Ground state solutions for the fractional Schrödinger-Poisson system involving doubly critical exponents

1.   Introduction and main results

In this paper, we study the following fractional Schrödinger-Poisson system with doubly critical exponent

where $s\in(0, 1)$, $2_{s}^{*} = \frac{6}{3-2s}$ is the critical fractional Sobolev exponent, the potential $V(x)$ and the nonlinearity $f$ satisfy the following assumptions:

$(V_1)$ $V\in C(\mathbb{R}^3)$ satisfies $\inf_{x\in\mathbb{R}^3}V(x)\geq V_0 > 0$;

$(V_2)$ There exists $h > 0$ such that $\lim_{|y|\rightarrow \infty}meas(\{x\in B_h(y)\mid V(x)\leq c\}) = 0$ for any $c > 0$, where $B_h(y)$ is the ball centered at point $y$ with radius $h$;

$(f_1)$ $f\in C(\mathbb{R}, \mathbb{R})$ and $f(t)\equiv0$ for $t\leq0$;

$(f_2)$ $\lim_{t\rightarrow0}\frac{f(t)}{t} = 0$;

$(f_3)$ $\lim_{t\rightarrow +\infty}\frac{f(t)}{t^{2^*_s-1}} = 0$;

$(f_4)$ There exist two constants $\rho\in (2, 2^*_s)$ and $0 < \delta < \frac{(\rho-2)V_0}{2}$ such that $\rho F(t)\leq f(t)t+\delta t^2$, where $F(t) = \int^t_0f(s)ds$;

$(f_5)$ $\lim_{t\rightarrow +\infty}\frac{F(t)}{t^{m+1}} = +\infty$, where $m > \max\{1, \frac{6s-3}{3-2s}\}.$

The non-local fractional Laplacian operator $(-\Delta)^{s}$ in $\mathbb{R}^3$ can be characterized as

P.V. represents the Cauchy principal value, and $C(s)$ is a positive constant only depending on $s$, see ^[1]. In the last several years, nonlinear equations involving the fractional Laplacian have been attracted a lot of attention by many scholars. One of the main reasons for this is that the fractional Laplacian operator naturally arises in many different areas, such as such as thin obstacle problem (see ^[2]), combustion (see ^[3]), financial mathematics (see ^[4]), minimal surfaces (see ^[5]), etc. Another main reason is that the fractional Laplacian $(-\Delta)^s$ is a non-local operator in contrast to the classical Laplacian $\Delta$, and previously developed methods may not be applied directly.

Technically, system (1.1) consists of a fractional Schrödinger equation coupled with a fractional Poisson equation. It can reduce to the following classical Schrödinger-Poisson system for $s = 1$,

Due to the real physical meaning, the classical Schrödinger-Poisson system has been studied extensively by many scholars. Here, we do not try to recall the details on this topic, we refer the interested readers to see ^[6,7,8,9] and the references therein.

From a physical standpoint, the fractional Schrödinger equation was discovered by Laskin as a result of extending Feynman path integral, from the Brownian-like to levy like quantum mechanical path. Since then, fractional Schrödinger-Poisson system has been attracted many scholars$^{\prime}$ dinterest, the existence and multiplicity of solutions have been established via applying the variational methods, please see ^{[10,11,12,13,14]}.

But to the best of our knowledge, there are few papers considering the frational Schrödinger-Poisson system with doubly critical exponent, such as ^[15,16]. More precisely, in ^[15], Feng and Yang used concentration-compactness principle to obtained a ground state solution for the following system

where $s\in(\frac{3}{4}, 1)$, $V\in L^{\frac{3}{2s}}(\mathbb{R}^3)$, $K\in L^{\infty}(\mathbb{R}^3)$. In ^[16], the author studied system (1.1), by limiting the order $s\in (\frac{3}{4}, 1)$, he obtained a positive solution with (AR) condition. Motivated by the previously mentioned works, in the present paper, we shall investigate the case of $s\in (0, 1)$ for doubly critical problem without (AR) condition, in other words, we will study the existence of positive ground state solution.

Then, our main result can be described as the follows.

