Consider the heat equation in a 2D rectangular region such that $0<x<L$ and $0<y<H$, $$\frac{\partial u}{\partial t} = k\bigg(\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2}\bigg)$$ subject to the initial condition $$u(x,y,t) = \alpha (x,y)$$ and boundary conditions $$u(0,y,t) =0, \frac{\partial u}{\partial x} (L,y,t)=0, \frac{\partial u}{\partial y} (x,0,t)=0, \frac{\partial u}{\partial y} (x,H,t)=0.$$ Find the solution to the problem.
My approach was the standard separation of variables (as all conditions are homogeneous) and then taking cases for the separation constants and ensuring that we don't get a trivial solution. Let $u(x,y,t) = f(x)g(y)h(t)$. Note: all $c_i$ are arbitrary constants.
$\frac{dh}{dt} + \lambda kh = 0 $
$\implies h(t) = c_1e^{-\lambda kt}\quad \ldots (1)$
$\frac{d^2f}{dx^2} + \mu f = 0$
Taking cases for $\mu \in \mathbb{R}$, we come to the conclusion that
for $\mu > 0$, $\mu = (\frac{2n+1}{2})^2\big(\frac{\pi}{L}\big)^2, n=0,1,2,\ldots$
And $f(x) = c_2 \sin(\sqrt{\mu} x) = c_2 \sin (\frac{2n+1}{2} \frac{\pi}{L} x) \quad \ldots (2)$
If $\mu =0$, $f(x) = c_3 \quad \ldots (3)$.
$\frac{d^2g}{dy^2} +(\lambda- \mu) g = 0$
Doing the same for this ODE, we get $\lambda _{mn} = \big(\frac{m\pi}{H}\big)^2 + \mu _n$.
My question is, before I apply the initial condition, what should my final expression for $u$ look like? Would it be $$u(x,y,t)=\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} A_n \sin\bigg(\big(n+\frac{1}{2}\big)\frac{\pi x}{L}\bigg)\cos \big(\frac{m\pi}{H}y\big) e^{-\Big(\big(\frac{m\pi}{H}\big)^2+\bigg(\big(n+\frac{1}{2}\big)\frac{\pi }{L}\bigg)^2\Big) kt}?$$ How should I incorporate the constant solution to find the unknown coefficients using the orthogonality of sines and cosines?