# Nodal solution for critical Kirchhoff-type equation with fast increasing weight in $$\mathbb{R}^{2}$$

## Abstract

In this paper, we investigate the existence of a least-energy sign-changing solutions for the following Kirchhoff-type equation:

$$- \biggl(1+b \int _{\mathbb{R}^{2}} K(x) \vert \nabla u \vert ^{2}\,dx \biggr) \operatorname{div} \bigl(K(x)\nabla u \bigr)=K(x)f(u),\quad x\in \mathbb{R}^{2},$$

where f has exponential subcritical or exponential critical growth in the sense of the Trudinger–Moser inequality. By using the constrained variational methods, combining the deformation lemma and Miranda’s theorem, we prove the existence of a least-energy sign-changing solution. Moreover, we also prove that this sign-changing solution has exactly two nodal domains.

## 1 Introduction and main results

In this present paper, we consider the existence of the least energy sign-changing solutions for the following equation:

$$- \biggl(1+b \int _{\mathbb{R}^{2}}K(x) \vert \nabla u \vert ^{2}\,dx \biggr) \operatorname{div} \bigl(K(x)\nabla u \bigr)=K(x)f(u), \quad x\in \mathbb{R}^{2},$$
(1.1)

where $$K(x)=\exp (|x|^{2}/4 )$$, b is a positive constant, and we assume that f satisfies:

$$(f_{0})$$:

$$f(t)\in C^{1} ( \mathbb{R},\mathbb{R} )$$;

$$(f_{1})$$:

$$f(t)=o{ ( |t| ) }$$ as $$|t|\to 0$$;

$$(f_{2})$$:

$$\lim_{|t|\to \infty}\frac{F(t)}{t^{4}}=\infty$$, where $$F(t)=\int _{0}^{t}f(s)\,ds$$;

$$(f_{3})$$:

$$\frac{f(t)}{t^{3}}$$ is an increasing function on $$\mathbb{R}\backslash \lbrace 0 \rbrace$$;

$$(f_{4})$$:

There exist

$$p>4 \quad \text{and}\quad \varrho _{0}> \biggl[4m_{p} \biggl( \frac{p-2}{p-4} \biggr) \frac{\alpha _{0}}{\pi} \biggr]^{ \frac{p-2}{2}} ,$$

such that

$$tf(t)\geq \varrho _{0} \vert t \vert ^{p},$$

for all $$t\in \mathbb{R}$$, where $$\alpha _{0}>0$$ and $$m_{p}$$ is attained in a ground state nodal energy of Eq. (1.1) when $$f(u)=|u|^{p-2}u$$.

As we all know, we call problems of type (1.1) nonlocal problems because there is an integral over $$\mathbb{R}^{2}$$. Such problems were first posed by G. Kirchhoff in [1] as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings.

Similar nonlocal problems also model several physical and biological systems, where u describes a process which depends on the average of itself, for example, the population density, see [2] and the references therein. After J.L. Lions [3] proposed the functional analysis method of the equation

$$u_{tt}- \biggl(a+b \int _{\Omega} \vert \nabla u \vert ^{2}\,dx \biggr) \Delta u=f(x,u),$$
(1.2)

where $$a,~b>0$$, and $$\Omega \subset \mathbb{R}^{N}$$ is a bounded domain, the steady-state form of the problem (1.2) has received a lot of attention. At the same time, many more results were obtained; we refer to [49] for bounded domains. In [6] the authors obtained sign-changing solutions to the nonlocal quasilinear elliptic boundary value problem using variational methods and invariant sets of descent flow in the subcritical case.

For the entire space $$\mathbb{R}^{N}(N\geq 3)$$, we know that the embedding $$H^{1}(\mathbb{R}^{N}) \hookrightarrow L^{q}(\mathbb{R}^{N})$$ $$( 2\leq q<2^{*} )$$ is not compact. In order to overcome the lack of compactness, many researchers introduced the potential function $$V(x)$$, to study the Kirchhof-type equation of the following form:

$$- \biggl(a+b \int _{\mathbb{R}^{N}} \vert \nabla u \vert ^{2}\,dx \biggr)\Delta u+V(x)u=f(x,u),$$
(1.3)

restoring spatial compactness by making different assumptions about $$V(x)$$. In [10], the author showed that problem (1.3) has sign-changing solutions, if we assume $$V \in C(\mathbb{R}^{3}, \mathbb{R})$$ satisfies $$\inf_{x\in \mathbb{R}^{3}} V(x)\geq a_{1} >0$$ and, for each $$A > 0$$, $$\mathrm{meas} \lbrace x\in \mathbb{R}^{3}: V(x)\leq A \rbrace <\infty$$, with $$a_{1}$$ being a constant and meas denoting the Lebesgue measure in $$\mathbb{R}^{3}$$. In [11], the author got a positive solution to the problem (1.3), considering $$V(x)$$ as a locally Hölder continuous function, and assuming there is a constant α such that $$V(x)\geq \alpha >0$$ for all $$x\in \mathbb{R}^{3}$$ and $$\inf_{x\in \Lambda}V(x)<\min_{x\in \partial \Lambda} V(x)$$, where Λ is an open bounded set. There are many diverse results for equations of type (1.3) in $$\mathbb{R}^{N}$$; we refer to [1215] and the references therein. In fact, by observation, we can see that our problem can be viewed as a generalization of the constant-coefficient Kirchhoff equation, when $$K(x)=1$$, it is exactly the Kirchhoff equation as in (1.3). At the same time, we use the properties of function $$K(x)$$ to avoid using potential function $$V(x)$$ to overcome the problem of lost space embedding compactness.

It is well known that the critical growth for nonlinear terms also leads to the loss of compactness for the embedding $$H^{1}(\mathbb{R}^{N})\hookrightarrow L^{2^{*}}(\mathbb{R}^{N})$$, where the critical Sobolev exponent is $$2^{*}=2N/(N-2)$$ ($$N>3$$). When $$N=2$$, the critical exponential growth is related to Trudinger–Moser inequality, which appears in the pioneer work [16, 17], that is,

$$\sup_{\lVert u\rVert _{H^{1}_{0}(\Omega )} \leq 1} \int _{\Omega}e^{ \alpha u^{2}}\leq C(\alpha )$$

for all $$\alpha \leq 4\pi$$ and $$\Omega \subset \mathbb{R}^{2}$$. Motivated by this inequality, de Figueiredo et al. [18] introduced the notion of subcritical and critical growth in the plane, i.e.,

$$(f_{5})$$:

$$f\in C ( \mathbb{R},\mathbb{R} )$$ and there exists $$\alpha _{0}\geq 0$$ such that

$$\lim_{|t|\to \infty}\frac{f(t)}{e^{\alpha |t|^{2}}}= \textstyle\begin{cases} 0,&\alpha >\alpha _{0}, \\ \infty ,&\alpha < \alpha _{0}. \end{cases}$$

If the above holds for all $$\alpha >0$$, we say that f has exponential subcritical growth at +∞, and if there exists $$\alpha _{0}>0$$ as above then f has exponential critical growth at +∞. When dealing with the entire space, we need a new version of the Trudinger–Moser inequality. It asserts that

$$\sup_{\lVert u\rVert _{H^{1}_{0}(\mathbb{R}^{2})} \leq 1} \int _{ \mathbb{R}^{2}} \bigl( e^{\alpha u^{2}}-1 \bigr) \leq C(\alpha )$$

for all $$\alpha \leq 4\pi$$; see [19, 20] and the references therein.

To obtain our results, we consider using the variational method in a weighted Sobolev space consisting of rapidly decaying functions at infinity, where the embedding of $$\mathbb{R}^{2}$$ is recovered in the weighted Sobolev space. This idea was first proposed by M. Escobedo and O. Kavian in [21], mainly used to find a self-similar solution of the heat equation in $$\mathbb{R}^{N}$$, more precisely, they define the weighting function

$$K(x):= \exp \biggl(\frac{ \vert x \vert ^{2}}{4} \biggr), \quad \text{for } x\in \mathbb{R}^{N}.$$

For scholars interested in weighted Sobolev spaces, we recommend [2228]. In [27], the author proves that the weighted semilinear elliptic problem has a sign-changing solution in the critical case, where the nonlinear term f satisfies the standard Ambrosetti–Rabinowitz superlinearity condition (namely, there exists $$\theta >2$$ such that $$tf(t)\geq \theta F(t)>0$$). In our paper, we directly use the Trudinger–Moser inequality in the weighted space considered in [29]; see Lemma 2.1.

Now, we introduce our work space. Consider $$C_{c}^{\infty}(\mathbb{R}^{2})$$, the space of infinitely differentiable functions with compact support, and denote by X the closure of $$C_{c}^{\infty}(\mathbb{R}^{2})$$ with respect to the norm

$$\rVert u\rVert = \biggl( \int _{\mathbb{R}^{2}}K(x) \vert \nabla u \vert ^{2}\,dx \biggr)^{\frac{1}{2}},$$

which is induced by the inner product

$$\langle u,v \rangle = \int _{\mathbb{R}^{2}} K(x)\nabla u \nabla v\,dx.$$

Define the weighted spaces for each $$s\geq 2$$ as

$$L_{K}^{s}\bigl(\mathbb{R}^{2}\bigr)= \biggl\{ u \text{ measurable in } \mathbb{R}^{2}: \int _{\mathbb{R}^{2}}K(x) \vert u \vert ^{s}\,dx< { \infty} \biggr\} .$$

By the results from [21, 22, 28] and Lemma 2.1 of [29], the space X is complete and the embedding $$\text{X}\hookrightarrow L_{K}^{s}(\mathbb{R}^{2})$$ is continuous and compact for all $$s\in [ 2,\infty )$$. Note that $$X\nsubseteq L_{K}^{\infty}(\mathbb{R}^{2})$$, thus we use the Trudinger–Moser inequality in $$\mathbb{R}^{2}$$ as a substitution of the Sobolev inequality.

