxn



part fpartxi

= limhrarr0


h




yprime = y =dydt



=d2ydt2︸︷︷︸

full notation

where y = y(t)


fx equivpart fpartx

fy equivpart fparty



part 2 fpartxparty


~uequiv uequiv u

for vectors






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1









dNdt

= rN





d f =n

sumi=1

part fpartxi

dxi =part fpartx1

dx+part fpartx2

dx2 + +part fpartxn

dxn



part fpartxi

= limhrarr0


h




yprime = y =dydt



=d2ydt2︸︷︷︸

full notation

where y = y(t)


fx equivpart fpartx

fy equivpart fparty



part 2 fpartxparty


~uequiv uequiv u

for vectors






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



part fpartxi

= limhrarr0


h




yprime = y =dydt



=d2ydt2︸︷︷︸

full notation

where y = y(t)


fx equivpart fpartx

fy equivpart fparty



part 2 fpartxparty


~uequiv uequiv u

for vectors






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1






dydx

+ r(x)y = f (x) (21)


yCF = Ay1 +By2 (22)


y = yCF + yPI (23)






ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



ad2ydx2 +b

dydx

+ cy = 0 (24)


ax2 d2ydx2 +bx

dydx

+ cy = 0 (25)



L =dn

dxn (26)


L[y] = microy (27)


y = emx (28)


L =ddx

(29)


dydx




radicmicrox into






am2 +bm+ c = 0 (213)

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

21 Introduction 11


m =minusbplusmn

radicb2minus4ac

2a (214)














2(218)


Exercise 21


Let y = xr then



General solution




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




Exercise 22






then


Differentiate wrt r

part

part rL[xr] = L

[part

part rxr]= L

[part

part rer logx

]= L [xr logx]

also

part

part rL[xr] =

part


part rxr




General solution










31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1







31 General remarks



F(

xyupartupartx

partuparty

part 2u


)= 0 (31)





















311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1












311 Examples

xpartupartx

+ ypartuparty


partupart t

+upartupartx


(partupartx

)2

+u3(

partuparty

)4

+partupart z


partupart t

+upartupartx


(x+3)partupartx

+ xy2 partuparty


xpart 2upart t2 + t

part 2uparty2 +u3

(partupartx

)2




part 2upartx2 +


partupart t









partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1








partupart t

= Dnabla2u (38)





nabla2 =

part 2

partx2 +part 2

party2 +part 2

part z2




of the medium




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




nabla2u = 0 (39)








potential


nabla2u =minus 1

c2part 2

part t2 u (310)




ux = 0


33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

33 PDEs vs ODEs 17


u =C



u = f (y)






R3 (Rn)


ux = g(x)


u =int

g(x)dx+ f (y) = w(x)+ f (y)




ux = f (y)x+g(y)


u = f (y)x22+g(y)x+w(y)



u(xy) =12


















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1
















1dudt

= u u(0) = 1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



dudt

= u2 u(0) = 1


dudt

=radic

u u(0) = 0



part


radicu








uxy = 0


ux = w(x)


u =int

w(x)dx+w2(y) = w1(x)+w2(y)







(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1






(a) uxx +u = 0 (b) uyy +u = 0


uprimeprime+u = 0








p = ux


py + p+ f (xy) = 0


micro = ey


part




by parts







intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




intC(x)dx

=minusx2


where D(x) =int




uuxyminusuxuy = 0


uyx

uy=

ux

u=rArr 1

uy

partuy

partx=

1u

partupartx



= lnu+ ln(b(y))

rArr uy = ub(y)


1u

partuparty

= b(y) rArr 1u

partu = b(y)party

rArr lnu =int


rArr u = E(x)D(y)














u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1













u = w(v(xy)

