On Schrödinger-Poisson equations with a critical nonlocal term

1.   Introduction

The Schrödinger-Poisson equation

arises in a physical context. It is introduced while describing the interaction of a charged particle with an electrostatic field. More details can be found in ^[3]. Also, it appears in other fields like semiconductor theory, nonlinear optics, and plasma physics. The readers may refer to ^[18] and the references therein for further discussion. When $V \equiv 1$, $\lambda = 1$, and $f(x, u) = |u|^{p-2}u$, problem (1.1) has been studied sufficiently. We refer to ^[9] for $p \le 2$ and $p \ge 6$, ^[7,8,10] for $4 \le p < 6$, ^[2] for $3 < p < 6$, and ^[22] for $2 < p < 6$. In ^[31], the authors obtained an axially symmetric solution of the following Schrödinger-Poisson equation in $\mathbb{R}^2$:

where $f \in C(\mathbb{R}, \mathbb{R})$, and $V$ and $K$ are both axially symmetric functions. In ^[4,5], the almost necessary and sufficient condition (Berestycki-Lions type condition) for the existence of ground state solutions of the problem

was given by ^[4] when $N = 2$ and ^[5] when $N \ge 3$. Precisely, they assumed $g$ satisfies the following conditions:

$(g_{1}) \;g(s) \in C(\mathbb{R}, \mathbb{R})$ is continuous and odd.

$(g_{2})$ $- \infty < \liminf_{s \rightarrow 0}\frac{g(s)}{s} \le \limsup_{s \rightarrow 0}\frac{g(s)}{s} = -a < 0$ for $N \ge 3$, and $\lim_{s \rightarrow 0}\frac{g(s)}{s} = -a < 0$ for $N = 2$.

$(g_{3})$ When $N \ge 3$, $\limsup_{s \rightarrow \infty}\frac{g(s)}{|s|^\frac{N+2}{N-2}} \le 0$; when $N = 2$, for any $\alpha > 0$ there exists $C_\alpha > 0$ such that $g(s) \le C_\alpha \mathrm{exp}(\alpha s^2)$ for all $s > 0$.

$(g_{4})$ There exists $\xi_0 > 0$ such that $G(\xi_0) > 0$, where $G(\xi_0) = \int_{0}^{\xi_0}g(s) \mathrm{d} s$.

When $g$ satisfies the above Berestycki-Lions type condition, the authors in ^[19] studied the problem

By using a truncation technique in ^[14], they proved that the problem admits a nontrivial positive radial solution for $q > 0$ small. For the critical case, the authors in ^[30] studied the existence of positive radial solutions of the problem

where $q \in (2, 4)$, $\mu > 0$, and $Q$ and $K$ are radial functions satisfying the following conditions:

$(h_1)$ $K \in C (\mathbb{R}^3, \mathbb{R})$, $\lim_{|x| \rightarrow \infty}K(x) = K_\infty \in (0, \infty)$ and $K(x) \ge K_\infty$ for $x \in \mathbb{R}^3$.

$(h_2)$ $Q \in C (\mathbb{R}^3, \mathbb{R})$, $\lim_{|x| \rightarrow \infty}Q(x) = Q_\infty \in (0, \infty)$ and $Q(x) \ge Q_\infty$ for $x \in \mathbb{R}^3$.

$(h_3)$ $|K(x)-K(x_0)| = o(|x-x_0|^\alpha)$, where $1 \le \alpha < 3$ and $K(x_0) = \max_{\mathbb{R}^3}K(x)$.

In ^[25], we studied (1.1) with $f$ satisfying the following Berestycki-Lions type condition with critical growth:

$(f_1)$ $f \in C (\mathbb{R}, \mathbb{R})$ is odd, $\lim_{u \rightarrow 0+}\frac{f(u)}{u} = 0$ and $\lim_{u \rightarrow +\infty}\frac{f(u)}{u^5} = K > 0$.

$(f_2)$ There exist $D > 0$ and $2 < q < 6$ such that $f(u) \ge K u^5+ D u^{q-1}$ for $u \ge 0$.

$(f_3)$ There exists $\theta > 2$ such that $\frac{1}{\theta}f(u)u-F(u)\ge 0$ for all $u \in \mathbb{R}^+$, where $F(u) = \int_0^u f(s) \mathrm{d}s$.

When $\lambda > 0$ is small, we obtained positive radial solutions for $q \in (4, 6)$, or $q \in (2, 4]$ with $D > 0$ large. In ^[29], the authors removed $(f_3)$ by using a local deformation argument in ^[6]. It should be pointed out that, in ^[25,29], the problems were considered in a radial setting.

When the nonlocal term is of critical growth, that is, $u^2$ is replaced by $u^5$, problem (1.1) is reduced to

These kind of equations are closely related with the Choquard-Pekar equation, which was proposed in ^[20] to study the quantum theory of a polaron at rest. Since the critical nonlocal term may cause the loss of compactness, problem (1.2) is quite different from the standard Schrödinger-Poisson equation. In ^[16], the authors considered the equation

where $b \ge 0$, $q \in \mathbb{R}$, and the subcritical nonlinearity $f$ satisfies the following conditions:

$(H_1)$ $f \in C (\mathbb{R}^+, \mathbb{R}^+)$ and $\lim_{u \rightarrow 0+}\frac{f(u)}{bu+u^5} = 0$.

$(H_2)$ $\lim_{u \rightarrow \infty}\frac{f(u)}{u^5} = 0$.

$(H_3)$ There is a function $z \in H_r^1(\mathbb{R}^3)$ such that $\int_{\mathbb{R}^3}F(z) > b \int_{\mathbb{R}^3}z^2$, where $F(z) = \int_0^z f(t) \mathrm{d}t$.

$(H_4)$ There exist $r \in (4, 6)$, $A > 0$, $B > 0$ such that $F(t) \ge A t^r-B t^2$ for $t \ge 0$.

For $q \ge 0$, they proved that there exists $q_0 > 0$ such that for $q \in [0, q_0)$, and problem (1.2) has at least one positive radially symmetric solution if $(H_1)$–$(H_3)$ hold. For $q = -1$, they proved that problem (1.2) has at least one positive radially symmetric solution if $(H_1)$–$(H_2)$ and $(H_4)$ hold. In ^[17], the authors studied the existence, nonexistence, and multiplicity of positive radially symmetric solutions of the equation

where $\lambda \in \mathbb{R}$, $\mu \ge 0$, and $p \in [1, 5]$. In ^[15], the author obtained positive solutions of the following equation with subcritical growth:

where $V$, $K$, and $f$ are asymptotically periodic functions of $x$. If the nonlinearity is of critical growth, the author in ^[12] studied ground state solutions of the equation

where $V(x) = 1+x_1^2+x_2^2$ with $x = (x_1, x_2, x_3) \in \mathbb{R}^3$ and $f$ is an appropriate nonlinear function.

