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1.   Introduction and main results

In the past decades, the following Kirchhoff problem

has attracted considerable attention. As we know, the following Dirichlet problem is one of the important deformations of Equation (1.1), which can be degenerated from (1.1). That is, if $V(x) = 0$ and $\Omega$ is a bounded subset of $\mathbb{R}^{3}$, then Equation (1.1) will become

Problem (1.2) corresponds to

which was advanced by Kirchhoff in ^[1]. As a generalization of the classical D'Alembert wave equation of the free vibration of elastic strings, the Kirchhoff model considers the changes of string length caused by lateral vibrations, which has important practical significance. For more mathematical and physical background about the Kirchhoff equations, we direct readers to ^[2,3] and the references quoted within them.

When $b = 0$, Equation (1.1) degenerates to the Schrödinger problem

which has been studied in the past few decades; see, for instance, ^{[4,5,6,7,8,9]} and the references therein. One interesting characteristic is the potential $V$ change sign on $\mathbb{R}^{N}$. In ^[6], Bahrouni, Ounaies, and Radulescu showed the existence of infinitely many solutions for Equation (1.3) with $f(x, u) = a(x)|u|^{q-1}u$ and $0 < q < 1$. Further, Furtado, Maia, and Medeiros ^[10] investigated the Schrödinger equation (1.3) with $a = 1$ and $f(x, u) = f(u)$. Concretely, the nonlinearity $f\in C^1(\mathbb{R}, \mathbb{R})$ is superlinear with subcritical growth. In addition, it verifies the Ambrosetti-Rabinowitz condition: for some $\theta > 2$, there is

Besides, $V$ satisfies the following assumptions:

$(V_0)$ $V\in L_{loc}^t(\mathbb{R}^3)$ for some $t > {\frac{3}{2}}$;

$(V_1)$ $0 < V_\infty: = \lim\limits_{|x|\rightarrow \infty}V(x) < +\infty;$

$(V_2)$ ${\int}_{\mathbb{R}^{3}}|V^-(x)|^{\frac{3}{2}}\mathrm{d}x < S^{\frac{3}{2}},$ where $V^-: = \mathrm{max}\{-V, 0\}$ and $S$ is the best constant for the Sobolev embedding, given by

$(V_3)$ $V(x)\le V_\infty$ for all $x \in \mathbb{R}^{3}$ and $V\not\equiv V_\infty$;

$(V_4)$ there exist $\gamma > 0$ and $C_V > 0$ such that

By using variational methods and the concentration-compactness principle, they obtained a positive ground state solution, and also a nodal solution. Therefore, we know that Equation (1.3) has a least energy sign-changing solution $v_0\in \mathcal{M}_0$ such that $\mathcal{I}_0(v_0) = c_{0, 2}: = \inf\limits_{u\in\mathcal{M}_0}\mathcal{I}_0(u)$, where $\mathcal{M}_0: = \big\{u\in X:u^{\pm}\neq0, \ \langle \mathcal{I}_0'(u), u^+\rangle = \langle \mathcal{I}_0'(u), u^-\rangle = 0\big\}$ with $\mathcal{I}_0(u) = \frac{1}{2} {\int}_{\mathbb{R}^{3}}(a|\nabla u|^2+V(x)u^2)\mathrm{d}x- {\int}_{\mathbb{R}^{3}}F(u)\mathrm{d}x$. For more results about problem (1.3) with indefinite potential, see ^{[10,11,12,13]} and the references therein.

When $b\ne 0$, due to the presence of ${\int}_{\mathbb{R}^3}|\nabla u|^2\mathrm{d}x\Delta u$, problem (1.1) becomes a nonlocal problem, which also brings some essential difficulties for our study. In the past, problems similar to (1.1) have attracted a lot of interest, and so there are many results. For instance, the existence of positive, ground state, and sign-changing solutions for problem (1.1) with various potential $V$ and nonlinearity $f$ has been extensively studied; see ^{[14,15,16,17,18,19,20,21]}. In a recent paper ^[22], Ni, Sun, and Chen also obtained the existence and multiplicity of normalized solutions for a Kirchhoff type problem by using minimization techniques and Lusternik-Schnirelmann theory.

In what follows, we are particularly interested in the case when the potential $V$ involved in problem (1.1) is indefinite. When $V$ can change its sign and satisfies conditions $(V_0)-(V_2)$ and $(V_4)$, Batista and Furtado ^[23] studied problem (1.1) with $f(x, u) = a(x)|u|^{p-2} u$, that is,

where $4 < p < 6$ and $a$ satisfies the following assumptions:

$(a_0)$ $a\in L^\infty (\mathbb{R}^3)$;

$(a_1)$ there exist $C_a, \ \theta_0 > 0$ such that

where

Via the constraint variational methods and the quantitative deformation lemma, the authors not only obtained a non-negative ground state solution, but also a sign-changing solution of Equation (1.4). Here, we point out that the proof of the existence of sign-changing solutions depends on the radial symmetry of $V$ in their paper. Indeed, once the potential $V$ is radial, one can overcome the lack of compactness by considering in the radial subspace. However, if $V$ is not radial, it makes no sense to restrict the problem to spaces of radial functions as the authors did in ^[23]. Besides, we know the embedding $H^1(\mathbb R^3)\hookrightarrow L^p(\mathbb R^3)$ is not compact for $2 < p < 6$. Therefore, one may ask if there will be a sign-changing solution for problem (1.4) or problem (1.1) when the indefinite potential $V$ is not radially symmetric and $f$ is a general nonlinearity.

Next, we will provide an answer to the questions raised above. Specifically, we are going to investigate the problem

where $a, b > 0$ and $V$ is a sign-changing potential.

Moreover, we shall impose that $f\in C(\mathbb{R}, \mathbb{R})$ is odd. In addition, $f$ also satisfies the following assumptions:

$(f_0)$ $\lim\limits_{t\to 0}\frac {f(t)}{t} = 0$;

$(f_1)$ $\lim\limits_{|t|\to\infty}\frac {f(t)}{|t|^5} = 0$;

$(f_2)$ $\lim\limits_{|t|\to\infty}\frac {F(t)}{|t|^{4}} = +\infty$, where $F(t): = {\int}_0^t f(s)\mathrm{d}s$;

$(f_3)$ $\frac{f(t)}{|t|^3} \ \mathrm{{is}}\ \mathrm{{a}}\ \mathrm{non}$-$\mathrm{decreasing} \ \mathrm{{function}}\ \mathrm{{for}}\ t\in \mathbb{R}\setminus\{0\};$

$(f_4)$ $F(t)\ge0$, for all $t\in \mathbb{R}$.

Through $(f_0)$ and $(f_1)$, it can be known that for any $\varepsilon > 0$, there is a positive constant $C_\varepsilon$ which makes

where $q\in(2, 6)$. Then, let $G(t): = \frac{1}{4}f(t)t-F(t)$. By using $(f_3)$, we can easily obtain

In fact, for any $t_2\ge t_1\ge 0$, from $(f_3)$, one has $\frac{f(t_1)}{t_1^3}\le\frac{f(t)}{t^3}\le\frac{f(t_2)}{t_2^3}$ for all $t_1\le t\le t_2$. Then,

Before presenting our main results, we first discussed the basic framework in the space

and the corresponding norm is given by

Next, we present the main results.

