Inverse problems for hyperbolic PDEs

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Inverse problems for hyperbolic PDEs Lauri Oksanen University College London

Example: inverse problem for the wave equation Let c be a smooth function on Ω R n and consider the wave equation t 2 u c 2 u = 0, in (0, T ) Ω, ν u (0,T ) Ω = f, u t=0 = t u t=0 = 0, and define the Neumann-to-Dirichlet map by Λf = u (0,T ) Ω. Theorem [Belishev 87]. Λ determines c uniquely for large enough T. Present challenges: Uniqueness for more general Lorentzian geometries Computational methods

More general Lorentzian geometries For a Lorentzian metric g on M = (0, T ) Ω, we define the wave operator g u = g 1/2 n ( x j g jk g 1/2 ) x k u. j,k=0 Writing x = (t, y) and g(x) = dt 2 + c 2 (y)dy 2, we have g u = 2 t u + c 2 (y) y u + lower order terms. Inverse problem for g. Determine g up to an isometry, given C(g) = {(u M, ν u M ); g u = 0, u C (M)}. The problem is open even when g is close to the Minkowski metric. The analogous elliptic inverse problem, with a Riemannian g close to the Euclidean metric, is also open.

The problem in a fixed conformal class Suppose now that g = ah where h is a known Lorentzian metric on M, and a is a strictly positive function on M. Inverse problem for a. Given h and the Cauchy data set C(g) = {(u M, ν u M ); g u = 0, u C (M)}. determine the conformal factor a. The gauge transformation, a (n+2)/4 g (a (n 2)/4 u) = h u + qu, q = a (n+2)/4 ( g a (n 2)/4 ), can be used to reduce determination of a conformal factor a to determination of a potential q.

Determination of a potential Suppose that h is of the product form h = dt 2 + g 0 (y), (t, y) M = (0, T ) Ω, where g 0 is a smooth Riemannian metric on Ω. Theorem [Kian-L.O. 17]. Suppose that (Ω, g 0 ) is simple. Then g 0 and C(q) = {(u M, ν u M ); h u + qu = 0, u C (M)}, determine the time-dependent potential q. Work in progress with Feizmohammadi, Ilmavirta and Kian: Non-smooth q Any g 0 such that the geodesic ray transform is injective on (Ω, g 0 ) Determination of a first order perturbation (i.e. magnetic potential)

Time independent problems Let (M 0, g 0 ) be a compact Riemannian manifold with boundary, and let E be a Hermitian vector bundle over M 0. Let be a connection on E and let Q C (M 0 ; End(E)). Suppose that and Q are symmetric: d u, v E = u, v E + u, v E, Qu, v E = u, Qv E for u, v C (M 0 ; E). Consider the local Cauchy data C(M 0, g 0, E,, Q) = {(u Γ, ν u Γ ); 2 t u u + Qu = 0, u M\Γ = 0}, where M = R M 0, Γ = R Γ 0, with Γ 0 M 0 open and non-empty, and is the adjoint of with respect to g 0. Theorem [Kurylev-L.O.-Paternain 18]. The set C(M 0, g 0, E,, Q) determines the structure (M 0, g 0, E,, Q) up to the gauge invariances.

Two proof strategies Theorem [Kian-L.O. 17]. Suppose that (M 0, g 0 ) is simple. Then g 0 and C(q) determine the time-dependent potential q. Use geometric optics to recover the light ray transform of q Reduce the light ray transform to the geodesic ray transform on (M 0, g 0 ) (uses the product structure) Invert the geodesic ray transform (uses simplicity of (M 0, g 0 )) Theorem [Kurylev-L.O.-Paternain 18]. The set C(M 0, g 0, E,, Q) determines the time-independent structure (M 0, g 0, E,, Q). The Boundary Control method [Belishev 87] based on hyperbolic unique continuation [Tataru 95] 1 1 Fails for smooth time-dependent coefficients [Alinhac 83]

The Minkowski space Consider R 1+3 with the Minkowski metric g = dt 2 + dy 2. As a matrix g = 1 1 1 1. y θ β y,θ A vector v = (v 0, v 1, v 2, v 3 ) is lightlike if the scalar product (v, v) g = v j g jk v k vanishes, and a line (i.e. a geodesic) is called a light ray if its tangent vector is lightlike. Writing S 2 = {θ R 3 ; θ = 1}, the light rays can be parametrized by β y,θ (s) = (s, y + sθ), (y, θ) R 3 S 2. Example: taking y = 0 and θ = (1, 0, 0) gives β(s) = (s, s, 0, 0).

