Boundary-Value Problems in Other Coordinates

CHAPTER 14

Ch14_2

Contents

14.1 Problems in Polar Coordinates14.2 Problems in Cylindrical Coordinates14.3 Problems in Spherical Coordinates

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14.1 Problems in Polar Coordinates

Laplacian in the Polar CoordinatesWe already know that

urr

uy

uyr

ru

yu

urr

ux

uxr

ru

xu

xy

yxrryrx

cossin

sincos

Besides,

tan , ,sin ,cos 222

Ch14_4

Thus

(1)

(2)

urr

ur

u

r

xu

rr

u

x

u

cossin2sinsin

cossin2cos

2

2

2

2

2

2

22

2

2

urr

ur

u

rxu

rr

u

x

u

cossin2sin

sincossin2cos

2

2

2

2

2

2

22

2

2

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Adding (1) and (2) we have

(3) 011

concern only wesection, In this

11

2

2

22

2

2

2

22

22

u

rru

rr

u

u

rru

rr

uu

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Example 1

Solve Laplace’s Equation (3) subject to u(c,) = f(), 0 < < 2.

Solution Since (r, + 2) is equivalent to (r, ), we must have u(r, ) = u(r, + 2). If we seek a product function u = R(r)(), then (r, + 2) = (r, ).

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Example 1 (2)

Introducing the separation constant , we have

We are seeking a solution of the form

(6)

(5) 0"

(4) 0'"

"'"

2

2

RrRRr

RrRRr

)2()0( ,0

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Example 1 (3)

Of the three possible general solutions of (5):(7)

(8)

(9)

we can dismiss (8) as an inherently non-periodic unless c1 = c2 = 0. Similarly (7) is non-periodic unless c2 = 0. The solution = c1 0 can be assigned any period and so = 0 is an eigenvalue.

0 hsinhcos 221 cc

0 )( 21 cc

0 sincos 221 cc

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Example 1 (4)

When we take = n, n = 1, 2, …, (9) is 2-periodic. The eigenvalues of (6) are then 0 = 0 and n = n2, n = 1, 2, …. If we correspond 0 = 0 with n = 0, the eigenfunctions are

When n = n2, n = 0, 1, 2, … the solutions of (4) are

,...2,1,sincos)(

;0,)(

21

1

nncnc

nc

(11) ,...2,1,)(

(10) 0,ln)(

43

43

nrcrcrR

nrccrRnn

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Example 1 (5)

Note we should define c4 = 0 to guarantee that the solution is bounded at he center of the plate (r = 0). Finally we have

(12) )sincos(),(

gives principleion superposit The

,...2,1,)sincos(

0,

10

00

nnn

n

nnn

n

BnArAru

nnBnAru

nAu

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Example 1 (6)

Applying the boundary condition at r = c, we get

(13) )( is,that

,,2

series,Fourier full ain ofexpansion an as

)sincos()(

2

00

00

10

dfA

bBcaAca

A

f

BnAcAf

nnn

nnn

nnn

n

(15) sin)(1

(14) cos)(1

2

0

2

0

dnfc

B

dnfc

A

nn

nn

Ch14_12

Example 2

Find the steady-state temperature u(r, ) shown in Fig 14.3.

Ch14_13

Example 2 (2)

Solution The boundary-value problem is

crruru

ucu

cru

rru

rr

u

0,0) ,( ,0)0 ,(

0 ,) ,(

0 ,0 ,011

0

2

2

22

2

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Example 2 (3)

and (16)

(17)

The boundary conditions translate into (0) = 0 and () = 0.

