PARTIALLY DELAYED STABILIZING FEEDBACKS ... - Project Euclid

Advances in Differential Equations

Volume 12, Number 1 (2007), 27–54

PARTIALLY DELAYED STABILIZING FEEDBACKS FOR MAXWELL’S SYSTEM Serge Nicaise Université de Valenciennes et du Hainaut Cambrésis MACS, Institut des Sciences et Techniques de Valenciennes 59313 Valenciennes Cedex 9 France Cristina Pignotti Dipartimento di Matematica Pura e Applicata, Universit` a di L’Aquila Via Vetoio, Loc. Coppito, 67010 L’Aquila, Italy (Submitted by: Viorel Barbu) Abstract. We consider, in a bounded and smooth domain, Maxwell’s equations with a delay term in the boundary or in the internal feedbacks. Under suitable assumptions we obtain exponential stability results. Some instability examples are also given.

1. Introduction In this paper we investigate the effect of time delay in boundary or internal stabilization of Maxwell’s equations in domains in R3 . Let Ω ⊂ R3 be an open and connected bounded set with a boundary Γ of class C ∞ consisting of a single connected component. In this domain Ω, we consider the initial-boundary-value problem Et (x, t) − curl H(x, t) = 0

in

Ω × (0, +∞)

(1.1)

Ht (x, t) + curl E(x, t) = 0

in Ω × (0, +∞)

(1.2)

div E(x, t) = div H(x, t) = 0

in

Ω × (0, +∞)

[(μ1 E(x, t) + μ2 E(x, t − τ )) × ν + H(x, t)] × ν = 0 E(x, 0) = E0 (x)

and H(x, 0) = H0 (x)

in

Ω

E(x, t − τ ) × ν(x) = F0 (x, t − τ ) in Γ × (0, τ ),

(1.3) on

Γ × (0, +∞) (1.4) (1.5) (1.6)

where E, H are the electric and magnetic vector fields and ν(x) denotes the outward unit normal vector to the point x ∈ Γ. Moreover, the constant τ > 0 Accepted for publication: August 2006. AMS Subject Classifications: 93D15, 93C20. 27

28

Serge Nicaise and Cristina Pignotti

is the time delay, μ1 and μ2 are positive real numbers and the initial datum (E0 , H0 , F0 ) belongs to a suitable space. We are interested in giving an exponential stability result for such a problem. Let us denote by v, w or, equivalently, by v · w the Euclidean inner product between two vectors v, w ∈ Rn . It is well known that if μ2 = 0, that is, in absence of delay, the energy of problem (1.1)–(1.6) is exponentially decaying to zero. See for instance [5, 4, 2, 9]. Assuming that μ2 < μ1 ,

(1.7)

we will obtain a stabilization result, by using a suitable observability estimate. Moreover, we show that if μ1 = μ2 then there exists a sequence of arbitrarily small (and large) delays such that instabilities occur. In the case μ2 > μ1 , we also obtain delays which destabilize the system. Furthermore, we study the problem for Maxwell’s equations with an internal feedback. In particular, we consider the system Et (x, t) − curl H(x, t) + σ(x)[μ1 E(x, t) + μ2 E(x, t − τ )] = 0 in Ω × (0, +∞) (1.8) Ht (x, t) + curl E(x, t) = 0

in

Ω × (0, +∞)

Ω × (0, +∞)

div H(x, t) = 0

in

E(x, t) × ν = 0

and H(x, t) · ν = 0

E(x, 0) = E0 (x) where σ ∈

(1.10) on

and H(x, 0) = H0 (x)

E(x, t − τ ) = G0 (x, t − τ ) L∞ (Ω)

(1.9)

Γ × (0, +∞) in

Ω

in Ω × (0, τ ),

(1.11) (1.12) (1.13)

is a function such that σ(x) ≥ 0

a. e. in

Ω,

(1.14)

and σ(x) > σ0 > 0, a. e. in ω, (1.15) where ω ⊂ Ω is an open neighborhood of Γ. Exponential stability results for the above problem in the case of μ2 = 0, that is, without delay, have been obtained in some papers. See for instance [9, 7]. In this paper, in the case μ2 < μ1 , we show that the energy is exponentially decaying to zero. This is done, as for the problem with boundary feedback, by using a suitable observability estimate. If μ2 ≥ μ1 , we obtain an explicit sequence of arbitrarily small delays that destabilize the system. Analogous analysis for scalar wave equations in both cases, boundary and internal feedbacks, has been recently carried out by the authors in [8].

Partially delayed stabilizing feedbacks

29

This problem has been inspired by [10] where similar results are proved, after a careful spectral analysis, for the scalar wave equation with boundary feedback in one space dimension. The paper is organized as follows. Well posedness of the problems is analyzed in section 2 using semigroup theory. In subsection 2.1 we study the well posedness of problem (1.1)–(1.6), while in subsection 2.2 we concentrate on problem (1.8)–(1.13). In section 3 and section 4 we prove the exponential stability of the problem with boundary and internal feedbacks respectively. Finally, in section 5 we give instability examples when μ2 ≥ μ1 in the case of boundary feedback (subsection 5.1) and in the case of internal feedback (subsection 5.2). 2. Well posedness of the problems In this section, assuming μ2 ≤ μ1 ,

(2.1)

we will give well–posedness results for problem (1.1)–(1.6) and for problem (1.8)–(1.13) using semigroup theory. 2.1. Boundary feedback. Let us set z(x, ρ, t) = E(x, t − τ ρ) × ν(x),

x ∈ Γ, ρ ∈ (0, 1), t > 0.

(2.2)

Then, problem (1.1)–(1.6) is equivalent to Et (x, t) − curl H(x, t) = 0

in

Ω × (0, +∞)

(2.3)

Ht (x, t) + curl E(x, t) = 0

in

Ω × (0, +∞)

(2.4)

div E(x, t) = div H(x, t) = 0 in Ω × (0, +∞)

(2.5)

τ zt (x, ρ, t) + zρ (x, ρ, t) = 0 in Γ × (0, 1) × (0, +∞)

(2.6)

[μ1 E(x, t) × ν + μ2 z(x, 1, t) + H(x, t)] × ν = 0

(2.7)

z(x, 0, t) = E(x, t) × ν(x) E(x, 0) = E0 (x)

on

Γ × (0, +∞)

Γ × (0, ∞)

and H(x, 0) = H0 (x)

z(x, ρ, 0) = F0 (x, −ρτ )

on

(2.8) in

Ω

in Γ × (0, 1).