Example: geometric optics in the Minkowski space Let g be the Minkowski metric on R 1+3 and consider the ansatz u(t, x) e iσ(t x1) (a(t, x) + iσ 1 b(t, x)), σ 1. The first order terms w.r.t. σ in g u + qu = 0 vanish when a satisfies the transport equation t a + x 1a = 0. Choosing a(t, x) = h(t x 1 )χ(x 2, x 3 ), with h δ and χ δ, focuses a on the light ray β(s) = (s, s, 0, 0). The zeroth order terms vanish when t b + x 1b = g a + qa. After eliminating the known term g a in t b + x 1b = g a + qa, we can read from b the integral (qa)(s, s t + x 1, x 2, x 3 )ds q(s, s, 0, 0)ds = q. R R β

Example: light ray transform in the Minkowski space Using the parametrization β y,θ (s) = (s, y + sθ), (y, θ) R 3 S 2, of light rays, the light ray transform is given by Lq(y, θ) = q = q(s, y + sθ)ds, (y, θ) R 3 S 2. β y,θ R Writing (s, y + sθ) = x = (x 0, x ), we have the Fourier slicing e iη y Lq(y, θ) dy = e iη (x x0θ) q(x) dx = q( η θ, η). R 3 R 4 Here η θ η, and choosing θ as a suitable linear combination of η and a vector orthogonal to it, we recover q(r η, η) for any r [ 1, 1]. The Fourier transform q(ξ) can be recovered for spacelike and lightlike ξ, that is, for ξ = (ξ 0, ξ ) satisfying ξ 0 ξ. If q is compactly supported, it can be recovered using analytic continuation [Stefanov 89].

Two proof strategies and convexity For time-dependent potential q. Recover the light ray transform Lq(β) = β q, with β a light ray. Then invert L. For time-independent structure (M 0, g 0, E,, Q). Use hyperbolic unique continuation that is sharp in time.

Inverse problems for non-linear wave equations Consider a 1 + 3-dimensional, globally hyperbolic Lorentzian manifold without boundary (M, g), and let V M be a bounded neighbourhood of a timelike curve µ : [0, 1] M. Moreover, let W be the intersection of the chronological future of µ(0) and past of µ(1). Theorem (IP with u 2 non-linearity) [Kurylev-Lassas-Uhlmann 17]. For any k N and r > 0 the data set D k,r (M, g) = {(u V, f ); g u + u 2 = f, u t<0 = 0, f C k 0 (V ), f r} determines (W, g) up to an isometry and a conformal rescaling.

Inverse problem for the Einstein-scalar field equations Theorem [Kurylev-Lassas-L.O.-Uhlmann]. Let M = R M 0 and ĝ = β(t, y)dt 2 + κ(t, y), where β is strictly positive and κ(t, ) is a Riemannian metric on M 0. Let ˆφ be scalar fields such that the Einstein-scalar field equations Ein(g) = T(g, φ) + f 1, g φ = f 2, (1) hold with (g, φ, f ) = (ĝ, ˆφ, 0), and suppose that d ˆφ l, l = 1, 2, 3, 4, are linearly independent near V. Then, for any k N and r > 0, and for a suitable pull-back to Fermi coordinates Φ g, the data set D k,r (M, ĝ, ˆφ) = Φ g {(g V, φ V, f ); (1) holds, g t<0 = ĝ, φ t<0 = ˆφ, f C k 0 (V ), f r} determines (W, ĝ) up to an isometry and a conformal rescaling.

