An inverse source problem in optical molecular imaging

Transcription

1 An inverse source problem in optical molecular imaging Plamen Stefanov 1 Gunther Uhlmann University of Washington

2 Formulation Direct Problem Singular Operators Inverse Problem Proof Conclusion Figure: The first and the last picture of this talk

3 Formulation Formulation of the problem The radiative transport equation in Ω is given by Z θ xu(x, θ)+σ(x, θ)u(x, θ) k(x, θ, θ )u(x, θ ) dθ = f (x), u SΩ = 0, S n 1 where σ is the absorption and k is the collision kernel. The source term f is assumed to depend on x only. Here, SΩ consists of x Ω and θ pointing inwards. The boundary measurements are modeled by Xf (x, θ) = u +SΩ, (x, θ) +SΩ, where u(x, θ) is a solution of the transport equation, and +SΩ denotes the points x Ω with direction θ pointing outwards.

4 Formulation Direct Problem Given f (and σ, k), find Xf. Inverse Problem Given Xf (and σ, k), find f. Clearly, it is a linear problem. Let σ = k = 0 first. Then X is just the X-ray transform: Xf (x, θ) = If (x, θ) := Z 0 τ (x,θ) f (x + tθ) dt, (x, θ) +SΩ (σ = k = 0),

5 Formulation Let k = 0 only. Then we get a weighted X-ray transform: Z Xf (x, θ) = I σf (x, θ) := E(x + tθ, θ)f (x + tθ) dt, (x, θ) +SΩ (k = 0), where E(x, θ) = exp Z 0 «σ(x + sθ, θ) ds. If σ = σ(x), then we get the attenuated X-ray transform, that we know how to invert. Without assuming that any one is zero, Bal and Tamasan proved injectivity when k = k(x, θ θ ), and k is small enough in a suitable norm. The main idea there is to treat k as a perturbation; then X is a perturbation of the attenuated X-ray transform. Also, results by Sharafutdinov on Riemannian manifolds, smallness conditions on the curvature k and σ. Our goal is to consider this problem for general (σ, k).

6 The Direct Problem: Generic Solvability The direct problem first We need assumptions, even for solvability of the direct problem! Assuming k 1 is enough. Also, R k(, θ, )dθ < σ suffices. Those conditions prevent a nuclear explosion, i.e., the corresponding time-dependent dynamics is bounded. They are not necessary conditions though. Theorem 1 (a) The direct problem is uniquely solvable for a dense open set of pairs (σ, k) in C 2, even for f = f (x, θ). (b) X : L 2 (Ω S n 1 ) L 2 ( +SΩ, dσ). Here, dσ = ν θds xdθ.

7 Sketch of the Proof Set T 0 = θ x, T 1 = T 0 + σ, T = T 0 + σ K, where σ and K are viewed as operators corresponding to the σ and to the k terms in the transport equation. Then T 0, T 1, T can be defined as closed operators acting on functions f satisfying f SΩ = 0. Then Z 0 [T 1 1 f ](x, θ) = exp Z 0 s «σ(x + τθ, θ) dτ f (x + sθ, θ) ds. Why did not we start with T 1 0? It turns out that T is a kind of relatively compact perturbation of T 1 but not of T 0. So, we have to think of σ as a part of the principal symbol but k is of low order. We now apply T 1 1 to both sides of Tu = f, u SΩ = 0. to get u = T 1 1 (Ku + f ) = (Id T 1 1 K)u = T 1 1 f. Therefore, if we can invert Id T 1 1 K, we can solve the direct problem.

8 Sketch of the Proof This is also equivalent to the existence of (Id KT 1 1 ) 1. Set [Jf ](x, θ) := f (x), Then (with f = f (x) independent of θ), and R +h = h +SΩ u = T 1 Jf = (Id T 1 1 K) 1 T 1 1 Jf, Xf = R +T 1 1 (Id KT 1 1 ) 1 Jf. Lemma 2 The operator KT 1 1 J : L 2 (Ω) L 2 (Ω S n 1 ) is compact. Proof. It is an integral operator of the form Z Σ x, x y, x y k x, θ, x y x y x y f (y) dy, x y n 1 where Ω Z 0 «Σ(x, s, θ ) = exp σ(x + τθ, θ ) dτ. s

9 Calderón-Zygmund Theorem Let K be an integral operator with kernel k(x, y) = φ(x, θ)r n, where θ = (x y)/ x y, r = x y, i.e., formally, [Kf ](x) = Z φ x, x y x y f (y) dy, x y n and the characteristic φ has mean value 0 as a function of θ, for any x. Then K is well defined, where the integral has to be understood in the principle value sense. Moreover, K L 2 (Ω) L 2 C sup φ(x, ) L 2 (S n 1 ), x The characteristic φ does not need to have zero mean value in θ but then the integral has to be considered as a convolution in distribution sense. The latter is well defined because the Fourier transform of the kernel w.r.t. the variable z = rθ is homogeneous of order 0, thus bounded.

10 Compactness of weakly singular operators Compactness of weakly singular operators Also, if B is an operator with a weakly singular kernel ψ(x, θ)r n+1, then xb is an integral operator with singular kernel x[β(x, θ)r n+1 ]. The latter, up to a weakly singular operator, has a singular kernel of the type φr n, and the integration is again understood in the principle value sense, see the next paragraph. In particular, the zero mean value condition is automatically satisfied. Informally speaking, operators with kernels of the kind x y n+1 map L s (Ω) into H 1, and are therefore compact as operators from L 2 (Ω) to itself. In particular, this proves the lemma.