Theorem 1.1. Assume that $(V_0)-(V_3)$ and $(f_0)-(f_3)$ are satisfied. Then, problem (1.5) has a positive ground state solution.

In the proof, we need to overcome the problems stemming from the lack of compactness of Sobolev embedding in the whole space $\mathbb{R}^3$. To solve this problem, we first analyze the relationship between the energy functional's minimax level and that of the limit problem. Subsequently, we obtain a positive ground state solution by applying a more general global compactness lemma (see Lemma 3.1).

In the second result, we mainly focused on researching the existence of least-energy sign-changing solutions. In order to obtain the result, we will use the method in ^[24].

Theorem 1.2. Assume that $f$ satisfies $(f_0)-(f_4)$. Assume also that $V$ satisfies $(V_0)-(V_2)$, and there exist positive constants $M, C,$ and $\gamma$ such that, for $|x|\ge M$,

Then, there exists $b^* > 0$ small enough such that Equation (1.5) possesses a least energy sign-changing solution for every $b\in (0, b^*)$.

Remark 1.3. Here, we would like to provide an example of nonlinearity $f$, which is odd and satisfies the conditions $(f_0)-(f_4)$, as shown below:

where $a_i > 0$ and $2 < b_i < 4$ for every $i\in\{1, 2, \ldots, l\}$.

In addition, we can find that a more obvious example that satisfies conditions $(V_0)-(V_2), (V_4^{'})$ is to take $V(x) = V_\infty-\frac{C}{1+|x|^\gamma}$, where $C$ and $\gamma$ are positive given by $({V_4^{'}})$. Alternatively, we can give another example below that satisfies our hypothesis. That is, the potential function $V$ is given by

where $\beta\in(0, 2)$ and $\alpha > 0$ is a sufficiently small.

Remark 1.4. Compared with ^[10], we apply weaker conditions $(f_2)$ and $(f_4)$ instead of the Ambrosetti-Rabinowitz condition to investigate the existence of sign-changing solutions. Moreover, condition $(f_4)$ can be used to construct a sign-changing $({PS})_{c_{b, 2}}$; please refer to Section 5 for the detailed process. We note that condition $(f_4)$ is only used here. Besides, it is not necessary for $f$ to be differentiable. By using the quantitative deformation lemma, we can prove that the minimizer on the Nehari manifold is a critical point of the energy functional $\mathcal{I}_b$ given in Section 2, please refer to Theorem 1.1 below.

Remark 1.5. In this result, we apply the weaker condition $(V_4^{'})$ instead of the condition $(V_4)$ given in ^[23], which makes our results applicable to a wider range of potential functions. Furthermore, the potential function $V$ does not need to be radial in our paper, which is different from ^[23]. Moreover, for the proof of Theorem 1.2, compared to ^[10], the essential problem we face is that the existence of the nonlocal term poses some difficulties to energy estimates. Here, the condition $(V_4^{'})$ and the restriction on the range of the parameter $b$ are crucial for estimating energy (see Lemma 5.1). After that, we can restore the compactness of a bounded Palais–Smale sequence at a certain level with the help of a global compactness lemma, thereby obtaining the conclusion of Theorem 1.2.

Theorem 1.6. Assume that $u_{b_n}$ are the least energy sign-changing solutions of Equation (1.5) obtained in Theorem 1.2. Then, for any sequence $\{b_n\}$ with $b_n\searrow 0$ as $n\to \infty$, there exists a subsequence, still denoted by $\{b_n\}$, such that $u_{b_n}\to u_0$ in $X$ as $n\to \infty$. Moreover, $u_0$ is a least energy sign-changing solution of Equation (1.3) with $f(x, u) = f(u)$.

Remark 1.7. In this paper, we not only weaken the potential condition, but also increase asymptotic behavior, which is different from ^[23]. In particular, our result regarding asymptotic behavior is new. The asymptotic research enriches the results of our paper, and at the same time, the sign-changing solution is closely related to the results of the Schrödinger equation in ^[10]. In practical terms, we study their asymptotic behavior as $b\searrow 0$ under assumptions $(V_0)-(V_2)$ and $(V_4)$, and we show that they converge to a least energy sign-changing solution of the Schrödinger equation (1.3) with $f(x, u) = f(u)$, $b\searrow 0$.

Now, we introduce the organizational structure of this paper. We first provide the notations and some necessary lemmas in Section 2. Then, in Section 3, our main job is to establish a more general global compactness result, which will be well applied in the proof of our main theorems. In Section 4, we first give energy estimates, then Theorem 1.1 is verified. Finally, we prove Theorem 1.2. After that, Theorem 1.6 is verifed in Section 5.

2.   Preliminaries

Next, we first give some notations and necessary lemmas, which are very helpful for us in proving the main theorems.

● $"\rightharpoonup"$ and $"\rightarrow "$ depict the weak and strong convergence, sequentially.

● $|u|_p = ( {\int}_{\mathbb{R}^{3}}|u|^{p}\mathrm{d}x)^{\frac{1}{p}}$ denotes the norm in $L^p({\mathbb{R}^{3}})$ for $p\in[1, +\infty)$.

● Let $H^1(\mathbb{R}^{3})$ be the Hilbert space with respect to the norm $\|u\|_{H^1}^2: = {\int}_{\mathbb{R}^{3}}(|\nabla u|^2+u^2)\mathrm{d}x$.

● We use $|\Omega|$ to represent the Lebesgue measure of the set $\Omega$.

● $o(1)$ denotes a quantity which goes to zero as $n\rightarrow \infty$.

● $C, \ C_i\ (i = 1, 2, \ldots)$ represent different positive constants.

● We denote $V = V^{+}-V^{-}$ with $V^\pm: = \mathrm{max}\{\pm V, 0\}$.

Lemma 2.1. Under the conditions of $(V_1)$ and $(V_2)$, the quadratic form

defines a norm in $H^1(\mathbb{R}^{3})$, which is equivalent to the usual one.

Proof. Due to $(V_1)$, one has that there is $R > 0$, which satisfies

where $B_R(0): = \{ x\in \mathbb{R}^3:|x| < R\}$. In general, we take $a = 1$. Then, using the Hölder inequality, the definition of $S$, and (2.2), one can ascertain

Furthermore, from [10, Lemma 2.1], one deduces that there is $C_1 > 0$, which holds the inequality ${\int}_{\mathbb{R}^{3}}(|\nabla u|^2+V^+u^2)\mathrm{d}x\ge C_1 {\int}_{\mathbb{R}^{3}}(|\nabla u|^2+u^2)\mathrm{d}x$. Moreover, from $(V_2)$, one has

Hence, combining with $V = V^{+}-V^{-}$, it can be obtained that

and the lemma is completed. □

Remark 2.2. From Lemma 2.1, it is obvious to see that the norm given by (2.1) is equivalent to the usual norm of $H^{1}(\mathbb{R}^{3})$. Moreover, we know that the embeddings $H^{1}(\mathbb{R}^{3})\hookrightarrow L^p(\mathbb{R}^3)$ (see [9, Theorem 1.8]) and $H^{1}(\mathbb{R}^{3})\hookrightarrow L_{loc}^q(\mathbb{R}^3)$ (see [9, Theorem 1.9]) are continuous and compact, respectively, where $p\in[2, 6]$ and $q\in[1, 6)$. Therefore, the embedding $X\hookrightarrow L^p(\mathbb{R}^3)$ is also continuous for all $p\in[2, 6]$. Besides, one can see that the continuity of the embedding mentioned above can be represented by the following inequalities:

in which $S_p^p > 0$ is a constant.

