Volume 2011, Article ID 879649, 14 pagesdoi:10.1155/2011/879649 Research Article Exponential Stability of Two Coupled Second-Order Evolution Equations Qian Wan and Ti-Jun Xiao Shanghai K
Trang 1Volume 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
Copyrightq 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 equationsone is an integro-differential equation decays exponentially if the convolution kernelk 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 equationsone
is an integro-differential equation in Hilbert space H
ut Aut αut −
t
0kt − sAusds β√Avt 0, 1.1
with initial data
u0 u0, v0 v0, u0 u1, v0 v1. 1.3
HereA : DA ⊂ H → H is a positive self-adjoint linear operator, α 0,β > 0, kt is a
nonnegative function on0, ∞ Moreover, the fractional power A1/2is defined as in the well known operator theorycf, e.g., 1,2
Trang 2An 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; see3, 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 system1.1–1.3 Nonlinear coupled systems with general decay rates will be discussed in a forthcoming paper
InSection 2, we present our exponential decay theorem and its proof An application
is given inSection 3
2 Exponential Decay Result
We start with stating our assumptions:
1 A is a self-adjoint linear operator in H, satisfying
wherea > 0.
2 α > 0, β > 0 are constants kt : 0, ∞ → 0, ∞ is locally absolutely continuous,
satisfying
k0 > 0, 1 −
∞
0 ktdt − a1α − β2 > 0, 2.2 and there exists a positive constantλ, such that
We define the energy of a mild solutionu of 1.1–1.3 as
Et E u t : 1
2ut21
2vt21−
t
0ksds
2
√Aut2
1 2
t
0kt − s√
Aus −√Aut2
ds
1 2
√Avt βut2
1 2
α − β2
ut 2.
2.4
The following is our exponential decay theorem
Trang 3Theorem 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
EtE0e1−Ct, ∀t 0, 2.5
for any mild solution of 1.1–1.3.
Proof We denote
wt
ut
vt
, w0t
u0t
v0t
, w1t
u1t
v1t
,
A
A α β√A
β√A A
ktA 0
.
2.6
Then,1.1–1.3 becomes
wt Awt −
t
0Bt − swsds 0 t ∈ 0, ∞,
w0 w0, w0 w1,
2.7
in H : H × H From the assumptions, one sees that—A is the generator of a strongly
continuous cosine function on H, and B· is bounded from DA into W1,1
loc0, ∞; H.
Therefore, we justify the assertioni cf., e.g 6
Suppose now thatu is a classical solution of 1.1–1.3 We observe
Et −1
2kt√
Aut2
1 2
t
0
kt − s√
Aus −√Aut2
ds
0,
2.8
by Assumption2 and so
Trang 4μ : 1 −
∞
0 ktdt − α − β2
and take 1< δ < 1 aμ/2β2 We have
Et 1
2ut21
2vt2μ
4
√Aut2
2− 1
2δ
√Avt2
1 − δβ2
2a
√Aut2
, t ∈ 0, ∞.
2.11
Furthermore, we need the following lemmas
Lemma 2.2 For any T S 0 and for any ε1 > 0, there exist positive numbers D1ε1, D2ε1
such that
T
S
√Avt βut2
dt
D1ES D2
T
S
t
0
kt − s√
Aus −√Aut2
ds dt
G1
T
S ut2
dt G2
1
ε1
T
S vt2
dt G3ε1 G4
T
S
√Aut2
dt,
2.12
for some positive constants G i 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 of1.1 with√Avt and integrate
overS, T Then, noticing 1.2, we obtain
T
S
ut,√Avt dt −
T
S
√
Aut, vt dt − β
T
S
√Aut2
dt
α
T
S
ut,√Avt dt
T
S
t
0kt − s√Ausds, vt
dt
β
T
S
t
0kt − s√Ausds,√Aut
dt β
T
S
√Avt2
dt 0.
2.13
Trang 5For the first item, integrating by parts, we have
T
S
ut,√Avt dt
T
S
ut,√Avt dt −
T
S
ut,√Avt dt. 2.14
The second and the fifth items can be treated similarly Therefore,
β
T
S
√Avt2
dt α
T
S
ut,√Avt dt
−
T
S
ut,√Avt dt
T
S
√
Aut, vt dt
−
T
S
t
0
kt − s√Ausds, vt
dt
β
T
S
1−
t
0
ksds
√Aut2dt
T
S
t
0
kt − s√
Aus −√Autds, vt
dt
− β
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
T
S kt √
Aut, vt dt.
2.15
Then, taking the inner product of both sides of1.1 with ut and integrating over S, T, we
obtain
β
T
S
√
Avt, ut dt α
T
S ut 2dt
−
T
S
ut, utdt
T
S ut2
dt
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
−
T
S
1−
t
0ksds
√Aut2dt.
2.16
Trang 6Equation2.15 × β/α 2.16 yields that
T
S
√Avt βut2
dt
−
T
S
ut, utdt
T
S ut2
dt
−α β
T
S
ut,√Avt dt β α
T
S
√
Aut, vt dt
−α β
T
S
t
0
kt − s√Ausds, vt
dt
β2− α
α
T
S
1−
t
0
ksds
√Aut2dt
β
α
T
S
t
0
kt − s√
Aus −√Autds, vt
dt
β
α
T
S kt √
Aut, vt dt
α − β α 2
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
−β2α − α
T
S
√Avt2
dt −α − β2 T
S ut 2dt.