Theorem 1.1. Assume that $V$, $f$ satisfy the assumptions $(V_1)-(V_2)$ and $(f_1)-(f_5)$, respectively. Then, system $(1.1)$ has at least one positive ground state solution.

Remark 1.1. As far as we know, there have not any works in the present literature for the system $(1.1)$ with $s\in(0, 1)$. The condition $0 < \mu F(t) = \mu\int^{t}_{0}f(\tau)d\tau\leq tf(t)$ for $\mu\in (4, 2_s^*)$ is well knowm as (AR) condition, which was first introduced to obtain a bounded Palais-Smale sequence. In fact, the (AR) condition implies that $F(t)$ is a 4-superlinear and subcritical nonlinearity, in order to cover the case where the degree of $F(t)$ is between (2, 4), we add a more weaker condition $(f_4)$.

Corollary 1.1. Assume that $V$ satisfy the assumptions $(V_1)-(V_2)$ and

then system $(1.1)$ has at least one positive ground state solution.

2.   Variational setting and preliminaries

In this section, firstly we will give the variational framework for the system $(1.1)$. Throughout this paper, we denote by $C, \; C_i > 0$ various positive constants which may vary from line to line and are not essential to the problem.

For any $s\in(0, 1)$, $D^{s, 2}(\mathbb{R}^3)$ is completion of the set $C^\infty_0(\mathbb{R}^3)$, which consists of infinitely differentiable functions $u:\mathbb{R}^3\rightarrow \mathbb{R}$ with compact support to the following norm

The fractional Sobolev space is defined by

is equipped with the norm

we denote by $\|\cdot\|_{p}$ the usual $L^{p}$-norm. Due to the appearance of potential function $V(x)$, we will work in the following space:

then, $E$ is equipped with the norm $\|u\|^2 = [u]^2_s+\int_{\mathbb{R}^3}V(x)u^2dx$. Let $S > 0$ be the best Sobolev constant for the embedding of $D^{s, 2}(\mathbb{R}^3)$ in $L^{2_s^*}(\mathbb{R}^3)$, which can be expressed as

As we all known, for $u\in E$, the Lax-Milgram theorem implies that Poisson equation $(-\Delta)^s\phi = |u|^{2^*_s-1}$ has a unique weak solution

where $C_s = \frac{\Gamma(\frac{3-2s}{2})}{2^{2s}\pi^{\frac{3}{2}}\Gamma(s)}$. Similar to the usual Schrödinger-Poisson system, we can insert $\phi_u$ into the first equation of the system $(1.1)$. Then system $(1.1)$ can be transformed in a single Schrödinger equation as follows

The Euler functional of Eq (2.2) is defined by $I:\; E\rightarrow \mathbb{R}$, that is,

Under the assumptions of $f(u)$, we can deduce that functional $I$ is well defined on $E$ and is of class $C^1(E, \mathbb{R})$. For each $u, v \in E$, we have

where $\langle\cdot, \cdot\rangle$ denotes the usual duality. It is easy to verify that if $u$ is a critical point of $I$, then the pair $(u, \phi_u)$ is a weak solution of system (1.1).

Lemma 2.1. If $s\in(0, 1)$, then for any $u\in E$, the following results hold

(1) $\phi_u\geq0;$

(2) $\phi_{tu} = t^{2^*_s-1}\phi_u$ for all $t > 0;$

(3) $\int_{\mathbb{R}^3}\phi_{u}|u|^{2^*_s-1}dx\leq S^{-1}\|u\|^{2(2^*_s-1)}_{2^*_s};$

(4) If $u_n\rightharpoonup u$ in $E$ as $n\rightarrow \infty$, then $\phi_{u_{n}}\rightharpoonup \phi_u$ in $D^{s, 2}(\mathbb{R}^3)$. Moreover,

Proof. The conclusions (1) and (2) are clear from simple calculation.

(3) Since $\phi_u(x)$ is a unique weak solution of $(-\Delta)^s\phi = |u|^{2^*_s-1}$, one has

Thus, it is easy to check that (3) holds.

