= Solution
The obstacle and wave pattern are stationary, so $\omega=\Omega=0$. With
$$
(k_1,k_2)=\kappa(\cos\phi,\sin\phi),
\qquad \kappa>0,
$$
the dispersion relation becomes
$$
0=\alpha\kappa^{1/2}-V\kappa\cos\phi.
$$
Hence $\cos\phi>0$, so $-\pi/2<\phi<\pi/2$, and
$$
\boxed{\kappa=\frac{\alpha^2}{V^2\cos^2\phi}.}
$$
Differentiating the dispersion relation with respect to the wavevector gives
$$
\begin{aligned}
\mathbf c_g
=\nabla_{\mathbf k}\Omega
&=\frac{\alpha}{2\sqrt\kappa}
(\cos\phi,\sin\phi)-(V,0)\\
&=\frac V2\left(\cos^2\phi-2,\,
\sin\phi\cos\phi\right),
\end{aligned}
$$
where the zero-frequency relation
$\alpha/\sqrt\kappa=V\cos\phi$ was used. Its first component is always negative, so the ray angle $\psi$ lies in
$$
\frac\pi2<\psi<\frac{3\pi}{2}.
$$
Moreover,
$$
\boxed{
\tan\psi
=\frac{\sin\phi\cos\phi}{\cos^2\phi-2}
=-\frac{\tan\phi}{1+2\tan^2\phi}.}
$$
Put $q=|\tan\phi|$. The magnitude of the slope relative to the negative $x_1$-axis is
$$
F(q)=\frac{q}{1+2q^2}.
$$
Its <derivative> is
$$
F'(q)=\frac{1-2q^2}{(1+2q^2)^2},
$$
so its <global maximum> occurs at $q=1/\sqrt2$ and equals
$$
F_{\max}=\frac1{2\sqrt2}=2^{-3/2}.
$$
Therefore all group-velocity rays lie in the <stationary square-root-dispersion wake>
$$
\boxed{
|\psi-\pi|\leq\tan^{-1}(2^{-3/2}),}
$$
a wedge of semi-angle $\tan^{-1}(2^{-3/2})$ extending in the negative $x_1$-direction.
Back to article page
