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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(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 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 $\partial D$. The point of this is to avoid any 'humps' in the coefficient functions (since this is the case for my Pde in edit 1)  .

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(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 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 $\partial D$. The point of this is to avoid any 'humps' in the coefficient functions (since this is the case for my Pde in edit 1).

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(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 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 $\partial D$. The point of this is to avoid any 'humps' in the coefficient functions.

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(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 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 $\partial D$. The point of this is to avoid any 'humps' in the coefficient functions (since this is the case for my Pde in edit 1) .

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(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'.

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(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 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 $\partial D$. The point of this is to avoid any 'humps' in the coefficient functions.