We start by changing the Laplacian operator in the 2-D heat equation from rectangular to cylindrical coordinates by the following definition:
D
:=
(
0
,
a
)
×
(
0
,
b
)
.
{\displaystyle D:=(0,a)\times (0,b)~.}
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
]
+
h
(
r
,
θ
,
t
)
,
where
(
r
,
θ
)
∈
D
,
t
∈
(
0
,
∞
)
.
{\displaystyle u_{t}=k\left[{\frac {1}{r}}\left(u_{r}+ru_{rr}\right)+{\frac {1}{r^{2}}}u_{\theta \theta }\right]+h(r,\theta ,t),{\text{ where }}(r,\theta )\in D,t\in (0,\infty )~.}
We choose for the example the Robin boundary conditions and initial conditions as follows:
{
|
u
(
0
,
θ
,
t
)
|
<
∞
α
1
u
(
a
,
θ
,
t
)
+
β
1
u
r
(
a
,
θ
,
t
)
=
0
α
2
u
(
r
,
0
,
t
)
−
β
2
u
θ
(
r
,
0
,
t
)
=
0
α
3
u
(
r
,
b
,
t
)
+
β
3
u
θ
(
r
,
b
,
t
)
=
0
u
(
r
,
θ
,
0
)
=
f
(
r
,
θ
)
{\displaystyle {\begin{cases}\left\vert u(0,\theta ,t)\right\vert <\infty \\\alpha _{1}u(a,\theta ,t)+\beta _{1}u_{r}(a,\theta ,t)=0\\\alpha _{2}u(r,0,t)-\beta _{2}u_{\theta }(r,0,t)=0\\\alpha _{3}u(r,b,t)+\beta _{3}u_{\theta }(r,b,t)=0\\u(r,\theta ,0)=f(r,\theta )\end{cases}}}
Step 1: Solve Associated Homogeneous Equation
edit
u
(
r
,
θ
,
t
)
=
R
(
r
)
Θ
(
θ
)
T
(
t
)
{\displaystyle u(r,\theta ,t)=R(r)\Theta (\theta )T(t)}
⇒
R
Θ
T
′
=
k
[
1
r
(
R
′
Θ
T
+
r
R
″
Θ
T
)
+
1
r
2
R
Θ
″
T
]
{\displaystyle \Rightarrow R\Theta T'=k\left[{\frac {1}{r}}(R'\Theta T+rR''\Theta T)+{\frac {1}{r^{2}}}R\Theta ''T\right]}
⇒
T
′
k
T
=
1
r
(
R
′
R
+
r
R
″
R
)
+
1
r
2
Θ
″
Θ
{\displaystyle \Rightarrow {\frac {T'}{kT}}={\frac {1}{r}}\left({\frac {R'}{R}}+r{\frac {R''}{R}}\right)+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}}
This means that a separation constant can be found that both sides will equal. Let's define it to be
λ
2
.
{\displaystyle \lambda ^{2}~.}
This yields:
T
′
+
k
λ
2
T
=
0
{\displaystyle {\color {Blue}T'+k\lambda ^{2}T=0}}
and multiplying the other side by
r
2
{\displaystyle r^{2}}
yields:
r
(
R
′
R
+
r
R
″
R
)
+
λ
2
r
2
=
−
Θ
″
Θ
{\displaystyle r\left({\frac {R'}{R}}+r{\frac {R''}{R}}\right)+\lambda ^{2}r^{2}=-{\frac {\Theta ''}{\Theta }}}
After defining another separation constant
μ
2
{\displaystyle \mu ^{2}}
, it yields:
Θ
″
+
μ
2
Θ
=
0
{\displaystyle {\color {Blue}\Theta ''+\mu ^{2}\Theta =0}}
Multiplying the other side by R yields:
r
2
R
″
+
r
R
′
+
(
λ
2
r
2
−
μ
2
)
R
=
0
{\displaystyle {\color {Blue}r^{2}R''+rR'+\left(\lambda ^{2}r^{2}-\mu ^{2}\right)R=0}}
We now have separate differential equations for each variable.
