## 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 L^{2}. The goal is to establish the existence, the uniqueness and the
regularity of the solution.

### 4.1 Introduction

Let Ω⊂R^{2} be the domain

Ω ={(t, x)∈R^{2} : 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)∂_{y}^{3}v(t, y) = g(t, y) in Ω,
v(0, y) = 0, y∈I_{0},

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

(4.1)

whereI_{0} =]ϕ_{1}(0), ϕ_{2}(0)[ and a(t), b(t) are given. g and ∂_{t}g are in L^{2}(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) +∂_{y}^{3}w(t^{′}, y) =h(t^{′}, y) (t^{′}, y)∈Ω^{′},
w(0, y) = 0,

w(t^{′}, ψ_{1}(t^{′})) =w(t^{′}, ψ_{2}(t^{′})) =∂_{y}w(t^{′}, ψ_{2}(t^{′})) = 0 t^{′} ∈(0, T^{′}),
where d(t^{′}) = a(t)

b(t), h(t^{′}, y) = g(t, y)

b(t) , Ω^{′} = {(t^{′}, x) ∈ R^{2}; 0 < t^{′} < T^{′}, x ∈ I_{t}^{′}} and

T

Z

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

∂tv(t, y) +c(t)v(t, y)∂yv(t, y) +∂_{y}^{3}v(t, y) = g(t, y) (t, y)∈Ω,
v(0, y) = 0 y∈I_{0},

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 c_{1} and c_{2}, 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} ∈L^{1}(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

|ϕ_{2}(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−ϕ_{1}(t)
ϕ2(t)−ϕ1(t)

, (4.7)

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

ϕ^{3}(t)∂_{x}^{3}u(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 ϕ^{3}(t)>0, we obtain

p(t)∂tu(t, x) +q(t)u(t, x)∂xu(t, x) +∂_{x}^{3}u(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) = ϕ^{3}(t), q(t) =ϕ^{2}(t)c(t) and r(t, x) = −ϕ^{2}(t)(xϕ^{′}(t) +ϕ^{′}_{1}(t)).

It is easy to see that by change of variables (4.7) we have
g ∈L^{2}(Ω) ⇔ f ∈L^{2}(R)

v ∈L^{∞}(0, T, L^{2}(I0)) ⇔ u∈L^{∞}(0, T, L^{2}(I))
v ∈L^{2}(0, T, H^{3}(I0)) ⇔ u∈L^{2}(0, T, H^{3}(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.

Theorem 4.2. If f, ∂tf are in L^{2}(R), and p, q, r satisfy assumption (4.10), Problem
(4.8) admits a unique solution such that

u∈L^{∞}(0, T, L^{2}(I))∩L^{2}(0, T, H^{3}(I)),

∂tu∈L^{∞}(0, T, L^{2}(I))∩L^{2}(0, T, H^{1}(I)).

Corollary 4.3. If g, ∂_{t}g+ν(t, y)∂yg are inL^{2}(Ω), and c(t), ϕ_{1}(t), ϕ_{2}(t)satisfy assump-
tions (4.3) and (4.4), Problem (4.1) admits a unique solution such that

v ∈L^{∞}(0, T, L^{2}(I0))∩L^{2}(0, T, H^{3}(I0)),

∂_{t}v ∈L^{∞}(0, T, L^{2}(I0))∩L^{2}(0, T, H^{1}(I0)).

where ν(t, y) = ϕ^{′}(t)y−ϕ_{1}ϕ^{′}_{2}+ϕ^{′}_{1}ϕ_{2}

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_{j}e_{j}, j ∈N^{∗},

e_{j}(0) =e_{j}(1) =e^{′′}_{j}(0) =e^{′}_{j}(1) = 0.

which create a basis in H^{4}(I) orthonormal in L^{2}(I)

Proof. For every f ∈L^{2}(I) there exists a unique solution satisfying

∂_{x}^{4}u=f,

u(0) =u(1) =u^{′′}(0) =u^{′}(1) = 0. (4.11)
Denote byT the operator f 7−→u considered as an operator from L^{2}(I) into L^{2}(I).

