 Solution

First remove the observation time from the spectral amplitude. For any fixed horizon $H>t$, replace $G_0(k,t)$ by $G_0(k,H)$ in the formula of part (d). The difference in the boundary integral is proportional toFor every $s>t$, this integrand is holomorphic inside $D_{+}$, where $\mathrm{Re}w<0$, and decays on its large closing arcs. The Cauchy integral theorem makes the difference zero. This finite-horizon extension of a boundary transform therefore gives the time-independent spectraAll observation-time dependence is now in $e^{ikx-wt}$.

When the boundary history is sufficiently decaying, the natural long-time version isThe integral converges in $\mathrm{Re}w<0$ for bounded boundary data and on its boundary for integrable histories; otherwise one must state an Abel prescription or a suitable temporal growth assumption. Arbitrary smooth $g_0$ need not have such an infinite-time transform. The finite-horizon formula remains exact without that extra assumption, and the source alone does not prescribe a universal long-time limit.

A useful further decomposition, if $g_0(s)=g_{\infty }+r(s)$ with integrable $r$, isThis separates the persistent forcing from the transient spectrum. The pole at $k=i$, where $w=0$, accounts for the decaying stationary profile $g_{\infty }{e}^{-x}$; that profile also follows directly from $q_{xxx}-q_x=0$, spatial decay and its boundary value. For the remaining oscillatory integrals the phase is $i[kx+(k^3+k)t]$. At a fixed ray $x/t=\xi$, its stationary points satisfy $3k^2+1+\xi =0$, or $k=\pm i\sqrt{(1+\xi )/3}$. These time-independent amplitudes expose precisely the poles and saddle points used in the method of steepest descent; allowable deformations must also respect the analyticity strip of the initial transform.