Singular semilinear elliptic equations in half-spaces

Phuong Le Phuong Le^1,2 (ORCID: 0000-0003-4724-7118)
¹Faculty of Economic Mathematics, University of Economics and Law, Ho Chi Minh City, Vietnam;
²Vietnam National University, Ho Chi Minh City, Vietnam phuongl@uel.edu.vn

Abstract.

We prove the monotonicity of positive solutions to the problem $-\Delta u=f(u)$ in $\mathbb{R}^{N}_{+}:=\{(x^{\prime},x_{N})\in\mathbb{R}^{N}\mid x_{N}>0\}$ under zero Dirichlet boundary condition with a possible singular nonlinearity $f$ . In some situations, we can derive a precise estimate on the blow-up rate of $\frac{\partial u}{\partial\eta}$ as $x_{N}\to 0^{+}$ , where $(\eta,e_{N})>0$ , and obtain a classification result. The main tools we use are the method of moving planes and the sliding method.

Key words and phrases:

semilinear elliptic equation, singular nonlinearity, half-space, monotonicity, rigidity

2020 Mathematics Subject Classification:

35J61, 35J75, 35B06, 35B09

1. Introduction

The monotonicity and symmetry of solutions to the semilinear elliptic problem

\begin{cases}-\Delta u=f(u)&\text{ in }\mathbb{R}^{N}_{+},\\ u>0&\text{ in }\mathbb{R}^{N}_{+},\\ u=0&\text{ on }\partial\mathbb{R}^{N}_{+},\end{cases}

(1)

where

\mathbb{R}^{N}_{+}:=\{x:=(x^{\prime},x_{N})\in\mathbb{R}^{N}\mid x_{N}>0\},

are well studied in the literature. Berestycki, Caffarelli, and Nirenberg [6, 4] demonstrated that if $f:[0,+\infty)\to\mathbb{R}$ is a Lipschitz function with $f(0)\geq 0$ , then any classical solution of (1) is monotone in the $x_{N}$ -direction. When $f$ is only locally Lipschitz continuous on $[0,+\infty)$ , a similar monotonicity result can be obtained for solutions that are bounded on all strips $\Sigma_{\lambda}:=\{(x^{\prime},x_{N})\in\mathbb{R}^{N}\mid 0<x_{N}<\lambda\}$ ( $\lambda>0$ ), as shown in [17, 27]. The case where $f(0)<0$ is more complex, and a complete proof of monotonicity for solutions in this scenario is currently only available for dimension $N=2$ , as detailed in [18, 19]. For results on symmetry of solutions, which is usually called rigidity in the literature, we refer to [7, 4, 2, 13, 5] and the references therein.

In this paper, we are mainly interested in problem (1) with singular nonlinearity at zero in the sense that $f:(0,+\infty)$ is locally Lipschitz continuous and $\lim_{t\to 0^{+}}f(t)=+\infty$ . A model problem is given by

\begin{cases}-\Delta u=\dfrac{1}{u^{\gamma}}+g(u)&\text{ in }\mathbb{R}^{N}_{+},\\ u>0&\text{ in }\mathbb{R}^{N}_{+},\\ u=0&\text{ on }\partial\mathbb{R}^{N}_{+},\end{cases}

(2)

where $\gamma>0$ and $g:[0,+\infty)$ is a locally Lipschitz continuous function. It’s well established that solutions to problem (2) are generally not smooth up to the boundary. In fact, it was shown in [23] and also in Theorem 2 below that the gradient of solutions becomes unbounded at the boundary. Given the natural regularity behavior of these solutions (as discussed in [14]), we focus on solutions $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ to (1). Consequently, the equation is well-defined in the classical sense within the domain’s interior.

Since the seminal paper [14], singular semilinear elliptic problems

\begin{cases}-\Delta u=\dfrac{1}{u^{\gamma}}+g(u)&\text{ in }\Omega,\\ u>0&\text{ in }\Omega,\\ u=0&\text{ on }\Omega,\end{cases}

(3)

where $\Omega\subset\mathbb{R}^{N}$ is a bounded domain, have been extensively studied from various perspectives. We specifically reference the works [3, 9, 12, 10, 21, 22, 23, 15], which are closely related to our research. A key focus in the study of these equations is understanding the behavior of solutions near the boundary, where they often lose regularity. A generalized version of the Höpf boundary lemma was obtained in [11]. The symmetry of solutions was studied in [15] (see also [16, 24] and the references therein).

As demonstrated in [11], to obtain the Höpf boundary lemma for (3), one may exploit a scaling argument near the boundary which leads to the study of a limiting problem in the half-space

\begin{cases}-\Delta u=\dfrac{1}{u^{\gamma}}&\text{ in }\mathbb{R}^{N}_{+},\\ u>0&\text{ in }\mathbb{R}^{N}_{+},\\ u=0&\text{ on }\partial\mathbb{R}^{N}_{+},\end{cases}

(4)

which is exactly problem (1) with $f\equiv 0$ . Solutions to problem (4) have been classified recently in elegant papers [25, 26]. These results reveal that all weak solutions to (4) with $\gamma>1$ must be either of the form

u(x)\equiv\frac{(\gamma+1)^{\frac{2}{\gamma+1}}}{(2\gamma-2)^{\frac{1}{\gamma+1}}}x_{N}^{\frac{2}{\gamma+1}}

or of the form

u(x)\equiv\lambda^{-\frac{2}{\gamma+1}}v(\lambda x_{N})

where $\lambda>0$ and $v\in C^{2}(\mathbb{R}_{+})\cap C(\overline{\mathbb{R}_{+}})$ is the unique solution to

\begin{cases}-v^{\prime\prime}=\dfrac{1}{v^{\gamma}},&t>0,\\ v(t)>0,&t>0,\\ v(0)=0,~{}\lim_{t\to+\infty}v^{\prime}(t)=1.\end{cases}

One notable feature of problem (4) is that its nonlinearity is decreasing on $(0,+\infty)$ . Hence the weak comparison principle holds in large subdomains of $\mathbb{R}^{N}_{+}$ and the monotonicity of solutions follows directly from this principle. In this paper, we consider a more general situation by studying the monotonicity and rigidity results for solutions $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ to (1) with a possible singular nonlinearity $f$ . In particular, we address the issue that $f$ is not decreasing in the whole $(0,+\infty)$ . The main assumption on $f$ we require is the following:

( $F$ )

for any $M>0$ , there exists $C(M)>0$ such that

$f(s)-f(t)\leq C(M)(s-t)\quad\text{ for all }0<t\leq s\leq M.$

Our first result is the following monotonicity result, which holds in a very general setting. Indeed, we require only the behavior of $f$ near its possible singular point.

Theorem 1.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function satisfying ( $F$ ) and there exist $c_{0},t_{0}>0$ such that

f(t)>c_{0}t\quad\text{ for all }0<t<t_{0}.

Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1) with $u\in L^{\infty}(\Sigma_{\lambda})$ for all $\lambda>0$ . Then

\frac{\partial u}{\partial x_{N}}>0\quad\text{ in }\mathbb{R}^{N}_{+}.

Theorem 1 applies not only to singular nonlinearities but also the superlinear ones. Since $f$ is not decreasing, we need to derive a weak comparison principle for the problem in narrow strips and exploit the moving plane method to prove Theorem 1. Similar results for singular problems in bounded domains were obtained in [15, 11]. Theorem 1 can be applied to problem (2) to yield the monotonicity of solutions. Furthermore, the inward derivatives of all such solutions must blow up near the boundary. Indeed, we can provide a precise estimate of the blow-up rate of derivatives as $x_{N}\to 0^{+}$ in our next result.

Theorem 2.

Assume that $\gamma>1$ and $g:[0,+\infty)$ is a locally Lipschitz continuous function. Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (2) with $u\in L^{\infty}(\Sigma_{\overline{\lambda}})$ for some $\overline{\lambda}>0$ . Then

\frac{\partial u}{\partial x_{N}}>0\quad\text{ in }\mathbb{R}^{N}_{+}.

Moreover, for each $\beta\in(0,1)$ , there exist $c_{1},c_{2},\lambda_{0}>0$ such that

c_{1}x_{N}^{\frac{1-\gamma}{\gamma+1}}<\frac{\partial u(x)}{\partial\eta}<c_{2}x_{N}^{\frac{1-\gamma}{\gamma+1}}\quad\text{ in }\Sigma_{\lambda_{0}}

(5)

for all $\eta\in\mathbb{S}^{N-1}_{+}$ with $(\eta,e_{N})\geq\beta$ , where $\mathbb{S}^{N-1}_{+}:=\mathbb{R}^{N}_{+}\cap\partial B_{1}(0)$ and $e_{N}:=(0,\dots,0,1)$ .

