Spiked solutions for Schrödinger systems with Sobolev critical exponent: the cases of competitive and weakly cooperative interactions

Abstract

In this paper we deal with the nonlinear Schrödinger system

$$\begin{aligned} -\Delta u_i =\mu _i u_i^3 + \beta u_i \sum _{j\ne i} u_j^2 + \lambda _i u_i, \qquad u_1,\ldots , u_m\in H^1_0(\Omega ) \end{aligned}$$

in dimension 4, a problem with critical Sobolev exponent. In the competitive case ($\beta <0$ fixed or $\beta \rightarrow -\infty$) or in the weakly cooperative case ($\beta \ge 0$ small), we construct, under suitable assumptions on the Robin function associated to the domain $\Omega$, families of positive solutions which blowup and concentrate at different points as $\lambda _1,\ldots , \lambda _m\rightarrow 0$. This problem can be seen as a generalization for systems of a Brezis–Nirenberg type problem.

Appendix A: Auxiliary estimates

We start by recalling here the notations

$$\begin{aligned}&U_{\delta ,\xi }(x)=c_4\frac{\delta }{\delta ^2+|x-\xi |^2},\qquad \psi ^0_{\delta ,\xi }=\delta \frac{\partial U_{\delta ,\xi }}{\partial \delta } \quad \text {and} \qquad \\&\quad \psi ^j_{\delta ,\xi }=\delta \frac{\partial U_{\delta ,\xi }}{\partial \xi _j},\ j=1,\ldots , 4. \end{aligned}$$

Lemma A.1

Given any $K\Subset \Omega \subset \mathbb {R}^4$, we have

$$\begin{aligned} PU_{\delta ,\xi }=U_{\delta ,\xi }-\delta A H(\cdot ,\xi )+R_{\delta ,\xi }, \end{aligned}$$

(A.1)

with

$$\begin{aligned} \Vert R_{\delta ,\xi }\Vert _\infty =\text {o}(\delta ),\quad \left\| \frac{\partial R_{\delta ,\xi }}{\partial \delta } \right\| _\infty =\text {o}(1),\quad \left\| \frac{\partial R_{\delta ,\xi }}{\partial \xi _j}\right\| _\infty =\text {o}(\delta ) \end{aligned}$$

as $\delta \rightarrow 0$, uniformly for $\xi \in K$, with

$$\begin{aligned} A:=\int _{\mathbb {R}^4} U_{1,0}^3=c_4^3\int _{\mathbb {R}^4}\frac{1}{(1+|y|^2)^3}=\frac{c_4}{\alpha _4}. \end{aligned}$$

In particular, we have

$$\begin{aligned} PU_{\delta ,\xi }=U_{\delta ,\xi }-\delta A H(\cdot ,\xi )+\text {o}(\delta ),\quad P\psi ^0_{\delta ,\xi }=\psi _{\delta ,\xi }^0-\delta A H(\cdot ,\xi )+\text {o}(\delta )\nonumber \\ \end{aligned}$$

(A.2)

and

$$\begin{aligned} P\psi _{\delta ,\xi }^j=\psi _{\delta ,\xi }^j-\delta ^2 A\frac{\partial H}{\partial \xi }(\cdot ,\xi )+\text {o}(\delta ^2), \quad j=1,\ldots , 4, \end{aligned}$$

(A.3)

as $\delta \rightarrow 0$, uniformly for $\xi \in K$.

Proof

See Proposition 1 in [27]. $\square$

Lemma A.2

Take $K\Subset \Omega \subset \mathbb {R}^4$. Then for every $0<p< 2$,

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^p=\text {O}(\delta ^p) \quad \text { as } \;\delta \rightarrow 0, \end{aligned}$$

while

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^2= c_4^2\delta ^2 \omega _3 |\ln \delta |+\text {O}(\delta ^2) \quad \text { as } \;\delta \rightarrow 0 \end{aligned}$$

and, for $2<p<4$,

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^p= \text {O}(\delta ^{4-p}) \quad \text { as } \;\delta \rightarrow 0, \end{aligned}$$

uniformly for $\xi \in K$, where $\omega _3$ denotes the measure of the unit sphere $S^3\subset \mathbb {R}^4$.

