chapter 13 partial differential equations

1

Chapter 13 Partial differential equations

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

Lecture 13 Laplace, diffusion, and wave equations

2

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

00,, 2

2

2

2

2

22

uz

uy

ux

zyxu

ex 2) Poisson’s equation:

ex 3) Diffusion or heat flow equation

zyxfu ,,2

y)diffusivit :( ,,,1

,,, 22

tzyxu

ttzyxu

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

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

3

ex 4) Wave equation 2

2

22 1

t

u

vu

ex 5) Helmholtz equation 022 FkF

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

4

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

In case of no heat source

0or 02

2

2

22

y

T

x

TT

s" variableof Separation"

. ofonly function a :)(

, ofonly function a :)( ,,

form, theofsolution a try weequation, thissolve To

yyY

xxXyYxXyxT

5

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

11 ,0

11 0

00

dy

Yd

Ydx

Xd

Xdy

Yd

Ydx

Xd

Xdy

YdX

dx

XdY

y

XY

x

XY

y

T

x

T

XY

kx

kx

kx

kx

XYTYkx

kxX

YkYXkX

kkconstdy

Yd

Ydx

Xd

X

ky

ky

ky

ky

ky

ky

cose

cose

sine

sine

,e

,e ,cos

,sin

. and

)0constant separation(.11

sides, separated- variablewith theequation above hesatisfy t To

22

22

2

2

2

6

kx

kx

kx

kx

XYT

ky

ky

ky

ky

cose

cose

sine

sine

:solutionGeneral

.0when10)

10/010sin .10 when 0)

cos discard .0 when 0 )

e discard . as 0)

yTiv

nkkxTiii

kxxTii

yTi ky

1) In the current problem, boundary conditions are

1

.10010

sin10

sin]10/exp[n

n

xnbT

xnynT

7

1.26 ,5 and 5 .

.10

3sin

3

1

10sin]10/exp[

400

.even ,0

, odd ,400

11200

10cos

1020

10sin100

10

2sin)(

2

series,Fourier theUsing

.10010

sin10

sin]10/exp[

10/3

10

0

10

00

1

TyxAtex

xe

xynT

n

nn

n

xn

ndx

xndx

l

xnxf

lb

xnbT

xnynT

y

n

l

n

nn

8

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

n odd

110

1

302130

2130

2130

21

10sin30

10sinh

3sinh

1400

.3sinh where,10

sin10

sin3sinh100

10sin30

10sinh

30sinh), is,that (30at 0

xny

n

nnT

nBbxn

bxn

nBT

xny

nBT

ykebeaeeyT

beaee

nnnn

nny

nn

kkykyk

kykyky

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

T=0 at 30 cm

9

.for ,

cose

cose

sine

sine

solution, generalanother have could We 2k

ky

ky

ky

ky

XYT

kx

kx

kx

kx

- 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.)

10

-. 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)

11

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

tkeTTkdt

dTFkF

kdt

dT

TkF

F

kdt

dT

TF

Fdt

dTFFT

tTzyxFu

t

uu

22

,

11 ,

1

.1111

s" variableof Separation" ,,

.1

: flowHeat -

2222

22

22

22

22

2

22

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

12

kxe

kxeu

kx

kxFFk

dx

FdkF

F

tk

tk

cos

sin

cos

,sin0

1

problem, D-1 thisof caseIn

22

22

22

222

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:

xl

ubaxu

dx

udu

100

0source)heat (no 0

00

20

2

02

13

nklkllxu

kxxu

0sin . when 0)2

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

.3

sin3

12sin

2

1sin

200

12001

12100

.100

sin

,0At

sinsin

222

22

/3/2/

11

01

0

/

1

/

l

xe

l

xe

l

xeu

nn

l

lb

xl

ul

xnbu

uut

l

xnebu

l

xneu

tltltl

nn

n

nn

tln

nn

tln

14

.sinsin

.sin

10

10

1

/ 2

nnff

nn

fn

tlnn

l

xnbuuu

l

xnbu

ul

xnebu

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.