From $$(f_{1})$$, for all $$\varepsilon >0$$, there exists $$\delta >0$$ such that, when $$|t|<\delta$$, we have

$$\bigl\vert f(t) \bigr\vert \leq \varepsilon \vert t \vert .$$
(1.4)

Let $$\alpha >\alpha _{0}$$ be given by $$(f_{5})$$ and $$q\geq 2$$. By using the critical growth of f, we obtain

$$\lim_{t\to +\infty} \frac{ \vert f(t) \vert }{ \vert t \vert ^{q-1}(e^{\alpha t^{2}}-1)}=0.$$
(1.5)

Therefore, for all $$\varepsilon >0$$, $$t\in \mathbb{R}$$, there exists $$C_{\varepsilon }$$ such that

$$\max \bigl\lbrace \bigl\vert f(t)t \bigr\vert , \bigl\vert F(t) \bigr\vert \bigr\rbrace \leq \varepsilon \vert t \vert ^{2}+C_{ \varepsilon} \vert t \vert ^{q} \bigl(e^{\alpha t^{2}}-1\bigr).$$
(1.6)

The problem (1.1) corresponds to the energy functional $$I:X\to \mathbb{R}$$ which can be constructed as

$$I_{b}(u)=\frac{1}{2} \int _{\mathbb{R}^{2}}K(x) \vert \nabla u \vert ^{2}\,dx+ \frac{b}{4} \biggl( \int _{\mathbb{R}^{2}} K(x) \vert \nabla u \vert ^{2}\,dx \biggr)^{2}- \int _{\mathbb{R}^{2}} K(x)F(u)\,dx.$$
(1.7)

By assumptions on f and a standard argument, we can affirm that $$I_{b}$$ is a well-defined $$C^{1}$$ functional, and its derivative can be computed as

$$\bigl\langle I^{\prime}_{b}(u),\varphi \bigr\rangle = \bigl(1+b \rVert u\rVert ^{2} \bigr) \int _{\mathbb{R}^{2}} K(x)\nabla u \nabla \varphi\,dx- \int _{\mathbb{R}^{2}} K(x)f(u)\varphi\,dx,$$
(1.8)

for all $$\varphi \in X$$. Furthermore, u is a sign-changing solution of system (1.1) if and only if u is a critical point of $$I_{b}$$ and $$u^{\pm}\neq 0$$, where

$$u^{+}:=\max (u,0),\qquad u^{-}:=\min (u,0).$$

Motivated by [5, 10], in order to find a sign-changing solution of equation (1.1), we make the following decompositions for $$u\in X$$:

\begin{aligned} &I_{b}(u)=I_{b}\bigl(u^{+} \bigr)+I_{b}\bigl(u^{-}\bigr)+\frac{b}{2} \int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{+} \bigr\vert ^{2}\,dx \int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{-} \bigr\vert ^{2}\,dx, \end{aligned}
(1.9)
\begin{aligned} &\bigl\langle I^{\prime}_{b}(u),u^{+} \bigr\rangle =\bigl\langle I^{\prime}_{b}\bigl(u^{+} \bigr),u^{+} \bigr\rangle +b \int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{+} \bigr\vert ^{2}\,dx \int _{ \mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{-} \bigr\vert ^{2}\,dx, \end{aligned}
(1.10)
\begin{aligned} &\bigl\langle I^{\prime}_{b}(u),u^{-} \bigr\rangle =\bigl\langle I^{\prime}_{b}\bigl(u^{-} \bigr),u^{-} \bigr\rangle +b \int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{+} \bigr\vert ^{2}\,dx \int _{ \mathbb{R}^{2}}K(x) \bigl\vert \nabla u^{-} \bigr\vert ^{2}\,dx. \end{aligned}
(1.11)

Meanwhile, we consider the Nehari manifold and Nehari nodal set associated to (1.7) defined respectively by

$$\mathcal{N}=\bigl\{ u\in X\backslash \{0\}:\bigl\langle I^{\prime}_{b}(u),u \bigr\rangle =0\bigr\}$$

and

$$\mathcal{M}=\bigl\{ u\in X; u^{\pm}\neq 0:\bigl\langle I^{\prime}_{b}(u),u^{+} \bigr\rangle =\bigl\langle I^{\prime}_{b}(u),u^{-}\bigr\rangle =0 \bigr\} .$$

In this paper, we have the following result.

### Theorem 1.1

(Subcritical case)

Assuming $$(f_{5})$$ with $$\alpha _{0}=0$$ and $$(f_{0})$$$$(f_{3})$$ hold, equation (1.1) has a least-energy sign-changing solution, which has precisely two nodal domains.

### Theorem 1.2

(Critical case)

Assuming $$(f_{5})$$ with $$\alpha _{0}>0$$ and $$(f_{0})$$$$(f_{4})$$ hold, equation (1.1) has a least-energy sign-changing solution, which has precisely two nodal domains.

We organize this paper as follows. In Sect. 2 we give some useful preliminary lemmas which pave the way for getting a least-energy sign-changing solution. Then Sect. 3 is devoted to proving Theorems 1.1 and 1.2.

## 2 Some preliminary lemmas

According to [29], the following version of the Trudinger–Moser inequality holds:

### Lemma 2.1

For any $$r\geq 0$$, $$u\in X$$, we have $$K(x)|u|^{r+2}\in L^{1}(\mathbb{R}^{2})$$. If $$\| u\| \leq M$$, $$\varsigma M^{2}<4\pi$$, then there exists $$C=C(M,r,\varsigma )>0$$ such that

$$\int _{\mathbb{R}^{2}}K(x) \vert u \vert ^{2+r}\bigl[\exp \bigl(\varsigma u^{2}\bigr)-1\bigr]\,dx\leq C(M,r, \varsigma ) \Vert u \Vert ^{r}.$$
(2.1)

### Proof

See [29, Theorem 1.1 and Corollary 1.2]. □

Next, we prove that the set $$\mathcal{M}$$ is nonempty. In this proof, we adopt in part the idea of Zhong and Tang [30].

### Lemma 2.2

Suppose that f satisfies $$(f_{0})$$$$(f_{3})$$. For any $$u\in X$$ with $$u^{\pm}\neq 0$$, there exists a unique pair of numbers $$s_{u},t_{u}>0$$ such that $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}$$ and $$I_{b}(s_{u}u^{+}+t_{u}u^{-})= \max_{s,t\geq 0}I_{b}(su^{+}+tu^{-})$$.

### Proof

Fix $$u\in X$$ with $$u^{\pm }\neq 0$$. We first verify the existence of $$( s_{u},t_{u})$$. Write

\begin{aligned} I_{b}\bigl(su^{+}+tu^{-} \bigr) ={}&\frac{1}{2} \bigl\Vert su^{+}+tu^{-} \bigr\Vert ^{2}+ \frac{b}{4} \bigl\Vert su^{+}+tu^{-} \bigr\Vert ^{4}- \int _{\mathbb{R}^{2}}K(x)F\bigl(su^{+}+tu^{-} \bigr)\,dx \\ ={}&\frac{1}{2}s^{2} \bigl\Vert u^{+} \bigr\Vert ^{2}+\frac{b}{4}s^{4} \bigl\Vert u^{+} \bigr\Vert ^{4}- \int _{ \mathbb{R}^{2}}K(x)F\bigl(su^{+}\bigr)\,dx+ \frac{b}{2}s^{2}t^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \\ & {}+\frac{1}{2}t^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}+\frac{b}{4}t^{4} \bigl\Vert u^{-} \bigr\Vert ^{4}- \int _{\mathbb{R}^{2}}K(x)F\bigl(tu^{-}\bigr)\,dx. \end{aligned}

Let $$\varPhi (s,t)=I_{b}(su^{+}+tu^{-})$$ and use $$\varPhi ^{u}$$ to represent the gradient at $$(s, t)$$, i.e., $$\varPhi ^{u}= ( \varPhi _{s}^{\prime}(s,t), \varPhi _{t}^{ \prime}(s,t) )= (I^{\prime}_{b}(su^{+}+tu^{-})u^{+}, I^{ \prime}_{b}(su^{+}+tu^{-})u^{-} )$$, and then

$$\textstyle\begin{cases} \varPhi _{s}^{\prime}(s,t)= s \Vert u^{+} \Vert ^{2}+bs^{3} \Vert u^{+} \Vert ^{4}+bst^{2} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}-\int _{\mathbb{R}^{2}}K(x)f(su^{+})(u^{+})\,dx, \\ \varPhi _{t}^{\prime}(s,t)=t \Vert u^{-} \Vert ^{2}+bt^{3} \Vert u^{-} \Vert ^{4}+bs^{2}t \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}-\int _{\mathbb{R}^{2}}K(x)f(tu^{-})(u^{-})\,dx. \end{cases}$$
(2.2)

