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A retarded Green function obeys $G(\tau )=0$ for $\tau <0$ andFor a bounded or suitably tempered response represented by the real-axis Fourier integral, analyticity in the upper frequency half-plane is required, with suitable behaviour at infinity. For negative $\tau$, $e^{-i\omega \tau }$ decays on a large upper semicircle. If that half-plane has no singularities, contour closure gives zero response before the source. Lower-half-plane poles give factors decaying for positive time. Real poles are approached from below for a retarded prescription; at $\epsilon =0$, for example, $G_R(\tau )=\Theta (\tau )\sin \tau$.

The causality versus stability of a Green function distinction matters: an unstable initial-value problem can still have a strictly causal response that grows exponentially. Such a response has no ordinary real-axis Fourier transform. Its inverse Laplace transform must run on a frequency contour above its growth poles. Thus “upper poles violate causality” is correct only with the stable/tempered real-axis Fourier assumptions, not as an unconditional theorem about differential equations.

A mathematician associated with Laplace expansion, the Laplace transform and Laplace's equation.

For a stable first-come first-served M/G/1 queue, write $\rho =\lambda \mathbb{E}S<1$ and $\tilde{S}(s)=\mathbb{E}{e}^{-sS}$. The stationary waiting-time Laplace transform isWith finite second service moment, differentiating at zero gives the Pollaczek-Khinchine mean formula. These are the transform and mean forms of the same queueing identity.

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 the whole-line equation $A_t+a_1{A}_{xx}+a_2{A}_{xxxx}+a_{3}A=0$ with $\mathrm{Re}{a}_2>0$, the dispersion relation is $s=a_1{k}^2-a_2{k}^4-a_3$. The temporal Laplace transform of the impulse response at zero is the real-k integral of $[s+a_3-a_1{k}^2+a_2{k}^4{]}^{-1}/(2\pi )$. Factor the denominator as $a_2(k^2+{\kappa }_1^2)(k^2+{\kappa }_2^2)$ with $\mathrm{Re}{\kappa }_j>0$ for large $\mathrm{Re}s$. Partial fractions give the displayed resolvent. Square-root sheets are chosen by continuation from this causal large-s region; choosing roots independently can introduce false spatial pinch points.

For $t>0$, this Laplace transform converges at zero because $dx/[x{\log }^{2}x]$ is integrable, and at infinity because of exponential damping. Its large-$t$ behavior is governed by a logarithmic zero-endpoint singularity and begins with $1/\log t$, rather than a power of $t$.

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 $2L/c$.

The set of complex numbers $s$ for which the displayed integral is finite is the region of absolute convergence of a Laplace transform. It depends only on the real part $\sigma =\mathrm{Re}s$. Once the condition holds for one $\sigma$, it holds for every larger real part. For the exponential function $e^{-2t}$ it is exactly $\mathrm{Re}s>-2$; in general, convergence on the boundary must be checked separately. An analytic continuation outside this set need not equal a convergent defining integral.

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.

This is the Mellin transform of the function extended by zero beyond the unit interval. The change of variables $\lambda =e^{-t}$ identifies it with the unilateral Laplace transform of $h(e^{-t})$. The Bromwich inversion formula therefore gives $h(\lambda )=(2\pi i{)}^{-1}{\int }_{\gamma -i\infty }^{\gamma +i\infty }{\lambda }^{-s}H(s)ds$ for $0<\lambda <1$, with $\gamma$ in the convergence half-plane and suitable inversion hypotheses. At a jump, symmetric inversion returns the mean of the two limiting values. In particular, for $0<c<1$, the cutoff $h(\lambda )={\lambda }^q{1}_{(0,c)}(\lambda )$ has transform $c^{s+q}/(s+q)$ when $\mathrm{Re}(s+q)>0$. Multiplication by $c^s$ shifts the support threshold in logarithmic coordinates, the Laplace transform time-shift rule expressed multiplicatively.

In a chain $N_1\to N_2\to N_3$ with distinct decay constants, the Bateman equation expresses each population as a linear combination of the exponential functions $e^{-{\lambda }_{i}t}$. 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.

