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Keep the coordinates and operator $D_k$ from part (a). The attenuation $f$ is known, and the source $g$ is to be reconstructed. Put $F(\rho ,\tau ,\theta )={\int }_{\tau }^{\infty }f(s\cos \theta -\rho \sin \theta ,s\sin \theta +\rho \cos \theta )ds$ and $S(\rho ,\theta )=\mathcal{R}f={\int }_{\mathbb{R}}fd\tau$. The directed forward attenuated Radon transform, for a detector at the positive end of the line, isThis weight accounts for absorption from each emission point to the detector. It is also derived by solving the real transport equation ${\partial }_{\tau }u+fu=g$ with zero incoming intensity: its outgoing limit is $A_f$. In particular reversing the line direction generally changes the measurement. The pair reconstructs $g$ from full directed data and known $f$; it does not infer both unknown functions from the same data.

We must specify the sign convention for the projectors. Use the Hilbert transform $Hv(\rho )={\pi }^{-1}\mathrm{PV}{\int }_{\mathbb{R}}v(s)/(\rho -s)ds$ and the signed Cauchy boundary operatorsThen $P^{+}-P^{-}=I$, and the positive idempotent projectors are $P^{+}$ and $-P^{-}$. This is the signed boundary-value convention in the given limits of $M$. To express everything with ordinary complementary projectors, one may instead write ${\Pi }_{+}=P^{+}$, ${\Pi }_{-}=-P^{-}$. The given limits becomeBoth $a_{\pm }$ depend on the transverse coordinate and direction but are constant along $\tau$.

The equation $D_{k}M=f$ supplies an integrating factor:where $T_k$ is the inverse of unattenuated complex transport, normalized to vanish at spatial infinity. To see the normalization and analyticity rather than assume them, let $w=x_1+i{x}_2$. The complexified transport Cauchy kernel isIn the rotated coordinates, $E_k=[2\pi |B|(\tau +iA\rho /B){]}^{-1}$. The identity ${\partial }_{\bar{z}}(1/(\pi z))=\delta$ and the Jacobian $A/|B|$ prove $D_k{E}_k=\delta$ with exactly this normalization. The denominator cannot vanish for nonzero real $w$ off the unit circle. The kernel is analytic separately inside and outside that circle, is $O(k)$ at zero and $O(k^{-1})$ at infinity. Thus $M=T_{k}f$ and the displayed $\mu$ have the corresponding spectral behavior for the smooth decaying data. In particular $\mu =O(k^{-1})$ at infinity and is bounded, indeed $O(k)$, at zero.

The limiting unattenuated inverse on any smooth decaying input $h$ isOne may verify this directly in the rotated Fourier variables: $D_k$ has multiplier $iA\xi -B\eta$, so its inverse has a pole at $\xi =-iB\eta /A$. As $B\to 0^{+}$, the positive transverse frequencies have the negative tail primitive and the negative transverse frequencies have the positive past primitive; for $B\to 0^{-}$ these allocations reverse. Splitting with the Fourier projectors ${\Pi }_{\pm }$ yields exactly the two formulas above. Applying them to $f$ reproduces the assumed limits of $M$ and checks the signs.

Now apply the same limiting inverse to $h=e^{M^{\pm }}g$. Because $a_{\pm }$ are independent of $\tau$, its line integral is $e^{a_{\pm }}{A}_f$, while its tail is $e^{a_{\pm }}{\int }_{\tau }^{\infty }{e}^{-F}gds$. HenceThe unknown incomplete line integral cancels in their difference. All remaining quantities are obtained from the measurements and known attenuation:Each projection acts in $\rho$ at fixed $\theta$, before evaluating $\rho =-x_1\sin \theta +x_2\cos \theta$ and $\tau =x_1\cos \theta +x_2\sin \theta$. Pulling a transverse-dependent exponential outside a projector would be incorrect.