Combining $$(f_{1})$$$$(f_{3})$$, it is easy to verify $$\varPhi _{s}^{\prime}(s,s)>0$$, $$\varPhi _{t}^{\prime}(s,s)>0$$ for $$s>0$$ small enough and $$\varPhi _{s}^{\prime}(t,t)<0$$, $$\varPhi _{t}^{\prime}(t,t)<0$$ for $$t>0$$ large enough. Then there exists $$0\leq r\leq R$$ such that

$$\varPhi _{s}^{\prime}(r,r)>0, \qquad \varPhi _{t}^{\prime}(r,r)>0; \qquad \varPhi _{s}^{\prime}(R,R)< 0, \qquad \varPhi _{t}^{\prime}(R,R)< 0.$$
(2.3)

By the monotonicity with respect to $$s>0$$ (resp. $$t>0$$) if $$t>0$$ (resp. $$s>0$$) is fixed, one has

\begin{aligned}& \varPhi _{s}^{\prime}(r,t)>0,\qquad \varPhi _{s}^{\prime}(R,t)< 0, \quad \text{for all } t\in [r,R] ,\\& \varPhi _{t}^{\prime}(s,r)>0,\qquad \varPhi _{t}^{\prime}(s,R)< 0, \quad \text{for all } s\in [r,R] . \end{aligned}

It follows from the Miranda’s theorem [31] that there exists a pair $$(s_{u},t_{u})\in [r,R] \times [r,R]$$ such that

$$\varPhi _{s}^{\prime}(s_{u},t_{u})=0, \qquad \varPhi _{t}^{\prime}(s_{u},t_{u})=0,$$

which implies that $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}$$, i.e., $$\mathcal{M}\neq \emptyset$$.

Next, we will prove that the positive number pair $$(s_{u},t_{u})$$ is unique. We suppose that there are two pairs of positive numbers $$(s_{u_{1}},t_{u_{1}})$$, $$(s_{u_{2}},t_{u_{2}})$$ satisfying $$\varPhi _{s}^{\prime}(s_{u_{i}},t_{u_{i}})=0$$, $$i=1,2$$. Without loss of generality, we assume $$s_{u_{1}}< s_{u_{2}}$$ and that there exists a unique $$s_{u}$$ such that $$\varPhi _{s}^{\prime}(s_{u},t_{u})=0$$. From $$\varPhi _{s}^{\prime}(s_{u_{i}},t_{u_{i}})=0$$ we derive that

$$s_{u_{1}} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u_{1}}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u_{1}}t_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}= \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u_{1}}u^{+} \bigr) \bigl(u^{+}\bigr)\,dx$$
(2.4)

and

$$s_{u_{2}} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u_{2}}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u_{2}}t_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}= \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u_{2}}u^{+} \bigr) \bigl(u^{+}\bigr)\,dx,$$
(2.5)

and then, combing (2.4) with (2.5), we have

\begin{aligned} & \biggl(\frac{1}{s_{u_{1}}^{2}}-\frac{1}{s_{u_{2}}^{2}} \biggr) \bigl\Vert u^{+} \bigr\Vert ^{2}+b \biggl( \frac{1}{s_{u_{1}}^{2}}-\frac{1}{s_{u_{2}}^{2}} \biggr)t_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \\ &\quad = \int _{\mathbb{R}^{2}}K(x) \biggl( \frac{f(s_{u_{1}}u^{+})}{(s_{u_{1}}u^{+})^{3}}- \frac{f(s_{u_{2}}u^{+})}{(s_{u_{2}}u^{+})^{3}} \biggr) \bigl(u^{+}\bigr)^{4}\,dx. \end{aligned}
(2.6)

We know that the left-hand side of the latter equality is positive due to assumption $$s_{u_{1}}< s_{u_{2}}$$. At the same time, using hypothesis $$(f_{3})$$, we can see that the right-hand side is negative, which leads to a contradiction. Therefore, we have $$s_{u_{1}}=s_{u_{2}}$$, so $$s_{u}$$ is unique. The proof of $$t_{u}$$ uniqueness is similar.

The existence of an extreme value of $$\varPhi (s, t)$$ at $$(s_{u}, t_{u})$$ is verified by using the sufficient condition for the existence of an extreme value of a binary function:

$$\textstyle\begin{cases} \varPhi _{ss}^{\prime \prime}(s,t)= \Vert u^{+} \Vert ^{2}+3bs^{2} \Vert u^{+} \Vert ^{4}+bt^{2} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}-\int _{\mathbb{R}^{2}}K(x)f^{\prime}(su^{+})(u^{+})^{2}\,dx, \\ \varPhi _{st}^{\prime \prime}(s,t)= 2bst \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}= \varPhi _{ts}^{\prime \prime}(s,t), \\ \varPhi _{tt}^{\prime \prime}(s,t)= \Vert u^{-} \Vert ^{2}+3bt^{2} \Vert u^{-} \Vert ^{4}+bs^{2} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}-\int _{\mathbb{R}^{2}}K(x)f^{\prime}(tu^{-})(u^{-})^{2}\,dx. \end{cases}$$
(2.7)

Substituting point $$(s_{u},t_{u})$$ into (2.7), we have

$$\textstyle\begin{cases} \varPhi _{ss}^{\prime \prime}(s_{u},t_{u})= \Vert u^{+} \Vert ^{2}+3bs_{u}^{2} \Vert u^{+} \Vert ^{4}+bt_{u}^{2} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}\\ \hphantom{\varPhi _{ss}^{\prime \prime}(s_{u},t_{u})=}{}-\int _{ \mathbb{R}^{2}}K(x)f^{\prime}(s_{u}u^{+})(u^{+})^{2}\,dx, \\ \varPhi _{st}^{\prime \prime}(s_{u},t_{u})= 2bs_{u}t_{u} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}=\varPhi _{ts}^{\prime \prime}(s_{u},t_{u}), \\ \varPhi _{tt}^{\prime \prime}(s_{u},t_{u})= \Vert u^{-} \Vert ^{2}+3bt_{u}^{2} \Vert u^{-} \Vert ^{4}+bs_{u}^{2} \Vert u^{+} \Vert ^{2} \Vert u^{-} \Vert ^{2}\\ \hphantom{\varPhi _{tt}^{\prime \prime}(s_{u},t_{u})=}{}-\int _{ \mathbb{R}^{2}}K(x)f^{\prime}(t_{u}u^{-})(u^{-})^{2}\,dx, \end{cases}$$
(2.8)

then, combing $$\varPhi _{s}^{\prime}(s_{u},t_{u})=0$$ with hypothesis $$(f_{3})$$, we obtain that

\begin{aligned}[b] \varPhi _{ss}^{\prime \prime}(s_{u},t_{u})&= \bigl\Vert u^{+} \bigr\Vert ^{2}+3bs_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{4}+bt_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}- \int _{ \mathbb{R}^{2}}K(x)f^{\prime}\bigl(s_{u}u^{+} \bigr) \bigl(u^{+}\bigr)^{2}\,dx \\ &< \bigl\Vert u^{+} \bigr\Vert ^{2}+3bs_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{4}+bt_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}-3 \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u}u^{+} \bigr) \frac{1}{s_{u}}\bigl(u^{+}\bigr)\,dx \\ &=-2 \bigl\Vert u^{+} \bigr\Vert ^{2}-2t_{u}^{2}b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \end{aligned}
(2.9)

and

\begin{aligned} \varPhi _{tt}^{\prime \prime}(s_{u},t_{u}) =& \bigl\Vert u^{-} \bigr\Vert ^{2}+3bt_{u}^{2} \bigl\Vert u^{-} \bigr\Vert ^{4}+bs_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}- \int _{ \mathbb{R}^{2}}K(x)f^{\prime}\bigl(s_{u}u^{-} \bigr) \bigl(u^{-}\bigr)^{2}\,dx \\ < & \bigl\Vert u^{-} \bigr\Vert ^{2}+3bt_{u}^{2} \bigl\Vert u^{-} \bigr\Vert ^{4}+bs_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}-3 \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u}u^{-} \bigr) \frac{1}{s_{u}}\bigl(u^{-}\bigr)\,dx \\ =&-2 \bigl\Vert u^{-} \bigr\Vert ^{2}-2s_{u}^{2}b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \end{aligned}
(2.10)

hold. Since, obviously,

$$\varPhi _{ss}^{\prime \prime}(s_{u},t_{u})< 0,$$

from (2.8)–(2.10) we get

\begin{aligned}[b] &\varPhi _{ss}^{\prime \prime}(s_{u},t_{u}) \varPhi _{tt}^{ \prime \prime}(s_{u},t_{u})- \varPhi _{st}^{\prime \prime 2}(s_{u},t_{u}) \\ &\quad > \bigl(2 \bigl\Vert u^{+} \bigr\Vert ^{2}+2t_{u}^{2}b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \bigr) \bigl(2 \bigl\Vert u^{-} \bigr\Vert ^{2}+2s_{u}^{2}b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \bigr)\\ &\qquad {}- \bigl(2bs_{u}t_{u} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \bigr)^{2} \\ &\quad >0. \end{aligned}
(2.11)

Thus we can get the maximum value of $$\varPhi (s,t)$$ at $$(s_{u},t_{u})$$. The proof is complete. □

### Lemma 2.3

Assume that $$(f_{0})$$$$(f_{3})$$ and $$(f_{5})$$ hold, as well as $$u\in X$$ and $$u^{\pm }\neq 0$$. Then we have:

1. (i)

If $$\varPhi _{s}^{\prime}(1,1)\leq 0$$, $$\varPhi _{t}^{\prime}(1,1)\leq 0$$, there is a unique positive number pair $$(s_{u},t_{u})$$ obtained in Lemma 2.2, satisfying $$0< s_{u},t_{u}\leq 1$$, such that $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}$$.

