Decay property of solutions near the traveling wave solutions for the second-order Camassa–Holm equation

Nonlinear Analysis 183 (2019) 230–258

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Nonlinear Analysis www.elsevier.com/locate/na

Decay property of solutions near the traveling wave solutions for the second-order Camassa–Holm equation✩ Danping Ding, Kai Wang ∗ Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu 212013, China

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Article history: Received 13 November 2018 Accepted 16 January 2019 Communicated by Enzo Mitidieri MSC: 35B30 35B35 35G20 35G25 47A50

abstract This paper studies the decay properties of solutions around the traveling wave solutions for Cauchy problem of the second-order Camassa–Holm equation. Applying the extended pseudo-conformal transformation methods, appearing the relevant works on the generalized Korteweg–de Vries equation (KdV) and nonlinear Schrödinger equation (NLS) from Martel and Merle, the solution is controlled by the decaying function with exponential speed, corresponding to the initial data and its second derivative with exponential decay. © 2019 Elsevier Ltd. All rights reserved.

Keywords: Second-order Camassa–Holm equation Decay Traveling wave solution Pseudo-conformal transformation

1. Introduction In 2003, when Constantin and Kolev [11] investigated the infinite-dimensional Lie group of all smooth orientation-preserving diffeomorphisms of the circle with a Riemannian structure, they obtained a geodesic equation as follows: ut = Bk (u, u), k ∈ N, (1.1) where u = u(t, x), (t, x) ∈ R+ × R, and Bk (u, u) := A−1 k Ck (u) − u∂x u, Ak (u) :=

k ∑

(−1)j ∂x2j u,

j=0

Ck (u) := −uAk (∂x u) + Ak (u∂x u) − 2∂x uAk (u). ✩ This work is supported by Natural Science Foundation of China (Grant No. 11371175). ∗ Corresponding author. E-mail addresses: [email protected] (D. Ding), [email protected] (K. Wang).

https://doi.org/10.1016/j.na.2019.01.018 0362-546X/© 2019 Elsevier Ltd. All rights reserved.

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

The operator A−1 k is given by the equivalent convolution form ∫ A−1 (f )(x) = P ∗ f = Pk (x − y)f (y)dy, k k

231

x ∈ R.

R

Denote by ∗ the convolution, Pˆk the Fourier transform of Pk , more precisely Pˆk (ξ) =

1+

ξ2

1 , + · · · + ξ 2k

ξ ∈ R,

here Pˆk as shown in [4]. The operator Ck (u) is a total derivative, that is, there exists a differential polynomial in u denoted by Γk such that Ck (u) = −∂x Γk (u). Therefore, by direct calculations, we have, −Γk (u) =

∫ x k k ∑ 1∑ (−1)j ∂x2j (u2 ) − (−1)j (u∂ξ2j ∂ξ u + 2∂ξ u∂ξ2j u)dξ. 2 j=0 −∞ j=0

For k = 0, Eq. (1.1) yields the inviscid Burgers equation ut + uux = −(u2 )x

or

ut + 3uux = 0.

For k = 1, Eq. (1.1) becomes the Camassa–Holm equation [2] ut − utxx + 3uux − 2ux uxx − uuxxx = 0.

(1.2)

) ( 1 ut + uux + P1 ∗ u2 + u2x = 0, 2 x

(1.3)

Eq. (1.2) can be rewritten as

where P1 (x) = 21 e−|x| , x ∈ R and Γ1 (u) = u2 + 12 u2x . There are many works on the Camassa–Holm equation, we do not repeat those works here and readers can refer to [1,3,6–10,12–14,17,18,27,28] and references therein. For k = 2, the second-order Camassa–Holm equation is ut − utxx + utxxxx + 3uux − 2ux uxx − uuxxx + 2ux uxxxx + uuxxxxx = 0.

(1.4)

Direct calculations show that √ P2 (x) =

3 − e 3

√

3 2 |x|

( sin

|x| π + 2 6

) ,

x ∈ R.

It is easy to verify that P2 (x) is a distributional solution to u − uxx + uxxxx = δ (here δ is the Dirac measure) and hence the third derivative has a jump at the origin. Since Γ2 (u) = u2 + 12 u2x − 12 u2xx − 3(ux uxx )x , then Eq. (1.4) is rewritten as ( ) 1 2 1 2 2 ut + uux + P2 ∗ u + ux − uxx − 3(ux uxx )x = 0. (1.5) 2 2 x Eq. (1.1) is called higher-order Camassa–Holm equation when k ≥ 2. There are several works on the well-posedness of Eq. (1.1) when k ≥ 2. In 2009, Coclite, Holden and Karlsen [4] first studied the Cauchy problem and obtained the following important results: Assume u0 ∈ Hk,p := {f ∈ H k (R)|∂xk f ∈ Lp (R)}, where 2 < p < +∞, then there exists a global weak solution u(t, x) ∈

232

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

( ) ( )⋂ C [0, +∞); C k−1 (R) L∞ [0, +∞); H k (R) with initial value u0 (x) = u(0, x). Moreover, if u0 ∈ H k+1 (R), then u ∈ L∞ ([0, T ]; H k+1 (R)), T > 0 and if u0 ∈ Hk,r , where 2 ≤ r < +∞, then u ∈ L∞ ([0, T ]; Hk,r ), T > 0. Furthermore, for k = 2, if there exists a map b ∈ L1 ([0, T ]), T > 0, such that ∥∂x2 u(t, ·)∥L∞ (R) ≤ b(t), for all t ≥ 0, then the solution u is unique. In 2010, Ding and Lv [16] proved that if u0 ∈ H k (R) and constant M > 0 such that p>M u0 = 0 (p is the smooth Littlewood–Paley projection operator), then there exists ) ( )⋂ ∞( L [0, +∞); H k (R) with initial value a unique global conservative solution u ∈ C [0, +∞); H k−1 (R) u0 (x). In 2017, Coclite and di Ruvo [5] investigated the convergence of the solution for Cauchy problem of (1.1) to the entropy ones of a scalar conservation law. The traveling wave solutions of the second-order Camassa–Holm equation play a crucial role in related problems. From [15], its form is ( ) ⎧ √ x − ct √ x − ct − 23 (x−ct) ⎪ ⎪ cos + 3 sin , x ≥ ct, ⎨ Ae 2 2 ( ) (1.6) QA (x − ct) = √ 3 ⎪ x − ct √ x − ct ⎪ ⎩ Ae 2 (x−ct) cos − 3 sin , x < ct, 2 2 √

where A > 0 is amplitude, c > 0 is velocity and A is related to c. Then P2 (ξ) = 6A3 QA (ξ) and √ QA − (QA )ξξ + (QA )ξξξξ = 2 3Aδ. For convenience, take A = 1 and let Q = Q1 in the remainder of this paper. There are the following conservation laws of Eq. (1.1) ∫ ∫ u(t, x)dx = u0 (x)dx, (1.7) R

and

k ∫ ∑ i=0

(∂xi u)2 (t, x)dx

R

R

=

k ∫ ∑ i=0

(∂xi u0 )2 (x)dx.

(1.8)

R

We consider the following Cauchy problem for the second-order Camassa–Holm equation in this paper { ut − utxx + utxxxx + 3uux − 2ux uxx − uuxxx + 2ux uxxxx + uuxxxxx = 0, (t, x) ∈ R+ × R, (1.9) u(0, x) = u0 (x), x ∈ R, for u0 ∈ H 5 (R). The purpose of this paper is to study the decay properties of the solution of (1.9) around the traveling wave solutions with time variable t(> 0) and space variable x(≥ 0). The discussion on this decay property can be traced back to the research on the critical generalized KdV equation. Martel and Merle [19,21] obtained the decay forms of the solution of Cauchy problem for the critical generalized KdV equation (see Proposition 1 in Section 2 [19], see Proposition 3 in Section 4 [21]). For completeness, we recall the relevant contents from [21]. Consider the following Cauchy problem for the critical generalized KdV equation { ut + (uxx + u5 )x = 0, (t, x) ∈ R+ × R, (1.10) u(0, x) = u0 (x), x ∈ R, for u0 ∈ H 1 (R). 31/4 d1/4 In 2001, Martel and Merle proved the instability of solitons Sd (x)(Sd (x) = ch1/2 , where d > 0) for (2d1/2 x) the equation in [21]. More precisely, use the translation and dilatation invariance of the equation, let 1/2

ω(t, x) = λ0 (t)u(t, λ0 (t)x + x0 (t)),

(t, x) ∈ R+ × R,

here λ0 (t) > 0, x0 (t) ∈ R are two C 1 functions, and u(t, x) is a solution of (1.10). Then, ω(t, x) is decomposed into ω(t, x) = κ(t, x) + S(x), (1.11)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

233

where κ is small, and S(x) := S1 (x). By the contradiction argument, suppose that S(x) is H 1 orbitally stable. Then, a contradiction is obtained at a finite time t(u0 ) from two facts. Therefore, S(x) is H 1 instable. Similarly, S(x) is L2 instable. By scaling arguments, they also obtained instability of Sd (x − dt), d > 0. For details refer to [19–26] and references therein. The second-order Camassa–Holm equation possesses similar invariances: translation invariance (if u(t, x) is a solution of (1.9), then u(t, x + x0 ) is also a solution) and dilatation invariance (if u(t, x) is a solution of 1/2 1/2 (1.9), then λ0 u(λ0 t, x) is also a solution). The similar invariance leads us to conjecture that it is possible to discuss the decay property of the solution of (1.9) by applying the approximate decomposition method. Let λ1/2 (t)u(t, y + x(t)) = ε(t, y) + Q(y),

(1.12)

where u(t, x) is a solution of (1.9), ε is small, Q is the traveling wave solution of the second-order Camassa– Holm equation (see (1.6)) and λ(t) > 0, x(t) ∈ R are geometrical parameters related to these invariances, and two C 1 functions which would be chosen later. Change the time variable as follows: t

∫ s= 0

dt′ , λ1/2 (t′ )

ds 1 = 1/2 . dt λ (t)

or equivalently,

(1.13)

Let β > 0, u ∈ H 2 (R), define the neighborhood of radius β around Q { Uβ =

}

2

u ∈ H (R); inf ∥u(·) − Q(· + r)∥H 2 ≤ β . r∈R

(1.14)

∑k ∫ Denote by ∥ · ∥H k the norm in Sobolev space H k (R), ∥f ∥2H k = j=0 R (∂xj f )2 dx, k ∈ N ; Denote by | · | ∫ absolute value and (, ) the scalar product in L2 (R), (f, g) = R f gdx, f, g ∈ L2 (R). We claim the following main results. Theorem 1.1.

Suppose u0 ∈ Uα0 , u0 = Q + ε0 , where ε0 (x) ∈ H 5 (R), and satisfies √

|ε0 (x)| + |ε0xx (x)| ≤ a0 e−

3 2 x

,

∀x ≥ 0,

(1.15)

for some a0 > 0, 0 < α0 ≪ α∗ , α∗ > 0. Then there exists constant θ > 0, such that ( |ε(t, x)| + |εxx (t, x)| ≤ θ

√

3 e− 4 t

)

+ 1 e−

√ 3 2 x

,

(1.16)

for all t > 0, and all x ≥ 0. By scaling arguments, Theorem 1.1 also gives the decay properties of solutions of (1.9) around the traveling wave solution QA (x − ct) for all A > 0, c > 0. This paper is organized as follows. Section 2 linearizes second-order Camassa–Holm equation around Q and studies the structure of operator A2 . Section 3 discusses the boundedness of geometrical parameters λ(t), x(t) and their derivatives by applying modulation theory and energy estimates to control nonlinear items in time. Section 4 investigates the uniform bound and the stability of ε(s) in H 2 (R). Section 5 proves the decay properties of ε(s) by the fixed point theory and contradictory argument, Theorem 1.1 is proved in this section. Appendices A and B give the detailed proof of some lemmas.

