Show solution to BVP has a saddle-point at $(0,0)$ This problem was from a PDE class I took years ago:

Let $\Omega=(-1,1)\times(-1,1)$ and consider the BVP
$$
\left\{\begin{aligned}
\Delta u&=0,~~x\in\Omega\\
u(-1,y)&=u(1,y)=0\\
u(x,-1)&=u(x,1)=f(x)
\end{aligned}\right.
$$
where $f:[-1,1]\to\mathbb{R}$ is even, strictly decreasing on $[0,1]$, and satisfies $f(\pm 1)=0$.  Show that $u$ has a saddle point at $u(0,0)$ and that $u(0,0)>0$.

I gave a rather hand-wavy argument about symmetry and the partial derivatives that I don't think is entirely accurate:

Note that $f$ is even and decreasing on $[0,1].$ Since $f:[-1,1]\to \mathbb{R},$ $f$ has a maximum at $x=0,$ i.e. $f$ has a critical point there (since $f$ has bounded variation, and is increasing on $[0,1]$, $f$ is differentiable a.e.). Since $u\equiv 0$ if $x=\pm 1,$ the directional derivative of $u$ pointing into $\Omega$ from the boundary is positive. In particular, $u$ has a critical point at $(0,0)$ by symmetry; from $x=\pm1$ it is increasing, but from $x=0,y=\pm1$ it is decreasing. Therefore $u(0,0)>0$ and $u$ has a saddle point there.

I also graphed the problem for $f(x)=\{|1-x|,1-x^2,\cos(\pi/2 x)\}$ and used Mathematica's NDSolve to make numerical plots of the solution; each appeared geometrically to have a saddle point at $(0,0)$. For example, with $f(x) = 1-x^2$ the solution looks like: 
I'd be interested in a proper solution, if only to satisfy my curiosity.
 A: Via the boundary conditions given, we can restrict the solution set of the PDE to a single function (which depends on $f(x)$). This function will always have a saddle point at $(x,y)=(0,0)$. Let $u = p(x)q(y)$ the PDE becomes
$$\Delta u = u_{xx}+u_{yy} = p''q + pq'' = 0 $$
which gives
$$-\frac{p''}{p} = \frac{q''}{q} = k$$
for some constant $k\in \mathbb{R}$ (or $\omega^2$). This gives two easy to solve ODE's, and recognizing that the boundary conditions 'want' the negative sign with the ODE in $x$, we arrive at solutions of the form
$$u = \left[A\cos(\omega x) + B\sin(\omega x)\right]\left[Ce^{\omega y} + De^{-\omega y} \right]$$
Since $f(x) = u(x,1) = u(x,-1)$ is even, we can restrict to only cosine functions. $u(1,y) = u(-1,y) = 0$ gives
\begin{align*}
A\cos(\omega x)\left[Ce^{\omega y} + De^{-\omega y} \right] &= 0\\
\cos(\omega x) &= 0\\
\text{(let) }\omega_n &= \frac{\pi}{2}+n\pi
\end{align*}
$u(x,1) = u(x,-1)$ gives
\begin{align*}
A\cos(\omega x)\left[Ce^{\omega} + De^{-\omega } \right]&= A\cos(\omega x)\left[Ce^{-\omega } + De^{\omega } \right]\\
(C-D)(e^\omega-e^{-\omega}) &= 0\\
C &= D
\end{align*}
So we have
$$u = \sum_{n=0}^{\infty}a_n\cos(\omega_n x)\cosh(\omega_n y)$$
using Fourier analysis we can establish the values for each constant $a_n$.
$$u(x,1) = \sum_{n=0}^{\infty}a_n\cos(\omega_n x)\cosh(\omega_n) = f(x)$$
So
\begin{align*}
a_n &= \frac{\langle f(x)/\cosh(\omega_n), \cos(\omega_n x)\rangle}{\langle \cos(\omega_n x), \cos(\omega_n x)\rangle}\\
&= \frac{\int_{-1}^{1}[f(x)/\cosh(\omega_n)]\cos(\omega_n x)\mathbf{d}x}{\int_{-1}^{1}\cos(\omega_n x)\cos(\omega_n x)\mathbf{d}x}\\
&= \frac{1}{\cosh(\omega_n)}\int_{-1}^{1}f(x)\cos(\omega_n x)\mathbf{d}x
\end{align*}
Now consider
\begin{align*}
\frac{\partial u}{\partial x}\Big|_{(0,0)} &= -\sum_{n=0}^{\infty}a_n\omega_n\sin(\omega_n x)\cosh(\omega_n y)\Big|_{(0,0)} = 0\\
\frac{\partial^2 u}{\partial x^2}\Big|_{(0,0)} &= -\sum_{n=0}^{\infty}a_n\omega_n^2\cos(\omega_n x)\cosh(\omega_n y)\Big|_{(0,0)} = -\sum_{n=0}^{\infty}a_n\omega_n^2\\
\frac{\partial u}{\partial y}\Big|_{(0,0)} &= \sum_{n=0}^{\infty}a_n\omega_n\cos(\omega_n x)\sinh(\omega_n y)\Big|_{(0,0)} = 0\\
\frac{\partial^2 u}{\partial y^2}\Big|_{(0,0)} &= \sum_{n=0}^{\infty}a_n\omega_n^2\cos(\omega_n x)\cosh(\omega_n y)\Big|_{(0,0)} = \sum_{n=0}^{\infty}a_n\omega_n^2\\
\end{align*}
After several failed attempts to show that $$u(0,0)= \sum_{n=0}^{\infty}a_n>0 $$ directly, I will instead use the maximum/minimum and the mean value property of Harmonic functions. The maximum property gives
\begin{align*}
\min_{\partial\Omega} u &\leq u(x,y)\leq \max_{\partial\Omega}u\text{ }(\forall (x,y)\in \Omega)\\
0&\leq u(x,y) \leq f(0)\text{ }(\forall (x,y)\in (-1,1)\times (-1,1))
\end{align*}
So $0\leq u(0,0) \leq f(0)$. Suppose $u(0,0)=0$ , then by the mean value property, $$u(0,0) = \frac{1}{\pi r^2}\int_{B_r(0,0)}u(x,y)\mathbf{d}x\mathbf{d}y = 0$$ for all $0<r<1$ (where $B_r(x,y)$ denotes the ball of radius $r$ centered at $(x,y)$), this implies either $u(x,y)=0$ for all $(x,y)\in B_r(0,0)$ or $(\exists \Omega'\subset B_r(0,0))$ such that $u(x,y) < 0$ for $(x,y)\in\Omega'$. The latter case is immediately a contradiction by the minimum property, and the former would imply that $u(x,y)=0$ on all of $\Omega$, which is of course a contradiction. Thus $u(0,0)>0$.
