Nonlinear Eigenvalue Problems and Bifurcation for Quasi-Linear Elliptic Operators

Abstract

In this paper, we analyze an eigenvalue problem for quasi-linear elliptic operators involving homogeneous Dirichlet boundary conditions in a open smooth bounded domain. We show that the eigenfunctions corresponding to the eigenvalues belong to $L^{\infty }$, which implies $C^{1,\alpha }$ smoothness, and the first eigenvalue is simple. Moreover, we investigate the bifurcation results from trivial solutions using the Krasnoselski bifurcation theorem and from infinity using the Leray–Schauder degree. We also show the existence of multiple critical points using variational methods and the Krasnoselski genus.

1 Introduction

We consider $\Omega \subset \mathbb {R}^N$ ($N\ge 2$) an open bounded domain with smooth boundary $\partial \Omega$. A classical result in the theory of eigenvalue problems guarantees that the problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u=\displaystyle \lambda u~~&{}\quad \text {in}\, \Omega ,\\ u = \displaystyle 0~~~~~~~~&{}\quad \text {on}\, \partial \Omega , \end{array} \right. \end{aligned}$$

(1.1)

possesses a nondecreasing sequence of eigenvalues and a sequence of corresponding eigenfunctions which define a Hilbert basis in $L^2(\Omega )$ (see, [16]). Moreover, it is known that the first eigenvalue of problem (1.1) is characterized in the variational point of view by

$$\begin{aligned} \lambda ^D_1:=\inf _{u\in W^{1,2}_0(\Omega )\backslash \{0\}}\left\{ \frac{\int _{\Omega }|\nabla u|^2~\mathrm{d}x}{\int _{\Omega }u^2~\mathrm{d}x}\right\} . \end{aligned}$$

Suppose that $p>1$ is a given real number and consider the nonlinear eigenvalue problem with Neumann boundary condition

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _p u=\displaystyle \lambda u~~&{}\quad \text {in}\, \Omega ,\\ \frac{\partial u}{\partial \nu } = \displaystyle 0~~~~~~~~&{}\quad \text {on}\, \partial \Omega , \end{array} \right. \end{aligned}$$

(1.2)

where $\Delta _p u:=\text {div}(|\nabla u|^{p-2}\nabla u)$ stands for the p-Laplace operator and $\lambda \in \mathbb {R}$. This problem was considered in [15], and using a direct method in calculus of variations (if $p>2$) or a mountain-pass argument (if $p\in (\frac{2N}{N+2},2)$) it was shown that the set of eigenvalues of problem (1.2) is exactly the interval $[0,\infty )$. Indeed, it is sufficient to find one positive eigenvalue, say $-\Delta _p u=\lambda u.$ Then a continuous family of eigenvalues can be found by the reparametrization $u=\alpha v,$ satisfying $-\Delta _p v=\mu (\alpha )v$, with $\mu (\alpha )=\frac{\lambda }{\alpha ^{p-2}}.$

In this paper, we consider the so-called (p, 2)-Laplace operator (see, [18]) with Dirichlet boundary conditions. More precisely, we analyze the following nonlinear eigenvalue problem:

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _p u-\Delta u=\displaystyle \lambda u \quad &{}\quad \text {in}\, \Omega ,\\ u = \displaystyle 0\quad \quad \quad \quad \quad \quad &{}\quad \text {on}\, \partial \Omega , \end{array} \right. \end{aligned}$$

(1.3)

where $p\in (1,\infty )\backslash \{2\}$ is a real number. We recall that if $1<p<q,$ then $L^q(\Omega )\subset L^p(\Omega )$ and as a consequence, one has $W^{1,q}_0(\Omega )\subset W^{1,p}_0(\Omega ).$ We will say that $\lambda \in \mathbb {R}$ is an eigenvalue of problem (1.3) if there exists $u\in W^{1,p}_0(\Omega )\backslash \{0\}$ (if $p>2$ ), $u\in W^{1,2}_0(\Omega )\backslash \{0\}$ (if $1<p<2$) such that

$$\begin{aligned} \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla v~\mathrm{d}x+\int _{\Omega }\nabla u\cdot \nabla v~\mathrm{d}x=\lambda \int _{\Omega }u~v~\mathrm{d}x, \end{aligned}$$

(1.4)

for all $v\in W^{1,p}_{0}(\Omega )$ (if $p>2$), $v\in W^{1,2}_{0}(\Omega )$ (if $1<p<2$). In this case, such a pair $(u,\lambda )$ is called an Eigenpair, and $\lambda \in \mathbb {R}$ is called an eigenvalue and $u\in W^{1,p}_0(\Omega )\backslash \{0\}$ is an eigenfunction associated with $\lambda .$ We say that $\lambda$ is a “first eigenvalue”, if the corresponding eigenfunction u is positive or negative.

The operator $-\Delta _p-\Delta$ appears in quantum field theory (see, [5]), where it arises in the mathematical description of propagation phenomena of solitary waves. We recall that a solitary wave is a wave which propagates without any temporal evolution in shape.

The operator $-\Delta _p-\Delta$ is a special case of the so called (p, q)-Laplace operator given by $-\Delta _p-\Delta _q$ which has been widely studied; for some results related to our studies, see, e.g., [6, 7, 10, 21, 25] .

The main purpose of this work was to study the nonlinear eigenvalue problem (1.3) when $p>2,$ and $1<p<2$, respectively . In particular, we show in section 2 that the set of the first eigenvalues is given by the interval $(\lambda _1^D,\infty )$, where $\lambda _1^D$ is the first Dirichlet eigenvalue of the Laplacian. We show that the first eigenvalue of (1.3) can be obtained variationally, using a Nehari set for $1<p<2$, and a minimization for $p>2.$ Also in the same section, we recall some results of [15, 22, 23].

In Sect. 3, we prove that the eigenfunctions associated with $\lambda$ belong to $L^{\infty }(\Omega )$: the first eigenvalue $\lambda _1^D$ of problem (1.3) is simple and the corresponding eigenfunctions are positive or negative. In addition, in Sect. 3.3 we show a homeomorphism property related to $-\Delta _p-\Delta$.

In Sect. 4, we prove that $\lambda _1^D$ is a bifurcation point for a branch of first eigenvalues from zero if $p>2,$ and $\lambda _1^D$ is a bifurcation point from infinity if $p<2.$ Also the higher Dirichlet eigenvalues $\lambda ^D_k$ are bifurcation points (from 0 if $p>2,$ respectively, from infinity if $1<p<2$ ), if the multiplicity of $\lambda ^D_k$ is odd. Finally in Sect. 5, we prove by variational methods that if $\lambda \in (\lambda ^D_k,\lambda ^D_{k+1}),$ then there exist at least k nonlinear eigenvalues using Krasnoselski’s genus. In what follows, we denote by $\Vert .\Vert _{1,p}$ and $\Vert .\Vert _2$ the norms on $W^{1,p}_0(\Omega )$ and $L^2(\Omega )$ defined, respectively, by

$$\begin{aligned} \Vert u\Vert _{1,p}=\left( \int _{\Omega }|\nabla u|^p~\mathrm{d}x\right) ^{\frac{1}{p}}~~\text {and}~~\Vert u\Vert _2=\left( \int _{\Omega }|u|^2~\mathrm{d}x\right) ^{\frac{1}{2}},~~\text {for all}~~u\in W^{1,p}_0(\Omega ),~~u\in L^2(\Omega ). \end{aligned}$$

We recall the Poincaré inequality, i.e., there exists a positive constant $C_p(\Omega )$ such that

$$\begin{aligned} \int _{\Omega }|u|^p~\mathrm{d}x\le C_p(\Omega )\int _{\Omega }|\nabla u|^p~\mathrm{d}x~~\text {for all} u\in W^{1,p}_0(\Omega ),~~1<p<\infty . \end{aligned}$$

(1.5)

2 The Spectrum of the Nonlinear Problem

We now begin with the discussion of the properties of the spectrum of the nonlinear eigenvalues problem (1.3).

Remark 2.1

Any $\lambda \le 0$ is not an eigenvalue of problem (1.3).

Indeed, suppose by contradiction that $\lambda =0$ is an eigenvalue of equation (1.3), then relation (1.4) with $v=u_{0}$ gives

$$\begin{aligned} \int _{\Omega }|\nabla u_{0}|^p~\mathrm{d}x+\int _{\Omega }|\nabla u_{0}|^2~\mathrm{d}x=0. \end{aligned}$$

Consequently, $|\nabla u_{0}|=0$; therefore, $u_0$ is constant on $\Omega$ and $u_0=0$ on $\Omega$. And this contradicts the fact that $u_0$ is a nontrivial eigenfunction. Hence $\lambda =0$ is not an eigenvalue of problem (1.3). Now it remains to show that any $\lambda <0$ is not an eigenvalue of (1.3). Suppose by contradiction that $\lambda <0$ is an eigenvalue of (1.3), with $u_{\lambda }\in W^{1,p}_0(\Omega )\backslash \{0\}$ the corresponding eigenfunction. The relation (1.4) with $v=u_{\lambda }$ implies

$$\begin{aligned} 0\le \int _{\Omega }\left| \nabla u_{\lambda }\right| ^p~\mathrm{d}x+\int _{\Omega }\left| \nabla u_{\lambda }\right| ^2~\mathrm{d}x=\lambda \int _{\Omega }u_{\lambda }^2~\mathrm{d}x<0, \end{aligned}$$

Which yields a contradiction and thus $\lambda <0$ cannot be an eigenvalue of problem (1.3).

Lemma 2.2

Any $\lambda \in (0,\lambda ^D_1]$ is not an eigenvalue of (1.3).

For the proof see also [15].

Proof

Let $\lambda \in (0,\lambda ^D_1)$, i.e., $\lambda ^D_1>\lambda .$ Let us assume by contradiction that there exists a $\lambda \in (0,\lambda ^D_1)$ which is an eigenvalue of (1.3) with $u_{\lambda }\in W^{1,2}_0(\Omega )\backslash \{0\},$ the corresponding eigenfunction. Letting $v=u_{\lambda }$ in relation (1.4), we have on the one hand,

$$\begin{aligned} \int _{\Omega }\left| \nabla u_{\lambda }\right| ^p~\mathrm{d}x+\int _{\Omega }\left| \nabla u_{\lambda }\right| ^2~\mathrm{d}x=\lambda \int _{\Omega }u_{\lambda }^2~\mathrm{d}x, \end{aligned}$$

and on the other hand,

$$\begin{aligned} \lambda ^D_1 \int _{\Omega }u_{\lambda }^2\mathrm{d}x\le \int _{\Omega }\left| \nabla u_{\lambda }\right| ^2~\mathrm{d}x. \end{aligned}$$

(2.1)

By subtracting both sides of (2.1) by $\lambda \displaystyle {\int _{\Omega }}u_{\lambda }^2~\mathrm{d}x$, we obtain

$$\begin{aligned} \left( \lambda ^D_1-\lambda \right) \int _{\Omega }u^2_{\lambda }~\mathrm{d}x\le \int _{\Omega }\left| \nabla u_{\lambda }\right| ^2~\mathrm{d}x-\lambda \int _{\Omega }u_{\lambda }^2~\mathrm{d}x, \end{aligned}$$

$$\begin{aligned} \left( \lambda ^D_1-\lambda \right) \int _{\Omega }u^2_{\lambda }~\mathrm{d}x\le \int _{\Omega }\left| \nabla u_{\lambda }\right| ^2~\mathrm{d}x-\lambda \int _{\Omega }u_{\lambda }^2~\mathrm{d}x+\int _{\Omega }\left| \nabla u_{\lambda }\right| ^p~\mathrm{d}x=0. \end{aligned}$$

Therefore, $(\lambda ^D_1-\lambda ) \displaystyle {\int _{\Omega }}u^2_{\lambda }~\mathrm{d}x\le 0,$ which is a contradiction. Hence, we conclude that $\lambda \in (0,\lambda ^D_1)$ is not an eigenvalue of problem (1.3). In order to complete the proof of the Lemma 2.2 we shall show that $\lambda =\lambda ^D_1$ is not an eigenvalue of (1.3).

By contradiction we assume that $\lambda =\lambda ^D_1$ is an eigenvalue of (1.3). So there exists $u_{\lambda ^D_1}\in W^{1,2}_0(\Omega )\backslash \{0\}$ such that relation (1.4) holds true. Letting $v=u_{\lambda ^D_1}$ in relation (1.4), it follows that

$$\begin{aligned} \int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^p~\mathrm{d}x+\int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^2~\mathrm{d}x=\lambda ^D_1 \int _{\Omega }u_{\lambda ^D_1}^2~\mathrm{d}x. \end{aligned}$$

But $\lambda ^D_1 \displaystyle {\int _{\Omega }}u_{\lambda ^D_1}^2~\mathrm{d}x\le \displaystyle {\int _{\Omega }}|\nabla u_{\lambda ^D_1}|^2~\mathrm{d}x;$ therefore

$$\begin{aligned} \int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^p~\mathrm{d}x+\int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^2~\mathrm{d}x\le \int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^2~\mathrm{d}x \Rightarrow \int _{\Omega }\left| \nabla u_{\lambda ^D_1}\right| ^p~\mathrm{d}x\le 0. \end{aligned}$$

Using relation (1.5), we have $u_{\lambda ^D_1}=0,$ which is a contradiction since $u_{\lambda ^D_1}\in W^{1,2}_0(\Omega )\backslash \{0\}.$ So $\lambda =\lambda ^D_1$ is not an eigenvalue of (1.3). $\square$

Theorem 2.3

Assume $p\in (1,2)$. Then the set of first eigenvalues of problem (1.3) is given by

$$\begin{aligned} (\lambda ^D_1,\infty ),~~\mathrm {where} \lambda ^D_1~\mathrm {denotes the first eigenvalue of} -\Delta \mathrm {on} \Omega . \end{aligned}$$

Proof

Let $\lambda \in (\lambda _1^D,\infty )$, and define the energy functional

$$\begin{aligned} J_{\lambda }: W^{1,2}_0(\Omega )\rightarrow \mathbb {R}~~ \text {by}~~J_{\lambda }(u)=\int _{\Omega }|\nabla u|^2~\mathrm{d}x+\frac{2}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x-\lambda \int _{\Omega }u^2~\mathrm{d}x. \end{aligned}$$

One shows that $J_{\lambda }\in C^1(W^{1,2}_0(\Omega ),\mathbb {R})$ (see, [18]) with its derivatives given by

$$\begin{aligned} \left\langle J'_{\lambda }(u),v\right\rangle= & {} 2\int _{\Omega }\nabla u\cdot \nabla v~\mathrm{d}x+2 \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla v~\mathrm{d}x-2\lambda \int _{\Omega }u~v~\mathrm{d}x~~ ,\\&~\forall ~v\in W^{1,2}_0(\Omega ). \end{aligned}$$

Thus we note that $\lambda$ is an eigenvalue of problem (1.3) if and only if $J_{\lambda }$ possesses a nontrivial critical point. Considering $J_{\lambda }(\rho e_1),$ where $e_1$ is the $L^2$-normalized first eigenfunction of the Laplacian, we see that

$$\begin{aligned} J_{\lambda }(\rho e_1)\le \lambda _1^D\rho ^2+C\rho ^p-\lambda \rho ^2\rightarrow -\infty ,~~~\text {as}~~\rho \rightarrow + \infty . \end{aligned}$$

