Control of evolution problems involving the

fractional Laplace operator

Umberto Biccarijoint work with Enrique Zuazua

BCAM - Basque Center for Applied MathematicsNUMERIWAVES group meeting

October 11, 2013

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Fractional laplacian

(−∆)su(x) := cn,sP.V .∫Rn

u(x)− u(y)|x − y |n+2s

dy , s ∈ (0, 1)

cn,s :=s2

2sΓ( n+2s2)

πn/2Γ(1−s)

We analyse the control problem for the fractional wave equation

utt + (−∆)s+1u = 0

on a bounded C 1,1 domain Ω of Rn. We focus on the control from aneighborhood of the boundary ∂Ω.

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Fractional laplacian

(−∆)su(x) := cn,sP.V .∫Rn

u(x)− u(y)|x − y |n+2s

dy , s ∈ (0, 1)

cn,s :=s2

2sΓ( n+2s2)

πn/2Γ(1−s)

We analyse the control problem for the fractional wave equation

utt + (−∆)s+1u = 0

on a bounded C 1,1 domain Ω of Rn. We focus on the control from aneighborhood of the boundary ∂Ω.

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Fractional Sobolev spaces

De�nition

Hs(Rn) :={u ∈ L2(Rn)

∣∣∣ |u(x)−u(y)||x−y |

n

2+s ∈ L2(Rn × Rn)

}De�nition

Ĥs(Rn) :={u ∈ L2(Rn)

∣∣(1 + |ξ|2)s/2û(ξ) ∈ L2(Rn)}Is an Hilbert space with norm ‖u(x)‖Hs := ‖(1 + |ξ|2)s/2û(ξ)‖L2

De�nition (Hs(Ω))

u ∈ Hs(Ω) if u = U|Ω with U ∈ Hs(Rn)

u ∈ Hs(Ω) if u ∈ L2(Ω) and∫∫

Ω×Ω

|u(x)− u(y)|2

|x − y |n+2sdxdy < +∞

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Fractional Sobolev spaces

De�nition

Hs(Rn) :={u ∈ L2(Rn)

∣∣∣ |u(x)−u(y)||x−y |

n

2+s ∈ L2(Rn × Rn)

}De�nition

Ĥs(Rn) :={u ∈ L2(Rn)

∣∣(1 + |ξ|2)s/2û(ξ) ∈ L2(Rn)}Is an Hilbert space with norm ‖u(x)‖Hs := ‖(1 + |ξ|2)s/2û(ξ)‖L2

De�nition (Hs(Ω))

u ∈ Hs(Ω) if u = U|Ω with U ∈ Hs(Rn)

u ∈ Hs(Ω) if u ∈ L2(Ω) and∫∫

Ω×Ω

|u(x)− u(y)|2

|x − y |n+2sdxdy < +∞

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Fractional Sobolev spaces

De�nition

Hs(Rn) :={u ∈ L2(Rn)

∣∣∣ |u(x)−u(y)||x−y |

n

2+s ∈ L2(Rn × Rn)

}De�nition

Ĥs(Rn) :={u ∈ L2(Rn)

∣∣(1 + |ξ|2)s/2û(ξ) ∈ L2(Rn)}Is an Hilbert space with norm ‖u(x)‖Hs := ‖(1 + |ξ|2)s/2û(ξ)‖L2

De�nition (Hs(Ω))

u ∈ Hs(Ω) if u = U|Ω with U ∈ Hs(Rn)

u ∈ Hs(Ω) if u ∈ L2(Ω) and∫∫

Ω×Ω

|u(x)− u(y)|2

|x − y |n+2sdxdy < +∞

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Proposition

Let Ω be a bounded C 0,1 domain; then, for any 0 < s < s1 we have

Hs1(Ω) ↪→ Hs(Ω)

with dense and compact embedding.

Proposition

Let s ∈ (0, 1); for any u ∈ Hs(Rn)

(−∆)su = F−1(|ξ|2sFu) ∀ξ ∈ Rn

Proposition

Let u, v be two Hs(Rn) function vanishing outside Ω; then∫Ω

v(−∆)sudx =∫Rn

(−∆)s/2u(−∆)s/2vdx =∫

Ω

u(−∆)svdx

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Proposition

Let Ω be a bounded C 0,1 domain; then, for any 0 < s < s1 we have

Hs1(Ω) ↪→ Hs(Ω)

with dense and compact embedding.

