Periodic solutions of an asymptotically linear Dirac equation

Abstract

Using the variational method, we investigate periodic solutions of a Dirac equation with asymptotically nonlinearity. The variational setting is established and the existence and multiplicity of periodic solutions are obtained.

1 Introduction and main results

Let us consider the following (stationary) Dirac equation

$$\begin{aligned} -i\sum ^3_{k=1}\alpha _k\partial _ku + a\beta u + V(x)u = G_u(x,u) \end{aligned}$$

(1.1)

for $x=(x_1,x_2,x_3)\in \mathbb {R}^3$, where $\partial _k=\partial /\partial x_k, a>0$ is a constant, $\alpha _1, \alpha _2, \alpha _3$ and $\beta$ are $4\times 4$ Pauli-Dirac matrices:

$$\begin{aligned} \beta = \left( \begin{array}{cc} I &{} 0\\ 0 &{} -I \end{array} \right) , \qquad \alpha _k=\left( \begin{array}{cc} 0 &{} \sigma _k\\ \sigma _k &{} 0 \end{array} \right) , \quad k=1,2,3 \end{aligned}$$

with

$$\begin{aligned} \sigma _1= \left( \begin{array}{cc} 0 &{} 1\\ 1 &{} 0 \end{array} \right) , \quad \sigma _2= \left( \begin{array}{cc} 0 &{} -i\\ i &{} 0 \end{array} \right) , \quad \sigma _3= \left( \begin{array}{cc} 1 &{} 0\\ 0 &{} -1 \end{array} \right) . \end{aligned}$$

This equation arises when one seeks for the standing wave solutions of the nonlinear Dirac equation (see [25])

$$\begin{aligned} -i\hbar \partial _t\psi =ic\hbar \sum ^3_{k=1}\alpha _k\partial _k \psi - mc^2\beta \psi - M(x)\psi + F_\psi (x,\psi ). \end{aligned}$$

(1.2)

Assuming that $F(x,e^{i\theta } \psi ) = F(x,\psi )$ for all $\theta \in [0,2\pi ]$, a standing wave solution of (1.2) is a solution of the form $\psi (t,x) = e^{\frac{i\mu t}{\hbar }}u(x)$. It is clear that $\psi (t,x)$ solves (1.2) if and only if u(x) solves (1.1) with $a=mc/\hbar , V(x)=M(x)/c\hbar +\mu I_4/\hbar$ and $G(x,u)=F(x,u)/c\hbar$.

For notational convenience, denoting

$$\begin{aligned} \alpha =(\alpha _1,\alpha _2,\alpha _3)\ \ \text {and}\ \ \alpha \cdot \nabla =\sum ^3_{k=1}\alpha _k\partial _k, \end{aligned}$$

we rewrite the Eq. (1.1) as

There are many papers studying the existence and multiplicity of standing wave of the equations under different assumptions on the potentials V and G, see, [3, 8–11, 14–18, 21, 23] and their references. Recall that, mathematically, the conditions that the potential functions depend periodically on x is used for describing a class of self-interaction of quantum electrodynamics in, e.g. [1, 2, 4, 5, 19, 20, 24, 26] for Schrödinger equations and [3] for Dirac equations. Note that if the potentials are periodic in x one may also study the existence and multiplicity of periodic solutions. Naturally, a periodic solution of $(D_V)$ may be referred as a standing periodic wave of (1.2). In recently paper [12], we have investigated periodic solutions of $(D_V)$ in both cases that the nonlinearity $G_u(x,u)$ is of superlinear and subcritical growth as $|u|\rightarrow \infty$. The case of concave and convex has been researched in the paper [13].

In the present paper, we are interested in the case that G(x, u) is asymptotically quadratic at 0 and $\infty$ and obtain the existence and multiplicity results of periodic solutions.

We make the following periodicity hypothesis on V(x) and G(x, u):

(V):

$V\in C(\mathbb {R}^3,\mathbb {R})$, and V(x) is 1-periodic in $x_k, k=1,2,3$.

$(G_0)$ :

$G\in C^1(\mathbb {R}^3\times \mathbb {C}^4,[0,\infty ))$, and G(x, u) is 1-periodic in $x_k, k=1,2,3$.

We are looking for periodic solutions of $(D_V)$: $u(x+z)=u(x)$ for any $z\in \mathbb {Z}^3$.

Setting $Q=[0,1]\times [0,1]\times [0,1]$, if u is a solution of $(D_V)$, its energy will be denoted by

$$\begin{aligned} \Phi (u)=\int _Q\left[ \frac{1}{2}(-i\alpha \cdot \nabla u + a\beta u + V(x) u)\cdot u-G(x,u)\right] dx, \end{aligned}$$

(1.3)

where (here and in the following) by $v\cdot w$ we denote the scalar product in $\mathbb {C}^4$ of v and w.

In order to state our results, let $A_0=-i\alpha \cdot \nabla +a\beta$ and $A_V=A_0+V$ denote the self-adjoint operators acting in $L^2(Q,\mathbb {C}^4)$. Let $\{\lambda _j\}_{j\in \mathbb {Z}}$ denote the sequence of all eigenvalues of $A_V$ counted by multiplicity:

$$\begin{aligned} \ldots \le \lambda _{-2}\le \lambda _{-1}<\lambda _0=0<\lambda _1\le \lambda _2\le \ldots , \end{aligned}$$

and let $\{e_j\}_{j\in \mathbb {Z}}$ be the associated sequence of eigenvectors of $A_V$:

$$\begin{aligned} A_Ve_j=\lambda _je_j,\quad |e_j|_{L^2}=1,\quad j=\pm 1,\pm 2,\dots . \end{aligned}$$

(1.4)

Remark 1.1

We can find out all eigenvalues and the associated eigenfunctions of $A_0$. Let

$$\begin{aligned} z=(k_1,k_2,k_3)\in \mathbb {N}^3,\quad x=(x_1,x_2,x_3)\in Q,\quad z x=k_1x_1+k_2x_2+k_3x_3, \end{aligned}$$

and $| z|=\sqrt{k^2_1+k^2_2+k^2_3}$. Note that

$$\begin{aligned} A_0= \left( \begin{array}{ll} aI &{}\quad -i(\sigma _1\partial _1+\sigma _2\partial _2+\sigma _3\partial _3)\\ -i(\sigma _1\partial _1+\sigma _2\partial _2+\sigma _3\partial _3) &{}\quad -aI \end{array}\right) \end{aligned}$$

and

$$\begin{aligned}-i(\sigma _1\partial _1{\text{ e }}^{2\pi zxi} +\sigma _2\partial _2{\text{ e }}^{2\pi zxi}+\sigma _3\partial _3{\text{ e }}^{2\pi zxi}) =2\pi {\text{ e }}^{2\pi zxi}W,\end{aligned}$$

where $W=\left( \begin{array}{ll} k_3 &{} \quad k_1-ik_2\\ k_1+ik_2 &{} \quad -k_3 \end{array}\right) .$ Setting $D=\left( \begin{array}{ll} aI &{} 2\pi W\\ 2\pi W &{} -aI \end{array}\right) ,$ one can verify that if $\lambda \ne 0$ is a eigenvalue of the matrix D and $\mathbf v$ is a eigenvector corresponding to $\lambda$, then $\lambda$ must be a eigenvalue of $A_0$ and ${\text{ e }}^{2\pi zxi}\mathbf v$ is a eigenfunction corresponding to $\lambda$. By $|\lambda I-D|=0$ we obtain

$$\begin{aligned} \begin{aligned}&\quad \ \left| \begin{array}{cc} (\lambda -a)I &{} -2\pi W \\ -2\pi W &{} (\lambda +a)I \end{array} \right| \\&\quad = \left| \begin{array}{llll} (\lambda -a) &{}\quad 0 &{} \quad -2\pi k_3 &{}\quad -2\pi (k_1-ik_2)\\ 0 &{}\quad (\lambda -a) &{}\quad -2\pi (k_1+ik_2) &{}\quad 2\pi k_3\\ -2\pi k_3 &{}\quad - 2\pi (k_1-ik_2) &{}\quad (\lambda +a) &{}\quad 0\\ -2\pi (k_1+ik_2) &{}\quad 2\pi k_3 &{} \quad 0 &{}\quad (\lambda +a) \end{array} \right| \\&\quad =(\lambda ^2-a^2-4\pi ^2|z|^2)^2=0, \end{aligned}\end{aligned}$$

and therefore

$$\begin{aligned} \lambda =\pm \sqrt{a^2+4\pi ^2|z|^2}. \end{aligned}$$

For $\mathbf{v}=(c_1,c_2,c_3,c_4)$, in virtue of $D\mathbf{v}^T=\lambda \mathbf{v}^T$ we get

$$\begin{aligned} \left\{ \begin{array}{l} 2\pi k_3c_3+2\pi (k_1-ik_2)c_4=(\lambda -a)c_1, \\ 2\pi (k_1+ik_2)c_3-2\pi k_3c_4=(\lambda -a)c_2, \end{array}\right. \end{aligned}$$

and so

$$\begin{aligned}\left\{ \begin{array}{l} \mathbf{v^{(1)}_\lambda }=(2\pi |z|^2,\ 0,\ (\lambda -a)k_3,\ (\lambda -a)(k_1+ik_2)),\\ \mathbf{v^{(2)}_\lambda }=(0,\ 2\pi |z|^2,\ (\lambda -a)(k_1-ik_2),\ (a-\lambda ) k_3). \end{array}\right. \end{aligned}$$

Put

$$\begin{aligned} {\bar{\mathbf{e}}_1}=(1,0,0,0), {\bar{\mathbf{e}}_2}=(0,1,0,0), {\bar{\mathbf{e}}_3}=(0,0,1,0), {\bar{\mathbf{e}}_4}=(0,0,0,1), \end{aligned}$$

then

$$\begin{aligned} \left. \begin{array}{l} \varphi ^{(1)}_\lambda (x):= {\text{ e }}^{2\pi i zx}[ 2\pi |z|^2{\bar{\mathbf{e}}_1}+ (\lambda -a) k_3{\bar{\mathbf{e}}_3}+(\lambda -a)(k_1+ik_2){\bar{\mathbf{e}}_4}],\\ \varphi ^{(2)}_\lambda (x):= {\text{ e }}^{2\pi i zx}[2\pi |z|^2 {\bar{\mathbf{e}}_2} +(\lambda -a)(k_1-ik_2) {\bar{\mathbf{e}}_3}-(\lambda -a) k_3 {\bar{\mathbf{e}}_4}] \end{array} \right. \end{aligned}$$

(1.5)

satisfy $A_0\varphi ^{(j)}_\lambda =\lambda \varphi ^{(j)}_\lambda , j=1,2$.