In this paper, we study the following Schrödinger-Poisson equation with a critical nonlocal term:

where $(u^+)^5$ is a critical term with $u^+: = \max\{u, 0\}$ and $\lambda > 0$ is a parameter. When we study (1.7) for the case $\lambda < 0$, the boundedness of the Palais-Smale sequence can be derived directly. However, for the case $\lambda > 0$, the problem is quite different. Since the term $\int_{\mathbb{R}^3}\phi_u |u|^5 \mathrm{d}x$ is homogeneous of degree $10$, the corresponding Ambrosetti-Rabinowitz condition on $f$ is the following:

$(f')$ There exists $\theta \ge 10$ such that $tf(t)- \theta F(t) \ge 0$ for any $t \in \mathbb{R}$.

Obviously, this condition is not suitable for the problem in dimension three. To solve the problem, the authors in ^[16] used a truncation technique in ^[14]. However, the argument is invalid when we study non-autonomous problems in a non-radial setting. Motivated by the above considerations, we first study the non-autonomous problem (1.7) in a non-radial setting, where the nonlinearity satisfies the Berestycki-Lions type condition with critical growth. We assume $V$ satisfies the following conditions:

$(V_{1})$ $V \in C^1(\mathbb{R}^3, \mathbb{R})$ and $\inf_{\mathbb{R}^3} V : = V_0 > 0$.

$(V_{2})$ $V(x) \le \lim_{|x| \rightarrow \infty}V(x): = V_\infty$ for all $x \in \mathbb{R}^3$ and the inequality is strict in a set of positive Lebesgue measure.

$(V_{3})$ There exists $\theta \in (0, 1)$ such that $\frac{t^3}{2}V(t x)-\frac{t^3}{2}V(x)-\frac{t^3-1}{6}(\nabla V(x), x) \le \frac{\theta(t-1)^2(t+2)}{24 |x|^2}$ for $x \in \mathbb{R}^3 \setminus \{0\}$ and $t \in \mathbb{R}^+$.

The result is as follows.

Theorem 1.1. Assume that $(V_1)$–$(V_3)$ and $(f_1)$–$(f_2)$ hold. Then there exists $\lambda_0 > 0$ such that for $\lambda \in (0, \lambda_0)$, problem (1.7) has a positive solution $(u_\lambda, \phi_\lambda)$. Moreover, as $\lambda \rightarrow 0$, $(u_\lambda, \phi_\lambda) \rightarrow (u, 0)$ in $H^1(\mathbb{R}^3) \times D^{1, 2}(\mathbb{R}^3)$, where $u$ is a ground state solution of the following limiting equation:

When $V \equiv 1$, problem (1.7) is reduced to the following equation:

Then we have the following result.

Corollary 1.1. Assume that $(f_1)$–$(f_2)$ hold. Then there exists $\lambda_0 > 0$ such that for $\lambda \in (0, \lambda_0)$, problem (1.9) has a positive solution $(u_\lambda, \phi_\lambda)$. Moreover, as $\lambda \rightarrow 0$, $(u_\lambda, \phi_\lambda) \rightarrow (u, 0)$ in $H^1(\mathbb{R}^3) \times D^{1, 2}(\mathbb{R}^3)$, where $u$ is a ground state solution of the following limiting equation:

Remark 1.1. Corollary 1.1 is still valid if we replace $(f_2)$ by $(H_3)$. So, we generalize the result in ^[16] to the critical case.

In the next, we consider the case that $\mathrm{int}V^{-1}(0)$ is contained in a spherical shell. We assume the following conditions.

$(V_{1}')$ $V \in C(\mathbb{R}^3, \mathbb{R})$ and $V(x) = V(|x|)$ for all $x \in \mathbb{R}^3$.

$(V_{2}')$ $V(x) = 0$ for $x \in \wedge_1$ and there exists $V_0 > 0$ such that $V(x) \ge V_0$ for $x \notin \wedge_2$, where $\wedge_1: = \{x \in \mathbb{R}^3: r_1 < |x| < r_2\}$ and $\wedge_2: = \{x \in \mathbb{R}^3: R_1 < |x| < R_2\}$ with $0 < R_1 < r_1 < r_2 < R_2$.

$(f_3')$ There exists $\theta > 2$ such that $\frac{f(u)}{u^{\theta-1}}$ is increasing for all $u > 0$.

To the best of our knowledge, there are no related results even for the case $\lambda = 0$. We must face several difficulties. A main difficulty is how to get the compactness. In ^[11], del Pino and Felmer developed a penalization approach to deal with singularly perturbed problems. Motivated by ^[11], instead of studying (1.7) directly, we turn to consider a modified problem. By studying the influence of the potential on the compactness and the behavior of positive solutions at infinity, we solve the problem. When $\lambda > 0$, we have to prove the boundedness of the Palais-Smale sequence for the modified problem. This is another difficulty. Now we state the result.

Theorem 1.2. Assume that $(V_1')$–$(V_2')$, $(f_1)$–$(f_2)$, and $(f_3')$ hold. Then there exists $R' > 0$ such that for $R_1 > R'$, there exists $\lambda' > 0$ such that problem (1.7) has a positive solution $(u_\lambda, \phi_\lambda)$ for $\lambda \in (0, \lambda')$. Moreover, as $\lambda \rightarrow 0$, $(u_\lambda, \phi_\lambda) \rightarrow (u, 0)$ in $H^1(\mathbb{R}^3) \times D^{1, 2}(\mathbb{R}^3)$, where $u$ is a positive solution of (1.8).

Notations.

$\bullet$ Denote $H^1: = H^1(\mathbb{R}^3)$ the Hilbert space with the norm $\|u\|_{H^1}^2: = \int_{\mathbb{R}^3}(|\nabla u|^2+|u|^2) \mathrm{d} x$.

