u ∈ W 1,p(x) (R N),

This work deals with the nonlocal p(x)-Laplacian equations in R N with nonvariational form ( A(u) − � p(x) u + j uj p(x)−2 u � = B(u)f(x, u) in R N , u 2 W 1,p(x) (R N ), and with the variational form 8 > > > > a � Z RN jr uj p(x) + j uj p(x) p(x) dx � (−� p(x) u + j uj p(x)−2 u)

The study on the problems of the nonlocal p(x)-Laplacian has attracted more and more interest in the recent years(e.g., see [1,2,3]), they mainly concerned the problems of the bounded domain, however the study on the existence of solutions for problems of nonlocal p(x)-Laplacian in R N is rare.We know that in the study of p-Laplacian equations in R N , a main difficulty arises from the lack of compactness.In this paper, we study the nonlocal p(x)-Laplacian equations in R N with non-variational form and with the variational form where A and B are two functionals defined on W 1,p(x) (R N ), F (x, t) = t 0 f (x, s)ds, and a is allowed to be singular at zero.To deal with the problems (1.1) and (1.2), we will overcome the difficulty caused by the absence of compactness through the method of weight function.
The variable exponent problems have been studied by many authors.We refer to [4,5] for applied background, to [6,7,8] for the variable exponent Lebesgue-Sobolev spaces and to [9,10,11,12] for the p(x)-Laplacian equations without nonlocal coefficient.This paper is organized as follows: in Section 2, we deal with the problem with nonvariational form; in Section 3, we deal with the problem with variational form.

The non-variational form
Denote by S(Ω) the set of all measurable real functions defined on Ω.Two measurable functions are considered as the same element of S(Ω) when they are equal almost everywhere.For p ∈ L ∞ + (Ω), define For some basic properties of the spaces L p(x) (Ω), W 1,p(x) (Ω) and W 1,p(x) 0 (Ω) we may refer to [6,7,8].
In this section we consider problem (1.1), the nonlocal p(x)-Laplacian equation in R N without the variational structure.
In what follows, for simplicity, we write X = W 1,p(x) (R N ) and c i , C, C i are positive constants.
Define the mapping T , G, L p(•) and N f : X → X * respectively by Proposition 2.3.Suppose that A satisfies the following condition: (A 1 ) A : X → [0, +∞) is continuous and bounded on any bounded subset of X, A(u) > 0 for all u ∈ X \ {0}, and for any bounded sequence {u n } ⊂ X for which A(u n ) → 0, u n must converge strongly to 0 in X. EJQTDE, 2012 No. 76, p. 3 Then the mapping T : X → X * is continuous and bounded, and is of type (S + ).
The proof is similar to [3], so omit it.Proposition 2.4.Suppose that the following conditions are satisfied: (B 1 ) The functional B : X → R is continuous and bounded on any bounded subset of X.
Then the mapping G : X → X * is completely continuous.
Proof.Under the condition (f 1 ), the mapping N f : X → X * is sequentially weaklystrongly continuous (see [9]).The continuity of G is obvious.Assume {u n } is bounded, then there exists a subsequence We know that the sum of an (S + ) type mapping and a completely continuous mapping is of type (S + ), so from Propositions 2.3 and 2.4 we have the following: Corollary 2.1.Let (A 1 ), (B 1 ) and (f 1 ) hold.Then the mapping T − G : X → X * is continuous and bounded, and is of type (S + ).
Theorem 2.1.Let (A 1 ), (B 1 ) and (f 1 ) hold.Suppose that the following conditions are satisfied: (A 2 )There are constants α ∈ R, M > 0 and c 1 > 0 such that Then problem (1.1) has at least one solution.If, in addition, α+p − > 1, then the mapping T − G : X → X * is subjective, and consequently for any h ∈ X * the operator equation T (u) − G(u) = h has at least one solution.
Proof.Under the hypotheses of Theorem 2.1, by Corollary 2.1, the mapping T − G : X → X * is continuous and bounded, and is of type (S + ).For sufficiently large u , we have that So, by the degree theory for (S + ) type mappings(see [13]), for R > 0 large enough, we have deg(T − G, B(0, R), 0) = 1, and consequently, there exists u ∈ B(0, R) such that T (u) − G(u) = 0, that is, (1.1) has at least one solution u ∈ B(0, R).If in addition, that is, the mapping T − G is coercive, and consequently, by the surjection theorem for the pseudomonotone mappings(see [14]), the mapping T − G is surjective.

