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The following nonlinear elliptic PDE arose in my research: $$\Delta f - e^f \partial_s f = E(s,t)\,,$$ where $f : \mathbb R(s) \times \mathbb R/\mathbb Z(t) \to \mathbb R$, $f = f(s,t)$, is the unknown function, $E(s,t)$ is a given nonnegative function which decays exponentially together with all derivatives as $|s| \to \infty$, and the Laplacian is $\Delta = \partial_s^2 + \partial_t^2$; $f$ is required to satisfy the following asymptotic conditions: $$\lim_{s \to \pm \infty} f(s,t) = a_\pm \in \mathbb R\,,$$ for some given constants $a_\pm$; it can be seen using integration over $s,t$ that these must satisfy $$e^{a-} - e^{a_+} = \int_{\mathbb R \times \mathbb R/ \mathbb Z}E(s,t)\,ds\,dt\,.$$

Question: does there always exist a unique solution of this equation for $f$ satisfying the given asymptotic conditions?

I checked standard sources such as Gilbarg and Trudinger, but even uniqueness doesn't seem to follow immediately from the form of the eqation, even though the maximum principle does hold.

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    $\begingroup$ Maybe you should clarify on which variables your Laplacian is taken. With a quick look your notation seems to indicate a parabolic problem, but then again this is not the case, I understand. (Even if this is evident since you say that the problem is elliptic). $\endgroup$ – Juhana Siljander Jun 5 '15 at 10:15
  • $\begingroup$ This PDE looks like you can shape into a Leray-Lions problem, and maybe some of the existence literature for that class of elliptic PDE could be helpful. $\endgroup$ – Arthur Suvorov Aug 30 '16 at 0:13
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This is not yet an answer. Yet, two comments and one information.

1- I don't think that the sign of $E$ is an issue. It might even happen that $\int E\,ds\,dt=0$, and you would be entitled to search for a solution such that $f\rightarrow0$ at infinity.

2- I do not see any problem with the maximum principle.

3- Let me denote $\bar g(s):=\int_{\mathbb R/\mathbb Z}g(s,t)\,dt$. Then $$\bar f''-\left(\overline{\exp f}\right)'=\bar E.$$ Let $h$ be the primitive of $\bar E$ such that $h(\pm\infty)=-e^{a_\pm}$. We obtain $$\bar f'-\overline{\exp f}=h.$$ Because of Jensen Inequality, $\overline{\exp f}\ge\exp\bar f$ and therefore $$\bar f'\ge h+\exp\bar f.$$ Let me denote $f_0=f_0(s)$ the solution (I leave the existence and uniqueness of $F_0$ as a calculus exercise) of $$f_0'=h+\exp f_0,\qquad f_0(\pm\infty)=a_\pm.$$ Then $\bar f\le f_0$ (argue by contradiction by considering otherwise a global maximum, a positive one, of $\bar f-f_0$). In other words, $$\int_{\mathbb R/\mathbb Z}f(s,t)\,dt\le f_0(s).$$

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This isn't an answer but I don't have enough points to make a comment. Is there a way to rescale the $s$ variable so as to make the equation look for like a standard parabolic equation $ \Delta g - g_s= h $ ?

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    $\begingroup$ I don't know, but I don't think so. Also, $s$ is one of the variables of the Laplacian, so I don't see how it's a parabolic equation. $\endgroup$ – Frol Zapolsky Apr 26 '15 at 9:53

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