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

234

2. The linearized equation and operator A2 In this section, we derive the equation for v(t, y) = λ1/2 (t)u(t, y + x(t)),

(2.1)

where u is a solution of (1.9), and λ(t) > 0, x(t) ∈ R are two C 1 functions. From (1.12), we have ε(t, y) = v(t, y) − Q(y) = λ1/2 (t)u(t, y + x(t)) − Q(y). Rewrite Eq. (1.5) as follows ) ( ∫ ∫ 3 1 2 1 2 2 P2zzz (x − z)u2z dz = 0. ut + uux + P2z (x − z) u + uz − uzz dz − 2 2 2 R R

(2.2)

(2.3)

There are the following lemmas. Lemma 2.1.

For all s ≥ 0, there is εs =

where

1 λs 1 λs Q + xs Qy + ε + xs εy − QQy − εεy − (Qε)y − R1 (ε), 2 λ 2 λ (

) ∫ 1 2 1 2 3 R1 (ε) = P2z (y − z) Q + Qz − Qzz dz − P2zzz (y − z)Q2z dz 2 2 2 R R ( ) ∫ ∫ 1 1 3 P2zzz (y − z)ε2z dz + P2z (y − z) ε2 + ε2z − ε2zz dz − 2 2 2 R R ∫ ∫ + P2z (y − z)(2Qε + Qz εz − Qzz εzz )dz − 3 P2zzz (y − z)Qz εz dz, ∫

(2.4)

2

R

(2.5)

R

and the third derivative of P2 is not continuous at the origin. Proof . Direct calculations show that 1 −1/2 λ λt u + λ1/2 ut + λ1/2 xt uy , 2 vy = λ1/2 uy , vyy = λ1/2 uyy . vt =

Therefore, ( ) 1 1 2 P2z (y − z) v 2 + vz2 − vzz dz 2 2 R ∫ 1 3 = λ−1/2 λt v + λ1/2 xt vy + P2zzz (y − z)vz2 dz. 2 2 R

λ1/2 vt + vvy +

∫

(2.6)

From (1.13) follows the equation (

) 1 2 1 2 vs + vvy + P2z (y − z) v + vz − vzz dz 2 2 R ∫ 1 λs 3 = v + x s vy + P2zzz (y − z)vz2 dz. 2 λ 2 R ∫

2

(2.7)

Due to v(s, y) = Q(y) + ε(s, y), we get εs =

1 λs 1 λs Q + xs Qy + ε + xs εy − QQy − εεy − (Qε)y − R1 (ε), 2 λ 2 λ

(2.8)

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where ) ( ∫ 1 3 1 P2zzz (y − z)Q2z dz P2z (y − z) Q2 + Q2z − Q2zz dz − 2 2 2 R ( ) ∫R ∫ 3 1 2 1 2 2 + P2zzz (y − z)ε2z dz P2z (y − z) ε + εz − εzz dz − 2 2 2 R R ∫ ∫ + P2z (y − z)(2Qε + Qz εz − Qzz εzz )dz − 3 P2zzz (y − z)Qz εz dz,

∫ R1 (ε) =

R

R

and P2xxx The lemma is proved.

⎧ √ (x π) 3 ⎪ ⎨ 2e− 2 x cos , + 2( 6 ) √ = ⎪ ⎩ − 2e 23 x cos x − π , 2 6

x ≥ 0, x < 0.

□

∑2 j Lemma 2.2. (Structure of A2 = j=0 (−1) ∂x2j ). The functions Qx and Qxx are two linearly independent eigenvectors of operator A2 in the sense of distributions. √ Proof . Let ϖ be an arbitrary Cc∞ function. Since Q − Qxx + Qxxxx = 2 3δ, then ∫ (A2 Qx , ϖ) = (Q − Qxx + Qxxxx )x ϖdx = 0 · Qx ,

(2.9)

R

and

∫ (Q − Qxx + Qxxxx )xx ϖdx = 0 · Qxx .

(A2 Qxx , ϖ) =

(2.10)

R

Therefore, Qx and Qxx are two eigenvectors of operator A2 corresponding to zero eigenvalue in the sense of distributions. Assuming there exists non-zero constant µ, such that Qx = µQxx , multiplying the equality by ∫ ∫ Qx and performing integral over R, one gets 0 ̸= R Q2x dx = µ R Qxx Qx dx = 0. This contradiction implies Qx and Qxx are linearly independent. The lemma is proved. □ On view of the structure, the difference between the critical generalized KdV equation and the secondorder Camassa–Holm equation mainly relies on the operator A2 . It is easy to verify that A2 is a self-adjoint linear differential operator, however, it acts on nonlinear terms Γ2 (u). This leads to the differences, such as the smoothness of the traveling wave solutions and other properties, between the second-order Camassa–Holm equation and the critical generalized KdV equation. 3. Selection of dilatation and translation parameters

∫

This section focuses on choosing λ(s) and x(s) such that ε(s)⊥Qx and ε(s)⊥Qxx , for all s ≥ 0, that is, ∫ ε(s)Qx dx = 0 and R ε(s)Qxx dx = 0, by using the implicit function theorem. R Suppose u(t) ∈ H 2 (R), λ1 > 0 and x 1 ∈ R, define ελ1 ,x1 (x) = λ 1 1/2 u(x + x1 ) − Q(x).

(3.1)

ˆ > 0, and a unique C 1 Proposition 3.1 (Choice of the Modulation Parameters). There exists α∗ > 0, λ ˆ 1 + λ) ˆ × R, such that if u(t) ∈ Uα∗ , and ελ ,x is defined as in (3.1), then map (λ1 , x1 ) : Uα∗ → (1 − λ, 1 1 ελ1 ,x1 ⊥Qx

and

ελ1 ,x1 ⊥Qxx .

(3.2)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

236

Moreover, if u(t) ∈ Uα , with 0 < α < α∗ , then there exists constant C1 > 0, such that ∥ελ1 ,x1 ∥H 2 ≤ C1 α Proof . Define the following functionals: ∫ fλ11 ,x1 (u) = ελ1 ,x1 Qx dx ,

fλ21 ,x1 (u) =

R

Since

|λ1 − 1| ≤ C1 α.

and

⏐ ∂ελ1 ,x1 ⏐⏐ 1 = u and ∂λ1 ⏐λ1 =1,x1 =0 2

(3.3)

∫ ελ1 ,x1 Qxx dx.

(3.4)

R

⏐ ∂ελ1 ,x1 ⏐⏐ = ux , ∂x1 ⏐λ1 =1,x1 =0

we obtain ⏐ ∂fλ11 ,x1 ⏐⏐ ⏐ ∂λ1 ⏐ and

1 = 2

∫

1 =− 2

∫

λ1 =1,x1 =0,u=Q

⏐ ∂fλ21 ,x1 ⏐⏐ ⏐ ∂λ1 ⏐

λ1 =1,x1 =0,u=Q

QQx dx = 0, R

Q2x dx

R

⏐ ∂fλ11 ,x1 ⏐⏐ ⏐ ∂x1 ⏐

⏐ ∂fλ21 ,x1 ⏐⏐ ̸ 0, = ⏐ ∂x1 ⏐

∫ =

Q2x dx ̸= 0,

R

λ1 =1,x1 =0,u=Q

∫ Qx Qxx dx = 0.

= λ1 =1,x1 =0,u=Q

R

The Jacobian matrix ⏐ ∂(fλ11 ,x1 , fλ21 ,x1 ) ⏐⏐ ⏐ ⏐ ∂(λ1 , x1 )

λ1 =1,x1 =0,u=Q

⏐ 1 ⏐⏐ ⏐ ∂fλ1 ,x1 ∂fλ11 ,x1 ⏐⏐ ⏐ ⏐⏐ (∫ )2 ⏐ ∂λ1 1 ∂x1 ⏐⏐⏐⏐ 2 ⏐ = =⏐ 2 Q dx ̸= 0. ⏐⏐ 2 2 R x ⏐ ∂fλ1 ,x1 ∂fλ1 ,x1 ⏐⏐ ⏐ ⏐⏐ ∂λ1 ∂x1 λ1 =1,x1 =0,u=Q

According to the implicit function theorem, there exists α > 0, a neighborhood Ω1,0 of (1,0) in R+ × R, and } { a unique C 1 map (λ1 , x1 ) : u ∈ H 2 (R); ∥u − Q∥H 2 < α → Ω1,0 , such that ∫ ∫ ελ1 ,x1 Qx dx = ελ1 ,x1 Qxx dx = 0. R

R

then (3.2) holds. Furthermore, if ∥u − Q∥H 2 < α ≤ α, then there is C1 > 0 such that |λ 1 − 1| ≤ C1 α. Since 1/2

1/2

∥ελ1 ,x1 ∥H 2 = ∥λ1 u(x + x1 ) − Q(x)∥H 2 ≤ |λ1

− 1|∥u∥H 2 + ∥u(x + x1 ) − Q(x)∥H 2 ,

then ∥ελ1 ,x1 ∥H 2 ≤ C1 α, for some C1 > 0. Extend the map (λ1 , x1 ) to the tube Uα . Using the implicit function theorem again, there exist α∗ < α, and a unique C 1 map r: Uα∗ → R, such that ∥u(·) − Q(· − r)∥H 2 = inf ∥u(·) − Q(· − r)∥H 2 < α1 < α, r∈R

(3.5)

for all u ∈ Uα∗ . Denote λ1 (u) = λ1 (u(· + r(u))), and x1 (u) = x1 (u(· + r(u))) + r(u) which satisfy (3.2), (3.3). The proposition is proved. □ Lemma 3.2. Suppose u(t) ∈ Uα∗ , α∗ > 0, for all t ≥ 0, with the initial value u0 (x), then there exists α0 > 0, such that u0 ∈ Uα0 .

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

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Proof . By the contradiction argument, assume u0 ̸∈ Uα0 , for any α0 > 0, but u(t) ∈ Uα∗ , u(t) is a solution of (1.9) corresponding to initial value u0 (x). Suppose u0 (x) ∈ H 2 (R), from the conservation law (1.8), u(t) ∈ L∞ ([0, +∞); H 2 (R)). By integrating by parts, we have √ ∥u(·) − Q(· + r)∥2H 2 =∥u0 − Q(· + r)∥2H 2 + 4 3[u(0, −r) − u(t, −r)]. Thus, √ (α∗ )2 ≥ inf ∥u(·) − Q(· + r)∥2H 2 = inf ∥u0 − Q(· + r)∥2H 2 + 4 3 inf [u(0, −r) − u(t, −r)] r∈R r∈R r∈R √ 2 > α0 − 4 3[∥u0 ∥H 2 + ∥u(t)∥H 2 ] √ = α02 − 8 3∥u0 ∥H 2 . √ Taking α02 = (α∗ )2 + 8 3∥u0 ∥H 2 + 1, we have √ (α∗ )2 > α02 − 8 3∥u0 ∥H 2 > (α∗ )2 + 1. This yields a contradiction. Then there is α0 > 0, such that u0 ∈ Uα0 . The lemma is proved.

