Stabilization and asymptotic behavior of a generalized telegraph - - PowerPoint PPT Presentation

Stabilization and asymptotic behavior of a generalized telegraph equation

Serge Nicaise

Université de Valenciennes et du Hainaut-Cambresis Laboratoire de Mathematiques et leurs Applications de Valenciennes, LAMAV

Serge Nicaise (LAMAV) Stabilization telegraph equation 18 June 2015, MIS 2015 1 / 32

Outline

1

Introduction

2

Well Posedness

3

Strong stability

4

Energy decay

5

A spectral analysis

6

Optimality

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Introduction

Motivation

Consider the evolution system:            Vt + gV + aIx + kW = 0 in Ω = (0, 1) × (0, +∞), It + rI + bVx = 0 in Ω × (0, +∞), Wt + cW = V in Ω × (0, +∞), V(0, ·) = V(1, ·) = 0,

• n (0, +∞),

V(x, 0) = V0(x), I(x, 0) = I0(x), W(x, 0) = W0(x), in Ω. (1) V electric potential, I electric current, W auxiliary variable (representing the non local effects) Coupling between the usual telegraph equation and a first order differential equation of parabolic type. Such problem was first introduced in

• S. Imperiale and P

. Joly. Error estimates for 1D asymptotic models in coaxial cables with non-homogeneous cross-section.

• Adv. Appl. Math. Mech., 4(6):647–664, 2012.

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Introduction

Particular cases

If r = k = 0, eliminating I, we find the wave equation with internal damping: Vtt − abVxx + gVt = 0, Exponential decay if ab > 0 and g > 0. If r = g = 0 by eliminating I and setting p = Wt, we arrive at        Vtt − abVxx + kp = 0, in Ω × (0, +∞), pt + cp = Vt in Ω × (0, +∞), V(0, ·) = V(L, ·) = 0,

• n (0, +∞),

V(x, 0) = V0(x), p(x, 0) = V0(x) − cW0(x), in Ω. Coupling between the wave equation in V with a first order differential equation in p for which a polynomial stability was established in

• S. Nicaise.

Stabilization and asymptotic behavior of dispersive medium models. Systems Control Lett., 61(5):638–648, 2012.

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Introduction

Main questions

Strong stability of the solution. Uniform Stability. Polynomial Stability. Optimality.

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Well Posedness

The energy space

We assume that all the coefficients involved in (1) are in L∞(Ω), real valued and non-negative. Moreover, we suppose that there exists a positive constant δ such that a ≥ δ, b ≥ δ, c ≥ δ, k + g ≥ δ, a. e. in Ω. (2) These assumptions are in agreement with physical settings. Consider the energy space H = L2(Ω)3, equipped with the inner product

• (V, I, W)⊤, (V ∗, I∗, W ∗)⊤

H =

• Ω

(αV ¯ V ∗ + βI¯ I∗ + γW ¯ W ∗) dx, (3) with α, β, γ ∈ L∞(Ω) fixed appropriately such that α(x) ≥ δ0, β(x) ≥ δ0, γ(x) ≥ δ0, a. e. in Ω, (4) for some δ0 > 0.

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Well Posedness

Cauchy problem/Maximal dissipativity

Problem (1) can be written as a Cauchy problem: ˙ u = Au, u(0) = u0, (5) with D(A) = H1

0(Ω) × H1(Ω) × L2(Ω),

and A   V I W   := −   aIx + gV + kW bVx + rI cW − V   , ∀   V I W   ∈ D(A). There exist α, β, γ ∈ L∞(Ω) satisfying (4) such that A is m-dissipative, i.e., λI − A is surjective for some λ > 0 and ℜ(AU, U) ≤ −1 2

• Ω

(αg|V|2 + 2βr|I|2 + γc|W|2) ≤ 0. Using Lumer-Phillips’ thm, A generates a C0-semigroup of contraction.

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Strong stability

Arendt-Batty’s thm

One simple way to prove the strong stability is to use the following theorem. Theorem (Arendt-Batty) Let X be a reflexive Banach space and (T(t))t≥0 be a semigroup generated by A on X. Assume that (T(t))t≥0 is bounded and that no eigenvalues of A lies on the imaginary axis. If σ(A) ∩ iR is countable, then (T(t))t≥0 is stable. Since the resolvent of our operator is not compact, we have to analyze the full spectrum on the imaginary axis.

• W. Arendt and C. J. K. Batty.

Tauberian theorems and stability of one-parameter semigroups.

• Trans. Amer. Math. Soc., 305(2):837–852, 1988.

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Strong stability

Eigenvalue problem

Lemma (Le 1) For all ξ ∈ R∗ := R \ {0}, we have ker(iξ − A) = {0}. If r is not identically equal to 0, we have ker A = {0}.

