Evolution problems involving the fractional Laplace operator: HUM control and Fourier analysis

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$Evolution problems involving the fractional Laplace$ joint work with Enrique Zuazua BCAM - Basque Center for

1 Evolution problems involving the fractional Laplace operator: HUM control and Fourier analysis Umberto Biccari joint work with Enrique Zuazua BCAM - Basque Center for Applied Mathematics NUMERIWAVES group meeting February 28, / 24

2 We analyse the control problem for the fractional Schrödinger equation and for the fractional wave equation iu t + ( ) s u = u tt + ( ) s+1 u = on a bounded C 1,1 domain Ω of R n. We focus on the control from a neighbourhood of the boundary Ω. Fractional laplacian ( ) s u(x) := c n,s P.V. R n u(x) u(y) dy, s (, 1) x y n+2s c n,s := s22s Γ( n+2s 2 ) π n/2 Γ(1 s) 2 / 24

3 We analyse the control problem for the fractional Schrödinger equation and for the fractional wave equation iu t + ( ) s u = u tt + ( ) s+1 u = on a bounded C 1,1 domain Ω of R n. We focus on the control from a neighbourhood of the boundary Ω. Fractional laplacian ( ) s u(x) := c n,s P.V. R n u(x) u(y) dy, s (, 1) x y n+2s c n,s := s22s Γ( n+2s 2 ) π n/2 Γ(1 s) 2 / 24

4 Fractional Schrödinger equation iu t + ( ) s u = in Ω [, T ] := Q u in (R n \ Ω) [, T ] u(x, ) = u (x) in Ω (1) Fractional wave equation u tt + ( ) s+1 u = in Ω [, T ] := Q u in (R n \ Ω) [, T ] u(x, ) = u (x) in Ω u t (x, ) = u 1 (x) in Ω (2) Thanks to Hille-Yosida theorem, both problems are well posed. 3 / 24

5 Fractional Schrödinger equation iu t + ( ) s u = in Ω [, T ] := Q u in (R n \ Ω) [, T ] u(x, ) = u (x) in Ω (1) Fractional wave equation u tt + ( ) s+1 u = in Ω [, T ] := Q u in (R n \ Ω) [, T ] u(x, ) = u (x) in Ω u t (x, ) = u 1 (x) in Ω (2) Thanks to Hille-Yosida theorem, both problems are well posed. 3 / 24

6 Internal control result - Schrödinger equation Let Ω be a bounded C 1,1 domain of R n. Let us consider the nonhomogeneous fractional Schrödinger equation iy t + ( ) s y = hχ ω [,T ] in Ω [, T ] := Q y on (R n \ Ω) [, T ] y(x, ) = y (x) in Ω where ω is a neighbourhood of the boundary of the domain and χ is the characteristic function. Theorem Let T > and ω Ω be a neighbourhood of the boundary of the domain. Then, for any y L 2 (Ω) there exists h L 2 (ω [, T ]) such that the solution of (3) satisfies y(x, T ) =. (3) 4 / 24

7 Proposition For any solution u of iu t + ( ) s u = in Ω [, T ] := Q u on (R n \ Ω) [, T ] u(x, ) = u (x) in Ω it holds Γ(1 + s) 2 Σ ( ) 2 u T δ s (x ν)dσdt = 2s ( ) s/2 u T + I ū(x u)dx Ω 2 L 2 (Ω)dt Σ := Ω [, T ] δ = δ(x) := d(x, Ω) 5 / 24

8 Proof. The identity is obtained by multiplying the equation by x ū + n 2 ū, taking the real part and integrating over Q by using the Pohozaev identity ( ) s u(x u)dx = 2s n u( ) s udx Ω 2 Ω (4) Γ(1 + ( s)2 u ) 2 (x ν)dσ 2 δ s X. ROS-OTON and J. SERRA - The Pohozaev identity for the Fractional laplacian Ω 6 / 24

9 Theorem For any u solution of (1) there exist two non negative constants A 1 and A 2, depending only on n, s, T and Ω, such that A 1 u 2 H s (Ω) Σ ( ) 2 u δ s (x ν)dσdt A 2 u 2 H s (Ω) Proof. The proof is merely technical. We use some interpolation results and some compactness-uniqueness arguments. 7 / 24