We will use the following hypotheses:

$(G_1)$ :

there is $b_0\ge 0$ such that and $G_u(x,u)-b_0u=o(|u|)$ as $u\rightarrow 0$ uniformly in $x\in Q$;

$(G_2)$ :

there is $b_\infty >0$ satisfying $G_u(x,u)-b_\infty u=o(|u|)$ as $|u|\rightarrow \infty$ uniformly in $x\in Q$;

$(G_3)$ :

either (i) $b_\infty \notin \sigma (A_V)$ or (ii) $G_u(x,u)-b_\infty u$ is bounded and $G(x,u)-\frac{1}{2}b_\infty |u|^2\rightarrow \infty$ as $|u|\rightarrow \infty$ uniformly in $x\in Q$.

Set

$$\begin{aligned} G^0(x,u):=G(x,u)-\frac{1}{2}b_0|u|^2,\ G^\infty (x,u):=G(x,u)-\frac{1}{2}b_\infty |u|^2,\end{aligned}$$

and define

$$\begin{aligned} b_0^+:=\min [\sigma (A_V)\cap (b_0,\infty )],\ \ b_\infty ^-:=\max [\sigma (A_V)\cap (b_0,b_\infty )].\end{aligned}$$

The first result reads as follows.

Theorem 1.2

Let $(V), (G_0)$ and $(G_1)-(G_3)$ be satisfied and $b_\infty >b^+_0$. Then

• (a) if $G^0(x,u)\ge 0$, then $(D_V)$ has at least one nontrivial periodic solution in $H^1(Q,\mathbb {C}^4)$;

• (b) if G is even in u, then $(D_V)$ has at least $d(b_0,b_\infty )$ pairs of periodic solutions, where $d(b_0,b_\infty )$ denotes the dimensionality of the eigenspace associated to $\sigma (A_V)\cap (b_0,b_\infty )$.

If $b_0\equiv 0$, then $b^+_0=\lambda _1$, we have

Corollary 1.3

Assume that $(V), (G_0)$ and $(G_1)-(G_3)$ hold with $b_0=0$. If $b_\infty >\lambda _1$, then $(D_V)$ has at least one nontrivial periodic solution in $H^1(Q,\mathbb {C}^4)$. If G is in addition even in u, then $(D_V)$ has at least $d(0, b_\infty )$ pairs of periodic solutions.

If $V(x)\equiv 0$, that is, $A_V=A_0$, then the equation $(D_V)$ becomes the following

We write $\{\mu _j\}$ the sequence of all eigenvalues of $A_0$ according to the size of order, not by multiplicity:

$$\begin{aligned} ...<\mu _{-2}<\mu _{-1}<\mu _0=0<\mu _1<\mu _2<.... \end{aligned}$$

Let $\sharp _{\mu _k}$ define the multiplicity of $\mu _k$, and $\lambda ^{(\mu _k)}_{j}$ the eigenvalues such that $\lambda ^{(\mu _k)}_{j}=\mu _k,\ j=1,\dots ,\ \sharp _{\mu _k}$.

Let N[j] denote the number of $z\in {\mathbb {N}}^3$ corresponding to $|z|^2=j$. For $0\le |z|^2\le 10$, we have:

$$\begin{aligned} N[0]=N[3]=1;\ N[j]=3,\ j=1,2,4,6,8; \end{aligned}$$

$$\begin{aligned} N[k]=6,\ k=5,9,10;\ \ N[7]=0, \end{aligned}$$

then by Remark 1.1,

$$\begin{aligned} \mu _j=\sqrt{a^2+4(j-1)\pi ^2},\ 1\le j\le 7;\ \mu _k=\sqrt{a^2+4k\pi ^2},\ k=8,9,10,\end{aligned}$$

and

$$\begin{aligned}\sharp _{\mu _1}=\sharp _{\mu _4}=1;\ \sharp _{\mu _j}=3,j=2,3,5,7,8;\ \sharp _{\mu _k}=6,k=6,9,10.\end{aligned}$$

Accordingly, we see

$$\begin{aligned}\begin{aligned}&\lambda ^{(\mu _1)}_1=\mu _1=a,\ \lambda ^{(\mu _2)}_1=\lambda ^{(\mu _2)}_2=\lambda ^{(\mu _2)}_3 =\sqrt{a^2+4\pi ^2},\\&\lambda ^{(\mu _3)}_1=\lambda ^{(\mu _3)}_2=\lambda ^{(\mu _3)}_3 =\sqrt{a^2+8\pi ^2},\ \lambda ^{(\mu _4)}_8=\sqrt{a^2+12\pi ^2},\\&\lambda ^{(\mu _5)}_1=\lambda ^{(\mu _5)}_2=\lambda ^{(\mu _5)}_3=\sqrt{a^2+16\pi ^2},\ \lambda ^{(\mu _6)}_1=\cdots =\lambda ^{(\mu _6)}_6=\sqrt{a^2+20\pi ^2},\\&\lambda ^{(\mu _7)}_1=\lambda ^{(\mu _7)}_2=\lambda ^{(\mu _7)}_3=\sqrt{a^2+24\pi ^2},\ \lambda ^{(\mu _8)}_1=\lambda ^{(\mu _8)}_2=\lambda ^{(\mu _8)}_3=\sqrt{a^2+32\pi ^2},\\&\lambda ^{(\mu _9)}_1=\cdots =\lambda ^{(\mu _9)}_6=\sqrt{a^2+36\pi ^2},\ \lambda ^{(\mu _{10})}_1=\cdots =\lambda ^{(\mu _{10})}_6=\sqrt{a^2+40\pi ^2}. \end{aligned}\end{aligned}$$

By (1.5), we can list the first 10 eigenvalues $\lambda _j$ and eigenfunctions $e_j$ corresponding to $\lambda _j$ as follows:

$$\begin{aligned}\begin{aligned} \lambda _1&=\lambda _2=\mu _1=a \ \ \text {with}\ \ z=(0,\ 0,\ 0),\\ e_1&=(1,\ 0,\ 0,\ 0),\ \ e_2=(0,\ 1,\ 0,\ 0);\\ \lambda _3&=\lambda _4=\mu _2=\sqrt{a^2+4\pi ^2} \ \ \text {with}\ \ z=(1,0,0),\\ e_3&=\Delta _1{\text{ e }}^{2\pi x_1i}(2\pi ,\ 0,\ 0,\ \mu _2-a),\\ e_4&=\Delta _1{\text{ e }}^{2\pi x_1i}(0,\ 2\pi ,\ \mu _2-a,\ 0);\\ \lambda _5&=\lambda _6=\mu _2 \ \ \text {with}\ \ z=(0,1,0),\\ e_5&=\Delta _1{\text{ e }}^{2\pi x_2i}(2\pi ,\ 0,\ 0,\ (\mu _2-a)i),\\ e_6&=\Delta _1{\text{ e }}^{2\pi x_2i}(0,\ 2\pi ,\ (a-\mu _2)i,\ 0);\\ \lambda _7&=\lambda _8=\mu _2 \ \ \text {with}\ \ z=(0,\ 0,\ 1),\\ e_7&=\Delta _1{\text{ e }}^{2\pi x_3i}(2\pi ,\ 0,\ \mu _2-a,0),\\ e_8&=\Delta _1{\text{ e }}^{2\pi x_3i}(0,\ 2\pi ,\ 0,\ a-\mu _2);\\ \lambda _9&=\lambda _{10}=\mu _3=\sqrt{a^2+8\pi ^2} \ \ \text {with}\ \ z=(1,\ 1,\ 0),\\ e_9&=\Delta _2{\text{ e }}^{2\pi (x_1+x_2)i}(4\pi ,\ 0,\ (\mu _3-a)(1+i),\ 0),\\ e_{10}&=\Delta _2{\text{ e }}^{2\pi (x_1+x_2)i}(0,\ 4\pi ,\ 0,\ (\mu _3-a)(1-i)), \end{aligned} \end{aligned}$$

where $\Delta _1=\frac{1}{\sqrt{4\pi ^2+(\mu _2-a)^2}},\ \Delta _2=\frac{1}{\sqrt{16\pi ^2+2(\mu _3 -a)^2} }.$

Now we have a special consequence corresponding to the equation $(D_0)$.

Corollary 1.4

Let $(G_0)$ and $(G_1)-(G_3)$ be satisfied with $b_0=0$. Then $(D_0)$ has at least one nontrivial periodic solution in $H^1(Q,\mathbb {C}^4)$, provided $b_\infty >a$. If moreover G is in even in u and $b_\infty ^-=\mu _k$ for some positive integer k, then $(D_0)$ has at least $l:=2(\sharp _{\mu _1}+\dots +\sharp _{\mu _k})$ pairs of periodic solutions.

A more general result can be obtained if $(G_1)$ is replaced by

$(G'_1)$ :

there is $b_0\in C(Q,[0,\infty ))$ such that $b_0(x)$ is 1-period with $b_0(x)\ge 0$ and $G_u(x,u)-b_0(x)u=o(|u|)$ as $|u|\rightarrow \infty$ uniformly in $x\in Q$,

$(G_2)$ is replaced by

$(G'_2)$ :

there is $b_\infty \in C(Q,(0,\infty ))$ such that $b_\infty (x)$ is 1-period and $G_u(x,u)-b_\infty (x)u=o(|u|)$ as $|u|\rightarrow \infty$ uniformly in $x\in Q$,

and $(G_3)$ is replaced by

$(G'_3)$ :

either (i) $0\notin \sigma (A_V-b_\infty )$ or (ii) ${\hat{G}}(x,u):= \frac{1}{2}\hat{G}_u(x,u)u- G(x,u)\ge 0$ and ${\hat{G}}(x,u)\rightarrow \infty$ as $|u|\rightarrow \infty$ uniformly in $x\in Q$.

