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Nova S´erie

DECAY OF SOLUTIONS OF SOME NONLINEAR EQUATIONS

Mohammed Aassila

Abstract: For a class of scalar partial differential equations that incorporate con- vection, diffusion, and possibly dispersion in one space and one time dimension, the stability of solutions is investigated.

1 – Introduction

The topic of this paper is the class of equations

(1.1) ut−α uxx+β uxxx+γ uxxxxt+ (g(u))x = 0, x∈R, t >0 where subscripts denote partial derivatives. The case g(u) = u22 and γ = 0 is typical and has received much attention. If α > 0, β = 0, this is known as Burgers equation. If α = 0, β > 0, this is essentially the Korteweg–de-Vries equation. The case α > 0, β >0 is thus refereed to as KdV-Burgers equation;

it also has been studied extensively as has been the case of general g. The case g(u) = up+1p+1 and γ = 1, β = 0 where p ≥ 1 is an integer is refereed to as the Rosenau–Burgers equation. Indeed, if α = 0 then we have the Rosenau equation proposed by Rosenau [8] for treating the dynamics of dense discrette systems in order to overcome the shortenings by the KdV equation, since the KdV equation describes unidimensional propagation of waves, but wave-wave and wave-wall interactions cannot be treated by it. Such a model were studied

Received: April 15, 2001; Revised: April 15, 2002.

AMS Subject Classification: 35Q53, 35B40.

Keywords: stability; convergence rates; asymptotic profile; traveling wave.

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by Park [9] and Chung and Ha [3] for the global existence of the solution to the IBVP. Equation (1.1) with α > 0, γ = 1, β = 0 is called the Rosenau–

Burgers equation and somehow corresponds to the KdV-Burgers equation and the Benjamin–Bona–Mahoney–Burgers equation, but it is given neither by Rosenau nor by Burgers. The Rosenau equation with the dissipative term−αuxx, or say, the Rosenau–Burgers equation arises in some natural phenomena as for example, in bore propagation and in water waves.

In section 2 of this paper, we study the problem (1.1) with β = 1, γ= 0:

(E1) ut−α uxx+uxxx+ (g(u))x = 0, x∈R, t >0

where α > 0 and g is a C2-class function. We give a general criterion for the existence of traveling wave solutions of the formu(x, t) =φ(x−ct).

In section 3, we study the asymptotic behaviour of the solution for the Rosenau–

Burgers equation (problem (1.1) withβ = 0, γ = 1):

(E2)

ut−α uxx+uxxxxt+ µup+1

p+1

x

= 0, x∈R, t >0, u(x,0) =u0(x)→0 as x→ ±∞ ,

whereα >0 and p≥1 is an integer.

The problem (E2) was studied by Mei [7]. He proved that if Z

Ru0(x)dx6= 0 then

ku(t)kL2 ≤ c

(1 +t)4 and ku(t)k≤ c

√1 +t, ∀t≥0. And, ifRRu0(x)dx= 0, then

ku(t)kL2 ≤ c

(1 +t)3/4 and ku(t)k≤ c

1 +t, ∀t≥0 .

Hence, it is proved in [7] that 0 is the asymptotic state of the solutionu(x, t) for the Rosenau–Burgers equation. In this paper, we prove that the solution of the nonlinear parabolic equationut−αuxx+³up+1p+1´

x= 0 is a better asymptotic profile for the Rosenau–Burgers equation. Furthermore, we prove that the convergence to this asymptotic profile is faster than the convergence to 0 proved in [7].

Before ending this section, we state and prove a general technical lemma which will be needed later.

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Lemma 1.1.

(i) If a >0, b >0, then we have for all t≥0 Z t

0 (1 +t−s)−a(1 +s)−bds ≤

c(1 +t)min(a,b) if max(a, b)>1, c(1 +t)min(a,b)log(2 +t) if max(a, b) = 1, c(1 +t)1−a−b if max(a, b)<1 . (ii) Let 0< a < bwithb >1. Let f: (0,∞)→R be bounded on [1,∞) and

integrable on (0,1). Then we have for allt≥0 Z t

0

f(t−s) (1 +t−s)−a(1 +s)−bds ≤ c t−a .