)


ux = wvvx uy = wvvy


wv(avx +bvy) = 0




c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



c = v(xy)




dydx

=b(xy)a(xy)


c = v(xy(x))


vx + vyyx = 0


vx

vy


minusvx

vy=

b(xy)a(xy)


dydx

=b(xy)a(xy)





dydx

=minusxy



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



12

y2 =minus12

x2 + c1


c = x2 + y2


v(xy) = x2 + y2


u = w(x2 + y2)




x = x(s) y = y(s) u = u(s) (44)


)






u = w(x2 + y2)


s = w(2s2)

Letr = 2s2

Then

s =radic

r2

So

w(r) =radic

r2



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



z =

radicx2 + y2

2


s = w(1+ s2)




z =radic

x2 + y2minus1 or x2 + y2 + z2 = 1



uxminusuy = 0


x = sy = 0 and u = s2



=minus1



c = x+ y



w(s) = s2


u = (x+ y)2





whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




whereη(xy) = v(xy)



x = x(η ξ ) y = y(η ξ )



)= u(η ξ )


ux =partupartx

=partupartη

partη

partx+

partupartξ

partξ


uy =partuparty

=partupartη

partη

party+

partupartξ

partξ



uy

]=

[ηx ξx

ηy ξy

][uη

uξ

]




avx +bvy = 0





J =

[ηx ξx

ηy xiy



J =


∣∣∣∣ 6= 0








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1








=ba=minus1

Solvec = x+ y


v = w(x+ y)


v = x+ y


η = x+ y


ξ = xminus y


x =12(s+ t)

y =12(ηminusξ )


J =part (η ξ )

part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣1 11 minus1

∣∣∣∣=minus2 6= 0




2uξ +u+(ξ +2) = 0




uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



uξ +12

u =minus12

ξ minus1


micro = expint 1

2dξ = eξ2

Proceed as usual

ddξ

(e12 ξ u) =minuse

12 ξ (1+

12

ξ )

ue12 ξ =minus2e

12 ξ minus

intξ d(e

12 ξ )


12 ξ +2e

12 ξ +C(η)







C(s) = 2se12 s





xux + yuyminusu = 0




xzx + yzy = 0


dydx

=ba=


rArr yxequiv elnC

rArr v(xy) =yx=C




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




η =yx ξ = xy



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣minus yx2 y

1x x


x

=minus2yx6= 0





minusyx

uη + xyuξ +yx



2ξ uξ = u

rArr duu

=1

2ξdξ

lnu =12

ln tξ + lnC(η)


u = c(η)radic

ξ = c(y

x

)radicxy


1 = c(tans)radic

cosssins


sins =rradic

1+ r2 coss =

1radic1+ r2

so

c(r) =

radic1+ r2

r=radic

r+ rminus1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



u =

radic(xy+

yx

)xy =

radicx2 + y2




x2zx + yzy = 0


dydx

=ba=


x+C



η = lny+1x ξ = x



part (xy)=

∣∣∣∣ηx ξx

ηy ξy

∣∣∣∣= ∣∣∣∣ηx 1ηy 0


ηy =1y6= 0

Then



uη


uξ +1ξ

eηminus1ξ u =1ξ






dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



dydx

=b(xy)a(xy)

(48)




dxdτ

= a(xy)dydτ

= b(xy)


dudx

=partupartx

+dydx

partuparty

=partupartx

+ba

partuparty


dudx

=c(xy)a(xy)


dudτ

=dudx

dxdτ

= c(xy)


dxdτ

= adydτ

= bdudτ

= c (49)



partupart t

+ cpartupartx

= 0 (410)




partu(ξ η)


44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

44 Linear waves 33


minus(xminus ct)2



dtdτ

= 1dxdτ

= cdudτ

= 0 (411)

Note this implies

dxdt










partupart t

+ cxpartupartx

= 0 (413)


dtdτ

= 1dxdτ

= cxdudτ

= 0 (414)







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1







s






dudτ

= c(xy)




44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

44 Linear waves 35




for c = 05




for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



for

c = 05





ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




ShocksTraffic Flow










(xyz(xy)

