• Aucun résultat trouvé

# Chapter 4 The KdV problem in domains that can be transformed into rectangles

N/A
N/A
Protected

Partager "Chapter 4 The KdV problem in domains that can be transformed into rectangles"

Copied!
24
0
0

Texte intégral

(1)

## The KdV problem in domains that can be transformed into rectangles

In this chapter, we study the Korteweg-de Vries equation subject to boundary condition in non-rectangular domain, with some assumptions on the boundary of the domain and the coefficients of equation. The right-hand side and its derivative with respect to t are in the Lebesgue space L2. The goal is to establish the existence, the uniqueness and the regularity of the solution.

(2)

### 4.1 Introduction

Let Ω⊂R2 be the domain

Ω ={(t, x)∈R2 : 0< t < T, x∈It}, It={x∈R:ϕ1(t)< x < ϕ2(t), t∈(0, T)},

we consider the semi-linear Korteweg-de Vries problem with time variables coefficients









tv(t, y) +a(t)v(t, y)∂yv(t, y) +b(t)∂y3v(t, y) = g(t, y) in Ω, v(0, y) = 0, y∈I0,

v(t, ϕ1(t)) = v(t, ϕ2(t)) =∂yv(t, ϕ2(t)) = 0 t ∈(0, T),

(4.1)

whereI0 =]ϕ1(0), ϕ2(0)[ and a(t), b(t) are given. g and ∂tg are in L2(Ω).

We are particularly interested in the question: What conditions on the functionsa(t), b(t) and (ϕi(t))i=1,2 guarantee that (4.1) admits a unique solution u.

Consider the case where a(t), b(t), a(t) and b(t) are positive and bounded functions for all t∈[0, T]. Let hbe defined by

h : [0, T] → [0, T] t 7→ h(t) =

t

Z

0

b(s)ds,

we put ψi = ϕi ◦h1 where i = 1,2. Using the change of variables t = h(t), w(t, y) = v(t, y), (4.1) can be written in the form









tw(t, y) +d(t)w(t, y)∂yw(t, y) +∂y3w(t, y) =h(t, y) (t, y)∈Ω, w(0, y) = 0,

w(t, ψ1(t)) =w(t, ψ2(t)) =∂yw(t, ψ2(t)) = 0 t ∈(0, T), where d(t) = a(t)

b(t), h(t, y) = g(t, y)

b(t) , Ω = {(t, x) ∈ R2; 0 < t < T, x ∈ It} and

T

Z

(3)

So, to solve (4.1), it is enough to solve the following problem









tv(t, y) +c(t)v(t, y)∂yv(t, y) +∂y3v(t, y) = g(t, y) (t, y)∈Ω, v(0, y) = 0 y∈I0,

v(t, ϕ1(t)) =v(t, ϕ2(t)) =∂yv(t, ϕ2(t)) = 0 t∈(0, T).

(4.2)

In the sequel, we assume that there exist positive constants c1 and c2, such that c1 ≤c(t)≤c2, for all t∈[0, T].

c1 ≤c(t)≤c2, for all t∈[0, T].

(4.3)

We also impose the hypothesis

ϕ1, ϕ2, ϕ′′1 , ϕ′′2 , ϕ′′′1 , ϕ′′′2 ∈L1(0, T) (4.4) to establish the existence and uniqueness of a solution to (4.1).

Then, for all t∈[0, T], we have

1(t)| ≤ |ϕ1(0)|+ Zt

0

1(s)|ds ≤ |ϕ1(0)|+ ZT

0

1(s)|ds ≤γ,

whereγ is a positive constant. Similarly, we have

2(t)| ≤γ, |ϕ1(t)| ≤γ, |ϕ2(t)| ≤γ,

′′1(t)| ≤γ, |ϕ′′2(t)| ≤γ,

for all t∈[0, T] (4.5)

therefore

|ϕ(t)| ≤2γ, |ϕ(t)| ≤2γ, |ϕ′′(t)| ≤2γ for all t∈[0, T], (4.6) whereϕ(t) = ϕ2(t)−ϕ1(t).

