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Submitted on 18 Nov 2018
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thermoelasticity and infinite memory
Mouhssin Bayoud, Khaled Zennir, Hocine Sissaoui
To cite this version:
Mouhssin Bayoud, Khaled Zennir, Hocine Sissaoui. Transmission problem with 1 – D mixed type
in thermoelasticity and infinite memory. Applied Sciences (Bucureş.., Balcan Society of Geometers,
2018. �hal-01926000�
thermoelasticity and infinite memory
Mouhssin Bayoud, Khaled Zennir and Hocine Sissaoui
Abstract. This paper describes a polynomial decay rate of solution for a transmission problem with 1 − D mixed type I and type II thermoelastic system with infinite memory acting in the first part. The main contri- butions here are to show that the infinite memory lets our problem still dissipative, and that the system is not exponentially stable, in spite of the kernel in the memory term is sub-exponential. Also we establish that the t
−1is the sharp decay rate. We extend the results in [27].
M.S.C. 2010: 35L05, 35B40, 93D20, 93C20.
Key words: Infinite memory; thermoelastic transmission problem; polynomial decay;
exponential stability; semigroup.
1 Introduction and position of problem
A qualitative studies for problems described the thermo-mechanical interactions in elastic materials has been increasing interest in recent years. The 1 − d linear model of the dynamical problems for classical thermoelastic systems is given by:
(1.1)
u
00− u
xx+ lθ
x= 0, x ∈ (0, L), t > 0, θ
0− θ
xx+ lu
0x= 0, x ∈ (0, L), t > 0,
where u(x, t) denotes the displacement of the rod at time t and θ(x, t) is the tem- perature difference with respect to a fixed reference temperature. This last system is so-called the type I thermoelastic, which is special case when we take k = 0 from the type III given by:
(1.2)
ρu
00− (au
x− lθ)
x= 0, cτ
00+ lu
0x− (βθ
x+ kτ
x)
x= 0.
When β = 0, the following thermoelastic system is named thermoelasticity without dissipation, that is, the energy is conservative (type II):
(1.3)
ρu
00− (au
x− lθ)
x= 0, cτ
00+ lu
0x− kτ
xx= 0.
A
pplied Sciences, Vol.20, 2018, pp. 18-35.c
Balkan Society of Geometers, Geometry Balkan Press 2018.
These last three types were introduced by Green an Naghdi [13]-[14].
In the present paper, we consider a transmission problem with 1 − D mixed type I and type II thermoelastic system and memory term for t > 0 in the following:
(1.4)
ρ
1u
00− a
1u
xx− R
t−∞
µ(t − s)u
xx(s)ds
+ β
1θ
x= 0, x ∈ (−L, 0), c
1w
001− lθ
xx+ β
1u
0x= 0, x ∈ (−L, 0), ρ
2v
00− a
2v
xx+ β
2q
x= 0, x ∈ (0, L), c
2w
002− kw
2,xx+ β
2v
x0= 0, x ∈ (0, L), u(0, t) = v(0, t),
θ(0, t) = q(0, t), w
1(0, t) = w
2(0, t), lθ
x(0, t) = kw
2,x(0, t),
a
1u
x(0, t) − a
2v
x(0, t) = β
1θ(0, t) + β
2q(0, t),
where u, v are the displacement of the system at time t in (−L, 0) and (0, L) and θ, q are respectively the temperature difference with respect to a fixed reference temper- ature, w
1, w
2are the so-called thermal displacement, which satisfies
w
1(., t) = Z
t0
θ(., s)ds + w
1(., 0) and
w
2(., t) = Z
t0
q(., s)ds + w
2(., 0).
The parameters a
1, a
2, ρ
1, ρ
2, β
1, β
2, c
1.c
2, k, l and L < ∞ are assumed to be positive constants.
The system (1.4) satisfies the Dirichlet boundary conditions:
(1.5)
u(−L, t) = v(L, t) = 0, t > 0, w
1(−L, t) = w
2(L, t) = 0, t > 0, and the following initial conditions:
(1.6)
u(.,0) =u0(x), u0(.,0) =u1(x), w1(.,0) =w01(x), θ(.,0) =θ0(x), x∈(−L,0) v(.,0) =v0(x), v0(.,0) =v1(x), w2(.,0) =w02(x), q(.,0) =q0(x), x∈(0, L).
We treat the infinite memory as Dafermos [6], adding a new variable η to the system which corresponds to the relative displacement history. Let us define the auxiliary variable
η = η
t(x, s) = u(x, t) − u(x, t − s), (x, s) ∈ (−L, 0) × R
+. By differentiation, we have
d
dt η
t(x, s) = − d
ds η
t(x, s) + d
dt u(x, t), (x, s) ∈ (−L, 0) × R
+.