15

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

1

1-

22222

2

22

2

2

2

22

2

vvk

TvkTXkXkconstdt

Td

Tvdx

Xd

X

tTxXy

t

y

vx

y

x=0 x=l

node

16

.cossin

sinsin

.,0at 0 :conditionBoundary

. where,

coscos

sincos

cossin

sinsin

,cos

,sin ,cos

,sin

l

vtn

l

xn

l

vtn

l

xn

y

lxxy

kv

tkx

tkx

tkx

tkx

XTykvt

kvtT

kx

kxX

17

l

vt

l

x

l

vt

l

xhy

xfl

xnby

l

vtn

l

xnby

lvtnyxfty

nn

nb

3cos

3sincossin

8

ts,coefficien seriesFourier thefindingAfter

.sin

.cossin

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

91

2

10

1

1) case 1

l

vtn

l

xn

l

vtn

l

xn

y

l

vtn

l

xn

l

vtn

l

xn

y

sinsin

cossin

cossin

sinsin

18

2) case 2

)expansion. seriesFourier (Use sinsin

.sinsin

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

110

1

0

xVl

xnb

l

xn

l

vnB

t

y

l

vtn

l

xnBy

lvtnyty

nn

nn

t

nn

.cossin

sinsin

l

vtn

l

xn

l

vtn

l

xn

y

19

3) Eigenfunctions

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

frequency sticcharacteri :2/

ioneigenfunctor function sticcharacteri a :sinsin

nn lvnl

vtn

l

xny

first harmonic, fundamentalsecond harmonic

third fourth

20



Lecture 14 Using Bessel equation

21

5. Steady-state temperature in a cylinder

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

011111

, with dividing and equation, theinto thisPutting

011

2

2

2

2

2

2

2

2

2

22

dz

Zd

Zd

d

rdr

dRr

dr

d

rR

ZRu

zZrRu

z

uu

rr

ur

rru

22

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

.,1 2

2

2

kz

kz

kz

euzu

e

eZk

dz

Zd

Z

011111

2

2

2

2

2

dz

Zd

Zd

d

rdr

dRr

dr

d

rR

22

2

2

2

2

11111k

dz

Zd

Zd

d

rdr

dRr

dr

d

rR

- 1st step

23

.cos

,sin,

1

Here,

01

01111

22

2

222

22

2

2

2

2

n

nn

d

d

krd

d

dr

dRr

dr

d

R

rk

d

d

rdr

dRr

dr

d

rRr

- 2nd step

. of zero a is where,cos

sin

.0 :conditionBoundary -

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

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

0: equation" Bessel"

0 00

1

22222

2222222222

nkrn

krn

nr

nn

nn

JkenkrJ

enkrJZRu

kJR

rkrNkrJR

krNkrJ

ypxyxyxypxyxx

RnrkRrRrRnrkdr

dRr

dr

drrkn

dr

dRr

dr

d

R

r

- 3rd step

24

.100

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

100

function, Bessel theofity orthogonal theusing Here,

.100 :conditionBoundary

1

0 0

1

0

20

1

0 01

0

1

0 0

100

drrkrJdrrkJrc

m

drrkrJdrrkJcrkrJ

rkJcu

mmm

nmm

nmmz

.

function) goscillatin:( 00 symmetry, azimuthalIn

10

0

n

zkmm

nm

merkJcu

JkJn

baaJaJaJ

badxbxJaxxJ

bJaJ

ppp

pp

pp

if ,

if ,0

)(0)(

212

1212

1212

1

1

0

.:left term 212

11

0

20 mm kJdrrkJr

25

mmmm

mm

m

mmmmmm

mm

mm

m

mm

m

mmmmm

m

kJkkJk

kJc

k

kJkJcdrrkrJdrrkJrc

kJk

rkrJk

drrkrJ

rkrJdr

d

krkrJ

rkrJkrkrJkdr

d

krkx

xxJxxJdx

d

12

1

1

1212

11

0 0

1

0

20

1

1

0

1

1

0 0

10

01

01

2002100

100100

results, twoCombining

.11

1

,1

,For

relation) (recusion :right term For the

.100 1

0 0

1

0

20 drrkrJdrrkJrc mmm

26

.sin,2

Similarly,

.

