Causal Green function of an overdamped oscillator ($G(t)=H(t)\frac{e^{-bt}-e^{-at}}{a-b},a>b>0$)

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 $G(t)=H(t)t{e}^{-bt}$.

For $a>b>0$ with $a,b\ne 1$, the solution of $(D_t+a)(D_t+b)y=e^{-|t|}$ that decays at both infinities isIts Fourier transform is $2/[(1+k^2)(a+ik)(b+ik)]$. The lower half-plane contains only the pole $-i$, while the upper half-plane contains $i,ia,ib$. Inverting by the residue theorem gives the two expressions. Direct substitution verifies the equation separately on each side, and $y,y^{\prime }$ match at zero, so there is no spurious impulsive source. The negative-time response is nonzero because the forcing already acts before zero, even though the Green function is causal. Coinciding rates $a=1$ or $b=1$ require limits with polynomial times exponential terms.