Ground state solutions of nonlocal equations with variable exponents and mixed criticality

Abstract

In this article, we use approximation techniques and variational methods to study a class of nonlocal equations with variable exponents and mixed criticality. We prove the existence of the ground state nontrivial solutions with the least energy. Our results are applied to a specific Schrödinger-Poisson type system.

Introduction

In this section, motivated by the work of Feng [1], we first develop a critical point framework for abstract functionals. The applications of this theory on Schrödinger-Poisson systems are discussed in the second subsection.

Abstract functional

The Schrödinger-Poisson system is one of the most important problems in nonlinear elliptic equations. One common form of this system is:

1.1

$-\mathrm{\Delta }v=|v{|}^{p-2}v+f(x,v),$ where v represents a wave function and $f(x,v)$ accounts for different interactions, including nonlocal effects. The complexity of these systems often lies in the interaction between local terms, such as $|v{|}^{p-2}v$, and nonlocal terms, particularly when critical growth is involved.

Recent interest in the Schrödinger-Poisson system has grown due to its role in modeling quantum particle interactions and electrostatic fields in nonlinear elliptic equations with subcritical and nonlocal terms. Guo [2] applied variational techniques and a refined Nehari manifold analysis to study a Schrödinger-Poisson system with a critical nonlocal term and a sublinear perturbation, and established the existence of multiple positive solutions, which broaden the understanding of such critical-type problems. Subsequent research, as exemplified by Chen, Qin, and Zhang [3], has focused on the existence of localized nodal solutions with higher topological complexity in the nonlinear Schrödinger-Poisson system. Using the symmetric mountain pass theorem, the authors constructed a sequence of localized sign-changing solutions that concentrates near the local minima of the potential, which effectively address the challenges posed by the nonlocal term. Research on Schrödinger-Poisson systems has also investigated how the symmetry of the potential influences the structure of the solutions. Ianni [4] examined the Schrödinger-Poisson-Slater problem with a radially symmetric potential, which revealed the presence of sign-changing radial solutions and provided an in-depth analysis through topological and nonlinear approaches. Meanwhile, Sun [5] studied a non-radially symmetric Schrödinger-Poisson system in three-dimensional space, presented new findings on the existence and stability of solutions, and broadened the theoretical understanding of such systems in asymmetric settings. Further research has investigated the impact of critical phenomena or mass constraints on the structure and existence of solutions. Zhang, Qin, Sahara, and Wu [6] investigated the existence of ground-state solutions for the Schrödinger-Poisson system with zero mass and the Coulomb critical exponent. The study analyzed the critical nonlinearity involved and established the existence of solutions under natural assumptions. Hu, Tang, and Jin [7] studied the normalized solutions of the Schrödinger-Poisson equation under mass constraints, and analyzed existence and multiplicity in Sobolev subcritical and critical cases. Their work overcame compactness issues caused by critical growth, which offer existence results under more general conditions.

It is worth noting that in addition to subcritical nonlocal terms, some researchers have also discussed the case of supercritical nonlocal terms. Earlier papers [8, 9] have examined the following elliptic equation with supercritical nonlinearity:

1.2

$\{\begin{array}{cc}-\mathrm{\Delta }v=|v{|}^{4+|x{|}^{\alpha }}v,\hfill & \mathrm{i}\mathrm{n}\mathrm{\Omega },\hfill \\ v=0,\hfill & \mathrm{o}\mathrm{n}\partial \mathrm{\Omega },\hfill \end{array}$ where $0<\alpha <1$ and $\mathrm{\Omega }\subset {\mathbb{R}}^3$ is the until ball. Based on the mountain pass lemma, María do Ó, Ruf and Ubilla [10] addressed (1.2) and demonstrated the existence of a positive radial solution. On this basis, Feng [1] considered a class of abstract functionals with supercritical terms, and obtained the non-trivial solution of this class of functionals.

Inspired by the above papers, denote by H the radially symmetric subspace of $H_0^1(B)$. Here, $B\subset {\mathbb{R}}^3$ represents the unit ball. In extending the supercritical case with a subcritical term, the computational steps remain largely consistent with those in [1]. We now examine the impact of this addition using the following abstract functional:

1.3

$J(v)=\frac{1}{2}{\int }_B|\mathrm{\nabla }v{|}^2+\lambda K(v)+{\int }_B\frac{1}{p}|v{|}^p-{\int }_B\frac{1}{6+|x{|}^{\alpha }}|v{|}^{6+|x{|}^{\alpha }},v\in H,$ where $\frac{8}{3}<p<6$ and $\lambda <0$ and $K\in C(H,\mathbb{R})$. Rather than analyzing J directly, we instead define a related functional $I:H\to \mathbb{R}$, which serves as an approximation and facilitates the search for a nontrivial critical point:

1.4

$I(v)=\frac{1}{2}{\int }_B|\mathrm{\nabla }v{|}^2+\lambda K(v)+{\int }_B\frac{1}{p}|v{|}^p-{\int }_B\frac{1}{6}|v{|}^6,v\in H.$

To derive some properties of the abstract functional in order to obtain the conclusions of this paper, we need to make the following assumptions regarding $K(v)$:

(K1) $K(v)\in C^1(H,{\mathbb{R}}^{+})$;

(K2) there exist $q>6$ and $C>0$ such that for $n>0$, $K(nv)=n^{q}K(v),K(v)\le C{\parallel v\parallel }^q,\mathrm{\forall }v\in H;$

(K3) $⟨K^{\prime }(v),v⟩=qK(v)$, $v\in H$;

(K4) For any $c>0$, if $\{v_m\}$ is a ${(PS)}_c$ sequence of J and $v_m$ converges weakly to v in H as $m\to \mathrm{\infty }$, then $J^{\prime }(v)=0$.

Based on the previous work, we focus on the existence of ground state for the functional J. Our main results are as follows.

Theorem 1.1

Assuming that conditions$(K1)-(K4)$are satisfied, the functionalJwill have a sequence that fulfills the${(PS)}_c$condition for some positive constant c. In addition, provided that the functionalImeets the${(PS)}_c$condition, there will be a nontrivial critical point for the functionalJ.

Theorem 1.2

Given thatKis an even function and weakly lower semicontinuous, and considering that Theorem 1.1holds, then the functionalJhas a critical point corresponding to the minimal energy.

Remark 1.3

The key distinction between J and I lies in the supercritical term $6+|x{|}^{\alpha }$. In fact, this term is responsible for the supercritical growth observed at all points other than the origin, highlighting the role of the perturbation in the system’s behavior.

Remark 1.4

For $\lambda <0$ and $q>6$, after adding the subcritical term, we can still construct the minimax level. We need to verify that this level is below the non-compactness level of functional I. However, accurately calculating the non-compactness level is quite challenging due to the variations in K.

Remark 1.5

Condition $(K4)$ is necessary as the proof approach changes with different K.

Applications

As an application of abstract functional theory, we consider the existence of nontrivial solutions for Schrödinger-Poisson systems with variable exponents of supercritical and subcritical combinations

1.5

$\{\begin{array}{cc}-\mathrm{\Delta }v-\varphi |v{|}^{3}v=|v{|}^{4+|x{|}^{\alpha }}v-|v{|}^{p-2}v\hfill & \text{in }B,\hfill \\ -\mathrm{\Delta }\varphi =|v{|}^5\hfill & \text{in }B,\hfill \\ v=\varphi =0\hfill & \text{on }\partial B,\hfill \end{array}$ where $\frac{8}{3}<p<6$, $0<\alpha <1$ and $B\subset {\mathbb{R}}^3$ is the until ball. The Schrödinger-Poisson system, which models the interplay between charged particles and electromagnetic fields, is frequently applied in mathematical physics (see [10] for more details). Several works have examined the system with nonlocal critical growth within bounded domains [11–13].

By using Theorems 1.1 and 1.2, we prove that there is a positive ground state solution for the system (1.5). In terms of technology, we have two problems in verifying our findings. First, the presence of supercritical nonlinearity in the system hinders the convergence of the bounded (PS) sequence. Second, because there are some key terms in this system, it is very hard to estimate the critical value of the mountain pass. To solve this problem, we use the method described in [8] to calculate the critical value of the mountain pass for (1.5), and then prove that it is lower than the noncompact level. As a result, we can conclude:

Theorem 1.6

The system described by equation (1.5) has at least one positive ground state solution.

Remark 1.7

It is clear that (1.5) lacks nontrivial solutions when $|x{|}^{\alpha }=0$ by applying Pohozaev’s identity from [11]. The undesirable result stems from the fact that the nonlinearity $|v{|}^{4+|x{|}^{\alpha }}v$ experiences supercritical growth across B, except at the origin, where it only shows critical growth.