(4) Since $u_n\rightharpoonup u$ in $E$ as $n\rightarrow \infty$, then $|u_n|^{2^*_s-1}\rightharpoonup |u|^{2^*_s-1}$ in $L^{\frac{2_s^*}{2_s^*-1}}(\mathbb{R}^3)$ as $n\rightarrow \infty$. Thus, for any $\varphi\in C_{0}^{\infty}(\mathbb{R}^3)$, the uniqueness of weak solution of Poisson equation implies that

that is, $\phi_{u_{n}}\rightharpoonup \phi_u$ in $D^{s, 2}(\mathbb{R}^3)$. It is now simple to conclude

Let $q = \frac{2_s^*}{2_s^*-1}$, using Hölder inequality, we have

notice that $u_n(x)\rightarrow u(x)$ a.e in $\mathbb{R}^3$ can be inferred from weak convergence of $u_n\rightharpoonup u$ in $E$, which implies that

Now, we combine (2.4) and (2.5), (2.3) holds.

Lemma 2.2. Suppose that $(V_1)$ and $(f_1)-(f_3)$ hold, then

(a) there exist $\alpha > 0, \beta > 0$ such that $I(u)\geq\alpha$, for all $\|u\| = \beta$,

(b) there exists $e\in H$ such that $\|e\| > \beta$ and $I(e) < 0$.

Proof. (a) By $(f_1)-(f_3)$, for all $\varepsilon > 0$ small enough, there exists $C_\varepsilon > 0$ such that

Thus, by Sobolev inequality and $(V_1)$, there holds

It follows from (2.6), Lemma 2.1(3) and Sobolev inequality that

Because $\varepsilon$ is small enough, we can assume $\varepsilon\in(0, V_0)$ and letting $\beta > 0$ small, $\|u\| = \beta$ implies that

(b) Fixed $u_0\in E$ and $u_0\neq0$, for any $t > 0$, we have

it is easy to check that $\lim_{t\rightarrow +\infty}I(tu_0) = -\infty.$ Therefore, there exists $t_0 > 0$ large enough such that $I(t_0u_0) < 0$ and $\|t_0u_0\| > \beta$. Thus, we complete the proof by taking $e = t_0u_0$.

Lemma 2.3. Assume that $(V_1)-(V_2)$ and $(f_1)-(f_4)$ hold, then the functional $I$ satisfies the $(PS)_c$ condition provided $c\in(0, c^*)$, where $c^* = \left(\frac{\sqrt{5}-1}{2}\right)^\frac{2}{2^*_s-2}\frac{(2^*_s-2)(22^*_s+1-\sqrt{5})}{4(2^*_s-1)2^*_s}S^{\frac{3}{2s}}.$

Proof. Let $\{u_n\}\subset E$ be a $(PS)_c$ sequence of $I$, that is,

Firstly, we claim that the sequence $\{u_n\}$ is bounded in $E.$ Indeed, by $(f_4)$ and (2.8), we get that

which implies that $\{u_n\}$ is bounded in $E$. Then we can extract a subsequence, still denoted $\{u_n\}\subset E$, that converges weakly to some $u_*\in E$. Under the conditions $(V_0)$ and $(V_1)$, we know from ^[10] that, the embedding $E\hookrightarrow L^p(\mathbb{R}^3)$ is continuous and compact for any $p\in(2, 2^*_s)$, thus we can make sure that

Next, we claim that $u_*$ is a solution of (2.2). It follows from $(f_3)$ and continuity of $f$ that for any $\varepsilon > 0$, there exists $C_\varepsilon > 0$ such that $f(u_n)\leq C_{\varepsilon}+\varepsilon u_n^{2^*_s-1}$. Set $0 < \theta < \varepsilon$ small enough, for $\Omega_{\theta} \subset \text{supp}\varphi$ with $meas(\Omega_\theta) < \theta$ and any $\varphi\in C_{0}^{\infty}(\mathbb{R}^3)$, there holds

due to $\{u_n\}$ is bounded in $L^{2_s^*}(\mathbb{R}^3)$, which implies that $\{f(u_n)\varphi\}$ is equiabsolutely continuous. Making use of Vitali theorem, we obtain

