Vtyjkyuk

Let $OmegasubsetmathbbR^2$ is bounded and convex and $partialOmega$ be smooth. Consider the second order elliptic PDE (1)
$$
begincases
Lu = f &text on Omega,,\ u=0 &text on partialOmega,,
endcases
$$
where $fin C^infty(Omega)$ 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(ScienceDirect):

Theorem 2: Let $Omega$ be a bounded, strictly convex domain in $mathbbR^2$ and $uin C^3(Omega)cap C^1(barOmega)$ a solution to the boundary value problem
beginalign*
Delta u &= f(u,nabla u)quadtext in Omega,,\
u=textconst,,& quadnabla uneq0,,quadtext on partialOmega,,
endalign*
where $fin C^1$, $f_ugeq0$. Then $u$ has exactly one critical point in $barOmega$ 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:
beginequation
(1+h^2y^2)partial_xxu +(1+h^2x^2)partial_yyu -2xyh^2partial_xyu-frach^2x(3+h^2rho^2)1+h^2rho^2partial_xu-frach^2y(3+h^2rho^2)1+h^2rho^2partial_yu = frac11+h^2rho^2
endequation
where $hin (0,1]$, $rho^2=x^2+y^2$ and the domain $Omega$ is convex, bounded and does not contain the origin. I have verified that this PDE is elliptic for $hin(0,1]$.

Edit 2: Unfortunately, I have found that it is not enough for $f$ to not have isolated critical points, so I am adding the condition that if $a(x,y)$ is a coefficient of $L$, then on any simply connected subset $D$ of $Omega$, $a$ attains its maximum and minimum on the boundary of $D$. The point of this is to avoid any 'humps' in the coefficient functions.