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

1

Z

(∂_{x}^{2}u)^{2}dx=

1

Z

f udx

and thus

k∂_{x}^{2}uk^{2}L^{2}(I) ≤ kfk^{2}L^{2}(I)kuk^{2}L^{2}(I)

≤ Ckfk^{2}L^{2}(I)

whereC is a positive constant. This can be written as

kT fk^{2}H^{2}(I)≤Ckfk^{2}L^{2}(I), ∀f ∈L^{2}(I).

Since the injection ofH^{1}(I) intoL^{2}(I) is compact (becauseI is bounded), we deduce that
T is a compact operator fromL^{2}(I) into L^{2}(I). Next, we show thatT is self-adjoint, i.e.,

Z1

0

(T f)gdx= Z1

0

f(T g) dx, ∀f, g∈L^{2}(I).

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

−∂_{x}^{4}u=f (4.12)

and

−∂_{x}^{4}v =g (4.13)

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

0

∂_{x}^{2}u∂_{x}^{2}vdx=
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

(∂_{x}^{2}u)^{2}dx≥0, ∀f ∈L^{2}(I).

Hence, assertions of Lemma 4.4 follow from Theorem 6.1 [10].

### 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) +∂_{x}^{3}uε(t, x)

+r(t, x)∂xuε(t, x)−ε∂_{x}^{2}uε+ε∂_{x}^{4}uε =f(t, x) (t, x)∈R,
u_{ε}(0, x) = 0 x∈(0,1),

uε(t,0) = uε(t,1) =∂_{x}^{2}uε(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 L^{2}(R), and p, q, r satisfy assumption (4.10), then (4.14)
admits a solution such that

uε ∈L^{∞}(0, T, H^{2}(I))∩L^{2}(0, T, H^{4}(I)),

∂tuε∈L^{∞}(0, T, L^{2}(I))∩L^{2}(0, T, H^{2}(I)).

Proof. To prove the existence of a solution to (4.14), we choose the basis (ej)j∈N⋆ ofH^{4}(I)
(see Lemma 4.4), orthonormal inL^{2}(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)

which satisfies the approximate problem

p(t)

1

Z

0

∂_{t}u_{εn}e_{j}dx+q(t)

1

Z

0

u_{εn}∂_{x}u_{εn}e_{j}dx+

1

Z

0

∂_{x}^{3}u_{εn}e_{j}dx

+

1

Z

0

r(t, x)∂xu_{εn}e_{j}dx−ε

1

Z

0

∂_{x}^{2}u_{εn}e_{j}dx+ε

1

Z

0

∂^{4}_{x}u_{εn}e_{j}dx

=

1

Z

0

f e_{j}dx,
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_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)ds≤K_{1}, (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

u^{2}_{εn}dx+ε

1

Z

0

(∂xu_{εn})^{2}dx+ε

1

Z

0

(∂_{x}^{2}u_{εn})^{2}dx

+ 1

2(∂xu(t,0))^{2}− 1
2

Z1

0

∂xr(t, x)u^{2}_{εn}dx=
Z1

0

f uεndx.

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

q(t)

1

Z

0

u^{2}_{εn}∂_{x}u_{εn}dx= q(t)
3

1

Z

0

∂_{x}(uεn)^{3}dx= 0,
Z1

0

∂_{x}^{3}uεnuεndx=−
Z1

0

∂_{x}^{2}uεn∂xuεndx= 1

2(∂xu(t,0))^{2},

1

Z

0

r(t, x)∂xu_{εn}u_{εn}dx=

1

Z

0

r(t, x)∂x

u^{2}_{εn}
2

dx=−1 2

1

Z

0

∂_{x}r(t, x)u^{2}_{εn}dx.

Then p(t)

2 d dt

Z1

0

u^{2}_{εn}dx+ε
Z1

0

(∂_{x}^{2}u_{εn})^{2}dx≤ 1
2

Z1

0

∂_{x}r(t, x)u^{2}_{εn}dx+
Z1

0

f u_{εn}dx, (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_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)ds

≤

t

Z

0

kf(s)k^{L}^{2}^{(I)}kuεn(s)k^{L}^{2}^{(I)}ds+β
2

t

Z

0

kuεn(s)k^{2}L^{2}(I)ds.

By the elementary inequality

|ab| ≤ ξ

2a^{2}+ b^{2}

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

α

2ku_{εn}k^{2}L^{2}(I)+ε
2

t

Z

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)ds

≤ 1 2

t

Z

0

kf(s)k^{2}L^{2}(I)ds+ 1
2

t

Z

0

ku_{εn}(s)k^{2}L^{2}(I)ds+ β
2

t

Z

0

ku_{εn}(s)k^{2}L^{2}(I)ds.

Hence

αku_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)ds

≤

t

Z

0

kf(s)k^{2}L^{2}(I)ds+ (β+ 1)

t

Z

0

kuεn(s)k^{2}L^{2}(I)ds,

and

αku_{εn}k^{2}L^{2}(I)+ε
Zt

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)ds

≤(β+ 1)

t

Z

0

ku_{εn}(s)k^{2}L^{2}(I)+ε

s

Z

0

k∂_{x}^{2}u_{εn}(τ)k^{2}L^{2}(I)dτ

ds

+

t

Z

0

kf(s)k^{2}L^{2}(I)ds.

As f is in L^{2}(R), there exists a positive constant C1 independent ofn such that

αkuεnk^{2}L^{2}(I)+ε
Zt

0

k∂_{x}^{2}uεn(s)k^{2}L^{2}(I)ds

≤C_{1}+ (β+ 1)

t

Z

0

ku_{εn}(s)k^{2}L^{2}(I)+ε

s

Z

0

k∂_{x}^{2}u_{εn}(τ)k^{2}L^{2}(I)dτ
ds,

then, by Gronwall’s inequality, we deduce that

αkuεnk^{2}L^{2}(I)+ε
Zt

0

k∂_{x}^{2}uεn(s)k^{2}L^{2}(I)ds≤C1exp((β+ 1)t).

TakingK1 =C1exp((β+ 1)T), we obtain

αkuεnk^{2}L^{2}(I)+ε
Zt

0

k∂_{x}^{2}uεn(s)k^{2}L^{2}(I)ds≤K1.

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

αk∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds≤K_{2}(ε),
where α is given in (4.10).

Proof. As ∂_{x}^{4}ej =λjej, we have

n

X

j=1

c_{j}(t)λ_{j}e_{j} =

n

X

j=1

c_{j}(t)∂_{x}^{4}e_{j} =∂_{x}^{4}u_{εn}(t).

Multiplying both sides of (4.16) by c_{j}λ_{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

(∂_{x}^{2}u_{εn})^{2}dx+ε

1

Z

0

(∂_{x}^{4}u_{εn})^{2}dx

=−q(t) Z1

0

uεn∂xuεn∂_{x}^{4}uεndx−
Z1

0

∂_{x}^{3}uεn∂_{x}^{4}uεndx

−

1

Z

0

r(t, x)∂xu_{εn}∂_{x}^{4}u_{εn}dx+ε

1

Z

0

∂_{x}^{2}u_{εn}∂_{x}^{4}u_{εn}dx+

1

Z

0

f ∂_{x}^{4}u_{εn}dx.

(4.20)

Using (4.10) and (4.19) with ξ= 3

ε, we obtain

1

Z

0

∂_{x}^{3}u_{εn}∂_{x}^{4}u_{εn}dx

≤ 3

2εk∂_{x}^{3}u_{εn}k^{2}L^{2}(I)+ ε

6k∂_{x}^{4}u_{εn}k^{2}L^{2}(I), (4.21)

Z1

0

f ∂_{x}^{4}uεndx

≤ 3

2εkfk^{2}L^{2}(I)+ ε

6k∂_{x}^{4}uεnk^{2}L^{2}(I), (4.22)

1

Z

0

r(t, x)∂xu_{εn}∂_{x}^{4}u_{εn}dx

≤ 3β^{2}

2ε k∂_{x}u_{εn}k^{2}L^{2}(I)+ ε

6k∂_{x}^{4}u_{εn}k^{2}L^{2}(I). (4.23)
Now (4.19) with ξ= 3, gives

ε

1

Z

∂_{x}^{2}u_{εn}∂_{x}^{4}u_{εn}dx≤ 3ε
2

1

Z

(∂_{x}^{2}u_{εn})^{2}dx+ ε
6

1

Z

(∂_{x}^{4}u_{εn})^{2}dx. (4.24)