□

Remark 3.3. Lemma 3.2 shows that there exists 0 < α0 ≪ α∗ , such that if u0 ∈ Uα0 , then u(t) ∈ Uα∗ , for all t ≥ 0, where u(t) is the solution of (1.9). Definition 3.4. Define ε(s) = ελ(s),x(s) which satisfies (ε(s), Qx ) = (ε(s), Qxx ) = 0, for all s ≥ 0 and some λ(s) > 0, x(s) ∈ R. To estimate the geometrical parameters

λs λ

and xs : we have the following lemmas

Lemma 3.5. Suppose u0 ∈ H 3 (R), and u(t) ∈ Uα1 , for all t ≥ 0, with 0 < α1 < α∗ , then λ(s), x(s) ∈ C 1 (R+ ). Moreover, ∫ ∫ ∫ ∫ ∫ 1 2 2 2 xs Qy dy = QQy dy − QQyy εdy − Qyy ε dy + R1 (ε)Qy dy, (3.6) 2 R R R R R and

1 λs 2 λ

∫

Q2y dy + xs

R

∫ Qyyy εdy = R

1 2

∫

Qyyy ε2 dy +

R

∫

∫ QQyyy εdy −

R

R1 (ε)Qyy dy,

(3.7)

R

where R1 (ε) is given by (2.5). Proof . Let χ(x) be a C 1 (R\{0}) function such that |χ(x)| + |χ′ (x)| ≤ Ce−

|x| 2

, ∀x ∈ R\{0},

(3.8)

and χ(0 − 0), χ(0 + 0) are finite. Suppose u0 ∈ H 2 (R), from (1.8), u ∈ L∞ ([0, +∞); H 2 (R)). Since u ∈ Uα1 , 0 < α1 < α∗ , by Proposition 3.1, λ(t), x(t) ∈ C 1 (R+ ). By using Eq. (1.5), we have ∫ ∫ d 1/2 d χu(s)dx = λ χu(s)dx ds R dt R ( ( ) ) ∫ 1 1 = λ1/2 χ −uux − P2 ∗ u2 + u2x − u2xx − 3(ux uxx )x dx 2 2 R x ( ( ) ) ∫ ∫ 1 1 1 = λ1/2 u2 χx dx − λ1/2 χ P2 ∗ u2 + u2x − u2xx − 3(ux uxx )x dx. 2 2 2 R R x

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238

Therefore, d ds

∫

d χv(s)dy = ds R

( ) ∫ 1/2 λ χ(x − x(s))u(s, x)dx R

∫ ∫ ∫ 1 λs 1 = χvdy − xs v 2 χy dy χy vdy + 2 λ R 2 R R ( ( ) ) ∫ 1 2 1 2 2 − χ P2 ∗ v + vy − vyy − 3(vy vyy )y dy. 2 2 R y

(3.9)

Noticing v(s, y) = Q(y) + ε(s, y), we obtain ∫ ∫ ∫ d 1 λs χε(s)dy = χ(Q + ε)dy − xs χy (Q + ε)dy ds R 2 λ R R (3.10) ∫ ∫ 1 χy (Q + ε)2 dy − χR1 (ε)dy. + 2 R R ∫ Let χ = Qy . Since (ε(s), Qy ) = R Qy εdy = 0, then ∫ ∫ ∫ ∫ ∫ ∫ d 1 0= Qy εdy = xs Q2y dy − QQ2y dy + QQyy εdy + Qyy ε2 dy − R1 (ε)Qy dy. ds R 2 R R R R R ∫ Similarly, let χ = Qyy . From (ε(s), Qyy ) = R Qyy εdy = 0, we have ∫ ∫ ∫ ∫ 1 λs 1 d Qyy εdy = − Q2y dy − xs Qyyy εdy + Qyyy ε2 dy 0= ds R 2 λ R 2 R R ∫ ∫ + QQyyy εdy − R1 (ε)Qyy dy, R

R

where R1 (ε) is defined in (2.5). Because of R Q2y dy > 0, it follows that both λλs terms of scalar products of (∂yi ε)j with (∂yi0 Q)j0 (i, i0 = 0, 1, 2; j, j0 = 0, 1, 2). Suppose u0 ∈ H 3 (R), take a sequence (u0n ) ∈ H 4 (R) such that u0n → u0 ∫

and xs are expressed in

in H 3 (R) as n → +∞, and denote by (un ) the corresponding solutions of (1.9). Let T > 0, from Theorem 2.4 in [4], we get that un ∈ L∞ ([0, T ]; H 3 (R)), for all 0 ≤ t ≤ T . It follows from the continuity of the solution with respect to the initial data that un → u in L∞ ([0, T ]; H 3 (R)) as n → +∞. Define εn , λn and xn in the same way. It follows from Proposition 3.1 that λn → λ and xn → x in ∞ L ([0, T ]) as n → +∞. ∫ ∫ Assume that χ satisfies (3.8), then R χ(∂yi εn )j dy → R χ(∂yi ε)j dy in L∞ ([0, T ]) as n → +∞, for all i, j = 0, 1, 2. Here, we only give the argument in detail for i = 2, j = 2, the discussion on other situations is 1/2 analogous. From the equality εn (t, y) = λn un (t, y + xn ) − Q(y) follows that ⏐∫ ⏐ ∫ ⏐ ⏐ 2 ⏐ χ(y)ε2nyy (s, y)dy − χ(y)εyy (s, y)dy ⏐⏐ ⏐ R ⏐∫R ⏐ ⏐∫ ⏐ ⏐ ⏐ ⏐ ⏐ 2 2 2 ⏐ ⏐ ⏐ ≤ ⏐ (λn χ(x − xn (s)) − λχ(x − x(s)))unxx (s, x)dx⏐ + λ ⏐ χ(x − x(s))(unxx − uxx )(s, x)dx⏐⏐ R R ⏐∫ ⏐ ⏐ ⏐ 1/2 ⏐ + 2λ ⏐ (χQxx )(x − x(s))(uxx − unxx )(s, x)dx⏐⏐ ⏐∫ ( R ⏐ ) ⏐ ⏐ 1/2 1/2 ⏐ + ⏐ 2 λ (χQxx )(x − x(s)) − λn (χQxx )(x − xn (s)) unxx (s, x)dx⏐⏐ R

=I1 + I2 + I3 + I4 . (i) Using the H¨ older inequality, we have ∫ I1 ≤ ∥λn χ(x − xn (s)) − λχ(x − x(s))∥L∞

R

u2nxx (s, x)dx.

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

239

By the Cauchy–Schwarz inequality, we get (∫ ( )1/2 )2 )1/2 (∫ 1/2 1/2 2 I4 ≤ 2 λ (χQxx )(x − x(s)) − λn (χQxx )(x − xn (s)) dx unxx dx . R

R 3

Since λn (s) → λ(s), xn (s) → x(s) as n → +∞, and un ∈ H (R), it follows from the decay properties of χ and Q that I1 → 0, I4 → 0 as n → +∞. (ii) Similarly, ∫ I2 ≤ λ∥un − u∥H 3 ∥un + u∥H 3 |χ|dx, R

and I3 ≤ 2λ1/2

)1/2 (χQxx )2 (x − x(s))dx ∥un − u∥H 3 .

(∫ R

∞

Due to the fact that un → u in L ([0, T ]; H 3 (R)) as n → +∞, and the decay properties of χ and Q, we have I2 → 0, I3 → 0 as n → +∞. ⏐ ⏐∫ ∫ By (i) and (ii), we obtain ⏐ R χ(y)ε2nyy (s, y)dy − R χ(y)ε2yy (s, y)dy ⏐ → 0 as n → +∞. Similar discussions show that ∫ ∫ χ(∂yi εn )j dy → χ(∂yi ε)j dy in L∞ ([0, T ]) as n → +∞, R

R

for all i, j = 0, 1, 2. Therefore, λn and xn are expressed as ∫s h1n (s′ )ds′ xn (s) = 0 ∫ , Q2 dy R y where

∫

QQ2y dy −

h1n (s) =

∫ QQyy εn dy −

R

and

R

∫ h2n (s) = xns

) ( ∫s 2 0 h2n (s′ )ds′ ∫ , λn (s) = exp Q2 dy R y

Qyy εny dy + R

1 2

∫

1 2

∫

Qyy ε2n dy +

∫ R1 (εn )Qy dy,

R

R

∫

Qyyy ε2n dy +

∫ QQyyy εn dy −

R

(3.11)

R

R1 (εn )Qyy dy. R

Taking limit as n → +∞ in the expressions of λn (s) and xn (s), we obtain λ(s) , x(s) and ε(s). In particular, λ(s) and x(s) are C 1 functions, and the formulas (3.6) and (3.7) are also true if u0 ∈ H 3 (R). The lemma is proved. □ Lemma 3.6.

Under the assumptions of Lemma 3.5, then there is constant C2 > 0, such that ⏐ ⏐ ⏐ λs ⏐ ⏐ ⏐ + |xs − 1| ≤ C2 ∥ε(s)∥ 2 , H ⏐λ⏐

for all s ≥ 0. Proof . From Lemma 3.5, we have xs = ∫ where

∫ h1 (s) = R

QQ2y dy −

h1 (s) , Q2 dy R y

λs 2h2 (s) =∫ , λ Q2 dy R y

∫ QQyy εdy − R

1 2

∫ R

Qyy ε2 dy +

∫ R1 (ε)Qy dy, R

(3.12)

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240

and

∫ h2 (s) = xs

Qyy εy dy + R

1 2

∫

Qyyy ε2 dy +

∫

∫ QQyyy εdy −

R

R

R1 (ε)Qyy dy. R

Let R1 (ε) = T1 + T2 , where ) ( ∫ ∫ 1 2 1 2 3 2 P2zzz (y − z)Q2z dz, T1 (y) = P2z (y − z) Q + Qz − Qzz dz − 2 2 2 R R ∫ ∫ T2 (s, y) = P2z (y − z)(2Qε + Qz εz − Qzz εzz )dz − 3 P2zzz (y − z)Qz εz dz R ) ( ∫R ∫ 1 3 1 P2zzz (y − z)ε2z dz. + P2z (y − z) ε2 + ε2z − ε2zz dz − 2 2 2 R R Since

√ P2 (x) =

3 Q(x), 6

√

Q(x) = 2e−

3 2 |x|

( sin

|x| π + 2 6

) , x ∈ R,

then by direct calculations, one has ⎧ √ √ √ √ 3 x ⎪ ⎨ − 2e− 2 x sin + e− 3x ( 3 + sin x − 3 cos x), 2 √ T1 (x) = √ √ √ ⎪ ⎩ − 2e 23 x sin x − e 3x ( 3 − sin x − 3 cos x), 2

x ≥ 0, (3.13) x < 0.