Proof:

(i) We give a brief outline of the proof. Let I =

Z 0

(1 +s)−a³1 +|t−s|´−b , wherea >0, b >0 and max(a, b)>1.

Assuming t >0, as is no essential loss of generality, since evidently I(−t)≤I(t), we may write

I = Z εt

0 + Z t

εt+ Z

t , where 0< ε <1, and εis held fixed. Now

Z εt

0

Z εt

0 (1 +s)−a³t(1−ε)´−bds = o(t−b) Z εt

0 (1 +s)−ads .

In casea >1, this expression iso(t−b). In casea= 1, it iso(logt·t−b), which is o(t−a) since b >1. In casea <1, it is

o(tb) Z εt

0

sads = o(tb)o(ta+1) = o(t1ab) ;

since 1 + min(a, b) ≤a+b by virtue of the assumption that max(a, b) >1, this is in turno(tmin(a,b)).

Similarly, Z t

εt ≤ o(t−a) Z t

εt

³1 +|t−s|´−bds = o(t−a)o(t−b+1)

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ifb6= 1, and so iso(tmin(a,b)). In caseb= 1, the integral is o(ta)o(logt) =o(tb) since in this casea >1. Ifb >1,

Z

t ≤ o(ta) Z

t

³1 +|t−s|´bds ,

which iso(t1−a−b). Finally, ifb≤1, Z

t ≤ o(t−b) Z

t

³1 +|t−s|´−a= o(t−b) .

(ii) We have Z t

0 f(t−s) (1 +t−s)−a(1 +s)−bds ≤

Z t

0 (1 +t−s)−a(1 +s)−bds + Z 1

0 f(s) (1 +s)−a(1 +t−s)−bds

Z t

0

(1 +t−s)a(1 +s)bds + c(1 +t)b . Now,

Z t

0 (1 +t−s)−a(1 +s)−bds =

= Z t

2

0

(1 +t−s)a(1 +s)bds + Z t

t 2

(1 +t−s)a(1 +s)bds . Ifa, b >1, we get

Z t

0 (1 +t−s)−a(1 +s)−bds ≤ 1 b−1

µ 1+t

2

a

+ (a−1) µ

1 + t 2

b

≤ c(1 +t)−a . Ifa <1< b, we get

Z t

0

(1 +t−s)a(1 +s)bds ≤ 1 b−1

µ 1 + t

2

−a

+ (1−a) (1 +t)1a µ

1 + t 2

−b

≤ c(1 +t)−a . Ifa= 1< b, we get

Z t

0 (1 +t−s)−a(1 +s)−bds ≤ 1 b−1

µ 1 + t

2

1

+ log µ

1 + t 2

¶ µ 1 + t

2

b

≤ c(1 +t)−1 .

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2 – Existence of traveling wave solutions

Consider the equation

(E1) ut−α uxx+uxxx+ (g(u))x = 0, x∈R, t >0 whereα >0 and gis aC2-class function.

Under certain conditions, the equation admits monotone traveling wave solu- tionsu(x, t) =φ(x−ct) with speedcthat connect the end states φ±= lim

r→±∞φ(r).

Such a wave profile must satisfy the third order ordinary differential equation

−c φ0+g(φ)0000−α φ00 = 0.

An example is g(x) = 2x(x−1) (b−x) with b≥ 2, which has the wave profile φ(x) = 1+e1x for the parameter α= 2b−1 and the speedc= 0. General profiles (non necessarily monotone) have been constructed in [1, 5]. It is known that monotone profiles exist forg(u) = up+1p+1 and α≥2√pc. More generally, we have the following criterion:

Theorem 2.1. Letg∈C2 be a strictly convex function. A monotone wave profileφfor (E1) exists if and only if:

c = g(φ+)−g(φ) φ+−φ

, (2.1)

α ≥ 2qg0)−c , (2.2)

φ+ < φ . (2.3)

The profile must therefore be monotonically decreasing.