coscos,

ity,orthogonal theof Because

.,sincos

.sincos

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

1

0

2

021

212

1

1

0

2

0

21

0

2

0

1 00

1 0

2

rdrdrkJrfkJ

b

kJa

rdrdrkJardrdrkJrf

rfnbnarkJu

enbnarkJu

nbnarf

m nmnmnmnnz

m n

zkmnmnmnn

nn

mn

27

- 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

28

xBNxAJy

p

xJxJpxN

ppdxexpx

pnnxJ

ypxyxyxypxyxx

pp

ppp

n

xppnn

p

solution general

or Weber)(Neumann sin

cos

.0 ,!1 where,211

1

:Solution -

0

00

12

22222

- Graph


29

2) Recursion relations

xJxJx

pxJxJ

x

pxJ

xJxJxJ

xJx

pxJxJ

xJxxJxdx

d

xJxxJxdx

d

ppppp

ppp

ppp

pp

pp

pp

pp

11

11

11

1

1

)5

2 )4

2 )3

)2

)1


30

3) Orthogonality

0 ,For 0 ,sinFor

,0,0,10sin,1

0,,,0sin,

of valueonefor just consider sinjust consider

,cos,sin

2222

ypxayxxaxJynyxny

bxJaxJxxnx

xJbaxxnxx

pxJx

xNxJxx

p

pp

p

p

pp

badxbxJaxxJmnxdxmxn pp for ,0 ,for ,0sinsin1

0

1

0

cf. Comparison


31

baaJaJaJ

badxbxJaxxJ

bJaJ

ppppp

pp

if ,

if ,0

)(0)(

212

1212

1212

1

1

0

0

0

222

22222

ypxayxx

ypxyxyxypxyxx


32

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

kvtn

kvtn

kvtn

kvtn

krJkvt

kvt

n

nkrJtTrRz

Rrkndr

dRr

dr

dr

nd

dn

d

d

kd

d

rdr

dRr

dr

d

Rr

FrRF

FkF

rr

Fr

rr

TvkTFkFtTyxFz

t

z

vz

nn

coscos

sincos

cossin

sinsin

cos

sin

cos

sin

equation) s(Bessel' 0

equation) harmonic (simple 01

01111

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

.011

,coordinatepolar In

0 and 0,

1

222

22

22

2

2

22

2

2

22

2

2

2222

2

2

22

33

2/ :modes normal theof freuqency sticCharacteri

series) (double of zeros :

.00)1( :conditionBoundary

coscos

sincos

cossin

sinsin

vk

Jk

kJrz

kvtn

kvtn

kvtn

kvtn

krJtTrRz

mnmn

nmn

n

n

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.

34

vtknrkJzex mnmnn coscos)

35

American Journal of Physics, 35, 1029 (1967)

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

36



Lecture 15 Using Legendre equation

37

7. Steady-state temperature in a sphere

.)polynomial (Legendre cos

)1equation Legendre d(Associate 0sin

sinsin

1

/1

1

.0sin

sinsin

111

cos

sin,

1

0sin

11sin

sin

111 ,For

0sin

1sin

sin

11

2

2

1

2

2

22

22

2

2

2

22

2

2

2222

22

ml

l

l

P

llkkm

d

d

d

d

r

rRk

dr

dRr

dr

d

R

m

d

d

d

d

dr

dRr

dr

d

R

m

mm

d

d

d

d

d

d

d

d

dr

dRr

dr

d

RrRu

u

r

u

rr

ur

rru

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

38

coscoscoscos100

.,16

7100 ,

4

3100 ,

2

1100

coscoscoscos100cos

)cos (Here,

lspolynomina Legendre of series ain thisexpandcan wecase, In this

.10 ,1

,01 ,0 where,100cosor

2/ ,0

2/0 ,100cos 3)

cos0 symmetry, azimuthal the toDue 2)

55

3211

33

167

143

021

210

53211

3167

143

021

0

53211

3167

143

021

01

01

0

PrPrrPPxu

ccc

PPPPPc

xxPxPxPxPxf

x

xxfxfPcu

Pcu

Prcum

lll

lllr

lllr

ll

ll

m

mPru

r

ml

l

l

cos

sincos

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

39

- Legendre’s equation

0 ,For

. 1 :Solution

01

121

equation) Legendre d(Associate 0 ii.