In this paper, we also exploit the techniques and ideas from [25] to establish one-dimensional symmetry of solutions to singular problems whose a model is problem (2), where $\gamma>1$ and $g:(0,+\infty)\to\mathbb{R}$ is a nonnegative locally Lipschitz continuous function such that $\limsup_{t\to+\infty}t^{\gamma}g(t)<+\infty$ . Notice that $g$ may have a singularity at zero such as $g(t)=\frac{1}{t^{\beta}}$ with $\beta\geq\gamma$ .

Theorem 3.

Assume that $\gamma>1$ and $f:(0,+\infty)$ is a positive locally Lipschitz continuous function satisfying ( $F$ ) and

(i)

there exists $c_{0}>0$ such that

$f(t)>\frac{c_{0}}{t^{\gamma}}\quad\text{ for all }t>0,$
(ii)

there exist $c_{1},t_{1}>0$ such that $f$ is nonincreasing on $(t_{1},+\infty)$ and

$f(t)<\frac{c_{1}}{t^{\gamma}}\quad\text{ for all }t>t_{1}.$

Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1) with $u\in L^{\infty}(\Sigma_{\overline{\lambda}})$ for some $\overline{\lambda}>0$ . Then $u(x)\equiv v(x_{N})$ , where $v$ is given by the formula

\int_{0}^{v(t)}\frac{ds}{\sqrt{M+F(s)}}=\sqrt{2}t\quad\text{ for all }t\geq 0

for some $M\geq 0$ , where $F(s)=\int_{s}^{+\infty}f(t)dt$ .

We will employ the sliding method, which was introduced by Berestycki and Nirenberg [8], to prove Theorem 3. We stress that in Theorem 3 we do not assume that $f$ is nonincreasing in the whole domain $(0,+\infty)$ . If this condition is granted, then we can show that solutions depend only on $x_{N}$ without the assumption of their boundedness on strips. This in turn yields a classification result. The proof for the following result is similar to the one in [26].

Theorem 4.

Assume that $f:(0,+\infty)$ is a nonincreasing positive locally Lipschitz continuous function. Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1). Then $u$ depends only on $x_{N}$ . Consequently, such a solution exists if and only if $\int_{1}^{+\infty}f(t)dt<+\infty$ . Moreover, when such a solution exists, it is given by $u(x)\equiv v(x_{N})$ , where $v$ is determined by the formula

\int_{0}^{v(t)}\frac{ds}{\sqrt{M+F(s)}}=\sqrt{2}t\quad\text{ for all }t\geq 0

for some $M\geq 0$ , where $F(s)=\int_{s}^{+\infty}f(t)dt$ .

The rest of this paper is devoted to the proofs of our main results. In Section 2 we prove a comparison principle for narrow strips and derive some a priori estimates for solutions. In Section 3, we prove the monotonicity and rigidity of solutions by means of the moving plane and sliding methods.

2. Preliminaries

2.1. Weak comparison principle for narrow strips

We prove a weak comparison principle which can be applied to problems with singular nonlinearities.

Proposition 5.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function such that ( $F$ ) holds and $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ satisfies

\begin{cases}-\Delta u\leq f(u),~{}u>0&\text{ in }\mathbb{R}^{N}_{+},\\ u=0&\text{ on }\partial\mathbb{R}^{N}_{+},\\ u\in L^{\infty}(\Sigma_{\overline{\lambda}})&\text{ for some }\overline{\lambda}>0.\end{cases}

Then there exists a small positive number $\lambda^{*}=\lambda^{*}(f,\|u\|_{L^{\infty}(\Sigma_{\overline{\lambda}})})<\overline{\lambda}$ such that: if $0<\lambda\leq\lambda^{*}$ and $v\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ satisfies

\begin{cases}-\Delta v\geq f(v),~{}v>0&\text{ in }\Sigma_{\lambda},\\ v>0&\text{ on }\partial\mathbb{R}^{N}_{+},\\ u\leq v&\text{ on }\{x_{N}=\lambda\},\\ \end{cases}

then $u\leq v$ in $\Sigma_{\lambda}$ .

Proof.

In what follows, we consider $\lambda<\overline{\lambda}$ . For $R>0$ , let $\varphi_{R}\in C^{\infty}(\mathbb{R}^{N-1})$ be such that

\begin{cases}0\leq\varphi_{R}\leq 1&\text{ in }\mathbb{R}^{N-1},\\ \varphi_{R}=1&\text{ in }B^{\prime}_{R},\\ \varphi_{R}=0&\text{ in }\mathbb{R}^{N-1}\setminus B^{\prime}_{2R},\\ |\nabla\varphi_{R}|\leq\frac{2}{R}&\text{ in }B^{\prime}_{2R}\setminus B^{\prime}_{R},\end{cases}

(6)

where $B_{r}^{\prime}$ denotes the ball in $\mathbb{R}^{N-1}$ of center $0^{\prime}\in\mathbb{R}^{N-1}$ with radius $r$ . We set

\varphi(x)=w^{+}(x)\varphi_{R}^{2}(x^{\prime})\chi_{\Sigma_{\lambda}}(x),

where $w^{+}:=\max\{u-v,0\}$ . Since the support of $\varphi$ is compactly contained in $\Sigma_{\lambda}\cup\{x_{N}=\lambda\}$ , we can use it as a test function in $-\Delta u\leq f(u)$ and $-\Delta v\geq f(v)$ . Then subtracting, we obtain

	$\displaystyle\int_{\Sigma_{\lambda}}\|\nabla w^{+}\|^{2}\varphi_{R}^{2}$	$\displaystyle\leq-2\int_{\Sigma_{\lambda}}(\nabla w^{+},\nabla\varphi_{R})w^{+}\varphi_{R}+\int_{\Sigma_{\lambda}}(f(u)-f(v))w^{+}\varphi_{R}^{2}$
		$\displaystyle\leq 2\int_{\Sigma_{\lambda}}\|\nabla w^{+}\|\|\nabla\varphi_{R}\|w^{+}\varphi_{R}+\int_{\Sigma_{\lambda}}(f(u)-f(v))w^{+}\varphi_{R}^{2}.$

In the set $\Sigma_{\lambda}\cap\{w^{+}>0\}$ we have

0<v<u\leq\|u\|_{L^{\infty}(\Sigma_{\overline{\lambda}})}.

Hence by exploiting Young’s inequality and using ( $F$ ), we have

	$\displaystyle\int_{\Sigma_{\lambda}}\|\nabla w^{+}\|^{2}\varphi_{R}^{2}$	$\displaystyle\leq\frac{1}{2}\int_{\Sigma_{\lambda}}\|\nabla w^{+}\|^{2}\varphi_{R}^{2}+2\int_{\Sigma_{\lambda}}\|\nabla\varphi_{R}\|^{2}(w^{+})^{2}$
		$\displaystyle\qquad+C(\\|u\\|_{L^{\infty}(\Sigma_{\overline{\lambda}})})\int_{\Sigma_{\lambda}}(w^{+})^{2}\varphi_{R}^{2}.$

This is equivalent to

\int_{\Sigma_{\lambda}}|\nabla w^{+}|^{2}\varphi_{R}^{2}\leq 4\int_{\Sigma_{\lambda}}|\nabla\varphi_{R}|^{2}(w^{+})^{2}+2C(\|u\|_{L^{\infty}(\Sigma_{\overline{\lambda}})})\int_{\Sigma_{\lambda}}(w^{+})^{2}\varphi_{R}^{2}.

(7)

By the classical Poincaré inequality in the interval $(0,\lambda)$ , we have

	$\displaystyle\int_{\Sigma_{\lambda}}\|\nabla w^{+}\|^{2}\varphi_{R}^{2}$	$\displaystyle\geq\int_{\mathbb{R}^{N-1}}\left(\int_{0}^{\lambda}\left(\frac{\partial w^{+}}{\partial x_{N}}\right)^{2}dx_{N}\right)\varphi_{R}^{2}(x^{\prime})dx^{\prime}$
		$\displaystyle\geq\frac{\pi^{2}}{\lambda^{2}}\int_{\mathbb{R}^{N-1}}\left(\int_{0}^{\lambda}(w^{+})^{2}dx_{N}\right)\varphi_{R}^{2}(x^{\prime})dx^{\prime}$
		$\displaystyle=\frac{\pi^{2}}{\lambda^{2}}\int_{\Sigma_{\lambda}}(w^{+})^{2}\varphi_{R}^{2}.$

Therefore, (7) leads to

\left(\frac{\pi^{2}}{\lambda^{2}}-2C(\|u\|_{L^{\infty}(\Sigma_{\overline{\lambda}})})\right)\int_{\Sigma_{\lambda}}(w^{+})^{2}\varphi_{R}^{2}\leq 4\int_{\Sigma_{\lambda}}|\nabla\varphi_{R}|^{2}(w^{+})^{2}.

Choosing $\lambda_{0}=\pi[4+2C(\|u\|_{L^{\infty}(\Sigma_{\overline{\lambda}})})]^{-\frac{1}{2}}$ , then for all $\lambda<\min\{\lambda_{0},\overline{\lambda}\}$ we have

\int_{\Sigma_{\lambda}}(w^{+})^{2}\varphi_{R}^{2}\leq\int_{\Sigma_{\lambda}}|\nabla\varphi_{R}|^{2}(w^{+})^{2}.

