[Retracted] Three Homoclinic Solutions for Second-Order p-Laplacian Differential System

Theorem A (see [5].)Assume that  f  and  V  satisfy the following conditions:

• (H1)

    f(t) = 0  and  ∇V(t, q(t)) = −L(t)q(t)+∇W(t, q(t));   

• (H2)

    L ∈ C(R, R^n×n)  is a symmetric and positive definite matrix for all  t ∈ R  and there is a continuous function  α : R → R  such that  α(t) > 0  for all  t ∈ R  and

  $$ ()
  as   | t | → ∞;   

• (H3)

  consider the following

  $$ ()
  where  m : R → R⁺  is a positive continuous function such that  m ∈ L^2/(2−γ)(R, R⁺)  and  1 < γ < 2, d ≥ 0, and p > 2  are constants.

Then (1) possesses infinitely many homoclinic solutions.

Moreover, Tang and Xiao [10] prove the existence of homoclinic solution of (1) as a limit of the  2kT-periodic solutions of the following extension of system (1):

and they established the following theorem.

Theorem B (see [10].)Assume that  f  and  V  satisfy the following conditions:

• (H4)

    V, f(t) ≠ 0  and V(t, x) = −K(t, x) + W(t, x), where V ∈ C¹(R × Rⁿ, R) is T-periodic with respect to t, and T > 0;

• (H5)

    ∇W(t, x) = o(|x|), as   | x | → 0 uniformly with respect to t;

• (H6)

  there is a constant  μ > 2  such that

  $$ ()

• (H7)

    f : R × Rⁿ   is a continuous and bounded function;

• (H8)

  there exist constants  b > 0 and γ ∈ (1,2]  such that

  $$ ()

• (H9)

  there is a constant ϱ ∈ [2, μ)  such that

  $$ ()

• (H10)

  consider the following

  $$ ()

Then system (1) possesses a nontrivial homoclinic solution.

Definition 1. A function

is said to be a 2kT-periodic solution of (Pk) if  u  satisfies the equation in (Pk).

Lemma 2. If  u ∈ E_(k)  is a critical point of  J_(k)(·, λ); then  u  is a 2kT-periodic solution of  (Pk).

Proof. Assume that  u ∈ E_(k)  is a critical point of  J_(k)(·, λ); then for all  v ∈ E_(k),  one has

It follows that By the definition of weak derivative, (18) implies that Thus$$and  u  satisfies the (Pk). Therefore,  u  is a solution of (Pk).

Definition 3. Let  X  be a Banach space and  f : X → (−∞, +∞].  f  is said to be sequentially weakly lower semi-continuous if  liminf _k→+∞f(x_k) ≥ f(x) as x_k⇀x in  X.

Definition 4. Suppose  E  is a real Banach space. For  ϕ ∈ C¹(E, Rⁿ), we say that ϕ  satisfies PS condition if any sequence  {u_k} ⊂ E  for which  ϕ(u_k)  is bounded and  ϕ^′(u_k) → 0  as  k → ∞  possesses a convergent subsequence.

Lemma 5 (see [16].)For  u ∈ E_(k), one then has$$, where

Lemma 6 (see [27].)Let  X  be a nonempty set, and  Φ, Ψ  are two real functions on  X. Assume that there are  r > 0, x₀, x₁ ∈ X  such that

Then, for each  ρ  satisfying one has

Lemma 7 (see [17].)Let  X  be a separable and reflexive real Banach space,  I⊆R  an interval, and  f : X × I → R  a function satisfying the following conditions:

• (i)

  for each  x ∈ X, the function  f(t, ·)  is continuous and concave;

• (ii)

  for each  λ ∈ I, the function  f(t, ·)  is sequentially weakly lower semicontinuous and Dâteaux differentiable, and  lim _{∥x∥→∞}f(x, λ) = +∞;

• (iii)

  there exists a continuous concave function  h : I → R  such that

  $$ ()

Then, there exist an open interval  J⊆I  and a positive real number  ρ, such that, for each  λ ∈ J, the equation has at least two solutions in  X  whose norms are less than  ρ. If, in addition, the function  f  is (strongly) continuous in  X × I, and, for each  λ ∈ I, the function  f(t, ·)  is  C¹  and satisfies the PS condition, then the above conclusion holds with “three” instead of “two.”

