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A frailty model uses an unobserved nonnegative random hazard multiplier to represent variation in individual risk. In a proportional frailty model, conditional on $U=u$ the cumulative hazard function is $u{H}_0(t)$, giving survivor $e^{-u{H}_0(t)}$. Averaging over the frailty random variable by the law of total probability yieldsthe Laplace transform of the frailty distribution evaluated at the baseline integrated hazard. This also applies to atoms by integration against the probability measure; the printed Dirac expression is not an ordinary continuous density.

In the given two-atom case, the Dirac delta function notation assigns probabilities $1/4$ and $3/4$ to $U=4/7$ and $U=8/7$. These have mean one, fixing the frailty scale relative to the baseline hazard. With $H_0(t)=7t/4$, the two conditional integrated hazards are $t$ and $2t$. ThereforeThe frailty distribution among survivors shifts toward the smaller multiplier $4/7$ over time, reproducing the declining population hazard and limit one without changing either individual's constant conditional hazard.

For a Poisson distribution of parameter $\lambda$, its probability generating function is $\exp \{\lambda (z-1)\}$. Applying the law of total expectation to the Poisson mixture givesThusFor $|z|\le 1$ the absolute value of the integrand is bounded by one, so this identity always exists as a Laplace transform of the positive mixing variable. An extension to $z>1$ requires the corresponding moment-generating function to be finite; the conditional computation itself does not guarantee positive exponential moments.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

Substitution of a normal mode gives the dispersion relationFor real $k$, $\sigma$ is the temporal growth rate. Since $\beta \ne 0$, its supremum is finite exactly whenFor $\beta <0$, arbitrarily short wavelengths grow arbitrarily fast. For $\beta >0$, the finite maximum temporal growth rate isThe Briggs-Bers criterion starts the inverse temporal Laplace transform above all temporal singularities and then deforms its contour downward while following the spatial roots. A finite growth bound supplies such an initial contour and a causal, high-frequency-controlled Green function. Unbounded temporal growth prevents that standard construction.

For absolute wave-packet instability, a candidate spatial pinch point must satisfy $D=0$, $D_k=0$, and $\mathrm{Im}\omega >0$. HeregivesThus candidate growing saddles require $\gamma <0$ at $k_0$, or $\gamma <{\alpha }^2/(4\beta )$ at $k_{\pm }$. Together with $\beta >0$, existence of at least one such candidate requires $\gamma <{\alpha }^2/(4\beta )$.

A growing double root is not sufficient: the roots must pinch the spatial inversion contour from opposite sides. Collisions of branches originating in the same spatial half-plane do not obstruct the relevant deformation. This distinction is part of the Briggs-Bers criterion; it is stated, for example, in the primary study doi.org/10.1017/jfm.2016.195.

An explicit false spatial saddle in quartic dispersion is $\alpha =-2$, $\beta =1$, $\gamma =1/2$. Its candidates $k=\pm i$ have $\omega =i/2$, but all real modes have $\sigma =-k^4-2k^2-1/2<0$. Each imaginary collision joins two branches in the same half-plane; neither is a relevant pinch. For $\omega =i\Omega$, the spatial roots obey $k^2=-1\pm \sqrt{1/2-\Omega }$, making those same-half-plane collisions transparent as $\Omega \downarrow 1/2$.

For this particular real, even dispersion relation one can also establish the actual threshold directly. At the origin its impulse Green function isLaplace method selects the real maximum, and gives a positive prefactor times $t^{-1/2}{e}^{{\sigma }_{\max }t}$ because $\alpha \ne 0$. Therefore the actual absolute wave-packet instability condition isThese are sufficient for this model as well as necessary. They follow after identifying relevant real saddles; the earlier algebraic double-root test alone lacks the pinch information. Equality is marginal, not exponential absolute growth.

