# Heat equation/Solution to the 3-D Heat Equation in Cylindrical Coordinates
## Definition

We are adding to the equation found in the 2-D heat equation in cylindrical coordinates, starting with the following definition:
$D:=(0,a)\times (0,b)\times (0,L)~.$
By changing the coordinate system, we arrive at the following nonhomogeneous PDE for the heat equation:
$u_{t}=k\left[{\frac {1}{r}}\left(u_{r}+ru_{rr}\right)+{\frac {1}{r^{2}}}u_{\theta \theta }+u_{zz}\right]+h(r,\theta ,z,t),{\text{ where }}(r,\theta ,z)\in D,t\in (0,\infty )~.$
We choose for the example the Robin boundary conditions and initial conditions as follows:
${\begin{cases}\left\vert u(0,\theta ,z,t)\right\vert <\infty \\\alpha _{1}u(a,\theta ,z,t)+\beta _{1}u_{r}(a,\theta ,z,t)=0\\\alpha _{2}u(r,0,z,t)-\beta _{2}u_{\theta }(r,0,z,t)=0\\\alpha _{3}u(r,b,z,t)+\beta _{3}u_{\theta }(r,b,z,t)=0\\\alpha _{4}u(r,\theta ,0,t)-\beta _{4}u_{z}(r,\theta ,0,t)=0\\\alpha _{5}u(r,\theta ,L,t)+\beta _{5}u_{z}(r,\theta ,L,t)=0\\u(r,\theta ,z,0)=f(r,\theta ,z)\end{cases}}$

## Solution

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
#### Separate Variables

$u(r,\theta ,z,t)=R(r)\Theta (\theta )Z(z)T(t)$
$\Rightarrow R\Theta ZT'=k\left(R''\Theta ZT+{\frac {1}{r}}R'\Theta ZT+{\frac {1}{r^{2}}}R\Theta ''ZT+R\Theta Z''T\right)$
${\frac {T'}{kT}}={\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+{\frac {Z''}{Z}}$
There is a separation constant $\gamma ^{2}$ that both sides of the equation are equivalent to. This yields:
${\frac {T'}{kT}}=-\gamma ^{2}\Rightarrow {\color {Blue}T'+k\gamma ^{2}T=0}$
${\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+\gamma ^{2}=-{\frac {Z''}{Z}}=\mu ^{2}$
The second equation yields the equations:
${\color {Blue}Z''+\mu ^{2}Z=0}$
${\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+\gamma ^{2}-\mu ^{2}=-{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}$
$\Rightarrow r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}+\left(\gamma ^{2}-\mu ^{2}\right)r^{2}=-{\frac {\Theta ''}{\Theta }}=\rho ^{2}$
This yields the following equations:
${\color {Blue}\Theta ''+\rho ^{2}\Theta =0}$
${\color {Blue}r^{2}R''+rR'+\left[\left(\gamma ^{2}-\mu ^{2}\right)r^{2}-\rho ^{2}\right]R=0}$

#### Translate Boundary Conditions

Just like in the 2-D heat equation, the boundary conditions yield:
${\begin{cases}\left\vert R(0)\right\vert <\infty \\\alpha _{1}R(a)+\beta _{1}R'(a)=0\\\alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0\\\alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0\\\alpha _{4}Z(0)-\beta _{4}Z'(0)=0\\\alpha _{5}Z(L)+\beta _{5}Z'(L)=0\\\end{cases}}$

#### Solve SLPs

$\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\$
$\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\$
$\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\$

#### Solve Time Equation

$T'+k\gamma _{kmn}^{2}T=0$
The solution to the equation is:
$T_{kmn}(t)=C_{kmn}e^{-k\left(\lambda _{kmn}^{2}+\mu _{k}^{2}\right)t},\;k,m,n=0,1,2,\cdots$

### Step 2: Satisfy Initial Condition

Define:
$u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)$
Applying the initial condition:
${\begin{aligned}u(r,\theta ,z,0)&=f(r,\theta ,z)\\&=\sum _{k,m,n=0}^{\infty }T_{kmn}(0)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)\\&=\sum _{k,m,n=0}^{\infty }C_{kmn}R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)\\\end{aligned}}$
This is the orthogonal expansion of $f$ in terms of $R_{kmn}\otimes \Theta _{m}(\theta )\otimes Z_{k}~.$ Hence,
$C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots$

### Step 3: Solve the Non-homogeneous Equation

Let:
$u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)$
$h(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }H_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z),\;H_{kmn}(t)={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}h(r,\theta ,z,t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots$
Substitute the expansions for u and h into the non-homogeneous equation:
$\sum T_{kmn}'R_{kmn}\Theta _{m}Z_{k}=k\left\{\sum T_{kmn}\left[\underbrace {\left(R_{kmn}''+{\frac {1}{r}}R_{kmn}'\right)} _{-\left[\left(\lambda _{kmn}^{2}-\mu _{k}^{2}\right)-{\frac {1}{r^{2}}}\rho _{m}^{2}\right]R_{kmn}}\Theta _{m}Z_{k}+{\frac {1}{r^{2}}}R_{kmn}\underbrace {\Theta _{m}''} _{-\rho _{m}^{2}\Theta _{m}}Z_{k}+R_{kmn}\Theta _{m}\underbrace {Z_{k}''} _{-\mu _{k}^{2}Z_{k}}\right]\right\}+\sum H_{kmn}R_{kmn}\Theta _{m}Z_{k}$
$\Leftrightarrow \sum T_{kmn}'\left[R_{kmn}\Theta _{m}Z_{k}\right]=\sum \left[(-k\gamma _{kmn}^{2})T_{kmn}+H_{kmn}\right]\left[R_{kmn}\Theta _{m}Z_{k}\right]$
From the linear independence of $R_{kmn}\otimes \Theta _{m}\otimes Z_{k}~$:
$T_{kmn}'(t)=k\gamma _{kmn}^{2}T_{kmn}(t)=H_{kmn}(t),\;k,m,n=0,1,2,\cdots$
$T_{kmn}(t)=e^{-k\gamma _{kmn}^{2}t}\int \limits _{0}^{t}e^{k\gamma _{kmn}^{2}s}H_{kmn}(s)ds+C_{kmn}e^{-k\gamma _{kmn}^{2}t}$
The undetermined coefficient satisfies the initial condition:
$C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots$