stability of some string-beam systems · 2017-05-19 · introduction feedback stabilization...

IntroductionFeedback stabilization

Thermoelastic case

Stability of some string-beam systems

Farhat Shel

Faculte des Sciences de Monastir

ContrOpt 2017

15-19 Mai 2017, Monastir, Tunisie

Farhat Shel Stability of some string-beam systems


Thermoelastic case



Thermoelastic case

Outline

1 Introduction

2 Feedback stabilizationAbstract settingAsymptotic behavior

3 Thermoelastic caseAbstract settingAsymptotic behavior



Thermoelastic case

Models

E. String E. Beamℓ2ℓ1 0

u1,tt − u1,xx = 0 , u2,tt − u2,xxxx = 0

Transmission conditions

u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case

Models

Energy

E (t) =

∫ `1

0|u1,t |2 dx+

∫ `1

0|u1,x |2 dx+

∫ `2

0|u2,t |2 dx+

∫ `2

0|u2,xx |2 dx

d

dtE (t) = 0.

The system is conservative.



Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0

θt + βvtx − κθxx = 0

, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1,x(`1, t) = −u1,t(`1, t),

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1,x(`1, t) = −u1,t(`1, t),

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is exponentially stable.Farhat Shel Stability of some string-beam systems


Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1(`1, t) = 0,

u2,x(`2, t) = 0, u2,xxx(`2, t) = u2,t(`2, t).

The system is



Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

Boundary conditions

u1(`1, t) = 0,

u2,x(`2, t) = 0, u2,xxx(`2, t) = u2,t(`2, t).

The system is polynomially stable.Farhat Shel Stability of some string-beam systems


Thermoelastic case

Models


u1,tt − u1,xx

+ βθx

= 0

θt + βutx − κθxx = 0

, u2,tt + u2,xxxx

+ βθx

= 0

θt + βutxx − κθxx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0,

θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

θx(0, t)

.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models

TE. String E. Beamℓ2ℓ1 0

u1,tt − u1,xx + βθx = 0θt + βutx − κθxx = 0

, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0,

θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

θx(0, t)

.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models



, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

θx(0, t)

.

Boundary conditions

u1(`1, t) = 0, θ(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models



, u2,tt + u2,xxxx

+ βθx

= 0



u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = 0.

θx(0, t)

.

Boundary conditions

u1(`1, t) = 0, θ(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is exponentially stable.Farhat Shel Stability of some string-beam systems


Thermoelastic case

Models

E. String TE. Beamℓ2ℓ1 0

u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx + βθx = 0θt + βutxx − κθxx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = θx(0, t).

Boundary conditions

u1(`1, t) = 0, θ(`2, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is



Thermoelastic case

Models

E. String TE. Beamℓ2ℓ1 0

u1,tt − u1,xx

+ βθx

= 0


, u2,tt + u2,xxxx + βθx = 0θt + βutxx − κθxx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ(0, t) = 0,

u2,xxx(0, t)− u1,x(0, t) = θx(0, t).

Boundary conditions

u1(`1, t) = 0, θ(`2, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.

The system is polynomially stable.Farhat Shel Stability of some string-beam systems


Thermoelastic case

References

Coupled string-beam systemAmmari, Jellouli, Mehrenberger, 2009.

Chain of beams and stringsAmmari et al, 2012.

String beams networkAmmari, Mehrenberger 2012



Thermoelastic case

Abstract settingAsymptotic behavior

Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={

f = (f1, f2) ∈ H1(0, `1)× H2(0, `2) | f satisfies (1)},

f2(`2) = 0,

δ ∈ {0, 1}

f1(0) = f2(0),∂x f2(0) = 0.

(1)

Energy space:H = V × L2(G),

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.



Thermoelastic case


Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={


f1(`1) = 0,

δ ∈ {0, 1}

f1(0) = f2(0),∂x f2(0) = 0.

(1)


〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.



Thermoelastic case


Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={


δf2(`2) = 0, (1− δ)f1(`1) = 0,

δ ∈ {0, 1}

f1(0) = f2(0),∂x f2(0) = 0.

(1)


〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.



Thermoelastic case


Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={


δf2(`2) = 0, (1− δ)f1(`1) = 0, δ ∈ {0, 1}f1(0) = f2(0),∂x f2(0) = 0.

