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

, 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

) ×(y

; y

) ×(z

; z

) in R

. 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

; x

) × (y

; y

) × (z

; z

). We consider the Laplace equation ∆u=0

subject to the boundary condition:

αu + β

= 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)

= ¡

¡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

= 0 on bnd(Ω)

The diﬀerential equation for Z(z) becomes:

¡λZ = 0:

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

∆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

Ω



∆u = 0

Ω

The ﬁrst problem has the solution of the form

u =

n;m=1

sinh ( λ

z)φ

(x; y);

where λ

= (n

+ m

)π

, and

sinh( λ

)

xy sin(nπx) sin(mπy) dydx =

4 cos( nπ) cos(mπ)

nmπ

sinh( λ

)

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

u =

n=1

sinh

+ 1

πx



sin(nπy) sin(πz);

where

sinh

+ 1



sin(nπy) dy =

2(1 ¡cos(nπ))

sinh

+ 1



n π

;

and ﬁnally

u(x; y; z) =

n;m=1

4 cos(nπ) cos(mπ)

nmπ

sinh

+ m

sinh

+ m

πz



sin(nπx) sin(mπy) +

n=1

2(1 ¡cos(nπ))

nπ sinh

+ 1

sinh

+ 1

πx



sin(nπy) sin(πz):

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

∆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)

∆u + 2@

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)

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

ZZZ

Ω

∆φφ = ¡λ

ZZZ

Ω

jφj

and by the integration by parts formula,

ZZZ

Ω

∆φφ =



bnd(Ω)

φ@

φ ¡

ZZZ

Ω

jrφj

;

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

eigenvalue problems



= ¡µX

X(0) = X(a) = 0

;



= ¡ηY

Y (0) = Y (b) = 0

;



= ¡νZ

Z(0) = Z(c) = 0

;

and ﬁnally

nmp

(x; y; z) = s in



nπ



sin



mπ



sin



pπ



;

with associated eigenvalues

nmp



nπ





mπ





pπ



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

∆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)

bnd(Ω)

;

for

1;2

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

w(x; y; z) =

n=1

sin(nπx)

1;2

(y; z)

By substituting the series into the equation, we o btain

n=1

¡(n

+ 5)π

sin(nπx)

1;2

(y; z) =

1;2

(y; z);

6.1 Cartesian coordinate 3

and thus

n=1

¡(n

+ 5)π

sin(nπx) = 1;

that results

¡2(1 ¡cos(nπ))

+ 5)nπ

;

and ﬁnally,

w(x; y; z) =

n=1

¡2(1 ¡cos(nπ))

+ 5)nπ

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)

(

¡5π

W = 1

W (0) = W (1) = 0

;

that is solved for

W (x) =

1 ¡cosh( 5

π )

5π

sinh( 5

π )

sinh( 5

πx) +

cosh( 5

πx) ¡1

5π

;

and thus

w(x; y; z) =



1 ¡cosh( 5

π)

5π

sinh( 5

π )

sinh( 5

πx) +

5π

(cosh( 5

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)

u = ∆u

Ω

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

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)

u = ∆u

Ω

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

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

= @

u +

θθ

u + @

u: (6.2)

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

tion

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^ =



;



= (cosθ; sinθ; 0)



dθ

;

dθ

;

dθ



= (¡sinθ; cosθ; 0)

z^=



;



= (0; 0; 1)

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

r= r^@

+ z^@

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

6.2.2 Laplace equatio n : Type I

Consider the following problem

∆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)

= 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

(r); J

(r)cos(nθ); J

(r) sin(nθ)g

n;p=1

; λ

(6.5)

The equation for Z becomes accordingly

¡λ

Z = 0;

and therefore

Z =



cosh



; sinh



Finally, the series solution u is

u(r; θ; z) =

n=0

p=1

cosh



(r)fA

cos(nθ) + B

sin(nθ)g+

n=0

p=1

sinh



(r)fC

cos(nθ) + D

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

∆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) =

p=1

cosh(z

z) + B

sinh(z

z)gJ

(r) cos(θ)

The condition at z = 0 determines A

= 0 for all p. Therefore we write

u(r; θ; z) =

p=1

sinh(z

z) J

(r)

cos(θ):

6 3D Linear Second-Order Equations

Substituting the condition at z = 1 determines B

sinh(z

)

(r) (1 ¡r) rdr;

where

r)j

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



= ¡λZ

Z(0) = Z(H) = 0

The solution is

(z) = sin



pπ



; λ

Therefore the equation for φ becomes

∆

(r;θ)

φ ¡

φ = 0:

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

+ rR

; (6.6)

and thus

(θ) = fcos(nθ); sin(nθ)g:

Substituting Θ

into (6.6) yields

+ rR



¡n



R = 0: (6.7)

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

. If

we take x = ¡

r = i

pπ

r, then the equation reduces to the standard Bessel equation

(x) + xR

(x) + (x

¡n

)R(x) = 0:

Therefore, the solution of Eq.(6.7) is

(r) = J



iπp



6.2 Problems in cylindrical domains 7

What is the series of J

(ix)? From the series of J

(x), we have

(ix) =

k=0

(¡1)

k!(k + n)!





2k+n

= (i)

k=0

k!(k + n)!





2k+n

The series

(x) =

k=0

k!(k + n)!





2k+n

;

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

πp



by I

(r) and write

(r; θ) = fI

(r) cos(nθ); I

(r) sin(nθ)g:

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

u(r; θ; z) =

n=0

p=1

sin



πp



(r)(A

cos(nθ) + B

sin(nθ)): (6.8)

The coeﬃcients A

; B

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

∆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) =

p=1

(r) sin(πpz)

cos(θ):

For r = 1, we have

z =

p=1

sin(πpz) I

(1);

and thus

2 (¡1)

pπI

(1)

The solution u is as follows

u(r; θ; z) =

p=1

2 (¡1)

(r)

pπI

(1)

sin(πpz)

cos(θ):

6.2.4 Eigenfu nctions of the Laplacian

Consider the following eigenvalue problem

(

∆φ = ¡λφ

φj

bnd(Ω)

; (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

, we

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

∆

(r;θ)

= ¡λ: (6.10)

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

(z) = sin



mπ



Substituting Z

into (6.10) results to

∆

(r;θ)

= ¡



λ ¡



: (6.11)

Take µ as µ = λ ¡

, 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

(r; θ) = fJ

(r) cos(nθ); J

(r) sin (nθ)g:

Therefore, the eigenfunctions of ∆ on a cylinder are

mnp

(r; θ; z) =

sin



mπ



(r) cos(nθ); sin



mπ



(r)sin (nθ)

; (6.13)

with eigenvalues λ

mnp

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

(θ; z) =

sin



mπ



cos(nθ); sin



mπ



sin (nθ)