Using (6), we deduce

\int_{\Sigma_{\lambda}^{R}}(w^{+})^{2}\leq\frac{4}{R^{2}}\int_{\Sigma_{\lambda}^{2R}}(w^{+})^{2},

where $\Sigma_{\lambda}^{R}:=B^{\prime}_{R}\times(0,\lambda)$ . Setting $h(R):=\int_{\Sigma_{\lambda}^{R}}(w^{+})^{2}$ , we have

h(R)\leq\frac{4}{R^{2}}h(2R)\quad\text{ and }\quad h(R)\leq C_{\lambda}R^{N-1}\quad\text{ for all }R>0.

Hence $h(R)\leq\frac{1}{2^{N}}h(2R)$ for all $R>2^{\frac{N+2}{2}}$ . By iteration of this inequality, we obtain

h(R)\leq\frac{1}{2^{Nk}}h(2^{k}R)\leq\frac{C_{\lambda}}{2^{Nk}}(2^{k}R)^{N-1}=\frac{C_{\lambda}}{2^{k}}R^{N-1}

for all $k\in\mathbb{N}$ and $R>2^{\frac{N+2}{2}}$ . Letting $k\to\infty$ , we deduce $h(R)\equiv 0$ . This implies $\int_{\Sigma_{\lambda}}(w^{+})^{2}=0$ , which means $u\leq v$ in $\Sigma_{\lambda}$ for all $\lambda<\lambda^{*}:=\min\{\lambda_{0},\overline{\lambda}\}$ . ∎

2.2. A priori estimates

We need the following property on solutions so that we can carry out the moving plane method to prove Theorem 1.

Lemma 6.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function such that ( $F$ ) holds. Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1) with $u\in L^{\infty}(\Sigma_{\overline{\lambda}})$ for some $\overline{\lambda}>0$ . Then

\lim_{x_{N}\to 0^{+}}u(x^{\prime},x_{N})=0\text{ uniformly in }x^{\prime}\in\mathbb{R}^{N-1}_{+}.

Proof.

Let $\lambda^{*}<\overline{\lambda}$ be defined as in Proposition 5 and choose some $\rho>\|u\|_{L^{\infty}(\Sigma_{\lambda^{*}})}$ . Let $h:(0,+\infty)\to\mathbb{R}$ be a $C^{1}$ function such that

	$\displaystyle h(t)>\max\{f(t),0\}$	$\displaystyle\quad\text{ in }(0,\rho),$
	$\displaystyle h(t)=\frac{c}{t^{2}}$	$\displaystyle\quad\text{ in }[\rho,+\infty)$

for some $c>0$ . We set $H(t)=\int_{\rho}^{t}h(s)ds$ for $t>0$ , then $H$ is strictly increasing in $(0,+\infty)$ and $H(t)<\int_{\rho}^{+\infty}h(s)ds=\frac{c}{\rho}$ . For each $\mu\geq\frac{c}{\rho}$ , one can check that

\int_{0}^{+\infty}\frac{ds}{\sqrt{\mu-H(s)}}>\int_{\rho}^{+\infty}\frac{ds}{\sqrt{\mu}}=+\infty

and

\int_{0}^{t}\frac{ds}{\sqrt{\mu-H(s)}}<\frac{t}{\sqrt{\mu-H(t)}}<+\infty\quad\text{ for }0<t<+\infty.

Hence the formula

\int_{0}^{v_{\mu}(t)}\frac{ds}{\sqrt{\mu-H(s)}}=\sqrt{2}t\quad\text{ for all }t\geq 0

uniquely determine a function $v_{\mu}\in C^{2}(\mathbb{R}_{+})\cap C(\overline{\mathbb{R}_{+}})$ , which is a solution to the ODE problem

\begin{cases}-v^{\prime\prime}=h(v)&\text{ in }\mathbb{R}_{+},\\ v^{\prime}(t)>0&\text{ in }\mathbb{R}_{+},\\ v(0)=0.\end{cases}

Moreover, $\lim_{\mu\to+\infty}v_{\mu}(t)=+\infty$ for all $t>0$ .

We fix some $\mu>0$ such that $v_{\mu}(\lambda^{*})>\rho$ . Then we choose $\lambda_{0}<\lambda^{*}$ satisfying $\|u\|_{L^{\infty}(\Sigma_{\lambda^{*}})}<v_{\mu}(\lambda_{0})<\rho$ . By abuse of notation, we will write $v_{\mu}(x^{\prime},x_{N}):=v_{\mu}(x_{N})$ . Then $0<v_{\mu}<\rho$ in $\Sigma_{\lambda_{0}}$ and $u<v_{\mu}$ on $\{x_{N}=\lambda_{0}\}$ .

For small $\varepsilon>0$ such that $v_{\mu}(\lambda_{0}+\varepsilon)<\rho$ , we define

v_{\mu,\varepsilon}(x):=v_{\mu}(x+\varepsilon e_{N}),

Then

\begin{cases}v_{\mu}(\varepsilon)<v_{\mu,\varepsilon}<v_{\mu}(\lambda_{0}+\varepsilon)<\rho&\text{ in }\Sigma_{\lambda_{0}},\\ -\Delta v_{\mu,\varepsilon}=h(v_{\mu,\varepsilon})>f(v_{\mu,\varepsilon})&\text{ in }\Sigma_{\lambda_{0}},\\ u\leq v_{\mu,\varepsilon}&\text{ on }\partial\Sigma_{\lambda_{0}}.\end{cases}

Now Proposition 5 implies $u\leq v_{\mu,\varepsilon}$ in $\Sigma_{\lambda_{0}}$ . Letting $\varepsilon\to 0$ , we have $u\leq v_{\mu}$ in $\Sigma_{\lambda_{0}}$ and the conclusion follows from that fact that $\lim_{t\to 0^{+}}v_{\mu}(t)=0$ . ∎

We prove some a priori estimates for solutions to (1) in what follows. The next lemma improves the upper bound on $u$ near the boundary in Lemma 6 when an explicit upper bound on $f$ is given.

Lemma 7.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function such that ( $F$ ) holds and $f(t)<\frac{c_{0}}{t^{\gamma}}$ for all $0<t<t_{0}$ , where $c_{0},t_{0}>0$ and $\gamma>1$ . Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1) with $u\in L^{\infty}(\Sigma_{\overline{\lambda}})$ for some $\overline{\lambda}>0$ . Then

u(x)\leq Cx_{N}^{\frac{2}{\gamma+1}}\quad\text{ in }\Sigma_{\lambda_{0}}

for some constants $C,\lambda_{0}>0$ .

Proof.

Let $\lambda^{*}<\overline{\lambda}$ be defined as in Proposition 5 and choose some $\rho>\|u\|_{L^{\infty}(\Sigma_{\lambda^{*}})}$ . By compactness, there exists $c_{\geq}c_{0}$ such that $f(t)<\frac{c_{1}}{t^{\gamma}}$ for all $0<t<\rho$ . Let

v(t)\equiv\frac{(\gamma+1)^{\frac{2}{\gamma+1}}}{(2\gamma-2)^{\frac{1}{\gamma+1}}}t^{\frac{2}{\gamma+1}},

then $v_{\mu}(x)=\mu v(x_{N})$ solves $-\Delta v_{\mu}=\frac{\mu^{\gamma+1}}{v_{\mu}^{\gamma}}$ in $\mathbb{R}^{N}_{+}$ . By abuse of notation, we will write $v_{\mu}(x_{N}):=v_{\mu}(x^{\prime},x_{N})$ . We choose $\mu$ large such that $\mu^{\gamma+1}>c_{1}$ and $v_{\mu}(\lambda^{*})>\rho$ . Then we choose $\lambda_{0}<\lambda^{*}$ satisfying $\|u\|_{L^{\infty}(\Sigma_{\lambda^{*}})}<v_{\mu}(\lambda_{0})<\rho$ . Now we have $0<v_{\mu}<\rho$ in $\Sigma_{\lambda_{0}}$ and $-\Delta v_{\mu}\geq f(v_{\mu})$ in $\Sigma_{\lambda_{0}}$ .

For small $\varepsilon>0$ such that $v_{\mu}(\lambda_{0}+\varepsilon)<\rho$ , we define

v_{\mu,\varepsilon}(x)=v_{\mu}(x+\varepsilon e_{N}).

Then

\begin{cases}-\Delta v_{\mu,\varepsilon}\geq f(v_{\mu,\varepsilon})&\text{ in }\Sigma_{\lambda_{0}},\\ v_{\mu,\varepsilon}=v_{\mu}(\varepsilon)>0&\text{ on }\{x_{N}=0\},\\ u<v_{\mu,\varepsilon}&\text{ on }\{x_{N}=\lambda_{0}\}.\end{cases}

Now Proposition 5 implies $u\leq v_{\mu,\varepsilon}$ in $\Sigma_{\lambda_{0}}$ . Letting $\varepsilon\to 0$ we conclude the proof. ∎

The next lemma concerns a lower bound on solutions.