The change of variables:

Ω → R

(t, y) 7→ (t, x) =

t, y−ϕ1(t) ϕ2(t)−ϕ1(t)

, (4.7)

(4)

transforms Ω into the rectangleR = (0, T)×I whereI = (0,1). Puttingv(t, y) =u(t, x) and g(t, y) = f(t, x), Problem (4.2) becomes

















tu(t, x) + c(t)

ϕ(t)u(t, x)∂xu(t, x) + 1

ϕ3(t)∂x3u(t, x)−xϕ(t) +ϕ1(t)

ϕ(t) ∂xu(t, x)

=f(t, x) (t, x)∈R, u(0, x) = 0 x∈I,

u(t,0) =u(t,1) =∂xu(t,1) = 0 t∈(0, T),

multiplying both sides of the equation of the last problem by ϕ3(t)>0, we obtain













p(t)∂tu(t, x) +q(t)u(t, x)∂xu(t, x) +∂x3u(t, x) +r(t, x)∂xu(t, x)

=f(t, x) (t, x)∈R, u(0, x) = 0 x∈I,

u(t,0) =u(t,1) =∂xu(t,1) = 0 t∈(0, T),

(4.8)

where

p(t) = ϕ3(t), q(t) =ϕ2(t)c(t) and r(t, x) = −ϕ2(t)(xϕ(t) +ϕ1(t)).

It is easy to see that by change of variables (4.7) we have g ∈L2(Ω) ⇔ f ∈L2(R)

v ∈L(0, T, L2(I0)) ⇔ u∈L(0, T, L2(I)) v ∈L2(0, T, H3(I0)) ⇔ u∈L2(0, T, H3(I))

(4.9)

Remark 4.1. According to (4.3), (4.5) and (4.6), the functions p, q and r satisfy the following assumptions









α < p(t)< β, α < p(t)< β ∀t∈[0, T], α < q(t)< β, α < q(t)< β ∀t∈[0, T],

|r(t, x)| ≤β, |∂xr(t, x)| ≤β, and |∂tr(t, x)| ≤β ∀(t, x)∈R

(4.10)

where α and β are positive constants.

(5)

Theorem 4.2. If f, ∂tf are in L2(R), and p, q, r satisfy assumption (4.10), Problem (4.8) admits a unique solution such that

u∈L(0, T, L2(I))∩L2(0, T, H3(I)),

tu∈L(0, T, L2(I))∩L2(0, T, H1(I)).

Corollary 4.3. If g, ∂tg+ν(t, y)∂yg are inL2(Ω), and c(t), ϕ1(t), ϕ2(t)satisfy assump- tions (4.3) and (4.4), Problem (4.1) admits a unique solution such that

v ∈L(0, T, L2(I0))∩L2(0, T, H3(I0)),

tv ∈L(0, T, L2(I0))∩L2(0, T, H1(I0)).

where ν(t, y) = ϕ(t)y−ϕ1ϕ21ϕ2

ϕ(t) .

We will prove Theorem 4.2 , and thanks to the change of variables (4.7) and (4.9) the Corollary 4.3 will be a direct result of Theorem 4.2 .

### 4.2 Preliminaries

Lemma 4.4. There exist eigenfunctions of the following problem

e(4)jjej, j ∈N,

ej(0) =ej(1) =e′′j(0) =ej(1) = 0.

which create a basis in H4(I) orthonormal in L2(I)

Proof. For every f ∈L2(I) there exists a unique solution satisfying

x4u=f,

u(0) =u(1) =u′′(0) =u(1) = 0. (4.11) Denote byT the operator f 7−→u considered as an operator from L2(I) into L2(I).

We claim thatT s self-adjoint and compact. First, the compactness. Because of (4.11) we have

1

Z

(∂x2u)2dx=

1

Z

f udx

(6)

and thus

k∂x2uk2L2(I) ≤ kfk2L2(I)kuk2L2(I)

≤ Ckfk2L2(I)

whereC is a positive constant. This can be written as

kT fk2H2(I)≤Ckfk2L2(I), ∀f ∈L2(I).

Since the injection ofH1(I) intoL2(I) is compact (becauseI is bounded), we deduce that T is a compact operator fromL2(I) into L2(I). Next, we show thatT is self-adjoint, i.e.,

Z1

0

(T f)gdx= Z1

0

f(T g) dx, ∀f, g∈L2(I).

Indeed, settingu=T f and v =T g, we have

−∂x4u=f (4.12)

and

−∂x4v =g (4.13)

Multiplying (4.12) by v and (4.13) by u and then integrating, we obtain Z1

0

x2u∂x2vdx= Z1

0

f vdx= Z1

0

gudx.

which is the desired conclusion.

