Multiple positive solutions for a slightly subcritical Choquard problem on bounded domains

Abstract

In this paper we study a slightly subcritical Choquard problem on a bounded domain $\Omega$. We prove that the number of positive solutions depends on the topology of the domain. In particular when the exponent of the nonlinearity approaches the critical one, we show the existence of $\hbox {cat}(\Omega )+1$ solutions. Here $\hbox {cat}(\Omega )$ denotes the Lusternik–Schnirelmann category.

Proof of Theorem 5.1

For the sake of completeness, we sketch here the proof of Theorem 5.1. We follow Lemma 2.3, Theorem 3.1 and Lemma 3.3 of [22].

First of all we recall that a (PS)-sequence for $I_*^\Omega$ is bounded. Hence, up to a subsequence, we may assume $v_n \rightharpoonup v$ in $H_0^1(\Omega )$ and v solves Problem (36). Then, if we consider $w_n=v_n-v$ we have $w_n\rightarrow 0$ in $L^2(\Omega )$ and, by Vitali’s convergence

$$\begin{aligned} \begin{aligned} \int _\Omega&(|\nabla (v_n)|^2-|\nabla w_n|^2)\mathop {}\!\mathrm {d}x=\int _\Omega \int _0^1 \frac{\mathop {}\!\mathrm {d}}{\mathop {}\!\mathrm {d}t}|\nabla v_n+(t-1)\nabla v|^2\mathop {}\!\mathrm {d}t \mathop {}\!\mathrm {d}x\\&=2\int _0^1\int _\Omega (\nabla v_n+(t-1)\nabla v)\nabla v\mathop {}\!\mathrm {d}x\mathop {}\!\mathrm {d}t\rightarrow 2\int _0^1\int _\Omega t|\nabla v |^2\mathop {}\!\mathrm {d}x \mathop {}\!\mathrm {d}t\\&=\int _\Omega |\nabla v|^2\mathop {}\!\mathrm {d}x \end{aligned} \end{aligned}$$

(80)

as $n\rightarrow +\infty$. Similarly we obtain

$$\begin{aligned} \int _\Omega (|v_n|^2-|w_n|^2)\mathop {}\!\mathrm {d}x\rightarrow \int _\Omega |v|^2 \mathop {}\!\mathrm {d}x \end{aligned}$$

(81)

as $n \rightarrow +\infty$. Then, by [11, Lemma 2.2] it holds

$$\begin{aligned} \begin{aligned} \int _\Omega \int _\Omega \frac{|v_n(x)|^{2_\mu ^*}|v_n(y)|^{2_\mu ^*}}{|x-y|^\mu }\mathop {}\!\mathrm {d}x\mathop {}\!\mathrm {d}y=&\int _\Omega \int _\Omega \frac{|w_n(x)|^{2_\mu ^*}|w_n(y)|^{2_\mu ^*}}{|x-y|^\mu }\mathop {}\!\mathrm {d}x\mathop {}\!\mathrm {d}y\\&+\int _\Omega \int _\Omega \frac{|v(x)|^{2_\mu ^*}|v(y)|^{2_\mu ^*}}{|x-y|^\mu }\mathop {}\!\mathrm {d}x\mathop {}\!\mathrm {d}y+o(1) \end{aligned} \end{aligned}$$

(82)

where $o(1)\rightarrow 0$ as $n\rightarrow +\infty$.

Therefore we obtain

$$\begin{aligned} I_*^\Omega (v_n)=I_*^\Omega (v)+I_*^\Omega (w_n)+o(1) \end{aligned}$$

and in the same way

$$\begin{aligned} (I_*^\Omega )'(v_n)=(I_*^\Omega )'(v)+(I_*^\Omega )'(w_n)+o(1)=(I_*^\Omega )'(w_n)+o(1) \end{aligned}$$

where $o(1)\rightarrow 0$ as $n\rightarrow +\infty$.

We need the following lemma:

Lemma A.1

[22, Lemma 3.3] Let $\{z_n\}_{n\in {\mathbb {N}}}$ be a (PS)-sequence for $I_*^\Omega$ in $H_0^1(\Omega )$ such that $z_n \rightharpoonup 0$. Then there exist two sequences $\{x_n\}_{n\in {\mathbb {N}}}\subset \Omega$ of points and $\{R_n\}_{n\in {\mathbb {N}}}$ of radii, $R_n \rightarrow +\infty$ as $n \rightarrow +\infty$, a non-trivial solution z to the limit problem (20) and a (PS)-sequence $\{w_n\}_{n\in {\mathbb {N}}}$ for $I_*^\Omega$ in $H_0^1(\Omega )$ such that for a subsequence $\{z_n\}_{n\in {\mathbb {N}}}$ there holds

$$\begin{aligned} z_n=w_n+R_n^{\frac{N-2}{2}}z(R_n(\cdot -x_n))+o(1) \end{aligned}$$

where $o(1) \rightarrow 0$ as $n\rightarrow +\infty$. In particular $w_n \rightharpoonup 0$ and

$$\begin{aligned} I_*^\Omega (w_n)=I_*^\Omega (z_n)-I_*(z)+o(1). \end{aligned}$$

Moreover

$$\begin{aligned} R_n{{\,\mathrm{dist}\,}}(x_n,\partial \Omega )\rightarrow \infty . \end{aligned}$$

