= Solution
Use normalized instantaneous modes $\varphi_p(x,t)=\sqrt{2/L}\sin(p\pi x/L)$ and expand $\Psi=\sum_p a_pe^{-i\vartheta_p}\varphi_p$. Differentiating at fixed $x$ gives
$$
\partial_t\varphi_p=-\frac{\dot L}{2L}\varphi_p-\sqrt{\frac2L}\frac{p\pi x\dot L}{L^2}\cos(p\pi x/L).
$$
For $n\ne p$, <integration> uses
$$
\int_0^1 u\sin(n\pi u)\cos(p\pi u)\,du
=-\frac{(-1)^{n+p}n}{\pi(n^2-p^2)},
$$
so the <moving-wall derivative coupling> is
$$
\langle\varphi_n|\dot\varphi_p\rangle
=\frac{\dot L}{L}(-1)^{n+p}\frac{2np}{n^2-p^2}.
$$
For $n=p$ the normalization <derivative> and the remaining <integral> cancel exactly, giving zero. The moving <integration> endpoint contributes no boundary term because every mode vanishes there. Therefore
$$
\boxed{\dot a_n=\frac{\dot L}{L}\sum_{p\ne n}(-1)^{n+p}\frac{2np}{p^2-n^2}\,a_p e^{i(\vartheta_n-\vartheta_p)}}.
$$
The real coupling <matrix> is antisymmetric, as required by conservation of total <probability>.
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