Finally, we note that

1

Z

0

(T f)fdx=

1

Z

0

ufdx=

1

Z

0

(∂x2u)2dx≥0, ∀f ∈L2(I).

Hence, assertions of Lemma 4.4 follow from Theorem 6.1 .

(7)

### 4.3 Parabolic regularization

To prove Theorem 4.2 , we consider in R= (0, T)×(0,1) the approximated problem













p(t)∂tuε(t, x) +q(t)uε(t, x)∂xuε(t, x) +∂x3uε(t, x)

+r(t, x)∂xuε(t, x)−ε∂x2uε+ε∂x4uε =f(t, x) (t, x)∈R, uε(0, x) = 0 x∈(0,1),

uε(t,0) = uε(t,1) =∂x2uε(t,0) = ∂xuε(t,1) = 0 t∈(0, T).

(4.14)

In the caser(t, x) = 0 andp(t),q(t) are constants, the equation (4.14) becomes the KdV- Burgers-Kuramoto equation. For other choice of coefficients, (4.14) is a model for several problems that appear in different physical situations.

We establish the existence of a solutionuεfor (4.14), using the Faedo-Galerkin method.

Then finding uniform estimates in ε we obtain Theorem 4.2 , as ε→0.

Theorem 4.5. If f, ∂tf are in L2(R), and p, q, r satisfy assumption (4.10), then (4.14) admits a solution such that

uε ∈L(0, T, H2(I))∩L2(0, T, H4(I)),

tuε∈L(0, T, L2(I))∩L2(0, T, H2(I)).

Proof. To prove the existence of a solution to (4.14), we choose the basis (ej)j∈N ofH4(I) (see Lemma 4.4), orthonormal inL2(I), defined as a subset of the eigenfunctions for the problem

e(4)jjej, j ∈N,

ej(0) =ej(1) =e′′j(0) =ej(1) = 0.

Then, we introduce the approximate solution uεn by uεn(t) =

n

X

j=1

cnj(t)ej, (4.15)

(8)

which satisfies the approximate problem

































 p(t)

1

Z

0

tuεnejdx+q(t)

1

Z

0

uεnxuεnejdx+

1

Z

0

x3uεnejdx

+

1

Z

0

r(t, x)∂xuεnejdx−ε

1

Z

0

x2uεnejdx+ε

1

Z

0

4xuεnejdx

=

1

Z

0

f ejdx, uεn(0) =u0n.

(4.16)

for all j = 1, ..., nand 0≤t ≤T.

Problem (4.16) is equivalent to a system ofn uncoupled nonlinear ordinary differential equations. So, this solution exists in some interval (0, T) with T ≤ T (see for example ). The a priori estimate (4.17) that will be given, shows thatT =T.

### 4.3.1 A priori estimates

Lemma 4.6. There exists a positive constant K1 independent of n and ε, such that for all t∈[0, T]

αkuεnk2L2(I)

t

Z

0

k∂x2uεn(s)k2L2(I)ds≤K1, (4.17)

where α is given in (4.10).

Proof. Multiplying (4.16) bycj, taking the summation with respect to j from 1 ton, and using (4.15), we get

p(t) 2

d dt

1

Z

0

u2εndx+ε

1

Z

0

(∂xuεn)2dx+ε

1

Z

0

(∂x2uεn)2dx

+ 1

2(∂xu(t,0))2− 1 2

Z1

0

xr(t, x)u2εndx= Z1

0

f uεndx.

(9)

Indeed, because of the boundary conditions, the integration by parts gives

q(t)

1

Z

0

u2εnxuεndx= q(t) 3

1

Z

0

x(uεn)3dx= 0, Z1

0

x3uεnuεndx=− Z1

0

x2uεnxuεndx= 1

2(∂xu(t,0))2,

1

Z

0

r(t, x)∂xuεnuεndx=

1

Z

0

r(t, x)∂x

u2εn 2

dx=−1 2

1

Z

0

xr(t, x)u2εndx.

Then p(t)

2 d dt

Z1

0

u2εndx+ε Z1

0

(∂x2uεn)2dx≤ 1 2

Z1

0

xr(t, x)u2εndx+ Z1

0

f uεndx, (4.18)

by integrating (4.18) with respect tot(t∈(0, T)), and according to (4.10), using Cauchy- Schwarz inequality, we find that

α

2kuεnk2L2(I)

t

Z

0

k∂x2uεn(s)k2L2(I)ds

t

Z

0

kf(s)kL2(I)kuεn(s)kL2(I)ds+β 2

t

Z

0

kuεn(s)k2L2(I)ds.

