Existence, regularity, asymptotic decay and radiality of solutions to some extension problems

TL;DR

This paper proves the radial symmetry, regularity, and decay properties of solutions to certain extension problems involving fractional Laplacians, and establishes existence of ground states under specific conditions.

Contribution

It introduces new symmetry and regularity results for solutions to extension problems with nonlinearities satisfying minimal growth conditions.

Findings

Solutions are radially symmetric in ^N.

Solutions exhibit exponential decay at infinity.

Existence of ground state solutions under Ambrosetti-Rabinowitz condition.

Abstract

Supposing only that $\displaystyle\lim_{t \to 0} \frac{f(t)}{t} = 0$ and $\displaystyle\lim_{t \to \infty} \frac{f(t)}{t^{p}} = 0$, for some $p \in \left(1,\frac{N+1}{N-1}\right)$, we prove that solutions to the extension problem \begin{equation*}\left\{ \begin{array}{rcll} -\Delta u+ m^2u &=& 0, &\mbox{in} \ \ \mathbb{R}^{N+1}_{+} \\ -\frac{\partial u}{\partial{x}} (0,y)& =& f(u(0,y)), & y \in \mathbb{R}^{N}, \end{array}\right. \end{equation*} and also to the extension Hartree problem \begin{equation*} \left\{\begin{aligned} -\Delta u +m^2u&=0, &&\mbox{in} \ \mathbb{R}^{N+1}_+,\\ -\displaystyle\frac{\partial u}{\partial x}(0,y)&=-V_\infty u(0,y)+\left(\frac{1}{|y|^{N-\alpha}}*F(u(0,y))\right)f(u(0,y)) &&\mbox{in} \ \mathbb{R}^{N}\end{aligned}\right. \end{equation*} are radially symmetric in $\mathbb{R}^N$. In the last problem, $V_\infty>0$ is a constant and $F$ the primitive of $f$. Under the same hypotheses, regularity and exponential decay of solutions to the first problem is also proved and, supposing the traditional Ambrosetti-Rabinowitz condition, also existence of a ground state solution.

Equations

$$\left\{\begin{array}[]{rcll}-\Delta u+m^{2}u&=&0,&\mbox{in}\ \ \mathbb{R}^{N+1}_{+}\\ -\frac{\partial u}{\partial{x}}(0,y)&=&f(u(0,y)),&y\in\mathbb{R}^{N},\end{array}\right.$$

$$\{\begin{array}{rlrl}-\Delta u+m^{2}u& =0,& & \text{mbox}in{\mathbb{R}}_{+}^{N+1},\\ -\frac{\partial u}{\partial x}(0,y)& =-V_{\infty }u(0,y)+\left(\frac{1}{|y{|}^{N-\alpha }}\ast F(u(0,y))\right)f(u(0,y))& & \text{mbox}in{\mathbb{R}}^N\end{array}$$

$$\left\{\begin{array}[]{rcll}-\Delta u+m^{2}u&=&0,&\mbox{in}\ \ \mathbb{R}^{N+1}_{+}\\ -\frac{\partial u}{\partial{x}}(0,y)&=&f(u(0,y)),&y\in\mathbb{R}^{N}.\end{array}\right.$$

$$\{\begin{array}{rlrl}-\Delta u+m^{2}u& =0,& & \text{mbox}in{\mathbb{R}}_{+}^{N+1},\\ -\frac{\partial u}{\partial x}(0,y)& =-V_{\infty }u(0,y)+\left(\frac{1}{|y{|}^{N-\alpha }}\ast F(u(0,y))\right)f(u(0,y))& & \text{mbox}in{\mathbb{R}}^N\end{array}$$

$$0<\theta F(t)<tf(t),\forall t>0,$$

$$\sqrt{-\Delta +m^2}u=f(u)\text{in }{\mathbb{R}}^N,$$

$$\sqrt{-\Delta +m^2}u+V_{\infty }u=\left(\frac{1}{|y{|}^{N-\alpha }}\ast F(u)\right)f(u)\text{in }{\mathbb{R}}^N.$$

$$v\in C^{1,\alpha }({\mathbb{R}}_{+}^{N+1})\cap C^2({\mathbb{R}}_{+}^{N+1})$$

$$0<v(x,y)\le C{e}^{-(m-\alpha )\sqrt{x^2+|y{|}^2}}{e}^{-\alpha x}$$

$$0<v(0,y)\le C{e}^{-\delta |y|},\text{for any}y\in {\mathbb{R}}^N.$$

$$H^1({\mathbb{R}}_{+}^{N+1})=\left\{u\in L^2({\mathbb{R}}_{+}^{N+1}):{\iint }_{{\mathbb{R}}_{+}^{N+1}}|\nabla u{|}^2\mathrm{d}x\mathrm{d}y<\infty \right\}$$

$$\Vert u{\Vert }^2={\iint }_{{\mathbb{R}}_{+}^{N+1}}\left(|\nabla u{|}^2+u^2\right)\mathrm{d}x\mathrm{d}y.$$

$$H^1({\mathbb{R}}_{+}^{N+1})$$

$$H^{1/2}({\mathbb{R}}^N)$$

$$2^{\ast }=\frac{2(N+1)}{N-1}\text{and}{2}^{\#}=\frac{2N}{N-1}.$$

$$[u{]}_{\Omega }^2={\int }_{\Omega }{\int }_{\Omega }\frac{|u(x)-u(y){|}^2}{|x-y{|}^{N+1}}\mathrm{d}x\mathrm{d}y$$