2. (ii)

If $$\varPhi _{s}^{\prime}(1,1)\geq 0$$, $$\varPhi _{t}^{\prime}(1,1)\geq 0$$, there is a unique positive number pair $$(s_{u},t_{u})$$ obtained in Lemma 2.2, satisfying $$s_{u},t_{u}\geq 1$$, such that $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}$$.

### Proof

(i) Assuming that $$s_{u}\geq t_{u}>0$$, in view of $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}$$, we have

\begin{aligned} s_{u} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \geq &s_{u} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u}t_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \\ =& \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u}u^{+} \bigr) \bigl(u^{+}\bigr)\,dx. \end{aligned}
(2.12)

From the hypothesis $$\varPhi _{s}^{\prime}(1,1)\leq 0$$, we have

$$\bigl\Vert u^{+} \bigr\Vert ^{2}+b \bigl\Vert u^{+} \bigr\Vert ^{4}+b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}\leq \int _{\mathbb{R}^{2}}K(x)f\bigl(u^{+}\bigr) \bigl(u^{+}\bigr)\,dx.$$
(2.13)

Combing (2.12) with (2.13), we get

$$\biggl(\frac{1}{s_{u}^{2}}-1 \biggr) \bigl\Vert u^{+} \bigr\Vert ^{2}\geq \int _{ \mathbb{R}^{2}}K(x) \biggl[\frac{f(s_{u}u^{+})}{(s_{u}u^{+})^{3}}- \frac{f(u^{+})}{(u^{+})^{3}} \biggr]\bigl(u^{+}\bigr)^{4}\,dx.$$
(2.14)

If $$s_{u}>1$$, then the left-hand side of this inequality is negative, but from $$(f_{3})$$ the right-hand side is positive, so (2.14) yields a contradiction. Therefore we conclude $$s_{u}\leq 1$$. Using a similar method, we can prove that $$t_{u}\leq 1$$.

(ii) Similarly, assuming that $$0< s_{u}\leq t_{u}$$ and using the fact that $$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M,}$$ we get

\begin{aligned} s_{u} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \leq& s_{u} \bigl\Vert u^{+} \bigr\Vert ^{2}+bs_{u}^{3} \bigl\Vert u^{+} \bigr\Vert ^{4}+bs_{u}t_{u}^{2} \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2} \\ =& \int _{\mathbb{R}^{2}}K(x)f\bigl(s_{u}u^{+} \bigr) \bigl(u^{+}\bigr)\,dx. \end{aligned}
(2.15)

From the assumption $$\varPhi _{s}^{\prime}(1,1)\geq 0$$, we have

$$\bigl\Vert u^{+} \bigr\Vert ^{2}+b \bigl\Vert u^{+} \bigr\Vert ^{4}+b \bigl\Vert u^{+} \bigr\Vert ^{2} \bigl\Vert u^{-} \bigr\Vert ^{2}\geq \int _{\mathbb{R}^{2}}K(x)f\bigl(u^{+}\bigr) \bigl(u^{+}\bigr)\,dx.$$
(2.16)

Now combing (2.15) with (2.16), we get

$$\biggl(\frac{1}{s_{u}^{2}}-1 \biggr) \bigl\Vert u^{+} \bigr\Vert ^{2}\leq \int _{ \mathbb{R}^{2}}K(x) \biggl[\frac{f(s_{u}u^{+})}{(s_{u}u^{+})^{3}}- \frac{f(u^{+})}{(u^{+})^{3}} \biggr]\bigl(u^{+}\bigr)^{4}\,dx.$$
(2.17)

If $$s_{u}<1$$, then the two sides of (2.17) are contradictory, therefore we conclude $$s_{u}\geq 1$$. Using a similar method, we can prove that $$t_{u}\geq 1$$. □

### Lemma 2.4

Assume that $$(f_{0})$$$$(f_{3})$$ and $$(f_{5})$$ hold. Then there exists $$\rho >0$$ such that $$\Vert u\Vert \geq \rho$$ for all $$u\in \mathcal{M}$$. Furthermore, $$m:=\inf \{I_{b}(u):u\in \mathcal{M}\}>0$$.

### Proof

Suppose, to the contrary, that there exists $$\lbrace u_{n} \rbrace \subset \mathcal{M}$$ such that $$\| u_{n}\|\to 0$$. Using (1.6), we obtain

\begin{aligned}[b] \Vert u_{n} \Vert ^{2}&\leq \Vert u_{n} \Vert ^{2}+b \Vert u_{n} \Vert ^{4} \\ &\leq \varepsilon \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{2}\,dx+C_{ \varepsilon} \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{q}\bigl(e^{\alpha u_{n}^{2}}-1\bigr)\,dx \\ &=\varepsilon \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{2}\,dx+C_{\varepsilon} \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{q} \bigl(e^{\alpha \Vert u_{n} \Vert ^{2} (\frac{u_{n}}{ \Vert u_{n} \Vert } )^{2}}-1 \bigr)\,dx. \end{aligned}
(2.18)

Using Sobolev embedding theorem and Hölder’s inequality with $$s',s>1$$, we get

$$\Vert u_{n} \Vert ^{2}\leq \varepsilon S_{2}^{-1} \Vert u_{n} \Vert ^{2}+C_{ \varepsilon} \biggl( \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{qs'}\,dx \biggr)^{ \frac{1}{s'}} \biggl( \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha \Vert u_{n} \Vert ^{2} (\frac{u_{n}}{ \Vert u_{n} \Vert } )^{2}}-1 \bigr)^{s}\,dx \biggr)^{\frac{1}{s}},$$

which after a rearrangement yields

$$\bigl(1-\varepsilon S_{2}^{-1}\bigr) \Vert u_{n} \Vert ^{2}\leq C_{\varepsilon} \biggl( \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{qs'}\,dx \biggr)^{\frac{1}{s'}} \biggl( \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha \Vert u_{n} \Vert ^{2} (\frac{u_{n}}{ \Vert u_{n} \Vert } )^{2}}-1 \bigr)^{s}\,dx \biggr)^{ \frac{1}{s}}.$$

Arguing as in the proof of Lemma 3.4 in [32], there exists $$\tilde{C_{\varepsilon}}$$ such that

$$\bigl(1-\varepsilon S_{2}^{-1}\bigr) \Vert u_{n} \Vert ^{2}\leq \tilde{C_{\varepsilon}} \biggl( \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{qs'}\,dx \biggr)^{ \frac{1}{s'}} \biggl( \int _{\mathbb{R}^{2}}K(x) \bigl(e^{s\alpha \Vert u_{n} \Vert ^{2} (\frac{u_{n}}{ \Vert u_{n} \Vert } )^{2}}-1 \bigr)\,dx \biggr)^{ \frac{1}{s}}.$$

Let $$v_{n}:=\frac{u_{n}}{\| u_{n}\|}$$, then $$\| v_{n}\|^{2}=1$$. Since $$\| u_{n}\|\to 0$$, there exists $$\beta <4\pi$$ such that $$s\alpha \| u_{n}\|^{2}<\beta$$ holds. For $$q>2$$, using Lemma 2.1 and the embedding theorem, there exists a constant $$\tilde{C_{\varepsilon}}$$ such that

$$\bigl(1-\varepsilon S_{2}^{-1}\bigr) \Vert u_{n} \Vert ^{2}\leq M\tilde{C_{\varepsilon}} \biggl( \int _{\mathbb{R}^{2}}K(x) \vert u_{n} \vert ^{qs'}\,dx \biggr)^{ \frac{1}{s'}}\leq M\tilde{C_{\varepsilon}} S_{qs'}^{-\frac{q}{2}} \Vert u_{n} \Vert ^{q}.$$

By simplifying we get

$$\frac{(1-\varepsilon S_{2}^{-1})}{M\tilde{C_{\varepsilon}} S_{qs'}^{-\frac{q}{2}}} \leq \Vert u_{n} \Vert ^{q-2}.$$

By arbitrariness of ε, there is a constant $$\rho = [ \frac{(1-\varepsilon S_{2}^{-1})}{M\tilde{C_{\varepsilon}} S_{qs'}^{-\frac{q}{2}}} ]^{\frac{1}{q-2}}>0$$ such that $$\| u_{n}\|\geq{\rho}>0$$.