Finally if $I_*^\Omega (z_n)\rightarrow m<m_*$, the sequence $\{z_n\}_{n\in {\mathbb {N}}}$ is relatively compact and hence $z_n\rightarrow 0$ in $H_0^{1}(\Omega )$, $I_*^\Omega (z_n)\rightarrow m=0$ as $n\rightarrow +\infty$.

Proof

We begin noticing that, being $m_*^\Omega :=\inf _{{\mathcal {N}}_*^\Omega }I_*^\Omega$, we can suppose that $I_*^\Omega (z_n)\rightarrow m \ge m_*^\Omega =\Big (\dfrac{2_\mu ^*-1}{2\cdot 2_\mu ^*}\Big )S_{H,L}^{\frac{2_\mu ^*}{2_\mu ^*-1}}$.

Then, being $(I_*^\Omega )'(z_n)\rightarrow 0$ we have

$$\begin{aligned} \Big (\dfrac{2_\mu ^*-1}{2\cdot 2_\mu ^*}\Big )\Vert z_n\Vert _\lambda ^2=I_*^\Omega (z_n)-\frac{1}{2\cdot 2_\mu ^*}\langle z_n, (I_*^\Omega )'(z_n)\rangle \rightarrow m \end{aligned}$$

and hence

$$\begin{aligned} \liminf _{n\rightarrow +\infty }\Vert z_n\Vert _\lambda ^2=m \Big (\frac{2\cdot 2_\mu ^*}{2_\mu ^*-1} \Big )\ge S_{H,L}^{\frac{2_\mu ^*}{2_\mu ^*-1}}. \end{aligned}$$

(83)

Let we call

$$\begin{aligned} Q_n(r):=\sup _{x\in \Omega }\int _{B_r(x)}(|\nabla z_n|^2+\lambda |z_n|^2)\mathop {}\!\mathrm {d}x, \end{aligned}$$

choose $x \in \Omega$ and consider

$$\begin{aligned} {\tilde{z}}_n:=R_n^{\frac{2-N}{2}}z_n\left( \frac{x}{R_n}+x_n\right) . \end{aligned}$$

It results

$$\begin{aligned} {\tilde{Q}}_n(1)=\sup _{\begin{array}{c} x\in {{\mathbb {R}}^N}\\ x/R_n+x_n \in \Omega \end{array}}\int _{B_1(x)}(|\nabla {\tilde{z}}_n|^2+\lambda |{\tilde{z}}_n|^2)\mathop {}\!\mathrm {d}x=\int _{B_1(0)}(|\nabla {\tilde{z}}_n|^2+\lambda |{\tilde{z}}_n|^2)\mathop {}\!\mathrm {d}x=\frac{1}{2L}S_{H,L}^{\frac{2_\mu ^*}{2_\mu ^*-1}} \end{aligned}$$

where $L\in {\mathbb {N}}$ is such that $B_2(0)$ is covered by L balls of radius 1. It is easy to see that, from (83), $R_n\ge R_0>0$ uniformly in n. Now denote with ${\tilde{\Omega }}_n:=\left\{ x\in {{\mathbb {R}}^N}:\frac{x}{R_n}+x_n\in \Omega \right\}$ so that we can regard ${\tilde{z}}_n\in H_0^1({\tilde{\Omega }}_n)\subset H^{1}({{\mathbb {R}}^N})$. It holds

$$\begin{aligned} \Vert {\tilde{z}}_n\Vert _\lambda ^2=\Vert z_n\Vert _\lambda ^2\rightarrow m \Big (\frac{2\cdot 2_\mu ^*}{2_\mu ^*-1} \Big )<\infty , \end{aligned}$$

so we can assume ${\tilde{z}}_n \rightharpoonup z$ in $H^{1}({{\mathbb {R}}^N})$.

Proceeding as in [21, Lemma 3.3], replacing $\beta ^*$ with $m_*^\Omega$ and $E_0$ with $I_*^\Omega$, we obtain that ${\tilde{z}}_n \rightarrow z$ in $H^1(\Omega ')$ for any $\Omega ' \subset \subset {{\mathbb {R}}^N}$.

Now we distinguish two cases:

• 1) $R_n {{\,\mathrm{dist}\,}}(x_n, \partial \Omega )\le c <\infty$ uniformly.