10 Control from a neighbourhood of the boundary We will use Hilbert Uniqueness Method Observability inequality T u 2 L 2 (Ω) C u 2 dxdt ω 8 / 24

11 Proof of the observability inequality 1 u 2 H s (Ω) C 1 2 u 2 H s (Ω) C 2 T T u 2 H s (Ω) C 3 3 u 2 H s (Ω) C 4 T u 2 dxdt ˆω (Ω ˆω) ω ( ) u t H s (ω) + u L 2 (ω) dt T u t 2 H s (ω) dt u 2 H s (ω) dt 4 u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)), u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)) 5 The proof is concluded by interpolation. J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

12 Proof of the observability inequality 1 u 2 H s (Ω) C 1 2 u 2 H s (Ω) C 2 T T u 2 H s (Ω) C 3 3 u 2 H s (Ω) C 4 T u 2 dxdt ˆω (Ω ˆω) ω ( ) u t H s (ω) + u L 2 (ω) dt T u t 2 H s (ω) dt u 2 H s (ω) dt 4 u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)), u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)) 5 The proof is concluded by interpolation. J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

13 Proof of the observability inequality 1 u 2 H s (Ω) C 1 2 u 2 H s (Ω) C 2 T T u 2 H s (Ω) C 3 3 u 2 H s (Ω) C 4 T u 2 dxdt ˆω (Ω ˆω) ω ( ) u t H s (ω) + u L 2 (ω) dt T u t 2 H s (ω) dt u 2 H s (ω) dt 4 u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)), u 2 H s (Ω) C 5 u 2 L 2 (,T ;H s (ω)) 5 The proof is concluded by interpolation. J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

14 Control result via HUM iu t + ( ) s u = u R n \Ω u(x, ) = u (x) BACKWARD SYSTEM iy t + ( ) s y = uχ ω [,T ] y R n \Ω y(x, T ) = The solution of the backward system is defined by transposition with a solution θ of iθ t + ( ) s θ = f θ R n \Ω θ(x, ) = θ J.L. LIONS and E. MAGENES - Problèmes aux limites non homogènes et applications / 24

15 Control result via HUM iu t + ( ) s u = u R n \Ω u(x, ) = u (x) BACKWARD SYSTEM iy t + ( ) s y = uχ ω [,T ] y R n \Ω y(x, T ) = The solution of the backward system is defined by transposition with a solution θ of iθ t + ( ) s θ = f θ R n \Ω θ(x, ) = θ J.L. LIONS and E. MAGENES - Problèmes aux limites non homogènes et applications / 24

16 We consider the operator Λ : L 2 (Ω) L 2 (Ω) defined by It is immediate to check the identity Λu := iy(x, ). Λu ; u = T ω u 2 dxdt Thus, thanks to the observability inequality, Λ is an isomorphism from L 2 (Ω) to L 2 (Ω). Hence, given y L 2 (Ω) we can choose the control function h := u ω where u is the solution of (1) with intial datum u = Λ 1 ( iy ). The proof is concluded. 11 / 24

17 We consider the operator Λ : L 2 (Ω) L 2 (Ω) defined by It is immediate to check the identity Λu := iy(x, ). Λu ; u = T ω u 2 dxdt Thus, thanks to the observability inequality, Λ is an isomorphism from L 2 (Ω) to L 2 (Ω). Hence, given y L 2 (Ω) we can choose the control function h := u ω where u is the solution of (1) with intial datum u = Λ 1 ( iy ). The proof is concluded. 11 / 24

18 We consider the operator Λ : L 2 (Ω) L 2 (Ω) defined by It is immediate to check the identity Λu := iy(x, ). Λu ; u = T ω u 2 dxdt Thus, thanks to the observability inequality, Λ is an isomorphism from L 2 (Ω) to L 2 (Ω). Hence, given y L 2 (Ω) we can choose the control function h := u ω where u is the solution of (1) with intial datum u = Λ 1 ( iy ). The proof is concluded. 11 / 24