Now assume that $$\{u_{n}\}\subset \mathcal{M}$$ is a minimizing sequence for m. Using hypothesis $$(f_{3})$$, we get

$$m = \lim_{n\to \infty} \inf \biggl[ I(u_{n})-\frac{1}{4} \bigl\langle I^{\prime}(u_{n}),u_{n} \bigr\rangle \biggr] \geq \frac{1}{4}\lim_{n\to \infty}\inf \Vert u_{n} \Vert ^{2} \geq \frac{1}{4}\rho ^{2} >0,$$
(2.19)

which completes the proof. □

Because $$\lbrace u_{n} \rbrace$$ is bounded in X, there exists $$u\in \text{X}$$ such that $$u_{n}^{\pm}\rightharpoonup u^{\pm }$$ in X. Since $$\{u_{n}\}\subset \mathcal{M}$$, one has $$\langle I^{\prime}(u_{n}),u_{n}^{\pm} \rangle =0$$, i.e.,

\begin{aligned} &\int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u_{n}^{\pm} \bigr\vert ^{2}\,dx+b \int _{ \mathbb{R}^{2}} K(x) \vert \nabla u \vert ^{2}\,dx \int _{\mathbb{R}^{2}}K(x) \bigl\vert \nabla u_{n}^{\pm} \bigr\vert ^{2}\,dx \\ &\quad = \int _{\mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{ \pm}\bigr)\,dx. \end{aligned}
(2.20)

Since $$\| u_{n}^{\pm}\|\geq \rho >0$$, using (1.6), we have

\begin{aligned}[b] \rho ^{2}\leq \bigl\Vert u_{n}^{\pm} \bigr\Vert ^{2}&\leq \int _{ \mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{\pm}\bigr)\,dx \\ &\leq \varepsilon \int _{\mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{2}\,dx+C_{ \varepsilon} \int _{\mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{q}\bigl(e^{\alpha (u_{n}^{ \pm})^{2}}-1\bigr)\,dx. \end{aligned}
(2.21)

By the boundedness of $$\lbrace u_{n} \rbrace$$ in X, there exists $$C_{1}$$ such that

\begin{aligned}[b] \rho ^{2}&\leq \varepsilon C_{1}+C_{\varepsilon} \int _{ \mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{q}\bigl(e^{\alpha (u_{n}^{\pm})^{2}}-1\bigr)\,dx \\ &\leq \varepsilon C_{1}+\tilde{C_{\varepsilon}} \biggl( \int _{ \mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{qs^{\prime}} \biggr)^{ \frac{1}{s^{\prime}}} \biggl( \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha s(u_{n}^{ \pm})^{2}}-1\bigr)\,dx \biggr)^{\frac{1}{s}}, \end{aligned}
(2.22)

from which we get

$$\rho ^{2}-\varepsilon C_{1}\leq M \tilde{C_{\varepsilon}} \biggl( \int _{ \mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{qs'}\,dx \biggr)^{\frac{1}{s'}}.$$
(2.23)

Choosing $$\varepsilon =\frac{\rho ^{2}}{2C_{1}}$$, we have

$$0< \frac{\rho ^{2}}{2M\tilde{C_{\varepsilon}}}\leq \biggl( \int _{ \mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{qs'}\,dx \biggr)^{\frac{1}{s'}}.$$
(2.24)

Since $$qs'>2$$, we conclude that $$u_{n}^{\pm}\rightarrow u^{\pm }$$ in $$L^{qs'} ( \mathbb{R}^{2} )$$. So, we have

$$\biggl( \int _{\mathbb{R}^{2}}K(x) \bigl\vert u^{\pm} \bigr\vert ^{qs'}\,dx \biggr)^{ \frac{1}{s'}}\geq \frac{\rho ^{2}}{2M\tilde{C_{\varepsilon}}}>0.$$
(2.25)

Therefore $$u^{\pm}\neq 0$$.

## 3 Proof of theorems

In this section, we will prove our main results. We first deal with the subcritical case $$( \alpha _{0}=0 )$$, it is related to the convergence of involved functions f and F, see $$(f_{0})$$$$(f_{5})$$.

### Lemma 3.1

Let $$\{u_{n}^{\pm}\}\subset \mathcal{M}$$ be a minimizing sequence for m. Then there exists $$u^{\pm }\in X$$ such that

$$\int _{\mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{\pm}\bigr)\,dx\rightarrow \int _{\mathbb{R}^{2}}K(x)f\bigl(u^{\pm}\bigr) \bigl(u^{\pm}\bigr)\,dx$$
(3.1)

and

$$\int _{\mathbb{R}^{2}}K(x)F\bigl(u_{n}^{\pm} \bigr)\,dx\rightarrow \int _{ \mathbb{R}^{2}}K(x)F\bigl(u^{\pm}\bigr)\,dx .$$
(3.2)

### Proof

According to Lemma 2.4, there exists $$M_{1} >0$$ such that

$$\bigl\Vert u_{n}^{\pm} \bigr\Vert ^{2}\leq M_{1},\quad \forall n\in \mathbb{N},$$
(3.3)

and there exists a function $$u\in X$$ such that $$u_{n}^{\pm}(x)\to u^{\pm}(x)$$ and $$f ( u_{n}^{\pm}(x) ) (u_{n}^{\pm}(x))\to f ( u^{\pm}(x) ) (u^{\pm}(x))$$ a.e. in $$\mathbb{R}^{2}$$. In order to prove the first limit, the generalized Lebesgue convergence theorem is used here. Letting $$g:\mathbb{R}\to \mathbb{R}$$ and $$g\in L^{1}(\mathbb{R}^{2})$$, and using (1.6), we have that

$$K(x)f \bigl( u_{n}^{\pm}(x) \bigr) \bigl(u_{n}^{\pm}(x) \bigr)\leq \varepsilon K(x) \bigl\vert u_{n}^{ \pm}(x) \bigr\vert ^{2}+C_{\varepsilon }K(x) \bigl\vert u_{n}^{\pm}(x) \bigr\vert ^{q} \bigl(e^{\alpha (u_{n}^{ \pm}(x))^{2}}-1\bigr):=g\bigl(u_{n}^{\pm}(x) \bigr).$$

We will prove that $$g(u_{n}^{\pm})$$ is convergent in $$L^{1}(\mathbb{R}^{2})$$. First, note that

$$\int _{\mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{2}\,dx\to \int _{\mathbb{R}^{2}}K(x) \bigl\vert u^{ \pm} \bigr\vert ^{2}\,dx .$$

Choosing $$s',s>1$$ such that $$\frac{1}{s}+\frac{1}{s'}=1$$, we have

$$K(x)^{\frac{1}{s'}} \bigl\vert u_{n}^{\pm} \bigr\vert ^{q}\to K(x)^{\frac{1}{s'}} \bigl\vert u^{\pm} \bigr\vert ^{q} \quad \text{in }L^{s'} \bigl(\mathbb{R}^{2}\bigr).$$
(3.4)

Using (3.3) and choosing $$\alpha <\frac{4\pi}{sM_{1}^{2}}$$, we conclude by Lemma 2.1 that

\begin{aligned}[b] \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha s(u_{n}^{\pm}(x))^{2}}-1 \bigr)\,dx&\leq \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha sM_{1}^{2} (\frac{u_{n}^{\pm}(x)}{\| u_{n}^{\pm}\|} )^{2}}-1 \bigr)\,dx \\ &\leq \int _{\mathbb{R}^{2}}K(x) \bigl(e^{4\pi ( \frac{u_{n}^{\pm}(x)}{\| u_{n}^{\pm}\|} )^{2}}-1 \bigr)\,dx\leq M_{2}. \end{aligned}
(3.5)

Because

$$K(x)e^{\alpha s|u_{n}^{\pm }(x)|^{2}}\to K(x)e^{\alpha s|u^{\pm }(x)|^{2}} \quad \text{a.e. in } \mathbb{R}^{2},$$

we can use Lemma 4.8 of [33] and conclude that

$$K(x)e^{\alpha s|u_{n}^{\pm}|^{2}}\rightharpoonup K(x)e^{\alpha s|u^{ \pm}|^{2}}.$$
(3.6)

Using (3.4) and (3.6), as well as Lemma 4.8 of [33] again, we conclude

$$\int _{\mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{\pm}\bigr)\,dx\rightarrow \int _{\mathbb{R}^{2}}K(x)f\bigl(u^{\pm}\bigr) \bigl(u^{\pm}\bigr)\,dx .$$

Analogously, $$\int _{\mathbb{R}^{2}}K(x)F(u_{n}^{\pm})\,dx\rightarrow \int _{ \mathbb{R}^{2}}K(x)F(u^{\pm})\,dx$$.

