In this article, we study a double phase variable exponents problem with mixed boundary value conditions of the form
{D(u)+|u|p(x)−2u+b(x)|u|q(x)−2u=f(x,u) in Ω,u=0 on Λ1,(|∇u|p(x)−2u+b(x)|∇u|q(x)−2u)⋅ν=g(x,u) on Λ2.
First of all, using the mountain pass theorem, we establish that this problem admits at least one nontrivial weak solution without assuming the Ambrosetti–Rabinowitz condition. In addition, we give a result on the existence of an unbounded sequence of nontrivial weak solutions by employing the Fountain theorem with the Cerami condition.
Citation: Mahmoud El Ahmadi, Mohammed Barghouthe, Anass Lamaizi, Mohammed Berrajaa. Existence and multiplicity results for a kind of double phase problems with mixed boundary value conditions[J]. Communications in Analysis and Mechanics, 2024, 16(3): 509-527. doi: 10.3934/cam.2024024
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In this article, we study a double phase variable exponents problem with mixed boundary value conditions of the form
{D(u)+|u|p(x)−2u+b(x)|u|q(x)−2u=f(x,u) in Ω,u=0 on Λ1,(|∇u|p(x)−2u+b(x)|∇u|q(x)−2u)⋅ν=g(x,u) on Λ2.
First of all, using the mountain pass theorem, we establish that this problem admits at least one nontrivial weak solution without assuming the Ambrosetti–Rabinowitz condition. In addition, we give a result on the existence of an unbounded sequence of nontrivial weak solutions by employing the Fountain theorem with the Cerami condition.
The study of solutions to superlinear problems driven by the double phase operator is a new and important topic, since it sheds light on a range of applications in the field of mathematical physics such as elasticity theory, strongly anisotropic materials, Lavrentiev's phenomenon, etc. (see [1,2,3]).
In the present paper, we study the existence and multiplicity of solutions for the following double–phase problems with mixed boundary conditions:
{D(u)+|u|p(x)−2u+b(x)|u|q(x)−2u=f(x,u) in Ω,u=0 on Λ1,(|∇u|p(x)−2u+b(x)|∇u|q(x)−2u)⋅ν=g(x,u) on Λ2, | (1.1) |
where Ω⊂RN(N≥2) is a bounded domain with Lipschitz boundary ∂Ω, Λ1,Λ2 are disjoint open subsets of ∂Ω such that ∂Ω=¯Λ1∪¯Λ2 and Λ1≠∅, 1<p(x)<q(x)<N for all x∈¯Ω, b:¯Ω↦[0,+∞) is Lipschitz continuous, ν denotes the outer unit normal of Ω at the point x∈Λ2, f:Ω×R→R and g:Λ2×R→R are Carathéodory functions, and D is the double phase variable exponents operator given by
D(u):=div(|∇u|p(x)−2∇u+b(x)|∇u|q(x)−2∇u), for u∈W1,D(Ω). | (1.2) |
Note that the differential operator defined above in (1.2) is called the double phase operator with variable exponents, which is a natural generalization of the classical double phase operator when p and q are constant functions
u↦div(|∇u|p−2∇u+b(x)|∇u|q−2∇u). |
From the physical point of view, while studying the behavior of strongly anisotropic materials, V.V. Zhikov [2] in 1986 discovered that their hardening properties changed radically point by point, what is known as the Lavrentiev phenomenon [3]. To describe this phenomenon, he initially introduced the functional
v⟼∫Ω(|∇v|p+b(x)|∇v|q)dx, | (1.3) |
where the integrand changes its ellipticity and growth properties according to the point in the domain. In the framework of mathematics, the functional (1.3) has been investigated by many authors with respect to regularity and nonstandard growth. For instance, we refer to the papers of P. Baroni et al. [4,5], P. Baroni et al. [6], G. Cupini [7], and the references therein.
Multiple authors have recently concentrated on the study of double phase problems in the case when the exponents p and q are constants, and a plethora of results have been obtained; see, for example, W. Liu and G. Dai [8], M. El Ahmadi et al. [9], L. Gasiński and P. Winkert [10], N. Cui and H.R. Sun [11], Y. Yang et al. [12], and the references therein. For example, N. Cui and H.R. Sun [11] considered the following problem in the particular case: p(x)=p, q(x)=q, and λ=1
{D(u)+|u|p−2u+b(x)|u|q−2u=f(x,u) in Ω,(|∇u|p−2u+b(x)|∇u|q−2u)⋅ν=g(x,u) on ∂Ω, |
where D(u):=div(|∇u|p−2∇u+b(x)|∇u|q−2∇u). The authors have proved the existence and multiplicity of nontrivial weak solutions for the above problem with superlinear nonlinearity. Their approach was based on critical point theory with Cerami condition.
Very recently, Y. Yang et al. [12] considered the problem (1.1) in the particular case of p(x)=p and q(x)=q. Based on the maximum principle and homological local linking, they proved the existence of at least two bounded nontrivial weak solutions.
The main novelty of the current paper is the combination of the double phase variable exponents operator with mixed boundary conditions, that is, the Dirichlet condition on Λ1 and the Steklov condition on Λ2, which is different from [13]. To the best of our knowledge, there are only a few results related to the study of such problems.
