Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2011, Article ID 879649, 14 pages
doi:10.1155/2011/879649
Research Article
Exponential Stability of Two Coupled
Second-Order Evolution Equations
Qian Wan and Ti-Jun Xiao
Shanghai Key Laboratory for Contemporary Applied Mathematics, School of Mathematical Sciences,
Fudan University, Shanghai 200433, China
Correspondence should be addressed to Ti-Jun Xiao, xiaotj@ustc.edu.cn
Received 30 October 2010; Accepted 21 November 2010
Academic Editor: Toka Diagana
Copyright q 2011 Q. Wan and T.-J. Xiao. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
By using the multiplier technique, we prove that the energy of a system of two coupled second
order evolution equations one is an integro-differential equation decays exponentially if the
convolution kernel k decays exponentially. An example is give to illustrate that the result obtained
can be applied to concrete partial differential equations.
1. Introduction
Of concern is the exponential stability of two coupled second-order evolution equations one
is an integro-differential equation in Hilbert space H
u t Aut αut −
t
√
kt − sAusds β Avt 0,
1.1
0
√
v t Avt β Aut 0,
1.2
with initial data
u0 u0 ,
v0 v0 ,
u 0 u1 ,
v 0 v1 .
1.3
Here A : DA ⊂ H → H is a positive self-adjoint linear operator, α
0, β > 0, kt is a
nonnegative function on 0, ∞. Moreover, the fractional power A1/2 is defined as in the well
known operator theory cf, e.g., 1, 2.
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An interesting and difficult point for it is to stabilize the whole system via the damping
effect given by only one equation 1.1. We remark that there is very few work concerning the
situation when the damping mechanism is given by memory terms; see 3, where a coupled
Timoshenko beam system is investigated.
On the other hand, the stability of the single integro-differential equation has been
studied extensively; see, for instance, 4, 5.
In this paper, through suitably choosing multipliers for the energy together with
other techniques, we obtain the desired exponential decay result for the system 1.1–1.3.
Nonlinear coupled systems with general decay rates will be discussed in a forthcoming paper.
In Section 2, we present our exponential decay theorem and its proof. An application
is given in Section 3.
2. Exponential Decay Result
We start with stating our assumptions:
1 A is a self-adjoint linear operator in H, satisfying
Au, u
a u 2,
u ∈ DA,
2.1
where a > 0.
2 α > 0, β > 0 are constants. kt : 0, ∞ → 0, ∞ is locally absolutely continuous,
satisfying
k0 > 0,
1−
∞
ktdt −
0
1 α − β2 > 0,
a
2.2
and there exists a positive constant λ, such that
k t −λkt,
for a.e. t
0.
2.3
We define the energy of a mild solution u of 1.1–1.3 as
t
2
2 1 2 1 − 0 ksds 1
√
Et Eu t :
u t v t Aut
2
2
2
√
2
√
1 t
kt − s Aus − Aut ds
2 0
2 1 1
√
Avt βut α − β2 ut 2 .
2
2
The following is our exponential decay theorem.
2.4
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Theorem 2.1. Let the assumptions be satisfied. Then,
√
i for any u0 , v0 ∈ D A and u1 , v1 ∈ H, problem 1.1–1.3 admits a unique mild
solution on 0, ∞.
√
The solution is a classical one, if u0 ∈ DA, v0 ∈ DA and u1 , v1 ∈ D A,
ii there exists a constant C > 0 such that the energy
Et E0e1−Ct ,
∀t
2.5
0,
for any mild solution of 1.1–1.3.
Proof. We denote
wt ut
vt
A
,
w0 t √
Aα β A
,
√
β A A
u0 t
v0 t
,
w1 t Bt ktA 0
0
0
u1 t
v1 t
,
2.6
.
Then, 1.1–1.3 becomes
w t Awt −
t
Bt − swsds 0 t ∈ 0, ∞,
0
w0 w0 ,
2.7
w 0 w1 ,
in H : H × H. From the assumptions, one sees that—A is the generator of a strongly
1,1
continuous cosine function on H, and B· is bounded from DA into Wloc
0, ∞; H.
Therefore, we justify the assertion i cf., e.g. 6.
Suppose now that u is a classical solution of 1.1–1.3. We observe
√
2 1 t
√
2
√
1
E t − kt Aut k t − s Aus − Aut ds
2
2 0
0,
2.8
by Assumption 2 and so
Et Es,
0 s t T.
2.9
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Let
μ : 1 −
∞
0
α − β2 ,
ktdt −
a
2.10
and take 1 < δ < 1 aμ/2β2 . We have
√
2
1
v t2 μ Aut
2
4
2
2
1 − δβ2 1 1
√
√
−
Avt Aut ,
2 2δ
2a
1
u t2 2
Et
2.11
t ∈ 0, ∞.