Hence, we cannot establish the coercivity of $J_{\lambda }$ on $W^{1,2}_0(\Omega )$ for $p\in (1,2)$, and consequently we cannot use a direct method in calculus of variations in order to determine a critical point of $J_{\lambda }.$ To overcome this difficulty, the idea will be to analyze the functional $J_{\lambda }$ on the so-called Nehari manifold defined by

$$\begin{aligned} \mathcal {N}_{\lambda } :=\left\{ u\in W^{1,2}_0(\Omega )\backslash \{0\}~:\int _{\Omega }|\nabla u|^2~\mathrm{d}x+\int _{\Omega }|\nabla u|^p~\mathrm{d}x=\lambda \int _{\Omega }u^2~\mathrm{d}x\right\} . \end{aligned}$$

Note that all non-trivial solutions of (1.3) lie on $\mathcal {N}_{\lambda }.$ On $\mathcal {N}_{\lambda }$ the functional $J_{\lambda }$ takes the following form

$$\begin{aligned} J_{\lambda }(u)= & {} \int _{\Omega }|\nabla u|^2~\mathrm{d}x+\frac{2}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x-\lambda \int _{\Omega }u^2~\mathrm{d}x\\= & {} \left( \frac{2}{p}-1\right) \int _{\Omega }|\nabla u|^p~\mathrm{d}x>0. \end{aligned}$$

We have seen in Lemma 2.2 that any $\lambda \in (0,\lambda _1^D]$ is not an eigenvalue of problem (1.3); see also [15]. It remains to prove the following:

Claim: Every $\lambda \in (\lambda ^D_1,\infty )$ is a first eigenvalue of problem (1.3). Indeed, we will split the proof of the claim into four steps follows:

• Step 1. Here we will show that $\mathcal {N}_{\lambda }\ne \emptyset$ and every minimizing sequence for $J_{\lambda }$ on $\mathcal {N}_{\lambda }$ is bounded. Since $\lambda >\lambda ^D_1$ there exists $v_{\lambda }\in W^{1,2}_0(\Omega )$ such that

  $$\begin{aligned} \int _{\Omega }\left| \nabla v_{\lambda }\right| ^2<\lambda \int _{\Omega }v^2_{\lambda }~\mathrm{d}x. \end{aligned}$$

  Then there exists $t>0$ such that $tv_{\lambda }\in \mathcal {N}_{\lambda } \Rightarrow$

  $$\begin{aligned}&\int _{\Omega }\left| \nabla \left( tv_{\lambda }\right) \right| ^2~\mathrm{d}x+\int _{\Omega }\left| \nabla \left( tv_{\lambda }\right) \right| ^p~\mathrm{d}x=\lambda \int _{\Omega }\left( tv_{\lambda }\right) ^2~\mathrm{d}x\Rightarrow \nonumber \\&t^2\int _{\Omega }\left| \nabla v_{\lambda }\right| ^2~\mathrm{d}x+t^p\int _{\Omega }\left| \nabla v_{\lambda }\right| ^p~\mathrm{d}x=t^2\lambda \int _{\Omega }v_{\lambda }^2~\mathrm{d}x \Rightarrow \nonumber \\&t=\left( \frac{\int _{\Omega }|\nabla v_{\lambda }|^p~\mathrm{d}x}{\lambda \int _{\Omega }v_{\lambda }^2~\mathrm{d}x- \int _{\Omega }|\nabla v_{\lambda }|^2~\mathrm{d}x}\right) ^{\frac{1}{2-p}}>0. \end{aligned}$$

  With such t we have $tv_{\lambda }\in \mathcal {N}_{\lambda }$ and $\mathcal {N}_{\lambda }\ne \emptyset .$ Note that for $u\in B_r(v_{\lambda }),$ $r>0$ small, the inequality $\lambda \int _{\Omega }|u|^2\mathrm{d}x>\int _{\Omega }|\nabla u|^2\mathrm{d}x$ remains valid, and then $t(u)u\in \mathcal {N}_{\lambda }$ for $u\in B_r(v_{\lambda }).$ Since $t(u)\in C^1$ we conclude that $\mathcal {N}_{\lambda }$ is a $C^1$-manifold. Let $\{u_k\}\subset \mathcal {N}_{\lambda }$ be a minimizing sequence of $J_{\lambda }|_{\mathcal {N}_{\lambda }}$, i.e., $J_{\lambda }(u_k)\rightarrow m=\displaystyle {\inf _{w\in \mathcal {N}_{\lambda }}}J_{\lambda }(w).$ Then

  $$\begin{aligned}&\lambda \int _{\Omega }u^2_k~\mathrm{d}x{-}\int _{\Omega }|\nabla u_k|^2~\mathrm{d}x\nonumber \\&\quad =\int _{\Omega }|\nabla u_k|^p~\mathrm{d}x{\rightarrow }\left( \frac{2}{p}-1\right) ^{-1}m~\mathrm {as}\,k{\rightarrow }\infty . \end{aligned}$$

  (2.2)

  Assume by contradiction that $\{u_k\}$ is not bounded, i.e., $\displaystyle \int _{\Omega }|\nabla u_k|^2~\mathrm{d}x\rightarrow \infty$ as $k\rightarrow \infty$. It follows that $\displaystyle {\int _{\Omega }}u^2_k~\mathrm{d}x \rightarrow \infty$ as $k\rightarrow \infty$, thanks to relation (2.2). We set $v_k=\frac{u_k}{\Vert u_k\Vert _2}.$ Since $\displaystyle {\int _{\Omega }}|\nabla u_k|^2~\mathrm{d}x<\lambda \displaystyle {\int _{\Omega }}u_k^2~\mathrm{d}x$, we deduce that $\displaystyle {\int _{\Omega }}|\nabla v_k|^2~\mathrm{d}x<\lambda ,$ for each k and $\Vert v_k\Vert _{1,2}<\sqrt{\lambda }.$ Hence $\{v_k\}\subset W^{1,2}_0(\Omega )$ is bounded in $W^{1,2}_0(\Omega ).$ Therefore, there exists $v_0\in W^{1,2}_0(\Omega )$ such that $v_k\rightharpoonup v_0$ in $W^{1,2}_0(\Omega )\subset W^{1,p}_0(\Omega )$ and $v_k\rightarrow v_0$ in $L^2(\Omega ).$ Dividing relation (2.2) by $\Vert u_k\Vert ^p_{2}$, we get

  $$\begin{aligned} \int _{\Omega }|\nabla v_k|^p~\mathrm{d}x=\frac{\lambda \displaystyle {\int _{\Omega }}u^2_k~\mathrm{d}x- \displaystyle {\int _{\Omega }}|\nabla u_k|^2~\mathrm{d}x}{\Vert u_k\Vert ^p_{2}}\rightarrow 0~~\mathrm {as}\,k \rightarrow \infty , \end{aligned}$$

  since $\lambda \displaystyle {\int _{\Omega }}u^2_k~\mathrm{d}x- \displaystyle {\int _{\Omega }}|\nabla u_k|^2~\mathrm{d}x\rightarrow \left( \frac{2}{p}-1\right) ^{-1}m<\infty$ and $\Vert u_k\Vert ^p_{2}\rightarrow \infty$ as $k\rightarrow \infty$. On the other hand, since $v_k\rightharpoonup v_0$ in $W^{1,p}_0(\Omega )$, we have $\displaystyle {\int _{\Omega }}|\nabla v_0|^p~\mathrm{d}x\le \lim _{k\rightarrow \infty }\inf \displaystyle {\int _{\Omega }}|\nabla v_k|^p~\mathrm{d}x=0$ and consequently $v_0=0$. It follows that $v_k\rightarrow 0$ in $L^2(\Omega ),$ which is a contradiction since $\Vert v_k\Vert _{2}=1$. Hence, $\{u_k\}$ is bounded in $W^{1,2}_0(\Omega ).$

• Step 2. $m= \displaystyle {\inf _{w\in \mathcal {N}_{\lambda }}}J_{\lambda }(w)>0.$ Indeed, assume by contradiction that $m=0$. Then, for $\{u_k\}$ as in step 1, we have

  $$\begin{aligned}&0<\lambda \int _{\Omega }u^2_k~\mathrm{d}x- \int _{\Omega }\left| \nabla u_k\right| ^2~\mathrm{d}x\nonumber \\&\quad =\int _{\Omega }\left| \nabla u_k\right| ^p~\mathrm{d}x\rightarrow 0, \text {as} k\rightarrow \infty . \end{aligned}$$

  (2.3)

  By Step 1, we deduce that $\{u_k\}$ is bounded in $W^{1,2}_0(\Omega ).$ Therefore there exists $u_0\in W^{1,2}_0(\Omega )$ such that $u_k \rightharpoonup u_0$ in $W^{1,2}_0(\Omega )$ and $W^{1,p}_0(\Omega )$ and $u_k\rightarrow u_0$ in $L^2(\Omega ).$ Thus $\displaystyle {\int _{\Omega }}|\nabla u_0|^p\le \lim _{k\rightarrow \infty }\inf \displaystyle {\int _{\Omega }}|\nabla u_k|^p~\mathrm{d}x=0.$ And consequently $u_0=0$, $u_k \rightharpoonup 0$ in $W^{1,2}_0(\Omega )$ and $W^{1,p}_0(\Omega )$ and $u_k\rightarrow 0$ in $L^2(\Omega ).$ Writing again $v_k=\frac{ u_k}{\Vert u_k\Vert _2}$ we have

  $$\begin{aligned} 0<\frac{\lambda \displaystyle {\int _{\Omega }}u^2_k~\mathrm{d}x-\int _{\Omega }\left| \nabla u_k\right| ^2~\mathrm{d}x}{\Vert u_k\Vert ^2_{2}}=\Vert u_k\Vert ^{p-2}_{2} \displaystyle {\int _{\Omega }}\left| \nabla v_k\right| ^p~\mathrm{d}x, \end{aligned}$$

  therefore,

  $$\begin{aligned} \int _{\Omega }\left| \nabla v_k\right| ^p~\mathrm{d}x= & {} \Vert u_k\Vert ^{2-p}_{2}\left( \frac{\lambda \Vert u_k\Vert ^2_2}{\Vert u_k\Vert ^2_{2}}-\frac{\displaystyle {\int _{\Omega }}\left| \nabla u_k\right| ^2~\mathrm{d}x}{\Vert u_k\Vert ^2_{2}}\right) \\= & {} \Vert u_k\Vert ^{2-p}_{2}\left( \lambda - \displaystyle {\int _{\Omega }}\left| \nabla v_k\right| ^2~\mathrm{d}x\right) \rightarrow 0~ \mathrm {as}\, k\rightarrow \infty , \end{aligned}$$

  since $\Vert u_k\Vert _{2}\rightarrow 0$ and $p\in (1,2)$, and $\{v_k\}$ is bounded in $W_0^{1,2}(\Omega ).$ Next since $v_k \rightharpoonup v_0$, we deduce that $\displaystyle {\int _{\Omega }}|\nabla v_0|^p~\mathrm{d}x\le \lim _{k\rightarrow \infty }\inf \displaystyle {\int _{\Omega }}|\nabla v_k|^p\mathrm{d}x=0$ and we have $v_0=0.$ And it follows that $v_k\rightarrow 0$ in $L^2(\Omega )$ which is a contradiction since $\Vert v_k\Vert _{2}=1$ for each k. Hence, m is positive.

• Step 3. There exists $u_0\in \mathcal {N}_{\lambda }$ such that $J_{\lambda }(u_0)=m.$ Let $\{u_k\}\subset \mathcal {N}_{\lambda }$ be a minimizing sequence, i.e., $J_{\lambda }(u_k)\rightarrow m$ as $k\rightarrow \infty .$ Thanks to Step 1, we have that $\{u_k\}$ is bounded in $W_0^{1,2}(\Omega ).$ It follows that there exists $u_0\in W_0^{1,2}(\Omega )$ such that $u_k\rightharpoonup u_0$ in $W_0^{1,2}(\Omega )$ and $W_0^{1,p}(\Omega )$ and strongly in $L^2(\Omega ).$ The results in the two steps above guarantee that $J_{\lambda }(u_0)\le \displaystyle {\lim _{k\rightarrow \infty }}\inf J_{\lambda }(u_k)=m.$ Since for each k we have $u_k\in \mathcal {N}_{\lambda }$, we have

  $$\begin{aligned} \int _{\Omega }\left| \nabla u_k\right| ^2~\mathrm{d}x+\int _{\Omega }\left| \nabla u_k\right| ^p~\mathrm{d}x=\lambda \int _{\Omega }u^2_k~\mathrm{d}x~~~\text {for all}\, k. \end{aligned}$$

  (2.4)

  Assuming $u_0\equiv 0$ on $\Omega$ implies that $\displaystyle {\int _{\Omega }}u^2_k~\mathrm{d}x\rightarrow 0$ as $k\rightarrow \infty$, and by relation (2.4) we obtain that $\displaystyle {\int _{\Omega }}|\nabla u_k|^2~\mathrm{d}x\rightarrow 0$ as $k\rightarrow \infty .$ Combining this with the fact that $u_k$ converges weakly to 0 in $W_0^{1,2}(\Omega )$, we deduce that $u_k$ converges strongly to 0 in $W_0^{1,2}(\Omega )$ and consequently in $W_0^{1,p}(\Omega )$. Hence we infer that

  $$\begin{aligned} \lambda \int _{\Omega }u^2_k~\mathrm{d}x- \int _{\Omega }\left| \nabla u_k\right| ^2~\mathrm{d}x=\int _{\Omega }\left| \nabla u_k\right| ^p~\mathrm{d}x\rightarrow 0, \text {as} k\rightarrow \infty . \end{aligned}$$

  Next, using similar argument as the one used in the proof of Step 2, we will reach to a contradiction, which shows that $u_0\not \equiv 0.$ Letting $k\rightarrow \infty$ in relation (2.4), we deduce that

  $$\begin{aligned} \int _{\Omega }|\nabla u_0|^2~\mathrm{d}x+\int _{\Omega }|\nabla u_0|^p~\mathrm{d}x\le \lambda \int _{\Omega }u_0^2~\mathrm{d}x. \end{aligned}$$

  If there is equality in the above relation, then $u_0\in \mathcal {N}_{\lambda }$ and $m\le J_{\lambda }(u_0)$. Assume by contradiction that

  $$\begin{aligned} \int _{\Omega }|\nabla u|^2~\mathrm{d}x+\int _{\Omega }|\nabla u|^p~\mathrm{d}x<\lambda \int _{\Omega }u^2~\mathrm{d}x. \end{aligned}$$

  (2.5)

  Let $t>0$ be such that $tu_0\in \mathcal {N}_{\lambda },$ i.e.,

  $$\begin{aligned} t=\left( \frac{\lambda \displaystyle {\int _{\Omega }}u_0^2~\mathrm{d}x- \displaystyle {\int _{\Omega }}\left| \nabla u_0\right| ^2~\mathrm{d}x}{\displaystyle {\int _{\Omega }}\left| \nabla u_0\right| ^p~\mathrm{d}x}\right) ^{\frac{1}{p-2}}. \end{aligned}$$

  We note that $t\in (0,1)$ since $1<t^{p-2}$ (thanks to (2.5)). Finally, since $tu_0\in \mathcal {N}_{\lambda }$ with $t\in (0,1)$ we have

  $$\begin{aligned}&0<m\le J_{\lambda }(tu_0)\\&\quad =\left( \frac{2}{p}-1\right) \int _{\Omega }\left| \nabla (tu_0)\right| ^p~\mathrm{d}x=t^p\left( \frac{2}{p}-1\right) \int _{\Omega }\left| \nabla u_0\right| ^p~\mathrm{d}x\\&\quad =t^p J_{\lambda }(u_0)\\&\quad \le t^p\lim _{k\rightarrow \infty }\inf J_{\lambda }(u_k)=t^p m<m~\text {for}\, t\in (0,1), \end{aligned}$$

  and this is a contradiction which assures that relation (2.5) cannot hold and consequently we have $u_0\in \mathcal {N}_{\lambda }$. Hence $m\le J_{\lambda }(u_0)$ and $m= J_{\lambda }(u_0)$.

