PORTUGALIAE MATHEMATICA
Vol. 56 Fasc. 4 – 1999
KIRCHHOFF–CARRIER ELASTIC STRINGS
IN NONCYLINDRICAL DOMAINS
J. Limaco Ferrel and L.A. Medeiros
In Memoriam Yukiyoshi Ebihara
Abstract: The existence and uniqueness of local and global solutions for the
Kirchhoff–Carrier nonlinear model for the vibrations of elastic strings in noncylindrical
domains are investigated by means of the Galerkin Method. The asymptotic behaviour
of the energy is also studied.
1 – Introduction
Let α : [0, T ] → R and β : [0, T ] → R, be two, twice continuously differentiable
functions such that:
α(t) < β(t)
for all 0 ≤ t ≤ T .
b contained in R2 , defined by:
We consider the noncylindrical domain Q,
n
b = (x, t) ∈ R2 : α(t) < x < β(t), for all 0 < t < T
Q
b of Q
b is given by
The lateral boundary Σ
b =
Σ
[
0<t<T
o
.
{α(t), β(t)} × {t} .
In this work we investigate the existence, uniqueness and the asymptotic behaviour of solutions of the following noncylindrical mixed problem, for Kirchhoff–
Carrier elastic strings:
Received : January 12, 1998; Revised : February 18, 1998.
AMS Subject Classification: 35F30, 35K35.
Keywords: Kirchhoff–Carrier, Strings, Noncylindrical, Global solutions, Asymptotic
behaviour.
466
(1.1)
J. LIMACO FERREL and L.A. MEDEIROS

¶ 2
¶
µZ β(t) µ

∂2u
∂ u
∂u 2

b

dx
−M
= f (x, t) in Q

2

∂t
∂x
∂x2

α(t)

b
 u = 0 on Σ




∂u

 u(x, 0) = u0 (x),
(x, 0) = u1 (x) in α(0) < x < β(0) .
∂t
In the bibliography at the end of this article a complete list of papers dealing
with the Kirchhoff–Carrier operator
∂2w
Lw =
−M
∂t2
µZ
2
Ω
¶
|∇w| dx ∆w
in a cylinder can be found. Regular global solutions are obtained in Arosio–
Spagnolo [1], Lions [13] and Pohozhaev [16]; local weak solutions in Ebihara–
Medeiros–Miranda [7] for the degenerate case, i.e. M (s) ≥ 0, cf. also Yamada
[17]. For perturbated Kirchhoff–Carrier operators see Arosio–Garivaldi [2].
We refer to Bainov and Minchev [6] for estimates on the interval of existence
of local solutions. Variational inequalities for Lu are discussed in Frota–Larkin
[8]. In these references complete information is given about the operator Lu.
For global existence in the cylindrical case we refer to Brito [4], Hosya–Yamada
[9] and Kouemou Patcheu [12]. Note that in this case initial data are chosen
inside a fixed ball. In the present work we also need to choose the initial data
satisfying analogous restrictions (see Section 6, Theorem 6.1, condition (6.10)).
In Nakao–Nakazaki [14], existence and decay for solutions of the nonlinear wave
equation, in noncylindrical domains for the d’Alembert operator 2 u = utt − ∆u
is investigated. They employed the penalty method as in Lions [14]. In the work
of Komornik–Zuazua [11], for d’Alembert operator with dissipative nonlinear
conditions on part of the boundary a method to study the asymptotic behaviour
of the energy was introduced. We adopt this method here to our case.
Some results for noncylindrical domains in dimensions n ≥ 2 will be published
elsewhere.
The plan of this work is as follows:
1. Introduction
2. Notations, Assumptions and Local Results
3. Approximations and Estimates
4. Proof of the Theorems
5. Applications
6. Global Solutions
7. Asymptotic Behaviour.
Bibliography
467
KIRCHHOFF–CARRIER ELASTIC STRINGS
2 – Notations, assumptions and local results
We consider real functions α(t), β(t) and M (λ) satisfying the following conditions:
(H1) α, β ∈ C 2 ([0, +∞]; R) with α(t) < β(t), α0 (t) ≤ 0, β 0 (t) ≥ 0 and
n
0
0
max |α (t)|, |β (t)|
o
≤
µ
m0
2
¶1/2
for all 0 ≤ t < ∞.
(H2) M ∈ C 1 ([0, ∞[; R) such that M (λ) ≥ m0 > 0 for all λ ≥ 0.
Remark 2.1. Note that the assumption α0 (t) ≤ 0 and β 0 (t) ≥ 0 means that
b is increasing in the sense that γ(t) = β(t) − α(t) is increasing.
Q
Remark 2.2. The condition
n
0
0
max |α (t)|, |β (t)|
o
≤
µ
m0
2
¶1/2
is equivalent to
¯
¯ µ m ¶1/2
0
¯ 0
¯
0
,
¯α (t) + y γ (t)¯ ≤
2
for all 0 ≤ y ≤ 1 .
In fact, if y = 0 and y = 1 in this last inequality we recover the hypothesis (H1).
Reciprocally, if (H1) is true we have:
0
|α (t)| ≤
µ
m0
2
¶1/2
and
0
|β (t)| ≤
µ
m0
2
¶1/2
.
By (H1) we have α0 (t) ≤ 0 and β 0 (t) ≥ 0. Then
α0 (t) + y γ 0 (t) ≤ y β 0 (t) ≤
µ
m0
2
¶1/2
for all 0 ≤ y ≤ 1
and
³
0
0
´
0
0
0
0
− α (t) + y β (t) ≤ −α (t) − β (t) + α (t) y ≤ −α (t) ≤
or
¯
¯ µ m ¶1/2
0
¯ 0
¯
0
¯α (t) + y γ (t)¯ ≤
2
µ
for all 0 ≤ y ≤ 1 .
m0
2
¶1/2
468
J. LIMACO FERREL and L.A. MEDEIROS
Remark 2.3.
In the investigation of global solutions of (1.1) we make a
³ m ´1/2
0
with
stronger assumption; namely max{|α0 (t)|, |β 0 (t)} ≤ C
2
1
6
C =
µ ¶1/2 µ
2
5
π
π+1
¶
.
x−α
varies in the
γ
b → Q, given by τ : (x, t) → (y, t) is
cylinder Q = ]0, 1] × ]0, T [ . The mapping τ : Q
a diffeomorphism. We transform system (1.1) by means of the change of variables
b the point (y, t), with y =
Observe that when (x, t) varies in Q
u(x, t) = v(y, t)
x−α
,
γ
with y =
which transforms the operator
b =
Lu
∂2u
−M
∂t2
µZ
β(t) µ ∂u ¶2
α(t)
∂x
dx
¶
∂2u
∂x2
in
into the operator
µ
Z µ
¶
µ
¶
b
Q
∂v
1
∂
1 1 ∂v 2
∂2v
∂2v
a(y, t)
−
M̌
−
dy
Ľv =
2
2
2
∂t
γ
γ 0 ∂y
∂y
∂y
∂y
2
∂ v
∂v
+ b(y, t)
+ c(y, t)
in Q .
∂t ∂y
∂y
¶
Here we have
• dx = γ dy ,
• M̌ (λ) = M (λ) −
• a(y, t) =
m0
−
2 γ2
µ
m0
m0
≥
>0,
2
2
µ
α0 + γ 0 y
γ
α0 + γ 0 y
• b(y, t) = −2
γ
µ
α00 + γ 00 y
• c(y, t) = −
γ
¶
¶
¶2
>0,
,
.
Then, the noncylindrical mixed problem (1.1) is transformed in the following
KIRCHHOFF–CARRIER ELASTIC STRINGS
469
cylindrical mixed problem:
(2.1)


Ľv(y, t) = g(y, t)





