Myint Tyn U., Debnath L. Linear Partial Differential Equations for Scientists and Engineers

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Higher-Dimensional Boundary-Value

Problems

“As long as a branch of knowledge oﬀers an abundance of problems, it is

full of vitality.”

David Hilbert

10.1 Introduction

The treatment of problems in more than two space variables is much more

involved than problems in two space variables. Here a number of multi-

dimensional problems involving the Laplace equation, wave and heat equa-

tions with various boundary conditions will be present ed. Included are the

Dirichlet problem for a cube, for a cylinder and for a sphere, wave and heat

equations in three dimensional rectangular, cylindrical polar and spherical

polar coordinates. The solution of the three-dimensional Schr¨odinger equa-

tion in a central ﬁeld with applications to hydrogen and helium atoms is

discussed. We also consider the forced vibration of a rectangular membrane

described by the three-dimensional, nonhomogeneous wave equation with

moving boundaries.

10.2 Dirichlet Problem for a Cube

The steady-state temperature distribution in a cube is described by the

Laplace equation

∇

u = u

+ u

= 0 (10.2.1)

for 0 <x<π,0<y<π,0<z<π. The faces are kept at zero degree

temperature except for the face z =0,thatis,

362 10 Higher-Dimensional Boundary-Value Problems

u (0,y,z)=u (π, y, z)=0

u (x , 0,z)=u (x, π, z)=0

u (x , y, π)=0,u(x, y, 0) = f (x, y) , 0 ≤ x ≤ π, 0 ≤ y ≤ π.

We assume a nontrivial separable solution in the form

u (x , y, z)=X (x) Y (y) Z (z) .

Substituting this in the Laplace equation, we obtain

′′

YZ+ XY

′′

Z + XY Z

′′

=0.

Division by XY Z yields

′′

= −

′′

Since the right side depends only on z and the left side is independent of

z, both terms must be equal to a constant. Thus, we have

′′

= −

′′

= λ.

Bythesamereasoning,wehave

′′

= λ −

′′

= µ.

Hence, we obtain the three ordinary diﬀerential equations

′′

− µX =0,

′′

− (λ − µ) Y =0,

′′

+ λZ =0.

Using the boundary conditions, the eigenvalue problem for X,

′′

− µX =0,

X (0) = X (π)=0,

yields the eigenvalues µ = −m

for m =1, 2, 3,... and the correspondin g

eigenfunctions sin mx.

Similarly, the eigenvalue problem for Y

′′

− (λ − µ) Y =0,

Y (0) = Y (π)=0,

gives the eigenvalues λ −µ = −n

where n =1, 2, 3,...and th e correspond-

ing eigenfunctions sin ny.

10.3 Dirichlet Problem for a Cylinder 363

Since λ is given by −



+ n



, it follows that the solution of Z

′′

+λZ =

0 satisfying the condition Z (π)=0is

Z (z)=C sinh



+ n

(π − z)

Thus, the solut ion of the Laplace equation satisfying the homogeneous

boundary conditions takes the form

u (x , y, z)=

∞



m=1

∞



n=1

sinh





+ n

(π − z)



sin mx sin ny.

Applying the nonhomogeneous boundary condition, we formally obtain

f (x, y)=

∞



m=1

∞



n=1

sinh





+ n



sin mx sin ny.

ThecoeﬃcientofthedoubleFourierseriesisthusgivenby

sinh





+ n





f (x, y)sinmx sin ny dx dy.

Therefore, the formal solution to the Dirichlet problem for a cube may be

writtenintheform

u (x , y, z)=

∞



m=1

∞



n=1

sinh



√

+ n

(π − z)



sinh



√

+ n



sin mx sin ny, (10.2.2)

where

= a

sinh





+ n



10.3 Dirichlet Problem for a Cylinder

Example 10.3.1. We consider the problem of determining the electric po-

tential u inside a charge-free cylin der. The potential u satisﬁes the Laplace

equation in cylindrical polar coordinates

∇

u = u

θθ

+ u

=0, (10.3.1)

for 0 ≤ r<a,0<z<l. Let the lateral surface r = a and the top z = l

be grounded, that is, zero potential. Let the potential on the base z =0be

given by

u (r, θ, 0) = f (r, θ) , (10.3.2)

where f (a, θ)=0.

364 10 Higher-Dimensional Boundary-Value Problems

We assume a nontrivial separable solution in the form

u (r, θ, z)=R (r) Θ (θ) Z (z) .

Substituting this in the Laplace equation yields

′′

′

′′

= −

′′

= λ.

It follows that

′′

+ rR

′

− r

λ = −

′′

= µ.

