In this paper, we studied a double-phase eigenvalue problem with large variable exponents. Let λ1(pn(⋅),qn(⋅)) be the first eigenvalues and un be the first eigenfunctions, normalized by ‖un‖Hn=1. Under some assumptions on the variable exponents pn(⋅) and qn(⋅), we showed that λ1(pn(⋅),qn(⋅)) converges to Λ∞, un converges to u∞ uniformly in the space Cα(Ω)(0<α<1) and u∞ is a nontrivial viscosity solution to a Dirichlet ∞-Laplacian problem. Even in the case where the variable exponents reduce to the constant exponents, our work is the first one dealing with a double-phase eigenvalue problem with large exponents.
Citation: Lujuan Yu, Beibei Wang, Jianwei Yang. An eigenvalue problem related to the variable exponent double-phase operator[J]. AIMS Mathematics, 2024, 9(1): 1664-1682. doi: 10.3934/math.2024082
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In this paper, we studied a double-phase eigenvalue problem with large variable exponents. Let λ1(pn(⋅),qn(⋅)) be the first eigenvalues and un be the first eigenfunctions, normalized by ‖un‖Hn=1. Under some assumptions on the variable exponents pn(⋅) and qn(⋅), we showed that λ1(pn(⋅),qn(⋅)) converges to Λ∞, un converges to u∞ uniformly in the space Cα(Ω)(0<α<1) and u∞ is a nontrivial viscosity solution to a Dirichlet ∞-Laplacian problem. Even in the case where the variable exponents reduce to the constant exponents, our work is the first one dealing with a double-phase eigenvalue problem with large exponents.
An expedient feature of the p-Laplacian eigenvalue problem is that the eigenfunctions may be multiplied by constant factors (in other words, the fact that if u is an eigenfunction, so is ku). Unfortunately, the p(x)-Laplacian eigenvalue problem does not possess this expedient property. It is important to stress that the loss of the property under consideration is not only a consequence of the dependence on x, but it can also occur in presence of unbalanced growth. For example, the double phase operator (that does not depend on x)
div(|∇u|p−2∇u+μ(x)|∇u|q−2∇u), | (1.1) |
loses this property. In this paper we are interested in considering that the operator has both peculiarities: It depends on x and it is unbalanced.
Let Ω⊂RN(N≥2) be a bounded domain with Lipschitz boundary ∂Ω. This article studies an eigenvalue problem coming from the minimization of the Rayleigh quotient:
||∇u||Hn||u||Hn, | (1.2) |
among all u∈W1,Hn0(Ω),u≢0. These functions belong to an appropriate Musielak-Orlicz Sobolev space with variable exponents; see its definition in section two. The function a:ˉΩ→[0,+∞) is a C1 differentiable function.
Put
Kn(u):=||∇u||Hn,kn(u):=||u||Hn,Sn(u):=∫Ω[pn(x)|∇uKn(u)|pn(x)−2+qn(x)a(x)|∇uKn(u)|qn(x)−2]dx∫Ω[pn(x)|ukn(u)|pn(x)−2+qn(x)a(x)|ukn(u)|qn(x)−2]dx,Hn:=tpn(x)+a(x)tqn(x) | (1.3) |
and define the first eigenvalue as
λ1(pn(⋅),qn(⋅))=infu∈W1,Hn0(Ω)∖{0}||∇u||Hn||u||Hn. | (1.4) |
By a similar proof of Proposition 3.1 in [1], we can show that the following equation
−div[(pn(x)|∇uKn(u)|pn(x)−2+qn(x)a(x)|∇uKn(u)|qn(x)−2)∇uKn(u)]=λ(pn(⋅),qn(⋅))Sn(u)ukn(u)(pn(x)|ukn(u)|pn(x)−2+qn(x)a(x)|ukn(u)|qn(x)−2),u∈W1,Hn0(Ω) | (1.5) |
is the Euler-Lagrange equation corresponding to the minimization of the Rayleigh quotient (1.2), where λ(pn(⋅),qn(⋅))=λ1(pn(⋅),qn(⋅)).
Here, we impose the following hypotheses on the variable exponents pn(x) and qn(x).
(H1): Assume that pn(x) and qn(x) are two sequences of C1 functions in ¯Ω, qn(⋅)>pn(⋅) for every n≥1 and
pn(x),qn(x)→+∞,uniformly for allx∈Ω, | (1.6) |
∇pn(x)pn(x)→ξ1(x),uniformly for allx∈Ω, | (1.7) |
∇qn(x)qn(x)→ξ2(x),uniformly for allx∈Ω. | (1.8) |
(H2): The following two quotients are bounded, namely,
lim supn→+∞p+np−n≤k1,lim supn→+∞q+nq−n≤k2, | (1.9) |
where for a function g we denote
g−=minx∈¯Ωg(x),g+=maxx∈¯Ωg(x). |
(H3): We also assume that
p−n>1,q−n>1,q+np−n<1+1N, | (1.10) |
then we can find a positive and continuous function θ(0<θ<+∞), such that
limn→∞qn(x)pn(x)=θ(x) | (1.11) |
uniformly for all x∈Ω.
The differential operator in (1.5) is the double-phase operator with variable exponents, which can be given by
div(|∇u|pn(x)−2∇u+μ(x)|∇u|qn(x)−2∇u). | (1.12) |
This operator is the classical double phase operator (1.1) when pn(x) and qn(x) are constants. Moreover, special cases of (1.12), studied extensively in the literature, occur when inf¯Ωμ>0 (the weighted (q(x),p(x))-Laplacian) or when μ≡0 (the p(x)-Laplacian).