  In this case, less than a rotation, we may suppose that the sequence ${\tilde{\Omega }}_n$ exhausts

  $$\begin{aligned} {\tilde{\Omega }}_\infty ={\mathbb {R}}_+^N=\{x=(x_1,\ldots , x_N); x_1>0 \}. \end{aligned}$$

• 2) $R_n {{\,\mathrm{dist}\,}}(x_n, \partial \Omega )\rightarrow \infty$ that implies ${\tilde{\Omega }}_n\rightarrow {\tilde{\Omega }}_\infty ={{\mathbb {R}}^N}.$

In both cases, for any $\varphi \in C_0^\infty ({\tilde{\Omega }}_\infty )$, we get $\varphi \in C_0^\infty ({\tilde{\Omega }}_n)$ for n large, so

$$\begin{aligned} \langle \varphi , (I_*^\Omega )'(z,{\tilde{\Omega }}_\infty )\rangle =\lim _{n\rightarrow \infty }\langle \varphi , (I_*^\Omega )'({\tilde{z}}_n,{\tilde{\Omega }}_n)\rangle =0 \end{aligned}$$

for all these $\varphi$. Hence $z\in H_0^{1}({\tilde{\Omega }}_\infty )$ and weakly solves (20) on ${\tilde{\Omega }}_\infty$. But in the case 1) by [11, Theorem 1.5] and [10, Theorem I.1.], $z\equiv 0$. Therefore it has to be $R_n {{\,\mathrm{dist}\,}}(x_n, \partial \Omega )\rightarrow \infty$.

We conclude the proof letting $\varphi \in C_0^\infty ({{\mathbb {R}}^N})$ such that $0\le \varphi \le 1$, $\varphi \equiv 1$ in $B_1(0)$, $\varphi \equiv 0$ outside $B_2(0)$ and

$$\begin{aligned} w_n(x)=z_n(x)-R_n^{\frac{N-2}{2}}z(R_n(x-x_n))\cdot \varphi ({\bar{R}}_n(x-x_n))\in H_0^{1}(\Omega ), \end{aligned}$$

where $\{R_n \}_{n\in {\mathbb {N}}}$ is chosen such that ${\tilde{R}}_n:=R_n({\bar{R}}_n)^{-1}\rightarrow \infty$ as $n \rightarrow +\infty$, i.e.

$$\begin{aligned} {\tilde{w}}_n=R_n^{\frac{2-N}{2}}w_n\Big (\frac{x}{R_n}+x_n\Big )={\tilde{z}}_n(x)-z(x)\varphi \Big (\frac{x}{{\tilde{R}}_n}\Big ). \end{aligned}$$

Then we may proceed as in [22, Proof of Lemma 3.3] obtaining ${\tilde{w}}_n={\tilde{z}}_n-z+o(1)$ with $o(1)\rightarrow 0$ as $n\rightarrow +\infty$ and using (80)–(82) we get

$$\begin{aligned} I_*^\Omega (w_n)=I_*({\tilde{w}}_n)=I_*({\tilde{z}}_n)-I_*^\Omega (z)+o(1) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert (I_*^\Omega )'(w_n,\Omega )\Vert _{H^{-1}}&=\Vert I_*'({\tilde{w}}_n,{\tilde{\Omega }}_n)\Vert _{H^{-1}}\le \Vert I_*'({\tilde{z}}_n,{\tilde{\Omega }}_n)\Vert _{H^{-1}}\\&\quad +\Vert (I_*^\Omega )'(z,{{\mathbb {R}}^N})\Vert _{H^{-1}}+o(1)=\Vert (I_*^\Omega )'(z_n,\Omega )\Vert _{H^{-1}}+o(1)\rightarrow 0 \end{aligned} \end{aligned}$$

(84)

as $n\rightarrow \infty$ and this conclude the proof. $\square$

Applying Lemma A.1 to sequence $z_n^1:=v_n-v$, $z_n^j:=~v_n-v-\sum _{i=1}^{j-1}{v_n^i}=z_n^{j-1}-v_n^{j-1}$ with $j>1$ and

$$\begin{aligned} v_n^i(x):=(R_n^i)^{\frac{N-2}{2}}v^i(R_n^i(x-x_n^i)), \end{aligned}$$

by induction we get

$$\begin{aligned} \begin{aligned} I_*^\Omega (z_n^i)&=I_*^\Omega (v_n)-I_*^\Omega (v)-\sum _{i=1}^{j-1}{I_*(v^i)}+o(1)\\&\le I_*^\Omega (v_n)-(j-1)m^*+o(1). \end{aligned} \end{aligned}$$

(85)

Note that for large j, the latter will be negative, so by Lemma A.1 the induction will stop after some index $k>0$. For this index we have

$$\begin{aligned} z_n^{k+1}=v_n-v-\sum _{j=1}^k{v_n^j}\rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} I_*^\Omega (v_n)-I_*^\Omega (v)-\sum _{j=1}^k{I_*(v^j)}\rightarrow 0, \end{aligned}$$

so we conclude the proof.