We can take as initial condition (t = 0)
η
0(x, s) = u
0(x) − u(x, −s), (x, s) ∈ (−L, 0) × R
+. Thus, the original memory term can be rewritten as follows
R
t−∞
µ(t − s)u
xx(s)ds = R
∞0
µ(s)u
xx(t − s)ds
= R
∞0
µ(t)dt
u
xx− R
∞0
µ(s)η
xxt(s)ds.
The problem (1.4) is transformed into the system
(1.7)
ρ
1u
00− a
1µ
0u
xx+ R
∞0
µ(s)η
txx(s)ds
+ β
1θ
x= 0, x ∈ (−L, 0), c
1w
100− lθ
xx+ β
1u
0x= 0, x ∈ (−L, 0), ρ
2v
00− a
2v
xx+ β
2q
x= 0, x ∈ (0, L), c
2w
200− kw
2,xx+ β
2v
x0= 0, x ∈ (0, L),
d
dt
η
t(x, s) +
dsdη
t(x, s) −
dtdu(x, t) = 0, x ∈ (−L, 0), u(0, t) = v(0, t),
θ(0, t) = q(0, t), w
1(0, t) = w
2(0, t), lθ
x(0, t) = kw
2,x(0, t),
a
1u
x(0, t) − a
2v
x(0, t) = β
1θ(0, t) + β
2q(0, t), η
0(x, s) = u
0(x, 0) − u
0(x, −s), s > 0, where µ
0= 1 − R
∞0
µ(t)dt.
The stability of various transmission problems on thermoelasticity have been con- sidered [8], [11], [20], [21], [22] and [25]. Without infinite memory, it is proved in [27]
that the energy of system (1.4) cannot achieve exponential decay rate. This paper is devoted to show that our system can achieve polynomial decay rate. That is, our main result here is to show that for these types of materials the dissipation produced by the viscoelastic part is not strong enough to produce an exponential decay of the solution despite that the infinite memory satisfies assumptions (3.1) and (3.2).
2 Previous results and stability
The transmission problem to hyperbolic equations was studied by Dautray and Lions [7], where the existence and regularity of solutions for the linear problem have been proved. In [21], the authors considered the transmission problem of viscoelastic waves (2.1)
ρ
1u
00− α
1u
xx= 0, x ∈ (0, L
0), ρ
2v
00− α
2v
xx+ R
t0
g(t − s)v
xx(s)ds = 0, x ∈ (L
0, L),
satisfying boundary conditions and initial conditions. The authors studied the wave
propagations over materials consisting of elastic and viscoelastic components. They
showed that the viscoelastic part produce exponential decay of the solution. In [18],
the authors investigated a 1D semi-linear transmission problem in classical thermoe-
lasticity and showed that a combination of the first, second and third energies of the
solution decays exponentially to zero. Marzocchi et al [19] studied a multidimen- sional linear thermoelastic transmission problem. An existence and regularity result has been proved. When the solution is supposed to be spherically symmetric, the authors established an exponential decay result similar to [18]. Next, Rivera and all [22], considered a transmission problem in thermoelasticity with memory. As time goes to infinity, they showed the exponential decay of the solution in case of radi- ally symmetric situations. We must mention the pioneer work by Rivera and all in [11], where a semilinear transmission problem for a coupling of an elastic and a ther- moelastic material is considered. The heat conduction is modeled by Cattaneo’s law removing the physical paradox of infinite propagation speed of signals. The damped, totally hyperbolic system is shown to be exponentially stable. In 2009, Mesaoudi and all [20] proposed and studied a 1D linear thermoelastic transmission problem, where the heat conduction is described by the theories of Green and Naghdi. By using the energy method, they proved that the thermal effect is strong enough to produce an exponential stability of the solution. The earliest result in this direction was established in [27], where the dynamical behavior of the system is described by
(2.2)
ρ
1u
001− a
1u
1,xx+ β
1θ
1,x= 0, x ∈ (−1, 0), c
1τ
100− bθ
1,xx+ β
1u
01,x= 0, x ∈ (−1, 0), ρ
2u
002− a
2u
2,xx+ β
2θ
2,x= 0, x ∈ (0, 1), c
2τ
200− kτ
2,xx+ β
2u
02,x= 0, x ∈ (0, 1).
The system consists of two kinds of thermoelastic components, one is of type I, another one is of type II. Under certain transmission conditions, these two components are coupled at the interface. The authors proved that the system is lack of exponential decay rate and they obtain the sharp polynomial decay rate.
3 Preliminaries
For simplicity reason denote u(x, t) = u, v(x, t) = v, w
i(x, t) = w
i, i = 1, 2, q(x, t) = q, when there is no confusion. Here u
0= du(t)/dt, v
0= dv(t)/dt and u
00= d
2u(t)/dt
2, v
00= d
2v(t)/dt
2, w
00i= d
2w
i(t)/dt
2, i = 1, 2.