The energy functional related to the double-phase operator (1.12) is given by
∫Ω|∇u|pn(x)+μ(x)|∇u|qn(x)dx, | (1.13) |
whose integrand switches two different elliptic behaviors. The integral functional (1.13) was first introduced by Zhikov [2,3,4,5], who obtained that the energy density changed its ellipticity and growth properties according to the point in order to provide models for strongly anisotropic materials. Moreover, double phase differential operators (1.12) and corresponding energy functionals (1.13) have several physical applications. We refer to the works of [6] on transonic flows, [7] on quantum physics and [8] on reaction diffusion systems. Finally, we mention a recent paper that is very close to our topic. For related works dealing with the double phase eigenvalue problems, we refer to the works of Colasuonno-Squassina [9], who proved the existence and properties of related variational eigenvalues. By using the Rayleigh quotient of two norms of Musielak-Orlicz space, the author of this paper has defined the eigenvalue, which has the same properties as the p-Laplace operator. Recently, Liu-Papageorgiou has considered an eigenvalue problem for the Dirichlet (p,q(⋅))−Laplacian by using the Nehari method (see [10]), a nonlinear eigenvalue problem for the Dirichlet (p,q)−Laplacian with a sign-changing Carathéodory reaction (see [11]) and a nonlinear eigenvalue problem driven by the anisotropic (p(⋅),q(⋅))−Laplacian (see [12]). Motivated by [9], Yu[13] discuss the asymptotic behavior of an eigenvalue for the double phase operator. However, to the author's knowledge, the eigenvalue problem for variable exponents double phase operator has remained open. Our article fits into this general field of investigation.
Assume that δ:Ω→[0,∞) is the distance function, which is given by
δ(x):=dist(x,∂Ω)=infy∈∂Ω|x−y|. |
This function is a Lipschitz continuous function. For all x∈Ω, we get |∇δ|=1. Define
Λ∞:=infφ∈W1,∞0(Ω)∖{0}‖∇φ‖L∞(Ω)‖φ‖L∞(Ω). | (1.14) |
It is known from the paper [1] that
Λ∞=‖∇δ‖L∞(Ω)‖δ‖L∞(Ω)=1maxx∈Ω{dist(x,∂Ω)}. | (1.15) |
Define
△∞u∞:=N∑i,j=1(u∞)xi(u∞)xj(u∞)xixj,k∞(u):=||u||L∞(Ω)=esssupx∈Ω|u|, | (1.16) |
K∞(u):=||∇u||L∞(Ω)=esssupx∈Ω|∇u|, | (1.17) |
and
k∞(u∞):=||u∞||L∞(Ω)=esssupx∈Ω|u∞|. |
The following are the main results of this paper.
Theorem 1.1. Let u∈C(Ω) be a weak solution of problem (1.5), then it is also a viscosity solution of the problem (3.2).
Theorem 1.2. Let hypotheses (H1)–(H3) be satisfied, λ1(pn(⋅),qn(⋅)) and Λ∞ be defined by (1.4) and (1.14), respectively. In addition, assume that un normalized by ‖un‖Hn=1 is the positive first eigenfunction, then,
(1)
limn→∞λ1(pn(⋅),qn(⋅))=Λ∞; | (1.18) |
(2) there exists a nonnegative function u∞ such that u∞∈Cα(Ω)∖{0} and ||u∞||L∞(Ω)=1;
(3) we can extract a subsequence, which is still denoted by un, such that
un→u∞, | (1.19) |
in the space Cα(Ω), where α(0<α<1) is a constant;
(4) we can obtain that the function u∞(x) is a nontrivial viscosity solution of the problem
{min{−Λ∞u∞+|∇u∞|,−Λ∞(u∞)θ(x0)+|∇u∞|,−△∞u∞−[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ2(x0)}=0,x∈Ω,u∞=0,x∈∂Ω. | (1.20) |
To the best of our knowledge, this is the first work dealing with the double phase eigenvalue problem (1.5). The rest of this paper is organized as follows. In section two, we collect some notations and facts about the Musielak-Orlicz space LH(Ω) and W1,H0(Ω), which will be used in this paper. Section three and section four are devoted to prove Theorems 1.1 and 1.2, respectively.
In this section, we recall some known results about the Musielak-Orlicz spaces LH(Ω) and W1,H0(Ω). For more detail, please see references [9,14,15,16,17].
We follow the notation of [9]. Let N(Ω) denote the set of all generalized N-functions. Let us introduce the nonlinear function H:Ω×[0,+∞)→[0,+∞) defined as
H(x,t):=tp(x)+a(x)tq(x),forall(x,t)∈Ω×[0,+∞), |
with 1<p(x)<q(x) and 0≤a(⋅)∈L1(Ω). It is clear that H∈N(Ω) is a locally integrable and generalized N-function. In addition, it fulfills the Δ2 condition, namely,
H(x,2t)≤2q+H(x,t). |
Therefore, in correspondence to H, we define the Musielak-Orlicz space LH(Ω) as
LH(Ω):={u:Ω→Rmeasurable:ρH(u)<+∞}, |
which can be equipped with the norm
‖u‖H:=inf{γ>0:ρH(u/γ)≤1}, |
where
ρH(u):=∫ΩH(x,|u|)dx, |
which is called H-modular.
