PHOTO-ACOUSTIC INVERSION IN CONVEX DOMAINS

. In photo-acoustics one has to reconstruct a function from its averages over spheres around points on the measurement surface. For special surfaces inversion formulas are known. In this paper we derive a formula for surfaces that bound smooth convex domains. It reconstructs the function mod- ulo a smoothing integral operator. For special surfaces the integral operator vanishes, providing exact reconstruction.


Introduction.
Recently there has been much interest in deriving inversion formulas for photo-acoustic imaging. One has to solve the following inverse problem: Let p be the solution of the intitial value problem ∂ 2 p ∂t 2 − ∆p = 0, (1) p(x, 0) = p 0 (x), ∂p ∂t (x, 0) = 0.
Let Ω be a domain in R 3 containing the support of p 0 and let ∂Ω be its boundary. The problem is to determine p 0 from the knowledge of p on ∂Ω × R 1 . In principle this problem can be solved by back-propagation. However there is much interest in explicit inversion formulas. Various special cases have been treated in the literature. Inversion formulas have been derived for special geometries: Ω a ball [1], various domains [5], balls, slabs and cylinders [8], cubes [2], ellipsoids [6].
In the present paper we consider the case of a bounded smooth convex domain Ω. We show that in such a domain the "universal back-projection formula" [8] reconstructs p 0 up to a smoothing integral operator. This operator disappears for special sets Ω, such as those mentioned above. Our derivation is in the spirit of [4].

The general inversion formula. Let us consider the integral
where ν is the exterior normal and σ the surface measure on ∂Ω. This is the inversion integral of [8]. It is shown there that I(x) = p 0 (x) for special measurement surfaces, such as spheres, cylinders and planes.
Let R be the Radon transform The derivative of (Rχ)(θ, s) with respect to s is denoted by (Rχ) (θ, s). For x, z ∈ Ω let θ(z, x) = (z − x)/|z − x|, s(z, x) = 1 2 (|z| 2 − |x| 2 )/|z − x|. (Rχ)(θ(z, x), s(z, x)) is the integral of χ over the plane of those points which have the same distance to z and x. Theorem: Assume that Ω is a convex bounded and smooth domain in R 3 . Let χ be the characteristic function of Ω. Then, This is the main result of our paper. It states that for Ω bounded, smooth and convex the inversion formula of [8] is correct up to a smoothing integral operator.
As Ω is bounded, smooth and convex, (Rχ)(θ(z, x), s(z, x)) is a smooth function of z, x on Ω for x = z.
Incidentally the Theorem extends the result of [7] to convex measurement geometries.
The proof of the theorem requires some preparations. We first state that according to Kirchhoff's formula the solution of problem (1,2) is given by Thus our problem calls for the reconstruction of p 0 from its averages over spheres centered on ∂Ω. Using the language of distributions we can write this as where δ is the Dirac δ function in R 1 . We state some well known facts about distributions. For the Dirac δ function in R 3 we have By direct calculation one shows that for the δ function in R 1 we have with the δ functions in R 3 and R 1 , respectively, on the left and the right hand side. This is easily seen by introducing polar coordinates.
Let for x, z ∈ Ω The key tool of the proof is the following Lemma: Let Ω be a bounded, smooth and convex domain in R 3 . With χ the characteristic function of Ω we have for x, z ∈ Ω, x = z I Ω (z, x) = − 1 |z − x| 2 (Rχ) (θ(z, x), s(z, x)).
Proof of the Lemma: In the integral we make the substitution y = (z+x)/2+t (5) to the t-integral we obtain we finally obtain the result. Now we proceed to the Proof of the Theorem: Applying the Gauss integral theorem we obtain Carrying out first ∇ y and then div x under the integral sign we get Since and, by (4), we conclude that For the evaluation of the integrals in (8), our lemma comes into play. First we note that due to (3), Since ∆p is the solution of (1,2) with p 0 replaced by ∆p 0 we also have and, correspondingly, Hence, From the lemma we get The second term on the right hand side of (10) is evaluated as follows: We have (see formula (II.1.5) of [3]) s(z, x)).