First we recall and make use the following assumptions on the functions µ:
We assume that the function µ : R
+−→ R
+is of class C
1satisfying:
(3.1) 1 −
Z
∞ 0µ(t)dt = µ
0> 0, ∀t ∈ R
+, and that there exists a constants k
1> 0 such that (3.2) µ
0(t) + k
1µ(t) ≤ 0 ∀t ∈ R
+.
We denote by A the unbounded operator in an appropriate Hilbert state space Let
V
k(0, L) = {h ∈ H
k(0, L); h(L) = 0}.
V
k(−L, 0) = {h ∈ H
k(−L, 0); h(−L) = 0},
H = V
1(−L, 0) × L
2(−L, 0) × L
2(−L, 0) × V
1(0, L) × L
2(0, L) × V
1(0, L) × L
2(0, L),
equipped, for (u, u
1, θ, v, v
1, w
2, q), (˜ u, u ˜
1, θ, ˜ ˜ v, v ˜
1, w ˜
2, q) ˜ ∈ H, with an inner product D (u, u
1, θ, v, v
1, w
2, q), (˜ u, u ˜
1, θ, ˜ ˜ v, v ˜
1, w ˜
2, q) ˜ E
H
= Z
0−L
h a
1µ
0u
x+ Z
t0
µ(s)η
xt(s)ds
˜
u
x+ ρ
1u
1u ˜
1+ c
1θ θ ˜ i dx
+ Z
L0
h
a
2v
xv ˜
x+ ρ
2v
1v ˜
1+ kw
2,xw ˜
2,x+ c
2q
xq ˜
xi dx.
with domain
D(A) = (u, u
1, θ, v, v
1, w
2, q) ∈ H :
u, θ ∈ H
2(−L, 0), u
1∈ H
1(−L, 0),
v ∈ H
2(0, L), v
1, q ∈ H
1(0, L), w
2∈ H
2(0, L), θ(−L) = q(L) = 0, lθ
x(0) = kw
2,x(0)
a
1µ
0u
x(0) − β
1θ(0) = a
2v
x(0) − β
2q(0) u(0) = v(0), θ(0) = q(0),
(3.3)
and
(3.4) A
u u
1θ v v
1w
2q
=
u
1ρ
−11a
1µ
0u
xx+ R
∞0
µ(s)η
txx(s)ds
− β
1θ
xc
−11− β
1u
1x+ lθ
xxv
1ρ
−12a
2v
xx− β
2q
xq
c
−12− β
2v
x1+ kw
2,xx
For U = (u, u
1, θ, v, v
1, w
2, q)
T, the problem (1.7) can be reformulated in the abstract from
(3.5) U
0= AU ,
where U (0) = (u
0, u
1, θ
0, v
0, v
1, w
02, q
0)
T∈ H is given.
We will use necessary and sufficient conditions for C
0-semigroups being exponentially stable in a Hilbert space. This result was obtained by Gearhart [12] and Huang [10]
Theorem 3.1. Let S(t) = e
Atbe a C
0-semigroup of contractions on Hilbert space.
Then S (t) is exponentially stable if and only if
ρ(A) ⊇ {iζ : ζ ∈ R } ≡ i R and
lim
|ζ|→∞
k(iζI − A)
−1k
L(H)< ∞.
4 Lack of Exponential Stability
Following the techniques in [2], it is easy to check that (H, k.k
H) is a Hilbert space.
In this section we prove the lack of exponential decay using Theorem 3.1, that is we show that there exists a sequence of values h
msuch that
(4.1) k(ih
mI − A)
−1k
L(H)→ ∞.
It is equivalent to prove that there exist a sequence of data F
m∈ H and a sequence of real numbers h
m∈ R , with kF
mk
H≤ 1 such that
(4.2) k(ih
mI − A)
−1F
mk
H= kU
mk
2H→ ∞.
Theorem 4.1. Assume that the kernel is of the form µ(s) = e
−hs, s ∈ R
+, with h > 1. The semi group S(t) on H is not exponentially stable.
Proof. As in [1], we will find a sequence of bounded functions
F
m= (f
1,m, f
2,m, f
3,m, f
4,m, f
5,m, f
6,m, f
7,m, f
8,m)
T∈ H, h ∈ R ,
for which the corresponding solutions of the resolvent equations is not bounded. This will prove that the resolvent operator is not uniformly bounded. We consider the spectral equation
ihU
m− AU
m= F
m.
and show that the corresponding solution U
mis not bounded when F
mis bounded in H. Rewriting the spectral equation in term of its components, we get
ihu − u
1= f
1mihρ
1u
1−
a
1µ
0u
xx+ R
∞0
µ(s)η
txx(s)ds
− β
1θ
x= ρ
1f
2mihc
1θ −
− β
1u
1x+ lθ
xx= c
1f
3mihv − v
1= f
4mihρ
2v
1−
a
2v
xx− β
2q
x= ρ
2f
5mihw
2− q = f
6mihc
2q −
− β
2v
x1+ kw
2,xx= c
2f
7mihη
t− u
1+ η
ts= f
8m.
