For constant width, <separation of variables> with $\phi=X(x)\sin(n\pi y/h_0)$ gives
$$
X''+k_n^2X=0,
\qquad
\boxed{k_n^2=k_0^2-\frac{n^2\pi^2}{h_0^2}.}
$$
Hence $X=A_ne^{ik_nx}+B_ne^{-ik_nx}$, with imaginary $k_n$ representing an evanescent mode.

For $X=\epsilon x$ and varying width,
$$
k_n(X)^2=k_0^2-\frac{n^2\pi^2}{h(X)^2},
\qquad
\Theta_n=\frac1\epsilon\int_0^Xk_n(\xi)d\xi.
$$
Projection of the next-order equation onto $\sin(n\pi y/h)$, whose squared norm is proportional to $h$, gives
$$
\boxed{2k_nA_n'+\left(k_n'+k_n\frac{h'}h\right)A_n=0,
\qquad
2k_nB_n'+\left(k_n'+k_n\frac{h'}h\right)B_n=0.}
$$
Thus the <WKB amplitude in a slowly varying duct> is
$$
\boxed{
A_n(X)=A_n(0)\left[\frac{k_n(0)h(0)}{k_n(X)h(X)}\right]^{1/2},
\quad
B_n(X)=B_n(0)\left[\frac{k_n(0)h(0)}{k_n(X)h(X)}\right]^{1/2}.}
$$
This applies for $n\geq1$ away from turning points $k_n=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 336 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-336/2/b/solution