I am considering the problem in the picture[2]. [2]: https://i.stack.imgur.com/krlS4.jpg'
I have so far got \begin{equation}\label{2.2.2.1} u(x,y,t) = X(x) \cdot Y(y) \cdot T(t) \end{equation} such that $$ T''(t) \cdot X(x) \cdot Y(y) = c^2 (T(t) \cdot X''(x) \cdot Y(y) + T(t) \cdot X(x) \cdot Y''(y)) $$ $$ \frac{1}{c^2} \frac{T''(t)}{T(t)} = \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = \lambda^2 $$ $$ T''(t) - c^2 \lambda^2 T(t) = 0 \hspace{5 mm} ; \hspace{5 mm} \frac{X''(x)}{X(x)} = - \frac{Y''(y)}{Y(y)} + \lambda^2. $$ Introducing $\mu$ and $\nu$ such that $\nu^2 = \lambda^2 + \mu^2$ we get \begin{equation}\label{2.2.2.2} X''(x) - \mu^2 X(x) = 0 \hspace{5 mm} , \hspace{5 mm} Y''(y) - \nu^2 Y(y) = 0 \hspace{5 mm} , \hspace{5 mm} T''(t) - c^2 \lambda^2 T(t) = 0. \end{equation}
By applying the periodic boundary conditions \eqref{2.2.0.3}, we obtain the following solutions to \eqref{2.2.2.2} \begin{equation}\label{2.2.2.3} X(x) = X_m(x) = \sin(\mu_m x) + \cos(\mu_m x), \end{equation} \begin{equation}\label{2.2.2.4} Y(y) = Y_n(y) = \sin(\nu_n y) + \cos(\nu_n y), \end{equation} \begin{equation}\label{2.2.2.5} T(t) = T_{mn}(t) = B^*_{mn}\sin(\lambda_{mn} t) + B_{mn}\cos(\lambda_{mn} t), \end{equation} where $$ \mu_m = m\pi \hspace{5 mm}, \hspace{5 mm} \nu_n = n\pi \hspace{5 mm}, \hspace{5 mm} \lambda_{mn} = c\sqrt{\mu_m^2 + \nu_n^2} = c\pi\sqrt{m^2 + n^2}. $$ By inserting \eqref{2.2.2.3}, \eqref{2.2.2.4} and \eqref{2.2.2.5} into \eqref{2.2.2.2}, we get \begin{equation}\label{2.2.2.6} u_{mn}(x,y,t) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} (\sin(\mu_m x) + \cos(\mu_m x)) \cdot (\sin(\nu_n y) + \cos(\nu_n y)) \cdot (B^*_{mn}\sin(\lambda_{mn} t) + B_{mn}\cos(\lambda_{mn} t)). \end{equation} \label{2.2.2.6} The initial conditions \eqref{2.2.0.2} gives that \begin{equation} \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} B_{mn}(\sin(m\pi x) + \cos(m\pi x))(\sin(n\pi y) + \cos(n\pi y)) = \cos(4\pi x)\sin(4\pi y) \end{equation} \begin{equation} \label{2.2.2.7} \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} B^*_{mn}(\sin(m\pi x) + \cos(m\pi x))(\sin(n\pi y) + \cos(n\pi y)) = 0. \end{equation} A plausible solution to \eqref{2.2.2.7} is $B^*_{mn} = 0$ for all $m,n \in \mathbb{N}$. $B_{mn}$ is given by \begin{equation} B_{mn} = \sum \end{equation}.
I am now stuck finding the fourier coefficient. Is what I wave done so far correct, or should I have another approach?