Multiple solutions and ground state solutions for a class of generalized Kadomtsev-Petviashvili equation

Theorem 1.1

Assume that ( h 1 ) – ( h 4 ) are satisfied, then equation (1.3) possesses infinitely many nontrivial solutions in Ω , where Ω ⊂ R N is a bounded domain.

Theorem 1.2

Assume that ( h 1 ) – ( h 2 ) and ( h 5 ) are satisfied, then equation (1.2) has a ground state solution.

Definition 2.1

[7] On Y = g x : g ∈ C 0 ∞ R N , define the inner product

and the norm is

where ∇ y = ∂ ∂ y 1 , … , ∂ ∂ y N − 1 and d V = d x d x .

If there exists a sequence u n ⊂ Y such that u n → u a.e. on R N , and ∥ u j − u k ∥ → 0 as j , k → ∞ , then we say that u : R N → R belongs to X .

Definition 2.2

[7] On Y = g x : g ∈ C 0 ∞ R N , define the inner product

and the norm is

where ∇ y = ∂ ∂ y 1 , … , ∂ ∂ y N − 1 and d V = d x d x .

If there exists a sequence u n ⊂ Y such that u n → u a.e. on R N , and ∥ u j − u k ∥ 0 → 0 as j , k → ∞ , then we say that u : R N → R belongs to X 0 .

Lemma 2.1

[7,11,12] The following continuous embeddings hold.

1. the embeddings X ↪ X 0 are continuous;

2. the embeddings X ↪ L q R N , for 1 ≤ q ≤ N ¯ are continuous;

3. the embeddings X ↪ L loc q R N , for 1 ≤ q < N ¯ are compact;

4. the embeddings X 0 ↪ L N ¯ R N are continuous.

Lemma 2.2

[14] Let X be an infinite dimensional Banach space, and there exists a finite dimensional space W such that X = W ⊕ V . I ∈ C 1 ( R ) satisfies the ( P S ) condition, and

1. I ( u ) = I ( − u ) for all u ∈ X , I ( 0 ) = 0 ;

2. there exist ρ > 0 , α > 0 such that I ∂ B ρ ∩ V ≥ α ;

3. for any finite dimensional subspace Y ⊂ X , there is R = R ( Y ) > 0 such that I ( u ) ≤ 0 on Y \ B R .

Lemma 2.3

[7] Assume that u n is a bounded sequence in X . If

then u n → 0 in L q R N for all q ∈ ( 2 , N ¯ ) .

Lemma 2.4

[13] Let ( X , ∥ ⋅ ∥ ) be a Banach space and T ⊂ R + be an interval. Consider a family of C 1 functionals on X of the form

with B ( u ) ≥ 0 and either A ( u ) → + ∞ or B ( u ) → + ∞ as ∥ u ∥ → + ∞ . If there are two points v 1 , v 2 ∈ X such that

where

Then, for almost every λ ∈ T , there exists a bounded ( P S ) c λ sequence in X , and the mapping λ → c λ is non-increasing and left continuous.

Lemma 3.1

Suppose h satisfies h 1 – h 4 . If u n ⊂ X satisfies

1. I u n is bounded;

2. I ′ u n , u n → 0 ,

Proof

If u n is unbounded in X , we can find a subsequence still denoted by u n such that u n → + ∞ . Let v n = u n ∥ u n ∥ , we have ∥ v n ∥ = 1 . Thus, we may assume that v n ⇀ v in X . As the embedding X ↪ L loc 2 R N is compact, we have v n → v in L 2 ( Ω ) . By h 4 and ( i ) , there exists c > 0 such that

as n → + ∞ , which implies 1 ≤ 2 κ μ − 2 lim n → + ∞ sup ∥ v n ∥ 2 2 . Therefore, v ≢ 0 . By h 3 and Fatou’s Lemma, one has

which is a contradiction. Hence, u n is bounded in X .□

Lemma 3.2

Suppose h satisfies ( h 1 ) – ( h 4 ) . Then the functional I satisfies the (PS) condition.

Proof

To prove that I satisfies the (PS) condition, we only need to prove { u n } ⊂ X has a convergent subsequence, where { u n } obtained by Lemma 3.1. As { u n } is bounded in X , there exists a subsequence still denoted by { u n } and u 0 ∈ X such that u n ⇀ u 0 in X and u n → u 0 in L q ( Ω ) for 1 ≤ q < N ¯ . From ( h 2 ) , we have

then

Applying the Hölder inequality, for 1 < p < N ¯ − 1 , one has

It follows from u n ⇀ u 0 in X and I ′ ( u 0 ) ∈ X * that I ′ u 0 , u n − u 0 → 0 . And as I ′ ( u n ) → 0 in X ∗ , it is easy to obtain

Therefore,

as n → + ∞ .

Thus, we have

as n → + ∞ .□

Proof

Proof of Theorem 1.1. We have verified that I satisfies the (PS) condition. It follows from ( h 5 ) that I is an even function. As X is a separable space, X has orthonormal basis { e i } . Define X j ≔ R e j , W k ≔ ⊕ j = 1 k X j , V k ≔ ⊕ j = k + 1 ∞ X j ¯ . Let W = W k , V = V k , clearly X = W ⊕ V and dim W < ∞ .

Next, we verify that I satisfies (ii) in Lemmas 2.2. By Lemma 2.1, for all u ∈ V , we have

Then there exists ρ > 0 small enough, α > 0 such that I ( u ) ≥ α > 0 as ∥ u ∥ = ρ .