One can easily verify that T1 and Qy are odd, Qyy is even. Direct calculations, √ √ ∫ ∫ ∫ T (y)Qyy dy = 0, R QQ2y dy = 4 3/21, R Q2y dy = 3/3. Obviously, we get R 1 √ ∫ ∫ ∫ 3 Q2y dy = QQ2y dy + T1 (y)Qy dy = . 3 R R R

∫ R

T1 (y)Qy dy =

√

3/7,

To estimate h1 (s) and h2 (s): Notice that ( ) ∫ ∫ √ √ 2 xs − 1 = 3 h1 (s) − QQy dy − T1 (y)Qy dy := 3h′1 (s). R

R

By the H¨ older inequality, we have ⏐ ⏐ ∫ ∫ ∫ ⏐ ⏐ 1 |h′1 (s)| = ⏐⏐− QQyy εdy − ε2 Qyy dy + T2 (y)Qy dy ⏐⏐ 2 R R √R ⏐∫ ⏐ ⏐ ⏐ 16 3 2 ≤ ∥ε∥H 2 + ∥ε∥H 2 + ⏐⏐ T2 (y)Qy dy ⏐⏐ . 3 R On the other hand, √ √ √ 3 3 4 3 2 2 ∥ε∥H 2 + ∥ε∥H 2 + ∥Q∥H 2 ∥ε∥H 2 + 3 ∥Q∥H 2 ∥ε∥H 2 |T2 (y)| ≤ 3√ 2√ 3 7 3 5 3 2 ∥ε∥H 2 + ∥Q∥H 2 ∥ε∥H 2 . = 6 3 √

Hence, ⏐∫ ⏐ ∫ ⏐ ⏐ 20 56 2 ⏐ T2 (y)Qy dy ⏐ ≤ ∥T2 ∥ ∞ |Qy |dy ≤ ∥ε∥H 2 + ∥Q∥H 2 ∥ε∥H 2 . L ⏐ ⏐ 3 3 R R √ ∫ ∫ From R QQ2y dy + R T1 (y)Qy dy = 3/3 follows that √ ⏐ √ ⏐ ⏐ ⏐ (16 3 + 56∥Q∥H 2 ) 3 23 ⏐≤ ⏐h1 (s) − ∥ε∥H 2 + ∥ε∥2H 2 . ⏐ 3 ⏐ 3 3

(3.14)

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241

Similarly,

√ ) ( √ (8 3 + 56) 23 8 3 |xs | + ∥Q∥H 2 ∥ε∥H 2 + ∥ε∥2H 2 . |h2 (s)| ≤ 3 3 3 Therefore, there exists a constant C2 > 0, such that ⏐ ⏐ ⏐ λs ⏐ ⏐ ⏐ + |xs − 1| ≤ C2 ∥ε(s)∥ 2 , s ≥ 0. H ⏐λ⏐ The lemma is proved.

(3.15)

□

4. Stability of ε(s) Consider the following Cauchy problems ⎧ ⎨ ε = 1 λs (Q + ε) + x (Q + ε) − QQ − εε − (Qε) − R (ε), (s, y) ∈ R+ × R, s s y y y y 1 2 λ ⎩ ε(0, y) = ε0 (y), y ∈ R, and

⎧ ⎨ εˆ = 1 λs (Q + εˆ) + x (Q + εˆ) − QQ − εˆεˆ − (Qˆ ε)y − R1 (ˆ ε), (s, y) ∈ R+ × R, s s y y y 2 λ ⎩ εˆ(0, y) = εˆ0 (y) = ε0 (y) + δ0 (y), y ∈ R,

(4.1)

(4.2)

where ε0 (y), δ0 (y) ∈ H 3 (R). Definition 4.1. Let ε, εˆ ∈ L∞ ([0, T ]; H 3 (R)) be solutions of (4.1), (4.2) respectively. Suppose for all ϕ ≥ 0, there is a β0 = β0 (ϕ) > 0, such that if ∥ˆ ε0 − ε0 ∥H 2 = ∥δ0 ∥H 2 ≤ β0 , then ∥ˆ ε(s) − ε(s)∥H 2 ≤ ϕ, for all 2 s ∈ [0, T ], T > 0. We claim that ε(s) is stable in H (R). Lemma 4.2 (Energy Relations). Suppose u0 = Q + ε0 , and ε0 (x) ∈ H 2 (R), then √ √ 2 2 ∥ε(s)∥H 2 = λ(s) ∥ε0 ∥H 2 + 2 3(λ(s) − 1) + 4 3(λ(s)ε0 (0) − ε(s, 0)),

(4.3)

for all s ≥ 0. Proof . From the conservation law (1.8) follows ∫ E(u(t)) = u2 + u2x + u2xx dx = E(u0 ). R

Thus,

∫ E(Q + ε(s)) = E(v(s)) = λ(s)

u2 + u2x + u2xx dx = λ(s)E(u) = λ(s)E(u0 ).

(4.4)

R

√ Using the fact ∥Q∥2H 2 = 2 3, and ∂x3 Q is not continuous at the origin, we get √ √ E(Q + ε(s)) = ∥ε(s)∥2H 2 + 2 3 + 4 3ε(s, 0), and

√ √ E(Q + ε0 ) = ∥ε0 ∥2H 2 + 2 3 + 4 3ε0 (0).

By (4.4), we have E(Q + ε(s)) = λ(s)E(Q + ε0 ). Therefore, √ √ 2 2 ∥ε(s)∥H 2 = λ(s) ∥ε0 ∥H 2 + 2 3(λ(s) − 1) + 4 3(λ(s)ε0 (0) − ε(s, 0)), The lemma is proved.

(4.5)

□

(4.6)

s ≥ 0.

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242

Lemma 4.3 (Uniform Bound of ε(s) in H 2 (R)). Assume u(t) ∈ Uα2 , with 0 < α2 ≤ α < α∗ and u0 = Q+ε0 , with ε0 ∈ H 2 (R), then there exists constant C3 > 0, such that 1/2

∥ε(s)∥H 2 ≤ C3 ∥ε0 ∥H 2 ,

s≥0

(4.7)

Proof . According to Lemma 4.2, we have √ √ 2 2 ∥ε(s)∥H 2 = λ(s) ∥ε0 ∥H 2 + 2 3(λ(s) − 1) + 4 3(λ(s)ε0 (0) − ε(s, 0)). Since u(t) ∈ Uα2 , it follows from Proposition 3.1 that ∥ε(s)∥H 2 + |λ(s) − 1| ≤ C1 α. Thus, there exists a constant C > 0, such that |λ(s) − 1| ≤ C ∥ε(s)∥H 2 , for all s ≥ 0. So, we obtain 2

∥ε(s)∥H 2 ≤ a1 ∥ε0 ∥H 2 + a2 ∥ε(s)∥H 2 ,

s ≥ 0,

(4.8)

√ ) ( where a1 = ∥ε0 ∥H 2 + 4 3 (1 + C1 α), 0 < a2 < +∞. Arguing by contradiction. Suppose there is a s0 > 0, such that ∥ε(s0 )∥H 2 ̸= 0 and for any 0 < C0 < +∞, we have 1/2

∥ε(s0 )∥H 2 > C0 ∥ε0 ∥H 2 , By (4.8), we get √ a22 + 4a1 ∥ε0 ∥H 2 a2 + , s ≥ 0. ∥ε(s)∥H 2 ≤ 2 2 According to the arbitrariness of the constant C0 , let √ a22 + 4a1 ∥ε0 ∥H 2 1 a2 + + . C0 = 1/2 1/2 2 2∥ε0 ∥H 2 2∥ε0 ∥H 2 Thus, √ √ 1/2 ∥ε0 ∥H 2 a22 + 4a1 ∥ε0 ∥H 2 a22 + 4a1 ∥ε0 ∥H 2 a2 a2 + + < ∥ε(s0 )∥H 2 ≤ + . 2 2 2 2 2 This leads to a contradiction. Therefore, there exists a constant C3 > 0, such that 1/2

∥ε(s)∥H 2 ≤ C3 ∥ε0 ∥H 2 ,

s ≥ 0.

The lemma is proved. □ Lemma 4.4 (Stability of ε(s) in H 2 (R)). Let β0 > 0, and ε0 (y), δ0 (y) ∈ H 3 (R). If ∥ˆ ε0 − ε0 ∥H 2 = ∥δ0 ∥H 2 ≤ β0 . Then there exists constant C4 > 0, independent of β0 , such that ∥ˆ ε(s) − ε(s)∥H 2 ≤ C4 β0 ,

s ∈ [0, T ],

(4.9)

for any 0 < T < +∞. Proof . From (4.2) and (4.1), we have ε¯s =

1 λs ε¯ + xs ε¯y − ε¯εˆy − ε¯y ε − (Q¯ ε)y − R2 (¯ ε), 2 λ

(4.10)

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243

where ε¯ = εˆ − ε, and ] [ 1 1 ε + ε)z − ε¯zz (ˆ ε + ε)zz dz P2z (y − z) ε¯(ˆ ε + ε) + ε¯z (ˆ 2 2 R ∫ ∫ 3 P2zzz (y − z) [¯ εz (ˆ ε + ε)z ] dz − 3 − P2zzz (y − z)Qz ε¯z dz 2 R R ∫ + P2z (y − z)(2Q¯ ε + Qz ε¯z − Qzz ε¯zz )dz.

∫ R2 (¯ ε) =

R

Multiply the equality (4.10) by ∂y2j ε (j = 0, 1, 2), and perform integral over R. Here, we only give the calculation detail in j = 2, the discussion on other situations is analogous. ∫ ∫ ∫ ∫ ∫ 1 λs 1 d ε¯yy ε¯εˆyyy dy ε¯yy ε¯y εˆyy dy − ε¯2yy εˆy dy − 2 ε¯2yy dy = ε¯2yy dy − 2 ds R 2 λ R R R R∫ ∫ ∫ 3 5 − ε¯2yy εy dy − ε¯yy ε¯y εyy dy − ε¯2 Qy dy 2 R 2 R yy R ∫ ∫ ∫ 3 + ε¯2y Qyyy dy − ε¯yy ε¯Qyyy dy − R2 (¯ ε)¯ εyyyy dy. 2 R R R Using the H¨ older inequality, Minkowski inequality and Cauchy–Schwarz inequality, it follows from Lemma 3.6 that there exists constant C ′ > 0, such that d 2 2 ∥¯ ε(s)∥H 2 ≤ C ′ (∥ε(s)∥H 3 + ∥ˆ ε(s)∥H 3 + ∥Q∥H 2 ) ∥¯ ε(s)∥H 2 . ds Since ε0 (y), δ0 (y) ∈ H 3 (R), from the Theorem 2.4 in [4], ε(s), εˆ(s) ∈ L∞ ([0, T ]; H 3 (R)), T > 0. Hence, by Gronwall inequality, we get ′ 2 2 ∥¯ ε(s)∥H 2 ≤ eC (∥ε0 ∥H 3 +∥ˆε0 ∥H 3 +∥Q∥H 2 )s ∥¯ ε0 ∥H 2 ,

s ≥ 0.

Due to ∥¯ ε0 ∥H 2 = ∥ˆ ε0 − ε0 ∥H 2 = ∥δ0 ∥H 2 ≤ β0 , there is C4 > 0, independent of β0 , such that ∥ˆ ε(s) − ε(s)∥H 2 ≤ C4 β0 ,

s ∈ [0, T ],

for any 0 < T < +∞. So, ε(s) is stable in H 2 (R). The lemma is proved.