Remark 1.8

In this paper, C is used to represent various positive constants, which may change depending on the context but do not affect the core of the problem.

Preliminary

This section introduces the lemmas and notations that will be utilized throughout the paper. Set $H=H_{0,rad}^1(B)=\{v\in H_0^1(B):v(x)=v(|x|)\}$ be the Sobolev space of radial symmetric functions. We can also define the corresponding norm as follows:

2.1

$\parallel v\parallel ={({\int }_B|\mathrm{\nabla }v{|}^2)}^{1/2}.$ We further define

2.2

$Z_{+}(\overline{B})=\{t:t\in \mathcal{C}(\overline{B}),t(x)>1,x\in \overline{B}\},t^{+}=\underset{x\in B}{sup}t(x),t^{-}=\underset{x\in B}{inf}t(x).$ Let $\theta \in Z_{+}(\overline{B})$. We define the space $L^{\theta (x)}(B)$, which is equipped with a variable exponent, as the set of measurable functions in B for which the following integral condition holds:

2.3

$L^{\theta (x)}(B)=\{{\int }_B|v(x){|}^{\theta (x)}dx<\mathrm{\infty }\}.$ The corresponding norm on this space is expressed as

2.4

${\parallel u\parallel }_{L^{\theta (x)}}=inf\{\mu >0,{\int }_B{\left|\frac{v}{\mu }\right|}^{\theta (x)}\le 1\}.$ This function space generalizes the traditional $L^{\theta }(B)$ spaces, where the exponent is constant, by allowing the exponent $\theta (x)$ to vary with position in B. The variation of $\theta (x)$ introduces a more refined structure to the space, offering greater flexibility in modeling functions with varying degrees of smoothness across the domain. Let $L^{{\theta }^{\prime }(x)}(B)$ denote the conjugate space of $L^{\theta (x)}(B)$, where $1/\theta (x)+1/{\theta }^{\prime }(x)=1$. For any $v\in L^{\theta (x)}(B)$ and $u\in L^{{\theta }^{\prime (x)}}(B)$, the Hölder inequality holds:

2.5

$\left|{\int }_{B}vu\right|\le (\frac{1}{{\theta }^{-}}+\frac{1}{{\theta }^{\prime -}}){\parallel v\parallel }_{L^{\theta (x)}}{\parallel u\parallel }_{L^{{\theta }^{\prime }(x)}}.$

Lemma 2.1

([6])

If$x\in B$and${\mu }_1>0$, ${\mu }_2>0$, We are able to obtain the following continuous embedding:

2.6

$H\hookrightarrow L^{6+{\mu }_1|x{|}^{{\mu }_2}}(B).$

By using Lemma 2.1 and the Hölder-type inequality, we can get

2.7

$⟨J^{\prime }(v),u⟩={\int }_B\mathrm{\nabla }v\cdot \mathrm{\nabla }u+{\int }_B|v{|}^{p-2}vu+\lambda ⟨K^{\prime }(v),u⟩-{\int }_B|v{|}^{4+|x{|}^{\alpha }}vu,v,u\in H.$ Next, we introduce the best embedding constant, denoted as S, which is defined as follows:

2.8

$S=\underset{v\in H\setminus \{0\}}{inf}\frac{{\int }_B|\mathrm{\nabla }v{|}^2}{{({\int }_B|v{|}^6)}^{1/3}}.$ We now define

2.9

$U_{\epsilon }(x)=\frac{{(3{\epsilon }^2)}^{1/4}}{{({\epsilon }^2+|x{|}^2)}^{1/2}},\epsilon >0,$ for which the equation holds:

2.10

$-\mathrm{\Delta }v=v^5\mathrm{i}\mathrm{n}{\mathbb{R}}^3.$ To proceed with the analysis, we define a cut-off function $\chi (x)$ with $\chi (x)=1$ on $B_{1/2}(0)$. Let $v_{\epsilon }=\chi (x)U_{\epsilon }(x)$, and applying standard techniques, we obtain the following estimates as $\epsilon \to 0^{+}$ (refer to [14]):

2.11

${\int }_B|\mathrm{\nabla }{v}_{\epsilon }{|}^2=S^{3/2}+O(\epsilon ),{\int }_B{v}_{\epsilon }^6=S^{3/2}+O({\epsilon }^3).$ As shown in [15], we have

2.12

$\begin{array}{r}{\int }_B|v_{\epsilon }{|}^p=\{\begin{array}{cc}O\left({\epsilon }^{\frac{p}{2}}\right),\hfill & \text{if }2<p<3;\hfill \\ O({\epsilon }^{\frac{3}{2}}|log\epsilon |),\hfill & \text{if }p=3;\hfill \\ O\left({\epsilon }^{\frac{6-p}{2}}\right),\hfill & \text{if }3<p<6.\hfill \end{array}\end{array}$ Next we set $O_p(\epsilon )=max{\int }_B|v_{\epsilon }{|}^{p}dx$, and for $p=2$, we have

2.13

${\int }_B|v_{\epsilon }{|}^2=C\epsilon .$

Lemma 2.2

([16])

If$\eta (v)={\int }_B|v(x){|}^{\theta (x)}$, $v\in L^{\theta (x)}(B)$, then:

$(1)$${\parallel v\parallel }_{L^{\theta (x)}}<1(=1;>1)\iff \eta (v)<1(=1;>1)$;

$(2)$${\parallel v\parallel }_{L^{\theta (x)}}>1\Rightarrow {\parallel v\parallel }_{L^{\theta (x)}}^{{\theta }^{-}}\le \eta (v)\le {\parallel v\parallel }_{L^{\theta (x)}}^{{\theta }^{+}}$, ${\parallel v\parallel }_{L^{\theta (x)}}<1\Rightarrow {\parallel v\parallel }_{L^{\theta (x)}}^{{\theta }^{+}}\le \eta (v)\le {\parallel v\parallel }_{L^{\theta (x)}}^{{\theta }^{-}}$.

Proof of Theorem 1.1

Lemma 3.1

Suppose that conditions$(K1)$and$(K2)$are satisfied.

$(1)$There exists${\eta }_0>0$, such that for all$\eta \in (0,{\eta }_0)$, one has${\delta }_{\eta }=inf\{J(v):v\in H,\parallel v\parallel =\eta \}>0$.

$(2)$One can find an element$e\in H$satisfying$\parallel e\parallel >\eta$and$J(e)<0$.

Proof

$(1)$ For $\eta >0$, let

3.1

${\mathrm{\Sigma }}_{\eta }=\{v\in H:\parallel v\parallel \le \eta \}.$ Invoking Sobolev’s inequality together with Lemma 2.2, we obtain that for each $v\in \partial {\mathrm{\Sigma }}_{\eta }$, a constant $C>0$ can be found such that:

3.2

$\begin{array}{rl}J(v)& =\frac{1}{2}{\parallel v\parallel }^2+{\int }_B\frac{1}{p}|v{|}^p+\lambda K(v)-{\int }_B\frac{1}{6+|x{|}^{\alpha }}|v{|}^{6+|x{|}^{\alpha }}\\ & \ge \frac{1}{2}{\parallel v\parallel }^2+C\lambda {\parallel v\parallel }^q-C({\parallel v\parallel }^6+{\parallel v\parallel }^7)\\ & =\frac{1}{2}{\eta }^2+C\lambda {\eta }^q-C{\eta }^6-C{\eta }^7.\end{array}$ By selecting ${\eta }_0$ sufficiently small, we can find ${\delta }_{\eta }>0$ such that part $(1)$ holds.