(1)


〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.



Thermoelastic case


Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={


δf2(`2) = 0, (1− δ)f1(`1) = 0, δ ∈ {0, 1}f1(0) = f2(0),∂x f2(0) = 0.

(1)


〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩

Hilbert space.



Thermoelastic case


Energy space


L2(G) = L2(0, `1)× L2(0, `2).

V ={


δf2(`2) = 0, (1− δ)f1(`1) = 0, δ ∈ {0, 1}f1(0) = f2(0),∂x f2(0) = 0.

(1)


〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.



Thermoelastic case


Evolution equation

Then the system (S) may be rewritten as the first order evolutionequation on H,

{y ′(t) = Ay(t), t > 0,y(0) = y0

(2)

where y = (u, ut), y0 = (u0, u1).

A

u1

u2

v1

v2

=

v1

v2

∂2xu1

−∂4xu2

.



Thermoelastic case


Evolution equation

D(A) ={

y = (u, v) ∈ V 2 | u1 ∈ H2(0, `1),u2 ∈ H4(0, `2),y satisfies (3)}

(1− δ)∂xu1(`1) = −(1− δ)v1(`1),

(1− δ)∂2xu2(`2) = 0,

δ∂3xu2(`2) = δv2(`2), δ∂xu2(`2) = 0,

∂xu1(0)− ∂3xu2(0) = 0.

(3)



Thermoelastic case


Theorem

The operator A generates a C0-semigroup S(t) = eAt ofcontraction on H.

For an initial datum y0 ∈ H there exists a unique solution

y ∈ C ([0,+∞),H)

of the Cauchy problem (2).Moreover if y0 ∈ D(A), then

y ∈ C ([0,+∞),D(A)) ∩ C 1([0,+∞),H).

Proof (of the theorem).

A is a dissipative operator on H.

]0,+∞) ⊂ ρ(A): the resolvent set of A.Conclusion: by Lumer phillips theorem.



Thermoelastic case


Theorem



y ∈ C ([0,+∞),H)

of the Cauchy problem (2).

Moreover if y0 ∈ D(A), then

y ∈ C ([0,+∞),D(A)) ∩ C 1([0,+∞),H).






Thermoelastic case


Theorem



y ∈ C ([0,+∞),H)


y ∈ C ([0,+∞),D(A)) ∩ C 1([0,+∞),H).






Thermoelastic case


Theorem



y ∈ C ([0,+∞),H)


y ∈ C ([0,+∞),D(A)) ∩ C 1([0,+∞),H).



]0,+∞) ⊂ ρ(A): the resolvent set of A.

Conclusion: by Lumer phillips theorem.



Thermoelastic case


Theorem



y ∈ C ([0,+∞),H)


y ∈ C ([0,+∞),D(A)) ∩ C 1([0,+∞),H).






Thermoelastic case


exponential stability ⇐⇒ S(t) = eAt is exponentially stable:

‖S(t)y0‖ ≤ Ce−wt ‖y0‖ ∀t > 0.

Lemma [Gearhard-Pruss-Huang]

A C0-semigroup of contraction etB is exponentially stable if, andonly if,

iR = {iβ | β ∈ R} ⊆ ρ(B) (4)

andlim sup|β|→∞

∥∥(iβ − B)−1∥∥ <∞. (5)



Thermoelastic case


polynomial stability ⇐⇒ S(t) = eAt is polynomially stable:

‖S(t)y0‖ ≤C

tα‖y0‖D(A) ∀t > 0.

Lemma [Borichev-Tomilov]

A C0-semigroup of contraction etB on a Hilbert space H satisfies∥∥etBy0

∥∥ ≤ C

t1α

‖y0‖D(B)

for some constant C > 0 and for α > 0 if, and only if, (4) holdsand

lim|β|→∞

sup1

βα∥∥(iβ − B)−1

∥∥ <∞ (6)



Thermoelastic case


Exponential stability

Theorem

If the feedback is applied at the exterior end of the string then, thesystem (S) is exponentially stable.



