Global Attractor for the Generalized Dissipative KDV Equation with Nonlinearity

Notice E_i, C, c, c_i denote for different positive constants.

Theorem 2.1. Let the generalized dissipative of four-order KDV equation with nonlinearity given by (2.1). Assume that ϕ(u), g(u) satisfy conditions (A1)–(A4) and, moreover, u₀ ∈ H², f ∈ H¹, then for α, β, γ > 0, there exists a global attractor 𝒜 of the problem (2.1), that is, there is a bounded absorbing set B ∈ H² in which sense the trajectories are attract to 𝒜, such that

where S(t) is semigroup operator generated by the problem (2.1).

Lemma 3.1. Assume that g(u) satisfied (A4), furthermore, u₀ ∈ H, f ∈ H, then for the solution u of the problem (2.1), one has the estimates

Proof. Taking the inner product of (2.1) with u, we have

where here, we apply Young’s inequality and the condition (A4).

Thus, from (3.4), we get

By virtue of Gronwall′s inequality and (3.6), one has (3.1) and which implies (3.2) and (3.3).

Lemma 3.2. In addition to the conditions of Lemma 3.1, one supposes that

then one has the estimate where

Proof. Taking the inner product of (2.1) with u_xx, we have

where

Noticing that

Using Nirberg′s interpolation inequality and the Sobolev embedding theory (see [11]), we have

Due to Lemma 3.1 and conditions of Lemma 3.2, we get that

From (3.10) and above inequalities, we get

Setting C − γ = c₈, then we can obtain that Thus, by Gronwall’s inequality and (3.15), we get that which implies Therefore, we prove Lemma 3.2.

Lemma 3.3. Suppose that ϕ(u), g(u) satisfy (A2), (A3) and, moreover, the following conditions hold true:

• (1)

  ϕ(u) ∈ C³, g(u) ∈ C²,

• (2)

  u₀ ∈ H², f ∈ H¹,

then for the solution u of the problem of (2.1), one has the following estimate furthermore,

Proof. Taking the inner product of (2.1) with u_xxxx, we have

where By Young’s inequality and Lemmas 3.1 and 3.2, thus from (3.22), we have Due to Lemmas 3.1 and 3.2 and (A3), we obtain

Using Young’s inequality, we have

By (3.21), (3.23) and (3.25), we get that is, where By virtue of Gronwall’s inequality, we have and (3.27) implies Therefore, we prove Lemma 3.3.

Lemma 3.4. Suppose that ϕ(u), g(u) satisfy (A2), (A3) and, moreover, the following conditions hold true:

• (1)

  ϕ(u) ∈ C³, g(u) ∈ C²,

• (2)

  u₀ ∈ H³, f ∈ H²,

then for the solution u of the problem of (2.1), we have the following estimates: where furthermore,

Proof. Taking the inner product of (2.1) with u_xxxxxx, we have

where

Using Nirberg’s interpolation inequality and Young’s inequality, from (3.36) and Lemmas 3.1–3.3, we have

Due to the condition (4.3), we get By direct calculations, it is easy to get that Due to (3.34)–(3.39), we have that is, where Using Gronwall′s inequality, we deduce that moreover, (3.41) implies Therefore, we prove Lemma 3.4.

Lemma 3.5. Suppose that the following conditions hold true:

• (1)

  u₀ ∈ H^m+1, f ∈ H^m,

• (2)

  ϕ(u) ∈ C^m+1, ∥ϕ(u)∥ ≤ A|u|^5−σ, (σ, A > 0),

• (3)

  g(u) ∈ C^m, |g₂(u)| ≤ K|u|⁵,

• (4)

  g(u) satisfies (A3), (A4) and g(u) is Lipschitz continuous, that is,

then there exists a unique global solution u for the problem (2.1) such that u ∈ L^∞(0, T; H^m(Ω)), and furthermore,the semigroup operator S(t) associated with the problem of (2.1) is continuous and there exists an absorbing set B ⊂ H²(Ω), where

Proof. Similar to the proof of Lemmas 3.1–3.4, we have

At the same time, we use the Galerkin method (see [11]) and Lemmas 3.1–3.4 to prove the existence of global solution for the problem (2.1). So, we omit them here.