(2.9) (2.10)

We need to introduce suitable space functions. As usual, we denote by H(div 0, Ω) the space H(div 0, Ω) := { Φ ∈ L2 (Ω)3 : div Φ(x) = 0 }

(2.11)

30


and by L2τ (Γ × (0, 1)) the space L2τ (Γ×(0, 1)) := { Ψ ∈ L2 (Γ×(0, 1))3 : ν(x)·Ψ(x, ρ) = 0, x ∈ Γ, ρ ∈ (0, 1) }. (2.12) Let us introduce the Hilbert space H := H(div 0, Ω) × H(div 0, Ω) × L2τ (Γ × (0, 1)),

(2.13)

equipped with the inner product ⎛ ⎞ ⎛ ⎞ ˜ 1 E E ˜ ˜ ⎝ H ⎠,⎝ H ˜ ⎠ = (E · E + H · H)dx + ξ z · z˜dρdΓ, (2.14) H Ω Γ 0 z z˜ where ξ is a positive real number such that τ μ2 ≤ ξ ≤ τ (2μ1 − μ2 ).

(2.15)

Note that, from (2.1), such a constant ξ exists. If we denote by U := (E, H, z)T , then

T U := (Et , Ht , zt )T = curl H, −curl E, −τ −1 zρ . Therefore, problem (2.3)–(2.10) can be rewritten as U = AU,

U (0) = (E0 , H0 , F0 (·, − · τ ))T ,

(2.16)

where the operator A is defined by ⎞ ⎛ ⎛ ⎞ curl H E A ⎝ H ⎠ := ⎝ −curl E ⎠ , z −τ −1 zρ with domain D(A) := (E, H, z)T ∈ V × V × L2 (Γ; H 1 (0, 1))3 : [μ1 E(x) × ν(x) + μ2 z(x, 1) + H(x)] × ν(x) = 0 on Γ;

(2.17)

E(x) × ν(x) = z(x, 0) on Γ ,

where, V := { Φ ∈ L2 (Ω)3 : curl Φ ∈ L2 (Ω)3 , div Φ = 0, Φ × ν ∈ L2τ (Γ) }, (2.18) Φ 2V =

(|Φ|2 + |curl Φ|2 )dx +

Ω

|ν × Φ|2 dΓ. Γ


31

Theorem 2.1. For any initial datum U0 ∈ H there exists a unique solution U ∈ C([0, +∞), H) of problem (2.16). Moreover, if U0 ∈ D(A), then U ∈ C([0, +∞), D(A)) ∩ C 1 ([0, +∞), H). Proof. Take U = (E, H, z)T ∈ D(A). Then, ⎞ ⎛ ⎞

⎛ curl H E (AU, U ) = ⎝ −curl E ⎠ , ⎝ H ⎠ z −τ −1 zρ H −1 = {Ecurl H − Hcurl E}dx − ξτ Ω

Γ

So, by Green’s formula (cfr. [2], page 150), −1 (AU, U ) = (E × ν) · HdΓ − ξτ Γ

Γ

1

1

zρ (x, ρ) · z(x, ρ)dρdΓ.

0

zρ (x, ρ) · z(x, ρ)dρdΓ.

(2.19)

0

Integrating by parts in ρ, we get 1 zρ (x, ρ) · z(x, ρ)dρdΓ Γ 0 1 =− zρ (x, ρ) · z(x, ρ)dρdΓ + {|z(x, 1)|2 − |z(x, 0)|2 }dΓ; Γ

that is,

0

Γ

1 zρ (x, ρ) · z(x, ρ)dρdΓ = {|z(x, 1)|2 − |z(x, 0)|2 }dΓ. (2.20) 2 Γ Γ 0 Therefore, from (2.19) and (2.20), ξτ −1 {|z(x, 1)|2 − |z(x, 0)|2 }dΓ (AU, U ) = − E · (H × ν)dΓ − 2 Γ Γ ξτ −1 {|z(x, 1)|2 − |z(x, 0)|2 }dΓ = E · [(μ1 (E × ν) + μ2 z(x, 1)) × ν]dΓ− 2 Γ Γ 2 = −μ1 |E × ν| dΓ − μ2 (z(x, 1) × ν) · (E × ν)dΓ Γ Γ −1 ξτ ξτ −1 |z(x, 1)|2 dΓ + |E × ν|2 dΓ, − 2 2 Γ Γ from which follows, using Cauchy-Schwarz’s inequality, μ μ2 ξτ −1 ξτ −1 2 2 (AU, U ) ≤ − μ1 + + |E × ν| dΓ + |z(x, 1)|2 dΓ. − 2 2 2 2 Γ Γ (2.21) 1

32


Then, from (2.15) we have that (AU, U ) ≤ 0, which means that the operator A is dissipative. Now, we want to prove the maximality of A showing that the operator I − A is surjective. For any given (F, G, w)T ∈ H we seek U = (E, H, z)T ∈ D(A), a solution of ⎞ ⎞ ⎛ ⎛ F E (I − A) ⎝ H ⎠ = ⎝ G ⎠ ; w z that is,

⎧ ⎨ E − curl H = F H + curl E = G ⎩ z + τ −1 zρ = w.

(2.22)

Suppose that we have found E and H with the appropriate regularity. Then we can determine z. Indeed, from (2.17), z(x, 0) = E(x) × ν(x),

for x ∈ Γ,

(2.23)

and, from (2.22), z(x, ρ) + τ −1 zρ (x, ρ) = w(x, ρ),

for x ∈ Γ, ρ ∈ (0, 1).

Then, by (2.23) and (2.24), we deduce −ρτ −ρτ z(x, ρ) = e E(x) × ν(x) + τ e

ρ

w(x, σ)eστ dσ,

x ∈ Γ, ρ ∈ (0, 1),

0

and, in particular, z(x, 1) = e−τ E(x) × ν(x) + τ e−τ

(2.24)

(2.25)

1

w(x, σ)eστ dσ

on Γ,

0

that we rewrite as z(x, 1) = e−τ E(x) × ν(x) + z0 (x) with z0 ∈ L2τ (Γ) defined by −τ

z0 (x) = τ e

1

on Γ,

x ∈ Γ.

w(x, σ)eστ dσ,

(2.26)

(2.27)

0

From (2.22), we get {(E − curl H)Φ + (H + curl E)Ψ}dx = (F Φ + GΨ)dx, Ω

Ω

(2.28)


33

for all (Φ, Ψ) ∈ V × L2 (Ω)3 . Integrating by parts, we obtain {EΦ − Hcurl Φ + (H + curl E)Ψ}dx + (H × ν) · ΦdΓ = (F Φ + GΨ)dx, Ω

Γ

Ω

and then, {EΦ − Hcurl Φ + (H + curl E)Ψ}dx − [(μ1 E × ν + μ2 z(x, 1)) × ν] · ΦdΓ = (F Φ + GΨ)dx. Ω

Γ

Ω

From this, putting Ψ = curl Φ, we obtain (EΦ + curl Ecurl Φ)dx − [(μ1 E × ν + μ2 z(x, 1)) × ν] · ΦdΓ Ω Γ (2.29) = (F Φ + Gcurl Φ)dx, ∀Φ ∈ V. Ω

Using (2.26) in (2.29) we have (EΦ + curl Ecurl Φ)dx + (μ1 + μ2 e−τ )(E × ν) · (Φ × ν)dΓ Ω Γ = (F Φ + Gcurl Φ)dx − μ2 z0 (Φ × ν)dΓ, ∀Φ ∈ V. Ω

