Symmetries of the Solution of a Laplace Boundary Value Problem Suppose that $\Psi (x,y)$ is a real valued function of the two real variables $x$ and $y$. Consider the following boundary value problem (BVP)
$$\begin{array}{c}
\begin{array}{*{20}{l}}
{{\nabla ^2}\Psi (x,y) = 0,}&{ - a \le x \le a}&{ - b \le y \le b}\\
{\Psi (a,y) = f(y)}&{}&{}\\
{\Psi ( - a,y) = f(y)}&{}&{}\\
{\Psi (x,b) = 0}&{}&{}\\
{\Psi (x, - b) = 0}&{}&{}
\end{array}\\
\end{array}\tag{1}$$ 
where $f(y)$ is a function of $y$ satisfying
$$\left\{ \matrix{
  f(y) = ( - y) \hfill \cr 
  f(b) = f( - b) = 0 \hfill \cr}  \right. \tag{2}$$
I want to prove the following theorem without directly obtaining $\Psi (x,y)$ from $(1)$.

Theorem. If $\Psi (x,y)$ satisfies the BVP in $(1)$ then $\Psi (x,y) = \Psi ( - x,y) = \Psi (x, - y) = \Psi ( - x, - y)$ over the rectangular domain $\left[ { - a,a} \right] \times \left[ { - b,b} \right]$.

I don't know how to start! Any hints and helps are appreciated. :)
 A: I just give this hint and you write this argument in formal language:
you have some defined Dirichlet's conditions. Think of the rectangle as being static in the $xy$ plane. Now invert the axes so that the same square is seen from this new $x'y'$ plane (think of it as taking the $x$ and $y$ axes and rotating them by a angle of $\pi$ to generate new axes $x'$ and $y'$). The conditions are invariant under this inversion, and the result follows.
A: I thought a little more and with the help of Michael and Vladimir Vargas I came up with something. The strategy is to first prove that $\Phi (x,y) = \Psi ( - x, - y)$ is also a solution of BVP $(1)$ in question above. Then I will apply the uniqueness theorem to conclude $\Phi (x,y) = \Psi (x,y)$. 
So let's just do the first part. At the beginning, we may notice by using chain-rule that
$$\left\{ \begin{array}{l}
\frac{{{\partial ^2}\Phi }}{{\partial {x^2}}}(x,y) = \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y)\\
\frac{{{\partial ^2}\Phi }}{{\partial {y^2}}}(x,y) = \frac{{{\partial ^2}\Psi }}{{\partial {y^2}}}( - x, - y)
\end{array} \right.\tag{3}$$
Hence, we can write
$$\begin{array}{l}
{\nabla ^2}\Phi (x,y) = \frac{{{\partial ^2}\Phi }}{{\partial {x^2}}}(x,y) + \frac{{{\partial ^2}\Phi }}{{\partial {y^2}}}(x,y)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y) + \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\nabla ^2}\Psi ( - x, - y) = 0
\end{array}\tag{4}$$
Also, the boundary conditions in $(1)$ become
$$\begin{array}{}
{\Phi (a,y) = \Psi ( - a, - y) = f( - y) = f(y)}& \to &{\Phi (a,y) = f(y)} \\
{\Phi ( - a,y) = \Psi (a, - y) = f( - y) = f(y)}& \to &{\Phi ( - a,y) = f(y)}&{}&{}\\
{\Phi (x,b) = \Psi ( - x, - b) = 0}& \to &{\Phi (x,b) = 0}\\
{\Phi (x, - b) = \Psi ( - x,b) = 0}& \to &{\Phi (x, - b) = 0}
\end{array}\tag{5}$$
it seems that we are done as we proved that $\Phi ( x, y)$ is a solution of BVP $(1)$ in question above. Other symmetries can be proved in a similar way. :)