Using the lower semicontinuity of convex functions, one has

$$\bigl\Vert u^{\pm} \bigr\Vert ^{2} \leq \liminf_{n\to \infty} \bigl\Vert u_{n}^{\pm} \bigr\Vert ^{2}.$$
(3.7)

Using (3.1), (3.2), and Lemma 2.3, there exists $$(s_{u},t_{u})\in (0,1]\times (0,1]$$ such that

$$\bar{u}:=s_{u}u^{+}+t_{u}u^{-}.$$

By $$(f_{3})$$, we have

\begin{aligned}[b] m\leq{}& I_{b}( \bar{u})=I_{b}(\bar{u})-\frac{1}{4}\bigl\langle I^{ \prime}_{b}(\bar{u}),\bar{u}\bigr\rangle \\ ={}&\frac{1}{4} \Vert \bar{u} \Vert ^{2}+ \frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f(\bar{u})\bar{u}-4F(\bar{u}) \bigr]\,dx \\ ={}&\frac{1}{4} \bigl\Vert s_{u}{u}^{+} \bigr\Vert ^{2}+\frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f\bigl(s_{u}{u}^{+} \bigr) \bigl(s_{u}{u}^{+}\bigr)-4F\bigl(s_{u}{u}^{+} \bigr) \bigr]\,dx \\ & {}+ \frac{1}{4} \bigl\Vert t_{u}{u}^{-} \bigr\Vert ^{2}+\frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f\bigl(t_{u}{u}^{-} \bigr) \bigl(t_{u}{u}^{-}\bigr)-4F\bigl(t_{u}{u}^{-} \bigr) \bigr]\,dx \\ \leq{}& \frac{1}{4} \Vert {u} \Vert ^{2}+ \frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f({u})u-4F({u}) \bigr]\,dx \\ \leq{}& \liminf_{n\to \infty} \biggl[ \Vert u_{n} \Vert ^{2}+\frac{1}{4} \int _{ \mathbb{R}^{2}}K(x) \bigl[f({u_{n}}){u_{n}}-4F({u_{n}}) \bigr]\,dx \biggr] \\ ={}&\liminf_{n\to \infty} \biggl(I(u_{n})- \frac{1}{4}\bigl\langle I^{\prime}(u_{n}),u_{n} \bigr\rangle \biggr)=m. \end{aligned}
(3.8)

Thus we conclude that $$s_{u}=t_{u}=1$$. So $$\bar{u}=u$$, $$I_{b}(u)=m$$. □

### Lemma 3.2

Assuming $$(f_{0})$$$$(f_{3})$$ and $$(f_{5})$$ hold, and $$u\in \mathcal{M}$$, one has $$\varPhi (s,t)<\varPhi (1,1)=I_{b}(u)$$ for all $$(s,t)\in C (\mathbb{R}^{+},\mathbb{R}^{+} )\backslash \lbrace (1,1) \rbrace$$. Furthermore, $$\det (\varPhi ^{u} )^{\prime}(1,1)>0$$.

### Proof

Letting $$u\in \mathcal{M}$$ and noting that $$\langle I^{\prime}_{b}(u), u^{\pm}\rangle = \langle I^{\prime}_{b}(u^{+}+u^{-}), u^{\pm}\rangle =0$$, we get that $$(1,1)$$ is a critical point of Φ, i.e.,

$$\varPhi ^{u}(1,1)= \biggl( \frac{\partial \varPhi}{\partial s}(1,1), \frac{\partial \varPhi}{\partial t}(1,1) \biggr) = ( 0,0 ).$$

According to Lemma 2.2, we know that $$\varPhi (s,t)$$ reaches its maximum at $$(s_{u},t_{u})$$, so from (3.8) we conclude that $$s_{u},t_{u}=1$$. To verify $$\det (\varPhi ^{u} )^{\prime}(1,1)>0$$, first note that

${\left({\Phi }^{u}\right)}^{\prime }\left(s,t\right)=\left(\begin{array}{cc}{g}_{1}^{\prime }\left(s\right)& 0\\ 0& {g}_{2}^{\prime }\left(t\right)\end{array}\right),$

where

\begin{aligned} &g_{1}(s):=\varPhi _{1}^{u} \bigl(su^{+}\bigr)u^{+}=s\bigl\lVert u^{+} \bigr\rVert ^{2}+bs^{3} \bigl\lVert u^{+} \bigr\rVert ^{4}- \int _{\mathbb{R}^{2}}K(x)f\bigl(su^{+}\bigr)u^{+} , \\ &g_{2}(s):=\varPhi _{2}^{u} \bigl(tu^{-}\bigr)u^{-}=t\bigl\lVert u^{-} \bigr\rVert ^{2}+bt^{3} \bigl\lVert u^{-} \bigr\rVert ^{4}- \int _{\mathbb{R}^{2}}K(x)f\bigl(tu^{-}\bigr)u^{-}. \end{aligned}

Because $$u^{+}\in \mathcal{N}$$, it follows from the definition of $$g_{1}(s)$$ and $$(f_{3})$$ that

\begin{aligned}[b] g_{1}^{\prime}(1)&= \bigl\lVert u^{+}\bigr\rVert ^{2}+3b\bigl\lVert u^{+} \bigr\rVert ^{4}- \int _{\mathbb{R}^{2}}K(x)f^{\prime}\bigl(u^{+} \bigr) \bigl(u^{+}\bigr)^{2} \\ &=-2\bigl\lVert u^{+}\bigr\rVert ^{2}+ \int _{\mathbb{R}^{2}}K(x) \bigl[ 3f\bigl(u^{+} \bigr)u^{+}-f^{ \prime}\bigl(u^{+}\bigr) \bigl(u^{+}\bigr)^{2} \bigr]\,dx< 0. \end{aligned}
(3.9)

Similarly, $$g_{2}^{\prime}(1)<0$$, and therefore we conclude that

$$\det \bigl(\varPhi ^{u} \bigr)^{\prime}(1,1)>0 .$$

□

### Lemma 3.3

Assume $$(f_{0})$$$$(f_{3})$$ and $$(f_{5})$$ hold. If $$u\in \mathcal{M}$$ and

$$I_{b}(u)=m:=\inf_{v\in \mathcal{M} }I(v),$$

then $$I^{\prime}_{b}(u)=0$$.

### Proof

Suppose to the contrary that the conclusion is not valid. Then there are $$\delta , \lambda >0$$ such that $$\lVert I^{\prime}_{b}(u)\rVert >\lambda$$ whenever $$\lVert u-v\rVert <3\delta$$. Let $$D\subset \mathbb{R}^{2}$$ be such that $$(1,1)\in D$$, and define a continuous mapping $$g:D\to X$$ by $$g(s,t)=su^{+}+tu^{-}$$. From Lemma 3.2, we conclude that

$$\alpha :=\max_{(s,t)\in \partial D}I_{b}\circ g< m .$$
(3.10)

For $$0<\varepsilon < \min \lbrace (m-\alpha )/2,\lambda \delta /8 \rbrace$$ and $$S:=B_{\delta}(v)$$, using Lemma 2.3 of [34], there exists $$\eta \in C ( [0,1]\times X,X )$$ verifying:

$$(a_{1})$$:

$$\eta ( 1,u ) =u$$, $$u\notin I^{-1}_{b} ( [ m-2\varepsilon ,m+2\varepsilon ] )$$;

$$(a_{2})$$:

$$\eta ( 1,I^{m+\varepsilon}_{b}\cap S )\subset I^{m- \varepsilon}_{b}$$;

$$(a_{3})$$:

$$I_{b} ( \eta (1,u ) )\leq I_{b}(u)$$, $$\forall u\in X$$.

By Lemma 3.2, $$(a_{2})$$, and $$(a_{3})$$, it follows that

$$\max_{(s,t)\in D}I_{b} \bigl( \eta \bigl( 1,g(s,t) \bigr) \bigr)< m.$$
(3.11)

It follows from the definition of $$\varPhi ^{u}$$ and $$u\in \mathcal{M}$$ that $$\varPhi ^{u}(s,t)=0$$ if and only if $$(s,t)=(1,1)\in D$$. Therefore, from the Brouwer degree theory and Lemma 3.2, we get

$$\deg \bigl(\varPhi ^{u},D,0 \bigr)= \operatorname{sgn}\det \bigl( \varPhi ^{u} \bigr)^{\prime} (1,1 )=1 .$$
(3.12)

Let $$h(s,t):=\eta (1,g(s,t) )$$ and

$$\Psi (s,t):= \bigl(s^{-1}I^{\prime}_{b} \bigl(h(s,t) \bigr)h(s,t)^{+},~t^{-1}I^{ \prime}_{b} \bigl(h(s,t) \bigr)h(s,t)^{-} \bigr) .$$
(3.13)

By the choice of $$\varepsilon >0$$, (3.10), and $$(a_{1})$$, we have $$g=h$$ in ∂D. Thus, the definition of $$\varPhi ^{u}$$ and (3.13) imply $$\varPhi ^{u}=\Psi$$ in ∂D, from which we get

$$\det (\Psi ,D,0 )=\det \bigl(\varPhi ^{u},D,0 \bigr)=1 .$$

So, there exists $$(s,t)\in D$$ such that $$h(s,t)\in \mathcal{M}$$, which is in contradiction with (3.11). Thus we get $$I ^{\prime}_{b} (u) =0$$. □

### Proof of Theorem 1.1

Letting $$\lbrace u_{n} \rbrace \subset \mathcal{M}$$ be a minimizing sequence for $$I_{b}$$ under the constraint set $$\mathcal{M}$$, we know that the sequence $$\lbrace u_{n} \rbrace$$ is bounded in X by Lemma 2.4. Also there exists $$u\in X$$ such that $$u_{n}\rightharpoonup u$$ in X. Combining (2.25), (3.8), and Lemma 3.3, we have $$I_{b}(u)=m$$, $$I^{\prime}_{b}(u)=0$$, and $$u^{\pm}\neq 0$$. Therefore, when $$\alpha _{0}=0$$, Eq. (1.1) has a least-energy sign-changing solution u.

