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 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
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
1
=0
+
u = 0
uj
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(x) sin(mπy) dydx =
4 cos( ) cos()
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(y) dy =
2(1 ¡cos())
sinh
¡
n
2
+ 1
p
π
n π
;
and finally
u(x; y; z) =
X
n;m=1
1
4 cos() cos()
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 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 := (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 := (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
(
φ = ¡λφ
φj
bnd(Ω)
=0
:
It is simply seen that λ > 0. In fact, we have
ZZZ
φφ = ¡λ
ZZZ
jφj
2;
and by the integration by parts formula,
ZZZ
φφ =
ZZ
bnd(Ω)
φ@
n
φ ¡
ZZZ
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
a
x
sin
b
y
sin
pπ
c
z
;
with associated eigenvalues
λ
nmp
=
a
2
+
b
2
+
c
2
:
Example 6.2. Let us solve th e following Poisson equation on := (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())
(n
2
+ 5)nπ
3
;
and finally,
w(x; y; z) =
"
X
n=1
1
¡2(1 ¡cos())
(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 := (0; π) ×(0; π) ×(0; π)
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
@
tt
u = u
uj
=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 := (0;
π) ×(0; π) ×(0; π)
8
<
:
@
t
u = u
@
n
uj
=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 denote the fo llowing c yl inder
= 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 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 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)
φ
φ
+
Z
00
Z
= 0; (6.4)
where
(r)
stands for the Laplacian on the disk. The associate eige nvalue of Eq.6.4 is
(
(r)
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(); J
np
(r) sin()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() + B
np
sin()g+
+
X
n=0
1
X
p=1
1
sinh
¡
λ
np
p
z
J
np
(r)fC
np
cos() + D
np
sin()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
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(); sin()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
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(); I
np
(r) sin()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() + B
np
sin()): (6.8)
The coefficients A
np
; B
np
are determined by the aid of the boundary data.
Example 6.4. Let denote a solid cylinder with the radius a = 1 and height H = 1. Let
us solve the following equation defined on
¯
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 denotes a solid cylinder with radius a and height H. Since φ =
(r)
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
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)
= ¡µ
(a; θ) = 0
: (6.12)
As we saw before, the eigenfunctions of the above problem are
np
(r; θ) = fJ
np
(r) cos(); J
np
(r) sin ()g:
Therefore, the eigenfunctions of on a cylinder are
φ
mnp
(r; θ; z) =
n
sin
H
z
J
np
(r) cos(); sin
H
z
J
np
(r)sin ()
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
H
z
cos(); sin
H
z
sin ()
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
¡
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()
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):