Furthermore, we need the following lemmas.
Lemma 2.2. For any T
such that
S
0 and for any ε1 > 0, there exist positive numbers D1 ε1 , D2 ε1 T 2
√
Avt βut dt
S
D1 ES D2
G1
T
S
T t
S
√
2
√
kt − s Aus − Aut ds dt
0
2
u t dt G2 1
ε1
T
S
2
v t dt G3 ε1 G4 2.12
T √
2
Aut dt,
S
for some positive constants Gi i 1, 2, 3, 4 which only depend on α, β, a, and k.
√
Proof. At first, let us take the inner product of both sides of 1.1 with Avt and integrate
over S, T. Then, noticing 1.2, we obtain
T
S
√
u t, Avt dt −
α
β
T
S
S
√
Aut, v t dt − β
√
ut, Avt dt T t
S
T
0
T t
S
T 2
√
Aut dt
S
√
kt − s Ausds, v t dt
0
T √
2
√
√
kt − s Ausds, Aut dt β
Avt dt 0.
S
2.13
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For the first item, integrating by parts, we have
T
S
T T
√
√
√
u t, Avt dt u t, Avt dt −
u t, Av t dt.
S
S
2.14
The second and the fifth items can be treated similarly. Therefore,
T
T √
2
√
ut, Avt dt
β
Avt dt α
S
−
−
S
S
S
√
kt − s Ausds, v t
dt
0
T
S
1−
t
√
2
ksds Aut dt
0
S
2.15
√
√
k t − s Aus − Aut ds, v t dt
T t
−β
T √
√
u t, Avt dt Aut, v t dt
T t
β
S
T 0
√
√
√
kt − s Aus − Aut ds, Aut dt
T t
S
T
S
0
√
Aut, v t dt.
kt
Then, taking the inner product of both sides of 1.1 with ut and integrating over S, T, we
obtain
β
T √
S
Avt, ut dt α
−
−
T
S
u t, ut
T t
S
S
ut 2 dt
dt T
S
2
u t dt
√
√
√
kt − s Aus − Aut ds, Aut dt
0
T
S
T
1−
t
0
√
2
ksds Aut dt.
2.16
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Equation 2.15 × β/α 2.16 yields that
T √
2
Avt βut dt
S
−
T
S
β
−
α
β
−
α
u t, ut
T S
β
α
√
kt − s Ausds, v t
dt
0
t
T
√
2
1 − ksds Aut dt
S
T t
S
S
S
2
u t dt
√
β T √
u t, Avt dt Aut, v t dt
α S
S
T
dt T
T t
β2 − α
α
β
α
0
2.17
√
√
k t − s Aus − Aut ds, v t dt
0
kt
α − β2
α
β2 − α
−
α
√
Aut, v t dt
T t
S
√
√
√
kt − s Aus − Aut ds, Aut dt
0
T √
2
T
2
ut 2 dt.
Avt dt − α − β
S
S
Next, we will estimate all the terms on the right side of 2.17. From 2.11, we have
the following estimate:
T
T √
√
2MES,
u t, Avt dt u t, Avt
S
S
2.18
T
where M is a positive constant. Those terms of the form S ·, · dt can be similarly treated.
Denote by J the sum of the other terms on the right of 2.17.
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7
Using Young’s inequality and noting 2.8, we get, for ε1 > 0,
T t
S
√
√
k t − s Aus − Aut ds, v t dt
0
ε1
2
T t
− ε21
S
√
√
k t − s
Aus − Autds
2
dt 0
T t
S
k sds
0
ε1 k0ES 2ε1
T
S
2
v t dt.
2
√
√
1 T
k t − s
v t2 dt.
Aus
−
Aut
ds
dt
2ε1 S
0
t
T
1
1
2ε1
S
2
v t dt.
2.19
The treatment of the other terms of J is similar, giving
√
2
√
k2 0β2 T ε1 T v t2 dt,
kt Aut, v t dt Aut dt 2
2 S
2ε1 α
S
S
2 T t
√
√
√
α−β
kt − s Aus − Aut ds, Aut dt
α
S
0
β
α
T
2 T t
t
√
√
2
2
√
α − β2
ε1 T ksds
kt
−
s
Aus
−
Aut
ds
dt
Aut
dt.
