Use a common factor $e^{i(kx-\omega t)}$, with $k=k_0>0$ and $\omega=\omega_0$. For the upward propagating branches with $0<\widehat\omega_2=\omega-kU_2<N_2$, define
$$
\boxed{\sin\theta_1=\frac{\omega}{N_1},\qquad
\sin\theta_2=\frac{\widehat\omega_2}{N_2},\qquad
m_j=-k\cot\theta_j,\quad 0<\theta_j<\frac\pi2.}
$$
These are the inclinations of upward-sloping <constant-phase lines of an internal gravity wave> to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as $A_0,A_r,A_t$. <Incompressibility> gives $(u_i,w_i)=A_0(\cos\theta_1,\sin\theta_1)$, $(u_r,w_r)=A_r(\cos\theta_1,-\sin\theta_1)$, and $(u_t,w_t)=A_t(\cos\theta_2,\sin\theta_2)$.

The <jump conditions for stratified inviscid shear flow> require a common interface displacement, not a common vertical velocity: $w=-i\widehat\omega\eta$ on each side. Since the background <mass density> is continuous, <pressure> is continuous without a hydrostatic jump. The horizontal momentum equation gives $p/\rho_0=(\widehat\omega/k)u$. Hence
$$
\frac{\sin\theta_1}{\omega}(A_0-A_r)
=\frac{\sin\theta_2}{\widehat\omega_2}A_t,\qquad
\omega\cos\theta_1(A_0+A_r)
=\widehat\omega_2\cos\theta_2A_t.
$$
These two matching equations determine <internal-wave transmission across a velocity jump>:
$$
\boxed{\frac{A_t}{A_0}=
\frac{2}{r\cos\theta_2/\cos\theta_1+r^{-1}\sin\theta_2/\sin\theta_1},\qquad
\frac{A_r}{A_0}=\frac{r\cos\theta_2/\cos\theta_1-r^{-1}\sin\theta_2/\sin\theta_1}{r\cos\theta_2/\cos\theta_1+r^{-1}\sin\theta_2/\sin\theta_1},\quad
r=\frac{\widehat\omega_2}{\omega}.}
$$
The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have $A_t/A_0=1$ and $A_r/A_0=0$, whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the <kinematic boundary condition> and pressure balance.

For equal <buoyancy frequencies> and the specified opposing current, $\theta_1=\pi/6$, $\theta_2=\arcsin(3/4)$, and
$$
\boxed{\frac{A_t}{A_0}=\frac8{4+\sqrt{21}},\qquad
\frac{A_r}{A_0}=\frac{\sqrt{21}-4}{\sqrt{21}+4}.}
$$
The incident phase lines have slope $1/\sqrt3$, the reflected lines slope $-1/\sqrt3$, and the transmitted lines slope $3/\sqrt7$: they steepen above the interface. For the other specified current, $\widehat\omega_2=0$. The transmitted vertical <wavenumber> then diverges and there is no regular propagating upper-layer wave of that frequency: this is the <critical level of an internal gravity wave> limit. Approaching it from $\widehat\omega_2>0$ gives $A_r/A_0\to-1$ and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 345 / 1 / c / Solution

Across a horizontal material interface with continuous background <mass density> but a velocity jump, an <internal gravity wave> preserves laboratory <frequency> and horizontal <wavenumber>. The matching conditions are continuity of displacement $w/\widehat\omega$ and <pressure>, not continuity of $w$. For velocity amplitudes and upward phase-line angles $\theta_1,\theta_2$, define $r=\widehat\omega_2/\widehat\omega_1$. Then the incident-to-transmitted amplitude ratio is
$$
\frac{A_t}{A_i}=\frac2{r\cos\theta_2/\cos\theta_1+r^{-1}\sin\theta_2/\sin\theta_1}.
$$
This follows by adding the displacement and pressure matching equations after eliminating the reflected amplitude. It assumes nonzero intrinsic frequencies and propagating outgoing branches.

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 Incoming links: Internal-wave transmission across a velocity jump