in Q ,
v(0, t) = v(1, t) = 0





 v(y, 0) = v0 (y),
on 0 < t < T ,
∂v
(y, t) = v1 (y)
∂y
on 0 < y < 1 .
We represent, as usual, by (( , )), k · k and ( , ), | · |, respectively, the scalar
product and norm in H01 (0, 1) and L2 (0, 1). By a(t, v, w) we denote the positive,
continuous, symmetric bilinear form in H01 (0, 1):
a(t, v, w) =
Z
1
a(y, t)
0
∂v ∂w
dy .
∂y ∂y
We have the following results:
Theorem 2.1. Let Ωt be the interval ]α(t), β(t)[ , 0 < t < T , and suppose
that (H1) and (H2) hold. Then, given
u0 ∈ H01 (Ω0 ) ∩ H 2 (Ω0 ) ,
u1 ∈ H01 (Ω0 ) ,
f ∈ L∞ ([0, T ]; H01 (Ωt )) ,
there exists 0 < T0 < T and a unique function
b0 → R ,
u: Q
satisfying the conditions:
b 0 = Ω × ]0, T0 [ ,
Q
u ∈ L∞ (0, T0 ; H01 (Ωt )∩H 2 (Ωt )), u0 ∈ L∞ (0, T0 ; H01 (Ωt )), u00 ∈ L∞ (0, T0 ; L2 (Ωt )) ,
b0.
solution of (1.1) in Q
Theorem 2.2. Under the assumptions of Theorem 2.1, given
v0 ∈ H01 (0, 1) ∩ H 2 (0, 1) ,
v1 ∈ H01 (0, 1)
and
g ∈ L∞ ([0, T ]; H01 (0, 1)) ,
there exists 0 < T0 < T and a unique function
v : Q0 → R
satisfying the conditions:
v ∈ L∞ (0, T0 ; H01 (0, 1) ∩ H 2 (0, 1)) ,
v 0 ∈ L∞ (0, T0 ; H01 (0, 1)) ,
which is solution of (2.1), in Q0 = ]0, 1[ × ]0, T0 [ .
470
J. LIMACO FERREL and L.A. MEDEIROS
x−α
b we obtain the term
in Q
γ
Remark 2.4. By the change of variables y =
³ α0 + γ 0 y ´2 ∂ 2 v
α + γ y γ 0 ∂v
+
in (2.1) that gives serious trouble when we
γ
γ ∂y
γ
∂y 2
∂ 2v0
multiply both sides of the equation (2.1) by − 2 and integrate in Q. However,
∂y
under the assumptions (H1) and (H2), the bad terms can be absorbed by the
positive terms that the nonlinearity M (λ) provides. Indeed, we incorporate the
³ m
m0 ´ ∂ 2 v
m0 m0
0
in Ľv, so that M̌ (λ) = M (λ) −
>
. The remaining
term − 2 + 2
2γ
2 γ ∂y 2
2
2
terms can be written in divergence form:
2
m0 ∂ 2 v
α0 + γ 0 y γ 0 ∂v
− 2
+
2
+
2 γ ∂y 2
γ
γ ∂y
µ
α0 + γ 0 y
γ
∂
=−
∂y
¶2
Ã"
∂2v
=
∂y 2
m0
−
2 γ2
µ
α0 + γ 0 y
γ
¶2 #
∂v
∂y
!
giving a positive symmetric bilinear continuous form a(t, v, w), which can be
handled easily when getting a priori estimates. Thus the operator Ľv is well
adapted to apply the energy method.
Remark 2.5. When getting estimates for v 0 in H01 (0, 1) and v in H 2 (0, 1)
terms of the form
·
¸
·
¸
(−α0 ) ∂ v 0 (0, t) 2
β 0 ∂ v 0 (1, t) 2
+
γ
∂y
γ
∂y
appear. In order to guarantee its positivity we need that α0 ≤ 0 and β 0 ≥ 0 or, in
b is increasing.
other words, that Q
3 – Approximations and estimates
Let {wj }, j = 1, 2, ..., be the solutions of the spectral problem:
((wj , v)) = λj (wj , v),
for all v ∈ H01 (0, 1) .
They can be chosen to constitute an orthonormal basis of L2 (0, 1).
We represent by Vm = {w1 , w2 , ..., wm } the subspace of H01 (0, 1) generated by
the first m eigenfunctions wj orthonormal in L2 (0, 1). Note that this is equivalent to say that −wj00 = λj wj , wj (0) = wj (1) = 0, for j = 1, 2, ..., i.e., they
are eigenfunctions of the Laplace operator with zero Dirichlet conditions on the
boundary.√In the one dimensional case we are working on, we obtain λj = (jπ)2
and wj = 2 sin jπx, j = 1, 2, ... .
471
KIRCHHOFF–CARRIER ELASTIC STRINGS
We look for vm (t) ∈ Vm solution of the system of ordinary differential equations:

(Ľvm (t), v) = (g(t), v)




(3.1)