Lemma 8.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function and $f(t)>\frac{c_{0}}{t^{\gamma}}$ for all $0<t<t_{0}$ , where $c_{0},t_{0}>0$ and $\gamma\geq 0$ . Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1). Then

u(x)\geq\min\{Cx_{N}^{\frac{2}{\gamma+1}},t_{0}\}\quad\text{ in }\mathbb{R}^{N}_{+}

for some constant $C>0$ independent of $t_{0}$ .

Proof.

Let $\lambda_{1}>0$ and $\phi_{1}\in C^{2}(\overline{B_{1}(0)})$ be the first eigenvalue and a corresponding positive eigenfunction of the Laplacian in $B_{1}(0)$ , namely,

\begin{cases}-\Delta\phi_{1}=\lambda_{1}\phi_{1}&\text{ in }B_{1}(0),\\ \phi_{1}>0&\text{ in }B_{1}(0),\\ \phi_{1}=0&\text{ on }\partial B_{1}(0).\end{cases}

Setting

w=\beta\phi_{1}^{\frac{2}{\gamma+1}},

where $\beta>0$ will be chosen later. Direct calculation yields that

-\Delta w=\frac{\alpha(x)}{w^{\gamma}}\quad\text{ in }B_{1}(0),

where

\alpha(x):=\frac{2\beta^{\gamma+1}}{\gamma+1}\left(\frac{\gamma-1}{\gamma+1}|\nabla\phi_{1}(x)|^{2}+\lambda_{1}\phi_{1}^{2}(x)\right).

Now we fix $\beta>0$ such that $\sup_{x\in B_{1}(0)}\alpha(x)=c_{0}$ and hence

-\Delta w\leq\frac{c_{0}}{w^{\gamma}}\text{ in }B_{1}(0).

Let $R_{0}>0$ be such that $R_{0}^{\frac{2}{\gamma+1}}w(0)=t_{0}$ . For any $0<R\leq R_{0}$ and $x_{0}=(x_{0}^{\prime},x_{0,N})\in\mathbb{R}^{N}$ with $x_{0,N}\geq R+\varepsilon$ , where $\varepsilon$ is sufficiently small, we set

w_{x_{0},R}(x):=R^{\frac{2}{\gamma+1}}w\left(\frac{x-x_{0}}{R}\right)\quad\text{ in }B_{R}(x_{0}).

Then

w_{x_{0},R}\leq t_{0}\quad\text{ and }\quad-\Delta w_{x_{0},R}\leq\frac{c_{0}}{w_{x_{0},R}^{\gamma}}\quad\text{ in }B_{R}(x_{0}).

On the other hand, since $w_{x_{0},R}=0<u$ on $\partial B_{R}(x_{0})$ , we can use $(w_{x_{0},R}-u)^{+}\chi_{B_{R}(x_{0})}$ as a test function in

-\Delta u=f(u)\quad\text{ and }\quad-\Delta w_{x_{0},R}\leq\frac{c_{0}}{w_{x_{0},R}^{\gamma}}

to obtain

\int_{B_{R}(x_{0})}\left|\nabla(w_{x_{0},R}-u)^{+}\right|^{2}\leq\int_{B_{R}(x_{0})}\left(\frac{c_{0}}{w_{x_{0},R}^{\gamma}}-f(u)\right)(w_{x_{0},R}-u)^{+}.

In $B_{R}(x_{0})\cap\{w_{x_{0},R}>u\}$ we have $f(u)\geq\frac{c_{0}}{u^{\gamma}}$ . Hence

\int_{B_{R}(x_{0})}\left|\nabla(w_{x_{0},R}-u)^{+}\right|^{2}\leq\int_{B_{R}(x_{0})}\left(\frac{c_{0}}{w_{x_{0},R}^{\gamma}}-\frac{c_{0}}{u^{\gamma}}\right)(w_{x_{0},R}-u)^{+} \leq 0.

This implies $u\geq w_{x_{0},R}$ in $B_{R}(x_{0})$ with $x_{0,N}\geq R+\varepsilon$ . Since $\varepsilon>0$ is arbitrary, we deduce

u\geq w_{x_{0},R}\text{ in }B_{R}(x_{0})\text{ for all }0<R\leq R_{0}\text{ and all }x_{0}\in\mathbb{R}^{N}\text{ with }x_{0,N}\geq R.

In particular, if $x_{0,N}=R<R_{0}$ , then

u(x_{0})\geq w_{x_{0},R}(x_{0})=w(0)R^{\frac{2}{\gamma+1}}=w(0)x_{0,N}^{\frac{2}{\gamma+1}}.

If $x_{0,N}\geq R=R_{0}$ , then

u(x_{0})\geq w_{x_{0},R}(x_{0})=w(0)R_{0}^{\frac{2}{\gamma+1}}=t_{0}.

The conclusion follows from the fact that $x_{0}$ is chosen arbitrarily in $\mathbb{R}^{N}$ . ∎

Under a weaker assumption on $f$ , we can still obtain a lower bound of $u$ , which is useful in many situations.

Lemma 9.

Assume that $f:(0,+\infty)$ is a locally Lipschitz continuous function and $f(t)>c_{0}t$ for all $0<t<t_{0}$ , where $c_{0},t_{0}>0$ . Let $u\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}})$ be a solution to (1). Then

u(x)\geq\min\{Cx_{N},t_{0}\}\quad\text{ in }\mathbb{R}^{N}_{+}.

for some constant $C>0$ .

An similar result was obtained by Berestycki et al. [7] for $C^{2}(\overline{\mathbb{R}^{N}_{+}})$ solutions and $f$ being locally Lipschitz continuous on $[0,+\infty)$ . We provide a proof that works in our more general situation.

Proof of Lemma 9.

Let $\lambda_{1}>0$ and $\phi_{1}\in C^{2}(\overline{B_{1}(0)})$ be the first eigenvalue and the corresponding positive eigenfunction of the Laplacian in $B_{1}(0)$ such that $\phi_{1}(0)=t_{0}$ . We take $R=\sqrt{\frac{\lambda_{1}}{c_{0}}}$ and set $\phi_{R}(x)=\phi_{1}\left(\frac{x}{R}\right)$ , then

\begin{cases}-\Delta\phi_{R}=c_{0}\phi_{R}\leq f(\phi_{R})&\text{ in }B_{R}(0),\\ \phi_{R}>0&\text{ in }B_{R}(0),\\ \phi_{R}=0&\text{ on }\partial B_{R}(0).\end{cases}

Since $\phi_{R}$ is radially symmetric and by abuse of notion, we may write $\phi_{R}(x)=\phi_{R}(|x|)$ . For each $x_{0}\in\mathbb{R}^{N}_{+}\setminus\Sigma_{R}$ we set

\phi_{R}^{x_{0}}(x)=\phi_{R}(x-x_{0})\quad\text{ for }x\in B_{R}(x_{0}).

We will show that

u\geq\phi_{R}^{x_{0}}\text{ in }B_{R}(x_{0})\quad\text{ for every }x_{0}\in\mathbb{R}^{N}_{+}\setminus\Sigma_{R}.

(8)

To this end, we let any $x_{0}:=(x_{0}^{\prime},x_{0,N})\in\mathbb{R}^{N}_{+}\setminus\Sigma_{R}$ .

We only consider the case $x_{0,N}>R$ since the case $x_{0,N}=R$ can be derived by continuity. We define the set

\Lambda:=\{s\in(0,1]\mid u>t\phi_{R}^{x_{0}}\text{ in }B_{R}(x_{0})\text{ for all }t\in(0,s)\}.

Since $u$ is positive on compact set $\overline{B_{R}(x_{0})}$ , we have $\Lambda\neq\emptyset$ . We denote $s_{0}=\sup\Lambda$ . To derive (8), we have to show that $s_{0}=1$ . Assume by contradiction that $s_{0}<1$ . We set $\tilde{\phi}:=s_{0}\phi_{R}^{x_{0}}$ . Then

u\geq\tilde{\phi}\quad\text{ in }B_{R}(x_{0})

u(\hat{x})=\tilde{\phi}(\hat{x})\quad\text{ for some }\hat{x}\in B_{R}(x_{0}).

(9)

From the boundary data of $\tilde{\phi}$ , we can choose $\varepsilon>0$ small such that

\hat{x}\in B_{R-\varepsilon}(x_{0})\quad\text{ and }\quad u>\tilde{\phi}\quad\text{ on }\partial B_{R-\varepsilon}(x_{0}).