By the elementary inequality

|ab| ≤ ξ

2a2+ b2

2ξ, ∀a, b∈R, ∀ξ >0, (4.19) with ξ= 1, we obtain

α

2kuεnk2L2(I)+ε 2

t

Z

0

k∂x2uεn(s)k2L2(I)ds

≤ 1 2

t

Z

0

kf(s)k2L2(I)ds+ 1 2

t

Z

0

kuεn(s)k2L2(I)ds+ β 2

t

Z

0

kuεn(s)k2L2(I)ds.

(10)

Hence

αkuεnk2L2(I)

t

Z

0

k∂x2uεn(s)k2L2(I)ds

t

Z

0

kf(s)k2L2(I)ds+ (β+ 1)

t

Z

0

kuεn(s)k2L2(I)ds,

and

αkuεnk2L2(I)+ε Zt

0

k∂x2uεn(s)k2L2(I)ds

≤(β+ 1)

t

Z

0

kuεn(s)k2L2(I)

s

Z

0

k∂x2uεn(τ)k2L2(I)

 ds

+

t

Z

0

kf(s)k2L2(I)ds.

As f is in L2(R), there exists a positive constant C1 independent ofn such that

αkuεnk2L2(I)+ε Zt

0

k∂x2uεn(s)k2L2(I)ds

≤C1+ (β+ 1)

t

Z

0

kuεn(s)k2L2(I)

s

Z

0

k∂x2uεn(τ)k2L2(I)dτ ds,

then, by Gronwall’s inequality, we deduce that

αkuεnk2L2(I)+ε Zt

0

k∂x2uεn(s)k2L2(I)ds≤C1exp((β+ 1)t).

TakingK1 =C1exp((β+ 1)T), we obtain

αkuεnk2L2(I)+ε Zt

0

k∂x2uεn(s)k2L2(I)ds≤K1.

(11)

Lemma 4.7. There exists a positive constant K2(ε) independent of n, such that for all t∈[0, T]

αk∂x2uεnk2L2(I)

t

Z

0

k∂x4uεn(s)k2L2(I)ds≤K2(ε), where α is given in (4.10).

Proof. As ∂x4ejjej, we have

n

X

j=1

cj(t)λjej =

n

X

j=1

cj(t)∂x4ej =∂x4uεn(t).

Multiplying both sides of (4.16) by cjλj, taking the summation with respect to j from 1 ton and using the previous relation, we obtain

p(t) 2

d dt

1

Z

0

(∂x2uεn)2dx+ε

1

Z

0

(∂x4uεn)2dx

=−q(t) Z1

0

uεnxuεnx4uεndx− Z1

0

x3uεnx4uεndx

1

Z

0

r(t, x)∂xuεnx4uεndx+ε

1

Z

0

x2uεnx4uεndx+

1

Z

0

f ∂x4uεndx.

(4.20)

Using (4.10) and (4.19) with ξ= 3

ε, we obtain

1

Z

0

x3uεnx4uεndx

≤ 3

2εk∂x3uεnk2L2(I)+ ε

6k∂x4uεnk2L2(I), (4.21)

Z1

0

f ∂x4uεndx

≤ 3

2εkfk2L2(I)+ ε

6k∂x4uεnk2L2(I), (4.22)

1

Z

0

r(t, x)∂xuεnx4uεndx

≤ 3β2

2ε k∂xuεnk2L2(I)+ ε

6k∂x4uεnk2L2(I). (4.23) Now (4.19) with ξ= 3, gives

ε

1

Z

x2uεnx4uεndx≤ 3ε 2

1

Z

(∂x2uεn)2dx+ ε 6

1

Z

(∂x4uεn)2dx. (4.24)

(12)

By the Ehrling inequality, there exists a positive constant η(ε, β), such that k∂x3uεnk2L2(I) ≤ ε2

9k∂x4uεnk2L2(I)+η(ε)kuεnk2L2(I). Using the previous inequality, (4.21) becomes

1

Z

0

x3uεnx4uεndx

≤ ε

3k∂x4uεnk2L2(I)+C2kuεnk2L2(I). (4.25) whereC2 =η(ε) 3

2ε.