Define the energy functional $\mathcal{I}_b: X\rightarrow \mathbb{R}$ by

One can see that $u\mapsto\int F(u)\mathrm{d}x$ is well defined on $X$, so $\mathcal{I}_b(u)$ is also well defined. Through discussions, one can deduce that $\mathcal{I}_b\in C^1(X, \mathbb R)$. Moreover, for any $v\in H^1(\mathbb{R}^{3})$,

The limit problem associated with (1.5) is the autonomous problem below:

The functional corresponding to Equation $(\mathcal{P}_\infty)$ is

Let

where

In addition, we define

where

Next, we will provide some important lemmas to prove Theorem 1.1 and Theorem 1.2.

Lemma 2.3. If $\{u_n\}\subset \mathcal{M}_b$ is a minimizing sequence for $\mathcal{I}_b$, then $C_1\le\|u_n^\pm\|_X\le C_2$ for some $C_1$, $C_2 > 0$.

Proof. Let $\{u_n\}\subset \mathcal{M}_b$ and $\mathcal{I}_b(u_n)\to m$ as $n\to \infty.$ On one hand, due to $(f_0)$, $(f_1)$, and the Sobolev inequality, one has

When $\varepsilon$ is sufficiently small to make $(1-\varepsilon{S_2^2}) > 0$, we can obtain that $\|{u_n^\pm}\|_X\ge C_1$ for some $C_1 > 0$. On the other hand, in light of $\{u_n\}\subset \mathcal{M}_b\subset \mathcal{N}_b$, we have $\langle \mathcal{I}_b'(u_n), u_n\rangle = 0$. In virtue of (1.7), one can arrive at

which means that $m > 0$, so one can deduce that $\{u_n\}$ is bounded in $X$. Namely, $\|u_n^\pm \|_X\le C_2$ for some $C_2 > 0$. Therefore, $C_1\le\|u_n^\pm \|_X\le C_2$. □

Remark 2.4. If $\{u_n\}\subset X$ be a $({PS})_{c}$ sequence, then by using (1.7) and similar arguments to Lemma 2.3, one can get $\|u_n\|_X\le C_2$ for some $C_2 > 0$.

Similar to the discussion in ^[25], one can derive Lemma 2.5 and Lemma 2.6. Here, we omit the proof process.

Lemma 2.5. Assume that $(V_0)-(V_2)$ and $(f_0)-(f_3)$ hold. If $u\in X$ with $u^{\pm}\neq 0$, then there exists a unique pair $(s_{u}, t_{u})\in (0, +\infty)\times(0, +\infty)$ such that $s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}_b$. Moreover,

Lemma 2.6. Assume that $(V_0)-(V_2)$ and $(f_0)-(f_3)$ hold. If $u\in X$ with $u\neq 0$, then there exists a unique $s_u > 0$ such that $s_uu\in\mathcal{N}_b$. Moreover,

Lemma 2.7. $c_{b, \infty}$ can be obtained by some positive and radially symmetric function $\bar{u}\in\mathcal{N}_{b, \infty}$, that corresponds to the ground state solution of $(\mathcal{P}_\infty)$ (see ^[26]). Furthermore, if $f$ is odd, then for every $0 < \delta < \sqrt{V_\infty}$, there is $C = C(\delta) > 0$ that satisfies

where $\alpha = \left(a+b {\int}_{\mathbb{R}^{3}}|\nabla \bar{u}|^2\mathrm{d}x\right)^\frac{1}{2}$.

Proof. The existence of the ground state solutions of $(\mathcal{P}_\infty)$ was proved in [26, Proposition 2.4]. For the properties of solutions of $(\mathcal{P}_\infty)$, see [27, Proposition 1.1]. □

3.   A global compactness result

Now, we will give a more general global compactness lemma, which is very useful.

Lemma 3.1. Let $\{u_n\}\subset X$ be a sequence such that

Then, there exist $u_0\in H^1(\mathbb{R}^3)$ and $A\in\mathbb{R}^+$, such that $\mathcal{J}'_{A}(u_0) = 0$, where

and either

${(i)}$ $u_n\rightarrow u_0$ in $X$, or

${(ii)}$ there exists a number $k\in\mathbb{N}^{+}$, $k$ sequences of points $\{y_n^j\}\subset \mathbb{R}^3$ with $|y_n^j|\rightarrow +\infty$, $1\le j\le k$, and $k$ functions $\left\lbrace u^1, u^2, \ldots, u^k\right\rbrace \subset H^1(\mathbb{R}^3)$, which are nontrivial weak solutions to

and

where

Proof. In view of Remark 2.4, we know $\{u_n\}$ is bounded in $X$. Then, there are $u_0\in X$ and $A\in\mathbb{R}^+$ that satisfy

Due to (3.4) and $\mathcal{I}_b'(u_n)\to0$ as $n\to\infty$, we arrive at

That is, $\mathcal{J}'_{A}(u_0) = 0$. On the other hand, it is clear to see that

Besides, for all $\varphi\in\mathcal{C}_0^\infty(\mathbb{R}^3)$, one has

It follows from (3.5) and (3.6) that

Next, we will demonstrate in three steps and provide detailed proof process.

Step 1. Letting $u_n^1: = u_n-u_0$, we can obtain

${(a_1)}$ $\mathcal{J}_A^\infty (u_n^1)\to c+\frac{bA^2}{4}-\mathcal{J}_A(u_0)$,

${(b_1)}$ $\langle(\mathcal{J}_{A}^\infty)' (u_n^1), u_n^1\rangle = \langle\mathcal{J}_{A} ' (u_n), u_n\rangle-\langle\mathcal{J}_{A} ' (u_0), u_0\rangle+o(1) = o(1).$

To prove ${(a_1)}$, we can use the weak convergence of $\{u_n\}$ and [28, Lemma 3] to conclude that

Moreover, by virtue of $(V_1)$, we know $\forall \varepsilon > 0$, $\exists R > 0$ such that $|V_\infty- V(x)| < \varepsilon$ on $\mathbb{R}^3\setminus B_R(0)$. Hence, it holds that

From the arbitrariness of $\varepsilon$, it can be concluded that

Using (3.7)-(3.9) and (3.11), we can show that

That is, ${(a_1)}$ is correct. As for ${(b_1)}$, by using a similar argument as before and (3.10), it is sufficient to get ${(b_1)}$. We omit the details here. Furthermore, by ${(a_1)}$, we can obtain that $\mathcal{J}_A^\infty (u_n^1)\ge 0$.