$\bullet$ Denote $D^{1, 2}: = D^{1, 2}(\mathbb{R}^3) = \left\{u \in L^{6}(\mathbb{R}^3): |\nabla u| \in L^2(\mathbb{R}^3)\right\}$ the Sobolev space with the norm $\|u\|^2_{D^{1, 2}}: = \int_{\mathbb{R}^3}|\nabla u|^2 \mathrm{d} x$.

$\bullet$ Denote the norm $\|u\|_s: = \big(\int_{\mathbb{R}^3}|u|^s\mathrm{d}x\big)^{\frac{1}{s}}$, where $2 \le s < \infty$.

$\bullet$ Denote $C$ a universal positive constant (possibly different).

2.   Proof of Theorem 1.1

Without loss of generality, we assume that $f(u) = 0$ for $u \le 0$. Define the best Sobolev constant

By the Lax-Milgram theorem, for any $u \in D^{1, 2}$ there exists a unique $\phi_u \in D^{1, 2}$ such that $- \Delta \phi_u = |u|^5$. The function $\phi_u$ has the following properties.

Lemma 2.1. (^[16])

($i$) $\phi_u \ge 0$, $\phi_{tu} = |t|^5 \phi_u$ and $\phi_{u(\frac{.}{t})} = t^2\phi_{u}(\frac{.}{t})$ for all $t > 0$.

($ii$) $\|\phi_u\|_{D^{1, 2}} \le S^{-\frac{1}{2}}\|u\|_6^5$.

($iii$) If $u_n \rightharpoonup u$ weakly in $L^6(\mathbb{R}^3)$ and $u_n \rightarrow u$ a.e. in $\mathbb{R}^3$, then $\phi_{u_n} \rightharpoonup \phi_u$ weakly in $D^{1, 2}$ up to a subsequence.

($iv$) Let $J(u) = \int_{\mathbb{R}^3}\phi_u|u|^5\mathrm{d} x$, where $u \in D^{1, 2}$. If $u_n \rightharpoonup u$ weakly in $L^6(\mathbb{R}^3)$ and $u_n \rightarrow u$ a.e. in $\mathbb{R}^3$, then

Define $X : = \left\{u \in H^{1}: \ \int_{\mathbb{R}^{3}}V(x)|u|^{2} \mathrm{d} x < \infty\right\}$ as the Hilbert space with the norm $\|u\| = \left(\int_{\mathbb{R}^{3}}|\nabla u|^{2} + V(x) |u|^{2} \mathrm{d} x\right)^{\frac{1}{2}}$. Define the functional on $X$ by

where $F(u): = \int_0^u f(s) \mathrm{d}s$. Obviously, the functional $I_\lambda$ is of class $C^1$ and critical points of $I_\lambda$ are weak solutions of (1.7). Let

If $I_0'(u) = 0$, by the arguments in ^[16,21,24] we can derive the Pohozǎev type identity $J_0(u) = 0$, where

When $V \equiv V_\infty$, problem (1.8) is reduced to the following equation:

The functional associated with (2.3) is

Define

Define

where $\Gamma: = \left\{\gamma \in C ([0, 1], H^1): \gamma(0) = 0, I_0^\infty(\gamma(1)) < 0\right\}$.

Lemma 2.2. Assume that $(V_1)$–$(V_3)$ hold. Then, for all $x \in \mathbb{R}^3 \setminus\{0\}$,

Proof. Let

By $(V_3)$, we get $g(0) \le 0$. Then $(\nabla V(x), x) \le \frac{\theta}{2 |x|^2}$ for all $x \in \mathbb{R}^3 \setminus\{0\}$. By $(V_2)$–$(V_3)$, we get $\lim_{t \rightarrow +\infty}\frac{g(t)}{t^3} \le 0$. Then $(\nabla V(x), x) \ge 3 V_\infty - 3 V(x)-\frac{\theta}{4 |x|^2}$ for all $x \in \mathbb{R}^3 \setminus\{0\}$.□

Theorem 2.1. (^[13]) Let $X$ be a Banach space equipped with a norm $\|.\|_{X}$ and let $J \subset \mathbb{R^{+}}$ be an interval. We consider a family $(I_{\mu})_{\mu \in J}$ of $C^{1}$ -functionals on $X$ of the form

where $B(u) \ge 0$ for all $u \in X$, and either $A(u) \rightarrow + \infty$ or $B(u) \rightarrow + \infty$ as $\|u\|_{X} \rightarrow \infty$. We assume there are two points $v_{1}$, $v_{2}$ in $X$ such that

where $\Gamma: = \{\gamma \in C ([0, 1], X); \gamma(0) = v_{1}, \gamma(1) = v_{2}\}$. Then, for almost every $\mu \in J$, there is a sequence $\{v_{n}\} \subset X$ such that $\{v_{n}\}$ is bounded, $I_{\mu} (v_{n}) \rightarrow c_{\mu}$, and $I_{\mu}' (v_{n}) \rightarrow 0$ in $X^{-1}$. Moreover, the map $\mu \rightarrow c_{\mu}$ is continuous from the left-hand side.

Lemma 2.3. Assume that $(V_1)$–$(V_3)$ and $(f_1)$–$(f_2)$ hold. Then $m_0 \in (0, m_0^\infty)$ is attained by a positive function.

Proof. Let $\mu_0 \in (0, 1)$. Define the functionals on $X$ by

where $\mu \in [\mu_0, 1]$. Similar to the argument in ^[27], we can use Theorem 2.1 to derive that for almost every $\mu \in [\mu_0, 1]$ there exists a positive function $u_\mu \in X$ such that $c_\mu = I_{0, \mu}(u_\mu)$ and $I_{0, \mu}'(u_\mu) = 0$.

Choose $\mu_n \uparrow 1$ such that $I_{0, \mu_n}(u_{\mu_n}) = c_{\mu_n}$ and $I_{0, {\mu_n}}'(u_{\mu_n}) = 0$. Then $u_{\mu_n}$ satisfies the following Pohozǎev type identity:

By (2.7), Lemma 2.2, and the Hardy inequality,

and

By (2.7)–(2.9) and $(f_1)$, we get that $\|u_{\mu_n}\|$ is bounded. Then $I_{0}(u_{\mu_n}) \rightarrow c_1$ and $I_0'(u_{\mu_n}) \rightarrow 0$. Similar to the argument in ^[27], we get that there exists a positive function $u_0 \in X$ such that $u_{\mu_n} \rightarrow u_0$ in $X$, $I_0(u_0) = c_1$, and $I_0'(u_0) = 0$. Moreover, $0 < m_0 \le c_1$ is attained. By ^[28], we get that $m_0^\infty = c_0^\infty$ is attained by a positive function $u_0^\infty$. Then by $(V_1)$-$(V_2)$ and a standard argument, we have $c_1 < c_0^\infty$.□

Let $S_0$ be the set of ground states of (1.8). By Lemma 2.3, we have $S_0 \ne \emptyset$.