Lemma 8. Let  u ∈ W^1,p(R, Rⁿ). Then for every  t ∈ R, the following inequality holds:

Proof. Fix  t ∈ R. For every  τ ∈ R,

Integrating (27) over  [t − 1/2, t + 1/2]  and using the Hölder inequality, we get

Remark 9. If there exist constant  μ ∈ [0, p)  and functions  τ₃(t) ∈ C([−kT, kT])  with  min _[−kT,kT]τ₃ > 0  such that

then (V2) holds.

In fact, (32) implies that there exists  c₂ > 0  such that

which combining the continuity of F_(k)(t, x) − τ₃(t) | x|^μ  on  [−kT, kT]×[−c₂, c₂]  yields that there exists constant  c₃ > 0  such that

Lemma 10. Assume that (V1) holds; then, for each  k ∈ N, there exists a continuous concave function  h_(k) : [0, +∞) → Rⁿ  such that

Proof. We define

It is clear that  u₁ ∈ E_(k). It follows from that

Let  g(x) = (1/p)x^p, x ≥ 0. It is clear that  g(x)  has the following properties: (1)  g(x) strictly increases for  x ≥ 0 and (2)  g(x) = w  has unique solution  Q(w)  for each  w > 0.

In view of (29), (38), and (1), one has

which yields that It follows from Lemma 5, (1), and (2) that Let G = MQ(r); then G/M is a solution of  g(x) = r.  From the definition of  g(x)  and  r, we have  g(c₁/M) = r. Thus, (2) implies  G = c₁, which combining (41) yields that

Therefore,

Since  F_(k)(t, 0) = 0, we obtain

It follows from$$ that$$, which combining  c₁, M > 0  yields that  E > 0. Therefore, in view of (V1) and (44), we get  Ω > 0. Thus, it follows from (38) and (40) that

From (43), (45), and (V1), we have

It is obvious that  ϕ₁(0) = −ϕ₂(0) = 0. Owing to Lemma 6, choosing  h(λ) = ρλ, we obtain which combining J_(k)(u, λ) = ϕ₁(u) + λϕ₂(u) implies the conclusion.

Lemma 11. If (V2) holds, then for each  k ∈ N,  lim _{∥u∥→∞}J_(k)(u, λ) = +∞  and  J_(k)(·, λ)  satisfies the  PS  condition.

Proof. Let$$be a sequence in  E_(k)  such that$$and$$is bounded, for each  k ∈ N.

Lemma 5 implies that

It follows from (V2) and (48) that which yields that for each  k ∈ N. Noting that  μ ∈ [0, p), the above inequality implies that  lim _{∥u∥→∞}J_(k)(u, λ) = +∞  and$$is bounded in  E_(k). Next, we will prove that$$converges strongly to some  u^(k)  in  E_(k). The proof is similar to [22]. Since$$is bounded in  E_(k), there exists a subsequence of$$for simplicity denoted again by$$ such that$$converges weakly to some  u^(k)  in  E_(k). Then$$converges uniformly to  u^(k)  on  [−kT, kT]  (see [28]). Therefore, as  n → +∞,  for each  k ∈ N. In view that$$ and$$converges weakly to some  u^(k), we get as n → +∞, for each  k ∈ N. Then, from (15), one has for each  k ∈ N. By [29], for each  k ∈ N, there exist  c_p, d_p > 0  such that

If  p ≥ 2, it follows from (51)–(54) that $$ as n → +∞.

If  1 < p < 2, by Holder’s inequality, we obtain

for each  k ∈ N. Similarly,

It follows from  1 < p < 2 and (54)–(56) that

In view of (51)–(53) and (57), we have$$as  n → +∞,  for each  k ∈ N.

Therefore,$$converges strongly to  u^(k)  in  E_(k), for each  k ∈ N. Thus, for each  k ∈ N, J_(k)(·, λ)  satisfies the  PS  condition.

Lemma 12. Assume that (V1) and (V2) hold; then there exist an open interval Λ⊆[0, +∞)  and a positive real number  σ, such that, for each  λ ∈ Λ  and  k ∈ N,  (Pk) has at least three 2kT-periodic solutions in  E_(k)  whose norms are less than  σ.

Proof. Let$$be a weakly convergent sequence to u^(k)  in  E_(k); then$$converges uniformly sequence to  u^(k)  on  [−kT, kT]. The continuity and convexity of$$ imply that$$is sequentially weakly lower continuous [28, Lemma 1.2], for each  k ∈ N, which combining the continuity of  f_(k)  yields that

Hence,  J_(k)(·, λ)  is sequentially weakly lower semi-continuous, for each  k ∈ N.