Proposition

Let s ∈ (0, 1); for any u ∈ Hs(Rn)

(−∆)su = F−1(|ξ|2sFu) ∀ξ ∈ Rn

Proposition

Let u, v be two Hs(Rn) function vanishing outside Ω; then∫Ω

v(−∆)sudx =∫Rn

(−∆)s/2u(−∆)s/2vdx =∫

Ω

u(−∆)svdx

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Proposition

Let Ω be a bounded C 0,1 domain; then, for any 0 < s < s1 we have

Hs1(Ω) ↪→ Hs(Ω)

with dense and compact embedding.

Proposition

Let s ∈ (0, 1); for any u ∈ Hs(Rn)

(−∆)su = F−1(|ξ|2sFu) ∀ξ ∈ Rn

Proposition

Let u, v be two Hs(Rn) function vanishing outside Ω; then∫Ω

v(−∆)sudx =∫Rn

(−∆)s/2u(−∆)s/2vdx =∫

Ω

u(−∆)svdx

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Fractional wave equation

utt + (−∆)s+1u = 0 in Ω× [0,T ] := Qu ≡ 0 on ∂Ω× [0,T ] := Σu(x , 0) = u0(x) in Ωut(x , 0) = u1(x) in Ω

In order to gurantee uniform velocity of propagation we need theexponent of the Fractional laplacian to be greater than 1.

Higher order Fractional laplacian

(−∆)s+1 := (−∆)(−∆)s = (−∆)s(−∆)

D((−∆)s+1) = H10 (Ω)× H2(s+1)(Ω)

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Fractional wave equation

utt + (−∆)s+1u = 0 in Ω× [0,T ] := Qu ≡ 0 on ∂Ω× [0,T ] := Σu(x , 0) = u0(x) in Ωut(x , 0) = u1(x) in Ω

In order to gurantee uniform velocity of propagation we need theexponent of the Fractional laplacian to be greater than 1.

Higher order Fractional laplacian

(−∆)s+1 := (−∆)(−∆)s = (−∆)s(−∆)

D((−∆)s+1) = H10 (Ω)× H2(s+1)(Ω)

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Energy

E (t) :=1

2

∫Ω

{(∇ut)2 +

((−∆)

s+22 u)2}

dx

dE (t)

dt= 0⇒ E (t) ≡ E (0) ≡ E0 ∀t ≥ 0

Pohozaev-type identity

∫Ω

(−∆)su(x · ∇u)dx =∫

Ω

u(−∆)sudx − Γ(1 + s)2

2

∫∂Ω

( uδs

)2(x · ν)dσ

X. ROS-OTON and J. SERRA - The Pohozaev identity for the Fractional laplacian

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Energy

E (t) :=1

2

∫Ω

{(∇ut)2 +

((−∆)

s+22 u)2}

dx

dE (t)

dt= 0⇒ E (t) ≡ E (0) ≡ E0 ∀t ≥ 0

Pohozaev-type identity

∫Ω

(−∆)su(x · ∇u)dx =∫

Ω

u(−∆)sudx − Γ(1 + s)2

2

∫∂Ω

( uδs

)2(x · ν)dσ

X. ROS-OTON and J. SERRA - The Pohozaev identity for the Fractional laplacian

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Proposition

For any solution of the fractional wave equation it holds

Γ(1 + s)2

2

∫Σ

(−∆uδs

)2(x · ν)dσdt = 2s − n

2

∫Q

((−∆)

s+22 u)2

dxdt

+2 + n

2

∫Q

(∇ut)2dxdt

+

∫Ω

ut(x · ∇(−∆u))dx∣∣∣∣T0

Proof.

The identity is obtained by multiplying the equation by x · ∇(−∆u) andintegrating by parts over Q by using the Pohozaev identity.

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Energy estimate

Theorem

Let E (t) de�ned as before; then, there exists two non negative constantsA1 and A2, depending only on s, T and Ω, such that

A1E0 ≤∫

Σ

(−∆uδs

)2(x · ν)dσdt ≤ A2E0

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Control from a neighborhood of the boundary

We will use Hilbert Uniqueness Method

Observability inequality

E0 ≤ C1∫ T0

∫ω

{(∇ut)2 +

((−∆)

s+22 u)2}

dxdt

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By using equipartition of the energy

E0 =∥∥∥(−∆) s+22 u0∥∥∥2

L2(Ω)+ ‖∇u1‖2L2(Ω) ≤ C2

∫ T0

∫ω

(∇ut)2dxdt (1)

