Electronic Journal of Qualitative Theory of Differential Equations

In this paper, we consider a mixed boundary value problem for nonuniformly elliptic equation in a variable exponent Sobolev space containing p(·)-Laplacian and mean curvature operator. More precisely, we are concerned with the problem with the Dirichlet condition on a part of the boundary and the Steklov boundary condition on an another part of the boundary. We show the existence of a nontrivial weak solution and at least two nontrivial weak solutions according to some hypotheses on given functions.


Introduction
In this paper, we consider the following problem (1.1) Here Ω is a bounded domain of R d (d ≥ 2) with a Lipschitz-continuous (C 0,1 for short) boundary Γ satisfying that and the vector field n denotes the unit, outer, normal vector to Γ.The function a(x, ξ) is a Carathéodory function on Ω × R d satisfying some structure conditions associated with an anisotropic exponent function p(x).Then the operator u → div [a(x, ∇u(x))] is more general than the p(•)-Laplacian ∆ p(x) u(x) = div [|∇u(x)| p(x)−2 ∇u(x)] and the mean curvature Email: aramaki@hctv.ne.jp operator div [(1 + |∇u(x)| 2 ) (p(x)−2)/2 ∇u(x)].These generalities bring about difficulties and requires some conditions.We impose the mixed boundary conditions, that is, the Dirichlet condition on Γ 1 and the Steklov condition on Γ 2 .The given data f : Ω × R → R and g : Γ 2 × R → R are Carathéodory functions satisfying some conditions.
However, since we can only find a few of papers associate with the problem with the mixed boundary condition in variable exponent Sobolev space as in (1.1).See Aramaki [1][2][3].We are convinced of the reason for existence of this paper.
In particular, the authors in [10] considered the problem (1.1) when p(x) = p = const.and Γ 2 = ∅, and derived the existence of a nontrivial weak solution to (1.1).This paper is an extension of the article [10] to the case of variable exponent and mixed boundary value problem.In the paper [10], the authors derived the weakly continuous differentiability of the corresponding energy functional and then applied a version of the Mountain-pass lemma introduced in Duc [9].However, in this paper we show that the corresponding energy functional is of class C 1 , and so it suffices to apply the standard Mountain-pass lemma.
The paper is organized as follows.Section 2 consists of two subsections.In Subsection 2.1, we recall some results on variable exponent Lebesgue-Sobolev spaces.In Subsection 2.2, we give the assumptions to the main theorems.In Section 3, we state the main theorems (Theorem 3.3 and Theorem 3.5) on the existence of at least one and two nontrivial weak solutions.The proofs of the main theorems are given in Section 4.

Preliminaries and the main theorems
Let Ω be a bounded domain in R d (d ≥ 2) with a C 0,1 -boundary Γ.Moreover, we assume that Γ satisfies (1.2).
Throughout this paper, we only consider vector spaces of real valued functions over R. For any space B, we denote B d by the boldface character B. Hereafter, we use this character to denote vectors and vector-valued functions, and we denote the standard inner product of vectors a = (a 1 , . . ., a d ) . Furthermore, we denote the dual space of B by B * and the duality bracket by ⟨•, •⟩ B * ,B .