Theorem 1.5

Suppose that $(V), (G_0), (G'_1)-(G'_3)$ are satisfied and $q_\infty >q_0^+$, where $q_\infty :=\min \limits _{x\in Q}b_\infty (x),\ q_0^+:=\min [\sigma (A_V)\cap (q_0,\infty )]$ and $q_0:= \max \limits _{x\in Q}b_0(x)$. Then

• (a) if $G(x,u)-\frac{1}{2} q_0|u|^2\ge 0$, then $(D_V)$ has at least one nontrivial periodic solution in $H^1(Q,\mathbb {C}^4)$;

• (b) if G is even in u, then $(D_V)$ has at least $d(q_0,q_\infty )$ pairs of periodic solutions.

This paper is organized as follows. In Sect. 2, we state the variational setting and establish a deformation theorem and abstract critical point theorems under the Cerami condition ($(C)_c$-condition). The proofs of the main results are given in Sect. 3.

2 Variational setting and abstract critical point theorems

To prove our main results, some preliminaries are first in order.

In what follows by $|\cdot |_q$ we denote the usual $L^q$-norm, and $(\cdot ,\cdot )_2$ the usual $L^2$-inner product. Let

$$\begin{aligned} L_T^q(Q):=\{u\in L_{loc}^q(\mathbb {R}^3,\mathbb {C}^4): u(x+{\hat{e}}_i)=u(x)\ a.e.\ ,\quad i=1,2,3\}, \end{aligned}$$

where ${\hat{e}}_1=(1,0,0), {\hat{e}}_2=(0,1,0), \hat{e}_3=(0,0,1)$. Let $A_0=-i\alpha \cdot \nabla +a\beta ,\, A_V=A_0+V$ denote the self-adjoint operators on $L^2(Q,{\mathbb {C}}^4)$ with domain

$$\begin{aligned}&\mathcal {D}(A_V)=\mathcal {D}(A_0)=H_T^1(Q)\\&\quad :=\{u\in H^1_{loc}(\mathbb {R}^3,\mathbb {C}^4):u(x+{\hat{e}}_i)=u(x)\ a.e.\ ,i=1,2,3\}. \end{aligned}$$

Set $E:=\mathcal {D}(|A_V|^{\frac{1}{2}})$ which is a Hilbert space with the inner product and norm, for $u=\sum _{j\in \mathbb {Z}} a_je_j$ and $v=\sum _{j\in \mathbb {Z}} b_je_j\in E$,

$$\begin{aligned} (u, v)=\sum _{j\not =0}|\lambda _j|a_j\cdot b_j+(u^0,v^0)_2 \quad \text {and}\quad \Vert u\Vert ^2=\sum _{j\not =0}|\lambda _j||a_j|^2+|u^0|^2_2, \end{aligned}$$

(2.1)

here $\{e_j\}_{j\in \mathbb {Z}}$ are the eigenvectors of $A_V$.

Then we have an orthogonal decomposition $E=E^-\oplus E^0\oplus E^+$ with $E^-:=\hbox {span}\{e_j: j<0\}, E^+:=\hbox {span}\{e_j: j>0\}$, and $E^0:=\ker (A_V)$. Note that if $0\not \in \sigma (A_V)$ then $E^0=\{0\}$.

The functional $\Phi$ defined by (1.3) can be rewritten by

$$\begin{aligned} \Phi (u)=\frac{1}{2}\big (\Vert u^+\Vert ^2-\Vert u^-\Vert ^2\big )-\int _{Q}G(x,u) \end{aligned}$$

for $u=u^-+u^0+u^+\in E$. Then $\Phi \in C^1(E,\mathbb {R})$ and critical points of $\Phi$ are solutions of $(D_V)$.

First we have the following (see [8, 11])

Lemma 2.1

$E=H^{1/2}(Q,\mathbb {C}^4)$ with equivalent norms, hence E embeds compactly into $L_T^s (Q)$ for all $s\in [1,3)$. In particular there is a constant $a_s>0$ such that

$$\begin{aligned} |u|_s\le a_s\Vert u\Vert \quad \text { for all }\;\; u\in E. \end{aligned}$$

(2.2)

We also use the following result, the proof is similar to that of Proposition B.10 in [22].

Lemma 2.2

Assume that

• (i) $G\in C^1(Q\times \mathbb {C}^4,\mathbb {R})$, and

• (ii) there are $k_1, k_2>0$ such that

  $$\begin{aligned} |G_u(x,u)|\le k_1+k_2|u|^s, \quad \forall (x,u)\in Q\times \mathbb {C}^4, \end{aligned}$$

  where $0\le s<3$.

Then

$$\begin{aligned} \psi (u):=\int _QG(x,u) \end{aligned}$$

(2.3)

is weakly continuous and $\psi '\in C(E, \mathbb {R})$ is compact.

Recall that a sequence $\{u_j\}$ in E is said to be a $(C)_c$-sequence of $\Phi$, if $\Phi (u_j)\rightarrow c$ and $(1+\Vert u_j\Vert )\Phi '(u_j)\rightarrow 0$ as $j\rightarrow \infty$. We say that $\Phi$ satisfies the $(C)_c$-condition if any $(C)_c$-sequence possesses a convergent subsequence ([6]).

Let X be a Banach space, and

$$\begin{aligned} \Phi ^b_a:=\Phi _a\cap \Phi ^b,\ \Phi _a:=\{u\in X: \Phi (u)\ge a \},\, \Phi ^b:=\{u\in X: \Phi (u)\le b \}. \end{aligned}$$

We first establish a deformation theorem which plays an important role in the multiplicity for $(D_V)$.

Theorem 2.3

Let $\Phi \in C^1(X,\mathbb {R})$ and satisfy the $(C)_c$-condition, $K_c=\{u\in X: \Phi (u)=c$ and $\Phi '(u)=0\}$. If ${\bar{\varepsilon }}>0$ and $\mathcal {O}$ is any neighborhood of $K_c$, then there exists an $\varepsilon \in (0,{\bar{\varepsilon }})$ and a deformation $\eta \in C([0,1]\times X,X)$ such that

$1^\circ \, \eta (0,u)=u$ for all $u\in X$.

$2^\circ \,\eta (t,u)=u$ for all $t\in [0,1]$ if $u\notin \Phi ^{c+\varepsilon }_{c-\varepsilon }$.

$3^\circ \,\eta (t,\cdot ): X\rightarrow X$ is homeomorphism for $t\in [0,1]$.

$4^\circ \,\Phi (\eta (\cdot ,u))$ is nonincreasing on [0, 1] for $u\in E$.

$5^\circ \,\eta (1,\Phi ^{c+\varepsilon }\setminus \mathcal {O})\subset \Phi ^{c-\varepsilon }$.

$6^\circ$ If $K_c=\emptyset , \eta (1,\Phi ^{c+\varepsilon })\subset \Phi ^{c-\varepsilon }$.

$7^\circ$ If $\Phi (u)$ is even in $u, \eta (t,u)$ is odd in u.

Proof

By the $(C)_c$-condition, $K_c$ is compact. Set $U_\delta =\{u\in X:\ d(u,K_c)<\delta \}$. Choosing $\delta$ suitably small ($\delta <1$), $U_\delta \subset \mathcal {O}$. Therefore it suffices to prove $5^\circ$ with $\mathcal {O}$ replaced by $U_\delta$. Note that $U_\delta =\emptyset$ when $K_c=\emptyset$, and so we get $6^\circ$ instead.

Let $M>0$ such that $\Vert u\Vert \le M$ for all $u\in U_\delta$.

One can easy to verify that there are ${\hat{\varepsilon }}>0$ and $\alpha >0$ such that

$$\begin{aligned} (1+\Vert u\Vert )\Vert \Phi '(u)\Vert \ge \alpha ,\ \ \text {for all}\ \ u\in \Phi ^{c+{\hat{\varepsilon }}}_{c-{\hat{\varepsilon }}}\setminus U_{\delta /2}. \end{aligned}$$

(2.4)

We may assume that

$$\begin{aligned} 0<{\hat{\varepsilon }}< \frac{3\delta }{8(1+M)}\min \left\{ {\bar{\varepsilon }},\alpha ^2,\frac{1}{4}\right\} . \end{aligned}$$

(2.5)

Let $\tilde{X}:=\{u\in X\, | \, \Phi '(u)\not =0\}$ and $V: \tilde{X}\rightarrow X$ be a pseudo gradient such that V is odd if $\Phi$ is even (see [22]). Choosing any $\varepsilon \in (0,{\hat{\varepsilon }})$, define

$$\begin{aligned} h(s)=\left\{ \begin{array}{ll} 1,\, &{}\quad {\text {if}}\, 0\le s\le 1,\\ \frac{1}{s},\, &{}\quad {\text {if}}\, s>1, \end{array}\right. \end{aligned}$$

$$\begin{aligned} f(u)=\frac{d(u,X\setminus \Phi ^{c+{\hat{\varepsilon }}}_{c-{\hat{\varepsilon }}})}{d(u,X\setminus \Phi ^{c+{\hat{\varepsilon }}}_{c-{\hat{\varepsilon }}})+d(u,\Phi ^{c+\varepsilon }_{c-\varepsilon })},\ g(u)=\frac{d(u, U_{\delta /8})}{d(u, U_{\delta /8})+d(u,X\setminus U_{\delta /4})}. \end{aligned}$$

Then

$$\begin{aligned}f |_{\Phi ^{c+\varepsilon }_{c-\varepsilon }}=g |_{X\setminus U_{\delta /4}}=1,\, \, f |_{X\setminus \Phi ^{c+{\hat{\varepsilon }}}_{c-{\hat{\varepsilon }}}}=g |_{U_{\delta /8}}=0.\end{aligned}$$

Let

$$\begin{aligned} W(u)=\left\{ \begin{array}{ll} -f(u)g(u)h((1+\Vert u\Vert )\Vert V(u)\Vert )(1+\Vert u\Vert )^2V(u),\, &{}\quad u\in {\tilde{X}},\\ 0,\, &{}\quad \text {otherwise}. \end{array}\right. \end{aligned}$$

It is easy to verify that

$$\begin{aligned} \Vert W(u)\Vert \le 1+\Vert u\Vert \qquad \text {for all }u. \end{aligned}$$

(2.6)

Then by construction, W is locally Lipschitz continuous on X and W is odd if $\Phi$ is even.