Proof:

(=⇒) Clearly we have

−c φ+g(φ) +φ00−α φ0 = constant and hence−cφ+g(φ) =−cφ++g(φ+) which implies (2.1).

Now, set ψ(z) =φ−φ(−z) and

(2.4) f(r) = g(φ)−c r−g(φ−r) .

Then,f is concave, and−αψ0−ψ00=f(ψ) which is the equation for a wave profile ψ of the Fisher–Kolmogorov–Petrovskii–Piskunov (F-KPP) equationvt−vxx = f(v) that travels to the right with speedα and has limitsψ−φ+, ψ+= 0.

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Thanks to [4], we know that such a monotone wave profile for concavef exists if and only ifα≥2pf0(0) = 2pg0)−c. In this case,ψ> ψ+ and therefore φ> φ+, sincef is positive betweenψ and ψ+. Thus (2.2)–(2.3) are true.

(⇐=) Definef as in (2.4), then by known results about the F-KPP equation, there exists a unique decreasing wave profile ψ with ψ(0) = φ−φ2 + that moves to the right with speedα. Then,φ(z) =φ−ψ(−z) is a monotone wave profile for (E1) withφ(±∞) =φ±.

3 – Asymptotic profile of Rosenau–Burgers equation Consider the Rosenau–Burgers equation in the form

(E2)

ut−α uxx+uxxxxt+ Ãup+1

p+ 1

!

x

= 0, x∈R, t >0, u(x,0) =u0(x)→0 as x→ ±∞ ,

whereα is any given constant,p≥1 is integer.

Consider the following scalings to the variables t→ t

ε2, x→ x

ε, u→ε1/pu , whereε¿1, then we obtain from (E2)

(3.1) ut−α uxx4uxxxxt+upux = 0.

Forε¿1, neglecting the small termε4uxxxxt leads to the asymptotic state equa- tion of the Rosenau–Burgers equation (E2) as follows

ut−α uxx+upux = 0 .

The solution of this parabolic equation should be a better asymptotic profile of equation (E2).

Concerning the parabolic equation (3.2)

(vt−α vxx+vpvx = 0 ,

v(x,0) =v0(x)→0 as x→ ±∞

we have the following result:

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Theorem 3.1 ([10, 11]). Suppose thatv0(x)∈H2(R)∩L1(R). Then, there exists a positive constantδ0 such that ifkv0kL1+kv0kH2 ≤δ0, then the problem (3.2) has a unique global solutionv(x, t) with

v ∈ C(R+, H2(R)∩L1(R)) ∩ L2(R+, H1(R)) and

(3.3) k∂xjv(t)kLq = O(1)³kv0kL1 +kv0kH2

´(1 +t)(j+1)q−12q , 1≤q ≤ ∞. Furthermore, ifv0 ∈L1(R)∩H6(R), then

(3.4) k∂xjvt(t)kL1 = O(1)³kv0kL1+kv0kH6

´(1 +t)−1−j2, j= 0,1,2,3,4.

The main result in this section is the following:

Theorem 3.2. Suppose that w0(x) =

Z x

−∞

³u0(y)−v0(y)´dy ∈ W3,1(R) and

v0(x) ∈ L1(R)∩H6(R) .

Letα: =kv0kL1+kv0kH6, then there exists a positive constant δ0 such that if kw0kW3,1+α ≤ δ0 ,

then the Cauchy problem (E2) has a unique global solution u(x, t) with u(x, t)−v(x, t)∈C(R+, H1(R)) and satisfies

(i) ifp= 1, for any η >0 we have

k(u−v)(t)kL2 ≤ c(1 +t)34 , (3.5)

k(u−v)x(t)kL2 ≤ c(1 +t)1+η , (3.6)

k(u−v)(t)kL ≤ c(1 +t)78 . (3.7)

(ii) If p≥2, then we have

k(u−v)(t)kL2 ≤ c(1 +t)34 , (3.8)

k(u−v)x(t)kL2 ≤ c(1 +t)54 , (3.9)

k(u−v)(t)kL ≤ c(1 +t)−1 . (3.10)