for all v ∈ Vm ,
vm (0) = v0m ,
0
vm
(0) = v1m .
By v0m and v1m we denote the projections of v0 and v1 over Vm . Note that
and
v0m → v0
in H01 (0, 1) ∩ H 2 (0, 1)
v1m → v1
in H01 (0, 1) ,
as m → ∞.
0 (t) in (3.1). We obtain:
Estimate I. Let us consider v = vm
(3.2)
µ
¶
1 d 0
1
1
d
|v (t)|2 +
M̌
kvm (t)k2
kvm (t)k2 +
2 dt m
2 γ2
γ
dt
¶ µ
¶
µ
0
∂vm 0
∂vm
0
0
0
, vm + c(y, t)
, vm = (g, vm
).
+ a(t, vm , vm ) + b(y, t)
∂y
∂y
If we set
we have
(3.3)
·
c(λ) =
M
¶¸
µ
λ
M̌ (s) ds ,
0
¶
µ
γ0
d 1 c 1
1
kvm (t)k2 +
M̌
kvm (t)k2 kvm (t)k2 +
M
3
dt 2 γ
γ
2γ
γ
µ
µ
¶
¶
γ0 c 1
1
d
1
2
2
+
M
kvm (t)k
M̌
kvm (t)k
kvm (t)k2 ,
=
2 γ2
γ
2 γ2
γ
dt
(3.4)
0
a(t, vm , vm
)=
1
1 d
a(t, vm , vm ) − a0 (t, vm , vm ) ,
2 dt
2
where
a0 (t, v, w) =
We also have:
(3.5)
Z
µ
b(y, t)
¶
Z
1
0
a0 (y, t)
∂v ∂w
dy .
∂y ∂y
¯
0
¯y=1
1
1
∂vm
0
0
, vm
= b(y, t) vm
(y, t)2 ¯¯
−
∂y
2
2
y=0
Z 1
1
∂b 0
=−
(v (t))2 dy .
2 0 ∂y m
Z
1 ∂b
0
∂y
0
(vm
(t))2 dy
472
J. LIMACO FERREL and L.A. MEDEIROS
Substituting (3.3), (3.4) and (3.5) in (3.2), we obtain:
µ
·
(3.6)
1
1
= a0 (t, vm , vm ) +
2
2
From (3.6), we obtain:
(3.7)
¶¸
1 d 0
1 d 1 c 1
1 d
M
|vm (t)|2 +
kvm (t)k2 +
a(t, vm , vm ) +
2 dt
2 dt γ
γ
2 dt
µ
¶
µ
¶
1
γ0 c 1
γ0
2
2
2
M̌
kv
(t)k
kv
(t)k
kv
(t)k
+
M
=
+
m
m
m
2 γ3
γ
2 γ2
γ
Z
1 ∂b
0
·
∂y
0
(vm
(t))2 dy
¶
µ
∂vm 0
0
).
+ c(y, t)
, vm + (g, vm
∂y
µ
¶¸
1 d 0
1 d 1 c 1
1 d
|vm (t)|2 +
kvm (t)k2 +
a(t, vm , vm ) +
M
2 dt
2 dt γ
γ
2 dt
¶
¶
µ
µ
γ0 c 1
γ0
1
2
2
2
kv
(t)k
+
≤
+
M̌
kv
(t)k
M
kv
(t)k
m
m
m
2 γ3
γ
2 γ2
γ
≤
1
0
|g(t)|2 + C1 |vm
(t)|2 + C2 kvm (t)k2 .
2
b is increasing, and
We have, by hypothesis, γ 0 ≥ 0 since Q
µ
¶
c 1 kvm k2 ≥ C kvm k2
M
(3.8)
γ
(3.9)
a(t, vm , vm ) > 0 .
Integrating (3.7) on [0, t[ contained in the interval of existence of vm (t) solution
of (3.1), we obtain
(3.10)
Z t³
0
|vm
(t)|2 + kvm (t)k2 ≤ K0 + K1
0
´
0
|vm
(s)|2 + kvm (s)k2 ds
where K0 , K1 are constants independent of m.
From (3.10) and Gronwall inequality, we obtain:
(3.11)
0
|vm
(t)|2 + kvm (t)k2 < C
on [0, T ] .
Estimate II. In the approximate system (3.1) we take v = −
gives:
¯
¶
µ
¯
0
∂ 2 vm
. This
∂y 2
1 d 0
1
d ¯¯ ∂ 2 vm ¯¯2
1
2
(3.12)
kvm (t)k2 +
M̌
kv
(t)k
+
m
2 dt
2 γ2
γ
dt ¯ ∂y 2 ¯
µ
µ
µ
0
0 ¶
0 ¶
0 ¶
∂vm
∂vm ∂ 2 vm
∂ 2 vm
∂ 2 vm
+
b(y,
t)
+
c(y,
t)
=
,
−
,
−
+ a t, vm , −
∂y 2
∂y
∂y 2
∂y
∂y 2
µ
∂ 2 vm
= g, −
∂y 2
¶
.
473
KIRCHHOFF–CARRIER ELASTIC STRINGS
We have:
µ
0
∂ 2 vm
a t, vm , −
∂y 2
¶
(3.13)
Note that:
¶
µ
Z
1 d 1
∂ 2 vm 2
dy
=
a(y, t)
2 dt 0
∂y 2
¶
µ 2
Z
1 1 0
∂ vm 2
−
a (y, t)
dy
2 0
∂y 2
·
¸
Z 1
0
0 ¯¯y=1
∂ ∂a ∂vm ∂vm
∂a ∂vm ∂vm
¯
dy +
−
.
∂y
∂y ∂y ∂y ¯y=0
0 ∂y ∂y ∂y
µ
0
∂ 2 vm
∂v 0
b(y, t) m , −
∂y
∂y 2
¶
=−
=
1
2
Z
0 ∂ 2v0
∂vm
m
dy
∂y ∂y 2
0
µ 0 ¶2
Z 1
1 ∂ ∂vm
= − b(y, t)
dy .
2 ∂y ∂y
0
1
b(y, t)
Integrating by parts we get:
(3.14)
µ
b(y, t)
0
0
∂ 2 vm
∂vm
,−
∂y
∂y 2
¶
Remark 3.1. Since b(y, t) = −2
µ
0
1
∂vm
− b(y, t)
2
∂y
Z
0
1 ∂b µ ∂v 0 ¶2
m
∂y
∂y
dy −
µ
0
∂vm
1
b(y, t)
2
∂y
¶2 ¯y=1
¯
¯
.
¯
y=0
α0 + γ 0 y
,
γ
¶2 ¯y=1
µ
µ
0 (1, t) ¶2
0 (0, t) ¶2
¯
β 0 ∂ vm
α 0 ∂ vm
¯
=
−
¯
γ
∂y
γ
∂y
y=0
which is non negative, by the hypothesis α0 ≤ 0, β 0 ≥ 0 on the non-decreasing
boundary. On the other hand,
(3.15)
µ
0
∂ 2 vm
∂vm
,−
c(y, t)
∂y
∂y 2
¶
Z
·
¸
0
∂
∂vm ∂vm
=
c(y, t)
dy
∂y
∂y
0 ∂y
0 ¯¯y=1
∂vm ∂vm
¯
− c(y, t)
.
∂y ∂y ¯y=0
1
Substituting (3.13), (3.14) and (3.15) in (3.12), we obtain:
µ
¶
¯
¯
1
1
d ¯¯ ∂ 2 vm ¯¯2
1 d 0
2
kvm (t)k2 +
M̌
kv
(t)k
+
m
2 dt
2 γ2
γ
dt ¯ ∂y 2 ¯
474
J. LIMACO FERREL and L.A. MEDEIROS
1 d
2 dt
+
(3.16)
−
Z
+
1
2
+
Z
1
0
0
1
a(y, t)
0
·
∂ 2 vm
∂y 2
¸
0
1 ∂b µ ∂v 0 ¶2
m
∂y
∂y
·
¶2
dy −
1
2
Z
1
a0 (y, t)
0
¯
µ
¸
µ
∂ 2 vm
∂y 2
¶2
dy
¶2 ¯y=1
¯
¯
¯
y=0
¯
0 ¯y=1
∂vm ∂vm
¯
=
dy − c(y, t)
∂y
∂y ∂y ¯y=0
0
∂vm
0
∂g ∂vm
,
∂y ∂y
¶
.
µ
¶
1
1
M̌
kvk2 , we have:
γ2
γ
Remark 3.2. Denoting by µ(t) =
µ
¶
µ
0
1
∂vm
b(y, t)
2
∂y
dy −
∂
∂vm
c(y, t)
∂y
∂y
=
µ
µ
0
0 ¯y=1
∂ ∂a ∂vm ∂vm
∂a ∂vm ∂vm
¯
dy +
∂y ∂y ∂y
∂y
∂y ∂y ∂y ¯y=0
Z
1
Z
¶
µ
¶
2
γ0
2 γ0
1
1
1
µ (t) = − 3 M̌
kvk2 + 3 M̌ 0
kvk2 ((v, v 0 )) − 4 M̌ 0
kvk2 kvk2
γ
γ
γ
γ
γ
γ
0
and, by Estimate I and the fact that M ∈ C 1 ([0, ∞[, R) we obtain, by the hypothesis (H1) on α, β:
0
|µ0 (t)| ≤ C + kvm
k.
From (3.16) and Remark 3.2, we have:
¯
(3.17)
¯
µZ
¶
¶
µ
1
∂vm 2
1 d 0
1
d ¯ ∂ 2 vm ¯¯2 1 d
dy +
a(y,
t)
kvm (t)k2 + µ(t) ¯¯
+
¯
2 dt
2
dt ∂y 2
2 dt 0
∂y
µ
µ
0 (1, t) ¶2
0 (0, t) ¶2
β 0 ∂ vm
(−α0 ) ∂ vm
+
+
=
γ
∂y
γ
∂y
=
1
2
Z
1
a0 (y, t)
0
Z
1
0
∂ 2 vm
∂y 2
¶2
0 ¯¯y=1
∂vm ∂vm
¯
dy +
1
−
¯
∂y y=0 2
∂a
−
∂y ∂y
−
µ
·
¸
Z
Z
1
0
1 ∂b
0
∂y
·
¸
0
∂ ∂a ∂vm ∂vm
dy
∂y ∂y ∂y
∂y
µ
0
∂vm
∂y
¶2
dy
¯
0
0 ¯y=1
∂
∂vm ∂vm
∂vm ∂vm
¯
dy + c(y, t)
c(y, t)
+
∂y
∂y
∂y
∂y ∂y ¯y=0
µ
0
∂g ∂vm
,
∂y ∂y
¶
Remark 3.3. We have the identity:
∂vm
(0, t) =
∂y
Z
1
0
·
¸
∂
∂vm
(y, t) dy =
(1−y)
∂y
∂y
Z
1
(1−y)
0
∂ 2 vm
dy −
∂y 2
Z
1 ∂v
0
m
∂y
dy
.
475
KIRCHHOFF–CARRIER ELASTIC STRINGS
or
¯
¯
¯ ∂vm
¯
¯
¯
(0,
t)
¯ ∂y
¯
by the Estimate II.
R
¯
¯
¯
¯
¯ ∂ 2 vm ¯
¯ ∂ 2 vm ¯
¯
¯
¯
+
kv
k
≤
C
+
,
≤ ¯¯
m
¯ ∂y 2 ¯ 2
∂y 2 ¯L2 (0,1)
L (0,1)
A similar estimate is true for
∂vm
(1, t).
∂y
Remark 3.4. We have:
a(y, t) =
m0
−
2 γ2
c(y, t) = −
