Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 67 (Fourier Transforms, Their Generalisations and the Imaging of the Brain)

For the Airy equation, let $P=e^{-ikx+\omega t}q$. Its local relation follows by arranging the flux so that the unwanted spatial derivatives cancel:ConsequentlyThus the dispersion relation and the required flux are given byThe calculation is polynomial in the complex spectral parameter, so the local relation holds for every $k\in \mathbb{C}$.

Use negative-exponential spatial Fourier transforms and finite-time spectral boundary transforms:Write $\mathcal{F}=k^2{F}_0-ik{F}_1-F_2$ and $\mathcal{G}=k^2{G}_0-ik{G}_1-G_2$. Integrate the local relation first over $0<x<L$ and then over time. The left and right endpoint fluxes have opposite signs, giving the finite-interval Airy global relationAll these spatial and temporal integral transforms are entire functions of $k$ for sufficiently smooth data on the finite intervals. Put $\alpha =e^{2\pi i/3}$ and $k_j={\alpha }^{j}k$, $j=0,1,2$. The dispersion relation obeys $\omega (k_j)=\omega (k)$; hence the same global relation holds at all three $k_j$. In particular, the missing traces satisfy the concrete linear systemwhere, with a common upper time limit $t$,The finite-time spectral boundary transforms $F_j,G_j$ are invariant under these rotations because their time kernel depends only on $k^3$. The coefficients multiplying them are not invariant; these differences make the three equations useful for eliminating the missing boundary traces.

LetThe upper domain is the sector $\pi /3<\arg k<2\pi /3$; the lower domain consists of the sectors $\pi <\arg k<4\pi /3$ and $5\pi /3<\arg k<2\pi$. Orient their boundaries with the corresponding domain to the left. For $0<x<L$ and $0<t<T$, the Fokas method gives the integral representationHere $\mathcal{F},\mathcal{G}$ are the combinations of all six finite-time spectral boundary transforms defined above. Initially, Fourier inversion of the global relation gives the first integral together with $-{\int }_{\mathbb{R}}{e}^{ikx-\omega t}\mathcal{F}/(2\pi )$ and $+{\int }_{\mathbb{R}}{e}^{ik(x-L)-\omega t}\mathcal{G}/(2\pi )$. For the left endpoint, deform through the upper complementary sectors, where the spatial exponential decays andis bounded. This replaces the real line by $\partial D_{+}$. For the right endpoint, deform in the lower half-plane; the real portions of $\partial D_{-}$ run from right to left. Thus its integral is the negative of the real-line integral, explaining the minus sign in the last displayed term. The arc contributions vanish by spatial exponential decay and integration by parts in the transforms. Oscillatory real-line integrals can equivalently be defined by a decaying cutoff and its limit.

Finally use the finite-horizon extension of a boundary transform. The difference between upper limits $t$ and $T$ contains ${\int }_t^T{e}^{\omega (s-t)}{f}_j(s)ds$, bounded in $D$, while the appropriate spatial factor decays inside $D_{+}$ or $D_{-}$. The Cauchy integral theorem makes this difference vanish. Keeping $T>t$ also avoids an artificial endpoint half-value in temporal inversion. The integral representation at this stage deliberately includes the unknown traces; the rotated global relations eliminate them in the next step.

The rotated global relations form a square system for the three missing boundary traces. Solve this system and substitute the resulting finite-time spectral boundary transforms into the integral representation. The pieces containing $Q(k_j,T)$ have zero contour contribution by analyticity, rotation of the relevant sectors and the Cauchy integral theorem. The remaining expression involves only $Q_0,F_0,G_0,G_1$. The finite-interval Airy spectral determinant has no nonremovable zeros in the contour domains, as established below, so this elimination introduces no extra spectral data there. Differentiating the representation, and using temporal inversion and compatibility at the corners, verifies the partial differential equation, initial value and the three prescribed boundary values.

For a direct Hadamard well-posedness check, the homogeneous difference satisfiesThis proves uniqueness and continuous dependence for homogeneous boundary data. More generally subtract a smooth quadratic boundary lift $b(x,t)$ satisfying the three prescribed endpoint conditions. The remaining problem has homogeneous boundary conditions and forcing $-b_t-b_{xxx}$. Its semigroup generator is $A=-d^3/d{x}^3$ on $H^3(0,L)$ with $v(0)=v(L)=v^{\prime }(L)=0$. It is a dissipative operator; for every real $\zeta >0$ the equation $(\zeta -A)v=h$ has a unique solution, since the homogeneous third-order boundary system is injective by the same energy identity and therefore invertible. Thus $A$ is maximal dissipative and generates a contraction semigroup. This gives existence and continuous dependence on the lifted data and forcing. Smooth compatible data therefore define a well-posed problem. This operator argument and the spectral elimination describe the same evolution.