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As a corollary, we obtain

Corollary 3.3. Under the hypotheses of theorem 3.2, we have for2≤q≤ ∞ the decay rates

(3.11) k(u−v)(t)kLq

c(1 +t)78+4q1 if p= 1, c(1 +t)−1+2q1 if p≥2 . The result in the corollary follows from the interpolation inequality

kfkLq ≤ kfk

q−2 q

L kfk

2 q

L2, 2≤q ≤ ∞. The rest of the paper is devoted to the proof of theorem 3.2.

From (E2) and (3.2), we have

(3.12) (u−v)t−α(u−v)xx−uxxxxt+

Ãup+1

p+ 1− vp+1 p+ 1

!

x

= 0.

Since v(±∞, t) = 0, and we expect u(±∞, t) = 0, ux(±∞, t) = 0, then after integrating (3.12) overR, we get

(3.13) d

dt Z

R

³u(x, t)−v(x, t)´dx = 0 .

Integration of (3.13) over [0, t] and thanks to the assumptions we get (3.14)

Z

R

³u(x, t)−v(x, t)´dx = Z

R

³u0(x)−v0(x)´dx = 0 .

Thus we have

wxt−α wxxx+wxxxxxt−vxxxxt+

Ã(v+wx)p+1

p+ 1 − vp+1 p+ 1

!

x

= 0 , where we set

(3.15) w(x, t) =

Z x

−∞

³u(y, t)−v(y, t)´dy

that iswx(x, t) =u(x, t)−v(x, t).

The integration over (−∞, x] with respect tox of the above equation yields (3.16)

(wt−α wxx+wxxxxt =Hp(w) , w(0, x) =w0(x),

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where

Hp(w) : = vxxxt− 1 p+ 1

³(v+wx)p+1−vp+1´

= vxxxt− 1 p+ 1

Xp

j=0

µ j p+1

vjwp+1−jx , p≥1 . We have

|Hp(w)| ≤ |vxxxt|+ 1 p+ 1

Xp

j=0

µ j p+1

|vjwxp+1−j|,

|∂xHp(w)| ≤ |vxxxxt|+|wpxwxx|

+ 1

p+1 Xp

j=1

µ j p+1

¶ ½

j|vj−1vxwp+1−jx |+ (p+1−j)|v j wxp−jwxx|

¾ . (3.17)

Taking the Fourier transform of (3.16), we get b

wt+ α ξ2

1 +ξ4wb = H\p(w) 1 +ξ4 which admits as solution

w(ξ, t) =b ea(ξ)twb0(ξ) + Z t

0

ea(ξ)(ts) H\p(w)(ξ, s) 1 +ξ4 ds , where we set a(ξ) : = 1+ξα ξ24.

The inverse Fourier transform of the above resultant equation yields

(3.18)

w(x, t) = 1 2π

Z

−∞

eiξxe−a(ξ)twb0(ξ)dξ + 1

Z t

0

Z

−∞

eiξxe−a(ξ)(t−s)H\p(w)(ξ, s)

1 +ξ4 dξ ds . The differentiation with respect tox of (3.18) gives

(3.19)

xjw(x, t) = 1 2π

Z

−∞

(iξ)jeiξxe−a(ξ)twb0(ξ)dξ

− 1 2π

Z t

0

Z

−∞

(iξ)jeiξxea(ξ)(ts)H\p(w)(ξ, s)

1 +ξ4 dξ ds . Now, we define the solution spaces as follows, for any positive integerp≥1 and givenδ >0:

Spδ : = nw∈C(R+, H2(R))|Ap(w)≤δo ,

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where

A1(w) = sup

0≤t≤∞

X1

j=0

(1+t)2j+14 ηk∂xjw(t)kL2 + (1+t)1ηkwxx(t)kL2

, Ap(w) = sup

0≤t≤∞

X2

j=0

(1+t)2j+14 k∂xjw(t)kL2, p≥2 .