Thermoelastic case


Proof

The operator A satisfies condition (4). It suffices to prove that (5)holds. Suppose the conclusion is false. Then there exists asequense (βn) of real numbers, without loss of generality, withβn −→ +∞, and a sequence of vectors (yn) = (un, vn) in D(A)with ‖yn‖H = 1, such that

‖(iβnI −A)yn‖H −→ 0.

We prove that this condition yields the contradiction ‖yn‖H −→ 0as n −→ 0.



Thermoelastic case



iβnu1,n − v1,n = f1,n −→ 0, in H1(0, `1),

iβnu2,n − v2,n = f2,n −→ 0, in H2(0, `2),

iβnv2,n − ∂2xu2,n = g2,n −→ 0, in L2(0, `1),

iβnv2,n + ∂4xu2,n = g2,n −→ 0, in L2(0, `2).

Then

−β2nu1,n − ∂2

xu1,n = g1,n + iβnf1,n, (7)

−β2nu2,n + ∂4

xu2,n = g2,n + iβnf2,n (8)

and‖vj ,n‖2 − β2

n ‖uj ,n‖2 −→ 0, j = 1, 2.



Thermoelastic case



I βnu1,n(`1) −→ 0, ∂xu1,n(`1) −→ 0.

I (7) ∗ q∂xu1,n : ‖∂xu1,n‖2 + β2n ‖u1,n‖2 −→ 0,

I βnu1,n(0), ∂xu1,n(0), Re (iβnf1,n(0)u1,n(0)) −→ 0,

I (8) ∗ q∂xu2,n :

−12

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

2

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

I1

2β2n ‖u2,n‖2 +

3

2

∥∥∂2xu2,n

∥∥2 → 0.

In conclusion ‖yn‖ converge to 0, which contradict the hypothesisthat ‖yn‖ = 1.



Thermoelastic case


Polynomial stability

Theorem

If no control is applied on the string then, the C0-semigroup ispolynomially stable. More precisely, there is C > 0 such that∥∥etAy0

∥∥ ≤ C

t‖y0‖D(A)

for every y0 ∈ D(A).

ProofIt suffices to prove that (6) holds for α = 1. Suppose theconclusion is false. There exists a sequence (βn) of real numbers,without loss of generality, with βn −→ +∞, and a sequence ofvectors (yn) = (un, vn) in D(A) with ‖yn‖H = 1, such that

‖βαn (iβnI −A)yn‖H −→ 0



Thermoelastic case



Lemma [Gagliardo-Nirenberg]

(1) There are two positive constants C1 and C2 such that for anyw in H1(0, `j),

‖w‖∞ ≤ C1 ‖∂xw‖1/2 ‖w‖1/2 + C2 ‖w‖ . (9)

(2) There are two positive constants C3 and C4 such that for anyw in H2(0, `j),

‖∂xw‖ ≤ C3

∥∥∂2xw∥∥1/2 ‖w‖1/2 + C4 ‖w‖ . (10)



Thermoelastic case


Non exponential stability

u1,tt − u1,xx = 0 in (0, π)× (0,∞),u2,tt + u2,xxxx = 0 in (0, π)× (0,∞),

u1(0, t) = u2(0, t), u2,x(0, t) = 0, u2,xxx(0, t) = u1,x(0, t),u1(π, t) = 0, u2,xxx(π, t) = u2,t(π, t), u2,x(π, t) = 0,

uj(x , 0) = u0j (x), uj ,t(x , 0) = u1

j (x), j = 1, 2.

The system is not exponentially stable.



Thermoelastic case


Proof

We prove that the corresponding semigroup etA is notexponentially stable. Let

I βn = n2 + 2√

n + 1n , βn → +∞

I fn = (0, 0,− sinβnx , 0), fn is in H and is bounded.

I yn = (u1,n, u2,n, v1,n, v2,n) ∈ D(A) such that (A− iβn)yn = fn.We will prove that yn → +∞.

Iu1,n = c1 sin(βnx) + (− x

2βn+ c2) cos(βnx),

u2,n = d1 sin(√βnx) + d2 cos(

√βnx)

+d3 sinh(√βnx) + d4 cosh(

√βnx).



Thermoelastic case


Proof

I

2β3/2n d1 ∼+∞

π2

2

√n.

