1

Solve $u_{xx}+u_{yy}+u_{zz}=0$ where $0<a<r<b$ and $u=A$ when $r=a$, $u=B$ when $r=b$

I know in $3$ dimensions I can rewrite this in polar coordinates as $u_{rr}+\frac{2}{r}u_r+\frac{1}{r^2}[u_{\theta\theta}+\cot(\theta)u_{\theta}+\frac{1}{\sin^2(\theta)}u_{\theta\theta}]$

The text says to look for solutions which only depend on $r$, so I want to solve $u_{rr}+\frac{2}{r}u_{r}=0$ with the given conditions.

But my problem is I'm not sure how to solve ODE's like this and the general solution that I've been using to solve other laplace equations was separation of variables but that seems not useful here.

AColoredReptile
  • 2,729
  • You solve $u_{rr}+\frac{2}{r}u_r=0$ for $u_r$ with the integrating factor method. Standard methods to solve ODEs should have been taught before learning about PDEs. – anon Dec 11 '21 at 22:27

1 Answers1

0

\begin{align*} u_{rr} + \frac{2}{r}u_r &= 0 \\ r^2u_{rr} +2ru_r &= 0 \\ (r^2u_r)_r &= 0 \\ r^2u_r &= c_1 \\ u_r &=\frac{c_1}{r^2} \\ u &= \frac{-c_1}{r} + c_2 \end{align*} Then, applying boundary conditions \begin{align*} aA &= -c_1 + ac_2 \\ bB &= - c_1 + bc_2 \\ c_2& = \frac{bB - aA}{b-a} \\ c_1 &= \frac{ab(B-A)}{b-a} \\ u &= \frac{r(bB-aA)-ab(B-A)}{r(b-a)} \end{align*}

Jorge
  • 1,187