It follows from Lemma 2.1(4) that

Using $u_n\rightharpoonup u_*$ weakly in $E$ again, we can prove that $\langle I'(u_*), \varphi\rangle = 0$ for $\forall\varphi\in C^\infty_0({\mathbb{R}^3})$. In the end, we claim that $u_n\rightarrow u_*$ strongly in $E$. In fact, we can define $v_n = {u_n}-u_*$, then $v_n\rightharpoonup0$ in $E$. For any $\varepsilon > 0$, there exists, by assumptions $(f_1), (f_2)$ and $(f_3)$, a $C_\varepsilon > 0$ such that

Consequently, by (2.9) and Lebesgue's dominated convergence theorem, we obtain

From Brézis-Lieb's lemma (see ^[17]), it holds that

Using Lemma 3.2 in ^[18], there holds

Summing up, the preceding equalities show that

and

Therefore, it follows from the hypotheses $I(u_n)\rightarrow c$ and $I'(u_n)\rightarrow0$ that

and

Now, we may assume that

Since the Lax-Milgram theorem implies that $(-\Delta)^s\phi = |v_n|^{2^*_s-1}$ has a sequence solution $\{\phi_{v_{n}}\}$, then we have

As $n\rightarrow \infty$ passing to the limit, it follows that

Using (2.13) and (2.14), we infer that

On the other hand, we obtain

It follows from (2.12)–(2.14), one has

The estimates (2.1) and (2.13) lead to

Thus, we get either

If $\ell\neq0$, then from (2.13), we infer that

which contradicts the fact that $c < c^*$. Hence $\ell = 0$ and we have that $u_n\rightarrow u$ in $E$.

As in ^[19], the extremal function $U_\varepsilon(x) = \varepsilon^{-\frac{3-2s}{2}}u^*(\frac{x}{\varepsilon})$ solves the equation $(-\Delta)^s\phi = |u|^{2^*_s-2}u$ in $\mathbb{R}^3$, where $u^*(x) = \frac{\widetilde{u}(x/S^\frac{1}{2s}_s)}{\|\widetilde{u}\|_{2^*_s}}$ and $\widetilde{u}(x) = k(\mu^2_0+|x|^2)^{-\frac{3-2s}{2}}$ with $k > 0$ and $\mu_0 > 0$ being fixed constants.

Let $\psi\in C^\infty_0(\mathbb{R}^3)$ is such that $0\leq\psi(x)\leq1$ in $\mathbb{R}^3, \psi(x) = 1$ in $B_1$ and $\psi(x) = 0$ in $\mathbb{R}^3\backslash B_2$, we define $v_\varepsilon(x) = \psi(x)U_\varepsilon(x)$. According to Propositions 21 and 22 in ^[19], we know that

Lemma 2.4. Under the assumptions of Theorem 1.1, then $0 < c < c^*$.

Proof. Since the Lax-Milgram theorem implies that $(-\Delta)^s\phi = |v_\varepsilon|^{2^*_s-1}$ has a unique solution $\{\phi_{\varepsilon}\}$, then we have

Let $Q(t) = \frac{t^2}{2}[v_\varepsilon]_s^2-\frac{t^{2(2^*_s-1)}}{2(2^*_s-1)}\int_{\mathbb{R}^3}\phi_{v_\varepsilon}|v_\varepsilon|^{2^*_s-1}dx-\frac{t^{2^*_s}}{2^*_s}\int_{\mathbb{R}^3} |v_\varepsilon|^{2^*_s}dx, t\geq0$, it follows from (2.16) that

as $\varepsilon\rightarrow0$. After computation, we have

As in Lemma 2.2, we see that $I(tv_\varepsilon) > 0$ for $t > 0$ small, and $I(tv_\varepsilon)\rightarrow -\infty$ as $|t|\rightarrow \infty$. According to the continuity of $I$, there exist $t_\varepsilon > 0$ such that

From $(f_5)$, given $M > 0$ large enough, there exists $R_M > 0$ large enough such that

This together with (2.6) implies that for all $M > 0$, there exists a constant $C_M > 0$ such that

In addition, we deduce that

Using the estimates (2.17) and (2.18), for $\varepsilon > 0$ small enough, we get

We distinguish three cases.

