POSITIVE SOLUTION FOR THE KIRCHHOFF-TYPE EQUATIONS INVOLVING GENERAL SUBCRITICAL GROWTH

. In this paper, the existence of a positive solution for the Kirchhoﬀ-type equations in R N is proved by using cut-oﬀ and monotonicity tricks, which unify and sharply improve the results of Li et al. [Existence of a positive solution to Kirchhoﬀ type problems without compactness conditions, J. Diﬀerential Equations 253 (2012) 2285–2294]. Our result cover the case where the nonlinearity satisﬁes asymptotically linear and superlinear at inﬁnity.

As well known, owing to the existence of the nonlocal term, usually it demands that the nonlinearity g satisfies A-R, monotonicity or supercubic growth conditions, which assures the boundedness of any (PS) or Cerami sequence. In Theorem A, the authors did not assume the above conditions. In the present paper, under the weaker assumptions, we consider equation (1.1). Our existence result reads as follows.
Thus (g 2 ) holds. By (g 6 ), there exists R > 0 such that for any s > R, one has g(s) > V + ab 2 s.

KIRCHHOFF-TYPE EQUATIONS INVOLVING GENERAL SUBCRITICAL GROWTH 447
So for any s ≥ V +ab V −ab R, we have since g(s) ≥ 0 for all s ≥ 0. Hence (g 3 ) holds and then we have the following corollary.
In [10], the authors claimed that it is not clear whether the result in Theorem A still holds for large λ > 0. In the present paper, we can give an answer. Suppose that the nonlinearity g satisfies (g 7 ) g ∈ C(R, R) and lim  The paper is organized as follows. Section 2 contains some preliminaries. In Section 3 we give the proof of theorems.

2.
Preliminaries. From now on, we will use the following notations.
• H 1 (R N ) denotes the usual Hilbert space endowed with the norm • L s (R N ) is the usual Lebesgue space endowed with the norm • C, C i denote various positive constants.
Because that we are looking for positive solution, we may assume that g(s) = 0 for all s < 0. The energy functional for equation (1.1) is defined by By (g 1 ) and (g 2 ), for any ε > 0, there exists C ε > 0 such that It is obvious that I λ ∈ C 1 (H 1 (R N ), R) and has the derivative given by Since equation (1.1) is autonomous, we can work in : u is radially symmetric}. We know that any critical point u ∈ H 1 r (R N ) is, by the principle of symmetric criticality of Palais (see Theorem 1.28 in [12]), also a critical point on To obtain a bounded Palais-Smale sequence for the functional I λ , following [7] and [10], we use a cut-off function ψ ∈ C ∞ (R + , R + ) satisfying and study the following modified functional I T λ : where for every T > 0, The functional I T λ satisfies the geometrical assumptions of the Mountain-pass theorem but we are not able to obtain the boundedness of the Palais-Smale sequences. Hence we use an indirect approach which developed by Jeanjean [6]. We apply the following Proposition 1 (See Theorem 1.1 and Lemma 2.3 in [6]). Proposition 1. Let X be a Banach space equipped with a norm · X and let J ⊂ R + be an interval. We consider a family {Φ µ } µ∈J of C 1 -functionals on X of the form where B(u) ≥ 0 for all u ∈ X and such that either A(u) → +∞ or B(u) → +∞, as u X → +∞. We assume that there are two points v 1 , v 2 in X such that Then, for almost every µ ∈ J, there is a sequence for {u n } ⊂ X, such that Moreover, the map µ → c µ is continuous from the left.
The following Pohozaev equality is important for proving the boundedness of the Palais-Smale sequence. For the proof, please see [10].

Proposition 2. If u is a nonzero solution of the equation
the following Pohozaev equality holds.
In our case, So the perturbed functional which we study is and its derivative is given by for any u, v ∈ H 1 r (R N ).
3. The proof of theorems. As a start, we give the proofs of some lemmas.
a.e. in R N . By (g 1 ) and (g 2 ), for any ε > 0, there exists C ε > 0 such that Thus combining the Hölder and Sobolev inequalities, we have Then we obtain Hence u n → u in H 1 r (R N ). We complete the proof.
The following lemma shows that u n ≤ T .
Proof. It follows from (I T λ,µn ) (u n ) = 0 and Proposition 2 that Combining (3.2) and I T λ,µn (u n ) = c µn , we have  (3.4) in which we use (3.1) and A 1 is independent of T . According the definition of ψ, we have and Then from (3.3) − (3.7), one has On the other hand, it follows from (2.1) with ε = ab 2 , (3.4), (3.5) (3.8) and We suppose by contradiction that there exists no subsequence of {u n } which is uniformly bounded by T . Without any loss of generality, we can suppose that u n > T for all n. Then combining (3.9), one has (3.10) We can find T 0 > 0 such that and λ * ∈ (0, a Hence for any λ ∈ [0, λ * ), (3.10) − (3.12) imply that which is a contradiction. We complete the proof.
Proof of Theorem 1.1. Let T and λ * be as in Lemma 3.5 and fix λ ∈ [0, λ * ). Suppose that (I T λ,µn ) (u n ) = 0 and I T λ,µn (u n ) = c µn . By Lemma 3.5, we may assume that u n ≤ T . Thus Since µ n → 1 − , I λ (u n ) → c 1 and I λ (u n ) → 0 in H 1 r (R N ). By similar argument as in the proof of Lemma 3.3, there exists u ∈ H 1 r (R N ) such that up to a subsequence, u n → u in H 1 r (R N ). Hence I λ (u) = c 1 and I λ (u) = 0. Define u ± := max{±u, 0}, then u = u + − u − . Multiplying equation (1.1) by u − , we have (a + λ u 2 ) u − 2 = 0, which deduces that u − = 0 and then u ≥ 0. By the strongly maximum principle, u is positive. We complete the proof.
Suppose that equation (1.1) has a nonzero solution u ∈ H 1 (R N ). Then one has If λ > C 1 , we obtain a contradiction.