Quasilinear equations with indefinite nonlinearity

Theorem 1.1.

Assume that condition (a) holds, r > 4 , 4 < s < 4 ⁢ N N - 2 , μ n > 0 , μ n → 0 , u n ∈ W 0 1 , p ⁢ ( Ω ) , D ⁢ I μ n ⁢ ( u n ) = 0 and I μ n ⁢ ( u n ) ≤ C . Then the following assertions hold:

1. There exists a function u ∈ H 0 1 ⁢ ( Ω ) ∩ L ∞ ⁢ ( Ω ) satisfying equation ( 1.1 ).

2. Up to a subsequence, there holds ∥ u n ∥ L ∞ ≤ C , u n → u for a.e. x ∈ Ω , u n → u in H 0 1 ⁢ ( Ω ) , and

   μ n ⁢ [ ∫ Ω ( 1 + u n 2 ) p 2 ⁢ | ∇ ⁡ u n | p ⁢ d x ] 2 p → 0   𝑎𝑛𝑑   I μ n ⁢ ( u n ) → I ⁢ ( u )    as  ⁢ n → ∞ .

Theorem 1.2.

Assume that condition (a) holds, r > 4 and 4 < s < 4 ⁢ N N - 2 . Then problem (1.1) has infinitely many solutions.

Theorem 1.3.

Assume that r > 4 , 4 < s < 4 ⁢ N N - 2 and conditions (a0)–(a4) hold. Then equation (1.3) has infinitely many solutions.

Lemma 2.1.

There holds that

Proof.

For φ ∈ W 0 1 . p ⁢ ( Ω ) , we know that

Since Ω ¯ - ∩ Ω ¯ + = ∅ , we can choose ψ ∈ C 0 ∞ ⁢ ( ℝ N ) such that ψ ≥ 0 , ψ ⁢ ( x ) = 1 for x ∈ Ω ¯ - , and ψ ⁢ ( x ) = 0 for x ∈ Ω ¯ + . Then

Taking ψ = u ⁢ ψ p as the test function in (2.1), we get

In view of the Sobolev inequality

it follows from (2.2) and (2.3) that

By choosing q ∈ ( 4 , s ) , we deduce that

and thus

Lemma 2.2.

Assume that μ n > 0 , μ n → 0 , u n ∈ W 0 1 , p ⁢ ( Ω ) , D ⁢ I μ n ⁢ ( u n ) = 0 and I μ n ⁢ ( u n ) ≤ C 0 . Then there exists a constant C > 0 independent of n such that

Moreover, there holds

Proof.

Assume that

By Lemma 2.1, we have

Let v n = ρ n - 1 ⁢ u n . Then

We have v n → v in L q ⁢ ( Ω ) , 1 ≤ q < 4 ⁢ N N - 2 , v n 2 ⇀ v 2 in H 0 1 ⁢ ( Ω ) , ∫ Ω a - ⁢ | v | r ⁢ d x = 0 and ∫ Ω a + ⁢ | v | s ⁢ d x = 0 . Thus, v ⁢ ( x ) = 0 for x ∈ Ω ¯ + ∩ Ω ¯ - and v 2 ∈ H 0 1 ⁢ ( Ω 0 ) ⊂ H 0 1 ⁢ ( Ω ) .

If p > N , then W 0 1 , p ⁢ ( Ω ) ↪ C α ⁢ ( Ω ) for some α ∈ ( 0 , 1 ) . Given ψ ≥ 0 and ψ ∈ C 0 ∞ ⁢ ( Ω 0 ) , we take

and have

We further obtain the estimates as

and

Hence, it gives

Since v 2 ∈ H 0 1 ⁢ ( Ω 0 ) , we have v 2 ≡ 0 in Ω. It follows from (2.4) and Lemma 2.1 that

and

In view of 1 ≤ q < 4 ⁢ N N - 2 , by taking v n → v = 0 in L q ⁢ ( Ω ) , we get

and

Let z n ∈ W 0 1 , p ⁢ ( Ω ) and

It is easy to see that

So we further deduce that

Consequently, the left-hand side of (2.5) converges to zero, which yields a contradiction. ∎

Proof of Theorem 1.1.