To state our results, we make the subsequent hypotheses on f and g:
(H0) There exist C1,C2>0, s1∈C+(Ω), and s2∈C+(Λ2) such that
(ⅰ) |f(x,t)|≤C1(1+|t|s1(x)−1) for all (x,t)∈Ω×R,
(ⅱ) |g(x,t)|≤C2(1+|t|s2(x)−1) for all (x,t)∈Λ2×R.
(H1) (ⅰ) 1<p+≤q+<s−1≤s+1<p∗(x) for all x∈Ω, $
(ⅱ) 1<p+≤q+<s−2≤s+2<p∗(x) for all x∈Λ2,
where
p∗(x):={Np(x)N−p(x) if p(x)<N,∞ if p(x)≥N,p∗(x):={(N−1)p(x)N−p(x) if p(x)<N,∞ if p(x)≥N.. |
(H2) (ⅰ) lim inf|t|→∞F(x,t)|t|q+=+∞ uniformly a.e. x∈Ω,
(ⅱ) lim inf|t|→∞G(x,t)|t|q+=+∞ uniformly a.e. x∈Λ2,
where F(x,t)=∫t0f(x,s)ds and G(x,t)=∫t0g(x,s)ds.
(H3) (ⅰ) There exist c1,r1≥0 and l1∈L∞(Ω) with l1(x)>Np− such as
|F(x,t)|l(x)≤c1|t|l(x)p−F(x,t), |
for all (x,t)∈Ω×R, |t|≥r1 and F(x,t):=1q+f(x,t)t−F(x,t)≥0.
(ⅱ) There exist c2,r2≥0 and l2∈L∞(Λ2) with l2(x)>N−1p−−1 such as
|G(x,t)|l(x)≤c2|t|l(x)p−G(x,t), |
for all (x,t)∈Λ2×R, |t|≥r2 and G(x,t):=1q+g(x,t)t−G(x,t)≥0.
(H4) (ⅰ) f(x,t)=∘(|t|p+−1) as t→0 uniformly for a.e. x∈Ω,
(ⅱ) g(x,t)=∘(|t|p+−1) as t→0 uniformly for a.e. x∈Λ2.
(H5) (ⅰ) f(x,−t)=−f(x,t) for all (x,t)∈Ω×R,
(ⅱ) g(x,−t)=−g(x,t) for all (x,t)∈Λ2×R.
Let us consider ϕ:X0→R the Euler functional corresponding to problem (1.1), which is defined as follows:
ϕ(u)=I(u)−φ(u), |
where
I(u)=∫Ω(1p(x)|∇u|p(x)+b(x)q(x)|∇u|q(x))dx+∫Ω(1p(x)|u|p(x)+b(x)q(x)|u|q(x))dx, |
and
φ(u)=∫ΩF(x,u)dx+∫Λ2G(x,u)dσ, |
with X0 will be defined in preliminaries and dσ is the measure on the boundary.
Then, it follows from the hypothesis (H0) that the functional ϕ∈C1(X0,R), and its Fréchet derivative is
⟨ϕ′(u),v⟩=∫Ω(|∇u|p(x)−2+b(x)|∇u|q(x)−2)∇u.∇v dx+∫Ω(|u|p(x)−2+b(x)|u|q(x)−2)u.v dx−∫Ωf(x,u)vdx−∫Λ2g(x,u)vdσ, |
for any u,v∈X0. It is clear that any critical point of ϕ is a weak solution to the problem (1.1).
Now, we present the main results of this paper.
Theorem 1. Suppose that (H0),(H1),(H2),(H3) and (H4) hold. Then problem (1.1) has at least one nontrivial weak solution.
Theorem 2. Suppose that (H0),(H1),(H2),(H3) and (H5) hold. Then problem (1.1) possesses a sequence of weak solutions (un) such that ϕ(un)→+∞ as n→+∞.
To study double phase problems, we need some definitions and basic properties of W1,D(Ω), which are called Musielak–Orlicz–Sobolev spaces. For more details, see [14,15,16,17,18,19] and references therein.
First, we recall the definition of variable exponent Lebesgue space. For p∈C+(¯Ω):={p∈C(¯Ω):p−:=infx∈¯Ωp(x)>1}, we designate the variable exponent Lebesgue space by
Lp(x)(Ω)={u:Ω→R is measurable and ∫Ω|u(x)|p(x)dx<+∞}, |
equipped with the Luxemburg norm
|u|p(x)=inf{μ>0:∫Ω|u(x)μ|p(x)dx≤1}. |
Proposition 1. [20]
1. The Sobolev space (Lp(x)(Ω),|.|p(x)) is defined as the dual space Lq(x)(Ω), where q(x) is conjugate to p(x), i.e., 1p(x)+1q(x)=1. For any u∈Lp(x)(Ω) and v∈Lq(x)(Ω), we have