; (6.14)

with eigenvalues λ

+ n

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



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 = 0

V (0) = 0; V (π) = 1

;

that is solved for

V (z) =

sinh(π)

sinh(z);

and thus v =

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



= z

W (0) = W (π) = 0

;

that is solved for W (z) =

z(z

¡π

). Finally, the solution is

u(θ; z) =

sinh(π)

sinh(z) cos(θ) +

z(z

¡π

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

function series as follows

w(θ; z) =

m=1

sin(mz); (6.15)

that leads to

w(z) =

m=1

2 cos(mπ)

sin(mz):

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

z(z

¡π

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

u = ∆u

u(t; θ; 0) = u(t; θ; 1) = 0

u(0; θ; z) = 0; @

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) =

m=1

(t) sin(mπz) sin(θ):

Substituting this into the wave eq uation , we d erive the following equation for U

(t)

(

(t) =¡(m

+ 1)U

(t)

(0) = 0; U

(0) =

2(1 ¡ cos(mπ))

mπ

;

which is solved fo r

(t) =

2(1 ¡cos(mπ))

mπ λ

sin( λ

t);

where λ

= (m

+ 1). Therefore we reach the solution as the following series

u(t; θ; z) =

m=1

2(1 ¡cos(mπ))

mπ λ

sin( λ

t) s in(mπz)

sin(θ):

6.2.6 Linear problems on a solid cylinder

We solve a few examples to show the method how linear problems can be solve by the

eigenfunction expansion method.

Example 6.7. Consider the the fol lowing problem in the solid cylinder of radius a = 1 and

height H = 1

∆u = r sin(πz)

u(1; θ; z) = 0

u(r; θ; 0) = u(r; θ; 1) = 0

Since the source term is independent of θ (the eigenfunction associated to n = 0), we write

the solution as follows

u(r; θ; z) = sin(πz)

p=1

(r):

Substituting u into the equation and using ∆u = ¡(z

+ π

)u, yields

p=1

¡(z

+ π

(r) = r:

The above equation determines U

as follows

¡1

+ π

)kJ

(r) r

dr;

6.2 Problems in cylindrical domains 11

where

r)j

rdr:

Example 6.8. Consider the the following problem in the solid cylinder D of radius a = 1

and height H = 1

u = ∆u

u(t; 1; θ; z) = 0

u(t; r; θ; 0) = 0

u(t; r; θ; 1) = sin(θ)

u(0; r; θ; z) = 0

First, we ﬁnd the steady state solution of the problem through solving the following one

∆v = 0

v(1; θ; z) = 0

v(r; θ; 0) = 0

v(r; θ; 1) = sin(θ)

Regarding the boundary data, the solut ion can be written in the following form

v(r; θ; z) =

p=1

sinh(z

z)J

(r) sin(θ);

where A

sinh(z

)

(r) rdr:

Now, we write u(t; r; θ; z) = v(r; θ; z) + w(t; r; θ; z) where w satisﬁes the following

homogeneous equation

(

w = ∆w

bnd(D)

It is simply seen that w has the series form

w(t; r; θ; z) =

n;m; p

¡λ

nmp

sin(mπz) J

(r) fA

nmp

cos(nθ) + B

nmp

sin(nθ)g:

Applying the initial condition of the problem leads to the following equality

p=1

sinh(z

z)J

(r) sin(θ) =

n;m; p

sin(mπz) J

(r)fA

nmp

cos(nθ) + B

nmp

sin(nθ)g;

that implies in turn A

nmp

= 0 and

¡A

sinh(z

z) =

m=1

1mp

sin(mπz) :

12 3D Linear Second-Order Equations

We obtain

1mp

= ¡2A

sinh(z

z) sin(mπz) dz;

and ﬁnally

u =

p=1

sinh(z

z) J

(r) sin(θ) +

m;p=1

1mp

¡λ

1mp

(r) sin(mπz) sin(θ):

Problems

Problem 6.6. Show that functions I

(r) are orthogonal with respect to σ(r) = r, that is,

; I

= 0; p =/ q:

Problem 6.7. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series s olution

to the following problem on Ω

∆u = 0

u(1; θ; z) = 0

u(r; θ; 0) = cosθ

u(r; θ; 1) = sinθ

Problem 6.8. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series s olution

to the following problem on Ω

∆u = 0

u(1; θ; z) = 0

u(r; θ; 0) = r

u(r; θ; 1) = sinθ + cosθ

Problem 6.9. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series s olution

to the following equation on Ω

∆u = 0

u(1; θ; z) = 0

u(r; θ; 0) = θ

u(r; θ; 1) = r

Problem 6.10. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series solution

to the following equation on Ω

∆u = 0

u(1; θ; z) = z cosθ

u(r; θ; 0) = 0

u(r; θ; 1) = 0

Draw solution at θ = π/4, z = 1/2 with respect to r.

Problem 6.11. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series solution

to the following equation on Ω

∆u = 0

u(1; θ; z) = sin(2πz) sinθ

u(r; θ; 0) = u(r; θ; 1 ) = 0

Draw solution at θ = π/4, z = 1/2 with respect to r.

6.2 Problems in cylindrical domains 13

Problem 6.12. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Find a series solution

to the following equation on Ω

∆u = 0

u(1; θ; z) = sin(2πz) θ

u(r; θ; 0) = cosθ; u(r; θ; 1) = 0

Problem 6.13. Let D denote t he side surface of a cylinder with radius a and height H. Note that the

cylinder does not include top and button caps. Solve the following equation on D for u = u(θ; z)



∆u = 0

u(θ; 0) = f(θ); u(θ; H) = g(θ)

Problem 6.14. Let D denote t he side surface of a cylinder with radius a and height H. Note that the

cylinder does not include top and button caps. Solve the following equation on D for u = u(θ; z)



∆u + 2@

u = 0

u(θ; 0) = f(θ); u(θ; H) = g(θ)

Problem 6.15. Let D denote the side surface of a cylinder with radius a = 1 and height H = 1. Note

that the cylinder does not include top and button caps. Find a closed for m solution to the following

problem on D



∆u = z

u(θ; 0) = cosθ; u(θ; 1) = 0

Problem 6.16. Let D denote the side surface of a cylinder with radius a = 1 and height H = 1. Note

that the c ylinder does not include top and button caps. Consider the following problem on D



∆u = z sin(2θ)

u(θ; 0) = u(θ; 1) = 0

a) Find a series solution to the problem.

b) Find a closed form solution to the problem and verify it is equal to the series solution obtained

in (a).