Case 1. $s\in(0, \frac{3}{4})$, then $\frac{3}{3-2s} < 2$. In this case, as we have seen in (2.17),

Thus, for $\varepsilon$ small enough, $O(\varepsilon^{2s})-O(\varepsilon^\frac{3(1-m)+2s(m+1)}{2}) < 0$ holds because of $3(1-m)+2s(m+1)-4s = (3-2s)(1-m) < 0$.

Case 2. $s = \frac{3}{4}$, then $\frac{3}{3-2s} = 2$. It is follows from (2.17) that

Due to $\lim_{\varepsilon\rightarrow 0}\frac{\varepsilon^{\frac{3}{2}}|log\varepsilon|}{\varepsilon^\frac{9-3m}{4}} = 0,$ as $\varepsilon\rightarrow0.$ Thus, we have $O(\varepsilon^{\frac{3}{2}}|log\varepsilon|)-O(\varepsilon^\frac{9-3m}{4}) < 0$ for $\varepsilon$ small enough.

Case 3. $s\in(\frac{3}{4}, 1)$, then $2 < \frac{3}{3-2s} < 3$. We remark that $m+1 > \frac{3}{3-2s}$ since $m > \frac{6s-3}{3-2s} > \frac{2s}{3-2s}$, we can obtain

Having observed $m > \frac{6s-3}{3-2s}$, it is easy to check that $O(\varepsilon^{3-2s})-O(\varepsilon^\frac{3(1-m)+2s(m+1)}{2}) < 0$ for $\varepsilon$ small enough.

Since $-C_1\|v_\varepsilon|^{m+1}_{m+1}+C_3\|v_\varepsilon\|^{2}_{2} < 0$ as $\varepsilon\rightarrow 0$, the result follows.

3.   Proof of Theorem 1.1

Proof. From Lemma 2.2, we know that the functional $I$ satisfies the mountain geometry structure. Thus we apply the Mountain-pass lemma, there exists a sequence $\{u_n\}\subset E$ satisfying

where

$\alpha$ and $e$ are defined by Lemma 2.2. By Lemmas 2.3 and 2.4, there exist a convergent subsequence $\{u_n\}\subset E$ (still denoted by itself) and $u_{**}\in E$ such that $u_n\rightarrow u_{**}$ in $E$. Thus, we conclude that

which implies that $(u_{**}, \phi_{u_{**}})$ is a nontrivial solution of system (1.1).

In the following, we claim that there exists a positive ground state solution $(v, \phi_v)$ of system (1.1). Define

we notice that $\mathcal{N}$ is nonempty because of $u_{**}\in \mathcal{N}$. For any $u\in \mathcal{N}$, It follows from (2.6), Lemma 2.1(3) and Sobolev inequality that

which implies that $\|u\|$ must be larger than some positive constant, thereby $0$ is not in $\partial M$. Meanwhile, we decuce that

for any $u\in \mathcal{N}$. By the fact $u\neq0$, we have $I(u) > 0$, thus we can get that $m > 0$. Let ${v_n}\subset \mathcal{N}$ be a minimizing sequence such that

Since $m\leq c < c^*$, after taking a subsequence, it follows from the proof of Lemma 2.3 that there exists $v\in E$ such that $v_n\rightarrow v$ in $E$. Hence, $v$ is a non-trivial critical point of $I$ with $I(v) = m$. Now, we define a new functional as

where $v^+: = \max\{v, 0\},$ $v^-: = \min\{v, 0\}.$ The condition $\langle I_{+}'(v), v^{-}\rangle = 0$ implies that $v\geq0$ in $\mathbb{R}^3$, which is a non-negation weak solution of system (1.1). By using the strong maximum principle and standard argument, $v$ is a positive ground state solution. This completes the proof of Theorem 1.1.

Acknowledgments

The work is supported by Natural Science Foundation of Sichuan Province (2022NSFSC1816) and Fundamental Research Funds of China West Normal University(17E089, 20A025).

Conflict of interest

The authors declare no conflict of interest.