(2.30)

Γ

Since the left-hand side of (2.30) is coercive on V, the Lax-Milgram lemma guarantees the existence and uniqueness of a solution E ∈ V of (2.30). Now, we set H = G − curl E ∈ L2 (Ω)3 . Then, (2.30) becomes (EΦ − Hcurl Φ)dx + (μ1 + μ2 e−τ )(E × ν)(Φ × ν)dΓ Ω Γ F Φdx − μ2 z0 (Φ × ν)dΓ, ∀Φ ∈ V, = Ω

Γ

and, integrating by parts, (E − curl H)Φdx − (H × ν) · ΦdΓ + (μ1 + μ2 e−τ )(E × ν)(Φ × ν)dΓ Ω Γ Γ F Φdx − μ2 z0 (Φ × ν)dΓ, ∀Φ ∈ V. (2.31) = Ω

Γ

From (2.31) we deduce curl H ∈ L2 (Ω)3 and E − curl H = F. Moreover, (H × ν) · ΦdΓ − (μ1 + μ2 e−τ )(E × ν)(Φ × ν)dΓ = μ2 z0 (Φ × ν)dΓ; Γ

Γ

Γ

34


that is, (H × ν) · ΦdΓ + [(μ1 + μ2 e−τ )(E × ν) × ν]ΦdΓ + μ2 (z0 × ν)ΦdΓ = 0. Γ

Γ

Γ

This implies H(x) × ν(x) = −(μ1 E(x) × ν(x) + μ2 z(x, 1)) × ν(x),

x ∈ Γ.

This proves the existence of a solution (E, H, z)T in D(A) of (2.22) and consequently the maximality of A. The well-posedness result follows from the Hille-Yosida theorem. 2.2. Internal feedback. Let us set z(x, ρ, t) = E(x, t − τ ρ),

x ∈ Ω, ρ ∈ (0, 1), t > 0.

(2.32)

Then, problem (1.8)–(1.13) is equivalent to Et (x, t) − curl H(x, t) + σ[μ1 E(x, t) + μ2 z(x, 1, t)] = 0 in Ω × (0, +∞) (2.33) Ht (x, t) + curl E(x, t) = 0 div H(x, t) = 0

in

(2.34)

in

(2.35)

in Ω × (0, 1) × (0, +∞)

and H(x, t) · ν = 0

z(x, 0, t) = E(x, t) E(x, 0) = E0 (x)

Ω × (0, +∞)

Ω × (0, +∞)

τ zt (x, ρ, t) + zρ (x, ρ, t) = 0 E(x, t) × ν = 0

in

on

Ω × (0, ∞)

and H(x, 0) = H0 (x)

z(x, ρ, 0) = G0 (x, −ρτ )

Γ × (0, +∞)

(2.36) (2.37) (2.38)

in

Ω

in Ω × (0, 1).

(2.39) (2.40)

We need to introduce suitable space functions. As usual, we denote H0 (curl , Ω) := { Ψ ∈ L2 (Ω)3 : curl Ψ ∈ L2 (Ω)3 ; Ψ × ν = 0 on Γ}, (2.41) H0 (div 0, Ω) := { Φ ∈ H(div 0, Ω) : Φ · ν = 0 on Γ }.

(2.42)

Let us introduce the Hilbert space H0 := L2 (Ω)3 × H0 (div 0, Ω) × L2 (Ω × (0, 1))3 ,

(2.43)

equipped with the inner product ⎞ ⎞ ⎛ ⎛ ˜ 1 E E ˜ ˜ ⎝ H ⎠,⎝ H ˜ ⎠ = (E · E + H · H)dx + ξ σz · z˜dρdΓ, (2.44) H0 Ω Ω 0 z z˜ where ξ is a positive real number satisfying (2.15).


35

If we denote U := (E, H, z)T , then

T U := (Et , Ht , zt )T = curl H − σμ1 E − σμ2 z(·, 1), −curl E, −τ −1 zρ . Therefore, problem (2.33)–(2.40) can be rewritten as U = A0 U, where the operator ⎛

U (0) = (E0 , H0 , G0 (·, − · τ ))T ,

(2.45)

A0

is defined by ⎞ ⎛ ⎞ curl H − σμ1 E − σμ2 z(·, 1) E ⎠, A0 ⎝ H ⎠ := ⎝ −curl E −1 z −τ zρ

with domain D(A0 ) := (E, H, z)T ∈ H0 (curl , Ω) × (H0 (div 0, Ω) ∩ H 1 (Ω)3 )

(2.46) × L2 (Ω; H 1 (0, 1))3 : E = z(·, 0) in Ω .

Theorem 2.2. For any initial datum U0 ∈ H0 there exists a unique solution U ∈ C([0, +∞), H0 ) of problem (2.45). Moreover, if U0 ∈ D(A0 ), then U ∈ C([0, +∞), D(A0 )) ∩ C 1 ([0, +∞), H0 ). Proof. Take U = (E, H, z)T ∈ D(A0 ). Then ⎞ ⎛ ⎞

⎛ curl H − σμ1 E − σμ2 z(·, 1) E 0

⎠,⎝ H ⎠ A U, U = ⎝ −curl E −1 z −τ zρ H0 =

{Ecurl H − Hcurl E − σμ1 |E|2 − σμ2 z(x, 1)E}dx Ω 1 −1 σzρ (x, ρ) · z(x, ρ)dρdΓ. −ξτ Ω

0

So, by Green’s formula,

0 σ(x)[μ1 |E|2 + μ2 z(x, 1)E]dx A U, U = − Ω 1 − ξτ −1 σ(x)zρ (x, ρ) · z(x, ρ)dρdx. Ω

(2.47)

0

Integrating by parts in ρ, we have 1 1 σ(x)zρ (x, ρ) · z(x, ρ)dρdx = − σ(x)zρ (x, ρ) · z(x, ρ)dρdx Ω 0 Ω 0 + σ(x)(|z(x, 1)|2 − |z(x, 0)|2 )dx; Ω

36


that is, 1 Ω

0

1 σ(x)zρ (x, ρ)·z(x, ρ)dρdx = 2

σ(x)(|z(x, 1)|2 −|z(x, 0)|2 )dx. (2.48) Ω

Therefore, from (2.47) and (2.48), 0

A U, U = − σ(x)[μ1 |E|2 + μ2 z(x, 1)E]dx Ω ξτ −1 − σ(x)(|z(x, 1)|2 − |z(x, 0)|2 )dx 2 Ω σ(x)μ2 2 (|z(x, 1)|2 + |E|2 )dx = − σ(x)μ1 |E| dx + 2 Ω Ω ξτ −1 ξτ −1 − σ(x)|z(x, 1)|2 dx + σ(x)|E|2 dx. 2 2 Ω Ω Therefore, 0

μ2 ξτ −1 A U, U ≤ −μ1 + σ(x)|E|2 dx (2.49) + 2 2 Ω μ2 ξτ −1 + σ(x)|z(x, 1)|2 dx. − 2 2 Ω Then, from (2.15) we deduce that the operator A0 is dissipative. Now, we want to prove the maximality of A0 showing that the operator I − A0 is surjective. For any given (F, G, w)T ∈ H0 we seek a solution U = (E, H, z)T ∈ D(A0 ) of ⎛ ⎞ ⎛ ⎞ E F (I − A0 ) ⎝ H ⎠ = ⎝ G ⎠ ; z w that is, ⎧ ⎨ E − curl H + σμ1 E + σμ2 z(·, 1) = F H + curl E = G (2.50) ⎩ z + τ −1 zρ = w. Suppose that we have found E and H with the appropriate regularity. Then we can determine z. Indeed, from (2.46), z(x, 0) = E(x),

for x ∈ Ω,

(2.51)

for x ∈ Ω, ρ ∈ (0, 1).

