On a class of Kirchhoff equations involving an anisotropic operator and potential

Abstract

In this work, we are concerned with a class of fractional equations of Kirchhoff type with potential. Using variational methods and a variant of quantitative deformation lemma, we prove the existence of a least energy sign-changing solution. Moreover, the existence of infinitely many solution is established.

1 Introduction and main result

Consider the following fractional Kirchhoff equation

$$\begin{aligned} \left( 1+b[u]_\alpha ^2\right) \left( (-\Delta _x)^\alpha u-\Delta _yu\right) +V(x,y)u=f(u), \quad (x,y)\in \mathbb {R}^N=\mathbb {R}^n\times \mathbb {R}^m, \end{aligned}$$

(1.1)

where $[u]_\alpha =\left( \int _{\mathbb {R}^N}\left( |(-\Delta _x)^{\frac{\alpha }{2}}u|^2+|\nabla _yu|^2\right) \mathrm{d}x\mathrm{d}y\right) ^{\frac{1}{2}}$, $\alpha \in (0,1),\,n,m\ge 1,\,(-\Delta _x)^\alpha$ denotes the fractional Laplacian in x which is defined via the Fourier transform by $\widehat{(-\Delta _x)^\alpha u}(X,Y)=|X|^{2\alpha }\widehat{u}(X,Y),$ where $\widehat{u}$ is the Fourier transform of u;  if u is sufficiently smooth, it can be expressed by

$$\begin{aligned} (-\Delta _x)^\alpha u(x,y)=C_{N,s} \mathrm{P.V.}\int _{\mathbb {R}^n}\frac{u(x,y)-u(z,y)}{|x-z|^{n+2\alpha }}\mathrm{d}z,\;(x,y)\in \mathbb {R}^N, \end{aligned}$$

with $C_{N,\alpha }$ is a normalization constant and P.V. stands for the Cauchy principal value; and V, f are a functions satisfying some conditions which will be specified later.

The presence of the nonlocal term $[u]_\alpha ^2$ in (1.1) causes some mathematical difficulties and so the study of such a class of equations is of much interest. Moreover, Eq. (1.1) is a fractional version related to the following hyperbolic equation

$$\begin{aligned} \rho \frac{\partial ^2u}{\partial t^2}-\left( \frac{\rho _0}{h}+\frac{E}{2L}\int _0^L\left| \frac{\partial u}{\partial x}\right| ^2\mathrm{d}x\right) \frac{\partial ^2u}{\partial x^2}=0, \end{aligned}$$

(1.2)

which was proposed by Kirchhoff [22] as an extension of the classical D’Alembert’s wave equation by considering the changes in the length of the strings produced by transverse vibrations.

On the other hand, when $b=0,$ $V\equiv 1$ and $f(u)=u^p$, equation (1.1) appears in the study of solitary waves of the generalized Benjamin–Ono–Zakharov–Kuznetsov equation

$$\begin{aligned} u_t+\partial _{x_1}\left( (-\Delta _x)^\alpha u-\Delta _yu+u^p\right) =0\,\, (x,y)\in \mathbb {R}^n\times \mathbb {R}^m, \end{aligned}$$

(1.3)

see [28] for some local and global well-posedness results on this equation with $m=n=1$. The anisotropic operator $(-\Delta _x)^\alpha u-\Delta _yu$ is observed in the study of toy models parabolic equations for which local diffusions occur only in certain directions and nonlocal diffusions. It models diffusion sensible to the direction in the Brownian and Lévy-Itô processes. For some regularity and rigidity properties of this operator, the readers can refer to [6, 17].

Recently, Esfahani [15] considered Eq. (1.1) with $b=0.$ Under suitable assumptions on $f\in C^1(\mathbb {R},\mathbb {R}),$ by adapting some arguments developed in [7,8,9] and using a variant of deformation lemma, the author shows that the equation admits a least energy sign-changing solution.

It is interesting to note that the fractional Laplacian (isotropic operator) problems have attracted a great attention by various study on the existence and multiplicity of solutions; see, e.g., [19, 20, 25, 33]. This interest comes from its significant applications in several areas such as physics, biology, chemistry and finance; see, e.g., [5, 11, 12, 18, 23]. We also refer to [10] for more applications specifically to an audience of mathematicians. We point out that the variational methods leading to existence theories in the fractional setting were introduced in [30].

In [18], Fiscella and Valdinoci first introduced a stationary Kirchhoff variational equation, which models the nonlocal aspect of the tension arising from nonlocal measurements of the fractional length of the string. After that, the fractional Kirchhoff problems have been investigated by many researchers, we cite here [2, 3, 13, 14, 21, 26, 27, 31]. However, to the best of our knowledge, there are no result about existence of solutions for a Kirchhoff-type equation with an operator containing local and nonlocal diffusions except [24]. Inspired by the results mentioned above, especially [1, 4, 15, 21], in the present paper, we will prove the existence of sign-changing solutions and infinitely many solutions for Eq. (1.1). To this aim, we need a compact embedding which is appropriate to our equation and some technical lemmas related to the minimization method on the nodal Nehari manifold.

We assume that $V:\mathbb {R}^N\rightarrow \mathbb {R}$ is continuous and satisfies:

$(V_0)$:

$V(x,y)>0$ for all $(x,y)\in \mathbb {R}^N$ and $V^{-1}\in L^\infty (\mathbb {R}^N).$

For the nonlinearity f,  we assume that $f\in C^1(\mathbb {R},\mathbb {R})$ and

$(\mathrm{f}_1)$:

$\lim _{t\rightarrow 0}\frac{f(t)}{t}=0;$

$(\mathrm{f}_2)$:

$\lim _{|t|\rightarrow \infty }\frac{f(t)}{|t|^{p^*-1}}=0,$ where $p^*=\frac{2(n+m\alpha )}{n+(m-2)\alpha };$

$(\mathrm{f}_3)$:

$\lim _{|t|\rightarrow \infty }\frac{F(t)}{t^4}=+\infty ,$ where $F(t)=\int _0^tf(\tau )\mathrm{d}\tau \ge 0$ for all $t\in \mathbb {R};$

$(\mathrm{f}_4)$:

the function $t\mapsto \frac{f(t)}{|t|^3}$ is increasing on $\mathbb {R}\backslash \{0\}.$

Before stating what we think our main result, we recall our working Sobolev space. Let $\mathcal {H}^\alpha (\mathbb {R}^N)$ be the fractional Sobolev–Liouville space

$$\begin{aligned} \mathcal {H}^\alpha (\mathbb {R}^N)=\left\{ u\in L^2(\mathbb {R}^N): \int _{\mathbb {R}^N}\left( |(-\Delta _x)^{\frac{\alpha }{2}}u|^2+|\nabla _yu|^2+u^2\right) \mathrm{d}x\mathrm{d}y<\infty \right\} \end{aligned}$$

with the norm

$$\begin{aligned} ||u||_{\mathcal {H}^\alpha }=\left( \int _{\mathbb {R}^N}\left( |(-\Delta _x)^{\frac{\alpha }{2}}u|^2+|\nabla _yu|^2+u^2\right) \mathrm{d}x\mathrm{d}y\right) ^{\frac{1}{2}}. \end{aligned}$$

The homogeneous fractional Sobolev–Liouville space $\mathcal {D}^\alpha (\mathbb {R}^N)$ is defined by

$$\begin{aligned} \mathcal {D}^{\alpha }(\mathbb {R}^N)=\{u\in L^2(\mathbb {R}^N):[u]_\alpha <\infty \}, \end{aligned}$$

where

$$\begin{aligned}{}[u]_\alpha =\left( \int _{\mathbb {R}^N}\left( |(-\Delta _x)^{\frac{\alpha }{2}}u|^2+|\nabla _yu|^2\right) \mathrm{d}x\mathrm{d}y\right) ^{\frac{1}{2}}. \end{aligned}$$

We search the solutions in the following subspace

$$\begin{aligned} E_\alpha =\left\{ u\in \mathcal {D}^{\alpha }(\mathbb {R}^N):\int _{\mathbb {R}^N} V(x,y)\mathrm{d}x\mathrm{d}y<\infty \right\} , \end{aligned}$$

endowed with the inner product

$$\begin{aligned} (u,v)_{E_\alpha }=\int _{\mathbb {R}^N}\left( (-\Delta _x)^{\frac{\alpha }{2}}u(-\Delta _x)^{\frac{\alpha }{2}}v+\nabla _yu\nabla _yv\right) \mathrm{d}x\mathrm{d}y+\int _{\mathbb {R}^N} V(x,y)uv\mathrm{d}x\mathrm{d}y \end{aligned}$$

and its related norm

$$\begin{aligned} ||u||_{E_\alpha }=(u,u)_{E_\alpha }^\frac{1}{2}. \end{aligned}$$

A function $u\in E_\alpha$ is called weak solution of (1.1) if

$$\begin{aligned}&\left( 1+b[u]_\alpha ^2\right) \int _{\mathbb {R}^N}\left( (-\Delta _x)^{\frac{\alpha }{2}}u(-\Delta _x)^{\frac{\alpha }{2}}v+\nabla _yu\nabla _yv\right) \mathrm{d}x\mathrm{d}y +\int _{\mathbb {R}^N}V(x,y)uv\mathrm{d}x\mathrm{d}y\\&\quad =\int _{\mathbb {R}^N} f(u)v\mathrm{d}x\mathrm{d}y, \text{ for } \text{ all } v\in E_\alpha . \end{aligned}$$

A sign-changing solution of (1.1) is a weak solution $u\in E_\alpha$ satisfying $u^\pm \ne 0,$ where

$$\begin{aligned} u^+=\max (u,0)\quad {and}\quad u^-=\min (u,0). \end{aligned}$$

The main result can be described by the following theorem.

