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

Outline

1 Introduction

2 Feedback stabilizationAbstract settingAsymptotic behavior

3 Thermoelastic caseAbstract settingAsymptotic behavior

Thermoelastic case

Outline

1 Introduction

Thermoelastic case

Outline

1 Introduction

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

dtE (t) = 0.

The system is conservative.

Thermoelastic case

Models

u1,tt − u1,xx

+ βθx

θt + βvtx − κθxx = 0

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

, u2,tt + u2,xxxx

+ βθx

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

θt + βutx − κθxx = 0

, u2,tt + u2,xxxx

+ βθx

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

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

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

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

, 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

, 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

References

Thermoelastic case

References

Thermoelastic case

Abstract settingAsymptotic behavior

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

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.

Energy space:H = V × L2(G),

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.

Thermoelastic case

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

f1(`1) = 0,

δ ∈ {0, 1}

f1(0) = f2(0),∂x f2(0) = 0.

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.

Thermoelastic case

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

δf2(`2) = 0, (1− δ)f1(`1) = 0,

δ ∈ {0, 1}

f1(0) = f2(0),∂x f2(0) = 0.

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.

Thermoelastic case

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

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

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

⟩Hilbert space.

Thermoelastic case

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

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

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 2

⟩+⟨g 1

1 , g21

⟩+⟨g 1

2 , g22

Hilbert space.

Thermoelastic case

Energy space

L2(G) = L2(0, `1)× L2(0, `2).

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

〈y1, y2〉H :=⟨∂x f 1

1 , ∂x f 21

⟩+⟨∂2x f 1

2 , ∂2x f 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

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

∂2xu1

−∂4xu2

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

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

∂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.

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).

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

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

‖y0‖D(B)

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

lim|β|→∞

βα∥∥(iβ − B)−1

∥∥ <∞ (6)

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

‖y0‖D(B)

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

lim|β|→∞

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

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

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).

−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

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).

−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 :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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

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 (8) ∗ q∂xu2,n :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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 (8) ∗ q∂xu2,n :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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 (8) ∗ q∂xu2,n :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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 (8) ∗ q∂xu2,n :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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 (8) ∗ q∂xu2,n :

∣∣∂2xu2,n(`2)

∣∣2 + 12β

2n ‖u2,n‖2 + 3

∥∥∂2xu2,n

∥∥2 → 0,

I (8) ∗ 1

β1/2n

e−β1/2n x : ∂2

xu2,n(`2)→ 0,

2β2n ‖u2,n‖2 +

∥∥∂2xu2,n

∥∥2 → 0.

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

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

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

I βn = n2 + 2√

n + 1n , βn → +∞

√βnx)

√βnx).

Thermoelastic case

I βn = n2 + 2√

n + 1n , βn → +∞

√βnx)

√βnx).

Thermoelastic case

2β3/2n d1 ∼+∞

−π2

∣∣∣∣− 1

2βn+ βnc1

∣∣∣∣2 − π

2|βnc2|2

= −1

∥∥u1n

∥∥2+∥∥∂xu1

∥∥2) + Re(

∫ π

0sin(βnx)(π − x)∂xu1

∥∥u1n

∥∥2+∥∥∂xu1

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

not bounded.

Thermoelastic case

2β3/2n d1 ∼+∞

−π2

∣∣∣∣− 1

2βn+ βnc1

∣∣∣∣2 − π

2|βnc2|2

= −1

∥∥u1n

∥∥2+∥∥∂xu1

∥∥2) + Re(

∫ π

∥∥u1n

∥∥2+∥∥∂xu1

not bounded.

Thermoelastic case

2β3/2n d1 ∼+∞

−π2

∣∣∣∣− 1

2βn+ βnc1

∣∣∣∣2 − π

2|βnc2|2

= −1

∥∥u1n

∥∥2+∥∥∂xu1

∥∥2) + Re(

∫ π

∥∥u1n

∥∥2+∥∥∂xu1

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

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

Remarks

fn = (β12−ε

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.

Thermoelastic case

Remarks

fn = (β12−ε

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.