µ
α0 + γ 0 y
γ
¶2
>0
and
∂a
α0 + γ 0 y γ 0
= −2
,
∂y
γ
γ
α00 + γ 00 y α0 + γ 0 y γ 0
−
.
γ
γ
γ
Then, since α0 + γ 0 = β 0 , we obtain:
¯
0 ¯y=1
0
2 β 0 γ 0 ∂vm
∂a ∂vm ∂vm
∂v 0
2 α0 γ 0 ∂vm
∂vm
¯
= + 2
−
(1, t) m (1, t) −
(0,
t)
(0, t) ,
¯
∂y ∂y ∂y y=0
γ
∂y
∂y
γ 2 ∂y
∂y
¯
¶
µ
0
0 ¯y=1
∂vm
∂vm ∂vm
β 0 γ 0 + γβ 00 ∂vm
¯
(1,
t)
(1, t)
c(y, t)
= −
¯
∂y ∂y y=0
γ2
∂y
∂y
µ 0 0
¶
α γ + α00 γ ∂vm
∂v 0
+
(0, t) m (0, t) .
2
γ
∂y
∂y
Let us consider, for example,
¯ 0
¯ 0 0¯¯
¯¯ 0
¯
¯2
¯2
¯ 0 0 ¯2 ¯
¯ β γ ¯ ¯ ∂vm
¯ ¯ ∂vm
¯
¯
¯
¯ β γ ¯ ¯ ∂vm
1 ¯¯ ∂vm
¯¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯ γ 2 ¯ ¯ ∂y (1, t)¯ ¯ ∂y (1, t)¯ ≤ λ ¯ γ 2 ¯ ¯ ∂y (1, t)¯ + 4 λ ¯ ∂y (1, t)¯ .
0
´2
1 β 0 ³ ∂vm
1 β0
= , then
(1, t) goes to the left side of (3.17) and it is compenλ γ
4 γ ∂y
´2
β 0 ³ ∂v 0
sated with
(1, t) which gives a positive contribution in the first member.
γ ∂y
¯ β 0 γ 0 ¯ ¯ ∂v
¯2
¯¯ m
¯
¯
The term ¯ 2 ¯ ¯
(1, t)¯ can be estimated as in Remark 3.3. We get
γ
∂y
If
¯ 0 0 ¯2
¯β γ ¯
¯
¯
¯ γ2 ¯
¯2
¯ 2
¯
¯
¯
¯ ∂ vm ¯ 2
¯ ∂vm
¯
¯
¯
¯
¯ ∂y (1, t)¯ ≤ C + ¯ ∂y ¯ ,
with a possibly different constant C.
The same argument is true for all the other terms above.
476
J. LIMACO FERREL and L.A. MEDEIROS
Remark 3.5. From the hypothesis on α and β, we can estimate all the other
¯ ∂ 2 v ¯2
m¯
¯
0 (t)k2 .
terms in the left side of (3.17) by C ¯
¯ and Ckvm
2
∂y
Then by the Remarks above, we modify (3.17) obtaining:
(3.18)
¯
¯
d ¯ ∂ 2 vm ¯¯2 d
d 0 2
+
kvm k + µ(t) ¯¯
dt
dt ∂y 2 ¯
dt
Z
1
0
µ
∂ 2 vm
a(y, t)
∂y
¶2
dy ≤
¯ 2
¯
¯ ∂ vm ¯ 2
¯ .
∂y 2 ¯
0
≤ C0 + C1 kvm
(t)k2 + ¯¯
·
¸
¯ ∂2v ¯
¯ 2
¯2
d
d ¯¯ ∂ 2 vm ¯¯2
m¯
¯
0 (t)¯ ∂ vm ¯ and by
µ(t)
−
µ
in
(3.18)
by
¯
¯
¯
¯
¯
¯
dt ∂y 2
dt
∂y 2
∂y 2
Remark 3.2, we obtain:
Substituting µ
(3.19)
"
¯
¯
¯ ∂ 2 vm ¯ 2
d
0
kvm
(t)k2 + µ(t) ¯¯
(t)¯¯ +
dt
∂y 2
Z
1
0
µ
∂vm
a(y, t)
∂y
¶2
¯ 2
¯ ∂ vm
0
hm (t) = kvm
(t)k2 + µ(t) ¯¯
we have from (3.19):
∂y 2
¯2
¯
(t)¯¯ +
Z
1
a(y, t)
0
dhm
≤ C0 + C1 hm + h3/2
m
dt
µ
≤
¯ 2
¯
¯ ∂ vm ¯ 2
¯
∂y 2 ¯
0
0
≤ C0 + C1 kvm
(t)k2 + kvm
(t)k ¯¯
If
dy
#
∂vm
∂y
¶2
in [0, T ] .
dy ,
in [0, T ] .
From this inequality, we get a number 0 < T0 < T , such that hm (t) is bounded
in [0, T0 ] independently of m. This gives the second estimate
¯
¯ 2
¯ ∂ vm ¯ 2
¯ <C
∂y 2 ¯
0
kvm
(t)k2 + ¯¯
(3.20)
on [0, T0 ] .
Note that µ(t) is strictly positive on [0, T ].
00 (t) in the approximate system (3.1) we obtain
Estimate III. Taking v = vm
00
|vm
(t)|2
µ
¶
∂ 2 vm 00
,v
− µ(t)
−
∂y 2 m
Z
1
·
¸
∂vm 00
∂
a(y, t)
vm dy +
∂y
0 ∂y
Z 1
Z 1
∂v 0
∂vm 00
00
+
b(y, t) m v 00 dy + c(y, t)
v dy = (g, vm
).
∂y
∂y m
0
0
From the first and second estimates and the hypothesis on α, β, we get:
(3.21)
00
|vm
(t)|2 < C
on [0, T0 ] .
477
KIRCHHOFF–CARRIER ELASTIC STRINGS
4 – Proof of the theorems
Proof of Theorem 2.2: In this step we prove that the estimates obtained
above are sufficient to take limits in the approximate equation (3.1). In view of
(3.11), (3.20) and (3.21) a subsequence represented by (vk ) can be extracted from
(vm ) such that:
(4.1)
vk * v
weak star in L∞ (0, T0 ; H01 (0, 1) ∩ H 2 (0, 1)) ,
(4.2)
vk0 * v 0
weak star in L∞ (0, T0 ; H01 (Ω)) .
By the classical compactness argument of Aubin–Lions, cf. Lions [13], it follows that:
(4.3)
strongly L2 (0, T0 ; H01 (0, 1)) .
vk → v
Because of the estimate (3.21) the subsequence satisfies also:
weak star in L∞ (0, T0 ; L2 (Ω)) .
vk00 * v 00
(4.4)
From hypothesis (H1), (H2) and the estimates (3.11) and (3.20), we obtain:
¯
µ
¶2 2 ¯
¯1
1
∂ v ¯¯
2
¯
<C
¯ γ 2 M̌ γ kvm (t)k
∂y 2 ¯
(4.5)
on [0, T0 [ .
Then, the sequence (vk ) is such that
µ
1
1
M̌
kvk (t)k2
2
γ
γ
(4.6)
¶2
∂ 2 vk
* χ
∂y 2
weak star in L∞ (0, T0 ; L2 (Ω)).
Lemma 4.1. χ =
µ
¶
1
∂2v
1
M̌
kv(t)k2
where v is the limit in (4.1).
γ
γ
∂y 2
Proof of Lemma 4.1: Let us introduce the notations µk (t) =
and µ(t) =
we have:
³1
´
1
2 . Because of (4.6), for every w ∈ L2 (0, T ; L2 (0, 1)),
M̌
kv(t)k
0
γ2
γ
Z
(4.7)
´
1 ³1
2
M̌
kv
(t)k
k
γ2
γ
0
T0µ
¶
∂2v
χ − µ(t) 2 , w dt =
∂y
Z
T0µ
¶
∂ 2 vk
χ − µk (t)
, w dt
∂y 2
0
µ 2
¶
Z T0
∂ vk
∂2v
µ(t)
+
−
,
w
dt
∂y 2
∂y 2
0
¶
Z T0h
i µ ∂2v
k
µk (t) − µ(t)
, w dt .
+
∂y 2
0
478
J. LIMACO FERREL and L.A. MEDEIROS
By (4.6) the first right side integral of (4.7) goes to zero when k → ∞ and the
second one, by (3.20), also goes to zero. To analyse the third member of the right
side of (4.7) we employ hypothesis (H2) on M (λ). Then, we have:
¯
¯
¯
¯
³
´
|µk (t) − µ(t)| ≤ c ¯ kvk (t)k2 − kv(t)k2 ¯ ≤ c kvk (t) − v(t)k kvk (t)k + kv(t)k .
It follows from (4.3), estimates (3.11), (3.20) and Lebesgue convergence theorem, that the last term of (4.7) goes to zero when k → ∞.
Integrating by parts we obtain that
and
a(t, vk , w) * a(t, v, w)
weakly in
L2 (0, T ; H 1 (0, 1))
(4.8)
µ
∂
∂vk
a(y, t)
∂y
∂y
¶
*
µ
∂
∂v
a(y, t)
∂y
∂y
¶
weakly in
L2 (0, T ; L2 (0, 1)) .
We also have, by the same argument,
(4.9)
b(y, t)
∂v 0
∂vk0
* b(y, t)
∂y
∂y
weakly in L2 (0, T ; L2 (0, 1)) ,
(4.10)
c(y, t)
∂vk
∂v
* c(y, t)
∂y
∂y
weakly in L2 (0, T ; L2 (0, 1)) .
Because of (4.4), (4.6), Lemma 1, (4.8), (4.9) and (4.10) we take m = k in the
