Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 67 / 2

For definiteness take $f$ in the Schwartz space; sufficiently smooth compactly supported data are also adequate. These hypotheses make the line integrals, Cauchy principal values and decay estimates below well defined. Write $\lambda =r{e}^{i\theta }$ with $r>0$, $r\ne 1$, and introduce orthonormal local coordinatesLet $F(\tau ,\rho ,\theta )=f(x_1,x_2)$, $a_r=(r+r^{-1})/2$ and $b_r=(r-r^{-1})/2$. Substitution into the directional transport equation givesNow make the invertible real-linear change of coordinatesSince $\mathrm{Im}z=-a_r\rho /b_r$, its Wirtinger derivative is ${\partial }_{\bar{z}}=({\partial }_{\tau }-i{b}_r{\partial }_{\rho }/a_r)/2$. Hence $L_{\lambda }=\nu (r){\partial }_{\bar{z}}$, as required. The change becomes singular at $r=1$, where the complex operator degenerates to a real directional derivative. Also the original equation contains ${\lambda }^{-1}$, so $\lambda =0$ is excluded from the differential equation; the normalized solution will extend analytically there.

The distributional identity ${\partial }_{\bar{z}}(1/(\pi z))=\delta$ gives the Cauchy-Green operator solutionwhere $\tilde{f}$ is $f$ in the changed real coordinates. The singularity $1/(z-z^{\prime })$ is locally integrable. For Schwartz functions, splitting the integral into a region near the origin and a rapidly decaying far region proves $\mu =O(|z{|}^{-1})$. If two solutions obey this decay, their difference has zero Wirtinger derivative and is an entire holomorphic function tending to zero; the Liouville theorem proves uniqueness.

The change-of-variables Jacobian has absolute value $a_r/|b_r|$. Therefore an equivalent form of the complexified transport Cauchy kernel isIn the original complex coordinate $w=x_1+i{x}_2$, it isFor $|\lambda |<1$ the denominator can be rewritten to give the kernel $\lambda /[\pi ((w-w^{\prime })-{\lambda }^2(\bar{w}-{\bar{w}}^{\prime }))]$. For $|\lambda |>1$ it is ${\lambda }^{-1}/[\pi ((\bar{w}-{\bar{w}}^{\prime })-{\lambda }^{-2}(w-w^{\prime }))]$. Consequently the spectral solution is holomorphic separately inside and outside the unit circle, with $\mu =O(\lambda )$ at zero and $\mu =O({\lambda }^{-1})$ at infinity. These normalizations will determine its jump reconstruction.

Define the Radon transform and its transverse Hilbert transform byUse the signed Cauchy boundary operators $P^{+}=(I+iH)/2$ and $P^{-}=(-I+iH)/2$. Thus $P^{+}$ and $-P^{-}$ are the complementary idempotent projections; $P^{-}$ itself carries the signed lower-boundary convention. The PDF defines the outside limit ${\lambda }^{-}=(1+\epsilon )e^{i\theta }$ but omits the definition of the inside limit in its displayed formula. We take ${\lambda }^{+}=(1-\epsilon )e^{i\theta }$, as required by that formula.

As $r\to 1$, apply the Sokhotski–Plemelj formula to the kernel from part (b):The principal value contribution is $i\mathrm{sgn}(b_r)HRf/2$. The delta contribution is the same from either side and equalsSince $b_r<0$ for the inside limit, the resulting boundary value isThe outside limit similarly has $P^{+}Rf$ in place of $-P^{-}Rf$. Naming the signed convention explicitly prevents a sign error from treating $P^{-}$ as a positive projection.

The two boundary values have jumpThe line-tail contributions cancel. Because the complexified transport Cauchy kernel is holomorphic on both sides of the unit circle and vanishes at spectral infinity, the Cauchy integral formula reconstructs it from this jump:There is no additional entire term: its inside and outside boundary values would agree, so it would define an entire function vanishing at infinity. Expand outside the circle as $\mu ={\mu }_1/\lambda +O({\lambda }^{-2})$. ThenIn the original coordinate $w=x_1+i{x}_2$, the transport equation is $(\lambda {\partial }_w+{\lambda }^{-1}{\partial }_{\bar{w}})\mu =f$. Its constant term at infinity is ${\partial }_w{\mu }_1=f$. Sincewe obtain the inverse Radon transform in filtered backprojection form:Opposite directions represent the same unoriented line. Reversing $\theta$ by $\pi$ reverses $\rho$; the Hilbert transform changes sign under this reversal, and its derivative changes it back. Thus the integrand agrees for opposite directions and the factor becomes $1/(2\pi )$ if the angular integral is over $[0,\pi )$. With the convention $\hat{h}(\kappa )=\int e^{-i\kappa \rho }h(\rho )d\rho$, the Hilbert transform has Fourier multiplier $-i\mathrm{sgn}\kappa$; the derivative makes the filter $|\kappa |$. This also identifies the formula with the standard ramp-filter reconstruction.

In ideal computed tomography, the unknown is an absorption coefficient $a(x)$. Along a straight ray the transmitted intensity satisfies $dI/d\tau =-aI$, soThus calibrated transmission measurements supply the Radon transform of the slice. The inversion above reconstructs the coefficient by filtered backprojection: filter each projection in its transverse variable and then accumulate it over the rays through each image point. This turns a collection of line-integral measurements into a spatial map, which is the key usefulness of the result.

The ideal derivation assumes known ray geometry, sufficiently complete angular sampling and a straight-ray attenuation model. The ramp Fourier multiplier grows at high frequency, so measurement noise is amplified. Practical reconstruction therefore uses discretization and regularization of an inverse problem, for example a window on the ramp filter. Limited angles or missing detector positions cannot be treated as if they were complete noiseless Radon transform data.

In single-photon emission computed tomography, the unknown is an emitting source, rather than the absorption coefficient reconstructed by transmission computed tomography. Photons emitted at different points on the same ray are weighted by their probability of surviving the path to the detector. If the known attenuation is $a$ and the detector is at the positive end of the ray, the data are an attenuated Radon transformThe directional transport equation is $\mathbf{d}\cdot \nabla u+au=f$ with zero incoming intensity. An integrating factor derives this attenuation weight. The same complexification, boundary-value and jump strategy gives spectral reconstruction of an attenuated Radon transform when attenuation is known. The unattenuated inverse from part (d) applies directly only when $a=0$; for nonzero attenuation the weighted jump and inverse must be used. This distinction allows emission imaging of the source without confusing it with reconstruction of the absorbing medium.