Now, we have to estimate the first term of the right-hand side of (4.20). An integration by parts gives

Z1

0

uεnxuεnx4uεndx=− Z1

0

x(uεnxuεn)∂x3uεndx

=−

1

Z

0

(∂xuεn)2x3uεndx−

1

Z

0

uεnx2uεnx3uεndx

= Z1

0

x(∂xuεn)2x2uεndx− 1 2

Z1

0

uεnx(∂x2uεn)2dx

= 5 2

1

Z

0

xuεn(∂x2uεn)2dx.

Then

q(t)

1

Z

0

uεnxuεnx4uεndx

≤ 5β

2 k∂xuεnkL(I)k∂x2uεnk2L2(I). (4.26) By integrating (4.20) between 0 and t, the estimates (4.22), (4.23), (4.24), (4.25) and (4.26) imply

α

2k∂x2uεnk2L2(I)+ ε 6

t

Z

0

k∂x4uεn(s)k2L2(I)ds

≤ 3 2ε

t

Z

0

kf(s)k2L2(I)ds+5β 2

t

Z

0

k∂xuεnkL(I)k∂x2uεnk2L2(I)ds

t t t

(13)

Using the injection of H01(I) in L(I) and Poincar´e’s inequality, we obtain α

2k∂x2uεnk2L2(I)+ ε 6

t

Z

0

k∂x4uεn(s)k2L2(I)ds

t

Z

0

k∂x2uεn(s)k2L2(I)

2 kuεn(s)k2H2(I)+3ε 2 + C2

4 + 3β2

ds

+ 3 2ε

t

Z

0

kf(s)k2L2(I)ds.

Observe that f ∈ L2(R). Then, from (4.17), there exists a positive constant C3(ε) such that

αk∂x2uεnk2L2(I)

t

Z

0

k∂x4uεn(s)k2L2(I)ds

≤C3(ε) +

t

Z

0

2 kuεn(s)k2H2(I)+ 3ε 2 + C2

4 + 3β2

k∂x2uεnk2L2(I)ds, so,

k∂x2uεnk2L2(I)

t

Z

0

k∂x4uεn(s)k2L2(I)ds

≤C3(ε) + Zt

0

2 kuεn(s)k2H2(I)+C4(ε)

k∂x2uεnk2L2(I)

s

Z

0

k∂x4uεn(τ)k2L2(I)

 ds.

whereC4(ε) = 3ε 2 + C2

4 + 3β2

4ε . Gronwall’s inequality shows that

αk∂x2uεnk2L2(I)

t

Z

0

k∂x4uεn(s)k2L2(I)ds

≤C3(ε) exp

t

Z

0

2 kuεn(s)k2H2(I)+C4(ε)

ds

.

(14)

From (4.17), there exists a positive constantK2(ε) independent of n, such that

αk∂x2uεnk2L2(I)

t

Z

0

k∂x4uεn(s)k2L2(I)ds≤K2(ε).

Lemma 4.8. There exists a positive constant K3(ε) independent of n, such that for all t∈[0, T]

αk∂tuεnk2L2(I)

t

Z

0

k ∂t(∂x2uεn(s)

k2L2(I)ds ≤K3(ε).

Proof. Differentiating (4.16) with respect to t, and replacing ej by∂tuεn, we get

1

Z

0

t(p(t)∂tuεn)∂tuεndx+

1

Z

0

t(q(t)uεnxuεn)∂tuεndx

+ Z1

0

tx3uεn

tuεndx+ Z1

0

t(r(t, x)∂xuεn)∂tuεndx

−ε

1

Z

0

tx2uεn

tuεndx+ε

1

Z

0

tx4uεn

tuεndx

= Z1

0

tf ∂tuεndx.

By the boundary conditions of (4.14), and (4.10) we obtain

αk∂tuεnk2L2(I)+ε Zt

0

k ∂t(∂x2uεn(s)

k2L2(I)ds

≤ 1

2k∂tfk2L2(R)

2kuεnkL(I)k∂xuεnk2L2(R)+βk∂xuεnk2L2(R)

+ 1

2 +3β

2 kuεnk2L(R)+ 3βk∂xuεnk2L(R)

k∂tuεnk2L2(R). Then, Gronwall’s inequality shows that

αk∂tuεnk2

t

Z

k ∂t(∂2uεn(s)

k2 ds ≤K3(ε).