approximate equation (3.11) and we let k go to +∞ obtaining:
for all w ∈ L2 (0, T0 ; L2 (0, 1))
(Ľv, w) = (g, w)
or, equivalently,
(4.11)
Ľv = g
in L2 (0, T0 ; L2 (0, 1)) .
From (4.1), (4.2) and (4.4), we obtain
(4.12)
v(0) = v0 ,
v 0 (0) = v1 on Ω .
Uniqueness
If v and v̂ are two solutions in the conditions of Theorem 2.2, then w = v − v̂
satisfies:
(4.13)
w 00 −
µ
¶
µ
¶
µ
2
2
1
1
1
∂w
1
∂
2 ∂ v
2 ∂ v̂
a(y, t)
M̌
kvk
+
M̌
kv̂k
−
2
2
2
2
γ
γ
∂y
γ
γ
∂y
∂y
∂y
+ b(y, t)
∂w0
∂w
+ c(y, t)
= 0
∂y
∂y
w(0) = w 0 (0) = 0 in Ω and w = 0 on ]0, 1[ × ]0, T0 [ .
¶
+
in L2 (0, T ; L2 (0, 1)) ,
479
KIRCHHOFF–CARRIER ELASTIC STRINGS
Multiplying (4.13) by w 0 and integrating we obtain:
(4.14)
µ
¶
1
1
d
1 d 0 2
M̌
|w (t)| +
kv(t)k2
kw(t)k2 +
2
2 dt
2γ
γ
dt
µ
¶
Z 1
Z
Z 1
∂w 2
∂w0 0
1 d 1
∂w 0
+
a(y, t)
dy + b(y, t)
w dy + c(y, t)
w dy =
2 dt 0
∂y
∂y
∂y
0
0
"
=
µ
µ
¶
1
1
1
1
M̌
kvk2 − 2 M̌
kv̂k2
γ2
γ
γ
γ
¶¸ Ã
∂ 2 v̂ 0
,w
∂y 2
¶
.
We have:
(4.15)
µ
¶
µ
¶
¶
µ
1
d
1 d 1
1
1
M̌
kvk2
kwk2 =
M̌
kvk2 kwk2
2
2
2γ
γ
dt
2 dt γ
γ
¶
µ
γ0
1
+ 3 M̌
kvk2
γ
γ
¶ µ
¶
µ
1
d 1
1
− 2 M̌ 0
kvk2
kvk2 kwk2
2γ
γ
dt γ
Z
(4.16)
1
0
∂w0 0
1
b(y, t)
w dy = −
∂y
2
Z
1 ∂b
∂y
0
(w0 )2 dy .
Substituting (4.4) and (4.5) in (4.3) we obtain:
(4.17)
Ã
µ
¶
1
d
1
kvk2 kwk2 +
|w0 (t)|2 + 2 M̌
dt
γ
γ
Z
1
0
µ
∂w
a(y, t)
∂y
³
¶2
dy
!
≤
´
≤ c |w0 (t)|2 + kw(t)k2 .
Integrating (4.17) over 0 ≤ t < T0 , we have:
0
2
2
|w (t)| + kw(t)k ≤ C0
Z t³
0
´
|w0 (s)|2 + kw(s)k2 ds .
This implies w = 0 by Gronwall’s inequality.
Proof of Theorem 2.1: If v is the solution of Theorem 2.2 we consider the
function
(4.18)
u(x, t) = v(y, t) ,
x=α+γy .
We also set
(4.19)
g(y, t) = f (α + γy, t) ,
480
J. LIMACO FERREL and L.A. MEDEIROS
³
(4.20)
´
v0 (y) = u(x, 0) = u0 α(0) + γ(0) y ,
³
v1 (y) = u0 (x, 0) = u1 α(0) + γ(0) y
(4.21)
³
´
³
´
´
+ α0 (0) + γ 0 (0) y u00 α(0) + γ(0) y .
The function u(x, t) defined by (4.18) is the solution of Theorem 2.1. To see
this it is sufficient to observe that the application
(x, t) →
µ
x−α
,t
γ
¶
b into ]0, 1[ × ]0, T0 [ is of class C 2 and we have:
from Q
1 ∂2v
∂2u
(x,
t)
=
(y, t) ,
∂x2
γ 2 ∂y
(4.22)
µ
¶
(4.23)
∂v
∂v
∂v 0
∂
(y, t)+c(y, t)
(y, t) ,
a(y, t)
+b(y, t)
u (x, t) = v (y, t)−
∂y
∂y
∂y
∂y
with y =
x−α
,
γ
00
00
Z
(4.24)
β(t) µ ∂u ¶2
α(t)
∂x
1
dx =
γ
Z 1µ
0
∂v
∂y
¶2
dy .
From (4.22)–(4.24) we obtain:
(4.25)
00
u (x, t) − M
µZ
β(t) µ ∂u ¶2
α(t)
∂x
dx
¶
∂2u
= Ľ v(y, t) ,
∂x2
x−α
. Then u solves (1.1), with initial conditions u0 and u1 .
γ
The regularity of v given by Theorem 2.2, implies the regularity of u claimed
in Theorem 2.1.
To prove the uniqueness observe that from (4.22)–(4.25) we have the equivalence between the mixed problems (1.1) and (2.1). Then, if u and û are two
solutions of (1.1) given by Theorem 2.1, then v and v̂ obtained by (4.18) are
solutions in the conditions of Theorem 2.2. Since we have uniqueness for v, i.e.,
v = v̂ it implies u = û.
with y =
481
KIRCHHOFF–CARRIER ELASTIC STRINGS
5 – Applications
We will give examples of functions α(t) and β(t) in C 1 ([0, ∞[ , R), such that
α(t) < β(t), α0 (t) ≤ 0 and β 0 (t) ≥ 0. If γ(t) = β(t) − α(t), we have γ 0 (t) ≥ 0 and,
by hypothesis, (H1), Remark 2.2, we must have:
µ
¶
¯
¯
m0 1/2
¯
¯ 0
0
,
¯α (t) + γ (t) y ¯ ≤
2
for all 0 ≤ t ≤ T and 0 ≤ y ≤ 1.
Let us consider the family of straight lines:
x = α0 (t) + γ 0 (t) y
depending of the parameter t ≥ 0.
We rewrite (H1) as:
−α0 (t) ≤
µ
m0
2
¶1/2
and
β 0 (t) ≤
µ
m0
2
¶1/2
Case 1. Let us consider in the (x, t) plane the lines:
α(t) = α0 − α1 t
with 0 ≤ α1 ≤
β(t) = β0 + β1 t
with 0 ≤ β1 ≤
µ
µ
m0
2
m0
2
¶1/2
¶1/2
,
,
with α0 < β0 positive constants.
The noncylindrical domains are cones with basis the intervals [α0 , β0 ].
Case 2. Let us consider the curves
1
α0 +β0
− (t + t0 ) 2k ,
α(t) =
2
1
α0 +β0
β(t) =
+ (t + t0 ) 2k ,
2
By the conditions of Lemma 1, we can choose
t0 = (2k)
2k
1−2k
k = 1, 2, ... .
k
( m20 ) 1−2k .
These curves could be written as:
t ≥ 0, with x0 =
α0 +β0
.
2
(x − x0 )2k = t + t0 ,
482
J. LIMACO FERREL and L.A. MEDEIROS
6 – Global solutions
In this section we prove that if we add some damping to the noncylindrical
Kirchhoff–Carrier model, with certain restrictions on the initial data, we obtain
a global solution in time. Note, however, that we also impose certain restrictions
on the boundary of the noncylindrical domain.
In fact, with the notation and hypothesis fixed in Section 2, we consider now,
for the sake of simplicity, the domains of the form
α(t) = −β(t)
for all t ≥ 0 .
Then
γ(t) = 2β(t)
for all t ≥ 0
and by Remark 2.2, we must have:
0
µ
1
C=
6
µ ¶1/2 µ
m0
0 < β (t) < C
2
(6.1)
with
2
5
¶1/2
.
¶
.
π
π+1
We suppose, in the present section, that
(6.2)
M (λ) = m0 + m1 λ ,
m0 , m1 > 0, λ ≥ 0 .
This is a particular case in which (H2) holds.
b into the cylinder Q is given by:
The mapping from Q
(6.3)
y=
x+β
2β
or
x = (2 y − 1) β ,
with 0 ≤ y ≤ 1.
We consider the perturbated system (6.4). The modified Kirchhoff–Carrier
model with damping is, for δ > 0 fixed, of the type:
(6.4)