2.17
Next, we will estimate all the terms on the right side of2.17 From 2.11, we have the following estimate:
T
S
ut,√Avt dt
ut,√Avt T
S
2MES, 2.18
whereM is a positive constant Those terms of the formS T ·, ·dt can be similarly treated.
Denote byJ the sum of the other terms on the right of 2.17
Trang 7Using Young’s inequality and noting2.8, we get, for ε1> 0,
T
S
t
0
kt − s√
Aus −√Autds, vt
dt
ε1
2
T
S
t
0
kt − s√Aus− √Autds
2
dt 1
2ε1
T
S vt2
dt.
−ε1
2
T
S
t
0
ksds
t
0kt − s√Aus− √Aut2ds dt 1
2ε1
T
S vt2
dt.
ε1k0ES 1
2ε1
T
S vt2
dt.
2.19 The treatment of the other terms ofJ is similar, giving
β
α
T
S
kt√Aut, vt dt
ε1
2
T
S
√Aut2
dt k20β2
2ε1α2
T
S vt2
dt,
α − β2
α
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
α − β22
2α2ε1
T
S
t
0
ksds
t
0
kt − s√
Aus −√Aut2
ds dt ε1
2
T
S
√Aut2
dt.
2.20 Thus, we obtain
J
ε1k0β
α ES
α − β22
2α2ε1
T
S
t
0
ksds
t
0
kt − s√
Aus −√Aut2
ds dt
α1 a1
β2− α ε
1
T
S
√Aut2
dt
k20β2
2ε1α2 β
2ε1α
T
S vt2
dt
T
S ut2
dt θ
T
S
√Avt2
dt,
2.21
whereθ max{α − β2/α, 0} Make use of the estimate
√Av2
1 ζ1√
Av βu2
1ζ1
1
β2
a√
Au2
Trang 8where ζ1 is a positive constant, small enough to satisfy1 ζ1θ < 1 We thus verify our
conclusion
Lemma 2.3 For any T S 0 and for any ε2 > 0, there exist positive numbers D3ε2, D4ε2,
such that
μ
2
T
S
√Aut2
dtD3ES D4
T
S
t
0
kt − s√
Aus −√Aut2
ds dt
1 β2
2ε2a
T
S ut2
dt ε2
2
T
S vt2
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 overS, T It follows that
−
T
S
√
Avt, ut dt
T
S
vt, wtdt −
T
S
vt, wtdt β
T
S ut 2dt. 2.24 Plugging this equation into2.16, we find
T
S
1−
t
0ksds
√Aut2dt
−
T
S
ut, utdt β
T
S
vt, wtdt
β2− α T
S ut 2dt
T
S ut2
dt
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
− β
T
S
vt, wtdt.
2.25
Observe
T
S
t
0kt − s√
Aus −√Autds,√Aut
dt
1
2δ3
T
S
t
0ksds
t
0kt − s√
Aus −√Aut2
ds dt δ3
2
T
S
√Aut2
dt,
2.26
Trang 9whereδ3 μ, and for ε2 > 0
β
T
S
vt, wtdt
ε2
2
T
S vt2
dt β2
2aε2
T
S ut2
The other items on the right of2.25 can be dealt with as in the proof ofLemma 2.2 Hence,
we get the conclusion
Lemma 2.4 For any T S 0, there exist positive numbers D5, D6such that
T
S ut2
dt
T
S vt2
dt
D5ES D6
T
S
t
0kt − s√
Aus −√Aut2
ds dt
3
2α − β2
a
T
S
√Aut2
dt
T
S
√Avt βut2
dt.
2.28
Proof Taking the inner product of both sides of1.2 with vt and integrating over S, T,
we see
T
S vt2
dt
T
S
vt, vtdt
T
S
√Avt2
dt β
T
S
√
Aut, vt dt. 2.29
Combining this equation and2.16 gives
T
S vt2
dt
T
S ut2
dt
T
S
ut, utdt
T
S
vt, vtdt
T
S
1−
t
0
ksds
√Aut2dt
−
T
S
t
0
kt − s√
Aus −√Autds,√Aut
dt
T
S
√Avt βut2
dt α − β2 T
S ut 2dt.
2.30 This yields the estimate as desired
Trang 10Lemma 2.5 Let S0> 0 be fixed For any T S S0and for any ε3> 0, there exist positive numbers
D7ε3, D8ε3 such that
T
S ut2
dtD7ES D8
T
S
t
0kt − s√
Aus −√Aut2
ds dt
ε3
T
S
√Aut2
dt ε3
T
S
√Avt βut2
dt.
