Groundstates and infinitely many high energy solutions to a class of nonlinear Schrödinger–Poisson systems

Abstract

We study a nonlinear Schrödinger–Poisson system which reduces to the nonlinear and nonlocal PDE

$$\begin{aligned} - \Delta u+ u + \lambda ^2 \left( \frac{1}{\omega |x|^{N-2}}\star \rho u^2\right) \rho (x) u = |u|^{q-1} u \quad x \in {{\mathbb {R}}}^N, \end{aligned}$$

where $\omega = (N-2)|{\mathbb {S}}^{N-1} |,$ $\lambda >0,$ $q\in (1,2^{*} -1),$ $\rho :{{\mathbb {R}}}^N \rightarrow {{\mathbb {R}}}$ is nonnegative, locally bounded, and possibly non-radial, $N=3,4,5$ and $2^*=2N/(N-2)$ is the critical Sobolev exponent. In our setting $\rho$ is allowed as particular scenarios, to either (1) vanish on a region and be finite at infinity, or (2) be large at infinity. We find least energy solutions in both cases, studying the vanishing case by means of a priori integral bounds on the Palais–Smale sequences and highlighting the role of certain positive universal constants for these bounds to hold. Within the Ljusternik–Schnirelman theory we show the existence of infinitely many distinct pairs of high energy solutions, having a min–max characterisation given by means of the Krasnoselskii genus. Our results cover a range of cases where major loss of compactness phenomena may occur, due to the possible unboundedness of the Palais–Smale sequences, and to the action of the group of translations.

Appendix A: Proof of the Pohozaev-type condition

Proof of Lemma 2.4

With the regularity remarks of Proposition 1 in place, we now multiply the first equation in (2.3) by $(x, \nabla u)$ and integrate on $B_R(0)$ for some $R>0$. We will compute each integral separately. We first note that

$$\begin{aligned} \begin{aligned} \int _{B_R}-\Delta u (x, \nabla u) \,\mathrm {d}x&= \frac{2-N}{2}\int _{B_R}|\nabla u|^2 \,\mathrm {d}x \\&\quad -\frac{1}{R}\int _{\partial B_R} |(x,\nabla u)|^2 \,\mathrm {d}\sigma +\frac{R}{2}\int _{\partial B_R}|\nabla u|^2 \,\mathrm {d}\sigma . \end{aligned} \end{aligned}$$

(5.20)

Fixing $i =1,\dots , N$, integrating by parts and using the divergence theorem, we then see that,

$$\begin{aligned} \int _{B_R} bu(x_i \partial _i u)\,\mathrm {d}x&=b\left[ -\frac{1}{2} \int _{B_R} u^2\,\mathrm {d}x +\frac{1}{2}\int _{B_R}\partial _i(u^2x_i)\,\mathrm {d}x \right] \\&=b\left[ -\frac{1}{2} \int _{B_R} u^2\,\mathrm {d}x + \frac{1}{2}\int _{\partial B_R} u^2 \frac{x_i^2}{|x|}\,\mathrm {d}\sigma \right] . \end{aligned}$$

So, summing over i, we get

$$\begin{aligned} \int _{B_R} bu(x, \nabla u) \,\mathrm {d}x= b\left[ -\frac{N}{2}\int _{B_R}u^2\,\mathrm {d}x +\frac{R}{2}\int _{\partial B_R}u^2 \,\mathrm {d}\sigma \right] . \end{aligned}$$

(5.21)