Theorem 1.1

Assume that $(V_0)$ and $(\mathrm{f}_1)$–$(\mathrm{f}_4)$ hold. Then, Eq. (1.1) admits a least energy sign-changing solution. Furthermore, if f is odd, then (1.1) has infinitely many solutions.

Remark 1.2

This result is known for the isotropic case with $N=3$ and the nonlinear term $f(x,u)=K(x)f(u),$ please refer to [21].

2 Auxiliary results and proof of main theorem

In this section, we give some preliminary results for the proof of Theorem 1.1.

Lemma 2.1

(Cf.[16]) The fractional Sobolev–Liouville space $\mathcal {H}^\alpha (\mathbb {R}^N)$ is continuously embedded into $L^q(\mathbb {R}^N)$ for $q\in [2,p^*];$ and is compactly embedded into $L^q(\mathbb {R}^N)$ for $q\in (2,p^*).$

According to Lemma 2.1, we have the next result (see [15, Lemma 2.1]).

Lemma 2.2

(Cf.[15]) Assume that $(V_0)$ holds. Then, $E_\alpha$ is continuously embedded into $L^q(\mathbb {R}^N)$ for $q\in [2, p^*];$ and compactly embedded into $L^q(\mathbb {R}^N)$ for $q\in [2, p^*);$

Obviously, the energy functional associated to (1.1) defined on $E_\alpha$ by

$$\begin{aligned} I(u)=\frac{1}{2}||u||_{E_\alpha }^2+\frac{b}{4}[u]_\alpha ^4-\int _{\mathbb {R}^N} F(u)\mathrm{d}x\mathrm{d}y, \end{aligned}$$

is of class $C^1$ and its critical points are solutions of (1.1).

To discus the existence of sign-changing solutions, we follow some arguments developed in [7,8,9, 32]. More precisely, we will minimize the functional I on the nodal set

$$\begin{aligned} \mathcal {M}=\left\{ u\in \mathcal {N}:u^\pm \ne 0,\,\langle I'(u),u^+\rangle =\langle I'(u),u^-\rangle =0\right\} , \end{aligned}$$

where

$$\begin{aligned} \mathcal {N}=\left\{ u\in E_\alpha \backslash \{0\}:\langle I'(u),u\rangle =0\right\} . \end{aligned}$$

Denote $S_1=\{u\in E_\alpha :||u||_{E_\alpha }=1\}.$ For all fixed $u\in E_\alpha \backslash \{0\},$ we set

$$\begin{aligned} g_u(t)=I(tu) \,\, \text{ for } \text{ all } \,\, t\ge 0. \end{aligned}$$

Lemma 2.3

Assume that $(V_0)$ and $(\mathrm{f}_1)$–$(\mathrm{f}_4)$ hold. Then

• (i) there exists an unique $t_u>0$ such that

  $$\begin{aligned} g_u'(t_u)=0,\quad g_u'>0 \text{ on } (0,t_u)\quad {and}\quad g_u'<0 \text { on } (t_u,\infty ); \end{aligned}$$

• (ii) $\rho :=\inf _{u\in S_1}g_u(t_u)>0.$ Furthermore, for each compact $\Lambda \subset S_1,$ there exists $C_\Lambda >0$ such that

  $$\begin{aligned} t_u\le C_\Lambda \,\, \text{ for } \text{ all } \,\, u\in \Lambda ; \end{aligned}$$

• (iii) let $\overline{\Phi }:E_\alpha \backslash \{0\}\rightarrow \mathcal {N}$ and $\Phi :S_1\rightarrow \mathcal {N}$ be the mappings defined by

  $$\begin{aligned} \overline{\Phi }(u)=t_uu\quad \text{ and }\quad \Phi =\overline{\Phi }|_{S_1}. \end{aligned}$$

  Then, $\overline{\Phi }$ is continuous and $\Phi$ is a homeomorphism. Moreover, the inverse of $\Phi$ is given by $\Phi ^{-1}(u)=\frac{u}{||u||_{E_\alpha }}.$

Proof

(i)   By $(\mathrm{f}_1)$–$(\mathrm{f}_2),$ for given $\varepsilon >0$ there exists $C_\varepsilon >0$ such that

$$\begin{aligned} F(t)\le \varepsilon t^2+C_\varepsilon |t|^{p^*} \text{ for } \text{ all } t\in \mathbb {R}. \end{aligned}$$

Therefore, by Lemma 2.2, for all $t>0,$

$$\begin{aligned} I(tu)\ge & {} \frac{t^2}{2}||u||_{E_\alpha }^2-\int _{\mathbb {R}^N}F(tu)\mathrm{d}x\mathrm{d}y\nonumber \\\ge & {} \frac{t^2}{2}||u||_{E_\alpha }^2-\varepsilon t^2\int _{\mathbb {R}^N}u^2\mathrm{d}x\mathrm{d}y-C_\varepsilon t^{p^*}\int _{\mathbb {R}^N}|u|^{p^*}\mathrm{d}x\mathrm{d}y\nonumber \\\ge & {} \left( \frac{1}{2}-\varepsilon ||V^{-1}||_\infty \right) t^2||u||_{E_\alpha }^2-CC_\varepsilon t^{p^*}||u||_{E_\alpha }^{p^*}. \end{aligned}$$

(2.1)

Choosing $\varepsilon <\frac{1}{2||V^-1||_\infty },$ for $t_*$ small enough,

$$\begin{aligned} I(tu)>0 \text{ for } \text{ all } 0<t<t_*. \end{aligned}$$

(2.2)

Let $\Omega _0=\{(x,y)\in \mathbb {R}^N:u(x,y)\ne 0\}$, we have

$$\begin{aligned} I(tu)\le \frac{t^2}{2}||u||_{E_\alpha }^2+\frac{bt^4}{4}||u||_{E_\alpha }^4-\int _{\Omega _0}F(tu)\mathrm{d}x\mathrm{d}y. \end{aligned}$$

It follows from Fatou’s lemma and $(\mathrm{f}_3),$

$$\begin{aligned} \limsup _{t\rightarrow \infty }\frac{I(tu)}{||tu||_{E_\alpha }^4}\le \frac{b}{4}-\int _{\Omega _0}\liminf _{t\rightarrow \infty }\frac{F(tu)}{(tu)^4}\frac{u^4}{||u||_{E_\alpha }^4}\mathrm{d}x\mathrm{d}y=-\infty . \end{aligned}$$

(2.3)

Hence, for some $T>0$ large enough,

$$\begin{aligned} I(tu)<0 \text{ for } \text{ all } t\ge T. \end{aligned}$$

(2.4)

By the continuity of $g_u,$ (2.2) and (2.4), there is $t_u>0$ such that $\displaystyle g_u(t_u)=\max _{t\ge 0}g_u(t).$ Then $t_uu\in \mathcal {N}.$ Moreover, using $(\mathrm{f}_4),$ it easy to see that $t_u$ is the unique number satisfying $g_u'(t_u)=0,$ and hence (i) follows.