(2.52)

and, from (2.50), z(x, ρ) + τ −1 zρ (x, ρ) = w(x, ρ),


Then, by (2.51) and (2.52), we deduce ρ w(x, σ)eστ dσ, z(x, ρ) = e−ρτ E(x) + τ e−ρτ

37

x ∈ Ω, ρ ∈ (0, 1),

(2.53)

0

and, in particular, z(x, 1) = e−τ E(x) + τ e−τ

1

w(x, σ)eστ dσ

in Ω,

0

that we rewrite as z(x, 1) = e−τ E(x) + z0 (x) with z0 ∈

L2 (Ω)3

defined by −τ

z0 (x) = τ e

1

w(x, σ)eστ dσ,

in Ω,

x ∈ Ω.

(2.54)

(2.55)

0

From (2.50), we see that {(E − curl H + σμ1 E + σμ2 z(·, 1))Φ + (H + curl E)Ψ}dx Ω = (F Φ + GΨ)dx, ∀ (Φ, Ψ) ∈ H0 (curl , Ω) × L2 (Ω)3 .

(2.56)

Ω

Integrating by parts and using the boundary condition satisfied by Φ, we obtain {EΦ − Hcurl Φ + σμ1 EΦ + σμ2 z(·, 1)Φ + (H + curl E)Ψ}dx Ω = (F Φ + GΨ)dx. Ω

From this, putting Ψ = curl Φ, we obtain {EΦ + curl Ecurl Φ + σμ1 EΦ + σμ2 z(·, 1)Φ}dx Ω = (F Φ + Gcurl Φ)dx, ∀ Φ ∈ H0 (curl , Ω).

(2.57)

Ω

Using (2.54) in (2.57) we have (EΦ + curl Ecurl Φ + σμ1 EΦ + σμ2 e−τ EΦ)dx Ω = (F Φ + Gcurl Φ)dx − σμ2 z0 Φdx, ∀Φ ∈ H0 (curl , Ω). Ω

(2.58)

Ω

Since the left-hand side of (2.58) is coercive on H0 (curl , Ω), the Lax-Milgram lemma guarantees the existence and uniqueness of a solution E ∈ H0 (curl , Ω).

38


Now, put H = G − curl E ∈ L2 (Ω)3 .

(2.59)

Then, from (2.58) we obtain curl H ∈ L2 (Ω)3

and

curl H = E + σμ1 E + σμ2 z(·, 1) − F.

Moreover, since G ∈ H0 (div 0, Ω), from (2.59) we have div H = div G = 0, and H · ν = −curl E · ν = 0 on Γ. Thus, H belongs to H0 (div 0, Ω). Moreover, H belongs to the space HT (curl , div , Ω) := {Φ ∈ L2 (Ω)3 : div Φ ∈ L2 (Ω), curl Φ ∈ L2 (Ω)3 , Φ · ν = 0 on Γ}, which is injected in H 1 (Ω)3 since the boundary of Ω is smooth. Therefore, (E, H, z)T belongs to D(A0 ). This concludes the proof of the maximality of A0 . 3. Boundary stability result In this section, we will prove the exponential stability of problem (1.1)– (1.6) assuming (1.7). Let us define the energy as 1 1 ξ E(t) := {|E(x, t)|2 + |H(x, t)|2 }dx + |E(x, t − τ ρ) × ν|2 dρdΓ, 2 Ω 2 Γ 0 (3.1) where ξ is a positive constant which satisfies τ μ2 < ξ < τ (2μ1 − μ2 ).

(3.2)

We have the following result. Proposition 3.1. For any strong solution of problem (1.1)−(1.6) the energy is decreasing and there exist two positive constants C and C such that (3.3) E (t) ≤ −C {|E(x, t) × ν|2 + |E(x, t − τ ) × ν|2 }dΓ Γ

and E (t) ≥ −C

{|E(x, t) × ν|2 + |E(x, t − τ ) × ν|2 }dΓ. Γ

(3.4)


39

Proof. Differentiating (3.1) we obtain 1 (E(x, t−τ ρ)×ν)·(Et (x, t−τ ρ)×ν)dρdΓ, E (t) = {E·Et +H·Ht }dx+ξ Ω

Γ

0

and then, using equations (1.1), (1.2) and integrating by parts, E (t) = − (H(x, t) × ν) · E(x, t)dΓ Γ 1 +ξ (E(x, t − τ ρ) × ν) · (Et (x, t − τ ρ) × ν)dρdΓ. Γ

(3.5)

0

Now, observe that Et (x, t − τ ρ) = −τ −1 Eρ (x, t − τ ρ). Then, we can rewrite 1 (E(x, t − τ ρ) × ν) · (Et (x, t − τ ρ) × ν)dρdΓ Γ 0 1 = −τ −1 (E(x, t − τ ρ) × ν) · (Eρ (x, t − τ ρ) × ν)dρdΓ Γ 0 τ −1 =− {|E(x, t − τ ) × ν|2 − |E(x, t) × ν|2 }dΓ. 2 Γ Thus, from (3.6), we obtain E (t) = − (H(x, t) × ν) · E(x, t)dΓ Γ ξτ −1 {|E(x, t) × ν|2 − |E(x, t − τ ) × ν|2 }dΓ. + 2 Γ

(3.6)

(3.7)

From the boundary condition (1.4), we have (H(x, t) × ν) · E(x, t)dΓ Γ (3.8) = (μ1 E(x, t) × ν + μ2 E(x, t − τ ) × ν) · (E(x, t) × ν)dΓ Γ 2 = μ1 |E(x, t) × ν| dΓ + μ2 (E(x, t − τ ) × ν) · (E(x, t) × ν)dΓ. Γ

Γ

So, by (3.7) and (3.8), we can estimate ξτ −1 |E(x, t) × ν|2 dΓ E (t) = − μ1 − 2 Γ

(3.9)