• Step 4. We conclude the proof of the claim. Let $u\in \mathcal {N}_{\lambda }$ be such that $J_{\lambda }(u)=m$ (thanks to Step 3). Since $u\in \mathcal {N}_{\lambda }$, we have

  $$\begin{aligned} \int _{\Omega }|\nabla u|^2~\mathrm{d}x+\int _{\Omega }|\nabla u|^p~\mathrm{d}x=\lambda \int _{\Omega }u^2~\mathrm{d}x, \end{aligned}$$

  and

  $$\begin{aligned} \int _{\Omega }|\nabla u|^2~\mathrm{d}x<\lambda \int _{\Omega }u^2~\mathrm{d}x. \end{aligned}$$

  Let $v\in \partial B_1(0)\subset W_0^{1,2}(\Omega )$ and $\varepsilon >0$ be very small such that $u+\delta v\ne 0$ in $\Omega$ for all $\delta \in (-\varepsilon ,\varepsilon )$ and

  $$\begin{aligned} \int _{\Omega }\left| \nabla (u+\delta v)\right| ^2~\mathrm{d}x<\lambda \int _{\Omega }\left( u+\delta v\right) ^2~\mathrm{d}x ; \end{aligned}$$

  this is equivalent to

  $$\begin{aligned}&\qquad \lambda \int _{\Omega }u^2~\mathrm{d}x-\int _{\Omega }|\nabla u|^2~\mathrm{d}x>\delta \left( 2\int _{\Omega }\nabla u\cdot \nabla v~\mathrm{d}x- 2\lambda \int _{\Omega }u v~\mathrm{d}x\right) \\&\qquad \quad +\delta ^2\left( \int _{\Omega }|\nabla v|^2~\mathrm{d}x-\lambda \int _{\Omega }v^2~\mathrm{d}x\right) , \end{aligned}$$

  which holds true for $\delta$ small enough since the left-hand side is positive while the function

  $$\begin{aligned} h(\delta ):= & {} |\delta |\left| 2\int _{\Omega }\nabla u\cdot \nabla v~\mathrm{d}x- 2\lambda \int _{\Omega }u v~\mathrm{d}x\right| \\&+\,\delta ^2\left| \int _{\Omega }|\nabla v|^2~\mathrm{d}x-\lambda \int _{\Omega }v^2~\mathrm{d}x\right| \end{aligned}$$

  dominates the term from the right-hand side and $h(\delta )$ is a continuous function (polynomial in $\delta$) which vanishes in $\delta =0$. For each $\delta \in (-\varepsilon ,\varepsilon ),$ let $t(\delta )>0$ be given by

  $$\begin{aligned} t(\delta )=\left( \frac{\lambda \displaystyle {\int _{\Omega }}\left( u+\delta v\right) ^2~\mathrm{d}x- \int _{\Omega }\left| \nabla (u+\delta v)\right| ^2~\mathrm{d}x}{\displaystyle {\int _{\Omega }}\left| \nabla (u+\delta v)\right| ^p~\mathrm{d}x}\right) ^{\frac{1}{p-2}}, \end{aligned}$$

  so that $t(\delta )\cdot (u+\delta v)\in \mathcal {N}_{\lambda }.$ We have that $t(\delta )$ is of class $C^1(-\varepsilon ,\varepsilon )$ since $t(\delta )$ is the composition of some functions of class $C^1$. On the other hand, since $u\in \mathcal {N}_{\lambda }$ we have $t(0)=1.$ Define $\iota :(-\varepsilon ,\varepsilon )\rightarrow \mathbb {R}$  by $\iota (\delta )=J_{\lambda }(t(\delta )(u+\delta v))$ which is of class $C^1(-\varepsilon ,\varepsilon )$ and has a minimum at $\delta =0.$ We have

  $$\begin{aligned}&\iota '(\delta )=\left[ t'(\delta )(u+\delta v)+v t(\delta )\right] J'_{\lambda }\left( t(\delta )(u+\delta v)\right) \Rightarrow \nonumber \\&0=\iota '(0)=J'_{\lambda }\left( t(0)(u)\right) \left[ t'(0)u+v t(0)\right] =\langle J'_{\lambda }(u),v\rangle \end{aligned}$$

  since $t(0)=1$ and $t'(0)=0.$ This shows that every $\lambda \in (\lambda ^D_1,\infty )$ is an eigenvalue of problem (1.3).

$\square$

In the next theorem we consider the case $p>2.$ For similar results for the Neumann case, (see, [22]).

Theorem 2.4

For $p>2$, the set of first eigenvalues of problem (1.3) is given by $(\lambda ^D_1,\infty ).$

The proof of Theorem 2.4 will follow as a direct consequence of the lemmas proved below:

Lemma 2.5

Let

$$\begin{aligned} \lambda _1(p):=\inf _{u\in W^{1,p}_0 \backslash \{0\}}\left\{ \frac{\frac{1}{p}\displaystyle {\int _{\Omega }}|\nabla u|^p~\mathrm{d}x+\frac{1}{2}\int _{\Omega }|\nabla u|^2~\mathrm{d}x}{\frac{1}{2}\displaystyle {\int _{\Omega }}u^2~\mathrm{d}x}\right\} . \end{aligned}$$

Then $\lambda _1(p)=\lambda ^D_1,$ for all $p>2.$

Proof

We clearly have $\lambda _1(p)\ge \lambda ^D_1$ since a positive term is added. On the other hand, consider $u_n=\frac{1}{n} e_1$ (where $e_1$ is the first eigenfunction of $-\Delta$), we get

$$\begin{aligned} \lambda _1(p)\le \frac{\frac{1}{2 n^2}\int _{\Omega }\left| \nabla e_1\right| ^2\mathrm{d}x+\frac{1}{p n^p}\int _{\Omega }\left| \nabla e_1\right| ^p \mathrm{d}x}{\frac{1}{2 n^2}\int _{\Omega }|e_1|^2\mathrm{d}x}\rightarrow \lambda ^D_1~~\text {as}\, n\rightarrow \infty . \end{aligned}$$

$\square$

Lemma 2.6

For each $\lambda >0$, we have

$$\begin{aligned} \lim _{\Vert u\Vert _{1,p}\rightarrow \infty }\left( \frac{1}{2}\int _{\Omega }|\nabla u|^2~\mathrm{d}x+\frac{1}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x-\frac{\lambda }{2}\int _{\Omega }u^2~\mathrm{d}x\right) =\infty . \end{aligned}$$

Proof

Clearly,

$$\begin{aligned} \frac{1}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x+\frac{1}{2}\int _{\Omega }|\nabla u|^2~\mathrm{d}x\ge & {} \frac{1}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x.\\ \end{aligned}$$

On the one hand, using Poincaré’s inequality with $p=2$, we have $\displaystyle {\int _{\Omega }}u^2~\mathrm{d}x\le C_2(\Omega )\displaystyle {\int _{\Omega }}|\nabla u|^2~\mathrm{d}x, \forall u\in W^{1,p}_0(\Omega )\subset W^{1,2}_0(\Omega )$ and then applying the Hölder inequality to the right-hand side term of the previous estimate, we obtain

$$\begin{aligned} \int _{\Omega }|\nabla u|^2~\mathrm{d}x \le |\Omega |^{\frac{p-2}{p}}\Vert u\Vert ^2_{1,p}, \end{aligned}$$

so $\displaystyle {\int _{\Omega }}u^2~\mathrm{d}x\le D \Vert u\Vert ^2_{1,p},$ where $D=C_2(\Omega ) |\Omega |^{\frac{p-2}{p}}.$ Therefore, for $\lambda >0,$

$$\begin{aligned} \frac{1}{2}\int _{\Omega }|\nabla u|^2~\mathrm{d}x+\frac{1}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x-\frac{\lambda }{2}\int _{\Omega }u^2~\mathrm{d}x\ge C\Vert u\Vert ^p_{1,p}-\frac{\lambda }{2} D \Vert u\Vert ^2_{1,p}, \end{aligned}$$

(2.6)

and the the right-hand side of (2.6) tends to $\infty$, as $\Vert u\Vert _{1,p}\rightarrow \infty$, since $p>2.$ $\square$

Lemma 2.7

Every $\lambda \in (\lambda ^D_1,\infty )$ is a first eigenvalue of problem (1.3).

Proof

For each $\lambda >\lambda ^D_1$ define $F_{\lambda }: W^{1,p}_0(\Omega )\rightarrow \mathbb {R}$ by

$$\begin{aligned} F_{\lambda }(u)=\frac{1}{2}\int _{\Omega }|\nabla u|^2~\mathrm{d}x+\frac{1}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x-\frac{\lambda }{2}\int _{\Omega }u^2~\mathrm{d}x~~,\forall u\in W^{1,p}_0(\Omega ). \end{aligned}$$

Standard arguments show that $F_{\lambda }\in C^1(W^{1,p}_0(\Omega ),\mathbb {R})$ (see, [18]) with its derivative given by

$$\begin{aligned} \left\langle F'_{\lambda }(u),\phi \right\rangle =\int _{\Omega }\left( |\nabla u|^{p-2}+1\right) \nabla u\cdot \nabla \phi ~\mathrm{d}x-\lambda \int _{\Omega }u\phi ~\mathrm{d}x, \end{aligned}$$

for all $u,\phi \in W^{1,p}_0(\Omega ).$ Estimate (2.6) shows that $F_{\lambda }$ is coercive in $W^{1,p}_0(\Omega ).$ On the other hand, $F_{\lambda }$ is also weakly lower semi-continuous on $W^{1,p}_0(\Omega )$ since $F_{\lambda }$ is a continuous convex functional (see [4], Proposition 1.5.10 and Theorem 1.5.3) . Then we can apply a calculus of variations result, in order to obtain the existence of a global minimum point of $F_{\lambda },$ denoted by $\theta _{\lambda },$ i.e., $F_{\lambda }(\theta _{\lambda })={\min _{W^{1,p}_0(\Omega )}}F_{\lambda }.$ Note that for any $\lambda > \lambda ^D_1$ there exists $u_{\lambda }\in W^{1,p}_0(\Omega )$ such that $F_{\lambda }(u_{\lambda })< 0$ . Indeed, taking $u_{\lambda }=re_1,$ we have

$$\begin{aligned} F_{\lambda }(re_1)=\frac{r^2}{2}\left( \lambda _1^D-\lambda \right) +\frac{r^p}{p}\int _{\Omega }|\nabla e_1|^p~\mathrm{d}x<0~~~\text {for}~~r>0~~\text {small}. \end{aligned}$$

But then $F_{\lambda }(\theta _{\lambda })\le F_{\lambda }(u_{\lambda })< 0$, which means that $\theta _{\lambda }\in W^{1,p}_0(\Omega )\backslash \{0\}.$ On the other hand, we have $\langle F'_{\lambda }(\theta _{\lambda }),\phi \rangle =0,\forall \phi \in W^{1,p}_0(\Omega )$ ($\theta _{\lambda }$ is a critical point of $F_{\lambda }$) with $\theta _{\lambda }\in W^{1,p}_0(\Omega )\backslash \{0\}\subset W^{1,2}_0(\Omega ) \backslash \{0\}.$ Consequently each $\lambda > \lambda ^D_1$ is an eigenvalue of problem (1.3). $\square$

A similar result of Theorem 3.1 was proved in [17] in the case of the p-Laplacian.

3 Properties of Eigenfunctions and the Operator $-\Delta _p-\Delta$

3.1 Boundedness of the Eigenfunctions

We shall prove boundedness of eigenfunctions and use this fact to obtain $C^{1,\alpha }$ smoothness of all eigenfunctions of the quasi-linear problem (1.3). The latter result is due to [17, Theorem 4.4], which originates from [13, 26].

Theorem 3.1

Let $(u,\lambda )\in W^{1,p}_0(\Omega )\times \mathbb {R}^{\star }_+$ be an eigensolution of the weak formulation (1.4). Then $u\in L^{\infty }(\Omega ).$

Proof

By Morrey’s embedding theorem it suffices to consider the case $p\le N.$ Let us assume first that $u>0.$ For $M\ge 0$ define $w_M(x)=\min \{u(x),M\}$. Letting

$$\begin{aligned} g(x)= \left\{ \begin{array}{l} x~~\text {if}~x\le M\\ M~~\text {if}~x>M. \end{array} \right. \end{aligned}$$

(3.1)

we have $g\in C(\mathbb {R})$ piecewise smooth function with $g(0)=0.$ Since $u\in W^{1,p}_0(\Omega )$ and $g'\in L^{\infty }(\Omega ),$ then $g\circ u\in W^{1,p}_0(\Omega )$ and $w_M\in W^{1,p}_0(\Omega )\cap L^{\infty }(\Omega )$ (see, Theorem B.3 in [17]). For $k>0,$ define $\varphi =w^{kp+1}_M$, then $\nabla \varphi =(kp+1)\nabla w_Mw^{kp}_M$ and $\varphi \in W^{1,p}_0(\Omega )\cap L^{\infty }(\Omega ).$

Using $\varphi$ as a test function in (1.4), one obtains

$$\begin{aligned}&(kp+1)\left[ \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla w_M w^{kp}_M~\mathrm{d}x+\int _{\Omega }\nabla u\cdot \nabla w_M w^{kp}_M~\mathrm{d}x\right] \\&\quad =\lambda \int _{\Omega }u~ w^{kp+1}_M~\mathrm{d}x. \end{aligned}$$

On the other hand, using the fact that $w^{kp+1}_M\le u^{kp+1}$, it follows that

$$\begin{aligned}&(kp+1)\left[ \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla w_M w^{kp}_M~\mathrm{d}x+\int _{\Omega }\nabla u\cdot \nabla w_M w^{kp}_M~\mathrm{d}x\right] \\&\quad \le \lambda \int _{\Omega }|u|^{(k+1)p}~\mathrm{d}x. \end{aligned}$$

We have $\nabla (w^{k+1}_M)=(k+1)\nabla w_M w^k_M\Rightarrow |\nabla w^{k+1}_M|^p=(k+1)^pw^{kp}_M|\nabla w_M|^p.$ Since the integrals on the left are zero on $\{x : u(x)>M\}$ we can take $u=w_M$ in the previous inequality, and it follows that

$$\begin{aligned} (kp+1)\left[ \int _{\Omega }\left| \nabla w_M\right| ^p w^{kp}_M~\mathrm{d}x+\int _{\Omega }\left| \nabla w_M\right| ^2 w^{kp}_M~\mathrm{d}x\right] \le \lambda \int _{\Omega }|u|^{(k+1)p}~\mathrm{d}x. \end{aligned}$$