(ii)   Let $\displaystyle \rho =\inf _{u\in S_1}g_u(t_u).$ Then $\rho >0.$ Indeed, by (2.1), we can find $\underline{t}>0$ such that

$$\begin{aligned} g_u(t_u)\ge \left( \frac{1}{2}-\varepsilon ||V^{-1}||_\infty \right) \underline{t}^2-CC_\varepsilon \underline{t}^{p^*}>0, \text{ for } \text{ all } u\in S_1. \end{aligned}$$

Thus $\rho >0.$

Now, let $\Lambda \subset S_1$ be a compact set. Suppose by contradiction that there exists $\{u_n\}\subset \Lambda$ such that $t_{u_n}\rightarrow +\infty .$ Then, up to subsequence, $u_n\rightarrow u$ for some $u\in \Lambda .$ By (2.3), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }I(t_{u_n}u_n)=-\infty . \end{aligned}$$

(2.5)

Exploiting $(\mathrm{f}_4),$ we can infer that the function $t\mapsto \frac{1}{4}tf(t)-F(t)$ is increasing on $(0,\infty )$ and decreasing on $(-\infty ,0).$ Therefore

$$\begin{aligned} \frac{1}{4}tf(t)-F(t)\ge 0 \text{ for } \text{ all } t\in \mathbb {R}. \end{aligned}$$

Since $\langle I'(t_{u_n}u_n),t_{u_n}u_n\rangle =0$ for all $n\in \mathbb {N},$

$$\begin{aligned} I(t_{u_n}u_n)= & {} I(t_{u_n}u_n)-\frac{1}{4}\langle I'(t_{u_n}u_n),t_{u_n}u_n\rangle \\= & {} \frac{1}{4}||t_{u_n}u_n||_{E_\alpha }^2+\int _{\mathbb {R}^N} \left( \frac{1}{4}t_{u_n}u_nf(t_{u_n}u_n)-F(t_{u_n}u_n)\right) \mathrm{d}x\mathrm{d}y\ge 0. \end{aligned}$$

This contradicts with (2.5).

(iii) By (i), $\overline{\Phi }$ is well defined. Let $v\in \mathcal {N},$ then equation $t_uu=v$ with $||u||_{E_\alpha }=1$ has an unique solution $u=\frac{v}{||v||_{E_\alpha }}$ with $t_u=||v||_{E_\alpha }.$ Therefore $\Phi$ is invertible and $\Phi ^{-1}(v)=\frac{v}{||v||_{E_\alpha }}.$ We notice that $\Phi ^{-1}$ is continuous. Now, we prove that $\overline{\Phi }$ is continuous. Let $u_n\rightarrow u$ in $E_\alpha \backslash \{0\}.$ By definition of $t_{u_n},$ we have $t_{\left( \frac{u_n}{||u_n||_{E_\alpha }}\right) }=t_{u_n}||u_n||_{E_\alpha }.$ Since $\left\{ \frac{u_n}{||u_n||_{E_\alpha }}\right\} \subset S_1,$ it follows from (ii) that $t_{u_n}||u_n||_{E_\alpha }\rightarrow l_0>0.$ Thereby $t_{u_n}\rightarrow \frac{l_0}{||u||_{E_\alpha }}.$ Recall that

$$\begin{aligned} t_{u_n}^2[u_n]_\alpha ^2+bt_{u_n}^4[u_n]_\alpha ^4+t_{u_n}^2\int _{\mathbb {R}^N} V(x,y)u_n^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}f(t_{u_n}u_n)t_{u_n}u_n\mathrm{d}x\mathrm{d}y. \end{aligned}$$

Letting n goes to infinity, we obtain

$$\begin{aligned} \frac{l_0^2}{||u||_{E_\alpha }^2}[u]_\alpha ^2+b\frac{l_0^4}{||u||_{E_\alpha }^4}[u]_\alpha ^4+\frac{l_0^2}{||u||_{E_\alpha }^2}\int _{\mathbb {R}^N} V(x,y)u^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}f\left( \frac{l_0u}{||u||_{E_\alpha }}\right) \frac{l_0u}{||u||_{E_\alpha }}\mathrm{d}x\mathrm{d}y, \end{aligned}$$

which yields that $\frac{l_0u}{||u||_{E_\alpha }}\in \mathcal {N}$ and $t_u=\frac{l_0}{||u||_{E_\alpha }}.$ Consequently $\overline{\Phi }(u_n)\rightarrow \overline{\Phi }(u),$ and so $\overline{\Phi }$ is continuous. The proof of Lemma 2.3 is completed. $\square$

Consider the functionals $\overline{\Psi }:E_\alpha \backslash \{0\}\rightarrow \mathbb {R}$ and $\Psi :S_1\rightarrow \mathbb {R}$ given by

$$\begin{aligned} \overline{\Psi }(u)=I(\overline{\Phi }(u))\quad \text{ and }\quad \Psi =\overline{\Psi }|_{S_1}. \end{aligned}$$

In view of Lemma 2.3 and with the help of [32], we have the following result.

Lemma 2.4

Assume that $(V_0)$ and $(\mathrm{f}_1)$–$(\mathrm{f}_4)$ hold. Then,

• (i) $\overline{\Psi }\in C^1(E_\alpha \backslash \{0\},\mathbb {R})$ and for all $u\in E_\alpha \backslash \{0\},\,v\in E_\alpha$

  $$\begin{aligned} \langle \overline{\Psi }'(u),v\rangle =\frac{||\overline{\Phi }(u)||_{E_\alpha }}{||u||_{E_\alpha }}\langle I'(\overline{\Phi }(u)),v\rangle ; \end{aligned}$$

• (ii) $\Psi \in C^1(S_1,\mathbb {R})$ and $\langle \Psi '(u),v\rangle =||\Phi (u)||_{E_\alpha }\langle I'(\Phi (u)),v\rangle$ for all $v\in T_uS_1$, where

  $$\begin{aligned} T_uS_1:=\{v\in E_\alpha :(u,v)_{E_\alpha }=0\}; \end{aligned}$$

• (iii) If $\{u_n\}$ is a (PS) sequence for $\Psi ,$ then $\{\Phi (u_n)\}$ is a (PS) sequence for I. Moreover, if $\{u_n\}\subset \mathcal {N}$ is a bounded (PS) for I,  then $\{\Phi ^{-1}(u_n)\}$ is a (PS) sequence for $\Psi ;$

• (iv) u is a critical point of $\Psi$ if and only if $\Phi (u)$ is a nontrivial critical point for I. Moreover, the corresponding critical values coincide and

  $$\begin{aligned} \inf _{u\in S_1}\Psi (u)=\inf _{u\in \mathcal {N}}I(u). \end{aligned}$$

Proof

The proof is similar to that of [32, Proposition 9, Corollary 10]. $\square$

Remark 2.5

We notice that

$$\begin{aligned} c_\infty ^\mathcal {N}:=\inf _{u\in \mathcal {N}}I(u)=\inf _{u\in E_\alpha \backslash \{0\}}\max _{t>0}I(tu)=\inf _{u\in S_1}\max _{t>0}I(tu)>0. \end{aligned}$$

(2.6)

For $u\in E_\alpha$ with $u^\pm \ne 0,$ we define the function $\varphi _u:[0,\infty )\times [0,\infty )\rightarrow \mathbb {R}$ by

$$\begin{aligned} \varphi _u(t,s)=I(t u^++su^-). \end{aligned}$$

Its gradient is given by

$$\begin{aligned} \nabla \varphi _u(t,s)= & {} (\langle I'(tu^++su^-),u^+\rangle ,\langle I'(tu^++su^-),u^-\rangle )\nonumber \\= & {} \left( \frac{1}{t}\langle I'(tu^++su^-),tu^+\rangle ,\frac{1}{s}\langle I'(tu^++su^-),su^-\rangle \right) ,\;t,s>0. \end{aligned}$$

(2.7)

Lemma 2.6

Assume that $(V_0)$ and $(\mathrm{f}_1)$–$(\mathrm{f}_4)$ hold. Then

• (i) The pair (t, s) with $t,s>0$ is a critical point of $\varphi _u$ if and only if $tu^++su^-\in \mathcal {M};$

• (ii) The function $\varphi _u$ has an unique critical point $(t_+,s_-)=(t_+(u),s_-(u))\in (0,\infty )^2,$ which is its unique global maximum;

Proof

(i)   By (2.7), for all $t,s>0,$ $\nabla \varphi _u(t,s)=0$ if and only if

$$\begin{aligned} \langle I'(tu^++su^-),tu^+\rangle =0\quad \text{ and } \quad \langle I'(tu^++su^-),su^-\rangle =0. \end{aligned}$$

j So that, $tu^++su^-\in \mathcal {M}.$

(ii)   Let $s\ge 0$ be fixed. Consider the function $\psi _1$ defined on $[0,\infty )$ by

$$\begin{aligned} \psi _1(t)=\varphi _u(t,s). \end{aligned}$$

Similar to Lemma 2.3, there exists an unique $t(u,s)>0$ satisfying

$$\begin{aligned} \psi _1'(t(u,s))=0,\quad \psi _1'(t)>0\, \text{ for } \, t\in (0,t(u,s))\quad {and}\quad \psi _1'(t)<0 \, \text{ for } \, t\in (t(u,s),\infty ). \end{aligned}$$

