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For $u_t+u_{xxx}=0$ on $x>0$, define $V(x,p)={\int }_0^{\infty }{R}_p(x-y)u_0(y)dy$ and let $G$ be the time Laplace transform of $u_x(0,t)$. The transformed solution with spatial decay isOnly one characteristic root has negative real part, so one scalar boundary derivative fixes the remaining homogeneous mode. The Bromwich inversion formula supplies an integral representation involving only the initial and boundary data. For general linear differential equations on a half-line, the number and form of the necessary boundary data depend on the decaying spatial roots; Fokas and Wang study the corresponding boundary maps for linear dispersive equations.

For $\mathrm{Re}p>0$, put $\lambda =p^{1/3}$ using the principal cube root and $r_{1,2}=\lambda e^{\pm i\pi /3}$. The decaying Green function for ${\partial }_s^3+p$ on the real line isThe identities among the three characteristic roots show that $R_p$ and $R_p^{\prime }$ are continuous, while $R_p^{\prime \prime }$ has jump one. Its distributional derivative therefore satisfies $({\partial }_s^3+p)R_p={\delta }_0$. This is the kernel of the resolvent operator $(p+{\partial }_s^3{)}^{-1}$, obtained by the time Laplace transform of the Airy equation.

Removing factorial growth from the coefficients of a formal inverse-power series produces its Borel transform. When continuation and exponential growth bounds allow it, ${\int }_{\mathcal{C}}{e}^{-zt}\mathcal{B}(t)dt$ reconstructs the series by Laplace transform. A singularity on the integration direction requires a lateral contour prescription. A simple pole contributes a residue when that prescription is changed, explaining how Stokes phenomenon arises from otherwise identical algebraic expansions.

A bounded function on $[0,\infty )$ has a Laplace transform satisfying the displayed bound, by integration of $|y(t)e^{-pt}|$. Consequently its Laplace transform cannot have a pole of order two or higher at $p=0$. This can eliminate the arbitrary constant introduced when multiplication by $t$ turns an ordinary differential equation into a first-order differential equation for its Laplace transform.

For example, transforming $t{y}^{\prime \prime }+(t-1)y^{\prime }-y=0$ with $y(0)=1$ gives $p(p+1)Y^{\prime }+(3p+2)Y=2$. An integrating factor $p^2(p+1)$ yieldsThe bound forces $C=0$, selecting $y=e^{-t}$. This selection can be justified without assuming unproved endpoint regularity for every solution: $e^{-t}$ and $t-1$ have nonzero Wronskian $t{e}^{-t}$ on $(0,\infty )$, so they form a fundamental system of solutions. Boundedness removes $t-1$, and the limit at zero fixes the remaining coefficient. The endpoint limit by itself does not select a unique solution.

Under suitable regularity and growth conditions, a function with Laplace transform $F$ is recovered by the inverse contour integralwhere the vertical line lies to the right of every singularity of $F$.

Extend a function given for nonnegative time to the real line by setting it to zero before the starting time. This turns unilateral convolution into whole-line convolution and identifies a Laplace transform slice with a Fourier transform after exponential function weighting.

A Laplace transform $G_T(p)={\int }_0^T{e}^{-ps}g(s)ds$ can grow exponentially as $\mathrm{Re}p\to -\infty$. At an evaluation time $t<T$, multiplying by $e^{pt}$ does not automatically permit closing the inverse contour to the left. Either retain the growth-bearing contour pieces, use a causal time convolution, or justify a deformation with the time-truncated transform. Resolvent-pole information alone is insufficient.

With inverse Fourier transform $G(t)=\int G(\omega )e^{-i\omega t}d\omega /(2\pi )$, a tempered retarded response vanishes for $t<0$ and is analytic in the upper frequency half-plane with suitable large-frequency growth bounds. Poles strictly below the real axis describe decaying modes; upper-half-plane poles obstruct a bounded retarded real-axis Fourier response. However causality alone does not imply stability: a retarded exponentially growing Green function can exist without an ordinary Fourier transform. Its Laplace transform inversion contour must pass above the growth poles in the frequency plane. Moving the contour to the real axis changes the boundary condition and can introduce a pre-response.

Suppose $X$ is a continuous local martingale starting at zero, with conditionally symmetric increments, and terminal conditional expectations have continuous martingale versions. ThenTo prove this, set $M_t=\mathbb{E}[e^{i\theta X_T}\mid {\mathcal{F}}_t]$. Symmetry makes $e^{-2i\theta X_t}{M}_t$ a martingale, and the Itô product rule gives $d⟨M,X⟩=i\theta Md⟨X⟩$. Consequently $e^{-i\theta X_t-{\theta }^2⟨X{⟩}_t/2}{M}_t$ is a bounded local martingale, hence a martingale. Evaluating at the endpoints proves the formula. If also $X_T\sim N(0,T)$ for all $T$, the values of the bracket Laplace transform at $1$ and $2$ force $⟨X{⟩}_T=T$ almost surely. Continuity and the Lévy characterization of Brownian motion then identify $X$ as Brownian motion.

The elapsed-time $h>0$ transition distribution of a driftless square-root diffusion starting at $x$ is the compound Poisson distributionwhere all variables on the right are independent and the empty sum is zero. Here the exponential distribution parameter is its rate. The Laplace transform of this sum iswhich is the Laplace transform of the driftless square-root diffusion; uniqueness of the Laplace transform proves the distributional identity. In particular, for $0\le v<2/h$, the exponential moment isFor $x>0$ this exponential moment is infinite when $v\ge 2/h$, since the event $N=1$ has positive probability and one exponential jump already has an infinite moment there. At $x=0$ the process remains zero and every such moment is one.

