0
$\begingroup$

I'm trying to solve a PDE using separation of variables. The heat eq. is given by: $$u_t = k(u_{rr}+ \frac{2}{r} u_r)$$ The boundary conditions are given by: $$ \lim_{r \to 0} \frac{\partial u}{\partial r} = 0; \quad \text{for} \quad t>0. $$ and $$ u(a,t) = u_{0}; \quad \text{for} \quad t>0. $$ where $u_0$ is a constant. The initial condition is given by: $$ u(r,0) = 0; \quad \text{for} \quad 0<r<a.$$.

Attempt at solution: Let: $u = R(r)T(t)$ be the solution to the PDE. From this, we have: $$\frac{\partial u}{\partial t} = R(r)T'(t)$$ $$\frac{\partial u}{\partial r} = R'(r)T(t)$$ $$\frac{\partial^2 u}{\partial r^2} = R''(r)T(t)$$ BY substituting back into the PDE, we have: $$\frac{T'(t)}{T(t)}=k(\frac{R''(r)}{R(r)} + \frac{2}{r}\frac{R'(r)}{R(r)})= -\lambda^2; \quad \text{where} \quad \lambda > 0.$$ Here, $\lambda$ is the separation constant. Then, we will get 2 ODEs given by: $$T'(t)+ \lambda T(t)=0$$ $$k(rR''(r)+2R'(r)+\lambda rR(r))= 0$$ The solution to the first ODE is $$ T = Ae^{- \lambda t}$$ How do I proceed from here and apply the BCs/ICs?

$\endgroup$

1 Answer 1

1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Lets $\ds{\varphi\pars{r,t} \equiv \on{u}\pars{r,t} - u_{0}}$ such that
$\left\{\begin{array}{rcl} \ds{\varphi_{t}} & \ds{=} & \ds{k\pars{\varphi_{rr} + {2 \over r}\,\varphi_{r}}} \\[2mm] &&\left.\begin{array}{rcl} \ds{\varphi_{r}\pars{0,t}} & \ds{=} & \ds{0} \\[1mm] \ds{\varphi\pars{a,t}} & \ds{=} & \ds{0} \end{array}\right\} \substack{\ds{Boundary} \\[0.5mm] \ds{Conditions}} \\[2mm] \ds{\varphi\pars{r,0}} & \ds{=} & \ds{-u_{0}}\ \pars{~Initial\ Condition~} \end{array}\right.$


  • Set $\ds{\varphi\pars{r,t} = \on{A}\pars{t}\on{B}\pars{r}}$. Then, \begin{align} \on{A}_{t}\on{B} & = k\pars{\on{A}\on{B}_{rr} + {2 \over r}\on{A}\on{B}_{r}} \\[2mm] {\on{A}_{t} \over \on{A}} & = k\pars{{\on{B}_{rr} \over \on{B}}+ {2 \over r}{\on{B}_{r} \over \on{B}}} = -q^{2}\,,\quad q \in \mathbb{R} \\[2mm] \on{A} &= -q^{2}\on{A}\,,\quad k\on{B}_{rr} + {2k \over r}\on{B}_{r} + q^{2}\on{B} = 0 \end{align} Lets $\ds{\on{B} \propto {\expo{\alpha r} \over r}}$ which yields $\ds{\alpha = \pm \ic q/\root{k}}$. In order to satisfy the boundary condition at $\ds{r = 0}$, $\ds{\on{B} \propto {\sin\pars{qr/\root{k}}}/r}$. At $\ds{r = a}$, $\ds{\sin\pars{qa/\root{k}} = 0}$ $\ds{\implies q_{n} = n\,{\pi\root{k} \over a}\,,\quad n = 1,2,3,\ldots}$ The general solution becomes \begin{align} \varphi\pars{r,t} & = \sum_{n = 1}^{\infty}\on{A}_{n}\pars{0}\exp\pars{-q_{n}^{2}\,t} {\sin\pars{q_{n}r/\root{k}} \over r} \\[5mm] -u_{0} & = \sum_{n = 1}^{\infty}\on{A}_{n}\pars{0} {\sin\pars{q_{n}r/\root{k}} \over r} \\[5mm] &-u_{0}\ \overbrace{\int_{0}^{a}r\sin\pars{{q_{n} \over \root{k}}\,r} \dd r} ^{\ds{-\,{\pars{-1}^{n}a^{2} \over n\pi}}} \\[2mm] & = \sum_{m = 1}^{\infty}\on{A}_{m}\pars{0}\ \underbrace{\int_{0}^{a}\sin\pars{{q_{m} \over \root{k}}\,r} \sin\pars{{q_{m} \over \root{k}}\,r}\,\dd r} _{\ds{{a \over 2}\,\delta_{mn}}} \end{align} $\ds{\implies \on{A}_{n}\pars{0} = {2\pars{-1}^{n}au_{0} \over n\pi}}$.
  • Finally $\ds{\pars{~\mbox{with}\ q_{n} = n\,{\pi\root{k} \over a}~}}$, $$ \bbx{\on{u}\pars{r,t} = u_{0} + {2au_{0} \over \pi}\,{1 \over r}\sum_{n = 1}^{\infty} {\pars{-1}^{n} \over n}\exp\pars{-q_{n}^{2}\,t} \sin\pars{{q_{n} \over \root{k}}\,r}}\\ $$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.