Stationary solutions for a generalized Kadomtsev-Petviashvili equation in bounded domain ∗

In this work, we are mainly concerned with the existence of stationary solutions for the generalized Kadomtsev-Petviashvili equation in bounded domain in R


Introduction
In this work, we shall investigate the stationary solutions for the generalized Kadomtsev-Petviashvili x ∆ y u(x, y), in Ω, where D −1 x h(x, y) := x −∞ h(s, y)ds denotes the inverse operator, (x, y) := (x, y 1 , . . ., . In this paper, we utilize variational methods and some critical point theroems to study the stationary solutions for the generalized Kadomtsev-Petviashvili equation (1.1).∂ ∂t w(t, x, y) + ∂ 3 ∂x 3 w(t, x, y) + ∂ ∂x f (w(t, x, y)) = D −1 x ∆ y w(t, x, y), ( where (t, x, y) := (t, x, y 1 , . . ., y n−1 x and ∆ y are as in (1.1).A solitary wave is a solution of the form w(t, x, y) = u(x − ct, y), where c > 0 is fixed.Substituting in (1.2), we have, In [1] and [2], by virtue of the constrained minimization method, De Bouard and Saut obtained the existence and nonexistence of solitary waves in the cases where power nonlinearities f (u) = u p , p = m/n, m, n are relatively prime, n is odd.In [3,4], Zou et al. established the existence of nontrivial solitary waves of problem (1.3) by a linking theorem.Wang and Willem [5] obtained multiple solitary waves for the generalized Kadomtsev-Petviashvili equation (1.2) in one-dimensional spaces by the Lyusternik-Schnirelman category theory.In [6], Liang and Su considered that the case that the nonconstant weight function for generalized Kadomtsev-Petviashvili equation, see [6, the problem (P) and the assumption (Q)].In [7][8][9], Xuan dealt with the case where N ≥ 2 and f (u) satisfies some superlinear conditions.Their main tool in [6][7][8][9] is the famous mountain pass theorem.
We also note Fountain and Dual Fountain theorems were established by Bartsch and Willem [12,13], and both theorems are effective tools for studying the existence of infinitely many large energy solutions and small energy solutions.For more details of recent development in the direction, we refer the reader to [14][15][16][17][18][19] and references cited therein.Meanwhile, Zou [20] established some variant fountain theorems and many people utilized these theorems to study nonlinear problems, for instance, see [21][22][23][24][25][26][27] and references therein.
It should be remarked that Chen and Tang [14] investigated the fractional boundary value problem of the following form In this paper, they adopted Fountain and Dual Fountain theorems to obtain the existence of infinite solutions under some adequate conditions.It is no doubt that the results in the literature are significantly improved.
In [21], by virtue of the variant fountain theorem established in [20], Sun considered the sublinear Schrödinger-Maxwell equations In that paper, his aim is to study the existence of infinitely many solutions for (1.4) when f (x, u) satisfies sublinear in u at infinity, see [21, (H 2 ) of Theorem 1.1].Motivated by paper [21], we also utilize the variant fountain theorem by [20] to investigate that the existence of infinitely many solutions for (1.1) with the nonlinearity f growing sublinearly in u, see Theorem 3.4 in Section 3.

Preliminaries
For Ω ∈ R n is a bounded domain with smooth boundary ∂Ω on where ∇ y := ∂ ∂y1 , . . ., ∂ ∂yn−1 , dV = dxdy, and the corresponding norm . (2.2) Note that the space X with inner product (2.1) and norm (2.2) is a Hilbert space, see [6, Definition] and [7, P12 and P13].We know the exponent p := 2(2n−1) 2n−3 > 2 is as critical as the critical Sobolev exponent p * := np n−p , i.e., there exists a constant C > 0 such that the estimate From the interpolation theorem, the boundedness of Ω and estimate (2. where In what follows, we shall establish the energy functional for problem (1.1).Note that we can rewrite (1.1) in the following form (see [7, (3) of Page 12]): (2.5) For each v ∈ X, multiply the both sides of the above equation in (2.5) by v(x, y) and integrate over and then we obtain by Green formula and integration by parts, Therefore, on X, define a functional ϕ as where F (u) := u 0 f (s)ds, Ψ(u) := Ω F (u)dV .For the nonlinearity f , we always assume that it satisfies the following conditions: (H1) f ∈ C(R, R), f (0) = 0, and for some 1 < p < p = 2(2n−1) 2n−3 , c 0 > 0, there holds for all u, v ∈ X, and critical points of ϕ on X are weak solutions of (1.1).