"'"

constant, separation and )()( Defining2

RrRRr

rRu

022 RRrRr

02

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Example 2 (4)

Together with (17) we have

(18)

The familiar problem possesses n = n2 and eigenfunctions () = c2 sin n, n = 1, 2, … Similarly, R(r) = c3rn and

un = R(r)() = An rn sin n

0)( 0,)( 0,

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Example 2 (5)

Thus we have

1

0

0

0 0

10

1

sin)1(12

),(

)1(12 ,sin

2

sin ,sin),(

n

nn

n

nnn

n

n

nn

n

nn

ncr

nu

ru

nc

uAdnucA

cAunrAru

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14.2 Problems in Polar Coordinates and Cylindrical Coordinates: Bessel Functions

Radial Symmetry The two-dimensional heat and wave equations expressed in polar coordinated are, in turn

(1)

where u = u(r, , t). The product solution is defined as u = R(r)()T(t). Here we consider a simpler problems that possesses radial symmetry, that is, u is independent of .

and 11

2

2

22

2

tuu

rru

rr

uk

2

2

2

2

22

22 11

t

uu

rru

rr

ua

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In this case, (1) take the forms, in turn,

(2)

where u = u(r, t).

and 1

2

2

tu

ru

rr

uk

2

2

2

22 1

t

uru

rr

ua

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Example 1

Find the displacement u(r, t) of a circular membrane of radius c clamped along its circumference if its initial displacement is f(r) and its initial velocity is g(r). See Fig 14.7.

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Example 1 (2)

Solution The boundary-value problem is

crrgtu

rfru

ttcu

tcrt

uru

rr

ua

t

0 ),( ),()0 ,(

0 ,0) ,(

0 ,0 ,1

0

2

2

2

22

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Example 1 (3)

Substituting u = R(r)T(t) into the PDE, then

(3)

The two equations obtained from (3) are(4)

(5)

This problem suggests that we use only = 2 > 0, > 0.

22

1

Ta

TR

Rr

R

02 rRRRr

022 TaT

Ch14_22

Example 1 (4)

Now (4) is the parametric Bessel differential equation of order v = 0, that is, rR” + R’ + 2rR = 0. The general solution is

(6)

The general solution of (5) is T = c3 cos at + c4 sin at

Recall that Y0(r) − as r 0+ and so the implicit assumption that the displacement u(r, t) should be bounded at r = 0 forces c2 = 0 in (6).

)()( 0201 rYcrJcR

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Example 1 (5)

Thus R = J0(r). Since the boundary condition u(c, t) = 0 implies R(c) = 0,we must have c1J0(c) = 0. We rule out c1 = 0, so

J0(c) = 0 (7)

If xn = nc are the positive roots of (7) then n = xn/c and so the eigenvalues are n = n

2 = xn2/c2 and the eigenfunc

tions are c1J0(nr). The product solutions are

(8)),()sincos( 0 rJtaBtaARTu nnnnnn

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Example 1 (6)

where we have done the useful relabeling of the constants. The superposition principle gives

(9)Setting t = 0 in (9) and using u(r, 0) = f(r) give

(10)This is recognized as the Fourier-Bessel expansion of f on the interval (0, c).

10 )()sincos(),(

nnnnnn rJtaBtaAtru

10 )()(

nnn rJArf

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Example 1 (7)

Now we have

(11)Next differentiating (9) with respect to t, set t = 0, and use ut(r, 0) = g(r):

c

nn

n drrfrrJcJc

A0 02

12 )()(

)(

2

(12) )()()(

2

then,)()(

0 021

2

10

c

nnn

n

nnnn

drrgrrJcJca

B

rJBarg

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Standing Waves

The solution (8) are called standing waves. For n = 1, 2, 3, …, they are basically the graph of J0(nr) with the time-varying amplitude

An cos nt + Bn sin nt The zeros of each standing wave in the interval (0, c) are the roots of J0(nr) = 0 and correspond to the set of points of a standing wave where there is no motion. This set is called a nodal line.

Ch14_27

As in Example 1, the zeros of standing waves are determined from

J0(nr) = J0(xnr/c) = 0Now from Table 5.2 and for n = 1, the first positive root of

J0(x1r/c) = 0 is 2.4r/c = 2.4 or r = cSince the desired interval is (0, c), the last result has

no nodal line. For n = 2, the roots of J0(x2r/c) = 0 are 5.5r/c = 2.4 and 5.5r/c = 5.5We have r = 2.4c/5.5 that has one nodal line. See Fig 14.8.