|∫Ωuv dx|≤(1p−+1q−)|u|p(x)|v|q(x)≤2|u|p(x)|v|q(x). |
2. If p1,p2∈C+(¯Ω), p1(x)≤p2(x), for all x∈¯Ω, then Lp2(x)(Ω)↪Lp1(x)(Ω) and the embedding is continuous.
Let p∈C+(∂Ω):={p∈C(∂Ω):p−:=infx∈∂Ωp(x)>1} and denote by dσ the Lebesgue measure on the boundary. We define
Lp(x)(∂Ω)={u:∂Ω→R is measurable and ∫∂Ω|u|p(x)dσ<+∞}, |
with the norm
|u|p(x),∂Ω=inf{α>0:∫∂Ω|u(x)α|q(x)dσ≤1}. |
Now, we give the main properties of the Musielak–Orlicz–Sobolev functional space that we will use in the rest of this paper. Denote by N(Ω) the set of all generalized N-functions. Let us denote by
D:Ω×[0,+∞)→[0,+∞) |
the function defined as
D(x,t)=tp(x)+b(x)tq(x), for all (x,t)∈Ω×[0,+∞), |
where the weight function b(.) and the variable exponents p(.),q(.)∈C+(¯Ω) satisfies the following hypothesis:
p(x)<q(x)<N,Nq(x)N+q(x)−1<p(x) for all x∈¯Ω and 0≤b(.)∈L∞(Ω). | (2.1) |
Note that the role of assuming the inequality Nq(x)N+q(x)−1<p(x) is to ensure that q(x)<p∗(x) and q(x)<p∗(x) for all x∈¯Ω, where p∗(x)=Np(x)N−p(x) and p∗(x)=(N−1)p(x)N−p(x).
It is clear that D is a generalized N-function, locally integrable, and
D(x,2t)≤2q+D(x,t), for all (x,t)∈Ω×[0,+∞), |
which is called condition (Δ2).
We designate the Musielak–Orlicz space by
LD(Ω)={u:Ω→R is measurable and ∫ΩD(x,|u|)dx<+∞}, |
equipped with the so-called Luxemburg norm
|u|D=inf{μ>0:∫ΩD(x,|uμ|)dx≤1}. |
The Musielak–Orlicz–Sobolev space W1,D(Ω) is defined as
W1,D(Ω)={u∈LD(Ω):|∇u|∈LD(Ω)}, |
endowed with the norm
∥u∥1,D=|u|D+|∇u|D. |
With such norms, LD(Ω) and W1,D(Ω) are separable, uniformly convex, and reflexive Banach spaces.
On LD(Ω), we consider the function ρ:LD(Ω)→R defined by
ρ(u)=∫Ω(|u|p(x)+b(x)|u|q(x))dx. |
The relationship between ρ and |.|D is established by the next result.
Proposition 2. (see [16]) For u∈LD(Ω), (un)⊂LD(Ω), and μ>0, we have
1. For u≠0, |u|D=μ⟺ρ(uμ)=1;
2. |u|D<1(=1,>1)⟺ρ(u)<1(=1,>1);
3. |u|D>1⟹|u|p−D≤ρ(u)≤|u|q+D;
4. |u|D<1⟹|u|q+D≤ρ(u)≤|u|p−D;
5. limn→+∞|un|D=0⇔limn→+∞ρ(un)=0 and limn→+∞|un|D=+∞⇔limn→+∞ρ(un)=+∞.
On W1,D(Ω), we introduce the equivalent norm by
‖u‖:=inf{μ>0:∫Ω[|∇uλ|p(x)+b(x)|∇uλ|q(x)+|uλ|p(x)+b(x)|uμ|q(x)]dx≤1}. | (2.2) |
Similar to Proposition (2), we have
Proposition 3. (see [16]) Let
ˆρ(u)=∫Ω(|∇u|p(x)+b(x)|∇u|q(x))dx+∫Ω(|u|p(x)+b(x)|u|q(x))dx. |
For u∈W1,D(Ω), (un)⊂W1,D(Ω), and μ>0, we have
1. For u≠0, ‖u‖=μ⟺ˆρ(uμ)=1;
2. ‖u‖<1(=1,>1)⟺ˆρ(u)<1(=1,>1);
3. ‖u‖>1⟹‖u‖p−≤ˆρ(u)≤‖u‖q+;
4. ‖u‖<1⟹‖u‖q+≤ˆρ(u)≤‖u‖p−;
5. limn→+∞‖un‖=0⇔limn→+∞ˆρ(un)=0 and limn→+∞‖un‖=+∞⇔limn→+∞ˆρ(un)=+∞.
We recall that problem (1.1) has a mixed boundary condition. For this, our Banach space workspace is given by
X0:={u∈W1,D(Ω):u=0 on Λ2}, |
endowed with the equivalent norm (2.2). Obviously, since X0 is a closed subspace of W1,D(Ω), then (X0,‖.‖) is a reflexive Banach space.
Proposition 4. (see [16]) Let hypothesis (2.1) be satisfied. Then the following embeddings hold:
1. There is a continuous embedding LD(Ω)↪Lr(x)(Ω) for r∈C(¯Ω) with 1≤r(x)≤p(x) for all x∈¯Ω.
2. There is a compact embedding W1,D(Ω)↪Lr(x)(Ω) for r∈C(¯Ω) with 1≤r(x)<p∗(x) for all x∈¯Ω.
3. If p∈C+(¯Ω)∩W1,γ(Ω)for someγ≥N. Then, there is a continuous embedding W1,D(Ω)↪Lr(x)(∂Ω) for r∈C(∂Ω) with 1≤r(x)≤p∗(x) for all x∈∂Ω.
4. There is a compact embedding W1,D(Ω)↪Lr(x)(∂Ω) for r∈C(∂Ω) with 1≤r(x)<p∗(x) for all x∈∂Ω.
It is important to note that when we replace W1,D(Ω) by X0 in Proposition 4, the embeddings 2 and 4 remain valid.