Variable exponent Lebesgue and Sobolev spaces
In this subsection, we recall some well-known results on variable exponent Lebesgue-Sobolev spaces.See Diening et al. [8], Fan and Zhang [12], Kováčik and Rákosník [16] and references therein for more detail.Throughout this paper, let Ω be a bounded domain in R d with a C 0,1 -boundary Γ and Ω is locally on the same side of Γ. Define P (Ω) = {p : Ω → [1, ∞); p is a measurable function}, and for any p ∈ P (Ω), put For any measurable function u on Ω, a modular ρ p(•) = ρ p(•),Ω is defined by The variable exponent Lebesgue space is defined by Then L p(•) (Ω) is a Banach space.We also define, for any integer m ≥ 0, The following three propositions are well known (see Fan et al. [14,22], Fan and Zhao [13], Zhao et al. [27], and [25]).Proposition 2.1.Let p ∈ P (Ω) and let u, u n ∈ L p(•) (Ω) (n = 1, 2, . ..)Then we have The following proposition is a generalized Hölder inequality.Proposition 2.2.Let p ∈ P + (Ω), where Here and from now on, p ′ (•) is the conjugate exponent of p(•), that is, For p ∈ P (Ω), define Let Ω be a bounded domain with C 0,1 -boundary and let p ∈ P + (Ω) and m ≥ 0 be an integer.Then we have the following: (i) The spaces L p(•) (Ω) and W m,p(•) (Ω) are separable, reflexive and uniformly convex Banach spaces.
We say that p ∈ P (Ω) belongs to P log (Ω) if p has the log-Hölder continuity in Ω, that is, p : Ω → R satisfies that there exists a constant for all x, y ∈ Ω.
We also write P log For the proof, see [8,Corollary 11.2.4].Next we consider the notion of trace.Let Ω be a domain of R d with a C 0,1 -boundary Γ and p ∈ P + (Ω).Since W 1,p(•) (Ω) ⊂ W 1,1 loc (Ω), the trace γ(u) = u Γ to Γ of any function u in W 1,p(•) (Ω) is well defined as a function in L 1 loc (Γ).We define , where the infimum can be achieved.Then (Tr W 1,p(•) )(Γ) is a Banach space.More precisely, see [8,Chapter 12].In the later we also write Moreover, we denote equipped with the norm where the infimum can also be achieved, so for any g ∈ (Tr Let q ∈ P + (Γ) := {q ∈ P (Γ); q − > 1} and denote the surface measure on Γ induced from the Lebesgue measure dx on Ω by dσ.We define and we also define a modular on L q(•) (Γ) by Proposition 2.5.We have the following properties.
In particular, the norm ∥∇u∥
Thus we can define the norm on the space X defined by (2.1) so that which is equivalent to ∥v∥ W 1,p(•) (Ω) from Lemma 2.8.

Assumptions to the main theorems
In this subsection, we state the assumptions to the main theorems.Let x ∈ Ω such that the following conditions hold.

For the function h
x ∈ Ω, we define a modular Define our basic space then Y is a Banach space (see Lemma 2.12 below).We note that Then we have the following lemma.
(i) Y → X and ∥v∥ X ≤ ∥v∥ Y for all v ∈ Y. (ii (iv When q ∈ P log + (Ω) satisfies q(x) ≤ p * (x) for all x ∈ Ω, define By Proposition 2.3 and Lemma 2.11, there exists a constant c > 0 such that ∥v∥ L q(•) (Ω) ≤ c∥v∥ X ≤ c∥v∥ Y for all v ∈ Y, so we can see that λ q > 0.
Proof.Since ∥v∥ Y = ∥h So there exists a subsequence {v n ′ } of {v n } such that ∇v n ′ (x) → ∇v(x) a.e. in Ω.By the Fatou lemma, Thereby v ∈ Y. Applying again the Fatou lemma, lim We continue to state the assumptions of f and g in (1.1).Let f is a real Carathéodory function on Ω × R having the following properties.
Let g be a real Carathéodory function on Γ 2 × R having the following properties.

Main theorems
In this section, we state the main theorems.
Definition 3.1.We say u ∈ Y is a weak solution of (1.1) if u satisfies that , then the equation (1.1) holds in the distribution sense.
Then we obtain the following two theorems.
Remark 3.4.This theorem extends the result of [10] in which the authors considered the case where p(x) = p = const.and Γ 2 = ∅.
We impose one more assumption.
Theorem 3.5.Addition to the hypotheses of Theorem 3.3, assume that (F4) also holds.Then the problem (1.1) has at least two nontrivial weak solutions.
Remark 3.6.The authors in [17] considered the equation and Γ 2 = ∅.The authors got the same result of Theorem 3.5 under stronger hypotheses than (A1) and (A4), that is, h 1 (x) ≡ 1.However, they use an inequality A(x, tξ) ≤ t p(x) A(x, ξ) for small t > 0 which does not hold for the function in Example 2.9 (ii).To overcome their mistake, we assume a stronger condition (F4).