Thermoelastic case

System

u1,tt − α1u1,xx

+ β1θ1,x

θ1,t + β1u1,tx − κ1θ1,xx = 0

, u2,tt + α2u2,xxxx

+ β2θ2,x

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

θ1,t + β1u1,tx − κ1θ1,xx = 0

, u2,tt + α2u2,xxxx

+ β2θ2,x

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)

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

∫ `1

(|u1,t |2 + α1 |u1,x |2 + |θ1|2

∫ `2

(|u2,t |2 + α2 |u2,xx |2 + |θ2|2

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

dtE (t) = −κ1 ‖∂xθ1‖2 − κ2 ‖∂xθ2‖2

and the system is dissipative.

Thermoelastic case

Let us consider

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

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)

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.

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

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

−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.

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

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

−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.

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

α1∂2xu1 − β1∂xθ1

−α2∂4xu2 + β2∂

−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

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

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

Thermoelastic case

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

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),

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

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

∣∣∂2xu2,n(0)

∣∣2 `2+1

∫ `2

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

∫ `2

∣∣∂2xu2,n

∣∣2 dx → 0

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

e−aw1/2n x

gives, with a = 1

α1/42

∂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

∣∣∂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

We take `1 = `2 = π, κ2 << α2.

Theorem

Thermoelastic case

We take `1 = `2 = π, κ2 << α2.

Theorem

Thermoelastic case

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

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

(√B2 + 4A3 − B

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

Thermoelastic case

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

(√B2 + 4A3 − B

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

Thermoelastic case

x1 =wn

2(a1 − a2) + i

2(a1 − a2)

x2 =wn

[−√

2a1 + i

2a1 + a2

x3 =wn

2a2 + i

(r − a1 −

Thermoelastic case

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

4(a1 − a2)2 + (r +

2(a1 − a2))2

2(a1 − a2),

2x ′2 =

1 + (r +1

2a1 + a2)2

−√

2x ′′2 =

2 + (r − a1 −1

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

wnR1, ±√

wnR2, ±√

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

Thermoelastic case

u2,n =3∑

(dke√wnRkx + bke−

√wnRkx).

Return back to (18),

β∂2xθ2,n = w 2

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

+αR2k )(dke

√wnRkx + bke−

√wnRkx) + a′x + b′.

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

Thermoelastic case

The transmission and boundary conditions are expressed as follow

3∑k=1

(dk + bk ) = c2,3∑

Rk (dk − bk ) = 0, (21)

w3/2n α

3∑k=1

(dk − bk ) = −1

2wn+ wnc1, (22)

3∑k=1

+ 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∑

(dk e√

wnRkπ + bk e−√wnRkπ) = 0, (24)

c1 sin(βnπ) + (−π

2βn+ c2) cos(βnπ) = 0. (25)

Thermoelastic case

After some calculus

[2a4e√

wn(R1+2R2)π + ...]

2βn+ βnc1) = w3/2

2βn− c1 tan(βnπ))

[a3e√

wn(R1+2R2)π + ...].

and then

2a4(− 1

2βn+ βnc1) + a3w

3/2n c1 tan(βnπ) ∼ π

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

n + ...

(− 1

2βn+ βnc1) ∼ π

βn=π√α1

√wn.

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

∣∣∣∣− 1

2wn+ wnc1

∣∣∣∣2 − π

2|wnc2|

2 = −1

∥∥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

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)

(iwnv1,n − α1∂

2xu1,n

)= g1,n −→ 0 in L2, (27)

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

(iwnv2,n + α2∂

4xu2,n − β2∂

2xθ2,n

)= g2,n −→ 0, in L2,(29)

(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

(w 2nu1,n + α1∂

2xu1,n

)= −g1,n − iwnf1,n, (31)

(θ2,n −

iwnκ2∂

2xθ2,n + β2∂

2xu2,n

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

⟨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

(w 2nu1,n + α1∂

2xu1,n

)= −g1,n − iwnf1,n, (31)

(θ2,n −

iwnκ2∂

2xθ2,n + β2∂

2xu2,n

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

⟨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

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

(|∂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

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

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

Thermoelastic case

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

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

Thermoelastic case

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

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

Thermoelastic case

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

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

Thermoelastic case

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

(|∂xu1,n(x)|2 + w 2

n |u1,n(x)|2)

dx −→ 0.

Thermoelastic case

THANKS!

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