)= 0




(x(s t)y(s t)z(s t)

)


x = x(τ)y = y(τ)z = z(τ)

(b) as a


)


F(xyz(xy)

)= 0Φ

(xyz(xy)

)= 0


~T (τ) =d~rdτ

=

(dxdτ

dydτ

dzdτ






Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1





Proof Omitted







F(x(τ)y(τ)z(τ)) = 0


partFpartx

partxpartτ

+partFparty

partypartτ

+partFpart z

part zpartτ

= 0

Or more concisely


where ~T =(

partxpartτ party

partτ part z

partτ








v2(xyu) = w(v1(xyu)

) (53)


dydx

=b(xyu)a(xyu)

dudx

=c(xyu)a(xyu)

dudy

=c(xyu)b(xyu)

(54)


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1


G(xyu(xy)) = 0


ux =minusGz

Gx uy =minus

Gz

Gy


aGx +bGy + cGz = 0



where

~A = (abc)

nablaG = (GxGyGu)


(x(τ)y(τ)z(τ)

)contained


(abc) =(

dxdτ

dydτ

dzdτ

)


= dτdyb

= dτdzc

= dτ


dydx

=b(xyu)a(xyu)

ux=

c(xyu)a(xyu)




c1 = c2


c2 = w(c1)


v2(xyu) = w(v1(xyu)

)




x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



x = x(s) y = y(s) u = u(s)



ux = 0

Identify~A = (100)


=ba= 0

dudx

=ca= 0


y = c1 u = c2


c2 = w(c1)


u(xy) = w(y)



yuxminus xuy = 2xyu


~A = (yminusx2xyu)


dydx

=minusxy

dudx

= 2xu

with solutions

y2

2=minusx2

2+ c0 lnu = x2 lnc2

v1(xyu) = x2 + y2 = c1 u = c2ex2



c2 = w(c1)




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




x = y = u = s


seminuss2= w(2s2)


r = 2s2 s =radic

r2


w(r) =radic

r2

eminusr2


u(xy) = ex2

radicx2 + y2

2eminus

x2+y22 =

radicx2 + y2

2e(x

2minusy2)2




Characteristic ODEs

dydx

=y+ux+u

dudx

= 0

with solutions


y+u = c1(x+u)

v1(xyu) =y+ux+u

= c1


c2 = w(c1)


u = w(

y+ux+u







xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1






xux + yuy = u (57)


u = w(x2 + y2)


u = w(v) v = x2 + y2



Solve

2vdwdv

= w

1w

dw =12v

dv

lnw =12

lnv+ lnc

w(v) = cradic

v


u = cradic

x2 + y2




2yux +uuy = 2yu






dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




dydx

=u2y

dudx

= u

with solutions


y2 = c2ex + c1 u = c2ex


= uminus y2 = c1






w(r) = eminusr






ux +uy = 0



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



dydx

= 1dudx

= 0



u = (xminus y)2







ut + c(x t)ux = 0


ut + c(x tu)ux = 0


ut + c(u)ux = 0 u(x0) = f (x) (58)


dtdτ

= 1dxdτ

= c(u)dudτ

= 0


x = tc(u)+ s






c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




c(u) = 1+u


u(x0) = f (x) =



u(x0) = f (s) =



u(x t) =



0 for xge 1+ t

(59)




ux = uprime0(s)sx




ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



ux =uprime0(s)

1+ tF prime(s)


sG(s)


xb = F(sb)tb + sb




t + s




Gprime(s) =(

1minus 12s2


2


radic2 The breaking

time is given by

tb = G(sb) =

radice2



radic2

54 Traffic Flow



aρ(x t)dx

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

54 Traffic Flow 47



partρ

part t=minuspartφ

partx


φ = ρu




φ = ρu = ρu(ρ) = f (ρ)


part t+

part


a quasilinear PDE




u = umax

(1minus ρ

ρc



ρc (510)