The sectionally analytic $\mu$ therefore solves an additive Riemann-Hilbert problem on the counterclockwise unit circle, with the plus value inside. The normalized Cauchy integral formula isThe Sokhotski–Plemelj theorem gives its jump; boundedness at zero and decay at infinity make it unique by the Liouville theorem. Its expansion at infinity isSubstitute into the original equation. The $k{\partial }_{x_1}-ik{\partial }_{x_2}$ part contributes at order one, the reciprocal derivative and $f\mu$ vanish at that order. This derives the projection formula for inverse attenuated Radon transform:Together with the boxed forward transform this is the requested pair. The derivation uses no small-attenuation approximation. For smooth compactly supported data the operations above are justified directly; sufficient integrable decay permits the same limiting construction.

As a normalization check, set $f=0$. Then $S=F=0$, $A_f=\mathcal{R}g$ and $J=(P^{-}+P^{+})\mathcal{R}g=iH\mathcal{R}g$. Since the line transform is independent of $\tau$ and $e^{i\theta }({\partial }_{x_1}-i{\partial }_{x_2})={\partial }_{\tau }-i{\partial }_{\rho }$, the formula reduces tothe full-angle inverse Radon transform. For the radial Gaussian $g=e^{-|x{|}^2}$, $\mathcal{R}g=\sqrt{\pi }{e}^{-{\rho }^2}$ and ${\partial }_{\rho }H\mathcal{R}g(0)=2$, so the reconstructed value at the origin is $1$, checking the $1/(4\pi )$ factor.

In single-photon emission computed tomography, $g$ represents the internal emission density, $f$ the known absorption coefficient and $A_f$ the directed detector counts. Each path integral weights a source point by its survival probability to the detector. Thus the attenuated Radon transform is the forward measurement model and its derived inverse reconstructs the emission distribution, providing the mathematical basis of SPECT.

On the active interval the excess temperature satisfies ${\theta }_2^{\prime }+a{\theta }_2=-1$. Multiplication by the integrating factor $e^{at}$ and integration from time one givesThere is no impulse in this equation, so continuity supplies ${\theta }_2(1)=(T_0-T_{\infty })e^{-a}$. ThereforeThe equivalent correction to the common unforced temperature is $-(1-e^{-a(t-1)})/a$, which vanishes continuously at the beginning of the interval.

Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.

On an interval where $y>0$, the Bernoulli differential equation is transformed by $u=y^{1-p}$:For $f_1=1$, $f_2=x$, put $s=1-p\ne 0$. An integrating factor $e^{-sx}$ solves $u^{\prime }-su=sx$, giving $u=C{e}^{sx}-x-s^{-1}$. Thus the positive branches areon intervals where the expression in parentheses is positive. The transformation omits the separate solution $y\equiv 0$. For $0<p<1$, the right-hand side need not be locally Lipschitz continuous at zero, so nonnegative solutions can also join zero and positive branches where the original equation permits; the formula describes every strictly positive segment.

For the autonomous case, $y^{\prime }=y-{\alpha }^2{y}^p$. Its two nonnegative equilibrium points areAt the positive equilibrium point, $f^{\prime }(y_{\ast })=1-p$. If $p>1$, $f(y)>0$ below $y_{\ast }$ and $f(y)<0$ above it; hence $0$ is unstable and $y_{\ast }$ is asymptotically stable. Every strictly positive solution tends to $y_{\ast }$ forward in time.

If $0<p<1$, those signs reverse: $0$ is stable from the nonnegative side and $y_{\ast }$ is unstable. The stability at zero follows from the phase-line signs, not a finite derivative there. For initial value $0<y_0<y_{\ast }$, the transformed equation is $u^{\prime }=s(u-{\alpha }^2)$ with $s>0$. The solution reaches zero atand remains zero thereafter as a nonnegative forward solution. This is finite-time extinction in a sublinear Bernoulli equation. Initial values above $y_{\ast }$ increase without bound; the sublinear negative term prevents neither that growth nor the instability of $y_{\ast }$.