Problem 6.17. Let D denote the side surface of a cylinder with radius a = 1 and height H = 1. Note that

the cylinder does not include top and button caps. Find a series solution to the following problem on D



∆u = zθ

u(θ; 0) = sinθ; u(θ; 1) = cosθ

Problem 6.18. Let D denote the side surface of a cylinder with radius a = 1 and height H = 1. Note

that the c ylinder does not include top and button caps. Consider the following problem on D



∆u = sin(πz)θ

u(θ; 0) = sin(θ); u(θ; 1) = 0

a) Find a series solution to the problem.

b) Find a closed form solution to the problem and verify it is equal to the series solution obtained

in (a).

Problem 6.19. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Solve the following

Poisson equation on Ω .

∆u = rz

u(1; θ; z) = 0

u(r; θ; 0) = 0 ; u(r; θ; 1) = r

14 3D Linear Second-Order Equations

Problem 6.20. Let D denote the side surface o f a cylinder with radius a = 1 and height H = 1. Solve

the following heat problem on D

u = ∆u

u(0; θ; z) = cos(θ) sin(πz)

u(t; θ; 0) = u(t; θ; 1) = 0

Problem 6.21. Let D denote the side surface o f a cylinder with radius a = 1 and height H = 1. Solve

the following heat problem on D

u = ∆u ¡t sinθ

u(0; θ; z) = 0

u(t; θ; 0) = cosθ; u(t; θ; 1) = sinθ

Problem 6.22. Let D denote the side surface o f a cylinder with radius a = 1 and height H = 1. Solve

the following heat problem on D

u = c

∆u

u(0; θ; z) = sinθ sin(πz); @

u(0; θ; z) = 0

u(t; θ; 0) = 0; u(t; θ; 1) = 0

Problem 6.23. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Solve the following

problem on Ω.

u = ∆u ¡e

¡t

u(t; 1; θ; z) = 0

u(t; r; θ; 0) = u(t; r; θ; 1) = 0

u(0; r; θ; z) = 0

Problem 6.24. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Solve the following

problem on Ω.

u = ∆u

u(t; 1; θ; z) = z sinθ

u(t; r; θ; 0) = u(t; r; θ; 1) = 0

u(0; r; θ; z) = 0

Problem 6.25. Let Ω denote a solid cylinder with radius a = 1 and height H = 1. Solve the following

problem on Ω.

u = c

∆u

u(t; 1; θ; z) = 0

u(t; r; θ; 0) = u(t; r; θ; 1) = 0

u(0; r; θ; z) = 0; @

u(0; r; θ; z) = (1 ¡r)z

6.3 Problems on sphe res

6.3.1 Spherical coordi n ate

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

distance of p to the origin, θ 2[0; π] is the angle it makes with z-axis, and φ 2[¡π; π] is the

angle the projection of p on the (x; y)-plane makes with x- a xi s.

6.3 Problems on spheres 15

If Ω denote a ball of radius a, then a closed solid ball is represented by the following set of

points

Ω = f(r; θ; φ); 0 ≤r ≤a; 0 ≤θ ≤π; ¡π ≤φ ≤π g:

The Laplacian operator ∆ in the spherical coordinate has the following form

∆u =

u) +

sinθ

(sinθ@

u) +

sin

φφ

u: (6.16)

To distinguish be tween a solid ball and a sphere, we use word sphere for the shell and denote

it by S

, and word ball for a solid ball. The inner product between two function f (r; θ; φ)

and g(r; θ; φ) in the spherical coordinate is as follows

hf ; gi=

¡π

f(r; θ; φ) g(r; θ; φ) r

sinθdrdθdφ: (6.17)

Remember that dV = r

sinθdrdθdφ is the volume element in the sph erical coordinate.

Remark 6.4. The spherical coordinate is speciﬁed by the following transformation

x = r sinθ cosφ

y = r sinθ sinφ

z = r c o sθ

In the ﬁrst chapter, we asked the reader to derive the form of nabla operator r by the aid

of unit vectors

r^ =

+ (@

x; @

y; @

z) = sinθ cosφ i

+ sinθ sinφ j

+ cosθ k

;

and similarly for θ

; φ

+ (@

x; @

y; @

z) = cosθ cosφ i

+ cosθ sinφ j

¡sinθ k

+ (@

x; @

y; @

z) = ¡sinφ i

+ cosφ j

16 3D Linear Second-Order Equations

It is simply seen that

r= r^@

r sin θ

;

and thus

∆ = r·r=



r^@

r sin θ





r^@

r sin θ



;

after after direct calculation gives the form of ∆ in the spherical coordinate as given in (6.16).

Problem 6.26. Show with the aid of following relations that the diﬀerential volume in spherical coor-

dinate is dV = r

sinθdrdθdφ

dx = sinθ cosφdr + r cosθ cosφdθ ¡r sinθ sinφdφ;

dy = sinθ sinφd r + r cosθ sinφdθ + r sinθ cosφdφ;

dz = cosθdr ¡r sinθdθ:

6.3.2 Laplace equatio n and Legendre polynomials

Let Ω denote a ball in R

of radius a. Consider the fo llowing e quation on Ω



∆u = 0

u(a; θ; φ) = f (θ; φ)

: (6.18)

First, assume that f is independent of φ, that is, f = f (θ). This leads to an important class

of special functions called the Legendre polynomials. Consider the equation



∆u = 0

u(a; θ; φ) = f(θ)

: (6.19)

Since the boundary data is independe nt of φ, (that is, f is axially symmetric about the z-

axes), we can assume that the solution is a xially sy mmetr ic in Ω, that is, u = u(r; θ). Due

to the form of ∆ in the spherical coordinate, the separated solution u(r; θ) = R(r) Θ(θ) leads

to the following equation

+ 2rR

Θ sinθ

dθ

(sinθΘ

) = 0: (6.20)

The equation for Θ leads to the f o llowing eigenvalue problem:

dθ

(sinθΘ

) = ¡λ

sinθΘ; (6.21)

for some eigenvalue λ

. It is observed that a singularity in Θ may happen at θ = 0; π, where

the coeﬃcient of the highest derivative Θ

become zero. To avoid the p o ssible blow up at

these points, we impose the following condition for the solution Θ(θ)

lim

θ!0;π

jΘ(θ)j< 1: (6.22)

If we take x = cos(θ), the eigenvalue problem (6.21) reduces to the following familiar form

which is called the Legendre's equation

[(1 ¡x

) Θ

] = ¡λ

Θ(x); ¡1 ≤x ≤1: (6.23)

6.3 Problems on spheres 17

The Legendre equation is discussed in our boo k on ordinary diﬀerential equations and

readers are referred to it. The solution of the Legendre equation can be derived by the

power series as follows

Θ(x) =

k=0

;

where c

satisfy the following recursive formula

k+2

k(k + 1) ¡λ

(k + 2)(k + 1)

Note that if λ

is an integer of th e form n(n + 1) for n = 0; 1; 2; ···, then o ne solution to (6.23)

is a polynomial of order n. This polynomial is called the Lege ndre polynomial of order n

after the French mathematician A.M. Legendre, a nd is denoted by P

(x). A closed form

representation of this polynomial is given by the O. Rodrigues formula

(x) =

(¡1)

(1 ¡x

)

: (6.24)

The following table shows P

(x) for n = 0; ···; 6.

n P

(x)

0 1

1 x

(3x

¡1)

x(5x

¡3)

(35x

¡30x

+ 3)

x(63x

¡70x

+ 1 5 )

(231x

¡315x

+ 1 0 5 x

¡5)

Table 6 .1.