40


ξτ −1 − μ2 (E(x, t − τ ) × ν) · (E(x, t) × ν)dΓ − |E(x, t − τ ) × ν|2 dΓ. 2 Γ Γ Therefore, by Schwarz’s inequality, μ2 ξτ −1 |E(x, t) × ν|2 dΓ E (t) ≤ − μ1 − − 2 2 Γ ξτ −1 μ 2 − |E(x, t − τ ) × ν|2 dΓ. − 2 2 Γ Since the constant ξ satisfies assumption (3.2), the upper bound (3.3) follows. The lower bound (3.4) directly follows from (3.7) and (3.8), and the wellknown estimate 2ab ≥ −a2 − b2 , valid for any real numbers a, b. Now we give a boundary observability inequality which will be useful to deduce the exponential decay of the energy. Proposition 3.2. There exists a time T > 0 such that for all times T > T there exists a positive constant C0 (depending on T) for which T E(T ) ≤ C0 {|E(x, t) × ν|2 + |E(x, t − τ ) × ν|2 }dΓ dt, (3.10) Γ

0

for any regular solution of problem (1.1) − (1.6). Proof. From [3] (Corollary 3.4), there exists a positive time T0 such that ˆ δ, ) for any δ, > 0 and any time T > T0 , there exists a constant Cˆ = C(T, for which T −δ T0 ES (t)dt − (ES (δ) + ES (T − δ)) δ (3.11) T 2 2 2 |E(t) × ν| + |H(t) × ν| dΓdt + l.o.t.(E, H) , ≤ Cˆ 0

Γ

where ES denotes the standard energy for Maxwell’s equations 1 (|E(t)|2 + |H(t)|2 )dx ES (t) := 2 Ω and l.o.t.(E, H) denotes lower order terms, namely l.o.t.(E, H) = (E, H) 2(H 1/2− (Ω×(0,T ))3 ) + (E, H) 2H −1+ (Γ×(0,T ))3 . The proof of this estimate is based on the multiplier method and requires some arguments from microlocal analysis.


41

Now, observe that, for δ > τ and T > 2δ, T −δ 1 |E(x, t − τ ρ) × ν|2 dρdΓdt Γ

δ

≤τ

−1

0 T −δ

Γ

δ

≤τ

−1

T −δ

δ

≤τ

−1

Γ T

t

|E(x, s) × ν|2 dsdΓdt

t−τ t+τ

|E(x, σ − τ ) × ν|2 dσdΓdt

t

|E(x, t − τ ) × ν|2 dΓdt.

T Γ

0

So, from (3.11), adding to both sides 1 ξ T −δ |E(x, t − τ ρ) × ν|2 dρdΓdt, 2 δ Γ 0 and using the boundary condition (1.4), we obtain T −δ T0 E(t)dt − (E(δ) + E(T − δ)) 2 δ T ≤ C 0 Γ (|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt + l.o.t.(E, H) ,

(3.12)

for a suitable positive constant C > 0 depending on T. Note that, since the energy E is decreasing, for T > 2δ, we have T −δ E(t)dt ≥ (T − 2δ)E(T ). E(T − δ) ≤ E(δ) and δ

Then, from (3.12), we obtain (T − 2δ)E(T ) (3.13) T (|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt + l.o.t.(E, H) + T0 E(δ). ≤C 0

Γ

Adding −T0 E(T ) to both sides of (3.13) and recalling (3.3), we obtain for T > T0 + 2δ, T C (|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt (3.14) E(T ) ≤ T − 2δ − T0 Γ 0 T C T0 + l.o.t.(E, H) + (|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt. T − 2δ − T0 0 Γ

42


Now, applying a compactness-uniqueness result as in [2] (page 156) we obtain, from (3.14), the observability inequality (3.10). We are ready to give the exponential stability result. Theorem 3.3. Let the assumption (1.7) be satisfied. Then, there exist positive constants C1 and C2 such that, for any regular solution of problem (1.1) − (1.6), E(t) ≤ C1 E(0)e−C2 t , ∀t ≥ 0. (3.15) Proof. From (3.3), E(T ) − E(0) ≤ −C

T

(|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt.

(3.16)

Γ

0

From (3.16) and the observability inequality (3.10), we obtain T (|E(t) × ν|2 + |E(t − τ ) × ν|2 )dΓdt ≤ C0 C −1 (E(0) − E(T )). E(T ) ≤ C0 0

Γ

Then ˜ E(T ) ≤ CE(0), with C˜ < 1. This easily implies the exponential stability estimate (3.15) since the system (1.1) – (1.6) is invariant under translation and the energy E is decreasing. 4. Internal stability result Here, under the assumption (1.7), we will prove that problem (1.8)–(1.13) is exponentially stable. Let us define the energy as 1 1 ξ F(t) := {|E(x, t)|2 + |H(x, t)|2 }dx + σ(x) |E(x, t − τ ρ)|2 dρdx, 2 Ω 2 Ω 0 (4.1) where ξ is a positive real number satisfying (3.2). We can give a first estimate on the derivative of the energy F. Proposition 4.1. For any strong solution of problem (1.8) − (1.13) the energy is decreasing and there exists a positive constant C such that F (t) ≤ −C σ(x){|E(x, t)|2 + |E(x, t − τ )|2 }dx. (4.2) Ω


Proof. Differentiating (4.1) we obtain σ(x) F (t) = {E · Et + H · Ht }dx + ξ Ω

Ω

1

43

E(x, t − τ ρ) · Et (x, t − τ ρ)dρdx,

0

and then, using (1.8), (1.9), the boundary conditions (1.11), and integrating by parts, 2 μ2 σ(x)E(x, t) · E(x, t − τ )dx F (t) = − μ1 σ(x)|E(x, t)| dx − Ω Ω 1 (4.3) σ(x) E(x, t − τ ρ) · Et (x, t − τ ρ)dρdx. +ξ Ω

0

Now, as in the proof of Proposition 3.1, we can compute 1 σ(x) E(x, t − τ ρ) · Et (x, t − τ ρ)dρdx Ω 0 1 −1 σ(x) E(x, t − τ ρ) · Eρ (x, t − τ ρ)dρdx = −τ 0 Ω −1 τ σ(x){|E(x, t − ρ)|2 − |E(x, t)|2 }dx. =− 2 Ω

(4.4)

Using (4.4) in identity (4.3), we have F (t) = −μ1 σ(x)|E(x, t)|2 dx − μ2 σ(x)E(x, t) · E(x, t − τ )dx Ω Ω ξτ −1 ξτ −1 2 + σ(x)|E(x, t)| dx − σ(x)|E(x, t − τ )|2 dx. (4.5) 2 2 Ω Ω From (4.5), applying Cauchy-Schwarz’s inequality, we obtain μ2 ξτ −1 F (t) = − μ1 − σ(x)|E(x, t)|2 dx − 2 2 Ω ξτ −1 μ 2 σ(x)|E(x, t − τ )|2 dx. − − 2 2 Ω

(4.6)

Therefore, recalling the assumption (3.2) on the constant ξ, estimate (4.2) immediately follows. Consider the homogeneous problem Eht (x, t) − curl Hh (x, t) = 0 in Ω × (0, +∞)

(4.7)

Hht (x, t) + curl Eh (x, t) = 0 in Ω × (0, +∞)

(4.8)

div Eh (x, t) = div Hh (x, t) = 0 Eh (x, t) × ν = 0

in Ω × (0, +∞)

and Hh (x, t) · ν = 0

on

Γ × (0, +∞)

(4.9) (4.10)

44


Eh (x, 0) = E0 (x)

and Hh (x, 0) = H0 (x)

in

Ω.

