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Ground states for planar axially Schrödinger–Newton system with an exponential critical growth
Boundary Value Problems volume 2020, Article number: 50 (2020)
Abstract
In this paper, we study the following planar Schrödinger–Newton system:
where V, f are axially symmetric about x, V is positive, and f is super-linear at zero and exponential critical at infinity. Using a weaker condition
with \(\theta\in[0,1)\) instead of the Nehari type monotonic condition on \(\frac{f(x,u)}{|u|^{3}}\), we obtain a ground state solution of the above problem via variational methods.
1 Introduction and main results
In the present paper, we are concerned with the wave solutions of the Schrödinger–Newton system
where \(\psi:\mathbb{R}^{d}\times\mathbb{R}\rightarrow\mathbb{C}\) is the wave function, \(W(x)\) is a real external potential, \(\lambda>0\) is a parameter. Problems of the type (1.1) arise in many problems from physics. We refer the readers to [15], therein (1.1) appears in a quantum mechanical context in the case \(d\leq3\).
A standing wave solution of (1.1) is a solution of the form \(\psi(x,t)=e^{-iEt}u(x)\) and its existence reduces (1.1) to the system
where \(V(x)=W(x)-E\), \(g(x,e^{-iEt}u)=f(x,u)e^{-iEt}\). For the case \(d=3\), it is called the Schrödinger–Poisson system and it has been well studied. For the existence, multiplicity, and concentration, we refer the readers to [2, 3, 9, 10, 13, 20] and the references therein. For Kirchhoff type equations involving subcritical and critical growth in three dimensions, please see [19] and the references therein. We also quote the paper [12] for Hardy–Schrödinger–Kirchhoff systems.
However, much less is known about the case \(d=2\). For \(\Delta\phi =u^{2}\), in \(\mathbb{R}^{2}\), one has
Substituting it into (1.2), we obtain the integro-differential equation
For simplicity, throughout this paper, let \(\lambda=2\pi\). The approach for \(d=3\) cannot be easily adapted to \(d=2\) since
which is the functional associated with the third term in (1.4), is sign-changing, and is neither bounded from above nor from below on \(H^{1}(\mathbb{R}^{2})\). This difficulty has been overcome recently in [7] or [16]. For
by introducing the following subspace of \(H^{1}(\mathbb{R}^{2})\)
endowed with the norm
Stubbe considered the \(L^{2}\)-constraint minimization problem and proved that (1.6) admits a ground state.
Soon afterwards, in [8], Cingolani and Weth processed successfully the two dimensional Schrödinger–Newton equations with nonlinear term \(|u|^{p-2}u\), \(p\geq4\). Du and Weth [11] provided some results about \(p>2\) and \(p\geq3\). The key tool is Pohozaev type identity (see [11, Lemma 2.4]). Chen, Shi, and Tang [4] used the same idea to obtain a ground state but they could deal with the general nonlinearity \(f(u)\). Simultaneously, Chen and Tang [5] investigated the existence of an axially symmetric Nehari type ground state and nontrivial solution for
where V, f is axially symmetric about x. Please see [6, 17] for further results about two dimensional Schrödinger–Newton equations with the axially symmetric assumptions. Recently, when \(V(x)=1\), Alves and Figueiredo [1] proved that (1.4) admits a positive ground state, where f is a continuous function with the exponential critical growth.
In this paper, motivated by the papers [1] and [5], we shall study the existence of ground state solutions of planar problem (1.1) with an exponential critical growth. In order to state our main result, we assume that
- (\(V_{1}\)):
\(V\in C(\mathbb{R}^{2},\mathbb{R})\), \(\inf_{x\in \mathbb{R}^{2}} V(x)>0\), \(V(x):=V(x_{1},x_{2})=V(|x_{1}|,|x_{2}|)\) for all \(x\in\mathbb{R}^{2}\).
- (\(V_{2}\)):
There exists a sequence \(\{t_{n}\}\subset(0,\infty )\) such that \(t_{n}\rightarrow\infty\) and
$$\sup_{x\in\mathbb{R}^{2},n\in\mathbb{N}}\frac {V(t^{-1}_{n}x)}{V(x)}< \infty. $$- (\(f_{1}\)):
\(f\in C^{1}(\mathbb{R}^{2}\times\mathbb{R},\mathbb {R})\), \(f(x,u):=f((x_{1},x_{2}),u)=f((|x_{1}|,|x_{2}|),u)\).