Rewriting (3.18) as the operational form w=Sw, we need to prove that S is a contraction mapping fromSpδ intoSpδ where δ >0 is a positive constant.

We have

Theorem 3.4. Under the hypotheses of theorem 3.2, there exists a positive constantδ1 such that if

kw0kW3,1+α < δ1

then Cauchy problem (3.16) has a unique global solution w(x, t)∈C(R+, H2(R)).

Furthermore, we have the following estimates:

(i) If p= 1, then for anyη >0we have (3.20)

X1

i=0

(1+t)2j+14 −ηk∂xjw(t)kL2 + (1+t)1−ηkwxx(t)kL2 ≤ c³kw0kW3,1´.

(ii) If p≥2, then we have (3.21)

X2

j=0

(1 +t)2j+14 k∂xjw(t)kL2 ≤ c³kw0kW3,1´ .

Sinceu(x, t)−v(x, t) =wx(x, t), once we prove theorem 3.1, then theorem 3.2 can be easily proved. Hence, we prove theorem 3.1 in the rest of the paper.

We need to this end two lemmas and two well-known estimates quoted from [7]:

Z

−∞

|ξ|je−ca(ξ)t

(1 +ξ4) (1 +|ξ|)j dξ ≤ c(1 +t)j+12 , j= 0,1,2,3,4 , (3.22)

°°

°° 1 2π

Z

−∞

(iξ)jeiξxe−a(ξ)twb0(ξ)dξ

°°

°°

L2 ≤ ckw0kWj+1,1(1 +t)2j+14 (3.23)

for j= 0,1,2.

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Lemma 3.5. Let w1(x, t), w2(x, t)∈ Spδ, then we have sup

ξ∈R

¯¯

¯H\1(w1)(ξ, s)−H\1(w2)(ξ, s)¯¯¯ ≤ c(α+δ)A1(w1−w2) (1 +s)−1+η , sup

ξ∈R

¯¯

¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯ ≤ c(α+δ)pAp(w1−w2) (1 +s)32, p≥2, sup

ξ∈R|ξ|¯¯¯H\1(w1)(ξ, s)−H\1(w2)(ξ, s)¯¯¯ ≤ c(α+δ)A1(w1−w2) (1 +s)54 , sup

ξR|ξ|¯¯¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯ ≤ c(α+δ)pAp(w1−w2) (1+s)−2, p≥2 . Lemma 3.6. Let w(x, t)∈ Spδ, then

(i) if p= 1 Z t

0

°°

°°

°° 1 2π

Z

−∞

(iξ)jeiξxe−a(ξ)(t−s)H\1(w)(ξ, s) 1 +ξ4

°°

°°

°°L2ds ≤

≤ c³α+ (α+δ)2´(1 +t)2j+14 , j = 0,1 , Z t

0

°°

°°

°° 1 2π

Z

−∞

(iξ)2eiξxea(ξ)(ts)H\1(w)(ξ, s) 1 +ξ4

°°

°°

°°

L2

ds ≤

≤ c³α+ (α+δ)2´(1 +t)1+η ; (ii) if p≥2, then we have

Z t

0

°°

°°

°° 1 2π

Z

−∞

(iξ)jeiξxe−a(ξ)(t−s)H\p(w)(ξ, s) 1 +ξ4

°°

°°

°°

L2

ds ≤

≤ c³α+ (α+δ)p+1´(1 +t)2j+14 , j= 0,1,2 . Proof of Theorem 3.1: Rewriting (3.18) in the form w = Sw, we need to prove that there exists a positive constant δ1 such that the operator S is a contraction mapping fromSpδ1 intoSpδ1.

We claim that S maps Spδ into itself. Indeed, for any w1(x, t) ∈ Spδ and denotingw =Sw1 we will prove that w ∈ Spδ for some small δ > 0. Thanks to lemma 3.6, for any positive integerpthere exists a constant c1 such that

Ap(w) ≤ c1

³kw0kW3,1 +α+ (α+δ)p+1´ .