Next we prove that Ψ ′ is weakly continuous.Suppose that L q (Ω) by (H1), (2.10) and (2.12).By Hölder inequality and Lemma 2.1, we get, Due to the form of ϕ ′ in (2.9), ϕ ′ is also continuous and hence ϕ ∈ C 1 (X, R).Furthermore, Ψ ′ is compact by the weak continuity of Ψ ′ since X is a Hilbert space.
As we have mentioned, we will utilize the critical point theory to prove our main results.Let us collect some definitions and lemmas that will be used below.One can refer to [10,28,29] for more details.
EJQTDE, 2012 No. 68, p. 5 Definition 2.1 Let X be a real Banach space, D an open subset of X. Suppose that a functional ϕ : D → R is Fréchet differentiable on D. If u 0 ∈ D and the Fréchet derivative satisfies ϕ ′ (u 0 ) = 0, then we say that u 0 is a critical point of the functional ϕ and ϕ(u 0 ) is a critical value of ϕ.
Let C 1 (X, R) denote the set of functionals that are Fréchet differentiable and their Fréchet derivatives are continuous on X.
Definition 2.2 Let X be a real Banach space and ϕ ∈ C 1 (X, R).We say that ϕ satisfies the Palais-Smale condition ((PS) condition for short) if for every sequence {u m } ⊂ X such that ϕ(u m ) is bounded and ϕ ′ (u m ) → 0 as m → ∞, there exists a subsequence of {u m } which is convergent in X.
Definition 2.3 Let X be a real Banach space, ϕ ∈ C 1 (X, R) and c ∈ R. We say that ϕ satisfies (PS) (2) For any u ∈ M , any sequence {u m } in M such that u m ⇀ u weakly in X, there holds: Then ϕ is bounded from below on M and attains its infimum in M .
Lemma 2.4(see [29,Theorem 9.12]) Let X be an infinite dimensional real Banach space.Let ϕ ∈ C 1 (X, R) be an even functional which satisfies the (PS) condition, and ϕ(0) = 0. Suppose that where Q 1 is infinite dimensional, and ϕ satisfies that (i) there exists α > 0 and ρ > 0 such that ϕ(u) ≥ α for all u ∈ Q 2 with u = ρ, Then ϕ has an unbounded sequence of critical values.
As X is a separable Hilbert space, there exist (see [30]) Clearly, X = j∈N X j with dimX j < ∞ for all j ∈ N. Lemma 2.5(see [12]) Let X be defined above.Suppose that (A4) ϕ satisfies the (PS) c condition for all c > 0.
Then ϕ has an unbounded sequence of critical values.
In the following, we shall introduce variant fountain theorems by Zou [20].Let X and the subspace X k , Y k and Z k are defined above.Consider the following C 1 -functional ϕ λ : X → R defined by (2.13) The following variant fountain theorem was established in [20].
Lemma 2.6 If the functional ϕ λ satisfies (T1) ϕ λ maps bounded sets to bounded sets uniformly for λ Then there exist Particularly, if {u λn } has a convergent subsequence for every k, then ϕ 1 has infinitely many nontrivial 3 Main Results Hence, there is a constant c such that u > c > lim inf n→∞ u n .Consequently, there exists a subsequence {u n k } ⊂ {u n } such that c > u n k , k = 1, 2, . ... From Hahn-Banach theorem, we know there exists f 0 ∈ X * (the dual space of X) such that f 0 = 1 and f 0 (u) = u .Therefore, on the other hand, note that u n k ⇀ u, and thus That is a contradiction.Secondly, we will discuss Ψ.By Lemma 2.1, there exists u ∈ X such that By integral mean value theorem, there is a number ξ = ξ(u m , u) between u m and u, we have Combining this and Hölder inequality, note that (2.10), we arrive at Therefore, Ψ(u m ) → Ψ(u) strongly in X.Hence, This completes the proof.
Proof of Theorem 3.1 The energy space X is a Hilbert space, so is reflexive.We easily verify the assumptions of Lemma 2.3 are true with M = X.Lemma 3.1 leads to ϕ is weak lower semi-continuous on X. Next, we will show ϕ is coercive on X.Indeed, by (2.7) and Lemma 2.1, there exists a constant and thus ϕ(u) → ∞ as u → ∞.Lemma 2.3 implies ϕ can attain its infimum in X, i.e., (1.1) has at least a weak solution.This completes the proof.
Theorem 3.2 Suppose that (H1) and the following two conditions are satisfied.