Let A:W1,D(Ω)→(W1,D(Ω))∗ be defined by
⟨A(u),v⟩=∫Ω(|∇u|p(x)−2+b(x)|∇u|q(x)−2)∇u.∇vdx+∫Ω(|u|p(x)−2+b(x)|u|q(x)−2)u.vdx, |
for all u,v∈W1,D(Ω), where (W1,D(Ω))∗ denotes the dual space of W1,D(Ω) and ⟨.,.⟩ stands for the duality pairing between W1,D(Ω) and (W1,D(Ω))∗.
Proposition 5. (see [16,Proposition 3.4]) Let hypothesis (2.1) be satisfied.
1. A:W1,D(Ω)→(W1,D(Ω))∗ is a continuous, bounded, and strictly monotone operator.
2. A:W1,D(Ω)→(W1,D(Ω))∗ satisfies the (S+)-property, i.e., if un⇀u in W1,D(Ω) and ¯limn→+∞⟨A(un)−A(u),un−u⟩≤0, then un→u in W1,D(Ω).
Definition 1. Let u∈X0. We say that u is a weak solution to the problem (1.1) if
∫Ω(|∇u|p(x)−2+b(x)|∇u|q(x)−2)∇u.∇v dx+∫Ω(|u|p(x)−2+b(x)|u|q(x)−2)u.v dx−∫Ωf(x,u)vdx−∫Λ2g(x,u)vdσ,=0, |
for all v∈X0.
Now, we give the definition of the Cerami condition that was first introduced by G. Cerami in [21].
Definition 2. Let (X,‖.‖) be a real Banach space and ϕ∈C1(X,R). We say that ϕ satisfies the Cerami condition (we denote (C)−condition) in X, if any sequence (un)⊂X such that (ϕ(un)) is bounded and ‖ϕ′(un)‖(1+‖un‖)→0 as n→+∞ has a strong convergent subsequence in X.
Remark 1. 1. It is clear from the above definition that if ϕ satisfies the (PS)-condition, then it satisfies the (C)-condition. However, there are functionals that satisfy the (C)-condition but do not satisfy the (PS)-condition (see [21]). Consequently, the (PS)-condition implies the (C)-condition.
2. The (C)-condition and the (PS)-condition are equivalent if ϕ is bounded below (see [22]).
Next, we present the following theorems, which will play a fundamental role in the proof of the main theorems.
Theorem 3. (see [23]) Let (X,‖.‖) be a real Banach space; ϕ∈C1(X,R) satisfies the (C)-condition; ϕ(0)=0, and the following conditions hold:
1. There exist positive constants ρ and α such that ϕ(u)≥α for any u∈X with ‖u‖=ρ.
2. There exists a function e∈X such that ‖e‖>ρ and ϕ(e)≤0.
Then, the functional ϕ has a critical value c≥α, that is, there exists u∈X such that ϕ(u)=c and ϕ′(u)=0 in X∗.
Let X be a real, reflexive, and separable Banach space. Then there exist {ej}j∈N⊂X and {e∗j}j∈N⊂X∗ such that
X=¯span{ej:j=1,2,...},X∗=¯span{e∗j:j=1,2,...}, |
and ⟨e∗i,ej⟩=1 if i=j, ⟨e∗i,ej⟩=0 if i≠j.
We denote Xj=span{ej}, Yk=⨁kj=1Xj, and Zk=¯⨁+∞j=kXj.
Theorem 4. (see [24]) Assume that X is a Banach space, and let ϕ:X→R be an even functional of class C1(X,R) that satisfies the (C)−condition. For every k∈N, there exists γk>ηk>0 such that
(A1) bk:=inf{ϕ(u):u∈Zk,‖u‖=ηk}→+∞ as k→+∞;
(A2) ck:=max{ϕ(u):u∈Yk,‖u‖=γk}≤0.
Then, ϕ has a sequence of critical values tending to +∞.
First of all, we are going to show that the functional ϕ fulfills the (C)-condition.
Lemma 3.1. If assumptions (H0),(H1),(H2) and (H3) hold, then the functional ϕ satisfies the (C)-condition.
Proof. Let (un)⊂X0 be a Cerami sequence for ϕ, namely,
(ϕ(un)) is bounded and ‖ϕ′(un)‖X∗0(1+‖un‖)→0, | (3.1) |
which implies that
sup|ϕ(un)|≤M and ⟨ϕ′(un),un⟩=∘n(1), | (3.2) |
where limn→+∞∘n(1)=0 and M>0.
We need to prove the boundedness of the sequence (un) in X0. To this end, assume to the contrary, that the sequence (un) is unbounded in X0. Without loss of generality, we can assume that ‖un‖>1. By virtue of (H3), for n large enough, we have
M+1≥ϕ(un)−1q+⟨ϕ′(un),un⟩=∫Ω(1p(x)|∇un|p(x)+b(x)q(x)|∇un|q(x))dx+∫Ω(1p(x)|un|p(x)+b(x)q(x)|un|q(x))dx−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ+1q+∫Ωf(x,un)undx+1q+∫Λ2g(x,un)undσ−1q+∫Ω(|∇un|p(x)+b(x)|∇un|q(x))dx−1q+∫Ω(|un|p(x)+b(x)|un|q(x))dx≥1q+∫Ω(|∇un|p(x)+b(x)|∇un|q(x))dx+1q+∫Ω(|un|p(x)+b(x)|un|q(x))dx−1q+∫Ω(|∇un|p(x)+b(x)|∇un|q(x))dx−1q+∫Ω(|un|p(x)+b(x)|un|q(x))dx+∫ΩF(x,un)dx+∫Λ2G(x,un)dσ, |
where F(x,un):=1q+f(x,un)un−F(x,un)≥0 and G(x,un)=1q+g(x,un)un−G(x,un)≥0.