.1!!2

1 :Solution0121

equation) (Legendre 0. i.

2/2

2

22

22

ml

l

mmm

l

ll

l

Plm

lmlxPdx

dxxP

yx

mllyxyx

m

xdx

d

lxPyllyxyx

m

1) Equation and solution

.equation) Legendre d(Associate 01sin

sinsin

1.

2

2

llm

d

d

d

dcf

Legendre’s equation 1

40

011

21

01cos1cos

cos2cos

cos1

01cos1cos

cos1cos

01sincos

sincos

01sin

cos

cossin

cos

cossin

1

equation) Legendre d(Associate 01sin

sinsin

1.

2

2

2

22

2

2

2

22

2

22

2

22

2

2

2

2

llx

m

dx

dx

dx

dx

llm

d

d

xd

d

llm

d

d

xd

d

llm

d

d

xd

d

llm

d

d

d

d

d

d

xd

d

llm

d

d

d

dcf


41

xPxPxcf

xxxxP

xxxxP

xxxP

xxxP

xxP

xxP

xP

313

2466

355

244

33

22

1

0

235

1 .

510531523116

1

1570638

1

330358

1

352

1

132

1

1

- Legendre polynomials - Associated Legendre polynomials

xP

ml

mlxP m

lmm

l !

!1

222

2/1212

202

12

12

22

22

2/1211

01

11

11

13

13

132

1

6

1

24

1

1

2

1

xxP

xxxP

xxP

xPxP

xPxP

xxP

xxP

xPxP


42

2) Orthogonality

ddxxcf

ml

ml

ldPdxxP

ldPdxxP

ml

ml

ll

sincos.

.!

!

12

2sincos

.12

2sincos

0

21

1

2

0

21

1

2


43

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

2

2

2

2

V

V

V

GVG

V

E

F

FFF

equationPoisson ofsolution :

equation) (Laplace 0

.,,

4

1

,,,, general,In

222

2

222

2

wufwuwu

w

zdydxdzzyyxx

zyxfu

zyxfzyxu

44

Example 1

grounded sphere

ll

llq

l

llm

ll

l

q

PrcVV

mm

r

Prm

mP

r

r

aarr

qV

azyx

qzdydxd

zzyyxx

zyxzyxV

V

cos

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

discard infiniteat zero 1)

coscos

sincosequation Laplace ofsolution Basic

.cos2

,coordinate sphericalIn

.,,4

4

1,,

4

1

1

1

22

222222

2

45

aRqaR

aRraRr

qaR

aarr

q

r

PaRaRq

aarr

q

a

PrRq

aarr

qV

a

qRc

a

qRRc

PRca

PRq

a

PRq

aaRR

q

PRcaaRR

qV

RrV

lll

l

ll

lll

l

l

ll

ll

l

ll

ll

ll

ll

ll

ll

ll

llRr

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

.cos/2/

/

cos2

cos/)/(

cos2

cos

cos2

.or relation, thisFrom

coscos

,cos

cos2 Using

.0coscos2

.for 0 :conditionBoundary

2

222222

1

2

22

1

112

22

1

12

11

11

122

1

22

‘Method of the images’

46

monopole dipole quadrupole octopole

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

cos21cos2

1

4

1

22222

0

r

r

r

rrrrrr

rdV

rr

rrr

r

322/1

16

5

8

3

2

11

11

11

cos2 where,1

rr

r

r

r

rr

rr

rr

cf. Electric multipoles

47

3322

cos216

5cos2

8

3cos2

2

11

11 r

r

r

r

r

r

r

r

r

r

r

r

rrr

expansion multipole cos1

4

1

01

0

n

nn

ndPr

rV

rr

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

lpolynomina Legendre cos1

2/cos3cos52/1cos3cos11

0

33

22

nn

n

Pr

r

r

r

r

r

r

r

r

r

chapter 13 partial differential equations

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