$$W^{1/2,2}(\Omega )$$

$$\Vert u{\Vert }_{W^{1/2,2}(\Omega )}=|u{|}_2+[u{]}_{\Omega }.$$

$$J(u)=\frac{1}{2}{\iint }_{{\mathbb{R}}_{+}^{N+1}}\left(|\nabla u{|}^2+m^2{u}^2\right)-\frac{1}{2}{\int }_{{\mathbb{R}}^N}F(\gamma (u))$$

$$J^{\prime }(u)\cdot \phi =$$

$$|u(0,y){|}^t={\int }_{\infty }^0\frac{\partial }{\partial x}|u(x,y){|}^t\mathrm{d}x={\int }_{\infty }^{0}t|u(x,y){|}^{t-2}u(x,y)\frac{\partial }{\partial x}u(x,y)\mathrm{d}x,$$

$${\int }_{{\mathbb{R}}^N}|\gamma (u){|}^t={\int }_{{\mathbb{R}}^N}|u(0,y){|}^t\mathrm{d}y$$

$$\le t{\left({\int }_{{\mathbb{R}}_{+}^{N+1}}|u{|}^{2(t-1)}\right)}^{1/2}{\left({\int }_{{\mathbb{R}}_{+}^{N+1}}|\nabla u{|}^2\right)}^{1/2}$$

$$\le t\Vert u{\Vert }_{2(t-1)}^{t-1}\Vert \nabla u{\Vert }_2.$$

$$2\le t\le \frac{2N}{N-1}=2^{\#}.$$

$$|\gamma (u){|}_t$$

$$\le \frac{t-1}{t}\Vert u{\Vert }_{2(t-1)}+\Vert \nabla u{\Vert }_2$$

$$\le C_t\Vert u\Vert ,$$

$$|\gamma (u)|\in L^t({\mathbb{R}}^N),\forall t\in [2,2^{\#}].$$

$$|\gamma (u){|}_2^2$$

$$\le \lambda {\iint }_{{\mathbb{R}}_{+}^{N+1}}{u}^2+\frac{1}{\lambda }{\iint }_{{\mathbb{R}}_{+}^{N+1}}|\nabla u{|}^2$$

$$|f(t)|\le \xi t+C_{\xi }{t}^{p-1},\forall t\ge 0$$

$$|F(t)|\le \xi t^2+C_{\xi }{t}^p\le C(t^2+t^p),\forall t\ge 0.$$

Full text

[TABLE]

and also to the extension Hartree problem

[TABLE]

are radially symmetric in $\mathbb{R}^{N}$. In the last problem, $V_{\infty}>0$ is a constant and $F$ the primitive of $f$. Under the same hypotheses, regularity and exponential decay of solutions to the first problem is also proved and, supposing the traditional Ambrosetti-Rabinowitz condition, also existence of a ground state solution.

H. Bueno is the corresponding author; Aldo H. S. Medeiros received research grants from CNPq/Brazil.

1. Introduction

We will denote the point of $\mathbb{R}^{N+1}_{+}$ by pair $(x,y)$, where $x\in(0,\infty)$ and $y=(y_{1},...,y_{N})\in\mathbb{R}^{N}$.

In this paper we recall various aspects of the extension problem

[TABLE]

Our main result consists in proving that solutions of (1.1) are radially symmetric in $\mathbb{R}^{N}$ with respect to a point $y_{0}\in\mathbb{R}^{N}$. We also prove the same result for the solutions to the pseudo-relativistic Hartree problem

[TABLE]

(where $V_{\infty}>0$ is a constant and $F$ is the primitive of $f$).

Our hypotheses on $f$ are very mild. We suppose that the $C^{1}$-nonlinearity $f$ satisfies

1. $(f_{1})$

$\displaystyle\lim_{t\to 0}\frac{f(t)}{t}=0$; 2. $(f_{2})$

$\displaystyle\lim_{t\to\infty}\frac{f(t)}{t^{p}}=0$, for some $p\in\left(1,\frac{N+1}{N-1}\right)$.

Under these hypotheses we also show that solutions of (1.1) are regular and have exponential decay. Although not new, we understand that the review of these results might be helpful. Consonant with this proposal, the proofs we present in this article are very detailed.

We also address a simple situation of existence of solutions to problem (1.1), supposing additionally that $f$ satisfies

1. $(f_{3})$

There exist $\theta>2$ such that

[TABLE]

where $F(t)=\displaystyle\int_{0}^{t}f(s)\mathrm{d}s$.

Of course, solutions of (1.1) can be obtained under much milder assumptions, see e.g., [2, 3]. Because we are looking for a positive solution, we suppose that $f(t)=0$ for $t<0$.

Of course, problem (1.1) results from the application of the Dirichlet to Neumann operator to the problem

[TABLE]

while (1.2) comes from

[TABLE]

Problems (1.3) and (1.4) have been extensively studied in recent years, see [6, 8, 9, 10, 11, 12, 16, 17, 18, 21, 22, 25]. See also the classical paper of Lieb [20].

Changing the operator $\sqrt{-\Delta+m^{2}}$ to its generalization $(-\Delta+m^{2})^{\sigma}$, $0<\sigma<1$, problems like (1.3) and (1.4) can be found in [1, 4, 14]. We summarize our results:

Theorem 1**.**

Suppose that conditions ($f_{1}$)-($f_{3}$), are valid. Then, problem (1.1) has a non-negative ground state solution $w\in H^{1}(\mathbb{R}^{N+1}_{+})$.

Theorem 2**.**

Assuming ($f_{1}$)-($f_{2}$), any solution $v$ of problem (1.1) satisfies

[TABLE]

and therefore is a classical solution of (1.1).

As a simple remark, we observe that if $f\in C^{\infty}$, then the solution is also $C^{\infty}$.

We also prove that the ground state solution has exponential decay:

Theorem 3**.**

Suppose that $v$ is a weak solution to (1.1).