I

−π2

∣∣∣∣− 1

2βn+ βnc1

∣∣∣∣2 − π

2|βnc2|2

= −1

2(β2

n

∥∥u1n

∥∥2+∥∥∂xu1

n

∥∥2) + Re(

∫ π

0sin(βnx)(π − x)∂xu1

ndx).

β2n

∥∥u1n

∥∥2+∥∥∂xu1

n

∥∥2must be not bounded. In conclusion yn is

not bounded.



Thermoelastic case


Remarks

Let ε > 0. By taking βn = n2 + 2n1−α + 1n2α with 0 < α < ε

and such that n1−α is integer and even and yn is such that

fn = (β12−ε

n (A− iβn))yn, then we can prove that yn is notbounded and then the polynomial stability of (S) can’t bebutter than 1

t2 .

If we replace the boundary conditions by the followings

δu1(`1, t) = 0, (1− δ)u1xx(`1, t) = 0,

(1− δ)u1,x(`1, t) = −(1− δ)u1,t(`1, t),

δu2,xx(`2, t) = −δu2,tx(`2, t), u2(`2, t) = 0.

then we obtain the same results.



Thermoelastic case


System


u1,tt − α1u1,xx

+ β1θ1,x

= 0

θ1,t + β1u1,tx − κ1θ1,xx = 0

, u2,tt + α2u2,xxxx

+ β2θ2,x

= 0

θ2,t + β2u2,txx − κ2θ2,xx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0,

θ1(0, t) = θ2(0, t),

α2u2,xxx(0, t)

− β2θ2,x(0, t)

= α1u1,x(0, t)

− β1θ1(0, t)

,

κ1κ2(κ1θ1,x(0, t) + κ2θ2,x(0, t)) = 0

.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case


System

TE. String TE. Beamℓ2ℓ1 0

u1,tt − α1u1,xx

+ β1θ1,x

= 0

θ1,t + β1u1,tx − κ1θ1,xx = 0

, u2,tt + α2u2,xxxx

+ β2θ2,x

= 0

θ2,t + β2u2,txx − κ2θ2,xx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0,

θ1(0, t) = θ2(0, t),

α2u2,xxx(0, t)

− β2θ2,x(0, t)

= α1u1,x(0, t)

− β1θ1(0, t)

,

κ1κ2(κ1θ1,x(0, t) + κ2θ2,x(0, t)) = 0

.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case


System


u1,tt − α1u1,xx + β1θ1,x = 0θ1,t + β1u1,tx − κ1θ1,xx = 0

, u2,tt + α2u2,xxxx + β2θ2,x = 0θ2,t + β2u2,txx − κ2θ2,xx = 0


u1(0, t) = u2(0, t), u2,x(0, t) = 0,

θ1(0, t) = θ2(0, t),

α2u2,xxx(0, t)

− β2θ2,x(0, t)

= α1u1,x(0, t)

− β1θ1(0, t)

,

κ1κ2(κ1θ1,x(0, t) + κ2θ2,x(0, t)) = 0

.

Boundary conditions

u1(`1, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case


System





u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ1(0, t) = θ2(0, t),α2u2,xxx(0, t)− β2θ2,x(0, t) = α1u1,x(0, t)− β1θ1(0, t),

κ1κ2(

κ1θ1,x(0, t) + κ2θ2,x(0, t)

)

= 0.

Boundary conditions

u1(`1, t) = 0, θ(`1, t) = 0, θ(`2, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case


System





u1(0, t) = u2(0, t), u2,x(0, t) = 0, θ1(0, t) = θ2(0, t),α2u2,xxx(0, t)− β2θ2,x(0, t) = α1u1,x(0, t)− β1θ1(0, t),κ1κ2(κ1θ1,x(0, t) + κ2θ2,x(0, t)) = 0.

Boundary conditions

u1(`1, t) = 0, θ(`1, t) = 0, θ(`2, t) = 0,

u2(`2, t) = 0, u2,xx(`2, t) = 0.



Thermoelastic case


System

For a solution (u, v , θ) of (S) the energy is defined as

E (t) =1

2

∫ `1

0

(|u1,t |2 + α1 |u1,x |2 + |θ1|2

)dx

+1

2

∫ `2

0

(|u2,t |2 + α2 |u2,xx |2 + |θ2|2

)dx .

