Existence of multiple solutions to Schrödinger–Poisson system in a nonlocal set up in $\mathbb {R}^3$

Abstract

The aim of this work is to study the following system:

$$\begin{aligned} {\begin{matrix} (-\Delta )^su+\alpha \phi u&{}=\beta u^{-\gamma }+g(u)+h(x)~\text {in}~\mathbb {R}^3\\ u&{}>0~\text {in}~\mathbb {R}^3\\ (-\Delta )^s\phi &{}=u^2~\text {in}~\mathbb {R}^3. \end{matrix}} \end{aligned}$$

under the Berestycki–Lions type condition. Here $\alpha , \beta >0$, $0<s,\gamma <1$, $g\in C(\mathbb {R},\mathbb {R})$, $h\in L^2(\mathbb {R}^3)$. We will prove the existence of at least two solutions using the Ekeland’s variational principle, Mountain pass theorem and a Pohožaev type identity.

Appendix

Lemma 4.2 guarantees the existence of a positive solution to (3.3) and Lemma 4.4 establishes that a solution to (1.1) is greater than or equal to the solution to (3.3).

Lemma 4.1

(Weak Comparison Principle) Let $u, v\in X$. Suppose, $(-\Delta )^{s}v-\frac{\mu }{v^{\gamma }}\ge (-\Delta )^{s}u-\frac{\mu }{u^{\gamma }}$ weakly in $\mathbb {R}^3$. Then $v\ge u$ in $\mathbb {R}^3.$

The proof follows verbatim of the Lemma 3.1 in [22].

Lemma 4.2

Let $\beta >0$. Then, the following problem

$$\begin{aligned}&(-\Delta )^{s}u+\alpha \phi _u u=\beta u^{-\gamma },~\text {in}~\mathbb {R}^3\nonumber \\&u>0,~\text {in}~\mathbb {R}^3 \end{aligned}$$

(4.1)

has a unique weak solution in $H^s(\mathbb {R}^3)$. This solution is denoted by $\underline{u}_{\beta }$, satisfies $\underline{u}_{\beta }\ge \epsilon _{\beta } v_0$ a.e. in $\Omega$, where $\epsilon _{\beta }>0$ is a constant.

Proof

We follow the proof in [8]. Firstly, we note that an energy functional on $H^s(\mathbb {R}^3)$ formally corresponding to (4.1) can be defined as follows.

$$\begin{aligned} \bar{E}(u)&=\frac{1}{2}[u]_2^2+\frac{\alpha }{4}\int \limits _{\mathbb {R}^3}\phi _u u^2\hbox {d}x-\frac{\beta }{1-\gamma }\int \limits _{\Omega }(u^+)^{1-\gamma }\hbox {d}x \end{aligned}$$

(4.2)

for $u\in X$. By the Poincaré inequality, this functional is coercive and continuous on $H^s(\mathbb {R}^3)$. It follows that $\bar{E}$ possesses a global minimizer $u_0\in H^s(\mathbb {R}^3)$. Clearly, $u_0\ne 0$ since $\bar{E}(0)=0>\bar{E}(\epsilon v_0)$ for sufficiently small $\epsilon$ and some $v_0>0$ in $\mathbb {R}^3$.

Secondly, we have the decomposition $u=u^+-u^-$. Thus if $u_0$ is a global minimizer for $\bar{E}$, then so is $|u_0|$, by $\bar{E}(|u_0|)\le \bar{E}(u_0)$. Clearly enough, the equality holds iff $u_0^-=0$ a.e. in $\mathbb {R}^3$. In other words we need to have $u_0\ge 0$, i.e. $u_0\in X^+$ where

$$\begin{aligned} X^+=\{u\in H^s(\mathbb {R}^3):u\ge 0~\text {a.e. in}~\mathbb {R}^3\} \end{aligned}$$

is the positive cone in $H^s(\mathbb {R}^3)$.

Third, we will show that $u_0\ge \epsilon v_0>0$ holds a.e. in $\mathbb {R}^3$ for small enough $\epsilon$. Observe that,

$$\begin{aligned} \begin{aligned} \frac{\mathrm{{d}}}{\hbox {d}t}\bar{E}(tv_0)|_{t=\epsilon }=&\epsilon [v_0]_2^2+\alpha \epsilon ^3\int \limits _{\mathbb {R}^3}\phi _u u^2\hbox {d}x-\beta \epsilon ^{-\gamma }\int \limits _{\Omega }v_0^{1-\gamma }\hbox {d}x<0 \end{aligned} \end{aligned}$$

(4.3)

whenever $0<\epsilon \le \epsilon _{\beta }$ for some sufficiently small $\epsilon _{\beta }$. We now show that $u_0\ge \epsilon _{\beta }v_0$. On the contrary, suppose $w=(\epsilon _{\beta }v_0-u_0)^+$ does not vanish identically in $\mathbb {R}^3$. Denote

$$\begin{aligned} \mathbb {R}^3_+=\{x\in \mathbb {R}^3:w(x)>0\}. \end{aligned}$$

We will analyse the function $\zeta (t)=E(u_0+tw)$ of $t\ge 0$. This function is convex owing to its definition over $X^+$ being convex. Further $\zeta '(t)=\langle E'(u_0+tw),w \rangle$ is nonnegative and nondecreasing for $t>0$. Consequently for $0<t<1$ we have

$$\begin{aligned} {\begin{matrix} 0\le \zeta '(1)-\zeta '(t)&{}=\langle E'(u_0+w)-E'(u_0+tw),w\rangle \\ &{}=\int \limits _{\mathbb {R}_+}E'(u_0+w)\hbox {d}x-\zeta '(t)\\ &{}<0 \end{matrix}} \end{aligned}$$

(4.4)

by inequality (4.3) and $\zeta '(t)\ge 0$ with $\zeta '(t)$ being nondecreasing for every $t>0$, which is a contradiction. Therefore $w=0$ in $\mathbb {R}$ and hence $u_0\ge \epsilon _{\beta }v_0$ a.e. in $\mathbb {R}$.

