= Backward four-wave-mixing boundary problem
{title2=$u'=i\kappa v^*,\ v'=-i\kappa u^*$}
Because one wave travels backward, its incident <boundary condition> belongs at the opposite end of the medium. With $u(0)=a$ and $v(L)=0$, differentiating once gives $u''=-\kappa^2u$ and the solution
$$
u(z)=a\frac{\cos[\kappa(L-z)]}{\cos\kappa L},\qquad v(z)=ia^*\frac{\sin[\kappa(L-z)]}{\cos\kappa L}.
$$
The invariant $|u|^2+|v|^2$ is constant along $z$, while the input/output powers obey transmission minus reflection equal to the incident power in this normalization. A proposed solution must satisfy both first-order equations, not just their second derivatives.
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