Now we consider the Cauchy problem:

$$\begin{aligned} \frac{{\text {d}}\eta }{{\text {d}}t}=W(\eta ), \, \, \eta (0,u)=u. \end{aligned}$$

(2.7)

By virtue of the locally Lipschitz continuity of W and (2.6), the basic existence uniqueness theorem for ordinary differentia equations implies that for each $u\in X$, (2.7) has a unique solution $\eta (t, u)$ defined for $t\in [0, \infty )$, and $\eta \in C([0,1]\times X, X)$. (2.7) implies that $1^\circ$ holds. Since $f(u)=0$ on $X\setminus \Phi ^{c+{\hat{\varepsilon }}}_{c-{\hat{\varepsilon }}}$, so $2^\circ$ is true. The semigroup property for solutions of (2.7) gives $3^\circ$. The oddness of W when $\Phi$ is even yields $7^\circ$.

If $W(u)\not =0,\ u\in {\tilde{X}}$ so V(u) is defined as is $V(\eta (t,u))$ and

$$\begin{aligned} \begin{aligned} \frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}&= \, \ (\Phi '(\eta (t,u)), W(\eta (t,u))) \\&= \, -f(\eta )g(\eta )h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )(1+\Vert \eta \Vert )^2(\Phi '(\eta ),V(\eta ))\\&\le \, -f(\eta )g(\eta )h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )(1+\Vert \eta \Vert )^2\Vert \Phi '(\eta )\Vert ^2\le 0. \end{aligned} \end{aligned}$$

(2.8)

It follows that $4^\circ$ holds.

Finally, we verify $\eta (1,\Phi ^{c+\varepsilon }\setminus U_\delta )\subset \Phi ^{c-\varepsilon }$. Let $u\in \Phi ^{c+\varepsilon }\setminus U_\delta$, then $\Phi (\eta (t,u))\le c+\varepsilon$ by $4^\circ$ and $1^\circ$. We need only prove that there exists $t_0\in [0,1]$ such that $\Phi (\eta (t_0,u))\le c-\varepsilon$, then $4^\circ$ gives $\Phi (\eta (1,u))\le c-\varepsilon$.

If otherwise, then $\Phi (\eta (t,u))>c-\varepsilon$ for all $t\in [0,1]$, and thus $\eta (t,u)\in \Phi ^{c+\varepsilon }_{c-\varepsilon }$, which implies

$$\begin{aligned} \Phi (\eta (0,u))-\Phi (\eta (t,u))\le 2\varepsilon <2{\hat{\varepsilon }},\ \forall t\in [0,1]. \end{aligned}$$

(2.9)

If $\eta (t,u)\in X\setminus U_{\delta /2}$ for all $t\in [0,1]$, we see $\eta (t,u)\in \Phi ^{c+\varepsilon }_{c-\varepsilon }\setminus U_{\delta /2}$. This shows $f(\eta (t,u))=g(\eta (t,u))=1$ and by (2.4),

$$\begin{aligned} (1+\Vert \eta (t,u)\Vert )\Vert \Phi '(\eta (t,u))\Vert \ge \alpha ,\ \forall t\in [0,1]. \end{aligned}$$

(2.10)

This yields

$$\begin{aligned} \begin{aligned} \frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}&= -h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )(1+\Vert \eta \Vert )^2(\Phi '(\eta ),V(\eta ))\\&\le -h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )(1+\Vert \eta \Vert )^2\Vert \Phi '(\eta )\Vert ^2,\ \ \forall t\in [0,1]. \end{aligned} \end{aligned}$$

(2.11)

If $(1+\Vert \eta \Vert )\Vert V(\eta )\Vert \le 1$, then $h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )=1$. It follows from (2.10) and (2.11) that

$$\begin{aligned} \frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}\le -\alpha ^2. \end{aligned}$$

(2.12)

If $(1+\Vert \eta \Vert )\Vert V(\eta )\Vert >1$, then

$$\begin{aligned} h((1+\Vert \eta \Vert )\Vert V(\eta )\Vert )=[(1+\Vert \eta \Vert )\Vert V(\eta )\Vert ]^{-1}, \end{aligned}$$

so (2.11) and the property of $V(\cdot )$ imply

$$\begin{aligned} \frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}\le -(1+\Vert \eta \Vert )\Vert V(\eta )\Vert \left[ \frac{\Vert \Phi '(\eta )\Vert }{\Vert V(\eta )\Vert }\right] ^2 \le -\frac{1}{4}. \end{aligned}$$

(2.13)

Consequently, by (2.12) and (2.13) we have

$$\begin{aligned} \frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}\le -\min \left\{ \alpha ^2,\frac{1}{4}\right\} \ \ \text {for all}\ t\in [0,1]. \end{aligned}$$

(2.14)

Integrating (2.14) and combing the result with (2.9) gives

$$\begin{aligned} \begin{aligned}&2{\hat{\varepsilon }}\ge \Phi (\eta (0,u))-\Phi (\eta (1,u))\\&\quad = \int _0^1-\frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}\ge \min \left\{ \alpha ^2,\frac{1}{4}\right\} , \end{aligned} \end{aligned}$$

(2.15)

this is contrary to (2.5). Consequently, we infer that there is ${\bar{t}}\in [0,1]$ such that $\eta ({\bar{t}},u)\in U_{\delta /2}$. Obviously, ${\bar{t}}>0$ since $\eta (0,u)=u \not \in U_\delta$. The continuity of $\eta (t,u)$ guarantees that there are $s_1,s_2\in [0,1]$ with $s_1\not =s_2$ such that $\eta (s_1,u)\in \partial U_{\delta /4},\ \eta (s_1,u)\in \partial U_\delta$ and $\eta (t,u)\in U_\delta \setminus \overline{U}_{\delta /4}$ for all $t\in (s_1,s_2)$ or $t\in (s_2,s_1)$, where $\overline{B}$ denotes the closure of B. This yields

$$\begin{aligned} \Vert \eta (s_1,u)-\eta (s_2,u)\Vert \ge 3\delta /4.\end{aligned}$$

(2.16)

By (2.6) we see $\Vert W(u)\Vert \le 1+ M$ for all $u\in U_\delta$, and so

$$\begin{aligned} \Vert \eta (s_2,u)-\eta (s_1,u)\Vert \le (1+M)|s_2-s_1| \end{aligned}$$

which together with (2.16) shows

$$\begin{aligned} |s_2-s_1|\ge \frac{3\delta }{4(1+M)}. \end{aligned}$$

We may assume that $s_1<s_2$.

On the other hand, similarly to (2.15) we get that

$$\begin{aligned}\begin{aligned} 2{\hat{\varepsilon }}&\ge \Phi (\eta (s_1,u))-\Phi (\eta (s_2,u))\\&= \int _{s_1}^{s_2}-\frac{{\text {d}}\Phi (\eta (t,u))}{{\text {d}}t}\\&\ge \min \left\{ \alpha ^2,\frac{1}{4}\right\} (s_2-s_1)\\&\ge \frac{3\delta }{4(1+M)}\min \left\{ \alpha ^2,\frac{1}{4}\right\} . \end{aligned} \end{aligned}$$

This, however, leads to a contradiction. The proof is complete. $\square$

Remark 2.4

In paper [12] (or [13]), we established a deformation theorem under the $(C)_c$-condition. However, it is difficult to use for the multiplicity. Therefore, Theorem 2.3 improves the corresponding result in [12].

In order to study the functional $\Phi$, we need certain abstract critical point theorems. In the following, we suppose that E is a real Hilbert space with $E=X\oplus Y.$

Theorem 2.5

Let $e\in X\setminus \{0\}$ and $\Omega =\{u=s e+v:\Vert u\Vert < R, s>0,v\in Y\}$. Suppose that

$(\Phi _1)$ :

$\Phi \in C^1(E,\mathbb {R})$, satisfies the $(C)_c$-condition for any $c\in \mathbb {R}$;

$(\Phi _2)$ :

there is a $r\in (0,R)$ such that $\rho :=\inf \Phi (X\cap \partial B_r)>\omega :=\sup \Phi (\partial \Omega )$, where $\partial \Omega$ refers to the boundary of $\Omega$ relative to $\hbox {span}\{e\}\oplus Y$, and $B_r=\{u\in E:\ \Vert u\Vert <r\}$.

Then $\Phi$ has a critical value $c\ge \rho$, with

$$\begin{aligned} c=\inf _{h\in \Gamma }\sup _{u\in \Omega }\,\Phi (h(u)), \end{aligned}$$

here

$$\begin{aligned} \begin{aligned} \Gamma =\{h\in C(E, E): h|_{\partial \Omega }=\text {id},\ \Phi (h(u))\le \Phi (u)\ \text {for}\ u\in \overline{\Omega }\}. \end{aligned} \end{aligned}$$

(2.17)

Proof

Put $S=X\cap \partial B_r$. We first show that for any $h\in \Gamma , h(\Omega )\cap S\ne \emptyset .$ We may assume $\Vert e\Vert =1$. Chose ${\hat{e}}\in Y$ with $\Vert {\hat{e}}\Vert =1$, and write $F:=\hbox {span}\{e,{\hat{e}}\}, \Omega _F:=F\cap \Omega .$ Let $\overline{\Omega }_F, \partial \Omega _F$ denote the closure and bound of $\Omega$ in F, respectively, P the project of E onto Y. For $u\in \overline{\Omega }_F, t\in [0,1],$ define

$$\begin{aligned} H(t,u)=t[\Vert (\text {id}-P)h(u)\Vert e+Ph(u)]+(1-t)u. \end{aligned}$$

Then $H:[0,1]\times \overline{\Omega }_F\rightarrow E$ is continuous. Obviously H is a compact operator. Since $h|_{\partial \Omega }=\text {id},$ if $u\in \partial \Omega _F$,

$$\begin{aligned} H(t,u)=t[\Vert u-Pu\Vert e+Pu]+(1-t)u=u, \end{aligned}$$

i.e., $H(t,\cdot )|_{\partial \Omega _F}=\text {id}$ for $t\in [0,1]$. In particular $H(t,u)\ne re$ for $t\in [0,1], u\in \partial \Omega _F$. By the property of Brouwer degree, we have

$$\begin{aligned} \deg (H(1,\cdot ),\Omega _F, re)=\deg (H(0,\cdot ),\Omega _F, re)=\deg (\text {id},\Omega _F, re)=1 \end{aligned}$$

which implies that there exists $u\in \Omega _F$ such that $H(1,u)=re\in S$. We find $Ph(u)=0, \Vert h(u)\Vert =r$, i.e. $h(u)\in S$, and therefore $c\ge \rho$.