The forcing vector has the eigenvector decompositionThere is no component in the zero eigenvalue direction. The scalar inhomogeneous linear differential equations in this basis areFor the first, a particular solution is $(\lambda +1)e^{2t}/4$. For the third, multiplying by the integrating factor $e^{-2t}$ gives $(e^{-2t}{z}_{+}{)}^{\prime }=\lambda -1$. Consequently for every real $\lambda$When $\lambda =1$, the resonant component vanishes and a particular solution is $\frac{1}{2}{v}_{-}{e}^{2t}$, of the stipulated constant-vector form. If $\lambda \ne 1$, substituting $c{e}^{2t}$ would require $(2I-M)c=2(-\lambda ,1,\lambda {)}^T$. In the eigenbasis the left side has zero $v_{+}$ component, while the right side has component $\lambda -1$. This is impossible. The factor $t$ is therefore unavoidable: it is resonance in a differential equation.

Let $\Delta M=M_1-M_0$ and retain only terms first order in $\mu /M_0$ and $\rho /M_0$. The outgoing event horizon equation expands toHere ${\kappa }_0$ is the initial Schwarzschild surface gravity. The integrating factor $e^{-{\kappa }_{0}v}$ gives $d(e^{-{\kappa }_{0}v}\rho )/dv=-e^{-{\kappa }_{0}v}\mu /(2M_0)$. Future stationarity imposes $\rho =2\Delta M$ for $v>v_0$ and removes the growing homogeneous solution. Integrating from $v$ to infinity yields the teleological response of a Vaidya event horizon:The integral is finite because $\mu$ is bounded. Integrating by parts also givesFor $v>v_0$, this reduces to $2\Delta M$. For $v<0$, the first form becomesConsequently $r_H\to 2M_0$ as $v\to -\infty$. The positive constant $C$ shows that this event horizon is already expanding in the initially vacuum region. For monotone accretion, $\rho -2\mu \ge 0$, so it lies outside the instantaneous apparent horizon to first order. All displayed radius corrections are understood up to $O((\Delta M{)}^2/M_0)$.

A differential one-form $Pdx+Qdy$ is an exact differential if there is a differentiable function $F$ with $P=F_x$ and $Q=F_y$, so the form equals $dF$. For continuously differentiable $P,Q$, equality of mixed partial derivatives gives the necessary conditionA local differential condition need not by itself give a global potential on a domain with holes; no such issue arises for the explicit primitives here.

For the first form, $P_y=2y=Q_x$ and direct differentiation givesFor the second, $P_y=2y$ but $Q_x=y^2$. These are not equal on any nonempty open subset of the plane, so that form is not an exact differential. Nevertheless,Thus one suitable pair is $f(x,y)=x{e}^y$, $g(y)=y^2{e}^{-y}$. This factorization is smooth even at $y=0$. On regions where $y\ne 0$, its reciprocal factor $e^y/y^2$ is an integrating factor which turns the original form into $df$.

A kinetic model of a dilute gas. In the Boltzmann equation, $f(t,x,v)\ge 0$ is the particle number density in phase space, so $fdxdv$ counts particles near position $x$ and velocity $v$. For unit particle mass in three dimensions, without external forces, the nonlinear equation isThe left side is the free transport equation. The quadratic right side describes binary elastic collisions occurring locally in space. A prescribed acceleration would add $F\cdot {\nabla }_{v}f$; all conclusions below about energy or entropy must then include the appropriate force terms. Assume enough smoothness, velocity decay and finite moments to justify the calculations, and either periodic space or decay that removes spatial boundary fluxes. These are assumptions on a classical solution, not a general existence theorem.

The dilute-gas approximation neglects simultaneous many-body collisions and treats the interaction region as small relative to macroscopic spatial scales. Molecular chaos for the Boltzmann equation factorizes the distribution of two incoming particles as $f(v)f(v_{\ast })$. It applies to incoming collision pairs; it does not assert that an interacting many-particle system stays exactly independent. This approximation closes the equation at the one-particle level and explains its nonlinearity.

