Some results about semilinear elliptic problems on half-spaces

All’amico Sandro con grande affetto e grande stima.

1.   Introduction

In this work we study some qualitative properties of the solutions to the elliptic boundary value problem

where ${\mathbb{R}}^N_+$ denotes the euclidean half-space $\bigl\{ x = (x', x_N) \in {\mathbb{R}}^{N-1}\times {\mathbb{R}} : x_N > 0 \bigr\}$, $N \geq2$. This type of problem naturally appears in the obtention of a priori bounds for positive solutions of nonlinear second order PDE's on smooth bounded domains (^[19]), in the study of semilinear problems with small diffusion on smooth bounded domains and in the study of regularity results for some free boundary problems (see e.g., ^[1,2,5,6]).

In the present work our focus is on the study of the growth of the solutions to (1.1) as well as on their monotonicity and symmetry properties. The situation is quite well understood in the two dimensional case (see ^[15,16] and also ^[4] when $u$ is bounded and positive) while, in the available results for $N\geq 3$ it is always assumed that $f$ is globally Lipschitz continuous (often with $f(0) \geq0$) and/or that the solution $u$ is positive and bounded (see ^{[2,3,4,5,7,8,9,10,11,12,13,17,18,21]}). For these reasons, in the present work, we concentrate on (possibly) unbounded solutions of (1.1) where $f$ is merely a locally Lipschitz continuous function. The paper is organized as follows. In section 2 we prove a comparison principle for solutions of semilinear problems on unbounded slab-type domains (see Theorem 2.1). By combining this result with the moving planes procedure we prove the monotonicity of the solutions which are bounded (only) on strips. See Theorem 3.1 and Corollary 3.4 in section 3. In section 4 we first establish some results about the growth of an arbitrary solution to (1.1) (see Theorem 4.1 and Theorem 4.4) and then we combine them with those of section 3 to get some new monotonicity and one-dimensional symmetry results (see Theorem 4.5 and Theorem 4.7). In particular, our results cover both the case of some superlinear and subcritical functions $f$ and the case of unbounded solutions with bounded gradient with a general nonlinearity $f$.

2.   A comparison principle

This section is devoted to the proof of a comparison principle for solutions of semilinear problems on unbounded slab-type domains. It is inspired by a result established in ^[14] and it will be used to obtain the main results of section 3 and 4.

Theorem 2.1 (Comparison principle in unbounded slabs of small width).

1) Let $N\ge 2$, $M > 0$ and assume that $f \in C^{0, 1}_{loc}([0, +\infty)).$ Then there exists $\vartheta\, = \, \vartheta(f, M) > 0$ such that, for any $(a, b) \subset {\mathbb{R}}$ with $0 < b-a < \vartheta$ and any $u, v\in C^2({\mathbb{R}}^{N-1}\times [a, b])$ satisfying

we have

2) Let $N\ge 2$ and assume that $f \in C^{0, 1}([0, +\infty)).$ Then there exists $\vartheta\, = \, \vartheta(f) > 0$ such that, for any $(a, b) \subset {\mathbb{R}}$ with $0 < b-a < \vartheta$ and any $u, v\in C^2({\mathbb{R}}^{N-1}\times [a, b])$, with at most polynomial growth at infinity and satisfying

we have

3) Let $N\ge 2$ and assume that $f \in C^0([0, +\infty))$ is a non-increasing function. Then, for any $(a, b) \subset {\mathbb{R}}$ and any $u, v\in C^2({\mathbb{R}}^{N-1}\times [a, b])$, with at most polynomial growth at infinity and satisfying

we have

Proof. Set $\Sigma_{a, b} : = {\mathbb{R}}^{N-1}\times (a, b).$ Testing the differential inequality with $w : = (u-v)^{+} \varphi^2$, $\varphi \in C^1_c({\mathbb{R}}^{N-1})$, we get

and so

Then

On the other hand, by the Poincaré inequality on the interval $(a, b)$ we have

and the combination of (2.5) and (2.6) yields

Now we distinguish the three cases.

In the case 1), from (2.7) we get

where $L(f, M)$ is the Lipschitz constant of $f$ on the interval $[-M, M]$.

