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)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/643410c38db432241aa104ccc1df886173026479)
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 )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea9714f124e86f7a661c2737c0288b59a3ac2580)
We choose for the example the Robin boundary conditions and initial conditions as follows:
![{\displaystyle u(r,\theta ,t)=R(r)\Theta (\theta )T(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca9bf9a9d6f6ac5cf48fe92d72f28cdb0ae25382)
![{\displaystyle \Rightarrow R\Theta T'=k\left[{\frac {1}{r}}(R'\Theta T+rR''\Theta T)+{\frac {1}{r^{2}}}R\Theta ''T\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5658baa9522d6d5ce6c0a53a8f63c14a3f872509)
![{\displaystyle \Rightarrow {\frac {T'}{kT}}={\frac {1}{r}}\left({\frac {R'}{R}}+r{\frac {R''}{R}}\right)+{\frac {1}{r^{2}}}{\frac {\Theta ''}{\Theta }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fb8143e143f35e1163405dfdd477c0e38d25b94)
This means that a separation constant can be found that both sides will equal. Let's define it to be
This yields:
![{\displaystyle {\color {Blue}T'+k\lambda ^{2}T=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cae950d4462772c993bc15c1f2d2545685884176)
and multiplying the other side by
yields:
![{\displaystyle r\left({\frac {R'}{R}}+r{\frac {R''}{R}}\right)+\lambda ^{2}r^{2}=-{\frac {\Theta ''}{\Theta }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/891d0667e9ad52760a97bfa853b814f2f16ea09e)
After defining another separation constant
, it yields:
![{\displaystyle {\color {Blue}\Theta ''+\mu ^{2}\Theta =0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/229d96ad13bdde0a5d25383a3072106490a252ce)
Multiplying the other side by R yields:
![{\displaystyle {\color {Blue}r^{2}R''+rR'+\left(\lambda ^{2}r^{2}-\mu ^{2}\right)R=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36052cd6496631fdcd1ca749442a22e082151298)
We now have separate differential equations for each variable.
![{\displaystyle \left\vert R(0)\Theta (\theta )T(t)\right\vert <\infty \Rightarrow \left\vert R(0)\right\vert <\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/938fc9d17668d8372a4ee4736de09f5dcda2e74e)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68c6785ca7d1b07943e199048f2d7850c18349da)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c358b3308ed5750da872218024d2e60edca1a83)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a91355469b21172e16726bde7b17f2f7e163e2f8)
The SLP for
is a singular Bessel type, whose eigenvalues
depends on
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/625b91b9dcf490c4a037237c59e25f283f1a8377)
and the eigenfunction is:
![{\displaystyle R_{mn}(r)=J_{\mu _{m}}(\lambda _{mn}r),\quad m,n=0,1,2,\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c858694109f49757ef5d0b9dabc9f539d1a123a)
where
is the Bessel function of the first kind of order
.
![{\displaystyle T'+k\lambda _{mn}^{2}T=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36d4028602465e09134ed8bdf1d79ee0e0fc31d8)
![{\displaystyle \Rightarrow T_{mn}(t)=C_{mn}e^{-k\lambda _{mn}^{2}t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f17d6fe00b8bf1e2620d02301d98662fc3948dd)
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 )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccddabe62d0d0ad5a40e7d61eff1ac1392dd4160)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/503f07763a5b5c151b6552aca152eb2ab966b600)
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 }}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/122f9870555273b38669818fbf51e7917a1b1cc3)
The weight function in the inner product
in integrals involving the Bessel functions. The Bessel functions
are orthogonal relative to the "weighted" scalar product
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 )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d50bbe15df5872315386ff714a03ec3636699b05)
![{\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 }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0363d8f9d2e020592f7ed89853b215868124900)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/476b1eb56deeca4f5d57393146c42bf2f9506f66)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86b8cfdf4a749d5265edf0815adee73f6d646626)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71fe15a33d6f86bc343d5d9da36d2617f7ed9735)
From the linear independence of
, it follows that:
![{\displaystyle T_{mn}'(t)+k\lambda _{mn}^{2}T_{mn}(t)=H_{mn}(t)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/179804e9f642677d7f33c1fa87687fa52ccd3345)
This first-order ODE can be solved with the following integration factor:
![{\displaystyle \mu (t)=e^{k\lambda _{mn}^{2}t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fae23bacc399e593ff9525c54ccbb4141d396d95)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08d52c71da7648a3469f8609e1781dc4747d4d98)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/103641475121b85fc23da8767bcc85d88ba8ff5b)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa62270149b92c750d5da6570c92726936070fad)