Let a function $u:\overline\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R}$ that satisfies $$\Delta u+f(u)=0 \ \ \ \mbox{in} \ \ \Omega,$$ and consider $$w(x_1,...,x_{n-1})=\lim_{x_n\rightarrow+\infty}u(x_1,...,x_n).$$ Why the function $u(x_1,...,x_n)$ converges uniformly in the $C^1$ sense in compact sets of $\mathbb{R}^{n-1}$ to the function $w$, and $w$ satisfies $$\Delta w+f(w)=0.$$

This argument is used in the paper: Further qualitative properties for elliptic equations in unbounded domains, by Berestycki, Caffarelli and Nirenberg. I didn't undesrtand this point. Someone can help me? Thanks.


This is not a general statement; it is a consequence of certain assumptions in the proposition where it is used (I assume you are on page 82 of the paper like I am).

In particular, $u$ is assumed to be a bounded monotone solution, in the sense that $\frac{\partial u}{\partial x_n} \geq 0$, and $f$ is assumed to be $C^1$ (this can be weakened to Lipschitz, probably, but since the paper says $C^1$, let's work with that).

Let $\Omega = \omega \times \mathbb{R}_+$, where $\omega$ is some compact subset with smooth boundary of $\mathbb{R}^{n-1}$.

Now since $u$ is uniformly bounded, consider $u_k(x) = u(x', x_n + k)$ where $x' \in \mathbb{R}^{n-1}$, $x \in \mathbb{R}_+$. In any compact subset of $\Omega$, this is uniformly bounded in the $C_0$ norm as $k \rightarrow \infty$, hence $f(u_k)$ is uniformly bounded as $f$ is $C^1$. Since $$ \Delta u_k + f(u_k) = 0$$, $u_k$ is uniformly bounded in the $C^1$ norm by elliptic estimates (see, e.g., Lemma 4.2 of Gilbarg-Trudinger), hence there is some subsequence that converges in the $C^\alpha$ norm for any $\alpha < 1$ (this is by the compact embedding of Holder spaces). Since we already know that they converge pointwise to $w$, that means that the whole sequence converges in $C^\alpha$.

Consider now the function $u_k - u_{k'}$ as $k,k'$ go to infinity. We consider the equation $$ -\Delta (u_k - u_{k'}) = f(u_k) - f(u_{k'}) $$ Since $f$ is Lipschitz, we have $$| f(u_k) - f(u_{k'}) | \leq \|f\|_{Lip} |u_k - u_{k'}| \leq \|f\|_{Lip} \|u_k - u_{k'}\|_{C^\alpha}$$ Applying elliptic estimates on the Poisson equation again, this tells us that the $u_k$ are a $C^1$ Cauchy sequence and hence their convergence to $w$ is in the $C^1$ norm.

Now let's examine how we know $w$ satisfies the equation. Morally speaking, $w$ satisfies $$\Delta w + f(w) = 0$$ because in the limit, the $x_n$ derivative of the $u$ disappears - it is constant in one direction, and hence does not contribute to the Laplacian. Making this rigorous requires a detour since we only have that the convergence is in $C^1$, and here I will invoke the theory of weak solutions. We have $$ \int \nabla u_k \nabla \phi dx = \int f(u_k) \phi dx$$ for test functions $\phi$. Over compact sets, we have that $\nabla u_k \rightarrow \nabla w$ and $f(u_k) \rightarrow f(w)$, and hence $w$ satisfies the equation in the weak sense. From here, you can invoke elliptic estimates to see that $w$ solves the equation in the classical sense, if you desire.

(Expanded update of the weak solution to classical, in response to question in comment). On second thought, it may be easier to use straightforward Schauder theory rather than estimates for weak solutions. Since $w$ is a $C^1$ function and so is $f$, that means $f(w)$ is $C^1$, in particular it is at least $C^\alpha$ (on any compact subset, that is). Hence the Schauder theory means that $$\Delta v = - f(w)$$ has a unique $C^{2,\alpha}$ solution $v$. Since $w$ is a weak solution of the same equation, the two coincide.

  • $\begingroup$ I have some questions about your answer. 1) when you said that u is uniformly bounded, means that u is bounded for the same constant in each subset of $\Omega$? 2) Why $u$ converge pointwise to $w$? 3) Could you explain better the step you said that $u$ converge to $w$ in the $C^1$ norm? 4) Wich estimate can i use to say that $w$ is a classical solution? Thank you very much!!! $\endgroup$ – José Carlos Feb 17 '13 at 8:03
  • $\begingroup$ 1) This follows because $u$ is globally bounded (that's what the assumption of a bounded solution means) 2) This is because $u$ is a bounded increasing function in the $x_n$ direction. Recall that any bounded increasing function always has a limit. For each $x' \in \mathbb{R}^{n-1}$ then, $u(x',x_n)$ is bounded and increasing as $x_n \rightarrow \infty$, and hence there is a limit for each $x'$. We call that limit $w$. 3) By elliptic estimates, the $C^1$ norm of $u_k - u_{k'}$ is bounded by the $L^\infty$ norm of $f(u_k) - f(u_{k'})$, which I showed goes to zero as $k,k'$ get large. $\endgroup$ – Ray Yang Feb 17 '13 at 15:37
  • $\begingroup$ 4 I have answered by updating the original answer. $\endgroup$ – Ray Yang Feb 17 '13 at 15:53
  • $\begingroup$ One last doubt, I promisse you!! How did you conclude that uk are a C1 Cauchy sequence and hence their convergence to w is in the C1 norm? Thank you very much! $\endgroup$ – José Carlos Feb 18 '13 at 20:26
  • $\begingroup$ Would you explain better the step of inequality: $$\|f\|_{Lip}|u_k-u_{k'}\leq\|f\|_{Lip}\|u_k-u_{k'}\|_{C^\alpha}(diam(K))^\alpha?$$ $\endgroup$ – José Carlos Feb 18 '13 at 20:48

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