- (\(f_{2}\)):
\(f(x,u)=o(|u|)\) as \(u\rightarrow0\), uniformly in \(x\in\mathbb{R}^{2}\).
- (\(f_{3}\)):
There exists \(\alpha_{0}>0\) such that
$$ \lim_{|u|\rightarrow\infty}\frac{f(x,u)}{\exp(\alpha u^{2})}=0 \quad\textrm {for } \alpha> \alpha_{0},\qquad \lim_{|u|\rightarrow\infty}\frac{f(x,u)}{\exp(\alpha u^{2})}=+\infty \quad\textrm{for } \alpha< \alpha_{0}. $$- (\(f_{4}\)):
There exists \(\theta\in[0,1)\) such that
$$\biggl[\frac{f(x,\tau)}{\tau^{3}}-\frac{f(x,t\tau)}{(t\tau)^{3}} \biggr]\operatorname{sign}(1-t)+ \theta V(x)\frac{ \vert 1-t^{2} \vert }{(t\tau)^{2}}\geq0,\quad \forall x \in\mathbb{R}^{2}, t>0, \tau\neq0; $$- (\(f_{5}\)):
\(\inf_{x\in\mathbb{R}^{2},u\neq0}\frac {F(x.u)}{u^{2}}>-\infty\), where \(F(u)=\int^{u}_{0}f(t)\,dt\).
Remark 1.1
A simple example of satisfying the hypotheses of (\(V_{1}\))–(\(V_{2}\)) is the function \(V(x)=1+|x_{2}|[1+\sin(\pi|x_{1}|)]\) with \(t_{n}=n\). Here we also give an example which satisfies (\(f_{1}\))–(\(f_{5}\)):
where \(K\in(\mathbb{R}^{2},\mathbb{R})\) is axially symmetric and \(\inf_{x\in\mathbb{R}^{2}} K(x)>0\), V satisfies (\(V_{1}\)) and (\(V_{2}\)). But it does not satisfy the Nehari type monotonic condition
Now we state our main result as follows.
Theorem 1
For\(d=2\), suppose that (\(V_{1}\)), (\(V_{2}\)) and (\(f_{1}\))–(\(f_{5}\)) are satisfied. Then, for any\(\alpha\in(0,\frac{\pi(1-\theta)}{m})\), wheremis the least energy (it will be defined in (2.22)), θis from (\(f_{3}\)), (1.7) possesses a ground state solution.
Remark 1.2
The condition \(\alpha\in(0,\frac{\pi(1-\theta)}{m})\) is used to prove the minimizing sequence of m is bounded, and please see Lemma 3.3. Up to now, we have not been able to remove it.
The paper is organized as follows. Section 2 is to establish the variational setting and to give some preliminaries. Section 3 is to prove the existence of ground states. Throughout the paper, we always assume that (\(V_{1}\)), (\(V_{2}\)) and (\(f_{1}\))–(\(f_{5}\)) hold and make use of the following notations:
C, \(C_{i}\) (\(i=0,1,2,\ldots\)) for positive constants (possibly different from line to line).
\(L^{s}(\mathbb{R}^{2}):=\{u:\mathbb{R}^{2}\rightarrow\mathbb {R}:\int_{\mathbb{R}^{2}}|u|^{s}\,dx<\infty\}\) and \(\|\cdot\|_{s}\) denotes the usual \(L^{s}\)-norm in \(L^{s}(\mathbb{R}^{2})\).
2 Variational setting and preliminaries
In this section, we begin our study by establishing the variational setting for (1.7). Let \(H^{1}(\mathbb{R}^{2})\) be the usual fractional Sobolev space with the usual norm
and
By (\(V_{1}\)) and (\(f_{1}\)), similar to [5], let E be defined as
endowed with the norm
Denote
According to [1, Lemma 2.1], we have the following.
Proposition 2.1
\(E\hookrightarrow L^{t}(\mathbb{R}^{2})\)is compact for all\(t\in [2,\infty)\).
We formally formulate problem (1.7) in a variational way as follows:
For simplicity of notations, denote
Similar to [8], using \(\ln(r)=\ln(1+r)-\ln(1+\frac{1}{r})\), \(\forall r>0\), it holds that
We give the following proposition which is used to estimate the nonlinearity.
