Damping by heat conduction in the Timoshenko system: · PDF fileIntroduction The Timoshenko...

Introduction The Timoshenko systems in thermoelasticity The Timoshenko with heat conduction damping Concluding Remarks

Damping by heat conduction in the Timoshenkosystem: Fourier and Cattaneo are the same

Belkacem Said-Houari

KAUST University, Saudi Arabia

December 3, 2012

Joint works with:

Aslan Kasimov (KAUST)

B. Said-Houari Timoshenko System December 3, 2012 1 / 40


Table of contents

1 IntroductionThe Fourier lawThe Cattaneo lawSingle-phase-lagging model

2 The Timoshenko systems in thermoelasticityThe initial boundary value problemThe Cauchy problem

The coupling with the Fourier lawThe coupling with the Cattaneo law

3 The Timoshenko with heat conduction dampingThe Cattaneo modelThe energy method in the Fourier spaceThe Fourier model

4 Concluding Remarks



The Fourier law

In 1807, the French mathematical physicist Joseph Fourier proposed aconstitutive relation of the heat flux of the form

q(x , t) = −κ∇θ(x , t), (1)

where x stands for the material point, t is the time, q is the heat flux, θis the temperature, ∇ is the gradient operator and κ is the thermalconductivity of the material, which is a thermodynamic state property.Equation (1) together with the energy equation of the heat conduction

ρθt + %divq = 0 (2)

yields the classical heat transport equation (of parabolic type)

ρθt − κ%∆θ = 0, (3)

that allows an infinite speed for thermal signals.B. Said-Houari Timoshenko System December 3, 2012 3 / 40


The Cattaneo law

To overcome the drawback in the Fourie law, a number of modifications ofthe basic assumption on the relation between the heat flux and thetemperature have been made, such as: Cattaneo law, Gurtin & Pipkintheory, Jeffreys law, Green & Naghdi theory and others. The commonfeature of these theories is that all lead to hyperbolic differential equationand permit transmission of heat flow as thermal waves at finite speed.The Cattaneo law

τqt+q + κ∇θ = 0, (τ > 0, relatively small) (4)

was proposed by Cattaneo in 1948. It is perhaps the most obvious, themost widely accepted and simplest generalization of Fourier’s law thatgives rise to a finite speed of propagation of heat.



The Cattaneo law

When the Fourier law (1) is replaced by the Cattaneo law (4) for the heatconduction, the equations of thermoelasticity become purely hyperbolic.Indeed, from the energy balance law

ρθt + %divq = 0 (5)

and (4), we obtain the telegraph equation

ρθtt −%κ

τ∆θ +

ρ

τθt = 0, (6)

which is a hyperbolic equation and predicts a finite signal speed equals to(%κ/ (ρτ))1/2.



Single-phase-lagging model

Note that the Cattaneo constitutive relation (4) is actually a first-orderapproximation of a more general constitutive relation (single-phase-laggingmodel).

q(x , t + τ) = −κ∇θ(x , t). (7)

The relation (7) states that the temperature gradient established at apoint x at time t gives rise to a heat flux vector at x at a later time t + τ .The delay time τ is interpreted as the relaxation time due to thefast-transient effects of thermal inertia (or small-scale effects of heattransport in time) and is called the phase-lag of the heat flux.It has been confirmed by many experiments that the Cattaneo lawgenerates a more accurate prediction than the classical Fourier law.However, some studies show that the Cattaneo constitutive relation hasonly taken account of the fast-transient effects, but not themicro-structural interactions.



The coupling of the heat conduction with elasticity

Racke (2002) has shown that the norm of the difference

between the solution U = (u, ut , θ, q) of the classical thermoelastic systemutt − αuxx + βθx = 0,

θt + qx + δutx = 0,

q + κθx = 0,

and the solution U =(u, ut , θ, q

)of the second sound thermoelastic

model utt − αuxx + βθx = 0,

θt + qx + δutx = 0,

τ qt + q + κθx = 0,

can be estimated as∥∥∥U(t, x)− U(t, x)

∥∥∥2≤ Cτ2. This last estimate

means that the difference will go to zero as τ → 0.B. Said-Houari Timoshenko System December 3, 2012 7 / 40


Coupling of the heat conduction with the Timoshenko system

The Timoshenko system consists of a coupled system of two waveequations describing the transverse vibration of a beam and it ignoresdamping effects of any nature. Specifically, Timoshenko derived thefollowing system{

ρϕtt = (K (ϕx − ψ))x , in (0, L)× (0,+∞),

Iρψtt = (EIψx)x + K (ϕx − ψ)), in (0, L)× (0,+∞),(8)

where t is time, x is the coordinate along the beam length. The functionϕ = ϕ(t, x) is the transverse displacement of the beam from anequilibrium state and ψ = ψ(t, x) is the rotation angle of the filament ofthe beam. The coefficients ρ, Iρ,E , I and K are, respectively, the density,the polar moment of inertia of a cross section, Young’s modulus ofelasticity, the moment of inertia of a cross section, and the shear modulus.



Some Results

Rivera and Racke (2002)→ Heat conduction through Fourier’s lawρ1ϕtt − σ(ϕx , ψ)x = 0, in (0, L)× (0,+∞),

ρ2ψtt − bψxx + k (ϕx + ψ) + γθx = 0, in (0, L)× (0,+∞),

ρ3θt − kθxx + γψtx = 0 in (0, L)× (0,+∞)

→ Exponential decay results for the linearized system and a non exponentialstability result for the case of different wave speeds.

Fernandez Sare and Racke (2009)→ Heat conduction through Cattaneo’slaw

ρ1ϕtt − K (ϕx + ψ)x = 0ρ2ψtt − bψxx + K (ϕx + ψ) + γθx = 0ρ3θt + κqx + γψtx = 0τ0qt + δq + κθx = 0,

(9)

→ No exponential decay even if the wave speeds are equal.

And additional damping of the form

∫ ∞0

g(t − s)ψxx(s)ds in the left-hand

side of (9) is not enough to stabilize the system exponentially.B. Said-Houari Timoshenko System December 3, 2012 9 / 40


The Cauchy problem (only few results)

Ide, Haramoto & Kawashima (2008) investigated the problemϕtt (t, x)− (ϕx − ψ)x (t, x) = 0, (t, x) ∈ R+ × R,

ψtt (t, x)−a2ψxx (t, x)− (ϕx − ψ) (t, x) +λψt (t, x) = 0, (t, x) ∈ R+ × R,

(ϕ,ϕt , ψ, ψt) (0, x) = (ϕ0, ϕ1, ψ0, ψ1) , x ∈ R,(10)

where a and λ are positive constants. They proved that if a = 1, then solutions of(10) decay as∥∥∂kxU (t)

∥∥2≤ C (1 + t)−1/4−k/2 ‖U0‖1 + Ce−ct

∥∥∂kxU0

∥∥2, (11)

where U = (ϕx − ψ,ϕt , aψx , ψt)T . If on the other hand, a 6= 1, then decay

property of system (10) is of regularity-loss type and solutions decay as∥∥∂kxU (t)∥∥

2≤ C (1 + t)−1/4−k/2 ‖U0‖1 + C (1 + t)−l/2 ∥∥∂k+l

x U0

∥∥2. (12)

The parameters k and l in (11) and (12) are non-negative integers, and C and c

are positive constants.B. Said-Houari Timoshenko System December 3, 2012 10 / 40


Racke & Said-Houari (2011) analyzed the semilinear problem

ϕtt − (ϕx − ψ)x = 0, (t, x) ∈ R+ × R,

ψtt − a2ψxx − (ϕx − ψ) +λψt = f (ψ) , (t, x) ∈ R+ × R,

(ϕ,ϕt , ψ, ψt) (0, x) = (ϕ0, ϕ1, ψ0, ψ1) , x ∈ R,

(13)

where f (ψ (t, x)) = |ψ (t, x)|p with p > 1. For the linear case (i.e. f = 0), they

improved the decay results obtained in (11) so that, for initial data

U0 ∈ Hs (R) ∩ L1,γ (R) with a suitably large s and γ ∈ [0, 1], solutions decay

faster than those given in (11) and (12). Further, we analyzed the asymptotic

behavior of the semilinear problem (13) with the power type nonlinearity |ψ|p

with p > 12.



The Cauchy problem: The coupling with the Fourier law

Said-Houari & Kasimov (2012)

We considered the Fourier-Timoshenko system:ϕtt − (ϕx − ψ)x = 0,

ψtt − a2ψxx − (ϕx − ψ) + λψt + βθx = 0,

θt − θxx + βψtx = 0,

(14)

where t ∈ (0,∞) denotes the time variable and x ∈ R is the space variable, thefunctions ϕ and ψ denote the displacements of the elastic material, the function θis the temperature difference and a, λ and β are certain positive constants.Initial conditions are of the following form,{

ϕ(., 0) = ϕ0(x), ϕt(., 0) = ϕ1(x), ψ(., 0) = ψ0(x),

ψt(., 0) = ψ1(x), θ(., 0) = θ0 (x) .(15)



Let us first write system (73)-(74) as a first order (in time) system of the form{Ut + AUx + LU = BUxx ,

U (x, 0) = U0.(16)

where A, L, B are matrices and U is a solution vector identified below. To obtain system (16), we introduce the followingvariables:

v = ϕx − ψ, u = ϕt , z = aψx , y = ψt .

Consequently, system (73) can be rewritten into the following first order system of hyperbolic-parabolic type

vt − ux + y = 0,

ut − vx = 0,

zt − ayx = 0,

yt − azx − v + λy + βθx = 0,

θt − θxx + βyx = 0,

(17)

System (75) is equivalent to system (16) with

U =

vuzyθ

, A =

0 −1 0 0 0−1 0 0 0 00 0 0 −a 00 0 −a 0 β0 0 0 β 0

, L =

0 0 0 1 00 0 0 0 00 0 0 0 0−1 0 0 λ 00 0 0 0 0

,

B=

0 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 −1

.B. Said-Houari Timoshenko System December 3, 2012 13 / 40


Fourier law: U0 ∈ Hs(R) ∩ L1(R)

Theorem 1

Let s be a nonnegative integer and assume that U0 ∈ Hs(R) ∩ L1(R). Then thesolution U of problem (16) satisfies the following decay estimates:

When a = 1,∥∥∂kxU (t)∥∥

2≤ C (1 + t)−1/4−k/2 ‖U0‖L1 + Ce−ct

∥∥∂kxU0

∥∥2. (18)

When a 6= 1,∥∥∂kxU (t)∥∥

2≤ C (1 + t)−1/4−k/2 ‖U0‖L1 + C (1 + t)−l/2 ∥∥∂k+l

x U0

∥∥2, (19)

where k and l are non-negative integers satisfying k + l ≤ s and C , c are twopositive constants.



The Fourier law: U0 ∈ Hs(R) ∩ L1,γ(R)

Theorem 2

Let γ ∈ [0, 1]. Let s be a nonnegative integer and assume thatU0 ∈ Hs(R) ∩ L1,γ(R). Then the solution U of problem (16) satisfies thefollowing decay estimates:

When a = 1, we have∥∥∂kxU (t)∥∥

2≤ C (1 + t)−1/4−k/2−γ/2 ‖U0‖L1,γ + Ce−ct

∥∥∂kxU0

∥∥2

+C (1 + t)−1/4−k/2

∣∣∣∣∫RU0 (x) dx

∣∣∣∣ . (20)

When a 6= 1, we have∥∥∂kxU (t)∥∥

2≤ C (1 + t)−1/4−k/2−γ/2 ‖U0‖L1,γ + C (1 + t)−l/2 ∥∥∂k+l

x U0

∥∥2

+C (1 + t)−1/4−k/2

∣∣∣∣∫RU0 (x) dx

∣∣∣∣ , (21)

where k and l are non-negative integers satisfying k + l ≤ s and C , c are twopositive constants.



The Cauchy problem: The coupling with the Cattaneo law

The second problem we investigate in the present work is the followingTimoshenko system in thermoelasticity of second sound

ϕtt − (ϕx − ψ)x = 0,

ψtt − a2ψxx − (ϕx − ψ) + βθx + λψt = 0,

θt + κqx + βψtx = 0,

τqt + δq + κθx = 0.

(22)

where t ∈ (0,∞) and x ∈ R and γ, τ, δ, κ, λ and β are positive constants and thefollowing initial conditions are assumed,{

ϕ(., 0) = ϕ0(x), ϕt(., 0) = ϕ1(x), ψ(., 0) = ψ0(x),

ψt(., 0) = ψ1(x), θ(., 0) = θ0 (x) , q(., 0) = q0 (x) .(23)



By introducing the following variables:

v = ϕx − ψ, u = ϕt , z = aψx , y = ψt , w = τ0q.

system (22) can be rewritten as the following first order hyperbolic system

vt − ux + y = 0,

ut − vx = 0,

zt − ayx = 0,

yt − azx − v + λy + βθx = 0,

θt +κ

τwx + βyx = 0,

wt +δ

τw + κθx = 0

(24)

and the initial conditions (31) take the form

(v, u, z, y, θ,w) (x, 0) = (v0, u0, z0, y0, θ0,w0) . (25)