$(2)$ According to [9], given a small $\epsilon >0$, there is a positive constant $C>0$ for which the following inequality holds:

3.3

$\begin{array}{rl}{\int }_B|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}& \ge {\int }_B|v_{\epsilon }{|}^6+C|log\epsilon |{\epsilon }^{\alpha }+O({\epsilon }^{2/3})+O(\epsilon )\\ & \ge S^{3/2}+C|log\epsilon |{\epsilon }^{\alpha }+O(\epsilon ).\end{array}$ By applying equation (2.11), we can show that for large enough n and whenever $\epsilon >0$ is sufficiently small:

3.4

$\begin{array}{rl}J(n{v}_{\epsilon })& =\frac{n^2}{2}{\parallel v_{\epsilon }\parallel }^2+{\int }_B\frac{n^p}{p}|v_{\epsilon }{|}^p+\lambda n^{q}K(v_{\epsilon })-{\int }_B\frac{n^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }}|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}\\ & \le \frac{n^2}{2}{\parallel v_{\epsilon }\parallel }^2+{\int }_B\frac{n^p}{p}|v_{\epsilon }{|}^p-\frac{n^6}{7}{\int }_B|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}\\ & \le S^{3/2}{n}^2+\frac{8n^p}{3}{\int }_B|v_{\epsilon }{|}^p-\frac{S^{3/2}}{14}{n}^6,\end{array}$ where $\frac{8}{3}<p<6$, it is easy to know that $J(n{v}_{\epsilon })\to -\mathrm{\infty }$ as $n\to +\mathrm{\infty }$. To construct a suitable path, we set $ħ(n)=nN{v}_{ϵ}$, where $N>0$, $ħ(n):[0,1]\to H$. If N is sufficiently large, then it follows that

3.5

${\int }_B|\mathrm{\nabla }ħ(1){|}^2>{\eta }^2,J(ħ(1))<0.$ With the choice $e=ħ(1)$, then $(2)$ holds, which completes the proof. □

The mountain pass geometry of the functional J is established by Lemma 3.1. It follows that there exists a ${(PS)}_c$ sequence $\{v_m\}\subset H$ corresponding to J, which satisfies the following condition:

3.6

$J(v_m)\to c,{\parallel J^{\prime }(v_m)\parallel }_{H^{-1}}\to 0,m\to \mathrm{\infty }.$ Here, the constant c is typically defined by

3.7

$c=\underset{\varsigma \in \tau }{inf}\underset{n\in [0,1]}{max}J(\varsigma (n)),$ and $\tau =\{\varsigma \in C([0,1],H):\varsigma (0)=0,J(\varsigma (1))<0\}$.

Lemma 3.2

Assume the assumption$(K3)$holds. If$\{v_m\}$is a${(PS)}_c$sequence, then$\{v_m\}$is bounded in H.

Proof

For m large enough, we can readily infer from assumption (K3) that

3.8

$\begin{array}{rl}c+o(1)\parallel v_m\parallel +1& \ge J(v_m)-\frac{1}{6}⟨J^{\prime }(v_m),v_m⟩\\ & =\frac{1}{3}{\parallel v_m\parallel }^2+{\int }_B(\frac{1}{p}-\frac{1}{6})|v_m{|}^p\\ & +\lambda (\frac{1}{q}-\frac{1}{6})⟨K^{\prime }(v_m),v_m⟩+{\int }_B(\frac{1}{6}-\frac{1}{6+|x{|}^{\alpha }})|v_m{|}^{6+|x{|}^{\alpha }}\\ & \ge \frac{1}{3}{\parallel v_m\parallel }^2.\end{array}$ Therefore, $\{v_m\}$ is bounded in H, and the proof is thereby completed. □

Lemma 3.3

([7])

For every$r>0$, any$v\in H$satisfies$|v(r)|\le r^{-1/2}\parallel v\parallel$.

Proof of Theorem 1.1

Base on Lemma 3.1, we find that there is a sequence $\{v_m\}\subset H$ such that $J(v_m)\to c$ and $J^{\prime }(v_m)\to 0$ as $m\to \mathrm{\infty }$, where c is defined by the minimax level (equation (3.7)). By Lemma 3.2, the sequence $\{v_m\}$ is bounded in H. Taking a subsequence, we assume $v_m\rightharpoonup v$ weakly in H, and $v_m(x)\to v(x)$, a.e. $x\in B$. Condition (K4) ensures that v is a nontrivial critical point of J whenever $v\ne 0$. Therefore, it remains to analyze the case $v=0$, which we will prove to be impossible. Since $H_r^1(B\mathrm{\setminus }{B}_{\rho })\hookrightarrow \hookrightarrow L^{\theta }(B\mathrm{\setminus }{B}_{\rho })$ for $\theta >1$ and $\rho \in (0,1)$, we conclude:

3.9

${\int }_{\rho }^1|v_m{|}^{6+r^{\alpha }}{r}^2\to 0\text{as }m\to \mathrm{\infty }$ and

3.10

${\int }_{\rho }^1|v_m{|}^6{r}^2\to 0\text{as }m\to \mathrm{\infty }.$ In order to conclude that $\{v_m\}$ forms a ${(PS)}_c$ for I, it suffices to verify that:

$(1)J(v_m)=I(v_m)+o(1)$;

$(2)⟨J^{\prime }(v_m),u⟩=⟨I^{\prime }(v_m),u⟩+o(1)\parallel u\parallel$, $u\in H$.

To prove that (1) is correct, we need only estimate

3.11

$\begin{array}{rl}& {\int }_B(\frac{1}{6}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^{6+|x{|}^{\alpha }})\\ & ={\int }_B(\frac{1}{6}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^6)+{\int }_B(\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^{6+|x{|}^{\alpha }}).\end{array}$ By (3.10) and coordinate transformation, there exist $\rho >0$ and $m_1\in N$ such that for any $\epsilon >0$, the following holds for all $m\ge m_1$,

3.12

$\begin{array}{rl}{\int }_B(\frac{1}{6}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^6)& \le \frac{\omega }{36}{\int }_0^{\rho }|v_m{|}^6{r}^{2+\alpha }+\frac{\omega }{36}{\int }_{\rho }^1|v_m{|}^6{r}^{2+\alpha }\\ & \le \frac{{\parallel v_m\parallel }^6}{36\alpha }\omega {\rho }^{\alpha }+\frac{\omega }{36}{\int }_{\rho }^1|v_m{|}^6{r}^2\\ & \le \frac{\epsilon }{2}.\end{array}$ Under the spherical coordinate transformation, ω denotes the surface area of the unit sphere in ${\mathbb{R}}^3$. In a similar manner, for any $\epsilon >0$, there exist a sufficiently small $\rho >0$ and $m_2\in N$, such that for any $m\ge m_2$, (3.9) and (3.10) imply that:

3.13

$\begin{array}{rl}& \left|{\int }_B(\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^{6+|x{|}^{\alpha }})\right|\\ & \le \frac{\omega }{6}{\int }_0^1|v_m{|}^6|1-|v_m{|}^{r^{\alpha }}|r^2+\frac{\omega }{6}|{\int }_{\rho }^1|v_m{|}^{6}1-|v_m{|}^{r^{\alpha }}{r}^2|\\ & \le \frac{\omega }{6}{\int }_0^{\rho }|v_m{|}^6{r}^2|exp[-\frac{r^{\alpha }}{2}log(Cr)]-1|+\frac{\omega }{18}{\rho }^3+\frac{\omega }{6}|{\int }_{\rho }^1|v_m{|}^6(|v_m{|}^{r^{\alpha }}-1)r^2|\\ & \le C\omega {\int }_0^{\rho }|v_m{|}^6{r}^2{r}^{\alpha }|logCr|+\frac{\omega }{18}{\rho }^3+\frac{\omega }{6}|{\int }_{\rho }^1|v_m{|}^6(|v_m{|}^{r^{\alpha }}-1)r^2|\\ & \le C\omega {\rho }^{\alpha }|logC\rho |+\frac{\omega }{18}{\rho }^3+\frac{\omega }{6}|{\int }_{\rho }^1|v_m{|}^6(|v_m{|}^{r^{\alpha }}-1)r^2|\le \frac{\epsilon }{2}.\end{array}$ Using (3.11), (3.12) and (3.13), we conclude that for every $\epsilon >0$, one can choose $m_0=max\{m_1,m_2\}$, so that for any $m\ge m_0$, the integral expression satisfies

3.14

$\left|{\int }_B(\frac{1}{6}|v_m{|}^6-\frac{1}{6+|x{|}^{\alpha }}|v_m{|}^{6+|x{|}^{\alpha }})\right|\le \epsilon ,$ which ensures that $(1)$ is valid.

