Double-phase problems and a discontinuity property of the spectrum

TL;DR

This paper investigates a nonlinear eigenvalue problem involving the sum of p- and q-Laplacians, demonstrating the spectrum's continuity and revealing a discontinuity as a parameter approaches a critical value.

Contribution

It establishes the continuity of the spectrum for the nonlinear eigenvalue problem and uncovers a discontinuity property related to the parametric (p,q)-differential operator.

Findings

The spectrum of the problem is continuous.

A discontinuity occurs in the spectrum as the parameter approaches 1 from below.

The study enhances understanding of spectral properties of nonlinear differential operators.

Abstract

We consider a nonlinear eigenvalue problem driven by the sum of $p$ and $q$-Laplacian. We show that the problem has a continuous spectrum. Our result reveals a discontinuity property for the spectrum of a parametric ($p,q$)-differential operator as the parameter $\beta\rightarrow 1^-$.

Equations

$$P_{\lambda }$$

$${\Delta }_{r}u=\mathrm{div}(|Du{|}^{r-2}Du)\text{mbox}forallu\in W_0^{1,r}(\Omega ).$$

$$J_{p,q}(u)={\int }_{\Omega }\left(|Du{|}^p+a(z)|Du{|}^q\right)dz0⩽a(z)⩽M,1<p<q,u\in W_0^{1,q}(\Omega ),$$

$$1<p<\infty ,q=2,p\ne q,\alpha =1-\beta ,\beta \in (0,1).$$

$$\hat{\sigma }(\beta )=(\beta {\hat{\lambda }}_1(q),+\infty )\text{mbox}for\beta \in (0,1).$$

$$-{\Delta }_{r}u(z)=\hat{\lambda }|u(z){|}^{r-2}u(z)\text{mbox}in\Omega ,u{|}_{\partial \Omega }=0$$

$${\hat{\lambda }}_1(r)=inf\left\{\frac{||Du|{|}_r^r}{||u|{|}_r^r}:u\in W_0^{1,r}(\Omega ),u\ne 0\right\}$$

$${\hat{u}}_1(r)\in C_{+}=\{u\in C_0^1(\overline{\Omega }):u(z)⩾0\text{mbox}forallz\in \overline{\Omega }\}.$$

$${\hat{u}}_1(r)\in \mathrm{int}{C}_{+}=\{u\in C_{+}:u(z)>0\text{mbox}forallz\in \Omega ,{\frac{\partial u}{\partial n}|}_{\partial \Omega }<0\},$$

$${\phi }_{\lambda }(u)=\frac{\alpha }{p}||Du|{|}_p^p+\frac{\beta }{q}||Du|{|}_q^q-\frac{\lambda }{q}||u|{|}_q^q\text{mbox}forallu\in W_0^{1,r}(\Omega ).$$

$$N_{\lambda }=\{u\in W_0^{1,r}(\Omega ):⟨{\phi }_{\lambda }^{\prime }(u),u⟩=0,u\ne 0\}.$$

$$u\to -\alpha {\Delta }_{p}u-\beta {\Delta }_{q}u\text{mbox}forallu\in W_0^{1,r}(\Omega ).$$

$$||u|{|}_{1,\tau }=||Du|{|}_{\tau }\text{mbox}forallu\in W_0^{1,\tau }(\Omega ).$$

$$⟨A_{\tau }(u),h⟩={\int }_{\Omega }|Du{|}^{\tau -2}(Du,Dh{)}_{{\mathbb{R}}^N}dt\text{mbox}forallu,h\in W_0^{1,\tau }(\Omega ).$$

$${\phi }_{\lambda }(u)$$

$$\Rightarrow$$

$${\phi }_{\lambda }({\hat{u}}_{\lambda })=inf\{{\hat{\phi }}_{\lambda }(u):u\in W_0^{1,p}(\Omega )\}.$$

$${\phi }_{\lambda }(t{\hat{u}}_1(q))$$

$${\phi }_{\lambda }(t{\hat{u}}_1(q))<0,$$

$${\phi }_{\lambda }^{\prime }({\hat{u}}_{\lambda })=0,$$

$$N_{\lambda }\ne \varnothing \Rightarrow \lambda >\beta {\hat{\lambda }}_1(q).$$