Similarly, we can define the Musielak-Orlicz Sobolev spaces. The space W1,H(Ω) is given by
W1,H(Ω)={u∈LH(Ω)suchthat|∇u|∈LH(Ω)}, |
with the norm
‖u‖1,H:=‖u‖H+‖∇u‖H. |
We denote by W1,H0(Ω) the completion of C∞0(Ω) in W1,H(Ω). With these norms, the spaces LH(Ω), W1,H(Ω) and W1,H0(Ω) are separable, reflexive and uniformly convex Banach spaces.
From Proposition 2.16 (ⅱ) in [18], if
q+p−<1+1N, |
then the following Poincarˊe-type inequality
‖u‖H≤C‖∇u‖H |
holds for all u∈W1,H0(Ω), where C is a positive constant independent of u. Therefore, in this paper, we equip W1,H0(Ω) with the equivalent norm ‖∇u‖H for all u∈W1,H0(Ω).
Proposition 2.1. [18] If u∈LH(Ω) and ρH(u) is the H-modular, then the following properties hold.
(1) Ifu≠0, then ‖u‖H=λ if, and only if, ϱH(uλ)=1;
(2) ‖u‖H<1 (=1;>1) if, and only if, ϱH(u)<1 (=1;>1);
(3) If ‖u‖H≤1, then ‖u‖q+H≤ρH(u)≤‖u‖p−H;
(4) If ‖u‖H≥1, then ‖u‖p−H≤ρH(u)≤‖u‖q+H;
(5) ‖u‖H→0 if, and only if, ρH(u)→0;
(6) ‖u‖H→0 if, and only if, ρH(u)→0.
Given u∈C(Ω)⋂W1,Hn0(Ω) and ϕ∈C2(Ω). Define
△pn(x)ϕ:=div(|∇ϕ|pn(x)−2∇ϕ)=|∇ϕ|pn(x)−4{|∇ϕ|2△ϕ+(pn(x)−2)△∞ϕ+|∇ϕ|2ln(|∇ϕ|)∇ϕ⋅∇pn}, |
△qn(x)ϕ:=div(|∇ϕ|qn(x)−2∇ϕ)=|∇ϕ|qn(x)−4{|∇ϕ|2△ϕ+(qn(x)−2)△∞ϕ+|∇ϕ|2ln(|∇ϕ|)∇ϕ⋅∇qn}, |
and
△∞ϕ:=N∑i,j=1∂ϕ∂xi∂ϕ∂xj∂2ϕ∂xi∂xj, |
where △∞ϕ is the ∞-Laplacian.
Here, we are now in a position to give the following definition of weak solutions to problem (1.5).
Definition 3.1. We call u∈W1,Hn0(Ω)∖{0} a weak solution of problem (1.5) if
∫Ω(pn(x)|∇uK(u)|pn(x)−2+qn(x)a(x)|∇uK(u)|qn(x)−2)∇u⋅∇vK(u)dx=λ(pn(⋅),qn(⋅))S(u)∫Ω(pn(x)|uk(u)|pn(x)−2+qn(x)a(x)|uk(u)|qn(x)−2)uvk(u)dx | (3.1) |
is satisfied for all test functions v∈W1,Hn0(Ω). If u≠0, we say that λ(pn(⋅),qn(⋅)) is an eigenvalue of (1.5) and that u is an eigenfunction corresponding to λ(pn(⋅),qn(⋅)).
In (1.5), we replace u by ϕ and keep Sn, Kn and kn unchanged, then
{−pn(x)(K(u))1−pn(x)△pn(x)ϕ−qn(x)a(x)(K(u))1−qn(x)△qn(x)ϕ−qn(x)(K(u))1−qn(x)|∇ϕ(x)|qn(x)−2∇ϕ(x)⋅∇a(x)−(K(u))1−pn(x)|∇ϕ|pn(x)−2∇ϕ(x)⋅∇pn(x)−a(x)(K(u))1−qn(x)|∇ϕ(x)|qn(x)−2∇ϕ(x)⋅∇qn(x)+pn(x)(K(u))1−pn(x)ln(K(u))|∇ϕ(x)|pn(x)−2∇ϕ(x)⋅∇pn(x)+qn(x)a(x)(K(u))1−qn(x)ln(K(u))|∇ϕ(x)|qn(x)−2∇ϕ(x)⋅∇qn(x)−λ(pn(⋅),qn(⋅))S(u)(pn(x)(k(u))1−pn(x)|ϕ|pn(x)−2ϕ+qn(x)a(x)(k(u))1−qn(x)|ϕ(x)|qn(x)−2ϕ(x))=0,x∈Ω,ϕ=0,x∈∂Ω. |
We first recall the definition of viscosity solutions. Assume we are given a continuous function
F:RN×R×RN×S(N)→R, |
where S(N) denotes the set of N×N symmetric matrices.
Consider the problem
F(x,u,∇u,D2u)=0, | (3.2) |
where
F(x,u,∇u,D2u)=−pn(x)(K(u))1−pn(x){|∇u|pn(x)−4[|∇u|2△u+(pn(x)−2)△∞u+|∇u|2ln(|∇u|)∇u⋅∇pn(x)]}−qn(x)a(x)(K(u))1−qn(x){|∇u|qn(x)−4[|∇u|2△u+(qn(x)−2)△∞u+|∇u|2ln(|∇u|)∇u⋅∇qn(x)]}−qn(x)(K(u))1−qn(x)|∇u|qn(x)−2∇u⋅∇a(x)−(K(u))1−pn(x)|∇u|pn(x)−2∇u⋅∇pn(x)−a(x)(K(u))1−qn(x)|∇u|qn(x)−2∇u⋅∇qn(x)+pn(x)(K(u))1−pn(x)ln(K(u))|∇u|pn(x)−2∇u⋅∇pn(x)+qn(x)a(x)(K(u))1−qn(x)ln(K(u))|∇u|qn(x)−2∇u⋅∇qn(x)−λ(pn(⋅),qn(⋅))S(u)(pn(x)(k(u))1−pn(x)|u|pn(x)−2u+qn(x)a(x)(k(u))1−qn(x)|u|qn(x)−2u). | (3.3) |
Definition 3.2. Assume that x0∈Ω, u∈C(Ω), ψ∈C2(Ω) and φ∈C2(Ω).