For positive $\epsilon$, the real-axis frequency response of the radiation-reaction oscillator pole structure excludes future runaway growth but retains an upper-half-plane pole. Closing its inverse Fourier transform upward at $t<0$ gives a nonzero response before the force. The fast pole contributes approximately $\epsilon e^{t/\epsilon }$ for negative $t$. A strictly causal retarded prescription instead includes an exponentially growing term for positive $t$ and requires a Laplace transform contour. The pre-response is a consequence of imposing boundedness at future infinity, not an unavoidable property of all initial-value solutions of the differential equation.

With packet arrival rate lambda, independent exponential payloads of mean one and byte-service rate C, let $\rho =\lambda /C<1$. The number N of unfinished packets has law $(1-\rho ){\rho }^n$. Conditional on N=n, the remaining payloads are n independent unit-mean exponentials, including the in-service residual by memorylessness. Summing their Laplace transforms gives $\mathbb{E}{e}^{-sW}=(1-\rho )(1+s)/(1+s-\rho )$. Thus W has mass $1-\rho$ at zero and, conditional on positivity, is exponential with rate $1-\rho$. The probability of at least r headers is instead ${\rho }^r$ for integer r.

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.

For thermal diffusivity $\kappa >0$ and thermal conductivity $k>0$, a semi-infinite conductor satisfies $u_t=\kappa u_{xx}$ for $x>0$. Take zero initial temperature and constant incoming heat flux $-k{u}_x(0,t)=Q$. Select the physical solution by boundedness at spatial infinity together with sufficient growth control for the time Laplace transform.

Writing the time transform as $U$ gives $U_{xx}=(p/\kappa )U$. The decaying branch is $U=A(p)e^{-x\sqrt{p/\kappa }}$, where $\mathrm{Re}\sqrt{p}>0$. The transformed flux condition gives $A=Q\sqrt{\kappa }/(k{p}^{3/2})$. Since $\mathcal{L}[2\sqrt{t}]=\sqrt{\pi }{p}^{-3/2}$, inversion gives the displayed surface temperature. More generally the full profile isDifferentiation verifies the heat equation and the applied flux; for every $x>0$ it tends to zero at initial time and decays at spatial infinity. Its surface temperature grows as $\sqrt{t}$, because the depth significantly heated by diffusion has the same scale.

A transfer function packages the response of a linear time-invariant system into a rule for turning a plant input into a plant output. Instead of solving a differential equation separately for every signal, take a Laplace transform, multiply by $G(s)$, then transform back. This also makes it possible to compare slow and fast oscillations through the continuous-time frequency response.

Here consider a continuous-time, causal, finite-dimensional linear time-invariant system with one plant input $u(t)$ and one plant output $y(t)$. With every initial state set to zero, its transfer function is defined byThe ratio $G(s)=Y(s)/U(s)$ is used where $U(s)\ne 0$ in a common Laplace convergence region; $G$ describes the system, independently of that particular plant input. Zero initial state matters: stored motion or another nonzero initial condition can contribute to the plant output even when the plant input vanishes. ^[ref]

The first-order low-pass filter below shows the same transfer function predicting both a step and a sustained sinusoid. Its pole gives a decay rate, its value at zero gives the eventual gain, and its values along the imaginary axis give the continuous-time frequency response. These descriptions make a useful bridge between a physical relaxation model and control theory calculations.

For $q_t-q_{xxx}=0$ on $x>0$, evaluate the global relation at $k=-ir/2$, where $\mathrm{Re}p>0$ and the principal cube root $r=p^{1/3}$ has positive real part. The time exponent becomes $-p$, giving the displayed Laplace transform identity. Thus the second-derivative trace is determined once the value and first-derivative traces are prescribed. The homogeneous transformed equation $u^{\prime \prime \prime }-pu=0$ has exactly two spatially decaying roots, so two independent admissible conditions are required. One condition leaves a free decaying mode; three independent conditions generally violate the global constraint.

An unbiased estimator $T$ is uniformly minimum-variance unbiased if its variance is no larger than that of every other unbiased estimator of the same parameter, for every parameter value. The Lehmann–Scheffé theorem gives such an estimator whenever an unbiased function of a complete sufficient statistic exists. Attaining the Cramér-Rao bound is sufficient for this property under its hypotheses, but is not necessary: for $T\sim \Gamma (n,\lambda )$ with $n>2$, $(n-1)/T$ is unbiased and has variance ${\lambda }^2/(n-2)$, while the information bound is ${\lambda }^2/n$. Completeness follows from uniqueness of the Laplace transform of $h(t)t^{n-1}$, so the estimator is still uniformly optimal among unbiased estimators.