Translate Boundary Conditions
edit
|
R
(
0
)
Θ
(
θ
)
T
(
t
)
|
<
∞
⇒
|
R
(
0
)
|
<
∞
{\displaystyle \left\vert R(0)\Theta (\theta )T(t)\right\vert <\infty \Rightarrow \left\vert R(0)\right\vert <\infty }
α
1
R
(
a
)
Θ
(
θ
)
T
(
t
)
+
β
1
R
′
(
a
)
Θ
(
θ
)
T
(
t
)
=
0
⇒
α
1
R
(
a
)
+
β
1
R
′
(
a
)
=
0
{\displaystyle \alpha _{1}R(a)\Theta (\theta )T(t)+\beta _{1}R'(a)\Theta (\theta )T(t)=0\Rightarrow \alpha _{1}R(a)+\beta _{1}R'(a)=0}
α
2
R
(
r
)
Θ
(
0
)
T
(
t
)
+
β
2
R
(
r
)
Θ
′
(
0
)
T
(
t
)
=
0
⇒
α
2
Θ
(
0
)
−
β
2
Θ
′
(
0
)
=
0
{\displaystyle \alpha _{2}R(r)\Theta (0)T(t)+\beta _{2}R(r)\Theta '(0)T(t)=0\Rightarrow \alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0}
α
3
R
(
r
)
Θ
(
b
)
T
(
t
)
+
β
2
R
(
r
)
Θ
′
(
b
)
T
(
t
)
=
0
⇒
α
3
Θ
(
b
)
+
β
3
Θ
′
(
b
)
=
0
{\displaystyle \alpha _{3}R(r)\Theta (b)T(t)+\beta _{2}R(r)\Theta '(b)T(t)=0\Rightarrow \alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0}
Θ
″
+
μ
2
Θ
=
0
α
2
Θ
(
0
)
−
β
2
Θ
′
(
0
)
=
0
α
3
Θ
(
b
)
+
β
3
Θ
′
(
b
)
=
0
}
⇒
Eigenvalues
μ
m
are solutions to
(
α
2
α
3
−
β
2
β
3
μ
2
)
sin
(
μ
B
)
+
(
α
2
β
3
+
α
3
β
2
)
μ
cos
(
μ
B
)
=
0
Θ
m
(
θ
)
=
β
2
μ
m
cos
(
μ
m
θ
)
+
α
2
sin
(
μ
m
θ
)
,
m
=
0
,
1
,
2
,
⋯
{\displaystyle \left.{\begin{aligned}\Theta ''+\mu ^{2}\Theta =0\\\alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0\\\alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0\end{aligned}}\right\}\Rightarrow {\begin{aligned}&{\text{Eigenvalues }}\mu _{m}{\text{ are solutions to }}(\alpha _{2}\alpha _{3}-\beta _{2}\beta _{3}\mu ^{2})\sin(\mu B)+(\alpha _{2}\beta _{3}+\alpha _{3}\beta _{2})\mu \cos(\mu B)=0\\&\Theta _{m}(\theta )=\beta _{2}\mu _{m}\cos(\mu _{m}\theta )+\alpha _{2}\sin(\mu _{m}\theta ),m=0,1,2,\cdots \end{aligned}}}
The SLP for
R
{\displaystyle R}
is a singular Bessel type, whose eigenvalues
λ
m
n
{\displaystyle \lambda _{mn}}
depends on
μ
m
{\displaystyle \mu _{m}}
and are non-negative solutions to the following equation:
(
α
1
a
+
β
1
μ
m
)
J
μ
m
(
λ
a
)
−
β
1
a
λ
J
μ
m
+
1
(
λ
a
)
=
0
{\displaystyle \left(\alpha _{1}a+\beta _{1}\mu _{m}\right)J_{\mu _{m}}(\lambda a)-\beta _{1}a\lambda J_{\mu _{m}+1}(\lambda a)=0}
and the eigenfunction is:
R
m
n
(
r
)
=
J
μ
m
(
λ
m
n
r
)
,
m
,
n
=
0
,
1
,
2
,
⋯
{\displaystyle R_{mn}(r)=J_{\mu _{m}}(\lambda _{mn}r),\quad m,n=0,1,2,\cdots }
where
J
ν
(
x
)
{\displaystyle J_{\nu }(x)}
is the Bessel function of the first kind of order
ν
{\displaystyle \nu }
.
T
′
+
k
λ
m
n
2
T
=
0
{\displaystyle T'+k\lambda _{mn}^{2}T=0}
⇒
T
m
n
(
t
)
=
C
m
n
e
−
k
λ
m
n
2
t
{\displaystyle \Rightarrow T_{mn}(t)=C_{mn}e^{-k\lambda _{mn}^{2}t}}
Step 2: Satisfy Initial Condition
edit
Let's define the solution as an infinite sum:
u
(
r
,
θ
,
t
)
:=
∑
m
,
n
=
0
∞
T
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
.
