# 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:

${\displaystyle D:=(0,a)\times (0,b)\times (0,L)~.}$

By changing the coordinate system, we arrive at the following nonhomogeneous PDE for the heat equation:

${\displaystyle u_{t}=k\left[{\frac {1}{r}}\left(u_{r}+ru_{rr}\right)+{\frac {1}{r^{2}}}u_{\theta \theta }+u_{zz}\right]+h(r,\theta ,z,t),{\text{ where }}(r,\theta ,z)\in D,t\in (0,\infty )~.}$

We choose for the example the Robin boundary conditions and initial conditions as follows:

${\displaystyle {\begin{cases}\left\vert u(0,\theta ,z,t)\right\vert <\infty \\\alpha _{1}u(a,\theta ,z,t)+\beta _{1}u_{r}(a,\theta ,z,t)=0\\\alpha _{2}u(r,0,z,t)-\beta _{2}u_{\theta }(r,0,z,t)=0\\\alpha _{3}u(r,b,z,t)+\beta _{3}u_{\theta }(r,b,z,t)=0\\\alpha _{4}u(r,\theta ,0,t)-\beta _{4}u_{z}(r,\theta ,0,t)=0\\\alpha _{5}u(r,\theta ,L,t)+\beta _{5}u_{z}(r,\theta ,L,t)=0\\u(r,\theta ,z,0)=f(r,\theta ,z)\end{cases}}}$

## 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

${\displaystyle u(r,\theta ,z,t)=R(r)\Theta (\theta )Z(z)T(t)}$

${\displaystyle \Rightarrow R\Theta ZT'=k\left(R''\Theta ZT+{\frac {1}{r}}R'\Theta ZT+{\frac {1}{r^{2}}}R\Theta ''ZT+R\Theta Z''T\right)}$

${\displaystyle {\frac {T'}{kT}}={\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+{\frac {Z''}{Z}}}$

There is a separation constant ${\displaystyle \gamma ^{2}}$ that both sides of the equation are equivalent to. This yields:

${\displaystyle {\frac {T'}{kT}}=-\gamma ^{2}\Rightarrow {\color {Blue}T'+k\gamma ^{2}T=0}}$

${\displaystyle {\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}+\gamma ^{2}=-{\frac {Z''}{Z}}=\mu ^{2}}$

The second equation yields the equations:

${\displaystyle {\color {Blue}Z''+\mu ^{2}Z=0}}$

${\displaystyle {\frac {R''}{R}}+{\frac {1}{r}}{\frac {R'}{R}}+\gamma ^{2}-\mu ^{2}=-{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}}$

${\displaystyle \Rightarrow r^{2}{\frac {R''}{R}}+r{\frac {R'}{R}}+\left(\gamma ^{2}-\mu ^{2}\right)r^{2}=-{\frac {\Theta ''}{\Theta }}=\rho ^{2}}$

This yields the following equations:

${\displaystyle {\color {Blue}\Theta ''+\rho ^{2}\Theta =0}}$

${\displaystyle {\color {Blue}r^{2}R''+rR'+\left[\left(\gamma ^{2}-\mu ^{2}\right)r^{2}-\rho ^{2}\right]R=0}}$

#### Translate Boundary Conditions

Just like in the 2-D heat equation, the boundary conditions yield:

${\displaystyle {\begin{cases}\left\vert R(0)\right\vert <\infty \\\alpha _{1}R(a)+\beta _{1}R'(a)=0\\\alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0\\\alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0\\\alpha _{4}Z(0)-\beta _{4}Z'(0)=0\\\alpha _{5}Z(L)+\beta _{5}Z'(L)=0\\\end{cases}}}$

#### Solve SLPs

{\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}}}

{\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}}}

{\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}}}

#### Solve Time Equation

${\displaystyle T'+k\gamma _{kmn}^{2}T=0}$

The solution to the equation is:

${\displaystyle T_{kmn}(t)=C_{kmn}e^{-k\left(\lambda _{kmn}^{2}+\mu _{k}^{2}\right)t},\;k,m,n=0,1,2,\cdots }$

### Step 2: Satisfy Initial Condition

Define:

${\displaystyle u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)}$

Applying the initial condition:

{\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 ${\displaystyle f}$ in terms of ${\displaystyle R_{kmn}\otimes \Theta _{m}(\theta )\otimes Z_{k}~.}$ Hence,

${\displaystyle C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }$

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

Let:

${\displaystyle u(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }T_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)}$

${\displaystyle h(r,\theta ,z,t)=\sum _{k,m,n=0}^{\infty }H_{kmn}(t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z),\;H_{kmn}(t)={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}h(r,\theta ,z,t)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }$

Substitute the expansions for u and h into the non-homogeneous equation:

${\displaystyle \sum T_{kmn}'R_{kmn}\Theta _{m}Z_{k}=k\left\{\sum T_{kmn}\left[\underbrace {\left(R_{kmn}''+{\frac {1}{r}}R_{kmn}'\right)} _{-\left[\left(\lambda _{kmn}^{2}-\mu _{k}^{2}\right)-{\frac {1}{r^{2}}}\rho _{m}^{2}\right]R_{kmn}}\Theta _{m}Z_{k}+{\frac {1}{r^{2}}}R_{kmn}\underbrace {\Theta _{m}''} _{-\rho _{m}^{2}\Theta _{m}}Z_{k}+R_{kmn}\Theta _{m}\underbrace {Z_{k}''} _{-\mu _{k}^{2}Z_{k}}\right]\right\}+\sum H_{kmn}R_{kmn}\Theta _{m}Z_{k}}$

${\displaystyle \Leftrightarrow \sum T_{kmn}'\left[R_{kmn}\Theta _{m}Z_{k}\right]=\sum \left[(-k\gamma _{kmn}^{2})T_{kmn}+H_{kmn}\right]\left[R_{kmn}\Theta _{m}Z_{k}\right]}$

From the linear independence of ${\displaystyle R_{kmn}\otimes \Theta _{m}\otimes Z_{k}~}$:

${\displaystyle T_{kmn}'(t)=k\gamma _{kmn}^{2}T_{kmn}(t)=H_{kmn}(t),\;k,m,n=0,1,2,\cdots }$

${\displaystyle T_{kmn}(t)=e^{-k\gamma _{kmn}^{2}t}\int \limits _{0}^{t}e^{k\gamma _{kmn}^{2}s}H_{kmn}(s)ds+C_{kmn}e^{-k\gamma _{kmn}^{2}t}}$

The undetermined coefficient satisfies the initial condition:

${\displaystyle C_{kmn}={\frac {\int \limits _{0}^{L}\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta ,z)R_{kmn}(r)\Theta _{m}(\theta )Z_{k}(z)rdrd\theta dz}{\int \limits _{0}^{a}R_{kmn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta \int \limits _{0}^{L}Z_{k}^{2}(z)dz}},\;k,m,n=0,1,2,\cdots }$