Next, we will prove the uniqueness of the global solution.

Assume that u, v are two solutions of the problem (2.1) and w = u − v, then we have

Taking the inner product in H of (2.1) with w, we have

where Due to the condition and from (3.49), we obtain that is, By application of Gronwall′s inequality, we get w = 0.

Finally, we recall some basic results in [11, 23] and by Lemmas 3.1–3.3, it is easy to prove that there exists an absorbing set

in H²(Ω). But as for the continuity of semigroup S(t), we can apply the following Lemmas 3.6, and 3.7 to prove the result.

Lemma 3.6. Suppose that u₀ ∈ H², f ∈ H¹, and ϕ(u), g(u) satisfy (A1)–(A4), there exists constant C > 0, such that

Proof. We take the scalar product in space H of (3.58) with u_η, we get

Due to (A3) and Young’s inequality, we get From (3.61), we obtain the following inequality: By Gronwall’s inequality, one has Hence, there exists C > 0, such that and implies

We take the scalar product in space H of (3.58) with u_ηxx and similar to the proof of Lemma 3.2, we have

By application of Gronwall′s inequality, we deduce that So, there exists C > 0, such that and implies

We take the scalar product in space H of (3.58) with u_ηxxxx and similar to the proof of Lemma 3.3, we have

It is easy to prove that

We take the scalar product in space H of (3.58) with u_ηxxxxxx and similar to the proof of Lemma 3.4, we have

that is,

Hence, by Gronwall’s inequality, we get

At the same time, we have and we omit them here.

Lemma 3.7. Under the conditions of Lemma 3.6, one has the following estimates

where C_i(η) > 0, (i = 1,2, 3).

Proof. We take the scalar product in space H of (3.60) with x²w_η and noticing that

it is easy to get that By Gronwall’s inequality, we have From (3.60), we obtain

We take the scalar product in space H of (3.61) with x²w_ηxx and noticing that

By Young’s inequality and the Sobolev embedding theory (see [11]) and (3.81)-(3.82), we deduce that

Using Gronwall’s inequality, we obtain The proof of Lemma 3.7 is completed.

Lemma 4.1 (see [23].)Assume that s > s₁, (s, s₁ ∈ N), then the following embedding $$into$$ is compact.

Proof. Let $$be a bounded set. It suffices to prove that B has a finite ε-net for any ε > 0. First, since

there exists an integer A > 0, such that Let Ω = {x||x| < A|}, then the imbedding $$ is compact. Thus, is relatively compact in $$ and has a finite $$-net $$, k = 1,2, …, m with $$,$$ and u_k ∈ B. We claim that $$ is an ε-net of B in $$.

Indeed, for any u ∈ B, $$, then there exists a $$ such that

Hence, This completes the lemma.

Lemma 4.2 (see [7], [11].)Let E be Banach space and {S(t), t ≥ 0} a set of semigroup operators, that is,  S(t) : E → E satisfy

where I is the identity operator and E is space H²(Ω). We also assume that

• (1)

  S(t) is bounded, that is, for each R > 0, there exists a constant C > 0 such that ∥u∥_E ≤ C implies ∥S(t)u∥_E ≤ R, (t ≥ 0),

• (2)

  there is an bounded absorbing set B₀ ⊂ E, that is, for any bounded set B ⊂ E, there exists a constant T, such that S(t)B ⊂ B₀, for t ≥ T,

• (3)

  S(t) is a continuous operator for t > 0, then S(t) has a compact global attractor

in the space E₀, such that

• (1)

  S(t)𝒜 = 𝒜, t ≥ 0,

• (2)

  dist (S(t)B,𝒜)_E → 0 as t → +∞, and dist (S(t)B,𝒜)_E denotes the Hausdorff semidistance defined as

for any bounded set B ⊂ H²(Ω) in which sense the trajectories are attracted to 𝒜 (see[9, 24]), using Kuratowski α-measure in order to overcome the non-compactness of the classical Sobolev embedding.

Definition 4.3 (see [11], [25].)Let {S(t)} _t≥0 be a semigroup in complete metric space E. For any subset B ⊂ E, the set ω(B) defined by $$ is called the ω-limit set of B.