By Lemmas 2.1 and 2.2, we have

Assume that u n → u in L q ⁢ ( Ω ) , 1 ≤ q < 4 ⁢ N N - 2 and u n 2 → u 2 in H 0 1 ⁢ ( Ω ) . We separate the proof into three steps.

Step 1. Moser’s iteration shows that the sequence { u n } is uniformly bounded.

Note that for p > N , the convergence W 0 1 , p ⁢ ( Ω ) ↪ C α ⁢ ( Ω ¯ ) holds for some α ∈ ( 0 , 1 ) . The function u n belongs to L ∞ ⁢ ( Ω ) . Here we prove that ∥ u n ∥ L ∞ ⁢ ( Ω ) is uniformly bounded. The term a + ⁢ | u | s - 2 ⁢ u in equation (1.2) is subcritical ( s < 4 ⁢ N N - 2 ), while the term of a - ⁢ | u | r - 2 ⁢ u in the equation does not cause any trouble to us at this step. The sequence { u n } is bounded in L 4 ⁢ N / ( N - 2 ) ⁢ ( Ω ) . Starting from this L 4 ⁢ N / ( N - 2 ) ⁢ ( Ω ) -bound, by Moser’s iteration we see the L ∞ ⁢ ( Ω ) -bound.

Step 2. Choose a suitable test function and show that the limit function u satisfies equation (1.2).

Let ψ ≥ 0 , ψ ∈ C 0 ∞ ⁢ ( Ω ) and φ = ψ ⁢ e - u n ∈ W 0 1 , p ⁢ ( Ω ) . Take φ as the test function. Then we have

We estimate each term in (2.7). From (2.6) we get

By the weak convergence, it gives

By the lower semi-continuity, we have

Using Lebesgue’s dominated convergence theorem leads to

and

It follows from (2.7) and the above estimates that

for ψ ≥ 0 and ψ ∈ C 0 ∞ ⁢ ( Ω ) . Moreover, we get

Given φ ≥ 0 and φ ∈ C 0 ∞ ⁢ ( Ω ) , we choose ψ n ∈ C 0 ∞ ⁢ ( Ω ) such that ψ n → φ ⁢ e u in H 0 1 ⁢ ( Ω ) , ∥ ψ n ∥ ≤ C and ψ n → φ ⁢ e - u for a.e. x ∈ Ω . Taking ψ n as the test function in (2.8) and letting n tend to infinity, we obtain

for φ ≥ 0 and φ ∈ C 0 ∞ ⁢ ( Ω ) .

Processing in a similar manner, one can also obtain an inequality with an opposite direction. Equation (1.2) holds for all φ ≥ 0 and φ ∈ C 0 ∞ ⁢ ( Ω ) . By the density argument, equation (1.2) also holds for all functions φ ∈ H 0 1 ⁢ ( Ω ) ∩ L ∞ ⁢ ( Ω ) .

Step 3. Since u ∈ H 0 1 ⁢ ( Ω ) ∩ L ∞ ⁢ ( Ω ) satisfies equation (1.2), we have

By D ⁢ I μ n ⁢ ( u n ) = 0 , we have 〈 D ⁢ I μ n ⁢ ( u n ) , u n 〉 = 0 . That is,

In view of s < 4 ⁢ N N - 2 and ∫ Ω a + ⁢ | u n | s ⁢ d x → ∫ Ω a + ⁢ | u | s ⁢ d x , as n tends to infinity, by the lower semi-continuity we deduce that

and

Hence, we obtain u n → u in H 0 1 ⁢ ( Ω ) and u n ⁢ ∇ ⁡ u n → u ⁢ ∇ ⁡ u in L 2 ⁢ ( Ω ) as n → ∞ . ∎