Proofs of Theorem 3.3 and Theorem 3.5
In this section, we give proofs of Theorem 3.3 and Theorem 3.5.In order to do so, we use the variational method.Define a functional on Y where The proof of Theorem 3.3 consists of several lemmas and propositions.
(v) There exists a constant c ) for all t ∈ R and a.e.x ∈ Γ 2 .
(v) and (vi) follow from the same arguments as (iii) and (iv), respectively.
Proposition 4.2.The functionals E, J, K ∈ C 1 (Y, R) and the Fréchet derivatives E ′ , J ′ and K ′ satisfy the following equalities.
for all u, v ∈ Y. Proof.
Step 2. We derive that E is Gateaux differentiable in Y. Let u, v ∈ Y and 0 < |t| ≤ 1.By the mean value theorem, From (A1), we have |a(x, ∇u(x) Here it follows from the Hölder inequality (Proposition 2.2), the last term of the above inequality is an integrable function independent of t.On the other hand, a(x, ξ) is a Carathéodory function, we have as t → 0. Using again the Lebesgue dominated convergence theorem, we have Thus E is Gateaux differentiable at u and the Gateaux derivative DE satisfies Clearly DE(u) is linear in Y.
Step 3. We show that for every u ∈ Y, we have We note that ∥v∥ Y = ∥h ).On the other hand, from (A1), . By the Hölder inequality (Proposition 2.2), we have Hence we see that DE(u) ∈ Y * and Step 4. We derive that the map Then (4.9) holds.So there exist a subsequence {u x ∈ Ω and all n ′ .By (4.10), In order to show that the right-hand side converges to zero, taking Proposition 2.1 into consideration, it suffices to derive that Since a(x, ξ) is a Carathéodory function, and ∇u n ′ (x) → ∇u(x) a.e.x ∈ Ω, we have As in the argument in Step 3, we have ). Step 5. We show that J and K belong to C 1 (Y, R) and (4.7) and (4.8) hold.By Lemma 4.1 (iii) and [2, Proposition 2.12], the Nemytskii operator By the Hölder inequality (Proposition 2.2), Since Y → X → L q(•) (Ω), we can see that J ∈ C 1 (Y, R) and (4.7) holds.Similarly, we can prove that K ∈ C 1 (Y, R) and (4.8) holds.(i) The functionals J and K are weakly continuous in Y, that is, if (ii) A(x, ξ) is a Carathéodory function on Ω × R d and A(x, ξ) ≥ 0 by (A4).Moreover, from (A2), A(x, ξ) is convex with respect to ξ for a.e.x ∈ Ω.If u n → u weakly in Y, then u n , u ∈ W 1,1 (Ω) and u n → u strongly in L 1 (Ω) and ∇u n → ∇u weakly in L 1 (Ω).Hence it follows from Struwe [21, Theorem 1.6, p. 9] that E(u) ≤ lim inf n→∞ E(u n ).
Therefore, we have Similarly, there exists there exist positive constants c 5 and c 6 such that On the other hand, from (A4), Thus we have where k 3 = k 0 /2 and c 3 = max{c 5 , c 6 } for all u ∈ Y with ∥u∥ Y < 1.
Step 1.The sequence {u n } is bounded in Y. Indeed, if {u n } is unbounded, there exists a subsequence {u n ′ } of {u n } such that ∥u n ′ ∥ ≥ n ′ for any n ′ ∈ N. By Lemma 4.5 (ii), This contradicts lim n ′ →∞ I(u n ′ ) = γ.
Step 2. Since {u n } is bounded in Y and Y is a reflexive Banach space, passing to a subsequence, we may assume that u n → u weakly in Y.By Proposition 4.4 (ii) and (iii), On the other hand, by Proposition 4.4 (iii) and the above equality, Thus by Proposition 4.4 (ii), we have lim n→∞ E(u n ) = E(u).
Step 3. We show that u n → u strongly in Y, that is, ρ p(•),h 1 (•) (∇u n − ∇u) → 0 as n → ∞.If this is not satisfied, there exist a subsequence {u n ′ } of {u n } and ε 0 > 0 such that Letting n ′ → ∞ and using Step 2, we have On the other hand, since This contradicts (4.12).

Remark 4 . 3 .
When p(•) = p = const.and Γ 2 = ∅, the authors of[10] only prove the weakly continuously differentiable on Y, and so they must use a version of the Mountain-pass lemma introduced in[9].However, since we derived that E belongs to C 1 (Y, R), it suffices to use the standard Mountain-pass lemma later.Proposition 4.4.