GLOBAL ATTRACTOR FOR WEAKLY DAMPED GKDV EQUATIONS IN HIGHER SOBOLEV SPACES

. Long time behavior of solutions for weakly damped gKdV equations on the real line is studied. With some weak regularity assumptions on the force f , we prove the existence of global attractor in H s for any s ≥ 1. The asymptotic compactness of solution semigroup is shown by Ball’s energy method and Goubet’s high-low frequency decomposition if s is an integer and not an integer, respectively.

The case k = 1 is the classical KdV equation, which models unidirectional propagation of nonlinear disperse long waves [12]. It is a completely integrable system, i.e., it can be solved by inverse scattering. The case k = 2 is called modified Korteweg-de Vries (mKdV) equation. The solutions of KdV and mKdV are connected by Miura transformation: KdV . Thus, mKdV is also a complete integrable system. But this does not hold for k−gKdV equations with k ≥ 3, see e.g. [13,17]. It is proved in [16,18] that the solutions of 4−gKdV associated with large initial data may blow up in finite time. Similar conclusions for k−gKdV with k > 4 are conjectured in [17]. Therefore, 3−gKdV is particular interesting as it is not a complete integrable and has global solutions. In fact, the well posedness, scattering and soliton stability of 3−gKdV are considered in [6,11,19,30]. Motivated by these works, in this paper we consider the long time dynamics of solutions for the following forced 3−gKdV equation with damping Here, λ > 0 is a constant, f is a given function independent of time, u 0 ∈ H s . We shall first prove the following well posedness of (1).
For an open interval I ⊂ R, it is convenient to introduce the restriction in time spaces X s,b I endowed with the norm { v X s,b , v(·) = u(·) on I}.
If I = (0, δ), we write X s,b δ instead u X s,b I for brevity. Grünrock proved the following multi-linear estimates, see Theorem 1 of [6].
Corollary 1. For any s ≥ 0, I ⊂ R, it holds that

MING WANG
Proof. The case s = 0 follows from Corollary Lemma 2.1 directly, so suppose s > 0. Let u i ∈ X s, 1 2 + and u i = u i on I, i = 1, 2, 3, 4, then ∂ x i=1 u i on I. It follows from Parseval identity and Lemma 2.1 that Combining this and the inequality Since the inequality holds for any u i , the Corollary follows. Lemma 2.2. Let s ∈ R, −1/2 < b < b ≤ 0 or 0 ≤ b < b < b < 1/2, δ ∈ (0, 1).
Proof. Let ψ : R → R be a smooth function of time with ψ = 1 on [0, 1] and suppψ ⊂ [−1, 2], and ψ δ (·) = ψ(δ −1 ·) for δ > 0. The desired conclusion follows from the following estimates in [9] ψ δ u X s,b See Lemma 6.3 of [34] for more details. Denote by W (t) the group generated by Airy equation, namely From the linear estimates in [9], we can easily obtain their variant version in terms of · X s,b δ . Lemma 2.3. Let s ∈ R, b > 1/2. Then the following inequalities hold We also need some space-time Lebesgue spaces. Let 1 ≤ p, q ≤ ∞, f : R 2 → R and I ⊂ R. Define With usual modification when p = ∞ or q = ∞. If I = R, we will omit I in the notation for brevity. Let Y be a Banach space, we also use · L ∞ (0,T ;Y ) to denote the norm sup t∈(0,T ) · Y . For α ≥ 0, we define the fractional derivatives by where f (x) denotes the partial Fourier transform of f in the x variable. Now we recall some inequalities between these spaces and Bourgain spaces. It is easy to check by definition that Moreover, we have It is easy to check that Lemma 2.4-2.5 still hold if we use the norms of L p x L q t (0, δ) and X 0, 1 2 + δ instead.
Lemma 2.6. It holds that Proof. It follows from Corollary 2.9 in [8] that Combining the inequality and Lemma 2.9 in [29] implies the desired conclusion.
Remark 3. The Lemma is concluded in Theorem A.8 of Kenig et al. [10]. It is obvious that the Lemma still holds if we use L p x L q t (0, δ) norms instead.

2.2.
Local well posedness. We consider a function such that Q be the solution of the stationary equation Proof. It's easy to see that | ξ 3 /(λ − iξ 3 )| ≤ λ −1 + 1, and then Q H s f H s−3 , so it suffices to estimate the norm in Bourgain space. Let χ(·) be a smooth function, In order to establish the well posedness of (1), we split u = w+Q, then w satisfies Here, constants c j are given by The Duhamel formulation of w reads Let 0 < δ < 1. It follows from (4) and Lemma 2.3 that 1 2 + δ and the last term of (5) can be controlled by By virtue of Corollary 1, we have Gathering (5)-(7), using Lemma 2.8 we arrive at Here and below, we use · X s,b (t,t+δ) to denote the norm · X s,b I with I = (t, t + δ).

Consider the set
Let w ∈ B, we define Γw as the right hand side of (4), then Therefore, ΓB ⊂ B. Moreover, one can check that on B with some α ∈ (0, 1). Thus, Γ is a contraction on B. Now we can state the main result in this subsection.
Moreover, the lifespan δ satisfies and for t ∈ (0, δ), the map w(0) → w(t) is continuous from H s to H s .

Existence of absorbing sets.
3.1. Absorbing sets in H 1 . In what follows, we derive some energy equalities for smooth solutions, which also hold for rough solutions after a limit process since the solution map is continuous from H s to H s . Lemma 3.1. Let u, w be a solution of (1) and (3), respectively. Then we have the following conserved quantities where ϕ(w) = 1 2
Proof. We first establish an L 2 bound of u. By (8) and Hölder inequality, we get d dt Then using Gronwall lemma we have for t ≥ 0 Let T 1 = λ −1 ln( u 0 2 L 2 + 2). It follows that u(t) L 2 ≤ C, t ≥ T 1 .
Since w = u − Q and Q is bounded in L 2 , we obtain Next, we obtain an L 2 bound of u x . It suffices to do this for w x , since Q is also bounded in H 1 . The main tools are (9) and the interpolation inequalities Thanks to (13), holds for any ε > 0 and some constant c ε depending only on ε. Moreover, by Hölder and Young's inequality, for j = 1, 2, 3, 4 Substituting these inequalities into (9), with proper ε = ε 0 , we have for By virtue of (12), we have It follows that Thus, for any t ≥ 0 This implies that for t ≥ 0 If t ≥ T 1 , we use (14) on the interval [T 1 , t] and apply Gronwall lemma to obtain Thanks to (15) Using (13) again and the L 2 bound of w, we obtain Note that u = w + Q and Q is bounded in H 1 , we get the desired estimate. It follows from (15) that here and below Φ(·) is a polynomial, which may change from line to line.