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Fig 14.8

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Laplacian in Cylindrical Coordinates

See Fig 14.10. We havex = r cos , y = r sin , z = z

and

2

2

2

2

22

22 11

z

uu

rru

rr

uu

Ch14_30

Fig 14.10

Ch14_31

Example 2

Find the steady-state temperature shown in Fig 14.11.

Ch14_32

Example 2 (2)

SolutionThe boundary conditions suggest that the temperature u has radial symmetry. Thus

20 ,)4 ,( ,0)0 ,(

40 ,0) ,2(

40 ,20 ,01

0

2

2

2

2

rururu

zzu

zrz

uru

rr

u

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Example 2 (3)

Using u = R(r)Z(z) and separation constant,

(13)

(14)(15)

For the choice = 2 > 0, > 0, the solution of (14) is R(r) = c1 J0(r) + c2 Y0(r)

Since the solution of (15) is defined on [0, 2], we haveZ(z) = c3 cosh z + c4 sinh z

2

1

ZZ

R

Rr

R

02 rRRRr 02 ZZ

Ch14_34

Example 2 (4)

As in Example 1, the assumption that u is bounded at r = 0 demands c2 = 0. The condition u(2, z) = 0 implies R(2) = 0. Then

J0(2) = 0 (16)defines the eigenvalues n = n

2. Last, Z(0) = 0 implies c3 = 0. Hence we have

R(r) = c1 J0(r), Z(z) = c4 sinh z,

10

0

)(sinh) ,(

and ),(sinh)()(

nnnn

nnnn

rzJAzru

rzJAzZrRu

Ch14_35

Example 2 (5)

2

0 021

20

100

0

)()2(2

24sinh

then,)(4sinh

thus,,4When

drrrJJ

uA

rJAu

uuz

nn

nn

nnnn

Ch14_36

Example 2 (6)

For the last integral, using t = nr and d[tJ1(t)]/dt = tJ0(t), then

10

10

1

0

1

02

0 121

20

)()2(4sinh

sinh),(

)2(4sinh

obtain we

,)2(

)()2(2

4sinh

nn

nnn

n

nnnn

nnnnnn

rJJz

uzru

Ju

A

Ju

dtttJdtd

J

uA n

Ch14_37

14.3 Problems in Spherical Coordinates: Legendre Polynomials

Laplacian in Spherical Coordinates See Fig 14.15. We knew that

(1)and

(2)We shall consider only a few of the simpler problems that are independent of the azimuthal angle .

cos ,sinsin ,cossin rzryrz

u

r

u

r

u

rru

rx

uu 22

2

22

2

222

22 cot1

sin

12

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Fig 14.15

Ch14_39

Example 1

Find the steady-state temperature u(r, ) shown in Fig 14.16.

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Example 1 (2)

Solution The problem is defined as

' cot"'2"

then),()(Let

0 ),() ,(

0 ,0 ,0cot1

2

222

2

22

2

RrRRr

rRu

fcu

cru

rr

uru

rr

u

Ch14_41

Example 1 (3)

and so (2)

(3)After letting x = cos , 0 , (3) becomes

(4)

This is a form of Legendre’s equation. Now the only solutions of (4) that are continuous and have continuous derivatives on [-1, 1] are the Legendre polynomials Pn(x) corresponding to 2 = n(n+1), n = 0, 1, 2, ….

02 22 RRrR

0sincossin 2

11,02)1( 22

22 x

dxd

xdx

dx

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Example 1 (4)

Thus we take the solutions of (3) to be = Pn(cos )

When = n(n + 1), the solution of (2) is R = c1 rn + c2 r –(n+1)

Since we again expect u to be bounded at r = 0, we define c2 = 0. Hence,

0

0

)(cos)( ,At

)(cos),( and ,)(cos

nn

nn

nn

nnn

nnn

PcAfcr

PrAruPrAu

Ch14_43

Example 1 (5)

Therefore Ancn are the coefficients of the Fourier-Legendre series (23) of Sec 12.5:

00

0

)(cossin)(cos)(2

12

),(

thus

,sin)(cos)(2

12

nn

n

n

nnn

Pcr

dPfn

ru

dPfc

nA