(1) Let u(x0)=ψ(x0) and suppose that u−ψ attains its strict maximum value at x0. If
F(x0,ψ(x0),∇ψ(x0),D2ψ(x0))≤0 |
for all of such x0, then the function u is said to be a viscosity subsolution of Eq (3.2).
(2) Let u(x0)=φ(x0) and suppose that u−φ attains its strict minimum value at x0. If
F(x0,φ(x0),∇φ(x0),D2φ(x0))≥0 |
for all of such x0, then the function u is said to be a viscosity supersolution of Eq (3.2).
(3) If u is both a subsolution and a supersolution of the problem (3.2), then u is a viscosity solution of the problem (3.2).
Proof of Theorem 1.1. Claim: u is a viscosity supersolution of (3.2).
Let x0∈Ω and φ∈C2(Ω). Assume that u(x0)=φ(x0) and the function u−φ obtains its strict minimum value at the point x0. Our goal is to show that
F(x0,u(x0),∇φ(x0),D2φ(x0))≥0. | (3.4) |
If
F(x0,u(x0),∇φ(x0),D2φ(x0))<0, |
then by continuity there exists a positive constant r such that B(x0,2r)⊂Ω, u>φ in this ball, except for the point x0 and
F(x,u(x),∇φ(x),D2φ(x))<0, |
for all x∈B(x0,2r). Thus, if x∈B(x0,r), we have
−div[(pn(x)|∇φ(x)Kn(u)|pn(x)−2+qn(x)a(x)|∇φ(x)Kn(u)|qn(x)−2)∇φ(x)Kn(u)]−λ(pn(⋅),qn(⋅))Sn(u)(pn(x)|u(x)kn(u)|pn(x)−2+qn(x)a(x)|u(x)kn(u)|qn(x)−2)u(x)kn(u)<0. |
If x∈∂B(x0,r), the minimum value of the function u−φ is defined as m. Let Φ(x):=φ(x)+m2. Note that m>0 and the above inequality still holds if the function φ(x) is replaced by Φ(x), namely,
−div[(pn(x)|∇Φ(x)Kn(u)|pn(x)−2+qn(x)a(x)|∇Φ(x)Kn(u)|qn(x)−2)∇Φ(x)Kn(u)]−λ(pn(⋅),qn(⋅))Sn(u)(pn(x)|u(x)kn(u)|pn(x)−2+qn(x)a(x)|u(x)kn(u)|qn(x)−2)u(x)kn(u)<0. | (3.5) |
Define η(x):=(Φ−u)+≥0, then if x∈∂B(x0,r), we have η(x)≡0.
Let
Ω1={x|x∈B(x0,r)andΦ(x)>u(x)}. |
We multiply (3.5) by the function η(x) and integrate over B(x0,r), then the inequality
∫Ω1(pn(x)|∇ΦKn(u)|pn(x)−2+qn(x)a(x)|∇ΦKn(u)|qn(x)−2)∇ΦKn(u)⋅∇(Φ−u)dx−∫Ω1λ(pn(⋅),qn(⋅))Sn(u)(pn(x)|ukn(u)|pn(x)−2+qn(x)a(x)|ukn(u)|qn(x)−2)ukn(u)(Φ−u)dx<0 | (3.6) |
is true.
If we define
η1(x)={(Φ−u)+,x∈B(x0,r),0,x∈Ω∖B(x0,r), |
and use η1(x) as a test function in (3.1), then we get
∫Ω1(pn(x)|∇uKn(u)|pn(x)−2+qn(x)a(x)|∇uKn(u)|qn(x)−2)∇uKn(u)⋅∇(Φ−u)dx−∫Ω1λ(pn(⋅),qn(⋅))Sn(u)(pn(x)|ukn(u)|pn(x)−2+qn(x)a(x)|ukn(u)|qn(x)−2)ukn(u)(Φ−u)dx=0. | (3.7) |
Subtracting (3.7) from (3.6), we arrive at
∫Ω1pn(x)(|∇ΦKn(u)|pn(x)−2∇ΦKn(u)−|∇uKn(u)|pn(x)−2∇uKn(u))⋅∇(Φ−u)dx+∫Ω1qn(x)a(x)(|∇ΦKn(u)|qn(x)−2∇ΦKn(u)−|∇uKn(u)|qn(x)−2∇uKn(u))∇(Φ−u)dx<0. | (3.8) |
The first integral is nonnegative due to the elementary inequality
⟨|a|p−2a−|b|p−2b,a−b⟩≥0, | (3.9) |
which holds for all p>1. Here, we take p=pn(x). We get a contradiction. Hence, (3.4) holds. Similarly, we conclude that u is a viscosity subsolution of (3.2) and we omit the details.
