I am struggling way too much with this problem, any help is highly appreciated. Consider the Pac-Man-like set described by
$$P=\left\{(\rho\cos\theta,\rho\sin\theta):\rho\in(0,1),\theta\in\left(\frac{\pi}{4},\frac{7\pi}{4}\right)\right\}$$
whose boundary can be described by $\partial P=\Gamma\cup\gamma_+\cup\gamma_-$ with $\Gamma=\{(\cos\theta,\sin\theta):\theta\in\left[\frac{\pi}{4},\frac{7\pi}{4}\right]\}$ and $\gamma_{\pm}=\{(x,\pm x):x\in[0,\frac{1}{\sqrt{2}}]\}$ as shown below (plotted with desmos).
I want to find a solution of the Laplace equation $\psi_{xx}+\psi_{yy}=0$ for $\psi:\overline{P}\to\mathbb{R}$ with boundary conditions
$$\boxed{\begin{cases}\psi\vert_{\Gamma}(\cos\theta,\sin\theta)=(\theta-\frac{\pi}{4})(\theta-\frac{7\pi}{4})\\ \psi\vert_{\gamma_+}(x,x)=\psi\vert_{\gamma_-}(x,-x)=0\end{cases}}$$
This is because I want it to be a non-zero polynomial on $\Gamma$, zero on $\gamma_{\pm}$ and continuous on $\partial P$. My question then is:
Does such a $\psi:\overline{P}\to\mathbb{R}$ exist? If so, how can I compute it?
Now I will show what I've managed to do. Suppose that we could express $\psi$ as a Fourier series like
$$\psi(\rho\cos\theta,\rho\sin\theta)=a_0+\sum_{n=1}^\infty\rho^n(a_n\cos(n\theta)+b_n\sin(n\theta))$$
Then, we have by the boundary conditions $\psi\vert_{\gamma_{\pm}}=0$ that
$$\begin{cases}0=\psi\left(\frac{\rho}{\sqrt{2}},\frac{\rho}{\sqrt{2}}\right)=a_0+\sum_{n=1}^\infty\rho^n(a_n\cos(n\pi/4)+b_n\sin(n\pi/4))\\ 0=\psi\left(\frac{\rho}{\sqrt{2}},-\frac{\rho}{\sqrt{2}}\right)=a_0+\sum_{n=1}^\infty\rho^n(a_n\cos(n\pi/4)-b_n\sin(n\pi/4))
\end{cases}$$
which immediately implies $a_n=0$ for $n\not\equiv2(\text{mod }4)$ and $b_n=0$ for $n\not\equiv0(\text{mod }4)$ by adding and subtracting both power series and equating the coeficients to zero. However, this means that
$$\psi(\rho\cos\theta,\rho\sin\theta)=a_2\rho^2\cos(2\theta)+b_4\rho^4\sin(4\theta)+a_6\rho^6\cos(6\theta)+…$$
And this is incompatible with the boundary condition for $\Gamma$ since, on $\Gamma$, $\psi$ isn't $\pi$-periodic while the right-hand side is. Note that $\rho^{-n}\cos(n\theta),\rho^{-n}\sin(n\theta)$ for $n\geq1$ and $\log\rho$ cannot appear in the polar form infinite series since these are not defined for $(x,y)=(0,0)$ and $(0,0)\in\overline{P}$. So does this mean that Laplace's equation has no solution on $\overline{P}$ with those boundary conditions?
Edit: In this question, user83387 states
Suppose we look for solutions to $\Delta u = 0$ on the circular sector $\{ (r, \theta) \mid 0 < r < a, 0 < \theta < \gamma \}$ with initial conditions $u(a, \theta) = g(\theta), \, u(r, 0) = u(r, \gamma) = 0$.
Using separation of variables, I showed that the solution is of the form
$$ u(r, \theta) = \sum_{n=1}^\infty \frac{2a^{-\frac{n\pi}{\gamma}}}{\gamma} \left(\int_0^\gamma g(\phi)\sin\left(\frac{n\pi \phi}{\gamma}\right) \, d\phi\right) \, r^{\frac{n\pi}{\gamma}}\sin\left(\frac{n\pi\theta}{\gamma}\right)$$
This is precisely what I needed shifting $\theta$ by $\pi/4$ radians and considering $a=1,\gamma=3\pi/2$. However, how did user83387 arrive to that expression?
Best Answer
Write the BCs in polar coordinates explicitly. Your domain is $[0,1] \times [\pi/4,7\pi/4]$ and the boundary conditions are:
$$u(1,\theta)=(\theta-\pi/4)(\theta-7\pi/4) \\ u(r,\pi/4)=0 \\ u(r,7\pi/4)=0$$
and the implicit boundary condition of regularity at $r=0$.
A separable solution to the equation is of the form $r^{\pm \sqrt{\lambda}} e^{\pm i\sqrt{\lambda} \theta}$ (all four sign combinations allowed). To keep regularity at $r=0$ you throw out the negative exponents for $r$. Now for any $\lambda>0$, you can combine the complex exponentials to get a solution that vanishes at $\theta=\pi/4$. In order for it to also vanish at $7\pi/4$, you must have one of $\tan(\sqrt{\lambda} \pi/4)=\tan(\sqrt{\lambda} 7\pi/4)$ or $\cot(\sqrt{\lambda} \pi/4)=\cot(\sqrt{\lambda} 7 \pi/4$).
Either way you conclude $\sqrt{\lambda}3\pi/2=n\pi$ for some nonnegative integer $n$. Hence $\lambda=4 n^2/9$ and $\sqrt{\lambda}=2n/3$.
Therefore the correct real Fourier series is of the form
$$a_0 + \sum_{n=1}^\infty r^{2n/3} \left ( a_n \cos(2n \theta/3) + b_n \sin(2n \theta/3) \right ).$$