We are adding to the equation found in the 2-D heat equation in cylindrical coordinates , starting with the following definition:
D
:=
(
0
,
a
)
×
(
0
,
b
)
×
(
0
,
L
)
.
{\displaystyle D:=(0,a)\times (0,b)\times (0,L)~.}
By changing the coordinate system, we arrive at the following nonhomogeneous PDE for the heat equation:
u
t
=
k
[
1
r
(
u
r
+
r
u
r
r
)
+
1
r
2
u
θ
θ
+
u
z
z
]
+
h
(
r
,
θ
,
z
,
t
)
,
where
(
r
,
θ
,
z
)
∈
D
,
t
∈
(
0
,
∞
)
.
{\displaystyle u_{t}=k\left[{\frac {1}{r}}\left(u_{r}+ru_{rr}\right)+{\frac {1}{r^{2}}}u_{\theta \theta }+u_{zz}\right]+h(r,\theta ,z,t),{\text{ where }}(r,\theta ,z)\in D,t\in (0,\infty )~.}
We choose for the example the Robin boundary conditions and initial conditions as follows:
{
|
u
(
0
,
θ
,
z
,
t
)
|
<
∞
α
1
u
(
a
,
θ
,
z
,
t
)
+
β
1
u
r
(
a
,
θ
,
z
,
t
)
=
0
α
2
u
(
r
,
0
,
z
,
t
)
−
β
2
u
θ
(
r
,
0
,
z
,
t
)
=
0
α
3
u
(
r
,
b
,
z
,
t
)
+
β
3
u
θ
(
r
,
b
,
z
,
t
)
=
0
α
4
u
(
r
,
θ
,
0
,
t
)
−
β
4
u
z
(
r
,
θ
,
0
,
t
)
=
0
α
5
u
(
r
,
θ
,
L
,
t
)
+
β
5
u
z
(
r
,
θ
,
L
,
t
)
=
0
u
(
r
,
θ
,
z
,
0
)
=
f
(
r
,
θ
,
z
)
{\displaystyle {\begin{cases}\left\vert u(0,\theta ,z,t)\right\vert <\infty \\\alpha _{1}u(a,\theta ,z,t)+\beta _{1}u_{r}(a,\theta ,z,t)=0\\\alpha _{2}u(r,0,z,t)-\beta _{2}u_{\theta }(r,0,z,t)=0\\\alpha _{3}u(r,b,z,t)+\beta _{3}u_{\theta }(r,b,z,t)=0\\\alpha _{4}u(r,\theta ,0,t)-\beta _{4}u_{z}(r,\theta ,0,t)=0\\\alpha _{5}u(r,\theta ,L,t)+\beta _{5}u_{z}(r,\theta ,L,t)=0\\u(r,\theta ,z,0)=f(r,\theta ,z)\end{cases}}}
All of the boundary conditions are homogeneous, so we don't have to partition the solution into a "steady-state" portion and a "variable" portion. Otherwise, that would be the way to solve this problem.
Step 1: Solve Associated Homogeneous Equation
edit
u
(
r
,
θ
,
z
,
t
)
=
R
(
r
)
Θ
(
θ
)
Z
(
z
)
T
(
t
)
{\displaystyle u(r,\theta ,z,t)=R(r)\Theta (\theta )Z(z)T(t)}
⇒
R
Θ
Z
T
′
=
k
(
R
″
Θ
Z
T
+
1
r
R
′
Θ
Z
T
+
1
r
2
R
Θ
″
Z
T
+
R
Θ
Z
″
T
)
{\displaystyle \Rightarrow R\Theta ZT'=k\left(R''\Theta ZT+{\frac {1}{r}}R'\Theta ZT+{\frac {1}{r^{2}}}R\Theta ''ZT+R\Theta Z''T\right)}
T
′
k
T
=
R
″
R
+
1
r
R
′
R
+
1
r
2
Θ
″
Θ
+
Z
″
Z
{\displaystyle {\frac {T'}{kT}}={\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+{\frac {Z''}{Z}}}
There is a separation constant
γ
2
{\displaystyle \gamma ^{2}}
that both sides of the equation are equivalent to. This yields:
T
′
k
T
=
−
γ
2
⇒
T
′
+
k
γ
2
T
=
0
{\displaystyle {\frac {T'}{kT}}=-\gamma ^{2}\Rightarrow {\color {Blue}T'+k\gamma ^{2}T=0}}
R
″
R
+
1
r
R
′
R
+
1
r
2
Θ
″
Θ
+
γ
2
=
−
Z
″
Z
=
μ
2