19 Fractional wave equation In order to guarantee uniform velocity of propagation we need the exponent of the Fractional laplacian to be greater than 1. Higher order Fractional laplacian ( ) s+1 := ( ) s ( ) D(( ) s+1 ) = H 3 (Ω) H 2(s+1) (Ω) 12 / 24

20 Fractional wave equation In order to guarantee uniform velocity of propagation we need the exponent of the Fractional laplacian to be greater than 1. Higher order Fractional laplacian ( ) s+1 := ( ) s ( ) D(( ) s+1 ) = H 3 (Ω) H 2(s+1) (Ω) 12 / 24

21 Internal control result Let Ω be a bounded C 1,1 domain of R n. Let us consider the nonhomogeneous fractional wave equation y tt + ( ) s+1 y = hχ ω [,T ] in Ω [, T ] := Q y on (R n \ Ω) [, T ] y(x, ) = y (x) in Ω y t (x, ) = y 1 (x) in Ω (5) where ω is a neighbourhood of the boundary of the domain and χ is the characteristic function. Theorem Let T > and ω Ω be a neighbourhood of the boundary of the domain. Then, for any couple of initial data (y, y 1 ) H 1 (Ω) L2 (Ω) there exists h L 2 (ω [, T ]) such that the solution of (5) satisfies y(x, T ) = y t (x, T ) =. 13 / 24

22 Proposition For any solution u of u tt + ( ) s+1 u = in Ω [, T ] := Q u on (R n \ Ω) [, T ] u(x, ) = u (x) in Ω u t (x, ) = u 1 (x) in Ω it holds Γ(1 + s) 2 Σ ( ) 2 u δ s (x ν)dσdt = 2s n ( ) 2 ( ) s+2 2 u dxdt 2 Q n ( u t ) 2 dxdt 2 Q T + u t (x ( u))dx Ω Σ := Ω [, T ] δ = δ(x) := d(x, Ω) 14 / 24

23 Proof. The identity is obtained by multiplying the equation by x ( u) and integrating over Q by using the Pohozaev identity for the fractional laplacian (4) applied to ( u). Theorem (Energy estimate) For any solution of (2) we define the energy as E(t) := 1 { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dx; 2 Ω for any T > there exists two non negative constants A 1 and A 2, depending only on s, T and Ω, such that A 1 E Σ ( ) 2 u (x ν)dσdt A 2 E δ s 15 / 24

24 Proof. The identity is obtained by multiplying the equation by x ( u) and integrating over Q by using the Pohozaev identity for the fractional laplacian (4) applied to ( u). Theorem (Energy estimate) For any solution of (2) we define the energy as E(t) := 1 { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dx; 2 Ω for any T > there exists two non negative constants A 1 and A 2, depending only on s, T and Ω, such that A 1 E Σ ( ) 2 u (x ν)dσdt A 2 E δ s 15 / 24

25 Control from a neighbourhood of the boundary We will use Hilbert Uniqueness Method Observability inequality E C T ω u 2 dxdt 16 / 24

26 Proof of the observability inequality T 1 E C 1 ˆω { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dxdt (Ω ˆω) ω 2 equipartition of the energy E C 2 T T 3 E C 2 u 2 H s+2 (ω) dt 4 The proof is concluded by interpolation. ( ) s u L 2 (ω) dt J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

27 Proof of the observability inequality T 1 E C 1 ˆω { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dxdt (Ω ˆω) ω 2 equipartition of the energy E C 2 T T 3 E C 2 u 2 H s+2 (ω) dt 4 The proof is concluded by interpolation. ( ) s u L 2 (ω) dt J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

28 Proof of the observability inequality T 1 E C 1 ˆω { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dxdt (Ω ˆω) ω 2 equipartition of the energy E C 2 T T 3 E C 2 u 2 H s+2 (ω) dt 4 The proof is concluded by interpolation. ( ) s u L 2 (ω) dt J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