Lemma 2.4. Assume that $(V_1)$–$(V_3)$ and $(f_1)$–$(f_2)$ hold. Then $S_0$ is compact in $X$.

Proof. By Lemma 2.3, for any $\{u_n\} \subset S_0$ we have $I_0(u_n) = m_0$, $I_0'(u_n) = 0$, and $J_0(u_n) = 0$. Moreover, $\|u_n\|$ is bounded. Assume that $u_{n} \rightharpoonup u_0$ weakly in $X$. Then $I_0'(u_0) = 0$. Let $v_n = u_n-u_0$. By $(V_1)$, $(f_1)$, and the Brezis-Lieb lemma in ^[24], we have

Since $v_n \rightharpoonup 0$ weakly in $X$, by the Lions Lemma in ^[24], $v_n \rightarrow 0$ in $L^t(\mathbb{R}^3)$ for any $t \in (2, 6)$, or there exists $\{y_n^1\} \subset \mathbb{R}^3$ with $|y_n^1| \rightarrow \infty$ such that $v_n^1: = v_n(.+y_n^1) \rightharpoonup v^1 \ne 0$ weakly in $X$. If $v_n \rightarrow 0$ in $L^t(\mathbb{R}^3)$ for any $t \in (2, 6)$, by $(f_1)$ we get $\int_{\mathbb{R}^3}F(v_n)\mathrm{d}x = o_n(1)$ and $\int_{\mathbb{R}^3}f(v_n)v_n\mathrm{d}x = o_n(1)$. Then

By $I_0'(u_0) = 0$, we have $J_0(u_0) = 0$. By Lemma 2.2 and the Hardy inequality, we get $I_0(u_0) \ge 0$. Assume that $\lim_{n \rightarrow \infty}\|v_n\|_6^6 = l$. If $l > 0$, by (2.11) and the definition of $S$, we get $l \ge S^{\frac{3}{2}}$. Then $m_0 \ge \frac{1}{3}S^{\frac{3}{2}}$, a contradiction. So, $l = 0$, from which we get $v_n \rightarrow 0$ in $X$. If there exists $\{y_n^1\} \subset \mathbb{R}^3$ with $|y_n^1| \rightarrow \infty$ such that $v_n^1: = v_n(.+y_n^1) \rightharpoonup v^1 \ne 0$ weakly in $X$, similar to the argument of Lemma 2.6 in ^[27] there exist $k \in \mathbb{N} \cup \{0\}$, $\{y_{n}^i\} \subset \mathbb{R}^3$ and $v^i \in X$ for $1 \leq i \leq k$ such that

Since $(I_0^{\infty})'(v^i) = 0$, we have $I_0^\infty(v^i) \ge m_0^\infty$. If $k \ge 1$, by $I_0(u_0) \ge 0$ and (2.12) we get $m_0 \ge m_0^\infty$, a contradiction. So, $k = 0$, from which we get $u_n \rightarrow u_0$ in $X$.□

Lemma 2.5. Assume that $(V_1)$–$(V_3)$ and $(f_1)$ hold. If $u \in S_0$, then $m_0 = I_0(u) > I_0(u(\frac{.}{t}))$ for all $t \in [0, 1) \cup (1, +\infty)$. Also, there exists $t_0 > 1$ independent of $u \in S_0$ such that $I_0(u(\frac{.}{t_0})) \le -2$.

Proof. By $u \in S_0$, we have $J_0(u) = 0$. Then

By $(V_3)$ and the Hardy inequality, we get $I_0(u) > I_0(u(\frac{.}{t}))$ for all $t \ne 1$. By Lemma 2.2 and the Hardy inequality,

Since $J_0(u) = 0$, by $(f_1)$ and (2.14) there exists $\varrho > 0$ independent of $u \in S_0$ such that $\|\nabla u\|_2^2 \ge \varrho$. So, by $(V_3)$, the Hardy inequality, and (2.13) we get there exists $t_0 > 1$ independent of $u \in S_0$ such that $I_0(u(\frac{.}{t_0})) \le -2$.□

Lemma 2.6. Assume that $(V_1)$–$(V_3)$ and $(f_1)$ hold. Then there exist $\lambda_1$, $M_0 > 0$ independent of $u \in S_0$ such that $I_\lambda(u(\frac{.}{t_0})) \le -1$, $\max_{t \in [0, 1]}\|u(\frac{.}{t t_0})\| \le M_0$ and $\|u\| \le M_0$ for all $\lambda \in [0, \lambda_1]$ and $u \in S_0$.

Proof. If $u \in S_0$, then $m_0 = I_0(u)$ and $J_0(u) = 0$. By the Hardy inequality and Lemma 2.2, we have $m_0 \ge \frac{1-\theta}{3}\|\nabla u\|_2^2$. Together with (2.14), $J_0(u) = 0$, and $(f_1)$, we derive that there exists $\sigma_1 > 0$ independent of $u \in S_0$ such that $\|u\|_{H^1} \le \sigma_1$. We note that

Together with $(V_1)$ and $\|u\|_{H^1} \le \sigma_1$, we get

By Lemma 2.1, we have

By Lemma 2.5 and (2.17), we derive that there exists $\lambda_1 > 0$ independent of $u \in S_0$ such that $I_\lambda(u(\frac{.}{t_0})) \le -1$ for $\lambda \in (0, \lambda_1)$ and $u \in S_0$.□

Choose $U_0 \in S_0$. Define

where $G_0: = \left\{g \in C ([0, 1], X): g(0) = 0, g(1) = U_0\left(\frac{.}{t_0}\right)\right\}$ and $\lambda \in (0, \lambda_1)$. Define

Lemma 2.7. $\lim_{\lambda \rightarrow 0}b_\lambda = \lim_{\lambda \rightarrow 0}B_\lambda = m_0$.