It turns out that if λ

=/ n(n+1), the inﬁnite series solution blows up at x = ±1. Therefore„

to keep the soluti o ns bounded, we assume λ

= n(n + 1), where n is a non-neg ative integer.

Consequently, Θ

(θ) =P

(cos θ) for n ≥0. Note that P

(cos θ) for n = 0; 1; :::, are solutions to

(6.21) for λ

= n(n+ 1) and therefore they are orthogonal with respect to the weight function

σ = sinθ

(cos θ) P

(cos θ) sin(θ) dθ = 0; n =/ m: (6.25)

In addition, the set fP

(cos θ)g

n=0

is a basis for smooth functions in θ 2[0; π]. At the end

of this chapter, we will prove the following proposition for P

(cos θ).

Proposition 6.1. The following relations hold for Legendre polynomials P

(cos θ)

(cos θ) sin(θ) dθ = 0;

(cos θ)j

sin(θ) dθ =

2n + 1

18 3D Linear Second-Order Equations

Now, we back to the equation for R. From (6.20), we reach

+ 2rR

¡n(n + 1)R = 0;

which is a Cauchy-Euler equation. The solution set of the above equation is R

(r) =

; r

¡n¡1

g where we ignore term r

¡n¡1

for p roblems deﬁned inside a ball. The reason is th at

¡n¡1

is unbounded at r = 0. For the exterior Laplace equations (equations deﬁned in the

exterior of a ball), we ignore the term r

and write R

(r) = r

¡n¡1

. Therefore, for problems

inside a ball, the solution of (6.19) has the following form

u(r; θ) =

n=0

(cos θ);

where U

are determined through the boundary condition and the following inner product

hf ; P

sinθ

2n + 1

f(θ) P

(cos θ) sin(θ) dθ:

Example 6.9. Consider the following problem



∆u = 0

u(1; θ; φ) = 1

Let us write the series solution as follows

u(r; θ; φ) =

n=0

(cos θ):

The boundary condition at r = 1 gives

1 =

n=0

(cos θ);

and since P

= 1, we d erive U

= 0 for n ≥1 and U

= 1 and thus u = 1 is the solution to the

problem. This solution conﬁrms the maximum principle for harmonic functions.

Example 6.10. We solve the f o llowing Laplace equation



∆u = 0

u(1; θ; φ) = cos(θ)

Since the boundary condition is independent of φ then we take u = u(r; θ) and write

u(r; θ) =

n=0

(cos θ):

Applying the boundary condition results to

cos θ =

n=0

(cos θ) )U



1 n = 1

0 n =/ 1

;

6.3 Problems on spheres 19

and ﬁnally u(r; θ) = r cos(θ).

6.3.3 Associated Legendre polynomials

Now we relax the restriction that f is axially symmetric. Let us write the separated solution

as u = R(r) Y (θ; φ) and substitute that into (6.18) to obtain

+ 2rR

∆

(θ;φ)

Y = 0; (6.26)

where ∆

(θ;φ)

Y stands for

∆

(θ;φ)

Y =

sinθ

(sinθ@

Y ) +

sin

φφ

Y : (6.27)

The above equat ion implies that

∆

(θ;φ)

Y is a constant. Let us denote this constant again

by λ

. Therefore

∆

(θ;φ)

Y = ¡λ

Y : (6.28)

To solve the a bove problem, we w rite Y (θ; φ) = Θ(θ)Φ(φ) and obtain

sinθ

dθ



sinθ

dΘ

dθ



dφ

= ¡λsin

θ: (6.29)

This equation implies in turn that

dφ

is constant. Let us denote this constant by λ

that is, Φ

= ¡λ

Φ. Notice that Φ(φ) must satisfy the periodicity condition

Φ(¡π) = Φ(π); Φ

(¡π) = Φ

(π):

Regarding this periodicity condition, we derive the following list o f solutions f o r Φ

(φ) 2f1; cos(mφ); sin(mφ)g

m=1

;

and λ

= m

. Substituting this constants into (6.29) gives

dθ

(sin(θ) Θ

) ¡

sin(θ)

Θ = ¡λ

sin(θ) Θ:

Observe that the derived equation is similar to a Legendre equation except the extra term

sin(θ)

Θ. Taking x = cos(θ), will transforms the equation into the foll owing one which is called

the associated Legendre equation:

[(1 ¡x

) Θ

] ¡

1 ¡x

Θ = ¡λ

Θ: (6.30)

The admissible values for λ

are aga in λ

= n(n + 1), n ≥ 0 to guarantee that the solution

remains bounded at x = ¡1; 1.

Proposition 6.2. If λ

= n(n + 1), then the solution of (6:30) are

n;m

(x) = (1 ¡x

)

(m)

(x); m = 0; :::; n;

20 3D Linear Second-Order Equations

where P

(m)

(x) =

(x).

Proof. Note that we have

(1 ¡x

¡2xP

= ¡n(n + 1)P

Diﬀerentiating the above equation m-times leads to the following one

(1 ¡x

(m+2)

¡2x(m + 1)P

(m+1)

= [¡n(n + 1) + m(m + 1)] P

(m)

: (6.31)

Now for the function P

n;m

(x) = (1 ¡x

)

(m)

(x), one needs to show



(1 ¡x

)

n;m



1 ¡x

n;m

= ¡n(n + 1) P

n;m

: (6.32)

A simple calculation shows



(1 ¡x

)

n;m



= (1 ¡x

)

f(1 ¡x

(m+2)

¡2(m + 1)xP

(m+1)



1 ¡x

¡m(m + 1)



n;m

Now, by the relation (6.31), we have

(1 ¡x

)

m/2

f(1 ¡x

(m+2)

¡2(m + 1)xP

(m+1)

g= [¡n(n + 1) + m(m + 1)] P

n;m

This implies in turn



(1 ¡x

)

n;m





¡n(n + 1) + m(m + 1) +



1 ¡x

¡m(m + 1)



n;m



¡n(n + 1) +

1 ¡x



;

that completes the proof. 