• therwise 0 is an eigenvalue of A whose associated eigenvector is

(0, 1, 0)⊤.

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Strong stability

Proof

Let ξ ∈ R and U = (V, I, W)⊤ ∈ D(A) be such that (iξ − A)U = 0, or equivalently satisfying    iξV + aIx + gV + kW = 0, iξI + bVx + rI = 0, iξW + cW − V = 0. (6) Hence by the dissipativeness of A, we get gV = rI = W = 0. Coming back to the third identity of (6), we obtain that V = 0 and by the first identity of (6) Ix = 0 (hence I is constant). By the second identity of (6) we obtain (iξ + r)I = 0. Hence ξ = 0 ⇒ I = 0; otherwise if ξ = 0, this identity reduces to rI = 0. If r = 0 ⇒ I = 0, otherwise if r ≡ 0: I = constant allowed.

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Strong stability

Surjectivity

Lemma (Le 2) For all ξ ∈ R∗, iξ − A is surjective. If r is not identically equal to 0, then A is surjective, otherwise the range R(A) of A is equal to H0 := {(V, I, W)⊤ ∈ H :

• Ω

βI = 0}. The proof is based on a compact perturbation argument and the use of the previous Lemma.

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Strong stability

Strong stability of A

In view of the previous result, if r = 0, we need to introduce the

• perator A0 from H0 into itself defined by D(A0) = D(A) ∩ H0 and

A0U = AU, ∀U ∈ D(A0). Corollary (Coro 3) If r is not identically equal to 0, A is strongly stable, otherwise A0 is strongly stable.

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Energy decay

Frequency domain approach: exponential decay

Lemma (Pruss, Huang) A C0 semigroup (etA)t≥0 of contractions on a Hilbert space H is exponentially stable, i.e., satisfies ||etAU0|| ≤ C e−ωt||U0||H, ∀U0 ∈ H, ∀t ≥ 0, for some positive constants C and ω if ρ(A) ⊃

• iβ
• β ∈ R
• ≡ iR,

(7) sup

β∈R

(iβ − A)−1L(H) < ∞. (8)

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Energy decay

Frequency domain approach: polynomial decay

Lemma (Borichev-Tomilov) A C0 semigroup (etA)t≥0 of contractions on a Hilbert space H satisfies ||etAU0|| ≤ C t− 1

l ||U0||D(A),

∀U0 ∈ D(A), ∀t > 1, for some constant C > 0 and for some positive integer l if (7) holds and if lim sup

|β|→∞

1 βl (iβ − A)−1L(H) < ∞. (9)

• A. Borichev and Y. Tomilov.

Optimal polynomial decay of functions and operator semigroups.

• Math. Ann., 347(2):455–478, 2010.

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Energy decay

Resolvent estimate

Since condition (7) was already treated, it remains to analyze the behaviour of the resolvent on the imaginary axis. Let us start with the exponential decay. Lemma (Le 4) In addition to the previous assumption (2), suppose that r ∈ W 1,∞(Ω) and that there exists δ1 > 0 such that r + g ≥ δ1 a. e. in Ω. (10) Then the resolvent of the operator of A satisfies condition (8), i.e., sup

β∈R

(iβ − A)−1L(H) < ∞.

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Energy decay

Proof

We use a contradiction argument, i.e., we suppose that (8) is false. Then there exist a sequence of real numbers ξn → +∞ and a sequence of vectors Un = (Vn, In, Wn)⊤ in D(A) such that UnH = 1, (11) (iξn + g)Vn + aIn,x + kWn → 0, (12) (iξn + r) In + bVn,x → 0, (13) (iξn + c)Wn − Vn → 0. (14) By the dissipativeness of A, we directly have √gVn → 0 in L2(Ω), (15) √ rIn → 0 in L2(Ω), (16) Wn → 0 in L2(Ω). (17)

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Energy decay

Proof ctd

Now we multiply (12) by a−1r ¯ Vn and integrate in Ω to find

• Ω
• (iξ + g)a−1r|Vn|2 + rIn,x ¯

Vn

• → 0.

Green’s formula: ⇒

• Ω
• (iξ + g)a−1r|Vn|2 − rxIn ¯

Vn − rIn ¯ Vn,x

• → 0.

