An inhomogeneous perturbation for a class of nonlinear scalar field equations

This paper deals with a class of nonlinear scalar field equations with an inhomogeneous perturbation. Two positive solutions were obtained using the variational methods.


Introduction and main result
In this paper, we consider the following nonlinear scalar field equations with an inhomogeneous perturbation u = g(u) + h(x), u ∈ H 1 R N , N ≥ 3. (1.1) Actually, g satisfies the Berestycki-Lions conditions: (g 1 ) g ∈ C(R, R); (g 2 ) -∞ < lim inf s→0 + g(s) s ≤ lim sup s→0 + g(s) s = -m < 0; (g 3  In [2,8,9], the authors studied Eq. (1.2) with a homogeneous perturbation and obtained a positive solution using the variational method. In this paper, we consider the effect of an inhomogeneous perturbation. In other words, we investigate the existence of positive solutions of Eq. (1.1). With regard to Eq. (1.1), there are some results, for example, [4,14,15]. Compared with those results, in the present paper, the nonlinearity g is almost optimal.
Set | · | s = ( R N | · | s dx) 1 s . Using the variational method, we get the following We have something to say about the perturbation h. The assumptions (h 1 ) and (h 2 ) are necessary, and (h 3 ) is to overcome the lack of compactness. Moreover, to prove the second positive solution, we need to use the Pohožaev identity, and then (h 4 ) seems appropriate.
Set f (s) = g(s) + ms, then Eq. (1.1) equals to the following equation We only need to prove the following where ω N denotes the volume of the unit ball in R N . By computing, we have h i ∈ L 2 (R N ), The rest of the paper is organized as follows: In Sect. 2, we introduce some preliminaries. In Sect. 3, we give the proof of the first positive solution. Section 4 is devoted to obtaining the second positive solution.

Preliminaries
From now on, C, C 1 , C 2 , . . . , denotes various positive constant, u ± = max{±u, 0} and (H, · ) is a Hilbert space, where To ensure the positivity of solutions and for simplicity, we always take f (s) = 0 for all s ≤ 0.
As is well known, the solutions of Eq. (1.3) correspond to the critical points of the following energy functional By Principle of symmetric criticality [13], we know that if u is a critical point of I restricted to H, then u is a critical point of I in H 1 (R N ). Set

The first positive solution of Eq. (1.3)
In this section, we prove that Eq. (1.3) has a local minimal solution. Proof From (2.1), it follows that Combining with Hölder's inequality and Sobolev's inequality, we get By (h 2 ), there exist L ∈ (0, θ ) and ϕ ∈ H such that R N hϕ dx > 0 and 0 ≤ ϕ(x) ≤ L for all x ∈ R N . Then we have which implies that there exists t 0 > 0 such that t 0 ϕ ≤ ρ and I(t 0 ϕ) < 0. Hence m < 0. It is obvious that m > -∞.

The second positive solution of Eq. (1.3)
In this section, we prove that Eq. (1.3) has another positive solution. In order to obtain a bounded Palais-Smale sequence, we use the following Jeanjean's theorem [7].

Theorem 4.1 Let X be a Banach space equipped with a norm · X and let J ⊂ R + be an interval. We consider a family { 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 {u n } ⊂ X such that

Moreover, the map μ → c μ is non-increasing and continuous from the left.
From [3], we know that Eq. (1.2) has a positive ground state solution ω ∈ H and Then there exists δ ∈ (0, 1) such that In Theorem 4.1, we set By Hölder's inequality and Sobolev's inequality, we have which implies A(u) → +∞ as u → +∞. Note that In the following, I 1 will always replace I. The next lemma is to verify the assumptions of Theorem 4.1.
The proof of the second positive solution We choose μ n ∈ J and μ n 1. Lemma 4.3 implies that there exists a positive sequence {u μ n } ⊂ H such that I μ n (u μ n ) = c μ n and I μ n (u μ n ) = 0. Note that (h 4 ) holds. Then we have (4.6) and the following Pohožaev identity Combining with Hölder's inequality and Sobolev's inequality, we get (4.7) By (2.1), (4.6), (4.7), Hölder's inequality, and Sobolev's inequality, we yield 1 2 . So R N u 2 μ n dx ≤ C for all n ∈ N.
By Sect. 3 and Sect. 4, we complete the proof of Theorem 1.2.