Let $\Omega\subset\mathbb{R}^2$ is bounded and convex and $\partial\Omega$ be smooth. Consider the second order elliptic PDE **(1)**
$$
\begin{cases}
Lu = f &\text{ on } \Omega\,,\\ u=0 &\text{ on } \partial\Omega\,,
\end{cases}
$$
where $f\in C^\infty(\Omega)$ satisfies $f(x) > 0$ and $f$ has no isolated critical points on $\Omega$. Then, does the solution $u$ of the above have only one critical point? I am looking for a result similar to Theorem 2 of [this paper][1](ScienceDirect):

> **Theorem 2:** Let $\Omega$ be a bounded, strictly convex domain in $\mathbb{R}^2$ and $u\in C^3(\Omega)\cap C^1(\bar{\Omega})$ a solution to the boundary value problem
\begin{align*}
\Delta u &= f(u,\nabla u)\quad\text{ in }\Omega\,,\\
u=\text{const}\,,& \quad\nabla u\neq0\,,\quad\text{ on }\partial\Omega\,,
\end{align*}
where $f\in C^1$, $f_u\geq0$. Then $u$ has exactly one critical point in $\bar{\Omega}$ and there $\det(D^2u)>0$ holds (i.e., a global maximum or minimum).

The paper claims that Theorem 2 is true even if we replace Laplace's operator by another elliptic operator. This means, for example, that if $f$ is constant, then the solution to (1) has a unique critical point. 

Edit 1: Here is the specific PDE I am interested in:
\begin{equation}
	(1+h^2y^2)\partial_{xx}u +(1+h^2x^2)\partial_{yy}u -2xyh^2\partial_{xy}u-\frac{h^2x(3+h^2\rho^2)}{1+h^2\rho^2}\partial_{x}u-\frac{h^2y(3+h^2\rho^2)}{1+h^2\rho^2}\partial_{y}u = \frac{1}{1+h^2\rho^2}
\end{equation}
where $h\in (0,1]$, $\rho^2=x^2+y^2$ and the domain $\Omega$ is convex, bounded and does not contain the origin. I have verified that this PDE is elliptic for $h\in(0,1]$.

Edit 2: Due to Mateusz's counterexample, I am adding the condition that the Hessian of the coefficients of $L$ has eigenvalues that do not change sign. This is to ensure that the coefficients don't have any 'humps'.


  [1]: https://www.sciencedirect.com/science/article/pii/S0022247X83711364