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1





ut +uux = 0

has a plus sign




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




F(xyu pq) = 0 (61)

where

p =partupartx

q =partuparty

(62)

Hence

part pparty

=partqpartx

(63)




partFpart p

part ppartx

+partFpartq

partqpartx


partFpartu

(65)

and

partFpart p

part pparty

+partFpartq

partqparty


partFpartu

(66)


partFpart p

part ppartx

+partFpartq

part pparty


partFpartu

(67)


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1


and

partFpart p

partqpartx

+partFpartq

partqparty


partFpartu

(68)


dxdτ

=partFpart p

dydτ

=partFpartq

(69)


dpdτ

=dux(x(τ)y(τ)

dτ=

part 2upartx2

dxdτ

+part 2u

partxpartydydτ

=part ppartx

dxdτ

+part pparty

dydτ


partFpartu

(610)

dqdτ

=duy(x(τ)y(τ)

dτ=

part 2uparty2

dydτ

+part 2u

partxpartydxdτ

=partqparty

dydτ

+partqpartx

dxdτ


partFpartu

(611)


dudτ

=partupartx

dxdτ

+partuparty

dydτ

= ppartFpart p

+qpartFpartq

(612)


dxdτ

=partFpart p

(613a)

dydτ

=partFpartq

(613b)

dpdτ


partFpartu

(613c)

dqdτ


partFpartu

(613d)

dudτ

= ppartFpart p

+qpartFpartq

(613e)


dxdτ

= adydτ

= bdudτ

= c (614)


63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

63 Boundary data 51

63 Boundary data


x = x0(s) y = y0(s) u = u0(s) (615)


du0

ds= p0

dx0

ds+q0

dy0

ds F(x0y0u0 p0q0) = 0 (616)


64 Examples


uxuy = u (617)


u = s2 on x = s y = s+1 (618)


2s = p0 +q0 p0q0 = s2 (619)




dxdτ

= qdydτ

= pdpdτ

= pdqdτ

= q (621a)

and

dudτ

= pq+qp = 2pq (621b)




u = x(yminus1)




u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



u2x +uy = 0 (623)


u = αs on x = s y = 0 (624)


p0 = α p20 +q0 = 0 (625)


F = p2 +q = 0 (626)


dxdτ

= 2pdydτ

= 1dpdτ

= 0dqdτ

= 0 (627a)

and

dudτ

= 2p2 +q (627b)



hence

dxdτ

= 2αdydτ

= 1 (629)

which give

x = 2ατ + s y = τ (630)


u = α2τ +αs (631)






65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

65 Sand Piles 53

N F

mg




u2x +u2

y +1 (634)


)2

+

(partuparty

)2

= micro2 (635)



|nablau|2 =(

partupartx

)2

+

(partuparty

)2

= micro2 (636)



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



)2

+

(partuparty

)2

= 1 (637)




dxdτ

= pdydτ

= qdpdτ

= 0dqdτ

= 0dudτ

= p2 +q2 = 1


p = p0(s) q = q0(s)




dx0

dsp0 +

dy0

dsq0 = 0 p2

0 +q20 = 1












u= τ

(638)

















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1
















andψxx =minusκ



minusκ2A(u2


2A = 0


u2x +u2

y = 1






Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


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Examples




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PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

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Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1


Examples






The part







J

(ξ η

xy

)=


ξy ηy

∣∣∣∣ 6= 0infin








and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1





and


2y



2y


2βuξ η +Φ = 0


a(

ξx

ξy

)2

+2b(

ξx

ξy

)+ c = 0

and

a(

ηx

ηy

)2

+2b(

ηx

ηy

)+ c = 0


ap2 +2bp+ c = 0




αγminusβ2 = J

(ξ η

xy

)2

(acminusb2)