Then, we obtain
M+1≥∫ΩF(x,un)dx+∫Λ2G(x,un)dσ, |
which implies
M+1≥∫ΩF(x,un)dx, | (3.3) |
and
M+1≥∫Λ2G(x,un)dσ. | (3.4) |
On the other hand, by Proposition 3, we have
M≥ϕ(un)=∫Ω(1p(x)|∇un|p(x)+b(x)q(x)|∇un|q(x))dx+∫Ω(1p(x)|un|p(x)+b(x)q(x)|un|q(x))dx−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ≥1q+(∫Ω(|∇un|p(x)+b(x)|∇un|q(x))dx+∫Ω(|un|p(x)+b(x)|un|q(x))dx)−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ≥1q+ˆρ(un)−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ. |
Because ‖un‖>1, we can obtain
M≥1q+‖un‖p−−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ. | (3.5) |
Since ‖un‖→+∞ as n→+∞, we deduce that
∫ΩF(x,un)dx+∫Λ2G(x,un)dσ≥1q+‖un‖p−−M→+∞ as n→+∞. | (3.6) |
Furthermore, using Proposition 3, we have
ϕ(un)=∫Ω(1p(x)|∇un|p(x)+b(x)q(x)|∇un|q(x))dx+∫Ω(1p(x)|un|p(x)+b(x)q(x)|un|q(x))dx−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ≤1p−‖un‖q+−∫ΩF(x,un)dx−∫Λ2G(x,un)dσ. |
Then, we obtain
ϕ(un)+∫ΩF(x,un)dx+∫Λ2G(x,un)dσ≤1p−‖un‖q+. | (3.7) |
In view of condition (H2), there exist T1,T2>0 such that
F(x,t)>|t|q+ for all x∈Ω and |t|>T1,G(x,t)>|t|q+ for all x∈Λ2 and |t|>T2. |
Since F(x,.) and G(x,.) are continuous functions on [−T1,T1] and [−T2,T2], respectively, there exist C0,C∗0>0 such that
|F(x,t)|≤C0 for all (x,t)∈Ω×[−T1,T1],|G(x,t)|≤C∗0for all (x,t)∈Λ2×[−T2,T2]. |
Then, there exist two real numbers K and K′, such that
F(x,t)≥K for all (x,t)∈Ω×R,G(x,t)≥K′ for all (x,t)∈Λ2×R. |
Hence,
F(x,un)−K1p−‖un‖q+≥0,G(x,un)−K′1p−‖un‖q+≥0, | (3.8) |
for all (x,n)∈¯Ω×N.
Put βn=un‖un‖, so ‖βn‖=1. Up to subsequences, for some β∈X0, we have
βn⇀β in X0,βn→β in Ls(x)(Ω),βn→β in Lr(x)(Λ2),βn(x)→β(x) a.e., in Ω,βn(x)→β(x) a.e., in Λ2, | (3.9) |
for s(x)<p∗(x) and r(x)<p∗(x).
Define the sets Ω0={x∈Ω:β(x)≠0} and Γ={x∈Λ2:β(x)≠0}.
Obviously, since ‖un‖→+∞ as n→+∞, we have
|un(x)|=|βn(x)|‖un‖→+∞, |
for any x∈Ω0∪Γ.
Therefore, due to (H2), for all x∈Ω0∪Γ, we deduce
F(x,un)1p−‖un‖q+=p−F(x,un)|un(x)|q+|βn(x)|q+→+∞,G(x,un)1p−‖un‖q+=p−G(x,un)|un(x)|q+|βn(x)|q+→+∞. | (3.10) |
Thus, |Ω0|=0 and |Γ|=0. In fact, suppose by contradiction that |Ω0|≠0 or |Γ|≠0. Using (3.6), (3.7), (3.10), and Fatou's lemma, we get
1=lim infn→+∞ ∫ΩF(x,un)dx+∫Λ2G(x,un)dσϕ(un)+∫ΩF(x,un)dx+∫Λ2G(x,un)dσ≥lim infn→+∞∫ΩF(x,un)dx+∫Λ2G(x,un)dσ1p−‖un‖q+≥lim infn→+∞[∫Ω0F(x,un)1p−‖un‖q+dx+∫ΓG(x,un)1p−‖un‖q+dσ]−lim supn→+∞∫Ω0K1p−‖un‖q+dx−lim supn→+∞∫ΓK′1p−‖un‖q+dσ≥lim infn→+∞∫Ω0F(x,un)1p−‖un‖q+dx+lim infn→+∞∫ΓG(x,un)1p−‖un‖q+dσ−lim supn→+∞∫Ω0K1p−‖un‖q+dx−lim supn→+∞∫ΓK′1p−‖un‖q+dσ=lim infn→+∞∫Ω0F(x,un)−K1p−‖un‖q+dx+lim infn→+∞∫Γg(x,un)−K′1p−‖un‖q+dσ≥∫Ω0lim infn→+∞F(x,un)−K1p−‖un‖q+dx+∫Γlim infn→+∞G(x,un)−K′1p−‖un‖q+dσ≥∫Ω0lim infn→+∞F(x,un)1p−‖un‖q+dx+∫Γlim infn→+∞G(x,un)1p−‖un‖q+dσ−∫Ω0lim supn→+∞K1p−‖un‖q+dx−∫Γlim supn→+∞K′1p−‖un‖q+dx=+∞, |
which is a contradiction. Therefore, β(x)=0 for a.e. x∈Ω and for a.e. x∈Λ2.