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Let n = maxn2 + 2p+1,c1

1

o, and choose δ2n c11, then for kv0kW3,1ncδ21, α≤ ncδ21 and δ < δ2 we have

Ap(w) ≤ c1 Ã δ2

n c1

+ δ2 n c1

+ µ δ2

n c1

2

p+1!

≤ c1(2 + 2p+12

n c1 ≤ δ2 . HenceS: Spδ→ Spδ for some smallδ < δ2.

Now, let us prove thatSis a contraction inSpδ. Suppose thatw1(x, t), w2(x, t)∈ Spδ (δ < δ2), then we have by (3.18): for j= 0,1,2 and p≥1

xj(Sw1−Sw2) = 1 2π

Z t

0

Z

−∞

(iξ)jeiξxe−a(ξ)(t−s)H\p(w1)(ξ, s)−H\p(w2)(ξ, s) 1 +ξ4 dξ ds . We estimate the termkSw1−Sw2kL2: we have

Z t

0

°°

°°

° 1 2π

Z

−∞

eiξxea(ξ)(ts) 1 +ξ4

³H\p(w1)(ξ, s)−H\p(w2)(ξ, s)´

°°

°°

°L2

ds =

= Z t

0

ÃZ

−∞

e−a(ξ)(t−s) (1 +ξ4)2

¯¯

¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯2

!12 ds

Z t

0 sup

ξ∈R

¯¯

¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯ ÃZ

−∞

e−a(ξ)(t−s) (1 +ξ4)2

!12 ds .

Using (3.22), lemma 1.1 and lemma 3.5, we obtain Z t

0

°°

°°

° 1 2π

Z

−∞

eiξxe−a(ξ)(t−s) 1 +ξ4

³H\1(w1)(ξ, s)−H\1(w2)(ξ, s)´

°°

°°

°L2

ds ≤

≤ c(α+δ)A1(w1−w2) Z t

0

(1 +s)(1η)(1 +t−s)14 ds

≤ c(α+δ)A1(w1−w2) (1 +t)−(1/4−η), 0< η ≤1/2, Z t

0

°°

°°

° 1 2π

Z

−∞

eiξxea(ξ)(ts) 1 +ξ4

³H\p(w1)(ξ, s)−H\p(w2)(ξ, s)´

°°

°°

°L2

ds ≤

≤ c(α+δ)pAp(w1−w2) Z t

0 (1 +s)32 (1 +t−s)14 ds

≤ c(α+δ)pAp(w1−w2) (1 +t)14 , p≥2.

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That is, we obtain kSw1−Sw2kL2

c(α+δ)A1(w1−w2) (1 +t)−(1/4−η), 0< η <1/2, for p= 1 , c(α+δ)pAp(w1−w2) (1 +t)14, for p≥2 .

Similarly, by using (3.22), lemma 1.1 and lemma 3.5, we have the estimates for k∂x(Sw1−Sw2)kL2 and k∂x2(Sw1−Sw2)kL2 as follows:

k∂x(Sw1−Sw2)kL2

c(α+δ)A1(w1−w2) (1 +t)(3/4η), 0< η <1/2, for p= 1 , c(α+δ)pAp(w1−w2) (1 +t)34, for p≥2 ,

and

k∂x2(Sw1−Sw2)kL2

c(α+δ)A1(w1−w2) (1 +t)−(1−η), 0< η <1/2, for p= 1 , c(α+δ)pAp(w1−w2) (1 +t)54 , for p≥2.

Hence, we deduce that, for some constantc1:

Ap(Sw1−Sw2) ≤ c1(α+δ)pAp(w1−w2) . Letn= maxnc2

1,2oand choose δ ≤δ3 < nc1

1, then for α < δ3 and Ap(w2)< δ3 we deduce that

Ap(Sw1−Sw2)< Ap(w1−w2) , that isS: Spδ → Spδ is a contraction for small δ < δ3.

Finally, letδ1 <min{δ2, δ3}, then we have proved thatSis a contraction from Spδ1 to Spδ1, and consequently by Banach’s fixed point theorem, S has a unique fixed point inSpδ1, and then we have the existence of a unique global solution.