ξx

ξy=minusbplusmn

radicb2minusac

a

ηx

ηy=minusbplusmn

radicb2minusac

a





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1





dxis

dξ

dx=

partξ

partx+

partξ

partydydx

= 0

solve to getdydx

=minusξx

ξy


dydx

=bplusmnradic

b2minusaca






u = F(s) ux = G(s) uy = H(s)

Thendux


duy




uxx

uxy

uyy

=

minusFHs

Gs


a(

dydx

)2

minus2b(

dydx

)+ c = 0




dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



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vectors


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examples


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Examples



Polar coordinates





fdrm0

fdrm1



dydx

=bminusradic

b2minusaca


dydx

=b+radic

b2minusaca



uξ η +1






dydx



uξ ξ +1α

Φ = 0


J

(ξ η

xy

)6= 0


dydx

=b+ i

radic|b2minusac|a







uξ ξ +uηη +1α

Φ = 0







dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1






dydx






xy

)∣∣∣∣= ∣∣∣∣ξx ηx

ξy ηy

∣∣∣∣= ∣∣∣∣minus1 01 1

∣∣∣∣=minus1 6= 0





dydx

=1plusmnradicminus4

1= 1plusmn2i



η = 2x


ξy ηy

∣∣∣∣= ∣∣∣∣minus1 21 0

∣∣∣∣ 6= 0







dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



dydx


1=plusmnix

y =plusmn ix2

2+C or yplusmn ix2

2=C






uξ ξ +uηη =1










bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



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examples


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Examples



Polar coordinates





fdrm0

fdrm1









bcs u(0 t) = 0u(` t) = 0 t ge 0



u(x t) = X(x)T (t)

ThenX primeprimeT =

1c2 XT primeprime

ieX primeprime

X=

1c2

T primeprime

T= constant λ




X = Aeradic

λx +Beminusradic

λx



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

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Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





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Nonlinear waves

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The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



A+B = 0 Aeradic

λ`+Beminusradic

λ` = 0


λ =minusp2







A = 0 Bsin(pl) = 0


pl = nπ n = 12

thus (C cos

(nπct`

)+Dsin

(nπct`

))Bsin

(nπx`


un =(

Cn cos(nπct

`

)+Dn sin

(nπct`

))sin(nπx

`


u =infin

sumn=1

(Cn cos

(nπct`

)+Dn sin

(nπct`

))sin(nπx

`


u(x0) =infin

sumn=1

Cn sin(nπx

`

)=U(x)

ut(x0) =infin

sumn=1

Dnnπc`

sin(nπx

`

)=V (x)


Cn =2`

int `

0U(xprime)sin

(nπxprime

`

)dxprime

nπc`

Dn =2`

int `

0V (xprime)sin

(nπxprime

`

)dxprime


u(x t)=infin

sumn=1

12

Cn

sin

nπ

`(x+ ct)+ sin

nπ

`(xminus ct)

+

infin

sumn=1

12

Dn

cos

nπ


nπ

`(x+ ct)



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



u(x t) = X(x)T (t)


1k

T prime

T=

X primeprime

X= const =minusω

2



2t)



utt = c2nabla

2u = c2(uxx +uyy)



T primeprime

c2T=minusω

2 =X primeprime

X+

Y primeprime

Y


andX primeprime

X=minusY primeprime

Yminusω

2


X=minusΩ


andY primeprime

Y= ω


2)Y = 0



utt = c2nabla

2u = c2(

1r

part

part r

(r

partupart r

)+

1r2

part 2upartθ 2

)utt = c2

nabla2u = c2

(urr +

ur

r+

uθθ

r2


T primeprime

T=minusω

2c2



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



R+

1r

Rprime

R+

1r2

Θprimeprime

Θ=minusω

2

or

r2 Rprimeprime

R+ r

Rprime

R+

Θprimeprime

Θ=minusω

2r2

r2 Rprimeprime

R+ r

Rprime

R+ω


Θ= Ω

2


Θ=minusΩ

2


r2 Rprimeprime

R+ r

Rprime

R+ω

2r2minusΩ2 = 0




nabla2φ =

part 2φ

partx2 +part 2φ

party2 +part 2φ

part z2 = 0




X+

Y primeprime

Y+

Zprimeprime

Z= 0

orX primeprime

X+

Y primeprime

Y=minusZprimeprime


X primeprime

X+

Y primeprime

Y=minusZprimeprime

Z= const = γ

2 (say)

