Toronto Math Forum

I shall expand upon this post later with the full list of steps I took to cancel out all my eigenfunctions and eigenvalues. For the moment, my final answer was u=-1, which is bounded and satisfies all boundary conditions (trivially, $u_x$ and $u_y$ are 0 as needed).

Update (apologies for the comparatively short response, I'm not feeling great at the moment): so we first write $u=XY$ so that the problem becomes $\frac{Y''}{Y}+\frac{X''}{X}=0$. Condition (2) implies periodicity in x, so we have associated problems $X''+\lambda X = 0$, and $Y''-\lambda Y=0$. Then $X=A_n \cos(\omega_n x) + B_n \sin(\omega_n x)$, but we can set $B_n=0$ immediately from (2). Additionally, $X'(0)=X'(\frac{\pi}{2}=0) $ so $\omega_n=2n, \lambda_n=4n^2$. Then  $Y = A_n e^{-2ny} + B_n e^{2ny} = A_n e^{-2ny} + 0$ , because u is bounded on $y>0$

So $u=A_0 + \sum_{n=1}^\infty e^{-2ny} \cos (2nx) \rightarrow (u_y-u)(x,0)=-A_0 - \sum_{n=1}^\infty (2n+1) A_n \cos (2nx) = 1 \rightarrow A_n(2n+1) = \frac{4}{\pi}\int_0^\frac{\pi}{2}\cos(2nx)$

In fact, this integral is 0 for all n. Therefore, we are just left with the condition $-A_0=1 \rightarrow u=-1$

First, we convert the Laplacian to polar coordinates: $\Delta u =u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}$. Write $u=RP$, so that we can solve the problem of form $\frac{r^2R''+rR'}{R}+\frac{P''}{P}=0$

Now, notice that boundary condition (4) becomes $u(r,0)=u(r,\pi)=0$ in polar coordinates. This gives us a condition for periodicity in P, so that we can associate a negative eigenvalue $-\lambda_n=-\omega_n^2$ to it. Solving $P''+\omega_n^2 P=0$ gives us $P=A_n \cos (\omega_n \theta) + B_n \sin (\omega_n \theta)$. We can immediately eliminate the $A_n$ term, because of the boundary condition on $\theta=0$, and we can also conclude that $\omega_n \pi = n \pi$ means that our $\omega_n=n$, $\lambda = n^2$.

Now, we expect some equations of the form $r^m$ for the solution to R. Solving the Euler problem $r^2R+rR'=\lambda R$ gives $m(m-1)+m=n^2\rightarrow m = \pm n$. So $R=A_n r^n+B_n r^{-n}$. Neither term can be negated, as r does not tend to 0 or infinity. However, $A_0$ and $B_0$ can be ignored as they do not obey the periodic behaviour we expect.

This gives general form $u(r,\theta) = \sum_{n=1}^\infty(A_n r^n+B_n r^{-n})\sin (n \theta)$. Now, we consider boundary conditions (2) and (3) in the cases $r=2, r=2^{-1}$ respectively.

Then

$u(2,\theta) = \sum_{n=1}^\infty(A_n 2^n+B_n 2^{-n})\sin (n \theta) = 1$ (2)

$u(2^{-1},\theta) = \sum_{n=1}^\infty(A_n 2^{-n}+B_n 2^{n})\sin (n \theta) = -1$ (3)

If we add these two summations together, this implies that $B_n=-A_n$. Use this simplification to find the fourier coefficients in (2):

$u(2,\theta) = \sum_{n=1}^\infty A_n (2^n - 2^{-n})\sin (n \theta) = 1$

Substitute $C_n=A_n(2^n-2^{-n})$. Then $C_n=\frac{2}{\pi}\int_0^\pi sin (n \theta) d\theta = \frac{4}{n\pi}$ for n odd, $0$ for n even.

So $u(r,\theta) = \sum_{n=1}^\infty \frac{4}{n\pi(2^n-2^{-n})} (r^n - r^{-n})\sin (n \theta)$

I went with a more brute force method, wherein I think I made a minor mistake (I did not have |k| at the end like we should have gotten) so it would be great to find out where.

I immediately plugged in the function $f(x)=\frac{1}{(x^2+a^2)}$ into the integral $F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-ikx}}{(x-ia)(x+ia)}$ (where I rewrote the denominator to make use of residue theorem ($\int f = 2\pi i Res (f;-ia)$)), and integrated over the contour:

Giving me $F(k)=\frac{\sqrt{\pi}}{\sqrt{2}a}e^{-ak}$

For the 2nd part, I simply applied the property that $g(x)=x*f(x)\rightarrow G(k)=ikF(x)$