A multiplicity result for asymptotically linear Kirchhoff equations

Theorem 1.1.

Assume that (A1), (A2) and (A3) hold. If l > μ with

then there exists b ~ > 0 such that for any b ∈ ( 0 , b ~ ) , problem (1.1) admits at least two nontrivial nonnegative solutions, in which one is a mountain pass type solution and the other is a ground state solution.

Theorem 2.1 ([4]).

Let E be a real Banach space with its dual space E * , and suppose that I ∈ C 1 ⁢ ( E , R ) satisfies

for some μ < η , ρ > 0 and e ∈ E with ∥ e ∥ > ρ . Let c ≥ η be characterized by

where Γ = { γ ∈ C ⁢ ( [ 0 , 1 ] , E ) : γ ⁢ ( 0 ) = 0 , γ ⁢ ( 1 ) = e } is the set of continuous paths joining 0 and e. Then there exists a sequence { u n } ⊂ E such that

Lemma 2.1.

Assume that (A1), (A2) and (A3) hold. Then there exist ρ > 0 , η > 0 such that

Proof.

By (A1) and (A2), for any ϵ > 0 , there exists C ϵ > 0 such that

and then

Moreover, by (A1), (A2) and (A3), there exists C 1 > 0 such that

So, from (2.2), (2.3) and the Sobolev inequality, for any u ∈ H , we have

Thus, one has

thanks to the fact that b 4 ( ∫ ℝ N | ∇ u | 2 d x ) 2 is nonnegative. Fixing ϵ ∈ ( 0 , C 1 - 1 ) and letting ∥ u ∥ = ρ > 0 be small enough, there exists η > 0 such that the desired conclusion holds. ∎

Lemma 2.2.

Assume that (A1), (A2) and (A3) hold. If l > μ , then there is b ~ > 0 such that for any b ∈ ( 0 , b ~ ) , there exists e ∈ H with ∥ e ∥ > ρ such that I b ⁢ ( e ) < 0 .

Proof.

According to the definition of μ and l > μ , there exists ϕ ∈ H , with ϕ ≥ 0 , such that

By (A2) and Fatou’s Lemma, we have

by taking e = t 0 ⁢ ϕ with t 0 large enough so that I 0 ⁢ ( e ) = I 0 ⁢ ( t 0 ⁢ ϕ ) < 0 and ∥ e ∥ = t 0 ⁢ ∥ ϕ ∥ > ρ . Since

is continuous and increasing in b ≥ 0 and I 0 ⁢ ( e ) < 0 , there exists b ~ > 0 sufficiently small such that I b ⁢ ( e ) < 0 for all b ∈ ( 0 , b ~ ) . ∎

Lemma 2.3.

Let (A1), (A2) and (A3) hold, and let l > μ . Then, for any b ∈ ( 0 , b ~ ) , where b ~ given by Lemma 2.2, the sequence { u n } defined in (2.4) is bounded in H.

Proof.

Assume on the contrary that ∥ u n ∥ → + ∞ as n → ∞ . Define ω n = u n ∥ u n ∥ . Clearly, { ω n } is bounded in H and there exists ω ∈ H such that, going if necessary to a subsequence,

We claim that ω ≢ 0 . Assume on the contrary that ω ≡ 0 . By (A3), there exists a constant θ ∈ ( 0 , 1 ) such that

For any n ∈ ℕ , this yields

Since the embedding H 1 ⁢ ( B R 0 ⁢ ( 0 ) ) ↪ L 2 ⁢ ( B R 0 ⁢ ( 0 ) ) is compact, we have ω n → ω in L 2 ⁢ ( B R 0 ⁢ ( 0 ) ) . According to [27, Lemma A.1], going if necessary to a subsequence, there exists g ∈ L 2 ⁢ ( B R 0 ⁢ ( 0 ) ) such that

By (A1) and (A2), there exists C > 0 such that

By (2.3), (2.7) and (2.8), for any n ∈ ℕ , we have

Noting that ω n → ω ≡ 0 a.e. in ℝ N , we obtain

It follows, from (2.9), (2.10) and the dominated convergence theorem, that

Thus, by (2.6) and (2.11), we get

Since ∥ u n ∥ → + ∞ as n → ∞ , it follows from (2.4) that

Therefore,

which contradicts (2.12), so ω ≢ 0 .

On the other hand, since ∥ u n ∥ → + ∞ as n → ∞ , it follows form (2.4) that

that is,

From (2.3) and (2.8), one has

From (2.13) and (2.14), it is clear that

Fatou’s Lemma yields

that is,

Since the embedding H ↪ D 1 , 2 ⁢ ( ℝ N ) is continuous, ω also belongs to D 1 , 2 ⁢ ( ℝ N ) . According to the definition of the norm of D 1 , 2 ⁢ ( ℝ N ) and (2.15), ω ≡ 0 . This is a contradiction. So, the sequence { u n } is bounded in H. ∎

Lemma 2.4.

Assume that (A1), (A2) and (A3) hold. Then, for any ϵ > 0 , there exists R ⁢ ( ϵ ) > R 0 and n ⁢ ( ϵ ) such that

for all R ≥ R ⁢ ( ϵ ) and n ≥ n ⁢ ( ϵ ) .

Proof.

Let ξ R : ℝ N → [ 0 , 1 ] be a smooth function such that

and, for some constant C 0 > 0 (independent of R),

Then, for all n ∈ ℕ and R ≥ R 0 , we have

This implies that

for all n ∈ ℕ and R ≥ R 0 . By (2.4), ∥ I ′ ⁢ ( u n ) ∥ H - 1 ⁢ ∥ u n ∥ → 0 as n → ∞ . So, for any ϵ > 0 , there exists n ⁢ ( ϵ ) > 0 such that

for all n ≥ n ⁢ ( ϵ ) . Hence, it follows from (2.17) and (2.18) that

for all R ≥ R 0 and n ≥ n ⁢ ( ϵ ) . Note that

For any ϵ > 0 , there exists R ⁢ ( ϵ ) > R 0 such that

By (2.21) and Young’s inequality, for all n ∈ ℕ and R ≥ R ⁢ ( ϵ ) , we get

By (A1), (A2), (A3) and (2.16), there exists η 1 ∈ ( 0 , 1 ) such that for all n ∈ ℕ and R ≥ R 0 ,

In virtue of the fact that b ∫ ℝ N | ∇ u n | 2 d x ∫ ℝ N | ∇ u n | 2 ξ R d x is nonnegative, together with (2.20), (2.22) and (2.23), for all n ∈ ℕ and R ≥ R ⁢ ( ϵ ) ≥ R 0 , we have

Since the sequence { u n } is bounded in H, it follows from (2.19) and (2.24) that there exists C 3 > 0 such that for all R ≥ R ⁢ ( ϵ ) and n ≥ n ⁢ ( ϵ ) ,

From η 1 ∈ ( 0 , 1 ) and (2.16), the desired conclusion easily follows. ∎

Proof of Theorem 1.1.

By Lemma 2.3, the sequence { u n } defined in (2.4) is bounded in H. Since H is a reflexive space, going if necessary to a subsequence, u n ⇀ u in H for some u ∈ H . In order to prove the theorem, we need to show that ∫ ℝ N | ∇ u n | 2 d x → ∫ ℝ N | ∇ u | 2 d x and that the sequence { u n } has a strong convergence subsequence in H, that is, ∥ u n ∥ → ∥ u ∥ as n → ∞ . Note that, by (2.4),

and

Since u n ⇀ u in H, we have

Since the embedding H ↪ D 1 , 2 ⁢ ( ℝ N ) is continuous, u n ⇀ u in D 1 , 2 ⁢ ( ℝ N ) , and thus

To show that

we first prove that

For any ϵ > 0 , by Lemma 2.4, (A3) and Hölder’s inequality, for n large enough, we have

This and the compactness of the embedding H 1 ⁢ ( ℝ N ) ↪ L loc 2 ⁢ ( ℝ N ) imply (3.5). Since { u n } is bounded in D 1 , 2 ⁢ ( ℝ N ) , we assume that ∫ ℝ N | ∇ u n | 2 d x → λ ≥ 0 . If λ = 0 , by Fatou’s Lemma,

so u ≡ 0 in H. From (3.1)–(3.6), it is easy to see that

But (3.6) and (3.7) invoke a contradiction with (2.4) in which c > 0 . So λ > 0 , and by Fatou’s Lemma we have

If λ = ∫ ℝ N | ∇ u | 2 d x , from (3.1)–(3.5), it is also easy to see that ∥ u n ∥ → ∥ u ∥ in H as n → ∞ , hence the proof is completed.

If λ > ∫ ℝ N | ∇ u | 2 d x , from (3.1)–(3.5), we obtain

By Fatou’s Lemma, it is easy to see from our assumption that

which is impossible. So,

Moreover, we get ∥ u n ∥ → ∥ u ∥ as n → ∞ . Now we show that the solution u is nonnegative. Multiplying equation (1.1) by u - and integrating over ℝ N , where u - = min ⁡ { u ⁢ ( x ) , 0 } , we find

Hence, u - = 0 and u is a nonnegative solution of problem (1.1).

To get a ground state solution, we denote by K the nontrivial critical set of I b . Set

It is easy to see that K is nonempty. For any u ∈ K , we have

Now we choose ϵ ∈ ( 0 , C 1 - 1 ) as in the proof of Lemma 2.1 and use (2.1), (2.3) and the Sobolev embedding theorem to get

Therefore, for any u ∈ K , we have

We recall that u ≠ 0 whenever u ∈ K , and (3.8) implies

Hence, any limit point of a sequence in K is different from zero.

We claim that I b is bounded from below on K, i.e., there exists M > 0 such that I b ⁢ ( u ) ≥ - M for all u ∈ K . Otherwise, there exists { u n } ⊂ K such that

It follows from (2.3) that

This and (3.10) imply that ∥ u n ∥ → + ∞ as n → ∞ . Let ω n = u n ∥ u n ∥ . There exists ω ∈ H such that (2.5) holds. Note that I b ′ ⁢ ( u n ) = 0 for u n ∈ K . As in the proof of Lemma 2.3, we obtain that ∥ u n ∥ → + ∞ is impossible. Then I b is bounded from below on K. So m ≥ - M . Let { u ¯ n } ⊂ K be such that I b ⁢ ( u ¯ n ) → m as n → ∞ . Then (2.4) holds for the sequence { u ¯ n } and m. Following almost the same procedures as in the proofs of Lemmas 2.3 and 2.4, and using the above arguments, we can show that { u ¯ n } is bounded in H and, going if necessary to a subsequence, ∥ u ¯ n ∥ → ∥ u ¯ ∥ , where u ¯ ∈ H ∖ { 0 } and ∫ ℝ N | ∇ u ¯ n | 2 d x → ∫ ℝ N | ∇ u ¯ | 2 d x as n → ∞ . There is only one difference in showing that (3.6) does not hold. Based on the aforementioned discussions, we know that the possible critical value c > 0 . Here we do not know if m > 0 , but (3.9) holds, and so λ > 0 . Moreover, I b ⁢ ( u ¯ ) = m and I b ′ ⁢ ( u ¯ ) = 0 . Therefore, u ¯ ∈ H ∖ { 0 } is a ground state solution of problem (1.1). Finally, we can also show that the ground state solution u ¯ is nonnegative. ∎