The next step is to confirm the correctness of $(2)$. Specifically, we need only estimate

3.15

$\begin{array}{rl}& |{\int }_0^1|v_m{|}^{4+r^{\alpha }}{v}_{m}u{r}^2-{\int }_0^1|v_m{|}^4{v}_{m}u{r}^2|\\ \le & {\int }_0^1|v_m{|}^5|u|||v_m{|}^{r^{\alpha }}-1|r^2\\ \le & {\int }_0^{\rho }|v_m{|}^5|u|||v_m{|}^{r^{\alpha }}-1|r^2+{\int }_{\rho }^1|v_m{|}^5|u||v_m{|}^{r^{\alpha }}{r}^2+{\int }_{\rho }^1|v_m{|}^5|u|r^2\\ =& A_1+A_2+A_3,\end{array}$ where $u\in H$, $0<\rho <1$. By [1], for any $\epsilon >0$, we can find

3.16

$A_1=|{\int }_0^{\rho }|v_m{|}^5|u|(|v_m{|}^{r^{\alpha }}-1)r^2|<\frac{\epsilon }{3}\parallel u\parallel .$ On the other hand, there exists $m_3\in N$ such that for $m>m_3$,

3.17

$A_2={\int }_{\rho }^1|v_m{|}^{5+r^{\alpha }}|u|r^2\le C{({\int }_{\rho }^1|v_m{|}^{6+r^{\alpha }}{r}^2)}^{5/7}\parallel u\parallel \le \frac{\epsilon }{3}\parallel u\parallel .$ In a similar manner, given any $\epsilon >0$, one can find $m_4\in \mathbb{N}$ such that for every $m>m_4$,

3.18

$A_3={\int }_{\rho }^1|v_m{|}^5|u|r^2\le C{({\int }_{\rho }^1|v_m{|}^6{r}^2)}^{5/6}\parallel u\parallel \le \frac{\epsilon }{3}\parallel u\parallel .$ Combining (3.16), (3.17), (3.18), we obtain for $\epsilon >0$, there exists $m_0^{\prime }=max\{m_3,m_4\}$, such that for any $m\ge m_0^{\prime }$,

3.19

$A_1+A_2+A_3\le \epsilon \parallel u\parallel ,$ which implies that $(2)$ is true. It can be concluded that $\{v_m\}$ constitutes a ${(PS)}_c$ sequence of I. Owing to the fact that I fulfills the ${(PS)}_c$ compactness criterion, the sequence necessarily converges strongly to the trivial solution in H, which leads to a contradiction. This completes the proof. □

Proof of Theorem 1.2

In what follows, the Nehari manifold method is employed to rigorously establish the existence of a ground state solution to the functional J, which is both nontrivial and nonnegative. To achieve this, we first introduce the Nehari manifold, defined as follows:

4.1

$\mathcal{N}=\{v\in H\setminus \{0\}:⟨J^{\prime }(v),v⟩=0\}.$

Lemma 4.1

Suppose that conditions$(K1)$and$(K2)$are satisfied. Given any$v\in H$with$v\ne 0$, there exists a positive number$n(v)$such that$n(v)v$lies exactly on$\mathcal{N}$, and among all positive scalar multiples ofv, this point yields the maximum value of the functional$J(n(v)v)={max}_{n\ge 0}J(nv)$.

Proof

The function $h(n)=J(nv)$, defined for $n>0$, is introduced to simplify the argument. It can be observed that $h^{\prime }(n)=⟨J^{\prime }(nv),v⟩=0$, and this derivative equals zero precisely when $nv\in \mathcal{N}$. Furthermore, a direct computation shows that when $\lambda <0$, $q>6$

4.2

$\begin{array}{rl}{h}^{\prime }(n)& =n{\parallel v\parallel }^2+{\int }_B{n}^{p-1}|v{|}^p+\lambda q{n}^{q-1}K(v)-{\int }_B{n}^{5+|x{|}^{\alpha }}|v{|}^{6+|x{|}^{\alpha }}\\ & =n({\parallel v\parallel }^2+{\int }_B{n}^{p-2}|v{|}^p+\lambda n^{q-2}K(v)-{\int }_B{n}^{4+|x{|}^{\alpha }}|v{|}^{6+|x{|}^{\alpha }})\\ & =n\zeta (n).\end{array}$ Obviously, ζ is a function having roots for $n>0$ and ${lim}_{n\to 0^{+}}\zeta (n)={\parallel v\parallel }^2>0$, ${lim}_{n\to \mathrm{\infty }}\zeta (n)=-\mathrm{\infty }$. Hence there exists a positive number $n(v)$ such that $n(v)v$ lies exactly on $\mathcal{N}$ and $J(n(v)v)={max}_{n\ge 0}J(nv)$. The proof is completed. □

Lemma 4.2

Assume the assumption$(K1)$-$(K3)$holds, thenJis lower bounded on the set$\mathcal{N}$.

Proof

From assumptions (K1) and (K2), together with the identity $⟨J^{\prime }(v),v⟩=0$, we conclude that

4.3

$\begin{array}{rl}{\parallel v\parallel }^2& =-\lambda qK(v)+{\int }_B|v{|}^p+{\int }_B|v{|}^{6+|x{|}^{\alpha }}\\ & \le C({\parallel v\parallel }^6+{\parallel v\parallel }^7+{\parallel v\parallel }^p+{\parallel v\parallel }^q),\end{array}$ which leads to the conclusion that $\parallel v\parallel$ cannot be smaller than some positive constant C. Conversely, we get

4.4

$\begin{array}{rl}J(v)& =J(v)-\frac{1}{6}⟨J^{\prime }(v),v⟩\\ & =\frac{1}{3}{\parallel v\parallel }^2+{\int }_B(\frac{1}{p}-\frac{1}{6})|v{|}^p+\lambda (\frac{1}{q}-\frac{1}{6})⟨K^{\prime }(v),v⟩\\ & +{\int }_B(\frac{1}{6}-\frac{1}{6+|x{|}^{\alpha }})|v{|}^{6+|x{|}^{\alpha }}\\ & \ge \frac{1}{3}{\parallel v\parallel }^2,v\in \mathcal{N}.\end{array}$ □

Based on Lemmas 4.1 and 4.2, we can introduce the following definitions:

4.5

$c^{\star }=\underset{v\in \mathcal{N}}{inf}J(v),c^{\star \star }=\underset{v\in H\setminus \{0\}}{inf}\underset{n\ge 0}{max}J(nv).$

Lemma 4.3

Under assumptions$(K1)$-$(K3)$, we have$c=c^{\star }=c^{\star \star }$.

Proof

Since $c^{\star }=c^{\star \star }$ by Lemma 4.1, we move forward to prove $c=c^{\star }$. To begin, let $v\in \mathcal{N}$. By applying Lemma 3.1, it can be shown that there exists some $n_0>1$ such that $J(n_{0}v)<0$. Thus, we have $c\le c^{\star }$.

In contrast, for every $v\in H$, we have that

4.6

$\begin{array}{rl}J(v)-\frac{1}{6}⟨J^{\prime }(v),v⟩& =\frac{1}{3}{\parallel v\parallel }^2+{\int }_B(\frac{1}{p}-\frac{1}{6})|v{|}^p+\lambda (\frac{1}{q}-\frac{1}{6})⟨K^{\prime }(v),v⟩\\ & +{\int }_B(\frac{1}{6}-\frac{1}{6+|x{|}^{\alpha }})|v{|}^{6+|x{|}^{\alpha }}\\ & \ge \frac{1}{3}{\parallel v\parallel }^2\ge 0.\end{array}$ Let $\varsigma \in \tau$, by (4.6), we already know that $⟨J^{\prime }(\varsigma (1)),\varsigma (1)⟩\le 6J(\varsigma (1))<0$. We introduce $s_1=inf\{s\in [0,1):⟨J^{\prime }(\varsigma (t)),\varsigma (t)⟩<0,t\in (s,1]\}$. Then $⟨J^{\prime }(\varsigma (s_1)),\varsigma (s_1)⟩=0$ and $\varsigma (t)\ne 0$ for all $t\in (s_1,1]$. We aim to prove that $\varsigma (s_1)\ne 0$. Suppose the contrary, that $\varsigma (s_1)=0$. Then, according to Lemma 3.1, $⟨J^{\prime }(\varsigma (t)),\varsigma (t)⟩>0$. As $s\to s_1^{+}$, we find an $\xi >0$ such that $s_1+\xi <1$ and $⟨J^{\prime }(\varsigma (s_1+$$\xi )),\varsigma (s_1+\xi )⟩>0$, which contradicts the assumption on $s_1$. Hence, $\varsigma (s_1)\ne 0$, and we conclude $\varsigma (s_1)\in \mathcal{N}$ and $c^{\star }\le c$. Thus the proof is completed. □

Lemma 4.4

Suppose that the conditions$(K1)$-$(K3)$are satisfied. When the value$c^{\star }$is achieved at$v\in \mathcal{N}$, vserves as a critical point of the functionalJwithin the spaceH.