29 Proof of the observability inequality T 1 E C 1 ˆω { ( ) } 2 ( u t ) 2 + ( ) s+2 2 u dxdt (Ω ˆω) ω 2 equipartition of the energy E C 2 T T 3 E C 2 u 2 H s+2 (ω) dt 4 The proof is concluded by interpolation. ( ) s u L 2 (ω) dt J.L. LIONS and E. MAGENES Problèmes aux limites non homogènes et applications / 24

30 Control result via HUM u tt + ( ) s+1 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) BACKWARD SYSTEM y tt + ( ) s+1 y = h(u) y Rn \Ω y(x, T ) = y (x, T ) = The solution of the backward system is defined by transposition: the function y is a solution of the problem if and only if for any solution θ of it holds Q θ tt + ( ) s+1 θ = f θ Rn \Ω θ(x, ) = θ (x) θ t (x, ) = θ 1 (x) ψfdxdt (ψ t ()θ ψ()θ 1 )dx = θhdxdt (6) Ω Q 18 / 24

31 Control result via HUM u tt + ( ) s+1 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) BACKWARD SYSTEM y tt + ( ) s+1 y = h(u) y Rn \Ω y(x, T ) = y (x, T ) = The solution of the backward system is defined by transposition: the function y is a solution of the problem if and only if for any solution θ of it holds Q θ tt + ( ) s+1 θ = f θ Rn \Ω θ(x, ) = θ (x) θ t (x, ) = θ 1 (x) ψfdxdt (ψ t ()θ ψ()θ 1 )dx = θhdxdt (6) Ω Q 18 / 24

32 We define the operator Λ(u, u 1 ) := (y t (x, ), y(x, )) by considering (6) with θ = u and choosing the control function h(u) = uχ {ω [,T ]} we obtain Λ(u, u 1 ); (u, u 1 ) = T ω u 2 dxdt. Observability inequality Λ : H 1 (Ω) L2 (Ω) H 1 (Ω) L 2 (Ω) is an isomorphism. For any couple of initial data (y, y 1 ) L 2 (Ω) H 1 (Ω) there exists a unique solution (u, u 1 ) H 1 (Ω) L2 (Ω) of Λ(u, u 1 ) = (y 1, y ) The control function h L 2 (, T ; L 2 (ω)) time T. drives the system in rest in 19 / 24

33 We define the operator Λ(u, u 1 ) := (y t (x, ), y(x, )) by considering (6) with θ = u and choosing the control function h(u) = uχ {ω [,T ]} we obtain Λ(u, u 1 ); (u, u 1 ) = T ω u 2 dxdt. Observability inequality Λ : H 1 (Ω) L2 (Ω) H 1 (Ω) L 2 (Ω) is an isomorphism. For any couple of initial data (y, y 1 ) L 2 (Ω) H 1 (Ω) there exists a unique solution (u, u 1 ) H 1 (Ω) L2 (Ω) of Λ(u, u 1 ) = (y 1, y ) The control function h L 2 (, T ; L 2 (ω)) time T. drives the system in rest in 19 / 24

34 Work in progress Pohozaev identity for the fractional laplacian ( ) s+1 u ( ) s+1 2(s + 1) n u (x u) dx = u( ) s+1 udx 2 Ω Γ(2 + s)2 2 Ω Ω ( u δ s+1 ) 2 (x ν)dσ (7) Starting from (7) it is possible to repeat the analysis just presented and prove the control result for the fractional wave equation. The observability inequality is proved with no need of interpolation techniques. 2 / 24

35 Work in progress Pohozaev identity for the fractional laplacian ( ) s+1 u ( ) s+1 2(s + 1) n u (x u) dx = u( ) s+1 udx 2 Ω Γ(2 + s)2 2 Ω Ω ( u δ s+1 ) 2 (x ν)dσ (7) Starting from (7) it is possible to repeat the analysis just presented and prove the control result for the fractional wave equation. The observability inequality is proved with no need of interpolation techniques. 2 / 24