Now we set $\vartheta : = \frac{\pi}{2 \sqrt{1 + L(f, M)}} > 0$ and thus, for any $(a, b) \subset {\mathbb{R}}$ with $b-a < \vartheta$ we have

In the case 2), from (2.7) we get

where $L_f$ is the Lipschitz constant of $f$. So that, for any $(a, b) \subset {\mathbb{R}}$ with $b-a < \vartheta : = \frac{\pi}{2 \sqrt{1+ L_f}} > 0$, we get (2.9) once again.

In the case 3), from (2.7) we get

since $f$ is non-increasing and so (2.9) is satisfied also in this case. Note that (2.9) holds true for any interval $(a, b) \subset {\mathbb{R}}$ (i.e., no smallness assumption on the lenght of $(a, b)$ is needed to treat the case 3)).

For $R > 0$ consider $\varphi = \varphi_R \in C^{1}_c ({\mathbb{R}}^{N-1})$ such that

where $B^{'}(0, R): = \left\{ x'\in \mathbb{R}^{N-1}: |x'| < R \right\}$ and define the set $\mathcal{C}(R): = \Sigma_{a, b} \cap ({B^{'}(0, R)\times \mathbb{R}}) = B^{'}(0, R)\times (a, b).$ Using $\varphi = \varphi_R$ in (2.9) we then obtain

For $R > 0$ we define the non-decreasing function $h(R) : = \int_{\mathcal{C}(R)} [(u-v)^+]^2$ and observe that $h$ has at most polynomial growth at infinity thanks to the (growth) assumptions on $u$ and $v$. Therefore $h$ satisfies

where $C$ and $k$ are positive constants.

From (2.14) we get $h(R) \leq 32 \frac{(b-a)^2}{\pi^2} C (1+2^kR^k)R^{-2}$ for $R > 0$ and thus, by iterating this procedure, we obtain $h(R) \leq \Big(32 \frac{(b-a)^2}{\pi^2}\Big)^m C (1 + 2^{mk} R^k) R^{-2m}$ for any $R > 0$ and any integer $m \geq1$. Now we fix $m > k$ and let $R \to +\infty$ to get $\lim_{R\to \infty} h(R) = 0$, which entails $h \equiv 0$. The latter implies $u \leq v$ on $\Sigma_{a, b}$ concluding the proof.

3.   The moving planes method for (possibly) unbounded solutions

Theorem 3.1. Assume $N\ge2$, $f \in C^{0, 1}_{loc}([0, +\infty))$ with $f(0) \ge 0$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of

Assume that $u$ is bounded on the slabs ${\mathbb{R}}^{N-1} \times [0, t]$, for every $t > 0$, i.e., for every $t > 0$ there is a constant $C(t) > 0$ such that $0 \leq u \leq C(t)$ on ${\mathbb{R}}^{N-1} \times [0, t].$

Then $u$ is monotone, i.e., $\frac{\partial u }{\partial x_N} > 0$ in ${\mathbb{R}}^N_+$.

Remark 3.2. When the space dimension $N = 2$, the above monotonicity result holds irrespective of the value of $f(0)$ and without the assumption of boundedness on slabs, see ^[15,16].

Proof. The proof is based on the moving planes procedure. For $t > 0$ we set

We aim at proving that

The monotonicity of $u$ will be then a consequence of (3.1) and the strong maximum principle. To prove (3.1) we shall show that

First we prove that $\Lambda$ is not empty. To this end we observe that, for every $t \in (0, 1)$, the functions $u$ and $u_t$ are bounded by $\Vert u \Vert_{L^{\infty}({ {\mathbb{R}}^{N-1} \times [0, 2])}} : = M > 0.$ Therefore, we can apply Theorem 2.1 to $u$ and $v: = u_t$ on $\Sigma_t$ to find that $u \leq u_t$ in $\Sigma_t$, for all sufficiently small $t > 0$.

Next we plan to prove that $\bar{t} : = \sup \Lambda$ is $+ \infty.$ Assume for contradiction that $\bar{t} < + \infty$ then we can prove the following

Proposition 3.3. For every $\delta \in (0, \frac{\bar t}{2})$ there is $\varepsilon(\delta) > 0$ such that

Proof of Proposition 3.3. If the claim were not true, there would exist $\delta \in (0, \frac{\bar t}{2})$ such that

Observe that the sequence $(x^k_N)$ is bounded and so, up to a subsequence, we may and do suppose that $x^k_N \to{ \bar x}_N \in [\delta, \bar t - \delta]$, as $k \to \infty$.

