Multiple solutions for a class of fourth-order elliptic equations with critical growth

Theorem 1.1

Let 1 < p < 2 , V ∈ C ( Ω ¯ ) be a positive function. Suppose that Q and V satisfy the hypotheses ( Q ) and ( V ) , respectively. Thus, there exists Γ * > 0 such that for μ ∈ ( 0 , Γ * ) , problem (1.1) has at least k nontrivial solutions.

Remark 1.2

Theorem 1.1 generalizes the results found in [7, 8]. In fact, the authors in [7] studied only the case that μ = 1 and subcritical case. Meanwhile, only one nontrivial solution is obtained in [8]. Hence, our result is new and interesting.

Lemma 2.1

Suppose that 1 < p < 2 and N > 4 , then, N μ ≠ ∅ . Furthermore, I μ is coercive and bounded below on N μ for all μ > 0 .

Proof

First, for u ∈ H 0 2 ( Ω ) , we define a firbering map J u : t → I μ ( t u ) by

from the above equation, we see that J u ( 0 ) = 0 and J u ( t ) → − ∞ as t → + ∞ . Besides, we have

therefore, there exists unique a t u > 0 such that

and then t u u ∈ N μ . Thus, N μ ≠ ∅ .

Second, by the Hölder and Sobolev inequalities, we know

So, for each u ∈ N μ , from (2.1) and the Young’s inequality, we obtain

Since 1 < p < 2 < 2 * * , I μ is coercive and bounded below on N μ for any μ > 0 .□

Proposition 2.2

For 1 ≤ i ≤ k , then β μ i > 0 for each μ ∈ ( 0 , Γ ) . And, ϕ ( t ε i v ε i ) → b i as ε → 0 + . Additionally, there exists ε 0 > 0 , such that for all 0 < ε < ε 0 , then, ϕ ( t ε i v ε i ) ∈ K r 0 2 for each 1 ≤ i ≤ k .

Proof

For every u ∈ N μ , by (2.1), we have

Since 1 < q < 2 < 2 * * , β μ i > 0 for all μ ∈ ( 0 , Γ ) .

By the definition of ϕ , we have

as ε → 0 + . This indicates that there exists ε 0 > 0 , such that ϕ ( t ε i v ε i ) ∈ K r 0 2 for any 0 < ε < ε 0 and 1 ≤ i ≤ k .□

Proposition 2.3

There exists Γ 0 > 0 , such that

for each 0 < μ < Γ 0 , and for all 1 ≤ i ≤ k .

Proof

We prove this proposition by contradiction. Assume that there exists a positive sequence { μ n } with μ n → 0 as n → ∞ , such that β μ i ˜ → m ˜ ≤ 2 N S N 4 for some 1 ≤ i ≤ k . Hence, there exists a sequence { u n } ⊂ ∂ O μ i such that I μ n ( u n ) → m ˜ as n → ∞ . For every 0 < μ n < Γ 0 , it follows that { u n } is bounded in H 0 2 ( Ω ) , and

Moreover,

From (2.4), fixed a constant l > 0 , by the Hölder and Sobolev inequalities, one obtains

By considering a subsequence of { u n } , it follows that

Moreover, one obtains

Then, we have l ≥ S N 4 . In addition, as n → ∞ , we have

It follows from (2.5) and (2.6) that l = S N 4 . Consequently, from (2.5), one obtains

Therefore, we know

Set ω n = u n ∣ u n ∣ 2 * * , then, ∣ ω n ∣ 2 * * = 1 , and

Hence, { ω n } is a minimizing sequence for S . According to [18], there exists a x 0 ∈ Ω ¯ such that

weakly in the sense of measure, here μ and ν are finite measures, and δ x 0 is the Dirac measure at x 0 . Since ω n ∈ ∂ O μ i , as n → ∞ , it follows that

which implies that x 0 ∈ ∂ O μ i , and x 0 ≠ b i for 1 ≤ i ≤ k . Additionally, from (2.7), we deduce

which yields a contradiction.□

Lemma 2.4

For u ∈ O μ i , there exist τ > 0 , and a differentiable functional ζ : B ( 0 ; τ ) ⊂ H 0 2 ( Ω ) → R + such that ζ ( 0 ) = 1 , ζ ( v ) ( u − v ) ∈ O μ i for any v ∈ B ( 0 ; τ ) , and

for every φ ∈ C 0 ∞ ( Ω ) .

Proof

For each u ∈ O μ i , define a function F u : R × H 0 2 ( Ω ) → R , given by

Then, F u ( 1 , 0 ) = ⟨ I μ ′ ( u ) , u ⟩ = 0 , and

According to the implicit function theorem, there exist τ > 0 , and a differentiable function ζ : B ( 0 ; τ ) ⊂ H 0 2 ( Ω ) → R + such that ζ ( 0 ) = 1

for all v ∈ B ( 0 ; τ ) , and

for all v ∈ B ( 0 ; τ ) , which is equivalent to

That is, ζ ( v ) ( u − v ) ∈ O μ i .□

Lemma 2.5

For each 1 ≤ i ≤ k , 0 < μ < Γ * = min { Γ , Γ 0 } , there is a ( P S ) β μ i sequence { u n } ⊂ O μ i for I μ .

Proof

First, we prove the following conclusion, suppose that ( Q ) and ( V ) hold, then

where 1 ≤ i ≤ k .

According to [19], we have the following results:

Hence,

then, I μ ( t v ε i ) → 0 as t → 0 , and I μ ( t v ε i ) → − ∞ as t → + ∞ . Moreover

then there exists a unique t ε > 0 such that I μ ( t v ε i ) achieves its maximum. Therefore, there exist two constants T 0 ¯ > 0 and T 0 ̲ > 0 , with T 0 ̲ < t ε < T 0 ¯ . Actually, from lim t → 0 + I μ ( t v ε i ) = 0 for all ε , we choose ξ = I μ ( t ε v ε i ) 4 > 0 , then, there exists T 0 ̲ > 0 such that