Evolution problems involving the fractional Laplace ...€¦ · Fractional Schrödinger equationFractional wave equationFourier analysis Evolution problems involving the fractional

Fractional Schrödinger equation Fractional wave equation Fourier analysis

Evolution problems involving the fractional Laplaceoperator: HUM control and Fourier analysis

Umberto Biccarijoint work with Enrique Zuazua

BCAM - Basque Center for Applied MathematicsNUMERIWAVES group meeting

February 28, 2014

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We analyse the control problem for the fractional Schrödinger equation

iut + (−∆)su = 0

and 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 aneighbourhood of the boundary ∂Ω.

Fractional laplacian

(−∆)su(x) := cn,sP.V .∫Rn

u(x)− u(y)|x − y |n+2s

dy , s ∈ (0, 1)

cn,s :=s22sΓ( n+2s2 )πn/2Γ(1−s)

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We analyse the control problem for the fractional Schrödinger equation

iut + (−∆)su = 0

and 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 aneighbourhood of the boundary ∂Ω.

Fractional laplacian

(−∆)su(x) := cn,sP.V .∫Rn

u(x)− u(y)|x − y |n+2s

dy , s ∈ (0, 1)

cn,s :=s22sΓ( n+2s2 )πn/2Γ(1−s)

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Fractional Schrödinger equation iut + (−∆)su = 0 in Ω× [0,T ] := Q

u ≡ 0 in (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ω

(1)

Fractional wave equationutt + (−∆)s+1u = 0 in Ω× [0,T ] := Qu ≡ 0 in (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ωut(x , 0) = u1(x) in Ω

(2)

Thanks to Hille-Yosida theorem, both problems are well posed.

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Fractional Schrödinger equation iut + (−∆)su = 0 in Ω× [0,T ] := Q

u ≡ 0 in (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ω

(1)

Fractional wave equationutt + (−∆)s+1u = 0 in Ω× [0,T ] := Qu ≡ 0 in (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ωut(x , 0) = u1(x) in Ω

(2)

Thanks to Hille-Yosida theorem, both problems are well posed.

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Internal control result - Schrödinger equation

Let Ω be a bounded C 1,1 domain of Rn.Let us consider the nonhomogeneous fractional Schrödinger equation iyt + (−∆)

sy = hχω×[0,T ] in Ω× [0,T ] := Qy ≡ 0 on (Rn \ Ω)× [0,T ]y(x , 0) = y0(x) in Ω

(3)

where ω is a neighbourhood of the boundary of the domain and χ is thecharacteristic function.

TheoremLet T > 0 and ω ⊂ Ω be a neighbourhood of the boundary of thedomain. Then, for any y0 ∈ L2(Ω) there exists h ∈ L2(ω × [0,T ]) suchthat the solution of (3) satisfies y(x ,T ) = 0.

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Proposition

For any solution u of iut + (−∆)su = 0 in Ω× [0,T ] := Q

u ≡ 0 on (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ω

it holds

Γ(1 + s)2∫

Σ

(|u|δs

)2(x · ν)dσdt = 2s

∫ T0

∥∥∥(−∆)s/2u∥∥∥2L2(Ω)

dt

+ =∫

Ω

ū(x · ∇u)dx∣∣∣∣T0

Σ := ∂Ω× [0,T ] δ = δ(x) := d(x , ∂Ω)

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Proof.The identity is obtained by multiplying the equation by x · ∇ū + n2 ū ,taking the real part and integrating over Q by using the Pohozaev identity∫

Ω

(−∆)su(x · ∇u)dx = 2s − n2

∫Ω

u(−∆)sudx

(4)

− 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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Theorem

For any u solution of (1) there exist two non negative constants A1 andA2, depending only on n, s, T and Ω, such that

A1‖u0‖2Hs(Ω) ≤∫

Σ

(|u|δs

)2(x · ν)dσdt ≤ A2‖u0‖2Hs(Ω)

Proof.The proof is merely technical. We use some interpolation results andsome compactness-uniqueness arguments.