{\displaystyle {\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+\gamma ^{2}=-{\frac {Z''}{Z}}=\mu ^{2}}
The second equation yields the equations:
Z
″
+
μ
2
Z
=
0
{\displaystyle {\color {Blue}Z''+\mu ^{2}Z=0}}
R
″
R
+
1
r
R
′
R
+
γ
2
−
μ
2
=
−
1
r
2
Θ
″
Θ
{\displaystyle {\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+\gamma ^{2}-\mu ^{2}=-{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}}
⇒
r
2
R
″
R
+
r
R
′
R
+
(
γ
2
−
μ
2
)
r
2
=
−
Θ
″
Θ
=
ρ
2
{\displaystyle \Rightarrow r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}+\left(\gamma ^{2}-\mu ^{2}\right)r^{2}=-{\frac {\Theta ''}{\Theta }}=\rho ^{2}}
This yields the following equations:
Θ
″
+
ρ
2
Θ
=
0
{\displaystyle {\color {Blue}\Theta ''+\rho ^{2}\Theta =0}}
r
2
R
″
+
r
R
′
+
[
(
γ
2
−
μ
2
)
r
2
−
ρ
2
]
R
=
0
{\displaystyle {\color {Blue}r^{2}R''+rR'+\left[\left(\gamma ^{2}-\mu ^{2}\right)r^{2}-\rho ^{2}\right]R=0}}
Translate Boundary Conditions
edit
Just like in the 2-D heat equation, the boundary conditions yield:
{
|
R
(
0
)
|
<
∞
α
1
R
(
a
)
+
β
1
R
′
(
a
)
=
0
α
2
Θ
(
0
)
−
β
2
Θ
′
(
0
)
=
0
α
3
Θ
(
b
)
+
β
3
Θ
′
(
b
)
=
0
α
4
Z
(
0
)
−
β
4
Z
′
(
0
)
=
0
α
5
Z
(
L
)
+
β
5
Z
′
(
L
)
=
0
{\displaystyle {\begin{cases}\left\vert R(0)\right\vert <\infty \\\alpha _{1}R(a)+\beta _{1}R'(a)=0\\\alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0\\\alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0\\\alpha _{4}Z(0)-\beta _{4}Z'(0)=0\\\alpha _{5}Z(L)+\beta _{5}Z'(L)=0\\\end{cases}}}
Z
″
+
μ
2
Z
=
0
α
4
Z
(
0
)
−
β
4
Z
′
(
0
)
=
0
α
5
Z
(
L
)
+
β
5
Z
′
(
L
)
=
0
}
Eigenvalues
μ
k
: solutions to equation
(
α
4
α
5
−
β
4
β
5
μ
2
)
sin
(
μ
L
)
+
(
α
4
β
5
+
α
5
β
4
)
μ
cos
(
μ
L
)
=
0
Z
k
(
z
)
=
β
4
μ
k
cos
(
μ
k
z
)
+
α
4
sin
(
μ
k
z
)
,
k
=
0
,
1
,
2
,
⋯
{\displaystyle \left.{\begin{aligned}&Z''+\mu ^{2}Z=0\\&\alpha _{4}Z(0)-\beta _{4}Z'(0)=0\\&\alpha _{5}Z(L)+\beta _{5}Z'(L)=0\end{aligned}}\right\}{\begin{aligned}&{\text{Eigenvalues }}\mu _{k}{\text{: solutions to equation }}(\alpha _{4}\alpha _{5}-\beta _{4}\beta _{5}\mu ^{2})\sin(\mu L)+(\alpha _{4}\beta _{5}+\alpha _{5}\beta _{4})\mu \cos(\mu L)=0\\&Z_{k}(z)=\beta _{4}\mu _{k}\cos(\mu _{k}z)+\alpha _{4}\sin(\mu _{k}z),\;k=0,1,2,\cdots \end{aligned}}}
Θ
″
+
ρ
2
Θ
=
0
α
2
Θ
(
0
)
−
β
2
Θ
′
(
0
)
=
0
α
3
Θ
(
b
)
+
β
3
Θ
′
(
b
)
=
0
}
Eigenvalues
ρ
m
: solutions to equation
(
α
2
α
3
−
β
2
β
3
ρ
2
)
sin
(
ρ
b
)
+
(
α
2
β
3
+
α
3
β
2
)
ρ
cos
(
ρ
b
)
=
0
Θ
m
(
θ
)
=
β
2
ρ
m
cos
(
ρ
m
θ
)
+
α
2
sin
(
ρ
m
θ
)
,
m
=
0
,
1
,
2
,
⋯