By the Ehrling inequality, there exists a positive constant η(ε, β), such that
k∂_{x}^{3}u_{εn}k^{2}L^{2}(I) ≤ ε^{2}

9k∂_{x}^{4}u_{εn}k^{2}L^{2}(I)+η(ε)ku_{εn}k^{2}L^{2}(I).
Using the previous inequality, (4.21) becomes

1

Z

0

∂_{x}^{3}u_{εn}∂_{x}^{4}u_{εn}dx

≤ ε

3k∂_{x}^{4}u_{εn}k^{2}L^{2}(I)+C_{2}ku_{εn}k^{2}L^{2}(I). (4.25)
whereC_{2} =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∂_{x}^{4}uεndx=−
Z1

0

∂x(uεn∂xuεn)∂_{x}^{3}uεndx

=−

1

Z

0

(∂xu_{εn})^{2}∂_{x}^{3}u_{εn}dx−

1

Z

0

u_{εn}∂_{x}^{2}u_{εn}∂_{x}^{3}u_{εn}dx

= Z1

0

∂x(∂xuεn)^{2}∂_{x}^{2}uεndx− 1
2

Z1

0

uεn∂x(∂_{x}^{2}uεn)^{2}dx

= 5 2

1

Z

0

∂_{x}u_{εn}(∂_{x}^{2}u_{εn})^{2}dx.

Then

q(t)

1

Z

0

u_{εn}∂_{x}u_{εn}∂_{x}^{4}u_{εn}dx

≤ 5β

2 k∂_{x}u_{εn}kL^{∞}(I)k∂_{x}^{2}u_{εn}k^{2}L^{2}(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∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ ε
6

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds

≤ 3 2ε

t

Z

0

kf(s)k^{2}L^{2}(I)ds+5β
2

t

Z

0

k∂_{x}u_{εn}k^{L}^{∞}^{(I}^{)}k∂_{x}^{2}u_{εn}k^{2}L^{2}(I)ds

t t t

Using the injection of H_{0}^{1}(I) in L^{∞}(I) and Poincar´e’s inequality, we obtain
α

2k∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ ε
6

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds

≤

t

Z

0

k∂_{x}^{2}u_{εn}(s)k^{2}L^{2}(I)

5β

2 ku_{εn}(s)k^{2}H^{2}(I)+3ε
2 + C_{2}

4 + 3β^{2}
4ε

ds

+ 3 2ε

t

Z

0

kf(s)k^{2}L^{2}(I)ds.

Observe that f ∈ L^{2}(R). Then, from (4.17), there exists a positive constant C3(ε) such
that

αk∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds

≤C_{3}(ε) +

t

Z

0

5β

2 ku_{εn}(s)k^{2}H^{2}(I)+ 3ε
2 + C_{2}

4 + 3β^{2}
4ε

k∂_{x}^{2}u_{εn}k^{2}L^{2}(I)ds,
so,

k∂_{x}^{2}uεnk^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{4}uεn(s)k^{2}L^{2}(I)ds

≤C3(ε) + Zt

0

5β

2 kuεn(s)k^{2}H^{2}(I)+C4(ε)

k∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ε

s

Z

0

k∂_{x}^{4}u_{εn}(τ)k^{2}L^{2}(I)dτ

ds.

whereC_{4}(ε) = 3ε
2 + C_{2}

4 + 3β^{2}

4ε . Gronwall’s inequality shows that

αk∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds

≤C_{3}(ε) exp

t

Z

0

5β

2 ku_{εn}(s)k^{2}H^{2}(I)+C_{4}(ε)

ds

.

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

αk∂_{x}^{2}u_{εn}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{4}u_{εn}(s)k^{2}L^{2}(I)ds≤K_{2}(ε).

Lemma 4.8. There exists a positive constant K_{3}(ε) independent of n, such that for all
t∈[0, T]

αk∂tuεnk^{2}L^{2}(I)+ε

t

Z

0

k ∂t(∂_{x}^{2}uεn(s)

k^{2}L^{2}(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})∂_{t}u_{εn}dx+

1

Z

0

∂_{t}(q(t)uεn∂_{x}u_{εn})∂_{t}u_{εn}dx

+ Z1

0

∂t ∂_{x}^{3}uεn

∂tuεndx+ Z1

0

∂t(r(t, x)∂xuεn)∂tuεndx

−ε

1

Z

0

∂_{t} ∂_{x}^{2}u_{εn}

∂_{t}u_{εn}dx+ε

1

Z

0

∂_{t} ∂_{x}^{4}u_{εn}

∂_{t}u_{εn}dx

= Z1

0

∂tf ∂tuεndx.

