Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 38A / Solution

Substitute $\varphi =e^{i(kx-\omega t)}$ into the mixed-derivative equation. Its left side is $k\omega \varphi$, so for $k\ne 0$ the dispersion relation, phase velocity and group velocity areThe graph of a function for each velocity is symmetric about $k=0$, singular at zero, and approaches the horizontal axis as $|k|$ grows. Energy packets travel right while their crests travel left.

At a distant fixed observer $x>0$, stationary components obey $x/t=c_g(k)$, so $|k|=\sqrt{\alpha t/x}$. As time increases their wavelength $2\pi /|k|$ decreases and their frequency magnitude $|\omega |=\sqrt{\alpha x/t}$ decreases; the crests pass toward decreasing $x$.

For the given initial spectrum the solution isThe value of a regular spectrum at the single point $k=0$ is irrelevant to this integral; a nonzero spatially constant mode is incompatible with the PDE. For asymptotic calculation assume the usual smooth localized spectrum, with sufficient integrability for the following stationary-phase steps.

At $x=Vt$, $V>0$, the phase $\Phi (k)=Vk+\alpha /k$ has stationary points $k=\pm k_0$, $k_0=\sqrt{\alpha /V}$. Their values are $\pm 2\sqrt{\alpha V}$ and their second derivatives are $\pm 2\alpha /k_0^3$. Expanding quadratically and using the oscillatory Gaussian integral gives the stationary phase leading termFor real data this is twice the real part of the positive-frequency term. If a stationary coefficient vanishes, the formula instead asserts that its contribution has smaller order. At $x=-Vt$, the phase derivative is $-V-\alpha /k^2<0$, so there is no real stationary point and no corresponding $t^{-1/2}$ wave packet. For smooth compactly supported spectra away from zero, repeated integration by parts gives decay faster than any power of $t$. More general spectra require endpoint/tail control; absence of stationary points alone does not justify that stronger rate for arbitrary $A$.