Collision geometry and the gain-loss operator. Let $g=v-v_{\ast }$ and $n\in S^2$. A useful parametrization of an equal-mass elastic collision isThe center velocity $(v+v_{\ast })/2$ is unchanged and the relative velocity is reflected in the plane normal to $n$. ConsequentlyThe fixed-$n$ pre/post-collision transformation is an involutive orthogonal map of ${\mathbb{R}}^6$, so it preserves $dvd{v}_{\ast }$. Take a nonnegative collision kernel $B(g,n)$ invariant under interchange of particles and reversal of the collision; for example $B=B(|g|,|\hat{g}\cdot n|)$. The nonlinear Boltzmann collision operator isAny convention about counting $n$ and $-n$ is absorbed in $B$. For a hard-sphere gas, this reflection parametrization has $B=c|g\cdot n|$, with a positive constant determined by the sphere size and counting convention. Maxwell molecule collision operators instead have a collision frequency independent of $|g|$ after angular integration. A Grad angular cutoff requires a finite angular integral; singular kernels need additional cancellation analysis and cannot be handled by every elementary gain-loss argument below.

Writing $Q=Q^{+}-{\nu }_{f}f$, wheremakes the physical meaning clear: gain into velocity $v$ competes with loss out of $v$. Along a free characteristic curve, put $F(t)=f(t,x+tv,v)$. Then $F^{\prime }+{\nu }_f(t,x+tv,v)F=Q^{+}(t,x+tv,v)$. The integrating factor givesFor finite collision integrals, this representation shows preservation of nonnegativity. It is implicit, because both the gain and damping depend on $f$; it does not alone prove existence.

The weak collision identity and conserved quantities. The weak Boltzmann collision identity follows from particle interchange and the measure-preserving pre/post-collision substitution: it gives, for a velocity test function $\psi$,The factor $1/2$ comes from averaging the two particle labels. Equivalently the right side isA collision invariant has zero square-bracketed change in every collision. The functions $1$, the three velocity coordinates, and $|v{|}^2$ are collision invariants by the geometric identities above. Hence the collision term conserves particle number, momentum and kinetic energy locally in position. Multiplying the full Boltzmann equation by these functions and integrating in velocity gives local conservation laws; integrating in space yields constant total mass, momentum and energy when boundary fluxes vanish.

The smooth collision invariants are exactly their linear span, provided the kernel allows all scattering directions. Here is a direct proof. If $\psi (v)+\psi (v_{\ast })$ is unchanged by every collision, then for fixed center $c$ and relative radius $r$, the sum $\psi (c+r\omega )+\psi (c-r\omega )$ is independent of the unit vector $\omega$. Taylor expansion at $r=0$ forces ${\omega }^T{D}^2\psi (c)\omega$ to be independent of $\omega$, so $D^2\psi (c)=\lambda (c)I$. Mixed derivatives vanish. Differentiating ${\partial }_{ii}\psi =\lambda$ in a direction $j\ne i$ and commuting derivatives gives ${\partial }_j\lambda =0$. Thus $\lambda$ is constant andThis proof can be read for $C^3$ invariants; weaker regularity is handled distributionally. The multidimensional scattering hypothesis matters: in one dimension equal-mass elastic collisions only exchange the two velocities, and every function is then a collision invariant.

Entropy dissipation and the H theorem. Define the Boltzmann H functional $H(f)=\int f\log fdxdv$, with $0\log 0=0$. The physical kinetic entropy has the opposite sign, up to constants and the particle-number normalization. Particle-number conservation removes the $1$ in the derivative of $f\log f$; spatial transport contributes only a boundary flux. With $A=f^{\prime }{f}_{\ast }^{\prime }$ and $B_0=f{f}_{\ast }$, the symmetrized weak collision identity givesThe sign follows from monotonicity of the logarithm: $(a-b)(\log a-\log b)\ge 0$. Zeros are treated by the usual extended nonnegative limit. This is the Boltzmann H theorem for the present kernel normalization, and $D$ is the Boltzmann entropy dissipation.

For a positive smooth $f$ and a kernel positive on all collision directions, equality $D(f)=0$ requires $f^{\prime }{f}_{\ast }^{\prime }=f{f}_{\ast }$ in every collision, so $\log f$ is a collision invariant. Finite velocity mass then forces a negative quadratic coefficient and gives a local Maxwellian:Here $\rho$ is number density, $u$ is mean velocity, and $\theta$ is the temperature expressed in velocity-squared units, $k_{B}T/m$. Completing the square proves the formula; conservation of momentum and kinetic energy makes $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$, so $Q(M,M)=0$. Vacuum is another collision equilibrium.