Remark 4.4. (1) It is easy to see that ψ ∈ ω(B) if and only if there exists a sequence of element ψ_n ∈ B and a sequence t_n → ∞, such that

 (2) If ω(B) is ω-limit compact set, then, for every bounded subset B of E and for any ε > 0, there exists a t₀ > 0, such that $$.

Definition 4.5 (see [11], [26].)Let {S(t)}_t≥0 be a semigroup in complete metric space E. A subset B₀ of E is called an absorbing set in E if, for any bounded subset B of E, there exists some t₀ ≥ 0, such that S(t)B ⊂ B₀, for all t ≥ t₀.

Definition 4.6 (see [11], [26].)Let {S(t)}_t≥0 be a semigroup in complete metric space E. A subset 𝒜 of E is called global attractor for the semigroup if 𝒜 is compact and enjoys the following properties:

• (1)

  𝒜 is a invariant set, that is, S(t)𝒜 = 𝒜, for any t ≥ t₀,

• (2)

  𝒜 attract all bounded set of E, that is, for any bounded subset B of E, dist (S(t)B, 𝒜) → 0, as t → ∞, where dist (B, A) is Hausdorff semidistance of two set B and A in space E: dist (B, A) = sup _x∈B  inf _y∈A d(x, y).

Definition 4.7 (see [12], [27].)Kuratowski α-measure of set B is defined by the formula

for every bounded set B of a Banach space X.

Remark 4.8. (1) If 𝒜 is compact set, then α(𝒜) = 0;

(2) α(𝒜 + ℬ) ≤ α(𝒜) + α(ℬ),

(3) α(𝒜 ∪ ℬ) ≤ Max {α(𝒜), α(ℬ)},

(4) if 𝒜 ≤ ℬ, α(𝒜) ≤ α(ℬ),

(5) $$.

Thirdly, we prove Theorem 2.1.

Proof. Using the result of [11], we have S(t) is ω-limit compact and B is bounded, for any ε > 0, there exists t ≥ 0 such that

Taking ε = 1/n, (n = 1,2, …), we can find a sequence {t_n}, t₁ < t₂ < ⋯t_n < ⋯, such that By $$, we get First, we prove that 𝒜 = ω(B) is variant. As a matter of fact, if ψ ∈ S(t)ω(B), then ψ = S(t)ϕ, for some ϕ ∈ ω(B). So, there exists a sequence ϕ_n ∈ B and t_n → ∞ such that S(t)ϕ_n → ϕ, that is, which implies that ψ ∈ ω(B) and S(t)ω(B) ⊂ ω(B).

Conversely, if ψ ∈ ω(B), by (4.9), we can find two sequences ϕ_n ∈ B and t_n → ∞ such that S(t_n)ϕ_n → ϕ. We need to prove that {S(t_n − t)ϕ_n} has a subsequence which converges in E. For any ε > 0, there exists a t_ε such that

which implies that Hence, there exists an integer N, such that Then, it follows that Notice that $$ contains only a finite number of elements, where N₀ is fixed such that t_n − t ≥ 0, as n ≥ N₀.

By properties (1)–(4) in Remark 4.8, we have

Let ε → 0, then we get that This implies that {S(t_n − t)ϕ_n} is relatively compact. So, there exists a subsequence $$ and ψ ∈ E, such that It is easy to see that ψ ∈ ω(B) and furthermore, ϕ ∈ S(t)ω(B).

Next, by virtue of Lemma 4.2 and the result of [11, 12], we prove that 𝒜 = ω(B) is an global attractor in E and attracts all bounded subsets of E.

Otherwise, then there exists a bounded subset B₀ of E such that dist (S(t)B₀, 𝒜) does not tend to 0 as t → ∞. Thus, there exists a δ > 0 and a sequence t_n → ∞ such that

For each n, there exist b_n ∈ B₀, (n = 1,2, …) satisfying Whereas B is an absorbing set, S(t_n)B₀ and S(t_n)b_n belong to B, for n sufficiently large. As in the discussion above, we obtain that S(t_n)b_n is relatively compact admits at least one cluster point γ, where t₁ follows S(t₁)B₀ ⊂ B. So, γ ∈ 𝒜 = ω(B) and this contradicts (4.24). The proof is complete.