□

5. Decay properties of the solution ε Lemma 5.1. Suppose ε is a solution of (2.4) with the initial value ε0 ∈ H 5 (R), then ε(t) ∈ L∞ ([0, T ]; H 5 (R)), for each T > 0. Proof . From (2.3), we get d u = −uux − dt

) ∫ 1 2 1 2 3 P2z (x − z) u + uz − uzz dz + P2zzz (x − z)u2z dz, 2 2 2 R R

∫

(

2

Multiply the equality by ∂x10 u, and perform integral over R, then ∫ ∫ ∫ ∫ 55 3 d 2 2 10 u dx = u uxxx dx − ∂ u(x) P2zzz (x − z)u2z dzdx dt R xxxxx 2 R xxxx 2 R x R ( ) ∫ ∫ ∫ 1 1 11 + ∂x10 u(x) P2z (x − z) u2 + u2z − u2zz dzdx − u2 ux dx. 2 2 2 R xxxxx R R

(5.1)

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244

Since ∂z3 P2 has a jump at the origin, integrating by parts and applying the H¨older inequality, Minkowski inequality and decay properties of P2 , there exists constant C > 0 such that ⏐ ⏐∫ ∫ ⏐ ⏐ 3 5 ⏐ ∂x5 u(x) ∂z P2 (z)(∂z u∂z ∂x u)(x − z)dzdx⏐⏐ ⏐ R⏐ ⏐ ∫ R ∫ ∫ ⏐ √ ⏐ 5 2 5 4 5 ⏐ ≤ ⏐2 3 (∂x u) ∂x udx + ∂x u ∂z P2 (z)∂z u∂x u(x − z)dzdx⏐⏐ R R ⏐∫ R ∫ ⏐ ⏐ ⏐ 5 3 2 5 + ⏐⏐ ∂x u ∂z P2 (z)∂z u∂x u(x − z)dzdx⏐⏐ R

R

≤ C∥u(t)∥2H 5 . Hence, d dt

∫ R

u2xxxxx dx ≤ C (∥u(t)∥H 2 + ∥u(t)∥H 4 ) ∥u(t)∥2H 5 .

To show that u(t) ∈ L∞ ([0, T ]; H 4 (R)), for all T > 0: Multiplying the equality (5.1) by ∂x2k u (k = 0, 1, 2, 3, 4), here, we only give the calculation detail in k = 4. ∫ ∫ ∫ ∫ ∫ 9 3 d u2xxxx dx = − u2xxxx ux dx + 5 u2xxx dx + ∂x8 u(x) P2zzz (x − z)u2z dzdx dt R 2 R 2 R ( R ) R ∫ ∫ 1 2 1 2 8 2 − ∂x u(x) P2z (x − z) u + uz − uzz dzdx. 2 2 R R Similar to above inequality, we have ∫ d u2 dx ≤ C (∥u(t)∥H 2 + ∥u(t)∥H 3 ) ∥u(t)∥2H 4 . dt R xxxx Similarly, 4

∑ d d ∥u(t)∥2H 4 = dt dt i=0

∫ R

(∂xi u)2 dx ≤ C (∥u(t)∥H 2 + ∥u(t)∥H 3 ) ∥u(t)∥2H 4 .

(5.2)

Since u0 ∈ H 3 (R), by Theorem 2.4 in [4], u(t) ∈ L∞ ([0, T ]; H 3 (R)), for each T > 0, i.e. there exists a constant C ′′ > 0, such that ∥u(t)∥H 3 ≤ C ′′ , for any t ∈ [0, T ]. By the Gronwall inequality, (5.2) yields ∫t ′′ C (∥u(t)∥H 2 +∥u(t)∥H 3 )dt 2 ∥u0 ∥2H 4 ≤ eC(∥u0 ∥H 2 +C )t ∥u0 ∥2H 4 . ∥u(t)∥H 4 ≤ e 0 Therefore, u(t) ∈ L∞ ([0, T ]; H 4 (R)), for any T > 0. Similar arguments show that ∥u(t)∥2H 5 ≤ eC(∥u0 ∥H 2 +∥u0 ∥H 4 )t ∥u0 ∥2H 5 ,

t ≥ 0.

So, u(t) ∈ L∞ ([0, T ]; H 5 (R)), for each T > 0. From ε(t, y) = λ1/2 (t)u(t, y + x(t)) − Q(y), one gets ε(t) ∈ L∞ ([0, T ]; H 5 (R)), T > 0. The lemma is proved. □ Introduce the following function η(s, x) = λ−1/2 (s)ε(s, x). Since

1 λs 1 λs Q + xs Qy + ε + xs εy − QQy − εεy − (Qε)y − R1 (ε), 2 λ 2 λ then by direct calculations, we obtain the following equation εs =

λ1/2 ηs − λ1/2 xs ηx = f1 (s, x) − f2 (s, x) − f3 (s, x),

(5.3)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

where

245

1 λs Q + xs Qx − QQx , 2 λ f2 (s, x) = εεx + (Qε)x ,

f1 (s, x) =

f3 (s, x) = R1 (ε). According to (1.13), we obtain ηt − xt ηx = g1 (t, x) − g2 (t, x) − g3 (t, x), where g1 (t, x) =

(5.4)

xt 1 λt Q + 1/2 Qx − λ−1 QQx , 2 λ3/2 λ

g2 (t, x) = ηηx + λ−1/2 (Qη)x , g3 (t, x) = λ−1 R1 (ε). The relation u0 = Q + ε0 implies λ(0) = 1, x(0) = 0, the initial condition, associated to (5.4), is η(0) = λ−1/2 (0)ε0 (x) = ε0 (x). Suppose ε0 (x) ∈ H 5 (R) and |ε0 (x)| + |ε0xx (x)| ≤ a0 e−

√ 3 2 x

,

∀x ≥ 0,

for some a0 > 0. To show that there exists constant θ1 > 0, such that ( √ ) √ 3 3 |η(t, x)| + |ηxx (t, x)| ≤ θ1 e− 4 t + 1 e− 2 x , t ≥ 0, x ≥ 0.

(5.5)

Choosing α < α1 < α∗ , by Proposition 3.1 and Lemma 3.6, there exists a constant C5 > 0, such that ⏐ ⏐ ⏐ λs ⏐ ⏐ ⏐ + |xs − 1| + ∥ε(s)∥ 2 ≤ C5 α, s ≥ 0. (5.6) H ⏐λ⏐ There is the following lemma, which will be proved in Appendix A. Lemma 5.2.

Suppose ε0 (x) ∈ H 5 (R) and √

|ε0 (x)| + |ε0xx (x)| ≤ a0 e−

3 2 x

,

∀x ≥ 0,

,

∀t ∈ (0, t0 ), ∀x ≥ 0,

for some a0 > 0. Then there exists constant θ2 > 0 such that √

|η(t, x)| + |ηxx (t, x)| ≤ θ2 (a0 + α)e−

3 2 (x+t)

for some t0 = t0 (a0 , α) > 0. Define ∗

{

t = sup t ≥ 0, ∀s ∈ (0, t), ∀x ≥ 0, |η| + |ηxx | ≤ θ2 (a0 + α +

√

3 1)(e− 4 t

√

+

3 1)e− 2 x

} ,

where η = η(t, x). For a0 > 0 small enough, take t∗ = t0 , so that the supremum is well defined. We claim that t∗ = +∞. By contradiction argument, assume t∗ < +∞. To control η(t, x) and ηxx (t, x) on t ∈ [0, t∗ ].

(5.7)

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246

Consider the following Cauchy problem { ηt − xt ηx = g1 (t, x) − g2 (t, x) − g3 (t, x), (t, x) ∈ R+ × R, η(0, x) = ε0 (x), x ∈ R, for ε0 (x) ∈ H 5 (R). Decompose the solution of (5.8) into the two solutions of the following Cauchy problems { (ηI )t − xt (ηI )x = 0, (t, x) ∈ R+ × R, ηI (0, x) = ε0 (x), x ∈ R, and

{

(ηII )t − xt (ηII )x = g1 (t, x) − g2 (t, x) − g3 (t, x), (t, x) ∈ R+ × R, ηII (0, x) = 0, x ∈ R,

where g1 (t, x) =

(5.8)

(5.9)

(5.10)

xt 1 λt Q + 1/2 Qx − λ−1 QQx , 2 λ3/2 λ

g2 (t, x) = ηII (ηII )x + λ−1/2 (QηII )x , and g3 (t, x) = λ−1 R1 (ε). That is, if ηI (t, x), ηII (t, x) are solutions of (5.9), (5.10) respectively, then η(t, x) = ηI (t, x) + ηII (t, x) is the solution of (5.8). From Proposition 3.1, |λ(s) − 1| ≤ C1 α, taking α ≤ (5C1 )−1 , we have 4/5 ≤ λ(s) ≤ 6/5,

∀s ≥ 0.

(5.11)

Since |xs − 1| ≤ C5 α by (5.6), and xs = λ1/2 xt , then |xs − 1| = |λ1/2 xt − 1| ≤ C5 α. Thus, xt ≥ λ−1/2 (1 − C5 α) ≥ (5/6)1/2 (1 − C5 α). ) ( Let α ≤ 1 − (1/2)(6/5)1/2 C5−1 , then xt (t) ≥ 1/2. So, we obtain x(t) − x(t′ ) ≥ (t − t′ )/2,

∀t, t′ ≥ 0.

(5.12)

There are the following lemmas. Lemma 5.3 (Linear Estimates). Let ηI ∈ L∞ ([0, t∗ ]; H 5 (R)) be a solution of (5.9), suppose ε0 (x) satisfies |ε0 (x)| + |ε0xx (x)| ≤ a0 e−

√ 3 2 x

,

∀x ≥ 0,

(5.13)

for some a0 > 0. Then there exists constant θ0 > 0, such that |ηI (t, x)| + |(ηI )xx (t, x)| ≤ θ0 a0 e− for all x ≥ 0, and all 0 ≤ t ≤ t∗ .

√ 3 4 (2x+t)

,

(5.14)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

247

Proof . For (5.9), we have (t, x) ∈ R+ × R.

ηI (t, x) = ε0 (x + x(t)), Since x(0) = 0, and from (5.12) and (5.13), we have √

|ηI (t, x)| = |ε0 (x + x(t))| ≤ a0 e−

3 4 (2x+x(t))

Similarly, one obtains

≤ a0 e−

√ 3 4 (2x+t)

, ∀x ≥ 0, ∀t ∈ [0, t∗ ].

√

|(ηI )xx (t, x)| ≤ a0 e−

3 4 (2x+t)

Thus, |ηI (t, x)| + |(ηI )xx (t, x)| ≤ θ0 a0 e−

, ∀x ≥ 0, ∀t ∈ [0, t∗ ].

√ 3 4 (2x+t)

∀x ≥ 0, ∀t ∈ [0, t∗ ],

,

for some constant θ0 , independent of a0 . The lemma is proved.

