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:

$D:=(0,a)\times (0,b).$

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

$u_t=k[\frac{1}{r}(u_r+r{u}_{rr})+\frac{1}{r^2}{u}_{\theta \theta }]+h(r,\theta ,t),\text{ where }(r,\theta )\in D,t\in (0,\mathrm{\infty }).$

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

$\{\begin{array}{c}|u(0,\theta ,t)|<\mathrm{\infty }\\ {\alpha }_{1}u(a,\theta ,t)+{\beta }_1{u}_r(a,\theta ,t)=0\\ {\alpha }_{2}u(r,0,t)-{\beta }_2{u}_{\theta }(r,0,t)=0\\ {\alpha }_{3}u(r,b,t)+{\beta }_3{u}_{\theta }(r,b,t)=0\\ u(r,\theta ,0)=f(r,\theta )\end{array}$

Solution

Step 1: Solve Associated Homogeneous Equation

Separate Variables

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

$\Rightarrow R\mathrm{\Theta }{T}^{\prime }=k[\frac{1}{r}(R^{\prime }\mathrm{\Theta }T+r{R}^{″}\mathrm{\Theta }T)+\frac{1}{r^2}R{\mathrm{\Theta }}^{″}T]$

$\Rightarrow \frac{T^{\prime }}{kT}=\frac{1}{r}(\frac{R^{\prime }}{R}+r\frac{R^{″}}{R})+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}$

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

$T^{\prime }+k{\lambda }^{2}T=0$

and multiplying the other side by $r^2$ yields:

$r(\frac{R^{\prime }}{R}+r\frac{R^{″}}{R})+{\lambda }^2{r}^2=-\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}$

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

${\mathrm{\Theta }}^{″}+{\mu }^2\mathrm{\Theta }=0$

Multiplying the other side by R yields:

$r^2{R}^{″}+r{R}^{\prime }+({\lambda }^2{r}^2-{\mu }^2)R=0$

We now have separate differential equations for each variable.

Translate Boundary Conditions

$|R(0)\mathrm{\Theta }(\theta )T(t)|<\mathrm{\infty }\Rightarrow |R(0)|<\mathrm{\infty }$

${\alpha }_{1}R(a)\mathrm{\Theta }(\theta )T(t)+{\beta }_1{R}^{\prime }(a)\mathrm{\Theta }(\theta )T(t)=0\Rightarrow {\alpha }_{1}R(a)+{\beta }_1{R}^{\prime }(a)=0$

${\alpha }_{2}R(r)\mathrm{\Theta }(0)T(t)+{\beta }_{2}R(r){\mathrm{\Theta }}^{\prime }(0)T(t)=0\Rightarrow {\alpha }_2\mathrm{\Theta }(0)-{\beta }_2{\mathrm{\Theta }}^{\prime }(0)=0$

${\alpha }_{3}R(r)\mathrm{\Theta }(b)T(t)+{\beta }_{2}R(r){\mathrm{\Theta }}^{\prime }(b)T(t)=0\Rightarrow {\alpha }_3\mathrm{\Theta }(b)+{\beta }_3{\mathrm{\Theta }}^{\prime }(b)=0$

Solve SLPs

$\begin{array}{c}{\mathrm{\Theta }}^{″}+{\mu }^2\mathrm{\Theta }=0\\ {\alpha }_2\mathrm{\Theta }(0)-{\beta }_2{\mathrm{\Theta }}^{\prime }(0)=0\\ {\alpha }_3\mathrm{\Theta }(b)+{\beta }_3{\mathrm{\Theta }}^{\prime }(b)=0\end{array}\}\Rightarrow \begin{array}{cc}& \text{Eigenvalues }{\mu }_m\text{ are solutions to }({\alpha }_2{\alpha }_3-{\beta }_2{\beta }_3{\mu }^2)\sin (\mu B)+({\alpha }_2{\beta }_3+{\alpha }_3{\beta }_2)\mu \cos (\mu B)=0\\ & {\mathrm{\Theta }}_m(\theta )={\beta }_2{\mu }_m\cos ({\mu }_m\theta )+{\alpha }_2\sin ({\mu }_m\theta ),m=0,1,2,\cdots \end{array}$

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

$({\alpha }_{1}a+{\beta }_1{\mu }_m)J_{{\mu }_m}(\lambda a)-{\beta }_{1}a\lambda J_{{\mu }_m+1}(\lambda a)=0$

and the eigenfunction is:

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

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

Solve Time Equation

$T^{\prime }+k{\lambda }_{mn}^{2}T=0$

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

Step 2: Satisfy Initial Condition

Let's define the solution as an infinite sum:

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

With the initial condition:

