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

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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.

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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 (ϕ_i(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 ψ_i = ϕ_i ◦h⁻¹ 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











∂_t^′w(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 ∈ I_t^′} and

T

Z

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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

ϕ^′₁, ϕ^′₂, 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

|ϕ1(t)| ≤ |ϕ1(0)|+ Zt

0

|ϕ^′₁(s)|ds ≤ |ϕ1(0)|+ ZT

0

|ϕ^′₁(s)|ds ≤γ,

whereγ is a positive constant. Similarly, we have

|ϕ₂(t)| ≤γ, |ϕ^′₁(t)| ≤γ, |ϕ^′₂(t)| ≤γ,

|ϕ^′′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)

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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) +ϕ^′₁(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) +ϕ^′₁(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],

(4.10)

where α and β are positive constants.

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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 assumptions (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−ϕ₁ϕ^′₂+ϕ^′₁ϕ₂

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⁾_j =λ_je_j, j ∈N^∗,

e_j(0) =e_j(1) =e^′′_j(0) =e^′_j(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

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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 [10].

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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, u_ε(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⁾_j =λjej, j ∈N^∗,

e_j(0) =e_j(1) =e^′′_j(0) =e^′_j(1) = 0.

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

n

X

j=1

c_nj(t)ej, (4.15)

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which satisfies the approximate problem









 p(t)

1

Z

0

∂_tu_εne_jdx+q(t)

1

Z

0

u_εn∂_xu_εne_jdx+

1

Z

0

∂_x³u_εne_jdx

+

1

Z

0

r(t, x)∂xu_εne_jdx−ε

1

Z

0

∂_x²u_εne_jdx+ε

1

Z

0

∂⁴_xu_εne_jdx

=

1

Z

0

f e_jdx, 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 [15]). 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]

αku_εnk²L²(I)+ε

t

Z

0

k∂_x²u_εn(s)k²L²(I)ds≤K₁, (4.17)

where α is given in (4.10).

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

p(t) 2

d dt

1

Z

0

u²_εndx+ε

1

Z

0

(∂xu_εn)²dx+ε

1

Z

0

(∂_x²u_εn)²dx

+ 1

2(∂xu(t,0))²− 1 2

Z1

0

∂xr(t, x)u²_εndx= Z1

0

f uεndx.

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

q(t)

1

Z

0

u²_εn∂_xu_εndx= q(t) 3

1

Z

0

∂_x(uεn)³dx= 0, Z1

0

∂_x³uεnuεndx=− Z1

0

∂_x²uεn∂xuεndx= 1

2(∂xu(t,0))²,

1

Z

0

r(t, x)∂xu_εnu_εndx=

1

Z

0

r(t, x)∂x

u²_εn 2

dx=−1 2

1

Z

0

∂_xr(t, x)u²_εndx.

Then p(t)

2 d dt

Z1

0

u²_εndx+ε Z1

0

(∂_x²u_εn)²dx≤ 1 2

Z1

0

∂_xr(t, x)u²_εndx+ Z1

0

f u_εndx, (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

2ku_εnk²L²(I)+ε

t

Z

0

k∂_x²u_εn(s)k²L²(I)ds

≤

t

Z

0

kf(s)k^L²^(I)kuεn(s)k^L²^(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

2ku_εnk²L²(I)+ε 2

t

Z

0

≤ 1 2

t

Z

0

kf(s)k²L²(I)ds+ 1 2

t

Z

0

ku_εn(s)k²L²(I)ds+ β 2

t

Z

0

ku_εn(s)k²L²(I)ds.

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Hence

αku_εnk²L²(I)+ε

t

Z

0

≤

t

Z

0

kf(s)k²L²(I)ds+ (β+ 1)

t

Z

0

kuεn(s)k²L²(I)ds,

and

αku_εnk²L²(I)+ε Zt

0

≤(β+ 1)

t

Z

0



ku_εn(s)k²L²(I)+ε

s

Z

0

k∂_x²u_εn(τ)k²L²(I)dτ



 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

ku_εn(s)k²L²(I)+ε

s

Z

0

k∂_x²u_εn(τ)k²L²(I)dτ ds,

then, by Gronwall’s inequality, we deduce that

0

k∂_x²uεn(s)k²L²(I)ds≤C1exp((β+ 1)t).

