Selected topics on wave equation with dynamic boundary conditions. Nicolas Fourrier

U NIVERSITY

OF

V IRGINIA

[email protected]

March 19, 2013

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

1 / 25

Overview

1

Introduction

2

Exponential Stability

3

Analyticity

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

2 / 25

Introduction

Overview

1

Introduction Presentation of the problem Mathematical model

2

Exponential Stability

3

Analyticity

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

3 / 25

Introduction

Nicolas Fourrier (U.V.A.)

Presentation of the problem

P.D.E. Seminar

March 19, 2013

4 / 25

Introduction

Presentation of the problem

Why this model ?

Understand the general dynamics of a strong damped wave equation with dynamic boundary conditions. Without strong damping in the interior, we expect to loose stability because of the inertial boundary term. Surface of local reaction vs. surface of extended reaction, i.e., do we assume that the acceleration is proportional to the displacement (simple harmonic oscillator) or to the relative displacement compared to its neighbours

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

5 / 25

Introduction

Mathematical model

 utt − ∆u − kΩ ∆ut = 0      utt − δη (u + kΩ ut ) − kΓ ∆Γ (αut + u) = 0 u=0   Initial conditions    u(0, x) = u0 , ut (0, x) = u1

in Ω in Γ1 in Γ0

(1)

on Ω

kΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or α > 0, allows to add a viscous boundary damper.

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

6 / 25

Introduction

Mathematical model

 utt − ∆u − kΩ ∆ut = 0      utt − δη (u + kΩ ut ) − kΓ ∆Γ (αut + u) = 0 u=0   Initial conditions    u(0, x) = u0 , ut (0, x) = u1

in Ω in Γ1 in Γ0

(1)

on Ω

kΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or α > 0, allows to add a viscous boundary damper.

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

6 / 25

Introduction

Mathematical model

 utt − ∆u − kΩ ∆ut = 0      utt − δη (u + kΩ ut ) − kΓ ∆Γ (αut + u) = 0 u=0   Initial conditions    u(0, x) = u0 , ut (0, x) = u1

in Ω in Γ1 in Γ0

(1)

on Ω

kΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or α > 0, allows to add a viscous boundary damper.

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

6 / 25

Introduction

Mathematical model

 utt − ∆u − kΩ ∆ut = 0      utt − δη (u + kΩ ut ) − kΓ ∆Γ (αut + u) = 0 u=0   Initial conditions    u(0, x) = u0 , ut (0, x) = u1

in Ω in Γ1 in Γ0

(1)

on Ω

kΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or α > 0, allows to add a viscous boundary damper.

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

6 / 25

Introduction

Mathematical model

Energy space For all kΓ ≥ 0, set B : L2 (Γ1 ) ⊃ D(B) → L2 (Γ1 ) to be: Bz = −kΓ ∆Γ z, D(B) = {z ∈ L2 (Γ1 ), kΓ ∆z ∈ L2 (Γ1 )} Z 1 kzk2 1 = kB 2 zk2L2 (Γ1 ) = kΓ |∇Γ z|2 d Γ1 = kΓ kukH 1 (Γ1 ) D(B 2 )

(2)

Γ1

Let (u, ut , u|Γ1 , u|Γ1 ) = (u1 , u2 , u3 , u4 ). The associated energy space is: 1

1

H = D(A 2 ) × L2 (Ω) × D(B 2 ) × L2 (Γ1 )

(3)

with associated norm: kukH = |∇u1 |2Ω + |u2 |2Ω + kΓ |∇Γ u3 |2Γ1 + |u4 |2Γ1 . Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

7 / 25

Introduction

Mathematical model

Evolutive problem With the definitions of A, B and N, we set:   0 I 0 0  ∆ kΩ ∆ 0 0   A=  0 0 0 I  −δη −kΩ δη −B −αB 1

1

(4)

1

1

D(A) ={[u1 , u2 , u3 , u4 ]T ∈ D(A 2 ) × D(A 2 ) × D(B 2 ) × D(B 2 ), such that ∆(u1 + kΩ u2 ) ∈ L2 (Ω), 1

1

1

δη (u1 + kΩ u2 ) + B 2 (B 2 u3 + αB 2 u4 ) ∈ L2 (Γ1 ), u1 |Γ1 = N ∗ Au1 = u3 , u2 |Γ1 = N ∗ Au2 = u4 }