Step 2. Define

Case 1 (Vanishing): $\delta^{(1)} = 0$. That is, as $n\to\infty$,

From the P.L. Lions lemma in ^[9], one has $u_n^1\to 0$ in $L^t(\mathbb{R}^3)$ for any $t\in (2, 6)$. Thus, we deduce that $\lim\limits_{n\rightarrow +\infty} {\int}_{\mathbb{R}^3}f(u_n^1)u_n^1\mathrm{d}x = 0$ by (1.6). Besides, it is easy to get ${\int}_{\mathbb{R}^3}(V(x)-V_\infty)(u_n^1)^2\mathrm{d}x = o(1)$. Hence, by ${(b_1)}$, we can conclude $\|u_n^1\|_X\to 0$ as $n\to\infty$.

Case 2 (Non-vanishing): $\delta^{(1)} > 0$. Assume that there exists $\{y_n^1\}\subset \mathbb{R}^3$ such that

We now define a new sequence $w_n^1: = u_n^1(\, \cdot\, +y_n^1)$. It is easy to get that $\{w_n^1\}$ is bounded in $X$. Moreover, we suppose that $w_n^1\rightharpoonup u^1$ in $X$. Since

it follows from the Sobolev embedding that $u^1\ne 0$. Moreover, $u_n^1\rightharpoonup 0$ in $X$ implies that $\{y_n^1\}$ is unbounded. That is, $|y_n^1|\to \infty$ as $n\to \infty$. Furthermore, we can show $(\mathcal{J}_{A}^\infty)'(u^1) = 0$.

Step 3. Setting $u_n^2: = u_n^1-u^1(\, \cdot\, -y_n^1)$, we can check that

${(a_2)}$ $\mathcal{J}_A^\infty (u_n^2)\to c+\frac{bA^2}{4}-\mathcal{J}_A(u_0)-\mathcal{J}_A^\infty (u^1)$,

${(b_2)}$ $\langle(\mathcal{J}_{A}^\infty)' (u_n^2), u_n^2\rangle = \langle\mathcal{J}_{A} ' (u_n), u_n\rangle-\langle\mathcal{J}_{A} ' (u_0), u_0\rangle-\langle(\mathcal{J}_{A}^\infty)' (u^1), u^1\rangle+o(1) = o(1).$

Similar to Step 2, define

If $\delta^{(2)} = 0$, we have $\|u_n^2\|_X\to 0$ as $n\to\infty$. That is, $\|u_n-u_0-u^1(\, \cdot\, -y_n^1)\|_X\to 0$ as $n\to\infty$. By (3.7) and ${(a_2)}$, we have $A = |\nabla u_0|_2^2+|\nabla u^1|_2^2$ and $c+\frac{bA^2}{4} = \mathcal{J}_A(u_0)+\mathcal{J}_A^\infty (u^1)$. Furthermore, we know that $\mathcal{J}_A^\infty (u_n^2) = o(1)$. If $\delta^{(2)} > 0$, as the arguments as above, we know that there exists $\{y_n^2\}\subset \mathbb{R}^3$ unbounded, and a sequence $w_n^2: = u_n^2(\, \cdot\, +y_n^2)$ that satisfies $w_n^2\rightharpoonup u^2$ in $X$ and $u^2\ne 0$. Besides, we can obtain $(\mathcal{J}_{A}^\infty)'(u^2) = 0$. Iterating the above process, we can show that

with $|y_n^j|\to \infty$ and

where $u^j$ is the nontrivial weak solution of Equation (3.1). Moreover, we can conclude that

Noticing that $u^i$ is the nontrivial weak solution of Equation (3.1), in view of (1.7), we can obtain $\mathcal{J}_A^\infty(u^i) > 0$. Besides, similar to the above discussion, we know that when $\delta^{(j)} = 0$, $\mathcal{J}_A^\infty(u_n^j) = o(1)$. Hence, there is some finite constant $k\in \mathbb{N}$. Moreover, the above process will stop after $k$ iterations. Namely, the proof is completed. □

Corollary 3.2. The functional $\mathcal{I}_b$ satisfies $(PS)_c$ condition for $c\in (0, c_{b, \infty})$.

Proof. Let $\{u_n\}\subset X$ be a $(PS)_c$ sequence for $c\in (0, c_{b, \infty})$. Then,

We only need to prove that $\{u_n\}$ has a convergent subsequence in $X$ next. From Remak 2.4, we first conclude that $\{u_n\}$ is bounded in $X$. Hence, there is a subsequence of $\{u_n\}$, still denoted as $\{u_n\}$. Besides, there is also $u_0\in H^1(\mathbb{R}^{3})$ that satisfies $u_n\rightharpoonup u_0$ in $X$. If it is strongly convergent, then the proof is completed. If $u_n\not\rightarrow u_0$ in $X$, from Lemma 3.1, there exist $A = |\nabla u_0|_2^2+\sum_{j = 1}^{k}|\nabla u^j|_2^2$, $k\in \mathbb{N}$ and $\{y_n^j\}\subset \mathbb{R}^3$ with $|y_n^j|\rightarrow +\infty$ for $j = 1, 2, \ldots, k$ and $u^j\in H^1(\mathbb{R}^{3})$ such that

where $u^j$ are nontrivial critical points of $\mathcal{J}_A^\infty$ for $j = 1, 2, \ldots, k$.

We first give the following two claims.

Claim 1: If $u_0\ne0$ and there exists $t_0 > 0$ such that $t_0u_0\in\mathcal{N}_b$, then we claim that $t_0\le 1$.

Since $t_0u_0\in\mathcal{N}_b$ and $\mathcal{J}'_{A}(u_0) = 0$, we can obtain that

and

It follows from (3.3), (3.12), and (3.13) that

From $(f_3)$ and (3.14), it is easy to obtain $t_0\le 1$. Therefore, the claim is proved.

Claim 2: If $u_0\ne0$, we claim that

Combining $t_0\le1$ and (1.7), we can arrive at

Hence, we can obtain

Therefore, the claim is proved. Similarly, since $u^j\ne 0$ and $(\mathcal{J}_A^\infty)'(u^j) = 0$, it is easy to see that

Now, we return to our proof. If $u_0\equiv0, k\ge1$, then $A = \sum_{j = 1}^{k}|\nabla u^j|_2^2$ and $c+\frac{bA^2}{4} = \sum_{j = 1}^{k}\mathcal{J}_A^\infty(u^j)\ge kc_{b, \infty}+\frac{bA^2}{4}$. Noticing that $c < c_{b, \infty}$, this is absurd. If $u_0\ne0, k\ge1$, then from (3.2), (3.15), and (3.16), we can obtain $c+\frac{bA^2}{4}\ge c_{b, 1}+kc_{b, \infty}+\frac{bA^2}{4}$. By using $(f_0)-(f_3)$ and similar discussions to those in [9, Theorem 4.2], one ascertains $c_{b, 1} > 0$. Combining $c < c_{b, \infty}$, we know this case cannot occur. Therefore, $k = 0$ and the proof is completed. □

4.   Positive ground state solution

Now, we prove Theorem 1.1. First, due to conditions $(f_0)-(f_2)$, one can easily deduce that $\mathcal{I}_b$ satisfies the mountain pass geometry. Then, by conditions $(f_0)-(f_3)$ and similar discussions to those in [9, Theorem 4.2], we have

where $\Gamma: = \left\{\gamma \in \mathcal{C}([0, 1], X):\gamma (0) = 0, \ \mathcal{I}_b(\gamma (1)) < 0\right\}.$ Then, we show the relationship between $c_{b, 1}$ and $c_{b, \infty}$. Note that $\bar{u}$ is a positive ground state solution of $(\mathcal{P}_\infty)$, so combined with Lemma 2.6, one can ascertain that there is $t_{\bar{u}} > 0$, which makes $t_{\bar{u}}\bar{u}\in \mathcal{N}_b$.