Replacing $|\nabla w_M|^p w^{kp}_M$  by $\frac{1}{(k+1)^p}|\nabla w^{k+1}_M|^p$, we have

$$\begin{aligned} \frac{kp+1}{(k+1)^p}\int _{\Omega }|\nabla w^{k+1}_M|^p~\mathrm{d}x+(kp+1)\int _{\Omega }|\nabla w_M|^2 w^{kp}_M~\mathrm{d}x\le \lambda \int _{\Omega }|u|^{(k+1)p}~\mathrm{d}x, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{kp+1}{(k+1)^p}\int _{\Omega }|\nabla w^{k+1}_M|^p~\mathrm{d}x\le \lambda \int _{\Omega }|u|^{(k+1)p}~\mathrm{d}x~ \end{aligned}$$

and then

$$\begin{aligned} \int _{\Omega }\left| \nabla w^{k+1}_M\right| ^p~\mathrm{d}x\le \left( \lambda \frac{(k+1)^p}{kp+1}\right) \int _{\Omega }|u|^{(k+1)p}~\mathrm{d}x. \end{aligned}$$

(3.2)

By Sobolev’s embedding theorem, there is a constant $c_1>0$ such that

$$\begin{aligned} \left\| w^{k+1}_M\right\| _{p^{\star }}\le c_1\left\| w^{k+1}_M\right\| _{1,p}, \end{aligned}$$

(3.3)

where $p^{\star }$ is the Sobolev critical exponent. Consequently, we have

$$\begin{aligned} \left\| w_M\right\| _{(k+1)p^{\star }}\le \left\| w^{k+1}_M\right\| ^{\frac{1}{k+1}}_{p^{\star }}, \end{aligned}$$

(3.4)

and, therefore,

$$\begin{aligned} \left\| w_M\right\| _{(k+1)p^{\star }}\le \left( c_1\left\| w^{k+1}_M\right\| _{1,p}\right) ^{\frac{1}{k+1}}=c_1^{\frac{1}{k+1}}\left\| w^{k+1}_M\right\| ^{\frac{1}{k+1}}_{1,p}. \end{aligned}$$

(3.5)

But by (3.2),

$$\begin{aligned} \left\| w^{k+1}_M\right\| _{1,p}\le \left( \lambda \frac{(k+1)^p}{kp+1}\right) ^{\frac{1}{p}}\Vert u\Vert ^{k+1}_{(k+1)p}, \end{aligned}$$

(3.6)

and we note that we can find a constant $c_2>0$ such that

$\left( \lambda \frac{(k+1)^p}{kp+1}\right) ^{\frac{1}{p\sqrt{k+1}}}\le c_2$, independently of k and consequently,

$$\begin{aligned} \left\| w_M\right\| _{(k+1)p^{\star }}\le c_1^{\frac{1}{k+1}}c_2^{\frac{1}{\sqrt{k+1}}}\Vert u\Vert _{(k+1)p}. \end{aligned}$$

(3.7)

Letting $M\rightarrow \infty$, Fatou’s lemma implies

$$\begin{aligned} \left\| u\right\| _{(k+1)p^{\star }}\le c_1^{\frac{1}{k+1}}c_2^{\frac{1}{\sqrt{k+1}}}\Vert u\Vert _{(k+1)p}. \end{aligned}$$

(3.8)

Choosing $k_1,$ such that $(k_1+1)p=p^{\star },$ then $\Vert u\Vert _{(k_1+1)p^{\star }}\le c_1^{\frac{1}{k_1+1}}c_2^{\frac{1}{\sqrt{k_1+1}}}\Vert u\Vert _{p^{\star }}$. Next we choose $k_2$ such that $(k_2+1)p=(k_1+1)p^{\star };$ then taking $k_2=k$ in inequality (3.8), it follows that

$$\begin{aligned} \Vert u\Vert _{(k_2+1)p^{\star }}\le c_1^{\frac{1}{k_2+1}}c_2^{\frac{1}{\sqrt{k_2+1}}}\Vert u\Vert _{(k_1+1)p^{\star }}. \end{aligned}$$

(3.9)

By induction we obtain

$$\begin{aligned} \Vert u\Vert _{(k_n+1)p^{\star }}\le c_1^{\frac{1}{k_n+1}}c_2^{\frac{1}{\sqrt{k_n+1}}}\Vert u\Vert _{(k_{n-1}+1)p^{\star }}, \end{aligned}$$

(3.10)

where the sequence $\{k_n\}$ is chosen such that $(k_n+1)p=(k_{n-1}+1)p^{\star }, k_0=0.$ One gets $k_n+1=(\frac{p^{\star }}{p})^n.$ As $\frac{p}{p^{\star }}<1,$ there is $C>0$ (which depends on $c_1$ and $c_2$) such that for any $n=1,2,\ldots$

$$\begin{aligned} \Vert u\Vert _{r_n}\le C\Vert u\Vert _{p^{\star }} \end{aligned}$$

(3.11)

with $r_n=(k_n+1)p^{\star }\rightarrow \infty$ as $n\rightarrow \infty .$ We note that (3.11) follows by iterating the previous inequality (3.10). We will indirectly show that $u\in L^{\infty }(\Omega ).$ Suppose $u\not \in L^{\infty }(\Omega ),$ then there exists $\varepsilon >0$ and a set A of positive measure in $\Omega$ such that $|u(x)|>C \Vert u\Vert _{p^{\star }}+\varepsilon =K,$ for all $x\in A.$ We then have,

$$\begin{aligned} \lim _{n\rightarrow \infty }\inf \Vert u\Vert _{r_n}{\ge } \lim _{n\rightarrow \infty }\inf \left( \int _{A} K^{r_n}\right) ^{1/r_n}{=}\lim _{n\rightarrow \infty }\inf K |A|^{1/r_n}=K>C \Vert u\Vert _{p^{\star }}, \end{aligned}$$

(3.12)

which contradicts (3.11). If u changes sign, we consider $u=u^+ -u^-$ where

$$\begin{aligned} u^+=\max \{u,0\}~~\text {and}~~u^-=\max \{-u,0\}. \end{aligned}$$

(3.13)

We have $u^+,u^-\in W^{1,p}_0(\Omega ).$ For each $M{>}0$ define $w_M{=}\min \{u^+(x),M\}$ and take again $\varphi =w^{kp+1}_M$ as a test function in (1.4). Proceeding the same way as above we conclude that $u^+\in L^{\infty }(\Omega )$. Similarly, we have $u^-\in L^{\infty }(\Omega )$. Therefore, $u=u^+ -u^-$ is in $L^{\infty }(\Omega ).$ $\square$

3.2 Simplicity of the Eigenvalues

We prove an auxiliary result which will imply uniqueness of the first eigenfunction.

Let

$$\begin{aligned} I(u,v)= & {} \left\langle -\Delta _p u,\frac{u^p-v^p}{u^{p-1}}\right\rangle +\left\langle -\Delta u,\frac{u^2-v^2}{u}\right\rangle \\&+\left\langle -\Delta _p v,\frac{v^p-u^p}{v^{p-1}}\right\rangle +\left\langle -\Delta v,\frac{v^2-u^2}{v}\right\rangle ,\nonumber \end{aligned}$$

for all $(u,v)\in D_I,$ where

$$\begin{aligned} D_I{=}\left\{ (u_1,u_2)\in W^{1,p}_0(\Omega )\times W^{1,p}_0(\Omega ):u_i> 0 ~\text {in}\, \Omega ~\text {and}\,u_i\in L^{\infty }(\Omega )~\text {for}\,~i=1,2\right\} ~\text {if}\, p>2, \end{aligned}$$

and

$$\begin{aligned} D_I{=}\left\{ (u_1,u_2){\in } W^{1,2}_0(\Omega )\times W^{1,2}_0(\Omega ):~u_i> 0 \text {in}\, \Omega ~\text {and}\,u_i\in L^{\infty }(\Omega )~\text {for}\, i=1,2\right\} ~\text {if}\, 1{<}p{<}2. \end{aligned}$$

Proposition 3.2

For all $(u,v)\in D_I$, we have $I(u,v)\ge 0.$ Furthermore, $I(u,v)=0$ if and only if there exists $\alpha \in \mathbb {R}^{\star }_+$ such that $u=\alpha v.$

Proof

We first show that $I(u,v)\ge 0.$ We recall that (if $2<p<\infty$)

$$\begin{aligned} \left\langle -\Delta _p u,w\right\rangle= & {} \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla w~\mathrm{d}x~~~\text {for all}\, w\in W^{1,p}_0(\Omega )\nonumber \\ \langle -\Delta u,w\rangle= & {} \int _{\Omega }\nabla u\cdot \nabla w~\mathrm{d}x~~~\text {for all}\, w\in W^{1,p}_0(\Omega ), \end{aligned}$$

and (if $1<p<2$)

$$\begin{aligned} \langle -\Delta _p u,w\rangle= & {} \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla w~\mathrm{d}x~~~\text {for all}\, w\in W^{1,2}_0(\Omega \nonumber \\ \langle -\Delta u,w\rangle= & {} \int _{\Omega }\nabla u\cdot \nabla w~\mathrm{d}x~~~\text {for all}\, w\in W^{1,2}_0(\Omega ). \end{aligned}$$

Let us consider $\beta =\frac{u^p-v^p}{u^{p-1}}$, $\eta =\frac{v^p-u^p}{v^{p-1}}$, $\xi =\frac{u^2-v^2}{u}$ and $\zeta =\frac{v^2-u^2}{v}$ as test functions in (1.4) for any $p>1$. Straightforward computations give

$$\begin{aligned} \nabla \left( \frac{u^p-v^p}{u^{p-1}}\right)= & {} \left\{ 1+(p-1)\left( \frac{v}{u}\right) ^p\right\} \nabla u-p\left( \frac{v}{u}\right) ^{p-1}\nabla v\nonumber \\ \nabla \left( \frac{v^p-u^p}{v^{p-1}}\right)= & {} \left\{ 1+(p-1)\left( \frac{u}{v}\right) ^p\right\} \nabla v-p\left( \frac{u}{v}\right) ^{p-1}\nabla u\nonumber \\ \nabla \left( \frac{u^2-v^2}{u}\right)= & {} \left\{ 1+\left( \frac{v}{u}\right) ^2\right\} \nabla u-2\left( \frac{v}{u}\right) \nabla v\\ \nabla \left( \frac{v^2-u^2}{v}\right)= & {} \left\{ 1+\left( \frac{u}{v}\right) ^2\right\} \nabla v-2\left( \frac{u}{v}\right) \nabla u. \end{aligned}$$

Therefore,

$$\begin{aligned}&\left\langle -\Delta _p u,\frac{u^p-v^p}{u^{p-1}}\right\rangle \\&\quad =\int _{\Omega }\left\{ -p\left( \frac{v}{u}\right) ^{p-1}|\nabla u|^{p-2}\nabla u\cdot \nabla v+\left( 1+(p-1)\left( \frac{v}{u}\right) ^p\right) |\nabla u|^p\right\} ~\mathrm{d}x\\&\quad = \int _{\Omega }\left\{ p\left( \frac{v}{u}\right) ^{p-1}|\nabla u|^{p-2}\left( |\nabla u||\nabla v|-\nabla u\cdot \nabla v\right) +\left( 1+(p-1)\left( \frac{v}{u}\right) ^p\right) |\nabla u|^p\right\} ~\mathrm{d}x\\&\qquad -\int _{\Omega } p\left( \frac{v}{u}\right) ^{p-1}|\nabla u|^{p-1}|\nabla v|~\mathrm{d}x \end{aligned}$$

and

$$\begin{aligned}&\left\langle -\Delta u,\frac{u^2-v^2}{u}\right\rangle \\&\quad =\int _{\Omega }\left\{ 2\left( \frac{v}{u}\right) \left( |\nabla u||\nabla v|-\nabla u\cdot \nabla v\right) +\left( 1+\left( \frac{v}{u}\right) ^2\right) |\nabla u|^2-2\left( \frac{v}{u}\right) |\nabla u||\nabla v|\right\} ~\mathrm{d}x. \end{aligned}$$

By symmetry we have

$$\begin{aligned}&\left\langle -\Delta _p v,\frac{v^p-u^p}{v^{p-1}}\right\rangle \\&\quad =\int _{\Omega }\left\{ -p\left( \frac{u}{v}\right) ^{p-1}|\nabla v|^{p-2}\nabla v\cdot \nabla u+\left( 1+(p-1)\left( \frac{u}{v}\right) ^p\right) |\nabla v|^p\right\} ~\mathrm{d}x\\&\quad =\int _{\Omega }\left\{ p\left( \frac{u}{v}\right) ^{p-1}|\nabla v|^{p-2}\left( |\nabla v||\nabla u|-\nabla v\cdot \nabla u\right) +\left( 1+(p-1)\left( \frac{u}{v}\right) ^p\right) |\nabla v|^p\right\} ~\mathrm{d}x\\&\qquad - \int _{\Omega }p\left( \frac{u}{v}\right) ^{p-1}|\nabla v|^{p-1}|\nabla u|~\mathrm{d}x \end{aligned}$$

and

$$\begin{aligned} \left\langle -\Delta v,\frac{v^2-u^2}{v}\right\rangle= & {} \int _{\Omega }\left\{ 2\left( \frac{u}{v}\right) \left( |\nabla v||\nabla u|{-}\nabla v{\cdot } \nabla u\right) \right. \\&\left. +\left( 1+\left( \frac{u}{v}\right) ^2\right) |\nabla v|^2-2\left( \frac{u}{v}\right) |\nabla v||\nabla u|\right\} ~\mathrm{d}x. \end{aligned}$$

Thus

$$\begin{aligned} I(u,v)= & {} \int _{\Omega }\left\{ p\left( \frac{v}{u}\right) ^{p-1}|\nabla u|^{p-2}\left( |\nabla u||\nabla v|-\nabla u\cdot \nabla v\right) +\left( 1+(p-1)\left( \frac{v}{u}\right) ^p\right) |\nabla u|^p\right\} ~\mathrm{d}x\\&- p\left( \frac{v}{u}\right) ^{p-1}|\nabla u|^{p-1}|\nabla v|~\mathrm{d}x\\&+\int _{\Omega }\left\{ p\left( \frac{u}{v}\right) ^{p-1}|\nabla v|^{p-2}\left( |\nabla v||\nabla u|-\nabla v\cdot \nabla u\right) +\left( 1+(p-1)\left( \frac{u}{v}\right) ^p\right) |\nabla v|^p\right\} ~\mathrm{d}x\\&- p\left( \frac{u}{v}\right) ^{p-1}|\nabla v|^{p-1}|\nabla u|~\mathrm{d}x\\&+\int _{\Omega }\left\{ 2\left( \frac{v}{u}\right) \left( |\nabla u||\nabla v|-\nabla u\cdot \nabla v\right) +\left( 1+\left( \frac{v}{u}\right) ^2\right) |\nabla u|^2-2\left( \frac{v}{u}\right) |\nabla u||\nabla v|\right\} ~\mathrm{d}x\\&+ \int _{\Omega }\left\{ 2\left( \frac{u}{v}\right) \left( |\nabla v||\nabla u|-\nabla v\cdot \nabla u\right) +\left( 1+\left( \frac{u}{v}\right) ^2\right) |\nabla v|^2-2\left( \frac{u}{v}\right) |\nabla v||\nabla u|\right\} ~\mathrm{d}x. \end{aligned}$$