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

8 / 25

Introduction

Mathematical model

Results The problem is well-posed, for kΩ , kΓ , α ≥ 0 A generates a C0 -semigroup of contractions provided some damping (kΩ or kΓ α is strictly positive) Under some conditions on kΩ , kΓ , α: {eAt }t≥0 is exponentially stable, i.e., ∃C, ω ≥ 0 such that keAt ukH ≤ Ce−ωt kukH Suppose |β| ≥ 1 so that (iβ − A)−1 ∈ L(H), then with F = (f1 , f2 , f3 , f4 )T ∈ H kR(iβ − A)kL(H) ≤

C |β|

It follows that {eAt }t≥0 is an analytic semigroup on H. Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

9 / 25

Exponential Stability

Overview

1

Introduction

2

Exponential Stability Proof Exp. Stability - Picture Comment

3

Analyticity

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

10 / 25

Exponential Stability

Proof

   

utt − kΩ ∆ut − ∆u = 0 u(x, t) = 0 u + δ (u + k u ) − k ∆ (αu  η t + u) = 0 Ω t Γ Γ   tt u(0, x) = u0 , ut (0, x) = u1

x x x x

∈ Ω, t > 0 ∈ Γ0 , t > 0 ∈ Γ1 , t > 0 ∈Ω

(5)

Steps Ep (t) = |ut |2Ω + |ut |2Γ1 : Potential Energy Ek (t) = |∇u|2Ω + kΓ |∇Γ u|2Γ1 : Kinetic Energy E (t) = Ep (t) + Ek (t). Multiply by ut the first equation of (5). multiply by u the first equation of (5) Case 1: α = 0 or kΓ = 0 Case 2: α > 0 Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

11 / 25

Exponential Stability

Proof

   

utt − kΩ ∆ut − ∆u = 0 u(x, t) = 0 u + δ (u + k u ) − k ∆ (αu  η t + u) = 0 Ω t Γ Γ   tt u(0, x) = u0 , ut (0, x) = u1

x x x x

∈ Ω, t > 0 ∈ Γ0 , t > 0 ∈ Γ1 , t > 0 ∈Ω

(5)

Steps Ep (t) = |ut |2Ω + |ut |2Γ1 : Potential Energy Ek (t) = |∇u|2Ω + kΓ |∇Γ u|2Γ1 : Kinetic Energy E (t) = Ep (t) + Ek (t). Multiply by ut the first equation of (5). multiply by u the first equation of (5) Case 1: α = 0 or kΓ = 0 Case 2: α > 0 Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

11 / 25

Exponential Stability

Proof

   

utt − kΩ ∆ut − ∆u = 0 u(x, t) = 0 u + δ (u + k u ) − k ∆ (αu  η t + u) = 0 Ω t Γ Γ   tt u(0, x) = u0 , ut (0, x) = u1

x x x x

∈ Ω, t > 0 ∈ Γ0 , t > 0 ∈ Γ1 , t > 0 ∈Ω

(5)

Steps Ep (t) = |ut |2Ω + |ut |2Γ1 : Potential Energy Ek (t) = |∇u|2Ω + kΓ |∇Γ u|2Γ1 : Kinetic Energy E (t) = Ep (t) + Ek (t). Multiply by ut the first equation of (5). multiply by u the first equation of (5) Case 1: α = 0 or kΓ = 0 Case 2: α > 0 Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

11 / 25

Exponential Stability

Proof

   

utt − kΩ ∆ut − ∆u = 0 u(x, t) = 0 u + δ (u + k u ) − k ∆ (αu  η t + u) = 0 Ω t Γ Γ   tt u(0, x) = u0 , ut (0, x) = u1

x x x x

∈ Ω, t > 0 ∈ Γ0 , t > 0 ∈ Γ1 , t > 0 ∈Ω

(5)

Steps Ep (t) = |ut |2Ω + |ut |2Γ1 : Potential Energy Ek (t) = |∇u|2Ω + kΓ |∇Γ u|2Γ1 : Kinetic Energy E (t) = Ep (t) + Ek (t). Multiply by ut the first equation of (5). multiply by u the first equation of (5) Case 1: α = 0 or kΓ = 0 Case 2: α > 0 Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

11 / 25

Exponential Stability

Exp. Stability - Picture

Imaginary Axis

kΩ = 0.12 kΓ = 0 α=0

5i

Real Axis −14

  utt − kΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (αut + u) = 0 u + δ(kΩδη  tt u=0

Nicolas Fourrier (U.V.A.)