(4.3)
We prove that there exists a sequence of real numbers h
mso that (4.3) verified. Let us consider f
1m= f
4m= f
6m= f
8m= 0. We eliminate the terms u
1, v
1. We can choose f
2m= f
3m= f
5m= f
6m= λ
mand we obtain u
1= ihu, v
1= ihv and q = ihw
2. Then, the system (4.3) takes the form
(4.4)
−h
2u − ρ
−11a
1µ
0u
xx+ R
∞0
µ(s)η
txx(s)ds
− β
1θ
x= λ
mihθ − c
−11− β
1u
1x+ lθ
xx= λ
m−h
2v − ρ
−12a
2v
xx− β
2ihw
2,x= λ
m−h
2w
2− c
−12− β
2v
1x+ kw
2,xx= λ
mihη
t− ihu + η
st= 0
We look for solutions of the form
u = aλ
m, v = bλ
m, θ = cλ
m, w
2= dλ
m, u
1= eλ
m, v
1= f λ
m, η
t(x, s) = γ(s)λ
mwith a, b, c, d, e, f ∈ C and γ(s) depend on h and will be determined explicitly in what follows. From (4.4), we get a, b, c, d, e and f satisfy
(4.5)
−h
2a − ρ
−11a
1h
mµ
0a + R
∞0
µ(s)γ(s)ds
− β
1ch
= 1, ihc − c
−11− β
1e + lh
mc
= 1,
−h
2b − ρ
−12a
2h
mb − β
2ihd
= 1, ihd − c
−12− β
2f + kh
md
= 1, γ
s+ ihγ − iha = 0.
From (4.5)
5we get
(4.6) γ(s) = a − ae
−ihs.
Then, from (4.6) we have Z
∞0
µ(s)γ(s) ds = Z
∞0
µ(s)(a − ae
−ihs)ds
= a
Z
∞ 0µ(s)ds − a Z
∞0
µ(s)ae
−ihsds
= a(1 − µ
0) − a Z
∞0
µ(s)e
−ihsds.
(4.7)
Now, we would like to find the parameters constants. To this end, we choose c
1ih = h
ml, c
2ih = kh
m,
(4.8)
and using the equations (4.5)
2and (4.5)
4, we obtain e = c
1β
1, (4.9)
f = c
2β
2. (4.10)
We choose −h
2ρ
2= a
2h
m. By equations (4.5)
1and (4.5)
3, we have
c =
1(−h2−ρ−11 hma1µ0)
1 + ρ
−11h
ma
1R
∞0
µ(s)γ(s)ds − ρ
−11h
mβ
1c , d =
βρ22ih
. Since c
2l = c
1k, recalling from (4.9), (4.10) that
u
1+ v
1= eλ
m+ f λ
m= c
1β
1λ
m+ c
2β
2λ
m,
we get
ku
1k
22+ kv
1k
22= h c
1β
1 2+ c
2β
2 2i h
2m.
Therefore we have
m→∞
lim kU
mk
2H≥ lim
m→∞
[ku
1k
22+ kv
1k
22]
= lim
m→∞
h c
1β
1 2+ c
2β
2 2i h
2m= +∞
which completes the proof.
5 Polynomial Stability
Our main result reads as follows.
Theorem 5.1. Assume that (3.1) and (3.2) hold. Then t
−1is the sharp decay rate.
Therefore there exists positive constant C such that the solution of our system satisfies
(5.1) E(t) ≤ C
t , ∀t ∈ R
+.
Proof. We will follow the idea for the proof of the corresponding results in [27]. We would show that
ζ→∞
lim k(iζI − A)
−1k < ∞ (5.2)
We prove that there exist a sequence
V
n= (u
n, u
1n, θ
n, v
n, v
1n, w
2,n, q
n) ∈ D(A), with kV
nk
H= 1, and a sequence ζ
n∈ R with ζ
n→ ∞ such that
n→∞
lim ζ
nk(iζ
nI − A)V
nk
H= 0
or
ζ
n(iζ
nu
n− u
1n) → 0, in H
1(−L, 0), (5.3)
ζ
niζ
nu
1n− ρ
−11a
1µ
0u
n,xx+ Z
∞0
µ(s)η
n,xxt(s)ds
− β
1θ
n,x→ 0, in L
2(−L, 0), (5.4)
ζ
niζ
nθ
n− c
−11− β
1u
1n,x+ lθ
n,xx→ 0, in L
2(−L, 0), (5.5)
ζ
n(iζ
nv
n− v
n1) → 0, in H
1(0, L), (5.6)
ζ
niζ
nv
1n− ρ
−12a
2v
n,xx− β
2q
n,x→ 0, in L
2(0, L), (5.7)
ζ
n(iζ
nw
2,n− q
n) → 0, in H
1(0, L), (5.8)
ζ
niζ
nq
n− c
−12− β
2v
n,x1+ kw
2,n,xx→ 0, in L
2(0, L), (5.9)
ihη
t− u
11,n+ η
ts= 0 (5.10)
Note that
Rehζ
n(iζ
n− A)V
n, V
ni
H= ζ
nk √
l θ
n,xk
2L2→ 0.