So

$$\begin{aligned} I(u,v)=\int _{\Omega }F\left( \frac{v}{u},\nabla v, \nabla u\right) ~\mathrm{d}x+\int _{\Omega }G\left( \frac{v}{u},|\nabla v|, |\nabla u|\right) ~\mathrm{d}x, \end{aligned}$$

where

$$\begin{aligned} F(t,S,R)= & {} p\left\{ t^{p-1}|R|^{p-2}\left( |R||S|-R\cdot S\right) +t^{1-p}|S|^{p-2}\left( |R||S|-R\cdot S\right) \right\} \\&+2\left\{ t\left( |R||S|-R\cdot S\right) \right\} + 2\left\{ t^{-1}\left( |R||S|-R\cdot S\right) \right\} \end{aligned}$$

and

$$\begin{aligned} G(t,s,r)= & {} \left( 1+(p-1)t^p\right) r^p+ \left( 1+(p-1)t^{-p}\right) s^p+(1+t^2)r^2~~~\\&+(1+t^{-2})s^2-pt^{p-1}r^{p-1}s-pt^{1-p}s^{p-1}r-2trs-2t^{-1}rs, \end{aligned}$$

for all $t=\frac{v}{u}>0, R=\nabla u, S=\nabla v\in \mathbb {R}^N$ and $r=|\nabla u|, s=|\nabla v|\in \mathbb {R}^+.$ We clearly have that F is non-negative. Now let us show that G is non-negative. Indeed, we observe that

$$\begin{aligned} G(t,s,0)=\left( 1+(p-1)t^{-p}\right) s^p+\left( 1+t^{-2}\right) s^2\ge 0 \end{aligned}$$

and $G(t,s,0)=0\Rightarrow s=0.$ If $r\ne 0$, by setting $z=\frac{s}{tr}$ we obtain

$$\begin{aligned} G(t,s,r)= & {} t^pr^p\left( z^p-pz+(p-1)\right) +r^p\left( (p-1)z^p-pz^{p-1}+1\right) \nonumber \\&+t^2r^2\left( z^2-2z+1\right) +r^2\left( z^2-2z+1\right) , \end{aligned}$$

and G can be written as

$$\begin{aligned} G(t,s,r)=r^p\left( t^pf(z)+g(z)\right) +r^2\left( t^2h(z)+k(z)\right) , \end{aligned}$$

with $f(z)=z^p-pz+(p-1),$ $g(z)=(p-1)z^p-pz^{p-1}+1,$ $h(z)=k(z)=z^2-2z+1$ $\forall p>1 .$ We can see that f, g, h and k are non-negative. Hence G is non-negative and thus $I(u,v)\ge 0$ for all $(u,v)\in D_I.$ In addition since f, g, h and k vanish if and only if $z=1$, then $G(t,s,r)=0$ if and only if $s=tr.$ Consequently, if $I(u,v)=0$ then we have

$$\begin{aligned} \nabla u\cdot \nabla v=|\nabla u||\nabla v|~~\text {and}~~u|\nabla v|=v|\nabla u| \end{aligned}$$

almost everywhere in $\Omega .$ This is equivalent to $\left( u\nabla v-v\nabla u\right) ^2=0,$ which implies that $u=\alpha v$ with $\alpha \in \mathbb {R}^{\star }_+.$ $\square$

Theorem 3.3

The first eigenvalues $\lambda$ of Eq. (1.3) are simple, i.e., if u and v are two positive first eigenfunctions associated to $\lambda ,$ then $u=v.$

Proof

By Proposition 3.2, we have $u=\alpha v.$ Inserting this into the equation (1.3) implies that $\alpha =1.$ $\square$

3.3 Invertibility of the Operator $-\Delta _p-\Delta$

To simplify some notations, here we set $X=W^{1,p}_0(\Omega )$ and its dual $X^{\star }=W^{-1,p'}(\Omega ),$ where $\frac{1}{p}+\frac{1}{p'}=1.$

For the proof of the following lemma, we refer to [19]:

Lemma 3.4

Let $p>2$. Then there exist two positive constants $c_1, c_2$ such that, for all $x_1,x_2\in \mathbb {R}^n,$ we have the following:

• (i) $(x_2-x_1)\cdot (|x_2|^{p-2}x_2-|x_1|^{p-2}x_1)\ge c_1|x_2-x_1|^p$

• (ii) $\left| |x_2|^{p-2}x_2-|x_1|^{p-2}x_1\right| \le c_2(|x_2|+|x_1|)^{p-2}|x_2-x_1|$

Proposition 3.5

For $p>2,$ the operator $-\Delta _p-\Delta$ is a global homeomorphism.

The proof is based on the previous Lemma 3.4.

Proof

Define the nonlinear operator $A:X\rightarrow X^{\star }$ by

$\langle Au,v\rangle =\displaystyle {\int _{\Omega }}\nabla u\cdot \nabla v~\mathrm{d}x+\displaystyle {\int _{\Omega }}|\nabla u|^{p-2}\nabla u\cdot \nabla v~\mathrm{d}x$ for all $u,v\in X.$

To show that $-\Delta _p-\Delta$ is a homeomorphism, it is enough to show that A is a continuous strongly monotone operator, (see [9, Corollary 2.5.10]).

For $p>2,$ for all $u, v\in X$, by (i), we get

$$\begin{aligned}&\langle Au-Av,u-v\rangle \\&\quad = \int _{\Omega }|\nabla (u-v)|^2\mathrm{d}x+\int _{\Omega }\left( |\nabla u|^{p-2}\nabla u-|\nabla v|^{p-2}\nabla v\right) \cdot \nabla (u-v)~\mathrm{d}x\\&\quad \ge \int _{\Omega }|\nabla (u-v)|^2\mathrm{d}x+c_1\int _{\Omega }|\nabla (u-v)|^p\mathrm{d}x\\&\quad \ge c_1\Vert u-v\Vert ^p_{1,p}. \end{aligned}$$

Thus A is a strongly monotone operator.

We claim that A is a continuous operator from X to $X^{\star }.$ Indeed, assume that $u_n\rightarrow u$ in X. We have to show that $\Vert Au_n-Au\Vert _{X^{\star }}\rightarrow 0$ as $n\rightarrow \infty .$ Indeed, using (ii) and Hölder’s inequality and the Sobolev embedding theorem, one has

$$\begin{aligned}&\left| \langle Au_n-Au,w\rangle \right| \\&\quad \le \int _{\Omega }\left| |\nabla u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right| |\nabla w|~\mathrm{d}x+\int _{\Omega }|\nabla (u_n-u)||\nabla w|~\mathrm{d}x\\&\quad \le c_2 \int _{\Omega }\left( |\nabla u_n|+|\nabla u|\right) ^{p-2}|\nabla (u_n-u)||\nabla w|~\mathrm{d}x+\int _{\Omega }|\nabla (u_n-u)||\nabla w|~\mathrm{d}x\\&\quad \le c_2\left( \int _{\Omega }\left( |\nabla u_n|+|\nabla u|\right) ^{p}\mathrm{d}x\right) ^{p-2/p}\left( \int _{\Omega }|\nabla (u_n-u)|^p \mathrm{d}x\right) ^{1/p}\\&\qquad \times \left( \int _{\Omega }|\nabla w|^p \mathrm{d}x\right) ^{1/p} + c_3\Vert u_n-u\Vert _{1,2}\Vert w\Vert _{1,2}\\&\quad \le c_4\left( \Vert u_n\Vert _{1,p}+\Vert u\Vert _{1,p}\right) ^{p-2}\Vert u_n-u\Vert _{1,p}\Vert w\Vert _{1,p}+c_5\Vert u_n-u\Vert _{1,p}\Vert w\Vert _{1,p}. \end{aligned}$$

Thus $\Vert Au_n-Au\Vert _{X^{\star }}\rightarrow 0$, as $n\rightarrow +\infty ,$ and hence A is a homeomorphism. $\square$

4 Bifurcation of Eigenvalues

In the next subsection we show that for Eq. (1.3) there is a branch of first eigenvalues bifurcating from $(\lambda _1^D, 0)\in \mathbb {R}^+\times W^{1,p}_0(\Omega ).$

4.1 Bifurcation from Zero: The Case $p>2$

By Proposition 3.5, Eq. (1.3) is equivalent to

$$\begin{aligned} u=\lambda (-\Delta _p-\Delta )^{-1}u~~~\text {for}~~~u\in W^{-1,p'}(\Omega ). \end{aligned}$$

(4.1)

We set

$$\begin{aligned} S_{\lambda }(u)=u-\lambda (-\Delta _p-\Delta )^{-1}u, \end{aligned}$$

(4.2)

$~u{\in } L^2(\Omega )\subset W^{-1,p'}(\Omega )~\text {and}~\lambda {>}0.$ By $\Sigma {=}\{(\lambda ,u)\in \mathbb {R}^+\times W^{1,p}_0(\Omega )/~u\ne 0~,S_{\lambda }(u)=0 \}$, we denote the set of nontrivial solutions of (4.1).

A bifurcation point for (4.1) is a number $\lambda ^{\star }\in \mathbb {R}^+$ such that $(\lambda ^{\star },0)$ belongs to the closure of $\Sigma .$ This is equivalent to say that, in any neighborhood of $(\lambda ^{\star },0)$ in $\mathbb {R}^+\times W^{1,p}_0(\Omega )$, there exists a nontrivial solution of $S_{\lambda }(u)=0$.

Our goal is to apply the Krasnoselski bifurcation theorem [see, [1]].

Theorem 4.1

(Krasnoselski, 1964)

Let X be a Banach space and let $T\in C^1(X,X)$ be a compact operator such that $T(0)=0$ and $T'(0)=0.$ Moreover, let $A\in \mathcal {L}(X)$ also be compact. Then every characteristic value $\lambda ^{*}$ of A with odd (algebraic) multiplicity is a bifurcation point for $u=\lambda Au+T(u).$

We state our bifurcation result.

Theorem 4.2

Let $p>2.$ Then every eigenvalue $\lambda _k^D$ with odd multiplicity is a bifurcation point in $\mathbb {R}^+\times W^{1,p}_0(\Omega )$ of $S_{\lambda }(u)=0,$ in the sense that in any neighbourhood of $(\lambda ^D_k,0)$ in $\mathbb {R}^+\times W^{1,p}_0(\Omega )$ there exists a nontrivial solution of $S_{\lambda }(u)=0$.

Proof

We write the equation $S_{\lambda }(u)=0$ as

$$\begin{aligned} u=\lambda Au+T_{\lambda }(u), \end{aligned}$$

where $Au=(-\Delta )^{-1}u$ and $T_{\lambda }(u)=[(-\Delta _p-\Delta )^{-1}-(-\Delta )^{-1}](\lambda u)$, where we consider

$$\begin{aligned} (-\Delta _p-\Delta )^{-1}: L^2(\Omega )\subset W^{-1,p'}(\Omega )\rightarrow W^{1,p}_0(\Omega )\subset \subset L^2(\Omega ) \end{aligned}$$

and $(-\Delta )^{-1} : L^2(\Omega )\subset W^{-1,2}(\Omega )\rightarrow W^{1,2}_0(\Omega )\subset \subset L^2(\Omega ).$

For $p>2,$ the mapping

$$\begin{aligned} (-\Delta _p-\Delta )^{-1}-(-\Delta )^{-1}: L^2(\Omega )\subset W^{-1,p'}(\Omega )\rightarrow W^{1,p}_0(\Omega )\subset \subset L^2(\Omega ) \end{aligned}$$

is compact thanks to Rellich–Kondrachov theorem. We clearly have $A\in \mathcal {L}(L^2(\Omega ))$ and $T_{\lambda }(0)=0$. Now we have to show that

• (1) $T_{\lambda }\in C^1$.

• (2) $T_{\lambda }'(0)=0$.

In order to show (1) and (2), it suffices to show that

• (a) $-\Delta _p-\Delta : W^{1,p}_0(\Omega )\rightarrow W^{-1,p'}(\Omega )$ is continuously differentiable in a neighborhood $u\in W^{1,p}_0(\Omega )$.

• (b) $(-\Delta _p-\Delta )^{-1}$ is a continuous inverse operator.

According to Proposition 3.5, $-\Delta _p-\Delta$ is a homeomorphism; hence $(-\Delta _p-\Delta )^{-1}$ is continuous and this shows (b). We also recall that in section 3.2, we have shown that $\lambda ^D_1$ is simple.

Let us show (a). We claim that $-\Delta _p : W^{1,p}_0(\Omega )\rightarrow W^{-1,p'}(\Omega )$ is Gâteaux differentiable. Indeed, for $\varphi \in W^{1,p}_0(\Omega )$ we have

$$\begin{aligned}&\left\langle -\Delta _p(u+\delta v), \varphi \right\rangle -\left\langle -\Delta _pu,\varphi \right\rangle \\&\quad = \left\langle |\nabla (u+\delta v)|^{p-2}\nabla (u+\delta v),\nabla \varphi \right\rangle -\left\langle |\nabla u|^{p-2}\nabla u,\nabla \varphi \right\rangle \\&\quad = \left\langle \left( |\nabla (u+\delta v)|^2\right) ^{\frac{p-2}{2}}\nabla (u+\delta v),\nabla \varphi \rangle -\langle |\nabla u|^{p-2}\nabla u,\nabla \varphi \right\rangle \\&\quad = \left\langle \left( |\nabla u|^{2}+2\delta \left\langle \nabla u, \nabla v\right\rangle +\delta ^2|\nabla v|^2\right) ^{\frac{p-2}{2}}\nabla (u+\delta v),\nabla \varphi \right\rangle \\&\qquad - \langle |\nabla u|^{p-2}\nabla u,\nabla \varphi \rangle \\&\quad = \langle [|\nabla u|^{p-2}+(p-2)|\nabla u|^{2(\frac{p-2}{2}-1)}\delta \langle \nabla u,\nabla v\rangle \\&\qquad + O(\delta ^2)]\nabla (u+\delta v),\nabla \varphi \rangle - \langle |\nabla u|^{p-2}\nabla u,\nabla \varphi \rangle \\&\quad = \langle [|\nabla u|^{p-2}+(p-2)|\nabla u|^{p-4}\delta \langle \nabla u,\nabla v\rangle \\&\qquad +O(\delta ^2)]\nabla (u+\delta v),\nabla \varphi \rangle \\&\qquad - \langle |\nabla u|^{p-2}\nabla u,\nabla \varphi \rangle \\&\quad = (p-2)\delta |\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u, \nabla \varphi \rangle \\&\qquad +\delta \langle |\nabla u|^{p-2}\nabla v,\nabla \varphi \rangle +O(\delta ^2)\\&\quad = \delta [(p-2)|\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u,\nabla \varphi \rangle \\&\qquad +\langle |\nabla u|^{p-2}\nabla v,\nabla \varphi \rangle +O(\delta )]. \end{aligned}$$

Define

$$\begin{aligned} \langle B(u)v,\varphi \rangle =(p-2)|\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u,\nabla \varphi \rangle +\langle |\nabla u|^{p-2}\nabla v,\nabla \varphi \rangle \end{aligned}$$

and let $(u_n)_{n\ge 0}\subset W^{1,p}_0(\Omega ).$ Assume that $u_n\rightarrow u,$ as $n\rightarrow \infty$ in $W^{1,p}_0(\Omega ).$ We have

$$\begin{aligned}&\langle B(u_n)v-B(u)v,\varphi \rangle \\&\quad =(p-2)\left[ |\nabla u_n|^{p-4}\langle \nabla u_n,\nabla v\rangle \langle \nabla u_n,\nabla \varphi \rangle -|\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u,\nabla \varphi \rangle \right] \\&\qquad + \langle |\nabla u_n|^{p-2}\nabla v,\nabla \varphi \rangle -\langle |\nabla u|^{p-2}\nabla v,\nabla \varphi \rangle . \end{aligned}$$