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Control from a neighbourhood of the boundary

We will use Hilbert Uniqueness Method

Observability inequality

‖u0‖2L2(Ω) ≤ C∫ T0

∫ω

|u|2dxdt

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Proof of the observability inequality

1 ‖u0‖2Hs(Ω) ≤ C1∫ T0

∫ω̂

|∇u|2dxdt (Ω ∩ ω̂) ⊂ ω

2 ‖u0‖2Hs(Ω) ≤ C2∫ T0

(‖ut‖H−s(ω) + ‖u0‖L2(ω)

)dt

⇒ ‖u0‖2Hs(Ω) ≤ C3∫ T0‖ut‖2H−s(ω)dt

3 ‖u0‖2H−s(Ω) ≤ C4∫ T0‖u‖2H−s(ω)dt

4 ‖u0‖2Hs(Ω) ≤ C5‖u‖2L2(0,T ;Hs(ω)), ‖u0‖

2H−s(Ω) ≤ C5‖u‖

2L2(0,T ;H−s(ω))

5 The proof is concluded by interpolation.

J.L. LIONS and E. MAGENES

Problèmes aux limites non homogènes et applications - 19689 / 24

1 ‖u0‖2Hs(Ω) ≤ C1∫ T0

∫ω̂

|∇u|2dxdt (Ω ∩ ω̂) ⊂ ω

2 ‖u0‖2Hs(Ω) ≤ C2∫ T0

(‖ut‖H−s(ω) + ‖u0‖L2(ω)

)dt

⇒ ‖u0‖2Hs(Ω) ≤ C3∫ T0‖ut‖2H−s(ω)dt

3 ‖u0‖2H−s(Ω) ≤ C4∫ T0‖u‖2H−s(ω)dt

4 ‖u0‖2Hs(Ω) ≤ C5‖u‖2L2(0,T ;Hs(ω)), ‖u0‖

2H−s(Ω) ≤ C5‖u‖

2L2(0,T ;H−s(ω))

1 ‖u0‖2Hs(Ω) ≤ C1∫ T0

∫ω̂

|∇u|2dxdt (Ω ∩ ω̂) ⊂ ω

2 ‖u0‖2Hs(Ω) ≤ C2∫ T0

(‖ut‖H−s(ω) + ‖u0‖L2(ω)

)dt

⇒ ‖u0‖2Hs(Ω) ≤ C3∫ T0‖ut‖2H−s(ω)dt

3 ‖u0‖2H−s(Ω) ≤ C4∫ T0‖u‖2H−s(ω)dt

4 ‖u0‖2Hs(Ω) ≤ C5‖u‖2L2(0,T ;Hs(ω)), ‖u0‖

2H−s(Ω) ≤ C5‖u‖

2L2(0,T ;H−s(ω))

Control result via HUM

iut + (−∆)su = 0u|Rn\Ω ≡ 0u(x , 0) = u0(x)

BACKWARD SYSTEMiyt + (−∆)sy = uχω×[0,T ]y |Rn\Ω ≡ 0y(x ,T ) = 0

The solution of the backward system is defined by transposition with asolution θ of

iθt + (−∆)sθ = fθ|Rn\Ω ≡ 0θ(x , 0) = θ0

J.L. LIONS and E. MAGENES - Problèmes aux limites non homogènes et applications - 1968

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iut + (−∆)su = 0u|Rn\Ω ≡ 0u(x , 0) = u0(x)

BACKWARD SYSTEMiyt + (−∆)sy = uχω×[0,T ]y |Rn\Ω ≡ 0y(x ,T ) = 0

The solution of the backward system is defined by transposition with asolution θ of

iθt + (−∆)sθ = fθ|Rn\Ω ≡ 0θ(x , 0) = θ0

J.L. LIONS and E. MAGENES - Problèmes aux limites non homogènes et applications - 1968

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We consider the operator Λ : L2(Ω)→ L2(Ω) defined by

Λu0 := −iy(x , 0).

It is immediate to check the identity

〈Λu0; u0〉 =∫ T0

∫ω

|u|2dxdt

Thus, thanks to the observability inequality, Λ is an isomorphism fromL2(Ω) to L2(Ω). Hence, given y0 ∈ L2(Ω) we can choose the controlfunction

h := u|ωwhere u is the solution of (1) with intial datum u0 = Λ−1(−iy0).

The proof is concluded.

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Λu0 := −iy(x , 0).

〈Λu0; u0〉 =∫ T0

∫ω

|u|2dxdt

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Λu0 := −iy(x , 0).