System (24)-(35) is equivalent to {Ut + AUx + LU = 0,

U (x, 0) = U0.(26)

where A is a real symmetric matrix and L is non-negative definite matrix, with

U =

vuzyθw

, A = −

0 1 0 0 0 01 0 0 0 0 00 0 0 a 0 00 0 a 0 −β 00 0 0 −β 0 κ/τ0 0 0 0 κ 0

, L =

0 0 0 1 0 00 0 0 0 0 00 0 0 0 0 0−1 0 0 λ 0 00 0 0 0 0 00 0 0 0 0 δ/τ

. (27)

and U0 = (v0, u0, z0, y0, θ0,w0)T .



System (26) can be seen as a particular case of a general hyperbolic system ofbalance laws. We point out that Shizuta & Kawashima introduced the so-calledalgebraic condition (SK), namely

(SK) ∀ξ ∈ R−{0} , Ker(L) ∩ {eigenvectors of (ξA)} = {0},

which is satisfied in many examples and sufficient to establish a general result ofglobal existence for small perturbations of constant-equilibrium state. Our system(26) satisfies the condition (SK), but the general theory on the dissipativestructure established by Shizuta & Kawashima (1985) is not applicable since thematrices A and L are not real symmetric.Beauchard & Zuazua (2011) have recently shown that the condition (SK) isequivalent to the classical Kalman rank condition in control theory for the pair(A, L), that is

rk[L, A(iξ)L, ..., A(iξ)N−1L] = N,

where A(iξ) = iξA is an N × N matrix.



The Cattaneo law: Decay results

Theorem 3 (U0 ∈ H s(R) ∩ L1(R))

Let s be a nonnegative integer and assume that U0 ∈ Hs(R) ∩ L1(R).Then the solution U = (v , u, z , y , θ, w)′ of problem (26) satisfies thefollowing decay estimates:∥∥∥∂kxU (t)

∥∥∥2≤ C (1 + t)−1/4−k/2 ‖U0‖L1 +C (1 + t)−l/2

∥∥∥∂k+lx U0

∥∥∥2, (28)

where k and l are non-negative integers satisfying k + l ≤ s and C , c arepositive constants.



Theorem 4 (U0 ∈ H s(R) ∩ L1,γ(R))

Let γ ∈ [0, 1], let s be a nonnegative integer, and assume thatU0 ∈ Hs(R)∩ L1,γ(R). Then the solution U = (v , u, z , y , θ, w)′ of problem(26) satisfies the following decay estimates:∥∥∥∂kxU (t)

∥∥∥2≤ C (1 + t)−1/4−k/2−γ/2 ‖U0‖L1,γ + C (1 + t)−l/2

∥∥∥∂k+lx U0

∥∥∥2

+C (1 + t)−1/4−k/2

∣∣∣∣∫RU0 (x) dx

∣∣∣∣ , (29)

where k and l are non-negative integers satisfying k + l ≤ s and C and care two positive constants.



The Timoshenko−Cattaneo without mechanical damping

We consider the system

ϕtt − (ϕx − ψ)x = 0,

ψtt − a2ψxx − (ϕx − ψ) + δθx = 0,

θt + qx + δψtx = 0,

τqt + βq + θx = 0,

(30)

where t ∈ (0,∞), x ∈ R, and a, τ, δ, β, β are positive constants. System(30) is supplied with the following initial conditions:{

ϕ(x , 0) = ϕ0(x), ϕt(x , 0) = ϕ1(x), ψ(x , 0) = ψ0(x),

ψt(x , 0) = ψ1(x), θ(x , 0) = θ0 (x) , q(x , 0) = q0 (x) .(31)



To rewrite the system as a first-order system, we define new variables, as follows:

v = ϕx − ψ, u = ϕt , z = aψx , y = ψt , w = q. (32)

Then, (30) takes the form:

vt − ux + y = 0,

ut − vx = 0,

zt − ayx = 0,

yt − azx − v + δθx = 0,

θt + wx + δyx = 0,

wt +β

τw +

1

τθx = 0

(33)

and, if we denoteU(x , t) = (v , u, z , y , θ,w)T , (34)

the initial conditions can then be written as

U(x , 0) = U0(x) = (v0, u0, z0, y0, θ0,w0)T , (35)

where v0 = ϕ0,x − ψ0, u0 = ψ1, z0 = aψ0,x , y0 = ψ1, w0 = q0.



Theorem 5 (Cattaneo model τ 6= 0)

Let s be a nonnegative integer and let α = (τ − 1)(1− a2

)− τδ2.

Assume that U0 ∈ Hs(R) ∩ L1(R). Then, the solution, U, of problem (33)satisfies the following decay estimates:

if α = 0,∥∥∥∂kxU (t)∥∥∥

2≤ C (1 + t)−1/12−k/6 ‖U0‖1 + Ce−ct

∥∥∥∂kxU0

∥∥∥2

; (36)

if α 6= 0,∥∥∥∂kxU (t)∥∥∥

2≤ C (1 + t)−1/12−k/6 ‖U0‖1 + C (1 + t)−l/6

∥∥∥∂k+lx U0

∥∥∥2,

(37)

where k and l are non-negative integers satisfying k + l ≤ s and C and care two positive constants.



Taking the Fourier transform of (33), we obtain the following ODE system:

vt − iξu + y = 0, (38a)

ut − iξv = 0, (38b)

zt − aiξy = 0, (38c)

yt − aiξz − v + δiξθ = 0, (38d)

θt + iξw + δiξy = 0, (38e)

wt +β

τw +

1

τiξθ = 0, (38f)

together with initial data, written in terms of the solution vector, U (ξ, t) =(v, u, z, y, θ, w

)T, as

U(ξ, 0) = U0(ξ). (39)

Proposition 1

Let U (ξ, t) be a solution of (38)-(39) and α = (τ − 1)(

1− a2)− τδ2. Then, there exist two positive constants, C and c,

such that, for all t ≥ 0 :

∣∣∣U (ξ, t)∣∣∣2 ≤ C

∣∣∣U (ξ, 0)∣∣∣2 e−cρ1(ξ)

, ρ1 (ξ) =ξ6(

1 + ξ2)3, if α = 0, (40)

∣∣∣U (ξ, t)∣∣∣2 ≤ C

∣∣∣U (ξ, 0)∣∣∣2 e−cρ2(ξ)

, ρ2 (ξ) =ξ6(

1 + ξ2)6, if α 6= 0. (41)



Lemma 6

Define the following energy functional:

E (t) =1

2(|v |2 + |u|2 + |z |2 + |y |2 + |θ|2 + τ |w |2). (42)

Then,

c1

∣∣∣U (ξ, t)∣∣∣2 ≤ E (ξ, t) ≤ c2

∣∣∣U (ξ, t)∣∣∣2 , (43)

where c1 = min (1/2, τ/2) and c2 = max (1/2, τ/2) , and

d

dtE (t) = −βτ |w |2, ∀t ≥ 0. (44)



In the following lemma, we show a dissipative term of |v |2.

Lemma 7

The functional

G (t) = Re

[−τaz∗u +

(τ +

1

δ2− a2

δ2

)δθ∗u − τ v y∗

]+

1− a2

δRe (−τ v∗w)

(45)satisfies

d

dtG (t) + τ |v |2 − τ |y |2 =

1− a2

δβRe (w v∗) +

1− a2

δRe (τ w∗y)

+αRe (iξy u∗) +α

δRe (iξw u∗) , (46)

whereα = (τ − 1)

(1− a2

)− τδ2. (47)



Lemma 8

Let K (t) be defined by

K (t) = Re(iξu∗v + aiξ¯zy − δaτ iξz∗w

). (48)

Then, for any ε1, ε2 > 0,

d

dtK (t) + (1− ε1) ξ2 |u|2 +

(a2 − ε2

)ξ2 |z |2

≤ C (ε1, ε2)(1 + ξ2

) (|v |2 + |y |2

)+ C (ε2)

(1 + ξ2

)|w |2 . (49)



Lemma 9

The following inequalities hold true:

d

dtRe(iξw∗θ

)+

(1

τ− ε3

)ξ2|θ|2 ≤ C

(ε3, ε

′3

) (1 + ξ2

)|w |2+ε′3

ξ4

1 + ξ2|y |2

(50)and

d

dtRe(iξθ∗y

)+ (δ − ε4) ξ2 |y |2 ≤ C

(ε′4) (

1 + ξ2)|θ|2 + ε′4

ξ4

1 + ξ2|z |2

+ε′4ξ2 |v |2 + C (ε4) ξ2 |w |2 , (51)

where ε3, ε′3, ε4, and ε′4 are arbitrary positive constants.



Proof of Proposition 1

To prove Proposition 1, we consider the two cases, α = 0 and α 6= 0, separately.Case 1. α = 0.

In this case, (46) takes the formdG (t)

dt+ (τ − ε0) |v|2 ≤ C (ε0)

(|w|2 + |y|2

). (52)

Define the functional

L1 (t) =ξ2

1 + ξ2

{γ0ξ

2G (ξ, t) + γ1ξ2

1 + ξ2K (ξ, t) + γ4Re

(iξθ∗ y

)}

+γ3Re(iξw∗θ

)+ N

(1 + ξ

2)E (ξ, t) , (53)

where N, γ0, ..., γ4 are positive constants to be fixed later. Taking the derivative of L (ξ, t) with respect to t and using (52),(49), (51, and (50), we get Consequently, we deduce that there exists a positive constant, η > 0, such that

d

dtL1 (t) + ηQ1 (t) ≤ 0, ∀t ≥ 0, (54)

where

Q1 (t) =ξ4

1 + ξ2|v|2 +

ξ4

1 + ξ2|y|2 +

ξ6(1 + ξ2

)2

(|u|2 + |z|2

)+ ξ

2|θ|2 +(

1 + ξ2)|w|2 . (55)

It is not difficult to see now that, for all t ≥ 0 and for all ξ ∈ R, we have

Q1 (t) ≥ξ6(

1 + ξ2)2

(|v|2 + |y|2 + |u|2 + |z|2 + |θ|2 + |w|2

)

=ξ6(

1 + ξ2)2

∣∣∣U (ξ, t)∣∣∣2 . (56)



On the other hand we can show that

β1

(1 + ξ

2)E (t) ≤ L1 (t) ≤ β2

(1 + ξ

2)E (t) . (57)

Combining (43), (54), (56), and (57), we deduce that there exists a positive constant, η1 > 0, such that for all t ≥ 0, we have

d

dtL1 (t) ≤ −η1

ξ6(1 + ξ2

)3L1 (t) . (58)

Integrating (58) over t and using once again (43) and (57), we deduce (40).

Case 2. α 6= 0.

In this case, the two last terms on the right-hand side of (46) can be estimated as

αRe(iξy u∗

)+α

δRe(iξw u∗

)≤ ε′0

ξ2

1 + ξ2|u|2 + C

(ε′0

) (1 + ξ

2) (|y|2 + |w|2

). (59)

Inserting (59) into (46) and applying Young’s inequality as before, we find that (52) takes the form

dG (t)

dt+ (τ − ε0) |v|2 ≤ C

(ε′0, ε0

) (1 + ξ

2) (|w|2 + |y|2

)+ ε′0

ξ2

1 + ξ2|u|2 . (60)

Now, we define

L2 (t) =ξ2(

1 + ξ2)2

{λ0

ξ2

1 + ξ2G (ξ, t) + λ1

ξ2(1 + ξ2

)2K (ξ, t) + λ4Re

(iξθ∗ y

)}

+λ3Re(iξw∗θ

)+ M

(1 + ξ

2)2

E (ξ, t) , (61)

where M, εj , ε′d , and λj , with j = 1, ..., 4 and d = 0, 3, 4, are positive constants to be fixed later.



Taking the time derivative of L2, we obtain

d

dtL2 (t) + λQ2 (t) ≤ 0, ∀t ≥ 0, (62)

for some λ > 0, where

Q2 =ξ4(

1 + ξ2)3|v|2 +

ξ4(1 + ξ2

)2|y|2 +

ξ6(1 + ξ2

)4

(|u|2 + |z|2

)+ ξ

2|θ|2 +(

1 + ξ2)2|w|2

≥ξ6(

1 + ξ2)4

∣∣∣U (ξ, t)∣∣∣2 . (63)

On the other hand, there exist two positive constants, β3 and β4, such that, for all t ≥ 0,

β3

(1 + ξ

2)2

E (t) ≤ L2 (t) ≤ β4

(1 + ξ

2)2

E (t) . (64)

Combining (43), (62), (63), and (64), we obtain (41), which completes the proof of Proposition 1.



Proof of Theorem 5

The proof is based on the pointwise estimates in Proposition 1.First, let us assume that α = 0. Then, ρ1(ξ) ≥ 1

8ξ6, for |ξ| ≤ 1,

ρ1(ξ) ≥ 18, for |ξ| ≥ 1.

(65)

Applying the Plancherel theorem and making use of the inequality in (40), we obtain∥∥∥∂kx U (t)∥∥∥2

2=

∫R|ξ|2k

∣∣∣U (ξ, t)∣∣∣2 dξ

≤ C

∫R|ξ|2k e−cρ1(ξ)t

∣∣∣U (ξ, 0)∣∣∣2 dξ

= C

∫|ξ|≤1

|ξ|2k e−cρ1(ξ)t∣∣∣U (ξ, 0)

∣∣∣2 dξ + C

∫|ξ|≥1

|ξ|2k e−cρ1(ξ)t∣∣∣U (ξ, 0)

∣∣∣2 dξ

= I1 + I2. (66)

That is, the integral is split into its low-frequency part, I1, and its high-frequency part, I2. For I1, we find, by using the firstinequality in (65), that

I1 ≤ C

∫|ξ|≤1

|ξ|2k e− 1

8ξ6t∣∣∣U (ξ, 0)

∣∣∣2 dξ ≤ C∥∥∥U0

∥∥∥2

∞

∫|ξ|≤1

|ξ|2k e− 1

8ξ6t

dξ. (67)

Then, using the inequality ∫ 1

0|ξ|2k e

− 18ξ6t

dξ ≤ (1 + t)−1/6−k/3,

we estimateI1 ≤ ‖U0‖

21 C (1 + t)−1/6−k/3

. (68)



For I2, using the second inequality in (65), we find

I2 ≤ C

∫|ξ|≥1

|ξ|2k e− 1

8t∣∣∣U (ξ, 0)

∣∣∣2 dξ ≤ Ce− 1

8t∥∥∥∂kx U0

∥∥∥2

2. (69)

Therefore, estimate (36) holds by adding (68) and (69). Now, we assume that α 6= 0. Then, ρ2(ξ) ≥ 164ξ6, for |ξ| ≤ 1,

ρ2(ξ) ≥ 164ξ−6, for |ξ| ≥ 1.

(70)

As before, using (41), we find that∥∥∥∂kx U (t)∥∥∥2

2≤ C

∫|ξ|≤1

|ξ|2k e−cρ2(ξ)t∣∣∣U (ξ, 0)

∣∣∣2 dξ + C

∫|ξ|≥1

|ξ|2k e−cρ2(ξ)t∣∣∣U (ξ, 0)

∣∣∣2 dξ

= J1 + J2.

The low-frequency part, J1, is estimated as I1 and results in

J1 ≤ ‖U0‖21 C (1 + t)−1/6−k/3

. (71)

The high-frequency part, J2, using the second inequality in (70), can be estimated as

J2 ≤ C

∫|ξ|≥1

|ξ|2k e− 1

64ξ−6t

∣∣∣U (ξ, 0)∣∣∣2 dξ.

Using the inequality

sup|ξ|≥1

{|ξ|−2l e−cξ−6t

}≤ C (1 + t)−l/3

,

we obtain

J2 ≤ C sup|ξ|≥1

{|ξ|−2l e−cξ−6t

}∫|ξ|≥1

|ξ|2(k+l)∣∣∣U (ξ, 0)

∣∣∣2 dξ

≤ C (1 + t)−l/3∥∥∥∂k+l

x U0

∥∥∥2

2. (72)

Combining the estimates of J1 and J2, estimate (37) is established. This completes the proof of Theorem 5.B. Said-Houari Timoshenko System December 3, 2012 33 / 40


The Timoshenko−Fourier without mechanical damping

We consider the model ϕtt − (ϕx − ψ)x = 0,

ψtt − a2ψxx − (ϕx − ψ) + δθx = 0,

θt − θxx + δψtx = 0,

(73)

These equations are supplemented with appropriate initial conditions,{ϕ(x, 0) = ϕ0(x), ϕt (x, 0) = ϕ1(x), ψ(x, 0) = ψ0(x),

ψt (x, 0) = ψ1(x), θ(x, 0) = θ0 (x) .(74)

As before, we introduce the following new variables:

v = ϕx − ψ, u = ϕt , z = aψx , y = ψt .

