The periodic Cauchy problem for a combined CH–mCH integrable equation

Nonlinear Analysis 143 (2016) 138–154

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Nonlinear Analysis www.elsevier.com/locate/na

The periodic Cauchy problem for a combined CH–mCH integrable equation Xingxing Liu Department of Mathematics, China University of Mining and Technology, Xuzhou, Jiangsu 221116, China

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Article history: Received 3 December 2015 Accepted 17 May 2016 Communicated by Enzo Mitidieri MSC: 35G25 35L05 Keywords: Combined CH–mCH equation Non-uniform dependence H¨ older continuity

This paper is concerned with the periodic Cauchy problem for a generalized Camassa–Holm integrable equation, which can be viewed as a generalization to both the Camassa–Holm (CH) and modified Camassa–Holm (mCH) equations. We mainly make a detailed presentation on the effects of varying the CH and mCH nonlocal nonlinearities on the non-uniform dependence and H¨ older continuity of the solution map. Using a Galerkin-type approximation method, we first establish the local well-posedness result in Sobolev spaces H s , s > 25 , with continuous dependence on the initial data. Then we prove that this dependence is sharp by showing that the data-to-solution map is not uniformly continuous, which is based on wellposedness estimates and the method of approximate solutions. Furthermore, we demonstrate that the solution map is H¨ older continuous in the H σ topology with 0 ≤ σ < s. © 2016 Elsevier Ltd. All rights reserved.

1. Introduction In this paper, we consider the following periodic Cauchy problem for a combined CH–mCH integrable equation with both quadratic and cubic nonlinearities:   1 1   mt = k1 (u2 − u2x )m x + k2 (umx + 2mux ), m = u − uxx , t > 0, x ∈ T, (1.1) 2 2 u(0, x) = u , x ∈ T, 0x

where k1 , k2 are arbitrary constants, and T denotes the unit circle R/Z. Eq. (1.1) was derived initially by Fokas [13,14], as an integrable generalization of the modified KdV equation. Integrability of Eq. (1.1), its biHamiltonian structure, and peakons, weak kinks and kink–peakon interactional solutions have shown in [36]. s Recently, we [31] have studied the non-periodic Cauchy problem for Eq. (1.1) with initial data u0 ∈ Bp,r , 1 5 1 for 1 ≤ p, r ≤ ∞, s > max{2 + p , 2 , 3 − p }. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.na.2016.05.013 0362-546X/© 2016 Elsevier Ltd. All rights reserved.

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Among Eq. (1.1) with various constants k1 , k2 , there are two: the Camassa–Holm equation and the modified Camassa–Holm equation, which have attracted much attention due to their integrability, physical derivation, wave breaking and solitary waves interacting like solitons. For k1 = 0, k2 = −2, Eq. (1.1) is reduced to the Camassa–Holm (CH) equation [4,15]: mt + umx + 2ux m = 0,

m = u − uxx ,

which was first derived as a bi-Hamiltonian system by Fokas and Fuchssteiner [15]. Then Camassa and Holm [4] independently re-derived it by approximating directly in the Hamiltonian for the Euler equations in the shallow water regime. u(t, x) stands for the fluid velocity and m(t, x) represents its potential density. In the past two decades, a large amount literature was devoted to the study of the CH equation. Here we only review some results about well-posedness. The local well-posedness of the non-periodic CH equation with initial data u0 ∈ H s (R), s > 23 , has been established in [11,26,30,32,37]. Moreover, it has both global strong solutions and finite time blow-up solutions [8,9]. After wave breaking the solutions can be continued uniquely as either global conservative [2] or dissipative solutions [3]. For the periodic case the initial value problem with u0 ∈ H s (T), s > 23 , is also locally well-posed [7]. It has been known that there exist global strong solutions for some certain class of initial data [10]. For k1 = −2, k2 = 0, Eq. (1.1) becomes the modified Camassa–Holm (mCH) equation:   mt + (u2 − u2x )m x = 0, m = u − uxx , which was derived independently by Fokas [13], Fuchssteiner [17], Olver and Rosenau [33], and Qiao [34]. Thus it is also called FORQ equation. The mCH equation regains attention due to its integrability and cuspon and peakon solutions[34]. Unlike the CH equation, the mCH equation admits not only new cusp solitons (cuspons), but also possesses weak kink solutions[35,36]. The Cauchy problem for the mCH equation in Besov spaces have been studied in [16]. In [18], the authors consider the formulation of the singularities of solutions and show that some solutions with certain initial date will blow up in finite time. Very recently, the blow-up mechanism to the mCH equation with varying linear dispersion has been presented in [5]. It is noticed that the nonlinear structure of Eq. (1.1) is similar to the one in dealing with the Gardner equation, or also called combined KdV–mKdV equation. In fact, Eq. (1.1) is a linear combination of the CH equation and mCH equation. Hence, we here called Eq. (1.1) as a combined CH–mCH equation. The purpose of our present paper is to study whether or not the solution map of Eq. (1.1) with two kinds of nonlinearities has similar properties of non-uniform dependence and H¨older continuity as the CH equation and the mCH equation. We first study the well-posedness result of the Cauchy problem to Eq. (1.1), which specifically reads as follows. Theorem 1.1. Let u0 ∈ H s (T), s > 52 . Then there exist a time T = T (∥u0 ∥H s (T) ) > 0 and a unique solution u to Eq. (1.1) such that u ∈ C([0, T ]; H s (T)). Moreover, the flow map u0 → u(t) is continuous from bounded sets of H s (T) into C([0, T ]; H s (T)). Furthermore, the estimate holds: ∥u(t)∥H s (T) ≤ max{4∥u0 ∥H s (T) , 1} , M,

for 0 ≤ t ≤ T ≤

1 , 32Cs ∥u0 ∥2H s (T)

(1.2)

where Cs is a constant depending only on s. To prove Theorem 1.1, it is convenient to rewrite Eq. (1.1) in the nonlocal form:    2  1  1 1 1 1 ut − k1 (3u2 ux − u3x ) − k2 uux = k1 Λ−2 (u3x ) + Λ−2 ∂x k1 uu2x + u3 + k2 u2 + u2x , 6 2 6 2 3 2 1

(1.3)

where the pseudo-differential operator Λ , (1 − ∂x2 ) 2 . The method used here is different from our approach to the non-periodic case in [31]. Inspired by [24], we also transform Eq. (1.3) into an ODE system, whose

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second equation is obtained from differentiating Eq. (1.3). That is, ∂t (ux ) =

k1 2 k1 2 k1 k1 k2 k2 k2 uu + u uxx − u2x uxx − u3 + u2x + uuxx − u2 2 x 2 2 3 4 2 2      k1 −2 1 2 1 + Λ ∂x (u3x ) + Λ−2 k1 uu2x + u3 + k2 u2 + u2x . 6 2 3 2

(1.4)