{\displaystyle u(r,\theta ,t):=\sum _{m,n=0}^{\infty }T_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )~.}
With the initial condition:
f
(
r
,
θ
)
=
u
(
r
,
θ
,
0
)
=
∑
m
,
n
=
0
∞
T
m
n
(
0
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
∑
m
,
n
=
0
∞
C
m
n
R
m
n
(
r
)
Θ
m
(
θ
)
{\displaystyle {\begin{aligned}f(r,\theta )&=u(r,\theta ,0)\\&=\sum _{m,n=0}^{\infty }T_{mn}(0)R_{mn}(r)\Theta _{m}(\theta )\\&=\sum _{m,n=0}^{\infty }C_{mn}R_{mn}(r)\Theta _{m}(\theta )\end{aligned}}}
where
C
m
n
=
∫
0
b
∫
0
a
f
(
r
,
θ
)
R
m
n
(
r
)
Θ
m
(
θ
)
r
d
r
d
θ
∫
0
a
R
m
n
2
(
r
)
r
d
r
∫
0
b
Θ
m
2
(
θ
)
d
θ
.
{\displaystyle C_{mn}={\frac {\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta )R_{mn}(r)\Theta _{m}(\theta )rdrd\theta }{\int \limits _{0}^{a}R_{mn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta }}~.}
The weight function in the inner product
w
(
r
)
=
r
{\displaystyle w(r)=r}
in integrals involving the Bessel functions. The Bessel functions
R
m
{\displaystyle R_{m}}
are orthogonal relative to the "weighted" scalar product
<
f
,
g
>
w
:=
∫
0
a
f
(
r
)
g
(
r
)
r
d
r
.
{\displaystyle <f,g>_{w}:=\int \limits _{0}^{a}f(r)g(r)rdr~.}
Step 3: Solve Non-homogeneous Equation
edit
Solving the non-homogeneous equation involves defining the following functions:
u
(
r
,
θ
,
t
)
:=
∑
m
,
n
=
0
∞
T
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
{\displaystyle u(r,\theta ,t):=\sum _{m,n=0}^{\infty }T_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )}
h
(
r
,
θ
,
t
)
:=
∑
m
,
n
=
0
∞
H
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
,
H
m
n
(
t
)
=
∫
0
b
∫
0
a
h
(
r
,
θ
,
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
r
d
r
d
θ
∫
0
a
R
m
n
2
(
r
)
r
d
r
∫
0
b
Θ
m
2
(
θ
)
d
θ
{\displaystyle h(r,\theta ,t):=\sum _{m,n=0}^{\infty }H_{mn}(t)R_{mn}(r)\Theta _{m}(\theta ),\quad H_{mn}(t)={\frac {\int \limits _{0}^{b}\int \limits _{0}^{a}h(r,\theta ,t)R_{mn}(r)\Theta _{m}(\theta )rdrd\theta }{\int \limits _{0}^{a}R_{mn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta }}}
Substitute the new definitions into the non-homogeneous equations:
∑
T
m
n
′
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
k
[
1
r
(
∑
T
m
n
(
t
)
R
m
n
′
(
r
)
Θ
m
(
θ
)
+
r
∑
T
m
n
(
t
)
R
m
n
″
(
r
)
Θ
m
(
θ
)
)
+
1
r
2
∑
T
m
n
(
t
)
R
m
n
(
r
)
Θ
m
″
(
θ
)
]
+
∑
H
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
{\displaystyle {\begin{aligned}\sum T_{mn}'(t)R_{mn}(r)\Theta _{m}(\theta )=&k\left[{\frac {1}{r}}\left(\sum T_{mn}(t)R_{mn}'(r)\Theta _{m}(\theta )+r\sum T_{mn}(t)R_{mn}''(r)\Theta _{m}(\theta )\right)+{\frac {1}{r^{2}}}\sum T_{mn}(t)R_{mn}(r)\Theta _{m}''(\theta )\right]\\&+\sum H_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )\end{aligned}}}
We will use the following substitutions in our equation above:
{
Θ
m
″
(
θ
)
=
−
μ
m
2
Θ
m
(
θ
)
r
R
m
n
″
(
r
)
+
R
m
n
′
(
r
)
=
1
r
(
μ
m
2
−
λ
m
n
2
r
2
)
R
m
n
(
r
)