ThenZprimeprime+ γ

2Z = 0

andX primeprime

Xminus γ



X primeprime

Xminus γ


Y= const = β

2 (say)




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




2)X = 0

orX primeprime+α

2X = 0

whereα

2 +β2 + γ

2 = 0



Here we have




φ = 0 for z = 0c y = 0b x = 0




Zn = An sinnπz

c (γ = γn

nπ

cn = 12 )

Ym = Bm sinmπy

b (β = βm

mπ

bm = 12 )


2X = 0


X(0) = 0rarr D = 0

ThenAnBmC sin

nπzc

sinmπy

bsinhλx



φ =infin

summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmnx


summ=1

infin

sumn=1

Cmn sinmπy

bsin

nπzc

sinhλmna = f (yz)






∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




∆ =part 2

part r2 +2r

part

part r+

1r2 sinθ

part

partθsinθ

part

partθ+

1r2 sin2

θ

part 2

partφ 2


urr +2r

ur +1


1r2 sin2

θuφφ = 0


Rprimeprime

R+

2r

Rprime

R+

1r2 sinθ


Θ+

1r2 sin2

θ

Φprimeprime

Φ= 0


Φprimeprime

Φ=minusm2

This gives



r2 Rprimeprime

R+2r

Rprime

R=minus 1

sinθ


Θ+

m2

sin2θ= λ (say)


d2Rdr2 +

2r

dRdrminus λ


and

1sinθ

ddθ

(sinθ

dΘ

dθ

)+

(λ minus m2

sin2θ

)Θ = 0 (82)


ddθ

=minussinθd

dmicro


ddmicro

((1minusmicro

2)dΘ

dmicro

)+

(λ minus m2

1minusmicro2

)Θ = 0 (83)




d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



d2Θ

dmicro2 minus2micro

1minusmicro2dΘ

dmicro+

(λ


(1minusmicro2)2

)Θ = 0



d2Rdr2 +

2r

dRdrminus n(n+1)

r2 R = 0 (84)



Pn(x) =1

2nndn



P0(x) = 1 P1(x) = x P2(x) =3x2minus1

2 P3(x) =

5x3minus3x2





(1minus x2)w(n+2)+

(n+1

n


(n+1nminus1


+2n(

xw(n+1)+

(n+1

n

)w(n)

)= 0







2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1



2nnPn(x) =dn


= (x+1)n dn

dxn [(xminus1)n]+

(n

nminus1





Pn(minusx) =1

2nndn


dn


as required





n(n+1)minus m2

1minus x2

)y = 0 (86)





Proof We have that





that is


)u(m) = 0 (lowast)




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1




((1minus x2)yprime

)=

ddx

((1minus x2)

ddx

((1minus x2)m2 u(m)

))=

ddx





(1minus mx2

1minus x2

)u(m)


= (1minus x2)m2u(m)


m2x2

1minus x2

)=minus

(n(n+1)minus m2

1minus x2

)y





dxm




Partial Derivatives



Partial derivatives

vectors


Introduction




General remarks

Examples




The wave equation

PDEs vs ODEs


Solution methods

Solution properties





Special Tricks


The truncated PDE



examples


Linear waves





The Cauchy Problem

Nonlinear waves

Shocks

Traffic Flow


The quadratic model


Introduction

Charpits equations

Boundary data

Examples

Sand Piles






Canonical forms

Examples



Polar coordinates





fdrm0

fdrm1

mas3111: partial differential equations with...

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