In comparison with the CH and mCH equations, the difficulty encountered by Eq. (1.1) is that it possesses both quadratic and cubic nonlinearities, which is required more delicate analysis on the approximate solutions. The second goal of our paper is to prove the following non-uniform dependence result. Theorem 1.2. If u0 ∈ H s (T), s > 25 , then the data-to-solution map for the Cauchy problem of Eq. (1.1) is not uniformly continuous from any bounded set of H s (T) into C([0, T ]; H s (T)). The investigation of non-uniform dependence on the initial data u0 ∈ H s (T), s ≥ 2 for the CH equation was initiated in [27] by using high frequency smooth traveling wave solutions. Then the result was sharpened to s > 32 on the periodic case [23] and non-periodic one [22]. Following that method of approximate solutions in conjunction with delicate commutator and multiplier estimates, the nonuniform dependence on the initial data for the Degasperis–Procesi (DP) equation (s > 32 ) [19], Novikov equation (s > 32 ) [20], Hunter–Saxton (HS) equation (s > 23 ) [28], and the mCH equation (s > 52 ) [24] has been established one after another. Our proof is also motivated by [23], which is based on the well-posedness estimates and the method of approximate solutions. However, it is noted that the constants k1 and k2 corresponding to the mCH cubic and CH quadratic nonlinearities respectively make a critical difference in the effects of varying nonlocal nonlinearities on the structure of the approximation solutions. More specifically, the pair of approximate solution sequences used for the CH, DP and HS equations [19,22,23,28] with quadratic nonlinearity has both low and high frequency components that symmetrically converge to zero at t = 0 and stay apart at any t > 0. In case of the equations with cubic nonlinearity including the Novikov and mCH equations [20,24], however, one of the approximate solution sequences has only high frequency part. Back to our Eq. (1.1), we find an interesting phenomenon appearing in Eq. (1.1) is the fact that we are required to use different approximate solutions, which depends on the existence of the quadratic nonlinearity or not. More precisely, when k2 ̸= 0, we define approximate solutions of the form uω,λ (t, x) = ωλ−1 + λ−s cos(λx + k22 ωt). When k2 = 0, the form of approximate 1 solutions is uω,λ (t, x) = ωλ− 2 + λ−s cos(λx + k21 ωt). Moreover, the pair of approximate solution sequences used for k2 ̸= 0 has both low and high frequency components. However, when k2 = 0, one of the sequences has no low frequency term, only with high frequency part. Now we have showed that the dependence of the solution map is continuous but not uniformly continuous. However, as stated in [6], if one studies the problem of Eq. (1.1) in a weaker H σ (T), 0 ≤ σ < s topology, it is reasonable to expect H¨ older continuity or even Lipschitz continuity of the solution map. Hence, we finally study the continuity properties of the solution map further and prove that it is actually H¨older continuous in the H σ (T) topology with 0 ≤ σ < s. To present our result, we first define the following function α as   1, (s, σ) ∈ Ω1 ,   s − σ, (s, σ) ∈ Ω2 , α= (1.5)  (s − σ)/2, (s, σ) ∈ Ω3 ,   (2s − 3)/(s − σ), (s, σ) ∈ Ω , 4 where the regions Ω1 , Ω2 , Ω3 , and Ω4 in the sσ-plane are given by Ω1 = {σ > 23 , σ ≤ s − 1} ∪ {0 ≤ σ ≤ 3 5 5 3 2 , 3 − s ≤ σ ≤ s − 2}, Ω2 = {(s, σ)|s > 2 , s − 1 ≤ σ < s}, Ω3 = {(s, σ)|s > 2 , s − 2 ≤ σ ≤ 2 }, and 5 Ω4 = {(s, σ)| 2 < s < 3, 0 ≤ σ ≤ 3 − s}. By the following result, the H¨older exponent α of Eq. (1.1) only

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with mixed nonlinearities is the same as the mCH equation with cubic nonlinearity [25]. Now we state the H¨ older (Lipschitz when α = 1) continuous result. Theorem 1.3. If s > 52 and 0 ≤ σ < s, then the data-to-solution map for the Cauchy problem of Eq. (1.1) is H¨ older continuous with the exponent α defined in (1.5) as a map from a ball B(0, M ) = {u| ∥u∥H s (T) ≤ M }, with H σ (T)-norm to C([0, T ]; H σ (T)). That is, for all initial data u0 , w0 ∈ B(0, M ), the corresponding solutions u(t), w(t) of Eq. (1.1) satisfy ∥u(t) − w(t)∥C([0,T ];H σ (T)) . ∥u0 − w0 ∥α H σ (T) . We now conclude the introduction by outlining the rest of our paper. In Section 2, we review and present some preliminary lemmas. In Section 3, we establish the local well-posedness result of Eq. (1.1) in the sense of Hadamard. In Section 4, we devote to proving that the date-to-solution map is not uniformly continuous. In Section 5, we complete the proof of H¨ older continuity. Notation. In the following, the symbols ., ≈ and & are used to express the corresponding inequality/equality that includes a universal constant. For example, f . g denotes that there exists a constant C > 0 such that f ≤ Cg. Since our discussion about Eq. (1.1) is on the circle T, for simplicity, we omit T in our notations of function spaces. 2. Preliminaries In this section, we recall and present some useful lemmas which will be used in the next section. Lemma 2.1 ([29]). If s > 0, then  s      Λ , f g  2 ≤ Cs ∥∂x f ∥L∞ ∥Λs−1 g∥L2 + ∥Λs f ∥L2 ∥g∥L∞ , L where Cs is a constant depending only on s. s

Lemma 2.2 ([1]). For any s ∈ R, let the pseudo-differential operator Λs = (1 − ∂x2 ) 2 be defined by  −ikx s f (k) , (1 + k 2 ) 2s f(k), where f(k) denotes the Fourier transform of f, f(k) , Λ e f (x)dx. Then T   12 1 s s 2 s 2  . In particular, the following for f ∈ H (T), we have ∥f ∥H s (T) , ∥Λ f ∥L2 (T) = √2π k∈Z (1 + k ) |f (k)| inequalities hold: ∥Λ−2 f ∥H s . ∥f ∥H s−2

and

∥Λ−2 ∂x f ∥H s . ∥f ∥H s−1 .

For brevity, we rewrite the ODE system (1.3)–(1.4) by denoting v = ∂x u  k k k  ut = 1 u2 v − 1 v 3 + 2 uv + F (u, v),   2 6 2  k1 2 k1 2 k1 k1 k2 k2 k2 vt = uv + u vx − v 2 vx − u3 + v 2 + uvx − u2 + G(u, v),   2 2 2 3 4 2 2    u(0, x) = u0 (x), v(0, x) = ∂x u0 (x) = v0 (x),

(2.1)

where     1 2  1  k  F (u, v) , 1 Λ−2 v 3 + Λ−2 ∂x k1 uv 2 + u3 + k2 u2 + v 2 , 6 2 3 2 (2.2)  G(u, v) , k1 Λ−2 ∂x (v 3 ) + 1 Λ−2 k1 uv 2 + 2 u3  + k2 u2 + 1 v 2 . 6 2 3 2 To get the existence of the solution for the above system, we consider the Cauchy problem of (2.1) by a mollified version

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  k1 k2 k1   ∂t uε = Jε (Jε uε )2 Jε vε − Jε (Jε vε )3 + Jε (Jε uε Jε vε ) + F (uε , vε ),   2 6 2     ∂ v = k1 J J u (J v )2  + k1 J (J u )2 ∂ (J v ) − k1 J (J v )2 ∂ (J v ) t ε ε ε ε ε ε ε ε ε x ε ε ε ε ε x ε ε 2 2 2   k2 k1 k2 k2     − u3ε + Jε (Jε vε )2 + Jε Jε uε ∂x (Jε vε ) − u2ε + G(uε , vε ),   3 4 2 2   uε (0, x) = u0 (x), vε (0, x) = v0 (x) = ∂x u0 (x),

(2.3)

where for each ε ∈ (0, 1], the mollifier Jε is defined by Jε f (x) = jε ∗ f (x). Here the periodic function jε is  1 ikx  , j(x) ∈ S(R). Thus, we have ∥Jε f ∥H s . ∥f ∥H s . given by jε (x) = 2π k∈Z j(εk)e Lemma 2.3. For all ε ∈ (0, 1], s > 52 , the mollified Cauchy problem of (2.3) has a unique solution (uε , vε ) ∈ C([0, T ]; H s−1 × H s−1 ) with lifespan T > 0. Moreover, we have the following size estimates: ∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ≤ M,

and

∥∂t uε ∥H s−2 + ∥∂t vε ∥H s−2 . M 3 ,

where the constant M , max{4∥u0 ∥H s , 1}.   Proof. For u0 ∈ H s , s > 25 , letting uε (0), vε (0) = (u0 , ∂x u0 ), (2.3) is a system of ODEs in H s−1 × H s−1 . Then by the Fundamental ODE Theorem in Banach spaces [12], there exists a unique solution (uε , vε ) ∈ C([0, Tε ]; H s−1 × H s−1 ) to the system (2.3) with lifespan Tε > 0. Next we shall derive energy estimates for solution (uε , vε ), and get a common lifespan T > 0 independent of ε. Applying the operator Λs−1 on the left of the second equation of system (2.3), multiplying by Λs−1 vε on the right, commuting Jε and Λs−1 and integrating over T, we have     s−1   1 d k1 k1 2 s−1 2 ∥vε (t)∥H s−1 = Λ Jε uε (Jε vε ) · Λ Jε vε dx + Λs−1 (Jε uε )2 ∂x (Jε vε ) · Λs−1 Jε vε dx 2 dt 2 T 2 T     s−1 k1 k1 s−1 2 Λ (Jε vε ) ∂x (Jε vε ) · Λ Jε vε dx − Λs−1 u3ε · Λs−1 vε dx − 2 T 3 T     k2 k2 + Λs−1 (Jε vε )2 · Λs−1 Jε vε dx + Λs−1 Jε uε ∂x (Jε vε ) · Λs−1 Jε vε dx 4 T 2 T   k2 k 1 + Λs−1 u2ε · Λs−1 vε dx + Λs−3 ∂x vε3 · Λs−1 vε dx 2 T 6 T     2  1  1 + Λs−3 k1 uε vε2 + u3ε + k2 u2ε + vε2 · Λs−1 vε dx 2 T 3 2 , F1 + F2 + F3 + · · · + F9 .

(2.4)

By the Cauchy–Schwarz inequality and the algebra property, we find F1 . ∥Jε uε (Jε vε )2 ∥H s−1 ∥Jε vε ∥H s−1 . ∥uε ∥H s−1 ∥vε ∥3H s−1 . Similarly, we deduce that F4 . ∥uε ∥3H s−1 ∥vε ∥H s−1 ,

F5 . ∥vε ∥3H s−1 ,

F7 . ∥uε ∥2H s−1 ∥vε ∥H s−1 ,

F8 . ∥vε ∥4H s−1 ,

and F9 . ∥uε ∥H s−1 ∥vε ∥3H s−1 + ∥uε ∥3H s−1 ∥vε ∥H s−1 + ∥uε ∥2H s−1 ∥vε ∥H s−1 + ∥vε ∥3H s−1 , where we applied Lemma 2.2 to F8 and F9 . By the Cauchy–Schwarz inequality and Lemma 2.1, we get     s−1  k1 2 s−1 2 s−1 s−1 Λ , (Jε uε ) ∂x (Jε vε ) · Λ Jε vε dx + (Jε uε ) Λ ∂x (Jε vε ) · Λ Jε vε dx F2 = 2 T  s−1T  . ∥Λ (Jε uε )2 ∥L2 ∥∂x (Jε vε )∥L∞ + ∥∂x (Jε uε )2 ∥L∞ ∥Λs−2 ∂x (Jε vε )∥L2 × ∥Λs−1 Jε vε ∥L2 + ∥Jε uε ∥2C 1 ∥Jε vε ∥2H s−1 . ∥Jε uε ∥2H s−1 ∥Jε vε ∥2H s−1 . ∥uε ∥2H s−1 ∥vε ∥2H s−1 ,

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where we used the algebra property and the Sobolev embedding H s−1 ↩→ C 1 for s > inequality. The third term F3 and the sixth term F6 can be calculated in a similar way, F3 . ∥vε ∥4H s−1

143

5 2

in the second

and F6 . ∥uε ∥H s−1 ∥vε ∥2H s−1 .

Combining the equality (2.4) and the estimates of F1 –F9 , we find that vε satisfies d ∥vε (t)∥H s−1 . (∥uε ∥H s−1 + ∥vε ∥H s−1 )3 + (∥uε ∥H s−1 + ∥vε ∥H s−1 )2 . dt Applying the similar computations to the first equation of system (2.3), for some Cs > 0 depending only on s, we then deduce that   d (∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ) ≤ Cs (∥uε ∥H s−1 + ∥vε ∥H s−1 )3 + (∥uε ∥H s−1 + ∥vε ∥H s−1 )2 . dt If ∥uε ∥H s−1 + ∥vε ∥H s−1 > 1, then we deduce from (2.5) that

(2.5)

d (∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ) ≤ 2Cs (∥uε ∥H s−1 + ∥vε ∥H s−1 )3 . dt Solving the above differential inequality, we get ∥u0 ∥H s−1 + ∥∂x u0 ∥H s−1 ∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ≤  . 1 − 4Cs t(∥u0 ∥H s−1 + ∥∂x u0 ∥H s−1 )2 By ∥u0 ∥H s−1 + ∥∂x u0 ∥H s−1 ≤ 2∥u0 ∥H s , setting T ,

1 32Cs ∥u0 ∥2H s

(2.6)

, we then find that (2.6) implies,

∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ≤ 4∥u0 ∥H s .

(2.7)

If ∥uε ∥H s−1 + ∥vε ∥H s−1 ≤ 1, this together with (2.7) gives ∥uε (t)∥H s−1 + ∥vε (t)∥H s−1 ≤ max{4∥u0 ∥H s , 1} , M,

for 0 ≤ t ≤ T.

(2.8)

Moreover, using (2.3) and similar argument about energy estimates, we obtain that for 0 ≤ t ≤ T , ∥∂t uε ∥H s−2 + ∥∂t vε ∥H s−2 . M 3 . This completes the proof of Lemma 2.3.

(2.9)



Lemma 2.4 ([38]). If σ +1 ≥ 0, then for f ∈ H ρ , g ∈ H σ , we have ∥[Λσ ∂x , f ]g∥L2 ≤ C∥f ∥H ρ ∥g∥H σ , provided that ρ > 23 and σ + 1 ≤ ρ. 3. Local well-posedness In this section, we briefly give the proof of Theorem 1.1, which is broken into the following Propositions 3.1–3.3. For simplicity, we denote the vector Uε , (uε , vε ) with the Sobolev norm defined by ∥Uε ∥H s = ∥uε ∥H s + ∥vε ∥H s , and the space (H s )2 , H s × H s . Proposition 3.1 (Existence). There exists a solution U , (u, v) ∈ C([0, T ]; (H s−1 )2 ) to the system (2.1), satisfying the size estimate ∥U (t)∥H s−1 = ∥u(t)∥H s−1 + ∥v(t)∥H s−1 ≤ M,

(3.1)

and the derivative size estimate ∥∂t U (t)∥H s−2 = ∥∂t uε ∥H s−2 + ∥∂t vε ∥H s−2 . M 3 , for 0 ≤ t ≤ T =

1 32Cs ∥u0 ∥2H s

, where the constant M is defined by max{4∥u0 ∥H s , 1}.

(3.2)

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Using the size estimates of the solution (uε , vε ) shown in Lemma 2.3. We see that the proof of Proposition 3.1 is similar to Theorem 1.1 in [24, pp. 505–509]. We omit it here. Proposition 3.2 (Uniqueness). The solution U = (u, v) ∈ C([0, T ]; (H s−1 )2 ) to the system (2.1) shown in Proposition 3.1 is unique. Proof. Suppose that there exists another solution W , (w, z) ̸= U in C([0, T ]; (H s−1 )2 ) for the system (2.1). Letting ϕ , u − w and ψ , v − z, then the Cauchy problem for (ϕ, ψ) is given by   k1  k1 2 k1 k2  k2   u − g + u ψ + (u + w) + ϕz + F (u, v) − F (w, z), ∂ ϕ =  t  2 6 2 2 2     ∂ ψ =  k1 u2 − k1 v 2 + k2 u∂ ψ +  k1 u(v + z) − k1 (v + z)∂ z + k2 (v + z)ψ t x x 2 2 2 2 2 4 (3.3)    k1 k2 k2 k1 2 k1    z + (u + w)∂x z − f + ∂x z − (u + w) ϕ + G(u, v) − G(w, z), +   2 2 3 2 2   ϕ(0, x) = ψ(0, x) = 0, where f , u2 + uw + w2 , g , v 2 + vz + z 2 and the functions F, G are given in (2.2). Let σ ∈ ( 21 , s − 2). Applying Λσ to the first equation of (3.3), multiplying by Λσ ϕ, and integrating on T, we find    k1 2 k1  k1 1 d k2  k2  ∥ϕ∥2H σ = Λσ u − g + u ψ · Λσ ϕdx + Λσ (u + w) + ϕz · Λσ ϕdx 2 dt 2 6 2 2 2 T T    + Λσ F (u, v) − F (w, z) · Λσ ϕdx. (3.4) T

By the Cauchy–Schwarz inequality, the algebra property and the size estimate (3.1), we obtain   k1 2 k1   k2  Λσ u − g + u ψ · Λσ ϕdx . ∥u∥2H σ + ∥g∥H σ + ∥u∥H σ ∥ψ∥H σ ∥ϕ∥H σ . (M 2 + M )∥ψ∥H σ ∥ϕ∥H σ . 2 6 2 T Similarly,  T

Λσ

 k1 k2  (u + w) + ϕz · Λσ ϕdx . (M 2 + M )∥ϕ∥2H σ . 2 2

Note that      2k1 k1 −2 1 k2 Λ ψg + Λ−2 ∂x k1 z 2 + f + k2 (u + w) ϕ + k1 u(v + z) + (v + z) ψ . 6 2 3 2 By Lemma 2.2, We have    Λσ F (u, v) − F (w, z) · Λσ ϕdx . (M 2 + M )(∥ψ∥H σ ∥ϕ∥H σ + ∥ϕ∥2H σ ). F (u, v) − F (w, z) =

T

Putting the above estimates together implies that d ∥ϕ∥H σ . (M 2 + M )(∥ψ∥H σ + ∥ϕ∥H σ ). (3.5) dt To get a similar estimate for ψ, we apply Λσ to the second equation of (3.3), multiply by Λσ ψ, and integrate on T, to obtain     k1 1 d k2  k2  2 σ k1 2 2 σ ∥ψ∥H σ = Λ (u − v ) + u ψx · Λ ψdx + Λσ (u − zx ) + (v + z)ψ · Λσ ψdx 2 dt 2 2 2 4 T T  k   k1  k2  1 + Λσ z 2 + (u + w)zx − f + zx − (u + w) ϕ · Λσ ψdx 2 3 2 T    + Λσ G(u, v) − G(w, z) · Λσ ψdx , Ψ1 + Ψ2 . (3.6) T

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Let us estimate the first term Ψ1 of the rhs of (3.6). Commuting ∂x with k21 (u2 − v 2 ) + k22 u, then using the Cauchy–Schwarz inequality, integrating by parts and the algebra property, we have   k1 2 k2  (u − v 2 ) + u ∂x ψ · Λσ ψdx Ψ1 = Λσ 2 2 T   σ  k1 2 k2  k2  k1 2 (u − v 2 ) + u Λσ ∂x ψ = Λ ∂x , (u − v 2 ) + u ψ · Λσ ψdx + 2 2 2 T T 2    k k 2 1 (u2 − v 2 ) + u ψ · Λσ ψdx × Λσ ψdx − Λσ ∂x 2 2 T   k1 2 k2 .  (u − v 2 ) + uH s−1 ∥ψ∥2H σ + (∥u∥2H s−1 + ∥v∥2H s−1 + ∥u∥H s−1 )∥ψ∥2H σ 2 2   k1 2 k2  + ∂x (u − v 2 ) + u H s−2 ∥ψ∥2H σ 2 2  k k2  1 (3.7) . (M 2 + M )∥ψ∥2H σ +  (u2 − v 2 ) + uH s−1 ∥ψ∥2H σ . (M 2 + M )∥ψ∥2H σ , 2 2 where we used Lemma 2.4 in the first inequality with ρ = s − 1. Note that      2k1 1 k2 k1 f + k2 (u + w) ϕ + k1 u(v + z) + (v + z) ψ . G(u, v) − G(w, z) = Λ−2 ∂x (ψg) + Λ−2 k1 z 2 + 6 2 3 2 Then the remaining terms Ψ2 on the rhs of (3.6) can be estimated similarly as the terms on the rhs of (3.4), we get Ψ2 . (M 2 + M )(∥ϕ∥H σ ∥ψ∥H σ + ∥ψ∥2H σ ).

(3.8)

Combining the inequalities (3.7)–(3.8) with (3.6), we obtain d ∥ψ∥H σ . (M 2 + M )(∥ϕ∥H σ + ∥ψ∥H σ ). dt

(3.9)

By (3.5) and (3.9), we have    d ∥ϕ∥H σ + ∥ψ∥H σ . (M 2 + M ) ∥ϕ∥H σ + ∥ψ∥H σ . dt An application of Gronwall’s inequality and the initial condition ϕ(0, x) = ψ(0, x) = 0 yield   2 ∥ϕ∥H σ + ∥ψ∥H σ . ∥ϕ(0)∥H σ + ∥ψ(0)∥H σ e(M +M )t = 0. Hence, we prove that the solution U = (u, v) to the system (2.1) is unique. This completes the proof of Proposition 3.2.  Now we relate the unique solution U = (u, v) obtained in Propositions 3.1–3.2 of the system (2.1) to the Cauchy problem of Eq. (1.1). Lemma 3.1. The Cauchy problem of (2.1) implies that v = ∂x u. Proof. Differentiating the first equation of the system (2.1), we have k1 k1 k2 k2 k1  2 2 3  k2  2 1 2  ∂t (∂x u) = k1 uux v + u2 vx − v 2 vx + ux v + uvx − uv + u − u + v + G(u, v), 2 2 2 2 2 3 2 2 where G is given in (2.2). Hence we can get the following initial value problem  k k k k k k k k  ∂t (∂x u) = k1 uux v − 1 uv 2 + 1 u2 vx − 1 v 2 vx − 1 u3 + 2 ux v − 2 v 2 + 2 uvx − 2 u2 + G(u, v),   2 2 2 3 2 4 2 2  k1 2 k1 2 k1 2 k1 3 k2 2 k2 k2 2 (3.10) ∂t v = uv + u vx − v vx − u + v + uvx − u + G(u, v),   2 2 2 3 4 2 2    ∂x u(0, x) = v(0, x) = ∂x u0 (x).

146

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Subtracting the second equation of (3.10) from the first one and setting ω = ∂x u − v, we obtain the following Cauchy problem of ω  ∂ ω = k uvω + k2 ωv, t 1 2  ω(0, x) = 0.  t  Solving the above differential equation yields ω(t, x) = ω(0, x) exp 0 (k1 uv + k22 v)dτ = 0. Therefore we prove that v = ∂x u. This completes the proof of Lemma 3.1.  Since v ∈ C([0, T ]; H s−1 ), then Lemma 3.1 implies that u ∈ C([0, T ]; H s ). A similar argument as the energy estimates for vε and uε , we can also get the size estimate and lifespan given in (1.2) for the solution u stated in Theorem 1.1. Now we have shown that there exists a unique solution u ∈ C([0, T ]; H s ) for the Cauchy problem of Eq. (1.1). Next we complete the proof of Theorem 1.1 by proving that the dependence of the solution on initial data is continuous. Proposition 3.3 (Continuous Dependence). The data-to-solution map u0 (x) → u(t, x) for the Cauchy problem of Eq. (1.1) is continuous from H s → C([0, T ]; H s ). Proof. Fix u0 ∈ H s and let {u0,n } ⊂ H s be a sequence such that limn→∞ u0,n = u0 in H s . Then it suffices to show that limn→∞ un (t, x) = u(t, x) in C([0, T ]; H s ), where u(t, x) and un (t, x) are the solutions to Eq. (1.1) with initial data u0 and u0,n , respectively. To overcome the difficulties of estimating the term involving u2 ∂x u, we introduce uε and uεn as the solutions to Eq. (1.1) with the mollified initial data Jε u0 = jε ∗ u0 and Jε u0,n , respectively. Moreover, due to the presence of the term (∂x u)3 , we will handle it in a similar way by transforming Eq. (1.1) into a system. We then find that the solutions (u, v), (un , vn ), (uε , v ε ) and (uεn , vnε ) satisfy the same system (2.1) with initial data (u0 (x), v0 (x)), (un (0, x), vn (0, x)), (uε (0, x), v ε (0, x)) and (uεn (0, x), vnε (0, x)), respectively. Using the triangle inequality, we have ∥u − un ∥C([0,T ];H s ) ≤ ∥u − uε ∥C([0,T ];H s ) + ∥uε − uεn ∥C([0,T ];H s ) + ∥uεn − un ∥C([0,T ];H s ) .

(3.11)

Thus, instead of estimating the difference u − un , we will calculate each of these terms on the rhs of (3.11). Due to the differences u − uε and un − uεn satisfying the same equations as u − ω in Proposition 3.2, we omit the details of the estimations. One can obtain that ∥uε − uεn ∥H s−1 + ∥v ε − vnε ∥H s−1 ≤ e

Cs T ε

∥u0 − u0,n ∥H s ,

(3.12)

and   ∥u − uε ∥H s−1 + ∥v − v ε ∥H s−1 ≤ et ∥u0 − uε0 ∥H s−1 + ∥v0 − v0ε ∥H s−1 − δ1 ,

(3.13)

where δ1 = δ1 (ε) → 0 as ε → 0. The same approach to the system satisfied by un − uεn yields,  ∥un − uεn ∥H s−1 + ∥vn − vnε ∥H s−1 ≤ et 2∥u0,n − u0 ∥H s−1 + ∥u0 − uε0 ∥H s−1 + 2∥v0,n − v0 ∥H s−1  + ∥v0 − v0ε ∥H s−1 − δ2 , with δ2 = δ2 (ε) → 0, as ε → 0. (3.14) Since ∥u0 − uε0 ∥H s−1 → 0 and ∥v0 − v0ε ∥H s−1 → 0 as ε → 0, we can choose ε sufficiently small such that for given η > 0, ∥u − uε ∥H s−1 + ∥v − v ε ∥H s−1 ≤

η . 3

(3.15)

By (3.12) and (3.14), for ε as specified earlier, we can choose N sufficiently large such that ∥uε − uεn ∥H s−1 + ∥v ε − vnε ∥H s−1 ≤

η , 3

∀n > N

(3.16)

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and η , ∀n > N. (3.17) 3 Therefore, the relations (3.15)–(3.17) together with (3.11) imply the proof of continuous dependence. This completes the proof of Proposition 3.3.  ∥un − uεn ∥H s−1 + ∥vn − vnε ∥H s−1 ≤

4. Non-uniform dependence In this section, we shall complete the proof of Theorem 1.2. More precisely, we show that there exists two sequences of solutions uλ (t) and vλ (t) in C([0, T ]; H s (T)) such that ∥uλ (t)∥H s . 1,

∥vλ (t)∥H s . 1,

lim ∥uλ (0) − vλ (0)∥H s = 0,

(4.1)

λ→∞

and  k2 t  . lim inf ∥uλ (t) − vλ (t)∥H s &  sin λ→∞ 2

(4.2)

Proof. Case 1: k2 ̸= 0. We break our proof into three steps. Step 1: Estimating the error of approximate solutions. We consider here approximate solutions of the following form with a low and a high frequency part  1 k2  ω + cos λx + ωt , λ λs 2 where ω is in a bounded subset of R and λ is in the set of positive integers Z+ . Then, we have uω,λ (t, x) =

∂t uω,λ (t, x) = −

k2 −s  k2  ωλ sin λx + ωt 2 2

 k2  and ∂x uω,λ (t, x) = −λ−s+1 sin λx + ωt . 2

(4.3)

(4.4)

Firstly, we compute the error of the approximate solution of (4.3). Substituting (4.4) into Eq. (1.3), we obtain 1 k2 F1ω,λ , ∂t uω,λ − k2 uω,λ ∂x uω,λ = λ−2s+1 sin(2λx + k2 ωt), 2 4  k1 ω,λ 2 k k2  k1 1 ω,λ F2 , − (u ) ∂x uω,λ = ω 2 λ−s−1 sin λx + ωt + ωλ−2s sin(2λx + k2 ωt) 2 2 2 2  k2  k2  k1 −3s+1   3 sin λx + ωt − sin λx + ωt , + λ 2 2 2   k1 k k 1 2 ω,λ F3 , (∂x uω,λ )3 = − λ3−3s sin3 λx + ωt , 6 6 2 k1 F4ω,λ , − Λ−2 (∂x uω,λ )3 = −Λ−2 F3ω,λ . 6 Since uω,λ (∂x uω,λ )2 = ωλ−2s+1 sin2 (λx +

k2 2 ωt)

+ λ−3s+2 sin2 (λx +

k2 2 ωt) cos(λx

+

k2 2 ωt),

  k1 −2  ω,λ k1 Λ ∂x u (∂x uω,λ )2 = − Λ−2 ωλ−2s+2 sin(2λx + k2 ωt) 2 2   k2  k2  + 2λ−3s+3 sin λx + ωt − 3λ−3s+3 sin3 λx + ωt , 2 2 k1 −2  ω,λ 3  , − Λ ∂x (u ) = 2Λ−2 F2ω,λ , 3   k2 , − Λ−2 ∂x (uω,λ )2 2   k2 k2  = − Λ−2 λ−2s+1 sin(2λx + k2 ωt) + 2ωλ−s sin λx + ωt , 2 2

(4.5)

(4.6) (4.7) (4.8)

then

F5ω,λ , −

F6ω,λ F7ω,λ

(4.9) (4.10)

(4.11)