Non-linearity in the Einstein-scalar field equations In the wave gauge, the Einstein-scalar field equations can be written as Ricĝ (g) dφ dφ = f 1, in M, g φ = f 2, in M, g = ĝ, φ = φ, in (, 0) M 0, where the reduced Ricci tensor Ricĝ (g) is given locally by Ricĝ (g) jk = 1 2 g pq p q g jk + g pq g rs (Γ prj Γ qsk + Γ prj Γ qks + Γ prk Γ qjs ) + j Ĥ k, Γ pkq = 1 2 ( pg kq + q g pk k g pq ), Ĥ k = g kr g pqˆγ r pq. Here j is the covariant derivative with respect to g in the coordinate direction j, and Γ r pq are the Christoffel symbols of ĝ.

Strategy of the proof Use again geometric optics but replace the WKB ansatz e iσφ a with conormal distributions. Linearize u(ɛ) = (g(ɛ), φ(ɛ)) with respect to a multi-dimensional family of sources f = J j=1 ɛ jf j. Here ɛ = (ɛ 1,..., ɛ J ). Choose each f j so that u j = ɛj u ɛ=0 is a geometric optics solution. Analyse the interaction of u j, j = 1,..., J, via the non-linearity, and show that singularities propagating back to V can be generated. Schematic. Geometric optics solutions u j focus near null geodesics (black curves). When they overlap, the non-linear terms cause new propagating singularities to appear (in red).

Example: quadratic non-linearity Consider the solution u(ɛ) = u of Then u j = ɛj u ɛ=0 satisfies and u 12 = ɛ1 ɛ2 u ɛ=0 satisfies g u u2 2 = ɛ 1f 1 + ɛ 2 f 2, u t<0 = 0. g u j = f j, u j t<0 = 0, g u 12 = u 1 u 2, u j t<0 = 0. How nice singularities in u 1 and u 2 behave under the product u 1 u 2?

Products of conormal distributions [Greenleaf-Uhlmann 93] Let K M be a submanifold. Recall [Hörmander 71]: a distribution 1/2-density u D (M) is in I (K) if it is smooth away from K and in local coordinates x = (x, x ) R N+(n N), with K = {x = 0}, it is of the form u, φ = R N iξ x e R n a(x, ξ )φ(x)dxdξ. We write σ[u] = a for the symbol, a 1/2-density on the conormal bundle, N K = {(x, ξ) T M; x K, ξv = 0 for v T x K} = {x = 0, ξ = 0}. Let K 1, K 2 M be transversal submanifolds, and let u j I (K j ). Then away from N K 1 N K 2 it holds that u 1 u 2 I (K 1 K 2 ) and σ[u 1 u 2 ](x, ξ) = σ[u 1 ](x, ξ (1) )σ[u 2 ](x, ξ (2) ), where ξ = ξ (1) + ξ (2) N x (K 1 K 2 ) = N x K 1 N x K 2.

Example: 4th order non-linearity Recall that we have assumed dim M = 1 + 3. Let K j M, j = 1, 2, 3, 4, be of codim 1. If x 4 j=1 K j and u j I (K j ), then generically T x M = N x K 1 N x K 4, σ[u 1 u 2 u 3 u 4 ](x, ξ) = The easiest non-linearity to analyse is 4 g u u4 4! = ɛ j f j, u t<0 = 0. j=1 Writing u j = ɛj u ɛ=0 and w = ɛ1... ɛ4 u ɛ=0 we have g u j = f j, g w = u 1 u 2 u 3 u 4. 4 σ[u j ](x, ξ (j) ). Near x, outside a small exceptional set, WF (u 1 u 2 u 3 u 4 ) = WF (δ x ) = T x M. j=1

Summary of uniqueness results The inverse problem to recover q given the Cauchy data C(q) = {(u M, ν u M ); g u + qu = 0, u C (M)} is open on (M, g) with M = (0, T ) Ω R 1+3 and g near the Minkowski metric. Time-independent problems are well-understood in theory, however, they are computationally challenging. The method based on interaction of singularities has proven to be flexible. It works for several non-linear wave equations, and it is under active further study. Non-linearity is used in an essential way, and there are (almost) no results for linear wave equations outside the case of product geometry.