Proof of Lemma 3.5: For p= 1, we have sup

ξR

¯¯

¯H\1(w1)(ξ, s)−H\1(w2)(ξ, s)¯¯¯

(3.24)

≤ kv(s)kL2k(w1x−w2x)(s)kL2

+ 1

2k(w1x+w2x)(s)kL2k(w1x−w2x)(s)kL2

≤ c³α(1 +s)14 +δ(1 +s)34´A1(w1−w2) (1 +s)34

≤ c(α+δ)A1(w1−w2) (1 +s)−1+η ,

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and

sup

ξR|ξ|¯¯¯H\1(w1)(ξ, s)−H\1(w2)(ξ, s)¯¯¯

(3.25)

Z

−∞

¯¯

¯H1x(w1)(x, s)−H1x(w2)(x, s)¯¯¯dx

≤ kvx(s)kL2k(w1x−w2x)(s)kL2 + kv(s)kL2k(w1xx−w2xx)(s)kL2

+kw1x(s)kL2k(w1xx−w2xx)(s)kL2 + kw2xx(s)kL2k(w1x−w2x)(s)kL2

≤ c½³α(1 +s)34 +δ(1 +s)−1+η´A1(w1−w2) (1 +s)34 +³α(1 +s)14 +δ(1 +s)34´A1(w1−w2) (1 +s)−1+η

¾

≤ c(α+δ)A1(w1−w2) (1 +s)54 .

For p≥2, we have

sup

ξR

¯¯

¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯

(3.26)

Z

−∞

¯¯

¯Hp(w1)(x, s)−Hp(w2)(x, s)¯¯¯dx

Z

−∞

1 p+1

Xp

j=0

µ j p+1

¶ ¯¯¯vjw1xp+1−j−vjwp+1−j2x ¯¯¯dx

≤ 1 p+1

p p+1

kv(s)kp−1L kv(s)kL2k(w1x−w2x)(s)kL2

+

p−1X

j=0

µ j p+1

kv(s)kjLk(w1x−w2x)(s)kL2 p−jX

i=0

kw1x(s)kiL2kw2x(s)kp−j−iL2 )

≤ c p+1

p p+1

αp+

p−1X

j=0

µ j p+1

(p−j)αjδp−j )

Ap(w1−w2) (1+s)p+12

≤ c(α+δ)pAp(w1−w2) (1 +s)32

(15)

and

sup

ξ∈R|ξ|¯¯¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯

Z

−∞

¯¯

¯xHp(w1)(x, s)−∂xHp(w2)(x, s)¯¯¯dx

°°°(w1xxwp1x−w2xxw2xp )(s)°°°

L1

+ 1

p+ 1 µ p

p+1

¶ ½

p°°°vp1(s)vx(s) (w1x−w2x)(s)°°°

L1

+°°°vp(s)(w1xx−w2xx)(s)°°°

L1

¾

+ 1

p+1

p1

X

j=1

µ j p+1

¶ ½

j°°°vj−1(s)vx(s) (wp+1−j1x −w2xp+1−j)(s)°°°

L1

+ (p+1−j)°°°vj(s) (wp−j1x w1xx−wp−j2x w2xx)(s)°°°

L1

¾ .

Since

°°

°(w1xxwp1x−w2xxw2xp )(s)°°°

L1

≤ kw1xx(s)kL2k(w1x−w2x(s)kL2 p−1X

i=0

kw1x(s)kiL2kw2x(s)kp−1−iL2 +kw2x(s)kpL2k(w1xx−w2xx)(s)kL2

≤ c δpAp(w1−w2) (p+ 1) (1 +s)−2

and 1 p+1

µ p p+1

¶ ½

p°°°vp−1(s)vx(s) (w1x−w2x)(s)°°°

L1+°°°vp(s) (w1xx−w2xx)(s)°°°

L1

¾

≤ c

½

p αpAp(w1−w2) (1 +s)p+22pAp(w1−w2) (1 +s)p+22

¾

≤ c αp(p+ 1)Ap(w1−w2) (1 +s)−2

(16)

and 1 p+1

p−1X

j=1

µ j p+1

j°°°vj1(s)vx(s) (w1xp+1−j−wp+1−j2x )(s)°°°

L1

≤ c p+1

p−1X

j=1

µ j p+1

j αj(1+s)j−12 1Ap(w1−w2)

