= Solution
Write $\mathbf X=\epsilon\mathbf x$, $\mathbf k=\nabla_{\mathbf X}\theta$, and denote the slow <density> and <velocity field> amplitudes by $a_\rho$ and $\mathbf a_u$. Acting on the rapid phase, $\partial_t+\mathbf U\cdot\nabla$ gives $i\Omega$, where $\Omega=\omega-\mathbf U\cdot\mathbf k$. Spatial differentiation gives $-i\mathbf k$, while derivatives of the amplitudes and the mean <shear flow> are $O(\epsilon)$. The leading <continuity equation> and <linearized Euler equations> therefore become
$$
\Omega a_\rho=\rho_0\mathbf k\cdot\mathbf a_u,\qquad
\rho_0\Omega\mathbf a_u=c_0^2\mathbf k a_\rho.
$$
Eliminating $\mathbf a_u$ on the nonzero acoustic branch gives the <acoustic eikonal equation in a shear flow>:
$$
\boxed{(\omega-\mathbf U\cdot\nabla\theta)^2=c_0^2|\nabla\theta|^2.}
$$
The mean-shear term is small here because $\nabla\mathbf U=O(\epsilon)$; retaining advection does not require retaining that gradient in the leading phase equation. The positive intrinsic-frequency acoustic <dispersion relation> is $\omega=Uk_x+c_0|\mathbf k|$, with <group velocity> $\mathbf v_g=U\mathbf e_x+c_0\mathbf k/|\mathbf k|$.
In a stationary wall <boundary layer>, horizontal homogeneity preserves $\omega$ and $k_x$ along a <Hamiltonian ray-tracing equations> trajectory. For a two-dimensional ray,
$$
k_y^2=\frac{(\omega-Uk_x)^2}{c_0^2}-k_x^2.
$$
Suppose $U$ increases from zero at the wall to a positive free-stream value. For a downstream-directed wave normal, $k_x>0$, increasing $U$ decreases $\Omega$ and $k_y$. The wave normal bends towards the wall, and the ray becomes more nearly parallel to it. If $U$ reaches $\omega/k_x-c_0$, then $k_y=0$: the ray turns and returns towards the wall rather than reaching the free stream. With launch angle $\beta_0$ above the wall and free-stream <Mach number> $M_\infty=U_\infty/c_0$, the escape condition is
$$
\boxed{\cos\beta_0\le\frac1{1+M_\infty}\qquad\text{for downstream launch}.}
$$
For upstream-directed wave normals, $k_x<0$, $\Omega$ and $|k_y|$ increase through the layer; there is no such downstream turning barrier on this branch. A <wavefront> normal is parallel to $\mathbf k$, while the energy ray follows the <group velocity>: advection makes their directions different. In particular, a wall-normal wave normal with $k_x=0$ can still have a ray drifting downstream. The ordinary <WKB method> needs a local turning-point continuation at $k_y=0$; the ray sketch depicts that qualitative continuation.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-82-shear-layer-rays.png]
{title=Acoustic energy rays through a wall shear layer, showing downstream escape, downstream turning and downstream drift of a wall-normal wave normal}
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