Hyperbolic inverse problems and exact controllability Lauri Oksanen University College London
An inverse initial source problem Let M R n be a compact domain with smooth strictly convex boundary, and let c C (R n ) be strictly positive. Suppose that c = 1 in R n \ M. Let φ C0 (M) and consider the wave equation, t 2 u c(x) 2 u = 0, on (0, ) R n, u t=0 = φ, t u t=0 = 0, on R n. Inverse source problem. Given c and u (0,T ) M find φ.
An inverse boundary value problem Consider the wave equation on the domain M t 2 u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M, and define the Dirichlet-to-Neumann map Λ c f = ν u (0,T ) M, f C0 ((0, T ) M). Inverse boundary value problem. Given Λ c find c. Contrary to the above inverse source problem, the inverse boundary value problem here is non-linear.
Uniqueness for the inverse source problem Inverse source problem. Given c and u (0,T ) M find φ C 0 (M), where 2 t u c(x) 2 u = 0, on (0, ) R n, u t=0 = φ, t u t=0 = 0, on R n. The inverse source problem is uniquely solvable for large enough T > 0. We get ν u (0,T ) M by solving an exterior problem. Uniqueness follows from Tataru s hyperbolic unique continuation.
Stability for the inverse source problem Given c and u (0,T ) M find φ C 0 (M), where 2 t u c(x) 2 u = 0, on (0, ) R n, u t=0 = φ, t u t=0 = 0, on R n. The stability estimate φ H 1 C(c) u (0,T ) M H 1 holds if the Riemannian (M, g) manifold is non-trapping [Stefanov-Uhlmann 09]. Here the metric tensor g = c 2 dx 2 gives the travel times of waves. Notice that the constant C(c) depends on the sound speed c.
Inverse source problem: is reconstruction of both f and c possible? This is not known in general. The scale of c is determined in the following sense: if c(x) is of the form αβ(x), where α a constant and β is a known function, then α is determined by u (0,T ) M [Hristova-Kuchment-Nguyen 08]. The linearized problem to reconstruct f and c from u (0,T ) M is not stable [Stefanov-Uhlmann 12] (if possible to solve at all). If f is known, f 0 and M is foliated by strictly convex hypersurfaces w.r.t. the Riemannian metric g = c 2 dx 2 [Stefanov-Uhlmann 13] showed that c is determined.
Stability with respect to the sound speed Given c close to c and u (0,T ) M find φ C 0 (M), where t 2 u c(x) 2 u = 0, on (0, ) R n, u t=0 = φ, t u t=0 = 0, on R n. Th. [L.O.-Uhlmann] Suppose that the Riemannian manifold (M, c 2 dx 2 ) has a strictly convex function with no critical points. Let C > 0 and let K M int be compact and let us consider the functions φ and c satisfying φ H 3 (K) C, c C 2 (K) C, supp(φ) K, c = c in Rn \ K. There are ɛ, T, C > 0 such that if c satisfies also c c C 1 (K) ɛ then we can reconstuct φ such that φ φ C c c H 1 L u (0,T ) M 1/2 H 1.
Uniqueness for the inverse boundary value problem Inverse boundary value problem. Given Λ c find c, where Λ c f = ν u (0,T ) M and 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. The inverse boundary value problem is uniquely solvable for large enough T > 0. This can be proven by using the Boundary Control method [Belishev 87] together with the unique continuation [Tataru 95] or by using the complex geometric optics (CGO) solutions [Sylvester-Uhlmann 87]. Both the methods are expected to be exponentially unstable.
Stability for the inverse boundary value problem Given Λ c find c, where Λ c f = ν u (0,T ) M and 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. If the Riemannian manifold (M, c 2 dx 2 ) is simple, then the inverse boundary value problem is locally Hölder stable. That is, there is C > 0 and a neighborhood U of c such that c c C Λ c Λ c 1/2, c U. This can be proven by using geometric optics solutions [Stefanov-Uhlmann 98, Bellassoued-Dos Santos Ferreira 11]. The method is not a global reconstruction method.