Differentiate formally the energy function with respect to time t,weget

d

dtE (t) = −κ1 ‖∂xθ1‖2 − κ2 ‖∂xθ2‖2

and the system is dissipative.



Thermoelastic case


Let us consider

V ={

f = (f1, f2) ∈ H1(0, `1)× H2(0, `2) | f satisfies (11)}

where

f1(`1) = 0, f2(`2) = 0, f1(0) = f2(0) and ∂x f2(0) = 0. (11)

Define the Hilbert space H

H = V ×(L2(0, `1)× L2(0, `2)

)×W

with W = L2(0, `1)× L2(0, `2) if e1 and e2 are thermoelastic,W = L2(0, `1)× {0} if only e1 is thermoelastic andW = {0} × L2(0, `2) if only e1 is purely elastic, and norm given by

‖z‖H := α1 ‖∂x f1‖2 + α2

∥∥∂2x f2

∥∥2+

2∑j=1

(‖gj‖2 + ‖hj‖2

)where z = (f = (f1, f2), g = (g1, g2), h = (h1, h2)) .



Thermoelastic case


D(A) =

{y = (u, v , θ) ∈ V ∩ (H2(0, `1)× H4(0, `2))× V ×W2 |

and y satisfies (12)

}with W2 = H2(0, `1)× H2(0, `2) if e1 and e2 are T.... and where

∂2xu2(`2) = 0, θ1(`1) = θ2(`2) = 0,θ1(0) = θ2(0),α2∂

3xu2(0)− β2∂xθ2(0) = α1∂xu1(0)− β1θ1(0),

κ1κ2 (κ1∂xθ1(0) + κ2∂xθ2(0)) = 0.

(12)

with βj = 0 and κj = 0 if ej is purely elastic, and

A

u1

u2

v1

v2

θ1

θ2

=

v1

v2

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

2xθ2

−β1∂xv1 + κ1∂2xθ1

−β2∂xxv2 + κ2∂2xθ2

.



Thermoelastic case


D(A) =

{y = (u, v , θ) ∈ V ∩ (H2(0, `1)× H4(0, `2))× V ×W2 |



∂2xu2(`2) = 0, θ1(`1) = θ2(`2) = 0,θ1(0) = θ2(0),α2∂

3xu2(0)− β2∂xθ2(0) = α1∂xu1(0)− β1θ1(0),

κ1κ2 (κ1∂xθ1(0) + κ2∂xθ2(0)) = 0.

(12)

with βj = 0 and κj = 0 if ej is purely elastic,

and

A

u1

u2

v1

v2

θ1

θ2

=

v1

v2

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

2xθ2

−β1∂xv1 + κ1∂2xθ1

−β2∂xxv2 + κ2∂2xθ2

.



Thermoelastic case


D(A) =

{y = (u, v , θ) ∈ V ∩ (H2(0, `1)× H4(0, `2))× V ×W2 |



∂2xu2(`2) = 0, θ1(`1) = θ2(`2) = 0,θ1(0) = θ2(0),α2∂

3xu2(0)− β2∂xθ2(0) = α1∂xu1(0)− β1θ1(0),

κ1κ2 (κ1∂xθ1(0) + κ2∂xθ2(0)) = 0.

(12)

with βj = 0 and κj = 0 if ej is purely elastic, and

A

u1

u2

v1

v2

θ1

θ2

=

v1

v2

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

2xθ2

−β1∂xv1 + κ1∂2xθ1

−β2∂xxv2 + κ2∂2xθ2

.



Thermoelastic case


Then, putting y = (u, ut , θ), we write the system (S) in the threecases, into the following first order evolution equation{

ddt y = Ayy(0) = y0

(13)

on the energy space H, where y0 = (u0, v 0, θ0).We have the following result,

Lemma

The operator A is the infinitesimal generator of a C0-semigroup ofcontraction S(t).



Thermoelastic case



Lemma

The semigroup S(t), generated by the operator A is asymptoticallystable.

Theorem

If the string is thermoelastic, then the system (S) is exponentiallystable.



