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:

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

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

$\{\begin{array}{c}|u(0,\theta ,z,t)|<\mathrm{\infty }\\ {\alpha }_{1}u(a,\theta ,z,t)+{\beta }_1{u}_r(a,\theta ,z,t)=0\\ {\alpha }_{2}u(r,0,z,t)-{\beta }_2{u}_{\theta }(r,0,z,t)=0\\ {\alpha }_{3}u(r,b,z,t)+{\beta }_3{u}_{\theta }(r,b,z,t)=0\\ {\alpha }_{4}u(r,\theta ,0,t)-{\beta }_4{u}_z(r,\theta ,0,t)=0\\ {\alpha }_{5}u(r,\theta ,L,t)+{\beta }_5{u}_z(r,\theta ,L,t)=0\\ u(r,\theta ,z,0)=f(r,\theta ,z)\end{array}$

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

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

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

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

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

$\frac{T^{\prime }}{kT}=-{\gamma }^2\Rightarrow T^{\prime }+k{\gamma }^{2}T=0$

$\frac{R^{″}}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}+{\gamma }^2=-\frac{Z^{″}}{Z}={\mu }^2$

The second equation yields the equations:

$Z^{″}+{\mu }^{2}Z=0$

$\frac{R^{″}}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+{\gamma }^2-{\mu }^2=-\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{″}}{\mathrm{\Theta }}$

$\Rightarrow r^2\frac{R^{″}}{R}+r\frac{R^{\prime }}{R}+({\gamma }^2-{\mu }^2)r^2=-\frac{{\Theta }^{″}}{\Theta }={\rho }^2$

This yields the following equations:

${\Theta }^{″}+{\rho }^2\Theta =0$

$r^2{R}^{″}+r{R}^{\prime }+[({\gamma }^2-{\mu }^2)r^2-{\rho }^2]R=0$

Translate Boundary Conditions

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

$\{\begin{array}{c}|R(0)|<\mathrm{\infty }\\ {\alpha }_{1}R(a)+{\beta }_1{R}^{\prime }(a)=0\\ {\alpha }_2\Theta (0)-{\beta }_2{\Theta }^{\prime }(0)=0\\ {\alpha }_3\Theta (b)+{\beta }_3{\Theta }^{\prime }(b)=0\\ {\alpha }_{4}Z(0)-{\beta }_4{Z}^{\prime }(0)=0\\ {\alpha }_{5}Z(L)+{\beta }_5{Z}^{\prime }(L)=0\end{array}$

Solve SLPs

$\begin{array}{cc}& Z^{″}+{\mu }^{2}Z=0\\ & {\alpha }_{4}Z(0)-{\beta }_4{Z}^{\prime }(0)=0\\ & {\alpha }_{5}Z(L)+{\beta }_5{Z}^{\prime }(L)=0\end{array}\}\begin{array}{cc}& \text{Eigenvalues }{\mu }_k\text{: solutions to equation }({\alpha }_4{\alpha }_5-{\beta }_4{\beta }_5{\mu }^2)\sin (\mu L)+({\alpha }_4{\beta }_5+{\alpha }_5{\beta }_4)\mu \cos (\mu L)=0\\ & Z_k(z)={\beta }_4{\mu }_k\cos ({\mu }_{k}z)+{\alpha }_4\sin ({\mu }_{k}z),k=0,1,2,\cdots \end{array}$

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

$\begin{array}{cc}& r^2{R}^{″}+r{R}^{\prime }+[({\gamma }^2-{\mu }^2)r^2-{\rho }^2]R=0\\ & |R(0)|<\mathrm{\infty }\\ & {\alpha }_{1}R(a)+{\beta }_1{R}^{\prime }(a)=0\end{array}\}\begin{array}{cc}& \text{Substitute }{\lambda }^2={\gamma }^2-{\mu }_k^2\text{ and }{\nu }^2={\rho }_m,k,m=0,1,2,\cdots \\ & \text{Eigenvals }{\lambda }_{kmn}\text{: solns to eqn }({\alpha }_1\lambda a+{\beta }_1{\rho }_m)J_{{\rho }_m}(\lambda a)-{\beta }_{1}a\lambda J_{{\rho }_m+1}(\lambda a)=0\\ & R_{kmn}(r)=J_{{\rho }_m}({\lambda }_{kmn}r),k,m,n=0,1,2,\cdots \\ & {\gamma }_{kmn}^2={\lambda }_{kmn}^2+{\mu }_k^2\end{array}$