Expanding the matrix determinant from part (b) givesThe finite-interval Airy spectral determinant obeys $\Delta (\alpha k)={\alpha }^2\Delta (k)$. At the origin,This zero comes from coalescence of the three rotated equations; it is removable in the underlying boundary system. Indeed, $v^{\prime \prime \prime }=0$ with $v(0)=v(L)=v^{\prime }(L)=0$ has only the zero quadratic solution.

For $k\ne 0$, a zero of $\Delta$ is equivalent to the existence of a nonzero combination $v(x)={\sum }_{j=0}^2{c}_j{e}^{i{\alpha }^{j}kx}$ satisfying the homogeneous endpoint conditions. This can be checked by expanding the determinant of those three endpoint equations. Such a function satisfies $v^{\prime \prime \prime }=\omega (k)v$, so it is an eigenfunction of $A$ with eigenvalue $-\omega (k)$. Integration by parts yieldsso every nonzero zero has $\mathrm{Re}\omega (k)\ge 0$. None lies on the boundary $\mathrm{Re}\omega =0$: by a cube-root rotation it suffices to take $K=kL$ real. With $u=\sqrt{3}K/2$, the determinant equation becomesThe right-hand side has squared modulus $1+4{\sinh }^{2}u>1$ for $K\ne 0$, whereas the left-hand side has modulus one. Therefore there are no nonremovable spectral poles in $\overline{D}$, and the contour representation has no discrete residue contribution from that domain.

Taken literally as a claim about the operator's discrete spectrum, however, the printed assertion is false. To see this constructively, put $k=-is/L$, $s>0$. ThenThus zeros occur whenBetween successive positive cosine maxima and adjacent negative minima the difference changes sign. There are consequently infinitely many positive roots, producing genuine negative eigenvalues $-s^3/L^3$ of $A$. The third-order resolvent maps into $H^3(0,L)$, whose inclusion into $L^2(0,L)$ is compact, so these are indeed discrete spectrum of an operator with compact resolvent. The valid conclusion is absence of poles in the reconstruction sectors, not absence of operator eigenvalues.

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.

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the modified Helmholtz equation, including its Laplace equation case $\lambda =0$. Equivalently,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.

Fix counterclockwise arclength on each side and take the outward normal derivative. WriteThe unit tangent is $i{c}_j$ and the outward unit normal is $c_j$. On side $j$ this gives the chain-rule identityPut $w=c_{j}k$. The exponential in the closed differential one-form pulls back toAlso $-\lambda qd\bar{z}/(ik)=(\lambda /w)q^{(j)}ds$. Consequently the contribution of side $j$ is ${\rho }_j(c_{j}k)$, where the required local side spectra areThese are spectral integral transforms in the local coordinate of each side. Keeping the tangential-derivative term is important: premature integration by parts in a single side integral would leave corner values, which cannot simply be discarded. At the special samples used below, those endpoint terms do vanish for the rotationally invariant solution.

The Generalized Stokes theorem applied to the closed differential one-form givesEquivalently, the global relation isThe identical outward Neumann boundary data make the three known transforms equal:To reduce the unknown traces to a single function, use rotation invariance and uniqueness. For $\lambda >0$ the difference $u$ of two solutions has zero outward derivative andby Green's first identity. Thus $u=0$. Rotation by $a$ preserves the equation, the domain and the oriented boundary data, so it preserves $q$; hence $q^{(1)}=q^{(2)}=q^{(3)}=q$ and ${\Phi }_j=\Phi$. The global relation becomes the single equation with three spectral unknownsThe three unknowns are evaluations of the same boundary transform, not three freely specified boundary data.

The printed allowance of all real $\lambda$ requires a qualification. At $\lambda =0$, existence requires $3{\int }_{-l/2}^{l/2}f(s)ds=0$ by the divergence theorem, and solutions differ by a constant. Fixing a rotation-invariant normalization gives uniqueness and the same rotational conclusion. In fact the unnormalized solution is already invariant: its rotation difference is constant, and three successive rotations make three times that constant zero. For $\lambda <0$, uniqueness fails precisely when $-4\lambda$ is an eigenvalue of the Neumann Laplacian. At these parameters identical data do not force an arbitrary solution to be rotationally invariant. Averaging any solution over the three rotations produces a solution with the common trace used here; the difference is a homogeneous Neumann eigenfunction, or a sum of them. Part (e) describes both solvability and this remaining freedom rather than assuming uniqueness for every real parameter.