Thermoelastic case


Proof

It suffices to prove that (5) holds. Suppose the conclusion is false.Then there exists a sequence (wn) of real numbers, withwn −→ +∞ and a sequence of vectors (yn) = (un, vn, θn) in D(A)with ‖yn‖H = 1, such that

‖(iwnI −A)yn‖HF−→ 0

which is equivalent to

iwnu1,n − v1,n = f1,n −→ 0, in H1(0, `1),

iwnv1,n − α1∂2xu1,n + β1∂xθ1,n = g1,n −→ 0, in L2(0, `1),

iwnθ1,n + β1∂xv1,n − κ1∂2xθ1,n = h1,n −→ 0, in L2(0, `1),

and

iw2,nu2,n − v2,n = f2,n −→ 0, in H2(0, `2),

iwnv2,n + α2∂4xu2,n = g2,n −→ 0, in L2(0, `2),



Thermoelastic case


We get

w 2nu1,n + α1∂

2xu1,n − β1∂xθ1,n = −g1,n − iwnf1,n, (14)

−w 2nu2,n + α2∂

4xu2,n = g2,n + iwnf2,n, (15)

and‖vj ,n‖2 − w 2

n ‖uj ,n‖2 −→ 0, j = 1, 2.

First, since

Re(〈(iwn −A)yn, yn〉H) = −κ1 ‖∂xθ1‖2

we obtain that ∂xθ1,n converges to 0 in L2(0, `2).As in [?] one can get

‖wnu1,n‖ , ‖∂xu1,n‖ , ‖θ1,n‖ −→ 0.

Moreoverwnu1,n(0), ∂xu1,n(0), θ1,n(0) −→ 0. (16)



Thermoelastic case


Proof

Taking the inner product of (15) with p = (`2 − x)∂xu2,n(x),

−1

2α2

∣∣∂2xu2,n(0)

∣∣2 `2+1

2

∫ `2

0w 2n |u2,n|2 dx+

3

2α2

∫ `2

0

∣∣∂2xu2,n

∣∣2 dx → 0

Now the inner product of the first member of (15) by 1

w1/2n

e−aw1/2n x

gives, with a = 1

α1/42

,

α2

w1/2n

∂3xu2,n(0) + α2a∂2

xu2,n(0) = o(1)

then∂2xu2,n(0) = o(1)

Return back to(4),∫ `2

0w 2n |u2,n|2 dx ,

∫ `2

0

∣∣∂2xu2,n

∣∣2 dx , converge to zero



Thermoelastic case


Lack of exponential stability

In this part the string is purely elastic.

We take `1 = `2 = π, κ2 << α2.

Theorem

If the string is purely elastic then the system (S) is not exponentialstable in the energy space H.



Thermoelastic case


Proof

We prove that the corresponding semigroup (S(t))t≥0 is notexponentially stable.For n ∈ N, let fn = (0, 0,−α1 sinβnx , 0, 0), with βn → +∞ and fnis in H and is bounded. Let yn = (u1,n, u2,n, v1,n, v2,n, θ2,n) ∈ D(A)such that (A− idn)yn = fn. We will prove that yn → +∞.We have

w 2nu1,n + α1∂

2xu1,n = α1 sinβnx

with wn =√α1βn, and

iw2,nu2,n − v2,n = 0, in H2(0, π), (17)

−w 2nu2,n + α2∂

4xu2,n − β2∂

2xθ2,n = 0, in L2(0, π), (18)

iwnθ2,n + iwnβ2∂2xu2,n − κ2∂

2xθ2,n = 0, in L2(0, π). (19)

Notations: α2 = α, β2 = β, κ2 = κ.



Thermoelastic case


Proof

The function u1,n is of the form

u1,n = c1 sin(wnx) + (− x

2wn+ c2) cos(wnx),

Using (18) and (19) we obtain that

ακ∂6xu2,n − iwn(α + β2)∂4

xu2,n − κw 2n∂

2xu2,n + iw 3

nu2,n = 0, (20)

By taking A = 3ακ2 + (α + β2)2,B = 9ακ2(α + β2) + 2(α + β2)3 − 27α2κ2,

a1 = 121/3

(√B2 + 4A3 + B

)1/3, b1 = 1

21/3

(√B2 + 4A3 − B

)1/3

and r = α + β2, the squares of the solutions of (20) are



Thermoelastic case


Proof

x1 =wn

3ακ

[√3

2(a1 − a2) + i

(r +

1

2(a1 − a2)

)]

x2 =wn

3ακ

[−√

3

2a1 + i

(r +

1

2a1 + a2

)],

x3 =wn

3ακ

[√3

2a2 + i

(r − a1 −

1

2a2

)]



Thermoelastic case


Proof

Let x2, x ′2 and x ′′2 the squares of the real parts of solutions of(20).