Next, it is proved that u has two nodal domains through contradictory assumptions. First, by Fatou’s lemma, one can easily observe that

$$\bigl\langle I^{\prime}_{b}(u),u^{\pm}\bigr\rangle \leq \liminf_{n\to \infty} \bigl\langle I^{\prime}_{b}(u_{n}),u_{n}^{\pm} \bigr\rangle =0.$$

Now, we assume

$$u=u_{1}+u_{2}+u_{3}$$
(3.14)

with $$u_{i}\neq 0$$, $$u_{1}>0$$, $$u_{2}<0$$, $$u_{3}\geq 0$$, $$\mathrm{supp}(u_{i})\cap \mathrm{supp}(u_{j})=\emptyset$$, $$i\neq j\ (i,j=1,2,3)$$, and

$$\bigl\langle I^{\prime}_{b}(u),u_{i}\bigr\rangle =0,\quad i=1,2,3.$$

Let $$v:=u_{1}+u_{2}$$, $$v^{+}=u_{1}$$ and $$v^{-}=u_{2}$$, as well as $$v^{\pm}\neq 0$$. Then, by Lemma 2.3(i), there exists $$(s_{v},t_{v})\in (0,1]\times (0,1]$$ such that

$$s_{v}v^{+}+t_{v}v^{-}=s_{v}u_{1}+t_{v}u_{2} \in \mathcal{M},\qquad I_{b}(s_{v}u_{1}+t_{v}u_{2}) \geq m.$$
(3.15)

Through direct calculation, we have

\begin{aligned}[b] I_{b} \bigl(s_{v}v^{+}+t_{v}v^{-} \bigr)={}&I_{b}\bigl(s_{v}v^{+} \bigr)+I_{b}\bigl(t_{v}v^{-}\bigr)+ \frac{bs_{v}^{2}t_{v}^{2}}{2}\rVert v^{+} \rVert ^{2}\rVert v^{-} \rVert ^{2} \\ ={}&\frac{s_{v}^{2}}{4} \Vert u_{1} \Vert ^{2}+ \frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f(s_{v}u_{1})s_{v}u_{1}-4F(s_{v}u_{1}) \bigr]\,dx \\ & {}+ \frac{t_{v}^{2}}{4} \Vert u_{2} \Vert ^{2}+ \frac{1}{4} \int _{ \mathbb{R}^{2}}K(x) \bigl[f(t_{v}u_{2})t_{v}u_{2}-4F(t_{v}u_{2}) \bigr]\,dx \\ \leq{}& \frac{1}{4} \Vert u_{1} \Vert ^{2}+\frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f(u_{1})u_{1}-4F(u_{1}) \bigr]\,dx \\ & {}+ \frac{1}{4} \Vert u_{2} \Vert ^{2}+ \frac{1}{4} \int _{\mathbb{R}^{2}}K(x) \bigl[f(u_{2})u_{2}-4F(u_{2}) \bigr]\,dx \\ ={}&I_{b}(u_{1})+I_{b}(u_{2})+ \frac{b}{2}\rVert u_{1} \rVert ^{2} \rVert u_{2} \rVert ^{2}+\frac{b}{4}\rVert u_{1} \rVert ^{2}\rVert u_{3} \rVert ^{2}\\ & {}+\frac{b}{4}\rVert u_{2}\rVert ^{2}\rVert u_{3}\rVert ^{2}. \end{aligned}
(3.16)

\begin{aligned}[b] 0&=\frac{1}{4}\bigl\langle I^{\prime}_{b}(u),u_{3}\bigr\rangle \\ &=\frac{1}{4} \Vert u_{3} \Vert ^{2}+ \frac{b}{4} \Vert u \Vert ^{2} \Vert u_{3} \Vert ^{2}- \frac{1}{4} \int _{\mathbb{R}^{2}}K(x)f(u_{3})u_{3}\,dx \\ &< I_{b}(u_{3})+\frac{b}{4}\rVert u_{1} \rVert ^{2}\rVert u_{3} \rVert ^{2}+\frac{b}{4}\rVert u_{2} \rVert ^{2}\rVert u_{3}\rVert ^{2}. \end{aligned}
(3.17)

From (3.15)–(3.17), we get the following contradiction:

\begin{aligned}[b] m&\leq I_{b}(s_{v}u_{1}+t_{v}u_{2}) \\ &< I_{b}(u_{1})+I_{b}(u_{2})+I_{b}(u_{3})+ \frac{b}{2}\rVert u_{1} \rVert ^{2}\rVert u_{2} \rVert ^{2}+\frac{b}{2}\rVert u_{1} \rVert ^{2} \rVert u_{3}\rVert ^{2}+\frac{b}{2}\rVert u_{2}\rVert ^{2}\rVert u_{3} \rVert ^{2} \\ &=I_{b}(u)=m. \end{aligned}
(3.18)

So $$u_{3}=0$$, and u exactly does have two nodal domains. □

In order to prove Theorem 1.2, we first introduce an auxiliary equation

$$- \biggl(1+b \int _{\mathbb{R}^{2}} K(x) \vert \nabla u \vert ^{2}\,dx \biggr) \operatorname{div} \bigl(K(x)\nabla u \bigr)=K(x) \vert u \vert ^{p-2}u,$$
(3.19)

where $$p>4$$ is given by $$(f_{4})$$. The energy functional corresponding to equation (3.19) is

$$I_{p}(u)=\frac{1}{2} \int _{\mathbb{R}^{2}}K(x) \vert \nabla u \vert ^{2}\,dx+ \frac{b}{4} \biggl( \int _{\mathbb{R}^{2}} K(x) \vert \nabla u \vert ^{2}\,dx \biggr)^{2}- \frac{1}{p} \int _{\mathbb{R}^{2}} K(x) \vert u \vert ^{p}\,dx.$$
(3.20)

The corresponding Nehari manifold and Nehari nodal set are

$$\mathcal{N}_{p}=\bigl\{ u\in X\backslash \{0\};u\neq 0:\bigl\langle I^{\prime}_{p}(u),u \bigr\rangle =0\bigr\}$$
(3.21)

and

$$\mathcal{M}_{p}=\bigl\{ u\in X; u^{\pm}\neq 0 :\bigl\langle I^{\prime}_{p}(u),u^{+} \bigr\rangle =\bigl\langle I^{\prime}_{p}(u),u^{-}\bigr\rangle =0\bigr\} .$$
(3.22)

When $$p>4$$, the embedding $$X\hookrightarrow L_{K}^{p} \bigl(\mathbb{R}^{2}\bigr)$$ is compact. We use the previous proof to establish the existence of $$w_{p}\in X$$ satisfying $$I_{p}(w_{p})=m_{p}$$, $$I^{\prime}_{p}(w_{p}) =0$$, and such that

$$m_{p}=\inf _{u\in \mathcal{M}_{p}}I_{p}(u)>0$$
(3.23)

holds.

For the critical case, we need to control m below the threshold to restore compactness, and now we estimate the value of m.

Let $$\lbrace u_{n} \rbrace \subset \mathcal{M}_{p}$$ be a minimizing sequence for $$I_{p}(u_{n})\rightarrow m_{p}$$.

### Lemma 3.4

For $$b>0$$, we have $$0< m<\frac{\pi}{2\alpha _{0}}$$.

### Proof

Let $$w=w^{+}+w^{-}$$ and $$w^{\pm}\neq 0$$ be the sign-changing solution of (3.19). Then we have

$$\bigl\langle I^{\prime}_{p}(w),w^{+} \bigr\rangle =\bigl\langle I^{\prime}_{p}(w),w^{-} \bigr\rangle =\bigl\langle I^{\prime}_{p}(w),w\bigr\rangle =0$$
(3.24)

and

$$m_{p}= I_{p}(w)=I_{p}(w)- \frac{1}{4}\bigl\langle I^{\prime}_{p}(w),w \bigr\rangle \geq \frac{p-4}{4p} \vert w \vert ^{p}_{p}.$$
(3.25)

Using $$(f_{4})$$ and (3.24), we have $$\langle I^{\prime}_{b}(w),w^{\pm}\rangle \leq 0$$, while using Lemmas 2.3 and 2.4, there is a unique number pair $$(s,t)\in (0,1]\times (0,1]$$ such that $$sw^{+}+tw^{-}\in \mathcal{M}$$. Combing $$(f_{4})$$, (3.24), (3.25), for $$(s,t)\in (0,1]\times (0,1]$$, we obtain