$$||Du|{|}_q^q<\frac{\lambda }{\beta }||u|{|}_q^q$$

$${\xi }_{\lambda }(t)$$

$${\xi }_{\lambda }(t)\to -\infty \text{mbox}ast\to +\infty$$

$${\xi }_{\lambda }(t)>0\text{mbox}see(\text{ref}eq5).$$

$${\xi }_{\lambda }(t_0)=0,$$

$$m_{\lambda }=inf\{{\phi }_{\lambda }(u):u\in N_{\lambda }\}.$$

$$\alpha ||Du|{|}_p^p+\beta ||Du|{|}_q^q=\lambda ||u|{|}_q^q.$$

$${\phi }_{\lambda }(u)$$

$$\Rightarrow$$

$$||u_n|{|}_{1,q}\to +\infty .$$

Full text

aa 2010 Mathematics Subject Classification. Primary: 35D30. Secondary: 35J60, 35P30

1. Introduction

Let $\displaystyle\Omega\subseteq\mathbb{R}^{N}$ be a bounded domain with a $\displaystyle C^{2}$-boundary $\displaystyle\partial\Omega$. In this paper we study the following nonlinear nonhomogeneous eigenvalue problem:

[TABLE]

For every $\displaystyle r\in(1,+\infty)$ by $\displaystyle\Delta_{r}$ we denote the $\displaystyle r$-Laplace differential operator defined by

[TABLE]

Equations driven by the sum of a $\displaystyle p$-Laplacian and of $\displaystyle q$-Laplacian, known as $\displaystyle(p,q)$-equations, arise in many problems of mathematical physics such as particle physics, see Benci, D’Avenia, Fortunato & Pisani [1] and nonlinear elasticity, see Zhikov [19]. Zhikov [19] introduced models for strongly anisotropic materials in the context of homogenization. For this purpose defined and studied the double phase functional

[TABLE]

where the modulating coefficient $\displaystyle a(z)$ dictates the geometry of the composite made of two different materials with hardening exponents $\displaystyle p$ and $\displaystyle q$ respectively.

Such problems were studied by Chorfi & Rădulescu [2], Gasinski & Papageorgiou [3, 4], Marano, Mosconi & Papageorgiou [9, 10], Mihailescu & Rădulescu [11], Papageorgiou & Rădulescu [12, 13], Papageorgiou, Rădulescu & Repovš [14, 15, 16], Rădulescu & Repovš [17], and Yin & Yang [18], under different conditions on the data of the problem.

In the present paper, we show that problem ($\displaystyle P_{\lambda}$) has a continuous spectrum which is the half line $\displaystyle(\beta\hat{\lambda}_{1}(q),+\infty)$, with $\displaystyle\hat{\lambda}_{1}(q)>0$ being the principal eigenvalue of $\displaystyle(-\Delta_{q},W^{1,q}_{0}(\Omega))$. So, for every $\displaystyle\lambda\in(\beta\hat{\lambda}_{1}(q),+\infty)$, problem ($\displaystyle P_{\lambda}$) admits a nontrivial solution. Our result reveals an interesting fact better illustration in the particular case where

[TABLE]

Let $\displaystyle L_{\beta}=-(1-\beta)\Delta_{p}u-\beta\Delta u$ and let $\displaystyle\hat{\sigma}(\beta)$ denote the spectrum of $\displaystyle L_{\beta}$. We have that

[TABLE]

The set function $\displaystyle\beta\rightarrow\hat{\sigma}(\beta)$ is $\displaystyle h$-continuous (Hausdorff continuous) on $\displaystyle\left(0,1\right]$, but at $\displaystyle\beta=0$ exhibits a discontinuity since $\displaystyle L_{0}=\Delta$ which has a discrete spectrum.

Our approach is based on the use of the Nehari manifold. So, we perform minimization under constraint.