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and F8ω,λ , −

 k2 −2  1 k2 Λ ∂x (∂x uω,λ )2 = − Λ−2 λ−2s+3 sin(2λx + k2 ωt). 2 2 4

(4.12)

To summarize, the error F ω,λ of the approximate solution of (4.3) is given by F ω,λ , F1ω,λ + F2ω,λ + · · · + F8ω,λ  k2 k1 k1  ω,λ 2 3(u ) ∂x uω,λ − (∂x uω,λ )3 − uω,λ ∂x uω,λ − Λ−2 (∂x uω,λ )3 = ∂t uω,λ − 6 2 6   ω,λ 2 1 1 −2   ω,λ 2 ω,λ 3  ω,λ 2 − Λ ∂x k1 u (∂x u ) + (u ) + k2 (u ) + (∂x uω,λ )2 , 2 3 2

(4.13)

where Fiω,λ , i = 1, 2, . . . , 8 are defined by (4.5)–(4.12). Next we shall estimate the H σ -norm of the error F ω,λ . We need the following useful lemma. Lemma 4.1 ([23]). Let σ ∈ R. If λ ∈ Z+ and λ ≫ 1, then ∥ cos(λx − α)∥H σ (T) ≈ λσ ,

α ∈ R.

(4.14)

Relation (4.14) is also true if cos(λx − α) is replaced by sin(λx − α). Finally, for any s ≥ 0 we have   k2  ∥uω,λ (t)∥H σ (T) = ωλ−1 + λ−s cos λx + ωt H σ (T) . λ−1 + λ−s+σ , 2

λ ≫ 1.

(4.15)

Now we can estimate the H σ -norm of the error F ω,λ by computing separately the H σ -norm of each term Fiω,λ , where we take σ ∈ ( 23 , s − 1), s > 52 , in the remainder of this proof. Using Lemmas 4.1, 2.2, (4.5)–(4.12) and the trigonometric identity sin3 t = 14 (3 sin t − sin 3t), we obtain ∥F1ω,λ ∥H σ . λ−2s+1+σ ,

∥F2ω,λ ∥H σ . λ−s−1+σ + λ−2s+σ + λ−3s+1+σ ,

∥F4ω,λ ∥H σ . ∥F3ω,λ ∥H σ−2 . λ−3s+1+σ ,

∥F3ω,λ ∥H σ . λ3−3s+σ ,

∥F5ω,λ ∥H σ . λ−2s+σ + λ−3s+1+σ ,

∥F6ω,λ ∥H σ . ∥F2ω,λ ∥H σ−2 . λ−s−3+σ + λ−2s−2+σ + λ−3s−1+σ , ∥F7ω,λ ∥H σ . λ−2s−1+σ + λ−s−2+σ ,

∥F8ω,λ ∥H σ . λ−2s+1+σ .

(4.16)

Since s + 1 − σ < s + 2 − σ, s + 3 − σ and 2s − 1 − σ < 2s − σ, 3s − 1 − σ, 2s + 1 − σ, 2s + 2 − σ, 3s + 1 − σ, then, by (4.13) and (4.16), we have ∥F ω,λ ∥H σ . λ−(s+1−σ) + λ−(2s−1−σ) .

1 , λs+1−σ

λ ≫ 1,

(4.17)

where we used the inequality s + 1 − σ < 2s − 1 − σ, for s > 52 . Step 2: Estimating the difference between approximate and actual solutions. Let us firstly define uω,λ (t, x) to be the actual solution for the Cauchy problem of Eq. (1.1) with initial data equal to the approximate solutions evaluated at time zero. More precisely, uω,λ (t, x) solves   k2 k1 k1  2  ∂t uω,λ − 3uω,λ ∂x uω,λ − (∂x uω,λ )3 − uω,λ ∂x uω,λ = Λ−2 (∂x uω,λ )3   6 2 6     1 −2   2 1 2 3 2 + Λ ∂x k1 uω,λ (∂x uω,λ ) + (uω,λ ) + k2 (uω,λ ) + (∂x uω,λ )2 ,   2 3 2   uω,λ (0, x) = uω,λ (0, x) = ωλ−1 + λ−s cos(λx), Note that, for any s ≥ 0, ω ∥ ∥H s (T) = |ω|λ−1 ∥1∥H s (T) = 2π|ω|λ−1 . λ

t > 0, x ∈ T, (4.18) x ∈ T.

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Thus, by Lemma 4.1, we find that the initial data uω,λ (0, x) ∈ H s for s ≥ 0, and ∥uω,λ (0, x)∥H s ≈ 1 for sufficiently large λ. Therefore, by Theorem 1.1, we conclude that there exists a T > 0 such that the Cauchy problem (4.18) has a unique solution uω,λ (t, x) ∈ C([0, T ]; H s ) and satisfying ∥uω,λ (t)∥H s . max{4∥uω,λ (0)∥H s , 1} . 1.

(4.19)

Next, we shall compute the difference between approximate and actual solutions. Subtracting (4.18) from (4.13) and letting v , uω,λ − uω,λ , we have  2 k k k k k  ∂t v − 1 ∂x (f v) + 1 gvx − 2 ∂x (hv) = F ω,λ + 1 Λ−2 (gvx ) + 1 Λ−2 ∂x f v   6 6 4 6 2 3   k2 −2   1 ω,λ 2 + (∂x u ) v + uω,λ ∂x h∂x v + Λ ∂x hv + ∂x h∂x v , t > 0, x ∈ T, (4.20)   2 2    v(0, x) = v0 (x) = 0, x ∈ T, where f , (uω,λ )2 + uω,λ uω,λ + (uω,λ )2 , g , (∂x uω,λ )2 + ∂x uω,λ ∂x uω,λ + (∂x uω,λ )2 and h , uω,λ + uω,λ . Applying the operator Λσ to both sides of (4.20), multiplying by Λσ v, and integrating over T, we obtain    k1 k1 k2 1 d 2 σ σ σ σ ∥v(t)∥H σ = Λ ∂x (f v) · Λ vdx − Λ gvx · Λ vdx + Λσ ∂x (hv) · Λσ vdx 2 dt 6 T 6 T 4 T    k1 k1 σ ω,λ σ σ−2 σ + Λ F · Λ vdx + Λ gvx · Λ vdx + Λσ−2 ∂x (f v) · Λσ vdx 6 3 T T T     σ   k1 k1 σ−2 ω,λ 2 σ−2 + Λ ∂x (∂x u ) v · Λ vdx + Λ ∂x uω,λ ∂x h∂x v · Λσ vdx 2 T 2 T   k2 k2 + Λσ−2 ∂x (hv) · Λσ vdx + Λσ−2 ∂x (∂x h∂x v) · Λσ vdx. (4.21) 2 T 4 T By (4.15) and (4.19), for σ ∈ ( 23 , s − 1), s >

5 2

and λ ≫ 1, we have

 ω,λ 2 ω,λ ω,λ 2 s s s ∥h∥H s . 1,  ∥f ∥H . ∥u ∥H s + ∥u ∥H ∥uω,λ ∥H + ∥u ∥H s . 1, ω,λ 2 ω,λ ω,λ 2 ∥g∥H σ . ∥u ∥H σ+1 + ∥u ∥H σ+1 ∥uω,λ ∥H σ+1 + ∥u ∥H σ+1 . 1,   ∥∂x h∥H σ . ∥uω,λ ∥H σ+1 + ∥uω,λ ∥H σ+1 . ∥uω,λ ∥H s + ∥uω,λ ∥H s . 1.

