Chapter 6

3D Linear Second-Order Equations

In this chapter, we delve into the study of linear problems, encompassing Poisson, heat,

and wave equations on Cartesian, cylindrical, and spherical domains. The cylindrical and

spherical domains are obtained through linear transformations from a cube in R

3

, enabling

us to employ the separation of variables technique for their solutions.

6.1 Cartesian coordinate

In this section, we initiate our exploration by solving Laplace's equations deﬁned on cubes

(x

0

; x

1

) ×(y

0

; y

1

) ×(z

0

; z

1

) in R

3

. This foundational step will en a ble us to tackle linear heat,

wave, and Poisson equations with nonhomogeneous boundary conditions.

6.1.1 Laplace equatio n

Ler Ω be the cube (x

0

; x

1

) × (y

0

; y

1

) × (z

0

; z

1

). We consider the Laplace equation ∆u=0

subject to the boundary condition:

αu + β

@u

@n

= f ;

on the boundary bnd(Ω). For this problem, we assume that the boundary condition is

homogeneous on the sides x and y. By applying the separation o f variables method, we

assume u(x; y; z) can be represented as φ(x; y)Z(z), which transforms the Laplace equation

into:

∆

(x;y)

φ

φ

= ¡

¡Z

Z

:

To satisfy this equality, we introduce a constant, denoted as ¡λ, which leads to the following

eigenvalue problem:

(

∆φ = ¡λφ on Ω

αφ + β

@φ

@n

= 0 on bnd(Ω)

:

The diﬀerential equation for Z(z) becomes:

Z

00

¡λZ = 0:

1

Example 6.1. Let's solve the following problem on a unit cube (0; 1) ×(0; 1) ×(0; 1)

8

>

>

>

>

>

>

<

>

>

>

>

>

>

:

∆u = 0

u(x; y; 1) = xy

u(1; y; z) = sin( πz)

u = 0 other sides

:

According to the boundary condition, we split the problem as

∆u = 0

uj

Ω

1

=0

+

∆u = 0

uj

Ω

3

=0

:

The ﬁrst problem has the solution of the form

u =

X

n;m=1

1

B

nm

sinh ( λ

nm

p

z)φ

nm

(x; y);

where λ

nm

= (n

2

+ m

2

)π

2

, and

B

nm

=

4

sinh( λ

nm

p

)

Z

0

1

Z

0

1

xy sin(nπx) sin(mπy) dydx =

4 cos( nπ) cos(mπ)

nmπ

2

sinh( λ

nm

p

)

:

For the s econd problem, we can write the solution as (why?)

u =

X

n=1

1

B

n

sinh

¡

n

2

+ 1

p

πx

sin(nπy) sin(πz);

where

B

n

=

2

sinh

¡

n

2

+ 1

p

π

Z

0

1

sin(nπy) dy =

2(1 ¡cos(nπ))

sinh

¡

n

2

+ 1

p

π

n π

;

and ﬁnally

u(x; y; z) =

X

n;m=1

1

4 cos(nπ) cos(mπ)

nmπ

2

sinh

¡

n

2

+ m

2

p

π

sinh

¡

n

2

+ m

2

p

πz

sin(nπx) sin(mπy) +

+

X

n=1

1

2(1 ¡cos(nπ))

nπ sinh

¡

n

2

+ 1

p

π

sinh

¡

n

2

+ 1

p

πx

sin(nπy) sin(πz):

Problem 6.1. Solve the following Laplace equation on Ω := (0; π) ×(0; π) (0; π)

8

<

:

∆u = 0

u(x; π; z) = sin(x) sin(z)

u = 0 other sides

:

Problem 6.2. Solve the following problem on Ω := (0; π) ×(0; π) (0; π)

8

<

:

∆u + 2@

x

u ¡u = 0

u(π; y; z) = sin(y) sin(z)

u = 0 other sides

2 3D Linear Second-Order Equations

6.1.2 Eigenfu nction of Laplacian i n Cartesian co ordinate

We solve the following problem on Ω

(

∆φ = ¡λφ

φj

bnd(Ω)

=0

:

It is simply seen that λ > 0. In fact, we have

ZZZ

Ω

∆φφ = ¡λ

ZZZ

Ω

jφj

2;

and by the integration by parts formula,

ZZZ

Ω

∆φφ =

ZZ

bnd(Ω)

φ@

n

φ ¡

ZZZ

Ω

jrφj

2

;

that gives λ > 0. Now, By the separation φ = X(x) Y (y) Z(z), we reach the following

eigenvalue problems

X

00

= ¡µX

X(0) = X(a) = 0

;

Y

00

= ¡ηY

Y (0) = Y (b) = 0

;

Z

00

= ¡νZ

Z(0) = Z(c) = 0

;

and ﬁnally

φ

nmp

(x; y; z) = s in

nπ

a

x

sin

mπ

b

y

sin

pπ

c

z

;

with associated eigenvalues

λ

nmp

=

nπ

a

2

+

mπ

b

2

+

pπ

c

2

:

Example 6.2. Let us solve th e following Poisson equation on Ω := (0; 1) ×(0; 1) ×(0; 1)

8

>

>

>

>

>

>

<

>

>

>

>

>

>

:

∆u = sin(πy) sin(2πz)

u(x; y; 1) = xy

u(1; y; z) = sin( πz)

u = 0 other sides

:

The soluti o n can be written as u = v + w, where v solves the Laplace equation with given

boundary conditions that we solved in the previous example. The equation for w reads

(

∆w =

1;2

(y; z)

wj

bnd(Ω)

=0

;

for

1;2

= sin(πy) sin(2πz). The solution can be written a s follows (why?)

w(x; y; z) =

"

X

n=1

1

A

n

sin(nπx)

#

1;2

(y; z)

By substituting the series into the equation, we o btain

X

n=1

1

¡(n

2

+ 5)π

2

A

n

sin(nπx)

1;2

(y; z) =

1;2

(y; z);

6.1 Cartesian coordinate 3

and thus

X

n=1

1

¡(n

2

+ 5)π

2

A

n

sin(nπx) = 1;

that results

A

n

=

¡2(1 ¡cos(nπ))

(n

2

+ 5)nπ

3

;

and ﬁnally,

w(x; y; z) =

"

X

n=1

1

¡2(1 ¡cos(nπ))

(n

2

+ 5)nπ

3

sin(nπx)

#

1;2

(y; z):

Remark 6.1. We could also assume the form of w as w(x; y; z) = W (x)

1;2

(y; z) for an

unknown function W . By this metho d, we obtain the following equation for W (x)

(

W

00

¡5π

2

W = 1

W (0) = W (1) = 0

;

that is solved for

W (x) =

1 ¡cosh( 5

p

π )

5π

2

sinh( 5

p

π )

sinh( 5

p

πx) +

cosh( 5

p

πx) ¡1

5π

2

;

and thus

w(x; y; z) =

1 ¡cosh( 5

p

π)

5π

2

sinh( 5

p

π )

sinh( 5

p

πx) +

1

5π

2

(cosh( 5

p

πx) ¡1)

1;2

(y; z):

This is the closed fo rm solution of the equation for w. It turns out tha t the closed form

solution is equivalent to the series solution obtained above.

Problem 6.3. Solve the following wave equation on Ω := (0; π) ×(0; π) ×(0; π)

8

>

>

>

>

>

>

<

>

>

>

>

>

>

:

@

tt

u = ∆u

uj

Ω

=0

u(0; x; y; z) = sin(x) sin(z)

@

t

u(0; x; y; z) = 0

Problem 6.4. Solve the following heat equation with Neumann boundary conditions on Ω := (0;

π) ×(0; π) ×(0; π)

8

<

:

@

t

u = ∆u

@

n

uj

Ω

=0

u(0; x; y; z) = cos(y) cos(2z)

6.2 Problems in cylindrica l domains

6.2.1 Cylindrical coordinate

A point p in spherical coordinate is represented by the triple (r; θ; z), where r ≥ 0 is the

distance of the projection of p in the (x; y)-plane to the origin, θ 2 [¡π; π] is the angle of

the projected point on the (x; y)-plane with respect to x-axis, and z is the height of point p.

4 3D Linear Second-Order Equations

y

x

z

p

z

θ

r

Let Ω denote the fo llowing c yl inder

Ω = f(r; θ; z); 0 ≤r < a; ¡ π ≤θ ≤π; 0 < z < H g; (6.1)

with boundary bnd(Ω) = f(r; θ; 0)g[f(r; θ; H)g[f(a; θ; z)g. The Laplacian operator in the

cylindrical coordinate is

∆u = ∆

(r;θ)

u + u

zz

= @

rr

u +

1

r

@

r

u +

1

r

2

@

θθ

u + @

zz

u: (6.2)

Note that the cube [0; a) ×[¡π; π] ×(0; H) is transformed to Ω by the following tr ansforma-

tion

8

<

:

x = r cosθ

y = r sinθ

z = z

;

that allows us to use the separation of variables technique for solving lin ear equations on a

cylinder.

Problem 6.5. The unit vectors r^; θ

^

; z^ in the spherical coordinate are respectively

r^ =

dx

dr

;

dy

dr

;

dz

dr

= (cosθ; sinθ; 0)

θ

^

=

1

r

dx

dθ

;

dy

dθ

;

dz

dθ

= (¡sinθ; cosθ; 0)

z^=

dx

dz

;

dy

dz

;

dz

dz

= (0; 0; 1)

a) Show that the nabla operator r in the coordinate is

r= r^@

r

+

1

r

θ

^

@

θ

+ z^@

z

:

b) Find the form of ∆ through the relation ∆ = r·r.

6.2.2 Laplace equatio n : Type I

Consider the following problem

8

<

:

∆u = 0

u(a; θ; z) = 0

u(r; θ; 0) = f(r; θ); u(r; θ; H) = g(r; θ)

: (6.3)

6.2 Problems in cylindrical domains 5

Note that u is homogeneous on the side surface of Ω and is equal to f unctions f ; g on the

bottom and upper caps respective ly . We relax this condition later and study the case u = h

on the side surface. By the separation of variable u(r; θ; z) = Φ(r; θ) Z(z), we reach the

following equation

∆

(r;θ)

φ

φ

+

Z

00

Z

= 0; (6.4)

where ∆

(r;θ)

stands for the Laplacian on the disk. The associate eige nvalue of Eq.6.4 is

(

∆

(r;θ)

φ = ¡λφ

φ(a; θ) = 0

:

As we saw in the last chapter, the set o f eigenpairs is as follows

Φ = fJ

0p

(r); J

np

(r)cos(nθ); J

np

(r) sin(nθ)g

n;p=1

1

; λ

np

=

z

np

2

a

2

(6.5)

The equation for Z becomes accordingly

Z

00

¡λ

np

Z = 0;

and therefore

Z =

cosh

¡

λ

np

p

z

; sinh

¡

λ

np

p

z

:

Finally, the series solution u is

u(r; θ; z) =

X

n=0

1

X

p=1

1

cosh

¡

λ

np

p

z

J

np

(r)fA

np

cos(nθ) + B

np

sin(nθ)g+

+

X

n=0

1

X

p=1

1

sinh

¡

λ

np

p

z

J

np

(r)fC

np

cos(nθ) + D

np

sin(nθ)g:

The boundary conditions at z = 0 and z = H determine con stants A; B; C ; D.

Example 6.3. Let a = 1 and H = 1. Consider the problem

8

>

>

>

>

>

>

<

>

>

>

>

>

>

:

∆u = 0;

u(1; θ; z) = 0

u(r; θ; 0) = 0

u(r; θ; 1) = (1 ¡r)cosθ

:

According to the boundary condition, it makes sense to w rite the solution as

u(r; θ; z) =

X

p=1

1

fA

p

cosh(z

1p

z) + B

p

sinh(z

1p

z)gJ

1p

(r) cos(θ)

The condition at z = 0 determines A

p

= 0 for all p. Therefore we write

u(r; θ; z) =

"

X

p=1

1

B

p

sinh(z

1p

z) J

1p

(r)

#

cos(θ):

6 3D Linear Second-Order Equations

Substituting the condition at z = 1 determines B

p

as

B

p

=

1

kJ

1p

k

2

sinh(z

1p

)

Z

0

1

J

1p

(r) (1 ¡r) rdr;

where

kJ

1p

k

2

=

Z

0

1

jJ

1

(z

1p

r)j

2

rdr:

6.2.3 Laplace equation: Type II

Now, we solve the Laplac e equation when u(a; θ; z) = f(θ; z) but u(r; θ; 0) = u(r; θ; H) = 0.