2. Preliminaries

Let $\displaystyle r\in(1,+\infty)$. We recall some basic facts about the spectrum of $\displaystyle(-\Delta_{r},W^{1,r}_{0}(\Omega))$. So, we consider nonlinear eigenvalue problem

[TABLE]

We say that $\displaystyle\hat{\lambda}$ is an eigenvalue of $\displaystyle(-\Delta_{r},W^{1,r}_{0}(\Omega))$ if problem (1) admits a nontrivial solution $\displaystyle\hat{u}\in W^{1,p}_{0}(\Omega)$, known as an eigenfunction corresponding to the eigenvalue $\displaystyle\hat{\lambda}$. From the nonlinear regularity theory (see, for example, Gasinski & Papageorgiou [5, pp. 737-738]), we have that $\displaystyle\hat{u}\in C^{1}_{0}(\overline{\Omega})=\left\{u\in C^{1}(\overline{\Omega}):u|_{\partial\Omega}=0\right\}$. There is a smallest eigenvalue $\displaystyle\hat{\lambda}_{1}(r)$ which has the following properties:

• •

$\displaystyle\hat{\lambda}_{1}(r)$ is isolated (that is, there exists $\displaystyle\epsilon>0$ such that the interval $\displaystyle(\hat{\lambda}_{1}(r),\hat{\lambda}_{1}(r)+\epsilon)$ contains no eigenvalue of $\displaystyle(-\Delta_{r},W^{1,r}_{0}(\Omega))$).

• •

$\displaystyle\hat{\lambda}_{1}(r)$ is simple (that is, if $\displaystyle\hat{u},\hat{v}$ are eigenfunction corresponding to $\displaystyle\hat{\lambda}_{1}(r)$, then $\displaystyle\hat{u}=\mu\hat{v}$ with $\displaystyle\mu\in\mathbb{R}\backslash\{0\}$).

• •

$\displaystyle\hat{\lambda}_{1}(r)>0$ and admits the following variational characterization

[TABLE]

The infimum in (2) is realized on the corresponding one dimensional eigenspace. The above properties imply that the elements of this eigenspace are in $\displaystyle C^{1}_{0}(\overline{\Omega})$ and do not change sign. By $\displaystyle\hat{u}_{1}(r)$ we denote the positive, $\displaystyle L^{r}$-normalized (that is, $\displaystyle||\hat{u}_{1}(r)||_{r}=1$) eigenfunction corresponding to $\displaystyle\hat{\lambda}_{1}(r)>0$. We have

[TABLE]

In fact the nonlinear maximum principle (see, for example, Gasinski & Papageorgiou [4, p. 738]), implies that

[TABLE]

with $\displaystyle\frac{\partial u}{\partial n}=(Du,n)_{\mathbb{R}^{N}}$ being the outward normal derivative of $\displaystyle u$. Note that if $\displaystyle\hat{u}$ an eigenfunction corresponding to an eigenvalue $\displaystyle\hat{\lambda}\neq\hat{\lambda}_{1}(r)$, then $\displaystyle\hat{u}$ is nodal (that is, sign changing). The Ljusternik-Schnirelmann minimax scheme gives in addition to $\displaystyle\hat{\lambda}_{1}(r)$ a whole strictly increasing sequence $\displaystyle\{\hat{\lambda}_{k}(r)\}_{k\in\mathbb{N}}$ of distinct eigenvalue such that $\displaystyle\hat{\lambda}_{k}(r)\rightarrow+\infty$. These are called “variational eigenvalues” and we do not know if they exhaust the spectrum of $\displaystyle(-\Delta_{r},W^{1,r}_{0}(\Omega))$. However, if $\displaystyle r=2$ (linear eigenvalue problem), then the spectrum is the sequence $\displaystyle\{\hat{\lambda}_{k}(2)\}_{k\in\mathbb{N}}$ of variational eigenvalues.

Let $\displaystyle r=max\{p,q\}$ and $\displaystyle\lambda>0$. The energy (Euler) functional for problem ($\displaystyle P_{\lambda}$) is defined by

[TABLE]

Evidently $\displaystyle\varphi_{\lambda}\in C^{1}(W^{1,r}_{0}(\Omega),\mathbb{R})$.

