A positive solution for an asymptotically cubic quasilinear Schrödinger equation

We consider the following quasilinear Schrodinger equation \begin{document} $ - \Delta u + V(x)u - \Delta ({u^2})u = q(x)g(u),\;\;\;\;x \in {\mathbb{R}^N}, $ \end{document} where \begin{document}$N≥ 1$\end{document} , \begin{document}$0 , \begin{document}$g∈ C(\mathbb{R}^+, \mathbb{R})$\end{document} and \begin{document}$g(u)/u^3 \to 1$\end{document} , as \begin{document}$u \to ∞.$\end{document} We establish the existence of a positive solution to this problem by using the method developed in Szulkin and Weth [ 27 , 28 ].

1. Introduction. In this paper, we are concerned with the following quasilinear Schrödinger equation This equation is related to the modified Schrödinger equation where ψ : R × R N → C, W = W (x), x ∈ R N is a given potential, κ is a positive constant and h is a real function. The form of (2) has been derived as models of several physical phenomena, see e.g. [19], [22]. We are interested in the standing wave solutions of the form: z(t, x) = u(x)e iλt . Observe that z(t, x) solves (2) if and only if u(x) satisfies (1) with V (x) = W (x) − λ and g(u) = h(u 2 )u. The existence of solutions of (1) has been studied widely. A positive ground state solution of problem (1) with q(x)g(u) = |u| p−2 u, 4 ≤ p < 2 · 2 * , N ≥ 3 has been obtained in [20] using a constrained minimization method. In [19], the quasilinear problem was reduced to a semilinear one by using a change of variables and an Orlicz space framework was used to prove the existence of a positive solution. This method was also employed by [9], but the usual Sobolev space H 1 (R N ) was used.
The semilinear equation (i.e., κ = 0 in (1)) with asymptotically linear nonlinearity has been extensively studied. The existence of a positive solution has been proved in [26] with assuming radial symmetry. In [10] and [16], for the non symmetric asymptotically linear case, they obtained a Mountain Pass positive solution. Later in [18], under restrictive conditions on the potential V , the existence of a 52 XIANG-DONG FANG positive solution corresponding to higher energy levels was shown, see also [17]. Subsequently, the result in [17] was extended to the quasilinear Schrödinger problem (see [5]). In a recent paper [8], they proved a positive bound state with the nonlinearity f ∈ C 3 . The existence of sign-changing solutions was proved for an asymptotically linear Schrödinger equation with deepening potential well in [21].
We consider the problem Setting G(u) := u 0 g(s)ds, we suppose that V , q and g satisfy the following assumptions: (V ) V ∈ C(R N ), lim |x|→∞ V (x) = V ∞ > 0, and inf x∈R N V (x) > 0. (Q) q ∈ C(R N ), lim |x|→∞ q(x) = q ∞ > 0, inf x∈R N q(x) > 0 and q(x) ≤ q ∞ , for all x ∈ R N , with the strict inequality holding on a subset of positive Lebesgue measure in R N .
and µ is the number in the right side of the inequality in Lemma 3.15.
and (g 1 )-(g 3 ) hold and the following hypothesis holds (H) the least energy level c ∞ of (7) is an isolated critical level for I ∞ . Then (3) has a positive solution.
Note that µ is independent of the particular choice of V (x) and q(x) by (11) and the definition ofc.
In [19], it is stated that V (x) ≤ V ∞ for all x ∈ R N (see (V 4) there) and the nonlinearity is autonomous, that is, q(x)g(u) := λ|u| p−1 u, 4 ≤ p < 2 · 2 * . In our paper (V ) is weaker. It may be possible that V (x) > V ∞ in our case. But we need an additional condition (R 2 ) to have delicate estimates for the energy of the functional. The existence of a ground state solution was proved in [19], while the minimal energy level need not be attained in our case (see Proof of Theorem 1.1).
To our knowledge, there is only a paper [5] concerned with the existence of solutions to quasilinear Schrödinger equations when the nonlinearity is asymptotically cubic at infinity and the potential q(x) ≤ q ∞ (or V (x) ≥ V ∞ equivalently). The conditions on the potential in [5] (see (a 1 ) − (a 6 ) there) are somewhat stronger than ours. For the nonlinearity, an example is the function g(u) = u 3 satisfying our assumptions but not the condition (g 2 ) in [5], since there Q(u) = 1 4 g(u)u − G(u) ≡ 0, for all u > 0.