2.31
Proof Take the inner product of both sides of1.1 with0t kt−sus−utds and integrate
overS, T This leads to
T
S
t
0ksds ut2
dt
−
T
S
ut,
t
0
kt − sus − utds
dt
T
S
ut,
t
0
kt − sus − utds
dt
−
T
S
1−
t
0
ksds
√
Aut,
t
0
kt − s√
Aus −√Autds
dt
− α
T
S
ut,
t
0
kt − sus − utds
dt
T
S
t
0
kt − s√Aus −√Autds
2
dt
β
T
S
√
Avt,
t
0
kt − sus − utds
dt.
2.32
Just as in the proofs of the above lemmas, using Young’s inequality and noting that
T
S
t
0
ksds ut2
dt
S0
0
ksds
T
S ut2
we prove the conclusion
Trang 11Proof of Theorem 2.1 (continued) From Assumption2 and 2.8, we have
T
S
t
0kt − s√
Aus −√Aut2
ds dt
−λ1
T
S
t
0
kt − s√
Aus −√Aut2
ds dt
−λ1
T
S Etdt
2
λ ES.
2.34
Now, fixS0> 0 Thanks to Lemmas2.2and2.3, we know that for anyT S S0and forη > 0,
μ
2
T
S
√Aut2
dt η
T
S
√Avt βut2
dt
D3 ηD1
D4 ηD2 2
λ
ES ηG3ε1 G4
T
S
√Aut2
dt.
ε2
2 ηG2 1
ε1
T
S vt2
dt
1 β2
2ε2a ηG1
T
S ut2
dt.
2.35
Moreover, by the use of Lemmas2.4and2.5, we have
μ
2
T
S
√Aut2
dt η
T
S
√Avt βut2
dt
p0ES p1
T
S
√Aut2
dt p2
T
S
√Avt βut2
dt,
2.36
where
p0
D3 ηD1 D5 ε2
2 ηG2 1
ε1
D7
1 β2
2ε2a ηG1
D4 ηD2 D6 ε2
2 ηG21
ε1
D8
1 β2
2ε2a ηG1
2
λ ,
p1 ηG3ε1 G4
3
2 α − β2
a
ε2
2 ηG2
1
ε1
ε3
1 β2
2ε2a ηG1
,
p2 ε2
2 ηG2
1
ε1 ε3
1 β2
2ε2a ηG1
.
2.37
Trang 12ε1 ε−1, η ε2 ε2, ε3 ε5. 2.38 Takingε small enough gives
p1< μ
Therefore, there is a constantN1 > 0 such that
T
S
√Aut2
dt
T
S
√Avt βut2
by2.36 UsingLemma 2.4and2.34, we deduce that for some N2> 0,
T
S ut2
dt
T
S vt2
dt
D5 2D6
1
λ
ES
3
2 α − β2
a
T
S
√Aut2
dt
T
S
√Avt βut2
dt
N2ES.
2.41
Next, define
Ht : ut2vt2√
Aut2
√
Avt βut2
t
0
kt − s√
Aus −√Aut2
ds.
2.42
It is easy to see that there existM1, M2> 0 such that M1EtHtM2Et Therefore,
T
S Htdt N1 N22λ
ES,
T
S Etdt
1
M1
T
S Htdt
N1 N2 2/λ
M1 ES.
2.43
Trang 13On the other hand, when 0SS0,
T
S Etdt
S0
S Etdt
T
S0
Etdt
S0− SES N1 N2 2/λ
M1 ES0
S0 N1 N2 2/λ
M1
ES,
2.44
that is,
T
S EtdtNES, ∀S 0. 2.45
By a standard approximation argument, we see that2.45 is also true for mild solutions From this integral inequality, we complete the proofcf., e.g., 7, Theorem 8.1
3 An Example
Example 3.1 Consider a coupled system of Petrovsky equations with a memory term
∂2
t ut, ξ Δ2ut, ξ αut, ξ −
t
0
kt − sΔ2ut, ξds βΔvt, ξ 0, t 0, ξ ∈ Ω,
∂2
t vt, ξ Δ2vt, ξ βΔut, ξ 0, t 0, ξ ∈ Ω, ut, ξ vt, ξ Δut, ξ Δvt, ξ 0, t 0, ξ ∈ ∂Ω, u0, ξ u0ξ, v0, ξ v0ξ, ∂ t u0, ξ u1ξ, ∂ t v0, ξ v1ξ, ξ ∈ Ω,
3.1
whereΩ is a bounded open domain inÊ
N, with sufficiently smooth boundary ∂Ω and α, β, k
as in Assumption2 Let H L2Ω 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.
DefineA : DA ⊂ H → H by
A Δ2, with DA H4Ω ∩ H2
Then, Assumption1 is satisfied Therefore, we claim in view ofTheorem 2.1that the energy
of the system decays exponentially at infinity
Trang 14The authors would like to thank the referees for their comments and suggestions This work was supported partially by the NSF of China10771202, 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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... NSF of China10771202, 11071042, the Research Fund for Shanghai Key Laboratory for Contemporary Applied Mathematics 08DZ2271900 and the Specialized Research Fund for the Doctoral Program of. .. vt2dt.
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 overS,... ut2
The other items on the right of 2.25 can be dealt with as in the proof ofLemma 2.2 Hence,
we get the conclusion
Lemma 2.4 For