Proof

By Lemma 4.1, $\mathcal{N}\ne \mathrm{\varnothing }$. Let $F(v)=⟨J^{\prime }(v),v⟩$, then $F\in C^1(H,\mathbb{R})$. It is easy to know that

4.7

$\begin{array}{rl}F(v)& ={\parallel v\parallel }^2+{\int }_B|v{|}^p+\lambda ⟨K^{\prime }(v),v⟩-{\int }_B|v{|}^{6+|x{|}^{\alpha }}\\ & \ge \frac{1}{2}{\parallel v\parallel }^2-C({\parallel v\parallel }^6+{\parallel v\parallel }^7)>0\end{array}$ as long as $v\in H$ and its norm $\parallel v\parallel$ is sufficiently small. For $\lambda <0$, $\frac{8}{3}<p<6$, $q>6$, we get,

4.8

$\begin{array}{rl}⟨F^{\prime }(v),v⟩& =⟨F^{\prime }(v),v⟩-6F(v)\\ & =-4{\parallel v\parallel }^2+(p-6){\int }_B|v{|}^p+\lambda q(q-6)K(v)-{\int }_B|x{|}^{\alpha }|v{|}^{6+|x{|}^{\alpha }}<0.\end{array}$ Therefore, for all $v\in \mathcal{N}$, $F^{\prime }(v)\ne 0$. It can be concluded from the implicit function theorem that $\mathcal{N}$ forms a $C^1$ manifold. Moreover, it is crucial to recall that v serves as the minimizer of the functional J on $v\in \mathcal{N}$. Consequently, the Lagrange multiplier method indicates that there exists ${\xi }_0\in \mathbb{R}$ satisfying

4.9

$J^{\prime }(v)={\xi }_0{F}^{\prime }(v).$ By making use of both (4.8) and (4.9), it follows that $J^{\prime }(v)=0$, this completes the proof. □

Proof of Theorem 1.2

The inequality $J\ge c^{\star }$ follows from Theorem 1.1. By invoking Lemma 4.2, along with Fatou’s lemma and weak semicontinuity of the norm, it is deduced that

4.10

$\begin{array}{rl}{c}^{\star }& =\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}inf}[J(v_m)-\frac{1}{6}⟨J^{\prime }(v_m),v_m⟩]\\ & =\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}inf}[\frac{1}{3}{\parallel v_m\parallel }^2+{\int }_B(\frac{1}{p}-\frac{1}{6})|v_m{|}^p+\lambda q(\frac{1}{q}-\frac{1}{6})K(v_m)\\ & +{\int }_B(\frac{1}{6}-\frac{1}{6+|x{|}^{\alpha }})|v_m{|}^{6+|x{|}^{\alpha }}]\\ & \ge \frac{1}{3}{\parallel v\parallel }^2+{\int }_B(\frac{1}{p}-\frac{1}{6})|v{|}^p+\lambda q(\frac{1}{q}-\frac{1}{6})K(v)+{\int }_B(\frac{1}{6}-\frac{1}{6+|x{|}^{\alpha }})|v{|}^{6+|x{|}^{\alpha }}\\ & =J(v)-\frac{1}{6}⟨J^{\prime }(v),v⟩=J(v).\end{array}$ This demonstrates that $J(v)=c^{\star }$. It is evident that $J(|v|)=J(v)=c^{\star }$. Consequently, the conclusions of Lemma 4.3 indicate that $|v|$ is a ground state of J. This completes the proof. □

Proof of Theorem 1.6

This part of the paper is primarily devoted to demonstrating the existence of positive ground state solutions to (1.5). As an application of the previous theory, we consider the case of $\lambda K(v)=-\frac{1}{10}{\int }_B{\varphi }_v|v{|}^5$. It follows from [11] that the following features hold.

Lemma 5.1

Let$v\in H$be fixed, the following holds.

$(1)$${\varphi }_v\ge 0$a.e. inB.

$(2)$${\varphi }_{nv}=n^5{\varphi }_v$for all$n>0$.

$(3)$$\parallel {\varphi }_v\parallel \le S^{-3}{\parallel v\parallel }^5$and

5.1

${\int }_B{\varphi }_v|v{|}^5\le S^{-6}{\parallel v\parallel }^{10}.$

$(4)$If$v_m\rightharpoonup v$inH, there exists a subsequence such that${\phi }_{v_m}\rightharpoonup {\phi }_v$inH.

Equation (1.5) can be derived from a variational principle, and its solutions are precisely the critical points of the associated functional defined in the space H:

5.2

$\tilde{E}(v)=\frac{1}{2}{\int }_B|\mathrm{\nabla }v{|}^2+{\int }_B\frac{1}{p}|v{|}^p-\frac{1}{10}{\int }_B{\varphi }_v|v{|}^5-{\int }_B\frac{1}{6+|x{|}^{\alpha }}|v{|}^{6+|x{|}^{\alpha }}.$ From equation (5.1) and Lemma 2.2, one can verify that Ẽ is properly defined in H, and moreover $\tilde{E}\in C^1(H,\mathbb{R})$. The corresponding derivative satisfies:

5.3

$\begin{array}{r}⟨{\tilde{E}}^{\prime }(v),u⟩={\int }_B\mathrm{\nabla }v\mathrm{\nabla }u+{\int }_B|v{|}^{p-2}vu-{\int }_B{\varphi }_v|v{|}^{3}vu-{\int }_B|v{|}^{4+|x{|}^{\alpha }}vu,v,u\in H.\end{array}$

Lemma 5.2

Let${\alpha }_0,{\beta }_0,{\gamma }_0>0$and define$f_0:[0,\mathrm{\infty })\to \mathbb{R}$as

5.4

$f_0(n)=\frac{{\alpha }_0}{2}{n}^2-\frac{{\beta }_0}{10}{n}^{10}-\frac{{\gamma }_0}{6}{n}^6.$Then

5.5

$\underset{n\in [0,\mathrm{\infty })}{sup}{f}_0(n)={\left(\frac{\sqrt{{\gamma }_0^2+4{\alpha }_0{\beta }_0}-{\gamma }_0}{2{\beta }_0}\right)}^{1/2}\frac{12{\alpha }_0{\beta }_0+{\gamma }_0^2-{\gamma }_0\sqrt{{\gamma }_0^2+4{\alpha }_0{\beta }_0}}{30{\beta }_0}.$

Proof

For $n\ge 0$, we have

5.6

$f_0^{\prime }(n)={\alpha }_{0}n-{\beta }_0{n}^9-{\gamma }_0{n}^5=n({\alpha }_0-{\beta }_0{n}^8-{\gamma }_0{n}^4).$ Let $h_1(n)={\alpha }_0-{\beta }_0{n}^8-{\gamma }_0{n}^4=0$, we write at

5.7

$n^4=\frac{\sqrt{{\gamma }_0^2+4{\alpha }_0{\beta }_0}-{\gamma }_0}{2{\beta }_0}.$ The result can be obtained by substituting it into $f_0(n)$. This completes the proof. □

Lemma 5.3

Let

5.8

$g_0(n)=\frac{n^2}{2}{\parallel v_{\epsilon }\parallel }^2-\frac{n^{10}}{10}{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-\frac{n^6}{6}{\int }_B|v_{\epsilon }{|}^6,$then, we have

5.9

$\underset{n\ge 0}{sup}{g}_0(n)\le \frac{13-\sqrt{5}}{30}{\left(\frac{\sqrt{5}-1}{2}\right)}^{1/2}{S}^{3/2}+O(\epsilon )=:\mathrm{\Lambda }+O(\epsilon ),\epsilon \to 0^{+}.$

Proof

By [1], we have, for $\epsilon >0$ small enough,

5.10

${\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5=S^{\frac{3}{2}}+O(\epsilon ).$ Combining this with the estimate (2.11) and Lemma 5.2 leads to the implication that

5.11

$\begin{array}{rl}{g}_0(n)& =\frac{n^2}{2}{\parallel v_{\epsilon }\parallel }^2-\frac{n^{10}}{10}{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-\frac{n^6}{6}{\int }_B|v_{\epsilon }{|}^6\\ & \le \frac{13-\sqrt{5}}{30}{\left(\frac{\sqrt{5}-1}{2}\right)}^{1/2}{S}^{3/2}+O(\epsilon ),\end{array}$ as long as $\epsilon >0$ is positive and sufficiently small. This completes the proof. □

Similarly, by Lemma 3.1, it can be concluded that the functional Ẽ also exhibits the mountain pass geometry. As a result, one can define a corresponding structure. In particular, there exists a ${(PS)}_{c_0}$ sequence $\{v_m\}\subset H$ for Ẽ, satisfying

5.12

$\tilde{E}(v_m)\to c_0,{\parallel {\tilde{E}}^{\prime }(v_m)\parallel }_{H^{-1}}\to 0,m\to \mathrm{\infty },$ in which $c_0$ is expressed as

5.13

$c_0=\underset{\varsigma \in \tilde{\tau }}{inf}\underset{n\in [0,1]}{max}\tilde{E}(\varsigma (n)),$ and $\tilde{\tau }=\{\varsigma \in C([0,1],H):\varsigma (0)=0,\tilde{E}(\varsigma (1))<0\}$.