On computational methods Let us consider again the basic wave equation 2 t u c 2 u = 0, in (0, T ) Ω, ν u (0,T ) Ω = f, u t=0 = t u t=0 = 0, and define the Neumann-to-Dirichlet map by Λf = u (0,T ) Ω. Λ determines c uniquely for large enough T [Belishev 87] Current computational methods in geophysical imaging are based on local techniques such as various linearizations together with least squares data fitting, not on the mathematical theory Convergence to local minima is a practical problem

Example: an acoustic lens The speed of propagation typically increases with the depth in geophysical imaging Low velocity regions cause problems for local data fitting methods due to focusing effects [Huang-Nammour-Symes 17] Speed of sound c(x). Surface is on top.

Example: an acoustic lens The speed of propagation typically increases with the depth in geophysical imaging Low velocity regions cause problems for local data fitting methods due to focusing effects [Huang-Nammour-Symes 17] Geodesics emanating from a point at the surface.

Reconstruction of the front face of the lens Reconstruction of c given simulated measurements corresponding to 241 point like sources on the top edge [de Hoop-Kepley-L.O.]. Our method is guaranteed to converge to the true solution We use ideas from the Boundary Control method [Belishev 87]

Reconstruction of the front face of the lens Reconstruction of c given simulated measurements corresponding to 241 point like sources on the top edge [de Hoop-Kepley-L.O.]. Our method is guaranteed to converge to the true solution We use ideas from the Boundary Control method [Belishev 87]

What about the back face of the lens? Due to the focusing, our method can not immediately recover the back face of the lens In theory, the data can be extended across the region where the speed of sound is already known by using unique continuation Then the speed of sound can be reconstructed deeper down by repeating the local reconstruction step We have developed stabilized finite element methods for unique continuation problems Heat equation [Burman-L.O. 18] Helmholtz equation [Burman-Nechita-L.O.] Time domain wave equation is a work in progress with Burman and Feizmohammadi

Unique continuation for the Helmholtz equation Unique continuation problem. Consider sets ω B Ω. Let u be a solution to u + k 2 u = 0 in Ω. Given u ω determine u B. Due to the lack of boundary conditions, the problem is ill-posed, and the optimal stability is of conditional Hölder type. B B ω ω Two computational solutions to the unique continuation problem. Left. Exact u. Middle. Solution in a convex geometry, that is, B belongs to the convex hull of ω. Right. Solution in a non-convex geometry.

Stabilized finite element method Using G(u, z) = ( u, z) L 2 (Ω) k 2 (u, z) L 2 (Ω), the weak formulation of the Helmholtz equation can be written: G(u, z) = 0 for all z H 1 0 (Ω). Our stabilized FEM is obtained as the equation for critical points of L(u h, z h ) = 1 2 u h u ω 2 L 2 (ω) + G(u h, z h ) + 1 2 s(u h, u h ) 1 2 s (z h, z h ). We consider L on V h W h, where V h is the H 1 conformal, P 1 finite element space over a quasi-uniform triangulation of polygonal Ω, and W h = V h H0 1 (Ω). Here h is the mesh size, and s(u h, u h ) = h n u h 2 F ds + hk 2 u h 2 L 2 (Ω), F F h s (z h, z h ) = z h 2 L 2 (Ω), F where n u h F is the jump of the normal derivative across F and F h is the set of internal faces of the triangulation.

Convergence estimate Theorem [Burman-Nechita-L.O.]. Let ω B Ω be as in the convex case. Let (u h, z h ) V h be the critical point of the Lagrangian L, and u the solution to the unique continuation problem. There are C > 0 and α (0, 1) such that for all k 0 and h > 0 satisfying kh 1, it holds that ( ) u h u L 2 (B) Chα k 2α 2 u H 2 (Ω) + k2 u L 2 (Ω). Contrary to the previous methods for unique continuation problems, e.g. the quasi-reversibility method [Lattès-Lions 67], we regularize on the discrete spaces, and h is the only asymptotic parameter In the non-convex case, we do not derive an explicit dependence on k but we show that if there is a line that intersects B but not ω, then the estimate must blow up faster than any polynomial in k

Convergence: the convex case Circles: H 1 -error, rate 0.64; squares: L 2 -error, rate 0.66; triangles : h 1 J(u h, u h ), rate 1; triangles : s (z h, z h ) 1/2, rate 1.3.

Convergence: the effect of noise in the convex case Perturbation O(h) Perturbation O(h 2 )