〈Λu0; u0〉 =∫ T0

∫ω

|u|2dxdt

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Fractional wave equation

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

Higher order Fractional laplacian

(−∆)s+1 := (−∆)s(−∆)

D((−∆)s+1) = H3(Ω) ∩ H2(s+1)0 (Ω)

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Fractional wave equation

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

Higher order Fractional laplacian

(−∆)s+1 := (−∆)s(−∆)

D((−∆)s+1) = H3(Ω) ∩ H2(s+1)0 (Ω)

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Internal control result

Let Ω be a bounded C 1,1 domain of Rn.Let us consider the nonhomogeneous fractional wave equation

ytt + (−∆)s+1y = hχω×[0,T ] in Ω× [0,T ] := Qy ≡ 0 on (Rn \ Ω)× [0,T ]y(x , 0) = y0(x) in Ωyt(x , 0) = y1(x) in Ω

(5)

where ω is a neighbourhood of the boundary of the domain and χ is thecharacteristic function.

TheoremLet T > 0 and ω ⊂ Ω be a neighbourhood of the boundary of thedomain. Then, for any couple of initial data (y0, y1) ∈ H10 (Ω)× L2(Ω)there exists h ∈ L2(ω × [0,T ]) such that the solution of (5) satisfiesy(x ,T ) = yt(x ,T ) = 0.

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Proposition

For any solution u ofutt + (−∆)s+1u = 0 in Ω× [0,T ] := Qu ≡ 0 on (Rn \ Ω)× [0,T ]u(x , 0) = u0(x) in Ωut(x , 0) = u1(x) in Ω

it holds

Γ(1 + s)2∫

Σ

(−∆uδs

)2(x · ν)dσdt = 2s − n

2

∫Q

((−∆)

s+22 u)2

dxdt

+2 + n2

∫Q

(∇ut)2 dxdt

+

∫Ω

ut(x · ∇(−∆u))dx∣∣∣∣T0

Σ := ∂Ω× [0,T ] δ = δ(x) := d(x , ∂Ω)

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Proof.

The identity is obtained by multiplying the equation by x · ∇(−∆u) andintegrating over Q by using the Pohozaev identity for the fractionallaplacian (4) applied to (−∆u).

Theorem (Energy estimate)

For any solution of (2) we define the energy as

E (t) :=12

∫Ω

{(∇ut)2 +

((−∆)

s+22 u)2}

dx ;

for any T > 0 there exists two non negative constants A1 and A2,depending only on s, T and Ω, such that

A1E0 ≤∫

Σ

(−∆uδs

)2(x · ν)dσdt ≤ A2E0

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Proof.

The identity is obtained by multiplying the equation by x · ∇(−∆u) andintegrating over Q by using the Pohozaev identity for the fractionallaplacian (4) applied to (−∆u).

Theorem (Energy estimate)

For any solution of (2) we define the energy as

E (t) :=12

∫Ω

{(∇ut)2 +

((−∆)

s+22 u)2}

dx ;

for any T > 0 there exists two non negative constants A1 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 neighbourhood of the boundary

We will use Hilbert Uniqueness Method

Observability inequality

E0 ≤ C∫ T0

∫ω

|u|2dxdt

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1 E0 ≤ C1∫ T0

∫ω̂

{(∇ut)2 +

((−∆)

s+22 u)2}

dxdt (Ω ∩ ω̂) ⊂ ω

2 equipartition of the energy ⇒ E0 ≤ C2∫ T0

∥∥∥(−∆) s+22 u∥∥∥2L2(ω)

dt

3 E0 ≤ C2∫ T0‖u‖2Hs+2(ω)dt

Problèmes aux limites non homogènes et applications - 1968

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1 E0 ≤ C1∫ T0

∫ω̂

{(∇ut)2 +

((−∆)

s+22 u)2}

∥∥∥(−∆) s+22 u∥∥∥2L2(ω)

dt

3 E0 ≤ C2∫ T0‖u‖2Hs+2(ω)dt

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1 E0 ≤ C1∫ T0

∫ω̂

{(∇ut)2 +

((−∆)

s+22 u)2}

∥∥∥(−∆) s+22 u∥∥∥2L2(ω)

dt

3 E0 ≤ C2∫ T0‖u‖2Hs+2(ω)dt

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1 E0 ≤ C1∫ T0

∫ω̂

{(∇ut)2 +

((−∆)

s+22 u)2}

∥∥∥(−∆) s+22 u∥∥∥2L2(ω)

dt

3 E0 ≤ C2∫ T0‖u‖2Hs+2(ω)dt

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

BACKWARD SYSTEMytt + (−∆)s+1y = h(u)y |Rn\Ω ≡ 0y(x ,T ) = y ′(x ,T ) = 0

The solution of the backward system is defined by transposition: thefunction y is a solution of the problem if and only if for any solution θ of

