Infinitely Many Solutions of Superlinear Elliptic Equation

Theorem 1. Assume that (S₁)–(S₃) hold and f(x, u) is odd in u. Then problem (1) possesses infinitely many solutions.

Remark 2. To show that our assumptions (S₂) and (S₃) are weaker than (f₄), we give two examples:

• (1)

  f(x, u) = 2uln u + u,

• (2)

  f(x, u) = γ|u|^γ−2u + (γ − 1)|u|^γ−3u sin²u + |u|^γ−1sin 2u,   u ∈ R∖{0},   γ > 2,

which do not satisfy (f₄). Example (2) can be found in [3]. So the case considered here cannot be covered by the cases mentioned in [6, 11].

Remark 3. Compared with papers [11, 12], we do not assume any superlinear conditions near zero. Compared with paper [2], we do not impose any kind of monotonic conditions. In addition, although we do not assume (f₄) holds, we are able to check the boundedness of P.S. (or P.S.*) sequences. So, our result is different from those in the literature.

Lemma 4. E embeds continuously into L^p, for all 0 < p ≤ 2^*, and compactly into L^p, for all 1 ≤ p < 2^*; hence there exists τ_p > 0 such that

where $$.

Lemma 5 (see [15] or [16].)Suppose that (S₁) is satisfied. Then Ψ ∈ C¹(E, R) and Ψ^′ : E → E^* is compact and hence I ∈ C¹(E, R). Moreover

for all u, v ∈ E, and critical points of I on E are solutions of (1).

Lemma 6 (see [17].)Assume that $$, and |f(x, u)| ≤ c(1 + |u|^p/r). Then for every u ∈ L^p(Ω),   f(x, u) ∈ L^r(Ω), and the operator A : L^p(Ω) ↦ L^r(Ω) : u ↦ f(x, u) is continuous.

Theorem 7 (see [3], Theorem 2.1.)Assume that the functional Φ_λ defined above satisfies the following:

•  

  (F₁) Φ_λ maps bounded sets to bounded sets uniformly for λ ∈ [1,2]; furthermore, Φ_λ(−u) = Φ_λ(u) for all (λ, u)∈[1,2] × E;

•  

  (F₂) B(u) ≥ 0 for all u ∈ E; moreover, A(u) → ∞ or B(u) → ∞ as ∥u∥ → ∞;

•  

  (F₃) there exists r_k > ρ_k > 0 such that

  $$ ()

•  

  Then

  $$ ()

•  

  where B_k = {u ∈ Y_k : ∥u∥ ≤ r_k} and $$. Moreover, for a.e. λ ∈ [1,2], there exists a sequence $$ such that

  $$ ()

Lemma 8. Assume that (S₁)-(S₂) hold. Then there exists a positive integer k₁ and two sequences r_k > ρ_k → ∞ as k → ∞ such that

where $$ and $$ for all k ∈ N.

Proof Step  1. We first prove (25).

By (16) and (23), for all λ ∈ [1,2] and u ∈ E, we have

where a₁ is the constant in (16). Let Then since E is compactly embedded into L^ν. Combining (14), (27), and (28), we have By (29), there exists a positive integer k₁ > 0 such that since ν > 2. Evidently, Combining (30) and (31), direct computation shows Step  2. We then verify (26).

We claim that for any finite-dimensional subspace F ⊂ E, there exists a constant ϵ > 0 such that

Here and in the sequel, m(·) always denotes the Lebesgue measure in R.

If not, for any n ∈ N, there exists u_n ∈ F∖{0} such that

Let v_n = u_n/∥u_n∥ ∈ F, for all n ∈ N. Then ∥v_n∥ = 1, for all n ∈ N, and Passing to a subsequence if necessary, we may assume v_n → v₀  in  E, for some v₀ ∈ F, since F is of finite dimension. Evidently, ∥v₀∥ = 1. In view of Lemma 4 and the equivalence of any two norms on F, we have and |v₀|_∞ > 0.