(10)

Since $u$ and $\tilde{\phi}$ are positive in the set $\overline{B_{R-\varepsilon}(x_{0})}$ and $f$ is locally Lipschitz continuous in $(0,+\infty)$ , the strong comparison principle can be applied in $B_{R-\varepsilon}(x_{0})$ to yield either $\tilde{\phi}<u$ in $B_{R-\varepsilon}(x_{0})$ or $\tilde{\phi}=u$ in $B_{R-\varepsilon}(x_{0})$ . However, the former contradicts (9), while the latter contradicts (10). Hence (8) holds. This implies

u(x)\geq\begin{cases}\phi_{R}(R-x_{N})&\text{ if }x_{N}<R,\\ \phi_{R}(0)&\text{ if }x_{N}\geq R.\end{cases}

The conclusion follows immediately from the fact that $\phi_{R}^{\prime}(R)<0$ and $\phi_{R}(0)=t_{0}$ . ∎

3. Qualitative properties of solutions

3.1. Monotonicity of solutions

As in the previous works, the main tool we use in proving the monotonicity of solutions is the method of moving planes, which was introduced by Alexandrov [1] in the context of differential geometry and by Serrin [29] in the PDE framework, for an overdetermined problem. We recall some familiar notions related to this method. For each $\lambda>0$ , we denote

x_{\lambda}=(x_{1},x_{2},\dots,2\lambda-x_{n}),

which is the reflection of $x$ through the hyperplane $\partial\Sigma_{\lambda}$ . Let $u$ be a solution to (1). We set

u_{\lambda}(x)=u(x_{\lambda}),

then $u_{\lambda}$ satisfies $-\Delta u_{\lambda}=f(u_{\lambda})$ in $\Sigma_{2\lambda}$ .

We are ready to prove the main results in this section.

Proof of Theorem 1.

Applying Proposition 5 with $v\equiv u_{\lambda}$ , we find $\lambda^{*}>0$ such that $u\leq u_{\lambda}$ in $\Sigma_{\lambda}$ for all $\lambda\leq\lambda^{*}$ . Hence the set

\Lambda=\{\lambda\in(0,+\infty)\mid u\leq u_{\mu}\text{ in }\Sigma_{\mu}\text{ for all }\mu\in(0,\lambda]\}

is not empty. Therefore, we can define

\lambda_{0}=\sup\Lambda.

We show that $\lambda_{0}=+\infty$ .

By contradiction, we assume that $\lambda_{0}<+\infty$ . By continuity, we know that

u\leq u_{\lambda_{0}}\quad\text{ in }\Sigma_{\lambda_{0}}.

(11)

By Lemmas 6 and 9, there exist $\tilde{\lambda},\tilde{\delta}>0$ sufficiently small such that

u+\tilde{\delta}<u_{\lambda}\text{ in }\Sigma_{\tilde{\lambda}}\quad\text{ for all }\lambda\geq\lambda_{0}.

(12)

We will reach a contradiction by showing that for some small $\varepsilon_{0}>0$ ,

u\leq u_{\lambda}\text{ in }\Sigma_{\lambda}\quad\text{ for all }\lambda\in[\lambda_{0},\lambda_{0}+\varepsilon_{0}].

If this is not true, then there exist $\lambda_{n}\searrow\lambda_{0}$ and $x_{n}:=(x_{n}^{\prime},x_{n,N})\in\Sigma_{\lambda_{n}}\setminus\Sigma_{\tilde{\lambda}}$ such that

u(x_{n})>u_{\lambda_{n}}(x_{n}).

(13)

Up to a subsequence, we may assume that $x_{n,N}\to y_{0}\in[\tilde{\lambda},\lambda_{0}]$ as $n\to\infty$ . Now we set

u_{n}(x^{\prime},x_{N}):=u(x^{\prime}+x_{n}^{\prime},x_{N}).

By Lemma 9, we know that

\min\{C\tilde{\lambda},t_{0}\}\leq u_{n}\leq\|u\|_{L^{\infty}(\Sigma_{\lambda})}\quad\text{ in }\Sigma_{\lambda}\setminus\Sigma_{\tilde{\lambda}}\text{ for }\lambda>0.

Hence $f(u_{n})$ are also bounded on each strip $\Sigma_{\lambda}\setminus\Sigma_{\tilde{\lambda}}$ . By standard regularity, Ascoli-Arzelà’s theorem and a diagonal process, we deduce that

u_{n}\to v\quad\text{ in }C^{2}_{\rm loc}(\mathbb{R}^{N}_{+}\setminus\Sigma_{\tilde{\lambda}})

up to a subsequence, where $v$ weakly solves $-\Delta u=f(u)$ in $\mathbb{R}^{N}_{+}\setminus\overline{\Sigma_{\tilde{\lambda}}}$ . Moreover, (11), (13) imply $v\leq v_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}\setminus\Sigma_{\tilde{\lambda}}$ and $v(0^{\prime},y_{0})\geq v_{\lambda_{0}}(0^{\prime},y_{0})$ . Hence

v(0^{\prime},y_{0})=v_{\lambda_{0}}(0^{\prime},y_{0}).

Therefore,

-\Delta(v_{\lambda_{0}}-v)+C(v_{\lambda_{0}}-v)=f(v_{\lambda_{0}})-f(v)+C(v_{\lambda_{0}}-v)\geq 0

(14)

in any compact set of $\{\tilde{\lambda}<x_{N}\leq\lambda_{0}\}$ with sufficiently large $C$ . By the strong maximum principle, we deduce $v<v_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}\setminus\overline{\Sigma_{\tilde{\lambda}}}$ . (The case $v\equiv v_{\lambda_{0}}$ in $\Sigma_{\lambda_{0}}$ cannot happen due to $v<v_{\lambda_{0}}$ on $\{x_{N}=\tilde{\lambda}\}$ deduced from (12).) This implies $y_{0}=\lambda_{0}$ . By the mean value theorem, there exists $\xi_{n}\in(x_{n,N},2\lambda_{n}-x_{n,N})$ such that

\frac{\partial u_{n}}{\partial x_{N}}(0^{\prime},\xi_{n})=\frac{u_{n}(0^{\prime},2\lambda_{n}-x_{n,N})-u_{n}(0^{\prime},x_{n,N})}{2\lambda_{n}-2x_{n,N}}=\frac{u_{\lambda_{n}}(x_{n})-u(x_{n})}{2\lambda_{n}-2x_{n,N}}<0.

Letting $n\to\infty$ , we obtain

\frac{\partial v}{\partial x_{N}}(0^{\prime},\lambda_{0})\leq 0.

Hence

\frac{\partial(v_{\lambda_{0}}-v)}{\partial x_{N}}(0^{\prime},\lambda_{0})=-2\frac{\partial v}{\partial x_{N}}(0^{\prime},\lambda_{0})\geq 0.

However, this contradicts the Höpf lemma [20] for (14) in $\Sigma_{\lambda_{0}}\setminus\Sigma_{\tilde{\lambda}}$ .

Therefore, $\lambda_{0}=+\infty$ . Hence $u\leq u_{\lambda}$ in $\Sigma_{\lambda}$ for all $\lambda>0$ . Exploiting the strong maximum principle and the Höpf lemma for $u_{\lambda}-u$ as above we deduce

\frac{\partial u}{\partial x_{N}}>0\quad\text{ in }\mathbb{R}^{N},

which is what we have to prove. ∎

Proof of Theorem 2.

Since $g:[0,+\infty)$ is a locally Lipschitz continuous, there exist $t_{0},c_{1},c_{2}>0$ such that the function $f(t)=\frac{1}{t^{\gamma}}+g(u)$ is decreasing on $(0,t_{0})$ and

\frac{c_{1}}{t^{\gamma}}<f(t)<\frac{c_{2}}{t^{\gamma}}\quad\text{ in }(0,t_{0}).

Hence Lemmas 7 and 8 imply the existence of $\lambda_{0}>0$ such that

cx_{N}^{\frac{2}{\gamma+1}}\leq u(x)\leq Cx_{N}^{\frac{2}{\gamma+1}}\quad\text{ in }\Sigma_{\lambda_{0}}.

(15)

The monotonicity of $u$ follows from Theorem 1. So we only prove (5). Our proof is motivated by an idea from [28].

Let any $A>a>0$ and a positive sequence $(\varepsilon_{n})$ such that $\varepsilon_{n}\to 0$ as $n\to\infty$ . We define

w_{n}(x):=\varepsilon_{n}^{-\frac{2}{\gamma+1}}u(\varepsilon_{n}x)\quad\text{ for }x\in\mathbb{R}^{N}_{+}.

For $n$ sufficiently large, we deduce from (15)

ca^{\frac{2}{\gamma+1}}\leq w_{n}(x)\leq CA^{\frac{2}{\gamma+1}}\quad\text{ in }\Sigma_{A}\setminus\Sigma_{a}

(16)

and

w_{n}(x)\leq Ca^{\frac{2}{\gamma+1}}\quad\text{ on }\{x_{N}=a\}.

