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I have the Laplace equation $$\Delta u=\frac{1}{r} \frac{\partial}{\partial r } \left(r \frac{\partial u}{\partial r} \right)+\frac{1}{r^2} \frac{\partial u }{\partial \theta^2}=0$$ on a unit disk $$0<r \leq 1$$

We note that we have the boundary condition for R $$|u(0,\theta)|<\infty \rightarrow |R(0)|<\infty$$ and the boundary conditions for $\theta$ are $$\Theta(- \pi)=\Theta(+ \pi)$$ and $$\Theta'(- \pi)=\Theta'(+ \pi)$$

I assume the solution is of the form $$u(r,\theta)=R(r)\Theta(\theta)$$ and subbing this into $$\Delta u$$

I get 2 ODE's, $$-\frac{r}{R(r)}(rR'(r))'=k$$ and $$\frac{\Theta''(\theta)}{\Theta(\theta)}=k$$

I first look at the case for when $$k=p^2>0$$ This gives $$\Theta=ae^{p \theta}+be^{-p \theta}$$

So by subbing in the boundary conditions for $\theta$, I get $$ae^{p \pi}+be^{-p \pi}=ae^{-p \pi}+be^{p \pi}$$ Why does this give me no solution?

When I look at the case $$k=0$$ the corresponding ODE for R(r) gives $$R=c_1 \ln r+c_2$$ Why does this being subject to the boundary condition $|R(0)|$ give me $R=c_2$? and why do I get the solution for this case being $u_0 (r,\theta)=1$?

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  • $\begingroup$ why $p$ must be a real number? $k$ can be negative. $\endgroup$
    – Yimin
    Commented Jun 8, 2015 at 16:58
  • $\begingroup$ I'd write that "solutionless" equations as $a(e^{p\pi}-e^{-p\pi})=b(e^{p\pi}-e^{-p\pi})$, which gives either $a=b$ or $e^{p\pi}=e^{-p\pi}$, the second option yielding $p=0$. $\endgroup$
    – MickG
    Commented Jun 8, 2015 at 17:05
  • $\begingroup$ So the result would be $\Theta(\theta)=a(e^{p\theta}+e^{-p\theta})=\frac a2\cosh(p\theta)$. $\endgroup$
    – MickG
    Commented Jun 8, 2015 at 17:07
  • $\begingroup$ For the last question, suppose $c_1\neq0$. Then for $r\to0$ we have $|R|\to\infty$, whereas the boundary condition states otherwise. $\endgroup$
    – MickG
    Commented Jun 8, 2015 at 17:10
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    $\begingroup$ $p=0,i,2i,3i,\cdots$ which gives $\Theta(\theta)=e^{in\theta}$ and $R(r)=r^{|n|}$ for $n=0,\pm 1,\pm 2,\cdots$ $\endgroup$ Commented Jun 8, 2015 at 17:36

1 Answer 1

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Why does this give me no solution?

You have $ae^{p\pi}+be^{-p\pi}=ae^{-p\pi}+be^{p\pi}$. That is equivalent to $a(e^{p\pi}-e^{-p\pi})=b(e^{p\pi}-e^{-p\pi})$, which implies either $a=b$ or $e^{p\pi}=e^{-p\pi}\iff p=ki$, but in our case $p^2=k>0$, so that $k$ must be 0. If $p\neq0$ you have no solution, with $p=0$ you get a constant function which is found also in the $k=0$ case, so nothing from this case that you don't get from other cases. But if $p\neq0$, the solution becomes $\Theta(\theta)=\frac{a}{2}\cosh(p\theta)$, which satisfies one boundary condition (the one on $\Theta$, being the $\cosh$ an even function), but not the other, as $\Theta'(\theta)=\frac{ap}{2}\sinh(p\theta)$, which is odd.

Why does this being subject to the boundary condition $|R(0)|$ give me $R=c_2$?

Well, suppose otherwise. If $c_1\neq0$, then for $r\to0$ we have $|R(r)|\to+\infty$, but the boundary condition states otherwise. So $c_1=0$ and $R(r)=c_2$.

And why do I get the solution for this case being $u_0(r,\theta)=1$?

I'd rather say you get a constant function. The $R$ part we have already shown to be constant, and the $\theta$ part has a zero second derivative and must thus be $\Theta(\theta)=a\theta+b$, which satisfies the boundary conditions iff $a=0$, and ends up being forced by the boundary conditions to be $\Theta(\theta)\equiv b$, a constant. $u_0(r,\theta)=R(r)\Theta(\theta)=b\cdot c_2$, so it is constant. Why it is one, I wouldn't know. Maybe there is some kind of normalization imposed by whatever you are following.

Hope this answers you.

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  • $\begingroup$ Your claim $e^{p\pi}=e^{-p\pi}\iff p=0$ is false. In general $p$ can be $ki$ for any $k\in\Bbb Z$, which is precisely the case in this scenario if we want non-constant solutions. $\endgroup$
    – GPerez
    Commented Jun 8, 2015 at 17:39
  • $\begingroup$ I was probably thinking real numbers. And since $p^2=k>0$ I have no reason to change my thinking. But yes, in general that does not hold :). $\endgroup$
    – MickG
    Commented Jun 8, 2015 at 18:01
  • $\begingroup$ The thing is, $k$ isn't positive. It is a fundamental property of the (positive) Dirichlet Laplacian (which is what we have in this case). Its eigenvalues are negative (and we know that $k$ is an eigenvalue from the equation $\Theta''-k\Theta = 0$). In this case $k=0$ is also an eigenvalue, because we are only prescribing periodicity in $\Theta$, as opposed to a particular boundary value. However this only gives a constant solution, and we want the full class of solutions, not just constant ones. In any case it is standard procedure to rule out $k > 0$. $\endgroup$
    – GPerez
    Commented Jun 8, 2015 at 19:04
  • $\begingroup$ Yes $k$ is positive. I am answering the OP's question, which is precisely about ruling out $k>0$. Standard procedure or not, it is the problem I am addressing. See my edit. $\endgroup$
    – MickG
    Commented Jun 8, 2015 at 19:13
  • $\begingroup$ I see what you mean now, with the edit it's much clearer :) $\endgroup$
    – GPerez
    Commented Jun 8, 2015 at 19:48

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