By the definition of norm | · |_∞, there exists a constant δ₀ > 0 such that

For any n ∈ N, let Set Λ₀ = {x ∈ Ω : |v₀(x)| ≥ δ₀}. Then for n large enough, by (36) and (38), we have Consequently, for n large enough, there holds This is in contradiction to (37). Therefore (34) holds.

Consequently, for any k ∈ N, there exists a constant ϵ_k > 0 such that

where $$, for all k ∈ N, and u ∈ Y_k∖{0}. By (S₂), for any k ∈ N, there exists a constant S_k > 0 such that Combining (23), (42), (43), and (S₂), for any k ∈ N and λ ∈ [1,2], we have with ∥u∥ ≥ S_k/ϵ_k. Now for any k ∈ N, if we choose then (44) implies ending the proof.

Proof of Theorem 1. It follows from (16), (23), and Lemma 5 that I_λ maps bounded sets to bounded sets uniformly for λ ∈ [1,2]. In view of the evenness of F(x, u) in u, it holds that I_λ(−u) = I_λ(u) for all (λ, u)∈[1,2] × E. Thus the condition (F₁) of Theorem 7 holds. Besides, A(u) = (1/2)∥u∥² → ∞ as ∥u∥ → ∞ and B(u) ≥ 0 since F(x, u) ≥ 0. Thus the condition (F₂) of Theorem 7 holds. And Lemma 8 shows that the condition (F₃) holds for all k ≥ k₁. Therefore, by Theorem 7, for any k ≥ k₁ and a.e. λ ∈ [1,2], there exists a sequence $$ such that

as m → ∞, where with B_k = {u ∈ Y_k : ∥u∥ ≤ r_k} and $$.

Furthermore, it follows from the proof of Lemma 8 that

where $$ and $$ by (32).

Claim  1.  $$ possesses a strong convergent subsequence in E, for ∀λ ∈ [1,2] and k ≥ k₁.

In fact, by the boundedness of the $$, passing to a subsequence, as m → ∞,  we may assume

By the Sobolev embedding theorem, Lemma 6 implies that Observe that By (47), it is clear that It follows from the Hölder inequality, (51), and (52) that as m → ∞. Thus by (53), (54), and (55), we have proved that that is, $$ in E.

Thus, for each k ≥ k₁, we can choose λ_n → 1 such that the sequence $$ obtained a convergent subsequence; passing again to a subsequence, we may assume

This together with (47) and (49) yields

Claim 2.   $$ is bounded in E for all k ≥ k₁.

For notational simplicity, we will set $$ for all n ∈ N throughout this paragraph. If {u_n} is unbounded in E, we define v_n = u_n/∥u_n∥. Since ∥v_n∥ = 1, without loss of generality we suppose that there is v ∈ E such that

Let Ω_≠ = {x ∈ Ω : v(x) ≠ 0}. If x ∈ Ω_≠, from (S₂)  it follows that On the other hand, after a simple calculation, we have We conclude that Ω_≠ has zero measure and   v ≡ 0  a.e.  in  Ω.

Moreover, from (49) and (58)

By (S₃), which contradicts (62). Hence {u_n} is bounded.

Claim 3.  $$ possesses a convergent subsequence with the limit u^k ∈ E for all k ≥ k₁.

In fact, by Claim 2, without loss of generality, we have assume

By virtue of the Riesz Representation theorem, $$ and Ψ^′ : E ↦ E^* can be viewed as $$ and Ψ^′ : E ↦ E, respectively, where E^* is the dual space of E. Note that that is By Lemma 5, Ψ^′ : E ↦ E is also compact. Due to the compactness of Ψ^′ and (64), the right-hand side of (66) converges strongly in E and hence $$ in E.

Now for each k ≥ k₁, by (58), the limit u^k is just a critical point of I₁ = I with $$. Since $$ as k → ∞ in (49), we get infinitely many nontrivial critical points of I. Therefore (1) possesses infinitely many nontrivial solutions by Lemma 5.