11 Another operator is compact! If f = f (x, θ), then Z Σ x, x y, x y k x, θ, x y [KT 1 x y x y 1 f ](x, θ) = f x y n 1 y, «x y dy. x y No enough integrations to make it compact! To invert Id KT 1 1 for generic k s, we want to apply the Analytic Fredholm Theorem. But KT 1 1 is not compact. Its square however is! Lemma 3 The operator KT 1 1 K : L 2 (Ω S n 1 ) L 2 (Ω S n 1 ) is compact. Sketch of the Proof: ZZΩ Sn 1 α x, y, x y, x y, θ, [KT 1 x y θ 1 Kf ](x, θ) = f (y, θ ) dy dθ x y n 1 Looks like a weakly singular operator but not exactly.

12 Generalized Calderón-Zygmund type of theorem This calls for a generalization of the Calderón-Zygmund arguments. Lemma 4 Let A be the operator [Af ](x) = Z α x, y, x y, x y x y f (y) dy x y n 1 with α(x, y, r, θ) compactly supported in x, y. Then (a) If α C 2, then A : L 2 H 1 is continuous with a norm not exceeding C α C 2. (b) Let α(x, y, r, θ) = α (x, y, r, θ)φ(θ). Then A L 2 H 1 C α C 2 φ H 1 (S n 1 ). Now we can prove the previous lemma.

13 End of the Proof of Thm 1 End of the Proof of Thm 1 Instead of trying to invert Id KT 1 1, we will invert its square (and then write P 1 = P(P 2 ) 1 ). Set A(λ) = Id `λkt1. By the Analytic Fredholm Theorem, for any fixed (σ, k), A(λ) is a meromorphic family. Thm 1 easily follows from this.

14 Inverse Problem The Inverse Problem Fix Ω 1 Ω. Define X 1 as X but in Ω 1. Theorem 5 For (σ, k) in an open and dense set of pairs in C 2 ( Ω S n 1 ) C 2 Ωx S n 1 θ ; C n+1 (S n 1 θ ), including (0, 0), the direct problem is solvable in Ω 1, and (a) the map X 1 is injective on L 2 (Ω), (b) the following stability estimate holds f L 2 (Ω) C X 1 X 1f H 1 (Ω 1 ), f L 2 (Ω), with a constant C > 0 locally uniform in (σ, k).

15 Sketch of the Proof Sketch of the Proof Denote X 1 = X again. Outline If k = 0, then X X = I σ I σ, that is an elliptic ΨDO, and the theorem is true (Frigyik, Uhlmann & S). k is responsible for a relative compact perturbation: X X = I σ I σ + L, with L : L 2 H 1 smoothing, i.e., L : L 2 H 2. Apply a parametrix Q of I σ I σ to X X : QX X = QI σ I σ + QL = Id + K with K one degree smoothing (and therefore compact). We get a Fredholm problem: invert Id + K, with K = K σ,k. For σ analytic, Id + K σ,0 is invertible. Consider now Id + K σ,λk for λ [0, 1]. Still invertible for generic λ s.

16 Sketch of the Proof Why is the theorem true for k = 0? Then Iσ I σ has a weakly singular kernel: Z [Iσ α(x, y) I σf ](x) = f (y) dy x y n 1 and is also an elliptic ΨDO of order 1. It is in the analytic class for σ analytic. If A is an elliptic ΨDO on Ω 1, if supp f Ω, and if Af = 0, then f is analytic in Ω 1. It also has compact support. Therefore, f = 0. Therefore, I σ I σ is injective on L 2 (Ω) for σ analytic. Apply a parametrix becomes a Fredholm problem. Therefore, stable under perturbations. Thus, I σ I σ is injective for an open dense set of σ s. The stability estimate also follows from the Fredholm Theory.

17 L : L 2 H 2 Proof of L : L 2 H 2 We write X = I σ + L, 1 L := R +T 1 1 KT 1 1 Id KT 1 1 J. Then X X = I σ I σ + L, L := I σ L + L I σ + L L. First (and most important) step: Can we prove that I σ L : L 2 H 2? A model operator A : h(x, θ) [Af ](x): [Ah](x) = ZZ α x, w, y, x w, w y, x w, w y x w w y w y h y, h (y) dy dw x w n 1 w y n 1 w y A : L 2 H 2? Not clear. Not so clear even when h is independent of θ...

18 L : L 2 H 2 [Ah](x) = ZZ α x, w, y, x w, w y, x w, w y x w w y x w h y, h (y) dy dw. x w n 1 w y n 1 w y If h = h(x) is independent of θ, and if α is of product type: «α x, w, y, x w, w y, x w, w y x w w y = α 1 x, w, x w, x w α x w 2 w, y, w y, w y, w y then we have a composition of two operators, each one of order 1, and we are done by our generalized C-Z theorem. This makes the statement believable. α is of product type, if, for example, k(x, θ, θ ) = Θ(θ)κ(x, θ ). In general, this is not true, but nothing prevents us from studying (infinite) linear combinations of terms like this.

19 L : L 2 H 2 We can always write k in the form k(x, θ, θ ) = X Θ j (θ)κ j (x, θ ). (1) j=1 For example, choose Θ j to be the spherical harmonics. Then the series converges in L 2 but for our analysis we need convergence in C 1 (almost). This explains the need for higher regularity in the θ variable. Now, each term in (1) contributes a compact operator; we have uniform convergence; so we are done. Well, we still have to deal with the fact that h depends on θ as well...

20 Conclusion If we want to recover the singularities of f only, up to order 1, then it is enough to apply to X Xf a parametrix of I σ I σf, that is explicit and depends on σ only. To actually find f from Xf = h, we still apply the explicit parametrix Q above, and get a Fredholm problem (Id + K)f = QX h, where Q and K are computable, and h is given. It seems strange that k plays no role in the recovery of the singularities of f (up to order 1). In reality, Z σ = σ true absorption + k(, θ, ) dθ. Therefore, k is involved in that step.