Proof

Take $R={{\mathrm{dist}}}(\xi ,\partial \Omega )/2$. Then

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^p&=c_4^p \int _{B_{R}(\xi )} \frac{\delta ^p}{(\delta ^2+|x-\xi |^2)^p}+\text {O}(\delta ^p)\\&=c_4^p \delta ^{4-p} \omega _3 \int _0^{R/\delta } \frac{t^3}{(1+t^2)^p}\, \mathrm{d}t+\text {O}(\delta ^p). \end{aligned}$$

We have $t^3/(1+t^2)^p\le t^{3-2p}$ for every $t\ge 0$. Thus, if $p\ne 3/2$,

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^p=\text {O}(\delta ^{4-p}) + c_4^p \delta ^{4-p} \omega _3 \left[ t^{4-2p}\right] _{t=1}^{t=R/\delta }+\text {O}(\delta ^p)=\text {O}(\delta ^{4-p})+\text {O}(\delta ^p). \end{aligned}$$

Thus $\int _\Omega U_{\delta ,\xi }^p=\text {O}(\delta ^p)$ if $0<p<2$, and $\int _\Omega U_{\delta ,\xi }^p=\text {O}(\delta ^{4-p})$ if $2<p<4$.

Likewise, we have that

$$\begin{aligned} \int _\Omega U_{\delta ,\xi }^2&= c_4^2 \delta ^2 \omega _3 \int _0^{R/\delta } \frac{t^3}{(1+t^2)^2}\, \mathrm{d}t + \text {O}(\delta ^2)\\&=\frac{c_4^2}{2}\delta ^2 \omega _3 \left[ \frac{1}{1+t^2}+\ln (1+t^2)\right] ^{t=R/\delta }_{t=0}+\text {O}(\delta ^2)\\&=-c_4^2 \delta ^2 \omega _3 \ln \delta +\text {O}(\delta ^2), \end{aligned}$$

as $\delta \rightarrow 0$. $\square$

Next we will present some asymptotic estimates related to the competition term of the system under consideration. For that, we will need the following simple pointwise estimates.

Lemma A.3

Let $F:\mathbb {R}\rightarrow \mathbb {R}$ be defined by $F(s)=(s^+)^4/4$. Then there exists $c>0$ such that, for every $a,b\in \mathbb {R}$,

$$\begin{aligned} |F(a+b)-F(a)-F'(a)b|\le & {} c (a^2b^2 +b^4), \\ |F'(a+b)-F'(a)-F''(a)b|\le & {} c(|a| b^2 +|b|^3), \end{aligned}$$

and

$$\begin{aligned} |F''(a+b)-F''(a)|\le c (|a| |b| + b^2). \end{aligned}$$

Moreover, for every $p>1$ there exists $C>0$ such that

$$\begin{aligned} \left| |a+b|^p-|a|^p\right| \le C \left( |a|^{p-1}|b| + |b|^p \right) , \end{aligned}$$

for every $a,b\in \mathbb {R}$.

Proof

The proof is very elementary, and follows simply from a Taylor’s expansion with Lagrange-type remainder. $\square$

Lemma A.4

Given $p,q>0$ and $\eta >0$ small, we have that

$$\begin{aligned} \int _\Omega U_{\delta _1,\xi _1}^p U_{\delta _2,\xi _2}^q \le \text {O}(\delta _2^q)\int _\Omega U_{\delta _1,\xi _1}^p + \text {O}(\delta _1^p) \int _\Omega U_{\delta _2,\xi _2}^q+\text {O}(\delta _1^p\delta _2^q). \end{aligned}$$

as $(\delta _1,\delta _2)\rightarrow (0,0)$, uniformly for all $\xi _1,\xi _2\in \Omega$ such that $|\xi _1-\xi _2|\ge 2\eta$, ${{\mathrm{dist}}}(\xi _1,\partial \Omega ),{{\mathrm{dist}}}(\xi _2,\partial \Omega )\ge 2\eta$.