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

αk∂tuεnk^{2}L^{2}(I)+ε
Zt

0

k ∂t(∂_{x}^{2}uεn(s)

k^{2}L^{2}(I)ds

≤ 1

2k∂_{t}fk^{2}L^{2}(R)+β

2ku_{εn}k^{L}^{∞}^{(I)}k∂_{x}u_{εn}k^{2}L^{2}(R)+βk∂_{x}u_{εn}k^{2}L^{2}(R)

+ 1

2 +3β

2 kuεnk^{2}L^{∞}(R)+ 3βk∂xuεnk^{2}L^{∞}(R)

k∂tuεnk^{2}L^{2}(R).
Then, Gronwall’s inequality shows that

αk∂tuεnk^{2} +ε

t

Z

k ∂t(∂^{2}uεn(s)

k^{2} ds ≤K3(ε).

### 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, H^{2}(I)), and ε∂_{x}^{4}u_{εn},∂_{t}u_{εn} are bounded in L^{2}(R).

So, it is possible to extract a subsequence from uεn , still denoted uεn, such that
uεn→uε weakly in L^{2} 0, T, H_{0}^{1}(I)

, (4.27)

uεn→uε strongly in L^{2} 0, T, L^{2}(I)

, (4.28)

ε∂_{x}^{4}uεn→ε∂_{x}^{4}uε weakly in L^{2} 0, T, L^{2}(I)

, (4.29)

∂tuεn→∂tuε, weakly in L^{2} 0, T, L^{2}(I)

. (4.30)

Note that (4.30) implies

T

Z

0 1

Z

0

p(t)∂_{t}u_{εn}wdxdt−→

T

Z

0 1

Z

0

p(t)∂_{t}u_{ε}wdxdt, ∀w∈L^{2}(R).

From (4.27) and (4.28)

u_{εn}∂_{x}u_{εn}−→u_{ε}∂_{x}u_{ε} weakly in L^{2}(R),
then

ZT

0

Z1

0

q(t)uεn∂xuεnwdxdt−→

ZT

0

Z1

0

q(t)uε∂xuεwdxdt, ∀w∈L^{2}(R).

ZT

0

Z1

0

∂_{x}^{3}uεnwdxdt−→

ZT

0

Z1

0

∂_{x}^{3}uεwdxdt, ∀w∈L^{2}(R),
ZT

0

Z1

0

r(t, x)∂_{x}u_{εn}wdxdt−→

ZT

0

Z1

0

r(t, x)∂_{x}u_{ε}wdxdt, ∀w∈L^{2}(R),
and

ε

T

Z

0 1

Z

0

(−∂_{x}^{2}uεn+∂_{x}^{4}uεn)wdxdt−→ε

T

Z

0 1

Z

0

(−∂_{x}^{2}uε+∂_{x}^{4}uε)wdxdt, ∀w∈L^{2}(R).

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

### 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

kuεk^{2}L^{2}(I)+ (α−ε)

t

Z

0

k∂xuε(s)k^{2}L^{2}(I)ds+ε

t

Z

0

k∂_{x}^{2}uε(s)k^{2}L^{2}(I)ds ≤K4,

where α is given in (4.10) and K_{4} 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

u^{2}_{ε}dx+ε

1

Z

0

(∂xu_{ε})^{2}dx+ε

1

Z

0

(∂_{x}^{2}u_{ε})^{2}dx

1

2(∂xu(t,0))^{2}− 1
2

Z1

0

∂xr(t, x)u^{2}_{ε}dx=
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_{ε}k^{2}L^{2}(I)+ε

t

Z

0

k∂_{x}^{2}u_{ε}(s)k^{2}L^{2}(I)ds ≤K_{1}. (4.31)

Now, we multiply both sides of (4.14) by e^{x}u_{ε} and integrate between 0 and 1, to get

p(t)