Next we prove there is a sequence $\{u_j\}$ in $\Omega$ such that

$$\begin{aligned} (1+\Vert u_j\Vert )\Vert \Phi '(u_j)\Vert \rightarrow 0\ \ \text {for}\ j\rightarrow \infty . \end{aligned}$$

(2.18)

Indeed otherwise there exist $\alpha _0>0$ and $\varepsilon _0>0$ such that

$$\begin{aligned} (1+\Vert u\Vert )\Vert \Phi '(u)\Vert \ge \alpha _0\ \ \ \text {for all}\ u\in \Omega \cap \Phi ^{c+\varepsilon _0}_{c-\varepsilon _0}. \end{aligned}$$

Set ${\bar{\varepsilon }}=\min \{\frac{1}{2}(\rho -\omega ),\varepsilon _0\}$. There is an $\varepsilon \in (0,{\bar{\varepsilon }})$ and $\eta \in C([0,1]\times E, E)$ given by Theorem 2.3 such that $1^\circ -4^\circ$ and $6^\circ$ are satisfied. Chose $h\in \Gamma$ such that $\sup \Phi (h(\Omega ))\le c+\varepsilon .$ Consequently

$$\begin{aligned} h(\Omega )\subset \Phi ^{c+\varepsilon }. \end{aligned}$$

(2.19)

Let $g(u):=\eta (1,h(u))$, then $g\in C(E,E)$. It follows from $3^\circ$ and $1^\circ$ that

$$\begin{aligned} \Phi (g(u))=\Phi (\eta (1,h(u)))\le \Phi (\eta (0,h(u)))=\Phi (h(u))\le \Phi (u) \end{aligned}$$

for all $u\in \overline{\Omega }$. For $u\in \partial \Omega , (\Phi _2)$ shows

$$\begin{aligned} \Phi (u)\le \omega <\rho -{\bar{\varepsilon }} \le c-{\bar{\varepsilon }}\le c-\varepsilon \end{aligned}$$

which, by $2^\circ$, implies $\eta (1,u)=u$, and so

$$\begin{aligned} g(u)=\eta (1,h(u))=\eta (1,u)=u. \end{aligned}$$

Thus $g\in \Gamma$. (2.19) and $6^\circ$ yield $g(\Omega )=\eta (1,h(\Omega ))\subset \Phi ^{c-\varepsilon }$ which leads to the contradiction $c\le \sup \Phi (g(\Omega ))\le c-\varepsilon$.

Now we find that there is a sequence $\{u_j\}$ in $\Omega$ satisfying (2.18). Since $\Phi$ satisfies $(\Phi _1)$ (the $(C)_c$-condition), there exists a convergent subsequence $\{u_{j_k}\}$ of $\{u_j\}$ such that $u_{j_k}\rightarrow {\bar{u}}$. The conclusion follows by $\Phi \in C^1(E,E)$. $\square$

Remark 2.6

In [22], Theorem 5.3], under the conditions that Y is finite dimensional and $\Phi$ satisfies the (PS)-condition, the same result was proved. Clearly, the conditions of Theorem 2.5 are weaker than that of Theorem 5.3.

Next, we consider a kind of pseudo-index (see [7]). Let $\Sigma$ denote the class of closed subsets of E symmetric with respect to the origin, and $\gamma : \Sigma \rightarrow \mathbb {N}\cup \{\infty \}$ the $\mathbb {Z}_2$ genus map (see [22]). Let $\Phi \in C(E,{\mathbb {R}}),\ J=(\sigma ,\infty )$,

$$\begin{aligned} {\mathcal {H}}=\{h\in C(E,E): \ h\ \text {is a homeomorphism and is odd}\}, \end{aligned}$$

$$\begin{aligned} {\mathcal {M}}_J=\{g\in {\mathcal {H}} : \ g|_{\Phi ^{-1}(\mathbb {R}\setminus J)}=\text {id} \ \text {and}\ \Phi (g(u))\le \Phi (u)\ \text {for}\ u\in E\}, \end{aligned}$$

and $\Lambda _*=\{h\in {\mathcal {M}}_J : \ h(B_1Y)\subset \Phi ^{-1}(J)\cup B_rY\}$.

Now we define the pseudo-index $(\Sigma , i^*)$ relative to ${\mathcal {M}}_J$ for the genus $\gamma$ as follows

$$\begin{aligned} i^*(A)=\inf \limits _{h\in \Lambda _*}\gamma (A\cap h(S_1Y)). \end{aligned}$$

One can verify the following

Lemma 2.7

Let $\Sigma ^*=\Sigma$, then $(\Sigma ^*, i^*)$ satisfies all properties for pseudo-index ([7]):

(P1):

$\Sigma ^*\subset \Sigma ,\ \overline{A\setminus B}\in \Sigma ^*$ and $\overline{g(A)}\in \Sigma ^*$ for all $A\in \Sigma ^*,\ B\in \Sigma$ and $g\in {\mathcal {M}}_J$;

(P2):

$A\subset B$ implies $i^*(A)\le i^*(B)$ for all $A, B\in \Sigma ^*$;

(P3):

$i^*(\overline{A\setminus B})\ge i^*(A)-\gamma (B)$ for all $A\in \Sigma ^*$ and $B\in \Sigma$;

(P4):

$i^*(\overline{g(A)})\ge i^*(A)$ for all $A\in \Sigma ^*$ and $g\in {\mathcal {M}}_J$.

Now, we give a abstract critical point theorem as follows.

Theorem 2.8

Assume that $\Phi$ is even and satisfies $(\Phi _1)$. If

$(\Phi _3)$ :

there exists $r>0$ with $\rho :=\inf \Phi (S_rY)>\Phi (0)=0$, where $S_r:=\partial B_r,\ AB:=A\cap B$;

$(\Phi _4)$ :

there exists a finite dimensional subspace $Y_0\subset Y$ and $R>r$ such that for $E_*:=X\oplus Y_0, M_*=\sup \Phi (E_*)<+\infty$ and $\sigma :=\sup \Phi (E_*\setminus B_R)<\rho$,

then $\Phi$ possesses at least m distinct pairs of critical points, where $m=\dim Y_0$.

Proof

Let

$$\begin{aligned} \Sigma _k=\{A\in \Sigma : i^*(A)\ge k\}, \quad k=1,2,\ldots , m.\end{aligned}$$

Define

$$\begin{aligned} c_k =\inf \limits _{A\in \Sigma _k }\sup \limits _{u\in A}\Phi (u), \quad k=1,2,\ldots ,m. \end{aligned}$$

(2.20)

We first show $\Sigma _k\not =\emptyset$. Set ${\tilde{A}}:=B_RE_*$. $(\Phi _4)$ implies $\Phi ^{-1}(J)\subset (E\setminus E_*)\cup B_R$, and hence

$$\begin{aligned} {\tilde{A}}\supset Y_0\cap (\Phi ^{-1}(J)\cup B_RY)\supset Y_0\cap h(B_1Y) \end{aligned}$$

for each $h\in \Lambda _*$, which yields

$$\begin{aligned} {\tilde{A}}\cap h(S_1Y)\supset Y_0\cap h(S_1Y)\supset \partial (Y_0\cap h(B_1Y)), \end{aligned}$$

and we get

$$\begin{aligned} \gamma ({\tilde{A}}\cap h(S_1Y))\ge \gamma (\partial (Y_0\cap h(B_1Y)))\ge m. \end{aligned}$$

Consequently, $\Sigma _k\not =\emptyset$, and $c_m\le M_*$ by $(\Phi _4)$. For any $A\in \Sigma _k$, by $h:=r\text {id}\in \Lambda _*$ one has

$$\begin{aligned} \gamma (A\cap S_rY)=\gamma (A\cap h(S_1Y)\ge i^*(A)\ge k \end{aligned}$$

which yields $c_k\ge \rho$ by $(\Phi _3)$. Noting that $\Sigma _1\supset \Sigma _2\supset \cdots \supset \Sigma _m$, we have

$$\begin{aligned} \sigma <\rho \le c_1\le c_2\le \cdots \le c_m\le M_*. \end{aligned}$$

It is obvious that $K_c:=\{u\in X: \Phi (u)=c$ and $\Phi '(u)=0\}\in \Sigma$, and $K_c$ is compact by the $(C)_c$-condition.

Finally, we claim:

$(P^*)$ :

If $1\le j,\ j+l\le m$, and $c_j=\cdots =c_{j+l}\equiv c$, then $\gamma (K_c)\ge l+1$.