Remark 2.
Since the solutions set of the limiting equation might be complicated, we need the hypothesis (H) which comes from [3] (see also [5]) in order to make sure that our minimax value will not coincide with the critical values of (7).
According to Theorems 1.2 and 1.3 in [2](take p = 3 there), it is easy to see that the ground state solution for the limiting problem (7) is unique and non-degenerate with g(u) = u 3 (see also [1,23]).
Notation. C, C 1 , C 2 , . . . will denote different positive constants whose exact value is inessential. B ρ (y) := {x ∈ R N : |x − y| < ρ}. The usual norm in the Lebesgue measure L p (R N ) is denoted by u p . E denotes the Sobolev space H 1 (R N ) with the norm and S is the unit sphere in E. For y ∈ R N , let y * u for the translate of u ∈ E, that is, (y * u)(x) := u(x − y).
2. Preliminary results. We observe that formally the problem (3) is the Euler-Lagrange equation associated with the energy functional Obviously the presence of the second order nonhomogeneous term ∆(u 2 )u prevents us to work directly with the functional J in E. To overcome this difficulty, we employ an argument developed in [9]. We make a change of variables v := f −1 (u), where f is defined by Let us collect some properties of f , which have been proved in [9,11,12,15,30]. (1) f is uniquely defined, C ∞ and invertible; (2) |f (t)| ≤ 1 for all t ∈ R; (3) |f (t)| ≤ |t| for all t ∈ R; (4) f (t)/t → 1 as t → 0; (9) there exists a positive constant C such that |f (t)| ≥ C|t| for |t| ≤ 1 and |f (t)| ≥ C|t| 1/2 for |t| ≥ 1; Therefore, after the change of variables, we have the following functional which is well defined on E and belongs to C 1 under the assumptions (V ), (Q), (g 1 ) and (g 2 ). Moreover, the critical points of I correspond to the weak solutions of the Euler-Lagrange equation According to [9], if v ∈ E is a critical point of the functional I, then u = f (v) ∈ E is a solution of (3). Define the Nehari manifold For the limiting problem − ∆v =g(v) , we define the functional and the corresponding Nehari manifold Note that (R 1 ) guarantees that (1.3) in [4] holds. It is easy to see thatg satisfies all assumptions of Theorem 1 in [4], so there exists a radially symmetric positive ground state solution u ∞ ∈ C 2 (R N ) associated to equation (7). Let By the condition (Q), we have E ⊂ E ∞ and E = ∅. Obviously, not all functions in E ∞ belong to E, while we can prove that the modified functions do (see Lemma 3.10). Under our assumptions, we can not make sure whether or not M is of class C 1 , so we take the methods developed by [27,28].
(1) For every u ∈ M, there is s > 0 such that su ∈ S ρ , then I(u) = I(t u u) ≥ I(su) by Lemma 3.2-(1). Hence inf M I ≥ inf Sρ I. We claim that there exist C 1 , ρ > 0 such that Arguing by contradiction, there is u n → 0 in E such that as n → ∞. Using the Hölder inequality, Hence v n = 1 and v n → 0 in E, a contradiction. By (Q), Lemma 3.1-(1), Lemma 2.1-(3),(7) and the Sobolev inequality, we obtain that Take ε small enough, such that for small ρ, I(u) ≥ C 4 u 2 −C 5 u p/2 and inf Sρ I > 0.
(3) Without loss of generality, we may assume that V ⊂ S. Suppose by contradiction that there exist u n ∈ V and w n = t n u n , such that u n → u, t n → ∞ and I(w n ) ≥ 0. It follows from (g 2 ) and Lemma 2.
The above follows from a similar argument as Lemmas 3.3 and 4.4 in [15], so we omit it.
Proof. Arguing indirectly, suppose u n → ∞ and v n := un un . If then v n → 0 in L p/2 (R N ) for 2 < p/2 < 2 * by P.L. Lions' lemma (cf. [29], Lemma 1.21). It follows from Lemma 3.1-(1) and Lemma 2.1-(3), (7) that G(f (tv n )) → 0 for every t > 0. Since (u n ) ⊂ M is a Palais-Smale sequence, there exists d > 0 such that d ≥ I(u n ) ≥ I(tv n ) for every t > 0 and using Lemma 2.1-(9) where o(1) → ∞, as n → ∞. We have a contradiction with t large enough. Then there is a sequence (y n ) ⊂ R N and δ > 0 such that If the sequence (y n ) is bounded in R N , going if necessary to a subsequence, v n v = 0 by the local compactness of the Sobolev embedding theorem. Note that Let x ∈ R N be such that v(x) = 0. Then u n (x) → ∞. It follows from (g 2 ) and Lemma 2.1- (5) that By the Lebesgue dominated convergence theorem, Then |y n | → ∞, we can assume that y n * v n w = 0. Similarly, we obtain that Proof. Taking the same argument as Lemma 4.6 in [15], we have that t n → ∞.