The Nehari manifold for the functional $\displaystyle\varphi_{\lambda}$ is the set

[TABLE]

In what follows, we denote by $\displaystyle\hat{\sigma}(\alpha,\beta)$ the spectrum of

[TABLE]

So, $\displaystyle\lambda\in\hat{\sigma}(\alpha,\beta)$ if and only if problem ($\displaystyle P_{\lambda}$) admits a nontrivial solution $\displaystyle\hat{u}\in C^{1}_{0}(\overline{\Omega})$. This solution is an eigenvector for the eigenvalue $\displaystyle\lambda$.

In what follows for every $\displaystyle\tau\in(1,+\infty)$ by $\displaystyle||\cdot||_{1,\tau}$ we denote the norm of $\displaystyle W^{1,\tau}_{0}(\Omega)$. On account of the Poincare inequality, we have

[TABLE]

Also, by $\displaystyle A_{\tau}:W^{1,p}_{0}(\Omega)\rightarrow W^{-1,\tau^{\prime}}_{0}(\Omega)=W^{1,p}_{0}(\Omega)^{*}\left(\frac{1}{\tau}+\frac{1}{\tau^{\prime}}=1\right)$ we denote the nonlinear operator defined by

[TABLE]

This operator is bounded (that is, maps bounded sets to bounded sets), continuous, strictly monotone (hence maximal monotone too).

3. The spectrum of ($\displaystyle P_{\lambda}$)

First we deal with the easy case where $\displaystyle 1<q<p$. As we will see in the sequel, for this case $\displaystyle\varphi_{\lambda}(\cdot)$ is coercive and so we can use the direct method of the calculus of variations.

Proposition 1**.**

If $\displaystyle 1<q<p$, then $\displaystyle\hat{\sigma}(\alpha,\beta)=(\beta\hat{\lambda}_{1}(q),+\infty)$ and the eigenvectors belong in $\displaystyle C^{1}_{0}(\overline{\Omega})$.

Proof.

Now $\displaystyle r=max\{p,q\}=p$. Evidently, if $\displaystyle\lambda\leqslant\beta\hat{\lambda}_{1}(q)$, then $\displaystyle\lambda\notin\hat{\sigma}(\alpha,\beta)$ ot other wise we violate (2).

Let $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$ and $\displaystyle u\in W^{1,p}_{0}(\Omega)$. We have

[TABLE]

Also, by the Sobolev embedding theorem $\displaystyle\varphi_{\lambda}(\cdot)$ is sequentially weakly lower semicontinuous. So, by Weierstrass-Tonelli theorem, we can find $\displaystyle\hat{u}_{\lambda}\in W^{1,p}_{0}(\Omega)$ such that

[TABLE]

For $\displaystyle t>0$ we have

[TABLE]

Since $\displaystyle q<p$, choosing $\displaystyle t\in(0,1)$ small we have

[TABLE]

From (3) we have

[TABLE]

∎

When $\displaystyle 1<p<q$, the energy functional is no longer coercive. So, the direct method of the calculus of the variations fails and we have to use a different approach. Instead we will minimize $\displaystyle\varphi_{\lambda}$ on the Nehair manifold $\displaystyle N_{\lambda}$.

First we show that $\displaystyle N_{\lambda}\neq\varnothing$.

Proposition 2**.**

$\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$* if and only if $\displaystyle N_{\lambda}\neq\varnothing$.*

Proof.

As before (see the proof of Proposition 1), using (2) we see that

[TABLE]

Now suppose that $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$. Then on account of (2) we can find $\displaystyle u\in W^{1,q}_{0}(\Omega)\ u\neq 0$ such that

[TABLE]

Consider the function $\displaystyle\xi_{\lambda}:(0,+\infty)\rightarrow\mathbb{R}$ defined by

[TABLE]

Since $\displaystyle q>p$, from (5) we see that

[TABLE]

On the other hand for $\displaystyle t\in(0,1)$ small, we have

[TABLE]

Therefore, by Balzano’s theorem, we can find $\displaystyle t_{0}>0$ such that

[TABLE]

∎

We define

[TABLE]

For $\displaystyle u\in N_{\lambda}$, we have

[TABLE]

Therefore

[TABLE]

From (8) we infer that $\displaystyle\varphi_{\lambda}|_{N_{\lambda}}$ is coercive on $\displaystyle W^{1,p}_{0}(\Omega)$.