\begin{aligned}[b] m\leq{}& I_{b} \bigl(sw^{+}+tw^{-}\bigr) \\ \leq{}& \frac{s^{2}}{2}\bigl\lVert w^{+}\bigr\rVert ^{2}+\frac{t^{2}}{2}\bigl\lVert w^{-} \bigr\rVert ^{2}+\frac{bs^{4}}{4}\bigl\lVert w^{+}\bigr\rVert ^{4}+\frac{bt^{4}}{4} \bigl\lVert w^{-}\bigr\rVert ^{4} \\ & {}+ \frac{bs^{2}t^{2}}{2}\bigl\lVert w^{+}\bigr\rVert ^{2}\bigl\lVert w^{-}\bigr\rVert ^{2}- \frac{\varrho _{0}s^{p}}{p}\bigl\lvert w^{+}\bigr\rvert ^{p}_{p}- \frac{\varrho _{0}t^{p}}{p}\bigl\lvert w^{-}\bigr\rvert ^{p}_{p} \\ ={}&\frac{s^{2}}{2} \biggl[ \int _{\mathbb{R}^{2}} K(x)\bigl\lvert w^{+} \bigr\rvert ^{p}\,dx-b\bigl\lVert w^{+}\bigr\rVert ^{4}-b\bigl\lVert w^{+}\bigr\rVert ^{2} \bigl\lVert w^{-}\bigr\rVert ^{2} \biggr]+ \frac{bs^{4}}{4}\bigl\lVert w^{+}\bigr\rVert ^{4} \\ & {}+ \frac{t^{2}}{2} \biggl[ \int _{\mathbb{R}^{2}}K(x)\bigl\lvert w^{-} \bigr\rvert ^{p}\,dx-b\bigl\lVert w^{-}\bigr\rVert ^{4}-b\bigl\lVert w^{-}\bigr\rVert ^{2} \bigl\lVert w^{-}\bigr\rVert ^{2} \biggr]+ \frac{bt^{4}}{4}\bigl\lVert w^{-}\bigr\rVert ^{4} \\ & {}+ \frac{bs^{2}t^{2}}{2}\bigl\lVert w^{+}\bigr\rVert ^{2}\bigl\lVert w^{-} \bigr\rVert ^{2}- \frac{\varrho _{0}s^{p}}{p}\bigl\lvert w^{+}\bigr\rvert ^{p}_{p}- \frac{\varrho _{0}t^{p}}{p}\bigl\lvert w^{-}\bigr\rvert ^{p}_{p} \\ \leq{}& \max_{\xi >0} \biggl(\frac{\xi ^{2}}{2}- \frac{\varrho _{0}\xi ^{p}}{p} \biggr) \lvert w\rvert ^{p}_{p}- \frac{bs^{2}}{4}\bigl\lVert w^{+}\bigr\rVert ^{4} \bigl( 2-s^{2} \bigr)- \frac{bt^{2}}{4}\bigl\lVert w^{-}\bigr\rVert ^{4} \bigl( 2-t^{2} \bigr) \\ & {}- \frac{s^{2}+t^{2}-s^{2}t^{2}}{2}b\bigl\lVert w^{+}\bigr\rVert ^{2} \bigl\lVert w^{-}\bigr\rVert ^{2} \\ \leq{}& \max_{\xi >0} \biggl(\frac{\xi ^{2}}{2}- \frac{\varrho _{0}\xi ^{p}}{p} \biggr) \lvert w\rvert ^{p}_{p}= \frac{p-2}{2p}\varrho _{0}^{-\frac{2}{p-2}}\lvert w\rvert ^{p}_{p} \\ \leq{}& \frac{2(p-2)}{p-4}\varrho _{0}^{-\frac{2}{p-2}}m_{p}. \end{aligned}
(3.26)

□

### Lemma 3.5

Suppose $$\lbrace u_{n} \rbrace \subset \mathcal{M}$$ is a minimizing sequence for m. Then

$$\limsup_{n\to \infty}\lVert u_{n}\rVert ^{2}< \frac{2\pi}{\alpha _{0}} .$$

### Proof

From the assumption, we have $$I_{b} (u_{n} )\to m$$, $$\langle I^{\prime}_{b}(u_{n}),u_{n} \rangle =0$$, when $$n\to +\infty$$. From $$( f_{3} )$$, we have

$$m+o(1)= I_{b}(u_{n})-\frac{1}{4}\bigl\langle I^{\prime}_{b}(u_{n}),u_{n} \bigr\rangle \geq \frac{1}{4}\lVert u_{n}\rVert ^{2} .$$

From Lemma 2.4, we have

$$\limsup_{n\to \infty}\lVert u_{n}\rVert ^{2}\leq 4m \leq \frac{8(p-2)}{p-4}\varrho _{0}^{-\frac{2}{p-2}}m_{p}.$$

Using $$(f_{4})$$, we get $$\limsup_{n\to \infty}\lVert u_{n}\rVert ^{2}< \frac{2\pi}{\alpha _{0}}$$. □

### Lemma 3.6

Assume $$\lbrace u_{n} \rbrace \subset \mathcal{M}$$ is a minimizing sequence for m. Then

$$\int _{\mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{\pm}\bigr)\,dx\to \int _{ \mathbb{R}^{2}}K(x)f\bigl(u^{\pm}\bigr) \bigl(u^{\pm}\bigr)\,dx$$

and

$$\int _{\mathbb{R}^{2}}K(x)F\bigl(u_{n}^{\pm} \bigr)\,dx\to \int _{\mathbb{R}^{2}}K(x)F\bigl(u^{ \pm}\bigr)\,dx .$$

### Proof

We only prove the first limit here, as the second is obtained similarly. By Lemma 3.5, we have $$\limsup_{n\to \infty}\lVert u_{n}\rVert ^{2}\leq \frac{2\pi}{\alpha _{0}}$$ and, up to a subsequence, $$u_{n}^{\pm}(x)\to u^{\pm}(x)$$ and

$$f \bigl( u_{n}^{\pm}(x) \bigr) \bigl(u_{n}^{\pm}(x) \bigr)\to f \bigl( u^{\pm}(x) \bigr) \bigl(u^{\pm}(x)\bigr) \quad \text{a.e. in}\ \mathbb{R}^{2} .$$

Arguing as in the proof of Lemma 3.1, introducing $$g:\mathbb{R}\to \mathbb{R}$$, $$g\in L^{1}(\mathbb{R}^{2})$$, and using (1.6), we have

$$K(x)f \bigl( u_{n}^{\pm}(x) \bigr) \bigl(u_{n}^{\pm}(x) \bigr)\leq \varepsilon K(x) \bigl\vert u_{n}^{ \pm}(x) \bigr\vert ^{2}+C_{\varepsilon }K(x) \bigl\vert u_{n}^{\pm}(x) \bigr\vert ^{q} \bigl(e^{\alpha (u_{n}^{ \pm}(x))^{2}}-1\bigr):=g\bigl(u_{n}^{\pm}(x) \bigr).$$

We will prove that $$g(u_{n}^{\pm})$$ converges in $$L^{1} (\mathbb{R}^{2})$$. First, note that

$$\int _{\mathbb{R}^{2}}K(x) \bigl\vert u_{n}^{\pm} \bigr\vert ^{2}\,dx\to \int _{\mathbb{R}^{2}}K(x) \bigl\vert u^{ \pm} \bigr\vert ^{2}\,dx .$$

Considering $$s',s>1$$ such that $$\frac{1}{s}+\frac{1}{s'}=1$$ and $$s\to 1^{+}$$, we obtain

$$K(x)^{\frac{1}{s'}} \bigl\vert u_{n}^{\pm} \bigr\vert ^{q}\to K(x)^{\frac{1}{s'}} \bigl\vert u^{\pm} \bigr\vert ^{q} \quad \text{in } L^{s'}\bigl(\mathbb{R}^{2}\bigr).$$
(3.27)

Now, choosing $$\alpha >\alpha _{0}$$ and close to $$\alpha _{0}$$, using Lemma 2.1, there exists $$M_{2}>0$$ such that

\begin{aligned}[b] \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha s(u_{n}^{\pm}(x))^{2}}-1 \bigr)\,dx&= \int _{\mathbb{R}^{2}}K(x) \bigl(e^{\alpha s\lVert u_{n}^{ \pm }\rVert ^{2} (\frac{u_{n}^{\pm}(x)}{\| u_{n}^{\pm}\|} )^{2}}-1 \bigr)\,dx \\ &\leq \int _{\mathbb{R}^{2}}K(x) \bigl(e^{4\pi ( \frac{u_{n}^{\pm}(x)}{\| u_{n}^{\pm}\|} )^{2}}-1 \bigr)\,dx\leq M_{2}. \end{aligned}
(3.28)

Since

$$K(x)e^{\alpha s|u_{n}^{\pm }(x)|^{2}}\to K(x)e^{\alpha s|u^{\pm }(x)|^{2}} \quad \text{a.e. in } \mathbb{R}^{2},$$

we use Lemma 4.8 of [33] and conclude that

$$K(x)e^{\alpha s|u_{n}^{\pm}|^{2}}\rightharpoonup K(x)e^{\alpha s|u^{ \pm}|^{2}} \quad \text{in } L^{s}\bigl(\mathbb{R}^{2}\bigr).$$
(3.29)

Using (3.27), (3.29), and Lemma 4.8 of [33] again, we conclude

$$\int _{\mathbb{R}^{2}}K(x)f\bigl(u_{n}^{\pm} \bigr) \bigl(u_{n}^{\pm}\bigr)\,dx\to \int _{ \mathbb{R}^{2}}K(x)f\bigl(u^{\pm}\bigr) \bigl(u^{\pm}\bigr)\,dx .$$

□

### Proof of Theorem 1.2

The proof is similar to that of Theorem 1.1. We conclude that in the critical case, $$I_{b}$$ has a least-energy sign-changing solution which has precisely two nodal domains. □

## Availability of data and materials

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## Funding

This research was funded by National Natural Science Foundation of China (No. 11661021; No. 11861021).

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Correspondence to Hongmin Suo.

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Qin, Q., Jie, G. & Suo, H. Nodal solution for critical Kirchhoff-type equation with fast increasing weight in $$\mathbb{R}^{2}$$. J Inequal Appl 2023, 40 (2023). https://doi.org/10.1186/s13660-023-02945-x