# Heat equation/Solution to the 2-D Heat Equation in Cylindrical Coordinates

## Definition

We start by changing the Laplacian operator in the 2-D heat equation from rectangular to cylindrical coordinates by the following definition:

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

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

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

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

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

## Solution

### Step 1: Solve Associated Homogeneous Equation

#### Separate Variables

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

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

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

This means that a separation constant can be found that both sides will equal. Let's define it to be ${\displaystyle \lambda ^{2}~.}$ This yields:

${\displaystyle {\color {Blue}T'+k\lambda ^{2}T=0}}$

and multiplying the other side by ${\displaystyle r^{2}}$ yields:

${\displaystyle r\left({\frac {R'}{R}}+r{\frac {R''}{R}}\right)+\lambda ^{2}r^{2}=-{\frac {\Theta ''}{\Theta }}}$

After defining another separation constant ${\displaystyle \mu ^{2}}$, it yields:

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

Multiplying the other side by R yields:

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

We now have separate differential equations for each variable.

#### Translate Boundary Conditions

${\displaystyle \left\vert R(0)\Theta (\theta )T(t)\right\vert <\infty \Rightarrow \left\vert R(0)\right\vert <\infty }$

${\displaystyle \alpha _{1}R(a)\Theta (\theta )T(t)+\beta _{1}R'(a)\Theta (\theta )T(t)=0\Rightarrow \alpha _{1}R(a)+\beta _{1}R'(a)=0}$

${\displaystyle \alpha _{2}R(r)\Theta (0)T(t)+\beta _{2}R(r)\Theta '(0)T(t)=0\Rightarrow \alpha _{2}\Theta (0)-\beta _{2}\Theta '(0)=0}$

${\displaystyle \alpha _{3}R(r)\Theta (b)T(t)+\beta _{2}R(r)\Theta '(b)T(t)=0\Rightarrow \alpha _{3}\Theta (b)+\beta _{3}\Theta '(b)=0}$

#### Solve SLPs

{\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 ${\displaystyle R}$ is a singular Bessel type, whose eigenvalues ${\displaystyle \lambda _{mn}}$ depends on ${\displaystyle \mu _{m}}$ and are non-negative solutions to the following equation:

${\displaystyle \left(\alpha _{1}a+\beta _{1}\mu _{m}\right)J_{\mu _{m}}(\lambda a)-\beta _{1}a\lambda J_{\mu _{m}+1}(\lambda a)=0}$

and the eigenfunction is:

${\displaystyle R_{mn}(r)=J_{\mu _{m}}(\lambda _{mn}r),\quad m,n=0,1,2,\cdots }$

where ${\displaystyle J_{\nu }(x)}$ is the Bessel function of the first kind of order ${\displaystyle \nu }$.

#### Solve Time Equation

${\displaystyle T'+k\lambda _{mn}^{2}T=0}$

${\displaystyle \Rightarrow T_{mn}(t)=C_{mn}e^{-k\lambda _{mn}^{2}t}}$

### Step 2: Satisfy Initial Condition

Let's define the solution as an infinite sum:

${\displaystyle u(r,\theta ,t):=\sum _{m,n=0}^{\infty }T_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )~.}$

With the initial condition:

{\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 ${\displaystyle C_{mn}={\frac {\int \limits _{0}^{b}\int \limits _{0}^{a}f(r,\theta )R_{mn}(r)\Theta _{m}(\theta )rdrd\theta }{\int \limits _{0}^{a}R_{mn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta }}~.}$

The weight function in the inner product ${\displaystyle w(r)=r}$ in integrals involving the Bessel functions. The Bessel functions ${\displaystyle R_{m}}$ are orthogonal relative to the "weighted" scalar product ${\displaystyle _{w}:=\int \limits _{0}^{a}f(r)g(r)rdr~.}$

### Step 3: Solve Non-homogeneous Equation

Solving the non-homogeneous equation involves defining the following functions:

${\displaystyle u(r,\theta ,t):=\sum _{m,n=0}^{\infty }T_{mn}(t)R_{mn}(r)\Theta _{m}(\theta )}$

${\displaystyle h(r,\theta ,t):=\sum _{m,n=0}^{\infty }H_{mn}(t)R_{mn}(r)\Theta _{m}(\theta ),\quad H_{mn}(t)={\frac {\int \limits _{0}^{b}\int \limits _{0}^{a}h(r,\theta ,t)R_{mn}(r)\Theta _{m}(\theta )rdrd\theta }{\int \limits _{0}^{a}R_{mn}^{2}(r)rdr\int \limits _{0}^{b}\Theta _{m}^{2}(\theta )d\theta }}}$

Substitute the new definitions into the non-homogeneous equations:

{\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:

${\displaystyle {\begin{cases}\Theta _{m}''(\theta )=-\mu _{m}^{2}\Theta _{m}(\theta )\\rR_{mn}''(r)+R_{mn}'(r)={\frac {1}{r}}\left(\mu _{m}^{2}-\lambda _{mn}^{2}r^{2}\right)R_{mn}(r)\end{cases}}}$

We can eliminate the derivatives by substituting:

{\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 ${\displaystyle R_{mn}\otimes \Theta _{m}}$, it follows that:

${\displaystyle T_{mn}'(t)+k\lambda _{mn}^{2}T_{mn}(t)=H_{mn}(t)~.}$

This first-order ODE can be solved with the following integration factor:

${\displaystyle \mu (t)=e^{k\lambda _{mn}^{2}t}}$

Thus, the equation becomes:

${\displaystyle \left[e^{k\lambda _{mn}^{2}t}T_{mn}(t)\right]'=e^{k\lambda _{mn}^{2}t}H_{mn}(t)}$

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

We satisfy the initial condition:

{\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}}}