Since P

(x) is a polynomial of order n, P

n;m

(x) are zero for m > n. It is evident also that

n;m

(1) = 0 for all n > 0 and P

n;0

(x) = P

(x).

Proposition 6.3. The function P

n;m

(cos θ) are orthogonal in t he following sense

n;m

(cos θ) P

(cos θ) sin(θ) dθ = 0; n =/ m;

and each list of the following class

(cos θ)g

n=0

; fP

n;1

(cos θ)g

n=1

; fP

n;2

(cos θ)g

n=2

; ···;

is a basis for smooth functions f (θ) deﬁned for θ 2[0; π].

The ﬁgur e (6.1), shows the convergence of a series in terms of the functions in fP

n;2

(x)g

n=2

to the functions f(x) = x

, ¡1 < x < 1 f o r N = 4; 10.

6.3 Problems on spheres 21

−

5 1

= 4

= 10

Figure 6.1.

The table (6.2) shows some associated Legendre fu nc tions P

n;m

(x).

m n = 0 n = 1 n = 2 n = 3 n = 4

0 1 P

(x) P

(x)

1 0 1 ¡x

3x 1 ¡x

1 ¡x

(5x

¡1)

x 1 ¡x

(7x

¡3)

2 0 0 3(1 ¡x

) 15x(1 ¡x

)

(1 ¡x

)(7x

¡1)

3 0 0 0 15(1 ¡x

)

3/2

105x(1 ¡x

)

3/2

4 0 0 0 0 105(1 ¡x

)

Table 6 .2.

Now back to (6.26), the equation for R is

+ 2rR

¡n(n + 1) = 0;

which is a Cauchy-Euler equation w ith the solution fr

; r

¡n¡1

g. Solving an equation

inside a ball, we reject th e solution r

¡n¡1

and keep r

, that is, R

(r) = r

. Finally, the

solution u in this case is written in the series form a s

u(r; θ; φ) =

n=0

m=0

n;m

(cosθ) (A

cos(mφ) + B

sin(mφ)):

The constants A

; B

are determined by the aid of the boundary condition.

Example 6.11. Let us solve the equation in the exterior of the u nit ball



∆u = 0

u(1; θ; φ) = sin(φ)

According to the boundary condition, we write the solution in the series fo rm as

u =

n=1

¡n¡1

n;1

(cosθ) sinφ:

Applying the boundary condition, we obtain

sinφ =

n=1

n;1

(cosθ) s inφ;

22 3D Linear Second-Order Equations

and thus

1 =

n=1

n;1

(cosθ):

The above equation determines constants U

n;1

(cosθ) s in(θ) dθ =

n;1

¡1

n;1

(x) dx:

6.3.4 Eigenvalue p roblem

In order to solve linear problems in a ball or on a sphere, one n eeds to solve the following

eigenvalue problem



∆u = ¡λu

u(a; θ; φ) = 0

The separation of variable u(r; θ; φ) = R(r) Y (θ; φ) leads to the eigenvalue problem

) +

∆

(θ;φ)

Y = ¡λr

: (6.33)

Thus

∆

(θ;φ)

Y is λ

= n(n + 1) (otherwise the solution goes unbounded), and hence

+ 2rR

+ (λr

¡n(n + 1))R = 0: (6.34)

The equation (6.34) is called the spherical Bessel equation (because of the term n(n + 1)

instead of n

and 2r instead of r). Since λ > 0, and we take x = λ

r to transform (6.34)

into the following standard form

+ 2xy

+ (x

¡n(n + 1))y = 0: (6.35)

The bounded solution to the above equation is called j-spherical Bessel functions and is

denoted by j

(x).

Proposition 6.4. The function j

(x) =

(x) is the bounded solution to the equation

(6:35).

The proof is left as an exercise to the reader. Applying the zero boundary condition,

determines λ as

( λ

a) = 0 )λ





We denote the zeroth of j

(x) by ζ

, that is , j

(ζ

) = 0. It is simply seen (and it is left a

an exercise to the reader again) that the foll owing orthogonality condition holds for spherical

Bessel functions

(r) j

(r) r

dr = 0; p =/ q;

where j

(r) stands for the functions j

(ζ

r / a). The table (6.3) shows the zeros of the

function j

(x).

6.3 Problems on spheres 23

p ζ

1 3.1415 4.4934 6.7635 6.9880

2 6.2832 7.7252 9.0950 10.4171

3 9.4248 10.9041 12.3230 13.6980

4 12.5664 14.0662 15.5146 16.9236

5 15.7080 17.2208 18.6890 20.1218

6 18.8496 20.3713 21.8539 23.3042

7 21.9911 23.5195 25.0129 26.4768

8 25.1327 26.6661 28.1678 29.6426

9 28.2743 29.8116 31.3204 32.8037

10 31.4159 32.9564 34.4705 35.9614

Table 6 .3.

The above discussion implies that the eigenfunctions of ∆ on a ball are:

nmp

(r; θ; φ) 2fj

(r)P

n;m

(cosθ)cos(mφ); j

(r)P

n;m

(cosθ)sin(mφ)g: (6.36)

In addition, we have h'

nmp

; '

i= 0 if at lea st one of indices are not equal. Note also that

∆'

nmp

= ¡

nmp

;

and in particular for m = 0 we have

∆fj

(r) P

(cos θ)g= ¡

(r) P

(cos θ):

Remark 6.5. The above discussion gives also the eigenvalues and eigenfunctions o f ∆

(θ;φ)

on S

. In fact, the solution to the problem

∆

(θ;φ)

Y = ¡λY ;

are

n;m

(θ; φ) 2fP

n;m

(cos θ) cos(mφ); P

n;m

(cosθ) sin(mφ)g: (6.37)

Notice that S

has no boundary, and thus ∆

(θ;φ)

is symmetric operator on C

). Further-

more, if f(θ; φ) is a smooth function deﬁned on S

, one can represent it as a series in terms

of functions in Y

(θ; φ). The appropriate inner product between two functions f (θ; φ) and

g(θ; φ) is

hf ; gi=

¡π

f(r; θ) g(r; θ) sinθdθdφ:

6.3.5 Linear problems on the shell

Let us solve the Poisson equation ∆u = f(θ; φ) on the shell of a ball. Since the solution

should be 2π-periodic with respect to φ, the solution should satisﬁed the following conditions

u(θ; ¡π) = u(θ; π); @

(θ; ¡π) = @

(θ; π):

24 3D Linear Second-Order Equations

But, according to the divergence theorem, we have

∆u =

bnd(S)

ru ·n^ dA = 0;

(since bnd(S) = ;) and thus the equation ∆u = f(θ; φ) has a solution only if

f(θ; φ) dA = 0:

Example 6.12. Let us solve the equation ∆u = φ on the unit s phere. Since the function φ

is odd in (¡π; π), we write the s o lution as

u(θ; φ) =

n=1

m=1

n;m

(cosθ) s in(mφ):

Substituting this into the equation and using the rel a tion

∆

(θ; φ)

n;m

(cosθ) sin(mφ)g= ¡n(n + 1)P

n;m

(cos θ) sin(mφ);

we obtain

n=1

m=1

¡n(n + 1)U

n;m

(cos θ) sin(mφ) = φ:

The above equation determines U

¡1

πn(n + 1)kP

n;m

¡π

φP

n;m

(cos θ) sin(mφ) sin(θ)dθdφ =

¡1

πn(n + 1)kP

n;m



¡1

n;m

(x)dx



¡π

φ sin(mφ) dφ



2(¡1)

mn(n + 1)kP

n;m



¡1

n;m

(x)dx



Example 6.13. Let us solve the f o llowing heat problem on the unit sphere

(

u = ∆

(θ;φ)

u ¡cos(θ)

u(0; θ; φ) = sin(θ)

We ﬁrst solve the Poisson equation ∆u = cosθ on the sphere. The solution is independent

of φ and thus we write

u(θ; φ) =

n=0

(cosθ):

Substitution into the Poisson equation gives

∆

(θ; φ)

u =

n=0

¡n(n + 1)U

(cosθ) = cos(θ):

6.3 Problems on spheres 25

Hence n = 1 and U

¡1

and consequently u(θ; φ) = ¡

cos(θ). Now, we write the solution

to the original equation as

u(t; θ; φ) = ¡

cos(θ) + w(t; θ; φ);

where w satisﬁes the homogeneous heat problem

(

w = ∆

(θ;φ)

w(0; θ; φ) = sin(θ) +

cos(θ)

We represent th e solution as the series

w(t; θ; φ) =

n=1

(t) P

(cosθ):

The equation for W

(t) is W

= ¡n(n + 1)W

and therefore

u(t; θ; φ) =

n=1

¡n(n+1)t

(cos θ):

Applying the initial condition gives

sin(θ) +

cos(θ) =

n=1

(cos θ):

This implies W

, W

= 0 for n = 2k + 1, k = 0; 1; 2; ···. For n even, we have

¡1

(x) 1 ¡x

dx:

Finally, we write the solution as

u(t; r; θ) = ¡

cos(θ) +

cos(θ) e

¡2t

n:even

¡n(n+1)t

(cos θ):

Example 6.14. We solve the f o llowing wave problem on a unit sphere

u = c

∆

(θ;φ)

u(0; θ; φ) = 0

u(0; θ; φ) = cos(θ) sin(2φ)

According to the initial condition, we write the solution as

u(t; θ; φ) =

n=2

(t)P

n;2

(cosθ) s in(2φ):

Substituting the series into the equation gives

n=2

(t)P

n;2

(cos θ) sin(2φ) =

n=2

¡c

n(n + 1)U

(t)P

n;2

(cos θ) sin(2φ):

26 3D Linear Second-Order Equations

Hence the following equation holds for U

(t)

(t) = ¡c

n(n + 1)U

(t):

Thus u has the series form

u(t; θ; φ) =

n=2

cos(c λ

t) + B

sin(c λ

t)]P

n;2

(cos θ) sin(2φ):

To determine A

; B

, we use the initial data. It is simply seen that A

= 0 and

cos(θ) =

n=2

c λ

n;2

(cos θ);

that gives

n;2

¡1

n;2

(x) dx:

6.3.6 Linear Problems on a solid ball

Example 6.15. Let us solve the f o llowing Poisson equation on a unit ball



∆u = cos(θ)

u(1; θ; φ) = 0

Note that the forcing term is independent of φ. Since P

(cosθ) =cos(θ), we write the solution

as the series

u(r; θ; φ) =

p=1

(r) cos(θ):

Regarding the relation

∆( j

(r) cos θ) = ¡ζ

(r) cos(θ);

we obtain

cos(θ)

p=1

¡ζ

(r) =cos(θ):

Constants U

are determined by the formula

¡1

(r) r

dr:

An alternative method is to write the solution as u(r; θ; φ) = U(r) cos(θ) and substitute that

into the equation to obtain

+ 2rU

¡2U = r

The above Cauchy-Euler equation is solved for U(r) = Ar +

. The boundary condition

u(1; θ; φ) = 0 implies A = ¡

and ﬁnally

u(r; θ; φ) =

r(r ¡1) cos(θ):

6.3 Problems on spheres 27

The series solution of the equation is just the series form of the later closed form solution.

Example 6.16. Let us solve the f o llowing heat problem

u = ∆u ¡e

¡t

u(t; 1; θ; φ) = cos(θ)

u(0; r; θ; φ) = 0

Since the boundary condition is nonzero, we write the solution as

u(t; r; θ; φ) = v(r; θ; φ) + w(t; r; θ; φ):

The equation for v is the following Laplace equation



∆v = 0

v(1; θ; φ) = cos(θ)

;

with the solution v = r cos(θ). The equation for w is

w = ∆w ¡e

¡t

w(t; 1; θ; φ) = 0

w(0; r; θ; φ) = ¡r cos(θ)

Note that the initial c o ndition for w is independent of φ and also the forcing term is inde-

pendent of φ and θ. According to this information, we write the solution as the series in

terms of fj

(r)P

(cos θ)g as

w(t; r; θ; φ) =

p=1

(t)j

(r) cos(θ):

Remember that P

(cosθ) = cos(θ). Substituting w into the equation gives

p=1

(t)j

(r) cos(θ) =

p=1

¡ζ

(t)j

(r) cos(θ) ¡e

¡t

In order to merge summations and ﬁnd the desired equation for W

, we have to expand the

source term e

¡t

in terms of eigenfunctions fj

(r) P

(cosθ)g. But this is impossible because

the source term is independent of θ. Fo r this, we split the equation of w into two sub-

problems

(a)

w = ∆w ¡e

¡t

w(t; 1; θ; φ) = 0

w(0; r; θ; φ) = 0

+ (b)

w = ∆w

w(t; 1; θ; φ) = 0

w(0; r; θ; φ) = ¡r cos(θ)

The solution to the sub-problem (b) is derived as

(t; r; θ; φ) =

p=1

¡ζ

(r)cos(θ);

28 3D Linear Second-Order Equations

where W

are determined as

¡1

(r) r

dr:

To solve the sub-problem (a), we expand the source terms as

¡t

= e

¡t

p=1

(r);

where

(r) r

dr:

We write the solution to (a) as

(t; r; θ; φ) =

p=1

¡1

¡t

¡e

¡ζ



(r):

