 Solution

Use the forward spatial transform $\hat{f}(k)=\int e^{-ikx}f(x)dx$ and inverse $f(x)=\int e^{ikx}\hat{f}(k)dk/(2\pi )$. The retarded transformed Green function obeysIt vanishes for negative $\tau$. Continuity at zero excludes an unwanted ${\delta }^{\prime }$ term; integrating across zero then fixes $G_k^{\prime }(0^{+})=c^2$. Define $\gamma =\alpha c^2$ and $d_k=\sqrt{{\gamma }^2-c^2{k}^2}$. The two homogeneous exponents are $-\gamma \pm d_k$. Imposing the two matching conditions gives the Fourier-mode Green function of the telegraph equation,For an explicit time-Fourier route, use $G_k(\tau )=\int e^{-i\omega \tau }{\mathcal{G}}_k(\omega )d\omega /(2\pi )$. ThenFor $k\ne 0$, both poles lie below the real axis. Closing above for $\tau <0$ gives zero; closing below for $\tau >0$ is clockwise, giving $-i$ times the residue sum. Since ${\mathcal{G}}_k=-c^2/[(\omega -{\omega }_{+})(\omega -{\omega }_{-})]$, this iswhich reduces to the same damped sine or hyperbolic sine above. The $k=0$ pole on the real axis is interpreted by the retarded limiting prescription, or directly by the initial-value matching above.

For $|k|<\alpha c$ the response is overdamped and nonoscillatory. For $|k|>\alpha c$, put ${\Omega }_k=\sqrt{c^2{k}^2-{\gamma }^2}$; analytic continuation gives $G_k=\Theta (\tau )c^2{e}^{-\gamma \tau }\sin ({\Omega }_k\tau )/{\Omega }_k$, a real damped oscillation. At equality the finite limit is $\Theta (\tau )c^2\tau e^{-\gamma \tau }$.

All nonzero Fourier modes decay. The uniform mode instead has $G_0=(1-e^{-2\gamma \tau })/(2\alpha )$ for positive time, leaving a constant temperature increment. At small wavenumber the slow decay exponent is $-k^2/(2\alpha )+O(k^4)$, so the diffusion limit of the damped heat-transmission equation has diffusivity $1/(2\alpha )$. The full hyperbolic equation has finite propagation speed $c$, unlike the limiting parabolic heat equation.