Finally, the functional E being strictly convex on $X^+$, we conclude that $u_0$ is the only critical point of E in $X^+$ with the property $\underset{V}{\text {ess}\inf }u_0>0$ for any compact subset $V{\subset }\mathbb {R}$. Therefore we choose $\underline{u}_{\beta }=u_0$ in the cutoff functional.

Remark 4.3

We will now conduct an apriori analysis on a solution (if it exists). Suppose u is a solution to (1.1), then we observe the following

• 1. $E(u)=E(|u|)$. This implies that $u^-=0$ a.e. in $\mathbb {R}^3$.

• 2. A solution to (1.1) can be taken to be positive, i.e. $u>0$ a.e. in $\mathbb {R}^3$ owing to being driven by the singular term.

Thus without loss of generality, we assume that the solution is positive.

We have the following result.

Lemma 4.4

(Apriori analysis) Fix a $\beta \in (0,\beta _0)$. Then a solution of (1.1), say $u>0$, is such that $u\ge \underline{u}_{\beta }$ a.e. in $\mathbb {R}^3$.

Proof

Fix $\beta \in (0,\beta _0)$ and let $u\in H^s(\mathbb {R}^3)$ be a positive solution to the system in (1.1) and $\underline{u}_{\beta }>0$ be a solution to (4.1). We will show that $u\ge \underline{u}_{\beta }$ a.e. in $\mathbb {R}^3$. Thus, we let $\underline{\mathbb {R}}^3=\{x\in \mathbb {R}^3:u(x)<\underline{u}_{\beta }(x)\}$ and from the equation satisfied by u, $\underline{u}_{\beta }$, we have

$$\begin{aligned} {\begin{matrix} 0\le &{}\int \limits _{\underline{\mathbb {R}}^3}\frac{|(\underline{u}_{\beta }-u)(x)-(\underline{u}_{\beta }-u)(y)|^2}{|x-y|^{3+2s}}\hbox {d}x\hbox {d}y\\ \le &{}\int \limits _{\underline{\mathbb {R}}^3}\frac{|(\underline{u}_{\beta }-u)(x)-(\underline{u}_{\beta }-u)(y)|^2}{|x-y|^{3+2s}}\hbox {d}x\hbox {d}y+\alpha \int \limits _{\underline{\mathbb {R}}^3}(\phi _{\underline{u}_{\beta }}\underline{u}_{\beta }-\phi _uu)(\underline{u}_{\beta }-u)\hbox {d}x\\ \le &{}\beta \int \limits _{\underline{\mathbb {R}}^3}(\underline{u}_{\beta }^{-\gamma }-u^{-\gamma })(\underline{u}_{\beta }-u)\hbox {d}x\\ \le &{} 0. \end{matrix}} \end{aligned}$$

(4.5)

Hence, from (4.5), we obtain $u\ge \underline{u}_{\beta }$ a.e. in $\mathbb {R}^3$. $\square$

We shall conclude the paper with the following observation.

Remark 4.5

$$\begin{aligned} {\begin{matrix} \lim \limits _{n\rightarrow +\infty }\int \limits _{\Omega }|u_n|^{-\gamma -1}uv\hbox {d}x=\int \limits _{\Omega }|u|^{-\gamma -1}uv\hbox {d}x~\text {for any}~v\in X_0. \end{matrix}} \end{aligned}$$

(4.6)

Indeed, let us denote the set $A_n=\{x\in \Omega :u_n(x)=0\}$. Since $|u_n|^{-\gamma } v\in L^1(\Omega )$, we have that the Lebesgue measure of $A_n$ is zero, i.e. $|A_n|=0$. Thus by the sub-additivity of the Lebesgue measure we have, $|\bigcup A_n|=0$. Let $x\in \Omega \setminus D$ such that $u(x)=0$. Here $|D|<\delta$—obtained from the Egorov’s theorem—where u is a uniform limit of (a subsequence of) $\{u_n\}$ in $\Omega$. In order to achieve (4.6), we shall show that the Lebesgue measure $|\{x\in \Omega \setminus D:~\text {as}~n\rightarrow \infty , u_n(x)\rightarrow 0\}|=0$. Further, define

$$\begin{aligned} A_{m,n}=\{x\in \Omega \setminus D:|u_n(x)|<\frac{1}{m}\}. \end{aligned}$$

Note that due to the uniform convergence, for a fixed n we have $|A_{m,n}|\rightarrow 0$ as $m\rightarrow \infty$. Now consider

$$\begin{aligned} \underset{m,n\in \mathbb {N}}{\bigcup }A_{m,n}=\underset{n\ge 1}{\bigcup }\underset{m\ge n}{\bigcap }A_{m,n}. \end{aligned}$$

Observe that, for a fixed n,

$$\begin{aligned} \left| \underset{m\ge n}{\bigcap }A_{m,n}\right| =\underset{m\rightarrow \infty }{\lim }A_{m,n}=0. \end{aligned}$$

The above argument is true for each fixed n and thus

$$\begin{aligned} \left| \underset{m,n\in \mathbb {N}}{\bigcup }A_{m,n}\right| =0. \end{aligned}$$

Therefore, $|\{x\in \Omega \setminus D:~\text {as}~n\rightarrow \infty , u_n(x)\rightarrow 0\}|=0$. Hence the claim follows. $\square$