Finally, the solution is

u(t; r; θ; φ) = r cos(θ) +

p=1

¡1

¡t

¡e

¡ζ



(r) +

p=1

¡ζ

(r)cos(θ):

Example 6.17. Let us solve the f o llowing wave equati on in a unit solid ball

u = ∆u ¡10rcos(θ)

u(1; θ; φ) = 0

u(0; r; θ; φ) = r(r

¡1)cos(θ); @

u(0; r; θ; φ) = sin(θ)sin(φ)

Write the solution as

u(t; r; θ; φ) = v(r; θ; φ) + w(t; r; θ; φ);

where v satisﬁes the Poisson equation



∆v = 10r cos(θ)

v(1; θ; φ) = 0

If we w rite v = V (r) cosθ, then V (r) satisﬁes the ordinary equation

(

+ 2rV

¡2V = 10r

V (1) = 0; V (0): bounded

This is a Cauchy-Euler equation and is solved for V (r) =

r(r

¡1). Therefore, we obtain

u(t; r; θ; φ) = r(r

¡1)cos(θ) + w(t; r; θ; φ);

where w satisﬁes the equation

w = ∆w

w(1; θ; φ) = 0

w(0; r; θ; φ) = 0; @

w(0; r; θ; φ) = sin(θ)sin(φ)

6.3 Problems on spheres 29

Note that P

(cosθ) =sinθ and thus the so lution of the above equation is written in the series

form as

w(t; r; θ; φ) =

p=1

(t)j

(r) P

1;1

(cosθ) s in(φ);

where W

(t) satisﬁes the equation W

= ¡ζ

. Since W

(0) = 0, the solution W (t) is

(t) = B

sin(ζ

t):

Therefore, w is

w(t; r; θ; φ) =

p=1

sin(ζ

t)j

(r) P

1;1

(cosθ) sin(φ);

where constants B

satisfy the relation

1 =

p=1

(r);

and

(r) r

dr:

Finally, the series solution to the prob lem is the series

u(t; r; θ; φ) = r(r

¡1)cos(θ) +

p=1

sin(ζ

t)j

(r) sin(θ) sin(φ):

Problems

Problem 6.27. Show that P

(x) a re orthogonal in the following sense

¡1

n;m

(x) P

(x) dx = 0; n =/ n

¡1

1 ¡x

n;m

(x) P

n;m

(x) dx = 0; m =/ m

Problem 6.28. Find series of the following fun ctions in terms of Legendre polynomials fP

(x)g for

x 2[¡1; 1] and draw the ﬁrst 5 terms of each series.

a) f(x) = 1 ¡x

b) f (x) = sin(πx)

Problem 6.29. Write the series representation of the following functions in terms of functions in the

list fP

(x)g for ¡1 ≤x ≤1. Draw each series for ﬁrst 5 terms.

a) f(x) = 1 ¡x

b) f (x) = x

Problem 6.30. Repeat the above problem in terms of fP

n;2

(x)g.

Problem 6.31. Solve the Laplace equation ∆u = 0 inside the unit ball with the following boundary

conditions

a) u(1; θ; φ) = 3 cos

θ ¡1

30 3D Linear Second-Order Equations

b) u(1; θ; φ) = θ + 1

c) u (1; θ; φ) = sin(3φ) + cos(2θ)

Problem 6.32. Solve the Laplace equation ∆u = 0 in the exterior of the unit ball with the following

bo undary conditions

a) u(1; θ; φ) = P

(cosθ)

b) u(1; θ; φ) = φ

Problem 6.33. Solve the Laplace equation ∆u = 0 deﬁned on the space 1 < r < 2 in R

(the space

between two sphere r = 1, r = 2) with the following boundary condition:

a) u(1; θ; φ) = ¡1 ; u(2; θ; φ) = 1 .

b) u(1; θ; φ) = sin(φ); u(2; θ; φ) = sin (φ).

Problem 6.34. Solve the following Laplace equation inside the unit ball



∆u = 0

u(1; θ; φ) = sin(φ)

Use a calculator to ﬁnd the coeﬃcients of ﬁrst 4 terms.

Problem 6.35. Verify that the spherical Bessel functions j

(x) =

n+1/2

(x) satisﬁes the spherical

Bessel equation

+ 2xy

+ (x

¡n(n + 1))y = 0;

where J

(x) is the so lution to the Bessel equation

+xy

+ (x

¡n

)y = 0:

Problem 6.36. Write the series representation of the following functions in terms of fj

(r)g in 0≤r ≤1

and plot each

a) f(r) = 1

b) f (r) = r

c) f (r) = r

Problem 6.37. Repeat the above problem for functions in the list {j

(r)}.

Problem 6.38. Solve the Poisson equation ∆

(θ; φ)

u = θ on the surface of the unit ball. Use a calculator

to determine the coeﬃcients of the ﬁrst 4 terms.

Problem 6.39. Solve the Poisson equation ∆

(θ; φ)

u = sin

φ on the surface of the unit ball.

Problem 6.40. Explain why the eigenfunction expansion method does not work for the Poisson

equation ∆

(θ;φ)

u = 1 deﬁned on the surface of the unit ball. Is there a unique solution to the problem?

Problem 6.41. Solve the following heat problem on the surface of the unit ball

(

u = ∆

(θ;φ)

u ¡P

(cosθ)

u(0; θ; φ) = 0

;

where k ≥1 is a ﬁxed integer. Draw the temperature of the circle (1; π /4; φ) in time fo r n = 2.

Problem 6.42. Solve the following heat problem on the surface of the unit ball

(

u = ∆

(θ;φ)

u(0; θ; φ) = P

(cosθ)

;

where n ≥0.

6.3 Problems on spheres 31

Problem 6.43. Solve the following wave problem on the surface of the unit sphere

(

u = ∆

(θ;φ)

u(0; θ; φ) = 0; @

u(0; θ; φ) = f (θ; φ)

Plot the solution at θ = π/4, if f (θ; φ) = 0.1 cosθ in terms of time t.

Problem 6.44. Even though the Dirac delta δ(x) is not a function in usual sense, let us write a

symbolic series of δ(x) in terms of P

(x) a s

δ(x) =

n=0

(x):

The coeﬃcients D

are determined by the aid of inner product as

(0)

2n + 1

(0):

a) Plot the series of δ(x) for 30 terms in the interval (¡1; 1).

b) Solve the following equation on the surface of the unit ball

(

u = ∆

(θ;φ)

u ¡δ(t ¡1) δ(θ ¡π /2)

u(0; θ; φ) = @

u(0; θ; φ) = 0

Problem 6.45. Consider the following Poisson equation inside the unit ball



∆u = 1

u(1; θ; φ) = 0

a) Find a series solution to the problem.

b) Find the solution in closed form.

