Ground state solution for some new Kirchhoff-type equations with Hartree-type nonlinearities and critical or supercritical growth

Theorem 1.1

If V satisfies ( V 1)–( V 2), f ∈ C 1 ( R , R ) verifies ( f 1)–( f 4), then problem (1) has a ground state solution.

Theorem 1.2

If V satisfies (V1)–(V3), then there exists some m 0 > 0 such that for m ∈ ( 0 , m 0 ] , problem (2) has a ground state solution.

Lemma 2.1

Assume that ( f 1)–( f 4) hold, then we have

1. for all ε > 0 and q ∈ ( 2 , 6 ) , there is a C ε > 0 such that ∣ f ( t ) ∣ ≤ ε ∣ t ∣ + C ε ∣ t ∣ q − 1 ,

2. for any s ≠ 0 , s f ( s ) > 2 F ( s ) and F ( s ) > 0 .

Proof

We could easily obtain the results by elementary calculation.□

Lemma 2.2

(Hardy-Littlewood-Sobolev inequality [16]). Let 0 < α < N , p , q > 1 and 1 ≤ r < s < ∞ be such that

1. For any f ∈ L p ( R N ) and g ∈ L q ( R N ) , one has

   ∫ R N ∫ R N f ( x ) g ( y ) ∣ x − y ∣ N − α d x d y ≤ C ( N , α , p ) ‖ f ‖ L p ( R N ) ‖ g ‖ L q ( R N ) .

2. For any f ∈ L r ( R N ) , one has

   1 ∣ ⋅ ∣ N − α ∗ f L s ( R N ) ≤ C ( N , α , r ) ‖ f ‖ L r ( R N ) .

Lemma 2.3

(Brezis-Lieb lemma [17]) Let s ∈ ( 1 , ∞ ) and { w n } be a bounded sequence in L s ( R N ) . If w n → w almost everywhere on R N , then for any q ∈ [ 1 , s ] ,

and

Lemma 2.4

(Nonlocal Brezis lemma [6]) Let α ∈ ( 0 , N ) , N ≥ 3 , p ∈ 1 , 2 N N + α and { u n } be a bounded sequence in L 2 N p N + α ( R N ) . If u n → u almost everywhere on R N , then

Lemma 2.5

[13] Let α ∈ ( 0 , N ) , N ≥ 3 , p ∈ 1 , 2 N N + α and { u n } be a bounded sequence in L 2 N p N + α ( R N ) . If { u n } → u almost everywhere on R N , then for any h ∈ L 2 N p N + α ( R N ) ,

Lemma 2.6

(Pohozaev identity [6,7,18,19]). Suppose V ( x ) satisfies ( V 1 )–( V 2 ) and let u ∈ X be a weak solution of Problem (1), then we have the following Pohozaev identity:

In particular, if V ≡ V ∞ , we have

Lemma 3.1

Let p ∈ 3 + α 3 , 3 + α , then I ∞ has no lower bounds.

Proof

For ∀ u ∈ X \ { 0 } and t > 0 , we have

as t → ∞ , since p + 3 + α > 4 , and then we can obtain the conclusion.□

Remark 3.2

For t > 0 , we set

Lemma 3.3

Let c 1 , c 2 , and c 3 be constants, which are positive and u ∈ X \ { 0 } . Then, the function

has only one positive critical point, which corresponds to its maximal value.

Proof

One can obtain the result by elementary calculation.□

Lemma 3.4

For any u ∈ X \ { 0 } , there exists only one t 0 > 0 such that u t 0 ∈ ℳ ∞ and I ∞ ( u t 0 ) = max t > 0 I ∞ ( u t ) .

Proof

I ∞ ( u t ) has the form of the function η ( t ) defined earlier. Since by Lemma 3.3, η ( t ) has only one critical point t 0 > 0 , which corresponds to its maximal value. Thus, η ( t 0 ) = max t > 0 η ( t ) and η ′ ( t 0 ) = 0 . It follows that G ∞ ( u t 0 ) = t 0 η ′ ( t 0 ) = 0 . This implies u t 0 ∈ ℳ ∞ and I ∞ ( u t 0 ) = max t > 0 I ∞ ( u t ) .□

Lemma 3.5

The functional I ∞ possesses the mountain-pass geometry, i.e.,

1. there exists ρ , δ > 0 such that I ∞ ≥ δ for all ‖ u ‖ = ρ ;

2. there exists e ∈ H 1 ( R 3 ) such that ‖ e ‖ > ρ and I ∞ ( e ) < 0 .

Proof

(1) By Lemmas 2.1(1) and 2.2, we have

Thus, there exists ρ , δ > 0 such that I ∞ ≥ δ for all ‖ u ‖ = ρ > 0 small enough.

(2) For any u ∈ X \ { 0 } , by the definition of I ∞ ( u t ) , we see I ∞ ( u t ) < 0 for t > 0 large. Note that

Taking e = u t 0 , with t 0 > 0 large, then we have ‖ e ‖ > ρ and I ∞ ( e ) < 0 .□

Lemma 3.6

Assume that ( f 1)–( f 4) hold, then m ∞ is obtained.

Proof

Let { u n } ⊂ ℳ ∞ be such that I ∞ ( u n ) → m ∞ . Since G ∞ ( u n ) = 0 , we have:

for n large enough. Therefore, { ∣ ∇ u n ∣ 2 2 } is bounded. So there exists A ≥ 0 such that ∫ R 3 ∣ ∇ u n ∣ 2 d x → A . We claim that { ∣ u n ∣ 2 2 } is also bounded. By Lemmas 2.1(1) and 2.2(1), the definition of G ∞ ( u n ) and Sobolev inequality, we have

which indicates { ∣ u n ∣ 2 2 } is bounded. Then let δ = lim n → ∞ sup y ∈ R 3 ∫ B 1 ( y ) ∣ u n ∣ 2 d x . Next we prove δ > 0 . If δ = 0 , then by Lions’ concentration compactness principle [19], we have u n → 0 in L q ( R 3 ) for q ∈ ( 2 , 6 ) . From Lemma 2.2, we can obtain that

since 2 < 6 p 3 + α < 6 . Together with G ∞ ( u n ) = 0 , we have u n → 0 in X . This conflicts with the fact that c ∞ > 0 . So δ > 0 and there exists { y n } ⊂ R 3 such that ∫ B 1 ( y n ) ∣ u n ∣ 2 d x ≥ δ 2 > 0 . We set v n ( x ) = u n ( x + y n ) , then