TakingK1 =C1exp((β+ 1)T), we obtain

0

k∂_x²uεn(s)k²L²(I)ds≤K1.

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

αk∂_x²u_εnk²L²(I)+ε

t

Z

0

k∂_x⁴u_εn(s)k²L²(I)ds≤K₂(ε), where α is given in (4.10).

Proof. As ∂_x⁴ej =λjej, we have

n

X

j=1

c_j(t)λ_je_j =

n

X

j=1

c_j(t)∂_x⁴e_j =∂_x⁴u_εn(t).

Multiplying both sides of (4.16) by c_jλ_j, 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

(∂_x²u_εn)²dx+ε

1

Z

0

(∂_x⁴u_εn)²dx

=−q(t) Z1

0

uεn∂xuεn∂_x⁴uεndx− Z1

0

∂_x³uεn∂_x⁴uεndx

−

1

Z

0

r(t, x)∂xu_εn∂_x⁴u_εndx+ε

1

Z

0

∂_x²u_εn∂_x⁴u_εndx+

1

Z

0

f ∂_x⁴u_εndx.

(4.20)

Using (4.10) and (4.19) with ξ= 3

ε, we obtain

1

Z

0

∂_x³u_εn∂_x⁴u_εndx

≤ 3

2εk∂_x³u_εnk²L²(I)+ ε

6k∂_x⁴u_εnk²L²(I), (4.21)

Z1

0

f ∂_x⁴uεndx

≤ 3

2εkfk²L²(I)+ ε

6k∂_x⁴uεnk²L²(I), (4.22)

1

Z

0

r(t, x)∂xu_εn∂_x⁴u_εndx

≤ 3β²

2ε k∂_xu_εnk²L²(I)+ ε

6k∂_x⁴u_εnk²L²(I). (4.23) Now (4.19) with ξ= 3, gives

ε

1

Z

∂_x²u_εn∂_x⁴u_εndx≤ 3ε 2

1

Z

(∂_x²u_εn)²dx+ ε 6

1

Z

(∂_x⁴u_εn)²dx. (4.24)

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By the Ehrling inequality, there exists a positive constant η(ε, β), such that k∂_x³u_εnk²L²(I) ≤ ε²

9k∂_x⁴u_εnk²L²(I)+η(ε)ku_εnk²L²(I). Using the previous inequality, (4.21) becomes

1

Z

0

∂_x³u_εn∂_x⁴u_εndx

≤ ε

3k∂_x⁴u_εnk²L²(I)+C₂ku_εnk²L²(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

uεn∂xuεn∂_x⁴uεndx=− Z1

0

∂x(uεn∂xuεn)∂_x³uεndx

=−

1

Z

0

(∂xu_εn)²∂_x³u_εndx−

1

Z

0

u_εn∂_x²u_εn∂_x³u_εndx

= Z1

0

∂x(∂xuεn)²∂_x²uεndx− 1 2

Z1

0

uεn∂x(∂_x²uεn)²dx

= 5 2

1

Z

0

∂_xu_εn(∂_x²u_εn)²dx.

Then

q(t)

1

Z

0

u_εn∂_xu_εn∂_x⁴u_εndx

≤ 5β

2 k∂_xu_εnkL^∞(I)k∂_x²u_εnk²L²(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∂_x²u_εnk²L²(I)+ ε 6

t

Z

0

k∂_x⁴u_εn(s)k²L²(I)ds

≤ 3 2ε

t

Z

0

kf(s)k²L²(I)ds+5β 2

t

Z

0

k∂_xu_εnk^L^∞^(I⁾k∂_x²u_εnk²L²(I)ds

t t t

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

2k∂_x²u_εnk²L²(I)+ ε 6

t

Z

0

≤

t

Z

0

k∂_x²u_εn(s)k²L²(I)