Proposition 2.2
([1, Lemma 2.5])
For every\(\alpha>0\)and for all\(u\in H^{1}(\mathbb{R}^{2})\), we have
Moreover, if\(\|\nabla u\|_{2}\leq1\), \(\|u\|_{2}\leq M\), and\(\alpha <4\pi\), then there exists\(C>0\)independent ofusuch that
Lemma 2.3
\(I\in C^{1}(E,\mathbb{R})\).
Proof
Noting that \(\ln(1+|x-y|)\leq\ln(1+|x|)-\ln(1+|y|)\), \(\forall x,y\in \mathbb{R}^{2}\), we get
In view of \(\ln(1+r)\leq r\), \(\forall r>0\), jointly with the Hardy–Littlewood–Sobolev inequality [14], we obtain
So \(I_{0}\) is well defined in E.
Using (\(f_{1}\))–(\(f_{3}\)), for each \(\varepsilon>0\), we have
where \(p>2\). Thus, using Hölder’s inequality with \(s>1\), \(\frac {1}{s}+\frac{1}{s'}=1\), we get
By Propositions 2.1 and 2.2, I is well defined in E. By [8, Lemma 2.2], \(I_{0}\in C^{1}(E,\mathbb{R})\). It is easy to check that \(\int_{\mathbb{R}^{2}}F(x,u)\,dx\) belongs to \(C^{1}(E,\mathbb{R})\). Thus, \(I\in C^{1}(E,\mathbb{R})\). □
Based on Lemma 2.3, we have
Lemma 2.4
For every\(u\in E\), we have
Proof
Since the proof is similar to [5, Lemma 2.3], we omit it here. □
Now, we define the Nehari manifold
Since the Nehari type monotonic condition on \(\frac{f(x,u)}{|u|^{3}}\) and super-cubic condition are not satisfied, we need to prove that \(\mathcal{N}\neq\emptyset\). To the end, we introduce the following new set:
Lemma 2.5
\(\mathcal{E}\neq\emptyset\).
Proof
Let \(u\in E\) with \(u\neq0\). \(u_{t}(x):=u(tx)\). By (\(V_{2}\)), there exists \(C_{1}>0\) such that
It follows that
In view of (\(f_{4}\)), \(t\geq0\), \(\tau\neq0\), it holds that
Taking \(t=0\), we obtain
By (\(f_{5}\)), one has
Thus, we get
So
which implies that \(\mathcal{E}\neq\emptyset\). □
The following lemma shows that \(\mathcal{N}\neq\emptyset\).
Lemma 2.6
For any\(u\in\mathcal{E}\), there exists unique\(t>0\)such that\(tu\in \mathcal{N}\).
Proof
Given \(u\in\mathcal{E}\), let \(\gamma_{u}(t):=\langle I'(tu),tu\rangle\) for \(t>0\). Then \(tu\in\mathcal{N}\) if and only if \(\gamma_{u}(t)=0\). Taking \(\varepsilon>0\) sufficiently small, jointly with Sobolev embedding, we obtain
Choosing \(t>0\) small such that \(\alpha s'\|tu\|^{2}<4\pi\), it follows from Proposition 2.2 that there exists \(\bar{t}>0\) small enough such that
Now, by (\(f_{4}\)), one has
which implies that
Therefore,
Thus, we have \(\gamma_{u}(t)\rightarrow-\infty\), as \(t\rightarrow\infty \). So there exists \(t_{0}>0\) such that \(\gamma_{u}(t_{0})=0\). Next, we shall prove that \(t_{0}\) is unique. Suppose to the contrary that there are \(t_{1}, t_{2}>0\) with \(t_{1}\neq t_{2}\) such that \(\gamma _{u}(t_{1})=\gamma_{u}(t_{2})=0\). For \(t_{1}u\in E\), using Lemma 2.4, for all \(t>0\), we have
Taking \(t=\frac{t_{2}}{t_{1}}\), it yields that
Similarly, one has
We obtain \(t_{1}=t_{2}\), so it is absurd. □
Since \(u\in\mathcal{N}\), by Lemma 2.4, one has
So we can define
Up to this stage, preparations have been made. We point out that we can define m without using the condition \(\alpha\in(0,\frac{\pi(1-\theta )}{m})\). In the next section, taking full advantage of the condition \(\alpha\in(0,\frac{\pi(1-\theta)}{m})\), we shall prove the existence of ground state solutions of (1.7).
3 Existence of ground states
In this section, with the additional condition \(\alpha\in(0,\frac{\pi (1-\theta)}{m})\), we are devoted to showing that m is achieved and the minimizer is a ground state solution of equation (1.7).