Absorbing sets in
Proof. Since u = w + Q, it suffices to obtain an H s bound of w. To this end, acting D s on both sides of (3), multiplying with D s w, and integrating over [0, δ] × R we have It amounts to deal with the integral on right hand side of (19). We first consider Estimates of I. If s is an integer, the proof is easier. So let s = k + α, k is an integer and 0 < α < 1.By Parseval identity, we rewrite I as where c j1,j2,··· ,j4 are constants. In the following, we will extensively use the fact that which can be proved following [5], see [ [21], p.26] for more details. For I 1 , we have where we have used Corollary 1. Then, by local well posedness and interpolation To estimate I 2 , we write the integral in I 2 as , using integration by parts we have By Hölder inequality, Thanks to Lemma (2.4), and (2), By local well posedness and interpolation inequality again, we find For I 22 , by Hölder inequality and fractional Leibniz rule (see Lemma 2.7) , we have By fractional Leibniz rule again, Note that By Lemma 2.4 and Lemma 2.5, we have and It follows from (27)-(32) that Therefore, by (23)-(26) and (33) we have Estimates of other terms. Since f ∈ H s−3 , by Lemma 2.8 we know Q ∈ X s, 1 2 + δ . One can proceed as above to obtain Since the case 2− 1 8(s−1) ≤ 0 is easier, we assume that 2− 1 8(s−1) > 0 in the following. One can show that for 0 < ε, σ < 1 2 2 Let x > 0, 0 < α < 2. Using Young's inequality Now let α = 2 − σ, x = w(0) H s , and note 0 < ε < 1, we find Thanks to Lemma 3.3, w is bounded in L ∞ (0, ∞; H 1 ). It follows from Corollary 2 that we can take w(δ) as a new data to derive an estimate of w(2δ). Repeat these process, after n steps, we find Note that for δ ∈ (0, 1) Thus, If t ≥ T 2 , it's convenient to regard w(t) as a solution of (3) starts from w(T 2 ). Then for t ≥ T 2 By virtue of (38), let T 3 = T 2 + ω −1 ln(Φ( w(0) H s )), we find for t ≥ T 3 Thus, problem (1) has a global solution in H s , the energy equation (19) holds on any interval [t 0 , t], Corollary 3. Let u 1 , u 2 be the solutions of (1) with initial data u 01 , u 02 ∈ H s , respectively. Then for 0 ≤ σ ≤ s Proof. Let U = u 1 − u 2 , then U solves the integral equation Similar to Section 2, we have It follows from (18) and Lemma 3.2 that u 1 are uniformly bounded with respect to j. Hence which of course implies Here, the implicit constants are independent of j. Therefore, Proof of Theorem 1.1. It follows from Lemma 3.3 and Corollary 3 directly.

4.
Global attractor in H s : Integral case. In this section, we assume that s = k, k is an integer. Since problem (1) has a bounded absorbing set B in H k , thanks to the classical existence result [31], {S(t)} t≥0 has a global attractor in H k if we can show that {S(t)} t≥0 is asymptotically compact. In other words, it amounts to check that for any sequence {u 0n } ⊂ B, and a positive real number sequence {t n }, t n → ∞ as n → ∞, then the sequence {S(t n )u 0n } is pre-compact in H k . Now let u n be the solutions of (1) associated with initial data u 0n , w n = u n − Q. For each T > 0, we only consider the sequence {w n (t n + ·)} n start with n large enough such that t n − T ≥ 0. It follows from section 3 that Thus, {w n (t n + ·)} n is precompact in C([−T, T ]; H k−3 loc ). By interpolation, we find that {w n (t n + ·)} n is precompact in C([−T, T ]; H k− loc ). Hence, there exists a subsequence, still denoted by {w n (t n + ·)} n , such that {w n (t n + ·)} n → w(·) strongly in C([−T, T ]; H k− loc ).
Moreover, {w n (t n + t)} n w(t) weakly in H k for every t ∈ R and w(t) is a solution of (3).
Using Lemma 4.1 and (40), it's easy to check that Now we are ready to prove the asymptotic compactness of {S(t)} t≥0 . The energy equation (39) for w n with t = t n and t 0 = t n − T reads as n → ∞. In view of (20), it suffices to show that as n → ∞.
For I 1 , by Hölder inequality, Lemma 2.4 we have Observe that j 1 ≤ k − 1, j 2 , j 3 , j 4 ≤ k, by Corollary 3 we find Thus, (45)holds, and then the claim (43) follows. Similar to the proof of the claim, one can show that Now, we pass to the limit as n → ∞ in (42) and obtain By virtue of energy equation for w, we find Since T > 0 can be arbitrary, we get Combining the weak convergence, we have This proves the asymptotic compactness of {S(t)} t≥0 , and, hence, the existence of global attractor in H k . Thus the proof of Theorem 1.2 in integral case is complete.

5.
Global attractor in H s : Fractional case.

5.1.
Splitting. Let P N and P N be Fourier projectors on higher frequency and lower frequency, respectively. Precisely, and P N = 1 − P N , where 1 |ξ|≤N is the characteristic function on {ξ : |ξ| ≤ N }. We split u = v + q such that where q, v satisfy the following equations: and Here and below, u 0 belongs to the absorbing set B s obtained in Section 3.