If $\gamma (K_c)\le l$, then there is a $\delta >0$ such that $\gamma (U_\delta (K_c))=\gamma (K_c)\le l$. Invoking Theorem 2.3 with $\mathcal {O}=U_\delta (K_c)$ and ${\bar{\varepsilon }}=\frac{\rho -\sigma }{2}$, there are $\varepsilon \in (0,{\bar{\varepsilon }})$ and $\eta \in C([0,1]\times E, E)$ such that $\eta (1,\cdot )$ satisfies the properties $1^\circ -7^\circ$ and

$$\begin{aligned} \eta (1,\Phi ^{c+\varepsilon }\setminus \mathcal {O})\subset \Phi ^{c-\varepsilon }. \end{aligned}$$

(2.21)

Choose ${\hat{A}}\in \Sigma _{j+l}$ such that $\sup \Phi ({\hat{A}})\le c+\varepsilon ,$ and hence

$$\begin{aligned} {\hat{A}}\subset \Phi ^{c+\varepsilon }. \end{aligned}$$

(2.22)

By (P3) one has

$$\begin{aligned} i^*(\overline{{\hat{A}}\setminus \mathcal {O}})\ge i^*(\hat{A})-\gamma (\mathcal {O})\ge j+l-l=j. \end{aligned}$$

(2.23)

Using $3^\circ$ and $7^\circ$ we get $\eta (1,\cdot )\in {\mathcal {H}}$. $4^\circ$ gives $\Phi (\eta (1,u))\le \Phi (u)$ for all $u\in E$. Since $\sigma <c-\varepsilon$, we have $\Phi ^{-1}(\mathbb {R}\setminus J)\subset E\setminus \Phi ^{c+\varepsilon }_{c-\varepsilon }$, and $2^\circ$ implies $\eta (1,\cdot )|_{\Phi ^{-1}(\mathbb {R}\setminus J)}=\text {id}$. Therefore $\eta (1,\cdot )\in {\mathcal {M}}_J$. Set $A_*:=\eta (1,\overline{\hat{A}\setminus \mathcal {O}})\in \Sigma$. It follows from (P4) and (2.23) that

$$\begin{aligned} i^*(A_*)=i^*\left( \overline{\eta (1,\overline{\hat{A}\setminus \mathcal {O}})}\right) \ge i^*\left( \overline{{\hat{A}}\setminus \mathcal {O}}\right) \ge j, \end{aligned}$$

and thus $A_*\in \Sigma _j$. Combing with (2.21), (2.22) and (2.20) we see

$$\begin{aligned} c\le \sup \Phi (A_*)\le c-\varepsilon <c, \end{aligned}$$

a contradiction. Therefore, the conclusion $(P^*)$ is valid and the proof is complete. $\square$

3 The proof of the main results

Throughout this section, we suppose that (V) and $(G_0)$ are satisfied.

Observe that, $(G_2)$ implies that for any $\varepsilon >0$ there is $R_\varepsilon >0$ such that

$$\begin{aligned} |G_u(x,u)-b_\infty u|\le \varepsilon |u| \hbox { whenever } |u|\ge R_\varepsilon , \end{aligned}$$

(3.1)

hence

$$\begin{aligned} |G_u(x,u)u-b_\infty |u|^2|\le |G_u(x,u)-b_\infty u||u|\le \varepsilon |u|^2 \end{aligned}$$

or

$$\begin{aligned} (b_\infty -\varepsilon )|u|^2\le G_u(x,u)u\le (b_\infty +\varepsilon )|u|^2 \quad \text {for all }|u|\ge R_\varepsilon . \end{aligned}$$

Fixed $s_0\in (0,1)$, in virtue of $G(x,u)\ge 0$ we get

$$\begin{aligned} \begin{aligned} G(x,u)&=\,G(x,s_0u)+\int ^1_{s_0}G_u(x,su)\cdot uds\\&\ge \,\int ^1_{s_0}\frac{1}{s}G_u(x,su)suds\\&\ge \,\frac{1}{2}(b_\infty -\varepsilon )(1-s^2_0)|u|^2 \end{aligned} \end{aligned}$$

for all $|u|\ge \frac{1}{s_0}R_\varepsilon$, and so

$$\begin{aligned} G(x,u)\ge \frac{1}{2}(b_\infty -\varepsilon )(1-s^2_0)|u|^2-C_{s_0}\quad \text {for all }(x,u). \end{aligned}$$

(3.2)

First, we have the following lemma.

Lemma 3.1

Suppose that $(G_1)$ and $(G_2)$ hold and $\{u_j\}$ is a bounded $(C)_c$-sequence of $\Phi$. Then there exists a critical point u of $\Phi$ such that $\Phi (u)=c$ and after passing to a subsequence, $u_j\rightarrow u$ strongly in E.

Proof

By Lemma 2.1, without loss of generality, we may assume that

$$\begin{aligned} u_n\rightharpoonup u\ \text {in}\ E\ \text {and}\ u_u\rightarrow u\ \text {in}\ L_T^s(Q)\ \text {for}\ s\in [1,3). \end{aligned}$$

(3.3)

Plainly, u is a critical point of $\Phi$. $(G_1)$ and $(G_2)$ yield that

$$\begin{aligned} |G_u(x,u)|\le C_1|u| \ \text {for}\ \text {all} \ (x,u) \end{aligned}$$

(3.4)

which shows that $\psi '$ is continuous and compact by Lemma 2.2, where $\psi$ is defined by (2.3). It follows from the representation of $\Phi '$, together with (3.3), the facts $\Phi '(u)=0$ and $\Phi '(u_n)\rightarrow 0$, and the compactness of $\psi '$, that

$$\begin{aligned} \ \ \Vert u_n^+-u^+\Vert ^2= & {} (\Phi '(u_n)-\Phi '(u),u_n^+-u^+)\\&+(\psi '(u_n)-\psi '(u),u_n^+-u^+ )\rightarrow 0 \ \ \text {as}\ \ n\rightarrow \infty . \end{aligned}$$

Similarly, $\Vert u^-_n-u^-\Vert \rightarrow 0$ as $n\rightarrow \infty$. It is clear that $\{u_j^0\}$ has a convergent subsequence since $E^0$ is finite dimensional. We have thus proved the lemma. $\square$

Lemma 3.2

If $b_\infty >\lambda _1$ and $(G_3)$ holds, then any $(C)_c$-sequence of $\Phi$ is bounded.

Proof

Let $\{u_j\}\subset E$ be such that $\Phi (u_j)\rightarrow c$ and $(1+\Vert u_j\Vert )\Phi '(u_j)\rightarrow 0$.

Defining

$$\begin{aligned} {\tilde{E}}^+:= & {} \left\{ u\in E:\ u=\sum _{\lambda _j>b_\infty }a_je_j\right\} ,\\ {\tilde{E}}^0:= & {} \left\{ u\in E:\ u=\sum _{\lambda _j=b_\infty }a_je_j\right\} ,\\ {\tilde{E}}^-:= & {} \left\{ u\in E:\ u=\sum _{\lambda _j<b_\infty ,\lambda _j\ne 0}a_je_j+u^0,u^0\in E^0\right\} , \end{aligned}$$

we have $E={\tilde{E}}^+\oplus {\tilde{E}}^0\oplus {\tilde{E}}^-$ and write $u=\tilde{u}^++{\tilde{u}}^0+{\tilde{u}}^-$ for $u\in E$ corresponding to this decomposition. Clearly, ${\tilde{E}}^0=\{0\}$ if $\hbox {b}_\infty \notin \sigma (A_V)$.

Let $P^{\pm }: E\rightarrow E^{\pm }$ be the orthogonal projections. One can see

$$\begin{aligned} (\Phi '(u),{\tilde{u}}^+)= & {} \Vert \tilde{u}^+\Vert ^2-b_\infty |{\tilde{u}}^+|^2_2 -\displaystyle \int _QG^\infty _u(x,u){\tilde{u}}^+,\nonumber \\ (\Phi '(u),{\tilde{u}}^-)= & {} (P^+u,{\tilde{u}}^-)-(P^-u,{\tilde{u}}^-)\nonumber \\&-b_\infty |{\tilde{u}}^-|^2_2-\displaystyle \int _QG^\infty _u(x,u){\tilde{u}}^-. \end{aligned}$$

(3.5)

For $u=\sum _{j\in \mathbb {Z},j\ne 0} a_je_j+u^0\in E\ (u^0\in E^0)$, we have

$$\begin{aligned} {\tilde{u}}^+=\sum _{\lambda _j>b_\infty }a_je_j,\ \ \ \ {\tilde{u}}^-=\sum _{\lambda _j<b_\infty ,\lambda _j\ne 0}a_je_j+u^0. \end{aligned}$$

By (2.1) one finds

$$\begin{aligned} \begin{aligned} \Vert {\tilde{u}}^+\Vert ^2-b_\infty |{\tilde{u}}^+|^2_2&= \sum _{\lambda _j>b_\infty }\lambda _j|a_j|^2-b_\infty \sum _{\lambda _j>b_\infty }|a_j|^2\\&\ge \, \left( 1-\frac{b_\infty }{\lambda '}\right) \Vert {\tilde{u}}^+\Vert ^2, \end{aligned} \end{aligned}$$

(3.6)

where $\lambda ':=\min (\sigma (A_V)\cap (b_\infty ,\infty ))$. Since $b_\infty >\lambda _1, \sigma (A_V)\cap (0,b_\infty )\ne \emptyset$. Setting $\lambda '':=\max (\sigma (A_V)\cap (0,b_\infty ))$, we obtain

$$\begin{aligned} \begin{aligned}&\,(P^+u,{\tilde{u}}^-)-(P^-u,{\tilde{u}}^-) -b_\infty |{\tilde{u}}^-|^2_2\\&\quad =\sum _{0<\lambda _j<b_\infty }\lambda _j|a_j|^2-\sum _{\lambda _j<0}|\lambda _j||a_j|^2 -b_\infty \sum _{\lambda _j<b_\infty ,\lambda _j\ne 0}|a_j|^2-b_\infty |u^0|^2_2\\&\quad \le \Vert {\tilde{u}}^-\Vert ^2 -2\sum _{\lambda _j<0}|\lambda _j| |a_j|^2 -\frac{b_\infty }{\lambda ''}\sum _{0<\lambda _j<b_\infty }\lambda _j|a_j|^2-(1+b_\infty )|u^0|^2_2, \end{aligned} \end{aligned}$$

and therefore

$$\begin{aligned} -(P^+u,{\tilde{u}}^-)+(P^-u,{\tilde{u}}^-)+b_\infty |\tilde{u}^-|^2_2\ge (w-1)\Vert {\tilde{u}}^-\Vert ^2, \end{aligned}$$