(15)

### 4.3.2 Existence of solution for the approximated problem

Lemmas 4.6, 4.7 and 4.8 show that the Galerkin approximation uεn is bounded in L(0, T, H2(I)), and ε∂x4uεn,∂tuεn are bounded in L2(R).

So, it is possible to extract a subsequence from uεn , still denoted uεn, such that uεn→uε weakly in L2 0, T, H01(I)

, (4.27)

uεn→uε strongly in L2 0, T, L2(I)

, (4.28)

ε∂x4uεn→ε∂x4uε weakly in L2 0, T, L2(I)

, (4.29)

tuεn→∂tuε, weakly in L2 0, T, L2(I)

. (4.30)

Note that (4.30) implies

T

Z

0 1

Z

0

p(t)∂tuεnwdxdt−→

T

Z

0 1

Z

0

p(t)∂tuεwdxdt, ∀w∈L2(R).

From (4.27) and (4.28)

uεnxuεn−→uεxuε weakly in L2(R), then

ZT

0

Z1

0

q(t)uεnxuεnwdxdt−→

ZT

0

Z1

0

q(t)uεxuεwdxdt, ∀w∈L2(R).

ZT

0

Z1

0

x3uεnwdxdt−→

ZT

0

Z1

0

x3uεwdxdt, ∀w∈L2(R), ZT

0

Z1

0

r(t, x)∂xuεnwdxdt−→

ZT

0

Z1

0

r(t, x)∂xuεwdxdt, ∀w∈L2(R), and

ε

T

Z

0 1

Z

0

(−∂x2uεn+∂x4uεn)wdxdt−→ε

T

Z

0 1

Z

0

(−∂x2uε+∂x4uε)wdxdt, ∀w∈L2(R).

Using these properties, we can pass to the limit in (4.16) as n −→ +∞. Consequently, there exists a solutionuε of (4.14).

(16)

### 4.4 Passage to the limit in ε

Our goal in this section is to find uniform estimates in ε, to pass to the limit in (4.14) when ε−→0.

Lemma 4.9. Under the hypotheses of Theorem 4.2 , for all ε ∈ (0, α), the solution of (4.14) satisfies the estimate

kuεk2L2(I)+ (α−ε)

t

Z

0

k∂xuε(s)k2L2(I)ds+ε

t

Z

0

k∂x2uε(s)k2L2(I)ds ≤K4,

where α is given in (4.10) and K4 is a positive constant independent of ε.

Proof. Multiplying both sides of (4.14) byuε and integrating between 0 and 1, we obtain

p(t) 2

d dt

1

Z

0

u2εdx+ε

1

Z

0

(∂xuε)2dx+ε

1

Z

0

(∂x2uε)2dx

1

2(∂xu(t,0))2− 1 2

Z1

0

xr(t, x)u2εdx= Z1

0

f uεdx.

Following the same steps as in lemma 4.6, we prove that there exists a constant K1

independent ofε, such that

αkuεk2L2(I)

t

Z

0

k∂x2uε(s)k2L2(I)ds ≤K1. (4.31)

Now, we multiply both sides of (4.14) by exuε and integrate between 0 and 1, to get

p(t)

1

Z

0

exuεtuεdx+

1

Z

0

exuεx3uεdx

−ε

1

Z

0

exuεx2uεdx+ε

1

Z

0

exuεx4uεdx

1 1 1

(4.32)

(17)

Taking into account the boundary conditions of Problem (4.14), an integration by parts gives

Z1

0

exuεtuεdx= 1 2

d dt

Z1

0

exu2εdx,

1

Z

0

exu∂x3uεdx=−

1

Z

0

exuεx2uεdx−

1

Z

0

exxu∂x2uεdx

=

1

Z

0

exuεxuεdx+

1

Z

0

ex(∂xuε)2dx− 1 2

1

Z

0

exx(∂xuε)2dx

=−1 2

1

Z

0

exu2εdx+ 3 2

1

Z

0

ex(∂xuε)2dx+1

2(∂xu(t,0))2,

1

Z

0

exuεx2uεdx=−

1

Z

0

exuεxuεdx−

1

Z

0

ex(∂xuε)2dx

= 1 2

Z1

0

exu2εdx− Z1

0

ex(∂xuε)2dx,

Z1

0

exuεx4uεdx=− Z1

0

exuεx3uεdx− Z1

0

exxuεx3uεdx

=

1

Z

0

exuεx2uεdx+

1

Z

0

exxuεx2uεdx+

1

Z

0

exxuε2xuεdx+

1

Z

0

ex(∂x2uε)2dx

=−

1

Z

0

exuεxuεdx−

1

Z

0

ex(∂xuε)2dx+

1

Z

0

exx(∂xuε)2dx+

1

Z

0

ex(∂x2uε)2dx

= 1 2

1

Z

0

exu2εdx−2

1

Z

0

ex(∂xuε)2dx+

1

Z

0

ex(∂x2uε)2dx−(∂xuε(t,0))2,

(18)