Estimates of q.
Using v = u − q, we rewrite (47) as for some different constants c j . Similar to Section 2, one can show that (49) has a unique solution q ∈ X 0, 1 2 + δ with the bound From the equation (49), we find that supp q ⊂ {ξ : |ξ| ≥ N }, ∀t ≥ 0. Thus, Multiplying (47) with q and integrating over [0, δ] × R yields that Using integration by parts, Hölder inequality and Lemma 2.4, we obtain Similarly, we have Now acting (49) with D s , multiplying D s q and integrating on [0, δ] × R implies that (52) Similar to Section 3, one can show that Since q(0) is bounded in H s , we choose N big enough, say N = N 0 , N 0 depends only on the absorbing set, such that the second term on right hand side is negative, thus Then an iteration argument yields that for some ω = Cλ

Estimates of v.
Since v = u − q, from previous discussion, we get the global existence of v in H s . Moreover, P N v = P N u is smooth and satisfies for all σ ≥ 0, t ≥ 0, δ ∈ (0, 1) . Thus, it suffices to focus on an estimate of P N v. To this end, we introduce Z = P N v − P N Q, which is a solution of Obviously, Z has the following bound Lemma 5.1. Let 0 < δ < 1. It holds that for some constants c j,k,l . By Lemma 2.8 and Corollary 2, we have and thus It is easy to see that In the case j = 1, k = 1, l = 2, it follows from Corollary 1, (55)-(56) that The other cases are similar. Thus the proof is complete.
and for j ≥ 2 and for j = 1, l ≥ 1 Now we write Z(t + t 0 ) as a solution of (54) starts from Z(t), namely Using Corollary 1 and Lemma 5.1, we have ).
Acting both sides of (54) with D s+ , multiplying D s+ Z and integrating over (t, t + δ) × R yields Now we need the following lemma.
Finally, we consider the most difficult case j = 1, l = 0, then k = 3. Let s+ = m+α, 0 ≤ α < 1. Then by Parseval identity On one hand, using integration by parts, we find It is easy to see that i ≤ m in both cases. Now by (66), using Hölder inequality and Lemma 2.4 we have It follows from (67) Now we need to bound the norms involved with g. Firstly, Moreover, observe that Thus, using Lemma 2.7 and Hölder inequality we have .
Proceed as the proof of (29), we get Since j 1 > m is impossible in the sum, use j 2 + j 3 ≤ m + 1 − n, we find Now if n = 0 or 1, using integration by parts, it follows from (73) that The integral in the sum of (78), using (21) and Corollary 1, is bounded by .
Thus, we obtain that v(t) H s+ 1, t ≥ 0. It's obvious that Kuratowski measure κ(A) depends on the metric of X. Sometimes we shall write κ X (A) instead, to emphasis the metric used, in the following. Some important properties of κ(A) are summarized as follows, see e.g. [7]. For the convenience of the reader, we recall the following criterion of the existence of global attractor, see [7,15,23]. Proposition 1. Let X be a Banach space and {S(t)} t≥0 be a continuous semigroup on X. Then {S(t)} t≥0 has a global attractor in X provided that the following conditions hold: (1) {S(t)} t≥0 has a bounded absorbing set in X ; (2) for any bounded subset B of X, we have κ(S(t)B) → 0, as t → ∞.
Let B be a bounded set in H s , it follows from the results in section 3 that S(t)B ⊂ B s for t ≥ T 3 (B), where B s is bounded in H s . Choose a smooth cut off function φ k such that 0 ≤ φ k ≤ 1, φ k = 0 if |x| ≤ k and φ k = 1 if |x| ≥ 2k. Thanks Lemma 4.1, we find for t ≥ T 4 = T 3 + T (ε) , k ≥ K(ε) φ k S(t)B L 2 ≤ ε. Now let u 0 ∈ B, split S(t)u 0 = u(t) = q(t) + v(t) := S 1 (t)u 0 + S 2 (t)u 0 .
Then Theorem 1.2 in fractional case follows from Proposition 1.