It is obvious that  J_(k)(u, ·)  is continuous and concave for each  u ∈ E_(k). In view of Lemmas 10 and 11, it follows from Lemma 7 that there exist an open interval  Λ⊆[0, +∞)  and a positive real number  σ, such that, for each  λ ∈ Λ   and  k ∈ N, J_(k)(·, λ)  has at least three critical points in  E_(k)  whose norms are less than  σ. Therefore, we can reach our conclusion by using Lemma 2.

Lemma 13. Assume that (V3) holds. Let$$be one of the three 2kT-periodic solutions of system  (Pk) obtained by Lemma 12 for each  k ∈ N. Then there exists a subsequence$$of$$convergent to a certain$$in$$.

Proof. From Lemma 12, we have

which combining Lemma 5 yields that there exists a positive constant  M₁  independent of  k  such that Thus, we obtain that$$ is a uniformly bounded sequence. Next, we will show that$$and$$are also uniformly bounded sequences. Since$$is a 2kT-periodic solutions of system (Pk) for every  t ∈ [−kT, kT),  we have

By (60), (61), and (V3), we get

which yields that Then, from (63), (V3), and the definition of  Φ_p(x), we obtain For  i = −k, −k + 1, …, k − 1, by the continuity of$$, we can choose$$such that

it follows that for  t ∈ [iT, (i + 1)T], i = −k, −k + 1, …, k − 1

Consequently,

Now we prove that the sequences$$and$$ are uniformly bounded and equicontinuous. In fact, for every  k ∈ N  and  t₁, t₂ ∈ R, we have by (67)

Similarly, from (64), we have Then, by application of the Arzelà-Ascoli Theorem, we obtain the existence of a subsequence$$of$$ and a function$$such that

Thus, Lemma 13 is proved.

Lemma 14. Let$$be determined by Lemma 13. Then$$is a nontrivial homoclinic solution of system (P).

Proof. The first step is to show that$$is a solution of system (P). By Lemma 13, one has

for  t ∈ [−k_jT, k_jT), j ∈ N. Take  a, b ∈ R  with  a < b. There exists  j₀ ∈ N  such that for all  j > j₀  one has Integrating (72) from  a  to  t ∈ [a, b], we obtain for  t ∈ [a, b]. Since (70) shows that$$uniformly on  [a, b]  and$$uniformly on  [a, b]  as  j → ∞. Let  j → ∞  in (73), we get for  t ∈ [a, b]. Since  a  and  b  are arbitrary, (74) yields that$$is a solution of system (P). It is easy to see that  u = 0  is not a solution of system (P) for  f(t, 0) ≠ 0  and so$$.

Secondly, we will prove that$$as  t → ±∞. By (59), we have

For every  l ∈ N, there exists  j₁ ∈ N  such that for  j > j₁   Let  j → ∞  in the above and use (70), and it follows that for each  l ∈ N, Let  l → ∞  in the above, and we get Thus Combining the above with (V3) we have By (26), we obtain Combining (80) with (81), we get$$as  t → ±∞.

Finally, we show that

From (60) and (70), one has From this and (64), we have If (82) does not hold, then there exist  ε₀ ∈ (0, 1/2)  and a sequence  {t_k}  such that which yield that for  t ∈ [t_k, t_k + ε₀/(1 + M₃)] It follows that which contradicts to (78) and so (82) holds. The proof is completed.

Theorem 15. Assume that (V1), (V2), and (V3) hold. Then system (P) possesses three nontrivial homoclinic solutions.

Example 1. Consider the following p-Laplacian problem:

where  λ ∈ [0, +∞),  kT = 2, and It is obvious that (V3) holds and for every  t ∈ [−2,2], Then, for each  t ∈ [−2,2].  Thus, there exists  c₄ > 0  such that which combining the continuity of F_(k)(t, x) − 2 | x|²  on  [−2,2]×[−c₄, c₄]  yields that there exists constant  c₅ > 0  such that Therefore, (V2) is satisfied. Furthermore, in view of Lemma 5,  M = 4. Let  η₁ = η₂ = 4, δ₁ = 1, δ₂ = 1, and$$;  then  K₁ = 0, K₂ = 6, K₃ = 1, E = 1.112 × 10⁻³, Ω = 1/2, and $$. Thus (V1) is satisfied. Moreover, f_(k)(t, 0) = 1 ≠ 0. In view of Theorem 15, we have that Example 1 possesses three nontrivial homoclinic solutions.