chapter 13 partial differential equations

Chapter 13 Partial differential equations

Mathematical methods in the physical sciences 3nd edition Mary L. Boas

Lecture 13 Laplace, diffusion, and wave equations

1. Introduction (partial differential equation)ex 1) Laplace equation

00,, 2

ex 2) Poisson’s equation:

ex 3) Diffusion or heat flow equation

zyxfu ,,2

y)diffusivit :( ,,,1

,,, 22

ttzyxu

: gravitational potential, electrostatic potential, steady-state temperature with no source

: with sources (=f(x,y,z))

ex 4) Wave equation 2

ex 5) Helmholtz equation 022 FkF

: space part of the solution of either the diffusion or the wave equation

2. Laplace’s equation: steady-state temperature in a rectangular plate (2D)

In case of no heat source

0or 02

s" variableof Separation"

. ofonly function a :)(

, ofonly function a :)( ,,

form, theofsolution a try weequation, thissolve To

xxXyYxXyxT

XYTYkx

YkYXkX

kkconstdy

,e ,cos

)0constant separation(.11

sides, separated- variablewith theequation above hesatisfy t To

:solutionGeneral

.0when10)

10/010sin .10 when 0)

cos discard .0 when 0 )

e discard . as 0)

nkkxTiii

kxxTii

yTi ky

1) In the current problem, boundary conditions are

.10010

sin]10/exp[n

1.26 ,5 and 5 .

10sin]10/exp[

.even ,0

, odd ,400

10sin100

2sin)(

series,Fourier theUsing

.10010

sin]10/exp[

TyxAtex

2) How about changing the boundary condition? Let us consider a finite plate of height 30 cm with the top edge at T=0.

302130

10sin30

10sinh

.3sinh where,10

sin3sinh100

10sin30

10sinh

30sinh), is,that (30at 0

ykebeaeeyT

kkykyk

kykyky

In this case, e^ky can not be discarded.

T=0 at 30 cm

.for ,

solution, generalanother have could We 2k

- To be considered I

- To be considered II

In case that the two adjacent sides are held at 100 (ex. C=D=100), the solution can be the combination of C=100 solutions (A, B, D: 0) and D=100 (A, B, C: 0) solutions.

This is correct, but makes the problem more complicated.(Please check the boundary condition.)

-. Summary of separation of variables.

1) A solution is a product of functions of the independent variables. 2) Separate partial equation into several independent ordinary equation. 3) Solve the ordinary differential eq. 4) Linear combination of these basic solutions 5) Boundary condition (boundary value problem)

3. Diffusion or heat flow equation; heat flow in a bar or slab

tkeTTkdt

tTzyxFu

s" variableof Separation" ,,

: flowHeat -

cf. “Why do we need to choose –k^2, not +k^2?”

problem, D-1 thisof caseIn

Let’s take a look at one example.

At t=0, T=0 for x=0 and T=100 x=l.From t=0 on, T=0 for x=l.

For T(x=0)=0 and T(x=l)=100 at t=0, the initial steady-state temperature distribution:

0source)heat (no 0

nklkllxu

0sin . when 0)2

cos discard .0 when 0 )1- Using Boundary condition

sinsin

tltltl

.sinsin

xnbuuu

For some variation, when T0, we need to consider uf.as the final state, maybe a linear function.

In this case, we can write down the solution simply like this.

4. Wave equation; vibrating string

Under the assumption that the string is not stretched,

number. wave22

wavelength

(radians)frequency angular 2 ),(secfrequency

.0 and 0.111

TvkTXkXkconstdt

x=0 x=l

.cossin

sinsin

.,0at 0 :conditionBoundary

. where,

coscos

sincos

cossin

sinsin

,sin ,cos

XTykvt

lvtnyxfty

3sincossin

ts,coefficien seriesFourier thefindingAfter

.cossin

discarded. be should /sin ,0' and )()0(For

1) case 1

sinsin

cossin

sinsin

2) case 2

)expansion. seriesFourier (Use sinsin

.sinsin

discarded. be should /cos ,0' and 0)0(For

lvtnyty

.cossin

sinsin

3) Eigenfunctions

. vibrationof mode normal thecalled is freuquency pure an with A vibratio

frequency sticcharacteri :2/

ioneigenfunctor function sticcharacteri a :sinsin

nn lvnl

first harmonic, fundamentalsecond harmonic

third fourth

Lecture 14 Using Bessel equation

5. Steady-state temperature in a cylinder

For this problem, cylindrical coordinate (r, , z) is more useful.

011111

, with dividing and equation, theinto thisPutting

In order to say that a term is constant, 1) function of only one variable 2) variable does not elsewhere in the equation.

.~ ,at 0~ Because

011111

11111k

- 1st step

- 2nd step

. of zero a is where,cos

.0 :conditionBoundary -

.0at because ),(~only take weHere,

.or Weber)(Neumann (Bessel), : equations Bessel theofSolution

0: equation" Bessel"

2222222222

JkenkrJ

enkrJZRu

rkrNkrJR

krNkrJ

ypxyxyxypxyxx

RnrkRrRrRnrkdr

- 3rd step

side.left in the termzero-non thehave we, ifOnly

function, Bessel theofity orthogonal theusing Here,

.100 :conditionBoundary

drrkrJdrrkJrc

drrkrJdrrkJcrkrJ

function) goscillatin:( 00 symmetry, azimuthalIn

merkJcu

baaJaJaJ

badxbxJaxxJ

.:left term 212

20 mm kJdrrkJr

mmmmmm

kJkkJk

kJkJcdrrkrJdrrkJrc

drrkrJ

rkrJdr

rkrJkrkrJkdr

xxJxxJdx

2002100

100100

results, twoCombining

relation) (recusion :right term For the

.100 1

20 drrkrJdrrkJrc mmm

.sin,2

Similarly,

coscos,

ity,orthogonal theof Because

.,sincos

.sincos

.sincosbut ,1 ,, iscylinder theof base theof eraturegiven temp theIf

rdrdrkJrfkJ

rdrdrkJardrdrkJrf

rfnbnarkJu

enbnarkJu

nbnarf

m nmnmnmnnz

zkmnmnmnn

- Bessel’s equation

1) Equation and solution

- named equation which have been studied extensively.- “Bessel function”: solution of a special differential equation.

