Diagonalization of the Finite Hilbert Transform on two adjacent intervals

We study the interior problem of tomography. The starting point is the Gelfand-Graev formula, which converts the tomographic data into the finite Hilbert transform (FHT) of an unknown function $f$ along a collection of lines. Pick one such line, call it the $x$-axis, and assume that the function to be reconstructed depends on a one-dimensional argument by restricting $f$ to the line. Let $Ω_1$ be the interval where $f$ is supported, and $Ω_2$ be the interval where the Hilbert transform of $f$ can be computed using the Gelfand-Graev formula. The equation we study is $H_1 f=g|_{Ω_2}$, where $H_1$ is the FHT that integrates over $Ω_1$ and gives the result on $Ω_2$, i.e. $H_1: L^2(Ω_1)\to L^2(Ω_2)$. In the case of the interior problem the tomographic data are truncated, and $Ω_1$ is no longer a subset of $Ω_2$. In this paper we consider the case when the intervals $Ω_1=(a_1,0)$ and $Ω_2=(0,a_2)$ are adjacent. Here $a_1 < 0 < a_2$. First we find a differential operator $L$ that commutes with $H_1$. Using the Titchmarsh-Weyl theory, we show that $L$ has only continuous spectrum and obtain two isometric transformations $U_1$, $U_2$, such that $U_2 H_1 U_1^*$ is the multiplication operator with the function $σ(λ)$, $λ\geq(a_1^2+a_2^2)/8$. Here $λ$ is the spectral parameter. Then we show that $σ(λ)\to0$ as $λ\to\infty$ exponentially fast. We also obtain the leading asymptotic behavior of the kernels involved in the integral operators $U_1$, $U_2$ as $λ\to\infty$. When the intervals are symmetric, i.e. $-a_1=a_2$, the operators $U_1$, $U_2$ are obtained explicitly in terms of hypergeometric functions.

No reviews yet.

No comments yet.