{\displaystyle \left.{\begin{aligned}&\Theta ''+\rho ^{2}\Theta =0\\&\alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0\\&\alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0\end{aligned}}\right\}{\begin{aligned}&{\text{Eigenvalues }}\rho _{m}{\text{: solutions to equation }}(\alpha _{2}\alpha _{3}-\beta _{2}\beta _{3}\rho ^{2})\sin(\rho b)+(\alpha _{2}\beta _{3}+\alpha _{3}\beta _{2})\rho \cos(\rho b)=0\\&\Theta _{m}(\theta )=\beta _{2}\rho _{m}\cos(\rho _{m}\theta )+\alpha _{2}\sin(\rho _{m}\theta ),\;m=0,1,2,\cdots \end{aligned}}}
r
2
R
″
+
r
R
′
+
[
(
γ
2
−
μ
2
)
r
2
−
ρ
2
]
R
=
0
|
R
(
0
)
|
<
∞
α
1
R
(
a
)
+
β
1
R
′
(
a
)
=
0
}
Substitute
λ
2
=
γ
2
−
μ
k
2
and
ν
2
=
ρ
m
,
k
,
m
=
0
,
1
,
2
,
⋯
Eigenvals
λ
k
m
n
: solns to eqn
(
α
1
λ
a
+
β
1
ρ
m
)
J
ρ
m
(
λ
a
)
−
β
1
a
λ
J
ρ
m
+
1
(
λ
a
)
=
0
R
k
m
n
(
r
)
=
J
ρ
m
(
λ
k
m
n
r
)
,
k
,
m
,
n
=
0
,
1
,
2
,
⋯
γ
k
m
n
2
=
λ
k
m
n
2
+
μ
k
2
{\displaystyle \left.{\begin{aligned}&r^{2}R''+rR'+\left[\left(\gamma ^{2}-\mu ^{2}\right)r^{2}-\rho ^{2}\right]R=0\\&\left\vert R(0)\right\vert <\infty \\&\alpha _{1}R(a)+\beta _{1}R'(a)=0\end{aligned}}\right\}{\begin{aligned}&{\text{Substitute }}\lambda ^{2}=\gamma ^{2}-\mu _{k}^{2}{\text{ and }}\nu ^{2}=\rho _{m},\;k,m=0,1,2,\cdots \\&{\text{Eigenvals }}\lambda _{kmn}{\text{: solns to eqn }}(\alpha _{1}\lambda a+\beta _{1}\rho _{m})J_{\rho _{m}}(\lambda a)-\beta _{1}a\lambda J_{\rho _{m}+1}(\lambda a)=0\\&R_{kmn}(r)=J_{\rho _{m}}(\lambda _{kmn}r),\;k,m,n=0,1,2,\cdots \\&\gamma _{kmn}^{2}=\lambda _{kmn}^{2}+\mu _{k}^{2}\end{aligned}}}
T
′
+
k
γ
k
m
n
2
T
=
0
{\displaystyle T'+k\gamma _{kmn}^{2}T=0}
The solution to the equation is:
T
k
m
n
(
t
)
=
C
k
m
n
e
−
k
(
λ
k
m
n
2
+
μ
k
2
)
t
,
k
,
m
,
n
=
0
,
1
,
2
,
⋯
{\displaystyle T_{kmn}(t)=C_{kmn}e^{-k\left(\lambda _{kmn}^{2}+\mu _{k}^{2}\right)t},\;k,m,n=0,1,2,\cdots }
Step 2: Satisfy Initial Condition
edit
Define:
u
(
r
,
θ
,
z
,
t
)
=
∑
k
,
m
,
n
=
0
∞
T
k
m
n
(
t
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
{\displaystyle u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)}
Applying the initial condition:
u
(
r
,
θ
,
z
,
0
)
=
f
(
r
,
θ
,
z
)
=
∑
k
,
m
,
n
=
0
∞
T
k
m
n
(
0
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
=
∑
k
,
m
,
n
=
0
∞
C
k
m
n
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
{\displaystyle {\begin{aligned}u(r,\theta ,z,0)&=f(r,\theta ,z)\\&=\sum _{k,m,n=0}^{\infty }T_{kmn}(0)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)\\&=\sum _{k,m,n=0}^{\infty }C_{kmn}R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)\\\end{aligned}}}
This is the orthogonal expansion of
f
{\displaystyle f}
in terms of
R
k
m
n
⊗
Θ
m
(
θ
)
⊗
Z
k
.