(4.11)

We recall the following result from [7]. Proposition 4.2. Let (Eh , Hh ) be the solution of the homogeneous problem (4.7) − (4.11) and let ω ⊂ Ω be an open neighborhood of Γ. Then, there exists a time T > 0 such that for every time T > T , T (|Eh (x, 0)|2 + |Hh (x, 0)|2 )dx ≤ C0 |Eh (x, t)|2 dxdt, (4.12) Ω

0

ω

for a suitable positive constant C0 depending on T and independent of the initial datum (E0 , H0 ). Using Proposition 4.2, we can prove an internal observability estimate for problem (1.8)–(1.13). Proposition 4.3. There exists a time T0 > 0 such that for all times T > T 0 there exists a positive constant Cˆ (depending on T ) for which T F(0) ≤ Cˆ σ(x){|E(x, t)|2 + |E(x, t − τ )|2 }dxdt, (4.13) 0

Ω

for any regular solution (E, H) of problem (1.8) − (1.13). Proof. As in [7] (cfr. Zuazua [11]), we can decompose the solution (E, H) of problem (1.8)–(1.13) as ˜ H), ˜ (E, H) = (Eh , Hh ) + (E, where (Eh , Hh ) solves the homogeneous problem (4.7)–(4.10) with initial condition Eh (x, 0) = E0 (x), Hh (x, 0) = H0 (x), in Ω, ˜ H) ˜ satisfies and (E, ˜ ˜t (x, t) − curl H(x, t) = −σ(x)[μ1 E(x, t) + μ2 E(x, t − τ )] in Ω × (0, +∞) E (4.14) ˜ t (x, t) + curl E(x, ˜ t) = 0 in Ω × (0, +∞) H (4.15) ˜ div H(x, t) = 0 in Ω × (0, +∞) ˜ t) × ν = 0 and H(x, ˜ E(x, t) · ν = 0 on ˜ 0) = 0 and H(x, ˜ E(x, 0) = 0 in Ω.

(4.16) Γ × (0, +∞)

(4.17) (4.18)


45

By definition (4.1), 1 ξ 1 2 2 {|E(x, 0)| + |H(x, 0)| }dx + σ(x) |E(x, −τ ρ)|2 dρdx F(0) = 2 Ω 2 Ω 0 1 1 ξ 2 2 = {|Eh (x, 0)| + |Hh (x, 0)| }dx + σ(x) |E(x, −τ ρ)|2 dρdx. 2 Ω 2 Ω 0 Then, for T > τ, using a change of variable in the last integral of the above identity, we obtain T 1 {|Eh (x, 0)|2 + |Hh (x, 0)|2 }dx + c σ(x) |E(x, t − τ )|2 dtdx, F(0) ≤ 2 Ω Ω 0 for some positive constant c. Now, let T be as in Proposition 4.2. Therefore, using the observability estimate (4.12) for the homogeneous problem, we have for T > T0 = max {τ, T }, T T 2 |Eh (x, t)| dxdt + c σ(x) |E(x, t − τ )|2 dtdx F(0) ≤ C0 0

≤ C

Ω

ω

T

0

+

0

˜ t)|2 )dxdt σ(x)(|E(x, t)|2 + |E(x, Ω

σ(x) Ω

T

(4.19)

|E(x, t − τ )|2 dtdx ,

0

for a suitable positive constant C . From (4.19) and standard energy estimates for problem (4.14)–(4.18), we easily obtain T σ(x){|E(x, t)|2 + |E(x, t − τ )|2 }dxdt. F(0) ≤ Cˆ 0

Ω

Now, using estimate (4.13), as in the case of boundary feedback we can deduce the exponential stability result. Theorem 4.4. Let the assumption (1.7) be satisfied. Then, there exist positive constants C1 and C2 such that , for any regular solution of problem (1.8) − (1.13), (4.20) F(t) ≤ C1 F(0)e−C2 t , ∀t ≥ 0. 5. Some instability examples In this section we will give some instability examples in the case μ2 ≥ μ1 .

46


5.1. Boundary feedback. In this subsection we consider the problem with boundary feedback (1.1)–(1.6), and prove the following result. Theorem 5.1. If (1.7) does not hold, then there exist a sequence of delays and solutions of problem (1.1) − (1.6), corresponding to these delays, such that their standard energy is larger than a positive constant. Proof. Let us consider the spectral problem for the system (1.1)–(1.4) by seeking a solution in the form E(x, t) = eλt e(x),

H(x, t) = eλt h(x),

λ ∈ C.

(5.1)

Then, (e, h) has to be a solution of the eigenvalue problem ⎧ λe(x) − curl h(x) = 0 in Ω ⎪ ⎪ ⎨ λh(x) + curl e(x) = 0 in Ω (5.2) div e(x) = div h(x) = 0 in Ω ⎪ ⎪ ⎩ ((μ1 + μ2 e−λτ )e × ν + h) × ν = 0 on Γ. Assuming that λ is different from 0, we can eliminate h, by the second equation, namely h = −λ−1 curl e and consequently e is a solution of ⎧ 2 ⎨ λ e(x) + curl curl e(x) = 0 in Ω div e(x) = 0 in Ω (5.3) ⎩ (λ(μ1 + μ2 e−λτ )e × ν − curl e) × ν = 0 on Γ. This problem can be reformulated, in a variational form, as 2 −λτ curl e·curl vdx+λ e·vdx+(μ1 +μ2 e )λ e×ν ·v ×νdΓ = 0, (5.4) Ω

Ω

Γ

for all v ∈ V, where V is defined in (2.18). As in [8] we want to find a solution for λ := ib, with b ∈ R. For this choice of λ the problem (5.4) can be rewritten as 2 −ibτ curl e·curl vdx−b e·vdx+(μ1 +μ2 e )ib e×ν ·v×νdΓ = 0, (5.5) Ω

Ω

Γ

for all v ∈ V. Assume that cos(bτ ) = −

μ1 . μ2

(5.6)

Note that, since we are considering the case μ2 ≥ μ1 , there exist b, τ such that (5.6) holds. Then, we choose μ2 sin(bτ ) = μ22 − μ21 . (5.7)


47

Under these assumptions, (5.5) becomes curl e · curl vdx − b2 e · vdx + b μ22 − μ21 e × ν · v × νdΓ = 0, (5.8) Ω

Ω

Γ

for all v ∈ V. In particular, for v = e, (5.8) gives 2 2 2 2 2 |curl e| dx − b |e| dx + b μ2 − μ1 |e × ν|2 dΓ = 0. Ω

Ω

(5.9)

Γ

Without loss of generality we can assume 2 e 2 := |e|2 dx = 1

(5.10)

and then, the identity (5.9) can be rewritten as 2 b − b μ22 − μ21 q0 (e) − q1 (e) = 0,

(5.11)

Ω

where

|ϕ × ν| dΓ, 2

q0 (ϕ) := Γ

|curl ϕ|2 dx.

q1 (ϕ) :=

(5.12)

Ω

Now we distinguish two cases. Case (a) μ2 > μ1 . In this case, from (5.11) we have 1 2 2 2 2 2 b= μ2 − μ1 q0 (e) ± (μ2 − μ1 )q0 (e) + 4q1 (e) . 2 Write for brevity l(w) =

μ22 − μ21 q0 (w) + (μ22 − μ21 )q02 (w) + 4q1 (w).