Then $v(x,y)>0$ in $[0,\infty)\times\mathbb{R}^{N}$ and, for any $0\leq\alpha<m$, there exists $C>0$ such that

[TABLE]

for any $(x,y)\in\mathbb{R}^{N+1}_{+}$.

In particular, there exists $\delta\in(0,m)$ such that

[TABLE]

Theorem 4**.**

Any solution $v$ to problem (1.1) is radially symmetric on $\mathbb{R}^{N}$ with respect to some $y_{0}\in\mathbb{R}^{N}$.

Theorem 5**.**

Any solution $v$ to problem (1.2) is radially symmetric on $\mathbb{R}^{N}$ with respect to some $y_{0}\in\mathbb{R}^{N}$.

The natural setting for problem (1.1) is the Sobolev space

[TABLE]

considered with the norm

[TABLE]

Notation. The norm in the space $\mathbb{R}^{N+1}_{+}$ will be denoted by $\|\cdot\|$. For all $q\in[1,\infty]$, we denote by $|\cdot|_{q}$ the norm in the space $L^{q}(\mathbb{R}^{N})$ and by $\|\cdot\|_{q}$ the norm in the space $L^{q}(\mathbb{R}^{N+1}_{+})$. From now on, integrals in $\mathbb{R}^{N+1}_{+}$ will be denoted without $\mathrm{d}x\mathrm{d}y$.

Traces of functions $H^{1}(\mathbb{R}^{N+1}_{+})$ are in $H^{1/2}(\mathbb{R}^{N})$ and every function in $H^{1/2}(\mathbb{R}^{N})$ is the trace of a function in $H^{1}(\mathbb{R}^{N+1}_{+})$, see [24]. Denoting $\gamma\colon H^{1}(\mathbb{R}^{N+1}_{+})\to H^{1/2}(\mathbb{R}^{N})$ the linear function that associates the trace $\gamma(v)\in H^{1/2}(\mathbb{R}^{N})$ of the function $v\in H^{1}(\mathbb{R}^{N+1}_{+})$, then $\ker\,\gamma=H^{1}_{0}(\mathbb{R}^{N+1}_{+})$.

The immersions

[TABLE]

are continuous for any $q\in[2,2^{*}]$ and $[2,2^{\#}]$ respectively, where

[TABLE]

For a bounded open set $\Omega\subset\mathbb{R}^{N}$ we have (see [13]), $H^{1/2}(\mathbb{R}^{N})=W^{1/2,2}(\mathbb{R}^{N})$. We recall the definition of and $W^{1/2,2}(\Omega)$. Let $u\colon\Omega\to\mathbb{R}$ a measurable function and $\Omega$ a bounded open set (that, in the sequel, we suppose to have Lipschitz boundary). Denoting

[TABLE]

and

[TABLE]

then $W^{1/2,2}(\Omega)$ is a reflexive Banach space (see, e.g., [13] and [15]) endowed with the norm

[TABLE]

The proof of the next result can be found in [13, Theorem 4.54].

Theorem 6**.**

The immersion $W^{1/2,2}(\Omega)\hookrightarrow L^{q}(\Omega)$ is compact for any $q\in\left[1,2^{\#}\right)$.

As usual, the immersion $W^{1/2,2}(\Omega)\hookrightarrow L^{2^{\#}}(\Omega)$ is continuous: see [13, Corollary 4.53]. We denote the norm in the space $L^{q}(\Omega)$ by $|\cdot|_{L^{q}(\Omega)}$.

Problem (1.1) is related to the energy functional

[TABLE]

and, since the derivative of the energy functional is given by

[TABLE]

for all $\varphi\in H^{1}(\mathbb{R}^{N+1}_{+})$, we see that critical points of $J$ are weak solutions (1.1).

2. Preliminaries

Let us suppose that $u\in H^{1}(\mathbb{R}^{N+1})\cap C^{\infty}_{0}(\mathbb{R}^{N+1}_{+})$ and $u(x,y)\geq 0$. Let us proceed heuristically: since

[TABLE]

it follows from Hölder’s inequality

[TABLE]

So, in order to apply the immersion $H^{1}(\mathbb{R}^{N+1}_{+})\hookrightarrow L^{q}(\mathbb{R}^{N+1}_{+})$ we must have $2\leq 2(t-1)\leq\frac{2(N+1)}{N-1}$, that is,

[TABLE]

By density of $H^{1}(\mathbb{R}^{N+1})\cap C^{\infty}_{0}(\mathbb{R}^{N+1}_{+})$ in $H^{1}(\mathbb{R}^{N+1}_{+})$, the estimate (2) is valid for all $u\in H^{1}(\mathbb{R}^{N+1}_{+})$.

Taking into account (1.5), Young’s inequality applied to (2) yields

[TABLE]

where $C_{t}$ is a constant. We summarize:

[TABLE]

The inequality (2.3) will also be valuable in the special case $t=2$:

[TABLE]

where $\lambda>0$ is a parameter, the last inequality being a consequence of Young’s inequality.

Remark 2.1**.**

It follows from ($f_{1}$) and ($f_{2}$) that, for any fixed $\xi>0$, there exists a constant $C_{\xi}$ such that

[TABLE]

and analogously

[TABLE]

Condition $(f_{3})$ also yields

[TABLE]

Observe also that $\gamma(u)\in L^{\theta}(\mathbb{R}^{N})$ and $\gamma(u)\in L^{2}(\mathbb{R}^{N})$ imply $F(\gamma(u))\in L^{1}(\mathbb{R}^{N})$.