□

Lemma 5.4 (Nonlinear Estimates). Suppose ηII ∈ L∞ ([0, t∗ ]; H 5 (R)) is a solution of (5.10). Then there exists constant C > 0, independent of θ2 , a0 , α, such that ) √ ( √ 3 3 (5.15) |ηII (t, x)| + |(ηII )xx (t, x)| ≤ Cθ22 (a0 + α + 1)2 e− 4 t + 1 e− 2 x , for all x ≥ 0, and all 0 ≤ t ≤ t∗ . The proof is detailed in Appendix B. For η(t, x) = ηI (t, x) + ηII (t, x), combining (5.14) and (5.15), we obtain ( √ ) √ 3 3 |η(t, x)| + |ηxx (t, x)| ≤ C0′ θ22 (a0 + α + 1)2 e− 4 t + 1 e− 2 x , ∀x ≥ 0, ∀t ∈ [0, t∗ ],

(5.16)

for some C0′ > 0 independent of θ2 , a0 , α. Proof of Theorem 1.1. Let 0 < t′0 < t∗ /2, define η˜(t, x) = η(t + t∗ − t′0 /2, x). By (5.16), we have

(5.17) √

|˜ η (0, x)| + |˜ ηxx (0, x)| ≤ Cθ22 (a0 + α + 1)e−

3 2 x

, ∀x ≥ 0.

By the same argument as in Lemma 5.2, there exists t′0 (depending on a0 , α) such that |˜ η (t, x)| + |˜ ηxx (t, x)| ≤ Cθ2 (a0 + α)e−

√ 3 2 (x+t)

, ∀t ∈ (0, t′0 ), ∀x ≥ 0.

Therefore, √

√

|η(t, x)| + |ηxx (t, x)| ≤ Cθ2 (a0 + α)e−

t′0 ∗ 3 2 (x+t+ 2 −t )

≤ Cθ2 (a0 + α)e−

t′0 3 2 (x+ 2 )

,

(5.18)

for all t∗ < t < t∗ + t′0 /2, and all x ≥ 0. To obtain a contradiction with the definition of t∗ , we control η(t, x) on [0, t∗ + δ ∗ ], for some δ ∗ > 0. Estimate η(t, x) by (5.16), (5.18) and find that ( √ ) √ √ ′ ∗ 3 3 ′′′ 2 2 − 43 t |η(t, x)| + |ηxx (t, x)| ≤ C θ2 (a0 + α + 1) e + 1 e− 2 x + Cθ2 (a0 + α)e− 2 (x+t0 /δ ) , (5.19) for all x ≥ 0, and all t ∈ [0, t∗ + δ ∗ ], with 0 < δ ∗ < t′0 /2.

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

248

Choosing a suitable θ2 > 0 and a0 , α > 0 small, for some δ ∗ > 0 small enough, we obtain a contradiction in the definition t∗ of (5.7). Thus, we have ( √ ) √ 3 3 |η(t, x)| + |ηxx (t, x)| ≤ θ1 e− 4 t + 1 e− 2 x , ∀t ≥ 0, ∀x ≥ 0, where θ1 = θ2 (a0 + α + 1). So, (5.5) is true. √ √ Due to 4/5 ≤ λ(s) ≤ 6/5, ds/dt = λ−1/2 (t), then 2 5s/5 ≤ t ≤ 30s/5, and since ε(s, x) = λ1/2 (s)η(s, x), modifying the constant θ1 , there exists constant θ = θ1 , such that ( √ ) √ 3 3 |ε(t, x)| + |εxx (t, x)| ≤ θ e− 4 t + 1 e− 2 x , ∀t ≥ 0, ∀x ≥ 0. Theorem 1.1 is proved. □ Acknowledgments The authors would like to warmly thank Professor Yvan Martel and Professor Frank Merle. Their series of works inspired the research. The authors thank the other researchers for their work and referees for their constructive comments and useful suggestions. The research is supported by National Natural Science Foundation of China (Grant No. 11371175). Appendix A. Proof of Lemma 5.2

Proof . Recall Eq. (2.3) ) ( ∫ 1 3 1 P2zzz (x − z)u2z dz = 0. P2z (x − z) u2 + u2z − u2zz dz − 2 2 2 R R

∫ ut + uux +

Let w = u − Q, that is, u = w + Q. By direct calculations and ε0 = u0 − Q, we get { wt + F1 + F2 = 0, (t, x) ∈ R+ × R, w(0, x) = ε0 (x), x ∈ R,

(A.1)

where F1 (t, x) = wwx + (wQ)x + QQx , and F2 (t, x) = R1 (w), here R1 is as shown in (2.5). For u0 (x) ∈ H 5 (R), from proof of Lemma 5.1, u(t) ∈ L∞ ([0, T ]; H 5 (R)), T > 0. There exists δ1 > 0 such that ∥w(t)∥H 5 ≤ δ1 , for all t ∈ [0, T ]. Define Ω T0 (f ) = sup ∥f ∥H 1 , (A.2) (0,T0 )

for any T0 > 0, f : [0, T0 ] × R → R. Let (∫ ∥f ∥Lq

T0

where 1 ≤ p, q ≤ +∞.

p

Lx

T0

(∫

p

|f (t, x)| dx

:= 0

R

)1/q

)q/p dt

,

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

249

We also define the following space Λ = {w ∈ L∞ ((0, t0 ); H 5 (R)), such that Ω t0 (w) ≤ M α √

and |w(t, x)| + |wxx (t, x)| ≤ θ2 (a0 + α)e−

3 2 (x+t)

, ∀t ∈ (0, t0 ), ∀x ≥ 0},

for θ2 > 0, M > 0, a0 > 0, 0 < α < α∗ and t0 > 0 which would be chosen later. It is easy to verify that Λ is Banach space equipped with the norm of L∞ ((0, t0 ); H 5 (R)). To prove that the solution w of (A.1) belongs to Λ, for t0 > 0 small enough. Apply a fixed point argument on w in the space Λ. We need to show that ∫ t Φ : w ∈ Λ ↦→ Φ(w)(t, x) = ε0 (x) + F (s, x)ds, (A.3) 0

maps Λ into itself and it is a contraction mapping under the norm Ω t0 , where F = −(F1 + F2 ). Divide the argument into two parts. The first part proves Φ is a contraction. The second part shows that |Φ(w)(t, x)| + |Φxx (w)(t, x)| ≤ θ2 (a0 + α)e−

√ 3 2 (x+t)

, ∀t ∈ (0, t0 ), ∀x ≥ 0,

for all w ∈ Λ, and t0 > 0 small enough. (1) The first part. Applying the Minkowski inequality and Cauchy–Schwarz inequality, for any w ∈ Λ, we have   ∫ t   t0  F (s, x)ds Ω (Φ(w)) = sup ε0 (x) +  (0,t0 )

H1

0

t

(∫ ≤∥ε0 ∥H 1 + sup (0,t0 )

≤∥ε0 ∥H 1 + sup

0 2 ∑

∫ ∥F (s)∥L2 ds +

1/2 t0

(0,t0 ) i=1

To estimate ∥Fi ∥L2

2 t0 Lx

and ∥Fix ∥L2

2 t 0 Lx

∥F1 ∥L2

2 t 0 Lx

0

t

) ∥Fx (s)∥L2 ds

(

) ∥Fi ∥L2

2 t 0 Lx

+ ∥Fix ∥L2

2 t0 Lx

.

(i = 1, 2):

=∥wwx + (wQ)x + QQx ∥L2

2 t0 Lx

t0

(∫ = 0 1/2

≤t0

)1/2 ∥wwx + (wQ)x + QQx ∥2L2 ds

((

) )2 Ω t0 (w) + 4(Ω t0 (w) + 1) .

Let F2 (s, x) = T1 + T2′ + T3 , where

( ) ∫ 1 1 3 P2z (x − z) Q2 + Q2z − Q2zz dz − P2zzz (x − z)Q2z dz, 2 2 2 R R ∫ ∫ ′ T2 (s, x) = P2z (x − z)(2Qw + Qz wz − Qzz wzz )dz − 3 P2zzz (x − z)Qz wz dz, ∫

T1 (x) =

R

R

(

) ∫ 1 2 1 2 3 2 T3 (s, x) = P2z (x − z) w + wz − wzz dz − P2zzz (x − z)wz2 dz. 2 2 2 R R ∫

Therefore, ∥F2 ∥L2

2 t0 Lx

(A.4)

= ∥T1 + T2′ + T3 ∥L2

2 t0 Lx

(∫ = 0

t0

)1/2 ∥T1 + T2′ + T3 ∥2L2 ds .

(A.5)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

250

(i) Since

⎧ √ √ √ √ 3 x ⎪ ⎨ − 2e− 2 x sin + e− 3x ( 3 + sin x − 3 cos x), x ≥ 0, 2 √ T1 (x) = √ √ √ ⎪ ⎩ − 2e 23 x sin x − e 3x ( 3 − sin x − 3 cos x), x < 0. 2 √ −√3|x|/2 Then |T1 (x)| ≤ (3 + 2 3)e , x ∈ R, and √ √ 2(3 + 2 3) . ∥T1 ∥L2 ≤ 31/4 √ (ii) Because P2 = 3Q/6, and ∂z3 P2 is not continuous at the origin, then ∫ ∫ x ∫ +∞ P2zzz (x − z)Qz wz dz = P2zzz (x − z)Qz wz dz + P2zzz (x − z)Qz wz dz R −∞ x ∫ ∫ P2zzzz (x − z)Qz wdz − P2zzz (x − z)Qzz wdz. = Qx w(s, x) + R

R

Hence, integrating by parts, we obtain √ ∫ √ ∫ 3 3 ′ T2 (s, x) = Qz (x − z)Qwdz − Qzzzz (x − z)Qz wdz 3 R 2 R √ ∫ √ ∫ 3 3 + Qzz (x − z)Qz wdz + Qzzz (x − z)Qzz wdz 6 R 3 R √ ∫ √ ∫ 3 3 + Qzz (x − z)Qzzz wdz − Qz (x − z)Qzz wdz 3 R 6 R √ ∫ 3 Qz (x − z)Qzzzz wdz − 3Qx w(s, x) − Qx w(s, 0). − 6 R ∫  Here, we only give the estimation details of  R Qz (x − z)Qwdz L2 .  ∫ ∫   4  Qz (x − z)Qwdz  ≤ ∥Qz (x − z)Qw∥L2 dz ≤ √ ∥Q∥L2 ∥w∥L2 ,  2  3 R R L In addition, ∥3Qx w(s, x) − Qx w(s, 0)∥L2 ≤ 4∥Qx ∥L2 ∥w∥H 1 . Thus, we have ∥T2′ (s)∥L2 ≤ (9∥Q∥L2 + 4∥Qx ∥L2 ) Ω t0 (w). (iii) T3 (s, x) =

√ ∫ √ ∫ 3 3 Qzz (x − z)wz wdz − Qzz (x − z)wzzz wdz 12 R 6 R √ ∫ √ ∫ 3 3 − Qz (x − z)wzz wdz − Qz (x − z)wzzzz wdz 12 R 12 R √ ∫ √ ∫ 3 3 + Qzzz (x − z)wzz wdz + Qz (x − z)w2 dz 6 R 6 R √ ∫ 3 3 1 − Qzzzz (x − z)wz wdz + wwx + Qx w(s, 0). 4 R 2 2

Since ∫     Qzz (x − z)wz wdz    R

and ∥Qx w(s, 0)∥L2 ≤ 4∥w∥L2 .