Proposition 3**.**

If $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$, then every minimizing sequence of (6) is bounded in $\displaystyle W^{1,q}_{0}(\Omega)$.

Proof.

We argue by contradiction. So, suppose that $\displaystyle\{u_{n}\}_{n\geqslant 1}\subseteq W^{1,q}_{0}(\Omega)$ is a minimizing sequence of (6) such that

[TABLE]

We have

[TABLE]

We set $\displaystyle y_{n}=\frac{u_{n}}{||u_{n}||_{q}}\ n\in\mathbb{N}$. Then $\displaystyle||y_{n}||_{q}=1\ n\in\mathbb{N}$. Also, from (3) we have

[TABLE]

So, we may assume that

[TABLE]

We multiply (9) with $\displaystyle\frac{1}{||u_{n}||^{q}_{q}}$. We obtain

[TABLE]

Recall that $\displaystyle\{u_{n}\}_{n\geqslant 1}\subseteq N_{\lambda}$ is a minimizing sequence for (6). So, we have

[TABLE]

Using (13) in (12), we infer that

[TABLE]

a contradiction since $\displaystyle||y_{n}||_{q}=1$ for all $\displaystyle n\in\mathbb{N}$.

Therefore we conclude that every minimizing sequence of (6) is bounded in $\displaystyle W^{1,q}_{0}(\Omega)$. ∎

We have already seen that $\displaystyle m_{\lambda}\geqslant 0$. We can say more.

Proposition 4**.**

If $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$, then $\displaystyle m_{\lambda}>0.$

Proof.

Arguing by contradiction, suppose that $\displaystyle m_{\lambda}=0$. Then we can find $\displaystyle\{u_{n}\}_{n\geqslant 1}\subseteq N_{\lambda}$ such that $\displaystyle\varphi_{\lambda_{-}}(u_{n})\rightarrow 0^{+}$. From (8) we have

[TABLE]

Then from (14) and Proposition 3, we infer that

[TABLE]

From (14), (15) and (7), it follows that

[TABLE]

Let $\displaystyle v_{n}=\frac{u_{n}}{||u_{n}||_{q}}\ n\in\mathbb{N}$. Then $\displaystyle||v_{n}||_{q}=1\ n\in\mathbb{N}$. We have

[TABLE]

Then from (3) and (18) it follows that

[TABLE]

From (18) and (19), we infer that

[TABLE]

a contradiction since $\displaystyle||v_{n}||_{q}=1\ n\in\mathbb{N}$. From this we conclude that $\displaystyle m_{\lambda}>0$. ∎

Proposition 5**.**

If $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$, then there exists $\displaystyle\hat{u}_{\lambda}\in N_{\lambda}$ such that $\displaystyle m_{\lambda}=\varphi_{\lambda}(\hat{u}_{\lambda}).$

Proof.

Let $\displaystyle\{u_{n}\}_{n\geqslant 1}\subseteq N_{\lambda}$ such that $\displaystyle\varphi_{\lambda}(u_{n})\rightarrow m_{\lambda}$. According to Proposition 3, $\displaystyle\{u_{n}\}_{n\geqslant 1}\subseteq W^{1,q}_{0}(\Omega)$ is bounded. So, we may assume that

[TABLE]

Since $\displaystyle u_{n}\in N_{\lambda}\ n\in\mathbb{N}$, we have

[TABLE]

Passing to the limit as $\displaystyle n\rightarrow\infty$ and using (20) and the weak lower semicontinuity of the functional in a Banach space, we obtain

[TABLE]

Note that $\displaystyle\lambda||\hat{u}_{\lambda}||^{q}_{q}-\beta||D\hat{u}_{\lambda}||^{q}_{q}\neq 0$ or otherwise from (18), we have

[TABLE]

Recall that

[TABLE]

So, it follows that

[TABLE]

which contradicts Proposition 4. Therefore

[TABLE]

Also, exploiting the sequential weak lower semicontinuity of $\displaystyle\varphi_{\lambda}(\cdot)$, we have

[TABLE]

If we show that $\displaystyle\hat{u}_{\lambda}\in N_{\lambda}$, then $\displaystyle\varphi_{\lambda}(\hat{u}_{\lambda})=m_{\lambda}$ and this will conclude the proof.