Define b :=

1 l(w). min 2 w∈V w 2 = 1

(5.13)

Let us first show that this minimum is positive. Indeed if this is not the case, there exists a sequence of wn ∈ V, n ∈ N, such that 1 (5.14) l(wn ) ≤ , n (5.15) wn 2 = 1, ∀n ∈ N. As

μ22 − μ21 q0 (w) ≤ l(w)

and 4q1 (w) ≤ l(w)2 ,

(5.16)

48


the above properties imply that there exists C > 0 such that wn V ≤ C, ∀n ∈ N. Consequently the sequence (wn )n is bounded in V . Since V is embedded into H 1/2 (Ω)3 (see Theorem 2 of [1]), by the Rellich-Kondracov theorem V is compactly embedded into L2 (Ω)3 . Therefore there exists a subsequence, still denoted by (wn )n for the sake of simplicity, and w ∈ V such that wn → w weakly in V and wn → w strongly in L2 (Ω)3 .

(5.17)

Coming back to (5.14)–(5.15) and using again (5.16), we deduce that wn → w strongly in V, with q0 (w) = q1 (w) = 0, w 2 = 1, ∀n ∈ N.

(5.18) (5.19)

This implies that w ∈ V satisfies curl w = 0, div w = 0 in Ω, w × ν = 0 on Γ, and due to our assumptions on Ω and its boundary we deduce that w = 0, which contradicts (5.19). This proves that b > 0. Let us now show that the minimum in (5.13) is hit at an element e ∈ V . Let us consider a minimizing sequence (wn )n , namely a sequence of wn ∈ V satisfying (5.15) and such that l(wn ) → 2b as n → ∞.

(5.20)

As before, using (5.16), the sequence (wn )n will be bounded in V. Therefore, there exist a subsequence, still denoted by (wn )n for the sake of simplicity, and e ∈ V such that wn → e weakly in V and wn → e strongly in L2 (Ω)3 . Now we remark that from the definition of b and (5.20), we have 2 2 2 b = lim b μ2 − μ1 q0 (wn ) + q1 (wn ) . n→∞

Indeed, if we denote bn :=

l(wn ) , 2

(5.21)

(5.22)


then bn satisfies b2n − bn

49

that is, b2n = bn

μ22 − μ21 q0 (wn ) − q1 (wn ) = 0;

μ22 − μ21 q0 (wn ) + q1 (wn ).

So, since bn → b as n → ∞, we have that b2 = lim [bn μ22 − μ21 q0 (wn ) + q1 (wn )]. n→∞

(5.23)

From (5.16) and (5.20) the sequence (q0 (wn ))n is bounded and therefore, since bn → b, (bn − b) μ22 − μ21 q0 (wn ) → 0. (5.24) By (5.23) and (5.24) we obtain (5.22). Moreover, the quantity 2 2 w ·w dx+b μ2 − μ1 w ×ν ·w ×νdΓ+ curl w ·curl w dx, ∀w, w ∈ V Ω

Γ

Ω

defines an inner product in V , whose associated norm is equivalent to the natural one. Consequently, from (5.21), we have e 22 + b μ22 − μ21 q0 (e) + q1 (e) ≤ lim wn 22 + b μ22 − μ21 q0 (wn ) + q1 (wn ) . n→∞

The strong convergence of wn to e in L2 (Ω)3 and (5.15) further imply that e 2 = 1, and the above property becomes 2 2 2 2 b μ2 − μ1 q0 (e) + q1 (e) ≤ lim b μ2 − μ1 q0 (wn ) + q1 (wn ) . n→∞

2 2 2 Due to (5.22), we conclude that b μ2 − μ1 q0 (e)+q 1 (e) ≤ b . This prove that 2 2 2 2 2 l(e) ≤ 2b, because b > 0 and μ2 − μ1 q0 (e)− (μ2 − μ1 )q0 (e) + 4q1 (e) ≤ 0. Since 2b ≤ l(e), we finally obtain l(e) = 2b. As in [8] we prove that if the minimum in the right–hand side of (5.13) is attained at e; that is, μ22 − μ21 q0 (e) + (μ22 − μ21 )q02 (e) + 4q1 (e) := min w ∈ V μ22 − μ21 q0 (w) + (μ22 − μ21 )q02 (w) + 4q1 (w) , (5.25)

w 2 = 1

50


then e is a solution of (5.8) with b as in (5.13). So, for such positive b, μ1 bτ = arccos − + 2lπ, l ∈ N, μ2 defines a sequence of time delays for which the problem (1.1)–(1.4) is not asymptotically stable, since the standard energy (|E(x, t)|2 + |H(x, t)|2 )dx ≥ 1, ∀t ≥ 0. Ω

Case (b) μ1 = μ2 . In this case, under our assumptions, (5.11) becomes b2 = q1 (e).

(5.26)

min q1 (w) w∈V w 2 = 1

(5.27)

But in this case the quantity

is equal to zero and therefore we cannot conclude as before. Consequently, we come back to (5.2) and remark that with the choice λ = ib, cos(bτ ) = −1, sin(bτ ) = 0, the factor (μ1 + μ2 e−λτ ) in front of e × ν is zero. This means that the boundary condition on Γ is simply h × ν = 0 on Γ, which is the standard “electric” boundary condition. We then eliminate e instead of h, in other words, we use the identity e = λ−1 curl h and consequently h is a solution of ⎧ 2 ⎨ λ h(x) + curl curl h(x) = 0 in Ω div h(x) = 0 in Ω (5.28) ⎩ h × ν = 0 on Γ. This problem is the classical eigenvalue problem for the Maxwell system with electric boundary condition. So, we can take a sequence {bn }n of positive real numbers defined by b2n = Λ2n , n ∈ N, where Λ2n , n ∈ N, are the eigenvalues for the above problem [6]. Then, putting bn τ = (2l + 1)π, l ∈ N, we obtain a sequence of delays (2l + 1)π τn,l = , l, n ∈ N, bn


51

which become arbitrarily small (or large) for suitable choices of the indices n, l ∈ N. Therefore, in the case μ1 = μ2 , we have found a set of time delays for which problem (1.1)–(1.4) is not asymptotically stable. Indeed, if h is a solution of (5.28) with the above choice of λ, b and τ , then, E(x, t) := −ib−1 eibt curl h(x) H(x, t) := eibt h(x)

(5.29)

is a solution of problem (1.1)–(1.4). Therefore, we have found a solution of our boundary-value problem whose energy is constant. Indeed, an easy computation shows that, for the pair (E, H) defined in (5.29), 2 2 (|E(x, t)| + |H(x, t)| )dx = 2 |h(x)|2 dx > 0, ∀t ≥ 0. Ω

The above examples prove Theorem 5.1.