Proof. By (2.17) and Lemmas 2.5–2.6, we get

Then $\limsup_{\lambda \rightarrow 0}b_\lambda \le \limsup_{\lambda \rightarrow 0}B_\lambda \le m_0$. On the other hand, for any $g \in G_0$,

where $b_0: = \inf_{g \in G_0} \max_{t \in [0, 1]}I_0(g (t))$. Then $b_\lambda \ge b_0$. By Lemma 2.6, there exists $\mu_0 \in (0, 1)$ such that $I_{0, \mu}(g(1)) \le -\frac{1}{2}$ for $\mu \in (\mu_0, 1)$. Define

By repeating the proof of Lemma 2.3, we get that $c_\mu$ is a critical value. Moreover, we can prove that $b_0$ is a critical value. Then $b_0 \ge m_0$. So, $\liminf_{\lambda \rightarrow 0}b_\lambda \ge m_0$.□

For $\eta$, $d > 0$, define $I_\lambda^\eta: = \{u \in X: I_\lambda(u) \le \eta\}$ and $S_0^d: = \{u \in X: \inf_{v \in S_0}\|u-v\| \le d\}$.

Lemma 2.8. Let $\{u_{\lambda_i}\} \subset S_0^d$ with $\lim_{i \rightarrow \infty}\lambda_i = 0$ be such that $\lim_{i \rightarrow \infty}I_{\lambda_i}(u_{\lambda_i}) \le m_0$ and $\lim_{i \rightarrow \infty}I_{\lambda_i}'(u_{\lambda_i}) = 0$. Then for $d > 0$ small, there exists $u_0 \in S_0$ such that $u_{\lambda_i} \rightarrow u_0$ in $X$ up to a subsequence.

Proof. By the proof of Lemma 2.5, there exists $\varrho > 0$ independent of $u \in S_0$ such that $\|u\|^2 \ge \varrho$ for $u \in S_0$. Since $\{u_{\lambda_i}\} \subset S_0^d$, by choosing $d > 0$ small we get $\|u_{\lambda_i}\|^2 \ge \frac{\varrho}{2}$. By Lemma 2.4, we have that $\|u_{\lambda_i}\|$ is bounded. Then $\lim_{i \rightarrow \infty}I_0(u_{\lambda_i}) \le m_0$ and $\lim_{i \rightarrow \infty}I_0'(u_{\lambda_i}) = 0$. By the argument of Lemma 2.4, there exists $u_0 \in X$ such that $u_{\lambda_i} \rightarrow u_0$ in $X$ up to a subsequence. So, $\|u_0\|^2 \ge \frac{\varrho}{2}$, $I_0(u_0) \le m_0$ and $I_0'(u_0) = 0$, which implies that $u_0 \in S_0$.□

Lemma 2.9. Let $d > 0$. Then there exists $\eta > 0$ such that for small $\lambda > 0$, $I_\lambda(\gamma(t)) \ge b_\lambda-\eta$ implies that $\gamma(t) \in S_0^{\frac{d}{2}}$, where $\gamma(0) = 0$ and $\gamma(t) = U_0(\frac{.}{t t_0})$ for $t \in (0, 1]$.

Proof. By Lemma 2.5, if $\gamma(t) \notin S_0^{\frac{d}{2}}$, then there exists $\delta > 0$ such that $|t t_0 -1| \ge \delta$. Moreover, there exists $\eta' > 0$ such that $I_0(\gamma(t))\le m_0-\eta'$. By Lemmas 2.1 and 2.6–2.7, there exists $\eta > 0$ such that for small $\lambda > 0$, it holds that $I_\lambda(\gamma(t)) < b_\lambda-\eta$.□

Proof of Theorem 1.1. Recall that if $u \in S_0$, then there exists $\varrho > 0$ independent of $u \in S_0$ such that $\|\nabla u\|_2^2 \ge \varrho$. So, we can choose $d > 0$ small such that $\|u\|^2 \ge \frac{\varrho}{2}$ for any $u \in S_0^d$. We use the idea in ^[6,29] to claim that for small $\lambda > 0$, there exists $\{u_n\} \subset S_0^d \cap I_\lambda^{B_\lambda}$ such that $I_\lambda'(u_n) \rightarrow 0$. Otherwise, there exists $a(\lambda) > 0$ such that $\|I_\lambda'(u)\| \ge a(\lambda)$ for $u \in S_0^d \cap I_\lambda^{B_\lambda}$. By Lemmas 2.7–2.8, there exists $\rho_0 > 0$ independent of $\lambda > 0$ small such that $\|I_\lambda'(u)\| \ge \rho_0$ for $u \in I_\lambda^{B_\lambda} \cap (S_0^d \setminus S_0^{\frac{d}{2}})$. We note that there exists a pseudo-gradient vector field $Q_\lambda$ on a neighborhood $Z_\lambda$ of $S_0^d \cap I_\lambda^{B_\lambda}$ for $I_\lambda$. Let $\eta_\lambda$ be a Lipschitz continuous function on $X$ such that $\eta_\lambda = 1$ on $S_0^d \cap I_\lambda^{B_\lambda}$, $\eta_\lambda = 0$ on $\mathrm{R}^3 \setminus Z_\lambda$, and $0 \le \eta_\lambda \le 1$ on $\mathrm{R}^3$. Let $\xi_\lambda$ be a Lipschitz continuous function such that $\xi_\lambda(t) = 1$ for $|t-b_\lambda| \le \frac{\eta}{2}$, $\xi_\lambda(t) = 0$ for $|t-b_\lambda| \ge \eta$, and $0 \le \xi_\lambda \le 1$ for $t \in \mathrm{R}^+$. Consider the initial value problem

Then (2.20) has a unique global solution $\psi_\lambda(u, t)$. Recall that $\lim_{\lambda \rightarrow 0}b_\lambda = \lim_{\lambda \rightarrow 0}B_\lambda = m_0$. Also, we have Lemma 2.9. By a standard argument, for any $t \in [0, 1]$ there exists $s(t) \ge 0$ such that $\psi_\lambda(\gamma(t), s(t))$ is continuous in $t \in [0, 1]$ and

where $\gamma$ is given in Lemma 2.9. Let $\gamma_0(.) = \psi_\lambda(\gamma(.), s(.))$. Then $\gamma_0 \in G_0$, from which we get

a contradiction. Since for $\lambda > 0$ small there exists $\{u_n\} \subset I_\lambda^{B_\lambda} \cap S_0^d$ such that $I_\lambda'(u_n) \rightarrow 0$, by Lemma 2.4 we get that $\|u_n\|$ is bounded. Assume that $u_n \rightharpoonup u_\lambda$ weakly in $X$. By Lemma 2.1, we have $I_\lambda'(u_\lambda) = 0$. Let $u_n = v_n+w_n$, where $v_n \in S_0$ and $\|w_n\| \le d$. By Lemma 2.4, there exists $v_\lambda \in S_0$ such that $v_n \rightarrow v_\lambda$ in $X$. Assume that $w_n \rightharpoonup w_\lambda$ in $X$. Then $\|w_\lambda\| \le d$. So, $u_\lambda \in S_0^d$. Moreover, $u_\lambda$ is positive. Together with Lemma 2.8, we get the result.□

