 Backward four-wave-mixing boundary problem

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 $u^{\prime \prime }=-{\kappa }^{2}u$ and the solutionThe 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.