0

on Ω on Γ1 on Γ0

P.D.E. Seminar

March 19, 2013

12 / 25

Exponential Stability

Comment

The proof enlights the necessity of interior damping, i.e. kΩ > 0. What happened if kΩ = 0 ?

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

13 / 25

Exponential Stability

Comment

Imaginary Axis

kΩ = 0 kΓ = 1 α = 0.0005

2.5e+02i

Real Axis −1.4

  utt − kΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (αut + u) = 0 u + δ(kΩδη  tt u=0

Nicolas Fourrier (U.V.A.)

0

on Ω on Γ1 on Γ0

P.D.E. Seminar

March 19, 2013

14 / 25

Analyticity

Overview

1

Introduction

2

Exponential Stability

3

Analyticity Proof Analyticity - Picture Comment

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

15 / 25

Analyticity

Proof

Proof Let β ≥ 1 so that (iβ − A)−1 ∈ L(H) F = (f1 , f2 , f3 , f4 )T ∈ H pre-image U = (u1 , u2 , u3 , u4 )T ∈ D(A) We consider the resolvent equation (iβ − A)U = F : (iβ − A)U = F     f1 u1   u2   = (iβ − A)−1 f2  = (iβ − A)−1 F U= f3  u3  f4 u4

(6)

We want to show |β|kUkH ≤ CkF kH .

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

16 / 25

Analyticity

Proof

Proof     Step I

  

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

(7)

Since the system is exponentially stable: kUkH ≤ CkF kH Multiply the third equation by u2 and make use of the fourth equation 1

Apply ∇ and B 2 on the first and second equations respectively i h 1 1 iβ |u2 |2Ω + |u4 |2Γ1 − |∇u1 |2Ω − |B 2 u3 |2Γ1 + kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1   E D 1 1 = (f2 , u2 )Ω + (f4 , u4 )Ω + ∇u1 , ∇f1 + B 2 u3 , B 2 f3 Ω

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

Γ1

March 19, 2013

17 / 25

Analyticity

Proof

Proof     Step I

  

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

(7)

Since the system is exponentially stable: kUkH ≤ CkF kH Multiply the third equation by u2 and make use of the fourth equation 1

Apply ∇ and B 2 on the first and second equations respectively i h 1 1 iβ |u2 |2Ω + |u4 |2Γ1 − |∇u1 |2Ω − |B 2 u3 |2Γ1 + kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1   E D 1 1 = (f2 , u2 )Ω + (f4 , u4 )Ω + ∇u1 , ∇f1 + B 2 u3 , B 2 f3 Ω

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

Γ1

March 19, 2013

17 / 25

Analyticity

Proof

Proof System       

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

(8)

i h 1 1 iβ |u2 |2Ω + |u4 |2Γ1 − |∇u1 |2Ω − |B 2 u3 |2Γ1 + kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1   E D 1 1 = (f2 , u2 )Ω + (f4 , u4 )Ω + ∇u1 , ∇f1 + B 2 u3 , B 2 f3 Ω

Γ1

Step II Take real part. 1

Use the first two equations:β(|∇u1|Ω + β|B 2 u3 |Γ1 ) Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

18 / 25

Analyticity

Proof

Proof System    

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

  

(8)

1

kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1 ≤ CkF kH kUkH ≤ CkF k2H Step II Take real part. 1

Use the first two equations:β(|∇u1|Ω + β|B 2 u3 |Γ1 ) Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

18 / 25

Analyticity

Proof

Proof System    

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

  

(8)

1

kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1 ≤ CkF kH kUkH ≤ CkF k2H Step II Take real part. 1

Use the first two equations:β(|∇u1|Ω + β|B 2 u3 |Γ1 ) Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

18 / 25

Analyticity

Proof

Proof System       

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

(9)

i h 1 1 iβ |u2 |2Ω + |u4 |2Γ1 − |∇u1 |2Ω − |B 2 u3 |2Γ1 + kΩ |∇u2 |2Ω + α|B 2 u4 |2Γ1   E D 1 1 2 2 = (f2 , u2 )Ω + (f4 , u4 )Ω + ∇u1 , ∇f1 + B u3 , B f3 Ω

Γ1

Step III Multiply by β Split terms on the right hand side Conclude Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

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Analyticity

Proof

Proof System       

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

h h i i 1 β 2 |u2 |2Ω + |u4 |2Γ1 ≤ β 2 |∇u1 |2Ω + |B 2 u3 |2Γ1 1

(9)