Proof

This is a simple consequence of the fact that

$$\begin{aligned}&\int _\Omega U_{\delta _1,\xi _1}^p U_{\delta _2,\xi _2}^q = \int _{B_\eta (\xi _1)} U_{\delta _1,\xi _1}^p U_{\delta _2,\xi _2}^q+ \int _{B_\eta (\xi _2)} U_{\delta _1,\xi _1}^p U_{\delta _2,\xi _2}^q\\&\quad +\int _{\Omega {\setminus } (B_\eta (\xi _1)\cup B_{\eta }(\xi _2))}U_{\delta _1,\xi _1}^p U_{\delta _2,\xi _2}^q \end{aligned}$$

together with the fact that $U_{\delta _i,\xi _i}\le C \delta _i$ on $\Omega {\setminus } B_\eta (\xi _i)$, for some $C>0$ independent of $\delta _i$. $\square$

Lemma A.5

Given $j=0,\ldots , 4$ and $\eta >0$ small, we have

$$\begin{aligned}&\left\| (PU_{\delta _2,\xi _2})^2(P\psi _{\delta _1,\xi _1}^j) \right\| _{4/3}=\text {O}(\delta _{1}\delta _2) \quad \text { and }\quad \\&\quad \left\| (PU_{\delta _1,\xi _1})(PU_{\delta _2,\xi _2})(P\psi _{\delta _1,\xi _1}^j)\right\| _{4/3}=\text {O}(\delta _{1}\delta _2) \end{aligned}$$

as $(\delta _1,\delta _2)\rightarrow (0,0)$, uniformly for all $\xi _1,\xi _2\in \Omega$ such that $|\xi _1-\xi _2|\ge 2\eta$, ${{\mathrm{dist}}}(\xi _1,\partial \Omega ),{{\mathrm{dist}}}(\xi _2,\partial \Omega )\ge 2\eta$.

Proof

To simplify notations, define $U_i:=U_{\delta _i,\xi _i}$ and $\psi _i^j:=\psi ^j_{\delta _i,\xi _i}$, for $i=1,2$, $j=0,\ldots , 4$. We have ($j=0$)

$$\begin{aligned} \left\| (PU_2)^2(P\psi _1^0) \right\| _{4/3}^{4/3}=\int _\Omega |U_{2}|^{8/3}|\psi ^0_1|^{4/3}+R_{\varvec{\delta },\varvec{\xi }}, \end{aligned}$$

with

$$\begin{aligned} R_{\varvec{\delta },\varvec{\xi }}= & {} \int _\Omega \left[ (PU_2)^{8/3}-U_2^{8/3}\right] \left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3} \right] \\&+\int _\Omega U_2^{8/3}\left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3} \right] +\;\int _\Omega |\psi _1^0|^{4/3}\left[ (PU_2)^{8/3}-U_2^{8/3}\right] , \end{aligned}$$

so that, by taking in consideration Lemmas A.1, A.3 and A.4, and the fact that $|\psi _1^0|\le C U_1$ for some $C>0$, we have

$$\begin{aligned} |R_{\varvec{\delta },\varvec{\xi }}|\le & {} \int _\Omega \left| U_2^{5/3}\text {O}(\delta _2) +\text {O}(\delta _2^{8/3}) \right| \left| |\psi _1^0|^{1/3}\text {O}(\delta _1)+\text {O}(\delta _1^{4/3})\right| \\&+\int _\Omega U_2^{8/3} \left| |\psi _1^0|^{1/3}\text {O}(\delta _1) + \text {O}(\delta _1^{4/3})\right| \\&+\;\int _\Omega |\psi _1^0|^{4/3}\left| U_2^{5/3}\text {O}(\delta _2)+\text {O}(\delta _2^{8/3})\right| =\text {O}(\delta _1^{4/3}\delta _2^{4/3}) \end{aligned}$$