For $x \in {\mathbb{R}}^N_+$ and $k \geq1$ let us set $u_k(x) : = u(x' + (x^k)^{'}, x_N)$. By the translation invariance of the equation satisfied by $u$, the boundedness of $u$ on every strip ${\mathbb{R}}^{N-1} \times [0, t]$ and standard elliptic estimates we have that the sequence of solutions $(u_k)$ is bounded in $C^{2, \alpha}_{loc}(\overline{\Sigma_t})$, for every $t > 0$ and some $\alpha \in (0, 1)$. Therefore, by the Ascoli-Arzelà theorem (via a diagonal procedure) we can extract a subsequence, still denoted $(u_k)$, which converges in $C^2_{loc}({\overline { {\mathbb{R}}^N_+}})$ to a limit $u^{\infty} \in C^2({\overline { {\mathbb{R}}^N_+}})$ satisfying

Furthermore, by the definition of $\Lambda$, (3.4) and the uniform convergence, we have that $u^{\infty} \le u^{\infty}_{\bar t}$ on $\Sigma_{\bar t}$ and $u^{\infty}(0', \bar x_N) \geq u_{\bar t}^{\infty}(0', \bar x_N)$ and so

Then,

where $C$ is the Lipschitz constant of $f$ on the interval $[0, \Vert u \Vert_{L^{\infty}({ {\mathbb{R}}^{N-1} \times [0, 2\bar t])}}]$ and so $u_{\bar t}^{\infty} \equiv u^{\infty}$ on $\Sigma_{\bar t}$ by (3.6) and the strong maximum principle. In particular $u_{\bar t}^{\infty} \equiv 0$ on the set $\{\, x_N = \bar x_N\}$ and so $u_{\bar t}^{\infty} \equiv 0$ on ${\mathbb{R}}^N_+$ thanks to (3.5) and the strong maximum principle (recall that $f(0) \geq 0$ is in force). We observe that $0 = - \Delta u^{\infty} = f(u^{\infty}) = f(0)$ and we set

so that $v_k(0', x_N^k) = 1$ for every $k \geq1.$ Then,

with $(c_k)_{k\geq1}$ uniformly bounded on every slab ${\mathbb{R}}^{N-1} \times [0, t]$, $t > 0$. We can therefore apply the Harnack inequality to $v_k$ to get, for every compact set $K_n : = \overline{B}(0', n) \times [0, n]$,

where in the latter we have used the fact that $(0', x_N^k) \in K_n$ for $k \geq 1$ and $n > \bar t$.

Moreover, by the definition of $\Lambda$, we know that $\frac{\partial u}{\partial x_N} > 0$ in $\Sigma_{\bar t}$ and so

Now we set $\alpha_k : = u_k(0', x_N^k)$, $f_k(t) : = \frac{f(\alpha_kt)}{\alpha_k}$, we rewrite (3.9) as

and we observe that the family $(f_k)_{k\geq1}$ is relatively compact in $C_{loc}^0([0, +\infty))$ since $f_k(0) = 0$ and

(the latter is satisfied with $C(\eta)$ being the Lipschitz constant of $f$ on the segment $[0, \eta \Vert u \Vert_{L^{\infty}({ {\mathbb{R}}^{N-1} \times [0, \bar t ])}} \, ]$). Thus, up to a subsequence, $f_k \to f^{\infty}$ in $C^0_{loc}([0, +\infty)$ with $f^{\infty} \in C^{0, 1}_{loc}([0, +\infty)$ and$f^{\infty}(0) = 0$.

In view of (3.11) and (3.12) we can use, once again, elliptic estimates and the Ascoli-Arzelà Theorem to find a subsequence (still denoted by $(v_k)$) which converges in $C^2_{loc}({\overline { {\mathbb{R}}^N_+}})$ to a limit $v^{\infty} \in C^2({\overline { {\mathbb{R}}^N_+}})$. By gathering together all those informations we finally get that

and

with $c^{\infty}$ locally bounded on ${\mathbb{R}}^N_+.$

The strong maximum principle and (3.13) imply that $v^{\infty} > 0$ in ${\mathbb{R}}^N_+$ while another application of the strong maximum principle to (3.14) yields $v_{\bar t}^{\infty} \equiv v^{\infty}$ in $\overline{\Sigma_{\bar t}}$ and so $v^{\infty}$ must vanish somewhere in ${\mathbb{R}}^N_+$. The latter contradicts $v^{\infty} > 0$ in ${\mathbb{R}}^N_+$ and concludes the proof of proposition 3.3.