We denote by $L^{q}_{w}(\mathbb{R}^{N})$ the weak $L^{q}(\mathbb{R}^{N})$ space and by $|\cdot|_{q_{w}}$ its usual norm (see [19]). The next result is a generalized version of the Hardy-Littlewood-Sobolev inequality and will be applied when considering the solutions of the Choquard equation (1.2):

Proposition 7** (Lieb [19]).**

Assume that $p,q,r\in(1,\infty)$ and

[TABLE]

Then, for some constant $N_{p,q,t}>0$ and for any $f\in L^{p}(\mathbb{R}^{N})$, $g\in L^{r}(\mathbb{R}^{N})$ and $h\in L^{q}_{w}(\mathbb{R}^{N})$, we have the inequality

[TABLE]

Lemma 8**.**

Let $A\subset\mathbb{R}^{N}$ be a measurable set. Then

[TABLE]

*Proof. * Of course, for any $y\in A$, we have

[TABLE]

Thus,

[TABLE]

$\hfill\Box$

3. Proof of Theorem 1

Lemma 9**.**

The functional $J$ satisfies the mountain pass theorem geometry. More precisely,

1. $(i)$

There exist $\rho,\delta>0$ such that $I|_{S}\geq\delta>0$ for all $u\in S$, where

[TABLE] 2. $(ii)$

For any $u_{0}\in H^{1}(\mathbb{R}^{N+1}_{+})$ such that $\gamma(u_{0})\neq 0$ we have $J(\tau u_{0})\to-\infty$ as $\tau\to\infty$.

*Proof. *Remark (2.1) and the Sobolev embedding yield

[TABLE]

Taking $0<\varepsilon<\frac{C}{2C_{1}}$, for some $a,A>0$ we obtain

[TABLE]

Since $p\in\left(1,\frac{N+1}{N-1}\right)$, condition ($i$) is proved.

In order to prove $(ii)$, fix $u_{0}\in H^{1}(\mathbb{R}^{N+1}_{+})$ with $\gamma(u_{0})\neq 0$. Thus,

[TABLE]

the last inequality being a consequence of the Ambrosetti-Rabinowitz condition ($f_{3}$). Therefore, since $\theta>2$ we obtain $J(\tau u_{0})\to-\infty$ when $\tau\to\infty$, completing the proof. $\hfill\Box$

The existence of a Palais-Smale sequence $(u_{n})\subset H^{1}(\mathbb{R}^{N+1}_{+})$ such that

[TABLE]

where

[TABLE]

is a consequence of the mountain pass theorem without the PS-condition. It is well-known an alternative characterization of the minimax value $c$, see [23] for details,

[TABLE]

Proof of Theorem 1. Let $(u_{n})\subset H^{1}(\mathbb{R}^{N+1}_{+})$ be a sequence such that $J(u_{n})\to c$ and $J^{\prime}(u_{n})\to 0$, with $c$ given by (3.1). Since we have, for all $n$ sufficiently large,

[TABLE]

and $\theta>2$, we conclude $(u_{n})$ is bounded in $H^{1}(\mathbb{R}^{N+1}_{+})$.

Thus, for a subsequence

[TABLE]

for all $q\in\left[2,2^{\#}\right]$.

As a consequence, for all $v\in H^{1}(\mathbb{R}^{N+1}_{+})$ we have

[TABLE]

Claim. There exists a sequence $(y_{n})$ in $\mathbb{R}^{N}$ and $\beta>0$ such that,

[TABLE]

In fact, assume that

[TABLE]

Thus $u_{n}\to 0$ in $L^{q}(\mathbb{R}^{N})$ for any $q\in(2,2^{*}_{s})$ (see [26, Lemma 1.2.1]). Since $(u_{n})$ is bounded in $L^{2}(\mathbb{R}^{N})$, there exists $M_{0}>0$ such that $|u_{n}|_{2}\leq M$ for all $n\in\mathbb{N}$. So, for any $\eta>0$, by taking $\epsilon=\eta/M_{0}$, we conclude that

[TABLE]

and since $p\in\left(1,2_{\#}\right)$ and $|u_{n}|_{p+1}\to 0$ we obtain

[TABLE]

Similarly,

[TABLE]

Consequently, when $n\to\infty$,

[TABLE]

from what follows

[TABLE]

reaching a contradiction that proves the Claim.

Therefore, there exists $\beta>0$ and a sequence $(y_{n})$ such that, for all $n\in\mathbb{N}$,

[TABLE]

Now, we define $w_{n}(x,y)=u_{n}(x,y+y_{n})$. Then $\|w_{n}\|=\|u_{n}\|$, $J(w_{n})=J(u_{n})$ and $J^{\prime}(w_{n})\to 0$ when $n\to\infty$. Passing to a subsequence if necessary, we can suppose that, for $q\in[2,2^{*}_{s})$, we have

[TABLE]

Note that

[TABLE]

that is, $\gamma(w)\neq 0$.

Furthermore, for all $v\in H^{1}(\mathbb{R}^{N+1}_{+})$, we have

[TABLE]

and we conclude that $J^{\prime}(w)=0$.

We now turn our attention to the positivity of $w$. Seeing that

[TABLE]

and choosing $v=w^{-}$, the left-hand side of the equality is positive, while the right-hand side is not positive. The proof is complete. $\hfill\Box$

4. Proof of Theorem 2

Following arguments in [11], we have:

Lemma 10**.**

For all $\theta\in\left(2,\frac{2N}{N-1}\right)$, we have $|\gamma(u)|^{\theta-2}\leq 1+g_{2}$, where $g_{2}\in L^{N}(\mathbb{R}^{N})$.