(17)

Moreover, $w_{n}$ solves

-\Delta w_{n}=\frac{1}{w_{n}^{\gamma}}+\varepsilon_{n}^{\frac{2\gamma}{\gamma+1}}g(\varepsilon_{n}^{\frac{2}{\gamma+1}}w_{n})\quad\text{ in }\mathbb{R}^{N}_{+}.

(18)

Since the right hand side of (18) is uniformly bounded in $\Sigma_{A}\setminus\Sigma_{a}$ and by the standard regularity [20], $(w_{n})$ is uniformly bounded in $C^{2,\alpha}(\overline{\Sigma_{A}\setminus\Sigma_{a}})$ , for some $0<\alpha<1$ . Since

|\nabla w_{n}(x)|=\varepsilon_{n}^{\frac{\gamma-1}{\gamma+1}}|\nabla u(\varepsilon_{n}x)|\geq\varepsilon_{n}^{\frac{\gamma-1}{\gamma+1}}\frac{\partial u(\varepsilon_{n}x)}{\partial\eta},

for $\varepsilon_{n}$ sufficiently small we get the estimate from above in (5).

Now we prove the estimate from below. Suppose by contradiction that there exist $\beta>0$ , a sequence of normal vectors $\eta_{n}\in\mathbb{S}^{N-1}_{+}$ with $(\eta_{n},e_{N})\geq\beta$ and a sequence of points $x_{n}=(x_{n}^{\prime},x_{n,N})\in\mathbb{R}^{N}_{+}$ such that

x_{n,N}^{\frac{\gamma-1}{\gamma+1}}\frac{\partial u(x_{n})}{\partial\eta_{n}}\to 0\text{ and }x_{n,N}\to 0\quad\text{ as }n\to\infty.

(19)

Passing to a subsequence, we may assume $\eta_{n}\to\eta\in\mathbb{S}^{N-1}_{+}$ with $(\eta,e_{N})\geq\beta$ as $n\to\infty$ . We define $w_{n}$ as above with $\varepsilon_{n}=x_{n,N}$ and $\tilde{w}_{n}(x^{\prime},x_{N})=w_{n}(x^{\prime}+\varepsilon_{n}^{-1}x^{\prime}_{n},x_{N})$ , namely,

\tilde{w}_{n}(x):=x_{n,N}^{-\frac{2}{\gamma+1}}u(x_{n,N}x^{\prime}+x_{n}^{\prime},x_{n,N}x_{N})\quad\text{ for }x\in\mathbb{R}^{N}_{+}.

Then (16), (17) and (18) still hold for $\tilde{w}_{n}$ . Since $(\tilde{w}_{n})$ is uniformly bounded in $C^{2,\alpha}(\overline{\Sigma_{A}\setminus\Sigma_{a}})$ , up to a subsequence, we have

\tilde{w}_{n}\to w_{a,A}\quad\text{ in }C^{2}_{\rm loc}(\overline{\Sigma_{A}\setminus\Sigma_{a}}).

Moreover, passing (18) to the limit, we get

-\Delta w_{a,A}=\frac{1}{w_{a,A}^{\gamma}}\quad\text{ in }\Sigma_{A}\setminus\Sigma_{a}.

Now we take $a=\frac{1}{j}$ and $A=j$ , for large $j\in\mathbb{N}$ and we construct $w_{\frac{1}{j},j}$ as above. For $j\to\infty$ , using a standard diagonal process, we can construct a limiting profile $w_{\infty}\in C^{2}_{\rm loc}(\mathbb{R}^{N}_{+})$ so that

-\Delta w_{\infty}=\frac{1}{w_{\infty}^{\gamma}}\quad\text{ in }\mathbb{R}^{N}_{+}

and $w_{\frac{1}{j},j}=w_{\infty}$ in $\Sigma_{j}\setminus\Sigma_{\frac{1}{j}}$ . Moreover, from (17) we know that

\lim_{x_{N}\to 0^{+}}w_{\infty}(x)=0\quad\text{ uniformly in }x^{\prime}\in\mathbb{R}^{N-1}.

Hence $w_{\infty}$ is a solution to (4). By [25, Theorem 1], $w_{\infty}$ depends only on $x_{N}$ and $w_{\infty}^{\prime}>0$ in $\mathbb{R}_{+}$ .

On the other hand, (19) gives $\frac{\partial\tilde{w}_{n}(e_{N})}{\partial\eta_{n}}=x_{n,N}^{\frac{\gamma-1}{\gamma+1}}\frac{\partial u(x_{n})}{\partial\eta_{n}}\to 0$ as $n\to\infty$ . This is a contradiction since $\frac{\partial\tilde{w}_{n}(e_{N})}{\partial\eta_{n}}\to\frac{\partial w_{\infty}(e_{N})}{\partial\eta}=w_{\infty}^{\prime}(1)\eta_{N}>0$ . ∎

Remark 1.

The proof indicates the following estimate which is stronger than the upper bound in (5)

|\nabla u(x)|<c_{2}x_{N}^{\frac{1-\gamma}{\gamma+1}}\quad\text{ in }\Sigma_{\lambda_{1}}

for some $\lambda_{1},c_{2}>0$ independent of $\beta$ .

3.2. Rigidity of solutions

In this subsection, we prove Theorem 3. We will make use of the following version of the maximum principle in unbounded domains which is due to Berestycki, Caffarelli and Nirenberg.

Lemma 10 (Lemma 2.1 in [7]).

Let $D$ be a domain (open connected set) in $\mathbb{R}^{N}$ , possibly unbounded. Assume that $\overline{D}$ is disjoint from the closure of an infinite open connected cone $\Sigma$ . Suppose there is a function $w\in C^{2}(D)\cap C(\overline{D})$ that is bounded above and satisfies for some continuous function $c(x)$ ,

	$\displaystyle-\Delta w-c(x)w\leq 0$	$\displaystyle\text{ in }D\text{ with }c(x)\leq 0,$
	$\displaystyle w\leq 0$	$\displaystyle\text{ on }\partial D.$

Then $w\leq 0$ in $D$ .

Lemma 10 allows us to derive a weak comparison principle. Notice that in the following result, we do not assume that $v$ is bounded from above.

Proposition 11.

Let $f:(0,+\infty)$ be a locally Lipschitz continuous function which is non-increasing on $(t_{1},+\infty)$ for some $t_{1}>0$ . Let $u,v\in C^{2}(\Omega)\cap C(\overline{\Omega})$ be solutions to

\begin{cases}-\Delta w=f(w)&\text{ in }\Omega,\\ w>0&\text{ in }\Omega,\end{cases}

where $\Omega\subset\mathbb{R}^{N}$ is an open connected set such that $\mathbb{R}^{N}\setminus\overline{\Omega}$ contains an infinite open connected cone. Assume that

u\leq v\text{ on }\partial\Omega,\quad v(x)\geq t_{1}\text{ in }\Omega

and

\sup_{\Omega}(u-v)<+\infty.

Then $u\leq v$ in $\Omega$ .

Proof.

Assume by contradiction that $u>v$ somewhere in $\Omega$ . Let $D$ be a connected component of the set where $u>v$ . Setting

w=u-v,

then

\begin{cases}-\Delta w-c(x)w=0&\text{ in }D,\\ w=0&\text{ on }\partial D,\end{cases}

where

c(x)=\frac{f(u(x))-f(v(x))}{u(x)-v(x)}.

Moreover, since $t_{1}\leq v<u$ in $D$ and $f$ is non-increasing on $(t_{1},+\infty)$ , we deduce $c(x)\leq 0$ in $D$ . Hence Lemma 10 applies to yield $w\leq 0$ in $D$ , a contradiction. Therefore, $u\leq v$ in $\Omega$ . ∎

We employ the technique from [25, Proposition 5] to show that solutions to problem (1) grow at most at a linear rate as $x_{N}\to+\infty$ .

Lemma 12.

Under the assumptions of Theorem 3, there exists a constant $C>0$ such that

u(x)\leq Cx_{N}\quad\text{ in }\mathbb{R}^{N}_{+}\setminus\Sigma_{\overline{\lambda}}.

Proof.

If $u$ is a solution to (1), then

v(x):=\left(\frac{\overline{\lambda}}{2}\right)^{-\frac{2}{\gamma+1}}u\left(\frac{\overline{\lambda}}{2}x\right)

is bounded in $\Sigma_{2}$ and $v$ satisfies

-\Delta v=\left(\frac{\overline{\lambda}}{2}\right)^{\frac{2\gamma}{\gamma+1}}f\left(\left(\frac{\overline{\lambda}}{2}\right)^{\frac{2}{\gamma+1}}v\right).

Moreover, the function $g(t):=\left(\frac{\overline{\lambda}}{2}\right)^{\frac{2\gamma}{\gamma+1}}f\left(\left(\frac{\overline{\lambda}}{2}\right)^{\frac{2}{\gamma+1}}t\right)$ still satisfies (i) and (ii) in Theorem 3 with possible different parameters $c_{0},c_{1},t_{1}$ . Therefore, without loss of generality, we may assume that our solution $u$ is bounded in the strip $\Sigma_{2}$ .