(4.22)

Now let us estimate each term of the rhs of (4.21). For the first term, by (4.22), the Sobolev inequality ∥∂x f ∥L∞ . ∥f ∥H s and Lemma 2.4 with ρ = s, we obtain     k1  k1 Λσ ∂x (f v) · Λσ vdx = [Λσ ∂x , f ]v · Λσ vdx + f Λσ ∂x v · Λσ vdx 6 T 6 T T . ∥f ∥H s ∥v∥2H σ . ∥v∥2H σ .

(4.23)

Similarly, one can obtain the estimate for the third term. For the second term, by Lemma 2.1 and (4.22), we have     −k1 −k1  Λσ gvx · Λσ vdx = [Λσ , g]∂x v · Λσ vdx + gΛσ ∂x v · Λσ vdx 6 T 6 T T   . ∥g∥H σ ∥∂x v∥L∞ + ∥∂x g∥L∞ ∥∂x v∥H σ−1 ∥v∥H σ + ∥∂x g∥L∞ ∥v∥2H σ . ∥g∥H σ ∥v∥2H σ . ∥v∥2H σ . For the fourth term, by (4.17), we see that  Λσ F ω,λ · Λσ vdx . ∥F ω,λ ∥H σ ∥v∥H σ . T

(4.24)

1 ∥v∥H σ , λs+1−σ

λ ≫ 1.

(4.25)

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For brevity, we denote Ψ as the remaining terms of the rhs of (4.21). Then, by the algebra property with 3 2 < σ < s − 1 < s, the inequalities (4.15), (4.19) and (4.22), we can deduce that  |Ψ | . ∥gvx ∥H σ−1 ∥v∥H σ−1 + ∥f v∥H σ−1 + ∥(∂x uω,λ )2 v∥H σ−1  + ∥uω,λ ∂x h∂x v∥H σ−1 + ∥hv∥H σ−1 + ∥∂x h∂x v∥H σ−1 ∥v∥H σ  . ∥g∥H σ−1 + ∥f ∥H σ−1 + ∥uω,λ ∥2H σ ∥v∥H σ−1 + ∥uω,λ ∥H σ−1 ∥h∥H σ ∥v∥H σ  + ∥h∥H σ−1 ∥v∥H σ−1 + ∥h∥H σ ∥v∥H σ ∥v∥2H σ . ∥v∥2H σ . (4.26) Combining inequalities (4.23)–(4.26) with (4.21), we get 1 d ∥v(t)∥H σ . ∥v(t)∥H σ + s+1−σ , dt λ

λ ≫ 1,

(4.27)

which implies that ∥v(t)∥H σ . λ−s−1+σ ,

λ ≫ 1.

(4.28)

Step 3: Completing the proof of Theorem 1.2. Let uλ , u1,λ and vλ , u−1,λ be the solutions to (4.18) with initial data u1,λ (0, x) and u−1,λ (0, x), respectively. Then, in order to prove Theorem 1.2, we only need to verify uλ and vλ , which satisfy (4.1)–(4.2). Firstly, we present the H s -norm of the difference between u±1,λ and u±1,λ . Since, for any integer k , [s] + 2, ∥u±1,λ (t) − u±1,λ (t)∥H k . ∥u±1,λ (0)∥H k , then (4.15) yields that ∥u±1,λ (t) − u±1,λ (t)∥H k . λk−s ,

λ ≫ 1.

(4.29)

Hence, applying interpolation with s1 = σ < s − 1 and s2 = k = [s] + 2 and using (4.28)–(4.29), we obtain k−s

s−σ

k−σ k−σ ±1,λ ∥u±1,λ (t) − u±1,λ (t)∥H s ≤ ∥u±1,λ (t) − u±1,λ (t)∥H (t) − u±1,λ (t)∥H σ ∥u k k−s

s−σ

k−s

1

. λ(−s−1+σ) k−σ · λ(k−s) k−σ = λ− k−σ ≤ λ− s+2 ,

λ ≫ 1.

(4.30)

Next, we show that uλ and vλ satisfy (4.1)–(4.2). By the definition of uλ and vλ , we have 2 ∥uλ (0) − vλ (0)∥H s = ∥u1,λ (0) − u−1,λ (0)∥H s = ∥ ∥H s ≈ λ−1 → 0, λ Using the triangle inequality and (4.30), we see that

as λ → ∞.

(4.31)

∥uλ (t) − vλ (t)∥H s = ∥u1,λ (t) − u−1,λ (t)∥H s 1

≥ ∥u1,λ (t) − u−1,λ (t)∥H s − Cλ− s+2     1 k2 t  k2 t   s − Cλ− s+2 = 2λ−1 + λ−s cos λx + − cos λx − H 2 2   1 k t 2 −s −1  − 2λ ∥1∥H s − Cλ− s+2 ≥ 2λ ∥ sin(λx)∥H s  sin . 2 Now letting λ ↗ 0 in the above inequality, by (4.14), we have  k2 t  lim inf ∥uλ (t) − vλ (t)∥H s &  sin . λ→∞ 2 Therefore, (4.31) and (4.32) prove Theorem 1.2 for Case 1: k2 ̸= 0.

(4.32)

Case 2: k2 = 0. When k2 = 0, Eq. (1.1) becomes  k1  2 (u − u2x )m x = 0, m = u − uxx . (4.33) 2 Thus, we find that the difference between the mCH equation and (4.33) is only the coefficient of equation. Therefore, the proof of Theorem 1.2 for Case 2: k2 = 0 is very similar to the mCH equation [24]. Here we omit the proof. This completes the proof of Theorem 1.2.  mt −

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151

5. H¨ older continuity In this section, we shall complete the proof of Theorem 1.3. For this purpose, the method used here is again required us to write Eq. (1.1) into a system as (2.1) in Section 2. We also denote by U = (u, v) ∈ C([0, T ]; (H s−1 )2 ) and W = (w, z) ∈ C([0, T ]; (H s−1 )2 ) are two solutions to (2.1) with initial data (u0 (x), v0 (x)) and (w0 (x), z0 (x)), respectively. If ϕ = u − w and ψ = v − z, then we find that (ϕ, ψ) satisfies the same equations in (3.3), but with different initial data ϕ(0, x) = u0 (x) − w0 (x) , ϕ0 (x) and ψ(0, x) = v0 (x) − z0 (x) = ∂x ϕ0 (x) , ψ0 (x). Now we finish our proof into two steps. Proof. Step 1: Lipschitz continuity in Ω1 . We begin the proof with the energy identities (3.4) and (3.6), and ′ ′ ′ denote by Ω1′ , Ω11 ∪ Ω12 ∪ Ω13 = {σ > 12 , σ ≤ s − 2} ∪ {− 12 < σ ≤ 21 , σ ≤ s − 3} ∪ {−1 ≤ σ ≤ − 12 , σ + s ≥ 2}. ′ In fact, for σ ∈ Ω11 , we obtain in Proposition 3.2 that    d ∥ϕ∥H σ + ∥ψ∥H σ . (M 2 + M ) ∥ϕ∥H σ + ∥ψ∥H σ . dt

(5.1)

′ Next we prove that the inequality (5.1) holds for the other subsets of Ω1′ . When σ ∈ Ω12 , we need the following lemma to get (5.1).