2x2 =

(3

4(a1 − a2)2 + (r +

1

2(a1 − a2))2

)1/2

+

√3

2(a1 − a2),

2x ′2 =

(3

4(a2

1 + (r +1

2a1 + a2)2

)1/2

−√

3

2a1,

2x ′′2 =

(3

4(a2

2 + (r − a1 −1

2a2)2

)1/2

+

√3

2a2.

2x2 > 2x ′′2 > 2x ′2.The equation (20) admits six simple solutions

±√

wnR1, ±√

wnR2, ±√

wnR3,

with0 < Re(R3) < Re(R2) < Re(R1).



Thermoelastic case


Proof

u2,n =3∑

k=1

(dke√wnRkx + bke−

√wnRkx).

Return back to (18),

β∂2xθ2,n = w 2

n

3∑k=1

(−1 + αR4k )(dke

√wnRkx + bke−

√wnRkx)

Then there exist two constants a′ and b′ such that

βθ2,n = wn

3∑k=1

(− 1

R2k

+αR2k )(dke

√wnRkx + bke−

√wnRkx) + a′x + b′.

Moreover, the equation (19) is verified if and only if a′ = b′ = 0.



Thermoelastic case


Proof

The transmission and boundary conditions are expressed as follow

3∑k=1

(dk + bk ) = c2,3∑

k=1

Rk (dk − bk ) = 0, (21)

w3/2n α

3∑k=1

1

Rk

(dk − bk ) = −1

2wn+ wnc1, (22)

3∑k=1

(−1

R2k

+ αR2k )(dk + bk ) = 0,

3∑k=1

(dk e√

wnRkπ + bk e−√wnRkπ) = 0, (23)

3∑k=1

R2k (dk e

√wnRkπ + bk e

−√wnRkπ) = 0,3∑

k=1

1

R2k

(dk e√

wnRkπ + bk e−√wnRkπ) = 0, (24)

c1 sin(βnπ) + (−π

2βn+ c2) cos(βnπ) = 0. (25)



Thermoelastic case


Proof

After some calculus

[2a4e√

wn(R1+2R2)π + ...]

(−1

2βn+ βnc1) = w3/2

n (π

2βn− c1 tan(βnπ))

[a3e√

wn(R1+2R2)π + ...].

and then

2a4(− 1

2βn+ βnc1) + a3w

3/2n c1 tan(βnπ) ∼ π

2

w3/2n

βn.

Hence, with βn = 2n + 1n , tan(βnπ) = π

n + ...

(− 1

2βn+ βnc1) ∼ π

4a4

w3/2n

βn=π√α1

4a4

√wn.

The real part of the inner product of (6) with (π − x)∂xu1,n gives

−π

2

∣∣∣∣− 1

2wn+ wnc1

∣∣∣∣2 − π

2|wnc2|

2 = −1

2(w2

n

∥∥u1,n∥∥2 +

∥∥∂xu1,n∥∥2) + Re(

∫ π0

sin(wnx)(π − x)∂xu1,ndx).

In conclusion yn is not bounded.Farhat Shel Stability of some string-beam systems


Thermoelastic case



Theorem

If the string is purely elastic, then the system (S) is polynomiallystable. More precisely, (for every γ < 2) there exists c > 0 suchthat

‖S(t)y0‖ ≤1

tγ‖y0‖D(A) .