From (i) and Lemma 8, we have

u(x)\geq\tilde{C}x_{N}^{\frac{2}{\gamma+1}}\quad\text{ in }\mathbb{R}^{N}_{+}.

(20)

From (ii) and the compactness, there exists $c_{2}>0$ such that

f(t)<\frac{c_{2}}{t^{\gamma}}\quad\text{ for all }t\geq\tilde{C}.

(21)

Let any $x_{0}=(x_{0}^{\prime},x_{0,N})\in\mathbb{R}^{N}_{+}$ with $x_{0,N}:=4R>2$ . We set

u_{R}(x):=R^{-\frac{2}{\gamma+1}}u(x_{0}+R(x-x_{0})),

then $u_{R}>0$ in $B_{4}(x_{0})$ and

-\Delta u_{R}(x)=R^{\frac{2\gamma}{\gamma+1}}f(u(x_{0}+R(x-x_{0})))=\frac{R^{2}f(u(x_{0}+R(x-x_{0})))}{u(x_{0}+R(x-x_{0})}u_{R}(x).

Remark that $x_{0}+R(x-x_{0})\in\mathbb{R}^{N}_{+}\setminus\Sigma_{2R}\subset\mathbb{R}^{N}_{+}\setminus\Sigma_{1}$ for $x\in B_{2}(x_{0})$ . Hence (20) and (21) give

\frac{R^{2}f(u(x_{0}+R(x-x_{0}))}{u(x_{0}+R(x-x_{0})}\leq\frac{c_{2}R^{2}}{u(x_{0}+R(x-x_{0}))^{\gamma+1}}\leq\frac{c_{2}}{4\tilde{C}^{\gamma+1}}\quad\text{ in }B_{2}(x_{0}).

By Harnack’s inequality, we have

\sup_{B_{1}(x_{0})}u_{R}\leq C_{H}\inf_{B_{1}(x_{0})}u_{R},

which implies

\sup_{B_{R}(x_{0})}u\leq C_{H}\inf_{B_{R}(x_{0})}u,

where $C_{H}>0$ is independent of $x_{0}$ . From this point, we can proceed as in the proof of [25, Proposition 5] to get the thesis. ∎

Given the previous asymptotic bound on $u$ , we can apply the scaling technique as in [25, Proposition 7] to establish a bound on the gradient.

Lemma 13.

Under the assumptions of Theorem 3, there exists a constant $C>0$ such that

|\nabla u(x)|\leq C\quad\text{ in }\mathbb{R}^{N}_{+}\setminus\Sigma_{\overline{\lambda}}.

Proof.

As in Lemma 12, we may assume $\overline{\lambda}=2$ . Let $x_{0}\in\mathbb{R}^{N}_{+}\setminus\Sigma_{2}$ and set $R=x_{0,N}\geq 2$ . We define

u_{R}(x):=\frac{u(Rx)}{R}\quad\text{ in }B_{\frac{1}{2}}\left(\frac{x_{0}}{R}\right).

By Lemma 12 we have $u_{R}\leq C$ . Moreover, from (20) and (21), we deduce

-\Delta u_{R}=Rf(u(Rx))\leq\frac{c_{2}R}{u(Rx)^{\gamma}}\leq\frac{4^{\frac{\gamma}{\gamma+1}}c_{2}}{\tilde{C}^{\gamma}}R^{-\frac{\gamma-1}{\gamma+1}}\leq\frac{2c_{2}}{\tilde{C}^{\gamma}}\quad\text{ in }B_{\frac{1}{2}}\left(\frac{x_{0}}{R}\right).

By the standard gradient estimate, we have $|\nabla u_{R}|\leq C^{\prime}$ in $B_{\frac{1}{4}}\left(\frac{x_{0}}{R}\right)$ . This indicates $|\nabla u|\leq C^{\prime}$ in $B_{\frac{R}{4}}(x_{0})$ . The thesis follows from the arbitrariness of $x_{0}$ . ∎

We are ready to prove Theorem 3 by employing the sliding method.

Proof of Theorem 3.

For each $\lambda>0$ and $\nu\in\mathbb{S}^{N-1}_{+}$ , we define

u_{\lambda}^{\nu}(x):=u(x+\lambda\nu).

We aim to show that

u\leq u_{\lambda}^{\nu}\,\text{ in }\mathbb{R}^{N}_{+}\quad\text{ for all }\lambda>0.

(22)

By Lemma 8, there exists $\lambda^{*}>0$ such that

u(x)>t_{1}\text{ for }x_{N}>\lambda^{*}.

Let $\lambda>\lambda_{\nu}^{*}$ , where $\lambda_{\nu}^{*}:=\frac{\lambda^{*}}{(\nu,e_{N})}$ , then $u_{\lambda}^{\nu}>t_{1}$ in $\mathbb{R}^{N}_{+}$ . Moreover, from Lemmas 12, 13 and the mean value theorem, we deduce

\sup_{\mathbb{R}^{N}_{+}}(u-u_{\lambda}^{\nu})<+\infty.

Since

\begin{cases}-\Delta u=f(u)&\text{ in }\mathbb{R}^{N}_{+},\\ -\Delta u_{\lambda}^{\nu}=f(u_{\lambda}^{\nu})&\text{ in }\mathbb{R}^{N}_{+},\\ u>0&\text{ in }\mathbb{R}^{N}_{+},\\ u_{\lambda}^{\nu}>t_{1}&\text{ in }\mathbb{R}^{N}_{+},\\ u\leq u_{\lambda}^{\nu}&\text{ on }\partial\mathbb{R}^{N}_{+},\end{cases}

we can apply Proposition 11 to derive

u\leq u_{\lambda}^{\nu}\,\text{ in }\mathbb{R}^{N}_{+}\quad\text{ for all }\lambda>\lambda_{\nu}^{*}.

Now that the set

\Lambda=\{\lambda>0\mid u\leq u_{\mu}^{\nu}\text{ in }\mathbb{R}^{N}_{+}\text{ for all }\mu>\lambda\}

is nonempty, we can define $\lambda_{0}=\inf\Lambda$ . We will show that

\lambda_{0}=0.

Assume on contrary that $\lambda_{0}>0$ . By continuity of $u$ , we have $u\leq u_{\lambda_{0}}^{\nu}$ in $\mathbb{R}^{N}_{+}$ . In order to reach a contradiction, we will search for some $\varepsilon_{0}$ small such that

u\leq u_{\lambda}^{\nu}\quad\text{ in }\mathbb{R}^{N}_{+}

(23)

for all $\lambda\in(\lambda_{0}-\varepsilon_{0},\lambda_{0}]$ .

$\circ$ Due to Lemmas 6 and 8, there exist $\tilde{\lambda},\tilde{\delta}>0$ sufficiently small such that

u+\tilde{\delta}\leq u_{\lambda}^{\nu}\quad\text{ in }\Sigma_{\tilde{\lambda}}

(24)

for all $\lambda>\lambda_{0}/2$ .

$\circ$ We claim that

u\leq u_{\lambda}^{\nu}\quad\text{ in }\Sigma_{\lambda^{*}}\setminus\Sigma_{\tilde{\lambda}}

(25)

for all $\lambda\in(\lambda_{0}-\varepsilon_{0},\lambda_{0})$ , where $\varepsilon_{0}>0$ is sufficiently small.

Assume that (25) does not hold. Then there exist two sequences $\lambda_{n}\nearrow\lambda_{0}$ and $x_{n}:=(x^{\prime}_{n},x_{n,N})\in\mathbb{R}^{N-1}\times[\tilde{\lambda},\lambda^{*})$ such that

u(x_{n})>u_{\lambda_{n}}^{\nu}(x_{n}).

(26)

Moreover, we may assume $x_{n,N}\to y_{0}\in[\tilde{\lambda},\lambda^{*}]$ . Now we set

u_{n}(x^{\prime},x_{N})=u(x^{\prime}+x^{\prime}_{n},x_{N}).