36 Fourier analysis In order to guarantee an uniform velocity of propagation we need s 1/2 in the Schrödinger equation and s 1 in the wave equation. 1-d fractional problems on ( 1, 1) (, T ) iu t + ( d 2 x ) 1/2 u = u R n \Ω u(x, ) = u (x) u tt + ( d 2 x ) 1/2 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) = u(x, t) = k 1 = u(x, t) = k 1 a k φ k (x)e iµ k t a k φ k (x)e i µ k t {µ k, φ k (x)} k 1 are the eigenvalues and the eigenfunctions of the square root of the laplacian in 1-d on the interval ( 1, 1). 21 / 24

37 Fourier analysis In order to guarantee an uniform velocity of propagation we need s 1/2 in the Schrödinger equation and s 1 in the wave equation. 1-d fractional problems on ( 1, 1) (, T ) iu t + ( d 2 x ) 1/2 u = u R n \Ω u(x, ) = u (x) u tt + ( d 2 x ) 1/2 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) = u(x, t) = k 1 = u(x, t) = k 1 a k φ k (x)e iµ k t a k φ k (x)e i µ k t {µ k, φ k (x)} k 1 are the eigenvalues and the eigenfunctions of the square root of the laplacian in 1-d on the interval ( 1, 1). 21 / 24

38 Fourier analysis In order to guarantee an uniform velocity of propagation we need s 1/2 in the Schrödinger equation and s 1 in the wave equation. 1-d fractional problems on ( 1, 1) (, T ) iu t + ( d 2 x ) 1/2 u = u R n \Ω u(x, ) = u (x) u tt + ( d 2 x ) 1/2 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) = u(x, t) = k 1 = u(x, t) = k 1 a k φ k (x)e iµ k t a k φ k (x)e i µ k t {µ k, φ k (x)} k 1 are the eigenvalues and the eigenfunctions of the square root of the laplacian in 1-d on the interval ( 1, 1). 21 / 24

39 Fourier analysis In order to guarantee an uniform velocity of propagation we need s 1/2 in the Schrödinger equation and s 1 in the wave equation. 1-d fractional problems on ( 1, 1) (, T ) iu t + ( d 2 x ) 1/2 u = u R n \Ω u(x, ) = u (x) u tt + ( d 2 x ) 1/2 u = u Rn \Ω u(x, ) = u (x) u t (x, ) = u 1 (x) = u(x, t) = k 1 = u(x, t) = k 1 a k φ k (x)e iµ k t a k φ k (x)e i µ k t {µ k, φ k (x)} k 1 are the eigenvalues and the eigenfunctions of the square root of the laplacian in 1-d on the interval ( 1, 1). 21 / 24

40 Spectral analysis for the square root of the Laplacian T. KULCZYCKI, M. KWA`SNICKI, J. MALECKI and E. STOS - Spectral properties of the Cauchy process on half-line and interval. µ k ( kπ 2 π ) = O 8 ( ) 1 k for k 1 (( kπ φ k sin 2 π ) (1 + x) + π ) ( ) 1 = O k 8 8 L 2 ( 1,1) for k 1 22 / 24

41 Gap between the eigenvalues Schrödinger equation [ (k + 1)π µ k+1 µ k π ] [ kπ π ] = π 8 2 Wave equation µk+1 µ k (k + 1)π 2 General case β (, 1) ( d 2 x ) β µ k = ( kπ 2 π 8 kπ 2 π 8 ) 2β (2 2β)π + O 8 ( ) 1 k 23 / 24

42 Gap between the eigenvalues Schrödinger equation [ (k + 1)π µ k+1 µ k π ] [ kπ π ] = π 8 2 Wave equation µk+1 µ k (k + 1)π 2 General case β (, 1) ( d 2 x ) β µ k = ( kπ 2 π 8 kπ 2 π 8 ) 2β (2 2β)π + O 8 ( ) 1 k 23 / 24

43 THANKS FOR YOUR ATTENTION! 24 / 24

Hilbert Uniqueness Method and regularity

Hilbert Uniqueness Method and regularity Sylvain Ervedoza 1 Joint work with Enrique Zuazua 2 1 Institut de Mathématiques de Toulouse & CNRS 2 Basque Center for Applied Mathematics Institut Henri Poincaré