Thermoelastic case


Proof

Let 1 > α > 12 . It suffices to prove that (5) holds. Suppose the

conclusion is false. Then there exists a sequence (wn) of realnumbers, with wn −→ +∞ and a sequence of vectors(yn) = (un, vn, θn) in D(A) with ‖yn‖H = 1, such that

‖wαn (iwnI −A)yn‖H −→ 0

which is equivalent to

wαn (iwnu1,n − v1,n) = f1,n −→ 0, in H1, (26)

wαn

(iwnv1,n − α1∂

2xu1,n

)= g1,n −→ 0 in L2, (27)

and

wαn (iwnu2,n − v2,n) = f2,n −→ 0, in H2,(28)

wαn

(iwnv2,n + α2∂

4xu2,n − β2∂

2xθ2,n

)= g2,n −→ 0, in L2,(29)

wαn

(iwnθ2,n + β2∂

2xv2,n − κ2∂

2xθ2,n

)= h2,n −→ 0, in L2.(30)



Thermoelastic case


Substituting (26) into (27) and (28) into (30) respectively to get

wαn

(w 2nu1,n + α1∂

2xu1,n

)= −g1,n − iwnf1,n, (31)

wαn

(θ2,n −

1

iwnκ2∂

2xθ2,n + β2∂

2xu2,n

)=

1

iwn(h2,n + ∂2

x f2,n)(32)

First, wα/2n ∂xθ2,n converge to 0 in L2(0, `2). Then w

α/2n θ2,n

converge to 0 in L2(0, `2) since θ2,n(0) = 0.

Multiplying (32) by 1

wα/2n

∂2xu2,n

β2wα/2n

∥∥∂2xu2,n

∥∥2+ w

α/2n

⟨θ2,n, ∂

2xu2,n

⟩(33)

−iκ2wα/2−1∂xθ2,n(0)∂2xu2,n(0)− iκ2w

α/2−1n

⟨∂xθ2,n, ∂

3xu2,n

⟩= 0.

Then we prove that

β2wα/2n

∥∥∂2xu2,n

∥∥2 −→ 0.



Thermoelastic case


Substituting (26) into (27) and (28) into (30) respectively to get

wαn

(w 2nu1,n + α1∂

2xu1,n

)= −g1,n − iwnf1,n, (31)

wαn

(θ2,n −

1

iwnκ2∂

2xθ2,n + β2∂

2xu2,n

)=

1

iwn(h2,n + ∂2

x f2,n)(32)

First, wα/2n ∂xθ2,n converge to 0 in L2(0, `2). Then w

α/2n θ2,n

converge to 0 in L2(0, `2) since θ2,n(0) = 0.Multiplying (32) by 1

wα/2n

∂2xu2,n

β2wα/2n

∥∥∂2xu2,n

∥∥2+ w

α/2n

⟨θ2,n, ∂

2xu2,n

⟩(33)

−iκ2wα/2−1∂xθ2,n(0)∂2xu2,n(0)− iκ2w

α/2−1n

⟨∂xθ2,n, ∂

3xu2,n

⟩= 0.

Then we prove that

β2wα/2n

∥∥∂2xu2,n

∥∥2 −→ 0.



Thermoelastic case


Proof

Using (29) we prove that wα/8n ‖v2,n‖2 → 0.

We built two sequences of positive numbers rm and sm suchthat

wrm/2n

∥∥∂2xu2,n

∥∥→ 0, wrm/2n ‖θ2,n‖ → 0, w

sm/2n ‖v2,n‖ → 0

and rm and sm converge to 1 + α.

α2∂3xu2,n(0)− β2∂xθ(0)→ 0.

wn ‖u2,n(0)‖ → 0.∫ `2

0

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

In summary, we have ‖yn‖H −→ 0. This resul contradicts thehypothesis that yn has the unit norm.



Thermoelastic case


Proof



wrm/2n

∥∥∂2xu2,n

∥∥→ 0, wrm/2n ‖θ2,n‖ → 0, w

sm/2n ‖v2,n‖ → 0


α2∂3xu2,n(0)− β2∂xθ(0)→ 0.

wn ‖u2,n(0)‖ → 0.

∫ `2

0

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.




Thermoelastic case


Proof



wrm/2n

∥∥∂2xu2,n

∥∥→ 0, wrm/2n ‖θ2,n‖ → 0, w

sm/2n ‖v2,n‖ → 0


α2∂3xu2,n(0)− β2∂xθ(0)→ 0.

wn ‖u2,n(0)‖ → 0.∫ `2

0

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.




Thermoelastic case


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stability of some string-beam systems · 2017-05-19 · introduction feedback stabilization...

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