The half-line solution of $C_t+U{C}_x=D{C}_{xx}$ with initially zero concentration, unit inlet concentration for positive time, and decay at infinity at fixed time. Both complementary error functions are needed to satisfy the inlet exactly. A Laplace transform gives $\tilde{C}=p^{-1}{e}^{(U-\sqrt{U^2+4Dp})x/(2D)}$. A finite layer requires an additional downstream boundary condition.

For a passive scalar on a half-line, impose constant total solute flux at its inlet, with initially zero concentration and constant positive advection speed $U$ and mass diffusivity $\mathcal{D}$. A Laplace transform in time givesWith $\xi =Ux/\mathcal{D}$, $\vartheta =U^{2}t/\mathcal{D}$ and $z_{\pm }=(\xi \pm \vartheta )/(2\sqrt{\vartheta })$, its inverse isThe complementary error function describes the smeared front. This is a flux boundary condition, not the constant-concentration inlet solution. At large axial Péclet number, both have the leading advancing front $\frac{1}{2}\mathrm{erfc}(z_{-})$.

The displayed kernel has residue one at zero and vanishes at the endpoints of a radius-$R$ imaginary-axis segment. On either radius-$R$ semicircle, $|1+z^2/R^2|=2|\mathrm{Re}z|/R$, cancelling the inverse real-part bound on a bounded function's Laplace transform tail. Use the full transform only on a right semicircle and a thin analytic leftward path; deform the finite-interval entire transform to the large left semicircle. This proves the Newman Tauberian theorem without assuming analytic continuation across an entire left half-disk.

For functions $f,g$ on $[0,\infty )$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ Lebesgue integrable functions, their unilateral convolution, defined almost everywhere if necessary, satisfiesTaking a causal extension by zero of both functions gives the existing whole-line convolution definition. Absolute convergence and Fubini's theorem justify the product rule: put $t=\tau +u$ in the double integral, so $e^{-st}=e^{-s\tau }{e}^{-su}$. For piecewise continuous functions of exponential order $a$ and $b$, the hypotheses hold for $\mathrm{Re}s>\max (a,b)$. The Laplace transform article works out the concrete example $h(t)=e^{-2t}$ and the Heaviside step function input, whose output is $(1-e^{-2t})/2$. ^[ref].

A small step of shear stress $s_{0}H(t)$ produces shear strain $\gamma (t)=s_{0}J(t)$ from a relaxed state. This time-dependent deformation under held stress is creep. The causal constitutive convolution gives $\hat{s}=p\hat{G}\hat{\gamma }$, proving the displayed Laplace transform relation. For a Maxwell fluid, $J(t)=1/G_0+t/(G_0\tau )$: an elastic jump followed by viscous flow.

A cure model mixes an individual class with identically zero hazard function and a susceptible class with survivor function $S_{\ast }$. For $0\le \pi <1$, its representation as a proportional frailty model with $\mathbb{E}U=1$ is $U=0$ with probability $\pi$ and $U=1/(1-\pi )$ otherwise, with baseline hazard $h_0=(1-\pi )h_{\ast }$. The frailty law is a mixture of two Dirac measures, rather than an ordinary density. Its Laplace transform evaluated at $H_0$ gives the displayed survivor function. The limiting survival equals $\pi$ when $S_{\ast }(t)\to 0$; otherwise the plateau also includes susceptible individuals who never experience the event.

A driftless square-root diffusion process is the nonnegative solution of $d{X}_t=\sqrt{X_t}d{W}_t$, with zero an absorbing boundary. For elapsed time $h>0$, starting from $x\ge 0$, its Laplace transform isTo prove this, fix the final time $h$ and set $F(t,x)=\exp [-xu/(1+u(h-t)/2)]$. Differentiation gives $F_t+\frac{1}{2}x{F}_{xx}=0$, while $F(h,x)=e^{-ux}$. By Itô formula, $F(t,X_t)$ is a local martingale; it takes values in $[0,1]$, so localization and bounded convergence make it a true martingale. Taking expectations at the two endpoints proves the formula. The Markov property gives the same formula conditionally at any starting time.

A dynamic boundary condition includes a time derivative of the boundary trace. For a half-line heat equation, $u_t(0,t)+\alpha u_x(0,t)+\beta u(0,t)=f(t)$ gives a separate boundary evolution coupled to the interior flux. Its Laplace transform contains the initial trace $u_0(0)$. The sign of $u_x$ differs from that of the outward normal derivative at a left endpoint.

A function has exponential order $a\in \mathbb{R}$ if some $M>0$ and $T\ge 0$ give the displayed bound. Bounded functions, including sine, have exponential order zero; $e^{-2t}$ has exponential order $-2$. The bound need not use the smallest possible $a$. Together with piecewise continuity, it guarantees a Laplace transform for $\mathrm{Re}s>a$.

With normal modes $e^{ikx-i\omega t}$, a finite upper bound on $\mathrm{Im}\omega$ allows the inverse temporal Laplace transform contour to start above all temporal singularities. An unbounded growth rate at high real wavenumber prevents the standard causal Green function construction used in the Briggs-Bers criterion. For $\mathrm{Im}\omega =\alpha k^2-\beta k^4-\gamma$ with nonzero real $\beta$, the bound holds exactly when $\beta >0$.