{\displaystyle R_{kmn}\otimes \Theta _{m}(\theta )\otimes Z_{k}~.}
Hence,
C
k
m
n
=
∫
0
L
∫
0
b
∫
0
a
f
(
r
,
θ
,
z
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
r
d
r
d
θ
d
z
∫
0
a
R
k
m
n
2
(
r
)
r
d
r
∫
0
b
Θ
m
2
(
θ
)
d
θ
∫
0
L
Z
k
2
(
z
)
d
z
,
k
,
m
,
n
=
0
,
1
,
2
,
⋯
{\displaystyle C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }
Step 3: Solve the Non-homogeneous Equation
edit
Let:
u
(
r
,
θ
,
z
,
t
)
=
∑
k
,
m
,
n
=
0
∞
T
k
m
n
(
t
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
{\displaystyle u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)}
h
(
r
,
θ
,
z
,
t
)
=
∑
k
,
m
,
n
=
0
∞
H
k
m
n
(
t
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
,
H
k
m
n
(
t
)
=
∫
0
L
∫
0
b
∫
0
a
h
(
r
,
θ
,
z
,
t
)
R
k
m
n
(
r
)
Θ
m
(
θ
)
Z
k
(
z
)
r
d
r
d
θ
d
z
∫
0
a
R
k
m
n
2
(
r
)
r
d
r
∫
0
b
Θ
m
2
(
θ
)
d
θ
∫
0
L
Z
k
2
(
z
)
d
z
,
k
,
m
,
n
=
0
,
1
,
2
,
⋯
{\displaystyle h(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }H_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z),\;H_{kmn}(t)={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}h(r,\theta ,z,t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }
Substitute the expansions for u and h into the non-homogeneous equation:
∑
T
k
m
n
′
R
k
m
n
Θ
m
Z
k
=
k
{
∑
T
k
m
n
[
(
R
k
m
n
″
+
1
r
R
k
m
n
′
)
⏟
−
[
(
λ
k
m
n
2
−
μ
k
2
)
−
1
r
2
ρ
m
2
]
R
k
m
n
Θ
m
Z
k
+
1
r
2
R
k
m
n
Θ
m
″
⏟
−
ρ
m
2
Θ
m
Z
k
+
R
k
m
n
Θ
m
Z
k
″
⏟
−
μ
k
2
Z
k
]
}
+
∑
H
k
m
n
R
k
m
n
Θ
m
Z
k
{\displaystyle \sum T_{kmn}'R_{kmn}\Theta _{m}Z_{k}=k\left\{\sum T_{kmn}\left[\underbrace {\left(R_{kmn}''+{\frac {1}{r}}R_{kmn}'\right)} _{-\left[\left(\lambda _{kmn}^{2}-\mu _{k}^{2}\right)-{\frac {1}{r^{2}}}\rho _{m}^{2}\right]R_{kmn}}\Theta _{m}Z_{k}+{\frac {1}{r^{2}}}R_{kmn}\underbrace {\Theta _{m}''} _{-\rho _{m}^{2}\Theta _{m}}Z_{k}+R_{kmn}\Theta _{m}\underbrace {Z_{k}''} _{-\mu _{k}^{2}Z_{k}}\right]\right\}+\sum H_{kmn}R_{kmn}\Theta _{m}Z_{k}}
⇔
∑
T
k
m
n
′
[
R
k
m
n
Θ
m
Z
k
]
=
∑
[
(
−
k
γ
k
m
n
2
)
T
k
m
n
+
H
k
m
n
]
[
R
k
m
n
Θ
m
Z
k
]
{\displaystyle \Leftrightarrow \sum T_{kmn}'\left[R_{kmn}\Theta _{m}Z_{k}\right]=\sum \left[(-k\gamma _{kmn}^{2})T_{kmn}+H_{kmn}\right]\left[R_{kmn}\Theta _{m}Z_{k}\right]}
From the linear independence of
R
k
m
n
⊗
Θ
m
⊗
Z
k
{\displaystyle R_{kmn}\otimes \Theta _{m}\otimes Z_{k}~}
:
T
k
m
n
′
(
t
)
=
k
γ
k
m
n
2
T
k
m
n
(
t
)
=
H
k
m
n
(
t
)
,
k
,
m
,
n
=
0
,
1
,
2
,
⋯
{\displaystyle T_{kmn}'(t)=k\gamma _{kmn}^{2}T_{kmn}(t)=H_{kmn}(t),\;k,m,n=0,1,2,\cdots }
T
k
m
n
(
t
)
=
e
−
k
γ
k
m
n
2
t
∫
0
t
e
k
γ
k
m
n
2
s
H
k
m
n
(
s
)
d
s
+
C
k
m
n
e
−
k
γ
k
m
n
2
t
{\displaystyle T_{kmn}(t)=e^{-k\gamma _{kmn}^{2}t}\int \limits _{0}^{t}e^{k\gamma _{kmn}^{2}s}H_{kmn}(s)ds+C_{kmn}e^{-k\gamma _{kmn}^{2}t}}
The undetermined coefficient satisfies the initial condition:
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{\displaystyle C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }