A nontrivial solution for a nonautonomous Choquard equation with general nonlinearity

. With the help of the monotonicity trick, a nonautonomous Choquard equations with general nonlinearity is studied and a nontrivial solution is obtained


Introduction and main result
In the paper, we explore nontrivial solutions for the following nonlocal problem where 1  |x| * u 2 = R 3 u 2 (y) |x−y| dy, the nonlinearity g satisfies general subcritical growth conditions (g 3 ) lim s→+∞ g(s) s 5 = 0; and the potential function V verifies (V 1 ) V ∈ C(R 3 , (−m, 0]) and lim |x|→∞ V(x) = 0; Eq. (1. 2) appeared at least as early as in 1954, in a work by S. I. Pekar describing the quantum mechanics of a polaron at rest [11].In 1976, P. Choquard used Eq.(1.2) to describe an electron trapped in its own hole in a certain approximation to Hartree-Fock theory of one component plasma [4].For more details in the physics aspects, please refer to [7].Therefore, many scholars have carried out in-depth research on Choquard equations and related problems.For recent results, we refer the readers to [6,8,9,12,14] and references therein.See also [10] for a broad survey of Choquard equations.
It is important to point out that Liu et al. in [6] considered the following special case of Eq. (1.1) Under the assumptions (g 1 )-(g 3 ), they investigated ground states of Eq. (1.3) by using the Pohožaev manifold method.In the present paper, we study Eq.(1.1) which can be regarded as the perturbation equation of Eq. (1.3).By using the monotonicity trick we obtain the following main result.
For the rest of this paper, we make the following marks.H := H 1 (R 3 ) is the usual Sobolev space endowed with the standard norm ∥ • ∥.L s (R 3 ), 2 ≤ s ≤ 6, denotes the usual Lebesgue space with the norm | • | s .C, C 1 , C 2 , . . .denote different positive constants whose exact value is inessential.For any u ∈ H, we define u t (•) := u(t −1 •) for t > 0.
It is widely known that the solutions of Eq. (1.4) correspond to the critical points of the functional defined by Using the Hardy-Littlewood-Sobolev inequality [5], one has Combining with ( f 1 )-( f 3 ) and (K 1 ) we know that I is well defined and I is of C 1 .But it is hard to obtain a bounded (PS) sequence for the functional I under the assumptions ( f 1 )-( f 3 ).
In addition, another difficulty we face is the lack of space compactness.

Preliminaries
In order to prove Theorem 1.2, we cannot directly apply the mountain pass theorem [1].Instead we use an indirect approach which dated to Struwe [13] and was developed by Jeanjean in [2].Exactly, we apply the following Proposition 2.1.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 {u n } ⊂ X such that Moreover, the map µ → c µ is non-increasing and continuous from the left.

The proof of Theorem 1.2
The following lemma is to verify the assumptions of Proposition 2.1.
Combining with the Hardy-Littlewood-Sobolev and Sobolev inequality, for any u ∈ H and µ ∈ J one has which implies that there exist α, ρ > 0 such that I µ (u) ≥ α for all µ ∈ J and ∥u∥ = ρ.Let ω ∈ H be a positive ground state solution of Eq. (2.2) with µ = 1.For any µ ∈ J, one has Combining with there exists t 0 > 0 such that ∥ω t 0 ∥ > ρ and Proof.According to the proof of Lemma 3.1, for any µ ∈ J, there exists t µ ∈ (0, t 0 ) such that Otherwise, there exists µ n ∈ J such that t µ n → 0 and then It is a contradiction.Note that K(x) ≤ m and K(x) ̸ ≡ m.Define . Suppose that (K 1 ) and ( f 1 )-( f 3 ) hold and that {u n } ⊂ H is a bounded (PS) c µ sequence for I µ .Then there exists u ∈ H, k ∈ N, v i ∈ H\{0}, y n,i ∈ R 3 for 1 ≤ i ≤ k such that up to a subsequence, where we agree that in the case k = 0 the above holds without v i , y n,i .
Proof.The proof is in the spirit of [3].Obviously, there exists u ∈ H such that up to a subsequence and Therefore, Define If β 1 = 0, one sees u n,1 → 0 in L p (R 3 ) with 2 < p < 6 from Lion's lemma [15].Then 3 and, up to a subsequence, we can assume that |y n,1 | → ∞.Thus ).If u n,2 → 0 in H, we are done.So we can assume that {u n,2 } does not converge strongly to 0 in H. Thus up to a subsequence u n,2 ⇀ 0 in H, u n,2 → 0 in L p loc (R 3 ) and u n,2 (x) → 0 a.e. in R 3 .Thus we have and Therefore, We replaced u n,1 by u n,2 and repeat the above arguments.If ).
Again we repeat the above arguments, then there exists k ∈ N, v i ∈ H\{0}, y n,i ∈ R 3 for 1 ≤ i ≤ k such that up to a subsequence, and Note that there exists α > 0 such that ∥v∥ ≥ α for any v ∈ {v ∈ H : v ̸ = 0 and (I ∞ µ ) ′ (v) = 0}.The iterations must stop after steps because {u n } is bounded in H.