Thus, we obtain the equations

′′

+ rR

′

−



λr

+ µ



R =0, (10.3.3)

′′

+ µΘ =0, (10.3.4)

′′

+ λZ =0. (10.3.5)

Using the periodicity conditions, the eigenvalue problem for Θ (θ),

′′

+ µΘ =0,

Θ (0) = Θ (2π) ,Θ

′

(0) = Θ

′

(2π) ,

yields the eigenvalues µ = n

for n =0, 1, 2,... with the corresponding

eigenfunctions sin nθ,cosnθ.Thus,

Θ (θ)=A cos nθ + B sin nθ. (10.3.6)

Suppose λ is real and negative and let λ = −β

where β>0. If the condition

Z (l) = 0 is imposed, then the solution of equation ( 10.3.5) can be written

in the form

Z (z)=C sinh β (l − z) . (10.3.7)

Next we introduce the new independent variable ξ = βr. Equation

(10.3.3) transforms into

dξ

+ ξ

dξ



− n



R =0

which is the Bessel equation of order n. The general solution is

(ξ)=DJ

(ξ)+EY

(ξ)

where J

and Y

are the Bessel functions of the ﬁrst and second kind

respectively. In terms of the original variable, we have

10.3 Dirichlet Problem for a Cylinder 365

(r)=DJ

(βr)+EY

(βr) .

Since Y

(βr) is u nb oun ded at r = 0, we choose E = 0. The condition

R (a) = 0 requires that

(βa)=0.

For each n ≥ 0, there exist positive zeros. Arranging these in an inﬁnite

increasing sequence, we have

0 <α

<α

<...<α

<....

Thus, we obtain

=(α

/a) .

Consequently,

(r)=DJ

(α

r/a) .

The solution u then ﬁnally takes the form

u (r, θ, z)=

∞



n=0

∞



m=1





cos nθ + b

sin nθ)

×sinh



(l − z)



To satisfy the nonhomogeneous boundary condition, it is required th at

f (r, θ)=

∞



n=0

∞



m=1





cos nθ + b

sin nθ)sinh





The coeﬃcients a

and b

are given by

πa

sinh





(α

)]



2π

f (r, θ) J





rdrdθ,

πa

sinh





n+1

(α

)]



2π

f (r, θ) J





×cos nθrdrdθ,

πa

sinh





n+1

(α

)]



2π

f (r, θ) J





×sin nθrdrdθ.

Example 10.3.2. We shall illustrate the same problem with diﬀerent bound-

ary conditions. Consider the problem

366 10 Higher-Dimensional Boundary-Value Problems

∇

u =0, 0 ≤ r<a, 0 <z<π,

u (r, θ, 0) = 0,u(r, θ, π)=0,

u (a , θ, z)=f (θ, z ) .

As before, by the separation of variables, we obtain

′′

+ rR

′

−



λr

+ µ



R =0,

′′

+ µΘ =0,

′′

+ λZ =0.

By the periodicity conditions, again as in the previous example, the Θ equa-

tion yields the eigenvalues µ = n

with n =0, 1, 2,...; the corresponding

eigenfunctions are sin nθ,cosnθ.Thus,wehave

Θ (θ)=A

cos n cos θ + B

sin nθ.

Now let λ = β

with β>0. Then, the boundary value problem

′′

+ β

Z =0

Z (0) = 0,Z(π)=0,

has the solution

Z (z)=C

sin mz, m =1, 2, 3,....

Finally, we have

′′

+ rR

′

−



+ n



R =0,

′′

′

−





R =0,

the general solution of which is

R (r)=DI

(mr)+EK

(mr) ,

where I

and K

are the modiﬁed Bessel functions of the ﬁrst and second

kind, respectively.

Since R must remain ﬁnite at r = 0, we set E =0.ThenR takes the

form

R (r)=DI

(mr) .

Applying the nonhomogeneous condition, we ﬁnd the solution

10.4 Dirichlet Problem for a Sphere 367

u (r, θ, z)=

∞



m=1





(mr)

(ma)

sin mz

∞



m=1

∞



n=1

cos nθ + b

sin nθ)

(mr)

(ma)

sin mz,

where



2π

f (θ, z)sinmz cos nθ dθ dz,



2π

f (θ, z)sinmz sin nθ dθ dz.

10.4 Dirichlet Problem for a Sphere

Example 10.4.3. To determine the potential in a sphere, we transform the

Laplace equation into spherical coordinates. It has the form

∇

u = u

θθ

cot θ

sin

ϕϕ

, (10.4.1)

where 0 ≤ r<a,0<θ<π,and0<ϕ<2π.

Let the prescribed potential on the sphere be

u (a , θ, ϕ)=f (θ, ϕ) . (10.4.2)

We assume a nontrivial separable solution in the form

u (r, θ, ϕ)=R (r) Θ (θ) Φ (ϕ) .

Substitution of u in the Laplace equation yields

′′

+2rR

′

− λR =0, (10.4.3)

sin

θΘ

′′

+sinθ cos θΘ

′



λ sin

θ − µ



Θ =0, (10.4.4)

′′

+ µΦ =0. (10.4.5)

The general solution of equation (10.4.5) is

Φ (ϕ)=A cos

√

µϕ+ B

sin

√

µϕ. (10.4.6)

The periodicity condition requires that

√

µ = m, m =0, 1, 2,....

Since equation (10.4.3) is of Euler typ e, the solution is of the form

R (r)=r

368 10 Higher-Dimensional Boundary-Value Problems

Inserting this is equation (10.4.3), we obtain

+ β − λ =0.

