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 ${\mathrm{\Omega }}_1$ be the interval where $f$ is supported, and ${\mathrm{\Omega }}_2$ be the interval where the Hilbert transform of $f$ can be computed using the Gelfand-Graev formula. The equation we study is $H_{1}f=g{\mathrm{\mid }}_{{\mathrm{\Omega }}_2}$, where $H_1$ is the FHT that integrates over ${\mathrm{\Omega }}_1$ and gives the result on ${\mathrm{\Omega }}_2$, i.e. $H_1:L^2({\mathrm{\Omega }}_1)\to L^2({\mathrm{\Omega }}_2)$. In the case of the interior problem the tomographic data are truncated, and ${\mathrm{\Omega }}_1$ is no longer a subset of ${\mathrm{\Omega }}_2$. In this paper we consider the case when the intervals ${\mathrm{\Omega }}_1=(a_1,0)$ and ${\mathrm{\Omega }}_2=(0,a_2)$ are adjacent. Here $a_1 < 0 < a_2$\mathrm{.}Firstwefindadifferentialoperator$L$thatcommuteswith$H_1$. Using the Titchmarsh-Weyl theory, we show that $L$ has only continuous spectrum and obtain two isometric transformations $U_1$, $U_2$, such that $U_2 H_1 U_1^*$isthemultiplicationoperatorwiththefunction$\sigma(\lambda)$,$\lambda\geq(a_1^2+a_2^2)/8$\mathrm{.}Here$\lambda$ is the spectral parameter. Then we show that $\sigma (\lambda )\to 0$ as $\lambda \to \mathrm{\infty }$ exponentially fast. We also obtain the leading asymptotic behavior of the kernels involved in the integral operators $U_1$, $U_2$ as $\lambda \to \mathrm{\infty }$. 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.