{\displaystyle {\begin{cases}\Theta _{m}''(\theta )=-\mu _{m}^{2}\Theta _{m}(\theta )\\rR_{mn}''(r)+R_{mn}'(r)={\frac {1}{r}}\left(\mu _{m}^{2}-\lambda _{mn}^{2}r^{2}\right)R_{mn}(r)\end{cases}}}
We can eliminate the derivatives by substituting:
∑
T
m
n
′
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
∑
{
T
m
n
(
t
)
k
[
1
r
(
R
m
n
′
(
r
)
+
r
R
m
n
″
(
r
)
)
⏟
1
r
(
μ
m
2
−
λ
m
n
2
r
2
)
R
m
n
(
r
)
Θ
m
(
θ
)
+
1
r
2
R
m
n
(
r
)
Θ
m
″
(
θ
)
⏟
−
μ
m
2
Θ
m
(
θ
)
]
}
+
∑
H
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
∑
{
T
m
n
(
t
)
k
[
1
r
2
(
μ
m
2
−
λ
m
n
2
r
2
)
−
1
r
2
μ
m
2
]
}
R
m
n
(
r
)
Θ
m
(
θ
)
+
∑
H
m
n
(
t
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
∑
[
(
−
k
λ
m
n
2
)
T
m
n
(
t
)
+
H
m
n
(
t
)
]
R
m
n
(
r
)
Θ
m
(
θ
)
{\displaystyle {\begin{aligned}\sum T_{mn}'(t)R_{mn}(r)\Theta _{m}(\theta )&=\sum \left\{T_{mn}(t)k\left[{\frac {1}{r}}\underbrace {\left(R_{mn}'(r)+rR_{mn}''(r)\right)} _{{\frac {1}{r}}\left(\mu _{m}^{2}-\lambda _{mn}^{2}r^{2}\right)R_{mn}(r)}\Theta _{m}(\theta )+{\frac {1}{r^{2}}}R_{mn}(r)\underbrace {\Theta _{m}''(\theta )} _{-\mu _{m}^{2}\Theta _{m}(\theta )}\right]\right\}+\sum H_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )\\&=\sum \left\{T_{mn}(t)k\left[{\frac {1}{r^{2}}}\left(\mu _{m}^{2}-\lambda _{mn}^{2}r^{2}\right)-{\frac {1}{r^{2}}}\mu _{m}^{2}\right]\right\}R_{mn}(r)\Theta _{m}(\theta )+\sum H_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )\\&=\sum \left[\left(-k\lambda _{mn}^{2}\right)T_{mn}(t)+H_{mn}(t)\right]R_{mn}(r)\Theta _{m}(\theta )\end{aligned}}}
From the linear independence of
R
m
n
⊗
Θ
m
{\displaystyle R_{mn}\otimes \Theta _{m}}
, it follows that:
T
m
n
′
(
t
)
+
k
λ
m
n
2
T
m
n
(
t
)
=
H
m
n
(
t
)
.
{\displaystyle T_{mn}'(t)+k\lambda _{mn}^{2}T_{mn}(t)=H_{mn}(t)~.}
This first-order ODE can be solved with the following integration factor:
μ
(
t
)
=
e
k
λ
m
n
2
t
{\displaystyle \mu (t)=e^{k\lambda _{mn}^{2}t}}
Thus, the equation becomes:
[
e
k
λ
m
n
2
t
T
m
n
(
t
)
]
′
=
e
k
λ
m
n
2
t
H
m
n
(
t
)
{\displaystyle \left[e^{k\lambda _{mn}^{2}t}T_{mn}(t)\right]'=e^{k\lambda _{mn}^{2}t}H_{mn}(t)}
⇒
T
m
n
(
t
)
=
e
−
k
λ
m
n
2
t
∫
0
t
e
k
λ
m
n
2
t
H
m
n
(
s
)
d
s
+
C
m
n
e
−
k
λ
m
n
2
t
{\displaystyle \Rightarrow T_{mn}(t)=e^{-k\lambda _{mn}^{2}t}\int \limits _{0}^{t}e^{k\lambda _{mn}^{2}t}H_{mn}(s)ds+C_{mn}e^{-k\lambda _{mn}^{2}t}}
We satisfy the initial condition:
u
(
r
,
θ
,
0
)
=
f
(
r
,
θ
)
=
∑
m
,
n
=
0
∞
T
m
n
(
0
)
R
m
n
(
r
)
Θ
m
(
θ
)
=
∑
m
,
n
=
0
∞
C
m
n
R
m
n
(
r
)
Θ
m
(
θ
)
,
C
m
n
=
∫
0
b
∫
0
a
f
(
r
,
θ
)
R
m
n
(
r
)
Θ
m
(
θ
)
r
d
r
d
θ
∫
0
a
R
m
n
2
(
r
)
r
d
r
∫
0
b
Θ
m
2
(
θ
)
d
θ
{\displaystyle {\begin{aligned}u(r,\theta ,0)&=f(r,\theta )\\&=\sum _{m,n=0}^{\infty }T_{mn}(0)R_{mn}(r)\Theta _{m}(\theta )\\&=\sum _{m,n=0}^{\infty }C_{mn}R_{mn}(r)\Theta _{m}(\theta ),\quad C_{mn}={\frac {\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta )R_{mn}(r)\Theta _{m}(\theta )rdrd\theta }{\int \limits _{0}^{a}R_{mn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta }}\end{aligned}}}