¶
¶
µ 0
¶ 2
µZ β(t) µ

β ∂u
∂ u
∂u 2

0
00

+δ
x
+u = 0
dx
u −M



∂x2
β ∂x
−β(t) ∂x

b ,

u = 0 on Σ





 u(x, 0) = u (x), u0 (x, 0) = u (x)
0
1
where β0 = β(0).
on − β0 < x < β0 ,
b ,
in Q
483
KIRCHHOFF–CARRIER ELASTIC STRINGS
In addition to (H1), (H2) we assume that
(H3) For all t ≥ 0, |β 00 (t)| ≤
(β 0 (t))2
.
β(t)
b into Q, the mixed problem (6.4) in
If we consider the mapping (6.3) from Q
b has the following form in Q:
the noncylindrical domain Q
(6.5)

µ
¶ 2
µ
¶
1
1
∂
∂v

00
2 ∂ v


v −
M̌
kvk
−
a(y, t)
+


4 β2
2β
∂y 2 ∂y
∂v





∂v 0
∂v

0

+ b(y, t)





v=0







∂y
+ c(y, t)
∂y
+δv = 0
in Q ,
on Σ ,
v(y, 0) = v0 (y), v 0 (y, 0) = v1 (y) ,
0<y<1.
Note that v(y, t) = u((2y−1)β, t) and Q = (0, 1) × (0, T ).
The coefficients of the operator Ľv(y, t) in (6.5) are given by:
• a(y, t) =
• b(y, t) =
• c(y, t) =
¸
·
´2
1 m0 ³ 0
,
−
β
(1
−
2
y)
4 β2 2
i
1h 0
β (1 − 2 y) ,
β
i
1 h 00
β (1 − 2 y) .
2β
We have:
¸
·
´2
1
m0 ³ 0
β0
.
− a0 (y, t) =
β 00 β (1 − 2 y)2 +
− β (1 − 2 y)
3
2
4β
2
Then, by (4.20), we obtain:
¯
¯
m0
¯ 00
¯
¯β β (1 − 2 y)2 ¯ ≤ |β 00 β| ≤ (β 0 )2 <
8
and
¯³
´2 ¯
m0
¯ 0
¯
.
¯ β (1 − 2 y) ¯ ≤ (β 0 )2 <
8
We have, consequently
(6.6)
m0 β 0
1
.
− a0 (y, t) ≥
2
16 β 3
484
J. LIMACO FERREL and L.A. MEDEIROS
Note that
v0 =
β 0 ∂u
x
+ u0
β ∂x
which gives in (6.5) the damping δv 0 .
Now we determine the condition on the initial data v0 , v1 which implies the
global existence of solution for the mixed problem (6.5).
Let us fix the number δ such that
δ
5
>
2
2 β0
(6.7)
µ
m0
2
¶1/2 µ
4 + π2
π2
¶
.
Let us define the function H(t) by:
(6.8)
where ∆v =
δ
δ2
(∇v 0 (t), ∇v(t)) +
kv(t)k2
4
8
µ
¶
Z 1
1
1
2
2
+
M̌
kv(t)k
a(y, t) (∆v)2 dy ,
|∆v(t)|
+
8 β2
2β
0
H(t) = kv 0 (t)k2 +
∂v
∂2v
, ∇v =
and v is a solution of the approximate problem for
2
∂y
∂y
(6.5).
Lemma 6.1. We have H(t) > 0, for all t ≥ 0.
Proof: In fact,
¯
¯
¯
¯
2
¯
¯δ
¯ (∇v, ∇v 0 )¯ ≤ δ ¯¯(∇v, ∇v 0 )¯¯ ≤ δ kvk2 + |v 0 |2 .
¯
¯4
4
32
We know that M̌ (λ) =
m0
m0
+ m1 λ ≥
, then
2
2
1
m0
M̌ (λ) ≥
2
8β
16 β 2
and a(y, t) ≥ 0. It implies:
H(t) ≥
1 0 2 3 δ2
m0
kv k +
kvk2 +
|∆v|2 .
2
32
16 β 2
We represent by C1 the constant:
(6.9)
µ
¶
1 c 1
1
M
kv0 k2 + a(0, v0 , v0 ) .
C1 = |v1 |2 +
2
4 β0
2 β0
485
KIRCHHOFF–CARRIER ELASTIC STRINGS
Theorem 6.1. Suppose (H1) and (H3) are true and α(t) = −β(t). Suppose
also that M = M (λ) is of the form (6.2). Then given
v0 ∈ H01 (0, 1) ∩ H 2 (0, 1) ,
v1 ∈ H01 (0, 1)
and δ > 0 satisfying (6.7), if v0 , v1 are such that
q
(6.10)
C1 H(0) <
√
µ
¶ µ
¶
2 m 0 m0 5
m0 1/2 4+π 2
16 m1 2 β0
2
π2
with C1 and H(0) as in (6.9) and (6.8), then the mixed problem (6.5) has a unique
weak solution v(y, t), defined for all t ≥ 0, y ∈ (0, 1).
Remark 6.1. Before going into the proof of Theorem 6.1, we analyse the
restriction (6.10) on the initial data u0 , u1 of (6.4).
We have
´
³
(i) v0 (y) = u0 (2 y − 1) β0 , with β0 = β(0) ,
´
³
(ii) v1 (y) = u1 (2 y − 1) β0 + (2 y − 1) β 0 (0)
Let us represent (−β0 , β0 ) by Ω0 . Then we have:
´
d ³
u0 (2 y − 1) β0 .
dx
(iii) kv0 k2 = 2 β0 ku0 k2H 1 (Ω0 ) ,
(iv) a(t, v0 , v0 ) ≤
m0
m0
kv0 k2 =
ku0 k2H 1 (Ω0 ) ,
2
4 β0
8 β0
(v) |4 v0 |2 = 8 β02 |4 u0 |2L2 (Ω0 ) ,
1
(β 0 (0))2
(vi) |v1 |2 ≤
|u1 |2L2 (Ω0 ) +
ku0 k2H 1 (Ω0 ) ,
0
β0
β0
(vii) kv1 k2 ≤ 4 β0 ku1 k2H 1 (Ω0 ) + 8 β0 β 02 (0) |∆u0 |2L2 (Ω0 ) + 8 β0 ku0 k2H 1 (Ω0 ) ,
(viii)
Z
1
0
0
2
0
a(y, 0, (∆v0 ) ) dy ≤
β0 m0 |∆u0 |2L2 (Ω0 )
.
Because of (iii)–(viii) and Remark 2.2, we obtain:
´
m0
1 c³
1
|u1 |2L2 (Ω0 ) +
kuk2H 1 (Ω0 ) +
M ku0 k2H 1 (Ω0 )
(ix) C1 ≤ Č0 =
0
0
2 β0
2 β0
4 β0
and
(x) H(0) ≤ G0 = 12 β0 ku1 k2H 1 (Ω0 ) + 13 β0 m0 |∆u0 |2L2 (Ω0 )
µ
¶
0
582
β0 ku0 k2H 1 (Ω0 )
0
16
³
´
2
+ β0 M̌ ku0 kH 1 (Ω0 ) |∆u0 |2L2 (Ω0 ) .
+
24 β0 +
0
486
J. LIMACO FERREL and L.A. MEDEIROS
Therefore, if we take
u0 ∈ H01 (Ω0 ) ∩ H 2 (Ω0 )
u1 ∈ H01 (Ω0 )
and
and fix δ satisfying (6.4), and u0 and u1 are restricted to
√
q
2 m 0 m0
δ,
Č0 G0 <
32 m1
then the mixed problem (6.5) has a unique solution u defined for all t > 0 and
x ∈ Ωt .
Proof of Theorem 6.1: We will employ the Faedo–Galerkin method choosing a Hilbertian basis in H01 (0, 1) ∩ H 2 (0, 1) (cf. Brezis [3]). We do only the a
priori estimates that imply the convergence of approximate solutions as we have
done in Section 3, employing classical compactness arguments as in Lions [13].
In fact, we use the basis of eigenvectors of the spectral problem
for all w ∈ H01 (0, 1) .
((v, w)) = λ(v, w)
Note that this basis can be obtained explicitly as in Section 3. Let us obtain the
a priori estimates.
Estimate I. Multiply both sides of (6.4)1 by v 0 and integrate on (0, 1).
We obtain:
(6.11)
µ
µ
1 d 0 2
d 1 c 1
|v (t)| +
kv(t)k2
M
2 dt
dt 4 β
2β
µ
¶¶
+
¶
µ
1
β0
β0 c 1
2
2
M̌
kv(t)k
M
kv(t)k2
+
kv(t)k
+
8 β2
2β
4 β2
2β
1
1
1 d
a(t, v, v) − a0 (t, v, v) −
+
2 dt
2
2
+
Z
1
c
0
with
−
Z
1 ∂b
0
∂y
(v 0 (t))2 dy
∂v 0
v dy + δ |v 0 |2 = 0
∂y
Z
1 ∂b
0
∂y
(v 0 )2 dy > 0 .
By (6.6) we obtain:
(6.12)
1
− a0 (t, v, v) = −
2
Z
11
0
µ
∂v
a (y, t)
2
∂y
0
¶2
dy ≥
m0 β 0
kvk2 .
16 β 3
¶
KIRCHHOFF–CARRIER ELASTIC STRINGS
Because of (H3), i.e., |β 00 | <
(β 0 )2
, we obtain:
β
¯Z 1
¯ Z 1 ¯ 00
¯
¯
¯
¯ β (1−2y) ∂v 0 ¯
¯ c(y, t) ∂v v 0 dy ¯ ≤
¯ dy
¯
v
¯
¯
¯
∂y
2β
∂y ¯
0
0
≤
≤
If we consider µ such that
Whence,
Z
Z
¯ ¯
¯ ∂v ¯
¯ ¯ · |v 0 | dy
¯ ∂y ¯
1
(β 0 )2
2 β2
1
µ(β 0 )4
4 β4
0
0
¯ ¯2
Z
¯ ∂v ¯
¯ ¯ dy +
¯ ∂y ¯
1
0
1 02
|v | dy .
4µ
µ(β 0 )4
m0 β 0
1
(β 0 )3
=
,
we
obtain
=
. Then:
4 β4
16 β 3
4µ
m0 β
¯Z 1
¯
µ
¶
0
¯
¯
m0 1/2 0 2
¯ c(y, t) ∂v v 0 dy ¯ ≤ m0 β kvk2 + 1
|v | .
¯
¯
∂y
16 β 3
16 β0
2
0
Z
(6.13)
1
c(y, 1)
0
√
m0
m0 β 0
∂v 0
2
√
|v 0 |2 .
v dy ≥ −
kvk
−
3
∂y
16 β
16 2 β0
From (6.12) and (6.13), we have:
(6.14)