According to Eq.6.4, we solve ﬁrst the following boundary value problem for Z

Z

00

= ¡λZ

Z(0) = Z(H) = 0

:

The solution is

Z

p

(z) = sin

pπ

H

z

; λ

p

=

p

2

π

2

H

2

:

Therefore the equation for φ becomes

∆

(r;θ)

φ ¡

p

2

π

2

H

2

φ = 0:

Using the separation of variables φ(r; θ) = R(r) Θ(θ), we reach the following equation

r

2

R

00

+ rR

0

R

+

Θ

00

Θ

=

p

2

π

2

H

2

r

2

; (6.6)

and thus

Θ

n

(θ) = fcos(nθ); sin(nθ)g:

Substituting Θ

n

into (6.6) yields

r

2

R

00

+ rR

0

+

¡

p

2

π

2

H

2

r

2

¡n

2

R = 0: (6.7)

Note that t he equation is similar to the Bessel equation except the negative sign of r

2

. If

we take x = ¡

p

2

π

2

H

2

q

r = i

pπ

H

r, then the equation reduces to the standard Bessel equation

x

2

R

00

(x) + xR

0

(x) + (x

2

¡n

2

)R(x) = 0:

Therefore, the solution of Eq.(6.7) is

R

np

(r) = J

n

iπp

H

r

:

6.2 Problems in cylindrical domains 7

What is the series of J

n

(ix)? From the series of J

n

(x), we have

J

n

(ix) =

X

k=0

1

(¡1)

k

k!(k + n)!

ix

2

2k+n

= (i)

n

X

k=0

1

1

k!(k + n)!

x

2

2k+n

:

The series

I

n

(x) =

X

k=0

1

1

k!(k + n)!

x

2

2k+n

;

is called the modiﬁed Bessel functions of the ﬁrst kind. For simplicity, we de note I

n

¡

πp

H

r

by I

np

(r) and write

φ

np

(r; θ) = fI

np

(r) cos(nθ); I

np

(r) sin(nθ)g:

Using the superposition principle, u(r; θ; z) can be represented as follows

u(r; θ; z) =

X

n=0

1

X

p=1

1

sin

πp

H

z

I

np

(r)(A

np

cos(nθ) + B

np

sin(nθ)): (6.8)

The coeﬃcients A

np

; B

np

are determined by the aid of the boundary data.

Example 6.4. Let Ω denote a solid cylinder with the radius a = 1 and height H = 1. Let

us solve the following equation deﬁned on Ω

¯

8

<

:

∆u = 0

u(1; θ; z) = z cos(θ)

u(r; θ; 0) = u(r; θ; 1) = 0

:

Based on the boundary condition, u has the series representation

u(r; θ; z) =

"

X

p=1

1

U

p

I

1p

(r) sin(πpz)

#

cos(θ):

For r = 1, we have

z =

X

p=1

1

U

p

sin(πpz) I

1p

(1);

and thus

U

p

=

2 (¡1)

p

pπI

1p

(1)

:

The solution u is as follows

u(r; θ; z) =

"

X

p=1

1

2 (¡1)

p

I

1p

(r)

pπI

1p

(1)

sin(πpz)

#

cos(θ):

6.2.4 Eigenfu nctions of the Laplacian

Consider the following eigenvalue problem

(

∆φ = ¡λφ

φj

bnd(Ω)

=0

; (6.9)

8 3D Linear Second-Order Equations

where Ω denotes a solid cylinder with radius a and height H. Since ∆φ = ∆

(r;θ)

φ + φ

zz

, we

use the separation of variables φ(r; θ; z) = (r; θ) Z(z), and write

∆

(r;θ)