Problem 6.46. Consider the following Poisson equation inside the unit ball



∆u = P

(cos θ)

u(1; θ; φ) = 0

a) Find a series solution to the problem.

b) Find the solution in closed form.

Problem 6.47. Solve the following Poisson equation inside the unit ball



∆u = φ

u(1; θ; φ) = 1 + cos(θ)

Problem 6.48. Solve the following heat equation inside the unit ball

u = ∆u ¡cos(θ)

u(t; 1; θ; φ) =

cos(θ)

u(0; r; θ; φ ) =

cos(θ)

and ﬁnd ﬁrst 3 terms of the transient solution.

Problem 6.49. Solve the following heat problem inside the unit ball

u = k ∆u ¡P

(cos θ) sin(φ)

u(t; 1; θ; φ) = 0

u(0; r; θ; φ ) = 0

;

32 3D Linear Second-Order Equations

and draw few terms of the series solution at the point r = 0.5; θ = π / 2,φ = π /2 with respect to time t

for k = 0.001.

Problem 6.50. (10 points) Let Ω denote the space between the spheres r = 1 and r = 2.

a) Solve the Laplace equation ∆u = 0 in Ω with t he boundary conditions

u(1; θ; φ) = 1 + cos(θ); u(2; θ; φ) = 2 + 2cos(θ)

b) Use P

(cosθ) and ﬁnd a closed form so lution to the following Poisson equation

∆u = 18r sin

θ sin(2φ)

u(1; θ; φ) = 1 + cos(θ)

u(2; θ; φ) = 2 + 2 cos(θ)

Problem 6.51. Solve the following heat problem inside the unit ball

u = 0.001∆u ¡δ(t ¡1)P

(cos θ) sin(φ)

u(t; 1; θ; φ) = 0

u(0; r; θ; φ) = 0

;

and draw few terms of the series solution at the point r = 0.5; θ = π/2,φ = π/2 with respect to time t.

Problem 6.52. Solve the following damped wave equation inside the unit ball

u + 2ξ@

u = c

∆u

u(t; 1; θ; φ) = 0

u(0; r; θ; φ ) = 0; @

u(0; r; θ; φ ) = φ

Assume tha t ξ > 0 is small enough and the system is in the under-damped mode.

Problem 6.53. Solve the following wave problem inside the unit ball

u = c

∆u ¡e

¡t

u(t; 1; θ; φ) = 0

u(0; r; θ; φ ) = r

Problem 6.54. Let D be the following section of the unit ball:

D =

(r; θ; φ); 0 < r < 1; 0 ≤θ ≤π; 0 ≤φ ≤

Solve the following heat equation on D:

u = ∆u

u(t; r; θ; 0) = u(t; r; θ; π/2) = 0

u(t; 1; θ; φ) = cos(θ) sin(2φ)

u(0; r; θ; φ) = 0

6.4 Legendre polynomials

The force ﬁeld generated by a mass m in R

= ¡

jrj

;

6.4 Legendre polynomials 33

where e

is the unit vector in the direction of r, the line connecting a point (x; y; z) to the

location of m. P. S. Laplace observed that the above force ﬁeld is the gradient of a scalar

function which is equal to

V (r) =

jrj

;

and moreover, V satisﬁes the equation ∆V = 0 if r =/ 0. The British Mathematician G.

Green was the ﬁrst person who called V a potential function for f

. For this reason, the

Laplace equation is sometimes called a potential equation. Th e French mathematician, J.

Legendre gave a series representati o n of the potential gen erated by a mass which is located

at the point (0; 0; R); see Fig.6.2.

Figure 6.2.

The potential (assuming m = 1 and omitting the constant G) i s equal to

V =

jrj

+ s

¡2sRcos(θ)

Substituting ρ =

and x = cos(θ), we reach

V =

1 + ρ

¡2ρx

Let us write the series of the right hand side of V (ρ) in terms of ρ as f o llows (omitting

)

1 + ρ

¡2ρx

n=0

(x) ρ

; (6.38)

for some functions P

(x). From (6.38) it is immediately obtained that P

is of th e following

form

(x) =

@ρ

(1 + ρ

¡2ρx)

¡1/2

ρ=0

34 3D Linear Second-Order Equations

On the other hand, it turns out that P

(x) satisfies a differential equation ca lled the

Legendre equation. In fact, since V is a potential, we have ∆V = 0 or equivalently

∆(P

(cosθ)ρ

) = 0. We have

∆(P

(cosθ) ρ

) =

n(n + 1)

n¡2

(cos θ) +

n¡2

sin(θ)

dθ

(sinθP

(cos θ)) = 0:

Simplifying the above relation g ives the fo llow ing equation:

n(n + 1) P

sin(θ)

dθ

(sinθP

) = 0:

For x = cosθ, the above equation reads



(1 ¡x

)



+ n(n + 1) P

= 0: (6.39)

From (6.38) we have

1 + ρ

¡2ρx

n=0

(x) ρ

n=/ m

(x) P

(x) ρ

n+m

;

and then

¡1

1 + ρ

¡2ρx

n=1

¡1

(x) dx:

But

¡1

1 + ρ

¡2ρx

log



ρ + 1

ρ ¡1



n=0

2n + 1

and we obtain

¡1

(x) dx =

2n + 1

Again, from (6.38) we obtain for x = 1

1 + ρ

¡2ρ

1 ¡ ρ

n=0

;

and thus P

(1) = 1 for all n ≥0. For x = ¡1 we h ave

1 + ρ

+ 2ρ

1 + ρ

n=0

(¡1)

and therefore P

(¡1) = (¡1)

. Also changing ρ to ¡ρ gives

n=0

(x) (¡ρ)

1 + ρ

+ 2ρx

1 + ρ

¡2ρ (¡x)

n=0

(¡x) ρ

Therefore P

(¡x) = (¡1)

(x). The list fP

(x)g

n=0

is an orthogonal basis for smooth

functions in [¡1; 1]. Therefore i f f(x), x 2[¡1; 1] is a smooth function, it can be represented

by the series

f(x) ∼

n=0

(x);

6.4 Legendre polynomials 35

where the coeﬃcients f

are calculated by the aid of inner product as

2n + 1

¡1

f(x) P

(x) dx: (6.40)

It is simply seen that the above series can be written in an integral form. In fact, we have

f(x) ∼

n=0



2n + 1

¡1

f(y) P

(y) dy



(x) =

¡1

f(y)

n=0

2n + 1

(x)P

(y)

and if we denote the kernel K(x; y) as

K(x; y) =

n=0

2n + 1

(x)P

(y); (6.41)

then the function f(x) can be represented as the integral

f(x) ∼

¡1

K(x; y)f(y) dy: (6.42)

36 3D Linear Second-Order Equations