Use a constant porosity $\varphi$ and let $v$ denote pore velocity. Darcy law gives the Darcy velocity $q=\varphi v$. If the velocity convention has already absorbed porosity, set $\varphi =1$ in the following formulas. Assume positive permeability of a porous medium throughout the layer, so $k_0>0$ and $k_0+k_1>0$.Here $\mu$ is dynamic viscosity. The flux-weighted residence time in a porous layer is the pore volume divided by throughput. Equivalently, weighting each streamline's transit time $L/v(y)$ by its inlet flux $v(y)dy$ givesThis is also the advective travel scale after cross-layer mixing has homogenized the concentration. For resident-weighted transit time in a layered flow, sampling the initial fluid uniformly by resident volume, while suppressing cross-stream exchange, is a different experiment: its mean iswith the continuous $k_1=0$ limit $\mu \varphi L^2/(\Delta P{k}_0)$. Specifying the sampling convention avoids confusing the reciprocal of the mean speed with the mean reciprocal speed.

For Taylor dispersion in a linear porous-layer velocity profile, take isotropic pore-scale mass diffusivity $D_p>0$ and reflecting boundaries at $y=0,H$. Write $v-U=\gamma (y-H/2)$, where $\gamma =\Delta P{k}_1/(\mu \varphi LH)$. Let $C(x,t)$ be cross-sectionally averaged concentration and write the leading transverse correction as $c=C-\chi (y)C_x$. Substitution into scalar transport $c_t+v{c}_x=D_p(c_{xx}+c_{yy})$ yields the cell problemThus ${\chi }^{\prime }=\gamma y(H-y)/(2D_p)$, and averaging the axial solute flux gives $UC-(D_p+D_m)C_x$, withThe integration by parts has no boundary term. It also proves nonnegativity, even when $k_1<0$. The total effective mass diffusivity is $\mathcal{D}=D_p+D_m$. For anisotropic pore dispersion, the denominator uses the transverse coefficient and the added molecular term uses the axial coefficient.

There are two separate tests for significance. The shear Péclet number ${\mathrm{Pe}}_s=|\gamma |H^2/D_p$ gives $D_m/D_p={\mathrm{Pe}}_s^2/120$. A large value implies substantial enhancement of axial spreading. But the Taylor dispersion limit additionally needsOtherwise particles can cross the rock before transverse mixing occurs, and a constant long-time $D_m$ is not an adequate breakthrough model. The relative front width is of order $\sqrt{\mathcal{D}/(UL)}$ once the long-time model applies.

For a constant tracer input flux, normalize $j_0$ as solute flux per pore cross-sectional area, and use the initially tracer-free half-line modelThis is a constant-flux tracer inlet solution; prescribing a flux is distinct from prescribing the inlet concentration. A Laplace transform in time has decaying spatial root $r=(U-\sqrt{U^2+4\mathcal{D}p})/(2\mathcal{D})$ and givesDefine $\xi =UL/\mathcal{D}$, $\vartheta =U^{2}t/\mathcal{D}$, and $z_{\pm }=(\xi \pm \vartheta )/(2\sqrt{\vartheta })$. Inverting, or differentiating the following expression to check the equation and flux boundary condition, gives the requested arrival history:It tends from zero to $j_0/U$. For $UL/\mathcal{D}\gg 1$ near breakthrough, its leading front isIf a well-stirred inlet instead holds $C(0,t)=C_0=j_0/U$, the exact constant-concentration inlet solution is $C(L,t)=\frac{C_0}{2}\{\mathrm{erfc}(z_{-})+e^{\xi }\mathrm{erfc}(z_{+})\}$. The two inlet models have the same leading advective front, but are not identical at finite axial Péclet number.

The printed differential in the transform integral is $dz$; it must be $dt$, since $z$ is the transform parameter. We prove the Newman Tauberian theorem in this corrected interpretation. Setalmost everywhere. The finite-interval transform $F_T$ is entire.