By Lemma 3.2, the mapping m is bijective and the inverse of m is given by m −1 : M → U, m −1 (u) = u u . Lemma 3.7. The mapping m −1 (u) is Lipschitz continuous.
Proof. Taking an argument as in [27], we have by Lemma 3.3, for all u, v ∈ M, So we have the following lemma. According to Lemma 4.8 in [15], the following lemma follows from Lemmas 3.2 and 3.3.
(2) If (w n ) is a Palais-Smale sequence for Ψ, then (m(w n )) is a Palais-Smale sequence for I. If (u n ) ⊂ M is a bounded Palais-Smale sequence for I, then (m −1 (u n )) is a Palais-Smale sequence for Ψ.  Proof. Let u ∞ ∈ M ∞ such that I ∞ (u ∞ ) = c ∞ . It is obvious that u ∞ ∈ E ∞ . In our case we cannot make sure that u ∞ ∈ E. But we prove the modified function does. Let u y ∞ := y * u ∞ . Using Lemma 2.1-(5),(10), (g 2 ) and the translation invariance, we obtain that as r → ∞. So there exist η < 0 and R > 0 such that J∞(ru y ∞ ) r 2 ≤ η, for r ≥ R. It follows from (V ), Lemma 2.1-(3) and the Lebesgue dominated convergence theorem that by (Q), (g 2 ), (g 3 ) and Lemma 2.1-(7). Then Note that J(ru y ∞ ) > 0 for r small enough, and the similar argument as in the proof of Lemma 3.2-(1) shows that J(ru y ∞ ) r 2 is strictly decreasing in r ∈ (0, ∞).

POSITIVE SOLUTION FOR SCHRÖDINGER EQUATION 59
Then, there exists a unique T y ∈ (0, R) such that J(T y u y ∞ ) = 0, i.e. T y u y ∞ ∈ M for |y| ≥ S (By Lemma 3.2-(2), we have u y ∞ ∈ E for |y| ≥ S). Therefore s is strictly increasing for s > 0, we get T y → 1, as |y| → ∞. Hence c 0 ≤ I(T y u y ∞ ) → c ∞ . In the following, we describe a splitting lemma on M.
Lemma 3.11. If there exists (u n ) ⊂ M such that then replacing {u n } if necessary by a subsequence, there exists a solution u 0 ∈ E of and k sequences (y j n ) ⊂ R N satisfying |y j n | → ∞, |y j n − y j n | → ∞, j = j , n → ∞,

Proof. By Lemma 3.4, (u n ) is bounded in E.
Going if necessary to a subsequence, u n u 0 in E. We claim that We prove the second. Note that [f (s)f (s)] ≤ [f (s)] 2 ≤ 1. For R > 0, using the mean value theorem and the Hölder inequality, Thus, for every ε > 0, there exists R > 0 such that

XIANG-DONG FANG
It follows from the Rellich theorem and Hölder inequality that The first is similar.
Let u 2 n := u 1 n − (−y 1 n ) * u 1 , we have u 2 n 0 in E and take the same argument as above except that I should be replaced by I ∞ . Since I ∞ (u) ≥ c ∞ > 0 for every nontrivial critical point u of I ∞ , the iteration must terminate after a finite number of steps with (I(u n )) is bounded.
where C 1 and C 2 depend on the bound on u n and V (x).
Proof. See Lemma 3.2 of [14] in detail. For the sake of completeness, we include a sketch of it. Obviously, we have by , Arguing by contradiction, assume u n = 0 such that Set A n := {x ∈ R N : |u n (x)| ≥ C} and B n := R N \ A n , for a given C > 0. Note that (u n ) is bounded in E, for ε > 0 there exists C large enough such that |A n | ≤ ε. It follows from Lemma 2.1-(6) that the function f (t) t is decreasing for t > 0. Hence Take ε small enough, then we have by Hölder inequality and (V ) a contradiction with (10).
has a nontrivial ground state solution.