Now we are ready to prove that $\bar t = +\infty$. By proposition 3.3 we know that for every $\delta \in (0, \frac{\bar t}{2})$ there is $\varepsilon(\delta) \in (0, \delta)$ such that

Now we set $M : = \Vert u \Vert_{L^{\infty}({\mathbb{R}}^{N-1} \times [0, 2 \bar t]) } > 0$ and choose $2 \delta < \min \{\frac{\bar t}{2}, \vartheta(M, f) \}$ so that we can apply Theorem 2.1 to $u$ and $u_{\bar t + \epsilon}$ on the sets ${\mathbb{R}}^{N-1} \times (0, \delta)$ and ${\mathbb{R}}^{N-1} \times (\bar t - \delta, \bar t + \epsilon)$. This implies

which clearly contradicts the definition $\bar t$. Therefore $\bar t = + \infty$ so that, for every $t > 0$,

with $c_t^{\infty}$ locally bounded on $\Sigma_t$. Again, as before, the maximum principle and the assumption $u > 0$ in ${\mathbb{R}}^N_+$ imply that

and the Hopf's lemma tell us that

The latter proves the desired conclusion.

An inspection of the first part of the proof of Theorem 3.1 immediately reveals that the moving planes procedure can always be started irrespectively of the value of $f(0)$ provided $u$ is bounded on a single slab ${\mathbb{R}}^{N-1} \times [0, t_0].$ More precisely we have the following

Corollary 3.4 (Starting the moving planes method). Assume $N \ge 2$, $f \in C^{0, 1}_{loc}([0, +\infty))$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of

Assume that there exists $t_0 > 0$ such that $u$ is bounded on the slab ${\mathbb{R}}^{N-1} \times [0, t_0]$. Then there exists $t_1 \in (0, t_0)$ such that

Furthermore, if $u \not \equiv 0$, there exists $t_2 \in (0, t_1)$ such that

Remark 3.5. When the space dimension $N = 2$, the above monotonicity result holds even without the assumption of boundedness on the slab ${\mathbb{R}}^{N-1} \times [0, t_0]$, see ^[15,16].

Proof. Just note that at the beginning of the proof of Theorem 3.1 we have never used anything about the value of $f(0)$ to prove that $\Lambda : = \{ t > 0 \, : \, u \leq u_{\theta} \quad {\text{in}} \, \, \Sigma_{\theta} \quad \forall \theta \leq t \}$ is not empty. This immediately yields (3.19) and (3.20). Let now suppose that $u \not \equiv 0$. Then, $u > 0$ in ${\mathbb{R}}^N_+$ if $f(0) \geq 0$ (by the strong maximum principle) and $u > 0$ in $\Sigma_{t_2}$, for some small $t_2 > 0$, if $f(0) < 0$ thanks to Theorem 6.1. of ^[15]. As before, this information and the strong maximum principle imply (3.21) and (3.22).

4.   Boundedness, monotonicity and symmetry

Next we prove a result which provides natural assumptions ensuring that all solutions $u$ of problem $(P)$ are automatically bounded on the slabs ${\mathbb{R}}^{N-1} \times [0, t]$, for every $t > 0$.

Theorem 4.1. Assume $N\geq 2$, $f \in C^{0}([0, +\infty))$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of (3.18). Then $u$ is bounded on the slabs ${\mathbb{R}}^{N-1} \times [0, t]$, for every $t > 0$, if one of the following assumptions holds true :

($H_1$) (Superlinear nonlinearities) $f$ satisfies $\lim_{t\to \infty} \frac{f(t)}{t} = \infty$ and $f(t) \le a_0(1 + t^r)$ for $t \geq 0$, where $r \in (1, \frac{N+1}{N-1})$ and $a_0 > 0$;

($H_2$) $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$;

($H_3$) $u$ has at most linear growth at infinity and $f(u(x)) \le 0$ for every $x \in \overline { {\mathbb{R}}^{N}_+}$.

When either $(H_1)$ or $(H_3)$ is in force, the bound on $u$ on the slab ${\mathbb{R}}^{N-1} \times [0, t]$ is independent of the solution $u$ (it actually depends on $f, N$ and $t$ only).

Remark 4.2. ⅰ) It will be clear from the proof that the conclusion of the theorem above holds true if ($H_2$) is replaced by : $\vert \nabla u \vert$ is bounded on the strips $\Sigma_t$, $t > 0$.