Then
p ζ
nθ
n,x→ 0, in L
2(−L, 0).
(5.11)
By Poincar´ e’s inequality, we get
p ζ
nθ
n→ 0, in L
2(−L, 0).
(5.12)
Thanks to the Gagliardo-Nirenberg inequality, we have k p
ζ
nθ
nk
L∞≤ C
1q
k p
ζ
nθ
n,xk
L2q k p
ζ
nθ
nk
L2+ C
2k p
ζ
nθ
nk
L2. (5.13)
Thus,
p ζ
nθ
n(0) → 0.
(5.14)
From (5.3), we have β
1(iζ
n)
−1u
1n,xis bounded in L
2(−L, 0). By (5.5) we have the boundedness of (iζ
n)
−1θ
n,xxin L
2(−L, 0).
Using again the Gagliardo-Nirenberg inequality, we have k qp
ζ
n−1θ
n,xk
L∞≤ d
1q
k(ζ
n)
−1θ
n,xxk
L2q k p
ζ
nθ
n,xk
L2+ d
2k qp ζ
n−1θ
n,xk
L2→ 0.
which gives
qp ζ
n−1θ
n,x(−L) → 0, qp ζ
n−1θ
n,x(0) → 0.
(5.15)
Multiplying (5.4) by p(x)u
n,xin L
2− norm for p(x) ∈ C
1[−L, 0], we get
−ζ
n2hu
n, p(x)u
n,xi
L2(−L,0)− ρ
−11a
1D
µ
0u
n,xx, p(x)u
n,xE
L2(−L,0)
−ρ
−11a
1D Z
∞ 0µ(s)η
tn,xx(s)ds, p(x)u
n,xE
L2(−L,0)
+ρ
−11β
1hθ
n,x, p(x)u
n,xi
L2(−L,0)→ 0.
(5.16)
Integration by parts gives
−ζ
n2hu
n, p(x)u
n,xi
L2(−L,0)= ζ
n2p(−L)|u
n(−L)|
2− ζ
n2p(0)|u
n(0)|
2+ ζ
n2D
p
x(x)u
n, u
nE
L2(−L,0)
−ρ
−11a
1µ
0D
u
n,xx, p(x)u
n,xE
L2(−L,0)
= −ρ
−11a
1µ
0p(0)|u
n,x(0)|
2+ ρ
−11a
1µ
0p(−L)|u
n,x(−L)|
2+ ρ
−11a
1µ
0D
p
x(x)u
n,x, u
n,xE
L2(−L,0)
and
−ρ
−11a
1D Z
∞ 0µ(s)η
n,xxt(s)ds, p(x)u
n,xE
L2(−L,0)
= −ρ
−11a
1p(0) Z
∞0
µ(s)η
n,xt(0, s)dsu
n,x(0)
+ ρ
−11a
1p(−L) Z
∞0
µ(s)η
n,xt(−L, s)dsu
n,x(−L) + ρ
−11a
1D
p
x(x) Z
∞0
µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
. Since
ρ
−11β
1hθ
n,x, p(x)u
n,xi
L2(−L,0)→ 0,
then by the above integrations, for p(x) = x ∈ C
1[−L, 0], (5.16) takes the form
−ζ
n2|u
n(−L)|
2+ ζ
n2D u
n, u
nE
L2(−L,0)
−ρ
−11a
1µ
0|u
n,x(−L)|
2+ ρ
−11a
1µ
0D
u
n,x, u
n,xE
L2(−L,0)
−ρ
−11a
1Z
∞ 0µ(s)η
tn,x(−L, s)dsu
n,x(−L) + ρ
−11a
1D Z
∞ 0µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
→ 0,
(5.17)
and hence, u
n,x(−L) and ζ
nu
n(−L) are bounded.
Similarly, taking p(x) = x + L ∈ C
1[−L, 0], (5.16) takes the form
−ζ
n2|u
n(0)|
2+ ζ
n2D u
n, u
nE
L2(−L,0)
−ρ
−11a
1µ
0|u
n,x(0)|
2+ ρ
−11a
1µ
0D
u
n,x, u
n,xE
L2(−L,0)
−ρ
−11a
1Z
∞ 0µ(s)η
tn,x(0, s)dsu
n,x(0)
+ ρ
−11a
1D Z
∞ 0µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
→ 0.
(5.18)
Then, we get boundedness of ζ
nu
n(0) and u
n,x(0).
Multiplying (5.5) by u
n,xand taking the integration, we get iζ
nD θ
n, u
n,xE
L2(−L,0)
+ c
−11β
1D
u
1,n,x, u
n,xE
L2(−L,0)
− c
−11l D
θ
n,xx, u
n,xE
L2(−L,0)
→ 0.
By (5.12), after dividing by i √
ζ
n, we have, where we have used ζ
n> 0 iζ
nD
θ
n, u
n,xE
L2(−L,0)
→ 0 Integrating by parts, we get
l(i p ζ
n)
−1θ
n,x(−L)u
n,x(−L) − θ
n,x(0)u
n,x(0) + l Dp
ζ
nθ
n,x, (iζ
n)
−1u
n,xxE
L2(−L,0)
+β
1p ζ
nD
u
1,n,x, u
n,xE
L2(−L,0)
→ 0 (5.19)
By (5.15) and the boundedness of u
n,x(−L) and u
n,x(0), we have l(i p
ζ
n)
−1θ
n,x(−L)u
n,x(−L) − θ
n,x(0)u
n,x(0)
→ 0
Moreover, from (5.4), we obtain that (iζ
n)
−1u
n,xxis bounded in L
2(−L, 0). Thus l( p
ζ
nθ
n,x, (iζ
n)
−1u
n,xx) → 0 Hence by (5.19), we get
(5.20)
q p
ζ
nu
n,x→ 0, in L
2(−L, 0).
Thanks to the Poincar´ e inequality, we have (5.21)
q p
ζ
nu
n→ 0, in L
2(−L, 0) By (5.20), (5.21) and Galiardo-Nirenberg inequality, we get
(5.22)
q p
ζ
nu
n(0) → 0.
From (5.4) and (5.11), using ζ
n> 0, we have iζ
nu
1,n− ρ
−11a
1µ
0u
n,xx+ Z
∞0
µ(s)η
n,xxt(s)ds
→ 0, in L
2(−L, 0), (5.23)
Multiplying the above by u
n, we get iζ
nD
u
1,n, u
nE
L2(−L,0)
− ρ
−11a
1D
µ
0u
n,xx+ Z
∞0
µ(s)η
tn,xx(s)ds , u
nE
L2(−L,0)
→ 0.
Integrating by parts, we get
− D
u
1,n, u
1,nE
L2(−L,0)
−ρ
−11a
1µ
0u
n,x(0)u
n(0) + ρ
−11a
1µ
0u
n,x(−L)u
n(−L) − ρ
−11a
1µ
0D
u
n,x, u
n,xE
L2(−L,0)
+ρ
−11a
1Z
∞ 0µ(s)η
n,xt(0, s)dsu
n(0) − ρ
−11a
1Z
∞ 0µ(s)η
n,xt(−L, s)dsu
n(−L) +ρ
−11a
1D Z
∞ 0µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
→ 0.
Since u
n,x(0), u
n,x(−L) are bounded, by (5.20) and u
n(−L) → 0, u
n(0) → 0, we have (5.24) u
1,n, ζ
nu
n→ 0, in L
2(−L, 0).
Multiplying (5.4) by (x + L)u
n,x, we get the real part as follows 2< h
− D
ζ
n2u
1,n, (x + L)u
n,xE
L2(−L,0)
−ρ
−11a
1D
µ
0u
n,xx+ Z
∞0
µ(s)η
n,xxt(s)ds
, (x + L)u
n,xE
L2(−L,0)
i
= −ζ
n2|u
n(0)|
2+ ζ
n2D u
n, u
nE
L2(−L,0)
− ρ
−11a
1µ
0|u
n,x(0)|
2+ ρ
−11a
1µ
0D
u
n,x, u
n,xE
L2(−L,0)
−ρ
−11a
1Z
∞ 0µ(s)η
n,xt(0, s)dsu
n,x(0) + ρ
−11a
1D Z
∞ 0µ(s)η
tn,x(s)ds, u
n,xE
L2(−L,0)
→ 0.
Hence, by (5.20) and (5.24), we get
(5.25) ζ
nu
n(0), u
n,x(0) → 0
Now, multiplying (5.4) by xu
n,x, we get the real part as follows 2< h
− D
ζ
n2u
1n, xu
n,xE
L2(−L,0)
(5.26)
−ρ
−11a
1D
µ
0u
n,xx+ Z
∞0
µ(s)η
n,xxt(s)ds , xu
n,xE
L2(−L,0)
i
= −ζ
n2|u
n(−L)|
2+ ζ
n2D u
n, u
nE
L2(−L,0)
− ρ
−11a
1µ
0|u
n,x(−L)|
2+ ρ
−11a
1µ
0D
u
n,x, u
n,xE
L2(−L,0)
−ρ
−11a
1Z
∞ 0µ(s)η
n,xt(−L, s)dsu
n,x(−L) + ρ
−11a
1D Z
∞ 0µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
→ 0.