2
2
2α ε1
S 0
0
S
2.20
Thus, we obtain
J
ε1 k0β
ES
α
2 T t
t
√
2
√
α − β2
ksds
kt
−
s
Aus
−
Aut
ds dt
2α2 ε1
S 0
0
T √
2
1 1 2
Aut dt
β − α ε1
α a
S
T
T √
T
2
2
2
2
2
k 0β
β
v t dt u t dt θ
Avt
dt,
2ε1 α
2ε1 α2
S
S
S
2.21
where θ max{α − β2 /α, 0}. Make use of the estimate
√ 2
Av
√
1 ζ1 2
1
Av βu 1 ζ1
β2 √ 2
Au ,
a
2.22
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where ζ1 is a positive constant, small enough to satisfy 1 ζ1 θ < 1. We thus verify our
conclusion.
Lemma 2.3. For any T
such that
μ
2
S
0 and for any ε2 > 0, there exist positive numbers D3 ε2 , D4 ε2 ,
T t
T √
2
√
2
√
Aut
dt
D
ES
D
kt − s Aus − Aut ds dt
3
4
S
S
β
1
2ε2 a
0
T
2
S
2
u t dt ε2
2
T
S
2
v t dt.
2.23
Proof. We denote w A−1/2 u and take the inner product of both sides of 1.2 with wt, and
integrate over S, T. It follows that
−
T
T
T √
T
Avt, ut dt v t, wt dt −
v t, w t dt β
ut 2 dt.
S
S
S
S
2.24
Plugging this equation into 2.16, we find
T
S
1−
t
√
2
ksds Aut dt
0
−
T
S
u t, ut
dt β
T
S
v t, wt dt
T
T
2
2
2
u t dt
β −α
ut dt T t
−β
S
S
S
2.25
√
√
√
kt − s Aus − Aut ds, Aut dt
0
T
S
v t, w t dt.
Observe
T t
S
√
√
√
kt − s Aus − Aut ds, Aut dt
0
2δ1
T t
3
S
0
ksds
t
√
√
2
2
√
δ3 T kt − s Aus − Aut ds dt Aut dt,
2
0
S
2.26
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where δ3 μ, and for ε2 > 0
β
T
ε2
v t, w t dt 2
S
T
S
2
2
v t dt β
2aε2
T
S
2
u t dt.
2.27
The other items on the right of 2.25 can be dealt with as in the proof of Lemma 2.2. Hence,
we get the conclusion.
Lemma 2.4. For any T
T
S
S
2
u t dt 0, there exist positive numbers D5 , D6 such that
T
S
2
v t dt
D5 ES D6
T t
S
3 α − β 2 2
a
√
2
√
kt − s Aus − Aut ds dt
2.28
0
T √
T √
2
2
Aut
dt
Avt βut dt.
S
S
Proof. Taking the inner product of both sides of 1.2 with vt and integrating over S, T,
we see
T
S
2
v t dt T
S
v t, vt dt T √
T √
2
Avt
dt
β
Aut, vt dt.
S
S
2.29
Combining this equation and 2.16 gives
T
S
2
v t dt T
S
−
T
S
2
u t dt
u t, ut dt T t
S
T
S
v t, vt dt t
T
√
2
1 − ksds Aut dt
S
0
√
√
√
kt − s Aus − Aut ds, Aut dt
0
T √
2
T
2
ut 2 dt.
Avt βut dt α − β
S
This yields the estimate as desired.
S
2.30
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Lemma 2.5. Let S0 > 0 be fixed. For any T
D7 ε3 , D8 ε3 such that
T
S
2
u t dt D7 ES D8
S
S0 and for any ε3 > 0, there exist positive numbers
T t
S
√
2
√
kt − s Aus − Aut ds dt
0
T √
T √
2
2
ε3
Aut dt ε3
Avt βut dt.
S
S
Proof. Take the inner product of both sides of 1.1 with
over S, T. This leads to
T t
S
2.31
t
0
kt−sus−utds and integrate
2
ksdsu t dt
0
T t
−
u t, kt − sus − utds
S
dt
0
T
t
u t, k t − sus − utds dt
S
0
t
t
T
√
√
√
1 − ksds
Aut, kt − s Aus − Aut ds dt
−
S
0
0
2.32
T
t
−α
ut, kt − sus − utds dt
S
0
2
T t
√
√
kt − s Aus − Autds dt
S 0
T √
t
β
Avt, kt − sus − utds dt.
S
0
Just as in the proofs of the above lemmas, using Young’s inequality and noting that
T t
S
we prove the conclusion.
0
2
ksdsu t dt
S0
0
ksds
T
S
2
u t dt,
2.33
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Proof of Theorem 2.1 (continued). From Assumption 2 and 2.8, we have
T t
S
−
1
λ
√
2
√
kt − s Aus − Aut ds dt
0
T t
S
− λ1
√
2
√
k t − s Aus − Aut ds dt
0
T
S
2.34
E tdt
λ2 ES.