ⅱ) Some control on the solution is however needed, even when $f(u(x)) \leq 0$ for every $x \in {\mathbb{R}}^N_+$. Indeed, the positive function $u(x) = x_N e^{x_1}$ solves $- \Delta u = - u \leq 0$ on ${\mathbb{R}}^N_+$, $u = 0$ on $\partial {\mathbb{R}}^N_+$, but it is unbounded on any slab $\Sigma_t$, $t > 0$.

Proof. When ($H_1$) is in force we use Theorem 2 of the recent work ^[23]. To this end we first observe that the assumptions on $f$ imply that $f(s) \geq -A$ for every $s \geq 0$ and some $A > 0$. Then, for $R > 1$ we set $\Omega : = B(0', 1) \times (0, 2R)$ and observe that, for any $z' \in {\mathbb{R}}^{N-1}$, the function $v(x): = u(x' + z', x_N)$ solves

Now, we fix $q > N$ and we apply Theorem 2 of ^[23] to $v$ with $A^{(1)} = A^{(2)} = Id$ (hence $\lambda = 1$ and $\Lambda = \Lambda(q, N, \Omega))$, $b \equiv 0$, $h = A$, $f(x, s) = f(s)+A \geq 0$, $g(x, s) = f^{+}(s), \; \xi(s) = s$, $\beta = 1$, $\Omega' = \Omega$ and $\omega = B(z_0, \frac{1}{4}),$ where $z_0 = (0', 1)$. This leads to

where $C$ is a positive constant depending only on $N, q, r, \Omega, T, f$. Since $z'$ is an arbitrary point of ${\mathbb{R}}^{N-1}$ we then have

where $C > 0$ depends only on $R, N, q, r, T, f$. The latter gives the desired conclusion since $R > 1$ is arbitrary.

When ($H_2$) holds true, the conclusion is clear thanks to the boundary condition satisfied by $u$ and the mean value theorem.

When ($H_3$) is satisfied we use the following consequence of the maximum principle. Hereafter, for $z \in \partial {\mathbb{R}}^N_+$ and $R > 0$, we set $B^+(z, R) : = B(z, R) \cap {\mathbb{R}}^N_+$.

Lemma 4.3. Assume $N\geq 2$ and let $v \in C^2(\overline {B^+(0, R)})$ be any solution of

Then

Proof of Lemma 4.3. If $x \in B^+(0, \frac{3R}{4})$ and $x_N \geq \frac{3R}{4}$, then (4.3) is clearly true. If $x = (x', x_N) \in B^+(0, \frac{3R}{4})$ and $x_N < \frac{3R}{4},$ we set $z = (x', 0) \in \partial {\mathbb{R}}^N_+$, $S: = \sup_{B^+(0, R)} v$, $r = R - \vert z \vert$ and observe that $0 < x_N < \frac{R}{4} < r < R$. Then, for $y \in B^+(z, r)$, we consider the harmonic function $H(y) : = S \Big (\frac{\vert y - z\vert^2}{r^2} + N (\frac{y_N}{r} - \frac{y_N^2}{r^2}) \Big)$, which also satisfies $H \geq v$ on $\partial B^+(z, r)$. Therefore, $0 \leq v \leq H$ on $B^+(z, r)$, by the maximum principle. In particular, for $y = x$, we get $0 \leq v(x) \leq S \Big (\frac{x_N^2}{r^2} + N (\frac{x_N}{r} - \frac{x_N^2}{r^2}) \Big) \leq S N (\frac{x_N}{r}) \leq 4N \Big(\sup_{B^+(0, R)} v \Big) \frac{x_N}{R}$. Which concludes the proof of the Lemma.

By ($H_3$) there is $a_0 > 0$ such that $u(x) \leq a_0(1 + \vert x \vert)$ for every $x \in {\mathbb{R}}^N_+$. Let $x \in {\mathbb{R}}^N_+$ and pick $R = 2 \vert x \vert + 1$ and observe that $x \in B^+(0, \frac{3R}{4})$. Thus, an application of the above Lemma 4.3 yields $0 \leq u(x) \leq 4N (a_0 (1 + R) \frac{x_N}{R} \leq (8a_0N) x_N.$ This concludes the proof of the Theorem.