L2

∫ ≤

(A.6)

4 ∥Qzz (x − z)wz w∥L2 dz ≤ √ ∥w∥H 2 ∥w∥L2 , 3 R

(A.7)

(A.8)

(A.9)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

251

Then, we get Due to ∥w(t)∥H 5

∥T3 (s)∥L2 ≤ (5∥w∥H 5 + 2) Ω t0 (w). (A.10) √ ≤ δ1 , for all 0 < t < t0 , and ∥Q∥2H 2 = 2 3, according to (A.6), (A.8) and (A.10), we obtain √ √ ∥F2 ∥L2 = ∥T1 ∥L2 + ∥T2′ ∥L2 + ∥T3 ∥L2 ≤ 5(δ1 + 6)Ω t0 (w) + 3 2 + 2 6,

Thus, ∥F2 ∥L2

2 t0 Lx

= ∥T1 + T2′ + T3 ∥L2

1/2

2 t0 Lx

≤ 10t0 ((δ1 + 6)Ω t0 (w) + 1).

(A.11)

Combining (A.5) and (A.11), one gets 2 ∑

∥Fi ∥L2

1/2

2 t0 Lx

i=1

≤ t0

(

) (Ω t0 (w))2 + 10(δ1 + 7)Ω t0 (w) + 14 .

Similarly, 2 ∑

1/2

∥Fix ∥L2

2 t0 Lx

i=1

≤ t0

(

) 1/2 (9δ1 + 44)Ω t0 (w) + 4δ1 (Ω t0 (w))1/2 + 36 .

By (5.6) and (A.4), we get 1/2

Ω t0 (Φ(w)) ≤ C5 α + t0 (M 2 α2 + M (17δ1 + 100)α + 4M 1/2 δ1 α1/2 + 48), 1/2

Let M > C5 , choosing t0 ≤ (M 2 α2 + M (17δ1 + 100)α + 4M 1/2 δ1 α1/2 + 48)−1 (M − C5 )α, we obtain Ω t0 (Φ(w)) ≤ M α. Similar discussions show that, for any w, w ˜ ∈ Λ, ( )1/2 Ω t0 (Φ(w) − Φ(w)) ˜ ≤ M1 t0 Ω t0 (w − w) ˜ ,

(A.12)

1/2

where M1 = C[(M α + 1)δ1 + (M α + δ1 + 1)(M α)1/2 ], for some C > 0. Therefore, by choosing t0 < M1−1 , Φ is a contraction mapping under the norm Ω t0 . (2) The second part. √ √ 3 3 Suppose |w(t, x)| + |wxx (t, x)| ≤ θ2 (a0 + α)e− 2 (x+t) , and |ε0 (x)| + |ε0xx (x)| ≤ a0 e− 2 x , ∀t ∈ (0, t0 ), ∀x ≥ 0. To show that √ 3 |Φ(w)(t, x)| + |Φxx (w)(t, x)| ≤ θ2 (a0 + α)e− 2 (x+t) , (A.13) for all t ∈ (0, t0 ) and all x ≥ 0. First to deal with Φ(w)(t, x): Taking t0 ≤ 1/2, we obtain √

|ε0 (x)| ≤ a0 e−

3 2 x

√

≤ θ2 a0 e−

3 2 (x+t)

,

(A.14)

√

for some θ2 ≥ e 3/4 , independent of a0 . To estimate F1 : ∫ t ∫ t F1 (s, x)ds = wwx + (wQ)x + QQx ds. 0

0

From (5.6), we have, ∥w(t)∥H 2 ≤ Cα, for some C > 0. Since |w(t, x)| + |wxx (t, x)| ≤ θ2 (a0 + α)e− x ≥ 0, t ∈ (0, t0 ), then ( ) √ √ 3 |F1 (s, x)| ≤ |wwx + (wQ)x + QQx | ≤ Cθ2 (a0 + α + 1) 1 + e− 2 s e− 3x .

√ 3 2 (x+t)

,

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

252

√

Due to t < e−

3t

on t ∈ [0, 2/5], thus ⏐ ⏐∫ ) ∫ t( √ √ ⏐ ⏐ − 23 s ⏐ F1 (s, x)ds⏐ ≤ θ2 (a0 + α + 1)e− 3x 1 + e ds ⏐ ⏐ R

0

≤ 2θ2 (a0 + α + 1)e √

≤ Cθ2 (a0 + α)e−

√ − 3x

(A.15)

t

√

3x − 23 t , e

for all x ≥ 0, 0 < t < t0 , with t0 ≤ min{2/5, (a0 + α)2 }, and some C > 0, independent of θ2 , a0 , α. To estimate F2 : ∫ t ∫ t ∫ t ∫ t ′ F2 (s, x)ds = T1 (x)ds + T2 (s, x)ds + T3 (s, x)ds. 0

√

√

0

0

0

3 3)e− 2 x ,

x ≥ 0, we have (i) By (3.13), |T1 (x)| ≤ (3 + 2 ⏐ ⏐∫ t √ √ √ √ ⏐ ⏐ ⏐ ≤ (3 + 2 3)e− 23 x t ≤ (3 + 2 3)(a0 + α)e− 23 (x+t) , ⏐ T (x)ds 1 ⏐ ⏐

(A.16)

0

where x ≥ 0, t ∈ (0, t0 ), with t0 ≤ min{2/5, (a0 + α)2 }. ∫t ∫ (ii) To estimate 0 T2′ (s, x)ds. Here, we only give the estimation details of R Qz (x−z)Qwdz, the discussion on other parts is analogous. ∫ Qz (x − z)Qwdz = L1 (s, x) + L2 (s, x) + L3 (s, x), R

where

0

∫

Qz (x − z)Qwdz,

L1 (s, x) = −∞ ∫ x

Qz (x − z)Qwdz,

L2 (s, x) = ∫

0 +∞

Qz (x − z)Qwdz.

L3 (s, x) = x

Let c1 = Cα, ∥w(t)∥H 2 ≤ c1 . Hence, for all x ≥ 0, ∫ ∫ 0 √ √ √ − 23 x − 23 (x−z) 23 z e dz ≤ 4c1 e |L1 | ≤ 4c1 e

√

e

3z

−∞

−∞ √

|L2 | ≤ 4θ2 (a0 +

0

8 = √ θ2 (a0 + α)e 3

e−

√ √ 3 3 3 2 (x−z) − 2 z − 2 z

e

0

√

√

3 2 x

,

√

x

∫

3 α)e− 2 s

4c1 dz ≤ √ e− 3

√ − 23 s − 23 x

(

e

√

1 − e−

e )

3 2 x

dz

,

and |L3 | ≤ 4θ2 (a0 + α)e−

√ 3 2 s

∫

+∞

e x

√ 8 ≤ √ θ2 (a0 + α)e− 3x e− 3 3

√ √ √ 3 3 3 2 (x−z) − 2 z − 2 z

√ 3 2 s

e

e

dz

.

Therefore, there exists a constant C > 0, independent of θ2 , a0 , α, such that ⏐ ⏐∫ ( ) √ √ ⏐ ⏐ ⏐ Qz (x − z)Qwdz ⏐ ≤ Cθ2 (a0 + α) 1 + e− 23 s e− 23 x , ⏐ ⏐ R

for all x ≥ 0 and all s ∈ (0, t0 ).

(A.17)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

Similarly, |T2′ (s, x)|

√

(

3 e− 2 s

≤ Cθ2 (a0 + α) 1 +

)

253

√

3 e− 2 x ,

∀x ≥ 0, ∀s ∈ (0, t0 ),

for some C > 0, independent of θ2√, a0 , α. Choosing t0 ≤ 1/2, thus t < e− 3t/2 on t ∈ [0, 1/2]. Then, ⏐ ⏐∫ t ) ∫ t( √ √ √ ⏐ ⏐ − 23 x ′ − 23 s − 23 (x+t) ⏐ ⏐ ≤ Cθ (a + α)e T (s, x)ds 1 + e ds ≤ Cθ (a + α)e , 2 0 2 0 2 ⏐ ⏐ 0

(A.18)

0

for all x ≥ 0, and all t ∈ (0, t0 ). ∫t (iii) To estimate 0 T3 (s, x)ds. ∫ 2 Qz (x − z)wzz dz = J1 (s, x) + J2 (s, x) + J3 (s, x), R

where

0

∫

2 Qz (x − z)wzz dz,

J1 (s, x) = −∞ ∫ x

2 Qz (x − z)wzz dz,

J2 (s, x) = ∫

0 +∞ 2 Qz (x − z)wzz dz.

J3 (s, x) = x

For all x ≥ 0, we obtain ∫

√

√

0

3 2 dz e− 2 (x−z) wzz

|J1 | ≤ 2

≤

3 2e− 2 x

0 2 wzz dz ≤ 2c21 e−

√ 3 2 x

,

−∞

−∞

|J2 | ≤

∫

2θ22 (a0

2

∫

+ α)

√

x

e−

√ √ 3 2 (x−z) − 3z − 3s

0 √ 4 = √ θ22 (a0 + α)2 e− 3s e− 3

√

e (

3 2 x

e

1−e

dz )

√ − 23 x

,

and |J3 | ≤ 2θ22 (a0 + α)2 e−

√

3s

∫

+∞

e

√ √ 3 2 (x−z) − 3z

x

Therefore,

e

√ √ 4 dz = √ θ22 (a0 + α)2 e− 3s e− 3x . 3

⏐∫ ⏐ ) √ ( √ ⏐ ⏐ 2 ⏐ Qz (x − z)wzz ⏐ ≤ Cθ22 (a0 + α)2 e− 3s + 1 e− 23 x , dz ⏐ ⏐

(A.19)

R

where x ≥ 0, s ∈ (0, t0 ). There exists a constant C > 0, independent of θ2 , a0 , α, such that ) √3 ( √ |T3 (s, x)| ≤ Cθ22 (a0 + α)2 e− 3s + 1 e− 2 x , for all x ≥ 0, and all s ∈ (0, t0 ), with t0 ≤ 1/2. Then, ⏐∫ t ⏐ ∫ t( √ √ √ ) ⏐ ⏐ 2 2 − 23 x − 3s 2 2 − 23 (x+t) ⏐ ⏐ T (s, x)ds ≤ Cθ (a + α) e e + 1 ds ≤ Cθ (a + α) e . 3 2 0 2 0 ⏐ ⏐ 0

(A.20)

0

According to (A.16), (A.18) and (A.20), we obtain ⏐∫ t ⏐ √ ⏐ ⏐ − 23 (x+t) ⏐ ⏐ . F (s, x)ds ≤ Cθ (a + α)e 2 0 2 ⏐ ⏐ 0

(A.21)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

254

By (A.15) and (A.21), we get ⏐ ⏐ ⏐∫ t ( √ ) √ 2 ⏐∫ t ⏐ ⏐ ∑ ⏐ ⏐ ⏐ ≤ Cθ2 (a0 + α) e− 23 x + 1 e− 23 (x+t) , ⏐ ⏐ ⏐ F (s, x)ds ≤ F (s, x)ds i ⏐ ⏐ ⏐ ⏐ 0

i=1

(A.22)

0

for x ≥ 0, t ∈ (0, t0 ), with t0 ≤ min{2/5, (a0 + α)2 }, and some C > 0, independent of θ2 , a0 , α. Therefore, combining (A.14) and (A.22), for t0 small enough, we obtain ( √ ) √ −1 23 (x+t) − 23 x (a0 + α) e |Φ(w)(t, x)| ≤ θ2 + Cθ2 e + 1 , ∀x ≥ 0, ∀t ∈ (0, t0 ),

(A.23)

Choosing a0 > 0 small enough, from (A.23), for any θ2 large enough, we have √

|Φ(w)(t, x)| ≤ θ2 (a0 + α)e−

3 2 (x+t)

∀x ≥ 0, ∀t ∈ (0, t0 ).