Let n∈N be large enough such that pn≥r>N, which results in W1,Hn0(Ω)↪W1,r0(Ω) (see Proposition 2.16 (1) of Blanco, Gasiński, Harjulehto and Winkert [18]). It follows that un are continuous functions. The reason is that the space W1,r0(Ω)↪↪Cα(Ω), 0<α<1. Moreover, it is known (see [9]) that for each n∈N fixed, we have un>0.
In order to prove Theorem 1.2, we only need to prove the following conclusions.
Lemma 4.1. Let h:¯Ω→(1,∞) be a given continuous function, then
||∇v|s|1sp(x)s≤||∇v||H, | (4.1) |
for all v∈W1,H0(Ω) and s∈(1,p−).
Proof. Since ‖∇v||∇v||H‖H=1, it follows from Proposition 2.1 that
∫Ω[(|∇v|||∇v||H)p(x)+a(x)(|∇v|||∇v||H)q(x)]dx=1. | (4.2) |
Thus,
∫Ω[(|∇v|||∇v||H)s]p(x)sdxp(x)s≤1. | (4.3) |
Invoking Proposition 2.1 again, we conclude that
|(|∇v|||∇v||H)s|p(x)s≤1, |
which implies (4.1).
Lemma 4.2. If u∈L∞(Ω), then we have
limn→∞kn(u)=k∞(u). | (4.4) |
Proof. Step1: To show that the following inequality holds,
lim supn→∞kn(u)≤k∞(u). | (4.5) |
If kn(u)≤k∞(u), the above inequality is true. Thus, we can assume that kn(u)>k∞(u), and since qn(x)>pn(x)>1, we have
1=(∫Ω|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1p−n≤[∫Ω(k∞(u)kn(u))pn(x)+a(x)(k∞(u)kn(u))qn(x)dx]1p−n≤[∫Ω(k∞(u)kn(u))p−n+a(x)(k∞(u)kn(u))p−ndx]1p−n=k∞(u)kn(u)(|Ω|+∫Ωa(x)dx)1p−n, |
which implies (4.5) holds.
Step2: To show that the following inequality holds,
lim infn→∞kn(u)≥k∞(u). | (4.6) |
Case1: k∞(u)=0. It is easy to find that (4.6) holds.
Case2: k∞(u)>0. Given ε>0, there exists a nonempty set Ωε⊂Ω such that, for all x∈Ωε, |u|>k∞(u)−ε. Ignoring those indices n that kn(u)≥k∞(u)−ε, we have
1=(∫Ω|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1p−n≥(∫Ωε|ukn(u)|pn(x)+a(x)|ukn(u)|qn(x)dx)1p−n≥(∫Ωε|k∞(u)−εkn(u)|pn(x)+a(x)|k∞(u)−εkn(u)|qn(x)dx)1p−n≥(∫Ωε|k∞(u)−εkn(u)|p−n+a(x)|k∞(u)−εkn(u)|p−ndx)1p−n=k∞(u)−εkn(u)(|Ωε|+∫Ωεa(x)dx)1p−n, |
which gives
lim infn→∞kn(u)≥k∞(u)−ε. |
The arbitrariness of ε implies that (4.6) is true. Consequently, (4.4) holds.
Remark 4.1. If |∇u|∈L∞(Ω), we can argue as Lemma 4.2 to obtain that
limn→∞Kn(u)=K∞(u). | (4.7) |
Lemma 4.3. If the assumptions of Theorem 1.2 hold, then
(1) (1.18) holds;
(2) there exists a nonnegative function u∞ such that u∞∈Cα(Ω)∖{0} and ||u∞||L∞(Ω)=1;
(3) we can extract a subsequence, which is still denoted by un, such that
un→u∞ |
in the space Cα(Ω), where α(0<α<1) is a constant.
Proof. Assume for simplicity that the following inequality holds
∫Ωdx=1. |
Step 1: To show that,
lim supn→∞λ1(pn(⋅),qn(⋅))≤Λ∞. | (4.8) |
Inserting u(x)=δ(x) into (1.4) gives
λ1(pn(⋅),qn(⋅))≤||∇δ||Hn||δ||Hn. |
Note that by Lemma 4.2 and Remark 4.1, we have
lim supn→∞λ1(pn(⋅),qn(⋅))≤‖∇δ‖L∞(Ω)‖δ‖L∞(Ω)=Λ∞. |
Step 2: We now claim that u∞∈W1,∞0(Ω).
Since (4.8) holds, for all n∈N sufficiently large, we can assume that λ1(pn(⋅),qn(⋅))≤Λ∞+1. Thus, we have
Λ∞+1≥λ1(pn(⋅),qn(⋅))=||∇un||Hn||un||Hn=||∇un||Hn. |
Note that the sequence {||∇un||Hn} is bounded.
Let r∈[1,∞) be arbitrary. We can find an integer nr, for all n≥nr, such that pn(⋅)≥r and
W1,Hn0(Ω)↪W1,r0(Ω)↪↪Lr(Ω). |
Hence, the sequence {un} is bounded in the reflexive Banach space W1,r0(Ω). We can find a subsequence, still defined by {un}, and a function u∞∈W1,r0(Ω), such that ∇un⇀∇u∞ in W1,r0(Ω) and un→u∞ in Lr(Ω).