(2.8)

Then, the function $\xi _1:[0,\infty )\rightarrow (0,\infty )$ given by $\xi _1(s)=t(u,s)$ is well defined. For all $s\ge 0,$ we have

$$\begin{aligned} \frac{\partial \varphi _u}{\partial t}(\xi _1(s),s)=\psi _1'(\xi _1(s))=0. \end{aligned}$$

(2.9)

Thus from (2.7),

$$\begin{aligned}&(1+b[\xi _1(s)u^++su^-]_\alpha ^2)\nonumber \\&\qquad \times \left( \int _{\mathbb {R}^{2n+m}}\frac{((\xi _1(s)u^++su^-)(x,y)-(\xi _1(s)u^++su^-)(z,y))\xi _1(s)(u^+(x,y)-u^+(z,y))}{|x-z|^{n+2\alpha }}\mathrm{d}x\mathrm{d}z\mathrm{d}y \right. \nonumber \\&\qquad +\left. \int _{\mathbb {R}^N}|\nabla _y(\xi _1(s)u^+)|^2\mathrm{d}x\mathrm{d}y\right) +\int _{\mathbb {R}^N}V(x,y)(\xi _1(s)u^+)^2\mathrm{d}x\mathrm{d}y\nonumber \\&\quad =\int _{\mathbb {R}^N}f(\xi _1(s)u^+)\xi _1(s)u^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.10)

Note that

$$\begin{aligned}{}[\xi _1(s)u^++su^-]_\alpha ^2=\xi _1(s)^2[u^+]_\alpha ^2+s^2[u^-]_\alpha ^2+2s\xi _1(s)Z(u^+,u^-) \end{aligned}$$

(2.11)

and

$$\begin{aligned}&\int _{\mathbb {R}^{2n+m}}\frac{((\xi _1(s)u^++su^-)(x,y)-(\xi _1(s)u^++su^-)(z,y))\xi _1(s)(u^+(x,y)-u^+(z,y))}{|x-z|^{n+2\alpha }}\mathrm{d}x\mathrm{d}z\mathrm{d}y\nonumber \\&\quad +\int _{\mathbb {R}^N}|\nabla _y(\xi _1(s)u^+)|^2\mathrm{d}x\mathrm{d}y =\xi _1(s)^2[u^+]_\alpha ^2+2s\xi _1(s)Z(u^+,u^-), \end{aligned}$$

(2.12)

where

$$\begin{aligned} Z(u^+,u^-)=-\int _{\mathbb {R}^{2n+m}}\frac{u^+(x,y)u^-(z,y)+u^-(x,y)u^+(z,y)}{|x-z|^{n+2\alpha }}\mathrm{d}x\mathrm{d}z\mathrm{d}y\ge 0. \end{aligned}$$

So, from (2.10) we entail

$$\begin{aligned}&\left( 1+b \left( \xi _1(s)^2[u^+]_\alpha ^2+s^2[u^-]_\alpha ^2+2s\xi _1(s)Z(u^+,u^-)\right) \right) \left( \xi _1(s)^2[u^+]_\alpha ^2+2s\xi _1(s)Z(u^+,u^-)\right) \nonumber \\&\quad +\int _{\mathbb {R}^N}V(x,y)(\xi _1(s)u^+)^2\mathrm{d}x\mathrm{d}y =\int _{\mathbb {R}^N}f(\xi _1(s)u^+)\xi _1(s)u^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.13)

$\square$

Claim 2.7

The function $\xi _1$ satisfies the following properties:

(a$_1$):

$\xi _1$ is continuous;

(a$_2$):

$\xi _1(s)< s$ for s large.

.

Proof of the Claim 2.7

(a$_1)$ Let $s_n\rightarrow s_0$ in $[0,\infty ).$ Then $\{\xi _1(s_n)\}$ is bounded. Suppose by contradiction up to a subsequence $\xi (s_n)\rightarrow +\infty .$ From (2.11)-(2.12), we can write

$$\begin{aligned}&(1+b[\xi _1(s_n)u^++s_nu^-]_\alpha ^2)[\xi _1(s_n)u^++s_nu^-]_\alpha ^2\nonumber \\&\quad \ge (1+b[\xi _1(s_n)u^++s_nu^-]_\alpha ^2)\nonumber \\&\quad \times \left( \int _{\mathbb {R}^{2n+m}}\frac{((\xi _1(s_n)u^++s_nu^-)(x,y)-(\xi _1(s_n)u^++s_nu^-)(z,y))\xi _1(s_n)(u^+(x,y)-u^+(z,y))}{|x-z|^{n+2\alpha }} \mathrm{d}x\mathrm{d}z\mathrm{d}y\right. \nonumber \\&\quad +\left. \int _{\mathbb {R}^N}|\nabla _y(\xi _1(s_n)u^+)|^2\mathrm{d}x\mathrm{d}y\right) . \end{aligned}$$

(2.14)

Combining (2.13) and (2.14), we get

$$\begin{aligned}&[\xi _1(s_n)u^++s_nu^-]_\alpha ^2+b[\xi _1(s_n)u^++s_nu^-]_\alpha ^4+\int _{\mathbb {R}^N}V(x,y)(\xi _1(s_n)u^+)^2\mathrm{d}x\mathrm{d}y\nonumber \\&\ge \int _{\mathbb {R}^N}f(\xi _1(s_n)u^+)\xi _1(s_n)u^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.15)

Thereby,

$$\begin{aligned}&\frac{[\xi _1(s_n)u^++s_nu^-]_\alpha ^2}{\xi _1(s_n)^4}+b\frac{[\xi _1(s_n)u^++s_nu^-]_\alpha ^4}{\xi _1(s_n)^4} +\frac{1}{\xi _1(s_n)^2}\int _{\mathbb {R}^N}V(x,y)(u^+)^2\mathrm{d}x\mathrm{d}y\nonumber \\&\quad \ge \int _{\mathbb {R}^N}\frac{f(\xi _1(s_n)u^+)}{(\xi _1(s_n)u^+)^3}(u^+)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.16)

According to $(\mathrm{f}_3)$–$(\mathrm{f}_4),$ Fatou’s lemma and taking account that $\xi _1(s_n)\rightarrow \infty ,$ and $s_n\rightarrow s_0<\infty ,$ passing to the limit as $n\rightarrow \infty$ in (2.16), we obtain

$$\begin{aligned} b[u^+]_\alpha ^4\ge \int _{\mathbb {R}^N}\liminf _{n\rightarrow \infty }\frac{f(\xi _1(s_n)u^+)}{(\xi _1(s_n)u^+)^3}(u^+)^4\mathrm{d}x\mathrm{d}y=+\infty , \end{aligned}$$

which is impossible and hence $\{\xi _1(s_n)\}$ is bounded. Then, up to a subsequence $\xi _1(s_n)\rightarrow t_0.$ By (2.13), we have

$$\begin{aligned}&\xi _1(s_n)^2[u^+]_\alpha ^2+2s_n\xi _1(s_n)Z(u^+,u^-)\\&\quad +b\left( \xi _1(s_n)^2[u^+]_\alpha ^2+s_n^2[u^-]_\alpha ^2+2s_n\xi _1(s_n)Z(u^+,u^-)\right) \left( \xi _1(s_n)^2[u^+]_\alpha ^2+2s_n\xi _1(s_n)Z(u^+,u^-)\right) \\&\quad +\int _{\mathbb {R}^N}V(x,y)(\xi _1(s_n)u^+)^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}f(\xi _1(s_n)u^+)\xi _1(s_n)u^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

Letting $n\rightarrow \infty ,$

$$\begin{aligned}&t_0^2[u^+]_\alpha ^2+2s_0t_0Z(u^+,u^-)\\&\quad + b\left( t_0^2[u^+]_\alpha ^2+s_0^2[u^-]_\alpha ^2+2s_0t_0Z(u^+,u^-)\right) \left( t_0^2[u^+]_\alpha ^2+2s_0t_0Z(u^+,u^-)\right) \\&\quad + \int _{\mathbb {R}^N}V(x,y)(t_0u^+)^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}f(t_0u^+)t_0u^+\mathrm{d}x\mathrm{d}y, \end{aligned}$$

which implies $\psi _1'(t_0)=\frac{\partial \varphi _u}{\partial t}(t_0,s_0)=0,$ and by the uniqueness of $\xi _1(s_0)$, we deduce $\xi _1(s_0)=t_0.$ The continuity of $\xi _1$ is achieved.