*Proof. *We have

[TABLE]

with $g_{2}=|\gamma(u)|^{\theta-2}\chi_{\{|\gamma(u)|>1\}}$. If $(\theta-2)N\leq 2$, then

[TABLE]

When $2<(\theta-2)N$, then $(\theta-2)N\in\left(2,\frac{2N}{N-1}\right)$ and $|\gamma(u)|^{\theta-2}\in L^{N}(\mathbb{R}^{N})$ as an outcome of (2.4). $\hfill\Box$

The proof of the next result adapts arguments in [5] and [11].

Proposition 11**.**

For all $p\in[2,\infty)$ we have $\gamma(v)\in L^{p}(\mathbb{R}^{N})$.

*Proof. * Choosing $\varphi=\varphi_{\beta,T}=vv^{2\beta}_{T}$ in (1.8), where $v_{T}=\min\{v_{+},T\}$ and $\beta>0$, we have $0\leq\varphi_{\beta,T}\in H^{1}(\mathbb{R}^{N+1}_{+})$ and

[TABLE]

Since $\nabla\varphi_{\beta,T}=v^{2\beta}_{T}\nabla v+2\beta vv^{2\beta-1}_{T}\nabla v_{T}$, the left-hand side of (4.1) is given by

[TABLE]

[TABLE]

where $D_{T}=\{(x,y)\in(0,\infty)\times\mathbb{R}^{N}\,:\,v_{T}(x,y)\leq T\}$.

Now we express (4.2) in terms of $\|vv^{\beta}_{T}\|^{2}$. For this, we note that $\nabla(vv^{\beta}_{T})=v^{\beta}_{T}\nabla v+\beta vv^{\beta-1}_{T}\nabla v_{T}$. Therefore,

[TABLE]

thus yielding

[TABLE]

where $C_{\beta}=\max\left\{m^{-2},\left(1+\frac{\beta}{2}\right)\right\}$. Gathering (4.1), (4.2) and (4), we obtain

[TABLE]

Since $|f(t)|\leq C_{1}(|t|+|t|^{\theta-1})$, it follows from (4.4)

[TABLE]

Applying Lemma 10, inequality (4) becomes

[TABLE]

where $g_{2}\in L^{N}(\mathbb{R}^{N})$.

Because $|\gamma(u)|_{2^{\#}}\leq C_{2^{\#}}\|u\|$ for all $u\in H^{1}(\mathbb{R}^{N+1}_{+})$, it follows then from (4.4) and (4) that

[TABLE]

Let us consider the last integral in the right-hand side of (4.7). For all $M>0$, define $A_{1}=\{g_{2}\leq M\}$ and $A_{2}=\{g_{2}>M\}$. Then, whereas $g_{2}\in L^{N}(\mathbb{R}^{N})$,

[TABLE]

and $\epsilon(M)=\left(\int_{A_{2}}g_{2}^{N}\right)^{1/N}\to 0$ when $M\to\infty$.

If $M$ is taken so that $\epsilon(M)C^{2}_{2^{\#}}C_{\beta}C_{1}<1/2$, we obtain

[TABLE]

for a positive constant $K$ depending on $\beta$. Now, since $vv^{\beta}_{T}\to v^{1+\beta}_{+}$ when $T$ goes to infinity, it follows

[TABLE]

Choosing $\beta_{1}+1:=(\theta/2)>1$, it follows from (2.4) that the right-hand side of (4.9) is finite. We conclude that $|\gamma(v_{+})|\in{L^{2\frac{\theta}{2}}}(\mathbb{R}^{N})<\infty$. Now, we choose $\beta_{2}$ so that $\beta_{2}+1=(\theta/2)^{2}$ and conclude that

[TABLE]

After $k$ iterations we obtain that

[TABLE]

from what follows that $\gamma(v_{+})\in L^{p}(\mathbb{R}^{N})$ for all $p\in[2,\infty)$. Since the same arguments are valid for $v_{-}$, we have $\gamma(v)\in L^{p}(\mathbb{R}^{N})$ for all $p\in[2,\infty)$. $\hfill\Box$

By simply adapting the proof given in [11], we present, for the convenience of the reader, the proof of our next result:

Proposition 12**.**

Let $v\in H^{1}(\mathbb{R}^{N+1}_{+})$ be a weak solution of (1.1). Then $\gamma(v)\in L^{p}(\mathbb{R}^{N})$ for all $p\in[2,\infty]$ and $v\in L^{\infty}(\mathbb{R}^{N+1}_{+})$.

*Proof. * We recall equation (4.4):

[TABLE]

where $C_{\beta}=\max\{m^{-2},(1+\beta^{2})\}$. Now $|f(t)|\leq C_{1}(|t|+|t|^{\theta-1})$ yields

[TABLE]

Since $|\gamma(v)|^{\theta-2}=|\gamma(v)|^{\theta-2}\chi_{\{|\gamma(v)\leq 1\}}+|\gamma(v)|^{\theta-2}\chi_{\{|\gamma(v)>1\}}$ and we know that $\gamma(v)\in L^{p}(\mathbb{R}^{N})$ for all $p\geq 2$, we have $|\gamma(v)|^{\theta-2}\chi_{\{|\gamma(v)>1\}}=:g_{3}\in L^{2N}(\mathbb{R}^{N})$. Thus,

[TABLE]

for a positive constant $C_{2}$ and a positive function $g_{3}\in L^{2N}(\mathbb{R}^{N})$ that depends neither on $T$ nor on $\beta$. Therefore,

[TABLE]

from what follows (when $T\to\infty$)

[TABLE]

From the inequality

[TABLE]

we conclude that

[TABLE]

and, by taking $\lambda>0$ so that

[TABLE]

we obtain

[TABLE]

Since

[TABLE]

for a positive constant $M$, it follows from (4) that

[TABLE]

We now apply an iteration argument, taking $2(1+\beta_{n+1})=2^{\#}(1+\beta_{n})$ and starting with $\beta_{0}=0$. This produces

[TABLE]

Because $(1+\beta_{n})=\left(\frac{2^{\#}}{2}\right)^{n}=\left(\frac{N}{N-1}\right)^{n}$, we have

[TABLE]

Thus,

[TABLE]

from what follows $|\gamma(v_{+})|_{p}<\infty$ for all $p\in[2,\infty]$. The same argument applies to $\gamma(v_{-})$, proving that $\gamma(v)\in L^{p}(\mathbb{R}^{N})$ for all $p\in[2,\infty]$.