1

+ |f2 |Ω |βu2 |Ω + |∇f1 |Ω |β∇u1 |Ω + |B 2 f3 |Γ1 |βB 2 u3 |Γ1 + |f4 |Γ1 |βu4 |Γ1 Step III Multiply by β Split terms on the right hand side Conclude Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

19 / 25

Analyticity

Proof

Proof System       

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

h h i i 1 β 2 |u2 |2Ω + |u4 |2Γ1 ≤ β 2 |∇u1 |2Ω + |B 2 u3 |2Γ1 1

(9)

1

+ |f2 |Ω |βu2 |Ω + |∇f1 |Ω |β∇u1 |Ω + |B 2 f3 |Γ1 |βB 2 u3 |Γ1 + |f4 |Γ1 |βu4 |Γ1 Step III Multiply by β Split terms on the right hand side Conclude Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

19 / 25

Analyticity

Proof

Proof System       

iβu1 − u2 = f1 iβu3 − u4 = f3 iβu2 − ∆u1 − kΩ ∆u2 = f2 iβu4 + δη (u1 + kΩ u2 ) + Bu3 + αBu4 = f4

h h i i 1 β 2 |u2 |2Ω + |u4 |2Γ1 ≤ β 2 |∇u1 |2Ω + |B 2 u3 |2Γ1 1

(9)

1

+ |f2 |Ω |βu2 |Ω + |∇f1 |Ω |β∇u1 |Ω + |B 2 f3 |Γ1 |βB 2 u3 |Γ1 + |f4 |Γ1 |βu4 |Γ1 Step III Multiply by β Split terms on the right hand side Conclude Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

March 19, 2013

19 / 25

Analyticity

Analyticity - Picture

Imaginary Axis

kΩ = 0 → 0.3 kΓ = 0 → 0.3 α=1

7.9i

Real Axis −30

  utt − kΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (αut + u) = 0 u + δ(kΩδη  tt u=0

Nicolas Fourrier (U.V.A.)

0

on Ω on Γ1 on Γ0

P.D.E. Seminar

March 19, 2013

20 / 25

Analyticity

Comment

The desired estimate relies on the value K1 where K = min{kΩ , α}. Thus, this proof works when both kΩ , α > 0. We have already seen the spectrum for kΩ = 0, which has a component along the imaginary axis ⇒ the analyticity is not possible. Is the boundary damping (α) really necessary ?

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

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Analyticity

Comment

Imaginary Axis

2

Im|λ| = 21.6(Re|λ|) − 58.3(Re|λ|) + 1.14

kΩ = 0.02 kΓ = 1 α=0

27i

Real Axis −18

  utt − kΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (αut + u) = 0 u + δ(kΩδη  tt u=0

Nicolas Fourrier (U.V.A.)

0

on Ω on Γ1 on Γ0

P.D.E. Seminar

March 19, 2013

22 / 25

Analyticity

Comment

Gevrey semigroup

As for characterization of analyticity, Gevrey’s regularity is described in terms of the bounds on all derivatives of the semigroup: Definition Let T (t) be a strongly continuous semigroup on a Banach space X and let δ > 1. We say that T (t) is of Gevrey class δ for t > t0 if T (t) is infinitely differentiable for t ∈ (t0 , ∞) and for every compact K ⊂ (t0 , ∞), and each θ > 0, there exists a constant C = Cθ,K such that ||T (n) (t)|| ≤ Cθ n (n!)δ , ∀t ∈ K and n ∈ {0, 1, 2, ...}

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

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Analyticity

Comment

Theorem Let T (t) be a strongly continuous semigroup satisfying ||T (t)|| ≤ Meωt . Assume one of the following holds: Suppose that for some γ satisfying 0 < γ ≤ 1: lim sup |β|γ ||R(iβ, A)|| = C∞ β→∞

Suppose that: lim t δ ||T ′ (t)|| = 0 t↓0

Then T (t) is of Gevrey class δ for t > 0 (for every δ > item).

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

1 γ

for the first

March 19, 2013

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Analyticity

Comment

Analytic ↓ Resolvent or time characterization (previous theorem) ↓ Gevrey ↓ Differentiable

(10)

Gevrey vs. Differentiable Let B ∈ H, then: A is Gevrey ⇒ A + B is Gevrey A is differentiable ⇒ A + B is not necessary differentiable.

Nicolas Fourrier (U.V.A.)

P.D.E. Seminar

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