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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
1 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
2 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
2 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
3 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
3 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
4 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
5 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
6 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
7 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Control from a neighbourhood of the boundary
We will use Hilbert Uniqueness Method
Observability inequality
‖u0‖2L2(Ω) ≤ C∫ T0
∫ω
|u|2dxdt
8 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
10 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
10 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
11 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
11 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
11 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
12 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
12 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
13 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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)
14 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
15 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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
15 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Control from a neighbourhood of the boundary
We will use Hilbert Uniqueness Method
Observability inequality
E0 ≤ C∫ T0
∫ω
|u|2dxdt
16 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Proof of the observability inequality
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
4 The proof is concluded by interpolation.
J.L. LIONS and E. MAGENES
Problèmes aux limites non homogènes et applications - 1968
17 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Proof of the observability inequality
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
4 The proof is concluded by interpolation.
J.L. LIONS and E. MAGENES
Problèmes aux limites non homogènes et applications - 1968
17 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Proof of the observability inequality
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
4 The proof is concluded by interpolation.
J.L. LIONS and E. MAGENES
Problèmes aux limites non homogènes et applications - 1968
17 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Proof of the observability inequality
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
4 The proof is concluded by interpolation.
J.L. LIONS and E. MAGENES
Problèmes aux limites non homogènes et applications - 1968
17 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Control result via HUM
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)
18 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Control result via HUM
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)
18 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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 .
19 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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 .
19 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
20 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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.
20 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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).
21 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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).
21 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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).
21 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
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).
21 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Spectral analysis for the square root of the Laplacian
T. KULCZYCKI, M. KWAS̀NICKI, J. MALECKI and E. STOS - Spectral properties of the
Cauchy process on half-line and interval.∣∣∣∣µk − (kπ2 − π8)∣∣∣∣ = O ( 1k
)for k ≥ 1
∥∥∥∥φk − sin((kπ2 − π8)
(1 + x) +π
8
)∥∥∥∥L2(−1,1)
= O(
1√k
)for k ≥ 1
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Fractional Schrödinger equation Fractional wave equation Fourier analysis
Gap between the eigenvalues
Schrödinger equation
µk+1 − µk →[
(k + 1)π2
− π8
]−[kπ2− π
8
]=π
2
Wave equation
√µk+1 −
√µk →
√(k + 1)π
2− π
8−√
kπ2− π
8→ 0
General case β ∈ (0, 1)
(−d2x )β ⇒ µk =(kπ2− (2− 2β)π
8
)2β+ O
(1k
)
23 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
Gap between the eigenvalues
Schrödinger equation
µk+1 − µk →[
(k + 1)π2
− π8
]−[kπ2− π
8
]=π
2
Wave equation
√µk+1 −
√µk →
√(k + 1)π
2− π
8−√
kπ2− π
8→ 0
General case β ∈ (0, 1)
(−d2x )β ⇒ µk =(kπ2− (2− 2β)π
8
)2β+ O
(1k
)
23 / 24
Fractional Schrödinger equation Fractional wave equation Fourier analysis
THANKS FOR YOUR ATTENTION!
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Fractional Schrödinger equationFractional wave equationFourier analysis