I'm studying for PDE qualifying exams and came across a problem that is giving me issues. The problem is:
Given the square $\Omega=\{|x|<1,|y|<1\}$, let $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$ be the solution to the boundary value problem:
\begin{array}{ll} &\Delta u(x,y)=0, & (x,y)\in\Omega\\ &u(x,y)=x^4-y^4, & (x,y)\in\partial\Omega. \end{array}
Find the value $u(0,0)$.
Note: I believe we are supposed to be able to answer this without explicitly solving for $u$.
What I've tried: My first thought was to use the Mean Value Property for harmonic functions. However, I have only ever applied it where $\Omega$ is a disk. The issue here is that we do not know the value of $u$ on any circle around the origin. I would not expect such a theorem to hold for integration over square domains, but I am curious if there is still a way to apply it to this specific problem.
I tried something slightly more contrived by noting that $u$ is a nonconstant function and furthermore, $u$ is harmonic in any open disk $D_\delta\subset\Omega$ of radius $0<\delta<1$, and hence the Maximum Modulus Principle gives us a strict inequality:
$$ \max_{(x,y)\in\partial D_{\delta_1}} u(x,y)<\max_{(x,y)\in\partial D_{\delta_2}} u(x,y) $$
where $0<\delta_1<\delta_2<1$. Then, one obtains a monotonically decreasing sequence $\left(\max_{(x,y)\in\partial D_{1/n}} u(x,y)\right)_{n=2}^\infty$ which is bounded from below and whose limit is $u(0,0)$. However, I failed to conclude anything from here.
At first, the problem seemed simple and maybe it is. Does anyone have any suggestions?
Edit: After some more thought, I think I figured out one approach. I was not previously using the symmetry of the boundary condition. Note that the two functions $u(x,y)$ and $v(x,y):=-u(y,x)$ are solutions to the original problem on $\Omega$, which by uniqueness, must mean they are identical. But then, $u(0,0)=-u(0,0)$ which implies $u(0,0)=0$. Does this seem correct?