Lemma 3.1
There exists\(C>0\)such that\(\|u\|\geq C\)for all\(u\in\mathcal{N}\); furthermore, \(m>0\).
Proof
Assume by contradiction that there is \(\{u_{n}\} \subset\mathcal{N}\) such that \(\|u_{n}\|\rightarrow0\). Obviously,
In view of \((f_{1})-(f_{3})\), combining Hölder’s inequality, it follows that
With Proposition 2.2 in hand, using the Sobolev embedding, it leads to
By direct calculation, it holds that
Thus, one has
Therefore, we obtain
That is,
Noting that \(\|u_{n}\|\rightarrow0\), using Proposition 2.2 again, we get
which is ridiculous. Combining with (2.21), we have \(m>0\). □
Next, we give the following lemma which shall be used later.
Lemma 3.2
For every\(u\in E\), it holds that\(I_{1}(u)\geq\frac{1}{16}\|u\| ^{2}_{2}\|u\|^{2}_{\ast}\).
Proof
The proof is similar to [5, Lemma 2.2]. Let
For any \((x,y)\in\varLambda_{1}\times\varLambda_{3}\), it holds that
Thus,
□
Let \(\{u_{n}\}\subset\mathcal{N}\) be a minimizing sequence of m. On the additional condition \(\alpha\in(0,\frac{\pi(1-\theta)}{m})\), we want to prove that \(\{u_{n}\}\) is bounded in E.
Lemma 3.3
If\(\alpha\in(0,\frac{\pi(1-\theta)}{m})\), we have\(\{u_{n}\}\)is bounded inE.
Proof
Similar to (2.21), \(\{\|u_{n}\|\}\) is bounded. Similar to (2.5), \(\{I_{2}(u_{n})\}\) is bounded. Next, we want to estimate the \(\{I_{1}(u_{n})\}\). Note that
For the second term on the right, using Hölder’s inequality with \(s'>1\) and \(s'\approx1\), it holds that
Taking into account \(\alpha\in(0,\frac{\pi(1-\theta)}{m})\), jointly with
for n large enough, we obtain \(\alpha s'\|u_{n}\|^{2}<4\pi\). So, by Proposition 2.2, we get
Since
which yields that \(\{I_{1}(u_{n})\}\) is bounded. And it follows from Lemma 3.2 that \(\{u_{n}\}\) is bounded in E. □
Next, we claim that there are \(R,\eta>0\) such that
If it is false, using Lions’ lemma (see [18, Lemma 1.21]), we get \(u_{n}\rightarrow0\) in \(L^{t}(\mathbb{R}^{2})\) for all \(t\in[2,\infty )\). Noting that
similar to (3.5), it holds that
which contradicts Lemma 3.1.
Lemma 3.4
mis achieved and the minimizer is a weak solution of (1.7).
Proof
Now, we can assume that \(u_{n}\rightharpoonup u_{0}\neq0\) in E, \(u_{n}\rightarrow u_{0}\) in \(L^{t}(\mathbb{R}^{2})\) for all \(t\in [2,\infty)\) and \(u_{n}(x)\rightarrow u_{0}(x)\) a.e. in \(\mathbb {R}^{2}\). By a standard argument, one can deduce that \(I'(u_{0})=0\). Obviously, we have
Here, we only check 3.12 since (3.11) is similar. We have already known that
Noting that \(\alpha\in(0,\frac{\pi(1-\theta)}{m})\) and (3.5), we obtain that \(\alpha\|u_{n}\|^{2}<4\pi\) for n large enough. By Proposition 2.2, there exists \(C>0\) independent of n such that
It follows from [18, Lemma A.1] and Lebesgue dominated convergence theorem that
Thus, we have
□
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The authors are grateful to the reviewer for her/his valuable comments upon which the paper was revised.
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This work is supported by the National Natural Science Foundation of China (11901514, 11861072, 11801153), the Honghe University Doctoral Research Programs (XJ17B11), the Yunnan Province Applied Basic Research for Youths (2018FD085), the Yunnan Province Local University (Part) Basic Research Joint Project (2017FH001-013), and the Yunnan Province Applied Basic Research for General Project (2019FB001).
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Wang, W., Li, Q. & Li, Y. Ground states for planar axially Schrödinger–Newton system with an exponential critical growth. Bound Value Probl 2020, 50 (2020). https://doi.org/10.1186/s13661-020-01349-w
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DOI: https://doi.org/10.1186/s13661-020-01349-w