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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
1 / 13
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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.
2 / 13
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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.
2 / 13
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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 < +∞
3 / 13
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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 < +∞
3 / 13
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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 < +∞
3 / 13
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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
4 / 13
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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
4 / 13
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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
4 / 13
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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)
5 / 13
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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)
5 / 13
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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
6 / 13
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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
6 / 13
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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.
7 / 13
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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
8 / 13
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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
9 / 13
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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)
10 / 13
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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)
10 / 13
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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)
10 / 13
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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)
11 / 13
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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
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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
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THANKS FOR YOUR ATTENTION!
13 / 13
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