Stability for a low frequency component Given Λ c find c, where Λ c f = ν u (0,T ) M and t 2 u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. Theorem [Liu-L.O.]. If the Riemannian manifold (M, c 2 dx 2 ) has a strictly convex function with no critical points, then the inverse problem is locally Lipschitz stable in low frequencies. That is, there are C, R > 0 and a neighborhood U of c such that F( c 2 c 2 )(ξ) Ce R ξ Λ c Λ c, c U. The method is a global reconstruction method.
Stability for the inverse source problem again Inverse source problem. Given c and u (0,T ) M find φ C 0 (M), where 2 t u c(x) 2 u = 0, on (0, ) R n, u t=0 = φ, t u t=0 = 0, on R n. The stability estimate φ H 1 C(c) u (0,T H ) M 1 holds if the Riemannian (M, c 2 dx 2 ) manifold is non-trapping. In the control theory literature, inequalities of the form initial data C boundary data are called continuous observability inequalities.
Exact controllability Control problem. Given c and φ L 2 (M) find f L 2 ((0, T ) M) such that u(t ) = φ where 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. The control problem can be solved stably if and only if the continuous observability inequality ψ L 2 C(c) ν v L (0,T ) M 2 holds, where t 2 v c(x) 2 v = 0, on (0, ) M, v x M = 0, on (0, ) M, v t=0 = 0, t v t=0 = ψ, on M.
The Boundary Control method Control problem using boundary data. Given Λ c and harmonic φ C (M) find f L 2 ((0, T ) M) such that u(t ) = φ where 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. The control problem can be solved stably if and only if the continuous observability inequality ψ L 2 C(c) ν v L (0,T ) M 2 holds, where t 2 v c(x) 2 v = 0, on (0, ) M, v x M = 0, on (0, ) M, v t=0 = 0, t v t=0 = ψ, on M.
About the BC method: Blagoveščenskĭı s identity The following identity is central to the Boundary Control method where u solves (u(t ), u(t )) L 2 (M;c 2 dx) = (f, Kf ) L 2 ((0,T ) M), 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M,, and K = K(Λ c ) = JΛ c RΛ c RJ. Here Rh(t) := h(t t) and Jh(t) := 1 2 2T t t h(s)ds.
About the BC method: a Blagoveščenskĭı type identity If φ C (M) is harmonic, then we have also the identity (u(t ), φ) L 2 (M;c 2 dx) = (f, Bφ) L 2 ( M (0,T )), where B = B(Λ c ) = RΛ c RI T 0 I T 1. Here Ih(t) := T t h(s)ds and T 0 and T 1 are the first two traces on M. Proof. Integration by parts gives 2 t (u(t), φ) L 2 (M;c 2 (x)dx) = ( u(t), φ) L 2 (M;dx) (u(t), φ) L 2 (M;dx) = (Λ c f (t), φ) L 2 ( M) (f (t), ν φ) L 2 ( M). By solving this differential equation with vanishing initial conditions at t = 0 we get the identity.
About the BC method: a Blagoveščenskĭı type identity If φ C (M) is harmonic, then we have also the identity (u(t ), φ) L 2 (M;c 2 dx) = (f, Bφ) L 2 ( M (0,T )), where B = B(Λ c ) = RΛ c RI T 0 I T 1. Here Ih(t) := T t h(s)ds and T 0 and T 1 are the first two traces on M. Proof. Integration by parts gives 2 t (u(t), φ) L 2 (M;c 2 (x)dx) = ( u(t), φ) L 2 (M;dx) (u(t), φ) L 2 (M;dx) = (Λ c f (t), φ) L 2 ( M) (f (t), ν φ) L 2 ( M). By solving this differential equation with vanishing initial conditions at t = 0 we get the identity.