Therefore,

$$\begin{aligned}&|\langle B(u_n)v-B(u)v,\varphi \rangle |\\&\quad \le (p-2) \left| |\nabla u_n|^{p-4}\langle \nabla u_n,\nabla v\rangle \langle \nabla u_n,\nabla \varphi \rangle -|\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u,\nabla \varphi \rangle \right| \\&\qquad + \left| |\nabla u_n|^{p-2}-|\nabla u|^{p-2}\right| |\langle \nabla v,\nabla \varphi \rangle |. \end{aligned}$$

By assumption, we can assume that, up to subsequences,

$(*)$:

$\nabla u_n\rightarrow \nabla u$ in $\left( L^p(\Omega )\right) ^N$ as $n\rightarrow \infty$ and

$(**)$:

$\nabla u_n(x)\rightarrow \nabla u(x)$ almost everywhere as $n\rightarrow \infty .$

Then $|\nabla u_n|^{p-4}\langle \nabla u_n,\nabla v\rangle \langle \nabla u_n,\nabla \varphi \rangle \rightarrow |\nabla u|^{p-4}\langle \nabla u,\nabla v\rangle \langle \nabla u,\nabla \varphi \rangle$ as $n \rightarrow \infty$ and consequently $\langle B(u_n)v,\varphi \rangle \rightarrow \langle B(u )v,\varphi \rangle$as $n\rightarrow \infty .$ Thus, we find that $-\Delta _p-\Delta \in C^1$ and thanks to the Inverse function theorem $(-\Delta _p-\Delta )^{-1}$ is differentiable in a neighborhood of $u\in W^{1,p}_0(\Omega )$. Therefore, according to the Krasnoselski bifurcation Theorem, we obtain that $\lambda _k^D$ is a bifurcation point at zero. $\square$

4.2 Bifurcation from Infinity: The Case $1<p<2$

We recall the nonlinear eigenvalue problem we are investigating

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _p u-\Delta u=\displaystyle \lambda u~~&{}\quad \text {in}\, \Omega ,\\ u = \displaystyle 0~~~~~~~~~~~~~~~~&{}\quad \text {on}\, \partial \Omega . \end{array} \right. \end{aligned}$$

(4.3)

Under a solution of (4.3) (for $1<p<2$), we understand a pair $(\lambda ,u)\in \mathbb {R}^+_{\star }\times W^{1,2}_0(\Omega )$ satisfying the integral equality

$$\begin{aligned} \int _{\Omega }|\nabla u|^{p-2}\nabla u\cdot \nabla \varphi ~\mathrm{d}x+\int _{\Omega }\nabla u\cdot \nabla \varphi ~\mathrm{d}x=\lambda \int _{\Omega }u\varphi ~\mathrm{d}x~~\text {for every}\, \varphi \in W^{1,2}_0(\Omega ). \end{aligned}$$

(4.4)

Definition 4.3

Let $\lambda \in \mathbb {R}.$ We say that the pair $(\lambda ,\infty )$ is a bifurcation point from infinity for problem (4.3) if there exists a sequence of pairs $\{(\lambda _n,u_n)\}_{n=1}^{\infty }\subset \mathbb {R}\times W^{1,p}_0(\Omega )$ such that Eq. (4.4) holds and $(\lambda _n,\Vert u_n\Vert _{1,2})\rightarrow (\lambda ,\infty ).$

We now state the main theorem concerning the bifurcation from infinity.

Theorem 4.4

The pair $(\lambda ^D_1,\infty )$ is a bifurcation point from infinity for the problem (4.3).

For $u\in W^{1,2}_0(\Omega ),~u\ne 0,$ we set $v=u/\Vert u\Vert _{1,2}^{2-\frac{1}{2}p}$. We have $\Vert v\Vert _{1,2}=\frac{1}{\Vert u\Vert _{1,2}^{1-\frac{1}{2}p}}$ and

$$\begin{aligned} |\nabla v|^{p-2}\nabla v=\frac{1}{\Vert u\Vert _{1,2}^{(2-\frac{1}{2}p)(p-1)}} |\nabla u|^{p-2}\nabla u. \end{aligned}$$

Introducing this change of variable in (4.4), we find that

$$\begin{aligned}&\Vert u\Vert _{1,2}^{\left( 2-\frac{1}{2}p\right) (p-2)}\int _{\Omega }|\nabla v|^{p-2}\nabla v\cdot \nabla \varphi ~\mathrm{d}x+\int _{\Omega }\nabla v\cdot \nabla \varphi ~\mathrm{d}x \nonumber \\&\quad =\lambda \int _{\Omega }v\varphi ~\mathrm{d}x~~\text {for every}\, \varphi \in W^{1,2}_0(\Omega ). \end{aligned}$$

(4.5)

But, on the other hand, we have

$$\begin{aligned} \Vert v\Vert ^{p-4}_{1,2}=\frac{1}{\Vert u\Vert _{1,2}^{\left( 1-\frac{1}{2}p\right) (p-4)}}=\frac{1}{\Vert u\Vert _{1,2}^{\left( 2-\frac{1}{2}p\right) (p-2)}}. \end{aligned}$$

Consequently, it follows that Eq. (4.5) is equivalent to

$$\begin{aligned} \Vert v\Vert _{1,2}^{4-p}\int _{\Omega }|\nabla v|^{p-2}\nabla v\cdot \nabla \varphi ~\mathrm{d}x+\int _{\Omega }\nabla v\cdot \nabla \varphi ~\mathrm{d}x=\lambda \int _{\Omega }v\varphi ~\mathrm{d}x~~\text {for every}\, \varphi \in W^{1,2}_0(\Omega ). \end{aligned}$$

(4.6)

This leads to the following nonlinear eigenvalue problem (for $1<p<2$):

$$\begin{aligned} \left\{ \begin{array}{ll} -\Vert v\Vert _{1,2}^{4-p}\Delta _p v-\Delta v =\lambda v~~&{}\quad \text {in}\,\, \Omega ,\\ v = \displaystyle 0~~~~~~~~~~~~~~~~&{}\quad \text {on}\,\, \partial \Omega . \end{array} \right. \end{aligned}$$

(4.7)

The proof of Theorem 4.4 follows immediately from the following remark, and the proof that $(\lambda _1^D,0)$ is a bifurcation of (4.7).

Remark 4.5

With this transformation, we have that the pair $(\lambda ^D_1,\infty )$ is a bifurcation point for the problem (4.3) if and only if the pair $(\lambda ^D_1,0)$ is a bifurcation point for the problem (4.7).

Let us consider a small ball $B_r(0) :=\{~w~\in W^{1,2}_0(\Omega )/~~~\Vert w\Vert _{1,2}< r~\},$ and consider the operator

$$\begin{aligned} T :=-\Vert \cdot \Vert _{1,2}^{4-p}\Delta _p-\Delta : W^{1,2}_0(\Omega )\rightarrow W^{-1,2}(\Omega ). \end{aligned}$$

Proposition 4.6

Let $1<p<2.$ There exists $r>0$ such that the mapping

$T : B_r(0)\subset W^{1,2}_0(\Omega )\rightarrow W^{-1,2}(\Omega )$ is invertible, with a continuous inverse.

Proof

In order to prove that the operator T is invertible with a continuous inverse, we again rely on [9, Corollary 2.5.10]. We show that there exists $\delta >0$ such that

$$\begin{aligned} \langle T(u)-T(v), u-v\rangle \ge \delta \Vert u-v\Vert ^2_{1,2}, ~~\text {for}~~u,v\in B_r(0)\subset W^{1,2}_0(\Omega ) \end{aligned}$$

with $r>0$ sufficiently small.

Indeed, using that $-\Delta _p$ is strongly monotone on $W^{1,p}_0(\Omega )$ on the one hand and the Hölder inequality on the other hand, we have

$$\begin{aligned} \nonumber&\langle T(u)-T(v), u-v\rangle \nonumber \\&\quad =\Vert \nabla u-\nabla v\Vert ^2_{2}+\left( \Vert u\Vert _{1,2}^{4-p}(-\Delta _pu)-\Vert v\Vert _{1,2}^{4-p}(-\Delta _pv), u-v \right) \nonumber \\&\quad = \Vert u-v\Vert ^2_{1,2}+\Vert u\Vert _{1,2}^{4-p}\left( (-\Delta _pu)-(-\Delta _p v), u-v\right) \nonumber \\&\qquad + \left( \Vert u\Vert _{1,2}^{4-p}-\Vert v\Vert _{1,2}^{4-p} \right) \left( -\Delta _pv, u-v\right) \nonumber \\&\quad \ge \Vert u-v\Vert ^2_{1,2}-\left| \Vert u\Vert _{1,2}^{4-p}-\Vert v\Vert _{1,2}^{4-p}\right| \Vert \nabla v\Vert _{p}^{p-1}\Vert \nabla (u-v)\Vert _{p}\nonumber \\&\quad \ge \Vert u-v\Vert ^2_{1,2}-\left| \Vert u\Vert _{1,2}^{4-p}-\Vert v\Vert _{1,2}^{4-p}\right| C\Vert v\Vert _{1,2}^{p-1}\Vert u-v\Vert _{1,2}. \end{aligned}$$

(4.8)

Now, we obtain by the Mean Value Theorem that there exists $\theta \in [0,1]$ such that

$$\begin{aligned} \left| \Vert u\Vert _{1,2}^{4-p}-\Vert v\Vert _{1,2}^{4-p}\right|= & {} \left| \frac{\mathrm{d}}{\mathrm{d}t}\left( \Vert u+t(v-u)\Vert ^2_{1,2}\right) ^{2-\frac{1}{2}p}|_{t=\theta } (v-u)\right| \\= & {} \left| (2-\frac{1}{2}p)\left( \Vert u+\theta (v-u)\Vert ^2_{1,2}\right) ^{1-\frac{1}{2}p}2\left( u+\theta (v-u),v-u\right) _{1,2}\right| \\\le & {} (4-p)\Vert u+\theta (v-u)\Vert ^{2-p}_{1,2}\Vert u+\theta (v-u)\Vert _{1,2}\Vert u-v\Vert _{1,2}\\= & {} (4-p)\Vert u+\theta (v-u)\Vert _{1,2}^{3-p}\Vert u-v\Vert _{1,2}\\\le & {} (4-p)\left( (1-\theta )\Vert u\Vert _{1,2}+\theta \Vert v\Vert _{1,2}\right) ^{3-p}\Vert u-v\Vert _{1,2}\\\le & {} (4-p) r^{3-p}\Vert u-v\Vert _{1,2}. \end{aligned}$$

Hence, continuing with the estimate of Eq. (4.8), we get

$$\begin{aligned}&\left\langle T(u)-T(v), u-v\right\rangle \ge \Vert u-v\Vert ^2_{1,2}\left( 1-(4-p) r^{3-p}Cr^{p-1}\right) \\&\quad =\Vert u-v\Vert ^2_{1,2}\left( 1-C'r^{2}\right) , \end{aligned}$$

and thus the claim, for $r>0$ small enough.

Hence, the operator T is strongly monotone on $B_r(0)$ and it is continuous, and hence the claim follows. $\square$

Clearly the mappings

$$\begin{aligned} T_{\tau }=-\Delta -\tau \Vert \cdot \Vert ^{\gamma }_{1,2}\Delta _p : B_r(0)\subset W^{1,2}_0(\Omega )\rightarrow W^{-1,2}(\Omega ),~~~0\le \tau \le 1 \end{aligned}$$

are also local homeomorphisms for $1<p<2$ with $\gamma =4-p>0$. Consider now the homotopy maps

$$\begin{aligned} H(\tau , y):=(-\tau \Vert \cdot \Vert _{1,2}^{\gamma }\Delta _p-\Delta )^{-1}(y),~~~y\in T_{\tau }(B_r(0))\subset W^{-1,2}(\Omega ). \end{aligned}$$

Then we can find a $\rho >0$ such that the ball

$$\begin{aligned} B_{\rho }(0)\subset \bigcap _{0\le \tau \le 1} T_{\tau }(B_r(0)) \end{aligned}$$

and

$$\begin{aligned} H(\tau ,\cdot ) : B_{\rho }(0)\cap L^2(\Omega )\mapsto W^{1,2}_0(\Omega )\subset \subset L^2(\Omega ) \end{aligned}$$

are compact mappings. Set now

$$\begin{aligned} \tilde{S}_{\lambda }(u)=u-\lambda \left( -\Vert u\Vert _{1,2}^{\gamma }\Delta _p-\Delta \right) ^{-1}u. \end{aligned}$$

Notice that $\tilde{S}_{\lambda }$ is a compact perturbation of the identity in $L^2(\Omega ).$ We have $0\notin H([0,1]\times \partial B_r(0) ).$ So it makes sense to consider the Leray–Schauder topological degree of $H(\tau ,\cdot )$ on $B_r(0).$ And by the property of the invariance by homotopy, one has

$$\begin{aligned} \deg (H(0,\cdot ),B_r(0),0)=\deg \left( H(1,\cdot ),B_r(0),0\right) . \end{aligned}$$

(4.9)

Theorem 4.7

The pair $(\lambda ^D_1,0)$ is a bifurcation point in $\mathbb {R}^+\times L^2(\Omega )$ of $\tilde{S}_{\lambda }(u)=0$, for $1<p<2.$

Proof

Suppose by contradiction that $(\lambda ^D_1,0)$ is not a bifurcation for $\tilde{S}_{\lambda }.$ Then, there exist $\delta _0>0$ such that for all $r\in (0,\delta _0)$ and $\varepsilon \in (0,\delta _0),$

$$\begin{aligned} \tilde{S}_{\lambda }(u)\ne 0~~~\forall ~|\lambda ^D_1-\lambda |\le \varepsilon ,~\forall ~u\in L^2(\Omega ),~\Vert u\Vert _{2}=r. \end{aligned}$$

(4.10)

Taking into account that (4.10) holds, it follows that it make sense to consider the Leray–Schauder topological degree $\deg (\tilde{S}_{\lambda }, B_r(0), 0)$ of $\tilde{S}_{\lambda }$ on $B_r(0).$

We observe that

$$\begin{aligned} \left( I-\left( \lambda _1^D-\varepsilon \right) H(\tau ,\cdot )\right) |_{\partial B_r(0)}\ne 0~~\text {for}~~ \tau \in [0,1]. \end{aligned}$$

(4.11)

Proving (4.11) guarantee the well posedness of $\deg (I{-}(\lambda ^D_1\pm \varepsilon )H(\tau ,\cdot ), B_r(0), 0)$ for any $\tau \in [0,1].$

Indeed, by contradiction suppose that there exists $v\in \partial B_r(0)\subset L^2(\Omega )$ such that

$v-\left( \lambda _1^D-\varepsilon \right) H(\tau ,v)=0,$ for some $\tau \in [0,1].$

One concludes that then $v\in W^{1,2}_0(\Omega ),$ and then that

$$\begin{aligned} -\Delta v-\tau \Vert v\Vert ^{\gamma }_{1,2}\Delta _pv=\left( \lambda ^D_1-\varepsilon \right) v. \end{aligned}$$