Proof. Though Ekeland's variational principle (cf: Theorem 4.8.1 in [7]) follows in a complete metric space, it still holds in U by Lemma 3.5 (the possibility that the limit reaches the boundary of U is ruled out). So there exists (w n ) ⊂ U satisfying Ψ(w n ) → c 0 and Ψ (w n ) → 0. Let u n := m(w n ). Then (u n ) is a Palais-Smale sequence for the functional I at level c 0 by Lemma 3.9, and moreover (u n ) is bounded in E by Lemma 3.4. According to Lemma 3.11 and c 0 < c ∞ , we have R N |∇(u n − u 0 )| 2 + f 2 (u n − u 0 ) → 0. Lemma 3.12 implies that u n − u 0 → 0. Using I(m(w n )) → I(u 0 ), m −1 (u 0 ) ∈ U and then u 0 ∈ M in view of Lemmas 3.5 and 3.8.
(4) Given y ∈ R N , β(y * u) = β(u) + y. Now we define b := inf u∈M, β(u)=0 I(u) = inf w∈U, β(m(w))=0 Ψ(w). It is easy to see that b ≥ c 0 = c ∞ . Proof. By the definition of b, we get a minimizing sequence (v n ) ⊂ U of Ψ with β(m(v n )) = 0 at level b. According to Ekeland's variational principle, we can find some (w n ) ⊂ U such that Ψ(w n ) → b, Ψ (w n ) → 0, w n − v n → 0. Put u n := m(w n ), then (u n ) ⊂ M is a Palais-Smale sequence for I at level b. Using Lemma 3.4, (u n ) is bounded and u n u 0 after passing to a subsequence. It follows from the property (1) of β that β(u n ) is bounded in R N .
According to the claim, we have u n → u 0 in E by Lemmas 3.11 and 3.12. Taking the same argument in Proposition 1, u 0 ∈ M.
Next, we consider the case b > c 0 = c ∞ . By the proof of Lemma 3.10, there exist S > 0 and T y ∈ (0, R) for some fixed R > 1 such that T y u y ∞ ∈ M, for |y| ≥ S. It is clear that T y → 1, as |y| → ∞ and is continuous with respect to y, and moreover sup |y|≥S T y < R is independent of the particular choice of V (x) and q(x). Now define the continuous operator Γ : R N → M as Γ[y] := T y u y ∞ ∈ M, for |y| ≥ S. It follows from the properties of β that β(Γ[y]) = y. Lemma 3.14. There exists some δ > 0 (with c ∞ + δ < b) such that β(u) = 0 for every u ∈ M ∩ I c∞+δ , where I c := {u ∈ E : I(u) ≤ c}.
Proof. According to the definition of b and b > c 0 = c ∞ , it is obvious that the conclusion holds.
Proof. In view of (g 2 ) and (g 3 ), G(s) ≤ 1 4 s 4 , for any s > 0. We have by Lemma 2.1-(3), (7), Proof of Theorem 1.1. By Lemma 3.10, we have c 0 ≤ c ∞ . If c 0 < c ∞ , we get a nontrivial solution of (3) in view of Proposition 1. In the case b = c 0 = c ∞ , Proposition 2 gives the desired conclusion. So we will consider the case b > c 0 = c ∞ . Arguing by contradiction, suppose the functional I does not have a critical value in (c ∞ ,c). By [24] P.86, there exists a decreasing flow η with η : Ψc \ Kc → Ψ c∞+δ (rule out the possibility that the flow η reaches at boundary of U by Lemmas 3.5 and 3.9), where Ψ c := {w ∈ U : Ψ(w) ≤ c}, K c := {w ∈ U : Ψ (w) = 0 and Ψ(w) = c} and δ is as in Lemma 3.14. Moreover, η(u) = u for every u ∈ Ψ c∞+δ . By Lemmas 3.13 and 3.14, there exists ρ 1 ≥ S > 0 such that for every ρ > ρ 1 Define a continuous map h : B ρ (0) → ∂B ρ (0) as According to (12), h(y) = y for y ∈ ∂B ρ (0), a contradiction with D.11 in [29]. Arguing as Lemma 2.4 in [8], it is easy to see that the solution u 0 does not change sign. The elliptic regularity theory implies that u 0 ∈ C 2 (R N ) (see Lemma 1.30 in [29]). It follows from the strong maximum principle that u 0 is positive.