1

Z

0

r(t, x)exuεxuεdx= 1 2

1

Z

0

r(t, x)exx(uε)2dx

=−1 2

Z1

0

xr(t, x)exu2εdx− 1 2

Z1

0

r(t, x)exu2εdx, and

1

Z

0

exu2εxuεdx=−1 3

1

Z

0

exu3εdx.

Using the previous calculations, (4.33) becomes p(t)

2 d dt

1

Z

0

exu2εdx+ 3

2 −ε Z1

0

ex(∂xuε)2dx

+ (1

2−ε)(∂xu(t,0))2

1

Z

0

ex(∂x2uε)2dx

=

1

Z

0

ex

2 (∂xr(t, x) +r(t, x)+)u2εdx+q(t) 3

1

Z

0

exu3εdx+

1

Z

0

f exuεdx.

As 1 ≤ ex ≤ e for all x ∈ [0,1], by integrating from 0 at t (t ∈ (0, T)) and using (4.10), we find thanks to Cauchy-Schwarz inequality

αkuεk2L2(I)+ (3α−2ε) Zt

0

k∂xuε(s)k2L2(I)ds+ε Zt

0

k∂x2uε(s)k2L2(I)ds

≤e(3β+ 2ε+ 1) Zt

0

kuε(s)k2L2(I)ds+e Zt

0

kf(s)k2L2(I)ds+2eβ 3

Zt

0

kuεk3L3(I)ds.

Now, we will get an estimate for the last term in the previous inequality. Since kuεk2L(I) ≤2kuεkL2(I)k∂xuεkL2(I),

then

2eβku k3 ≤ 2eβ

ku k ku k2

(19)

By Young’s inequality |AB| ≤ |A|p

p +|B|p

p with 1< p <∞and p = p

p−1, choosing p= 4 (then p = 4

3)

A=√

14k∂xuεk

1 2

L2(I), B = eβ

√2α14kuεk

5 2

L2(I), we obtain

eβk∂xuεk

1 2

L2(I)kuεk

5 2

L2(I) ≤αk∂xuεk2L2(I)+3 4

√2α14 43

kuεk

10 3

L2(I).

On the other hand, f ∈L2(R), then there exists a constantC5 independent of ε such that

αkuεk2L2(I)+ 2(α−ε)

t

Z

0

k∂xuε(s)k2L2(I)ds+ε

t

Z

0

k∂2xuε(s)k2L2(I)ds

≤C5+e(3β+ 2ε+ 1)

t

Z

0

kuε(s)k2L2(I)ds

+3 4

√2α14 43 Zt

0

kuεk

4 3

L2(I)kuεk2L2(I)ds.

As ε goes to 0 we can chooseε∈(0, α). Then, we deduce that

ϕ(t) = αkuεk2L2(I)+ (α−ε)

t

Z

0

k∂xuε(s)k2L2(I)ds+ε

t

Z

0

k∂x2uε(s)k2L2(I)ds

verifies the following inequality

ϕ(t)≤C5+

t

Z

0

C6kuεk

4 3

L2(I)+C7

ϕ(s) ds,

whereC6 = 3 4

√2α14 43

and C7 =e(3β+ 2ε+ 1).

By the Gronwall’s inequality, we obtain

ϕ(t)≤C5exp

t

Z

C6k∂xuεk

4 3

L2(I)ds+C7T

.

(20)

From (4.31) we have a bound for

t

R

0 kuεk4/3L2(I)ds, then there exists a positive constant K4 independent ofε such that

kuεk2L2(I)+ (α−ε) Zt

0

k∂xuε(s)k2L2(I)ds+ε Zt

0

k∂2xuε(s)k2L2(I)ds≤K4

Lemma 4.10. Under the hypotheses of Theorem 4.2 , for all ε >0, the solution of (4.14) satisfies the estimate

αk∂tuεk2L2(I)+C Zt

0

k∂t(∂xuε(s))k2L2(I)ds+ε Zt

0

k∂t(∂x2uε(s))k2L2(I)ds ≤K5

where C and K5 are a positive constants independent of ε.