However, we get the contradiction,

$$\begin{aligned} \left( \lambda ^D_1-\varepsilon \right) \Vert v\Vert _2^2=\Vert \nabla v\Vert ^2_2+\tau \Vert v\Vert ^{\gamma }_{1,2}\Vert \nabla v\Vert ^p_p\ge \Vert \nabla v\Vert ^2_2\ge \lambda ^D_1\Vert v\Vert ^2_2. \end{aligned}$$

By the contradiction assumption, we have

$$\begin{aligned} \deg \left( I-(\lambda ^D_1+\varepsilon )H(1,\cdot ), B_r(0), 0\right) =\deg \left( I-(\lambda ^D_1-\varepsilon )H(1,\cdot ), B_r(0), 0\right) . \end{aligned}$$

(4.12)

By homotopy using (4.9), we have

$$\begin{aligned} \begin{aligned} \deg \left( I-(\lambda ^D_1-\varepsilon )H(1,\cdot ), B_r(0), 0\right)&=\deg \left( I-(\lambda ^D_1-\varepsilon )H(0,\cdot ), B_r(0), 0\right) \\&=\deg \left( I{-}(\lambda ^D_1-\varepsilon )({-}\Delta )^{-1}, B_r(0), 0\right) =1 \end{aligned} \end{aligned}$$

(4.13)

Now, using (4.13) and (4.12), we find that

$$\begin{aligned} \deg (I-(\lambda ^D_1+\varepsilon )H(1,\cdot ), B_r(0), 0)=\deg (I-(\lambda ^D_1-\varepsilon )H(0,\cdot ), B_r(0), 0)=1 \end{aligned}$$

(4.14)

Furthermore, since $\lambda ^D_1$ is a simple eigenvalue of $-\Delta ,$ it is well known [see [1]] that

$$\begin{aligned} \deg \left( I-(\lambda ^D_1+\varepsilon )(-\Delta )^{-1}, B_r(0\right) , 0)=\deg \left( I-(\lambda ^D_1+\varepsilon )H(0,\cdot ), B_r(0), 0\right) =-1 \end{aligned}$$

(4.15)

In order to get contradiction (to relation (4.14)), it is enough to show that

$$\begin{aligned} \deg \left( I-(\lambda ^D_1+\varepsilon )H(1,\cdot ), B_r(0\right) , 0)=\deg \left( I-(\lambda ^D_1+\varepsilon )H(0,\cdot ), B_r(0), 0\right) , \end{aligned}$$

(4.16)

$r>0$ sufficiently small. We have to show that

$$\begin{aligned} \left( I-(\lambda _1^D+\varepsilon )H(\tau ,\cdot )\right) |_{\partial B_r(0)}\ne 0~~~\text {for}~~\tau \in [0,1]. \end{aligned}$$

Suppose by contradiction that there is $r_n\rightarrow 0$, $\tau _n\in [0,1]$ and $u_n\in \partial B_{r_n}(0)$ such that

$$\begin{aligned} u_n-\left( \lambda _1^D+\varepsilon \right) H\left( \tau _n,u_n\right) =0 \end{aligned}$$

or equivalently

$$\begin{aligned} -\tau _n\Vert u_n\Vert ^{\gamma }_{1,2}\Delta _p u_n-\Delta u_n=\left( \lambda _1^D+\varepsilon \right) u_n. \end{aligned}$$

(4.17)

Dividing the Eq. (4.17) by $\Vert u_n\Vert _{1,2}$, we obtain

$$\begin{aligned} -\tau _n\Vert u_n\Vert ^{\gamma +p-1}_{1,2}\Delta _p\left( \frac{u_n}{\Vert u_n\Vert _{1,2}}\right) -\Delta \left( \frac{u_n}{\Vert u_n\Vert _{1,2}}\right) =(\lambda _1^D+\varepsilon )\frac{u_n}{\Vert u_n\Vert _{1,2}}, \end{aligned}$$

and by setting $v_n=\frac{u_n}{\Vert u_n\Vert _{1,2}},$ it follows that

$$\begin{aligned} -\tau _n\Vert u_n\Vert ^{\gamma +p-1}_{1,2}\Delta _p v_n-\Delta v_n=(\lambda _1^D+\varepsilon )v_n. \end{aligned}$$

(4.18)

But since $\Vert v_n\Vert _{1,2}=1,$ we have $v_n\rightharpoonup v$ in $W^{1,2}_0(\Omega )$ and $v_n\rightarrow v$ in $L^2(\Omega ).$ Furthermore, the first term in the left-hand side of Eq. (4.18) tends to zero in $W^{-1,p'}(\Omega )$ as $r_n\rightarrow 0$ and hence in $W^{-1,2}(\Omega ).$ Equation (4.17) then implies that $v_n\rightarrow v$ strongly in $W^{1,2}_0(\Omega )$ since $-\Delta : W^{1,2}_0(\Omega )\rightarrow W^{-1,2}(\Omega )$ is a homeomorphism and thus v with $\Vert v\Vert _{1,2}=1$ solves $-\Delta v=(\lambda _1^D+\varepsilon )v$, which is impossible because $\lambda _1^D+\varepsilon$ is not the first eigenvalue of $-\Delta$ on $W^{1,2}_0(\Omega )$ for $\varepsilon >0.$

Therefore, by homotopy it follows that

$$\begin{aligned} \deg \left( I-(\lambda ^D_1+\varepsilon )H(1,\cdot ), B_r(0\right) , 0)=\deg \left( I-(\lambda ^D_1+\varepsilon )H(0,\cdot ), B_r(0), 0\right) . \end{aligned}$$

Now, thanks to (4.15), we find that

$$\begin{aligned} \deg \left( I-(\lambda ^D_1+\varepsilon )H(1,\cdot ), B_r(0), 0\right) =-1, \end{aligned}$$

which contradicts Eq. (4.14). $\square$

Theorem 4.8

The pair $(\lambda ^D_k,0)$ ($k>1$) is a bifurcation point of $\tilde{S}_{\lambda }(u)=0$, for $1<p<2$ if $\lambda ^D_k$ is of odd multiplicity.

Proof

Suppose by contradiction that $(\lambda ^D_k,0)$ is not a bifurcation for $\tilde{S}_{\lambda }.$ Then, there exist $\delta _0>0$ such that for all $r\in (0,\delta _0)$ and $\varepsilon \in (0,\delta _0),$

$$\begin{aligned} \tilde{S}_{\lambda }(u)\ne 0~~~\forall ~|\lambda ^D_k-\lambda |\le \varepsilon ,~\forall ~u\in L^2(\Omega ),~\Vert u\Vert _{2}=r. \end{aligned}$$

(4.19)

Taking into account that (4.19) holds, it follows that it make sense to consider the Leray–Schauder topological degree $\deg (\tilde{S}_{\lambda }, B_r(0), 0)$ of $\tilde{S}_{\lambda }$ on $B_r(0).$

We show that

$$\begin{aligned} \left( I-\left( \lambda _k^D-\varepsilon \right) H(\tau ,\cdot )\right) |_{\partial B_r(0)}\ne 0~~\text {for}~~ \tau \in [0,1]. \end{aligned}$$

(4.20)

Proving (4.20) garantees the well posedness of $\deg (I-(\lambda ^D_k\pm \varepsilon )H(\tau ,\cdot ), B_r(0), 0)$ for any $\tau \in [0,1].$ Indeed, consider the projections $P^-$ and $P^+$ onto the spaces $\text {span}\{e_1,\dots ,e_{k-1}\}$ and $\text {span}\{e_k,e_{k+1},\dots \}$, respectively, where $e_1\dots ,e_k,e_{k+1},\dots$denote the eigenfunctions associated with the Dirichlet problem (1.1).

Suppose by contradiction that relation (4.20) does not hold. Then there exists $v\in \partial B_r(0)\subset L^2(\Omega )$ such that $v-(\lambda ^D_k-\varepsilon )H(\tau ,v)=0,$ for some $\tau \in [0,1].$ This is equivalent of having

$$\begin{aligned} -\Delta v-\left( \lambda ^D_k-\varepsilon \right) v=\tau \Vert v\Vert ^{\gamma }_{1,2}\Delta _p v. \end{aligned}$$

(4.21)

Replacing v by $P^+v+P^-v,$ and multiplying equation (4.21) by $P^+v-P^-v$ in the both sides, we obtain

$$\begin{aligned}&\left\langle [-\Delta -(\lambda ^D_k-\varepsilon )](P^+v+P^-v),P^+v-P^-v\right\rangle \\&\quad =\tau \Vert P^+v+P^-v\Vert ^{\gamma }_{1,2}\left\langle \Delta _p[P^+v+P^-v], P^+v-P^-v\right\rangle . \end{aligned}$$

$$\begin{aligned} \Updownarrow \end{aligned}$$

$$\begin{aligned}&\qquad -\left[ \Vert \nabla P^-v\Vert ^2_2-(\lambda ^D_k-\varepsilon )\Vert P^-v\Vert ^2_2\right] \\&\qquad +\Vert \nabla P^+v\Vert ^2_2-(\lambda ^D_k-\varepsilon )\Vert P^+v\Vert ^2_2 = \tau \Vert P^+v+P^-v\Vert ^{\gamma }_{1,2}\\&\qquad \times \langle \Delta _p[P^+v+P^-v], P^+v-P^-v\rangle . \end{aligned}$$

But

$$\begin{aligned}&\left\langle \Delta _p[P^+v+P^-v], P^+v-P^-v\right\rangle \\&\quad =-\int _{\Omega }\left| \nabla (P^+v+P^-v)\right| ^{p-2}\nabla (P^+v+P^-v)\cdot \nabla \left( P^+v-P^-v\right) ~\mathrm{d}x, \end{aligned}$$

and using the Hölder inequality, the embedding $W^{1,2}_0(\Omega )\subset W^{1,p}_0(\Omega )$ and the fact that $P^+v$ and $P^-v$ do not vanish simultaneously, there is some positive constant $C'>0$ such that $\Vert P^+v-P^-v\Vert _{1,2}\le C'(\Vert P^+v\Vert ^2_{1,2}+\Vert P^-v\Vert ^2_{1,2})=C'\Vert P^+v-P^-v\Vert ^2_{1,2},~~\text {since}~\left( P^+v,P^-v\right) _{1,2}=0,$ we have

$$\begin{aligned} \begin{aligned}&\left| \langle \Delta _p[P^+v+P^-v], P^+v-P^-v\rangle \right| \\&\quad \le \Vert P^+v+P^-v\Vert ^{p-1}_{1,p}\Vert P^+v-P^-v\Vert _{1,p}\\&\quad \le C'\Vert P^+v+P^-v\Vert ^{p-1}_{1,2}\Vert P^+v-P^-v\Vert ^{2}_{1,2}\\&\quad \le C'\Vert P^+v+P^-v\Vert ^{p+1}_{1,2},~~\text {since}~~\Vert P^+v-P^-v\Vert ^{2}_{1,2}\\&\quad =\Vert P^+v+P^-v\Vert ^{2}_{1,2}. \end{aligned} \end{aligned}$$

On the other hand, thanks to the Poincaré inequality as well as the variational characterization of eigenvalues we find

$$\begin{aligned} -\left[ \Vert \nabla P^-v\Vert ^2_2-(\lambda ^D_k-\varepsilon )\Vert P^-v\Vert ^2_2\right] \ge 0 \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla P^+v\Vert ^2_2-(\lambda ^D_k-\varepsilon )\Vert P^+v\Vert ^2_2\ge 0, \end{aligned}$$

we can bound from below these two inequalities together by $\Vert \nabla P^+v\Vert ^2_{2}+\Vert \nabla P^-v\Vert ^2_{2}.$

Finally, we have

$$\begin{aligned} \Vert v\Vert ^2_{1,2}=\Vert \nabla P^+v\Vert ^2_{2}+\Vert \nabla P^-v\Vert ^2_{2}\le \tau C'\Vert P^+v+P^-v\Vert ^{\gamma +p+1}_{1,2},~~\text {with}~~\gamma =4-p, \end{aligned}$$

$$\begin{aligned} \Updownarrow \end{aligned}$$

$$\begin{aligned} \Vert v\Vert ^2_{1,2}\le C''\Vert v\Vert ^{\gamma +p+1}_{1,2}\Leftrightarrow 1\le C'' r^{3}\rightarrow 0, \end{aligned}$$

for r taken small enough. This shows that (4.20) holds.

By the contradiction assumption, we have

$$\begin{aligned} \deg \left( I-(\lambda ^D_k+\varepsilon )H(1,\cdot ), B_r(0), 0\right) =\deg \left( I-(\lambda ^D_k-\varepsilon )H(1,\cdot ), B_r(0), 0\right) . \end{aligned}$$

(4.22)

By homotopy using (4.20), we have

$$\begin{aligned}&\deg \left( I-\left( \lambda ^D_k-\varepsilon \right) H(1,\cdot ), B_r(0), 0\right) \nonumber \\&\quad =\deg \left( I-\left( \lambda ^D_k-\varepsilon \right) H(0,\cdot ), B_r(0), 0\right) \nonumber \\&\quad =\deg \left( I-(\lambda ^D_k-\varepsilon )(-\Delta )^{-1}, B_r(0), 0\right) =(-1)^{\beta }, \end{aligned}$$

(4.23)

where $\beta$ is the sum of algebraic multiplicities of the eigenvalues $\lambda _k^D-\varepsilon < \lambda .$ Similarly, if $\beta '$ denotes the sum of the algebraic multiplicities of the characteristic values of $(-\Delta )^{-1}$ such that $\lambda >\lambda ^D_k+\varepsilon ,$ then

$$\begin{aligned} \deg \left( I-\left( \lambda ^D_k+\varepsilon \right) H(1,\cdot ), B_r(0), 0\right) =(-1)^{\beta '} \end{aligned}$$

(4.24)

But since $[\lambda ^D_k-\varepsilon ,\lambda ^D_k+\varepsilon ]$ contains only the eigenvalue $\lambda ^D_k,$ it follows that $\beta '=\beta +\alpha ,$ where $\alpha$ denotes the algebraic multiplicity of $\lambda ^D_k.$ Consequently, we have

$$\begin{aligned}&\deg \left( I-\left( \lambda ^D_k+\varepsilon \right) H(1,\cdot ), B_r(0), 0\right) \\&\quad =(-1)^{\beta +\alpha }\\&\quad =(-1)^{\alpha }\deg \left( I-\left( \lambda ^D_k+\varepsilon \right) H(1,\cdot ), B_r(0), 0\right) \\&\quad = -\deg \left( I-\left( \lambda ^D_k+\varepsilon \right) H(1,\cdot ), B_r(0), 0\right) , \end{aligned}$$

since $\lambda ^D_k$ is with odd multiplicity. This contradicts (4.22). $\square$

5 Multiple Solutions

In this section we prove multiplicity results by distinguishing again the two cases $1<p<2$ and $p>2$. We recall the following definition which will be used in this section. Let X be a Banach space and $\Omega \subset X$ an open bounded domain which is symmetric with respect to the origin of X,  that is, $u\in \Omega \Rightarrow -u\in \Omega .$ Let $\Gamma$ be the class of all the symmetric subsets $A\subseteq X\backslash \{0\}$ which are closed in $X\backslash \{0\}.$

Definition 5.1

(Krasnoselski genus) Let $A\in \Gamma$. The genus of A is the least integer $p\in \mathbb {N}^*$ such that there exists $\Phi : A\rightarrow \mathbb {R}^p$ continuous, odd and such that $\Phi (x)\ne 0$ for all $x\in A.$ The genus of A is usually denoted by $\gamma (A)$.