Lemma 5.4

Given the definition of$c_0$in (5.13), it holds that$0<c_0<\mathrm{\Lambda }$.

Proof

According to (3.3), for sufficiently small ε, it holds that

5.14

$\begin{array}{rl}\tilde{E}(n{v}_{\epsilon })& =\frac{n^2}{2}{\int }_B|\mathrm{\nabla }{v}_{\epsilon }{|}^2+{\int }_B\frac{n^p}{p}|v_{\epsilon }{|}^p-\frac{n^{10}}{10}{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-{\int }_B\frac{n^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }}|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}\\ & \le \frac{n^2}{2}{\parallel v_{\epsilon }\parallel }^2+{\int }_B\frac{n^p}{p}|v_{\epsilon }{|}^p-\frac{n^6}{7}{\int }_B|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}\\ & \le S^{3/2}{n}^2+{\int }_B\frac{n^p}{p}|v_{\epsilon }{|}^p-\frac{S^{3/2}}{14}{n}^6:=\phi (n).\end{array}$ As a result, there exists a sufficiently large $Q>0$, independent of ε ensuring $\phi (Q)=0$ and $\tilde{E}(Q{v}_{\epsilon })\le 0$ for sufficiently small ε. Therefore, we can identify some $0<n_{\epsilon }<Q$ such that

5.15

$0<\delta \le c_0\le \underset{n\in [0,Q]}{max}\tilde{E}(n{v}_{\epsilon })=\tilde{E}(n_{\epsilon }{v}_{\epsilon }).$ Let $\frac{d}{dn}\tilde{E}(n{v}_{\epsilon }){|}_{n=n_{\epsilon }}=0$, we have

5.16

$n_{\epsilon }{\parallel v_{\epsilon }\parallel }^2=n_{\epsilon }^9{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-{\int }_B{n}_{\epsilon }^{p-1}|v_{\epsilon }{|}^p+{\int }_B{n}_{\epsilon }^{5+|x{|}^{\alpha }}|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}.$ Next, combining the ideas in literature by [9], we can estimate $n_{\epsilon }^{p-2}{\int }_B|v_{\epsilon }{|}^p$. Let θ be such that $|n_{\epsilon }{v}_{\epsilon }|\le 1$ for $r\ge \theta$,

5.17

$\begin{array}{rl}{A}_{\epsilon }& =n_{\epsilon }^{p-2}{\int }_B|v_{\epsilon }{|}^p\\ & =\omega {\int }_0^{\theta }|v_{\epsilon }{|}^p{r}^2{n}_{\epsilon }^{p-2}+\omega {\int }_{\theta }^1|v_{\epsilon }{|}^p{r}^2{n}_{\epsilon }^{p-2},\end{array}$

5.18

$\begin{array}{rl}{A}_{\epsilon }& >w{\int }_0^{\theta }|v_{\epsilon }{|}^2{r}^2+w{\int }_{\theta }^1|n_{\epsilon }{v}_{\epsilon }{|}^4|v_{\epsilon }{|}^2{r}^2\\ & >w{\int }_0^{\theta }|v_{\epsilon }{|}^2{r}^2\\ & =w{\int }_0^{\epsilon }|v_{\epsilon }{|}^2{r}^2+w{\int }_{\epsilon }^{\theta }|v_{\epsilon }{|}^2{r}^2\\ & >C{\int }_0^{\epsilon }{\epsilon }^{-1}{r}^2+C{\int }_{\epsilon }^{\theta }{\epsilon }^1\\ & >C{\epsilon }^2,\end{array}$ and

5.19

$\begin{array}{rl}{A}_{\epsilon }& <w{\int }_0^{\theta }|n_{\epsilon }{v}_{\epsilon }{|}^4|v_{\epsilon }{|}^2{r}^2+w{\int }_{\theta }^1|v_{\epsilon }{|}^2{r}^2\\ & =w{\int }_0^{\theta }{n}_{\epsilon }^4|v_{\epsilon }{|}^6{r}^2+w{\int }_{\theta }^1|v_{\epsilon }{|}^2{r}^2\\ & <w{\int }_0^{\epsilon }{n}_{\epsilon }^4|v_{\epsilon }{|}^6{r}^2+w{\int }_{\epsilon }^{\theta }{n}_{\epsilon }^4|v_{\epsilon }{|}^6{r}^2+w{\int }_{\theta }^1|v_{\epsilon }{|}^2{r}^2\\ & <C{\int }_0^{\epsilon }{n}_{\epsilon }^4{\epsilon }^{-4}{r}^2+C{\int }_{\epsilon }^{\theta }{n}_{\epsilon }^4{\epsilon }^2{r}^{-4}+C{\int }_{\theta }^1{\epsilon }^{-1}\\ & <C{\epsilon }^{-1}-C{\epsilon }^2\\ & <C{\epsilon }^{-1}.\end{array}$ Thus, we have

5.20

$\begin{array}{rl}{S}^{\frac{3}{2}}+O(\epsilon )& =n_{\epsilon }^8{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5+n_{\epsilon }^{p-2}{\int }_B|v_{\epsilon }{|}^p\\ & +n_{\epsilon }^4{\int }_B|v_{\epsilon }{|}^6+n_{\epsilon }^4{\int }_B(n_{\epsilon }^{|x{|}^{\alpha }}|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}-|u_{\epsilon }{|}^6)\\ & =n_{\epsilon }^8{\int }_B{\varphi }_{v_{\epsilon }}|u_{\epsilon }{|}^5+n_{\epsilon }^4[S^{\frac{3}{2}}+O({\epsilon }^2)+A_{\epsilon }+B_{\epsilon }]\\ & =n_{\epsilon }^8{\int }_B{\varphi }_{v_{\epsilon }}|u_{\epsilon }{|}^5+n_{\epsilon }^4[S^{\frac{3}{2}}\\ & +O({\epsilon }^3)+O({\epsilon }^{-1})+O({\epsilon }^2)+O({\epsilon }^{\alpha }|log\epsilon |)+O({\epsilon }^{3/2})].\end{array}$ As shown in [9], $B_{\epsilon }=O({\epsilon }^{\alpha }|log\epsilon |)+O({\epsilon }^{3/2})$. For simplicity, we define $A=S^{\frac{3}{2}}+O(\epsilon )$, $B={\int }_B{\varphi }_{u_{\epsilon }}|u_{\epsilon }{|}^5$, $C=S^{\frac{3}{2}}+O({\epsilon }^3)+O({\epsilon }^{-1})+O({\epsilon }^2)+O({\epsilon }^{\alpha }|log\epsilon |)+O({\epsilon }^{3/2})$. Consequently, equation (5.20) takes the form $A=B{n}_{\epsilon }^8+C{n}_{\epsilon }^4$. For ε small, it is easy to see that

5.21

$n_{\epsilon }^4=\frac{\sqrt{C^2+4AB}-C}{2B}.$ Thus, for ε small enough, we get

5.22

$0<n_{\epsilon }^2<1.$ Next, we are going to deal with the supercritical term

5.23

$\begin{array}{rl}& {\int }_B\frac{n_{\epsilon }^6}{6}|v_{\epsilon }{|}^6-{\int }_B\frac{n_{\epsilon }^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }}|v_{\epsilon }{|}^{6+|x{|}^{\alpha }}\\ =& {\int }_B(\frac{n_{\epsilon }^6}{6}-\frac{n_{\epsilon }^6}{6+|x{|}^{\alpha }})|v_{\epsilon }{|}^6+{\int }_B\frac{n_{\epsilon }^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }}(|v_{\epsilon }{|}^6-|v_{\epsilon }{|}^{6+|x{|}^{\alpha }})\\ & +{\int }_B(\frac{n_{\epsilon }^6}{6+|x{|}^{\alpha }}-\frac{n_{\epsilon }^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }})|v_{\epsilon }{|}^6\\ =& B_1+B_2+B_3.\end{array}$ By [9] and (5.22), we can find

5.24

$B_1={\int }_B(\frac{n_{\epsilon }^6}{6}-\frac{n_{\epsilon }^6}{6+|x{|}^{\alpha }})|v_{\epsilon }{|}^6\le C{\epsilon }^{\alpha },$

