For the initial data, the two weighted integrals equal $(\phi_0/c)\operatorname{sgn}(x)\int_0^Le^{\mp fy/c}dy$. Following their opposite characteristics to large time at fixed $x$ gives the additional final-state conditions
$$
\boxed{
\int_0^Le^{-fy/c}\left(-\frac{\Phi_y}{f}+\frac\Phi c\right)dy
=-\frac{\phi_0}{c}\int_0^Le^{-fy/c}dy,}
$$
$$
\boxed{
\int_0^Le^{fy/c}\left(-\frac{\Phi_y}{f}+\frac\Phi c\right)dy
=\frac{\phi_0}{c}\int_0^Le^{fy/c}dy.}
$$
These two scalar constraints determine the two coefficients $A,B$ in the nullspace from part v and therefore make the adjusted solution unique.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 333 / 1 / viii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-333/1/viii/solution