Fix $R>0$. Because the analytic domain contains the whole imaginary axis, compactness supplies a $\delta >0$ such that the thin rectangle $-\delta \le \mathrm{Re}z\le 0$, $|\mathrm{Im}z|\le R$ lies in the domain. Let $C_{+}$ be the right semicircle of radius $R$, oriented from $-iR$ to $iR$. Join its endpoints by a leftward path $L$ along the other three sides of that rectangle. This is a closed positively oriented contour.

Use contour damping for bounded Laplace transforms withThe residue theorem givesOn $C_{+}$, with $x=\mathrm{Re}z>0$,Thus the integrand has modulus at most $2M/R^2$, and this half-circle contributes at most $M/R$ after division by $2\pi$.

On $L$, the $F$ term tends to zero as $T\to \infty$: every interior point of the path has negative real part, $F$ is bounded on this fixed compact path, and the remaining kernel factor is bounded because the path avoids zero. Dominated convergence applies, with the two endpoints irrelevant to the path integral.

For the $F_T$ term, deform $L$ to the left semicircle $C_{-}$ of radius $R$. This deformation uses only the entire function $F_T$; it does not demand that $F$ extend across a large left half-disk. Both paths lie to the left of zero and their enclosed deformation region avoids the kernel pole. On $C_{-}$, for $x<0$,The same circle factor gives another bound $M/R$. ConsequentlyThe radius $R$ is arbitrary, soThis proves convergence of the ordinary improper integral, not merely a damped limit.

For the weakened domain hypothesis, take $f(t)=1$. Its Laplace transform is $F(z)=1/z$, analytic on the open right half-plane, but ${\int }_0^{T}f(t)dt=T$ diverges. Analyticity only in the open right half-plane is insufficient.

For $\mathrm{Re}s>0$, integration by parts recursively in the Laplace transform givesThe initial exponential boundary term vanishes, and $\mathcal{L}[1]=1/s$ starts the recursion. Here $H$ is the Heaviside step function.

Transform the displacement in time, writing $Y(x,s)={\mathcal{L}}_t[y](x,s)$. Zero initial displacement and velocity give $s^{2}Y=c^2{Y}_{xx}+g/s$. A particular solution is $g/s^3$, and the spatial homogeneous solutions are $e^{\pm sx/c}$. The prescribed far-field behavior excludes the growing exponential. The fixed boundary sets the remaining coefficient to $-g/s^3$, henceSince $\mathcal{L}[t^2/2]=1/s^3$, the Laplace transform time-shift rule givesEquivalently, $y=g{t}^2/2$ for $t\le x/c$, and $y=gxt/c-g{x}^2/(2c^2)$ for $t\ge x/c$. This solves the forced wave equation in each region and has continuous displacement and first derivatives across the characteristic $t=x/c$, so it also solves the equation weakly there. It satisfies the fixed-end and initial conditions and equals $g{t}^2/2$ sufficiently far away at each fixed time. The end's influence travels at speed $c$.

The small root and the root near $-2$ are regular perturbation roots: they stay finite as the parameter vanishes. Substitution of power series givesFor the third root, dominant balance for algebraic roots requires $\lambda =\mu /\epsilon$. The largest terms give ${\mu }^2(\mu +1)=0$, and the missing branch is $\mu =-1$. The sum of the three roots then yieldsThus the leading roots are $-\epsilon /2$, $-2$, and $-1/\epsilon$; only the last is a divergent perturbation root. The first regular branch happens to have a zero limit, so retaining its first nonzero term is essential on the long spatial scale.

First take $\epsilon \to 0^{+}$, the decaying half-line regime. If the three exact roots are $r_j$, the exact initial value problem solution isThis follows either by solving the three initial-value equations or from the Laplace transform $\hat{y}(s)=\epsilon /(\epsilon s^3+s^2+2s+\epsilon )$. The coefficient formula has sums $\sum A_j=\sum r_j{A}_j=0$ and $\sum r_j^2{A}_j=1$.