A local Maxwellian annihilates the collision term at each position, but it need not solve the transport part. A spatially uniform time-independent Maxwellian is a full equilibrium on periodic space. On all of ${\mathbb{R}}_x^3$ it has infinite total mass unless it is vacuum. Spatially varying local Maxwellians must satisfy their additional transport equations; collision equilibrium alone is insufficient.

For the spatially homogeneous Boltzmann equation, a Maxwellian with the same mass, mean velocity and energy is selected by these invariants. The relative entropy $\int f\log (f/M)dv$ is nonnegative. Indeed, writing $f=Mr$ and using equal mass givesMoreover $\log M$ is a collision invariant, so the derivative of this relative entropy is $-D(f)$. Thus $M$ minimizes $H$ at fixed conserved moments. Entropy decrease and the characterization of its equality case are structural information; convergence to equilibrium or a decay rate additionally needs compactness or a quantitative coercive estimate.

Moment equations and fluid closure. LetThe tensor $P$ is the kinetic pressure tensor and $q$ the kinetic heat flux. Put $e=(2\rho {)}^{-1}\int |v-u{|}^{2}fdv$. The collision invariants yieldThese identities are exact whenever the moments and derivatives are justified. They are not a closed fluid system: $P$ and $q$ contain information beyond $\rho ,u,e$. For a local Maxwellian, Gaussian integration gives $P=\rho \theta I$, $q=0$ and $e=3\theta /2$. Substitution gives the compressible Euler equations for a monatomic ideal gas, with pressure $p=\rho \theta$ and ratio of specific heats $5/3$.

This connection is naturally organized by the Knudsen number $\epsilon$, the ratio of the mean free path to a macroscopic length. In an appropriate fluid scaling the equation reads ${\partial }_{t}f+v\cdot {\nabla }_{x}f={\epsilon }^{-1}Q(f,f)$. Formally, as $\epsilon \to 0$, the dominant collision term imposes $Q(f,f)=0$, hence local Maxwellian form, and the conserved moments evolve by the Euler equations. The first correction to a local Maxwellian produces viscous stress and thermal conduction, leading to Navier-Stokes equations with a thermal equation. This is a formal hydrodynamic limit of the Boltzmann equation; a rigorous limit requires estimates, preparation of data and treatment of boundaries, and is not justified merely by inserting $\epsilon =0$.

Linearization and its dissipative structure. Around a fixed positive Maxwellian $M$, write $f=M(1+h)$. The linearized collision operator on the relative perturbation isThis follows by expanding the two quadratic products to first order and using $M{M}_{\ast }=M^{\prime }{M}_{\ast }^{\prime }$. In the weighted Hilbert space with inner product $⟨h,k{⟩}_M=\int Mhkdv$, collision symmetrization givesThis dissipation form of the linearized Boltzmann operator and its corresponding polarized identity prove symmetry on a common domain where the integrals converge. Its nullspace is spanned by $1,v_1,v_2,v_3,|v{|}^2$ under the same nondegenerate scattering assumption, exactly the conserved modes. For unbounded collision frequencies, the domain and closed realization must be specified before calling the operator self-adjoint. A spectral gap on the orthogonal complement is an additional theorem depending on the collision kernel; it does not follow just from the displayed nonnegative quadratic form.

Analytic scope and irreversibility. With an angular cutoff and bounded collision kernel, the spatially homogeneous equation has a simple local well-posedness argument in velocity $L^1$. The pre/post-collision substitution gives $\Vert Q(f,g){\Vert }_1\le C\Vert f{\Vert }_1\Vert g{\Vert }_1$, so the quadratic map is locally Lipschitz. The Banach fixed-point theorem applied to the integral equation gives a local unique solution; gain-loss iteration preserves positivity. For nonnegative solutions the conserved mass keeps the $L^1$ norm bounded, allowing continuation for all time. This bounded-kernel homogeneous argument is a precise special case, not a proof of global classical well-posedness for arbitrary inhomogeneous hard-sphere data.