Now, we verify that I satisfies (iii) in Lemma 2.2. For any finite dimensional subspace Y ⊂ X , since lim ∣ t ∣ → + ∞ H ( t ) ∣ t ∣ 2 = + ∞ , for u ≠ 0 ,

as r → + ∞ . Thus, there exists r 0 > 0 such that I r u < 0 for all r ≥ r 0 > 0 . So we can conclude that there exists a R ( Y ) > 0 such that I ( u ) ≤ 0 on Y \ B R ( Y ) .

Hence, according to Lemma 2.2, equation (1.3) possesses infinitely many nontrivial solutions.□

Lemma 4.1

Suppose that h satisfies ( h 1 ) – ( h 2 ) and ( h 5 ) . Then

1. there exists v ∈ X ⧹ { 0 } such that I λ ( v ) < 0 for all λ ∈ 1 2 , 1 ;

2. c λ = inf γ ∈ Γ max t ∈ [ 0 , 1 ] I λ ( γ ( t ) ) > max I λ ( 0 ) , I λ ( v ) for all λ ∈ 1 2 , 1 , where

Proof

(i) By ( h 5 ) , we have lim s → + ∞ H ( s ) s 2 = + ∞ . Furthermore, for some u ∈ X

Thus, there exists t 0 > 0 such that I λ t 0 u < 0 . By taking v = t 0 u , we have I λ ( v ) < 0 .

(ii) By virtue of ( h 2 ) , for any ε > 0 and some p ∈ ( 1 , N ¯ − 1 ) , there exists C ε > 0 such that

By Lemma 2.1, we have

Then there exists ρ > 0 small enough such that

Therefore, c λ > max I λ ( 0 ) , I λ ( v ) .□

Lemma 4.2

Suppose h satisfies ( h 1 ) – ( h 2 ) and ( h 5 ) . For almost every λ ∈ 1 2 , 1 , there is a bounded sequence { v m } , such that I λ ( v m ) → c λ in X and I ′ λ ( v m ) → 0 in the dual X ∗ of X .

Lemma 4.3

If v m is a bounded sequence in X and lim m → + ∞ sup ( x , y ) ∈ R N ∫ B 1 ( ( x , y ) ) v m 2 d V = 0 , then lim m → + ∞ ∫ R N G v m = 0 , where G v m = 1 2 h v m v m − H v m .

Proof

On one hand, by simple calculations, we derive

On the other hand, by Lemma 2.3, we have v m → 0 in L q ( R N ) for all q ∈ ( 2 , N ¯ ) . Hence, we can conclude that

Thus, lim m → + ∞ ∫ R N G v m d V = 0 .□

Lemma 4.4

If v m ⊂ X is the sequence obtained by Lemma 4.2, then for a.e. λ ∈ 1 2 , 1 , there exists a sequence of points x m , y m ⊂ R × R N − 1 , u m ( x , y ) ≔ v m x − x m , y − y m , such that

1. u m ⇀ u λ ≠ 0 in X ;

2. I λ ′ u λ = 0 in X * ;

3. I λ u λ ≤ c λ in X ; and

4. there exists M > 0 such that I λ u λ ≥ M .

Proof

By Lemma 4.2, we know that for almost every λ ∈ 1 2 , 1 , there exists a bounded sequence v m that satisfy I λ v m → c λ in X and I λ ′ v m → 0 in X * as m → + ∞ . Furthermore,

By Lemma 4.3, there exist a sequence of points x m , y m ⊂ R × R N − 1 and α > 0 , such that

Let u m ( x , y ) ≔ v m x − x m , y − y m . By the invariance translations of I λ , as m → + ∞ , we have that I λ u m → c λ in X and I λ ′ u m → 0 in X * . Since u m is bounded, there exists u λ ∈ X such that u m ⇀ u λ in X .

In the following, we complete the proof of this lemma.

1. It follows from Lemma 2.1 that

   C ∥ u λ ∥ 2 ≥ ∥ u λ ∥ 2 2 ≥ ∫ B 1 ( 0 ) u λ 2 d V = lim m → + ∞ ∫ B 1 ( 0 ) u m 2 d V ≥ α > 0 ,

   and thus obtain u λ ≢ 0 in X .

2. As C 0 ∞ R N is dense in X , we only need to check that I λ ′ u λ , φ = 0 for any φ ∈ X . We have

   I λ ′ u m , φ − I λ ′ u λ , φ = u m − u λ , φ − λ ∫ R N h u m − h u λ φ d V → 0 ,

   since u m ⇀ u λ in X , u m → u λ in L loc p R N for 1 ≤ p ≤ N ¯ . It follows from I λ ′ u m → 0 that I λ ′ u λ = 0 .

3. By ( h 5 ) and Fatou’s Lemma, we get

   c λ = lim m → + ∞ I λ u m − 1 2 I λ ′ u m , u m = λ lim m → + ∞ ∫ R N G u m d V ≥ λ ∫ R N G u λ d V = I λ u λ − 1 2 I λ ′ u λ , u λ = I λ u λ .

4. Combining (ii) with h 2 and Lemma 2.1, we obtain that for any ε > 0 , there exists C ε > 0 such that

   ∥ u λ ∥ 2 = λ ∫ R N h u λ u λ d V ≤ ∫ R N h u λ u λ d V ≤ C ε ∥ u λ ∥ 2 + C ε ∥ u λ ∥ p + 1 .

   Then there exists β > 0 such that ∥ u λ ∥ ≥ β > 0 . Therefore,

   I λ u λ = I λ u λ − 1 μ I λ ′ u λ , u λ = 1 2 − 1 μ ∥ u λ ∥ 2 + ∫ R N 1 μ h u λ u λ − H u λ d V ≥ 1 2 − 1 μ ∥ u λ ∥ 2 ≥ 1 2 − 1 μ β 2 ≔ M > 0 .