From (3.5) and (3.9), respectively, we can deduce that
βn→0 in Ls(x)(Ω),βn→0 in Lr(x)(Λ2),βn(x)→0 a.e. in Ω,βn(x)→0 a.e. in Λ2, | (3.11) |
for s(x)<p∗(x), r(x)<p∗(x), and
0<1q+≤lim supn→+∞[∫Ω|F(x,un)|‖un‖p−dx+∫Λ2|G(x,un)|‖un‖p−dσ]≤lim supn→+∞∫Ω|F(x,un)|‖un‖p−dx+lim supn→+∞∫Λ2|G(x,un)|‖un‖p−dσ. | (3.12) |
Using (H0) and (H1), we obtain
∫{0≤|un(x)|≤r1}|F(x,un)|‖un‖p− dx≤C1∫{0≤|un(x)|≤r1}|un|+1s1(x)|un|s(x)‖un‖p− dx≤C1|un|1‖un‖p−+C1s−1∫{0≤|un(x)|≤r1}|un|s1(x)−p−|βn|p−dx≤C1C3‖un‖‖un‖p−+Crs−p−1s−|βn|p−p−≤CC3‖un‖p−−1+C1C4rs−p−1s−1‖βn‖p−→0, as n→+∞, | (3.13) |
where C3,C4>0, s is either s+1 or s−1 and r1 comes from (H3).
Put l′1(x)=l(x)l(x)−1. Since l1∈L∞(Ω) with l1(x)>Np−, it follows that l′1(x)p−<p∗(x).
On the other hand, by virtue of hypothesis (H3)(i), (3.3), and (3.11), we deduce
∫{|un(x)|≥r1}|F(x,un)|‖un‖p−dx≤2[∫{|un(x)|≥r1}(|F(x,un)||un|p−)l1(x)dx]1l1(x)[∫{|un(x)|≥r1}|βn|l′1(x)p−dx]1l′1(x)≤2c1l1(x)1[∫{|un(x)|≥r1}F(x,un)dx]1l1(x)[∫{|un(x)|≥r1}|βn|l′1(x)p−dx]1l′1(x)≤2c1l1(x)1[∫ΩF(x,un)dx]1l1(x)[∫Ω|βn|l′1(x)p−dx]1l′1(x)≤2c1l1(x)1(M+1)1l(x)[∫Ω|βn|l′(x)p− dx]1l′(x)→0 as n→+∞. |
Combining this with (3.13), we obtain
∫Ω|F(x,un)|‖un‖p−dx=∫{0≤|un(x)|≤r1}|F(x,un)|‖un‖p−dx+∫{|un(x)|≥r1}|F(x,un)|‖un‖p−dx⟶0, as n→+∞. | (3.14) |
Similarly, let l′2(x)=l2(x)l2(x)−1. Since l2∈L∞(Ω) with l2(x)>N−1p−−1, it follows that l′2(x)p−<p∗(x). Then, by (H3)(ii), (3.4), and (3.11), we can prove in a similar way that
∫Λ2|G(x,un)|‖un‖p−dσ⟶0, as n→+∞. | (3.15) |
Consequently, combining (3.14) with (3.15), we obtain
∫Ω|F(x,un)|‖un‖p−dx+∫Λ2|G(x,un)|‖un‖p−dσ⟶0, as n→+∞, |
which is a contradiction to (3.12). Thus, (un) is bounded in X0.
Finally, we need to prove that any (C)-sequence has a convergent subsequence. Let (un)⊂X0 be a (C)-sequence. Then, (un) is bounded in X0. Passing to the limit, if necessary, to a subsequence, from Proposition 4, we have
un⇀u in X0, un→u in Ls1(x)(Ω), un→u in Ls2(x)(Λ2),un(x)→u(x) a.e. x∈Ω, un(x)→u(x) a.e. x∈Λ2, | (3.16) |
for 1≤s1(x)<p∗(x) and 1≤s2(x)<p∗(x). It is easy to check from (H0), (3.16) and Hölder's inequality that
|∫Ωf(x,un)(un−u)dx|≤C1|1+|un|s1(x)−1|s′1(x)|un−u|s1(x)⟶0 as n→+∞, | (3.17) |
and
|∫Λ2g(x,un)(un−u)dσ|≤C2|1+|un|s2(x)−1|s′2(x)|un−u|s2(x)⟶0 as n→+∞, | (3.18) |
where 1s1(x)+1s′1(x)=1 and 1s2(x)+1s′2(x)=1.
Next, since un⇀u, from (3.1), we have
⟨ϕ′(un),un−u⟩⟶0, as n→+∞. | (3.19) |
Then
⟨ϕ′(un),un−u⟩=⟨A(un),un−u⟩−∫Ωf(x,un)(un−u)dx−∫Λ2g(x,un)(un−u)dσ⟶0 as n→+∞, |
where A is given in Proposition 5.
Finally, the combination of (3.17), (3.18), and (3.19) implies
⟨A(un),un−u⟩⟶0 as n→+∞. |
Since the operator A satisfies the (S+) property in view of Proposition 5, we can obtain that un→u in X0. The proof is complete.
