= Solution
Take the spatial <Fourier transform> of the causal <Green function>. The transformed equation is $\widehat G_t=(\mu-iUk-\gamma k^2)\widehat G+\delta(t)$, with zero response for $t<0$. Integration across the impulse sets $\widehat G(k,0^+)=1$, so
$$
\widehat G(k,t)=H(t)e^{\mu t-iUkt-\gamma k^2t}.
$$
Inverting the <Gaussian Fourier transform> gives
$$
\boxed{G(x,t)=\frac{H(t)}{\sqrt{4\pi\gamma t}}\exp\!\left(\mu t-\frac{(x-Ut)^2}{4\gamma t}\right).}
$$
This solves the homogeneous equation for $t>0$ and tends to the <Dirac delta distribution> as $t\downarrow0$. Its spatial integral is $e^{\mu t}$, as required by the zero-wavenumber growth law.
Along a ray $x=vt$, the exponential <growth rate> is
$$
s(v)=\mu-\frac{(v-U)^2}{4\gamma}.
$$
The impulse grows in the moving interval of rays $U-2\sqrt{\gamma\mu}<v<U+2\sqrt{\gamma\mu}$, while its peak at $x=Ut$ grows as $t^{-1/2}e^{\mu t}$. At a fixed position,
$$
G(x,t)\sim\frac{e^{Ux/(2\gamma)}}{\sqrt{4\pi\gamma t}}e^{(\mu-U^2/(4\gamma))t}.
$$
Thus, under the printed assumption $\mu>0$, \b[$0<\mu<U^2/(4\gamma)$ gives convective wave-packet instability], because the growing disturbance is swept away from each fixed observation point. \b[$\mu>U^2/(4\gamma)$ gives absolute wave-packet instability], with exponential growth at fixed position. At equality the fixed-position exponential rate is zero and the impulse decays algebraically as $t^{-1/2}$, even though its moving peak grows. The two bounding rays have the same algebraic prefactor.
These statements classify the <Green function> and localized initial disturbances propagated by its convolution. They do not assert that every possible initial condition decays at fixed position in the convective case: spatially uniform initial data instead produce $e^{\mu t}$. Persistent external forcing can likewise change the observed late-time response. The distinction between <convective wave-packet instability> and <absolute wave-packet instability> concerns localized disturbances in a specified frame.
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