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Neglect radiation in cosmology and take pressureless matter plus a separately conserved dark energy component with $p_{\mathrm{DE}}=w{\rho }_{\mathrm{DE}}{c}^2$, where $\rho$ here denotes mass-equivalent energy density. The Friedmann acceleration equation givesThus $w<-1/3$ for a positive-density component is necessary for that component to accelerate expansion, but the total must also overcome pressureless matter. For a cosmological constant, $w=-1$ and $q_0={\Omega }_{m,0}/2-{\Omega }_{\Lambda ,0}$, so acceleration requires ${\Omega }_{\Lambda ,0}>{\Omega }_{m,0}/2$. Spatial curvature of an FLRW universe enters the distance relation but has no separate term in this acceleration criterion. A Hubble diagram alone does not identify a microscopic dark energy mechanism.

For an isotropic harmonic oscillator with mass $m$, restoring constant $k$ and linear drag coefficient $b$, put $\omega =\sqrt{k/m}$ and $\gamma =b/(2m)$. Choose initial position ${\mathbf{r}}_0=(c,c)$ and tangential velocity ${\mathbf{v}}_0=\omega (-c,c)$, which would give anticlockwise circular motion without damping. For underdamping, $\nu =\sqrt{{\omega }^2-{\gamma }^2}$ givesTaking the squared norm and multiplying by $k/2$ gives the exact potential energyOnly to leading order in $\gamma /\omega \ll 1$ does this reduce to $k{c}^2{e}^{-2\gamma t}$. The exact expression has $\dot{U}(0)=0$, consistent with the initially tangential velocity, while the leading approximation does not reproduce this short-time derivative. Angular momentum obeys $L_z(t)=L_z(0)e^{-2\gamma t}$ by angular momentum decay under uniform linear drag.

For $\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0$, an initial velocity $v_0$ at zero displacement givesAll three cases begin with slope $v_0$. Only underdamping produces repeated crossings through equilibrium; critical damping and overdamping remain on the initial side of equilibrium for this particular initial condition. The overdamped long-time tail decays at rate $\gamma -s$, which becomes slow for strong damping.

For $L=(D_t+a)(D_t+b)$ with $a>b>0$, $H$ denotes the Heaviside step function. This Green function vanishes at negative time and satisfies $LG=\delta (t)$. Its positive-time part solves the homogeneous ordinary differential equation, is zero at the origin, and has right derivative one. Thus it is continuous with a unit derivative jump, which gives precisely the Dirac delta function in its second distributional derivative.

With the negative-exponential Fourier transform convention,Both poles lie in the upper half-plane. Contour integration closes above for positive time and below for negative time, giving the title expression. When the integral converges, convolution $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$ supplies the response with no freely added homogeneous transient growing toward negative infinity. The continuous limit $a\to b$ is $G(t)=H(t)t{e}^{-bt}$.