,

Second to deal with Φxx (w)(t, x): Similar discussions show that: |Φxx (w)(t, x)| ≤ θ2 (a0 + α)e−

√ 3 2 (x+t)

, ∀x ≥ 0, ∀t ∈ (0, t0 ),

for t0 small enough. Thus, (A.13) holds. It follows from (1) and (2) that the solution w of (A.1) belongs to the space Λ. On the other hand, η(t, x) = λ−1/2 ε(t, x) = u(t, x + x(t)) − λ−1/2 Q(x) = w(t, x + x(t)) + Q(x + x(t)) − λ−1/2 Q(x), and ηxx (t, x) = wxx (t, x + x(t)) + Qxx (x + x(t)) − λ−1/2 Qxx (x). Since λ(0) = 1, x(0) = 0, and |λs /λ| + |xs − 1| ≤ C5 α , then |λ(t) − 1| ≤ Cαt, |x(t)| ≤ (Cα + 1)t. Therefore, taking a smaller t0 , and using the decay properties of Q, we have √

|η(t, x)| + |ηxx (t, x)| ≤ θ2 (a0 + α)e− Lemma 5.2 is proved.

3 2 (x+t)

, ∀0 < t < t0 , ∀x ≥ 0.

□

Appendix B. Proof of Lemma 5.4

Proof . For (5.10), we have ηII (t, x) = η1 (t, x) − η2 (t, x) − η3 (t, x), ∫t

where ηi (t, x) = 0 gi√(s, x + x(t) − x(s))ds. Since Q(x) = 2e− 3x/2 sin(x/2 + π/6), x ≥ 0, then 4 √ ∑ ⏐ j ⏐ ⏐∂x Q(x)⏐ ≤ 8e− 23 x ,

∀x ≥ 0.

(B.1)

j=0

To estimate ηi (t, x)(i = 1, 2, 3) respectively: (i) From (B.1) and (5.6), there exists a constant C > 0, independent of α, such that √

|g1 (s, x)| ≤ C(α + 1)e−

3 2 x

,

∀x ≥ 0,

(B.2)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

By (5.12), for all x ≥ 0, and all t ∈ [0, t∗ ], we have ⏐ ⏐∫ t ⏐ ⏐ g1 (s, x + x(t) − x(s))ds⏐⏐ |η1 (t, x)| = ⏐⏐ 0 ∫ t √ 3 ≤ C(α + 1) e− 2 (x+x(t)−x(s)) ds

255

(B.3)

0

( √ ) √ 3 3 ≤ C(α + 1) e− 4 t + 1 e− 2 x . (ii) Since η(s, x) = λ−1/2 (s)ε(s, x), and ∥ε(s)∥H 2 ≤ C5 α, ∥ηII (s)∥H 2 ≤ Cα, for some C > 0. Then, ( √ ) √ 3 |g2 (s, x)| ≤ Cθ22 (a0 + α + 1)2 e− 4 s + 1 e− 3x , ∀x ≥ 0, ∀s ∈ [0, t∗ ], (B.4) for some C > 0, independent of θ2 , a0 , α. Therefore, ⏐ ⏐∫ t ⏐ ⏐ ⏐ g2 (s, x + x(t) − x(s))ds⏐⏐ |η2 (t, x)| = ⏐ 0 ( √ ) √ 3 ≤ Cθ22 (a0 + α + 1)2 e− 4 t + 1 e− 3x ,

(B.5)

where x ≥ 0, and t ∈ [0, t∗ ]. (iii) Let g3 (s, x) = λ−1 R1 = G1 (s, x) + G2 (s, x) + G3 (s, x), where

(

) ∫ 1 2 1 2 3 λG1 (s, x) = P2z (x − z) Q + Qz − Qzz dz − P2zzz (x − z)Q2z dz, 2 2 2 R R ∫ ∫ λ1/2 G2 (s, x) = P2z (x − z)(2QηII + Qz (ηII )z − Qzz (ηII )zz )dz − 3 P2zzz (x − z)Qz (ηII )z dz, ∫

2

R

R

(

) ∫ 1 1 3 2 2 2 G3 (s, x) = P2z (x − z) ηII + (ηII )z − (ηII )zz dz − P2zzz (x − z)(ηII )2z dz. 2 2 2 R R ∫

From (3.13) and (5.11) follows √ √ 5(3 + 2 3) − 3 x e 2 , 4

|G1 (s, x)| ≤ λ−1 |T1 (x)| ≤

∀x ≥ 0.

Thus,

⏐∫ t ⏐ ) √ ( √ ⏐ ⏐ 3 − 43 t ⏐ ⏐ G1 (s, x + x(t) − x(s))ds⏐ ≤ 9 e + 1 e− 2 x , ∀x ≥ 0, ∀t ∈ [0, t∗ ]. ⏐ 0 ∫t ∫ To estimate 0 G2 ds: We only give the estimation details of R Qz (x − z)QηII dz. Let ∫ Qz (x − z)QηII dz = L′1 (s, x) + L′2 (s, x) + L′3 (s, x), R

where

Qz (x − z)QηII dz,

L′2 (s, x) = L′3 (s, x) =

0

∫

L′1 (s, x) =

−∞ ∫ x

Qz (x − z)QηII dz,

∫

0 +∞

Qz (x − z)QηII dz. x

(B.6)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

256

Let c2 = Cα, ∥ηII (s)∥H 2 ≤ c2 . Hence, for all x ≥ 0, we have ∫ ∫ 0 √ √ √ − 23 x ′ − 23 (x−z) 23 z e dz = 4c2 e |L1 | ≤ 4c2 e

0

√

e

3z

−∞

−∞

4c2 dz = √ e− 3

√

3 2 x

,

( √ )∫ x √ √ √ 3 3 3 − 43 s e 2 z e− 2 z e− 2 z dz ≤ 4θ2 (a0 + α + e +1 ( √ ) ( √0 ) √ 3 3 8 − 43 s ≤ √ θ2 (a0 + α + 1) e +1 e− 2 x + 1 e− 2 x , 3 √

|L′2 |

3 1)e− 2 x

and ( √ ) √ ∫ +∞ √ 3 3 3 3 e− 2 z dz |L′3 | ≤ 4θ2 (a0 + α + 1) e− 4 s + 1 e 2 x ( √ ) √x 3 8 ≤ √ θ2 (a0 + α + 1) e− 4 s + 1 e− 3x . 3 3 Then, there exists constant C > 0, independent of θ2 , a0 , α, such that ⏐ ⏐∫ ( √ ) √ ⏐ ⏐ ⏐ Qz (x − z)QηII dz ⏐ ≤ Cθ2 (a0 + α + 1) e− 43 s + 1 e− 23 x , ⏐ ⏐ R

for all x ≥ 0, and all s ∈ [0, t∗ ]. Similarly,

( √ ) √ 3 3 |G2 (s, x)| ≤ Cθ2 (a0 + α + 1) e− 4 s + 1 e− 2 x , ∀x ≥ 0, ∀s ∈ [0, t∗ ],

Therefore, ⏐∫ t ⏐ ( √ ) ∫ t √ ⏐ ⏐ − 23 (x+x(t)−x(s)) − 43 s ⏐ ⏐ G2 (s, x + x(t) − x(s))ds⏐ = Cθ2 (a0 + α + 1) e e + 1 ds ⏐ 0 0 ) √ ( √ 3 3 ≤ Cθ2 (a0 + α + 1) e− 4 t + 1 e− 2 x . To estimate

∫t 0

G3 ds: Let ∫

Qz (x − z)(ηII )2zz dz = J1′ (s, x) + J2′ (s, x) + J3′ (s, x),

R

where

J2′ (s, x) = J3′ (s, x) =

0

∫

J1′ (s, x) =

Qz (x − z)(ηII )2zz dz,

−∞ ∫ x

Qz (x − z)(ηII )2zz dz,

∫

0 +∞

Qz (x − z)(ηII )2zz dz.

x

Then, for any x ≥ 0, we have ∫ 0 ∫ √ √ ′ − 23 (x−z) 2 − 23 x |J1 | ≤ 2 e (ηII )zz dz ≤ 2e −∞

|J2′ |

√

0

(ηII )2zz dz ≤ 2c22 e−

−∞

(

)2 ∫

√

3 e− 4 s

x

√ 3

√

e− 2 (x−z) e− 3z dz 0 ( √ )2 ( √ ) √ 3 3 3 4 ≤ √ θ22 (a0 + α + 1)2 e− 4 s + 1 e− 2 x + 1 e− 2 x , 3 ≤

2θ22 (a0

+ α + 1)

2

+1

3 2 x

,

(B.7)

(B.8)

D. Ding and K. Wang / Nonlinear Analysis 183 (2019) 230–258

257

and )2 ∫ +∞ √ ( √ √ 3 3 e 2 (x−z) e− 3z dz |J3′ | ≤ 2θ22 (a0 + α + 1)2 e− 4 s + 1 x ( √ )2 √ 3 4 ≤ √ θ22 (a0 + α + 1)2 e− 4 s + 1 e− 3x . 3 3 Thus, for all x ≥ 0, and all s ∈ [0, t∗ ], we obtain ⏐ ⏐∫ ( √ ) √ ⏐ ⏐ ⏐ Qz (x − z)(ηII )2zz dz ⏐ ≤ Cθ22 (a0 + α + 1)2 e− 43 s + 1 e− 23 x , ⏐ ⏐ R

for some C > 0, independent of θ2 , a0 , α. Similar discussions show that ) √ ( √ 3 3 |G3 (s, x)| ≤ Cθ22 (a0 + α + 1)2 e− 4 s + 1 e− 2 x ,

∀x ≥ 0, ∀s ∈ [0, t∗ ],

(B.9)

So, ⏐∫ t ⏐ ∫ t √ √ ⏐ ⏐ 2 2 − 23 (x+x(t)−x(s)) − 43 s ⏐ ⏐ G3 (s, x + x(t) − x(s))ds⏐ = Cθ2 (a0 + α + 1) e (e + 1)ds ⏐ 0 0 ( √ ) √ 3 3 ≤ Cθ22 (a0 + α + 1)2 e− 4 t + 1 e− 2 x .

(B.10)

Combining (B.6), (B.8), and (B.10), there exists constant C > 0 independent of θ2 , a0 , α, such that ) √ ( √ 3 2 2 − 43 t + 1 e− 2 x , ∀x ≥ 0, ∀t ∈ [0, t∗ ]. (B.11) |η3 (t, x)| ≤ Cθ2 (a0 + α + 1) e According to (B.3), (B.5), and (B.11), we obtain ( √ ) √ 3 3 |ηII (t, x)| ≤ C6 θ22 (a0 + α + 1)2 e− 4 t + 1 e− 2 x , ∀x ≥ 0, ∀t ∈ [0, t∗ ]. Similarly, |(ηII )xx (t, x)| ≤

C7 θ22 (a0

+ α + 1)

2

(

√

3 e− 4 t

)

+ 1 e−

√ 3 2 x

, ∀x ≥ 0, ∀t ∈ [0, t∗ ].

(B.12)

(B.13)

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