Define
sn(x):=pn(x)pn(x)−r,x∈Ω, |
and it follows that
s+n=p−np−n−r,s−n=p+np+n−r |
and
|1|sn(x)≤max{|Ω|1s+n,|Ω|1s−n}. | (4.9) |
Using Hölder's inequality and the above inequality, we have
(∫Ω|∇un|rdx)1r≤(1s−n+rp−n)|1|1rsn(x)||∇un|r|1rpn(x)r≤(1s−n+rp−n)max{|Ω|1s+n,|Ω|1s−n}1r||∇un|r|1rpn(x)r. | (4.10) |
Thus, (4.1) and (4.10) ensure that
‖∇un‖Lr(Ω)≤2(1+|Ω|)||∇un||Hn≤2(1+|Ω|)Λ∞+1. | (4.11) |
We choose an arbitrary positive real number r1 such that B(x,r1)⊂Ω, where the point x∈Ω is a Lebesgue point such that |∇u∞|∈L1(Ω), then we find that
1|B(x,r1)|∫B(x,r1)|∇u∞(y)|dy≤lim infn→∞1|B(x,r1)|∫B(x,r1)|∇un(y)|dy≤lim infn→∞|B(x,r1)|−1r||∇un||Lr(Ω)≤|B(x,r)|−1r2(1+|Ω|)(Λ∞+1). | (4.12) |
Passing to the limit as r→+∞ in the above inequality, gives
1|B(x,r1)|∫B(x,r1)|∇u∞(y)|dy≤2(1+|Ω|)(Λ∞+1). |
Letting r1→0+ in the above inequality, gives
|∇u∞(x)|≤2(1+|Ω|)(Λ∞+1), |
for a.e. x∈Ω, which implies that ∇u∞∈L∞(Ω), as claimed.
Step 3: We want to prove that un→u∞ in Cα(Ω) (0<α<1) and ‖u∞‖L∞(Ω)=1.
Keeping in mind that r∈[1,∞) is an arbitrary constant, we can assume that r>N. Therefore, this combined with the fact that W1,r0(Ω)↪↪Cα(Ω)(0<α<1) implies that there exists a nonnegative function u∞∈Cα(Ω)∖{0}, such that un→u∞ in Cα(Ω) and un converges uniformly to u∞ in Ω. Given ε∈(0,1), we can find a constant Nε∈N such that
|un(x)−u∞(x)|<ε, | (4.13) |
for all x∈Ω,n≥Nε. It follows that
[ρHn(un−u∞)]1p−n=[∫Ω|un−u∞|pn(x)+a(x)|un−u∞|qn(x)dx]1p−n≤[∫Ωεpn(x)+a(x)εqn(x)dx]1p−n≤ε[∫Ω(1+a(x))dx]1p−n≤[∫Ω(1+a(x))dx]1p−n | (4.14) |
and
[ρHn(un−u∞)]1q+n≤εp−nq+n[∫Ω(1+a(x))dx]1q+n≤[∫Ω(1+a(x))dx]1q+n, | (4.15) |
for all n≥Nε. Letting n→∞ in (4.14) and (4.15) yields
limn→∞[ρHn(un−u∞)]1p−n=limn→∞[ρHn(un−u∞)]1q+n=0. | (4.16) |
Thus, the inequality
|||un||Hn−‖u∞‖L∞(Ω)|≤|||un||Hn−||u∞||Hn|+|||u∞||Hn−‖u∞‖L∞(Ω)|≤||un−u∞||Hn+|||u∞||Hn−‖u∞‖L∞(Ω)|≤{[ρHn(un−u∞)]1p−n+[ρHn(un−u∞)]1q+n}+|||u∞||Hn−‖u∞‖L∞(Ω)|.≤{[ρHn(un−u∞)]1p−n+[ρHn(un−u∞)]1q+n}+|kn(u∞)−‖u∞‖L∞(Ω)| |
holds. In view of Lemma 4.2 and (4.16), we can get
‖u∞‖L∞(Ω)=limn→∞||un||Hn=1. | (4.17) |
Step 4: To show that lim infn→∞λ1(pn(⋅),qn(⋅))≥Λ∞.
Since ∇un⇀∇u∞ in W1,r0(Ω), ||un||Hn=1 and the inequality (4.11) holds, we have
‖∇u∞‖Lr(Ω)≤lim infn→∞‖∇un‖Lr(Ω)≤lim infn→∞‖∇un‖Hn=lim infn→∞λ1(pn(⋅),qn(⋅)). |
Letting r→∞ and using Proposition 7 in [19] and equality (4.17), we get
Λ∞≤‖∇u∞‖L∞(Ω)‖u∞‖L∞(Ω)≤lim infn→∞λ1(pn(⋅),qn(⋅)). | (4.18) |
Thus, (4.8) and (4.18) imply that (1.18) holds. The proof is complete.
Remark 4.2. We can again argue with Step 3 to obtain
‖∇u∞‖L∞(Ω)=limn→∞||∇un||Hn. | (4.19) |
The function u∞(x) also has the following property.
Lemma 4.4. If the assumptions of Theorem 1.2 hold, we can deduce that u∞(x) is a nontrivial viscosity solution of the problem (1.20).
Proof. For the first part we only need to show that u∞ is a viscosity subsolution of (1.20). Let x0∈Ω and ψ∈C2(Ω). Assume that u∞−ψ attains its strict maximum value of zero at x0, namely, u∞(x0)−ψ(x0)=0.
Claim: We want to show that
max{Λ∞ψ(x0)−|∇ψ(x0)|,(ψ(x0))θ(x0)K∞(u∞)−|∇ψ(x0)|,△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ2(x0)}≤0. | (4.20) |
By Lemma 4.3, we know that the convergence of un to u∞ in Ω is uniform. Therefore, there exists a sequence {xn}⊂Ω such that xn→x0 (as n→∞), un(xn)=ψ(xn) and un−ψ attains its strict maximum value at xn.
