= Solution
Substitute $\phi=e^{i(kx-\omega t)}$ into the mixed-derivative equation. Its left side is $k\omega\phi$, so for $k\ne0$ the <dispersion relation>, <phase velocity> and <group velocity> are
$$
\boxed{\omega=-\frac\alpha k,\qquad c(k)=-\frac\alpha{k^2},\qquad
c_g(k)=\frac\alpha{k^2}.}
$$
The <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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-3-wave-velocities.png]
{title=Opposite phase and group velocities for omega=-alpha/k, shown with alpha=1}
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 is
$$
\boxed{\phi(x,t)=\int_{\mathbb R}A(k)\exp\left[i\left(kx+\frac{\alpha t}k\right)\right]dk,\qquad
A(-k)=\overline{A(k)}\text{ for real }\phi.}
$$
The 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 $\pm2\sqrt{\alpha V}$ and their second <derivatives> are $\pm2\alpha/k_0^3$. Expanding quadratically and using the oscillatory <Gaussian integral> gives the <stationary phase> leading term
$$
\boxed{\phi(Vt,t)\sim\sqrt{\frac{\pi k_0^3}{\alpha t}}
\left[A(k_0)e^{i(2\sqrt{\alpha V}t+\pi/4)}
+A(-k_0)e^{-i(2\sqrt{\alpha V}t+\pi/4)}\right].}
$$
For 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 $A$.
Back to article page