+

Z

00

Z

= ¡λ: (6.10)

The ordinary diﬀerential equation for Z is solved f o r sine functions

Z

m

(z) = sin

mπ

H

z

:

Substituting Z

m

into (6.10) results to

∆

(r;θ)

= ¡

λ ¡

m

2

π

2

H

2

: (6.11)

Take µ as µ = λ ¡

m

2

π

2

H

2

, and write the problem as foll ows

(

∆

(r;θ)

= ¡µ

(a; θ) = 0

: (6.12)

As we saw before, the eigenfunctions of the above problem are

np

(r; θ) = fJ

np

(r) cos(nθ); J

np

(r) sin (nθ)g:

Therefore, the eigenfunctions of ∆ on a cylinder are

φ

mnp

(r; θ; z) =

n

sin

mπ

H

z

J

np

(r) cos(nθ); sin

mπ

H

z

J

np

(r)sin (nθ)

o

; (6.13)

with eigenvalues λ

mnp

=

m

2

π

2

H

2

+

z

np

2

a

2

.

Remark 6.2. Fo r problems deﬁned on the side surface of a cylinder (when r = a), we solve

the eigenvalue problem ∆φ = ¡λφ on th e following set

D := f(θ; z); ¡π ≤θ ≤π; 0 ≤z ≤H g:

It is simply seen that in this case, th e eigenfunctions are

φ

mn

(θ; z) =

n

sin

mπ

H

z

cos(nθ); sin

mπ

H

z

sin (nθ)

o

; (6.14)

with eigenvalues λ

nm

=

m

2

π

2

H

2

+ n

2

.

6.2.5 Linear problems on the surface of a cylinder

Let D denote the side surface of a cylinder of radius a and h eight H (without top and bottom

caps).

Example 6.5. Let us solve th e following Poisson equation

∆u = z

u(θ; 0) = 0; u(θ; π) = cos(θ)

;

6.2 Problems in cylindrical domains 9

where u is deﬁned in the side surface of a cylinder of radius 1 and height π. By superposition

principle, the solution consists two terms, the solution that is cont ributed by the boundary

data, and the solution that is contr ibuted by the source term. For the contribution from

boundary, we solve the following equation

∆v = 0

v(θ; 0) = 0; v(θ; π) = co s(θ)

:

Since cos(θ) is a part of the eigenfunction s in

¡

mπ

H

z

cos(θ), we write v = V (z) cos(θ), for some

unknown function V (z). Substituting this into the equation leads to the following one

V

00

¡V = 0

V (0) = 0; V (π) = 1

;

that is solved for

V (z) =

1

sinh(π)

sinh(z);

and thus v =

1

sinh(π )

sinh(z)cos(θ). For the contribution by the source z, we solve the following

Poisson equation

∆w = z

w(θ; 0) = 0; w(θ; π) = 0

:

Note that the source term is independent f θ and thus we can write w = W (z). Substituting

w into the equation leads to the following equation

W

00

= z

W (0) = W (π) = 0

;

that is solved for W (z) =

1

6

z(z

2

¡π

2

). Finally, the solution is

u(θ; z) =

1

sinh(π)

sinh(z) cos(θ) +

1

6

z(z

2

¡π

2

):

Remark 6.3. In the above example, we could solve the Poisson equation for w by eigen-

function series as follows

w(θ; z) =

X

m=1

1

W

m

sin(mz); (6.15)

that leads to

w(z) =

X

m=1

1

2 cos(mπ)

m

3

sin(mz):

It is simply veriﬁed that the above series is th e series expansion of

1

6

z(z

2

¡π

2

).

Example 6.6. Let us solve the following wave problem on the side surface of a cylinder o f

radius a = 1 and height H = 1

8

<

:

@

tt

u = ∆u

u(t; θ; 0) = u(t; θ; 1) = 0

u(0; θ; z) = 0; @

t

u(0; θ; z) = sin(θ)

:

10 3D Linear Second-Order Equations

According to the initial data, the solution can be written in the following form

u(t; θ; z) =

X

m=1

1

U

m

(t) sin(mπz) sin(θ):