To this end, let

[TABLE]

Evidently $\displaystyle\hat{\xi}_{\lambda}(\cdot)$ is a continuous function. Arguing indirectly, suppose that $\displaystyle\hat{u}_{\lambda}\notin N_{\lambda}$. Then since $\displaystyle u_{n}\in N_{\lambda}$ for all $\displaystyle n\in\mathbb{N}$, from (20) we infer that

[TABLE]

On the other hand, note that since $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$, we have

[TABLE]

From (23), (24) and Bolzano’s theorem, we see that there exists $\displaystyle t^{*}\in(0,1)$ such that

[TABLE]

Then using (8) we have

[TABLE]

a contradiction. Therefore $\displaystyle\hat{u}_{\lambda}\in N_{\lambda}$ and this finishes the proof. ∎

So, we can state the following theorem concerning problem ($\displaystyle P_{\lambda}$).

Theorem 6**.**

If $\displaystyle\lambda>\beta\hat{\lambda}_{1}(q)$ then $\displaystyle\lambda$ is an eigenvalue of problem ($\displaystyle P_{\lambda}$) with eigenfunction $\displaystyle\hat{\lambda}\in C^{1}_{0}(\overline{\Omega})$.

Proof.

For $\displaystyle 1<q<p$, this follows from Proposition 1.

For $\displaystyle 1<q<p$, let $\displaystyle h\in W^{1,q}_{0}(\Omega)$. Choose $\displaystyle\epsilon>0$ such that $\displaystyle\hat{u}_{\lambda}+sh\not\equiv 0$ for $\displaystyle s\in(-\epsilon,\epsilon)$. We set

[TABLE]

Then we have that $\displaystyle s\rightarrow t(s)$ is a curve in $\displaystyle N_{\lambda}$ and it is differentiable. Let $\displaystyle\hat{\xi}_{\lambda}:(-\epsilon,\epsilon)\rightarrow\mathbb{R}$ be defined by

[TABLE]

Evidently $\displaystyle s=0$ is a minimizer of $\displaystyle\hat{\xi}_{\lambda}(\cdot)$ and so

[TABLE]

Then nonlinear regularity theory of Lieberman [8, p. 320], implies that $\displaystyle\hat{u}_{\lambda}\in C^{1}_{0}(\overline{\Omega})$. ∎

Remark 1**.**

In the terminology of critical point theory, the above proof shows that the Nehari manifold, is a natural constant for the functional $\displaystyle\varphi_{\lambda}$ (see Gasinski & Papageorgiou [5, p. 812]).

Now suppose that $\displaystyle\alpha=(1-\beta),\beta\in(0,1)$. Let $\displaystyle L_{\beta}=-(1-\beta)\Delta_{p}-\beta\Delta_{q}$ and let $\displaystyle\hat{\sigma}(\beta)$ be the spectrum of $\displaystyle L_{\beta}$. From Theorem 6, we know that

[TABLE]

Evidently $\displaystyle\hat{\sigma}(\cdot)$ is Hausdorff and Vietoris continuous on $\displaystyle(0,1)$ (see Hu & Papageorgiou [6]), but at $\displaystyle\beta=1$, it exhibits a discontinuity since

[TABLE]

and from Section 2, we know that $\displaystyle\hat{\lambda}_{1}(q)>0$ is isolated and so $\displaystyle\hat{\sigma}(1)\neq(\hat{\lambda}_{1}(q),+\infty)$. This is more emphatically illustrated when $\displaystyle q=2$. Then

[TABLE]

but at $\displaystyle\beta=1$, we have

[TABLE]

Acknowledgments. This research was supported by the Slovenian Research Agency grants P1-0292, J1-8131, J1-7025, N1-0064, and N1-0083. V.D. Rădulescu acknowledges the support through a grant of the Romanian Ministry of Research and Innovation, CNCS-UEFISCDI, project number PN-III-P4-ID-PCE-2016-0130, within PNCDI III.