Lemma 4.1. Assume that $(V_0)-(V_3)$ and $(f_0)-(f_3)$ hold. Then,

Proof. We first claim that $t_{\bar{u}} < 1$.

Since $\bar{u}\in\mathcal{N}_{b, \infty}$, one has that

Due to $t_{\bar{u}}\bar{u}\in\mathcal{N}_b$, we ascertain that

Combining (4.1), (4.2), and $(V_3)$, we deduce that

If $t_{\bar{u}}\ge 1$, the above equation does not hold by $(f_3)$. Thus, we can obtain that $t_{\bar{u}} < 1.$

It follows from (1.7), $(V_3)$, Lemma 2.6, and the above claim that

The proof is completed. □

Proof of Theorem 1.1. Through using [9, Theorem 1.15], one knows that there exists a sequence $\{u_n\}\subset X$ such that

In view of Corollary 3.2 and Lemma 4.1, the sequence $\{u_n\}$ has a subsequence which strongly converges to $u\in X$. Besides, the function $u$ satisfies $\mathcal{I}_b(u) = c_{b, 1} > 0$ and $\mathcal{I}_b'(u) = 0$. We can easily get that $u\ne 0$. This indicates that $u$ is a ground solution of Equation (1.5). Next, we prove that Equation (1.5) has a positive ground solution. First, based on $f$ being an odd function, we can see $\mathcal{I}_b(|u|) = \mathcal{I}_b(u) = c_{b, 1}$ and $|u|\in \mathcal{N}_b$. We can claim that $\mathcal{I}_b'(|u|) = 0$ by using the deformation lemma, where $f$ does not require differentiability. For convenience, let us note $w = |u|$. Then, we only need to prove $\mathcal{I}_b'(w) = 0$.

By contradiction, we assume $\mathcal{I}_b'(w)\ne 0$. Then, there are $\varrho > 0$ and $\delta > 0$ that satisfy

Let $D: = (1-\sigma, 1+\sigma)$, where $\sigma\in(0, \min{\{{\frac{1}{2}}, {\frac{\delta}{\sqrt{2}\|w\|_X}}\}})$. By using the fact that $w\in \mathcal{N}_b$ and the condition $(f_3)$, we have

and

Hence, we can take $t_1, t_2\in D\setminus\{1\}$, such that

It follows from Lemma 2.6 that

For $\varepsilon: = \mathrm{min}\left\lbrace (c_{b, 1}-\bar{c})/{3}, \delta\varrho/{8}\right\rbrace$ and $S_\delta: = \{u\in X:\|u-w\|_X\le\delta\}$, due to the deformation lemma in [9, Lemma 2.3], one deduces that there exists $\eta\in\mathcal{C}([0, 1]\times X, X)$ that satisfies

$(a)$ $\eta(1, u) = u$ if $u\notin \mathcal{I}_b^{-1}([c_{b, 1}-2\varepsilon, c_{b, 1}+2\varepsilon])\cap S_{2\delta}$;

$(b)$ $\eta(1, \mathcal{I}_b^{c_{b, 1}+\varepsilon}\cap S_\delta)\subset\mathcal{I}_b^{c_{b, 1}-\varepsilon}$;

$(c)$ $\mathcal{I}_b(\eta(1, u))\le \mathcal{I}_b(u)$, for any $u\in X$.

From this deformation, with the help of Lemma 2.6, we can claim that

Indeed, on the one hand, from Lemma 2.6, we know $\forall t\in D$,

In addition, based on the definition of $\sigma$, it can be concluded that

which means $tw\in S_\delta$. Hence, $tw\in \mathcal{I}_b^{c_{b, 1}+\varepsilon}\cap S_\delta$. According to $(b)$, it is easy to obtain (4.5).

In what follows, we can first claim that $\eta(1, tw)\cap\mathcal{N}_b\neq\emptyset$ for some $t\in D$. We define

From (4.4), the definition of $\varepsilon$ and $(a)$, we have $\eta(1, t_1w) = t_1w$ and $\eta(1, t_2w) = t_2w$. From (4.3), one has

In view of (4.6) and the continuity of $\Phi$, there exists $t_0\in[t_1, t_2]\subset D$ such that $\Phi(t_0) = 0$. From the definition of $\Phi$, one has $\eta(1, t_0w)\in \mathcal{N}_b$. Namely, $\eta(1, tw)\cap\mathcal{N}_b\neq\emptyset$ for some $t\in[t_1, t_2]$. Then, one gets $c_{b, 1}\le\mathcal{I}_b(\eta(1, t_0w))$, which is clearly contradictory to (4.5). Hence, the assumption is not valid and we ascertain that $\mathcal{I}_b'(w) = 0$. Therefore, $w$ is a non-negative solution of Equation (1.5). Finally, according to the maximum principle [29, Theorem 3.5], one can obtain $w > 0$ in $\mathbb{R}^3$. Then, we say Equation (1.5) has a positive ground solution $w$. So far, the proof is completed. □

Remark 4.2. Let $u_1\in X$ be the solution obtained from Theorem 1.1. From the theory of classical Schrödinger equations (see [30, Theorem 3.1]), it is easy to conclude that $u_1$ decays exponentially as $|x|\to \infty$. Namely, for every $\delta > 0$, there exist $C = C(\delta) > 0$ and $R > 0$ such that

5.   Existence and asymptotic behavior of least energy sign-changing solutions

In this section, we will prove the existence and asymptotic behavior of least-energy sign-changing solutions of Eq. (1.5). We first give the relationship among $c_{b, 1}$, $c_{b, \infty}$ and $c_{b, 2}$. Then, inspired by ^[24], we can construct a sign-changing $({PS})_{c_{b, 2}}$ sequence for $\mathcal{I}_b$. After that, we can use Lemma 3.1 to prove Theorem 1.2. Finally, we prove Theorem 1.6.

Lemma 5.1. Assume that $(V_0)-(V_2)$, $(V_4^{'})$, and $(f_0)-(f_3)$ hold. Then, there exists $b^* > 0$ small enough such that for $0 < b < b^*$ we have

Proof. Define $\bar{u}_n(x): = \bar{u}(x-ne_1)$ and $e_1: = (1, 0, 0)$, where $\bar{u}$ is given in Lemma 2.7. In what follows, $u_1$ represents the positive ground state solution of Equation (1.5), which is obtained from Theorem 1.1.

Claim 1: There exist $s_0, t_0 > 0$ such that $s_0u_1-t_0\bar{u}_{n_0}\in\mathcal{M}_b$ for some $n_0\in \mathbb{N}$ large enough.