Employing Theorem 1.1, it turns out that for any n∈N large enough, un are continuous viscosity solutions of (1.5) with λ(pn(⋅),qn(⋅))=λ1(pn(⋅),qn(⋅)). Thus, we have
−pn(xn)(Kn(un))1−pn(xn)|∇ψ(xn)|pn(xn)−4{|∇ψ(xn)|2△ψ(xn)+(pn(xn)−2)△∞ψ(xn)+[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇pn(xn)}−qn(xn)a(xn)(Kn(un))1−qn(xn)|∇ψ(xn)|qn(xn)−4{|∇ψ(xn)|2△ψ(xn)+(qn(xn)−2)△∞ψ(xn)+[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇qn(xn)}−qn(xn)(Kn(un))1−qn(xn)|∇ψ(xn)|qn(xn)−2∇ψ(xn)⋅∇a(xn)−(Kn(un))1−pn(xn)|∇ψ(xn)|pn(xn)−2∇ψ(xn)⋅∇pn(xn)−a(xn)(Kn(un))1−qn(xn)|∇ψ(xn)|qn(xn)−2∇ψ(xn)⋅∇qn(xn)−λ1(pn(⋅),qn(⋅))Sn(un)pn(xn)(kn(un))1−pn(xn)|ψ(xn)|pn(xn)−2ψ(xn)−λ1(pn(⋅),qn(⋅))Sn(un)qn(xn)a(xn)(kn(un))1−qn(xn)|ψ(xn)|qn(xn)−2ψ(xn)≥0. | (4.21) |
Case 1: ψ(x0)=u∞(x0)>0.
Continuing (4.21), for n∈N sufficiently large, we have |∇ψ(xn)|>0. Let us assume the assertion is not true, then by (4.21) and continuity, we have ψ(x0)≤0. This leads to a contradiction.
Dividing both sides of (4.21) by
pn(xn)(pn(xn)−2)(Kn(un))1−pn(xn)|∇ψ(xn)|pn(xn)−4, |
we see that the following inequality holds
−|∇ψ(xn)|2△ψ(xn)pn(xn)−2−△∞ψ(xn)−[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇pn(xn)pn(xn)−2−qn(xn)pn(xn)|∇ψ(xn)Kn(un)|qn(xn)−pn(xn){a(xn)|∇ψ(xn)|2△ψ(xn)pn(xn)−2+a(xn)(qn(xn)−2pn(xn)−2)△∞ψ(xn)+|∇ψ(xn)|2∇ψ(xn)⋅∇a(xn)pn(xn)−2+a(xn)|∇ψ(xn)|2pn(xn)−2∇ψ(xn)⋅∇qn(xn)qn(xn)+a(xn)[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇qn(xn)pn(xn)−2}−|∇ψ(xn)|2pn(xn)∇ψ(xn)⋅∇pn(xn)pn(xn)−2≥(λ1(pn(⋅),qn(⋅)))3Sn(un)|λ1(pn(⋅),qn(⋅))ψ(xn)∇ψ(xn)|pn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2+(λ1(pn(⋅),qn(⋅)))3Sn(un)qn(xn)pn(xn)a(xn)|ψ(xn)|2ψ(xn)pn(xn)−2[(|ψ(xn)|kn(un))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|]pn(xn)−4≥0. | (4.22) |
Now, letting n→∞, we deduce that
−|∇ψ(xn)|2△ψ(xn)pn(xn)−2−△∞ψ(xn)→−△∞ψ(x0),−[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇pn(xn)pn(xn)−2→−[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ1(x0),−qn(xn)pn(xn){a(xn)|∇ψ(xn)|2△ψ(xn)pn(xn)−2+a(xn)(qn(xn)−2pn(xn)−2)△∞ψ(xn)+|ψ(xn)|2∇ψ(xn)⋅∇a(xn)pn(xn)−2+a(xn)[ln(|∇ψ(xn)|)−ln(Kn(un))]|∇ψ(xn)|2∇ψ(xn)⋅∇qn(xn)pn(xn)−2+a(xn)|∇ψ(xn)|2pn(xn)−2∇ψ(xn)⋅∇qn(xn)qn(xn)}→−θ2(x0)a(x0){△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ2(x0)},−|∇ψ(xn)|2pn(xn)∇ψ(xn)⋅∇pn(xn)pn(xn)−2→0. |
Taking the lower limit in inequality (4.22) and employing the limits above, we have
−|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))θ2(x0)a(x0)⋅{△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ2(x0)}−{△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ1(x0)}=−(|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))θ2(x0)a(x0)+1)△∞ψ(x0)−[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)(ξ1(x0)+θ2(x0)a(x0)|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))ξ2(x0))≥(Λ∞)3lim infn→∞Sn(un)|λ1(pn(⋅),qn(⋅))ψ(xn)∇ψ(xn)|pn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2+(Λ∞)3θ(x0)a(x0)lim infn→∞Sn(un)|ψ(xn)|2ψ(xn)pn(xn)−2[(|ψ(xn)|kn(un))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|]pn(xn)−4≥0. | (4.23) |
Note that by (4.17), (4.19) and u∞(x0)=ψ(x0)>0, we have
|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))=|∇ψ(x0)Λ∞k∞(u∞)|lim infn→∞(qn(xn)−pn(xn))≤|∇ψ(x0)Λ∞u∞(x0)|lim infn→∞(qn(xn)−pn(xn))=|∇ψ(x0)Λ∞ψ(x0)|limn→∞(qn(xn)−pn(xn)) | (4.24) |
and
|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))=|∇ψ(x0)Λ∞(k∞(u∞))θ(x0)|lim infn→∞(qn(xn)−pn(xn))≤|∇ψ(x0)Λ∞(u∞(x0))θ(x0)|lim infn→∞(qn(xn)−pn(xn))=|∇ψ(x0)Λ∞(ψ(x0))θ(x0)|lim infn→∞(qn(xn)−pn(xn)). | (4.25) |
Claim:
Λ∞ψ(x0)−|∇ψ(x0)|≤0. | (4.26) |
Assume that Λ∞ψ(x0)>|∇ψ(x0)|, then (4.24) and (1.11) imply