Then
(5.27) ζ
nu
n(−L), u
n,x(−L) → 0.
Taking again (5.4), multiplying by u
n, we have p ζ
nD
iζ
nu
11, n, u
nE
L2(−L,0)
+ ρ
−11p ζ
nβ
1D θ
n,x, u
nE
L2(−L,0)
−ρ
−11p ζ
na
1µ
0D
u
n,xx, u
nE
L2(−L,0)
−ρ
−11p ζ
na
1D Z
∞ 0µ(s)η
n,xxt(s)ds, u
nE
L2(−L,0)
→ 0.
(5.28)
By (5.20) and (5.25), we have
−ρ
−11p
ζ
na
1µ
0D
u
n,xx, u
nE
L2(−L,0)
= −ρ
−11a
1µ
0p
ζ
nu
n,x(0)u
n(0) + ρ
−11a
1µ
0p
ζ
nu
n,x(−L)u
n(−L) +ρ
−11a
1µ
0p ζ
nD
u
n,x, u
n,xE
L2(−L,0)
→ 0 and
−ρ
−11p ζ
na
1D Z
∞ 0µ(s)η
tn,xx(s)ds, u
nE
L2(−L,0)
= −ρ
−11a
1µ
0p ζ
nZ
∞ 0µ(s)η
n,xt(0, s)dsu
n(0)
+ρ
−11a
1µ
0p ζ
nZ
∞ 0µ(s)η
n,xt(−L, s)dsu
n(−L) +ρ
−11a
1µ
0p
ζ
nD Z
∞ 0µ(s)η
tn,x(s)ds, u
n,xE
L2(−L,0)
→ 0.
(5.29)
Thus, by(5.29) and (5.11), we go to (5.30)
q p
ζ
nu
1n→ 0, in L
2(−L, 0).
Multiplying (5.4) by (x + L)u
n,x, we have D
i p
ζ
nζ
nu
1n, (x + L)u
n,xE
L2(−L,0)
+ ρ
−11p ζ
nβ
1D
θ
n,x, (x + L)u
n,xE
L2(−L,0)
−ρ
−11p ζ
na
1µ
0D
u
n,xx, (x + L)u
n,xE
L2(−L,0)
−ρ
−11p ζ
na
1D Z
∞ 0µ(s)η
n,xxt(s)ds, (x + L)u
n,xE
L2(−L,0)
→ 0.
(5.31)
Integrating by parts and using (5.11) and the boundedness of u
n,xin L
2(−L, 0), we get
− p
ζ
n|u
1n(0)|
2+ p ζ
nD
u
1n, u
1nE
L2(−L,0)
− ρ
−11a
1µ
0p
ζ
n|u
n,x(0)|
2+ρ
−11a
1µ
0p ζ
nD
u
n,x, u
n,xE
L2(−L,0)
−ρ
−11a
1p ζ
nZ
∞ 0µ(s)η
tn,x(0, s)dsu
n,x(0)
−ρ
−11a
1p ζ
nZ
0−L
D Z
∞ 0µ(s)η
tn,x(s)ds, u
n,xE
L2(−L,0)
→ 0.
(5.32)
Thus by (5.20) and (5.30), we go to (5.33)
q p
ζ
nu
1n(0), q p
ζ
nu
n,x(0) → 0 Multiplication of (5.23) by u
n,xyields
iζ
nD u
1n, u
n,xE
L2(−L,0)
− ρ
−11a
1µ
0D
u
n,xx, u
n,xE
L2(−L,0)
−ρ
−11a
1D Z
∞ 0µ(s)η
n,xxt(s)ds, u
n,xE
L2(−L,0)
→ 0.
(5.34)
Due to (5.25)and (5.27), we get
−ρ
−11a
1µ
0D
u
n,xx, u
n,xE
L2(−L,0)
− ρ
−11a
1D Z
∞ 0µ(s)η
n,xxt(s)ds, u
n,xE
L2(−L,0)
= 1
2 (−ρ
−11a
1µ
0)|u
n,x(0)|
2+ ρ
−11a
1µ
0)|u
n,x(−L)|
2−ρ
−11a
1Z
∞ 0µ(s)η
n,xt(0, s)dsu
n,x(0) + ρ
−11a
1Z
∞ 0µ(s)η
n,xt(−L, s)dsu
n,x(−L) +ρ
−11a
1D Z
∞ 0µ(s)η
n,xt(s)ds, u
n,xE
L2(−L,0)
→ 0.
(5.35)
Thus, it follows from (5.34) that
(5.36) (iζ
nu
1n, u
n,x) → 0.