Lemma 5.1 ([21]). If σ > − 12 , then ∥f g∥H σ ≤ Cσ ∥f ∥H σ+1 ∥g∥H σ . For the first term on the rhs of (3.4), by Lemma 5.1, the algebra property with σ + 1 > 21 , and the size estimate (3.1) with σ + 1 ≤ s − 2, we have   k1 2 k1  k1 2 k1 k2  k2  u − g + u ψ · Λσ ϕdx ≤ ∥ u − g + u ψ∥H σ ∥ϕ∥H σ Λσ 2 6 2 2 6 2 T   2 2 . ∥u∥H σ+1 + ∥g∥H σ+1 + ∥u∥H σ+1 ∥ψ∥H σ ∥ϕ∥H σ . (M + M )∥ψ∥H σ ∥ϕ∥H σ . (5.2) Similarly,  T

Λσ

 k1 k2  (u + w) + ϕz · Λσ ϕdx . (M 2 + M )∥ϕ∥2H σ , 2 2

(5.3)

and for the third term, we obtain    Λσ F (u, v) − F (w, z) · Λσ ϕdx T

   k2   2k1 . ∥gψ∥H σ ∥ϕ∥H σ +  k1 z 2 + f + k2 (u + w) ϕ + (v + z) k1 u + ψ H σ ∥ϕ∥H σ 3 2 . ∥g∥H σ+1 ∥ψ∥H σ ∥ϕ∥H σ + (∥z∥2H σ+1 + ∥f ∥H σ+1 + ∥u + w∥H σ+1 )∥ϕ∥2H σ + (∥u(v + z)∥H σ+1 + ∥v + z∥H σ+1 )∥ψ∥H σ ∥ϕ∥H σ . (M 2 + M )(∥ψ∥H σ ∥ϕ∥H σ + ∥ϕ∥2H σ ).

(5.4)

Putting the estimates (5.2)–(5.4) together implies that d ∥ϕ∥H σ . (M 2 + M )(∥ϕ∥H σ + ∥ψ∥H σ ). dt

(5.5)

To get a similar estimate for ψ, let us consider the energy identity (3.6). Indeed, for the first term Ψ1 on the rhs of (3.6), we can estimate it as handled in (3.7) except the third term of Ψ1 , because Lemma 2.4 is always true for σ ∈ Ω1′ . Then using Lemma 5.1, we get the estimate of the third term of Ψ1 as follows.   k1 2  k1 2 k2  k2  (u − v 2 ) + u ψ · Λσ ψdx ≤ ∥∂x (u − v 2 ) + u ψ∥H σ ∥ψ∥H σ Λσ ∂x 2 2 2 2 T

X. Liu / Nonlinear Analysis 143 (2016) 138–154

152

 k1 2 k2  k2 k1 (u − v 2 ) + u ∥H σ+1 ∥ψ∥H σ ∥ψ∥H σ . ∥ (u2 − v 2 ) + u∥H σ+2 ∥ψ∥2H σ 2 2 2 2 2 2 . (M + M )∥ψ∥H σ . . ∥∂x

(5.6)

The remaining terms on the rhs of (3.6) can be estimated similarly as (5.2)–(5.4) by Lemma 5.1. Thus, we have d ∥ψ∥H σ . (M 2 + M )(∥ψ∥H σ + ∥ϕ∥H σ ). dt

(5.7)

′ ′ Hence, (5.5) and (5.7) yield that (5.1) is right for Ω12 . When σ ∈ Ω13 , one can also obtain (5.1) with the help of the following lemma.

Lemma 5.2 ([25]). If −1 ≤ σ ≤ 0, s >

5 2

and σ + s ≥ 2, then ∥f g∥H σ ≤ Cσ,s ∥f ∥H s−2 ∥g∥H σ .

′ Since the procedure is very similar to the case of σ ∈ Ω12 , we only need to use Lemma 5.2 instead of Lemma 5.1 to estimate the product terms. For example,   k1 2 k1 k2  k1 k1 k2 u − g + u ψ · Λσ ϕdx . ∥ u2 − g + u∥H s−2 ∥ψ∥H σ ∥ϕ∥H σ Λσ 2 6 2 2 6 2 T   2 . ∥u∥H s−2 + ∥g∥H s−2 + ∥u∥H s−2 ∥ψ∥H σ ∥ϕ∥H σ . (M 2 + M )∥ψ∥H σ ∥ϕ∥H σ ,

and the term as handled in (5.6),   k1 2  k1 2 k2  k2  (u − v 2 ) + u ψ · Λσ ψdx . ∥∂x (u − v 2 ) + u ∥H s−2 ∥ψ∥H σ ∥ψ∥H σ Λσ ∂x 2 2 2 2 T k1 k2 . ∥ (u2 − v 2 ) + u∥H s−1 ∥ψ∥2H σ . (M 2 + M )∥ψ∥2H σ . 2 2 The remaining terms can be calculated similarly. To summarize, we prove that (5.1) holds for all σ ∈ Ω1′ . Then an application of Gronwall’s inequality to (5.1) yields   2 ∥ϕ∥H σ + ∥ψ∥H σ . ∥ϕ0 (x)∥H σ + ∥ψ0 (x)∥H σ e(M +M )t . By the definition of ϕ, ψ, and the result of v = ∂x u, z = ∂x w, shown in Lemma 3.1, we find   2 ∥u − w∥H σ + ∥u − w∥H σ+1 . ∥u0 (x) − w0 (x)∥H σ + ∥u0 (x) − w0 (x)∥H σ+1 e(M +M )t , from which it follows that ∥u − w∥H σ+1 . ∥u0 (x) − w0 (x)∥H σ+1 e(M

2

+M )T

.

Thus, by lowering the Sobolev index from σ + 1 to σ, we get ∥u − w∥H σ . ∥u0 (x) − w0 (x)∥H σ e(M

2

+M )T

,

(5.8)

for σ ∈ Ω1 , Ω11 ∪ Ω12 ∪ Ω13 = {σ > 23 , σ ≤ s − 1} ∪ { 12 < σ ≤ 32 , σ ≤ s − 2} ∪ {0 ≤ σ ≤ 12 , σ + s ≥ 3}. Step 2: H¨ older continuity in Ω2 ∪Ω3 ∪Ω4 . For (s, σ) ∈ Ω2 , by interpolation between the H s−1 and H s -norms, we have s−σ σ−s+1 ∥u − w∥H σ ≤ ∥u − w∥H , s−1 ∥u − w∥H s

which combines with the size estimate (3.1) and the Lipschitz continuity in Ω11 , we find s−σ σ−s+1 (M ∥u − w∥H σ . M σ−s+1 ∥u − w∥H e s−1 . M

. M σ−s+1 e(M

2

+M )T

s−σ ∥u0 − w0 ∥H σ .

2

+M )T

s−σ ∥u0 − w0 ∥H s−1

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153

For (s, σ) ∈ Ω3 , a similar argument by interpolation between the H s−2 and H s -norms, and the Lipschitz continuity in Ω12 , one can easily prove that ∥u − w∥H σ . M

σ−s+2 2

e(M

2

+M )T

s−σ

∥u0 − w0 ∥H2s−2 . M

σ−s+2 2

e(M

2

+M )T

s−σ

∥u0 − w0 ∥H2σ .

For (s, σ) ∈ Ω4 , by the Lipschitz continuity in Ω13 , and interpolation between the H σ and H s -norms, we obtain ∥u − w∥H σ ≤ ∥u − w∥H 3−s ≤ e(M . e(M

2

+M )T

2

∥u0 − w0 ∥

This completes the proof of Theorem 1.3.

+M )T

2s−3 s−σ Hσ

∥u0 − w0 ∥H 3−s 3−s−σ

∥u0 − w0 ∥Hs−σ .M s

3−s−σ s−σ

e(M

2

+M )T

2s−3

∥u0 − w0 ∥Hs−σ σ .



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