By taking $\lambda=1$ and $|\gamma(v^{1+\beta}_{+})|_{p}<C_{4}$ for all $p$ in (4), we obtain for any $\beta>0$,

[TABLE]

But $\|v_{+}\|^{1+\beta}_{2^{*}(1+\beta)}=\|v_{+}^{1+\beta}\|_{2^{*}}\leq C_{2^{*}}\|v_{+}^{1+\beta}\|$ and for a positive constant $\tilde{c}$ results from (4.12) that

[TABLE]

Thus,

[TABLE]

and the right-hand side of the last inequality is uniformly bounded for all $\beta>0$. We are done. $\hfill\Box$

We now state a result obtained by Coti Zelati and Nolasco [11, Proposition 3.9]:

Proposition 13**.**

Suppose that $v\in H^{1}(\mathbb{R}^{N+1}_{+})\cap L^{\infty}(\mathbb{R}^{N+1}_{+})$ is a weak solution of

[TABLE]

where $h\in L^{p}(\mathbb{R}^{N})$ for all $p\in[2,\infty]$.

Then $v\in C^{\alpha}([0,\infty)\times\mathbb{R}^{N})\cap W^{1,q}((0,R)\times\mathbb{R}^{N})$ for all $q\in[2,\infty)$ and $R>0$.

In addition, if $h\in C^{\alpha}(\mathbb{R}^{N})$, then $v\in C^{1,\alpha}([0,\infty)\times\mathbb{R}^{N})\cap C^{2}(\mathbb{R}^{N+1}_{+})$ is a classical solution of (4.13).

Proof of Theorem 2. In the proof of Proposition 13 (see [11, Proposition 3.9]), defining

[TABLE]

taking the odd extension of $h$ and $\rho$ to the whole $\mathbb{R}^{N+1}$ (which we still denote simply by $h$ and $\rho$), in [11] is obtained that $\rho$ satisfies the equation

[TABLE]

and $\rho\in C^{1,\alpha}(\mathbb{R}^{N+1})$ for all $\alpha\in(0,1)$ by applying Sobolev’s embedding. Therefore, $v(x,y)=\frac{\partial\rho}{\partial x}(x,y)\in C^{\alpha}(\mathbb{R}^{N})$.

In our case

[TABLE]

We now rewrite equation (4.14) as

[TABLE]

Since $f\in C^{1}$ and $\frac{\partial\rho}{\partial x}(x,y)$ is bounded, the right-hand side of the last equality belongs to $C^{\alpha}(\mathbb{R}^{N+1})$. Thus, classical elliptic boundary regularity yields

[TABLE]

Hence, by applying classical interior elliptic regularity directly to $v$, we deduce that $v\in C^{1,\alpha}(\mathbb{R}^{N+1}_{+})\cap C^{2}(\mathbb{R}^{N+1}_{+})$ is a classical solution of problem (1.1). $\hfill\Box$

5. Proof of Theorem 3

Let us consider a critical point $v\in H^{1}(\mathbb{R}^{N+1}_{+})$ of $J$. Then

[TABLE]

Considering the Fourier transform with respect to the variable $y\in\mathbb{R}^{N}$ we obtain

[TABLE]

and hence

[TABLE]

Since we have $\gamma(u)\in L^{q}(\mathbb{R}^{N})$ for any $q\in[2,\infty)$ as a consequence of Proposition 12, we have that $u(x,y)\to 0$ as $|y|\to\infty$ for any $x$ and conclude that $u(x,y)e^{\lambda x}\to 0$, as $x+|y|\to\infty$, for any $0<\lambda<m$.

Proof of Theorem 3. Applying the strong maximum principle and Hopf lemma, we have $u(x,y)>0$ for any $(x,y)\in\mathbb{R}^{N+1}_{+}$.

For $R>0$ let us define

[TABLE]

and the auxiliary function

[TABLE]

with $0\leq\alpha<m$ and $C_{R}>0$ a constant to be fixed later.

We have

[TABLE]

Moreover,

[TABLE]

Let us define $w(x,y)=f_{R}(x,y)-v(x,y)$, for $(x,y)\in\overline{\Omega}_{R}^{+}$. Thus,

[TABLE]

and choosing $C_{R}=e^{mR}\displaystyle\max_{\partial B_{R}^{+}}u$ we get for $(x,y)\in\partial B_{R}^{+}$,

[TABLE]

In addition, since we already know that $u(x,y)\to 0$ when $x+|y|\to\infty$ and the same is true for $f_{R}(x,y)$, we conclude that $w(x,y)\to 0$ as $x+|y|\to\infty$.

Claim. We have $w(x,y)\geq 0$ in $\Omega_{R}^{+}$.

In fact, suppose that

[TABLE]

By the strong maximum principle, there exists $(0,y_{0})\in\Gamma_{R}$ such that

[TABLE]

Now, we define $W(x,y)=w(x,y)e^{\lambda x}$, with $\lambda\in(0,m)$. Thus,

[TABLE]

and, as before, we have $W(x,y)\to 0$ as $x+|y|\to\infty$ and $W(x,y)\geq 0$ in $\partial B_{R}^{+}$.