In fact, denote $\chi(a) = \frac{1}{a}u_1-\bar{u}_n$ with $a > 0$, and define $a_1, a_2$ by

By Lemma 2.5, there exists $(s(\chi(a)), t(\chi(a)))$ such that $s(\chi(a))\chi^+(a)+t(\chi(a))\chi^-(a)\in \mathcal{M}_b$. Because $u_1$ is positive and $\bar{u}$ is radial, we can prove that $a_1 = +\infty$. Indeed, we first notice that $\bar{u}$ is radially symmetric, as the same arguments presented in [31, Lemma 3.1.2] (see also [32, Radial Lemma 1]), one has that there exists a constant $C > 0$ such that

We can obtain that for every $x\in B_R(0)$, there is $|x-ne_1|\ge n-|x|\ge n-R$. By using (5.1), one can ascertain

Then, for fixed $x\in B_R(0)$, we will have

Therefore, we can take $\varepsilon = \frac{u_1}{2aC_2} > 0$ and $n_0\in\mathbb{N}$, such that $\frac{1}{n-R} < \varepsilon$ for every $n\ge n_0$. Combining with (5.2), we know that for all $n\ge n_0$ and $a\in(0, +\infty)$, there is

Finally, we obtain that $a_1 = +\infty$ by the definition of $a_1$.

If $a\to a_1 = +\infty$, then $\frac{1}{a}u_1\to 0$ in $X$ and $\chi^+(a)\to 0$. Similar to [33, Lemma 2.2 $(ii)$], one concludes that $s(\chi(a))\to +\infty$ and $\{t(\chi(a))\}$ is bounded in $\mathbb{R}^+$. Hence, as $a\to a_1$, one has

Similarly, if $a\to a_2^+$, $\chi^-(a)\to 0$, one has that $t(\chi(a))\to +\infty$ and $\{s(\chi(a))\}$ is bounded in $\mathbb{R}^+$. So, as $a\to a_2^+$, we have

From [33, Lemma 2.2 $(i)$], we can ascertain the continuity of $s$ and $t$. Combining (5.3) and (5.4), we obtain that there exists $a_0\in (a_2, a_1)$ such that $s(\chi(a_0)) = t(\chi(a_0))$ for $n_0$. Hence, let $s_0 = \frac{1}{a_0}s(\chi(a_0))$ and $t_0 = t(\chi(a_0))$, it is easy to show that

Claim 2: There exist $b^* > 0$ small enough and $n\in\mathbb{N}^+$ large enough such that for $0 < b < b^*$ we have

Obviously, $\mathcal{I}_b(su_1-t\bar{u}_n) < 0$ for $s$ or $t$ large enough. Next, we only need to consider this problem in a bounded interval. That is, we consider the case that $s, t\in (0, C)$, where $C > 0$. Moreover, one can check that there is $t_n > 0$ that satisfies $t_n\to 1$ as $n\to \infty$ and $t_n\bar{u}_n\in\mathcal{N}_b$. Then, by a direct calculation, we can obtain

where

First, similar to the conclusion presented in ^[10], we can show that

for any $q\in [1, 5]$.

For $A_n$, in view of $(V_4^{'})$, we can arrive at

As for $D_n$, due to (1.6), one has

In what follows, we estimate $B_n$. Since $\langle\mathcal{I}_{b, \infty} '(\bar{u}_n), u_1\rangle = 0$, we deduce that

From (1.6), (5.7), and (5.11), we derive that

Since $\langle\mathcal{I}_b'(u_1), \bar{u}_n\rangle = 0$, we can get that

Let us go back to the term $B_n$ now. Combining (1.6), (5.8), (5.12), and the above equation, we can obtain that

By (5.12), it is easy to conclude that

Then, due to (5.9), (5.10), (5.13), and (5.14), we can ascertain that

Therefore, choosing $n_0\in \mathbb{N}^+$ large enough, we can obtain $-\frac{C_7}{n_0^\gamma}+C_{18}e^{{-\frac{\delta}{6\alpha}n_0}} +C_{19}e^{{-\frac{\delta}{6}n_0}} < 0$. Hence, we can take $b^* = \frac{C_7\frac{1}{n_0^\gamma}-C_{18}e^{ -{\frac{\delta}{6\alpha}n_0}}-C_{19}e^{ -{\frac{\delta}{6}n_0}} }{2C_{17}} > 0$, and then, $\forall b\in(0, b^*)$, we have

for all $n\ge n_0$. Noticing that (5.6), we deduce that

for all $n\ge n_0$. Therefore, the claim is proved. Finally, combining Claim 1 and Claim 2, for any $b\in (0, b^*)$, we can show that

The proof is completed. □

In order to construct a sign-changing $({PS})_{c_{b, 2}}$ sequence of the functional $\mathcal{I}_b$, we follow the method in ^[24]. Define

First, since $f$ being an odd function, we can obtain $g(u, v) > 0$ if $u\neq 0$. Besides, we can see that $u\in\mathcal{M}_b$ if and only if $g(u^{+}, u^{-}) = g(u^{-}, u^{+}) = 1$. Moreover, we can construct the following set $U$, which is larger than the set $\mathcal{M}_b$. Namely, we define

We know $\mathcal{M}_b\subset U$. Furthermore, from Lemma 2.5, it can be concluded that $\mathcal{M}_b\ne \emptyset$. That is, $\mathcal{M}_b\subset U\ne \emptyset$. Now, we use $P$ to represent the cone of the non-negative functions in $X$, $D: = [0, 1]$, $E: = D\times D$, and $\Sigma$ to represent the set of continuous maps $\sigma$, so that for all $s, t \in D$,

$(i)$ $\sigma\in C(E, X)$;

$(ii)$ $\sigma(s, 0) = 0, \, \sigma(0, t)\in P$ and $\sigma(1, t)\in -P$;

$(iii)$ $(\mathcal{I}_b\circ\sigma)(s, 1)\leq 0$ and $\frac{ {\int}_{\mathbb{R}^{3}}f(\sigma(s, 1))\sigma(s, 1)\, \, dx} {\|\sigma(s, 1)\|_X^2+b( {\int}_{\mathbb{R}^{3}}|\nabla (\sigma(s, 1))|^2\, \, dx)^2}\geq 2$.

We can claim that $\Sigma\neq\emptyset$. In fact, for every $u\in X$ and $u^{\pm}\neq 0$, we can set $\sigma(s, t) = \mu (1-s)tu^{+}+\mu stu^{-}$, where $\mu > 0$ and $s, t\in D$. Then, through simple calculations, one can conclude $\sigma(s, t)\in\Sigma$ for some $\mu > 0$. Moreover, we can also obtain the following lemmas.