Now, fix S0 > 0. Thanks to Lemmas 2.2 and 2.3, we know that for any T
for η > 0,
μ
2
S
S0 and
T √
T √
2
2
Aut dt η
Avt βut dt
S
S
T √
2
2
D3 ηD1 D4 ηD2 λ ES ηG3 ε1 G4 Aut dt.
S
T
T
2
2
β2
ε2
1
u t dt.
v t dt 1 ηG2
ηG1
2
ε1
2ε
a
2
S
S
2.35
Moreover, by the use of Lemmas 2.4 and 2.5, we have
μ
2
T √
T √
2
2
Aut dt η
Avt βut dt
S
S
T √
T √
2
2
p0 ES p1 Aut dt p2 Avt βut dt,
S
2.36
S
where
β2
ηG1
D7 1 p0 D3 ηD1 D5
2ε2 a
β2
2
1
ε2
ηG2
ηG1
,
D8 1 D4 ηD2 D6
2
ε1
2ε2 a
λ
β2
1
3 α − β2 ε2
p1 ηG3 ε1 G4 ηG2
ηG1 ,
ε3 1 2
a
2
ε1
2ε2 a
β2
1
ε2
ηG2 ε3 1 ηG1 .
p2 2
ε1
2ε2 a
1
ε2
ηG2
2
ε1
2.37
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Let
ε1 ε−1 ,
η ε2 ε2 ,
ε3 ε5 .
2.38
Taking ε small enough gives
μ
,
2
p1 <
p2 < η.
2.39
Therefore, there is a constant N1 > 0 such that
T √
T √
2
2
Aut dt Avt βut dt N1 ES
S
S
2.40
by 2.36. Using Lemma 2.4 and 2.34, we deduce that for some N2 > 0,
T
S
T
2
u t dt S
2
v t dt
1
ES
λ
T T 2
2
3 α − β2 √
√
Aut dt Avt βut dt
2
a
S
S
D5 2D6
2.41
N2 ES.
Next, define
2 √
2
2 2 √
Ht : u t v t Aut Avt βut
t
√
2
√
kt − s Aus − Aut ds.
2.42
0
It is easy to see that there exist M1 , M2 > 0 such that M1 Et Ht M2 Et. Therefore,
T
T
S
S
Htdt Etdt 1
M1
T
S
N1 N2 Htdt 2
ES,
λ
N1 N2 2/λ
ES.
M1
2.43
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On the other hand, when 0 S S0 ,
T
S
S0
Etdt S
Etdt T
S0
Etdt
2 2/λ
S0 − SES N1 N
ES0 M
2.44
1
S0 N1 N2 2/λ
ES,
M1
that is,
T
S
Etdt NES,
∀S
2.45
0.
By a standard approximation argument, we see that 2.45 is also true for mild solutions.
From this integral inequality, we complete the proof cf., e.g., 7, Theorem 8.1.
3. An Example
Example 3.1. Consider a coupled system of Petrovsky equations with a memory term
∂2t ut, ξ Δ2 ut, ξ αut, ξ −
t
kt − sΔ2 ut, ξds βΔvt, ξ 0,
∂2t vt, ξ Δ2 vt, ξ βΔut, ξ 0,
ut, ξ vt, ξ Δut, ξ Δvt, ξ 0,
u0, ξ u0 ξ,
t
0, ξ ∈ Ω,
0
v0, ξ v0 ξ,
∂t u0, ξ u1 ξ,
t
0, ξ ∈ Ω,
t
3.1
0, ξ ∈ ∂Ω,
∂t v0, ξ v1 ξ,
ξ ∈ Ω,
where Ω is a bounded open domain in ÊN , with sufficiently smooth boundary ∂Ω and α, β, k
as in Assumption 2. Let H L2 Ω with the usual inner product and norm. Here, we denote
by ∂t u the time derivative of u and by Δu the Laplacian of u with respect to space variable ξ.
Define A : DA ⊂ H → H by
A Δ2 ,
with DA H 4 Ω ∩ H02 Ω.
3.2
Then, Assumption 1 is satisfied. Therefore, we claim in view of Theorem 2.1 that the energy
of the system decays exponentially at infinity.
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Acknowledgments
The authors would like to thank the referees for their comments and suggestions. This
work was supported partially by the NSF of China 10771202, 11071042, the Research Fund
for Shanghai Key Laboratory for Contemporary Applied Mathematics 08DZ2271900 and
the Specialized Research Fund for the Doctoral Program of Higher Education of China
2007035805.
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