3.   Proof of Theorem 1.2

Define $X_r : = \left\{u \in H_r^{1}(\mathbb{R}^3): \ \int_{\mathbb{R}^{3}}V(x)|u|^{2} \mathrm{d} x < \infty\right\}$ as the Hilbert space with the norm $\|u\| = \left(\int_{\mathbb{R}^{3}}|\nabla u|^{2} + V(x) |u|^{2} \mathrm{d} x\right)^{\frac{1}{2}}$. By $(V_2')$, we derive that for all $u \in X_r$,

Then the imbedding $X_r \hookrightarrow H_r^1(\mathbb{R}^3)$ is continuous. Define $g(u) = 0$ for $u \le 0$ and $g(u) = \min\left\{f(u)+(u^+)^5, \frac{V_0u}{\kappa}\right\}$ for $u > 0$, where $\kappa > 2$. Let $\chi$ be the characteristic function such that $\chi(x) = 1$ for $x \in \wedge_2$ and $\chi(x) = 0$ for $x \in \mathbb{R}^3 \setminus \wedge_2$. Consider the truncated problem of (1.8) as

where $h(x, u) = \chi(x)\left[f(u)+ (u^+)^5\right]+ (1-\chi(x))g(u)$. The functional associated with (3.2) is

where $H(x, u) = \int_0^u h(x, s) \mathrm{d}s = \chi(x) \left[F(u)+\frac{1}{6}(u^+)^6\right]+ (1-\chi(x))G(u)$ with $G(u) = \int_0^u g(s) \mathrm{d}s$. In what follows, we look for critical points of $\hat{I}_0$. Define

where $\Gamma_0: = \left\{\gamma \in C ([0, 1], X_r): \gamma(0) = 0, \hat{I}_0(\gamma(1)) < 0\right\}$.

Lemma 3.1. There exists a bounded sequence $\{u_n\} \subset X_r$ such that $\hat{I}_0(u_n) \rightarrow \hat{c}_0 \in \left(0, \frac{1}{3}S^{\frac{3}{2}}\right)$ and $\hat{I}_0'(u_n) \rightarrow 0$.

Proof. By $(f_1)$, for any $\varepsilon > 0$ there exists $C_\varepsilon > 0$ such that

Then there exist $\rho$, $\varrho > 0$ such that $\hat{I}_0(u) \ge \varrho$ for $\|u\| = \rho$, in view of the definition of $S$. Also, $\hat{I}_0(0) = 0$ and $\lim_{t \rightarrow +\infty}\hat{I}_0(t \varphi) = -\infty$ for any $\varphi \in C_0^\infty(\wedge_2) \setminus \{0\}$. By the mountain pass theorem in ^[1], there exists a sequence $\{u_n\} \subset X_r$ such that $\hat{I}_0(u_n) \rightarrow \hat{c}_0 \ge \varrho$ and $\hat{I}_0'(u_n) \rightarrow 0$. By $(f_3')$, we get $\frac{1}{\theta}f(u)u-F(u) \ge 0$ for all $u \in \mathbb{R}$. Then

So, $\|u_n\|$ is bounded. By ^[24], the function $U(x): = \frac{3^{\frac{1}{4}}}{\big(1 + |x|^{2}\big)^{\frac{1}{2}}}$ is a minimizer for $S$. Define $U_\varepsilon(x): = \varepsilon^{-\frac{1}{2}}U(\frac{x}{\varepsilon})$. Let $x_0 \in \wedge_1$. Choose $r > 0$ such that $B_{2r}(x_0) \subset \wedge_1$. Define $u_\varepsilon(x) : = \psi(x) U_\varepsilon(x)$, where $\psi \in C_{0}^{\infty} (B_{2r}(x_0))$ such that $\psi(x) = 1$ for $x \in B_{r}(x_0)$, $\psi(x) = 0$ for $x \in \mathbb{R}^3 \setminus B_{2r}(x_0)$, $0 \le \psi(x) \le 1$, and $|\nabla \psi(x)| \le C$. By the definition of $\hat{c}_0$, we get $\hat{c}_0 \le \sup_{t \ge 0}\hat{I}_0(t u_\varepsilon)$. Moreover, by Lemma 2.1 in ^[28], we get $\hat{c}_0 < \frac{1}{3}S^{\frac{3}{2}}$.□

Lemma 3.2. $\hat{I}_0$ admits a positive critical point $u_0$ with $\hat{I}_0(u_0) = \hat{c}_0$.

Proof. By Lemma 3.1, there exists a bounded sequence $\{u_n\} \subset X_r$ such that $\hat{I}_0(u_n) \rightarrow \hat{c}_0 \in \left(0, \frac{1}{3}S^{\frac{3}{2}}\right)$ and $\hat{I}_0'(u_n) \rightarrow 0$. Assume that $u_n \rightharpoonup u_0$ weakly in $X_r$. Then $\hat{I}_0'(u_0) = 0$. For $R > R_2$, define $\psi_R \in C_0^\infty(\mathbb{R}^3)$ such that $\psi_R(x) = 0$ for $|x| \le R$, $\psi_R(x) = 1$ for $|x| \ge 2R$, and $0 \le \psi_R \le 1$ and $|\nabla \psi_R| \le \frac{C}{R}$. By $\left(\hat{I}_0'(u_n), \psi_R u_n\right) = o_n(1)$,

Then, for any $\delta > 0$, there exists $R_\delta > 0$ such that for $R > R_\delta$,

Since $h(x, u)u \le \frac{V_0}{\kappa}u^2$ for $x \in \mathbb{R}^3 \setminus \wedge_2$, by the Lebesgue dominated convergence theorem

By the argument of Lemma 2.1 in ^[26], we obtain that

Combining (3.6)–(3.8), we have

Let $v_n = u_n-u_0$. Then

from which we derive that $u_n \rightarrow u_0$ in $X_r$, $\hat{I}_0(u_0) = \hat{c}_0$ and $\hat{I}_0'(u_0) = 0$. By $\left(\hat{I}_0'(u_0), u_0^-\right) = 0$, we get $u_0 \ge 0$. The maximum principle implies that $u_0$ is positive.□

Let $\hat{m}_0: = \inf\{\hat{I}_0(u): u \in X_r, \hat{I}_0'(u) = 0\}$.