Lemma 5.2. Assume that $(V_0)-(V_2)$ and $(f_0)-(f_4)$ hold. Then,

Proof. On one hand, for all $u\in\mathcal{M}_b$ and $s, t\in D$, there is $\sigma(s, t) = \mu (1-s)tu^{+}+\mu stu^{-}$ such that when $\mu > 0$ is sufficiently large, $\sigma(s, t)\in \Sigma$. In light of Lemma 2.5, one concludes that $\mathcal{I}_b(u) = \max_{s, t\geq0}\mathcal{I}_b(su^{+}+tu^{-})$. Hence, one obtains that

which implies that

On the other hand, we can see that for every $\sigma\in\Sigma$, there is $u_{\sigma}\in \sigma(E)\cap\mathcal{M}_b$. Therefore,

Then, one can arrive at

Hence, combining (5.17) and (5.18), we can conclude that

Now it remains to prove the claim. Indeed, due to the definition of $\Sigma$, we can know $\sigma(0, t)\in P$ and $\sigma(1, t)\in -P$, which holds for every $\sigma\in\Sigma$ and $t\in D$. Moreover, by using conditions $(f_4)$ and (1.7), it is easy to ascertain that for every $t\in\mathbb{R}$, there is $\frac{1}{4}f(t)t\ge F(t)\ge 0$. Next, for convenience, one can define $l^\pm (s, t): = g(\sigma^{+}(s, t), \sigma^{-}(s, t))\pm g(\sigma^{-}(s, t), \sigma^{+}(s, t))$. Then, one can deduce

and

Then, we can use property $(iii)$ and the inequality $\frac{b}{a}+\frac{d}{c}\geq\frac{b+d}{a+c}$, where $a, b, c, d > 0$, to obtain that

Thus,

In addition, we can verify $l^+(s, 0) = 0$. Hence,

Therefore, using Miranda's theorem ^[34] and (5.19)-(5.22), there is $(s_{\sigma}, t_{\sigma})\in E$ that satisfies

As a result, one has

Namely, there exists $u_{\sigma} = \sigma(s_\sigma, t_\sigma)\in{\sigma(E)}\cap{\mathcal{M}_b}$ for any $\sigma\in\Sigma$. The proof is completed. □

Lemma 5.3. Assume that $(V_0)-(V_2)$ and $(f_0)-(f_4)$ hold, then there exists a sequence $\{u_n\}\subset U$ satisfying $\mathcal{I}_b (u_n)\rightarrow c_{b, 2}$ and $\mathcal{I}_b' (u_n)\rightarrow 0$ as $n\to\infty$.

Proof. Let $\{{w_n}\}\subset{\mathcal{M}_b}$ be a minimizing sequence and $\sigma(s, t) = \mu (1-s)tw_n^{+}+\mu stw_n^{-}\in\Sigma$, so

We can claim that there is a sequence $\{u_n\}\subset X$, which satisfies, as $n\rightarrow \infty$,

Suppose there is a contradiction, then, there exists $\delta > 0$ such that $\sigma_n(E)\cap V_\delta = \emptyset$ for $n$, which is sufficiently large, where

Through using ^[35], due to Hofer ^[36], there is $\eta\in \mathcal{C}(D\times X, X)$, which satisfies the following properties for some $\varepsilon\in(0, \frac{c_{b, 2}}{2})$ and every $t\in D$.

$(i)$ $\eta(0, u) = u$, $\eta(t, -u) = -\eta(t, u)$;

$(ii)$ $\eta(t, u) = u$ for any $u\in \mathcal{I}_b^{c_{b, 2}-\varepsilon}\cup (X\setminus \mathcal{I}_b^{c_{b, 2}+\varepsilon})$, where $\mathcal{I}_b^c = \{u\in X:\mathcal{I}_b(u)\le c\}$;

$(iii)$ $\eta(1, \mathcal{I}_b^{c_{b, 2}+\frac{\varepsilon}{2}}\setminus V_\delta)\subset \mathcal{I}_b^{c_{b, 2}-\frac{\varepsilon}{2}}$;

$(iv)$ $\eta(1, (\mathcal{I}_b^{c_{b, 2}+\frac{\varepsilon}{2}}\cap P)\setminus V_\delta)\subset \mathcal{I}_b^{c_{b, 2}-\frac{\varepsilon}{2}}\cap P$.

By (5.23), select $n$ to be sufficiently large so that

Define $\tilde{\sigma}_n(s, t): = \eta(1, \sigma_n(s, t))$ for all $(s, t)\in E$. Then, it is clear to see that $\tilde{\sigma}_n\in \Sigma$. From (5.25) and property $(iii)$, it can be inferred that $\tilde{\sigma}_n(E)\subset \mathcal{I}_b^{c_{b, 2}-\frac{\varepsilon}{2}}$. Therefore,

which is absurd, so the claim is valid.

Now, we begin to prove $\{u_{n}\}\subseteq U$ when $n$ is large enough. Indeed, in light of $\mathcal{I}_b'(u_n)\rightarrow0$ as $n\to\infty$, one can ascertain $\langle \mathcal{I}_b'(u_n), u_n^\pm\rangle = o(1)$. After that, it is sufficient to prove $u_n^\pm\neq0$, which means $g(u_n^+, u_n^-)\rightarrow1$ and $g(u_n^-, u_n^+)\rightarrow1$. Therefore, one gets $\{u_n\}\subset U$ for $n$ large enough. Using (5.24), there is a sequence $\{\nu_n\}$ that satisfies

To get $u_{n}^{\pm}\neq0$, we only need to show that $s_nw_n^+\neq0$ and $t_nw_n^-\neq0$ for $n$ large enough. It derives from Lemma 2.3 and the fact of $\{w_n\}\subset \mathcal{M}_b$ that $C_{1}\le\|w_{n}^\pm\|\le C_{2}$ for some $C_{1}, C_{2} > 0$. Next, we only need to prove $s_n\nrightarrow0$ and $t_n\nrightarrow0$ as $n\rightarrow \infty$. By contradiction, if $s_n\rightarrow0$ as $n\to\infty$, in light of (5.26) and $\mathcal{I}_b$ being continuous, one can get

Choosing $0 < \varepsilon < \frac{1}{2S_2^2}$ such that

leads to a contradiction. Then, the above assumption is not valid. Therefore, $\{u_{n}\}\subseteq U$ for $n$ large enough. □

Based on the previous lemma, we will now focus on the proof of Theorem 1.2.

Proof of Theorem 1.2. First, there is a $(PS)_{c_{b, 2}}$ sequence $\{u_n\}\subseteq U$ by Lemma 2.3 and Lemma 5.3, which is bounded. Then, we assume that there exists a subsequence, which satisfies $u_n\rightharpoonup u_0$ in $X$. We can claim $u_n\to u_0$ in $X$, thus we deduce that

From Lemma 2.3, by $u_n\to u_0$ in $X$, we get $\|u_0^{\pm}\|_X\geq C_1 > 0$, namely $u_0\in\mathcal{M}_b$. Hence, $u_0\in \mathcal{M}_b$ is a least energy sign-changing solution of Equation (1.5).

It remains to verify the above claim. In fact, due to Lemma 3.1, if case ${(i)}$ occurs, the proof is completed. If case ${(ii)}$ occurs, since $c_{b, 1} < c_{b, \infty}$, it follows from (3.2) that $k\le 1$. Hence, $k = 0$ or $k = 1$.

