1
$\begingroup$

I have two BVP's and for both of them the solution is similar, however there is one thing I can't explain. First of all let me state the BBVP's. The first one is

$$(1)= \begin{cases} &u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,1<r<2,|\theta|\leq \pi,\\ &u(1,\theta) = 0&&, \theta\leq \pi\\ &u(2,\theta) = 1-\frac{\theta^2}{\pi^2}&&, \theta\leq \pi \end{cases} $$

And the second one is

$$(2)= \begin{cases} u_{rr}+r^{-1}u_r+r^{-2}u_{\theta\theta} = 0&&,0<r<1,|\theta|\leq \pi,\\ u(1,\theta) = \sin^2\theta + \cos\theta&&,\theta\leq \pi\\ \end{cases} $$

In both cases, one can let $u(r,\theta)=R(r)\Theta(\theta)$ and after plugging this into the PDE, one can write

$$r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''}{\Theta}=\lambda, \ \text{where} \ \lambda=\text{constant.}\tag3$$

Since the $\Theta-$part is easier we start with that one. Since we are working in disks/annuluses we need the solution to be $2\pi-$periodic. Thus the solution has the form (up to constant multiples) $\Theta=e^{in\theta},$ where $n\in\mathbb{Z}.$ Plugging this onto the ODE, $\Theta''=-\lambda\Theta$ gives $\lambda=n^2.$ Thus

$$\Theta_n(\theta)=e^{in\theta}. \tag4$$

Using $(3)$ we can solve for $R(r)$ since $r^2R''+rR'=\lambda R=n^2R,$ re arranging we simply have the Euler equation

$$r^2R''+RR'-n^2R=0.$$

Thus far, both $(1)$ and $(2)$ can be solve thid way. But here comes the differents. In the notes for $(1)$, my professor says that:

We need to examine the solutions of the Euler equation for two cases: $n=0$ and $n\neq 0.$

But in the notes for $(2)$ he does not do this and only considers $n\neq 0.$

What in the problem statement makes this difference?

$\endgroup$
1
$\begingroup$

There is no difference. Perhaps he just neglected to mention $n=0$ for (2), or thought you remembered it from (1). But problem (2) also needs a boundary condition at $r=0$, which you didn't mention. Assuming the boundary condition is that $\lim_{r \to 0} u(r,\theta)$ exists, your solution should end up as

$$ u(r,\theta) = \frac{1}{2} + r \cos(\theta) - \frac{r^2}{2} \cos(2\theta) $$

and the term $1/2$ comes from $n=0$.

$\endgroup$
  • $\begingroup$ I don't think he neglected to mention it because then the full solution he provided would have differed. I did not mention the boundary condition because there is none :S This is what my prof got as his final answer for $(1)$ and $(2)$ respectively: $$u(r,\theta)=\frac{2\ln(r)}{3\ln(2)}+\sum_{n\in\mathbb{Z}\setminus {0}}e^{in\theta}a_n(r^n+r^{-n}),$$ where $$a_n=\frac{1}{2\pi(2^n-2^{-n})}\int_{-\pi}^{\pi}\left(1-\frac{\theta^2}{\pi^2}\right)e^{-in\theta} \ d\theta.$$ $\endgroup$ – Parseval Mar 4 at 22:27
  • $\begingroup$ And for $(2)$ it's $$u(r,\theta)=\sum_{n\in Z}c_nr^{|n|}e^{in\theta},$$ where $$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}(\sin^2(\theta)+\cos(\theta))e^{in\theta} d\theta.$$ Both of these seem to differ from your answer. $\endgroup$ – Parseval Mar 4 at 22:28
  • $\begingroup$ @Parseval The given solution for (2) and Robert's solution are in fact the same. $\endgroup$ – Dylan Mar 5 at 9:24

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.