# An eigenvalue problem on a half circle

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

• Dirichlet boundary condition imposed on where? The diameter plus the half-circle or just the diameter or just the half-circle? Dec 26, 2022 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. Dec 26, 2022 at 11:32
• @CheeHan The entire boundary if you want a well-posed problem. Dec 26, 2022 at 11:33
• Please make the title more descriptive. Dec 26, 2022 at 19:06

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

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 $$k = 1, 2, 3, \text{etc}$$.
• Thanks for this. So the solution is $$u(r,\theta)=J_n\bigg(\frac{\alpha_{n,k}}{J_n(R)}\bigg)\sin n\phi$$ Dec 26, 2022 at 11:52

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.

• Thanks Chee. What about the initial condition $J(R)=0$? Dec 26, 2022 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)? Dec 26, 2022 at 11:31