5β

2 ku_εn(s)k²H²(I)+3ε 2 + C₂

4 + 3β² 4ε

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

t

Z

0

≤C₃(ε) +

t

Z

0

5β

2 ku_εn(s)k²H²(I)+ 3ε 2 + C₂

4 + 3β² 4ε

k∂_x²u_εnk²L²(I)ds, so,

k∂_x²uεnk²L²(I)+ε

t

Z

0

k∂_x⁴uεn(s)k²L²(I)ds

≤C3(ε) + Zt

0

5β

2 kuεn(s)k²H²(I)+C4(ε)



k∂_x²u_εnk²L²(I)+ε

s

Z

0

k∂_x⁴u_εn(τ)k²L²(I)dτ



 ds.

whereC₄(ε) = 3ε 2 + C₂

4 + 3β²

4ε . Gronwall’s inequality shows that

t

Z

0

≤C₃(ε) exp





t

Z

0

5β

2 ku_εn(s)k²H²(I)+C₄(ε)

ds



.

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

t

Z

0

k∂_x⁴u_εn(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)+ε

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)∂tu_εn)∂_tu_εndx+

1

Z

0

∂_t(q(t)uεn∂_xu_εn)∂_tu_εndx

+ Z1

0

∂t ∂_x³uεn

∂tuεndx+ Z1

0

∂t(r(t, x)∂xuεn)∂tuεndx

−ε

1

Z

0

∂_t ∂_x²u_εn

∂_tu_εndx+ε

1

Z

0

∂_t ∂_x⁴u_εn

∂_tu_εndx

= 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)+β

2ku_εnk^L^∞^(I)k∂_xu_εnk²L²(R)+βk∂_xu_εnk²L²(R)

+ 1

2 +3β

2 kuεnk²L^∞(R)+ 3βk∂xuεnk²L^∞(R)

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

αk∂tuεnk² +ε

t

Z

k ∂t(∂²uεn(s)

k² ds ≤K3(ε).

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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∂_x⁴u_εn,∂_tu_εn 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)∂_tu_εnwdxdt−→

T

Z

0 1

Z

0

p(t)∂_tu_εwdxdt, ∀w∈L²(R).

From (4.27) and (4.28)

u_εn∂_xu_εn−→u_ε∂_xu_ε weakly in L²(R), then

ZT

0

Z1

0

q(t)uεn∂xuεnwdxdt−→

ZT

0

Z1

0

q(t)uε∂xuεwdxdt, ∀w∈L²(R).

ZT

0

Z1

0

∂_x³uεnwdxdt−→

ZT

0

Z1

0

∂_x³uεwdxdt, ∀w∈L²(R), ZT

0

Z1

0

r(t, x)∂_xu_εnwdxdt−→

ZT

0

Z1

0

r(t, x)∂_xu_εwdxdt, ∀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 solutionu_ε of (4.14).

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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

u²_εdx+ε

1

Z

0

(∂xu_ε)²dx+ε

1

Z

0

(∂_x²u_ε)²dx

1

2(∂xu(t,0))²− 1 2

Z1

0

∂xr(t, x)u²_εdx= 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

αku_εk²L²(I)+ε

t

Z

0

k∂_x²u_ε(s)k²L²(I)ds ≤K₁. (4.31)

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

p(t)

1

Z

0

e^xuε∂tuεdx+

1

Z

0

e^xuε∂_x³uεdx

−ε

1

Z

0

e^xu_ε∂_x²u_εdx+ε

1

Z

0

e^xu_ε∂_x⁴u_εdx

1 1 1

(4.32)