5.25

$B_2={\int }_B\frac{n_{\epsilon }^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }}(|v_{\epsilon }{|}^6-|v_{\epsilon }{|}^{6+|x{|}^{\alpha }})\le -C{\epsilon }^{\alpha }|log\epsilon |,$

5.26

$B_3={\int }_B(\frac{n_{\epsilon }^6}{6+|x{|}^{\alpha }}-\frac{n_{\epsilon }^{6+|x{|}^{\alpha }}}{6+|x{|}^{\alpha }})|v_{\epsilon }{|}^6\le C{\int }_B(1-n_{\epsilon }^{|x{|}^{\alpha }})|v_{\epsilon }{|}^6\le C{\epsilon }^{\alpha }.$ Integrating (5.24)-(5.26) with Lemma 5.3, we conclude

5.27

$\begin{array}{rl}\tilde{E}(n_{\epsilon }{v}_{\epsilon })& =\frac{n_{\epsilon }^2}{2}{\parallel v_{\epsilon }\parallel }^2+{\int }_B\frac{n_{\epsilon }^p}{p}|v_{\epsilon }{|}^p-\frac{n_{\epsilon }^{10}}{10}{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-{\int }_B\frac{n_{\epsilon }^{6+|x{|}^a}}{6+|x{|}^a}|v_{\epsilon }{|}^{6+|x{|}^a}\\ & =\frac{n_{\epsilon }^2}{2}{\parallel v_{\epsilon }\parallel }^2-\frac{n_{\epsilon }^{10}}{10}{\int }_B{\varphi }_{v_{\epsilon }}|v_{\epsilon }{|}^5-{\int }_B\frac{n_{\epsilon }^6}{6}|v_{\epsilon }{|}^6+{\int }_B\frac{n_{\epsilon }^p}{p}|v_{\epsilon }{|}^p+B_1+B_2+B_3\\ & \le \underset{n\ge 0}{sup}g(n)+{\int }_B\frac{n_{\epsilon }^p}{p}|v_{\epsilon }{|}^p+B_1+B_2+B_3\\ & \le \frac{13-\sqrt{5}}{30}{\left(\frac{\sqrt{5}-1}{2}\right)}^{1/2}{S}^{3/2}+O(\epsilon )+C{O}_p(\epsilon )+C{\epsilon }^{\alpha }-C{\epsilon }^{\alpha }|log\epsilon |,\end{array}$ where $O_p(\epsilon )=max{\int }_B|v_{\epsilon }{|}^p$. Thus, for ε small enough, by (5.27), we get:

5.28

$c_0\le \tilde{E}(n_{\epsilon }{v}_{\epsilon })<\mathrm{\Lambda }.$ The proof is finished. □

Lemma 5.5

For a${(PS)}_{c_0}$sequence$\{v_m\}$inẼ, then there exists$v\in H$such that, up to subsequence, $\widetilde{E^{\prime }}=0$and$v_m\rightharpoonup v$.

Proof

That the sequence $\{v_m\}$ is bounded in H is established by Lemma 3.2. Thus, up to a subsequence, we can assume that $\{v_m\}$ converges weakly to v in H and that $v_m\to v$ a.e. in B. By selecting $\phi \in C_0^{\mathrm{\infty }}(B)$, we obtain

5.29

$⟨{\tilde{E}}^{\prime }(v_m),\phi ⟩={\int }_B\mathrm{\nabla }{v}_m\mathrm{\nabla }\phi +{\int }_B|v_m{|}^{p-2}{v}_m\phi -{\int }_B{\varphi }_{v_m}|v_m{|}^3{v}_m\phi -{\int }_B|v_m{|}^{4+|x{|}^{\alpha }}{v}_m\phi .$ We deduce from Lemma 5.1 that ${\varphi }_{v_m}\rightharpoonup {\varphi }_v$ holds in both H and $L^6(B)$. Thus

5.30

${\int }_B({\varphi }_{v_m}-{\varphi }_v)|v{|}^{3}v\phi \to 0,m\to \mathrm{\infty }.$ Since $v_m\to v$ a.e. in B and

5.31

${\int }_B|{\varphi }_{v_m}(|v_m{|}^3{v}_m-|v{|}^{3}v){|}^{\frac{6}{5}}\le C(|{\varphi }_{v_m}{|}_6^{\frac{6}{5}}|v_m{|}_6^{\frac{24}{5}}+|{\varphi }_{v_m}{|}_6^{\frac{6}{5}}|v{|}_6^{\frac{24}{5}})\le C,$ we have ${\varphi }_{v_m}(|v_m^3{v}_n-|v{|}^{3}v)\rightharpoonup 0$ in $L^{\frac{6}{5}}(B)$ and

5.32

${\int }_B{\varphi }_{v_m}(|v_m{|}^3{v}_m-|v{|}^{3}v)\phi \to 0,m\to \mathrm{\infty },$ which together with (5.30) ensures that

5.33

${\int }_B{\varphi }_{v_m}|v_m{|}^3{v}_m\phi \to {\int }_B{\varphi }_v|v{|}^{3}v\phi ,m\to \mathrm{\infty }.$ By Hölder inequality, for all measurable subset $D\subset B$, we get

5.34

$\begin{array}{rl}|{\int }_D(|v_m{|}^{4+|x{|}^a}{v}_m-|v{|}^{4+|x{|}^a}v)\phi |& \le {\int }_D(|v_m{|}^{5+|x{|}^a}+|v{|}^{5+|x{|}^a})|\phi |\\ & \le {\parallel |v_m{|}^{5+|x{|}^a}+|v{|}^{5+|x{|}^a}\parallel }_{L^{\frac{p(x)}{p(x)-1}}(D)}{\parallel \phi \parallel }_{L^{p(x)}(D)}.\end{array}$ Thus, by Vitali’s theorem ([17]), we have

5.35

${\int }_B|v_m{|}^{4+|x{|}^{\alpha }}{v}_m\phi \to {\int }_B|v{|}^{4+|x{|}^{\alpha }}v\phi ,m\to \mathrm{\infty }.$ Similarly, we have

5.36

${\int }_B|v_m{|}^{p-2}{v}_m\phi \to {\int }_B|v{|}^{p-2}v\phi ,m\to \mathrm{\infty }.$ Combining (5.33), (5.35) and (5.36), we conclude

5.37

$⟨{\tilde{E}}^{\prime }(v),\phi ⟩=\underset{m\to \mathrm{\infty }}{lim}⟨{\tilde{E}}^{\prime }(v_m),\phi ⟩=0.$ Hence, ${\tilde{E}}^{\prime }(v_m)=0$. The proof is completed. □

Lemma 5.6

Let$\tilde{I}(v)=\frac{1}{2}{\parallel v\parallel }^2+\frac{1}{p}{\int }_B|v{|}^p-\frac{1}{10}{\int }_B{\varphi }_v|v{|}^5-\frac{1}{6}{\int }_B|v{|}^6$, thenĨsatisfies the${(PS)}_{c_0}$condition and$0<c_0<\mathrm{\Lambda }$.

Proof

Given a sequence $\{v_m\}$ with

5.38

$\tilde{I}(v_m)\to c_0,{\tilde{I}}^{\prime }(v_m)\to 0\mathrm{a}\mathrm{s}m\to \mathrm{\infty }.$ Using the same reasoning as in Lemma 3.2, we immediately obtain that $\{v_m\}$ is bounded in H. We may assume, after taking a subsequence, that $v_m\to v$ in H for some $v\in H$. In view of Lemma 5.5, it follows without difficulty that ${\tilde{I}}^{\prime }(u)=0$, hence

5.39

$\begin{array}{rl}\tilde{I}(v)& =\tilde{I}(v)-\frac{1}{6}⟨{\tilde{I}}^{\prime }(v),v⟩\\ & =\frac{1}{3}{\parallel v\parallel }^2+(\frac{1}{p}-\frac{1}{6}){\int }_B|v{|}^p+\frac{1}{15}{\int }_B{\varphi }_v|v{|}^5\ge 0.\end{array}$ Next, let $u_m=v_m-v$, we have

5.40

${\parallel v_m\parallel }^2={\parallel u_m\parallel }^2+{\parallel v\parallel }^2+o(1).$ As a consequence of the Brézis-Lieb Lemma in [18, 19], one obtains