Proof. Differentiating (4.14) with respect tot, multiplying both sides by extuε and inte- grating between 0 and 1, we obtain

p(t)

1

Z

0

extuεt2uεdx+p(t)

1

Z

0

ex(∂tuε)2dx+

1

Z

0

ext(q(t)uεxuε)∂tuεdx

+

1

Z

0

extx3uε

tuεdx+

1

Z

0

ext(r(t, x)∂xuε)∂tuεdx

−ε

1

Z

0

extx2uε

tuεdx+ε

1

Z

0

extx4uε

tuεdx

=

1

Z

0

extf ∂tuεdx.

(4.33)

By the boundary conditions of (4.14), an integration by parts gives Z1

0

extuεt2uεdx= 1 2

d dt

Z1

0

ex(∂tuε)2dx,

Références

Documents relatifs

This type of flow field has an advantage over the serpentine shaped channels due to the lower pressure losses; on the other hand, a major drawback is that different paths

When applying the database ‘standard’ and an X calibration set (Figure 4 ), a different tendency arises: for each etching temperature different from the stan- dard etching

Rosier, Exact boundary controllability for the Korteweg–de Vries equation on a bounded domain. Rosier, Exact boundary controllability for the linear Korteweg–de Vries equation on

A methodology to follow physical aging effects on their thermal, mechanical and dielectric properties is applied to a commercial epoxy adhesive.. The analytic description, using

Influence of parameters settings: maximal overlapping ratio effect on the number of extracted segments (a), the overall complexity (b) and quality of the full detection process

In this paper we will follow a different strategy: we first discretize the equation (1.5) with respect to time and space and then derive the suitable artificial boundary conditions

• Design and implementation of the Mental Model Browser web application, a system for enhancing the experience of searching for information on the web through recording a

With the exception of AZA3, all toxins displayed good thermal stability, although at higher temperatures (&gt;130 °C) each toxin investigated was degraded to some extent. In spite

R´ esum´ e - Nous continuons l’´etude du probl`eme de Cauchy et aux limites pour l’´equation de Korteweg-de Vries initi´ee dans .. On obtient des effets r´egularisants

For smooth and decaying data, this problem (the so-called soliton resolution conjecture) can be studied via the inverse transform method ((mKdV) is integrable): for generic such

Sanz-Serna and Christi di- vis d a modified Petrov-Galerkin finite elem nt method, and compared their method to the leading methods that had been developed to date:

En effet, lorsque l’induction de la rachi-anesthésie s’effectue chez un patient en décubitus dorsal, table à l’horizontale, il existe une corrélation entre

Tandis que le deuxième chapitre a eu pour objectif principal l’élaboration de la relation hauteur- débit ainsi que celle du coefficient de débit par une approche théorique en

For sake of completeness, we give also the limit behaviour of the previous Ginzburg-Landau equation with homogeneous Dirichlet boundary condition (for scalar problems with

This paper concerns the inverse problem of retrieving the principal coefficient in a Korteweg-de Vries (KdV) equation from boundary measurements of a single solution.. The

However, in our work, using another method, we prove a more general result concerning the existence, uniqueness and regularity of a solution to the Burgers problem with

In this chapter, we study the Burgers equation with time variable coefficients, subject to boundary condition in a non-parabolic domain.. Some assumptions on the boundary of the

The …rst approach will be illustrated in this work by the study of a parabolic equation with Cauchy-Dirichlet boundary conditions in a non-regular domain of R 3 , in Chapter 3, and

Indeed it has been shown using experimental study  that the first force peak of the dynamic edge impact is due to fiber loading until the compressive fiber failure, while the

For example, in  Beauchard and Coron proved, for a time large enough, the local exact controllability along the ground state trajectory of a Schrödinger equation and Coron proved

Zhang, Boundary smoothing properties of the Korteweg–de Vries equation in a quarter plane and applications, Dynamics Partial Differential Equations 3 (2006) 1–70..

In this paper, we study a class of Initial-Boundary Value Problems proposed by Colin and Ghidaglia for the Korteweg-de Vries equation posed on a bounded domain (0 , L ).. We show

Lastly it should be more realist to set boundary conditions on the value or the rst (spatial) derivative of the function y rather than on the second derivative y xx , if we have in