Again, fixing $i =1,\dots , N$, integrating by parts and using the divergence theorem, we find that,

$$\begin{aligned} \int _{B_R}c\rho \phi _u u x_i (\partial _i u) \,\mathrm {d}x&= c\bigg [ -\frac{1}{2}\int _{B_R} \rho \phi _u u^2 \,\mathrm {d}x -\frac{1}{2}\int _{B_R} \phi _u u^2x_i (\partial _i\rho ) \,\mathrm {d}x \\&\quad -\frac{1}{2}\int _{B_R} \rho u^2 x_i (\partial _i \phi _u) \,\mathrm {d}x+\frac{1}{2}\int _{B_R}\partial _i (\rho \phi _u u^2 x_i)\,\mathrm {d}x \bigg ] \\&=c\bigg [ -\frac{1}{2}\int _{B_R}\rho \phi _u u^2 \,\mathrm {d}x -\frac{1}{2}\int _{B_R} \phi _u u^2x_i (\partial _i\rho ) \,\mathrm {d}x \\&\quad -\frac{1}{2}\int _{B_R} \rho u^2 x_i (\partial _i \phi _u) \,\mathrm {d}x+\frac{1}{2}\int _{\partial B_R}\rho \phi _u u^2 \frac{x_i^2}{|x|}\,\mathrm {d}\sigma \bigg ]. \end{aligned}$$

Thus, summing over i, we get

$$\begin{aligned} \int _{B_R}c\rho \phi _u u (x, \nabla u) \,\mathrm {d}x&= c\bigg [ -\frac{N}{2}\int _{B_R}\rho \phi _u u^2 \,\mathrm {d}x -\frac{1}{2}\int _{B_R} \phi _u u^2 (x, \nabla \rho ) \,\mathrm {d}x \nonumber \\&\quad -\frac{1}{2}\int _{B_R} \rho u^2 (x, \nabla \phi _u) \,\mathrm {d}x+\frac{R}{2}\int _{\partial B_R} \rho \phi _u u^2 \,\mathrm {d}\sigma \bigg ]. \end{aligned}$$

(5.22)

Finally, once more fixing $i=1,\dots , N$, integrating by parts and using the divergence theorem, we find that,

$$\begin{aligned} \int _{B_R}d |u|^{q-1}u (x_i \partial _i u) \,\mathrm {d}x = d \left[ \frac{-1}{q+1} \int _{B_R} |u|^{q+1} \,\mathrm {d}x+ \frac{1}{q+1} \int _{\partial B_R} |u|^{q+1} \frac{x_i^2}{|x|} \,\mathrm {d}\sigma \right] , \end{aligned}$$

and so, summing over i, we see that

$$\begin{aligned} \begin{aligned} \int _{B_R}d|u|^{q-1}u (x, \nabla u) \,\mathrm {d}x&=d\bigg [\frac{-N}{q+1}\int _{B_R}|u|^{q+1}\,\mathrm {d}x\\&\quad +\frac{R}{q+1}\int _{\partial B_R}|u|^{q+1}\,\mathrm {d}\sigma \bigg ]. \end{aligned} \end{aligned}$$

(5.23)

Putting (5.20), (5.21), (5.22) and (5.23) together, we see that

$$\begin{aligned} \begin{aligned}&\frac{2-N}{2}\int _{B_R}|\nabla u|^2 \,\mathrm {d}x -\frac{1}{R}\int _{\partial B_R} |(x,\nabla u)|^2 \,\mathrm {d}\sigma +\frac{R}{2}\int _{\partial B_R}|\nabla u|^2 \,\mathrm {d}\sigma \\&\quad +b\bigg [-\frac{N}{2}\int _{B_R}u^2\,\mathrm {d}x +\frac{R}{2}\int _{\partial B_R}u^2 \,\mathrm {d}\sigma \bigg ] \\&\quad + c\bigg [ -\frac{N}{2}\int _{B_R}\rho \phi _u u^2 \,\mathrm {d}x-\frac{1}{2}\int _{B_R} \phi _u u^2 (x, \nabla \rho ) \,\mathrm {d}x\\&\quad -\frac{1}{2}\int _{B_R} \rho u^2 (x, \nabla \phi _u) \,\mathrm {d}x +\frac{R}{2}\int _{\partial B_R} \rho \phi _u u^2 \,\mathrm {d}\sigma \bigg ]\\&\quad -d\left[ \frac{-N}{q+1}\int _{B_R}|u|^{q+1}\,\mathrm {d}x+\frac{R}{q+1}\int _{\partial B_R}|u|^{q+1}\,\mathrm {d}\sigma \right] =0. \end{aligned} \end{aligned}$$