(3.7)

here $w:=\min \{1+b_\infty ,2,\frac{b_\infty }{\lambda ''}\}$. For $\delta >0$ small, it follows from (3.1) that

$$\begin{aligned} |G^\infty _u(x,u)|<\delta |u|+C_\delta ,\ \text {for all } (x,u). \end{aligned}$$

(3.8)

Putting $u_j={\tilde{u}}^+_j+{\tilde{u}}^-_j+{\tilde{u}}^0_j$, by (3.5) we know

$$\begin{aligned} \left. \begin{array}{ll} &{}\ \Vert {\tilde{u}}^+_j\Vert ^2-b_\infty |\tilde{u}^+_j|^2_2 =(\Phi '(u_j),{\tilde{u}}^+_j) +\displaystyle \int _QG^\infty _u(x,u_j){\tilde{u}}^+_j,\\ &{}\,-(P^+u_j,{\tilde{u}}^-_j)+(P^-u_j,{\tilde{u}}^-_j)+b_\infty |{\tilde{u}}^-_j|_2^2\\ &{}\qquad \qquad \qquad =-(\Phi '(u_j),{\tilde{u}}^-_j)- \displaystyle \int _QG^\infty _u(x,u_j){\tilde{u}}^-_j. \end{array}\right. \end{aligned}$$

(3.9)

(3.6)–(3.9) and (2.2) yield

$$\begin{aligned} \begin{aligned} \xi \Vert {\tilde{u}}^+_j+{\tilde{u}}^-_j\Vert ^2&\le \Vert \Phi '(u_j)\Vert \Vert {\tilde{u}}^+_j+{\tilde{u}}^-_j\Vert \\&\quad \,\, + \delta C'\Vert u_j\Vert \Vert {\tilde{u}}^+_j+{\tilde{u}}^-_j\Vert +C'_\delta \Vert {\tilde{u}}^+_j+{\tilde{u}}^-_j\Vert \end{aligned} \end{aligned}$$

(3.10)

with $\xi =\min \{1-\frac{b_\infty }{\lambda '},w-1\}$.

If (i) of $(G_3)$ holds, then $u_j={\tilde{u}}^+_j+{\tilde{u}}^-_j$. (3.10) implies that

$$\begin{aligned} \xi \Vert u_j\Vert \le \Vert \Phi '(u_j)\Vert + \delta C'\Vert u_j\Vert +C'_\delta , \end{aligned}$$

and so $\{u_j\}$ is bounded.

Next let (ii) of $(G_3)$ be satisfied. (3.6), (3.7) and (3.9) yield that $\{{\tilde{u}}^+_j+{\tilde{u}}^-_j\}$ is bounded. We claim that $\{{\tilde{u}}^0_j\}$ is bounded.

Assume by contradiction that $\Vert {\tilde{u}}^0_j\Vert \rightarrow \infty$ as $j\rightarrow \infty$. Since ${\tilde{E}}^0$ is finite dimensional, we have: along a subsequence, there exists $Q_0\subset Q$ satisfying $|Q_0|>0$ such that $|{\tilde{u}}^0_j(x)|\rightarrow \infty$ as $j\rightarrow \infty$ uniformly in $x\in Q_0$. Here, we write |W| for the Lebesgue measure of $W\subset {\mathbb {R}}^3$. It follows from the hypotheses that $G^\infty (x,\tilde{u}^0_j)\rightarrow \infty$ as $j\rightarrow \infty$ uniformly in $x\in Q_0$, and thus

$$\begin{aligned} \begin{aligned} G^\infty (x,u_j)&=G^\infty (x,{\tilde{u}}^0_j) +\int _0^1G^\infty _u(x,s(u_j-{\tilde{u}}^0_j))({\tilde{u}}^+_j+{\tilde{u}}^-_j)ds\\&\ge G^\infty (x,{\tilde{u}}^0_j)-K_1\Vert {\tilde{u}}^+_j+{\tilde{u}}^-_j\Vert \rightarrow \infty \end{aligned} \end{aligned}$$

(3.11)

as $j\rightarrow \infty$ uniformly in $x\in Q_0$.

By (3.2) and $G^\infty (x,u)\rightarrow \infty$ as $|u|\rightarrow \infty$ we obtain that there exists $m_0>0$ such that

$$\begin{aligned} G^\infty (x,u)\ge -m_0\ \ \text {for all }\ (x,u). \end{aligned}$$

(3.12)

Noting that

$$\begin{aligned} \begin{aligned}&\, \Vert u^+\Vert ^2-\Vert u^-\Vert ^2-b_\infty |u|^2_2\\&\quad =\,\Vert {\tilde{u}}^+\Vert ^2+\sum _{0<\lambda _j<b_\infty }\lambda _j|a_j|^2 -\sum _{\lambda _j<0}|\lambda _j||a_j|^2 -b_\infty |{\tilde{u}}^++{\tilde{u}}^-|^2_2, \end{aligned} \end{aligned}$$

we get by (2.2)

$$\begin{aligned} \left| \Vert u^+\Vert ^2-\Vert u^-\Vert ^2-b_\infty |u|^2_2\right| \le (1+a_2b_\infty )(\Vert {\tilde{u}}^+\Vert ^2+\Vert {\tilde{u}}^-\Vert ^2). \end{aligned}$$

(3.13)

On account of (3.13), (3.12) and (3.11) we see that

$$\begin{aligned} \begin{aligned} |\Phi (u_j)|&=\, \left| \frac{1}{2}(\Vert u_j^+\Vert ^2-\Vert u_j^-\Vert ^2-b_\infty |u_j|^2_2) -\int _QG^\infty (x,u_j)\right| \\&\ge \, \left| \int _QG^\infty (x,u_j)\right| -\frac{1}{2}(1+a_2b_\infty )(\Vert {\tilde{u}}^+_j\Vert ^2+\Vert {\tilde{u}}^-_j\Vert ^2)\\&\ge \, \int _{Q_0}G^\infty (x,u_j)-m_0-\frac{1}{2}(1+a_2b_\infty )(\Vert \tilde{u}^+_j\Vert ^2+\Vert {\tilde{u}}^-_j\Vert ^2) \rightarrow \infty \end{aligned} \end{aligned}$$

as $j\rightarrow \infty$, a contradiction. Consequently $\{x^0_j\}$ is bounded and the proof is complete. $\square$

We need to introduce another orthogonal decomposition: $E=\hat{E}^+\oplus {\hat{E}}^0\oplus {\hat{E}}^-$, where

$$\begin{aligned} \left\{ \begin{array}{l} {\hat{E}}^+:=\left\{ u\in E:\ u=\displaystyle \sum _{\lambda _j>b_0}a_je_j\right\} ,\\ {\hat{E}}^0:=\left\{ u\in E:\ u=\displaystyle \sum _{\lambda _j=b_0}a_je_j\right\} , \\ {\hat{E}}^-:=\left\{ u\in E:\ u=\displaystyle \sum _{\lambda _j<b_0,\lambda _j\ne 0}a_je_j+u^0,u^0\in E^0\right\} ,\end{array}\right. \end{aligned}$$

(3.14)

One can verify that there is $\xi _0\in (0,1)$ such that

$$\begin{aligned} \begin{aligned}&\, \Vert {\hat{u}}^+\Vert ^2-b_0|{\hat{u}}^+|^2_2 \ge \xi _0 \Vert {\hat{u}}^+\Vert ^2,\\&\ (P^+u,{\hat{u}}^-)-(P^-u,{\hat{u}}^-)-b_0|{\hat{u}}^-|^2_2\le -\xi _0 \Vert {\hat{u}}^-\Vert ^2 \end{aligned} \end{aligned}$$

(3.15)

for any $u={\hat{u}}^++{\hat{u}}^0+{\hat{u}}^-\in E$, the proof is similar to that of (3.6) and (3.7).

Lemma 3.3

Suppose that $(G_1)$ and $(G_2)$ hold, then there exist $r>0$ and $\rho >0$ such that $\inf \Phi ({\hat{E}}^+\cap B_r)\ge 0$ and $\inf \Phi (\hat{E}^+\cap \partial B_r)\ge \rho$.

Proof

Choosing $q\in (2,3)$, we have that, for any $\varepsilon >0$, there is $C_\varepsilon >0$ such that

$$\begin{aligned} G^0(x,u)\le \varepsilon |u|^2+C_\varepsilon |u|^q,\, \hbox { for all }\, (x,u). \end{aligned}$$

This implies

$$\begin{aligned}\begin{aligned} \Phi ({\hat{u}}^+)&=\, \frac{1}{2}(\Vert {\hat{u}}^+\Vert ^2-b_0|{\hat{u}}^+|^2_2)-\int _QG^0(x,{\hat{u}}^+)\\&\ge \, \frac{1}{2}\xi _0\Vert {\hat{u}}^+\Vert ^2-\varepsilon C_1'\Vert {\hat{u}}^+\Vert ^2-C'_2C_\varepsilon \Vert {\hat{u}}^+\Vert ^p \end{aligned} \end{aligned}$$

via (3.15) for ${\hat{u}}\in {\hat{E}}^+$, which follows that the conclusion is valid. $\square$

Lemma 3.4

Let $(G_2)$ be satisfied. If $b_\infty >b_0^+$, then for any $n\in {\mathbb {N}}$ with $b^-_\infty =\lambda _n$, there exists $R_n>r$ such that $\sup \Phi ( E_n\setminus B_{R_n})<0$ and $\sup \Phi (E_n)<\infty$, where r is as in Lemma 3.3, $E_n:=E^-\oplus E^0\oplus \hbox {span}\{e_1,...,e_n\}$.