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

Z1

0

e^xuε∂tuεdx= 1 2

d dt

Z1

0

e^xu²_εdx,

1

Z

0

e^xu∂_x³uεdx=−

1

Z

0

e^xuε∂_x²uεdx−

1

Z

0

e^x∂xu∂_x²uεdx

=

1

Z

0

e^xu_ε∂_xu_εdx+

1

Z

0

e^x(∂_xu_ε)²dx− 1 2

1

Z

0

e^x∂_x(∂_xu_ε)²dx

=−1 2

1

Z

0

e^xu²_εdx+ 3 2

1

Z

0

e^x(∂xuε)²dx+1

2(∂xu(t,0))²,

1

Z

0

e^xu_ε∂_x²u_εdx=−

1

Z

0

e^xu_ε∂_xu_εdx−

1

Z

0

e^x(∂xu_ε)²dx

= 1 2

Z1

0

e^xu²_εdx− Z1

0

e^x(∂xuε)²dx,

Z1

0

e^xuε∂_x⁴uεdx=− Z1

0

e^xuε∂_x³uεdx− Z1

0

e^x∂xuε∂_x³uεdx

=

1

Z

0

e^xu_ε∂_x²u_εdx+

1

Z

0

e^x∂_xu_ε∂_x²u_εdx+

1

Z

0

e^x∂_xu_ε∂²_xu_εdx+

1

Z

0

e^x(∂_x²u_ε)²dx

=−

1

Z

0

e^xuε∂xuεdx−

1

Z

0

e^x(∂xuε)²dx+

1

Z

0

e^x∂x(∂xuε)²dx+

1

Z

0

e^x(∂_x²uε)²dx

= 1 2

1

Z

0

e^xu²_εdx−2

1

Z

0

e^x(∂_xu_ε)²dx+

1

Z

0

e^x(∂_x²u_ε)²dx−(∂_xu_ε(t,0))²,

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1

Z

0

r(t, x)e^xu_ε∂_xu_εdx= 1 2

1

Z

0

r(t, x)e^x∂_x(uε)²dx

=−1 2

Z1

0

∂xr(t, x)e^xu²_εdx− 1 2

Z1

0

r(t, x)e^xu²_εdx, and

1

Z

0

e^xu²_ε∂_xu_εdx=−1 3

1

Z

0

e^xu³_εdx.

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

2 d dt

1

Z

0

e^xu²_εdx+ 3

2 −ε Z¹

0

e^x(∂xuε)²dx

+ (1

2−ε)(∂_xu(t,0))²+ε

1

Z

0

e^x(∂_x²u_ε)²dx

=

1

Z

0

e^x

2 (∂xr(t, x) +r(t, x)+)u²_εdx+q(t) 3

1

Z

0

e^xu³_εdx+

1

Z

0

f e^xu_εdx.

As 1 ≤ e^x ≤ 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

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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=√

2α¹⁴k∂_xu_εk

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

eβ

√2α¹⁴ ⁴₃

kuεk

10 3

L²(I).

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

αku_εk²L²(I)+ 2(α−ε)

t

Z

0

k∂_xu_ε(s)k²L²(I)ds+ε

t

Z

0

k∂²_xu_ε(s)k²L²(I)ds

≤C₅+e(3β+ 2ε+ 1)

t

Z

0

ku_ε(s)k²L²(I)ds

+3 4

eβ

√2α¹⁴ ⁴₃ Z^t

0

ku_εk

4 3

L²(I)ku_εk²L²(I)ds.

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

ϕ(t) = αku_εk²L²(I)+ (α−ε)

t

Z

0

k∂_xu_ε(s)k²L²(I)ds+ε

t

Z

0

k∂_x²u_ε(s)k²L²(I)ds

verifies the following inequality

ϕ(t)≤C₅+

t

Z

0

C₆ku_εk

4 3

L²(I)+C₇

ϕ(s) ds,

whereC6 = 3 4

eβ

√2α¹⁴ ⁴₃

and C7 =e(3β+ 2ε+ 1).

By the Gronwall’s inequality, we obtain

ϕ(t)≤C₅exp





t

Z

C₆k∂_xu_εk

4 3

L²(I)ds+C₇T



.

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From (4.31) we have a bound for

t

R

0 ku_εk^4/3_L²_(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∂²_xuε(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 e^x∂_tu_ε and integrating between 0 and 1, we obtain

p(t)

1

Z

0

e^x∂_tu_ε∂_t²u_εdx+p^′(t)

1

Z

0

e^x(∂tu_ε)²dx+

1

Z

0

e^x∂_t(q(t)uε∂_xu_ε)∂_tu_εdx

+

1

Z

0

e^x∂t ∂_x³uε

∂tuεdx+

1

Z

0

e^x∂t(r(t, x)∂xuε)∂tuεdx

−ε

1

Z

0

e^x∂_t ∂_x²u_ε

∂_tu_εdx+ε

1

Z

0

e^x∂_t ∂_x⁴u_ε

∂_tu_εdx

=

1

Z

0

e^x∂tf ∂tuεdx.

(4.33)

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

0

e^x∂tuε∂_t²uεdx= 1 2

d dt

Z1

0

e^x(∂tuε)²dx,