Proof of Theorem 1
Let us check that the functional ϕ satisfies the geometric conditions of the mountain pass in Theorem 3. By Lemma 3.1, ϕ satisfies the (C)−condition. According to the definition of ϕ, we have ϕ(0)=0. Then, to apply Theorem 3, it remains to prove that
(ⅰ) There exist positive constants ρ and α such that ϕ(u)≥α for any u∈X0 with ‖u‖=ρ.
(ⅱ) There exists a function e∈X0 such that ‖e‖>ρ and ϕ(e)≤0.
For (i), let ‖u‖<1. Then, by Proposition 3, we have
ϕ(u)≥1p+ˆρ(u)−∫ΩF(x,u)dx−∫Λ2G(x,u)dσ≥1p+‖u‖q+−∫ΩF(x,u)dx−∫Λ2G(x,u)dσ. | (3.20) |
Using (H0) and (H4), for ε>0 be small enough, there exist C1(ε),C2(ε)>0 such that
F(x,t)≤ε|t|p++C1(ε)|t|s1(x),∀(x,t)∈Ω×R,G(x,t)≤ε|t|p++C2(ε)|t|s2(x),∀(x,t)∈Λ2×R. | (3.21) |
Since p+<s1(x)<p∗(x) and p+<s2(x)<p∗(x) for all x∈¯Ω and for all x∈¯Λ2 in view of condition (H1), we have from Proposition 4 that
X0↪Lp+(Ω),X0↪Ls1(x)(Ω),X0↪Lp+(Λ2),X0↪Ls2(x)(Λ2). |
So, there exist ci>0(i=3,...6) such that
|u|p+≤c3‖u‖,|u|s1(x)≤c4‖u‖, ∀u∈X0|u|p+,Λ2≤c6‖u‖,|u|s(x)≤c7‖u‖, ∀u∈X0. |
Therefore, by (3.20) and (3.21), for ‖u‖<1 sufficiently small, we obtain
ϕ(u)⩾1q+‖u‖q+−ε∫Ω|u|p+dx−C1(ε)∫Ω|u|s1(x)dx−ε∫Λ2|u|p+dσ−C2(ε)∫Λ2|u|s2(x)dσ⩾1q+‖u‖q+−εcp+3‖u‖p+−C1(ε)cs−14‖u‖s−1−εcp+5‖u‖p+−C2(ε)cs−26‖u‖s−2. |
Since s−1>p+ in view of condition (H1) and ‖u‖<1, then ‖u‖s−1<‖u‖p+. Thus, we obtain
ϕ(u)⩾1q+‖u‖q+−εcp+3‖u‖p+−C1(ε)cs−14‖u‖p+−εcp+5‖u‖p+−C2(ε)cs−26‖u‖s−2≥1q+‖u‖q+−(εcp+3+C1(ε)cs−14+εcp+5)‖u‖p+−C2(ε)cs−26‖u‖s−2. |
Since s−2>q+≥p+, then by the standard argument, there exist positive constants ρ and α such that ϕ(u)≥α for any u∈X0 with ‖u‖=ρ.
Next, we affirm that there exists e∈X0 with ‖u‖>ρ such that
ϕ(e)<0. | (3.22) |
In fact, from (H2), it follows that for every k>0, there exist constants Tk and T∗k such that
F(x,t)>k|t|q+ for all x∈Ω and |t|>Tk,G(x,t)>k|t|q+ for all x∈Λ2 and |t|>T∗k. |
Since F(x,.) and G(x,.) are continuous functions on [−Tk,Tk] and [−T∗k,T∗k], respectively, there exist constants C0,C∗0>0 such that
|F(x,t)|≤C0 for all (x,t)∈Ω×[−Tk,Tk],|G(x,t)|≤C∗0 for all (x,t)∈Λ2×[−T∗k,T∗k]. |
Thus,
F(x,t)≥k|t|q+−C0, for all (x,t)∈Ω×R,G(x,t)≥k|t|q+−C∗0, for all (x,t)∈Λ2×R. | (3.23) |
Let w∈X0∖{0} such that ‖w‖=1 and l>1 be large enough. Using the above inequality, we obtain
ϕ(lw)=∫Ω(1p(x)|∇lw|p(x)+b(x)q(x)|∇lw|q(x))dx+∫Ω(1p(x)|lw|p(x)+b(x)q(x)|lw|q(x))dx−∫ΩF(x,lw)dx−∫Λ2G(x,lw)dσ≤lq+p−ˆρ(w)−klq+∫Ω|w|q+dx−klq+∫Λ2|w|q+dσ+C0|Ω|+C∗0|Λ2|≤lq+p−−klq+∫Ω|w|q+dx−klq+∫Λ2|w|q+dσ+C0|Ω|+C∗0|Λ2|=lq+(1p−−k∫Ω|w|q+dx−k∫Λ2|w|q+dσ)+C0|Ω|+C∗0|Λ2|. |
As
1p−−k∫Ω|w|q+dx−k∫Λ2|w|q+dσ<0, |
for k large enough, we deduce
ϕ(lw)→−∞, as l→+∞. |
Thus, there exist t0>1 and e=t0w∈X0∖¯Bρ(0) such that ϕ(e)<0.