Since also

$$\begin{aligned} \int _\Omega (U_2)^{8/3}|\psi _1^0|^{4/3}&\le C \int _\Omega (U_2)^{8/3}(U_1)^{4/3}\le \text {O}(\delta _1^{4/3})\int _\Omega (U_2)^{8/3}+\text {O}(\delta _2^{8/3})\\&\quad \times \int _\Omega (U_1)^{4/3}+\text {O}(\delta _1^{4/3}\delta _2^{4/3})=\text {O}(\delta _1^{4/3}\delta _2^{4/3}), \end{aligned}$$

we have

$$\begin{aligned} \left\| (PU_2)^2(P\psi _1^0)\right\| _{4/3}^{4/3}=\text {O}(\delta _1^{4/3}\delta _2^{4/3}). \end{aligned}$$

For $j\ge 1$, we can reason in an analogous way: by using the fact that $|\psi _1^j|\le C U_1^2$ for some $C>0$, we obtain

$$\begin{aligned} \Vert (PU_2)^2(P\psi _1^j)\Vert _{4/3}^{4/3}=\int _\Omega (U_{2})^{8/3}|\psi ^j_1|^{4/3}+\text {O}(\delta _1^{4/3}\delta _2^{4/3})=\text {O}(\delta _1^{4/3}\delta _2^{4/3}). \end{aligned}$$

As for the second conclusion of the lemma, reasoning in the same line, we write ($j=0$)

$$\begin{aligned} \Vert (PU_1)(PU_2)(P\psi _1^0)\Vert _{4/3}^{4/3}=\int _\Omega (U_{1})^{4/3} (U_2)^{4/3}|\psi ^0_1|^{4/3}+{\tilde{R}}_{\varvec{\delta },\varvec{\xi }}, \end{aligned}$$

with

$$\begin{aligned} {\tilde{R}}_{\varvec{\delta },\varvec{\xi }}= & {} \int _\Omega \left[ (PU_1)^{4/3}-U_1^{4/3}\right] \left[ (PU_2)^{4/3}-U_2^{4/3}\right] \left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3}\right] \\&+\;\int _\Omega U_1^{4/3} U_2^{4/3} \left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3}\right] \\&+ \int _\Omega U_1^{4/3} \left[ (PU_2)^{4/3}-U_2^{4/3}\right] |\psi _1^0|^{4/3}\\&+\;\int _\Omega \left[ (PU_1)^{4/3}-U_1^{4/3}\right] U_2^{4/3} |\psi _1^0|^{4/3}+\int _\Omega \left[ (PU_1)^{4/3}-U_1^{4/3}\right] \\&\times \left[ (PU_2)^{4/3}-U_2^{4/3}\right] |\psi _1^0|^{4/3}\\&+\;\int _\Omega \left[ (PU_1)^{4/3}-U_1^{4/3}\right] U_2^{4/3} \left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3}\right] \\&+\;\int _\Omega U_1^{4/3} \left[ (PU_2)^{4/3}-U_2^{4/3}\right] \left[ |P\psi _1^0|^{4/3}-|\psi _1^0|^{4/3}\right] \end{aligned}$$

By using once again Lemmas A.1, A.3 and A.4, we can prove that ${\tilde{R}}_{\varvec{\delta },\varvec{\xi }}=\text {O}(\delta _1^{4/3}\delta _2^{4/3})$. Since moreover (by Lemma A.4)

$$\begin{aligned} \int _\Omega U_{1}^{4/3} U_2^{4/3}|\psi ^0_1|^{4/3}\le C\int _\Omega U_1^{8/3}U_2^{4/3}=\text {O}(\delta _1^{4/3}\delta _2^{4/3}), \end{aligned}$$

we conclude, as wanted, that

$$\begin{aligned} \Vert (PU_1)(PU_2)(P\psi _1^0)\Vert _{4/3}^{4/3}=\text {O}(\delta _1^{4/3}\delta _2^{4/3}). \end{aligned}$$

The fact that

$$\begin{aligned} \Vert (PU_1)(PU_2)(P\psi _1^j)\Vert _{4/3}^{4/3}=\text {O}(\delta _1^{4/3}\delta _2^{4/3}) \quad \text { for } j=1,\ldots ,4, \end{aligned}$$

follows in an analogous way, using this time the estimate: $|\psi _1^j|\le C U_1^{2}$, for some $C>0$. $\square$