θtt + (−∆)s+1θ = fθ|Rn\Ω ≡ 0θ(x , 0) = θ0(x)θt(x , 0) = θ1(x)

it holds ∫Qψfdxdt −

∫Ω

(ψt(0)θ0 − ψ(0)θ1)dx =∫

Qθhdxdt (6)

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

BACKWARD SYSTEMytt + (−∆)s+1y = h(u)y |Rn\Ω ≡ 0y(x ,T ) = y ′(x ,T ) = 0

The solution of the backward system is defined by transposition: thefunction y is a solution of the problem if and only if for any solution θ of

θtt + (−∆)s+1θ = fθ|Rn\Ω ≡ 0θ(x , 0) = θ0(x)θt(x , 0) = θ1(x)

it holds ∫Qψfdxdt −

∫Ω

(ψt(0)θ0 − ψ(0)θ1)dx =∫

Qθhdxdt (6)

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We define the operator

Λ(u0, u1) := (yt(x , 0),−y(x , 0))

by considering (6) with θ = u and choosing the control function

h(u) = uχ{ω×[0,T ]}

we obtain

〈Λ(u0, u1); (u0, u1)〉 =∫ T0

∫ω

|u|2dxdt.

Observability inequality ⇒Λ : H10 (Ω)× L2(Ω)→ H−1(Ω)× L2(Ω) is an isomorphism.

For any couple of initial data (y0, y1) ∈ L2(Ω)× H−1(Ω) there exists aunique solution (u0, u1) ∈ H10 (Ω)× L2(Ω) of Λ(u0, u1) = (y1,−y0)

The control function h ∈ L2(0,T ; L2(ω)) drives the system in rest intime T .

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We define the operator

Λ(u0, u1) := (yt(x , 0),−y(x , 0))

by considering (6) with θ = u and choosing the control function

h(u) = uχ{ω×[0,T ]}

we obtain

〈Λ(u0, u1); (u0, u1)〉 =∫ T0

∫ω

|u|2dxdt.

Observability inequality ⇒Λ : H10 (Ω)× L2(Ω)→ H−1(Ω)× L2(Ω) is an isomorphism.

For any couple of initial data (y0, y1) ∈ L2(Ω)× H−1(Ω) there exists aunique solution (u0, u1) ∈ H10 (Ω)× L2(Ω) of Λ(u0, u1) = (y1,−y0)

The control function h ∈ L2(0,T ; L2(ω)) drives the system in rest intime T .

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Work in progress

Pohozaev identity for the fractional laplacian (−∆)s+1u∫Ω

(−∆)s+1u (x · ∇u) dx = 2(s + 1)− n2

∫Ω

u(−∆)s+1udx

(7)

− Γ(2 + s)2

2

∫∂Ω

( uδs+1

)2(x · ν)dσ

Starting from (7) it is possible to repeat the analysis just presented andprove the control result for the fractional wave equation.

The observability inequality is proved with no need of interpolationtechniques.

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Work in progress

Pohozaev identity for the fractional laplacian (−∆)s+1u∫Ω

(−∆)s+1u (x · ∇u) dx = 2(s + 1)− n2

∫Ω

u(−∆)s+1udx

(7)

− Γ(2 + s)2

2

∫∂Ω

( uδs+1

)2(x · ν)dσ

Starting from (7) it is possible to repeat the analysis just presented andprove the control result for the fractional wave equation.

The observability inequality is proved with no need of interpolationtechniques.

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Fourier analysis

In order to guarantee an uniform velocity of propagation we need s ≥ 1/2in the Schrödinger equation and s ≥ 1 in the wave equation.

1-d fractional problems on (−1, 1)× (0,T )iut + (−d2x )1/2u = 0u|Rn\Ω ≡ 0u(x , 0) = u0(x)utt + (−d2x )1/2u = 0u|Rn\Ω ≡ 0u(x , 0) = u0(x)ut(x , 0) = u1(x)

=⇒ u(x , t) =∑k≥1

akφk(x)e iµk t

=⇒ u(x , t) =∑k≥1

akφk(x)e i√µk t

{µk , φk(x)}k≥1 are the eigenvalues and the eigenfunctions of the squareroot of the laplacian in 1-d on the interval (−1, 1).

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Fourier analysis

=⇒ u(x , t) =∑k≥1

akφk(x)e iµk t

=⇒ u(x , t) =∑k≥1

akφk(x)e i√µk t

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