System (73) can then be rewritten as

vt − ux + y = 0,

ut − vx = 0,

zt − ayx = 0,

yt − azx − v + δθx = 0,

θt − θxx + δyx = 0,

(75)

and initial conditions (74), using V (x, t) = (v, u, z, y, θ)T to denote the solution vector, take the form

V (x, 0) = V0(x) = (v0, u0, z0, y0, θ0) , (76)

where v0 = ϕ0,x − ψ0, u0 = ψ1, z0 = aψ0,x , and y0 = ψ1.



Theorem 10

Let s be a nonnegative integer and assume that V0 ∈ Hs(R) ∩ L1(R). Then, thesolution, V , of problem (75)-(76) satisfies the following decay estimates:

when a = 1,∥∥∂kxV (t)∥∥

2≤ C (1 + t)−1/12−k/6 ‖V0‖1 + Ce−ct

∥∥∂kxV0

∥∥2

; (77)

when a 6= 1,∥∥∂kxV (t)∥∥

2≤ C (1 + t)−1/12−k/6 ‖V0‖1 + C (1 + t)−l/2 ∥∥∂k+l

x V0

∥∥2, (78)

where k and l are non-negative integers satisfying k + l ≤ s, and C and c are twopositive constants.



Taking the Fourier transform of (75), we obtain

vt − iξu + y = 0, (79a)

ut − iξv = 0, (79b)

zt − aiξy = 0, (79c)

yt − aiξz − v + δiξθ = 0, (79d)

θt + ξ2θ + δiξy = 0, (79e)

with the initial data written in terms of V (ξ, t) = (v, u, z, y, θ)T :

V (ξ, 0) = V0(ξ, 0) = (v0, u0, z0, y0, θ0)T . (80)

Proposition 2

Let V (ξ, t) be the solution of (79)-(80). Then, there exist two positive constants, C and c, such that, for all t ≥ 0, thefollowing estimates hold true:if a = 1, ∣∣∣V (ξ, t)

∣∣∣2 ≤ C∣∣∣V (ξ, 0)

∣∣∣2 e−c%1(ξ), %1 (ξ) =

ξ6(1 + ξ2

) (1 + ξ2 + ξ4

) ; (81)

if a 6= 1, ∣∣∣V (ξ, t)∣∣∣2 ≤ C

∣∣∣V (ξ, 0)∣∣∣2 e−c%2(ξ)

, %2 (ξ) =ξ6(

1 + ξ2)2 (1 + ξ2 + ξ4

) . (82)



We proved here that there is essentially no difference between theTimoshenko-Cattaneo model and the Timoshenko-Fourier model interms of decay rates and regularity assumptions on the initial data. Inthe regularity-loss case and in both systems, the decay rate isobtained only for low-order derivatives, but not for all derivatives upto order s. Indeed, for the Cattaneo model, we see from (37) that ourestimates can be obtained for all k satisfying0 ≤ k ≤ [s/2− 1/4] = s1. For the Fourier model, we find from (78)that our estimates hold for 0 ≤ k ≤ [3s/4− 1/8] = s2. We see thats1 < s2 which is the only difference between the two models.



By comparing our present result with other known results on thisproblem, we see that in bounded domains and for appropriateassumptions on the coefficients, the following models all lead to anexponential decay rate:

only linear mechanical damping ψt present as in (10).

only viscoelastic damping present, i.e.,∫ t

0g(s)ψxx(t − s)ds instead of

ψt in (10) with an exponentially decaying g(s));Fourier heat conduction with no mechanical damping.Cattaneo heat conduction with no mechanical damping.

However, this is not true for the Cauchy problem. Indeed, withfrictional damping as in (10), the decay rate of the L2-norm is(1 + t)−1/4 (Ide & Kawashima), with the viscoelastic damping, thedecay rate is (1 + t)−1/8 (Liu & Kawashima), whereas as we haveshown here, the decay rate with heat dissipation alone is (1 + t)−1/12.As we see, the heat dissipation is slowest of all. The fact that bothFourier and Cattaneo models lead to the same decay rate smears thedistinction between them that prior studies have highlighted.



As in Theorem 2, it can be shown that if one assumes that∫R U0(x)dx = 0, then the decay rate in Theorem 5 can be improved.

Nevertheless, the total decay rate is still polynomial. We recall thatthe assumption

∫R U0(x)dx = 0 is equivalent to U0(0) = 0. An

exponential decay rate can be obtained by assuming that U0(ξ) = 0for |ξ|2 ≤ α0, with some α0 > 0.

The decay rates can be further improved by assuming that the highermomenta of the initial data are zeros.



Thank you for your attention


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