|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))=0 | (4.27) |
and
limn→∞|λ1(pn(⋅),qn(⋅))ψ(xn)∇ψ(xn)|(pn(xn)−4)∖(qn(xn)−4)=(Λ∞ψ(x0)|∇ψ(x0)|)1θ(x0)>1. | (4.28) |
Thus, choosing ε>0 small enough, we have
|λ1(pn(⋅),qn(⋅))ψ(xn)∇ψ(xn)|(pn(xn)−4)∖(qn(xn)−4)≥1+ε, | (4.29) |
for all n∈N sufficiently large. By (4.29), we get
lim infn→∞|λ1(pn,qn)ψ(xn)∇ψ(xn)|pn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2=lim infn→∞(|λ1(pn,qn)ψ(xn)∇ψ(xn)|(pn(xn)−4)∖(qn(xn)−4))qn(xn)−4qn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2qn(xn)−4≥Rψ(x0)3limn→∞(1+ε)qn(xn)−4qn(xn)−4=+∞. | (4.30) |
From (4.23), (4.27) and (4.30), we see that
−{△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ1(x0)}≥+∞, | (4.31) |
which is a contradiction. Hence, (4.26) holds.
Claim:
(ψ(x0))θ(x0)K∞(u∞)−|∇ψ(x0)|≤0. | (4.32) |
Suppose that the above inequality is not true, then we have
limn→∞[(ψ(xn)kn(un))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|](pn(xn)−4)/(qn(xn)−4)=limn→∞[(ψ(xn))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|](pn(xn)−4)/(qn(xn)−4)=[(ψ(x0))θ(x0)K∞(u∞)|∇ψ(x0)|]1θ(x0)>1. |
Thus, choosing ε1>0 small enough, we have
[(ψ(xn)kn(un))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|](pn(xn)−4)/(qn(xn)−4)≥1+ε1, | (4.33) |
for all n∈N sufficiently large. We are led to
lim infn→∞[(|ψ(xn)|kn(un))(qn(xn)−4)/(pn(xn)−4)Kn(un)|∇ψ(xn)|]pn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2≥lim infn→∞(1+ε1)qn(xn)−4qn(xn)−4|ψ(xn)|2ψ(xn)pn(xn)−2qn(xn)−4=θ(x0)ψ(x0)3limn→∞(1+ε1)qn(xn)−4qn(xn)−4=+∞. | (4.34) |
In view of (ψ(x0))θ(x0)K∞(u∞)−|∇ψ(x0)|>0 and (4.25),
|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))=0. |
Therefore, this fact along with (4.23) shows that (4.31) holds. This is a contradiction. Thus we deduce that (4.32) holds.
Claim:
△∞ψ(x0)+[ln(|∇ψ(x0)|)−ln(K∞(u∞))]|∇ψ(x0)|2∇ψ(x0)⋅ξ2(x0)≤0. | (4.35) |
Taking (4.24) and (4.26) into account, we have
|∇ψ(x0)K∞(u∞)|lim infn→∞(qn(xn)−pn(xn))=+∞. | (4.36) |
At the same time, by (4.25) and (4.32), we also deduce that (4.36) holds. If we assume that inequality (4.35) does not hold, then by (4.23) and (4.36), there is a contradiction. Thus, we deduce that (4.35) holds.
Case 2: ψ(x0)=u∞(x0)=0.
Note that if |∇ψ(x0)|=0 (in this case, we have △∞ψ(x0) = 0), the inequality (4.20) trivially holds. Hence, let us assume that |∇ψ(x0)|>0, then |∇ψ(xn)|>0 for n∈N large enough. We can use very similar arguments as Case 1 to conclude that (4.20) holds. The same argument can be used in order to show that u∞ is a viscosity supersolution of (1.20).
By Lemmas 4.3 and 4.4, it follows that Theorem 1.2 holds.
Remark 4.3. In the particular case where pn(x)=np(x) and qn(x)=nq(x), Theorems 1.1 and 1.2 are also true.
In this paper, we studied a double-phase eigenvalue problem with large variable exponents. As we know, for p-Laplace operator eigenvalue problems, there is an important feature that if u is an eigenfunction, so is ku, where k is an arbitrary constant. However, the double-phase operator with variable exponents looses this property. To overcome the above mentioned shortcoming, we defined the eigenvalue by using the Rayleigh quotient of two norms of Musielak-Orlicz space. Moreover, in the particular case where pn(⋅)=pn and qn(⋅)=qn, Theorems 1.1 and 1.2 are also true (see [13]).
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported by the National Natural Science Foundation of China (No.12001196) and the Natural Science Foundation of Henan (No. 232300421143).
The authors declare that they have no competing interests.
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