Using (13) and (11) we get

• Ω
• (iξ + g)a−1r|Vn|2 − rxIn ¯

Vn + b−1r(−iξn + r)|In|2 → 0. Dividing by iξn and again thanks to (11) we obtain

• Ω

(a−1r|Vn|2 − b−1r|In|2) → 0. (16) ⇔ √ rIn → 0 in L2(Ω) ⇒ √ rVn → 0 in L2(Ω). (15) ⇔ √gVn → 0 in L2(Ω) and reminding the assumption r + g ≥ δ1: Vn → 0 in L2(Ω). (18)

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Energy decay

Proof ctd

We multiply (13) by 1

b¯

In and integrate in Ω to find thanks to (11)

• Ω

iξn + r b |In|2 + Vn,x¯ In

• → 0.

Applying Green’s formula in the last term, we find that

• Ω

iξn + r b |In|2 − Vn¯ In,x

• → 0.

The properties (12), (17) and (11) yields

• Ω

iξn + r b |In|2 + −iξn + g a |Vn|2 → 0. (19) By dividing by iξn and taking into account (18), we obtain In → 0 in L2(Ω). (20)

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Energy decay

Resolvent estimate ctd

Lemma (Le 5) Under the assumption (2), the resolvent of the operator of A always satisfies condition (9) with l = 2, i.e., lim sup

|β|→∞

1 β2 (iβ − A)−1L(H) < ∞.

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Energy decay

Exponential stability results

Theorem (Thm 6) Let the assumption (2) be satisfied. Then

• 1. If r ∈ W 1,∞(Ω), r is not identically equal to 0 and (10) holds, then

problem (5) is exponentially stable in H, i.e., there exist two positive constants C and ω such that E(t) ≤ C e−ωtE(0), ∀t ≥ 0, U0 ∈ H.

• 2. If r ≡ 0 and (10) holds, then problem (5) is exponentially stable in

H0, i.e., there exist two positive constants C and ω such that U(t) − κ(0, 1, 0)⊤2

H ≤ C e−ωtE(0),

∀t ≥ 0, for all U0 ∈ H, with κ = (

• Ω β)−1U0, (0, 1, 0)H.

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Energy decay

Polynomial stability results

Theorem (Thm 7) Let the assumption (2) be satisfied. Then

• 1. if r is not identically equal to 0, problem (5) is polynomially stable in

H, more precisely there exists a positive constant C such that E(t) ≤ C t−1||U0||2

D(A), ∀t > 1, U0 ∈ D(A).

(21)

• 2. If r ≡ 0, problem (5) is polynomially stable in H0, in other words,

there exists a positive constant C such that U(t) − κ(0, 1, 0)⊤2

H ≤ C t−1||U0||2 D(A), ∀t > 1, U0 ∈ D(A).

(22)

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A spectral analysis

Roots of a family of polynomials

In order to give optimal decay rate of the energy, we perform a careful spectral analysis of the operator A. This analysis is made under the assumptions that the functions a, b, c, k, r, g are constant and that k > 0. The general case is an open question. Under these assumptions, we note that (2) reduces to the positiveness of a, b and

• c. For simplicity, we also assume that r = c.

The discrete spectrum of A is related to the roots of a family of

• polynomials. More precisely for any ℓ ∈ N∗ introduce the third order

polynomial pℓ(λ) = (λ + r)

• (λ + g)(λ + c) + k
• + ℓ2π2ab(λ + c).

(23) Denotes by λℓ,i, i = 1, 2, 3 its roots repeated according to their multiplicity.

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A spectral analysis

Properties of the roots

Lemma (Le 8) The polynomial pℓ has always a real eigenvalue, say λℓ,3. λℓ,3 belongs to the interval (− max{r, c}, − min{r, c}) and neither −r nor −c is a root of pℓ. There exists ℓ0 ∈ N∗ (depending on the coefficients) such that for ℓ ≥ ℓ0, ℑλℓ,1 > 0 (hence the roots of pℓ are simple) with lim

ℓ→∞ ℜλℓ,1 = −r + g

2 , lim

ℓ→∞

ℑλℓ,1 ℓ = κ, lim

ℓ→∞ ℓ2(λℓ,3 + c) = ρ,

with κ = π √ ab and ρ = k(c−r)

π2ab .

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A spectral analysis

Illustration

0.9 0.8 0.7 0.6 0.5 60 40 20 20 40 60

Figure: Roots of pℓ for r = 0 and a = b = c = g = k = 1.

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A spectral analysis

Spectrum of A

Theorem (Thm 9) For all ℓ ∈ N∗, the operator A has three eigenvalues λℓ,i, i = 1, 2, 3, which are roots of the polynomial pℓ. If the roots of pℓ are simple, the eigenvector associated with λℓ,i, i = 1, 2 or 3, is given by (up to a multiplicative factor) Uℓ,i =    sin(ℓπ·) − bℓπ

λℓ,i+r cos(ℓπ·) 1 λℓ,i+c sin(ℓπ·)

   . (24)

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A spectral analysis

Proof

Let λ ∈ C and U = (V, I, W)⊤ ∈ D(A) be such that (λ − A)U = 0, or    (λ + g)V + aIx + kW = 0, (λ + r)I + bVx = 0, (λ + c)W − V = 0. (25) Assuming that λ = −c and λ = −r, we can eliminate W and I to get Vxx = λ + r ab

• λ + g +

k λ + c

• V.

Reminding that V(0) = V(1) = 0, a non-zero V exists ⇔ ∃ℓ ∈ N∗ s. t. λ + r ab

• λ + g +

k λ + c

• = −ℓ2π2,

(26) and V = cℓ sin(ℓπ·), with cℓ ∈ C. Multiplying (26) by λ + c, we see that λ (different from −c) is a root of (26) if and only if it is a root of pℓ.

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A spectral analysis

Spectrum of A ctd

Theorem (Thm 10) The complex number −c always belongs to σess(A), while −r is an eigenvalue of A, its associated eigenvector is (0, 1, 0)⊤. We can show that the range of c + A is not closed and hence the

• perator c + A cannot be a Fredholm operator.

Corollary (Coro 11) σ(A) = σd(A) ∪ σess(A), with σess(A) = {−c} and σd(A) = {−r} ∪

• ℓ∈N∗

{λℓ,i : i = 1, 2, 3}.

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A spectral analysis

Riesz basis for H

Denote by U∗

ℓ,i = cℓ,iUℓ,i, i ∈ I = {1, 2, 3}, where the normalization

factor cℓ,i ∈ (0, ∞) is chosen such that U∗

ℓ,i = 1.

Theorem (Thm 12) The set {U∗

ℓ,i}i∈I,ℓ∈N∗ {(0, 1, 0)⊤} is a Riesz basis of H.

The main steps are:

• 1. The set {U∗

ℓ,i}i∈I,ℓ∈N∗ is a Riesz basis of the closed subspace of H2

spanned by these eigenvectors,

• 2. H = Span {(0, 1, 0)⊤} ⊕ H2.

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Optimality

Optimal polynomial decay rate

Lemma In addition to (2), assume that the functions a, b, c, k are constant and that r = g = 0. Then there exists ℓ0 ∈ N large enough such that the sequence of eigenvalues λℓ,1 of A, for all ℓ ≥ ℓ0 satisfies λℓ,1 = iℓπ √ ab + ik 2ℓπ √ ab − c 2ℓ2π2ab + o( 1 ℓ2 ), ∀ℓ ≥ ℓ0. (27) Further Uℓ,1 belongs to D(A0). By Thm 9, λ ∈ C is an eigenvalue of A ⇔ λ is a root of λ ab

• λ +

k λ + c

• = −ℓ2π2.

A careful analysis of the behavior of the roots of this eq. leads to (27). As ℜλℓ,1 ∼ −

c 2ℓ2π2ab, the decay in t−1 is optimal ([Littman Markus 87]).

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Optimality

Optimal exponential decay rate

If r > 0, Theorem 4.6 garantees an exponential decay of the energy in

• H. Hence we can define the optimal decay rate as

ωopt = inf{ω : ∃C(ω) > 0 s. t. E(t) ≤ C(ω)e2ωtE(0), ∀t ≥ 0}. Our goal is here to prove that the decay rate is equal to the spectral abscissa of A defined by µ = sup{ℜλ : λ ∈ σ(A)}. It is clear that µ ≤ ωopt, therefore it remains to prove the converse inequality. Theorem If r > 0, we have ωopt ≤ µ, (28) and consequently ωopt = µ.

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Optimality

Proof

Take an initial datum U0 ∈ H, then by Thm 12, it can be written as U0 =

• ℓ∈N∗
• i∈I

cℓ,iU∗

ℓ,i + d(0, 1, 0)⊤,

with cℓ,i, d ∈ C such that U02

H ∼ ℓ∈N∗

• i∈I |cℓ,i|2 + |d|2.

Hence the solution U of (5) is given by U(t) = etAU0 =

• ℓ∈N∗
• i∈I

cℓ,ietλℓ,iU∗

ℓ,i + de−tr(0, 1, 0)⊤, ∀t > 0.

⇒ U(t)2

H

• ℓ∈N∗
• i∈I

(1 + t4)e2tℜλℓ,i|cℓ,i|2 + |d|2e−2tr, ∀t > 0, since we recall that some eigenvalues λℓ,i could be triple. Again by Thm 12, we deduce that U(t)2

H (1 + t4)e2tµU02 H Cεe2(µ+ε)t, ∀t > 0.

for all ε > 0. Hence ωopt ≤ µ + ε, ∀ε > 0.

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Optimality

Thank you for your attention.

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