$\begin{array}{cc}f(r,\theta )& =u(r,\theta ,0)\\ & =\sum _{m,n=0}^{\mathrm{\infty }}{T}_{mn}(0)R_{mn}(r){\mathrm{\Theta }}_m(\theta )\\ & =\sum _{m,n=0}^{\mathrm{\infty }}{C}_{mn}{R}_{mn}(r){\mathrm{\Theta }}_m(\theta )\end{array}$

where $C_{mn}=\frac{\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}f(r,\theta )R_{mn}(r){\mathrm{\Theta }}_m(\theta )rdrd\theta }{\underset{0}{\overset{a}{\int }}{R}_{mn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta }.$

The weight function in the inner product $w(r)=r$ in integrals involving the Bessel functions. The Bessel functions $R_m$ are orthogonal relative to the "weighted" scalar product $<f,g{>}_w:=\underset{0}{\overset{a}{\int }}f(r)g(r)rdr.$

Step 3: Solve Non-homogeneous Equation

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

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

$h(r,\theta ,t):=\sum _{m,n=0}^{\mathrm{\infty }}{H}_{mn}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta ),H_{mn}(t)=\frac{\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}h(r,\theta ,t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )rdrd\theta }{\underset{0}{\overset{a}{\int }}{R}_{mn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta }$

Substitute the new definitions into the non-homogeneous equations:

$\begin{array}{cc}\sum T_{m^{\prime }n}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )=& k[\frac{1}{r}(\sum T_{mn}(t)R_{m^{\prime }n}(r){\mathrm{\Theta }}_m(\theta )+r\sum T_{mn}(t)R_{m^{″}n}(r){\mathrm{\Theta }}_m(\theta ))+\frac{1}{r^2}\sum T_{mn}(t)R_{mn}(r){{\mathrm{\Theta }}_m}^{″}(\theta )]\\ & +\sum H_{mn}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )\end{array}$

We will use the following substitutions in our equation above:

$\{\begin{array}{c}{{\mathrm{\Theta }}_m}^{″}(\theta )=-{\mu }_m^2{\mathrm{\Theta }}_m(\theta )\\ r{R}_{m^{″}n}(r)+R_{m^{\prime }n}(r)=\frac{1}{r}({\mu }_m^2-{\lambda }_{mn}^2{r}^2)R_{mn}(r)\end{array}$

We can eliminate the derivatives by substituting:

$\begin{array}{cc}\sum T_{m^{\prime }n}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )& =\sum \{T_{mn}(t)k[\frac{1}{r}\underset{\frac{1}{r}({\mu }_m^2-{\lambda }_{mn}^2{r}^2)R_{mn}(r)}{\underbrace{(R_{m^{\prime }n}(r)+r{R}_{m^{″}n}(r))}}{\mathrm{\Theta }}_m(\theta )+\frac{1}{r^2}{R}_{mn}(r)\underset{-{\mu }_m^2{\mathrm{\Theta }}_m(\theta )}{\underbrace{{{\mathrm{\Theta }}_m}^{″}(\theta )}}]\}+\sum H_{mn}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )\\ & =\sum \{T_{mn}(t)k[\frac{1}{r^2}({\mu }_m^2-{\lambda }_{mn}^2{r}^2)-\frac{1}{r^2}{\mu }_m^2]\}{R}_{mn}(r){\mathrm{\Theta }}_m(\theta )+\sum H_{mn}(t)R_{mn}(r){\mathrm{\Theta }}_m(\theta )\\ & =\sum [(-k{\lambda }_{mn}^2)T_{mn}(t)+H_{mn}(t)]R_{mn}(r){\mathrm{\Theta }}_m(\theta )\end{array}$

From the linear independence of $R_{mn}\otimes {\mathrm{\Theta }}_m$, it follows that:

$T_{m^{\prime }n}(t)+k{\lambda }_{mn}^2{T}_{mn}(t)=H_{mn}(t).$

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

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

Thus, the equation becomes:

$[e^{k{\lambda }_{mn}^{2}t}{T}_{mn}(t)]=e^{k{\lambda }_{mn}^{2}t}{H}_{mn}(t)$

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

We satisfy the initial condition:

$\begin{array}{cc}u(r,\theta ,0)& =f(r,\theta )\\ & =\sum _{m,n=0}^{\mathrm{\infty }}{T}_{mn}(0)R_{mn}(r){\mathrm{\Theta }}_m(\theta )\\ & =\sum _{m,n=0}^{\mathrm{\infty }}{C}_{mn}{R}_{mn}(r){\mathrm{\Theta }}_m(\theta ),C_{mn}=\frac{\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}f(r,\theta )R_{mn}(r){\mathrm{\Theta }}_m(\theta )rdrd\theta }{\underset{0}{\overset{a}{\int }}{R}_{mn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta }\end{array}$