Retain the three leading roots but compute their coefficients without expanding their denominators. Set $a=\epsilon /2$, $b=2$, $d=1/\epsilon$. A convenient composite asymptotic expansion isIt satisfies all three initial conditions exactly and retains the fast transient, the ordinary decay, and the slow decay. Its first two coefficients are $\epsilon /2+O({\epsilon }^2)$ and $-\epsilon /2+O({\epsilon }^2)$; the fast coefficient is ${\epsilon }^2+O({\epsilon }^3)$. Consequently the simpler bulk expression is $\frac{\epsilon }{2}(e^{-\epsilon x/2}-e^{-2x})$, but that expression alone does not reproduce the initial derivative layer.

The absolute error estimate isTo see its uniformity, the slow-root error is $O({\epsilon }^2)$, while its magnitude is $O(\epsilon )$ and its coefficient is $O(\epsilon )$. The uniform error bound for nearby decaying exponentials therefore gives an $O({\epsilon }^2)$ contribution even for $x=O({\epsilon }^{-1})$. The middle-root error is $O(\epsilon )$ with coefficient $O(\epsilon )$, again giving $O({\epsilon }^2)$. The fast-root error is $O(1)$ with magnitude $O({\epsilon }^{-1})$ and coefficient $O({\epsilon }^2)$, giving only $O({\epsilon }^3)$. Coefficient errors are $O({\epsilon }^2)$ or smaller. This estimate concerns $y$ itself, rather than asserting the same uniform order for all derivatives or a relative error at its zero.

For the sketch, on $x=\epsilon X$ one obtainsThus $y\sim x^2/2$ at the origin and rises smoothly after the fast layer. For $x=O(1)$ it rises towards a plateau of height approximately $\epsilon /2$, then decays on the much longer scale $x=O({\epsilon }^{-1})$. The bulk maximum lies near $x=\log (4/\epsilon )/(2-\epsilon /2)$ and has height asymptotic to $\epsilon /2$.

The sign of the parameter matters on an infinite interval. If $\epsilon \to 0^{-}$ is allowed, the singular mode grows rather than decays. For each fixed $x>0$, its dominant contribution is $y(x)\sim {\epsilon }^2\exp [(-1/\epsilon +2)x]$. The extra $2$ in that exponent is needed for relative leading accuracy at fixed $x$. The positive-parameter uniform absolute bound and decaying sketch do not extend to that regime.

The imposed pressure gradient and Darcy's law give a linear velocity profile. Use $U$ as the mean pore velocity in the contaminant advection-diffusion equation; thenThe last conditions express zero transverse solute flux under the standard closed-layer interpretation. Transverse exchange through the layer boundaries would change this cell problem and requires additional boundary data. If the prescribed mean speed is a Darcy velocity, its corresponding pore velocity is that speed divided by porosity.

The given parameter is the ratio of transverse mixing time to travel time:Thus transverse dispersion mixes the contaminant across the layer many times during its passage along the long layer. Different fluid speeds still matter: repeated exchange between slow and fast portions produces Taylor dispersion, rather than transport solely at each parcel's original speed. The averaged description applies after $t\gg H^2/D_T$ and on longitudinal scales large compared with the distance travelled during that mixing time.

Write $C=\overline{c}$ and, to leading correction order, $c=C+\chi (y)C_x+\cdots$, with $\overline{\chi }=0$. Taking $C_t\simeq -U{C}_x$ in the local equation gives the cell problemWith $\eta =y/H$, its solution isAveraging the local advection-diffusion equation now yields $C_t+U{C}_x=[D_L-\overline{(u-U)\chi }]C_{xx}$. An integration by parts in the cell problem gives $-\overline{(u-U)\chi }=D_T\overline{({\chi }^{\prime }{)}^2}$. Since ${\int }_0^1({\eta }^2-\eta {)}^{2}d\eta =1/30$, the effective equation isThis is Taylor dispersion in a linear porous-layer velocity profile. For a localized pulse away from the inlet and outlet, its mean position advances at $U$ and its longitudinal variance grows as $2D_{\mathrm{eff}}t$.