Proof of Theorem 2
To prove Theorem 2, we need the following auxiliary lemmas:
Lemma 3.2. (see [25,26]) For s∈C+(¯Ω) and r∈C+(¯Λ2) such that s(x)<p∗(x) for all x∈¯Ω and r(x)<p∗(x) for all x∈¯Λ2. Let
δk=sup{|u|s(x):‖u‖=1,u∈Zk},δ′k=sup{|u|s(x),Λ2:‖u‖=1,u∈Zk}. |
Then, limk→+∞δk=limk→+∞δ′k=0.
Lemma 3.3. (see [27]) For all s∈C+(¯Ω) (r∈C+(¯Λ2)) and u∈Ls(x)(Ω) (v∈Lr(x)(Λ2)), there exists y∈Ω (z∈Λ2) such that
∫Ω|u|s(x)dx=|u|s(y)s(x),∫Λ2|u|r(x)dσ=|u|r(z)r(x),Λ2. | (3.24) |
Now, we return to the proof of Theorem 2. To this end, based on Fountain Theorem 4, we will show that the problem (1.1) possesses infinitely many weak solutions with unbounded energy. Evidently, according to (H5), ϕ is an even functional. By Lemma 3.1, we know that ϕ satisfies the (C)-condition. Then, to prove Theorem 2, it only remains to verify the following assertions:
(A1) bk:=inf{ϕ(u):u∈Zk,‖u‖=ηk}→+∞ as k→+∞,
(A2) ck:=max{ϕ(u):u∈Yk,‖u‖=γk}≤0.
(A1) For any u∈Zk such that ‖u‖=ηk>1. It follows from (H0), Proposition 3, and Lemma 3.3 that
ϕ(u)=∫Ω(1p(x)|∇u|p(x)+b(x)q(x)|∇u|q(x))dx+∫Ω(1p(x)|u|p(x)+b(x)q(x)|u|q(x))dx−∫ΩF(x,u)dx−∫Λ2G(x,u)dσ≥1q+‖u‖p−−C1∫Ω|u|dx−C1∫Ω|u|s1(x)s1(x)dx−C2∫Λ2|u|dσ−C2∫Λ2|u|s2(x)s2(x)dσ≥1q+‖u‖p−−C1c1‖u‖−C1s−1|u|s1(y)s1(x)−C2c2‖u‖−C2s−2|u|s2(z)s2(x),λ2≥1q+‖u‖p−−c3‖u‖−C1s−1|u|s1(y)s1(x)−C2s−2|u|s2(z)s2(x),λ2, |
where c3=max{C1c1,C2c2}.
Then, it follows that
ϕ(u)≥{1q+‖u‖p−−c3‖u‖−C1s−1−C2s−2 if |u|s1(x)≤1,|u|s2(x)≤11q+‖u‖p−−c3‖u‖−C1s−1(δk‖u‖)s+1−C2s−2(δ′k‖u‖)s+2 if |u|s1(x)>1,|u|s2(x)>11q+‖u‖p−−c3‖u‖−C1s−1−C2s−2(δ′k‖u‖)s+2 if |u|s1(x)≤1,|u|s2(x)>11q+‖u‖p−−c3‖u‖−C1s−1(δk‖u‖)s+1−C2s−2 if |u|s1(x)>1,|u|s2(x)≤1≥1q+‖u‖p−−c3‖u‖−2C1s−1(δk‖u‖)s+1−2C2s−2(δ′k‖u‖)s+2−2C1s−1−2C2s−2≥12q+‖u‖p−+(14q+‖u‖p−−2C1s−1δs+1k‖u‖s+1)+(14q+‖u‖p−−2C1s−2(δ′k)s+2‖u‖s+2)−c3‖u‖−2C1s−1−2C2s−2. | (3.25) |
Let us consider the following equations:
14q+tp−−2C1s−1δs+1kts+1=0, | (3.26) |
and
14q+tp−−2C1s−2(δ′k)s+2ts+2=0. | (3.27) |
Let ak and dk be the two non-zero solutions of (3.26) and (3.27), respectively. Then, we obtain
ak=(4q+2C1s−1δs+1k)1p−−s+1→+∞ and dk=(4q+2C2s−2(δ′k)s+2)1p−−s+2→+∞ as k→+∞. |
We fix ηk as follows
ηk=min{ak,dk}. |
Then, by Lemma 3.2, (3.25) and s+1,s+2>q+>p−, we obtain
ϕ(u)≥12q+ηp−k−c3ηk−C6→+∞ as k→+∞, |
where C6>0. Hence, (A1) holds.
(A2) In view of Proposition 3 and (3.23), for u∈Yk with ‖u|>1, we have
ϕ(u)≤1p−‖u‖q+−∫ΩF(x,u)dx−∫Λ2G(x,u)dσ≤1p−‖u‖q+−k∫Ω|u|q+dx−k∫Λ2|u|q+dσ+C0|Ω|+C∗0|Λ2|≤1p−‖u‖q+−k(|u|q+q++|u|q+q+,Λ2)+C0|Ω|+C∗0|Λ2|. |
Since dimYk<∞, then all norms are equivalent in Yk. Therefore, as 1p−<1, for k large enough, we obtain
ϕ(u)→−∞ as ‖u‖→+∞. |
Finally, the assertion (A2) is also valid.
This completes the proof.
Mahmoud El Ahmadi: Writing-original draft, Writing-review & editing; Mohammed Barghouthe: Formal Analysis, Methodology; Anass Lamaizi: Formal Analysis; Mohammed Berrajaa: Supervision, Validation.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the referees for their valuable comments and suggestions, which have improved the quality of this paper.
The authors declare there is no conflict of interest.
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