Question about 2nd order PDE. In one paper, I have read the following:

Let $c\in\mathbb{R}$. Consider the differential equation: 
  $$\nabla^2\phi-\phi+x=F(\phi-cx),$$
where $\phi:\mathbb{R}^2\to\mathbb{R}$ and $F:\mathbb R\to \mathbb R$
  is some function. If we assume that $\phi\to 0$ for $x,y$ large
  enough, if we fix $x$ and let $y\to\infty$ then $F$ is actually a linear function and $F(z)=-\frac{1}{c}z.$

Is the above true? 
I have tried the following: 
Fix $x$. Let $$\varphi(x)=\lim_{y\to\infty}\phi(x,y).$$ 
Take $y\to\infty$ in both sides of the differential equation then 
$$\nabla^2\varphi(x)+x=F(\varphi(x)-cx).$$
Of course all functions are smooth enough to justify the interchanging of limit sign with $\nabla^2$ and $F$. 
But then I got stuck here. 
EDIT
I just realized I cannot interchange the limit sign with the $\nabla^2$ operator. An easy counter example for this is in a comment in the first answer to the question. So actually we are left with:
$$ \lim_{y\to\infty}\nabla^2\phi(x,y)+x=F(-cx), $$
so $F$ is not necessarily linear?
 A: What I would do is to multiply both sides by $\psi(x,y)$, where $\psi(x,y)$ is a compactly supported, smooth, bump function.  Use integration by parts twice on the term involving $\nabla^2$.  Now choose a suitable sequence of such functions $\psi_n(x,y)$, where the support goes to infinity in the $y$ direction as $n \to \infty$, and where the suppose in the $x$ direction stays the same.  Use the Lebesgue dominated convergence theorem to justify taking the limits under the integral sign.
To give more detail, let $\psi_n(x,y) = \theta_1(x) \theta_2(y-y_n)$ where $\theta_1$ and $\theta_2$ are smooth, compactly supported, bump functions.  Performing the operation above, we get
$$ 0 = \int\int (\theta_1''(x)\theta_2(y-y_n) + \theta_1(x)\theta_2''(y-y_n) + \theta_1(x)\theta_2(y-y_n)) \phi(x,y) + \theta_1(x)\theta_2(y-y_n) x - \theta_1(x)\theta_2(y-y_n) F(\phi(x,y) -cx) \, dx \, dy \\= \int\int (\theta_1''(x)\theta_2(y) + \theta_1(x)\theta_2''(y) + \theta_1(x)\theta_2(y)) \phi(x,y+y_n) + \theta_1(x)\theta_2(y) x - \theta_1(x)\theta_2(y) F(\phi(x,y+y_n) -cx) \, dx \, dy.$$
Let $y_n \to \infty$ and use Lebesgue dominated convergence.  We will add an extra hypothesis that F is continuous.  Then we get
$$ 0 = \int\int \theta_1(x)\theta_2(y) x - \theta_1(x)\theta_2(y) F(-cx) \, dx \, dy = \int \theta_2(y) \, dy \int \theta_1(x)( x -  F(-cx)) \, dx .$$
Since $\theta_1$ can be chosen arbitrarily, it follows that
$$ F(-cx) = x .$$
I do wonder if one can do without the hypothesis that $F$ is continuous.
Added later: Suppose we don't know if $F$ is continuous, but we do know that $\phi$ is $C^2$.  I have some incomplete ideas about this.  For example, suppose that for some $x^*$ that the function $y \mapsto f_{x^*}(y) = \phi(x^*,y)$ has a root $y^*$ such that $f_{x^*}(y^*) = 0$ and $f_{x^*}'(y^*) \ne 0$.  Then looking at a set $\{x^*\}\times(y^*-\epsilon,y^*+\epsilon)$ for $\epsilon>0$ small enough, and looking that the differential equation, we can see that $F$ has to be continuous in a neighborhood of $x^*$.
But there are many other cases to consider, and you can see that it quickly becomes a mess.