Lemma 3.3. $\hat{m}_0 \in \left(0, \frac{1}{3}S^{\frac{3}{2}}\right)$ is attained.

Proof. By Lemmas 3.1–3.2, we get $\hat{m}_0 \le \hat{I}_0(u_0) = \hat{c}_0 < \frac{1}{3}S^{\frac{3}{2}}$. By the definition of $\hat{m}_0$, there exists $\{u_n\} \subset X_r$ such that $\hat{I}_0(u_n) \rightarrow \hat{m}_0$ and $\hat{I}_0'(u_n) = 0$. By $\left(\hat{I}_0'(u_n), u_n\right) = 0$, (3.4), and the definition of $S$, there exists $C_1 > 0$ such that $\|u_n\|^2 \ge C_1 S^{\frac{3}{2}}$. Similar to (3.5), we get $\hat{m}_0 > 0$. Also, there exists $C_2 > 0$ such that $\|u_n\|^2 \le C_2 S^{\frac{3}{2}}$. Assume that $u_n \rightharpoonup u_0$ weakly in $X_r$. Then $\hat{I}_0'(u_0) = 0$. Similar to the argument of Lemma 3.2, we get $u_n \rightarrow u_0$ in $X_r$. So $\hat{m}_0 = \hat{I}_0(u_0)$ and $\hat{I}_0'(u_0) = 0$, that is, $\hat{m}_0$ is attained.□

Define by $\hat{S}_0$ the set of ground states of (3.2). By Lemma 3.3, we get $\hat{S}_0 \ne \emptyset$.

Lemma 3.4. $\hat{S}_0$ is compact and there exist $C_1$, $C_2 > 0$ such that $C_1 S^{\frac{3}{2}} \le \|u\|^2 \le C_2 S^{\frac{3}{2}}$ for all $u \in \hat{S}_0$.

Proof. Similar to the argument of Lemma 3.3, we get $C_1 S^{\frac{3}{2}} \le \|u\|^2 \le C_2 S^{\frac{3}{2}}$ for all $u \in \hat{S}_0$. For any $\{u_n\} \subset \hat{S}_0$, since $\|u_n\|^2 \le C_2 S^{\frac{3}{2}}$, we assume that $u_n \rightharpoonup u$ weakly in $X_r$. By Lemma 3.3, we get $\hat{I}_0(u_n) = \hat{m}_0 \in \left(0, \frac{1}{3}S^{\frac{3}{2}}\right)$. Similar to the argument of Lemma 3.2, we obtain that $u_n \rightarrow u$ in $X_r$. So, $\hat{S}_0$ is compact.□

Lemma 3.5. (^[23]) There exists a constant $C_0 > 0$ such that for all $u \in H_r^1(\mathrm{R}^3)$, there holds $|u(x)| \le \frac{C_0}{|x|^{\frac{1}{2}}}\|u\|_{H^1}$ for any $x \ne 0$.

By $(f_1)$, there exists $C' > 0$ such that

Choose $R' > 0$ such that for $R_1 > R'$,

Lemma 3.6. If $u \in \hat{S}_{0}$, then $\hat{m}_{0} = \hat{I}_{0}(u) > \hat{I}_0(t u)$ for all $t \ne 1$. Also, there exists $t_0 > 1$ independent of $u \in \hat{S}_{0}$ such that $\hat{I}_0(t_0 u) \le -2$.

Proof. We claim that

Otherwise, there exists $u \in \hat{S}_0$ such that $\left|\mathrm{supp} u \cap \{x \in \mathrm{R}^3: \chi(x) > 0\}\right| = 0$. By $(\hat{I}_0'(u), u) = 0$,

a contradiction. Let $l(t) = \hat{I}_0(t u)$, where $t \ge 0$ and $u \in \hat{S}_0$. Then $l'(t) = t y(t)$, where

Since $l'(1) = 0$, we have $y(1) = 0$. By $(f_3')$, we get that $y(t)$ is strictly decreasing on $t > 0$. Then $l'(t) > 0$ for $t \in (0, 1)$ and $l'(t) < 0$ for $t > 1$, from which we get $\hat{I}_{0}(u) > \hat{I}_0(t u)$ for all $t \ne 1$. By $\left(\hat{I}_0'(u), u\right) = 0$, (3.4), and the definition of $S$, there exists $\delta_0 > 0$ independent of $u \in \hat{S}_{0}$ such that $\int_{\mathrm{R}^3}\chi(x)|u|^6\mathrm{d}x \ge \delta_0$. Together with Lemma 3.4, we derive that there exists $t_0 > 1$ independent of $u \in \hat{S}_{0}$ such that $\hat{I}_0(t_0 u) \le -2$.□

We consider the following truncated problem of (1.7):

The functional associated with (3.13) is as follows:

Lemma 3.7. There exists $\lambda_1' > 0$ independent of $u \in \hat{S}_0$ such that $\hat{I}_{\lambda}(t_0 u) \le -1$ for $\lambda \in (0, \lambda_1')$.

Proof. By Lemma 2.1, we have

By Lemma 3.4, Lemma 3.6, and (3.14), we derive that there exists $\lambda_1' > 0$ independent of $u \in \hat{S}_0$ such that $\hat{I}_{\lambda}(t_0 u) \le -1$.□

Choose $V_0 \in \hat{S}_0$. Define

where $\Gamma: = \left\{\gamma \in C ([0, 1], X_r): \gamma(0) = 0, \gamma(1) = t_0V_0\right\}$ and $\lambda \in (0, \lambda_1')$. Define

Lemma 3.8. $\lim_{\lambda \rightarrow 0}d_\lambda = \lim_{\lambda \rightarrow 0}D_\lambda = \hat{m}_0$.

