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Solve the eigenvalue problem for the half-circle $x^2+y^2\le R^2$ $y\ge 0$ with homogeneous Dirichlet conditions as boundary conditions.

This is what I did:

Let $u=u(r,\phi)$

$$\Delta u=-\lambda u$$

$$R_{rr}\Phi+\frac{1}{r}R_r\Phi+\frac{1}{r^2}R\Phi_{\phi\phi}=-\lambda R\Phi$$

This gives $$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\frac{\Phi_{\phi\phi}}{\Phi}=-\lambda$$

Rearranging and setting both equations equal to some constant $\mu$:

$$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\lambda=-\frac{\Phi_{\phi\phi}}{\Phi}=\mu$$

Gives the two ODEs:

$$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\lambda=\mu$$ $$-\frac{\Phi_{\phi\phi}}{\Phi}=\mu$$

The second gives with Dirichlet conditions:

$$-\frac{\Phi_{\phi\phi}}{\Phi}=\mu \longrightarrow \Phi(\phi)=C_1\sin\sqrt{\mu}\phi$$

The former gives the Bessel equation,:

$$R_{rr}+rR_r+(\lambda-\mu)\frac{R}{r^2}=0 $$

But this should rather be in the form:

$$R_{rr}+rR_r+(\lambda-\frac{\mu}{r^2})R=0 $$

What went wrong, and what about the fact that with Dirichlet conditions we should also include $J(R)=0$? That gives a trivial solution.

Is this procedure incomplete?

Thanks

Luthier415Hz
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  • Dirichlet boundary condition imposed on where? The diameter plus the half-circle or just the diameter or just the half-circle? – Chee Han Dec 26 '22 at 11:23
  • General remark, you introduced the additional parameter $\mu$. The solution for the angular component depends on $\mu$. You should now also get a general solution for the radial component depending on both $\mu$ and $\lambda$. Then you plug in the boundary conditions and only for discrete values of $\lambda$ a finite number of non-trivial solutions (that is finitely many values of $\mu$) exist. – quarague Dec 26 '22 at 11:32
  • @CheeHan The entire boundary if you want a well-posed problem. – quarague Dec 26 '22 at 11:33
  • Please make the title more descriptive. – mr_e_man Dec 26 '22 at 19:06

2 Answers2

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You forget to multiply $\lambda$ by $r^2$. It should be $$ \frac{r^2R_{rr}}{R} + \frac{rR_r}{R} + \frac{\Phi_{\phi\phi}}{\Phi} = -\lambda r^2. $$ Following your notation, you would arrive at $$ r^2R_{rr} + rR_r + (\lambda r^2 - \mu)R = 0. $$ This is a Bessel differential equation after you do a change of variable $s = \sqrt{\lambda} r$, which you can write down the solution.

Chee Han
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  • Thanks Chee. What about the initial condition $J(R)=0$? – Luthier415Hz Dec 26 '22 at 11:29
  • I added a comment above right after I realised your problem is on the upper half-disk. Where is the Dirichlet boundary condition being imposed? Is it on the half-circle only (without the diameter)? – Chee Han Dec 26 '22 at 11:31
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[...] $$R_{rr}\Phi+\frac{1}{r}R_r\Phi+\frac{1}{r^2}R\Phi_{\phi\phi}=-\lambda R\Phi$$

This gives $$\frac{r^2R_{rr}}{R}+\frac{rR_r}{R}+\frac{\Phi_{\phi\phi}}{\Phi}=-\lambda$$ [...]

What went wrong[?]

A factor of $r^2$ has been dropped from the right-hand side at this step.

But this should rather be in the form: $$R_{rr}+rR_r+(\lambda-\frac{\mu}{r^2})R=0 $$

The coefficient of the $R_r$ term here should be $1/r$.

what about the fact that with Dirichlet conditions we should also include $J(R)=0$? That gives a trivial solution.

For the azimuthal equation, the zero-value condition at $\phi = \pi$ implies that $\sin (\sqrt{\mu} \pi) = 0$, and hence $\sqrt{\mu} = n$, where $n = 1, 2, 3, \text{etc}$.

Eventually you should get $$R_{rr} + \frac{R_r}{r} + \left( \lambda-\frac{n^2}{r^2} \right) R = 0, $$ whose fundamental solutions are Bessel $J_n (r \sqrt{\lambda})$ and $Y_n (r \sqrt{\lambda})$.

  • To get the solution to vanish at $r = 0$, you only want $J_n (r \sqrt{\lambda})$ (since Bessel-Y has a singularity).
  • To get the solution to vanish at $r = r_\text{max}$ (I've avoided $R$ which clashes with the radial dependent variable), you need $r_\text{max} \sqrt{\lambda} = \alpha_{n,k}$, where $\alpha_{n,k}$ is the $k$th root of $J_n$, with $k = 1, 2, 3, \text{etc}$.
yawnoc
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