The variational form
In this section we consider problem (1.2) with variational form, where f satisfies condition (f 1 ), k and g are two real functions satisfying the following conditions.(k 1 ) k : (0, +∞) → (0, +∞) is continuous and k ∈ L 1 (0, t) for any t > 0.
Note that the function k satisfying (k 1 ) may be singular at t = 0. Define Proposition 3.1.Let (f 1 ), (k 1 ), (g 1 ) hold.Then the following statements hold: Thus u ∈ X \ {0} is a (weak) solution of (1.2) if and only if u is a nontrivial critical point of E.
(3) The functional J : X → R is sequentially weakly lower semi-continuous, Φ : X → R is sequentially weakly continuous, and thus E is sequentially weakly lower semi-continuous.
Proof.The proof of statements ( 1) and ( 2) is obvious.Since the function k(t) is increasing and the functional I 1 is sequentially weakly lower semi-continuous, we can see that the functional J : X → R is sequentially weakly lower semi-continuous.Moreover, under the condition (f 1 ), Φ and Φ ′ are sequentially weakly-strongly continuous.Note let D ∈ X \ {0}, it is clear that the mapping J ′ and E ′ : D → X * are bounded.In order to prove that J ′ : D → X * is of type (S + ), assuming that {u n } ⊂ D, u n ⇀ u in X and lim n→∞ J ′ (u n )(u n − u) ≤ 0, then there exist positive constants c 1 and c 2 such that c 1 ≤ dx ≤ c 2 and so there exist positive constants c 3 and c 4 such that is of type (S + ), we obtain u n k → u in X.This shows that the mapping J ′ : D → X * is of type (S + ).Moreover, since Φ ′ is sequentially weakly-strongly continuous, the mapping E ′ : D → X * is of type (S + ).
Remark 3.1.To verify that E satisfies (P.S) condition on E, it is enough to verify that any (P.S) sequence is bounded.Theorem 3.1.Let (f 1 ), (k 1 ), (g 1 ) and the following conditions hold.(k 2 ) There are positive constants α 1 , M and C such that k(t) ≥ Ct α 1 for t ≥ M. (g 2 ) There are positive constants β 1 and Then the functional E is coercive and obtain its infimum in X at some u 0 ∈ X.Thus u 0 is a solution of (1.2) if E is differentiable at u 0 , and in particular, if u 0 = 0.
Proof.For u large enough, by (f 1 ), (k 2 ), (g 2 ) and (E 2 ), we have that hence E is coercive.Since E is sequentially weakly lower semi-continuous and X is reflexive, E attains its infimum in X at some u 0 ∈ X.In the case where E is differential at EJQTDE, 2012 No. 76, p. 6 u 0 , u 0 is a solution of (1.2).The proof is complete.
As X is a separable and reflexive Banach space, there exist (see [15,Section 17]) Proposition 3.2.Assume that Φ : X → R is weakly-strongly continuous and Φ(0) = 0, γ > 0 is a given positive constant.Set The proof of the Proposition 3.2 is similar to [9], here we omit it.
As b 0 ≡ 0 and b 0 ≥ 0, we can find a bounded open set Ω ⊂ R N , such that b 0 (x) > 0 for x ∈ Ω.The space W 1,p(x) 0 (Ω) is a subspace of X.For any k, we can choose a k dimensional linear subspace E k of W 1,p(x) 0 (Ω) such that E k ⊂ C ∞ 0 (Ω).As the norms on E k are equivalent each other, there exists ρ k ∈ (0, 1) such that u ρ (k) and sufficiently small λ > 0 we have As q + 0 β 2 < α 2 p − , we can find λ k ∈ (0, 1) and ǫ k > 0 such that By the genus theory (see [16]), each c k is a critical value of E, hence there is a sequence of solutions {±u By the coerciveness of E, there exists a constant γ > 0 such that E(u) > 0 when u ≥ γ.Taking arbitrarily A ∈ Σ k , then γ(A) ≥ k.Let Y k and Z k be the subspaces of X as mentioned in (3.1), according to the properties of genus we know that A ∩ Z k = ∅.Let EJQTDE, 2012 No. 76, p. 8 Proposition 3.3.Let (f 1 ), (k 1 ), (g 1 )and the following conditions be satisfied: 4 ) There exists λ > 0 such that λ k(t) ≥ tk(t) for t > 0. (g 4 ) There exists ν > 0 such that νĝ(t) ≤ g(t)t for t > 0. (f 4 ) There exists µ > 0, such that 0 < µF (x, t) ≤ f (x, t)t, t = 0 and ∀x ∈ R N .(E 4 ) λp + < νµ.Then E satisfies condition (P.S) c for any c = 0.