Ω

5.2. Internal feedback. In this subsection we will give instability examples for the problem with internal feedback (1.8)–(1.13), proving the following result. Theorem 5.2. If (1.7) does not hold, there exist a sequence of arbitrarily small (or large) delays and solutions of problem (1.8) − (1.13), corresponding to these delays, such that their standard energy does not tend to 0. Proof. Let us consider the spectral problem for the system (1.8)–(1.11) in the case σ(x) ≡ 1 in Ω. Namely we seek a solution of (1.8)–(1.11) in the form (5.1). Then, the pair (e, h) has to solve the eigenvalue problem ⎧ λe(x) − curl h(x) + [μ1 e(x) + μ2 e−λτ e(x)] = 0 in Ω ⎪ ⎪ ⎨ λh(x) + curl e(x) = 0 in Ω (5.30) div e(x) = div h(x) = 0 in Ω ⎪ ⎪ ⎩ e(x) × ν = 0 and h(x) · ν = 0 on Γ. As before assuming that λ is different from 0, we can eliminate h, by the second equation, namely h = −λ−1 curl e and consequently e is a solution of ⎧ ⎨ curl curl e = −λ(λ + μ1 + μ2 e−λτ )e in Ω (5.31) div e = 0 in Ω ⎩ e × ν = 0 on Γ. Let us consider the standard problem for the Maxwell system with electric boundary condition [6] ⎧ ⎨ curl curl e = −μ2 e in Ω div e = 0 in Ω (5.32) ⎩ e × ν = 0 on Γ.

52


Following [8], we show that for any Λ2 eigenvalue of problem (5.32), there exists a solution λ ∈ C of the equation λ2 + (μ1 + μ2 e−λτ )λ = −Λ2 .

(5.33)

We seek a solution λ = α + iβ, α, β ∈ R, with βτ = (2l + 1)π,

l ∈ N.

Under this assumption the equation (5.33) becomes 2 α + β 2 = Λ2 μ2 e−ατ = 2α + μ1 .

(5.34)

(5.35)

Now, we distinguish two cases. Case (a) μ1 = μ2 . In this case, from (5.35) we have α = 0,

β 2 = Λ2 .

Therefore, for any Λ2n eigenvalue of problem (5.32), if βn ∈ R satisfies βn2 = Λ2n , then for λ = iβn problem (5.30) admits a non–zero solution. Take βn positive. From our assumption (5.34) τn,l =

(2l + 1)π , βn

n, l ∈ N,

is a set of time delays that become arbitrarily small (or large) for suitable choices of the indices n, l ∈ N. For such delays the problem (1.8)–(1.13) admits solutions in the form E(x, t) = eiβt e(x),

H(x, t) = eiβt h(x),

whose standard energy is constant and strictly positive. So, system (1.8)– (1.13) is not asymptotically stable. Case (b) μ2 > μ1 . For a fixed α > 0, from the second equation of (5.35), we obtain μ2 1 τ (α) = ln , (5.36) α μ1 + 2α and so, in order to have τ (α) > 0, we consider 0 < α < 12 (μ2 − μ1 ). From (5.34), the first equation of (5.35) becomes α2 + where τ (α) is given by (5.36).

(2l + 1)2 π 2 = Λ2 , τ 2 (α)

(5.37)


53

It is easy to verify (see [8]) that for any fixed Λ2 eigenvalue of problem (5.32) there exists α (0 < α < (μ2 − μ1 )/2) such that (5.37) is satisfied. Therefore, for such α there exists a delay τ (α) (defined by (5.36)) such that a function of the form eα+iβ (e(x), h(x)) solves problem (1.8)–(1.13). Since α ≥ 0 the energy of such a solution is not decaying to zero. So, this solution is not asymptotically stable. Note that, for any Λ2n eigenvalue of problem (5.32) and for any l ∈ N there exist αn,l and a delay τn,l = τ (αn,l ) such that (5.35) is satisfied with βn,l =

(2l + 1)π . τn,l

From the first equation of (5.35), (2l + 1)2 π 2 ≤ Λ2n . 2 τn,l Then, for a fixed l ∈ N, if n → +∞, then τn,l → 0+ . On the contrary, for a fixed n ∈ N, for l → +∞, then τn,l → +∞. Therefore, we have instability phenomena for a sequence of arbitrarily small or large time delays. The examples of case (a) and case (b) prove Theorem 5.2. References [1] M. Costabel, A remark on the regularity of solutions of Maxwell’s equations on Lipschitz domains, Math. Methods Appl. Sc., 12 (1990), 365–368. [2] M. Eller, J. Lagnese, and S. Nicaise, Decay rates for solutions of Maxwell system with nonlinear boundary damping, Comput. Appl. Math., 21 (2002), 135–165. [3] M. Eller and J. Masters, Exact boundary controllability of electromagnetic fields in a general region, Appl. Math. Optim., 45 (2002), 99–123. [4] B. V. Kapitonov, Stabilization and exact boundary controllability for Maxwell’s equations, SIAM J. Control Optim., 32 (1994), 408–420. [5] V. Komornik, Boundary stabilization, observation and control of Maxwell’s equations, PanAm. Math. J., 4 (1994), 47–61. [6] P. Monk, “Finite Element Methods for Maxwell’s Equations,” Numer. Math. Scientific Comp., Oxford Univ. Press, New York, 2003. [7] S. Nicaise and C. Pignotti, Internal stabilization of Maxwell’s equations in heterogeneous media, Abstr. Appl. Anal., 7 92005), 791–811. [8] S. Nicaise and C. Pignotti, Stability and instability results of the wave equation with a delay term in the boundary or internal feedbacks, Siam J. Control Opt., to appear. [9] K. D. Phung, Contrˆ ole et stabilisation d’ondes ´ electromagnétiques, ESAIM: Control Optim. Calc. Var., 5 (2000), 87–137. [10] G. Q. Xu, S. P. Yung, and L. K. Li. Stabilization of wave systems with input delay in the boundary control, ESAIM: Control Optim. Calc. Var., to appear.

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[11] E. Zuazua, Exponential decay for the semi-linear wave equation with locally distributed damping, Comm. Partial Differential Equations, 15 (1990), 205–235.