× (1+s)34

pXj

i=0

δp−j(1 +s)3(p−j)4

≤ c Ap(w1−w2) (1+s)−2 1 p+1

p−1X

j=1

µ j p+1

j(p−j+1)αjδp−j

and 1 p+1

p−1X

j=1

µ j p+1

(p+1−j)°°°vj(s)(w1xp−jw1xx−wp−j2x w2xx)(s)°°°

L1

≤ c p+1

p−1X

j=1

µ j p+1

(p+1−j)αj(1+s)j2

×

½

δpjAp(w1−w2) (1 +s)3p+5−3j4

+ δp−jAp(w1−w2) (1 +s)−2(1 +s)3(p−j−1)4 (p−j)

¾

≤ c Ap(w1−w2)(1 +s)2 1 p+1

p−1X

j=1

µ j p+1

(p+1−j)2αjδpj ,

we obtain that sup

ξ∈R|ξ|¯¯¯H\p(w1)(ξ, s)−H\p(w2)(ξ, s)¯¯¯

(3.27)

≤ c Ap(w1−w2) (1 +s)−2 (

δp(p+1) +αp(p+1)

+ 1

p+1

p−1X

j=1

õ j p+1

j(p+1−j)αjδp−j + µ j

p+1

(p+1−j)2αjδp−j

!)

≤ c(α+δ)pAp(w1−w2) (1 +s)2 .

Thanks to (3.24)–(3.27), we deduce the result of lemma 3.5.

(17)

Proof of Lemma 3.6: Since for f ∈H1 we have kfkL ≤√

2kfk

1 2

L2 · kfxk

1 2

L2

we easily deduce that (i) for p= 1,

(3.28) kw(t)kL ≤ √

2kw(t)kL122kwx(t)kL122 ≤ √

2δ(1 +t)12 , kwx(t)kL ≤ √

2kwx(t)k

1 2

L2kwxx(t)k

1 2

L2 ≤ √

2δ(1 +t)78 ; (ii) for p≥2:

(3.29) kw(t)kL ≤ √

2kw(t)k

1 2

L2kwx(t)k

1 2

L2 ≤ √

2δ(1 +t)12 , kwx(t)kL ≤ √

2kwx(t)k

1 2

L2kwxx(t)k

1 2

L2 ≤ √

2δ(1 +t)−1 . Thanks to (3.17) we have

sup

ξ∈R|H\p(w)(ξ, s)| ≤ Z

−∞|Hp(w)(x, s)|dx

Z

−∞

(

|vxxxt|+ 1 p+1

Xp

j=0

µ j p+1

|vjwxp+1−j| )

dx

≤ kvxxxt(s)kL1 + 1 p+1

µ p p+1

kv(s)kp−1L kv(s)kL2kwx(s)kL2

+ 1

p+1

pX1

j=1

µ j p+1

kv(s)kjLkwx(s)kpL1jkwx(s)k2L2 . Now, because of (3.28)–(3.29) we have

sup

ξR|H\1(w)(ξ, s)| ≤ c

½

α(1+s)522(1+s)−(3/2−2η)+α δ(1+s)−(1−η)

¾

≤ cnα+ (α+δ)2o(1+s)−(1−η) , (3.30)

by choosingη such that 0< η≤1/2, and

(3.31)

sup

ξR|H\p(w)(ξ, s)| ≤ c (

α(1+s)52 + µ p

p+1

αpδ(1+s)p+12

+

p−1X

j=0

µ j p+1

αjδp+1−j(1+s)2p−j+12 )

≤ cnα+ (α+δ)p+1o(1+s)32 , p≥2,

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