Let $\Phi$ denote the common trace transform of the unique solution, or of the rotational average at an exceptional parameter. Because $q$ and $f$ are real and $\lambda$ is real, $\overline{\Phi (\bar{k})}=\Phi (k)$ and $\overline{\Psi (\bar{k})}=\Psi (k)$. Taking the Schwarz conjugate of the global relation means replacing $k$ by $\bar{k}$ and conjugating the whole equation. It yields the second of the conjugate global relations for the modified Helmholtz equation:In this conjugation the rotated arguments interchange; after putting terms in the order $k,\bar{a}k,ak$, the coefficients have the displayed signs and arguments. DefineThe elementary cube-root identities giveMultiply the first global relation by $E(iak)$ and the conjugate one by $E(-iak)$. Then the coefficient of $\Phi (ak)$ is one in both equations:where all the quantities on the right are known,Subtracting eliminates $\Phi (ak)$. Ifthe resulting solution for the requested transform isAt a zero of the denominator this identity is interpreted in its undivided form. The transforms themselves remain finite for $k\ne 0$; consequently the numerator must vanish. That cancellation, rather than evaluation of a singular quotient, determines the Fourier coefficients below.

For $\lambda >0$, choose a sample $k_n$ satisfying $\chi (k_n)=i\pi n$ for each integer $n$. A convenient branch isThus $k_n+\lambda /k_n=2\pi in/l$. The undivided identity in part (d) givesRotation invariance makes all three vertex values equal, so $q(l/2)=q(-l/2)$. The exponential has the same endpoint value $(-1{)}^n$ at both ends. Integration by parts in the definition of $\Phi$ therefore givesThese are the Fourier coefficients with the positive-exponential convention. Hence the computed boundary trace isEvery quantity in this rotationally invariant Neumann reconstruction in an equilateral triangle is determined by the prescribed $f$. To check that no denominator was silently divided by zero, put $k_n=i{\kappa }_n$, ${\kappa }_n>0$. ThenNeither factor vanishes for $\lambda >0$. For sufficiently smooth compatible traces, the periodic extension of $q$ is continuous and piecewise smooth; the Fourier series recovers $q$, with the usual endpoint convention. Reality implies $Q_{-n}=\overline{Q_n}$ even though the sampling formula uses complex arguments.

For $\lambda =0$, first impose the zero-flux compatibility from part (c). For $n\ne 0$ take the nonzero root $k_n=2\pi in/l$ and use the same coefficient formula with $\lambda =0$. The coefficient $Q_0$ is an arbitrary real constant times $l$, expressing the additive constant for this problem with a Neumann boundary condition. The zero sample cannot determine it. For example, $f=0$ admits every constant solution, so no uniquely specified $q(s)$ is possible without a normalization.

For completeness, negative parameters can be handled without assuming away resonances. Let ${\varphi }_j$ be a real orthonormal eigenbasis of the Neumann Laplacian, with $-\Delta {\varphi }_j={\mu }_j{\varphi }_j$, and defineusing the corresponding oriented $f$ on each side. The weak equation and Green's first identity give the explicit coefficient relationAt a nonresonant parameter this determines the unique solution. Its common trace is the continuation of the displayed spectral Fourier coefficients from $\lambda >0$; choose either nonzero root of $k_n^2-(2\pi in/l)k_n+\lambda =0$. If an individual quotient is $0/0$ at a nonresonant parameter, use its removable limit in $\lambda$, rather than declare a new eigenvalue. For example, a double sampling root can make $k_n-\lambda /k_n=0$ without making the boundary problem singular. The weak coefficient relation proves that the solution is analytic in the parameter wherever no ${\mu }_j+4\lambda$ vanishes, so these nonresonant limits exist.

At a genuine resonance $\lambda =-{\mu }_{\ast }/4$, the Fredholm solvability condition is precisely $F_j=0$ for every ${\mu }_j={\mu }_{\ast }$. Necessity follows directly from the coefficient relation. For compatible smooth data, sufficiency follows by dividing all nonresonant coefficients and leaving the resonant ones free. The boundary functional is continuous on $H^1(D)$, so ${\sum }_j|F_j{|}^2/(1+{\mu }_j)<\infty$; away from the finite resonant eigenspace this implies ${\sum }_j(1+{\mu }_j)|F_j/({\mu }_j-{\mu }_{\ast }){|}^2<\infty$. Thus the resulting series converges in $H^1(D)$ and satisfies the weak equation, not just its individual coefficient identities. The full solution and its boundary values are thuswith convergence in the weak solution space and the associated trace sense. Rotation averaging selects the common-side component; a nonsymmetric homogeneous addition can have different traces on the three sides. Incompatible resonant data give no solution. For arbitrary real $\lambda$, compatibility and the free homogeneous components are essential; the source does not specify conditions that would select a unique trace at resonance.