1

Z

0

e^{x}uε∂tuεdx+

1

Z

0

e^{x}uε∂_{x}^{3}uεdx

−ε

1

Z

0

e^{x}u_{ε}∂_{x}^{2}u_{ε}dx+ε

1

Z

0

e^{x}u_{ε}∂_{x}^{4}u_{ε}dx

1 1 1

(4.32)

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

Z1

0

e^{x}uε∂tuεdx= 1
2

d dt

Z1

0

e^{x}u^{2}_{ε}dx,

1

Z

0

e^{x}u∂_{x}^{3}uεdx=−

1

Z

0

e^{x}uε∂_{x}^{2}uεdx−

1

Z

0

e^{x}∂xu∂_{x}^{2}uεdx

=

1

Z

0

e^{x}u_{ε}∂_{x}u_{ε}dx+

1

Z

0

e^{x}(∂_{x}u_{ε})^{2}dx− 1
2

1

Z

0

e^{x}∂_{x}(∂_{x}u_{ε})^{2}dx

=−1 2

1

Z

0

e^{x}u^{2}_{ε}dx+ 3
2

1

Z

0

e^{x}(∂xuε)^{2}dx+1

2(∂xu(t,0))^{2},

1

Z

0

e^{x}u_{ε}∂_{x}^{2}u_{ε}dx=−

1

Z

0

e^{x}u_{ε}∂_{x}u_{ε}dx−

1

Z

0

e^{x}(∂xu_{ε})^{2}dx

= 1 2

Z1

0

e^{x}u^{2}_{ε}dx−
Z1

0

e^{x}(∂xuε)^{2}dx,

Z1

0

e^{x}uε∂_{x}^{4}uεdx=−
Z1

0

e^{x}uε∂_{x}^{3}uεdx−
Z1

0

e^{x}∂xuε∂_{x}^{3}uεdx

=

1

Z

0

e^{x}u_{ε}∂_{x}^{2}u_{ε}dx+

1

Z

0

e^{x}∂_{x}u_{ε}∂_{x}^{2}u_{ε}dx+

1

Z

0

e^{x}∂_{x}u_{ε}∂^{2}_{x}u_{ε}dx+

1

Z

0

e^{x}(∂_{x}^{2}u_{ε})^{2}dx

=−

1

Z

0

e^{x}uε∂xuεdx−

1

Z

0

e^{x}(∂xuε)^{2}dx+

1

Z

0

e^{x}∂x(∂xuε)^{2}dx+

1

Z

0

e^{x}(∂_{x}^{2}uε)^{2}dx

= 1 2

1

Z

0

e^{x}u^{2}_{ε}dx−2

1

Z

0

e^{x}(∂_{x}u_{ε})^{2}dx+

1

Z

0

e^{x}(∂_{x}^{2}u_{ε})^{2}dx−(∂_{x}u_{ε}(t,0))^{2},

1

Z

0

r(t, x)e^{x}u_{ε}∂_{x}u_{ε}dx= 1
2

1

Z

0

r(t, x)e^{x}∂_{x}(uε)^{2}dx

=−1 2

Z1

0

∂xr(t, x)e^{x}u^{2}_{ε}dx− 1
2

Z1

0

r(t, x)e^{x}u^{2}_{ε}dx,
and

1

Z

0

e^{x}u^{2}_{ε}∂_{x}u_{ε}dx=−1
3

1

Z

0

e^{x}u^{3}_{ε}dx.

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

2 d dt

1

Z

0

e^{x}u^{2}_{ε}dx+
3

2 −ε
Z^{1}

0

e^{x}(∂xuε)^{2}dx

+ (1

2−ε)(∂_{x}u(t,0))^{2}+ε

1

Z

0

e^{x}(∂_{x}^{2}u_{ε})^{2}dx

=

1

Z

0

e^{x}

2 (∂xr(t, x) +r(t, x)+)u^{2}_{ε}dx+q(t)
3

1

Z

0

e^{x}u^{3}_{ε}dx+

1

Z

0

f e^{x}u_{ε}dx.

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

αkuεk^{2}L^{2}(I)+ (3α−2ε)
Zt

0

k∂xuε(s)k^{2}L^{2}(I)ds+ε
Zt

0

k∂_{x}^{2}uε(s)k^{2}L^{2}(I)ds

≤e(3β+ 2ε+ 1) Zt

0

kuε(s)k^{2}L^{2}(I)ds+e
Zt

0

kf(s)k^{2}L^{2}(I)ds+2eβ
3

Zt

0

kuεk^{3}L^{3}(I)ds.