(5.24)

We now multiply the second equation in (2.3) by $(x, \nabla \phi _u)$ and integrate on $B_R(0)$ for some $R>0$. By a simple calculation we see that

$$\begin{aligned} \begin{aligned} \int _{B_R} \rho u^2 (x, \nabla \phi _u)\,\mathrm {d}x&= \int _{B_R} -\Delta \phi _u (x, \nabla \phi _u) \,\mathrm {d}x \\&= \frac{2-N}{2}\int _{B_R}|\nabla \phi _u|^2 \,\mathrm {d}x -\frac{1}{R}\int _{\partial B_R} |(x,\nabla \phi _u)|^2 \,\mathrm {d}\sigma \\&\quad +\frac{R}{2}\int _{\partial B_R}|\nabla \phi _u|^2 \,\mathrm {d}\sigma . \end{aligned} \end{aligned}$$

Substituting this into (5.24) and rearranging, we get

$$\begin{aligned} \begin{aligned}&\frac{N-2}{2}\int _{B_R}|\nabla u|^2 \,\mathrm {d}x +\frac{Nb}{2}\int _{B_R}u^2\,\mathrm {d}x +\frac{(N+k)c}{2}\int _{B_R}\rho \phi _u u^2 \,\mathrm {d}x \\&\qquad +\frac{c(2-N)}{4}\int _{B_R}|\nabla \phi _u|^2\,\mathrm {d}x -\frac{Nd}{q+1}\int _{B_R}|u|^{q+1}\,\mathrm {d}x \\&\quad \le \frac{N-2}{2}\int _{B_R}|\nabla u|^2 \,\mathrm {d}x +\frac{Nb}{2}\int _{B_R}u^2\,\mathrm {d}x +\frac{Nc}{2}\int _{B_R}\rho \phi _u u^2 \,\mathrm {d}x \\&\qquad +\frac{c}{2}\int _{B_R} \phi _u u^2 (x, \nabla \rho ) \,\mathrm {d}x +\frac{c(2-N)}{4}\int _{B_R}|\nabla \phi _u|^2\,\mathrm {d}x-\frac{Nd}{q+1}\int _{B_R}|u|^{q+1}\,\mathrm {d}x \\&\quad =-\frac{1}{R}\int _{\partial B_R} |(x,\nabla u)|^2 \,\mathrm {d}\sigma +\frac{R}{2}\int _{\partial B_R}|\nabla u|^2 \,\mathrm {d}\sigma +\frac{bR}{2}\int _{\partial B_R}u^2 \,\mathrm {d}\sigma \\&\qquad +\frac{cR}{2}\int _{\partial B_R} \rho \phi _u u^2 \,\mathrm {d}\sigma +\frac{c}{2R}\int _{\partial B_R} |(x,\nabla \phi _u)|^2 \,\mathrm {d}\sigma \\&\qquad -\frac{cR}{4}\int _{\partial B_R}|\nabla \phi _u|^2 \,\mathrm {d}\sigma -\frac{dR}{q+1}\int _{\partial B_R}|u|^{q+1}\,\mathrm {d}\sigma , \end{aligned} \end{aligned}$$

(5.25)

where we have used the assumption $k\rho (x)\le (x,\nabla \rho )$ for some $k\in {{\mathbb {R}}}$ to obtain the first inequality. We now call the right hand side of (5.25) $I_R$, namely

$$\begin{aligned} \begin{aligned} I_R&{:}{=}-\frac{1}{R}\int _{\partial B_R} |(x,\nabla u)|^2 \,\mathrm {d}\sigma +\frac{R}{2}\int _{\partial B_R}|\nabla u|^2 \,\mathrm {d}\sigma +\frac{bR}{2}\int _{\partial B_R}u^2 \,\mathrm {d}\sigma \\&\quad +\frac{cR}{2}\int _{\partial B_R} \rho \phi _u u^2 \,\mathrm {d}\sigma +\frac{c}{2R}\int _{\partial B_R} |(x,\nabla \phi _u)|^2 \,\mathrm {d}\sigma \\&\quad -\frac{cR}{4}\int _{\partial B_R}|\nabla \phi _u|^2 \,\mathrm {d}\sigma -\frac{dR}{q+1}\int _{\partial B_R}|u|^{q+1}\,\mathrm {d}\sigma . \end{aligned} \end{aligned}$$