Note that,

[TABLE]

thus yielding $-\Delta W+2\lambda\frac{\partial W}{\partial x}+(m^{2}-\lambda^{2})W\geq 0$ on $\Omega_{R}^{+}$.

By the strong maximum principle,

[TABLE]

Therefore,

[TABLE]

It follows from Hopf’s lemma that

[TABLE]

But

[TABLE]

Since $|y_{0}|\to\infty$ as $R\to\infty$ and $v(0,y)\to 0$ as $|y|\to\infty$, follow from $(f_{1})$ and $(f_{2})$ that $f(v(0,y_{0}))\to 0$ as $R\to\infty$. Thus, for any $0<\lambda<\alpha<m$ and $R$ large enough we have

[TABLE]

a contradiction with (5.1).

Therefore, $w(x,y)\geq 0$ in $\overline{\Omega}_{R}^{+}$ and thus

[TABLE]

for all $(x,y)\in\overline{\Omega}_{R}{+}$.

In particular, for $\delta=m-\alpha>0$ we finally obtain

[TABLE]

and we are done. $\hfill\Box$

6. Radial solution

In this section we will prove that two different problems have radially symmetric solutions. The proof of our results adapt ideas of Choi and Seok [7, Proposition 4.2]. We initially consider the problem (1.1).

Proof of Theorem 4. By applying Theorems 2 and 3, any solution $v$ of (1.1) is regular and satisfies

[TABLE]

We now apply the moving planes method together with the maximum principle. For any $\lambda>0$ we define

[TABLE]

Note that $\Sigma_{\lambda}$ is the projection of $R_{\lambda}$ on $\mathbb{R}^{N}$. Denoting $w_{\lambda}=v_{\lambda}-v$, we have

[TABLE]

Claim. For $\lambda>0$ large enough, we have $w_{\lambda}\geq 0$ in $R_{\lambda}$.

In fact, define $w_{\lambda}^{-}=\min\{0,w_{\lambda}\}$ and consider

[TABLE]

As a consequence of ($f_{1}$) we have $f^{\prime}(t)\to 0$ as $t\to 0$ and thus $c^{\lambda}(y)$ converges to [math] uniformly on $R_{\lambda}$ when $\lambda\to\infty$.

Taking $w^{-}_{\lambda}$ as a test-function in (6.1) yields

[TABLE]

Now, observe that the change variable $y\to y^{\lambda}$ yields

[TABLE]

and also

[TABLE]

Substituting the last two equalities in (6.2) we obtain

[TABLE]

Since $c^{\lambda}(y)\to 0$ uniformly in $\Sigma_{\lambda}$ when $\lambda\to\infty$, it follows that, for $\lambda$ large enough, $|c~{}{\lambda}(y)|\leq\varepsilon$ for any $\varepsilon>0$. Thus, it follows from Lemma 8 that

[TABLE]

allowing us to conclude that $w^{-}_{\lambda}=0$ in $R_{\lambda}$ for $\lambda$ large enough, that is, $w_{\lambda}\geq 0$ in $R_{\lambda}$, proving our claim.

Now we define

[TABLE]

We start considering the case $\nu>0$ and claim that, in this case, we have $w_{\nu}\equiv 0$ on $R_{\lambda}$. If not, it follows from the continuity of $w_{\nu}$ and the strong maximum principle that $w_{\nu}>0$ on the set

[TABLE]

We now assert that $w_{\nu}>0$ on the set $\{y\in\mathbb{R}^{N}\,:\,y_{1}>\nu\}$. Otherwise, there exists $\bar{y}\in\mathbb{R}^{N}$ such that $w_{\nu}(\bar{y})=0$, with its first coordinate greater than $\nu$ . By the Hopf lemma we have $-\frac{\partial}{\partial x}w_{\nu}(0,\bar{y})>0$ and we have reached a contradiction, since $v_{\nu}(0,\bar{y})=v(0,\bar{y})$ and

[TABLE]

Thus $w_{\nu}>0$ on the set $\{y\in\mathbb{R}^{N}\,:\,y_{1}>\nu\}$.

In order to reach a contradiction with the definition of $\nu$ if $\nu>0$, consider a sequence $\lambda_{j}<\nu$ such that $\lambda_{j}\to\nu$ when $j\to\infty$. Since $c^{\lambda_{j}}\to 0$ uniformly in $\Sigma_{\lambda}$ for $\lambda$ large enough, we have that $|c^{\lambda_{j}}(y)|\leq\varepsilon<\frac{1}{\gamma_{0}}$, for a positive constant $\varepsilon$ and any $|y|>r_{0}>0$.

Let $D=\|c_{\lambda_{j}}\|_{L^{\infty}(\mathbb{R}^{N})}<\infty$ and $B_{r_{0}}(p_{j})$ be the open ball with center $p_{0}=(\lambda_{j},0,\ldots,0)\in\mathbb{R^{N}}$ and radius $r_{0}>0$. Then, according to (6.3) and Lemma 8,

[TABLE]

[TABLE]

and we conclude that

[TABLE]

Denote by $E_{j}$ the set $\textrm{supp}\,w^{-}_{\lambda_{j}}(0,y)$ in $B_{r_{0}}(p_{j})$. Since $w^{-}_{\nu}>0$ in $\Sigma_{\nu}$ and $\lambda_{j}\to\nu$, the continuity of $w_{\nu}$ yields that $|E_{j}|$ converges to [math] as $j\to\infty$, since $w^{-}_{\lambda_{j}}(0,y)\to w^{-}_{\nu}(0,y)=0$ in $\Sigma_{\nu}$. Thus, the dominated convergence theorem and Hölder’s inequality imply that

[TABLE]

Consequently, $w^{-}(x,y)=0$ in $R_{\lambda_{j}}$, that is, $w_{\lambda_{j}}\geq 0$ in $R_{\lambda_{j}}$, contradicting the definition of $\nu$. Thus, obtain $w_{\nu}=0$ in $R_{\nu}$ and we obtain the symmetry in the $y_{1}$ direction with respect to $y_{1}=\nu$.