Now, we will get an estimate for the last term in the previous inequality. Since
kuεk^{2}L^{∞}(I) ≤2kuεkL^{2}(I)k∂xuεkL^{2}(I),

then

2eβku k^{3} ≤ 2eβ

ku k ku k^{2}

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=√

2α^{1}^{4}k∂_{x}u_{ε}k

1 2

L^{2}(I),
B = eβ

√2α^{1}^{4}kuεk

5 2

L^{2}(I),
we obtain

eβk∂xuεk

1 2

L^{2}(I)kuεk

5 2

L^{2}(I) ≤αk∂xuεk^{2}L^{2}(I)+3
4

eβ

√2α^{1}^{4}
^{4}_{3}

kuεk

10 3

L^{2}(I).

On the other hand, f ∈L^{2}(R), then there exists a constantC_{5} independent of ε such
that

αku_{ε}k^{2}L^{2}(I)+ 2(α−ε)

t

Z

0

k∂_{x}u_{ε}(s)k^{2}L^{2}(I)ds+ε

t

Z

0

k∂^{2}_{x}u_{ε}(s)k^{2}L^{2}(I)ds

≤C_{5}+e(3β+ 2ε+ 1)

t

Z

0

ku_{ε}(s)k^{2}L^{2}(I)ds

+3 4

eβ

√2α^{1}^{4}
^{4}_{3} Z^{t}

0

ku_{ε}k

4 3

L^{2}(I)ku_{ε}k^{2}L^{2}(I)ds.

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

ϕ(t) = αku_{ε}k^{2}L^{2}(I)+ (α−ε)

t

Z

0

k∂_{x}u_{ε}(s)k^{2}L^{2}(I)ds+ε

t

Z

0

k∂_{x}^{2}u_{ε}(s)k^{2}L^{2}(I)ds

verifies the following inequality

ϕ(t)≤C_{5}+

t

Z

0

C_{6}ku_{ε}k

4 3

L^{2}(I)+C_{7}

ϕ(s) ds,

whereC6 = 3 4

eβ

√2α^{1}^{4}
^{4}_{3}

and C7 =e(3β+ 2ε+ 1).

By the Gronwall’s inequality, we obtain

ϕ(t)≤C_{5}exp

t

Z

C_{6}k∂_{x}u_{ε}k

4 3

L^{2}(I)ds+C_{7}T

.

From (4.31) we have a bound for

t

R

0 ku_{ε}k^{4/3}_{L}^{2}_{(I)}ds, then there exists a positive constant K_{4}
independent ofε such that

kuεk^{2}L^{2}(I)+ (α−ε)
Zt

0

k∂xuε(s)k^{2}L^{2}(I)ds+ε
Zt

0

k∂^{2}_{x}uε(s)k^{2}L^{2}(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∂tuεk^{2}L^{2}(I)+C
Zt

0

k∂t(∂xuε(s))k^{2}L^{2}(I)ds+ε
Zt

0

k∂t(∂_{x}^{2}uε(s))k^{2}L^{2}(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}∂_{t}u_{ε} and inte-
grating between 0 and 1, we obtain

p(t)

1

Z

0

e^{x}∂_{t}u_{ε}∂_{t}^{2}u_{ε}dx+p^{′}(t)

1

Z

0

e^{x}(∂tu_{ε})^{2}dx+

1

Z

0

e^{x}∂_{t}(q(t)uε∂_{x}u_{ε})∂_{t}u_{ε}dx

+

1

Z

0

e^{x}∂t ∂_{x}^{3}uε

∂tuεdx+

1

Z

0

e^{x}∂t(r(t, x)∂xuε)∂tuεdx

−ε

1

Z

0

e^{x}∂_{t} ∂_{x}^{2}u_{ε}

∂_{t}u_{ε}dx+ε

1

Z

0

e^{x}∂_{t} ∂_{x}^{4}u_{ε}

∂_{t}u_{ε}dx

=

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∂_{t}^{2}uεdx= 1
2

d dt

Z1

0

e^{x}(∂tuε)^{2}dx,