−
1 0
a (t, v, v) +
2
Z
1
c(y, t)
0
∂v 0
v dy + δ |v 0 |2 ≥
∂y
µ
¶ ¶
µ
1
m0 1/2
≥ δ−
|v 0 |2 .
16 β0
2
By (6.14) we modify (6.11), obtaining:
(6.15)
µ
µ
¶¶
1 d 0 2
d 1 c 1
|v (t)| +
kv(t)k2
M
+
2 dt
dt 4 β
2β
µ
¶ ¶
µ
1
m0 1/2
1 d
|v 0 |2 ≤ 0 .
a(t, v, v) + δ −
+
2 dt
8 β0
2
The parameter δ was fixed satisfying the condition (6.7), what implies:
1
δ
>
2
8 β0
µ
m0
2
¶1/2
.
Then, from (6.15) we get:
·
µ
d 1 c 1
1 d 0 2
|v (t)| +
kv(t)k2
M
2 dt
dt 4 β
2β
¶¸
+
δ
1 d
a(t, v, v) + |v 0 (t)|2 ≤ 0 .
2 dt
2
487
488
J. LIMACO FERREL and L.A. MEDEIROS
Integrating from 0 to t, we get:
µ
¶
Z
1 0 2
1 c 1
δ t 0
1
M
|v (t)| +
kv(t)k2 + a(t, v, v) +
|v (s)|2 ds ≤
2
4β
2β
2
2 0
µ
¶
1 c 1
1
1
2
2
M
kv0 k + a(0, v0 , v0 )
≤ |v1 | +
2
4 β0
2β0
2
for all t ≥ 0.
If we set
¶
µ
1
1 c 1
|v1 |2 +
M
kv0 k2 + a(0, v0 , v0 )
2
4 β0
2 β0
C1 =
we obtain the first estimate
1 0 2
m0
|v (t)| +
kv(t)k2 < C1
2
16 β 2
(6.16)
for all t > 0. Note that C1 does not depend on t, but depends on v0 , v1 , β0 ,
a(y, 0).
Estimate II. We multiply both sides of the equation (6.5)1 by −∆v 0 = −
∂ 2v0
∂y 2
and integrate on (0, 1).
We obtain:
(6.17)
µ
¶
1
1
d
1 d 0
kv (t)k2 +
M̌
kv(t)k2
|∆v(t)|2 +
2
2 dt
8β
2β
dt
1 d
+
2 dt
1
+
2
−
+
Z
Z
Z
1
0
1
a(y, t) (∆v) dy −
2
1 ∂b
0
∂y
µ
2
µ
∂v 0
∂y
¶2
¶
Z
1
0
µ
a0 (y, t) (∆v)2 dy
¶¯
∂v 0 ¯¯y=1
1
dy − b(y, t)
2
∂y ¯y=0
¯
∂a ∂v ∂v 0 ¯¯y=1
∂v 0
dy +
∂y
∂y ∂y ∂y ¯y=0
1
∂ ∂a ∂v
∂y ∂y ∂y
1
∂v ∂v 0
∂v ∂v 0 ¯¯y=1
∂
+ δ kv 0 (t)k2 = 0 .
c(y, t)
dy − c(y, t)
∂y
∂y ∂y
∂y ∂y ¯y=0
0
0
Z
·
¸
¯
The inequalities below for the various terms of this identity are true. In fact,
we have:
(6.18)
−
1
2
Z
1
0
a0 (y, t) (∆v)2 dy ≥
m0 β 0
|∆v|2 ,
32 β 3
489
KIRCHHOFF–CARRIER ELASTIC STRINGS
(6.19)
¯Z
¯
¯
¯
1
0
·
¯ Z
¯1
¯
¯2
(6.20)
(6.21)
¯
¸
µ
2 m0
∂ ∂a ∂v ∂v 0 ¯¯ m0 β 0
dy ≤
|∆v|2 +
∂y ∂y ∂y ∂y ¯
64 β 3
β0 2
0
1 ∂b µ ∂v 0 ¶2
∂y
∂y
¶1/2 µ
¯
µ
¶
¯
1 m0 1/2 0 2
¯
dy ¯ ≤
kv k ,
2 β0 2
¯Z 1 ·
¯
µ
¸
¶ µ
¶
¯
∂
∂v ∂v 0 ¯¯
1 m0 1/2 4+π 2
m0 β 0
2
¯
kv 0 k2 ,
¯ ∂y c(y, t) ∂y ∂y dy ¯ ≤ 64 β 2 |∆v| + 2β
2
π2
0
0
µ
¶2 ¯y=1
µ
¶
µ
¶
¯
β 0 ∂ v 0 (1, t) 2 β 0 ∂ v 0 (0, t) 2
¯
=
+
,
¯
2β
∂y
2β
∂y
y=0
(6.22)
∂v 0
−b(y, t)
∂y
(6.23)
¯
¯
¶2
¶2
µ
µ
¯ ∂a ∂v ∂v 0 ¯¯y=1 ¯
β 0 ∂v 0
β 0 ∂v 0
¯
¯
¯
+
(1,
t)
(0,
t)
≤
¯
¯
¯ ∂y ∂y ∂y ¯y=0 ¯
8 β ∂y
8 β ∂y
µ 0 ¶3 µ
¶2
β
β
+4
(6.24)
¶
4 + π2
kv 0 k2 ,
π2
π+1
π
|∆v|2 ,
¯
¯ ¯
¶2
¶2
µ
µ
¯
β 0 ∂v 0
∂v ∂v 0 ¯¯1 ¯¯
β 0 ∂v
¯
(1, t) +
(0, t)
¯ c(y, t)
¯ ≤
¯
∂y ∂y ¯0 ¯
8 β ∂y
8 β ∂y
µ 0 ¶3 µ
¶2
β
β
+
π+1
π
|∆v|2 .
Using (6.18)–(6.24) the following inequality can be obtained from (6.17):
(6.25)
¶
µ
1 d 0
1
d
1
kv (t)k2 +
M̌
kv(t)k2
|∆v(t)|2 +
2 dt
8 β2
2β
dt
1 d
+
2 dt
+
β0
Z
µ
1
0
2
a(y, t) (∆v) dy −
∂v 0
(1, t)
4 β ∂y
·
µ
¶2
5 m0
+ δ−
2 β0 2
+
µ
µ
β0
β
¶3 µ
β 0 ∂v 0
(0, t)
4 β ∂y
¶1/2 µ
4+π 2
π2
¶¸
π+1
π
¶2
¶2
kv 0 (t)k2 ≤ 0 .
The coefficient of |∆v|2 is assumed to be positive, i.e.,
µ
β 03 π+1
m0 β 0
−
32 β 3
β3
π
|∆v|2 +
¶2
>0.
m0 β 0
|∆v|2
32 β 2
490
J. LIMACO FERREL and L.A. MEDEIROS
From the condition (6.7) on δ, it follows from (6.25):
µ
¶
1
1 d 0
1
d
kv (t)k2 +
M̌
kv(t)k2
|∆v(t)|2 +
2
2 dt
8β
2β
dt
Z
1 d 1
δ
+
a(y, t) (∆v(t))2 dy + kv 0 (t)k2 ≤ 0 .
2 dt 0
2
(6.26)
Estimate III. Multiply both sides of (6.7)1 by w = −∆v and integrate on
(0, 1). We have:
µ
¶
1
1
M̌
kvk2 |∆v|2 − a(t, v, ∆v) −
−(v , ∆v) +
2
4β
2β
µ
¶ µ
¶
∂v 0
∂v
− δ(v 0 , ∆v) + b
, −∆v + c , −∆v = 0 .
∂y
∂y
00
Whence,
(6.27)
µ
¶
1
1
d
(∇v 0 , ∇v) − kv 0 k2 +
M̌
kvk2 |∆v|2 −
dt
4 β2
2β
µ
¶ µ
¶
∂v 0
∂v
δ d
2
kvk + b
, −∆v + c , −∆v = 0 .
− a(t, v, ∆v) −
2 dt
∂y
∂y
We obtain the inequalities:
(6.28)
(6.29)
(6.30)
a(t, v, −∆v) ≥
m0
8 β2
µ
¶
3 1
|∆v|2 ,
−
4 π
¯µ
¶¯
0
¯
¯
¯ b(y, 1) ∂v , −∆v ¯ ≤ m0 |∆v|2 + kv 0 k2 ,
¯
¯
∂y
32 β 2
¯µ
¶¯
¯
¯
¯ c(y, t) ∂v , −∆v ¯ ≤ m0 |∆v|2 .
¯
¯
∂y
16 πβ 2
By (6.28), (6.29), (6.30), we modify (6.27) obtaining:
(6.31)
δ d
m0
d
(∇v 0 , ∇v) +
kvk2 +
|∆v|2 − 2 kv 0 k2 ≤ 0 ,
dt
2 dt
8 β2
m0
> 0 and the coefficient of |∆v|2 is also positive. Note that δ is
2
δ
a fixed parameter. Multiply (6.31) by and adding to (6.26) we cancel −2 kv 0 k2 ,
4
since M̌ (λ) >
491
KIRCHHOFF–CARRIER ELASTIC STRINGS
obtaining:
"
µ
¶
δ
1 d
δ2 0 2
1
1
kv 0 k2 + (∇v 0 , ∇v) +
kv k +
kvk2 ) |∆v|2 +
M̌
2
2 dt
4
8
8β
2β
(6.32)
1
+
2
≤
·
µ
Z
1
2
#
a(y, t) (∆v) dy +
0
1
1
M̌
kv(t)k2
2
8β
2β
¶¸0
m0 δ
|∆v|2 ≤
32 β 2
|∆v|2 .
m0
Note that M̌ (λ) =
+ m1 λ with m0 , m1 > 0. Then M̌ 0 (λ) = m1 is constant.
2
We obtain:
·
µ
1
1
M̌
kv(t)k2
8 β2
2β
¶¸0
µ
¶
1
β0
m1
= − 2 M̌
kvk2 +
(∇v, ∇v 0 )
4β
2β
8 β3
β0
kvk2 ,
with − β 0 = α0 < 0 ,
−
16 β 4
or
·
(6.33)
µ
1
1
M̌
kvk2
2
8β
2β
¶¸0
≤
m1 kvk 0
kv k .
8 β2 β
From Estimate I, we have:
µ
C1
kvk
≤4
β2
m0
(6.34)
with
C1 =
¶1/2
¶
µ
1 c 1
1
|v1 |2 +
M
kv0 k2 a(0, v0 , v0 ) .
2
4 β0
2 β0
Then, from (6.31) and (6.32) it follows:
(6.35)
·
µ
1
1
M̌
kvk2
2
8β
2β
¶¸0
µ
m 1 C1
|∆v| ≤
2 β m0
2
¶1/2
kv 0 k |∆v|2 .
Then, from (6.31) and (6.34), we have:
(6.36)
"
µ
¶
1
δ2 0 2
1
δ
1 d
kv k +
M̌
kvk2 |∆v|2 +
kv 0 k2 + (∇v 0 , ∇v) +
2
2 dt
2
8
8β
2β
+
Z
1
2
#
a(y, t) (∆v) dy +
0
µ
µ
m0
m 1 C1
δ−
32
2 m0
¶1/2
kv 0 k
¶
|∆v|2
≤ 0.
β2
492
J. LIMACO FERREL and L.A. MEDEIROS
With the notations of Theorem 6.1, we obtain from (6.35):
µ
¶
1 0
m0
|∆v|2
H (t) ≤ −
δ + γ(t)
2
32
β2
(6.37)
where
γ(t) =
and kv 0 (t)k <
p
µ
C1
m0
¶1/2
m1 0
kv (t)k
2
H(t).
m0 δ
and let us see that this leads us to a contradiction.
Suppose that γ(t) ≥
32
In fact, we have:
1 m1 q
C1 H(t) ,
γ(t) < √
m0 2
then
1 m1 q
C1 H(0) .
γ(0) ≤ √
m0 2
(6.38)
By assumption (6.10) of Theorem 6.1 about v0 , v1 , we obtain, from (6.37):
γ(0) <
m0 δ
.
32
mδo
. This minimum exists and it is
32
greater than zero, by continuity of γ(t). We have:
n
Let us consider t∗ = min t > 0, γ(t) =
(6.39)