Proof

It will suffice to show that for $u\in E_n$

$$\begin{aligned} \Phi (u)\rightarrow -\infty \ \ \ \text {as} \ \ \Vert u\Vert \rightarrow \infty . \end{aligned}$$

(3.16)

Choose $s_0\in \left( 0,\sqrt{1-\frac{b^-_\infty }{b_\infty }}\right)$ in (3.2). Noting that $u^+=\sum \nolimits _{j=1}^n s_{j}e_j$ for $u\in E_n$, by (3.2), for $\varepsilon =\frac{1}{2} (b_\infty -\frac{b^-_\infty }{1-s^2_0})$, we find

$$\begin{aligned} \begin{aligned} 2\Phi (u)&= \Vert u^+\Vert ^2-\Vert u^-\Vert ^2-2\int _QG(x,u)\\&\le \Vert u^+\Vert ^2-\Vert u^-\Vert ^2- \alpha _0(|u^+|_2^2+|u^0|_2^2+|u^-|^2_2)+2C_{s_0}, \end{aligned} \end{aligned}$$

(3.17)

where $\alpha _0:=(b_\infty -\varepsilon )(1-s^2_0)>b^-_\infty$. Since

$$\begin{aligned}\begin{aligned} \alpha _0|u^+|_2^2-\Vert u^+\Vert ^2 \ge (\alpha _0-\lambda _n)\sum \limits _{j=1}^n|s_{j}|^2 \ge \frac{\alpha _0-b^-_\infty }{\lambda _1}\Vert u^+\Vert ^2_2, \end{aligned} \end{aligned}$$

by (3.17) we get

$$\begin{aligned} \begin{aligned} 2\Phi (u)&\le \,-\frac{\alpha _0-b^-_\infty }{\lambda _1}\Vert u^+\Vert ^2 -C_2(\Vert u^-\Vert ^2+\Vert u^0\Vert ^2)+2C_{s_0}\\&\le -{\hat{C}}\Vert u\Vert ^2+2C_{s_0} \end{aligned} \end{aligned}$$

which implies that (3.16) is valid and $\sup \Phi (E_n)<\infty$. $\square$

As a consequence, we have

Lemma 3.5

Under the conditions of Lemma 3.4, if $G^0(x,u)\ge 0$, then there is $R_0>r$ such that $\sup \Phi (\partial \Omega )\le 0$, where

$$\begin{aligned} \Omega :=\{u={\hat{u}}^-+{\hat{u}}^0+se_m: {\hat{u}}^-+{\hat{u}}^0\in {\hat{E}}^-\oplus {\hat{E}}^0, s>0, \Vert u\Vert <R_0\} \end{aligned}$$

with $A_Ve_m=b^+_0e_m$, and $\partial \Omega$ refers to the boundary of $\Omega$ relative to $\hbox {span}\{e_m\}\oplus {\hat{E}}^-\oplus {\hat{E}}^0$.

Proof

Since ${\hat{E}}^-\oplus {\hat{E}}^0\oplus {\mathbb {R}}^+e_m\subset E_m$ and $\lambda _m=b_0^+\le b^-_\infty$, by Lemma 3.4 we find that $\Phi (u)<0$ for $u={\hat{u}}^-+{\hat{u}}^0+se_m$ with $\Vert u\Vert =R_0$ and $s>0$ when $R_0>r$ large.

Let $u={\hat{u}}^-+{\hat{u}}^0$ with $\Vert u\Vert \le R_0$. By $G^0(x,u)\ge 0$ and (3.15) one has

$$\begin{aligned} \begin{aligned} 2\Phi (u)&= (P^+u,{\hat{u}}^-)-(P^-u,{\hat{u}}^-)-b_0|{\hat{u}}^-|^2_2- 2\int _QG^0(x,u)\\&\le -\xi _0\Vert {\hat{u}}^-\Vert ^2-2\int _QG^0(x,u) \le 0 \end{aligned} \end{aligned}$$

which yields that the result is valid. $\square$

Now, with the above arguments, we are ready to prove Theorem 1.2.

Proof of Theorem 1.2

(Existence) Let us verify the conditions of Theorem 2.5. Let $X=\hat{E}^+,\ Y={\hat{E}}^-\oplus {\hat{E}}^0,\ r>0$ be from Lemma 3.3. Lemma 3.1 and 3.2 imply that $(\Phi _1)$ is true. Lemma 3.3 yields $\inf \Phi (X\cap \partial B_r)\ge \rho$, and Lemma 3.5 gives $\Phi |_{\partial \Omega }<\sigma _0$ for $\sigma _0\in (0,\rho )$. Therefore $(\Phi _2)$ holds. It follows from Theorem 2.5 that $\Phi$ possesses a critical value $c\ge \rho$, with

$$\begin{aligned} c=\inf _{h\in \Gamma }\sup _{u\in \Omega }\,\Phi (h(1,u)), \end{aligned}$$

where $\Gamma$ is defined as (2.17).

Next, we proceed to prove the multiplicity. Since G is even in $u, \Phi$ is even. Using Lemma 3.3 we know that the condition $(\Phi _3)$ holds with $X={\hat{E}}^-\oplus Y^0$ and $Y=\hat{E}^+$. Let $\hbox {span}\{e_m,\dots ,e_n\}$ be the eigenspace associated to $\sigma (A_V)\cap (b_0,b_\infty )$, and $\lambda _j$ the eigenvalue corresponding to $e_j$ (i.e., $A_Ve_j=\lambda _je_j),\ j=m,\dots ,n$, then $b^+_0=\lambda _m,\ b^-_\infty =\lambda _n$ and $d(b_0,b_\infty )=n-m$. It follow from Lemma 3.4 that $\Phi$ satisfies $(\Phi _4)$ with $Y_0=\hbox {span}\{e_m,\dots , e_n\}, R=R_n,\ M_*=M_n$ and $\sigma \in (0,\rho )$. Therefore, $\Phi$ has at least $n-m$ pairs of nontrivial critical points by Theorem 2.8. $\square$

We are now in a position to give the proof of Theorem 1.5.

Proof

The main difference to the proof of Theorem 1.2 lies in the boundedness of the (C)c-sequences.

Claim 1. Any (C)c-sequence is bounded.

Let $\{u_j\}\subset E$ be such that

$$\begin{aligned} \Phi (u_j)\rightarrow c,\ (1+\Vert u_j\Vert )\Phi '(u_j)\rightarrow 0 \ \text {as}\ \ j\rightarrow \infty . \end{aligned}$$

We then have

$$\begin{aligned} \int _Q{\hat{G}}(x,u_j)=\Phi (u_j)-\frac{1}{2}\Phi '(u_j)\cdot u_j\le C_0. \end{aligned}$$

(3.18)

Assume by contradiction that $\Vert u_j\Vert \rightarrow \infty$. Then the normalized sequence $v_j=u_j/\Vert u_j\Vert$ satisfies (up to a subsequence) $v_j\rightharpoonup v$ in E. Lemma 2.1 guarantees $v_j\rightarrow v$ in $L_T^s(Q)$ for $s\in [1,3)$ and $|v_j|_s\le a_s$ for all $s\in [1,3]$. We write $\tilde{u}_j=u^-_j+u^+_j, \tilde{v}_j=v^-_j+v^+_j$. Then

$$\begin{aligned}\Phi '(u_j)(u^+_j-u^-_j) =\Vert u_j\Vert ^2\left( \Vert \tilde{v}_j\Vert ^2-\int _Q \frac{G_u(x,u_j)(v^+_j-v^-_j)|v_j|}{|u_j|}\right) , \end{aligned}$$

and therefore

$$\begin{aligned} o(1) =\Vert \tilde{v}_j\Vert ^2-\int _Q \frac{G_u(x,u_j)(v^+_j-v^-_j)|v_j|}{|u_j|}. \end{aligned}$$

(3.19)

We distinguish the two cases: $v=0$ or $v\ne 0$.

let $v=0$. $(G'_1)$ and $(G'_2)$ yield that (3.4) is true, this implies

$$\begin{aligned}\int _{Q}\frac{|G_u(x,u_j)|}{|u_j|}|v^+_j-v^-_j|\,|v_j| \le C_1 |v_j|^2_2 \end{aligned}$$

which jointly with (3.19) shows $\Vert \tilde{v}_j\Vert ^2\rightarrow 0$, and so $|\tilde{v}_j|_2\rightarrow 0$. $|v_j|_2\rightarrow 0$ yields $|v^0_j|_2\rightarrow 0$. We obtain $1=\Vert v_j\Vert =\Vert \tilde{v}_j\Vert +|v^0_j|_2\rightarrow 0$, a contradiction.

Assume $v\ne 0$. First let (i) of $(G_3)$ hold. Since $|u_j(x)|=|v_j(x)|\ \Vert u_j\Vert \rightarrow \infty$, by (3.4) and Lebesgue dominated convergence theorem we obtain

$$\begin{aligned} \int _{Q}\frac{G_u(x,u_j)v_j\varphi }{|u_j|}\rightarrow \int _{Q}b_\infty (x)v\varphi \end{aligned}$$

for any $\varphi \in C^\infty [Q,{\mathbb {C}}^4]$, hence $A_V v= b_\infty v$, which contradicts $0\notin \sigma (A_V-b_\infty )$.

Suppose that (ii) of $(G_3)$ is satisfied. $v_j\rightarrow v$ in $L_T^s(Q)$ guarantees (up to a subsequence) $v_j(x)\rightarrow v(x)$ a.e. on Q. Since $v\ne 0$, there exists $Q_0\subset Q$ with $|Q_0|>0$ such that

$$\begin{aligned} v_j(x)\rightarrow v(x)\ \text { as }\ j\rightarrow \infty \ \ \text {uniformly on }\ Q_0 \end{aligned}$$

and $|v_j(x)|\ge \varepsilon _0>0$ for large j. Observe that $|u_j(x)|=\Vert u_j\Vert |v_(x)|\ge \varepsilon _0\Vert u_j\Vert \rightarrow \infty$ for $x\in Q_0$. By (ii) of $(G_3)$ we have

$$\begin{aligned} \int _{Q_0}\hat{G}(x,u_j)\rightarrow \infty , \end{aligned}$$

which contradicts (3.18).

Next we have

Claim 2. The conclusions of Lemmas 3.3–3.5 are true where $(G_1)$ and $(G_2)$ are replaced by $(G'_1)$ and $(G'_2)$ respectively, and $b_0$ is replaced by $q_0$ in (3.14).

Since (3.2) and (3.4) are satisfied, where $b_\infty$ is replaced by $q_\infty$, one can prove as before.

Finally, repeating the arguments of the proof of Theorem 1.2, we obtain the desired results. $\square$