Theorem 5.2

Let $1<p<2$ or $2<p<\infty$, and suppose that $\lambda \in (\lambda ^D_k,\lambda ^D_{k+1})$ for any Then Eq. (1.3) has at least k pairs of nontrivial solutions.

Proof

Case 1: $1<p<2$. In this case we will avail of [ [1], Proposition 10.8]. We consider the energy functional $I_{\lambda }: W^{1,2}_0(\Omega )\backslash \{0\}\rightarrow \mathbb {R}$ associated with the problem (1.3) defined by

$$\begin{aligned} I_{\lambda }(u)=\frac{2}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x+\int _{\Omega }|\nabla u|^2~\mathrm{d}x-\lambda \int _{\Omega }u^2~\mathrm{d}x. \end{aligned}$$

The functional $I_{\lambda }$ is not bounded from below on $W^{1,2}_0(\Omega )$, so we consider again the natural constraint set, the Nehari manifold on which we minimize the functional $I_{\lambda }.$ The Nehari manifold is given by

$$\begin{aligned} \mathcal {N}_{\lambda }:=\left\{ u\in W^{1,2}_0(\Omega )\backslash \{0\right\} :~\langle I'_{\lambda }(u),u\rangle =0\}. \end{aligned}$$

On $\mathcal {N}_{\lambda },$ we have $I_{\lambda }(u)=(\frac{2}{p}-1)\displaystyle {\int _{\Omega }}|\nabla u|^p~\mathrm{d}x>0.$ We clearly have that $I_{\lambda }$ is even and bounded from below on $\mathcal {N}_{\lambda }.$

Now, let us show that every (PS) sequence for $I_{\lambda }$ has a converging subsequence on $\mathcal {N}_{\lambda }.$ Let $(u_n)_n$ be a (PS) sequence, i.e., $|I_{\lambda }(u_n)|\le C$, for all n, for some $C>0$ and $I'_{\lambda }(u_n)\rightarrow 0$ in $W^{-1,2}(\Omega )$ as $n\rightarrow +\infty .$ We first show that the sequence $(u_n)_n$ is bounded on $\mathcal {N}_{\lambda }$. Suppose by contradiction that this is not true, then $\displaystyle {\int _{\Omega }}|\nabla u_n|^2~\mathrm{d}x\rightarrow +\infty$ as $n\rightarrow +\infty .$ Since $I_{\lambda }(u_n)=(\frac{2}{p}-1)\displaystyle {\int }_{\Omega }|\nabla u_n|^p~\mathrm{d}x$ we have $\displaystyle {\int }_{\Omega }|\nabla u_n|^p~\mathrm{d}x\le c.$ On $\mathcal {N}_{\lambda }$, we have

$$\begin{aligned} 0<\int _{\Omega }|\nabla u_n|^p~\mathrm{d}x=\lambda \int _{\Omega }u_n^2~\mathrm{d}x-\int _{\Omega }|\nabla u_n|^2~\mathrm{d}x, \end{aligned}$$

(5.1)

and hence $\displaystyle {\int _{\Omega }}u_n^2~\mathrm{d}x\rightarrow +\infty .$ Let $v_n=\frac{u_n}{\Vert u_n\Vert _2};$ then $\displaystyle {\int _{\Omega }}|\nabla v_n|^2~\mathrm{d}x\le \lambda$ and hence $v_n$ is bounded in $W^{1,2}_0(\Omega ).$ Therefore, there exists $v_0\in W^{1,2}_0(\Omega )$ such that $v_n\rightharpoonup v_0$ in $W^{1,2}_0(\Omega )$ and $v_n\rightarrow v_0$ in $L^2(\Omega ).$ Dividing (5.1) by $\Vert u_n\Vert ^p_2,$ we have

$$\begin{aligned} \frac{\lambda \displaystyle {\int _{\Omega }}u_n^2~\mathrm{d}x-\int _{\Omega }|\nabla u_n|^2~\mathrm{d}x}{\Vert u_n\Vert ^p_2}=\int _{\Omega }|\nabla v_n|^p~\mathrm{d}x\rightarrow 0, \end{aligned}$$

since $\lambda \displaystyle {\int _{\Omega }}u_n^2~\mathrm{d}x-\int _{\Omega }|\nabla u_n|^2~\mathrm{d}x=(\frac{2}{p}-1)^{-1}I_{\lambda }(u_n)$, $|I_{\lambda }(u_n)|\le C$ and $\Vert u_n\Vert ^p_2\rightarrow +\infty .$ Now, since $v_n\rightharpoonup v_0$ in $W^{1,2}_0(\Omega )\subset W^{1,p}_0(\Omega ),$ we infer that

$$\begin{aligned} \int _{\Omega }|\nabla v_0|^p~\mathrm{d}x\le \liminf _{n\rightarrow +\infty }\int _{\Omega }|\nabla v_n|^p~\mathrm{d}x=0, \end{aligned}$$

and consequently $v_0=0.$ So $v_n\rightarrow 0$ in $L^2(\Omega )$ and this is a contradiction since $\Vert v_n\Vert _2=1.$ So $(u_n)_n$ is bounded on $\mathcal {N}_{\lambda }$.

Next, we show that $u_n$ converges strongly to u in $W^{1,2}_0(\Omega ).$

To do this, we will use the following vector inequality for $1<p<2$

$$\begin{aligned} \left( |x_2|^{p-2}x_2-|x_1|^{p-2}x_1\right) \cdot (x_2-x_1)\ge C'\left( |x_2|+|x_1|\right) ^{p-2}|x_2-x_1|^2, \end{aligned}$$

for all $x_1,x_2\in \mathbb {R}^N$ and for some $C'>0,$ (see [19]).

We have $\displaystyle {\int _{\Omega }}u_n^2~\mathrm{d}x\rightarrow \displaystyle {\int _{\Omega }}u^2~\mathrm{d}x$ and since $I_{\lambda }'(u_n)\rightarrow 0$ in $W^{-1,2}(\Omega ),$ $u_n\rightharpoonup u$ in $W^{1,2}_0(\Omega ),$ we also have $I_{\lambda }'(u_n)(u_n-u)\rightarrow 0$ and $I_{\lambda }'(u)(u_n-u)\rightarrow 0$ as $n\rightarrow +\infty .$ On the other hand, one has

$$\begin{aligned} \langle I'_{\lambda }(u_n)- I'_{\lambda }(u), u_n-u\rangle= & {} 2\left[ \int _{\Omega }\left( |\nabla u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right) \cdot \nabla (u_n-u)~\mathrm{d}x\right] \\&+ 2\int _{\Omega }|\nabla (u_n-u)|^2~\mathrm{d}x-2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x\\\ge & {} C'\int _{\Omega }\left( |\nabla u_n|+|\nabla u|\right) ^{p-2}|\nabla (u_n-u)|^2~\mathrm{d}x\\&+ 2\int _{\Omega }|\nabla (u_n-u)|^2~\mathrm{d}x-2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x\\\ge & {} 2\int _{\Omega }|\nabla (u_n-u)|^2~\mathrm{d}x-2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x\\\ge & {} \Vert u_n-u\Vert ^2_{1,2}-\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x. \end{aligned}$$

Therefore, $\Vert u_n-u\Vert _{1,2}\rightarrow 0$ as $n\rightarrow +\infty$ and $u_n$ converges strongly to u in $W^{1,2}_0(\Omega ).$

Let $\Sigma '=\{A\subset \mathcal {N}_{\lambda }:~A~\text {closed}~\text {and}~-A=A\}$ and $\Gamma _j=\{A\in \Sigma ':~\gamma (A)\ge j\},$ where $\gamma (A)$ denotes the Krasnoselski’s genus. We show that $\Gamma _j\ne \emptyset .$

Set $E_j=\text {span}\{e_i,~~i=1,\dots ,j\},$ where $e_i$ are the eigenfunctions associated with the problem (1.1). Let $\lambda \in (\lambda ^D_j,\lambda ^D_{j+1}),$ and consider $v\in S_{j}:=\{ v\in E_j:~\int _{\Omega }|v|^2~\mathrm{d}x=1\}.$ Then set

$$\begin{aligned} \rho (v)=\left[ \frac{\int _{\Omega }|\nabla v|^p~\mathrm{d}x}{\lambda \int _{\Omega }v^2~\mathrm{d}x-\int _{\Omega }|\nabla v|^2~\mathrm{d}x}\right] ^{\frac{1}{2-p}}. \end{aligned}$$

Then $\lambda \int _{\Omega }v^2~\mathrm{d}x-\int _{\Omega }|\nabla v|^2~\mathrm{d}x\ge \lambda \int _{\Omega }v^2~\mathrm{d}x-\sum \limits _{i=1}^{j}\int _{\Omega }\lambda _i| e_i|^2~\mathrm{d}x\ge (\lambda -\lambda _j)\int _{\Omega }|v|^2~\mathrm{d}x > 0$. Hence, $\rho (v)v\in \mathcal {N}_{\lambda },$ and then $\rho (S_j)\in \Sigma ',$ and $\gamma (\rho (S_j))=\gamma (S_j)=j$ for $1\le j\le k,$ for any .

It is then standard (see [1], Proposition 10.8) to conclude that

yields k pairs of nontrivial critical points for $I_{\lambda },$ which gives rise to k nontrivial solutions of problem (1.3).

Case 2: $p>2$.

In this case, we will rely on the following theorem:

Theorem (Clark, [11]) .

Let X be a Banach space and $G\in C^1(X,\mathbb {R})$ satisfying the Palais–Smale condition with $G(0)=0.$ Let $\Gamma _k =\{~A\in \Sigma ~:~\gamma (A)\ge k~\}$ with $\Sigma = \{~A\subset X~;~A=-A~\text {and}~A~\text {closed}~ \}.$ If $c_k=\inf \nolimits _{A\in \Gamma _k}\sup \nolimits _{u\in A}G(u)\in (-\infty , 0),$ then $c_k$ is a critical value.

Let us consider the $C^1$ energy functional $I_{\lambda }: W^{1,p}_0(\Omega )\rightarrow \mathbb {R}$ defined as

$$\begin{aligned} I_{\lambda }(u)=\frac{2}{p}\int _{\Omega }|\nabla u|^p~\mathrm{d}x+\int _{\Omega }|\nabla u|^2~\mathrm{d}x-\lambda \int _{\Omega }|u|^2~\mathrm{d}x. \end{aligned}$$

We want to show that

$$\begin{aligned} -\infty <\sigma _j=\inf _{\left\{ A\in \Sigma ', \gamma (A)\ge j\right\} }\sup _{u\in A}I_{\lambda }(u) \end{aligned}$$

(5.2)

is a critical point for $I_{\lambda }$, where $\Sigma '=\{A\subseteq S_j\},$ where $S_j=\{v\in E_j~:~\int _{\Omega }|v|^2 \mathrm{d}x=1\}.$

We clearly have that $I_{\lambda }(u)$ is an even functional for all $u\in W^{1,p}_0(\Omega )$, and also $I_{\lambda }(u)$ is bounded from below on $W^{1,p}_0(\Omega )$ since $I_{\lambda }(u)\ge C\Vert u\Vert _{1,p}^p-C'\Vert u\Vert ^2_{1,p}$.

We show that $I_{\lambda }(u)$ satisfies the (PS) condition. Let $\{u_n\}$ be a Palais–Smale sequence, i.e., $|I_{\lambda }(u_n)|\le M$ for all n,  $M>0$ and $I_{\lambda }'(u_n)\rightarrow 0$ in $W^{-1,p'}(\Omega )$ as $n\rightarrow \infty .$ We first show that $\{u_n\}$ is bounded in $W^{1,p}_0(\Omega ).$ We have

$$\begin{aligned} M\ge & {} |C\Vert u_n\Vert _{1,p}^p-C'\Vert u_n\Vert ^2_{1,p}|\ge \left( C\Vert u_n\Vert _{1,p}^{p-2}-C'\right) \Vert u_n\Vert _{1,p}^2, \end{aligned}$$

and so $\{u_n\}$ is bounded in $W^{1,p}_0(\Omega ).$

Therefore, $u\in W^{1,p}_0(\Omega )$ exists such that, up to subsequences that we will denote by $(u_n)_n$ we have $u_n\rightharpoonup u$ in $W^{1,p}_0(\Omega )$ and $u_n\rightarrow u$ in $L^2(\Omega ).$

We will use the following inequality for $v_1,v_2\in \mathbb {R}^N:$ there exists $R>0$ such that

$$\begin{aligned} |v_1-v_2|^p\le R\left( |v_1|^{p-2}v_1-|v_2|^{p-2}v_2\right) (v_1-v_2), \end{aligned}$$

for $p>2$ (see [19]). Then we obtain

$$\begin{aligned} \begin{aligned} \langle I'_{\lambda }(u_n)-I'_{\lambda }(u),u_n-u\rangle&=2\int _{\Omega }\left( |\nabla u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right) \cdot \nabla (u_n-u)~\mathrm{d}x\\&\quad +2\int _{\Omega }|\nabla u_n-\nabla u|^2~\mathrm{d}x\\&\quad -2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x\\&\ge \frac{2}{R}\int _{\Omega }|\nabla u_n-\nabla u|^p~\mathrm{d}x \\&\quad +2 \int _{\Omega }|\nabla u_n-\nabla u|^2~\mathrm{d}x-2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x\\&\ge \frac{2}{R}\Vert u_n-u\Vert ^p_{1,p}-2\lambda \int _{\Omega }|u_n-u|^2~\mathrm{d}x. \end{aligned} \end{aligned}$$

Therefore, $\Vert u_n-u\Vert _{1,p}\rightarrow 0$ as $n\rightarrow +\infty$, and so $u_n$ converges to u in $W^{1,p}_0(\Omega ).$

Next, we show that there exists sets $A_j$ of genus $j=1,\dots ,k$ such that $\sup \limits _{u\in A_j}I_{\lambda }(u)<0.$

Consider $E_j=\text {span}\{e_i,~i=1,\dots ,j\}$ and $S_j=\{v\in E_j:~\int _{\Omega }|v|^2~\mathrm{d}x=1\}.$ For any $s\in (0,1)$, we define the set $A_j(s):=s(S_j\cap E_j)$ and so $\gamma (A_j(s))=j$ for $j=1,\dots ,k.$ We have, for any $s\in (0,1)$

$$\begin{aligned} \sup \limits _{u\in A_j}I_{\lambda }(u)= & {} \sup \limits _{v\in S_j\cap E_j}I_{\lambda }(sv)\nonumber \\\le & {} \sup \limits _{v\in S_j\cap E_j}\left\{ \frac{s^p}{p}\int _{\Omega }|\nabla v|^p\mathrm{d}x+\frac{s^2}{2}\int _{\Omega }|\nabla v|^2\mathrm{d}x-\frac{\lambda s^2}{2}\int _{\Omega }|v|^2\mathrm{d}x\right\} \\\le & {} \sup \limits _{v\in S_j\cap E_j}\left\{ \frac{s^p}{p}\int _{\Omega }|\nabla v|^p\mathrm{d}x+\frac{s^2}{2}(\lambda _j-\lambda )\right\} <0 \end{aligned}$$

for $s>0$ sufficiently small, since $\displaystyle {\int _{\Omega }}|\nabla v|^p~\mathrm{d}x\le c_j$, where $c_j$ denotes some positive constant.

Finally, we conclude that $\sigma _{\lambda ,j}$ $(j=1,\dots ,k)$ are critical values thanks to Clark’s Theorem. $\square$