m0


on 0 ≤ t < t∗ ,
 γ(t) < 32 δ
m


 γ(t∗ ) = 0 δ .
32
We obtain:
∗
(6.40)
H(t ) − H(0) =
Z
t∗
H 0 (s) ds .
0
By (6.34) we have H 0 (s) < 0 on (0, t∗ ), then H(t∗ ) < H(0) or H(t∗ )1/2 < H(0)1/2 .
Consequently,
¶
¶
µ
µ
1
1
m0
m1 q
m1 q
∗
C1 H(t ) < √
C1 H(0) <
γ(t ) < √
δ
m0 2
m0 2
32
∗
which is in contradiction with (6.39)2 .
KIRCHHOFF–CARRIER ELASTIC STRINGS
493
Then, from (6.37) we have H 0 (t) ≤ 0 for all t ≥ 0 and, therefore, the global
estimate
kv 0 (t)k2 + |∆v(t)|2 < c,
(6.41)
for all t ≥ 0 .
The estimate for v 00 (t) and the uniqueness can be proved as in the local case.
7 – Asymptotic behaviour
In this section we investigate the behaviour, when t goes to infinity, of a
perturbed energy associated to the global solution v obtained in Theorem 6.1.
In fact we consider
(7.1)
µ
¶
1 c 1
1
1
kv(t)k2 + a(t, v(t), v(t)) ,
M
E(v, t) = |v 0 (t)|2 +
2
4β
2β
2
which we still call the energy associated to the solution v of the mixed problem
(6.5).
m0
c(λ) = m0 λ + m1 λ2 and (7.1)
Remember that M̌ (λ) =
+ m1 λ, then M
2
2
2
takes the form
(7.2)
E(v, t) =
m0
m1
1
1 0 2
|v (t)| +
kv(t)k2 +
kv(t)k4 + a(t, v(t), v(t)) .
2
16 β 2
32 β 3
2
Observe that E(v, t) depends on the boundary β = β(t) of the noncylindrical
b and that −α(t) = β(t).
domain Q,
Theorem 7.1. If v is the global solution of the mixed problem (6.5), then
we have
R
E(v, t) ≤ C1 e
−C2
t ds
0 β(s)2
,
for all t > 0.
Remark 7.1. Note that C1 = C0 β02 E(v0 , 0), C2 =
C0 =
with k such that 20 k =
δ
.
4
1 + 2 k 16 (k + δ)
+
m0
β02
δ
and
60 C0
494
J. LIMACO FERREL and L.A. MEDEIROS
Proof of Theorem 7.1: Let us consider the perturbation of E(v, t) given
by:
h
F (v, t) = E(v, t) + k 2 (v 0 (t), v(t)) + δ |v 0 (t)|2
(7.3)
i
where k is a constant to be fixed; see Remark 7.1. This method of working with a
perturbed energy F (v, t) was employed by Komornik–Zuazua [11]. We also refer
to Nakao–Narazaki [15] and Hosya–Yamada [9].
Since
¯
¯
k 0 2
¯
¯
|v (t)| + k δ |v(t)|2 ,
¯2 k (v 0 (t), v(t))¯ ≤
(7.4)
δ
we obtain:
k
2 k (v 0 (t), v(t)) + k δ |v(t)|2 ≥ − |v 0 (t)|2 .
δ
(7.5)
Then, from (7.4) and (7.5) we modify (7.3), obtaining:
F (v, t) ≥
µ
¶
1 k
m0
m1
1
|v 0 (t)|2 +
−
kv(t)k2 +
kv(t)k4 + a(t, v, v) .
2
3
2 δ
16 β
32 β
2
If we take k > 0 such that
1
k
< we obtain:
δ
4
(7.6)
F (v, t) ≥
1
E(v, t) .
2
From (7.3) we also have
(7.7)
F (v, t) ≤ E(v, t) + k |v 0 (t)|2 + (k + δ) |v(t)|2 ,
since
|v 0 (t)|2 ≤ 2 E(v, t)
and
|v(t)|2 ≤
16 2
β (t) E(v, t) .
m0
From (7.7) we obtain:
µ
¶
k+δ
F (v, t) ≤ (1 + 2 k) E(v, t) +
16 β 2 E(v, t)
m0
¶
µ
k+δ
1 + 2k 2
16 β 2 E(v, t) ,
β E(v, t) +
≤
β0
m0
since β(t) ≥ β(0) = β0 . Then:
(7.8)
F (v, t) ≤ C0 β 2 E(v, t) ,
495
KIRCHHOFF–CARRIER ELASTIC STRINGS
with
C0 =
1 + 2 k 16 (k+δ)
.
+
m0
β02
Taking the time derivative of F (v, t) we have:
(7.9)
h
i
dE
dF
=
+ k 2 (v 00 (t), v(t)) + 2 |v 0 (t)|2 + 2 δ (v 0 (t), v(t)) .
dt
dt
From the first estimate in Section 7, we have:
dE
δ
≤ − |v 0 (t)|2 .
dt
2
(7.10)
Multiplying both sides of (6.5)1 by v and integrating on (0, 1), we obtain:
(7.11)
¶
µ
1
1
M̌
kv(t)k2 kv(t)k2 + a(t, v(t), v(t)) +
4 β2
2β
µ
¶ µ
¶
∂v 0
∂v
+ b(y, t)
, v(t) + c(y, t) , v(t) + δ(v 0 (t), v(t)) = 0 .
∂y
∂y
(v 00 (t), v(t)) +
We verify that:
(7.12)
(7.13)
(7.14)
¯µ
µ 0 ¶2
¶¯
0
¯
¯
¯ b(y, t) ∂v , v(t) ¯ ≤ 1 β
kv(t)k2 + 9 |v 0 (t)|2 ,
¯
¯
∂y
4 β
¯µ
µ 0 ¶2
¶¯
¯
¯
¯ c(y, t) ∂v , v(t) ¯ ≤ 1 β
kv(t)k2 ,
¯
¯
∂y
2 β
¶
µ
a(t, v(t), v(t)) ≥
1
4 β2
m0
− β 02 kv(t)k2 .
2
Then, from (7.12), (7.13) and (7.14) we obtain:
(7.15)
¶
µ
µ
¶
∂v
∂v 0
, v(t) + c(y, t)
, v(t) ≥
a(t, v(t), v(t)) + b(y, t)
∂y
∂y
¶
µ
1
m0
02
≥
kv(t)k2 − 9 |v 0 (t)|2
−
4
β
4 β2 2
≥ −9 |v 0 (t)|2 ,
m0
− 4β 02 > 0.
2
Then, from (7.11), (7.15) and by definition of M̌ (λ), we have:
since
(7.16)
(v 00 (t), v(t)) + δ(v 0 (t), v(t)) ≤ −
m1
m0
kv(t)k2 −
kv(t)k4 + 9 |v 0 (t)|2 .
2
8β
8 β3
496
J. LIMACO FERREL and L.A. MEDEIROS
Then, substituting (7.10) and (7.16) in (7.9), we obtain:
"
δ
m0
m1
dF
≤ − |v 0 (t)|2 + k 2 |v 0 (t)|2 −
kv(t)k2 −
kv(t)k4 + 18 |v 0 (t)|2
2
dt
2
4β
4 β3
or
dF
+
dt
µ
¶
δ
k m0
k m1
− 20 k |v 0 (t)|2 +
kv(t)k2 +
kv(t)k2 ≤ 0 .
2
2
4β
4 β2
Taking k such that 20 k =
(7.17)
δ
, we get
4
·
¸
dF
δ
δ m0
m1
+ |v 0 (t)|2 +
kv(t)k2 +
kv(t)k2 ≤ 0 .
2
dt
4
20 16 β
16 β 3
We have:
0 < a(t, v(t), v(t)) ≤
m0
kv(t)k2
8 β2
which substituted in (7.2) gives:
E(v, t) ≤
(7.18)
m1
1 0 2 3 m0
|v (t)| +
kv(t)k2 +
kv(t)k4 ,
2
2
16 β
32 β 3
µ
δ
δ
m1
m0
δ
E(v, t) ≤ |v 0 (t)|2 +
kv(t)k2 +
kv(t)k4
2
20
4
20 16 β
16 β 3
Substituting (7.18) in (7.17) we obtain:
(7.19)
δ
dF
+
E(v, t) ≤ 0
dt
60
for all t ≥ 0 .
From (7.8) and (7.19) we obtain:
F (v, t) ≤ F (v0 , 0) e
− 60δC
0
Rt
.
Rt
.
ds
0 β(s)2
From (7.6), this last inequality implies:
E(v, t) ≤ 2 F (v0 , 0) e
− 60δC
0
ds
0 β(s)2
Applications
Case 1. Suppose that m > 2, is an integer and
1
β(t) = (t + t0 ) m ,
t, t0 > 0 .
¶
.
#
497
KIRCHHOFF–CARRIER ELASTIC STRINGS
We have:
m C2
E(v, t) ≤ Cm e− m−2 (t+t0 )
m−2
m
which gives an exponential decay.
Case 2. Suppose that m = 2. Then
β(t) = (t + t0 )1/2 .
We obtain:
E(v, t) ≤ C1 (t + t0 )−C2 ,
C2 > 0 ,
which gives an algebraic decay.
Remark 7.2. We have:
0 2
Z µ
2 β y = x+β ,
2 β dy = dx ,
Z
¶
1 ∂v 2
m0
dy
(v ) dy +
16 β 2 0 ∂y
0
µZ 1µ ¶2 ¶2
µ ¶2
Z
m1
∂v
∂v
1 1
+
a(y,
t)
dy
dy .
+
32 β 3 0 ∂y
2 0
∂y
1
E(v, t) =
2
1
If
∂v
∂u
=
,
∂x
∂y
2β
then the above quantity, in the cylinder Q, is transformed, into the noncylindrical
b in the following one
domain Q,
" Z
1 1
E(u, t) =
2β 2
m1
+
4
+βµ β 0 x
∂u
+ u0
β ∂x
−β
µZ
+βµ ∂u ¶2
∂x
−β
dx
¶2
¶2
+
c(λ) = m0 λ + m1 λ2 . Then
We have M
2
2
(7.20)
" Z
1 1
E(u, t) =
2β 2
+
Z
+βµ β 0 x
−β
0
2
Z
+βµ m
−
µ
β0x
β
0
2
−β
∂u
+ u0
β ∂x
−β
+βµ m
m0
dx +
4
¶2
¶2 ¶ µ
+βµ ∂u ¶2
∂x
−β
−
µ
β0x
β
∂u
∂x
1 b
E(u, t)
2β
¶2
#
dx .
dx
¶2 ¶ µ
1 c
dx + M
2
The last integral is positive, then:
E(u, t) ≥
Z
µZ
∂u
∂x
¶2
dx .
+βµ ∂u ¶2
−β
∂x
#
dx
¶
498
J. LIMACO FERREL and L.A. MEDEIROS
where
1
b
E(u,
t) =
2
Z
+βµ β 0 x
∂u
+ u0
β ∂x
−β
¶2
µZ
1 c
dx + M
2
+βµ ∂u ¶2
∂x
−β
dx
¶
.
b
The behaviour of E(u,
t) depends essentially on β because
Rt
1 b
C2
E(u, t) ≤ β C1 e
2
ds
0 β(s)2
.
For example, if β(t) = (1+t)1/2 , then
1 b
E(u, t) ≤ C1 (1 + t)2−C2 ,
2
and we need C2 > 2 to have a decay of algebraic type. Note that C2 depends on
m0 , δ and β0 .
Remark 7.3. We found in (7.20)
· Z
µ
¶¸
Z
µ
¶
2
1 +β β 0 ∂u 2
1 1 +β β 0 ∂u
dx
x
+ u0
x
dx +
E(u, t) =
2 β 2 −β β ∂x
2 −β β ∂x
µZ +βµ
¶ Z +β ·
¶
µ
¶
¶ ¸µ
∂u 2
m0 3 β 0 2 ∂u 2
1 c
dx +
dx .
+ M
−
x
2
∂x
2
2 β
∂x
−β
−β
Observe that
0 2
(u ) =
µ
β 0 ∂u β 0 ∂u
u + x
− x
β ∂x
β ∂x
0
Then,
1
E(u, t) ≥
2β
ÃZ
+β
−β
If we represent by
we have
b
b
E(u,
t) =
Z
β
¶2
1 c
(u ) dx + M
2
0 2
(u0 )2 dx +
−β
E(u, t) ≤
µ
β 0 ∂u
1
u0 + x
≤
2
β ∂x
µZ
1 c
M
2
1 bb
E(u, t),
2β
µZ
−β
µ
∂x
∂u
∂x
¶2
1
δ
120 C0 β
for all t ≥ 0
dx
dx
for all t ≥ 0 .
Now suppose that β 0 is strictly positive and
0 < β 0 (t) <
1
+
2
+βµ ∂u ¶2
−β
β
¶2
¶
µ
β 0 ∂u
x
β ∂x
¶!
.
¶2
.
KIRCHHOFF–CARRIER ELASTIC STRINGS
with
µ
m0
δ
<C
120 C0 β0
2
¶1/2
499
.
Note that δ, m0 , C0 , C are the constants fixed above.
We obtain
Z t
Z t 0
ds
β
δ
−
≤ −2
ds
2
60 C0 0 β (s)
0 β
or
µ
Z t ¶
δ
β02
ds
exp −
<
.
60 C 0 β 2
β(t)2
Substituting in the inequality of Theorem 7.1, we obtain:
E(v, t) ≤
bb
But E(u,
t) < 2 β E(v, t). This gives
b
b
E(u,
t) ≤
C0 β 0
E(v, 0) .
β 2 (t)
2 C0 β04 E(v0 , 0)
.
β(t)
ACKNOWLEDGEMENT – We take this opportunity to express our appreciation to
E. Zuazua who read the first version of this work, proposing substantial modifications.
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J. Limaco Ferrel,
Instituto de Matemática UFF,
Rua São Paulo s/n, CEP 24210-110 Niterói RJ – BRASIL
and
L.A. Medeiros,
Instituto de Matemática UFRJ,
C.P. 68530 – CEP 21944-970, Rio de Janeiro, RJ – BRASIL