(a$_2)$ For s large enough, we have $\xi _1(s)< s.$ If not, there is $s_n\rightarrow \infty$ such that $\xi _1(s_ n)\ge s_n$ for all n. Then $\xi _1(s_ n)\rightarrow \infty .$ Therefore, using (2.16), we arrive at a contradiction and consequently (a$_2)$ follows. The claim is proved. $\square$

In the same way, for $t\ge 0$ fixed, we define $\psi _2$ on $[0,\infty )$ by $\psi _2(s)=\varphi _u(t,s),$ and so we can find $\xi _2:[0,\infty )\rightarrow (0,\infty )$ satisfying $(\mathrm{a}_1){-}(\mathrm{a}_2)$ and

$$\begin{aligned} \psi _2'(\xi _2(t))=\frac{\partial \varphi _u}{\partial s}(t,\xi _2(t))=0 \text{ for } \text{ all } t\ge 0. \end{aligned}$$

(2.17)

In view of $(\mathrm{a}_2),$ there exists $A_1>0$ such that for all $t,s\ge A_1,$

$$\begin{aligned} \xi _1(s)\le s\quad \text{ and }\quad \xi _2(t)\le t. \end{aligned}$$

Set

$$\begin{aligned} A_2=\max \left\{ \max _{s\in [0,A_1]}\xi _1(s),\max _{t\in [0,A_1]}\xi _2(t)\right\} \end{aligned}$$

and $A=\max \{A_1,A_2\}.$ Let $\xi :[0,A]\times [0,A]\rightarrow (0,\infty )^2$ defined by

$$\begin{aligned} \xi (t,s)=(\xi _1(s),\xi _2(t)). \end{aligned}$$

Obviously, $\xi$ is continuous since $\xi _1$ and $\xi _2$ are continuous by Claim 2.7. Moreover, we have

$$\begin{aligned} \xi ([0,A]\times [0,A])\subset [0,A]\times [0,A]. \end{aligned}$$

Applying Brouwer’s fixed point theorem, there exists $(t_+,s_-)\in [0,A]^2$ such that

$$\begin{aligned} \xi (t_+,s_-)=(\xi _1(s_-),\xi _2(t_+))=(t_+,s_-) \end{aligned}$$

Since $\xi _1$ and $\xi _2$ are positive functions, $t_+,s_->0.$ In addition, $\nabla \varphi _u(t_+,s_-)=0,$ hence $(t_+,s_-)$ is a critical point of $\varphi _u.$

Let us now prove the uniqueness of the critical points of $\varphi _u$.

Claim 2.8

For all $v\in \mathcal {M},$ we have (1, 1) is a critical point of $\varphi _v.$ Moreover, if $(t_0,s_0)$ with $t_0,s_0>0$ is a critical point of $\varphi _v,$ then $t_0=s_0=1.$

Proof of the Claim 2.8

Since $v=v^++v^-\in \mathcal {M},$ $\nabla \varphi _u(1,1)=0.$ Thus (1, 1) is a critical point of $\varphi _v.$ Let $(t_0,s_0)$ be a critical point of $\varphi _v.$ Without loss of generality, we may assume that $0<t_0\le s_0.$ Then

$$\begin{aligned} \langle I'(t_0v^++s_0v^-),t_0v^+\rangle =0\quad \text{ and } \quad \langle I'(t_0v^++s_0v^-),s_0v^-\rangle =0. \end{aligned}$$

(2.18)

The first part of (2.18) can be written

$$\begin{aligned}&\left( t_0^2[v^+]_\alpha ^2+2t_0s_0Z(v^+,v^-)\right) \left( 1+b \left( t_0^2[v^+]_\alpha ^2+s_0^2[v^-]_\alpha ^2+2t_0s_0Z(v^+,v^-)\right) \right) \nonumber \\&\quad +\int _{\mathbb {R}^N}V(x,y)(t_0v^+)^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}f(t_0v^+)t_0v^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

Dividing by $t_0^4,$ we obtain

$$\begin{aligned}&\frac{1}{t_0^2}[v^+]_\alpha ^2+b[v^+]_\alpha ^4+b\frac{s_0^2}{t_0^2}[v^+]_\alpha ^2[v^-]_\alpha ^2 +2\left( \frac{s_0}{t_0^3}+2b\frac{s_0}{t_0}[v^+]_\alpha ^2+b\frac{s_0^3}{t_0^3}[v^-]_\alpha ^2\right) Z(v^+,v^-)\\&\quad + \, 4b\frac{s_0^2}{t_0^2}Z(v^+,v^-)^2+\frac{1}{t_0^2}\int _{\mathbb {R}^N}V(x,y)(v^+)^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}\frac{f(t_0v^+)}{t_0^3}v^+\mathrm{d}x\mathrm{d}y. \end{aligned}$$

We have $0<t_0\le s_0$ and $Z(v^+,v^-)\ge 0,$ thus

$$\begin{aligned}&\frac{1}{t_0^2}[v^+]_\alpha ^2+b[v^+]_\alpha ^4+b[v^+]_\alpha ^2[v^-]_\alpha ^2 +2\left( \frac{1}{t_0^2}+2b[v^+]_\alpha ^2+b[v^-]_\alpha ^2\right) Z(v^+,v^-)\nonumber \\&\quad + \, 4bZ(v^+,v^-)^2 +\frac{1}{t_0^2}\int _{\mathbb {R}^N}V(x,y)(v^+)^2\mathrm{d}x\mathrm{d}y\le \int _{\mathbb {R}^N}\frac{f(t_0v^+)}{(t_0v^+)^3}(v^+)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.19)

On the other hand, $v^++v^-\in \mathcal {M}$ implies

$$\begin{aligned}&[v^+]_\alpha ^2+b[v^+]_\alpha ^4+b[v^+]_\alpha ^2[v^-]_\alpha ^2+2\left( 1+2b[v^+]_\alpha ^2+b[v^-]_\alpha ^2\right) Z(v^+,v^-)\nonumber \\&\quad + \, 4bZ(v^+,v^-)^2+\int _{\mathbb {R}^N}V(x,y)(v^+)^2\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}\frac{f(v^+)}{(v^+)^3}(v^+)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.20)

From (2.19) to (2.20), we deduce

$$\begin{aligned}&\left( 1-\frac{1}{t_0^2}\right) \left( [v^+]_\alpha ^2+\int _{\mathbb {R}^N}V(x,y)(v^+)^2\mathrm{d}x\mathrm{d}y\right) +2\left( 1-\frac{1}{t_0^2}\right) Z(v^+,v^-)\nonumber \\&\quad \ge \int _{\mathbb {R}^N}\left( \frac{f(v^+)}{(v^+)^3}-\frac{f(t_0v^+)}{(t_0v^+)^3}\right) (v^+)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.21)

If $t_0<1,$ by $(\mathrm{f}_4)$ and (2.21) we derive a contradiction. Then $s_0\ge t_0\ge 1.$

Similarly, using the second part of (2.18), we get

$$\begin{aligned}&\left( \frac{1}{s_0^2}-1\right) \left( [v^-]_\alpha ^2+\int _{\mathbb {R}^N}V(x,y)(v^-)^2\mathrm{d}x\mathrm{d}y\right) +2\left( \frac{1}{s_0^2}-1\right) Z(v^+,v^-)\nonumber \\&\quad \ge \int _{\mathbb {R}^N}\left( \frac{f(s_0v^-)}{(s_0v^-)^3}-\frac{f(v^-)}{(v^-)^3}\right) (v^-)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.22)

This and $(\mathrm{f}_4)$ imply $t_0\le s_0\le 1.$ So, we conclude that $t_0=s_0=1.$ The claim is proved. $\square$

Proof

Now, we turn to prove (ii). Let $(t_1,s_1)$ and $(t_2,s_2)$ be two critical points of $\varphi _u$ with $t_1,s_1,t_2,s_2>0.$ By (i), we have

$$\begin{aligned} v_1:=t_1u^++s_1u^-\in \mathcal {M}\quad \text{ and } \quad v_2:=t_2u^++s_2u^-\in \mathcal {M}. \end{aligned}$$

Then $v_1^\pm \ne 0$ and

$$\begin{aligned} \frac{t_2}{t_1}v_1^++\frac{s_2}{s_1}v_1^-=v_2\in \mathcal {M}. \end{aligned}$$

Therefore, as a consequence of (i), $\left( \frac{t_2}{t_1},\frac{s_2}{s_1}\right)$ is a critical point of $\varphi _{v_1}.$ From Claim 2.8, we obtain $\frac{t_2}{t_1}=\frac{s_2}{s_1}=1,$ that is $t_1=t_2$ and $s_1=s_2,$ and hence the uniqueness is achieved.

