Is that the general solution of the Helmholtz equation? Helmholtz equation:
$$u_{xx}+u_{yy}+k^2u=0$$
$$0 \leq x \leq L$$
$$0 \leq y \leq L$$
The solution is in the form $u=X(x)Y(y)$
Replacing this at the equation we get the following problems:
$$X''+k_x^2X=0, X(0)=X(L)=0$$
$$Y''+k_y^2Y=0, Y(0)=Y(L)=0$$
We get that:
$$X_m=\sin{(\frac{m \pi x}{L})}, Y_n=\sin{(\frac{n \pi y}{L})}$$
So $u_{m,n}=X_m(x)Y_n(y)$
$$u_{m,n}(x,y)=\sin{(\frac{m \pi x}{L})} \sin{(\frac{n \pi y}{L})}, m,n=1,2,3, \dots$$
where $k_{m,n}^2=\frac{(m^2+n^2) \pi^2}{L^2}$
Could you tell me if $u_{m,n}(x,y)=\sin{(\frac{m \pi x}{L})} \sin{(\frac{n \pi y}{L})}, m,n=1,2,3, \dots$ is the general solution of the Helmhotz equation, or does the general solution contain also a sum??

Since the equation is linear, you must sum up your solutions.