In order to obtain that ϕ satisfies (PS) condition, we need the following lemma.Note that if ϕ satisfies (PS) condition, then ϕ satisfies (PS) c condition for all c ∈ R by Remark 2.1.
Proof.Let {u k } be a sequence in X such that {ϕ(u k )} is bounded and ϕ(u k ) → 0 as k → ∞.We first prove {u k } is bounded.From the definition of functional ϕ, there exists C > 0 such that Consequently, by (H2), there holds which implies {u k } is bounded in X. Going if necessary to a subsequence, we can assume that there exists u ∈ X such that Hence, (ϕ ′ (u k ) − ϕ ′ (u))(u k − u) → 0, and note that (H1) leads to f ∈ L q (Ω) (see (2.10)), where p −1 + q −1 = 1, hence, by Hölder inequality, we get Therefore, It is easy to see that Consequently, the second inequality of (3.6) and (3.7), we get Choosing In view of (H1) and (H3), it is obvious ϕ(u) is even and ϕ(0) = 0.By Lemma 3.2, ϕ(u) satisfies (PS) condition.
We first prove ϕ satisfies (i) of Lemma 2.4.For any u ∈ X and u ≤ τ −1 have by the first inequality of (3.6) and Lemma 2.1 Therefore, Therefore, we can choose ρ > 0 small enough such that ϕ(u) ≥ β > 0 with u = ρ.
Finally, we show ϕ satisfies (ii) in Lemma 2.4.Let W ⊂ X is a finite dimensional subspace.For every r ∈ R \ {0} and u ∈ W \ {0} with u = 1, we obtain by (3.8) and Lemma 2.1 Note α > 2, the above inequality leads to there exists r 0 such that ru > ρ and ϕ(ru) < 0 for each r ≥ r 0 > 0. Since W is a finite dimensional subspace, there exists R(W ) > 0 such that ϕ(u) ≤ 0 on Lemma 2.4 yields that ϕ(u) has infinitely many critical points, i.e., (1.1) has infinitely many weak solutions.This completes the proof.
For any u ∈ Y k , let and it is easy to verify that • defined by (3.9) is a norm of Y k .Since all the norms of a finite dimensional normed space are equivalent, so there exists positive constant c 4 such that c 4 u ≤ u * .
Up until now, we have proved the functional ϕ satisfies all the conditions of Lemma 2.5, then ϕ has an unbounded sequence of critical values c n = ϕ(u n ).We only need to show u n → ∞ as n → ∞.
Indeed, going to a subsequence if necessary, we may assume that there is a constant M > 0 such that u n ≤ M. By this, there exist ξ n between u n and 0, integral mean value theorem and the definition of ϕ(u n ) enable us to obtain Note Ω is a bounded domain, we easily know Ω f (u n )u n dV and Ω f (ξ n )u n dV are bounded from the continuity of f and the boundedness of u n and ξ n .This contradicts the unboundness of c n .This completes the proof.
In the following, we shall prove (1.1) has infinitely many weak solutions by variant fountain theorems by Zou [20].To facilitate computations for the following proof, without loss of generality, we only consider a special case of (H1), i.e., f satisfies the following condition: (H6) f (u) = µ|u| µ−1 , where 1 < µ < 2 is a constant.
In order to apply Lemma 2.6 to prove the result, we first define the functionals A, B and ϕ λ on our working space X by for all u ∈ X and λ ∈ [1,2].By Lemma 2.2, we know ϕ is determined by (2.6).
The following three lemmas play some important roles in our Theorem 3.4.
Proof.By simple computation, we have It yields B(u) ≥ 0. We will prove there exists ε > 0 such that There exists otherwise a sequence {u n } n∈N ⊂ X \{0} such that Passing to a subsequence if necessary, we may assume v n → v 0 in X for some v 0 ∈ X since X is of finite dimension.We easily find v 0 = 1.Consequently, there exists a constant σ 0 > 0 such that meas(|v 0 | µ ≥ σ 0 ) ≥ σ 0 .
(3.18) Indeed, if not, then we have which implies This leads to v 0 = 0, contradicting to v 0 = 1.In view of Lemma 2.1 and the equivalence of any two norms on X , we have For every n ∈ N, denote and N 0 := {|v 0 | µ ≥ σ 0 }, where σ 0 is defined by (3.18).Then for n large enough, by (3.18), we see Consequently, for n large enough, we arrive at immediately This contradicts to (3.20).Therefore, (3.15) holds.For the ε given in (3.15), we let Then by (3.15), we find meas(N u ) ≥ ε, ∀u ∈ X \{0}.
For every k ∈ N, we can choose ρ k := 8