Next, we will show that $\varphi _u$ admits a global maximum. Let $\Omega _0^\pm :=\{(x,y)\in \mathbb {R}^N:u^\pm (x,y)\ne 0\}.$ Then $|\Omega _0^\pm |>0$ and

$$\begin{aligned} \varphi _u(t,s)= & {} \frac{1}{2}||tu^++su^-||_{E_\alpha }^2+\frac{b}{4}[tu^++su^-]_\alpha ^4-\int _{\mathbb {R}^N}F(tu^++su^-)\mathrm{d}x\mathrm{d}y\\= & {} \frac{t^2}{2}||u^+||_{E_\alpha }^2+\frac{s^2}{2}||u^-||_{E_\alpha }^2+tsZ(u^+,u^-)+\frac{b}{4}[tu^++su^-]_\alpha ^4-\int _{\mathbb {R}^N}F(tu^++su^-)\mathrm{d}x\mathrm{d}y\\= & {} \frac{t^2}{2}||u^+||_{E_\alpha }^2+\frac{s^2}{2}||u^-||_{E_\alpha }^2+tsZ(u^+,u^-)+\frac{b}{4}\left( t^2[u^+]_\alpha ^2+s^2[u^-]_\alpha ^2+2tsZ(u^+,u^-)\right) ^2\\&- \, \int _{\Omega _0^+}F(tu^+)\mathrm{d}x\mathrm{d}y-\int _{\Omega _0^-}F(su^-)\mathrm{d}x\mathrm{d}y\\\le & {} (t^2+s^2)\left( \frac{1}{2}||u^+||_{E_\alpha }^2+\frac{1}{2}||u^-||_{E_\alpha }^2+Z(u^+,u^-)\right) +\frac{b(t^2+s^2)^2}{4}\left( [u^+]_\alpha ^2 +[u^-]_\alpha ^2+Z(u^+,u^-)\right) ^2\\&- \, \int _{\Omega _0^+}F(tu^+)\mathrm{d}x\mathrm{d}y-\int _{\Omega _0^-}F(su^-)\mathrm{d}x\mathrm{d}y.\\ \end{aligned}$$

Thus

$$\begin{aligned} \frac{\varphi _u(t,s)}{(t^2+s^2)^2}\le & {} \frac{1}{t^2+s^2}\left( \frac{1}{2}||u^+||_{E_\alpha }^2+\frac{1}{2}||u^-||_{E_\alpha }^2+Z(u^+,u^-)\right) +\frac{b}{4}\left( [u^+]_\alpha ^2+[u^-]_\alpha ^2+Z(u^+,u^-)\right) ^2\nonumber \\&- \, \frac{t^4}{(t^2+s^2)^2}\int _{\Omega _0^+}\frac{F(tu^+)}{(tu^+)^4}(u^+)^4\mathrm{d}x\mathrm{d}y -\frac{s^4}{(t^2+s^2)^2}\int _{\Omega _0^-}\frac{F(su^-)}{(su^-)^4}(u^-)^4\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.23)

Observing that

$$\begin{aligned} \frac{t^4}{(t^2+s^2)^2}\ge \frac{1}{4} \text{ if } t\ge s\quad \text{ and }\quad \frac{s^4}{(t^2+s^2)^2}\ge \frac{1}{4} \text{ if } t\le s, \end{aligned}$$

it follows from (2.23), $(\mathrm{f}_3)$ and Fatou’s Lemma that

$$\begin{aligned} \limsup _{\begin{array}{c} |(t,s)|\rightarrow \infty \\ t\ge s \end{array}}\frac{\varphi _u(t,s)}{(t^2+s^2)^2}\le & {} \frac{b}{4}\left( [u^+]_\alpha ^2+[u^-]_\alpha ^2+Z(u^+,u^-)\right) ^2\\&- \, \frac{1}{4}\int _{\Omega _0^+}\liminf _{t\rightarrow +\infty }\frac{F(tu^+)}{(tu^+)^4}(u^+)^4\mathrm{d}x\mathrm{d}y=-\infty \end{aligned}$$

and

$$\begin{aligned} \limsup _{\begin{array}{c} |(t,s)|\rightarrow \infty \\ s\ge t \end{array}}\frac{\varphi _u(t,s)}{(t^2+s^2)^2}\le & {} \frac{b}{4}\left( [u^+]_\alpha ^2+[u^-]_\alpha ^2+Z(u^+,u^-)\right) ^2\\&- \, \frac{1}{4}\int _{\Omega _0^-}\liminf _{s\rightarrow +\infty }\frac{F(su^-)}{(su^-)^4}(u^-)^4\mathrm{d}x\mathrm{d}y=-\infty . \end{aligned}$$

Hence, $\varphi _u(t,s)\rightarrow -\infty$ as $|(t,s)|\rightarrow \infty .$ So, by the continuity, we conclude that $\varphi _u$ admits a global maximum $(\hat{t},\hat{s})\in [0,\infty )^2.$ We notice that $[tu^+]_\alpha ^2+[su^-]_\alpha ^2\le [tu^++su^-]_\alpha ^2$ and

$$\begin{aligned} \int _{\mathbb {R}^N}F(tu^++su^-)\mathrm{d}x\mathrm{d}y=\int _{\mathbb {R}^N}F(tu^+)\mathrm{d}x\mathrm{d}y+\int _{\mathbb {R}^N}F(su^-)\mathrm{d}x\mathrm{d}y. \end{aligned}$$

Therefore

$$\begin{aligned} \varphi _u(t,0)+\varphi _u(0,s)=I(tu^+)+I(su^-)\le I(tu^++su^-)=\varphi _u(t,s) \text{ for } \text{ all } t,s\ge 0. \end{aligned}$$

This together with the fact that $\max _{t\ge 0} \varphi _u(t,0),\max _{s\ge 0}\varphi _u(0,s)>0$ leads to

$$\begin{aligned} \max _{t\ge 0} \varphi _u(t,0)< \max _{t,s\ge 0}\varphi _u(t,s)=\varphi _u(\hat{t},\hat{s})\quad \text{ and }\quad \max _{s\ge 0}\varphi _u(0,s)<\max _{t,s\ge 0}\varphi _u(t,s)=\varphi _u(\hat{t},\hat{s}), \end{aligned}$$

which shows that $\hat{t},\hat{s}>0.$ Consequently $(\hat{t},\hat{s})$ is a critical point with positive coordinates. By the uniqueness of $(t_+,s_-),$ we deduce that $(\hat{t},\hat{s})=(t_+,s_-),$ and hence $(t_+,s_-)$ is an unique global maximum. This finishes the proof of Lemma 2.6. $\square$

Lemma 2.9

Let $\{u_n\}\subset \mathcal {M}.$ If $u_n\rightharpoonup u$ weakly in $E_\alpha ,$ then $u^\pm \ne 0.$

Proof

For all $v\in \mathcal {M},$ we have

$$\begin{aligned} ||v^\pm ||_{E_\alpha }^2\le \int _{\mathbb {R}^N}f(v^\pm )v^\pm \mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.24)

Therefore, using $(\mathrm{f}_1)$–$(\mathrm{f}_2)$ and Lemma 2.2, for $\varepsilon >0,$ there exists $C_\varepsilon ,$ such that

$$\begin{aligned} ||v^\pm ||_{E_\alpha }^2\le \varepsilon ||V^{-1}||_\infty ||v^\pm ||_{E_\alpha }^2+CC_\varepsilon ||v^\pm ||_{E_\alpha }^{p^*}. \end{aligned}$$

Letting $\varepsilon <\frac{1}{||V^{-1}||_\infty },$ we get

$$\begin{aligned} ||v^\pm ||_{E_\alpha }\ge \delta :=\left( \frac{1-\varepsilon ||V^{-1}||_\infty }{CC_\varepsilon }\right) ^{\frac{1}{p^*-2}}. \end{aligned}$$

Let $u_n\rightharpoonup u$ weakly in $E_\alpha .$ It follows from (2.24) that

$$\begin{aligned} \int _{\mathbb {R}^N}f(u_n^\pm )u_n^\pm \mathrm{d}x\mathrm{d}y \ge \delta \text{ for } \text{ all } n\in \mathbb {N}. \end{aligned}$$

By $(\mathrm{f}_1)$–$(\mathrm{f}_2)$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\mathbb {R}^N}f(u_n^\pm )u_n^\pm \mathrm{d}x\mathrm{d}y= & {} \int _{\mathbb {R}^N}f(u^\pm )u^\pm \mathrm{d}x\mathrm{d}y, \end{aligned}$$