By gathering together the previous results we can deduce various consequences. We start with

Theorem 4.4. Assume $N\geq 2$, $f \in C^0([0, +\infty))$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of (3.18).

i) If $f$ satisfies $\lim_{t\to \infty} \frac{f(t)}{t^r} = \ell \in (0, +\infty)$ for some $r \in (1, \frac{N+1}{N-1})$, then $u$ is bounded on ${\mathbb{R}}^N_+$.

ii) If $f$ satisfies $\, t^r -t \leq f(t) \leq \Lambda(t^r +1)$ for $t \geq 0$, where $r \in (1, \frac{N+1}{N-1})$ and $\Lambda > 1,$ then $u$ is bounded on ${\mathbb{R}}^N_+$.

In both cases the bound on $u$ is universal, i.e, it depends on $f$ and $N$ only.

iii) If $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$ and $f$ satisfies $\lim_{t\to \infty} \frac{f(t)}{t^p} = \ell \in (0, +\infty)$ for some $p \in (1, \overline p(N))$, then $u$ is bounded on ${\mathbb{R}}^N_+$.

Here $\overline p(N)$ is the Sobolev exponent, i.e., $\overline p(N) = \frac{N+2}{N-2}$ if $N \geq 3$ and $p_S(2) = +\infty.$

iv) If $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$ and $f$ satisfies $\, t^p -t \leq f(t) \leq \Lambda(t^p +1)$ for $t \geq 0$, where $p \in (1, \underline p(N))$ and $\Lambda > 1,$ then $u$ is bounded on ${\mathbb{R}}^N_+$.

Here $\underline p(N)$ is the Serrin exponent, i.e., $\underline p(N) = \frac{N}{N-2}$ if $N \geq 3$ and $p_S(2) = +\infty.$

Proof. If $f$ satisfies the assumption of item i), then $f$ also satisfies the assumption $(H_1)$ of Theorem 4.1. Thus $u$ is bounded on the slab ${\mathbb{R}}^{N-1} \times [0, 1]$ by a constant depending only on $N$ and $f$. On the other hand, by Theorem 2.1 of ^[20], applied with $\Omega = {\mathbb{R}}^N_+$, we have that $u(x) \leq C(N, f)(1 + dist^{-\frac{2}{r-1}}(x, \partial \Omega))$ for every $x \in \Omega = {\mathbb{R}}^N_+$. Hence $u$ is bounded on the set ${\mathbb{R}}^{N-1} \times [1, +\infty)$ by the universal constant $2C(N, f)$. This gives the conclusion.

If $f$ satisfies the assumption of item ii), then $f$ also satisfies the assumption $(H_1)$ of Theorem 4.1 and so, as before, $u$ is bounded on the slab ${\mathbb{R}}^{N-1} \times [0, 4]$ by a constant depending only on $N$ and $f$. On the other hand the following standard integral estimate holds true for $u$

for all $x_0$ such that $\overline{B(x_0, 2)} \subset {\mathbb{R}}^N_+$. Here $C(N, r)$ is a positive constant independent on $x_0$ and $u$ (it actually depends on $N$ and $r$ only). To this end, we first observe that the functions $u_{x_0}(x) : = u(x+x_0)$ satisfy $- \Delta u_{x_0} \geq u^r_{x_0} -u_{x_0}$ on $B(0, 2)$ and then we multiply the previous differential inequality by $\phi_1$ (a positive first eigenfunction of $-\Delta$ with homogeneous Dirichlet boundary conditions in $B(0, 2)$) and integrate by parts to get

where $\lambda_1 > 0$ is the first eigenvalue of $-\Delta$ with homogeneous Dirichlet boundary conditions in $B(0, 2).$ From the latter, after an application of Holder inequality, we obtain

and so

as claimed. From (4.4) we then get

hence, for all $x_0$ such that $\overline{B(x_0, 2)} \subset {\mathbb{R}}^N_+$,

where $C'(N, r)$ is a positive constant independent on $x_0$ and $u$.

Combining (4.5) with the Harnack inequality (see e.g. item (b) of Theorem 4.1 and item (b) of Theorem 4.3 of ^[22]), applied to every ball $B(x_0, 1)$ where $x_0 \in {\mathbb{R}}^{N-1} \times [3, +\infty),$ we obtain

where $C''(N, r)$ is a positive constant independent on $x_0$ and $u$. The desired conclusion then follows.

The cases ⅲ) and iv) are treated as the cases ⅰ) and ⅱ) with the only difference that we use that $(H_2)$ of Theorem 4.1 is now in force.