About the BC method: the control problem again Control problem using boundary data. Given Λ c and harmonic φ C (M) find f L 2 ((0, T ) M) such that u(t ) = φ where 2 t u c(x) 2 u = 0, on (0, ) M, u x M = f, on (0, ) M, u t=0 = 0, t u t=0 = 0, on M. We can find a solution f by minimizing u(t ) φ 2 L 2 (M;c 2 (x)dx) = (f, Kf ) 2(f, Bφ) + φ 2 L 2 (M;c 2 (x)dx). Let ξ, η R n satisfy ξ = η and ξ η. Then the functions φ(x) := e i(ξ+iη) x/2, ψ(x) := e i(ξ iη) x/2 are harmonic. If f is a solution to the control problem u(t ) = φ then F(c 2 )(ξ) = (φ, ψ) L 2 (M;c 2 (x)dx) = (f, Bψ).
Stability for a low frequency component To show the stability estimate F( c 2 c 2 )(ξ) Ce R ξ Λ c Λ c, c U, we need to show that the constant C(c) in the continuous observability inequality, ψ L 2 C(c) ν v (0,T ) M L 2, is uniform with respect to the sound speed c. Here v solves 2 t v c(x) 2 v = 0, on (0, ) M, v x M = 0, on (0, ) M, v t=0 = 0, t v t=0 = ψ, on M.
Continuous observability Let us consider the continuous observability inequality w H 1 0 (M) L 2 (M) C(c) νv L 2 ( M (0,T )), (1) where w = (v(0), t v(0)) and v solves 2 t v c(x) 2 v = 0, on (0, ) M, v x M = 0, on (0, ) M. The inequality (1) holds if and only if (M, c 2 dx 2 ) is non-trapping [Bardos-Lebeau-Rauch 92, Burq-Gérard 97]. The proof is based on Microlocal analysis (propagation of singularities) to obtain w H 1 0 (M) L 2 (M) C νv L 2 ( M (0,T )) + Rw L 2 ( M (0,T )), where and R is a compact operator. Unique continuation to improve to (1). The second step loses all control over C(c).
Trying to perturb the continuous observability inequality Let v and ṽ be the solutions of the wave equations 2 t v c 2 v = 0, 2 t ṽ c 2 ṽ = 0, with the initial data w and vanishing boundary conditions. Then ṽ v satisfies the wave equation ( 2 t c(x) 2 )(ṽ v) = ( c 2 c 2 ) ṽ, on M (0, T ), with vanishing initial and boundary conditions. In particular, w H 1 0 (M) L 2 (M) C(c) νv L 2 ( M (0,T )) C(c) ν ṽ L 2 ( M (0,T )) + C(c) νv ν ṽ L 2 ( M (0,T )) C(c) ν ṽ L 2 ( M (0,T )) + δ ṽ L 1 (0,T ;L 2 (M)) C(c) ν ṽ L 2 ( M (0,T )) + δ w H 2 (M) H 1 0 (M), where δ > 0 is small if c is close to c. However, δ w on the RHS can not be absorbed in the LHS as we have lost one degree of smoothness.
Continuous observability inequality with explicit constants Let (M, g) be a smooth Riemannian manifold with a boundary and consider the wave equation 2 t u(t, x) µ u(t, x) = 0, (t, x) (0, ) M, u(t, x) = 0, where µ is the weighted Laplace-Beltrami operator, and µ C (M) is strictly positive. µ u = µ 1 div(µ u), (t, x) (0, ) M, Notice that if g = c(x) 2 dx 2 and µ(x) = c(x) n 2, then µ = c(x) 2, where is the Euclidean Laplacean.
Continuous observability inequality with explicit constants Theorem [Liu-L.O.]. Suppose that there is a strictly convex function l C 3 (M) with no critical points on (M, g). Let ρ, r > 0 satisfy D 2 l(x, X ) ρ X 2, l(x) r, for all X T x M and x M, and define Γ := {x M; ( l, ν) > 0}. Suppose that ( ) T > 2(C 3 τ + C 4 )e (B l β l 3 )τ τ, where τ = max ρ, C 1 2ρr 2. Let u C 2 ([0, T ] M) be a solution of the wave equation with homogeneous Dirichlet boundary condition. Then T u(0) 2 L 2 (M) + tu(0) 2 L 2 (M) C(T ) ( ν u) 2 µdsdt, where C(T ) = C 2 e (B l β l )τ τ T 2(C 3 τ+c 4 )e (B l β l )τ τ. 0 Γ