- being something like damped sines and cosines.- many applications. ex) problems involving cylindrical symmetry (cf. cylinder function); motion of pendulum whose length increases steadily; small oscillations of a flexible chain; railway transition curves; stability of a vertical wire or beam; Fresnel integral in optics; current distribution in a conductor; Fourier series for the arc of a circle.

022222 ypxyxyxypxyxx

Bessel’s equation 1

xBNxAJy

xJxJpxN

ppdxexpx

ypxyxyxypxyxx

solution general

or Weber)(Neumann sin

.0 ,!1 where,211

:Solution -

- Graph

2) Recursion relations

xJxJxJ

xJxxJxdx

3) Orthogonality

0 ,For 0 ,sinFor

,0,0,10sin,1

0,,,0sin,

of valueonefor just consider sinjust consider

,cos,sin

ypxayxxaxJynyxny

bxJaxJxxnx

xJbaxxnxx

xNxJxx

badxbxJaxxJmnxdxmxn pp for ,0 ,for ,0sinsin1

cf. Comparison

baaJaJaJ

badxbxJaxxJ

ypxayxx

ypxyxyxypxyxx

6. Vibration of a circular membrane (just like drum)

krJkvt

nkrJtTrRz

Rrkndr

TvkTFkFtTyxFz

coscos

sincos

cossin

sinsin

equation) s(Bessel' 0

equation) harmonic (simple 01

, with dividing then and equation, above theinto form thisPutting .

,coordinatepolar In

0 and 0,

2/ :modes normal theof freuqency sticCharacteri

series) (double of zeros :

.00)1( :conditionBoundary

coscos

sincos

cossin

sinsin

krJtTrRz

cf. They are not integral multiples of the fundamental as is true for the string (characteristics of the bessel function). This is why a drum is less musical than a violin.

vtknrkJzex mnmnn coscos)

American Journal of Physics, 35, 1029 (1967)

(m,n)=(1,0) (2,0) (1,1)

Lecture 15 Using Legendre equation

7. Steady-state temperature in a sphere

.)polynomial (Legendre cos

)1equation Legendre d(Associate 0sin

sinsin

111 ,For

- Sphere of radius 1 where the surface of upper half is 100, the other is 0 degree.

coscoscoscos100

7100 ,

3100 ,

coscoscoscos100cos

)cos (Here,

lspolynomina Legendre of series ain thisexpandcan wecase, In this

.10 ,1

,01 ,0 where,100cosor

2/0 ,100cos 3)

cos0 symmetry, azimuthal the toDue 2)

PrPrrPPxu

PPPPPc

xxPxPxPxPxf

xxfxfPcu

sincos

./1 discard Therefore, .divergence no sphere, theofinterior For the 1) 1

- Legendre’s equation

0 ,For

. 1 :Solution

equation) Legendre d(Associate 0 ii.

1 :Solution0121

equation) (Legendre 0. i.

lmlxPdx

mllyxyx

lxPyllyxyx

1) Equation and solution

.equation) Legendre d(Associate 01sin

sinsin

Legendre’s equation 1

01cos1cos

cos2cos

01cos1cos

cos1cos

01sincos

sincos

cossin

equation) Legendre d(Associate 01sin

sinsin

xPxPxcf

510531523116

1570638

330358

- Legendre polynomials - Associated Legendre polynomials

mlxP m

2/1212

2/1211

2) Orthogonality

ddxxcf

ldPdxxP

sincos.

2sincos

8. Poisson’s equation

00 cf.

.unit)Gaussian (in 44 force, ticelectrostaFor 2)

equation Laplace :00 cf.

equationPoisson :44 force, nalgravitatioFor 1)

ve)conservati :( 0

equationPoisson ofsolution :

equation) (Laplace 0

,,,, general,In

wufwuwu

zdydxdzzyyxx

zyxfzyxu

Example 1

grounded sphere

qzdydxd

zzyyxx

zyxzyxV

1cos,0 . of indep. issolution axis-zabout symmetric 2)

discard infiniteat zero 1)

coscos

sincosequation Laplace ofsolution Basic

,coordinate sphericalIn

222222

aRraRr

PaRaRq

PRcaaRR

/0,0, :position ,/:charge term,second In the

.cos/2/

cos/)/(

.or relation, thisFrom

coscos

cos2 Using

.0coscos2

.for 0 :conditionBoundary

222222

‘Method of the images’

monopole dipole quadrupole octopole

2) Expansion for the potential of an arbitrary localized charge distribution

cos21cos2

rrrrrr

cos2 where,1

cf. Electric multipoles

cos216

expansion multipole cos1

- n = 0 : monopole contribution- n = 1 : dipole- n = 2 : quadrupole- n = 3 : octopole

lpolynomina Legendre cos1

2/cos3cos52/1cos3cos11

chapter 13 partial differential equations

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