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The zero initial condition turns the time Laplace transform of the heat equation intoDecay as $x\to \infty$ selectsThe boundary value givesso

A proportional frailty model specifiesIts population survival function is the Laplace transform of the frailty density,If $m=\mathbb{E}U<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m{h}_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb{E}U=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

Taking the Laplace transform in $t$ and using the zero initial condition givesThe solution bounded as $x\to \infty$ isThe transformed boundary condition isso this is the sinusoidally forced heat equation on a half-line. Integrating in $x$ gives

Use the principal square root, with a branch cut along the negative real axis. The Bromwich contour is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contributeHence

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt{p}$ are $i\sqrt{s}$ and $-i\sqrt{s}$. Their jump and the opposite contour orientations combine to giveThis is precisely Bromwich inversion with a square-root branch cut: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.

Taking the Laplace transform of the differential equation and usinggivesHenceWhenever $s$ is not an eigenvalue of $A$, the matrix $sI-A$ is invertible, and thusThis is the Laplace-transform solution of a constant-coefficient vector ODE.

Set $g=0$. The given homogeneous solution is $y(t)=e^{tA}{y}_0$, so the preceding formula saysfor every initial vector $y_0$. Equality on every vector gives the matrix identityon the common domain of convergence and in particular away from the eigenvalues of $A$. This is the Laplace transform of a matrix exponential.

Under the Cameron-Martin theorem for a linear drift, the probability that Brownian motion with drift $-\mu$ reaches $x$ isUsing the supplied Laplace transform with $\lambda ={\mu }^2/2$ givesand henceThis is the survival function of the exponential distribution with rate $2\mu$.

For $x,y>0$, the Laplace transform identityexhibits $k_1$ as an inner product of the feature functions $t\mapsto e^{-tx}$ in $L^2(0,\infty )$. More explicitly, for any real $c_i$ and positive $x_i$,Therefore $k_1$ is a positive-semidefinite kernel.

The laplace transform is $\hat{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t),dt$, and inversion iswhere the vertical Bromwich contour lies to the right of all singularities. For $t>0$ it is closed leftwards, and for $t<0$ rightwards, where the exponential decays.

Transforming the PDE and imposing boundedness as $x\to \infty$ giveswith the square root cut conventionally along $(-\infty ,0]$. ThereforeThere are poles at $p=0,-1$ (with cancellation assessed in the complete expression) and a branch point at $0$; close left for $t>0$ and wrap the cut while including the relevant residues.

A proportional frailty model has conditional hazard $h(t\mid U=u)=u{h}_0(t)$. Its conditional survival function is $e^{-u{H}_0(t)}$, so its population survival function is the Laplace transform of the frailty distribution evaluated at $H_0(t)$.

For $y_{tt}=c^2{y}_{xx}$ on $0<x<L$, initially at rest with zero displacement, apply $y(0,t)=a$ for $t>0$ and keep $y(L,t)=0$. A geometric series in the Laplace transform givesHere $H$ is the Heaviside step function, $L,c>0$, and only finitely many terms contribute at any finite time. The second step in each pair is a sign-reversed reflection from the fixed endpoint. The formula solves the wave equation classically away from the fronts and distributionally across them; the midpoint is a square wave of period $2L/c$.

For $L>0$, constant $\beta >0$ and $\mathrm{Re}p>0$, on $(0,L)$ put $b=\beta /2$, $\kappa =\sqrt{p+b^2}$ with the principal square root, andThese satisfy $(p-{\partial }_x^2-\beta {\partial }_x)u_p=(p-{\partial }_x^2-\beta {\partial }_x)v_p=0$ and the left and right homogeneous Neumann boundary conditions, respectively. The weighted Wronskian givesThe first derivative jumps by $-e^{-\beta y}$, so the resolvent integrates against $e^{\beta y}dy$. Its time Laplace transform interpretation is ${\mathcal{R}}_p={\int }_0^{\infty }{e}^{-pt}{K}_{\beta }(t)dt$. The factor $p$ records the zero eigenvalue and constant stationary eigenfunction.

In a chain $N_1\to N_2\to N_3$ with distinct decay constants, each population is a linear combination of the exponentials $e^{-{\lambda }_{i}t}$. When two consecutive constants coalesce to $\lambda$, the repeated Laplace-transform pole produces a term proportional to $t{e}^{-\lambda t}$.

For a time Laplace transform parameter with $\mathrm{Re}p>0$, the spatial characteristic roots satisfy $r^3+r+p=0$. Exactly one root $r_{-}$ has negative real part and two have positive real part. Indeed an imaginary root would force $p=-i(k-k^3)$ to be imaginary. The count is therefore constant on the connected right half-plane; for real $p>0$, the cubic has one negative real root and a conjugate pair with positive real parts. Repeated roots would satisfy $3r^2+1=0$, which again puts $p$ on the imaginary axis. Thus $r_{-}(p)$ is a nonzero holomorphic function in the right half-plane. Only $e^{r_{-}x}$ decays at positive spatial infinity.

If $X$ is a Lévy process and $T$ an independent subordinator, then $Y_t=X_{T_t}$ is a Lévy process. Conditional on the clock, increments of $X$ over disjoint clock intervals are independent; averaging over the independent stationary increments of the clock gives the same properties for $Y$. If $\mathbb{E}{e}^{iu{X}_s}=e^{-s\Psi (u)}$, its characteristic function is $\mathbb{E}{e}^{-T_t\Psi (u)}$. For standard Brownian motion, $\Psi (u)=u^2/2$, so this is the clock's Laplace transform at $u^2/2$.

A subordinator is a Lévy process with nondecreasing paths. Its nonnegative stationary increments have a Laplace transform of the formwhere $\Phi$ is called its Laplace exponent. Its standard path convention is càdlàg. Subordinators provide random clocks for subordination of a Lévy process.