Theorem 4.5. Assume $N\geq 2$, $f \in C^{0, 1}_{loc}([0, +\infty))$ with $f(0) \geq 0$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of $(P)$. If either the condition $(H_1)$ or $(H_2)$ or $(H_3)$ of Theorem 4.1 is satisfied, then $u$ is monotone, i.e., $\frac{\partial u }{\partial x_N} > 0$ in ${\mathbb{R}}^N_+$.

Remark 4.6. In the case $N = 2$ the conclusion of the theorem above was already known to hold under the sole assumption that $f$ is locally lipschitz continuous, see ^[15,16].

Proof. Theorem 4.1 implies that $u$ is bounded on every slab. The conclusion then follows by applying Theorem 3.1.

Theorem 4.7. Assume $f \in C^{0, 1}_{loc}([0, +\infty))$ and let $u \in C^2(\overline{ {\mathbb{R}}^N_+})$ be a solution of (3.18).

a) Assume $N = 2, 3$ and let us suppose that one of the following assumptions holds true :

i) $f(0) \geq 0$ and $\lim_{t\to \infty} \frac{f(t)}{t^r} = \ell \in (0, +\infty)$ for some $r \in (1, \frac{N+1}{N-1})$.

ii) $t^r -t \leq f(t) \leq \Lambda(t^r +1)$ for $t \geq 0$, where $r \in (1, \frac{N+1}{N-1})$ and $\Lambda > 1$.

iii) $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$ and $f(0) \geq 0, \; \lim_{t\to \infty} \frac{f(t)}{t^p} = \ell \in (0, +\infty)$ for some $p \in (1, \overline p(N))$.

iv) $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$ and $\, t^p -t \leq f(t) \leq \Lambda(t^p +1)$ for $t \geq 0$, where $p \in (1, \underline p(N))$ and $\Lambda > 1$.

Here $\underline p$ and $\overline p$ are as in Theorem 4.4. \end{itemize}

Then, either $u \equiv 0$ and $f(0) = 0$, or $u$ is positive, bounded, monotone and one-dimensional on ${\mathbb{R}}^N_+$.

b) Assume $N \geq 2$.

i) If $\, t^r \leq f(t) \leq \Lambda t^r$ for $t \geq 0$, where $r \in (1, \frac{N+1}{N-1})$ and $\Lambda > 1,$ then $u \equiv 0$ in ${\mathbb{R}}^N_+$.

ii) if $\nabla u \in L^{\infty}({\mathbb{R}}^N_+)$ and $\, t^p \leq f(t) \leq \Lambda t^p$ for $t \geq 0$, where $p \in (1, \underline p(N))$ and $\Lambda > 1$, then $u \equiv 0$ in ${\mathbb{R}}^N_+$.

Remark 4.8. For $N = 2$ : item a) ⅰ) holds true for any $r > 1$ (see ^[16]), item a) ⅲ) holds true for any locally Lipschitz function $f$ satisfying $f(0) \geq 0$ (see ^[15]) while item a) iv) and item b) ⅱ) hold true for any locally Lipschitz function $f$ (see ^[15]).

Proof. Note that $f(0) \geq 0$ in any case. Then, by the strong maximum principle, either $u \equiv 0$ and so $f(0) = 0$, or $u > 0$ on ${\mathbb{R}}^N_+$. Then, to conclude the proof of item a) we just need to treat the case $u > 0$. By Theorem 4.4 and Theorem 4.5 $u$ is bounded and monotone. Since a solution for $N = 2$ can be seen as a solution for $N = 3$, the one-dimensional symmetry of $u$ then follows from Theorem 1.5 of ^[18] (or from Theorem 1.5 of ^[4] if $f \in C^1$). Let us now turn to item b) and suppose for contradiction that $u > 0$. If $N = 2, 3$ then, thanks to item a), $u$ would be a 1D, bounded and monotone increasing solution to $-u'' = f(u)$ on ${\mathbb{R}}^+$, which is clearly impossible. If $N\ge 4$, $u$ would be bounded and monotone increasing by Theorem 4.5. But this is in contradiction with the last sentence of item (a) of Theorem IV of ^[22] (which implies that $u \to 0$ as $x_N \to +\infty$). Thus $u\equiv 0$ on ${\mathbb{R}}^N_+$, as claimed.

Conflict of interest

The authors declare no conflict of interest.