5.41

${\int }_B|v_m{|}^{6}dx={\int }_B|u_m{|}^{6}dx+{\int }_B|v{|}^{6}dx+o(1),$

5.42

${\int }_B|v_m{|}^{p}dx={\int }_B|u_m{|}^{p}dx+{\int }_B|v{|}^{p}dx+o(1),$

5.43

${\int }_B{\varphi }_{v_m}|v_m{|}^5={\int }_B{\varphi }_{u_m}|u_m{|}^5+{\int }_B{\varphi }_v|v{|}^5+o(1).$ These four equalities imply that

5.44

$\begin{array}{rl}{c}_0-\tilde{I}(v)& =\tilde{I}(v_m)-\tilde{I}(v)+o(1)\\ & =\frac{1}{2}{\parallel v_m\parallel }^2-\frac{1}{2}{\parallel v\parallel }^2-\frac{1}{10}{\int }_B{\varphi }_{v_m}|v_m{|}^5+\frac{1}{10}{\int }_B{\varphi }_v|v{|}^5\\ & +\frac{1}{p}{\int }_B|v_m{|}^p-\frac{1}{6}{\int }_B|v{|}^p+o(1)-\frac{1}{6}{\int }_B|v_m{|}^6+\frac{1}{6}{\int }_B|v{|}^6+o(1)\\ & =\frac{1}{2}{\parallel u_m\parallel }^2+\frac{1}{p}{\int }_B|u_m{|}^p-\frac{1}{10}{\int }_B{\varphi }_{u_m}|u_m{|}^5-\frac{1}{6}{\int }_B|u_m{|}^6+o(1),\end{array}$ and similarly,

5.45

$\begin{array}{rl}o(1)& =⟨{\tilde{I}}^{\prime }(v_m),v_m⟩-⟨{\tilde{I}}^{\prime }(v),v⟩\\ & ={\parallel v_m\parallel }^2-{\parallel v\parallel }^2+{\int }_B|v_m{|}^p-{\int }_B|v{|}^p-{\int }_B{\varphi }_{v_m}|v_m{|}^5\\ & +{\int }_B{\varphi }_v|v{|}^5-{\int }_B|v_m{|}^6+{\int }_B|v{|}^6\\ & ={\parallel u_m\parallel }^2+{\int }_B|u_m{|}^p-{\int }_B{\varphi }_{u_m}|u_m{|}^5-{\int }_B|u_m{|}^6+o(1).\end{array}$ We will prove that $\parallel u_m\parallel \to 0$. If this is not true, then there exists a subsequence, still labeled as $\{u_m\}$, such that ${\parallel u_m\parallel }^2\to l>0$. For simplicity, let $a_m={\int }_B{\varphi }_{u_m}|u_m{|}^5$ and $b_m={\int }_B|u_m{|}^6$ and $d_m={\int }_B|u_m{|}^p$. Without loss of generality, assume $a_m\to a_0$ and $b_m\to b_0$ and $d_m\to d_0$ as $m\to \mathrm{\infty }$. Observe that

5.46

$\begin{array}{rl}{\int }_B|u_m{|}^6& ={\int }_B\mathrm{\nabla }{\varphi }_{u_m}\mathrm{\nabla }|u_m|\\ & \le \frac{{\epsilon }^2}{2}{\int }_B|\mathrm{\nabla }|u_m|{|}^2+\frac{1}{2{\epsilon }^2}{\int }_B|\mathrm{\nabla }{\varphi }_{u_m}{|}^2\\ & =\frac{1}{2{\epsilon }^2}{\int }_B{\varphi }_{u_m}|u_m{|}^5+\frac{{\epsilon }^2}{2}{\int }_B|\mathrm{\nabla }{u}_m{|}^2,\end{array}$ then as $m\to \mathrm{\infty }$, we conclude that

5.47

$b_0\le \frac{1}{2{\epsilon }^2}{a}_0+\frac{{\epsilon }^2}{2}l.$ Combining with (5.45), and taking ${\epsilon }^2=\frac{\sqrt{5}-1}{2}$, we get

5.48

$a_0\ge \frac{3-\sqrt{5}}{2}l+\frac{\sqrt{5}-1}{\sqrt{5}}{d}_0,$ from which, using (5.39), (5.44), and (5.45), we obtain that

5.49

$\begin{array}{rl}{c}_0\ge c_0-\tilde{I}(u)& =\frac{2}{5}{a}_0+\frac{1}{3}{b}_0+(\frac{1}{p}-\frac{1}{2})d_0+o(1)\\ & =\frac{1}{3}l+\frac{1}{15}{a}_0+(\frac{1}{p}-\frac{1}{6})d_0+o(1)\\ & \ge \frac{13-\sqrt{5}}{30}l+(\frac{1}{p}-\frac{1}{6}+\frac{\sqrt{5}-1}{15\sqrt{5}})d_0+o(1).\end{array}$ On the other hand, by (5.1) and (5.45), we have

5.50

$l<S^{-6}{l}^5+S^{-3}{l}^3.$ Therefore, $l^2>\frac{-1+\sqrt{5}}{2}{S}^3$. Combining this with (5.16) leads to $c_0\ge \mathrm{\Lambda }$, which results in a contradiction. Hence, $u_m\to 0$ strongly in H, or equivalently, $v_m\to v$ in H as $m\to \mathrm{\infty }$. This completes the proof. □

Lemma 5.7

([18])

Let$g:B\times \mathbb{R}\to \mathbb{R}$be a Carathéodory function that satisfies the following condition for almost every$x\in B:$

5.51

$|g(x,v)|\le a(x)(1+|v|).$If$a\in L^{\frac{3}{2}}(B)$with$a\ge 0$, and$v\in H_0^1(B)$is a weak solution to the equation$-\mathrm{\Delta }v=$$g(x,v)$inB, then$v\in L^{\theta }(B)$for every$\theta <\mathrm{\infty }$.

Proof of Theorem 1.6

Based on the results of Lemmas 5.4 and 5.5, we can conclude that the functional $\tilde{E}(v)$ satisfies the ${(PS)}_{c_0}$ condition. Combining this with the application of Theorem 1.2, we can obtain that equation (1.5) admits a nonnegative, nontrivial ground state solution $v\in H$, satisfying

5.52

$-\mathrm{\Delta }v={\varphi }_v|v{|}^{3}v+v^{5+|x{|}^{\alpha }}-v^{p-1}\text{in}B.$ We define:

5.53

$\tilde{g}(v(x))={\varphi }_v|v{|}^{3}v+v^{5+|x{|}^{\alpha }}-v^{p-1},x\in B.$ By applying Lemma 2.2 and considering $\frac{8}{3}<p<6$, we deduce that ${\int }_B|v{|}^{-3+\frac{3}{2}p}\le C$ and ${\int }_B{v}^{6+\frac{3}{2}|x{|}^a}\le C$. Given that ${\varphi }_v\in D^{1,2}(B)$, it follows that ${\varphi }_v\in L^6(B)$. Furthermore, we can easily verify that $|{\varphi }_v{|}^{\frac{3}{2}}\in L^4(B)$ and $|v{|}^{\frac{9}{2}}\in L^{\frac{4}{3}}(B)$. Consequently, by applying Hölder’s inequality, we arrive at the conclusion that ${\varphi }_v|v{|}^3\in L^{\frac{3}{2}}(B)$, which leads to the equation:

5.54

$a=\frac{\tilde{g}(v)}{1+|v|}\in L^{\frac{3}{2}}(B).$ Moving forward, Lemma 5.7 allows us to establish that $v\in L^{\theta }(B)$ for all $1<q<\mathrm{\infty }$. As a result, $\tilde{g}(v)\in L^{\theta }(B)$ holds for any $1<\theta <\mathrm{\infty }$. By applying the Calderón-Zygmund inequality and the $L^{\theta }$ estimate provided in [20, 21], we deduce that $v\in W^{2,\theta }(B)$, which implies v∈ $C^{1,\gamma }(B)$ by the Sobolev embedding theorem for any $\gamma \in (0,1)$. Additionally, we infer that $v(x)>0$ for all $x\in B$ by the Harnack inequality [22]. This completes the proof. □

Acknowledgements

The authors would like to thank the referees for their useful suggestions which have significantly improved the paper.

Author contributions

Y.L. , X.C. and Q.Z. provided equal contribution to this research article. All authors read and approved the final manuscript.

Funding information

This work is supported by the National Natural Science Foundation of China (No. 11961014) and Guangxi Natural Science Foundation (2021GXNSFAA196040).

Data availability

No datasets were generated or analysed during the current study.

Declarations

Ethics approval and consent to participate

Not applicable.

Competing interests

The authors declare no competing interests.

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