We note that $|(x,\nabla u)|\le R|\nabla u |$ and $|(x,\nabla \phi _u)|\le R|\nabla \phi _u |$ on $\partial B_R$, so it holds that

$$\begin{aligned} |I_R|&\le \frac{3R}{2}\int _{\partial B_R}|\nabla u|^2 \,\mathrm {d}\sigma +\frac{bR}{2}\int _{\partial B_R}u^2 \,\mathrm {d}\sigma \\&\quad + \frac{cR}{2}\int _{\partial B_R}\rho \phi _u u^2 \,\mathrm {d}\sigma +\frac{3cR}{4}\int _{\partial B_R}|\nabla \phi _u|^2 \,\mathrm {d}\sigma +\frac{dR}{q+1}\int _{\partial B_R}|u|^{q+1}\,\mathrm {d}\sigma . \end{aligned}$$

Now, since $|\nabla u|^2$, $u^2 \in L^1({{\mathbb {R}}}^N)$ as $u\in E ({{\mathbb {R}}}^N)\subseteq H^1({{\mathbb {R}}}^N)$, $\rho \phi _u u^2$, $|\nabla \phi _u|^2 \in L^1({{\mathbb {R}}}^N)$ because $\int _{{{\mathbb {R}}}^N}\rho \phi _u u^2\,\mathrm {d}x=\int _{{{\mathbb {R}}}^N}|\nabla \phi _u|^2\,\mathrm {d}x$ and $\phi _u \in D^{1,2}({{\mathbb {R}}}^N)$, and $|u|^{q+1} \in L^1({{\mathbb {R}}}^N)$ because $E({{\mathbb {R}}}^N) \hookrightarrow L^s({{\mathbb {R}}}^N)$ for all $s \in [2,2^{*}]$, then it holds that $I_{R_n} \rightarrow 0$ as $n\rightarrow +\infty$ for a suitable sequence $R_n \rightarrow +\infty .$ Moreover, since (5.25) holds for any $R>0$, it follows that

$$\begin{aligned}&\frac{N-2}{2}\int _{{{\mathbb {R}}}^N}|\nabla u|^2 \,\mathrm {d}x +\frac{Nb}{2}\int _{{{\mathbb {R}}}^N}u^2\,\mathrm {d}x +\frac{(N+k)c}{2}\int _{{{\mathbb {R}}}^N}\rho \phi _u u^2 \,\mathrm {d}x \\&\quad +\frac{c(2-N)}{4}\int _{{{\mathbb {R}}}^N}|\nabla \phi _u|^2\,\mathrm {d}x -\frac{Nd}{q+1}\int _{{{\mathbb {R}}}^N}|u|^{q+1}\,\mathrm {d}x \le 0, \end{aligned}$$

and so, we obtain

$$\begin{aligned}&\frac{N-2}{2}\int _{{{\mathbb {R}}}^N}|\nabla u|^2 \,\mathrm {d}x +\frac{Nb}{2}\int _{{{\mathbb {R}}}^N}u^2\,\mathrm {d}x +\frac{(N+2+2k)c}{4}\int _{{{\mathbb {R}}}^N}\rho \phi _u u^2 \,\mathrm {d}x\\&\quad -\frac{Nd}{q+1}\int _{{{\mathbb {R}}}^N}|u|^{q+1}\,\mathrm {d}x \le 0, \end{aligned}$$

using the fact that $\int _{{{\mathbb {R}}}^N}|\nabla \phi _u|^2\,\mathrm {d}x=\int _{{{\mathbb {R}}}^N}\rho \phi _u u^2\,\mathrm {d}x$. This completes the proof. $\square$