For the inlet step, use the semi-infinite inlet approximation $x>0$, $C(0,t)=1$, $C(x,0)=0$ and $C\to 0$ as $x\to \infty$ at fixed $t$. The resulting constant-concentration inlet solution isOne derivation is to Laplace transform in time: $\tilde{C}(x,p)=p^{-1}\exp [(U-\sqrt{U^2+4D_{\mathrm{eff}}p})x/(2D_{\mathrm{eff}})]$. Inverting gives the two complementary error functions. Their sum is one at $x=0$; for fixed $x>0$ both terms vanish as $t\downarrow 0$, and substitution verifies the averaged equation. Retaining only the first term gives the familiar moving error-function front far from the inlet when longitudinal advection dominates diffusion, but does not satisfy the inlet condition exactly.

For a literal finite layer $0<x<L$, a downstream boundary condition is also needed once the outlet influences the solution. The PDF's step-inlet request specifies only the initially clean region $x>0$, so the formula above is the semi-infinite/long-layer solution, not a claim of a unique finite-interval solution with unspecified outlet data.

For $a>0$, the Heaviside step function vanishes before $a$ and is one afterwards. For $\mathrm{Re}p>0$,The value assigned at the single jump point does not affect this Laplace transform. The positive real part ensures convergence at infinity.

Use the Bromwich inversion formula in the real variable $\omega$, with $p=\gamma +i\omega$:When the relevant Laplace transforms and spatial integral transforms are explicit, every sample of this integrand can be evaluated directly, without a time-stepping approximation to the partial differential equation. Split the initial condition integral at $y=x$: the two pieces of $R_q$ are linear combinations of $e^{\pm qy}$, so truncated spatial exponential transforms suffice. For real data the negative-frequency half is the complex conjugate of the positive-frequency half.

A practical numerical integration is to truncate to $|\omega |\le \Omega$, apply an adaptive quadrature rule, and independently increase $\Omega$ and refine the quadrature. Choose $\gamma >0$ large enough to remain to the right of all singularities, but avoid an unnecessarily large $e^{\gamma t}$. Evaluate hyperbolic function ratios in scaled form; for example,This prevents overflow when $\mathrm{Re}qL$ is large. The boundary contributions at an interior point are damped by $e^{-x\mathrm{Re}q}$ or $e^{-(L-x)\mathrm{Re}q}$, with $\mathrm{Re}q\sim \sqrt{|\omega |/2}$. Near an endpoint more frequencies are needed; at the endpoint itself use the prescribed Dirichlet boundary condition. The initial condition resolvent has a generally only $O(1/\omega )$ tail. Treat its oscillatory tail accurately, subtract a known transform with the same leading term, or evaluate that contribution separately with the heat kernel.

An equally useful check is the eigenfunction expansion. DefineThe causal modal formula isIt follows either from the heat kernel formula or directly from integration by parts against a sine eigenfunction. Exponential-transform formulas evaluate the time integral explicitly when available. The initial condition contribution has a Gaussian function cutoff in $n$ for each positive time. Nonzero endpoint forcing produces a more slowly convergent sine series tail. A boundary lifting $\ell (x,t)=(1-x/L)g_0(t)+(x/L)e^{aL}h(t)$ gives $w=\ell +z$, where $z$ has zero Dirichlet boundary conditions and forcing $-{\ell }_t-a^2\ell$; expanding $z$ gives better convergence and imposes the endpoint values exactly. At very small times the method of images is often more efficient than retaining many modes.

Check convergence by increasing both the frequency cutoff and quadrature resolution, and compare with a separately truncated causal modal or image-kernel evaluation. Causality in finite-time Laplace contour inversion means a leftward contour deformation must respect both resolvent operator poles and growth of the transformed data. In particular, the transform of data extended by zero after $T$ can grow like $e^{-pT}$ in the left half-plane: for $t<T$ one cannot close that contour indiscriminately and discard its large arc. The modal forcing integral with upper limit $t$ avoids this causality error.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.