Taking the product of (5.23) with θ
n, yields iζ
nD u
1,n, θ
nE
L2(−L,0)
− ρ
−11a
1µ
0D
u
n,xx, θ
nE
L2(−L,0)
−ρ
−11a
1D Z
∞ 0µ(s)η
tn,xx(s)ds, θ
nE
L2(−L,0)
→ 0 in L
2(−L, 0).
(5.37)
Due to (5.11),(5.14) and (5.25), we have
−ρ
−11a
1µ
0D
u
n,xx, θ
nE
L2(−L,0)
= −ρ
−11a
1µ
0u
n,x(0)θ
n(0) + ρ
−11a
1µ
0u
n,x(−L)θ
n(−L) +ρ
−11a
1µ
0D
u
n,x, θ
n,xE
L2(−L,0)
→ 0
(5.38)
and
−ρ
−11a
1D Z
∞ 0µ(s)η
tn,xx(s)ds, θ
nE
L2(−L,0)
= −ρ
−11a
1Z
∞0
µ(s)η
n,xt(0, s)dsθ
n(0) +ρ
−11a
1Z
∞ 0µ(s)η
tn,x(−L, s)dsθ
n(−L) +ρ
−11a
1D Z
∞ 0µ(s)η
tn,x(s)ds, θ
n,xE
L2(−L,0)
→ 0.
(5.39)
Then from (5.37), we obtain iζ
nD u
1n, θ
nE
L2(−L,0)
→ 0 (5.40)
Multiplying (5.5) by u
1n, we have (5.41) D
iζ
nθ
n, u
1nE
L2(−L,0)
−c
−11l D
θ
n,xx, u
1nE
L2(−L,0)
+c
−11β
1D
u
1n,x, u
1nE
L2(−L,0)
→ 0.
By (5.36), (5.40), we have
(5.42) D
θ
n,xx, u
1nE
L2(−L,0)
→ 0.
Integrating by parts
(5.43) θ
n,x(0)u
1n(0) − θ
n,x(−L)u
1n(−L) − D
θ
n,x, u
1n,xE
L2(−L,0)
→ 0.
Due to (5.15)and (5.33), we get
(5.44) θ
n,x(0)u
1n(0) − θ
n,x(−L)u
1n(−L) → 0.
From (5.43) we have
(5.45) D
θ
n,x, u
1n,xE
L2(−L,0)
→ 0.
Multiplying (5.5) by (x + L)θ
n,xand integrating, we get
< hD
iζ
nθ
n, (x + L)θ
n,xE
L2(−L,0)
− c
−11D
(lθ
n,xx− β
1u
1n,x), (x + L)θ
n,xE
L2(−L,0)
i → 0 (5.46)
By (5.11) and (5.12), we obtain
(5.47) D
iζ
nθ
n, (x + L)θ
n,xE
L2(−L,0)
→ 0.
Thus by (5.46) and (5.11), we have
(5.48) −c
−11lθ
n,x(0)θ
n,x(0) + 2<[c
−11β
1(u
1n,x, (x + L)θ
n,x)] → 0.
Then, by (5.45), we get
(5.49) θ
n,x(0) → 0
Hence, by (5.35),(5.25),(5.14)and (5.49), we have
(5.50) u
n,x(0), u
n(0), θ
n(0), θ
n,x(0) → 0.
Taking the product of (5.9) with (x − L)w
2,n,x, yields
< h iζ
nD
q
n, (x − L)w
2,n,xE
L2(0,L)
+ c
−12β
2D
v
1n,x, (x − L)w
2,n,xE
L2(0,L)
−c
−12k D
w
2,n,xx, (x − L)w
2,n,xE
L2(0,L)
i → 0.
(5.51)
Using the transmission conditions in (1.4), we get (5.52) (q
n, q
n) + c
−12k(w
2,n,x, w
2,n,x) − 2< h
c
−12β
2D
v
n,x, (x − L)q
n,xE
L2(0,L)
i → 0.
Taking the product of (5.7) with (x − L)v
n,x, we obtain iζ
nD
v
1n, (x − L)v
n,xE
L2(0,L)
− ρ
−12a
2D
v
n,xx, (x − L)v
n,xE
L2(0,L)
+ρ
−12β
2D
q
n,x, (x − L)v
n,xE
L2(0,L)
→ 0.
(5.53)
Integrating (5.53) by parts we have D
v
1n, v
1nE
L2(0,L)
+ ρ
−12a
2D
v
n,x, v
n,xE
L2(0,L)
+2< h ρ
−12β
2D
q
n,x, (x − L)q
n,xE
L2(0,L)
i → 0.
(5.54)
Thus by (5.52) and (5.54), we obtain a
2D
v
n,x, v
n,xE
L2(0,L)
+ D
ρ
2v
1n, v
1nE
L2(0,L)
+ k D
w
2,n,x, w
2,n,xE
L2(0,L)
+c
2D q
n, q
nE
L2(0,L)