Take φ(x , t) :=

∫ t0

u(x , s)ds − (−∆)−(s+1)u1(x); this φ satis�es the

fractional wave equation with initial data

φ0 := φ(0) = −(−∆)−(s+1)u1 φ1 := φt(0) = u0

Thus (u0, u1) ∈ L2(Ω)× H−(s+1)(Ω)⇒ (φ0, φ1) ∈ Hs+1(Ω)× L2(Ω)

Considering (1) for the function φ we have

‖∇u0‖2L2(Ω) + ‖u1‖2H−s(Ω) ≤ C2

∫ T0

∫ω

(∇u)2dxdt (2)

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By using equipartition of the energy

E0 =∥∥∥(−∆) s+22 u0∥∥∥2

L2(Ω)+ ‖∇u1‖2L2(Ω) ≤ C2

∫ T0

∫ω

(∇ut)2dxdt (1)

Take φ(x , t) :=

∫ t0

u(x , s)ds − (−∆)−(s+1)u1(x); this φ satis�es the

fractional wave equation with initial data

φ0 := φ(0) = −(−∆)−(s+1)u1 φ1 := φt(0) = u0

Thus (u0, u1) ∈ L2(Ω)× H−(s+1)(Ω)⇒ (φ0, φ1) ∈ Hs+1(Ω)× L2(Ω)

Considering (1) for the function φ we have

‖∇u0‖2L2(Ω) + ‖u1‖2H−s(Ω) ≤ C2

∫ T0

∫ω

(∇u)2dxdt (2)

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By using equipartition of the energy

E0 =∥∥∥(−∆) s+22 u0∥∥∥2

L2(Ω)+ ‖∇u1‖2L2(Ω) ≤ C2

∫ T0

∫ω

(∇ut)2dxdt (1)

Take φ(x , t) :=

∫ t0

u(x , s)ds − (−∆)−(s+1)u1(x); this φ satis�es the

fractional wave equation with initial data

φ0 := φ(0) = −(−∆)−(s+1)u1 φ1 := φt(0) = u0

Thus (u0, u1) ∈ L2(Ω)× H−(s+1)(Ω)⇒ (φ0, φ1) ∈ Hs+1(Ω)× L2(Ω)

Considering (1) for the function φ we have

‖∇u0‖2L2(Ω) + ‖u1‖2H−s(Ω) ≤ C2

∫ T0

∫ω

(∇u)2dxdt (2)

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utt + (−∆)s+1u = 0u|Σ ≡ 0u(x , 0) = u0(x)ut(x , 0) = u1(x)

RETROGRADE SYSTEM ψtt + (−∆)s+1ψ = h(u)

ψ|Σ ≡ 0ψ(x ,T ) = ψ′(x ,T ) = 0

The solution of the retrograde system is de�ned by transposition: thefunction ψ is a solution of the problem if and only if for any solution θ of

θtt + (−∆)s+1θ = fθ|Σ ≡ 0θ(x , 0) = θ0θt(x , 0) = θ1

∫Q

ψfdxdt −∫

Ω

(ψt(0)θ0 − ψ(0)θ1)dx =∫Q

θhdxdt (3)

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We de�ne the operator

Λ(u0, u1) := (ψt(x , 0),−ψ(x , 0))

by considering (3) with θ = u and choosing the control function

h(u) = div(ρ(t)u)χ{ω×(0,T )}

we obtain

〈Λ(u0, u1); (u0, u1)〉 =∫ T0

∫ω

|∇u|2 dxdt.

Thus, thanks to the observability inequality, for any given couple of initialdata (ψ0, ψ1) ∈ H(s+1)(Ω)× L2(Ω) there exists a unique solution(u0, u1) ∈ L2(Ω)× H−(s+1)(Ω) of the problem Λ(u0, u1) = (ψ1,−ψ0)and we have found a control

h ∈ L2(0,T ;H−1(ω))

that drives the system in rest in time T .12 / 13

We de�ne the operator

Λ(u0, u1) := (ψt(x , 0),−ψ(x , 0))

by considering (3) with θ = u and choosing the control function

h(u) = div(ρ(t)u)χ{ω×(0,T )}

we obtain

〈Λ(u0, u1); (u0, u1)〉 =∫ T0

∫ω

|∇u|2 dxdt.

Thus, thanks to the observability inequality, for any given couple of initialdata (ψ0, ψ1) ∈ H(s+1)(Ω)× L2(Ω) there exists a unique solution(u0, u1) ∈ L2(Ω)× H−(s+1)(Ω) of the problem Λ(u0, u1) = (ψ1,−ψ0)and we have found a control

h ∈ L2(0,T ;H−1(ω))

that drives the system in rest in time T .12 / 13

THANKS FOR YOUR ATTENTION!

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