More difficult collision kernels and spatial dependence require different solution frameworks. One should distinguish classical solutions, weak solutions and renormalized solutions rather than infer smoothness from conservation and entropy alone. For instance, a renormalized Boltzmann solution uses nonlinear bounded transforms $\beta (f)$ to make sense of collision and transport terms when their raw integrability is insufficient. No general smooth global existence conclusion for unrestricted large inhomogeneous data follows from the elementary calculations above.

Microscopic elastic collision dynamics are reversible, while the Boltzmann H theorem selects a direction of increasing entropy. The step that introduces this direction is the incoming-pair factorization: time reversal creates correlations in incoming pairs that the same factorization discards. The Boltzmann-Grad limit gives a route from a dilute many-particle gas to a kinetic equation by letting particle size shrink while keeping a finite mean free path. Propagation of an appropriate chaos hypothesis and control of recollisions are substantive parts of that limit; the collision geometry alone is not a derivation.

Retain the derivative marks and coefficients from the printed system. In matrix form it is $M\dot{u}+Ku=b(t)$, withSince $\det M=11$, multiplication by its inverse givesThe linear transformation $u=x+y$, $v=x-y$ diagonalizes this system:Both start at zero. An integrating factor gives $u=v=e^{-t}-e^{-3t}$, soDirect substitution into the two original equations confirms the different forcing coefficients; the exact cancellation in $y$ depends on those coefficients.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

For $c\ne 0$ and on an interval where $x\ne 0$, the logarithmic derivative substitution givesThe quadratic terms cancel. Multiplying by $cx$ gives the linearization of a Riccati equation:Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order linear ordinary differential equation, solvable by an integrating factor.

For the particular equation, $c=1$ and the linear equation is $x^{\prime \prime }+x^{\prime }/t-{\lambda }^{2}x/t^2=0$. Use the PDF's logarithmic coordinate $\tau =\log t$ and write $X(\tau )=x(e^{\tau })$. The chain rule gives $x^{\prime }=X^{\prime }/t$ and $x^{\prime \prime }=(X^{\prime \prime }-X^{\prime })/t^2$, leaving $X^{\prime \prime }-{\lambda }^{2}X=0$. Thus $x=A{t}^{\lambda }+B{t}^{-\lambda }$, as also follows from the Euler-Cauchy equation. Recovering the Riccati equation solution yieldsThe initial value imposes $A=-3B$; choose $A=3,B=-1$. ThereforeThe denominator vanishes only at $t_{\ast }>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim 1/(t-t_{\ast })$. The maximal interval containing $t=1$ is $(t_{\ast },\infty )$; the same expression on $(0,t_{\ast })$ is a separate solution branch. These are poles of a Riccati solution from zeros of its linearizing solution.

Assume for this spectral construction that the source and known attenuation are sufficiently smooth and decaying, for instance compactly supported, and that $\mu \ge 0$ for the physical interpretation. The following steps identify both the forward attenuated Radon transform and the route to its inversion.

First write the complex transport operator asFor $k=e^{i\theta }$ this is the real directional derivative ${\mathbf{d}}_{\theta }\cdot \nabla$, with ${\mathbf{d}}_{\theta }=(\cos \theta ,\sin \theta )$. Put ${\mathbf{n}}_{\theta }=(-\sin \theta ,\cos \theta )$ and $x=p{\mathbf{n}}_{\theta }+s{\mathbf{d}}_{\theta }$. The spectral equation on the unit circle becomesIts minus sign fixes the appropriate endpoint condition: use $F(p,+\infty )=0$. The integrating factor givesThus the measured quantity at the opposite end of the line isThe exponent is the attenuation accumulated between the source point and the detector at the negative end. The common convention with detector at the positive end is the same transform after reversing the direction, ${\mathcal{R}}_{\mu }^{+}f(p,\theta )={\mathcal{R}}_{\mu }^{-}f(-p,\theta +\pi )$. Using an incoming zero condition at the negative end while retaining the printed minus sign would instead give a growing integrating factor, not physical attenuation.