(2.25)

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\mathbb {R}^N}F(u_n^\pm )\mathrm{d}x\mathrm{d}y= & {} \int _{\mathbb {R}^N}F(u^\pm )\mathrm{d}x\mathrm{d}y. \end{aligned}$$

(2.26)

Then

$$\begin{aligned} \int _{\mathbb {R}^N}f(u^\pm )u^\pm \mathrm{d}x\mathrm{d}y \ge \delta >0, \end{aligned}$$

which implies that $u^\pm \ne 0.$ $\square$

Remark 2.10

We have

$$\begin{aligned} c_\infty ^\mathcal {M}:=\inf _{u\in \mathcal {M}}I(u)\ge c_\infty ^\mathcal {N}, \end{aligned}$$

being $\mathcal {M}\subset \mathcal {N}.$

Proof of Theorem 1.1

Let $\{u_n\}\subset \mathcal {M}$ be a minimizing sequence of I,  that is

$$\begin{aligned} \lim _{n\rightarrow \infty }I(u_n)=\inf _{v\in \mathcal {M}}I(v)=c_\infty ^{\mathcal {M}}>0. \end{aligned}$$

(2.27)

Let us first prove that $\{u_n\}$ is bounded in $E_\alpha .$ Suppose that up to a subsequence $||u_n||_{E_\alpha }\rightarrow \infty$ and set $v_n=\frac{u_n}{||u_n||_{E_\alpha }}.$ Then $\{v_n\}$ is bounded in $E_\alpha .$ So, in view of Lemma 2.2, we may assume that

$$\begin{aligned} v_n\rightharpoonup v \text{ in } {E_\alpha },\;v_n\rightarrow v \text{ a.e. } \mathbb {R}^N \text{ and } v_n\rightarrow v \text{ in } L^q(\mathbb {R}^N) \text{ for } q\in [2,p^*). \end{aligned}$$

(2.28)

Since $||u_n||_{E_\alpha }v_n^++||u_n||_{E_\alpha }v_n^-=u_n\in \mathcal {M},$ by Lemma 2.6,

$$\begin{aligned} t_+(v_n)=s_-(v_n)=||u_n||_{E_\alpha }. \end{aligned}$$

Therefore, for all $t>0$ and $n\in \mathbb {N},$

$$\begin{aligned} I(u_n)= & {} I(||u_n||_{E_\alpha }v_n)=I(t_+(v_n)v_n^++s_-(v_n)v_n^-)=\varphi _{v_n}(t_+(v_n),s_-(v_n))\nonumber \\\ge & {} \varphi _{v_n}(t,t)= I(tv_n)=\frac{t^2}{2}+\frac{bt^4}{4}[v_n]_\alpha ^4-\int _{\mathbb {R}^N}F(tv_n)\mathrm{d}x\mathrm{d}y\nonumber \\\ge & {} \frac{t^2}{2}-\int _{\mathbb {R}^N}F(tv_n)\mathrm{d}x\mathrm{d}y \end{aligned}$$

(2.29)

If $v=0,$ then (2.27)–(2.29) imply

$$\begin{aligned} c_\infty ^{\mathcal {M}}\ge \frac{t^2}{2} \text{ for } \text{ all } t>0, \end{aligned}$$

(2.30)

which is a contradiction. Consequently $v\ne 0.$ Then, $|\Omega _0|>0,$ where $\Omega _0=\{(x,y):v(x,y)\ne 0\}.$

Dividing $I(u_n)$ by $||u_n||_{E_\alpha }^4$ and using (2.28), $(\mathrm{f}_3)$ and Fatou’s lemma, we get

$$\begin{aligned} 0=\lim _{n\rightarrow \infty }\frac{I(u_n)}{||u_n||_{E_\alpha }^4}\le \frac{b}{4}-\int _{\Omega _0}\liminf _{n\rightarrow \infty } \frac{F(||u_n||_{E_\alpha }v_n)}{(||u_n||_{E_\alpha }v_n)^4}v_n^4\mathrm{d}x\mathrm{d}y=-\infty . \end{aligned}$$

This is impossible and hence $\{u_n\}$ is bounded in $E_\alpha .$ Then $u_n\rightharpoonup u$ in ${E_\alpha }.$ From Lemma 2.9, we infer that $u^\pm \ne 0.$ Applying Lemma 2.6, there exist $t_+,s_->0$ such that $t_+u^++s_-u^-\in \mathcal {M}$ and

$$\begin{aligned} \langle I'(t_+u^++s_-u^-),u^\pm \rangle =0. \end{aligned}$$

Since $u_n\rightharpoonup u$ and $u_n\in \mathcal {M}$, it follows from (2.25) and Fatou’s lemma that

$$\begin{aligned} \langle I'(u),u^\pm \rangle \le \liminf _{n\rightarrow \infty }\langle I'(u_n),u_n^\pm \rangle =0. \end{aligned}$$

Similar to the proof of Claim 2.8, we see that $t_+,s_-\le 1.$ It follows from $(\mathrm{f}_4)$ and (2.25)–(2.26) that

$$\begin{aligned} c_\infty ^\mathcal {M}\le & {} I(t_+u^++s_-u^-)\\= & {} I(t_+u^++s_-u^-)-\frac{1}{4}\langle I'(t_+u^++s_-u^-),t_+u^++s_-u^-\rangle \\= & {} \frac{1}{4}||t_+u^++s_-u^-||_{E_\alpha }^2+\int _{\mathbb {R}^N}\left( \frac{1}{4}f(t_+u^+)t_+u^+-F(t_+u^+)\right) \mathrm{d}x\mathrm{d}y\\&+ \, \int _{\mathbb {R}^N}\left( \frac{1}{4}f(s_-u^-)s_-u^--F(s_-u^-)\right) \mathrm{d}x\mathrm{d}y\\\le & {} \frac{1}{4}||u||_{E_\alpha }^2+\int _{\mathbb {R}^N}\left( \frac{1}{4}f(u^+)u^+-F(u^+)\right) \mathrm{d}x\mathrm{d}y\\&+ \, \int _{\mathbb {R}^N}\left( \frac{1}{4}f(u^-)u^--F(u^-)\right) \mathrm{d}x\mathrm{d}y\\= & {} \frac{1}{4}||u||_{E_\alpha }^2+\int _{\mathbb {R}^N}\left( \frac{1}{4}f(u)u-F(u)\right) \mathrm{d}x\mathrm{d}y\\\le & {} \liminf _{n\rightarrow \infty }\left[ \frac{1}{4}||u_n||_{E_\alpha }^2+\int _{\mathbb {R}^N}\left( \frac{1}{4}f(u_n)u_n-F(u_n)\right) \mathrm{d}x\mathrm{d}y\right] \\= & {} \liminf _{n\rightarrow \infty }\left[ I(u_n)-\frac{1}{4}\langle I'(u_n),u_n\rangle \right] =c_\infty ^\mathcal {M}. \end{aligned}$$

This yields $I(t_+u^++s_-u^-)=c_\infty ^\mathcal {M}.$ Arguing as in the proof of Claim 2.8, we derive that $t_+=s_-=1,$ and so

$$\begin{aligned} I(u)=\inf _{v\in \mathcal {M}}I(v). \end{aligned}$$

Now, we show that $I'(u)=0.$ Suppose by contradiction that $I'(u)\ne 0.$ Then, by the continuity of $I',$ there exist $\nu ,\varrho >0$ such that

$$\begin{aligned} ||I'(v)||\ge \nu \text{ for } \text{ all } v\in {E_\alpha } \text{ with } ||v-u||_{E_\alpha }\le 3\varrho . \end{aligned}$$

(2.31)

Let $0<\mathrm{a}_1<1<\mathrm{a}_2$ with $\mathrm{a}_2-\mathrm{a}_1<\frac{\varrho }{\sqrt{2}||u||_{E_\alpha }},$ and $D=(\mathrm{a}_1,\mathrm{a}_2)\times (\mathrm{a}_1,\mathrm{a}_2).$ Define the function $\gamma _u:\overline{D}\rightarrow {E_\alpha }$ by

$$\begin{aligned} \gamma _u(t,s)=tu^++su^-. \end{aligned}$$

Then $I(\gamma _u(1,1))=c_\infty ^\mathcal {M},$ and by Lemma 2.6(ii), $I(\gamma _u(t,s))<c_\infty ^\mathcal {M}$ for all $(t,s)\in \overline{D}\backslash \{(1,1)\}.$ Therefore

$$\begin{aligned} \varpi :=\max _{(t,s)\in \partial D}I(\gamma _u(t,s))<c_\infty ^\mathcal {M}. \end{aligned}$$