Since $C\tilde{\lambda}^{\frac{2}{\gamma+1}}\leq u_{n}\leq\|u\|_{L^{\infty}(\Sigma_{\lambda})}$ in $\Sigma_{\lambda}\setminus\Sigma_{\tilde{\lambda}}$ , we have that $f(u_{n})$ is bounded in $\Sigma_{\lambda}\setminus\Sigma_{\tilde{\lambda}}$ for each $\lambda>\tilde{\lambda}$ . The standard regularity gives $\|u_{n}\|_{C^{2,\alpha}(\overline{\Sigma_{\lambda}\setminus\Sigma_{\tilde{\lambda}}})}<C_{\lambda}$ for some $0<\alpha<1$ . By the Arzelà–Ascoli theorem, via a standard diagonal process, we have

u_{n}\to v\quad\text{ in }C^{2}_{\rm loc}(\overline{\mathbb{R}^{N}_{+}\setminus\Sigma_{\tilde{\lambda}}})

up to a subsequence. Moreover, $v$ weakly solves $-\Delta v=f(v)$ in $\mathbb{R}^{N}_{+}\setminus\overline{\Sigma_{\tilde{\lambda}}}$ . Using the definition of $\lambda_{0}$ and passing (26) to the limit, we have

	$\displaystyle v\leq v_{\lambda_{0}}^{\nu}\quad\text{ in }\mathbb{R}^{N}_{+}\setminus\Sigma_{\tilde{\lambda}},$
	$\displaystyle v(x_{0})=v_{\lambda_{0}}^{\nu}(x_{0}),$

where $x_{0}=(0^{\prime},y_{0})$ . On the other hand, by (24) we have $v+\tilde{\delta}\leq v_{\lambda_{0}}^{\nu}$ on $\{x_{N}=\tilde{\lambda}\}$ . Hence the strong comparison principle implies $v<v_{\lambda_{0}}^{\nu}$ in $\mathbb{R}^{N}_{+}\setminus\Sigma_{\tilde{\lambda}}$ . This contradicts the fact that $v(x_{0})=v_{\lambda_{0}}^{\nu}(x_{0})$ . Therefore, (25) must hold.

$\circ$ Next, we show that

u\leq u_{\lambda}^{\nu}\quad\text{ in }\mathbb{R}^{N}_{+}\setminus\Sigma_{\lambda^{*}}

(27)

for all $\lambda\in(\lambda_{0}-\varepsilon_{0},\lambda_{0})$ .

From (25) and the continuity, we already have $u\leq u_{\lambda}^{\nu}$ on $\{x_{N}=\lambda^{*}\}$ . Moreover, $u_{\lambda}^{\nu}(x)\geq t_{1}$ for each $x\in\mathbb{R}^{N}_{+}\setminus\Sigma_{\lambda^{*}}$ . Hence (27) follows by applying Proposition 11 with $u$ and $v:=u_{\lambda}^{\nu}$ on $\mathbb{R}^{N}_{+}\setminus\Sigma_{\lambda^{*}}$ .

Combining (24), (25) and (27), we obtain (23). This contradicts the definition of $\lambda_{0}$ and hence (22) is proved.

Therefore, $u$ is monotone increasing in direction $\nu$ for all $\nu\in\mathbb{S}^{N-1}_{+}$ . That is,

\frac{\partial u}{\partial\nu}:=(\nabla u,\nu)\geq 0\quad\text{ in }\mathbb{R}^{N}_{+}.

To deduce the one-dimensional symmetry of $u$ , we take $\zeta$ be any direction in $\{x\in\partial B_{1}(0)\mid x_{N}=0\}$ . Let $\nu_{n}\in\mathbb{S}^{N-1}_{+}$ be a sequence converging to $\zeta$ , we have $\frac{\partial u}{\partial\nu_{n}}\geq 0$ . By sending $n\to\infty$ , we deduce $\frac{\partial u}{\partial\zeta}\geq 0$ in $\mathbb{R}^{N}_{+}$ . Similarly, let another sequence $\tau_{n}\in\mathbb{S}^{N-1}_{+}$ converging to $-\zeta$ , we obtain $\frac{\partial u}{\partial\zeta}\leq 0$ in $\mathbb{R}^{N}_{+}$ . Therefore, $u$ is constant in direction $\zeta$ . Since $\zeta$ is arbitrary, we deduce that $u$ does not depend on $x^{\prime}$ . Hence $u$ depends only on $x_{N}$ and monotone increasing in $x_{N}$ .

By the Höpf lemma [20], we have actually $\frac{\partial u}{\partial x_{N}}>0$ in $\mathbb{R}^{N}_{+}$ . By writing $v(x_{N})=u(x)$ , problem (1) reduces to

\begin{cases}-v^{\prime\prime}=f(v)&\text{ in }\mathbb{R}_{+},\\ v^{\prime}(t)>0&\text{ in }\mathbb{R}_{+},\\ v(0)=0.\end{cases}

Hence for every $t>0$ , we have

\frac{1}{2}(v^{\prime})^{2}-F(v)=M,

(28)

which is a constant. Letting $t\to+\infty$ and noticing that $v(t)\to+\infty$ by Lemma 8, we deduce $M\geq 0$ . By integrating (28) and using $v(0)=0$ , this gives

\int_{0}^{v(t)}\frac{ds}{\sqrt{M+F(s)}}=\sqrt{2}t\quad\text{ for all }t\geq 0.

(29)

Conversely, for every $M\geq 0$ we have

\int_{0}^{+\infty}\frac{ds}{\sqrt{M+F(s)}}=+\infty\text{ and }\int_{0}^{t}\frac{ds}{\sqrt{M+F(s)}}<+\infty\text{ for all }t>0.

Therefore, for each $M\geq 0$ , formula (29) uniquely determines a function $v:=v_{M}$ which is a solution to (1). It is also clear from (28) that each solution $v_{M}$ is characterized by the property $\lim_{t\to+\infty}v_{M}^{\prime}(t)=\sqrt{2M}$ . ∎

Finally, we discuss the case that $f$ is nonincreasing in the whole domain $(0,+\infty)$ .

Proof of Theorem 4.

The proof is almost the same as that of [26, Theorem 6], so we only comment on the difference. By employing the Kelvin transform

\hat{u}(x):=\frac{1}{|x|^{N-2}}u\left(\frac{x}{|x|^{2}}\right),

we deduce that $\hat{u}\in C^{2}(\mathbb{R}^{N}_{+})\cap C(\overline{\mathbb{R}^{N}_{+}}\setminus\{0\})$ and $\hat{u}$ is a solution to

\begin{cases}-\Delta\hat{u}=\dfrac{1}{|x|^{N+2}}f(|x|^{N-2}\hat{u})&\text{ in }\mathbb{R}^{N}_{+},\\ \hat{u}>0&\text{ in }\mathbb{R}^{N}_{+},\\ \hat{u}=0&\text{ on }\partial\mathbb{R}^{N}_{+}\setminus\{0\}.\end{cases}

As in [26] we denote $\Sigma_{\lambda}=\{(x_{1},x^{\prime})\in\mathbb{R}^{N}_{+}\mid x_{1}<\lambda\}$ , $x_{\lambda}=(2\lambda-x_{1},x^{\prime})$ and $\hat{u}_{\lambda}(x)=\hat{u}(x_{\lambda})$ . Then for test functions of type $w=(\hat{u}-\hat{u}_{\lambda}-\tau)^{+}\psi\chi_{\Sigma_{\lambda}}$ with compact support in $\Sigma_{\lambda}$ and $\tau>0$ we have

\int_{\Sigma_{\lambda}}(\nabla(\hat{u}-\hat{u}_{\lambda}),\nabla w)=\int_{\Sigma_{\lambda}}\left(\frac{1}{|x|^{N+2}}f(|x|^{N-2}\hat{u})-\frac{1}{|x_{\lambda}|^{N+2}}f(|x_{\lambda}|^{N-2}\hat{u}_{\lambda})\right)w\leq 0

since $\hat{u}\geq\hat{u}_{\lambda}$ on the support of $w$ and $|x|\geq|x_{\lambda}|$ in $\Sigma_{\lambda}$ . From this inequality, we can argue as in the proof of [26, Theorem 6] and repeat the arguments there to get $u(x)=v(x_{N})$ , where $v$ is a solution to

\begin{cases}-v^{\prime\prime}=f(v)&\text{ in }\mathbb{R}_{+},\\ v(t)>0&\text{ in }\mathbb{R}_{+},\\ v(0)=0.\end{cases}

By Theorem 1 we have $v^{\prime}(t)>0$ in $\mathbb{R}_{+}$ . Moreover, since $-v^{\prime\prime}\geq 0$ we have that $v^{\prime}$ is nondecreasing and hence $\lim_{t\to+\infty}v(t)=+\infty$ . From $-v^{\prime\prime}=f(v)$ we deduce

\frac{1}{2}(v^{\prime})^{2}+F_{1}(v)=M_{1},

where $F_{1}(s)=\int_{1}^{s}f(t)dt$ and $M_{1}$ is a constant. Sending $t\to+\infty$ , we obtain $M_{1}\geq\int_{1}^{+\infty}f(t)dt$ . In particular, $\int_{1}^{+\infty}f(t)dt$ is finite. Hence also $F(s)=\int_{s}^{+\infty}f(t)dt$ is finite for all $s>0$ . Similar to the proof of Theorem 3 we deduce

\int_{0}^{v(t)}\frac{ds}{\sqrt{M+F(s)}}=\sqrt{2}t\quad\text{ for all }t\geq 0\text{ and some }M\geq 0

(30)

and (30) indeed provides a solution to our problem. ∎

Conflict of interest The author declares no conflict of interest.

Data Availability Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

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