Proof. By (3.14), Lemma 3.4, and 3.6, we get

Then $\limsup_{\lambda \rightarrow 0}d_\lambda \le \limsup_{\lambda \rightarrow 0}D_\lambda \le \hat{m}_0$. By Lemma 3.6, for any $\gamma \in \Gamma$,

from which we get $d_\lambda \ge \hat{c}_0$. By Lemma 3.2, we have $\hat{c}_0 \ge \hat{m}_0$, which implies that $\liminf_{\lambda \rightarrow 0}d_\lambda \ge \hat{m}_0$.□

For $\eta$, $d > 0$, define $\hat{I}_\lambda^\eta: = \{u \in X_r: \hat{I}_\lambda(u) \le \eta\}$ and $\hat{S}_0^d: = \{u \in X_r: \inf_{v \in S_0}\|u-v\| \le d\}$. By Lemma 3.4, we can choose $d > 0$ small such that $\frac{C_1}{2}S^{\frac{3}{2}} \le \|u\|^2 \le 2 C_2 S^{\frac{3}{2}}$ for all $u \in \hat{S}_0^d$.

Lemma 3.9. Let $\{u_{\lambda_i}\} \subset \hat{S}_0^d$ with $\lim_{i \rightarrow \infty}\lambda_i = 0$ be such that $\lim_{i \rightarrow \infty}\hat{I}_{\lambda_i}(u_{\lambda_i}) \le \hat{m}_0$ and $\lim_{i \rightarrow \infty}\hat{I}_{\lambda_i}'(u_{\lambda_i}) = 0$. Then, for $d > 0$ small, there exists $u_0 \in \hat{S}_0$ such that $u_{\lambda_i} \rightarrow u_0$ in $X_r$ up to a subsequence.

Proof. Since $\{u_{\lambda_i}\} \subset \hat{S}_0^d$, we have $\frac{C_1}{2}S^{\frac{3}{2}} \le \|u_{\lambda_i}\|^2 \le 2 C_2S^{\frac{3}{2}}$. Moreover, $\lim_{i \rightarrow \infty}\hat{I}_0(u_{\lambda_i}) \le \hat{m}_0$ and $\lim_{i \rightarrow \infty}\hat{I}_0'(u_{\lambda_i}) = 0$. Similar to the argument of Lemma 3.2, we derive that there exists $u_0 \in X_r$ such that $u_{\lambda_i} \rightarrow u_0$ in $X_r$. So, $\|u_0\|^2 \ge \frac{C_1}{2}S^{\frac{3}{2}}$, $\hat{I}_0(u_0) \le \hat{m}_0$, and $\hat{I}_0'(u_0) = 0$, from which we get $u_0 \in \hat{S}_0$.□

Lemma 3.10. Let $d > 0$. Then there exists $\eta > 0$ such that for small $\lambda > 0$, $\hat{I}_\lambda(\gamma(t)) \ge d_\lambda-\eta$ implies that $\gamma(t) \in \hat{S}_0^{\frac{d}{2}}$, where $\gamma(t) = t t_0 V_0$ for $t \in [0, 1]$.

Proof. By Lemma 3.6, if $\gamma(t) \notin \hat{S}_0^{\frac{d}{2}}$, then there exists $\delta > 0$ such that $|t t_0 -1| \ge \delta$. Moreover, there exists $\eta' > 0$ such that $\hat{I}_0(\gamma(t))\le m_0-\eta'$. By Lemma 2.1, Lemma 3.4, and Lemma 3.8, there exists $\eta > 0$ such that for small $\lambda > 0$, it holds that $\hat{I}_\lambda(\gamma(t)) < d_\lambda-\eta$.□

Proof of Theorem 1.2. Similar to the proof of Theorem 1.1, we can use Lemmas 3.8–3.10 to derive that, for small $\lambda > 0$, there exists $\{u_n\} \subset \hat{S}_0^d \cap \hat{I}_\lambda^{D_\lambda}$ such that $\hat{I}_\lambda'(u_n) \rightarrow 0$. Then $\frac{C_1}{2}S^{\frac{3}{2}} \le \|u_n\|^2 \le 2 C_2 S^{\frac{3}{2}}$. Assume that $u_n \rightharpoonup u_\lambda$ weakly in $X_r$. Then $\hat{I}_\lambda'(u_\lambda) = 0$. Let $u_n = v_n+w_n$, where $v_n \in \hat{S}_0$ and $\|w_n\| \le d$. By Lemma 3.4, there exists $v_\lambda \in \hat{S}_0$ such that $v_n \rightarrow v_\lambda$ in $X_r$. Assume that $w_n \rightharpoonup w_\lambda$ in $X_r$. Then $\|w_\lambda\| \le d$. So, $u_\lambda \in \hat{S}_0^d$. Moreover, $\frac{C_1}{2}S^{\frac{3}{2}} \le \|u_\lambda\|^2 \le 2 C_2 S^{\frac{3}{2}}$. Together with (3.1) and Lemma 3.5, we have

By (3.11), we get $\max_{x \in \overline{\wedge_2}}u_\lambda(x) \le \sqrt[4]{\frac{V_0}{2\kappa C'}}$. Let $\varphi = (u_\lambda-\sigma)^+$, where $\sigma = \sqrt[4]{\frac{V_0}{2\kappa C'}}$. By $\left(\hat{I}_\lambda'(u_\lambda), \varphi\right) = 0$,

Since $V(x) \ge V_0$ for $x \in \mathrm{R}^3 \setminus \wedge_2$, by (3.18), we get $u_\lambda(x) \le \sigma$ for $x \in \mathrm{R}^3 \setminus \wedge_2$. Then $h(x, u_\lambda) = f(u_\lambda)+u_\lambda^5$, from which we get $I_\lambda'(u_\lambda) = 0$. Together with Lemma 3.9, we get the result.□

4.   Conclusions

In this paper, we study the existence and asymptotic behavior of positive solutions of a non-autonomous Schrodinger-Poisson equation with critical growth. First, we consider the case that the nonlinearity satisfies the Berestycki-Lions type condition with critical growth. To the best of our knowledge, existing results on Schrodinger-Poisson equations are about radial solutions. However, the problem is quite different when we consider the problem in a non-radial setting. Second, we consider the case that the zero set of the potential is contained in a spherical shell. To the best of our knowledge, there are no results on this question. By developing some techniques in variational methods, we solve the problem successfully.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This project is supported by Natural Science Foundation of Shandong Province(ZR2023MA037) and NSFC(No. 12101192). The authors would like to thank the editors and referees for their useful suggestions and comments.

Conflict of interest

All authors declare no conflict of interest in this paper.