Solve Time Equation

$T^{\prime }+k{\gamma }_{kmn}^{2}T=0$

The solution to the equation is:

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

Step 2: Satisfy Initial Condition

Define:

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

Applying the initial condition:

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

This is the orthogonal expansion of $f$ in terms of $R_{kmn}\otimes {\mathrm{\Theta }}_m(\theta )\otimes Z_k.$ Hence,

$C_{kmn}=\frac{\underset{0}{\overset{L}{\int }}\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}f(r,\theta ,z)R_{kmn}(r){\mathrm{\Theta }}_m(\theta )Z_k(z)rdrd\theta dz}{\underset{0}{\overset{a}{\int }}{R}_{kmn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta \underset{0}{\overset{L}{\int }}{Z}_k^2(z)dz},k,m,n=0,1,2,\cdots$

Step 3: Solve the Non-homogeneous Equation

Let:

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

$h(r,\theta ,z,t)=\sum _{k,m,n=0}^{\mathrm{\infty }}{H}_{kmn}(t)R_{kmn}(r){\mathrm{\Theta }}_m(\theta )Z_k(z),H_{kmn}(t)=\frac{\underset{0}{\overset{L}{\int }}\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}h(r,\theta ,z,t)R_{kmn}(r){\mathrm{\Theta }}_m(\theta )Z_k(z)rdrd\theta dz}{\underset{0}{\overset{a}{\int }}{R}_{kmn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta \underset{0}{\overset{L}{\int }}{Z}_k^2(z)dz},k,m,n=0,1,2,\cdots$

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

$\sum T_{k^{\prime }mn}{R}_{kmn}{\mathrm{\Theta }}_m{Z}_k=k\{\sum T_{kmn}[\underset{-[({\lambda }_{kmn}^2-{\mu }_k^2)-\frac{1}{r^2}{\rho }_m^2]R_{kmn}}{\underbrace{(R_{k^{″}mn}+\frac{1}{r}{R}_{k^{\prime }mn})}}{\mathrm{\Theta }}_m{Z}_k+\frac{1}{r^2}{R}_{kmn}\underset{-{\rho }_m^2{\mathrm{\Theta }}_m}{\underbrace{{{\mathrm{\Theta }}_m}^{″}}}{Z}_k+R_{kmn}{\mathrm{\Theta }}_m\underset{-{\mu }_k^2{Z}_k}{\underbrace{{Z_k}^{″}}}]\}+\sum H_{kmn}{R}_{kmn}{\mathrm{\Theta }}_m{Z}_k$

$\iff \sum T_{k^{\prime }mn}\left[R_{kmn}{\mathrm{\Theta }}_m{Z}_k\right]=\sum [(-k{\gamma }_{kmn}^2)T_{kmn}+H_{kmn}]\left[R_{kmn}{\mathrm{\Theta }}_m{Z}_k\right]$

From the linear independence of $R_{kmn}\otimes {\mathrm{\Theta }}_m\otimes Z_k$:

$T_{k^{\prime }mn}(t)=k{\gamma }_{kmn}^2{T}_{kmn}(t)=H_{kmn}(t),k,m,n=0,1,2,\cdots$

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

The undetermined coefficient satisfies the initial condition:

$C_{kmn}=\frac{\underset{0}{\overset{L}{\int }}\underset{0}{\overset{b}{\int }}\underset{0}{\overset{a}{\int }}f(r,\theta ,z)R_{kmn}(r){\mathrm{\Theta }}_m(\theta )Z_k(z)rdrd\theta dz}{\underset{0}{\overset{a}{\int }}{R}_{kmn}^2(r)rdr\underset{0}{\overset{b}{\int }}{\mathrm{\Theta }}_m^2(\theta )d\theta \underset{0}{\overset{L}{\int }}{Z}_k^2(z)dz},k,m,n=0,1,2,\cdots$