(2.32)

In view of (2.31) and [34, Lemma 2.3], for $\varepsilon <\min \left\{ \frac{\nu \varrho }{8},\frac{c_\infty ^\mathcal {M}-\varpi }{2}\right\} ,$ there exists $\eta \in C([0,1]\times {E_\alpha },{E_\alpha })$ verifying

$\eta _1)$:

$\eta (t,v)=v$ if $v\not \in I^{-1}([c_\infty ^\mathcal {M}-2\varepsilon ,c_\infty ^\mathcal {M}+2\varepsilon ]);$

$\eta _2)$:

$I(\eta (1,v))\le c_\infty ^\mathcal {M}-\varepsilon$ for all $v\in E_\alpha$ with $||v-u||_{E_\alpha }\le \varrho$ and $I(v)\le c_\infty ^\mathcal {M}+\varepsilon ;$

$\eta _3)$:

$I(\eta (1,v))\le I(\eta (0,v))=I(v)$ for all $v\in {E_\alpha }.$

We claim that

$$\begin{aligned} \max _{(t,s)\in \overline{D}}I(\eta (1,\gamma _u(t,s))) <c_\infty ^\mathcal {M}. \end{aligned}$$

(2.33)

Indeed, by Lemma 2.6,

$$\begin{aligned} I(\gamma _u(t,s))<c_\infty ^\mathcal {M}+\varepsilon \text{ for } \text{ all } (t,s)\in \overline{D}. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} ||\gamma _u(t,s)-u||_{E_\alpha }^2=||(t-1)u^++(s-1)u^-||_{E_\alpha }^2\\ \le 2(\mathrm{a}_2-\mathrm{a}_1)^2||u||_{E_\alpha }^2 <\varrho ^2, \end{aligned}$$

thus $||\gamma _u(t,s)-u||_{E_\alpha }<\varrho$ for all $(t,s)\in \overline{D}.$ Therefore, $(\eta _2)$ implies $I(\eta (1,\gamma _u(t,s)))\le c_\infty ^\mathcal {M}-\varepsilon ,$ and hence (2.33) holds.

Now, we show that

$$\begin{aligned} \eta (1,\gamma _u(D))\cap \mathcal {M}\ne \emptyset . \end{aligned}$$

(2.34)

Let

$$\begin{aligned} \zeta _u(t,s)=\eta (1,\gamma _u(t,s)), \end{aligned}$$

$$\begin{aligned} \phi _0(t,s)= & {} \left( \langle I'(\gamma _u(t,s)),u^+\rangle ,\langle I'(\gamma _u(t,s)),u^-\rangle \right) \\=: & {} \left( \phi _0^1(t,s),\phi _0^2(t,s)\right) \end{aligned}$$

and

$$\begin{aligned} \phi _1(t,s)=\left( \frac{1}{t}\langle I'(\zeta _u(t,s)),(\zeta _u(t,s))^+\rangle ,\frac{1}{s}\langle I'(\zeta _u(t,s)),(\zeta _u(t,s))^-\rangle \right) . \end{aligned}$$

By direct computation

$$\begin{aligned} \frac{\partial \phi _0^1}{\partial t}(1,1)= & {} ||u^+||_{E_\alpha }^2+3b[u^+]_\alpha ^4+8bZ(u^+,u^-)[u^+]_\alpha ^2+b[u^+]_\alpha ^2[u^-]_\alpha ^2+4bZ(u^+,u^-)^2\nonumber \\&- \, \int _{\mathbb {R}^N}f'(u^+)(u^+)^2\mathrm{d}x\mathrm{d}y \end{aligned}$$

(2.35)

and

$$\begin{aligned} \frac{\partial \phi _0^1}{\partial s}(1,1)= & {} 4bZ(u^+,u^-)[u^+]_\alpha ^2+6bZ(u^+,u^-)[u^-]_\alpha ^2+2b[u^+]_\alpha ^2[u^-]_\alpha ^2+8bZ(u^+,u^-)^2\nonumber \\&+2Z(u^+,u^-). \end{aligned}$$

(2.36)

From $\langle I'(u^++u^-),u^+\rangle =0,$ we entail

$$\begin{aligned} \int _{\mathbb {R}^N}f(u^+)u^+\mathrm{d}x\mathrm{d}y= & {} ||u^+||_{E_\alpha }^2+b[u^+]_\alpha ^4+4bZ(u^+,u^-)[u^+]_\alpha ^2+2bZ(u^+,u^-)[u^-]_\alpha ^2\nonumber \\&+ \, b[u^+]_\alpha ^2[u^-]_\alpha ^2+4bZ(u^+,u^-)^2+2Z(u^+,u^-). \end{aligned}$$

(2.37)

By $(\mathrm{f}_4),$ we see that

$$\begin{aligned} f'(t)t^2>3f(t)t \text{ for } \text{ all } t\ne 0. \end{aligned}$$

(2.38)

Combining (2.35) and (2.37)–(2.38), we obtain

$$\begin{aligned} -\frac{\partial \phi _0^1}{\partial t}(1,1)> & {} 2||u^+||_{E_\alpha }^2+4bZ(u^+,u^-)[u^+]_\alpha ^2+6bZ(u^+,u^-)[u^-]_\alpha ^2+2b[u^+]_\alpha ^2[u^-]_\alpha ^2\nonumber \\&+ \, 8bZ(u^+,u^-)^2+6Z(u^+,u^-). \end{aligned}$$

(2.39)

Similarly, we have

$$\begin{aligned} \frac{\partial \phi _0^2}{\partial t}(1,1)= & {} 6bZ(u^+,u^-)[u^+]_\alpha ^2+4bZ(u^+,u^-)[u^-]_\alpha ^2+2b[u^+]_\alpha ^2[u^-]_\alpha ^2+8bZ(u^+,u^-)^2\nonumber \\&+2Z(u^+,u^-) \end{aligned}$$

(2.40)

and

$$\begin{aligned} -\frac{\partial \phi _0^2}{\partial s}(1,1)> & {} 2||u^-||_{E_\alpha }^2+6bZ(u^+,u^-)[u^+]_\alpha ^2+4bZ(u^+,u^-)[u^-]_\alpha ^2+2b[u^+]_\alpha ^2[u^-]_\alpha ^2\nonumber \\&+8bZ(u^+,u^-)^2+6Z(u^+,u^-). \end{aligned}$$

(2.41)

From (2.36) and (2.39)–(2.41), we deduce

$$\begin{aligned} -\frac{\partial \phi _0^1}{\partial t}(1,1)>\frac{\partial \phi _0^1}{\partial s}(1,1)\quad \text{ and }\quad -\frac{\partial \phi _0^2}{\partial s}(1,1)>\frac{\partial \phi _0^2}{\partial t}(1,1). \end{aligned}$$

Therefore,

$$\begin{aligned} \text{ Jac }(\phi _0)(1,1)>0. \end{aligned}$$

Note that (1, 1) is the unique pre-image of (0, 0) by $\phi _0,$ thus (1, 1) is the unique regular point of $\phi _0.$ By the Brouwer degree theory,

$$\begin{aligned} \deg (\phi _0,D,(0,0))=\text{ sign } (\text{ Jac }(\phi _0)(1,1))=1. \end{aligned}$$

(2.42)

By (2.32) for all $(t,s)\in \partial D$,

$$\begin{aligned} I(\gamma _u(t,s))\le \varpi <c_\infty ^{\mathcal {M}}-2\varepsilon . \end{aligned}$$

This together with $(\eta _1)$ gives $\eta (1,\gamma _u(t,s)))=\gamma _u(t,s)$ for all $(t,s)\in \partial D.$ Hence

$$\begin{aligned} \deg (\phi _1,D,(0,0))=\deg (\phi _0,D,(0,0))=1, \end{aligned}$$

(2.43)

which implies $\phi _1(t_0,s_0)=(0,0)$ for some $(t_0,s_0)\in D.$ Consequently

$$\begin{aligned} \eta (1,\gamma _u(t_0,s_0))=\zeta _u(t_0,s_0)\in \mathcal {M}, \end{aligned}$$

but this contradicts (2.33). We conclude that u is a critical point of I and so it is a sign-changing solution of (1.1).

If f is odd, then I is even. In view of Lemma 2.4 and Remark 2.5, $\Psi$ is bounded below in $S_1.$ By standard arguments and Lemma 2.3, we see that $\Psi$ satisfies (PS) condition on $S_1.$ According to [29, Theorem 8.10] and Lemma 2.4, we deduce that I has infinitely many critical points, which completes the proof of Theorem 1.1. $\square$