If $u_0\equiv 0$, in light of $c_{b, 2} > 0$, one can deduce for $k = 1$ and $A = |\nabla u^1|_2^2$,

Since $|y_n^1|\rightarrow +\infty$ and $\{u_n\}\subset U$, (5.27) and Lemma 2.3 imply that $(u^1)^\pm\ne 0$. Besides, due to $(\mathcal{J}_{A}^\infty)'(u^1) = 0$, we get $\langle(\mathcal{J}_{A}^\infty)' (u^1)^\pm, (u^1)^\pm\rangle = \langle(\mathcal{J}_{A}^\infty)' (u^1), (u^1)^\pm\rangle = 0$. Hence, we have $\mathcal{J}_A^\infty((u^1)^\pm)\ge c_{b, \infty}+\frac{bA}{4} {\int}_{\mathbb{R}^{3}}|\nabla (u^1)^\pm|^2\mathrm{d}x$ by (3.16). Therefore,

which contradicts with $c_{b, 1} < c_{b, \infty}$. Thus, $u_0\ne 0$, combining $c_{b, 2} < c_{b, 1}+c_{b, \infty}$, (3.2), (3.15), and (3.16), we can get that

Therefore, we can show that $k = 0$ and $u_n\to u_0$ in $X$. The proof is completed. □

From now on, we will study the asymptotic behavior of the above sign-changing solutions with respect to $b$. In order to facilitate research, we set $c_{0, 1}: = \inf\limits_{u\in\mathcal{N}_0}\mathcal{I}_0(u)$, $c_{0, \infty}: = \inf\limits_{u\in\mathcal{N}_{0, \infty}}\mathcal{I}_{0, \infty}(u)$ and $c_{0, 2}: = \inf\limits_{u\in\mathcal{M}_0}\mathcal{I}_0(u)$ with $\mathcal{N}_0: = \big\{u\in X\setminus\{0\}:\langle \mathcal{I}_0'(u), u\rangle = 0\big\}$, $\mathcal{N}_{0, \infty}: = \big\{u\in X\setminus\{0\}:\langle\mathcal{I}_{0, \infty}'(u), u\rangle = 0\big\}$, and $\mathcal{M}_0: = \big\{u\in X:u^{\pm}\neq0, \ \langle \mathcal{I}_0'(u), u^+\rangle = \langle \mathcal{I}_0'(u), u^-\rangle = 0\big\}$, where $\mathcal{I}_0(u)$ and $\mathcal{I}_{0, \infty}(u)$ respectively represent $\mathcal{I}_{b}(u)$ and $\mathcal{I}_{b, \infty}(u)$ with $b = 0$. Besides, from Theorem 1.2, we have obtained that Equation (1.5) has a least energy sign-changing solution for all $b\in(0, b^*)$ under hypothesis $(V_0)-(V_2)$ and $(V_4)$. Next, we denote it as $u_b$ and consider the problem with $b\in(0, \mathrm{min}\{b^*, 1\})$.

Proof of Theorem 1.6. First of all, in virtue of Theorem 1.2 and Lemma 5.1, we immediately know that $c_{b, 2}$ is achievable by some $u_b$ for all $b\in (0, b^*)$ and satisfies $\mathcal{I}_b(u_b) = c_{b, 2} < c_{b, 1}+c_{b, \infty}$. Hence, there is a sequence $\{u_{b_n}\}$ that satisfies $\mathcal{I}_{b_n}(u_{b_n}) = c_{b_n, 2} < c_{b_n, 1}+c_{b_n, \infty}$ and $\mathcal{I}'_{b_n}(u_{b_n}) = 0$. Similar to the proof in ^[37], we can get that $\mathcal{I}_b(u_b) = c_{b, 2}\le M_0$, where $M_0$ is a positive constant. Due to $(f_3)$, we have that

and we can easily infer that $\{u_{b_n}\}$ is bounded. Thus, there is a subsequence of $\{u_{b_n}\}$ and $u_0\in X$ that satisfies $u_{b_n} \rightharpoonup u_0$ in $X$. Besides that, we can easily check that the sequence $\{ {\int}_{\mathbb{R}^{3}}|\nabla u_{b_n}|^2\mathrm{d}x {\int}_{\mathbb{R}^{3}}\nabla u_{b_n}\cdot\nabla \varphi \mathrm{d}x\}$ is bounded for all $\varphi\in\mathcal{C}_0^\infty(\mathbb{R}^3)$. Therefore, combining $b_n\to 0$ as $n\to +\infty$, we can arrive at

Through simple calculations, one has

for all $\varphi\in\mathcal{C}_0^\infty(\mathbb{R}^3)$, which implies that $\mathcal{I}_0'(u_0) = 0$.

In the following, we deduce that $\mathcal{I}_0(u_0)\le c_{0, 2}$. Because of $(f_2)$, we know that there is $A_0 > 0$ large enough such that $\mathcal{I}_{b_n}(sv_0^++tv_0^-) < 0$ for all $s+t\ge A_0$. Due to Lemma 2.5, we get that there is $(s_{b_n}, t_{b_n})\in (0, +\infty)\times(0, +\infty)$ that satisfies $s_{b_n}v_0^++t_{b_n}v_0^-\in \mathcal{M}_{b_n}$, where $v_0\in \mathcal{M}_0$ satisfies $\mathcal{I}_0(v_0) = c_{0, 2}$. Hence, we have $\mathcal{I}_{b_n}(s_{b_n}v_0^++t_{b_n}v_0^-)\ge 0$ by $(f_3)$, and then $0 < s_{b_n}, t_{b_n} < A_0$. Therefore, we can conclude that

Hence,

In light of $(f_3)$ and Fatou's Lemma, one gets

Next, we show that $u_0\in \mathcal{M}_0$ and $\mathcal{I}_0(u_0)\ge c_{0, 2}$. First, combining the boundedness of $\{u_{b_n}\}$ and (5.28), one deduces that

Besides, it is easy to check that

for all $\varphi\in\mathcal{C}_0^\infty(\mathbb{R}^3)$, which means that

Hence, owing to (5.30) and (5.31), we know that there is a subsequence of $\{u_{b_n}\}$, still denoted by $\{u_{b_n}\}$, which satisfies

In virtue of [10, Proposition 4.1], we get that $c_{0, 2} < c_{0, 1}+c_{0, \infty}$. According to ^[10], we know that $\mathcal{I}_0$ satisfies the $(PS)_{c^*}$ condition, where $c^* < c_{0, 1}+c_{0, \infty}$. Hence, by (5.32), we get that the $({PS})_c$ sequence $\{u_{b_n}\}$ has a convergent subsequence, still denoted by $\{u_{b_n}\}$. Then, one has $u_{b_n}\to u_0$ in $X$. Moreover, in light of $\{u_{b_n}\}\subset \mathcal{M}_{b_n}$ and Lemma 2.3, we can arrive at $\|u_{b_n}^{\pm}\|_X\geq C_1 > 0$, and then $\|u_0^{\pm}\|_X\geq C_1 > 0$. Noting that $\mathcal{I}_0'(u_0) = 0$, we can show that $u_0\in \mathcal{M}_0$. Therefore, we can obtain that $\mathcal{I}_0(u_0)\ge c_{0, 2}$. Combining with (5.29), we conclude that $u_0\in \mathcal{M}_0$ and $\mathcal{I}_0(u_0) = c_{0, 2}$. Namely, $u_0$ is a least energy sign-changing solution of Equation (1.3). The proof is completed. □

Author contributions

Yan-Fei Yang: Writing-original draft, Writing-review & editing; Chun-Lei Tang: Formal analysis, Methodology, Supervision, Writing-review & editing.

Use of AI tools declaration

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

Acknowledgments

This study was funded by the National Natural Science Foundation of China (No. 12371120).

Conflict of interest

All authors declare no conflicts of interest in this paper.