If $\nu=0$, we repeat the previous arguments for $\lambda<0$ and $w_{\lambda}=v_{\lambda}-v$ defined on

[TABLE]

Thus, as before, we conclude that $w_{\lambda}\geq 0$ when $\lambda\to-\infty$. Define

[TABLE]

If $\nu<0$, the preceding discussion applies and we obtain the symmetry with respect to $y_{1}=\nu^{\prime}$. If $\nu^{\prime}=0$, we have

[TABLE]

and since $\nu=0$ we have also

[TABLE]

From (6.5) and (6.6) follows that

[TABLE]

Consequently, replacing $y_{1}$ by $-y_{1}$ in (6.7), we obtain the symmetry with respect to $y_{1}=0$:

[TABLE]

To conclude the proof we apply the same procedure with respect to the other directions $y_{i}$, for $i=2,...,N$. $\hfill\Box$

We now consider the problem (1.2). For $y\in\mathbb{R}^{N}$ we denote

[TABLE]

Proof of Theorem 5. We maintain the notation introduced in the proof of Theorem 4 and define $g_{\lambda}(y)=g(y^{\lambda})$. Observe that, as before, any solution of (1.2) is regular and satisfies

[TABLE]

see e.g. [3].

We now apply the moving planes method in integral form. As before, $w_{\lambda}$ stands for $v_{\lambda}-v$. Then we have, for $y\in\mathbb{R}^{N}$,

[TABLE]

We claim that $w_{\lambda}\leq 0$ on $R_{\lambda}$ for $\lambda$ large enough. To prove our claim, we define $w^{+}_{\lambda}(x,y)=\max\{0,w_{\lambda}\}$ and consider

[TABLE]

Observe that, when $\lambda\to\infty$, both $c^{\lambda}_{1}(x,y)$ and $c^{\lambda}_{2}(x,y)$ converge to [math] uniformly on $R_{\lambda}$.

Taking $w^{+}_{\lambda}$ as a test-function in (1.2), the same argument applied to obtain (6.3) yields

[TABLE]

[TABLE]

respectively.

We now consider the integral $I_{1}$ in right-hand side of (6). Since $g\in L^{\infty}(\mathbb{R}^{N})$ and $c^{\lambda}_{1}\to 0$ uniformly when $\lambda\to\infty$, we have $c^{\lambda}_{1}(0,y)g_{\lambda}(y)=\epsilon(\lambda)$, where $\epsilon(\lambda)\to 0$ when $\lambda$ is large enough. Thus

[TABLE]

We now consider $I_{2}$. Since $F$ is increasing, we have

[TABLE]

since the second integral in (6.11a) is negative and can be ignored, while the first integral in (6.11a) is bounded by the integral in (6.11b). So,

[TABLE]

as a consequence of the Hardy-Littlewood inequality, with

[TABLE]

the constant $K$ depending on $q$. Observe that we have $q<2$.

Thus, it follows from Hölder’s inequality that

[TABLE]

where

[TABLE]

and $\delta=2q/(2-q)$.

Returning to (6), we conclude that

[TABLE]

[TABLE]

as a consequence of Lemma 8. Thus, $w^{+}_{\lambda}\equiv 0$ on $R_{\lambda}$ for $\lambda$ large enough, proving our claim.

Now we define

[TABLE]

Let us start considering the case $\nu>0$. We claim that $w_{\nu}\equiv 0$ on $R_{\lambda}$. If not, as a consequence of the strong maximum principle, we have $w_{\nu}<0$ on the set

[TABLE]

since $w_{\nu}\leq 0$ on $R_{\nu}$ is valid by continuity. We assert that $w_{\nu}<0$ on the set $\{y\in\mathbb{R}^{N}\,:\,y_{1}>\nu\}$. Otherwise, there exists $\bar{y}\in\mathbb{R}^{N}$ such that $w_{\nu}(\bar{y})=0$, with $\bar{y}_{1}>\nu$. By the Hopf Lemma, we have $-\frac{\partial}{\partial x}w_{\nu}(0,\bar{y})>0$ and we have reached a contradiction, because

[TABLE]

since $v_{\nu}(0,\bar{y})=v(0,\bar{y})$ and $1/|x-y^{\lambda}|\leq 1/|x-y|$.

In order to reach a contradiction with the definition of $\nu$, consider a sequence $\lambda_{j}<\nu$ such that $\lambda_{j}\to\nu$ when $j\to\infty$. Let $B_{r_{0}}(p_{j})$ be the ball with center $p_{0}=(\lambda_{j},0,\ldots,0)\in\mathbb{R^{N}}$ and radius $r_{0}>0$. Then, according to (6),

[TABLE]

[TABLE]

Since $w^{+}_{\lambda_{j}}$ has exponential decay, by taking $r_{0}$ large enough we have that

[TABLE]

Thus,

[TABLE]

[TABLE]

The same arguments applied to obtain (6) yield $w_{\lambda_{j}}\leq 0$ in $R_{\lambda_{j}}$, contradicting the definition of $\nu$. Thus $w_{\nu}=0$ on $R_{\nu}$ and the symmetry in the $y_{1}$ direction follows.

If $\nu=0$, we also repeat the arguments in the proof of Theorem 4 to conclude that $u$ is symmetric in the $y_{1}$ direction. $\hfill\Box$

Acknowledgements: Aldo H. S. Medeiros received a grant by CNPq - Brasil.