The roots are β =



−1+

√

1+4λ



/2and−(1 + β). Hence, the general

solution of equation (10.4.3) is

R (r)=Cr

+ Dr

−(1+β)

. (10.4.7)

The variable ξ =cosθ transforms equation (10.4.4) into the form



1 − ξ



′′

− 2ξΘ

′



β (β +1)−

1 − ξ



Θ = 0 (10.4.8)

which is Legendre’s associated equation. The general solution with β = n

for n =0, 1, 2,... is

Θ (θ)=EP

(cos θ)+FQ

(cos θ) .

Continuity of Θ (θ)atθ =0,π corresponds to continuity of Θ (ξ)at

ξ =+

1. Since Q

(ξ) has a logarithmic singularity at ξ = 1, we choose

F = 0. Thus, the solution of equation (10.4.8) becomes

Θ (θ)=EP

(cos θ) .

Consequently, the solution of the Laplace equation in spherical coordinates

u (r, θ, ϕ)=

∞



n=0



m=0

(cos θ)(a

cos mϕ + b

sin mϕ) .

In order for u to satisfy the prescribed fu nction on th e boundary, it is

necessary that

f (θ, ϕ)=

∞



n=0



m=0

(cos θ)(a

cos mϕ + b

sin mϕ)

for 0 ≤ θ ≤ π,0≤ ϕ ≤ 2π. By the orthogonal properties of the functions

(cos θ)cosmϕ and P

(cos θ)sinmϕ, the coeﬃcients are given by

(2n +1)

2πa

(n − m)!

(n + m)!



2π



f (θ, ϕ) P

(cos θ)cosmϕ sin θdθdϕ,

(2n +1)

2πa

(n − m)!

(n + m)!



2π



f (θ, ϕ) P

(cos θ)sinmϕ sin θdθdϕ,

for m =1, 2,...,andn =1, 2,...,and

(2n +1)

4πa



2π



f (θ, ϕ) P

(cos θ)sinθdθdϕ,

for n =0, 1, 2,....

10.4 Dirichlet Problem for a Sphere 369

Example 10.4.4. Determine the potential of a grounded conducting sphere

in a uniform ﬁeld that satisﬁes the p rob lem

∇

u =0, 0 ≤ r<a, 0 <θ<π, 0 <φ<2π,

u (a , θ)=0,u→−E

r cos θ, as r →∞.

Let the ﬁeld be in the z direction so that the potential u will be inde-

pendent of φ. Then, the Laplace equation takes the form

θθ

cot θ

=0.

We assume a nontrivial separable solution in the form

u (r, θ)=R (r) Θ (θ) .

Substitution of this in the Laplace equation yields

′′

+2rR

′

− λR =0,

sin

θΘ

′′

+sinθ cos θΘ

′

+ λ sin

θΘ =0.

If we set λ = n (n +1)withn =0, 1, 2,..., then the second equation is the

Legendre equation. The general solution of this equation is

Θ (θ)=A

(cos θ)+B

(cos θ) ,

where P

and Q

are the Legendre functions of the ﬁrst and second kind

respectively. In order for the solution not to be singular at θ = 0 and θ = π,

we set B

=0.Thus,Θ (θ) becomes

Θ (θ)=A

(cos θ) .

ThesolutionoftheR-equation is obtained in the form

R (r)=C

+ D

−(n+1)

Thus, the potential function is

u (r, θ)=

∞



n=0



+ b

−(n+1)



(cos θ) .

To satisfy the condition at inﬁnity, we must have

= −E

, and a

=0, for n ≥ 2

and hence,

u (r, θ)=−E

r cos θ +

∞



n=1

n+1

(cos θ) .

370 10 Higher-Dimensional Boundary-Value Problems

The condition u (a, θ) = 0 yields

0=−E

a cos θ +

∞



n=1

n+1

(cos θ) .

Using the orthogonality of the Legendre functions, we ﬁnd that b

are given

(2n +1)

n+2



−π

cos θP

(cos θ) d (cos θ)=E

since the integral vanishes for all n except n = 1. Hence, the potential

function is given by

u (r, θ)=−E

r cos θ + E

cos θ.

Example 10.4.5. A dielectric sphere of radius a is placed in a uniform elec-

tric ﬁeld E

. Determine the potentials inside and outside the sphere.

The problem is to ﬁnd p otentials u

and u

that satisfy

∇

= ∇

=0,

∂u

∂r

∂u

∂r

= u

, on r = a,

→−E

r cos θ as r →∞,

where u

and u

are the potentials inside and outside the sphere, respec-

tively, and K is the dielectric constant.

As in the preceeding example, the potential function is

u (r, θ)=

∞



n=0



+ b

−(n+1)



(cos θ) . (10.4.9)

Since u

must be ﬁnite at the origin, we take

(r, θ)=

∞



n=0

(cos θ)forr ≤ a. (10.4.10)

For u

, which must approach inﬁnity in the prescribed manner, we choose

(r, θ)=−E

r cos θ +

∞



n=0

−(n+1)

(cos θ) . (10.4.11)

From the two continuity conditions at r = a, we obtain

= −E

,Ka

= −E

−

= b

=0,n≥ 2.