Solving 2D heat equation with separation of variables Solve:
$$
\begin{cases}
u_t = Ku_{xx}\\
u(0,y,t) = u(\pi,y,t) = 0\\
u_y(x,0,t) = u_y(x,\pi,t) = 0\\
u(x,y,0) = 1
\end{cases}$$
for $0 < x < \pi$, $0 < y < \pi$, $t>0$:
This was the problem given to me, but I don't believe it has a nontrivial solution (correct me if I'm wrong). Instead, I think the problem was meant to say:
$$u_t = K(u_{xx} + u_{yy})$$
In this case,
Let $u(x,y,t) = f(x)g(y)h(t)$:
$$\frac{h'(t)}{h(t)} = \frac{f"(x)}{f(x)} + \frac{g"(y)}{g(y)}$$
Let:
$$\frac{f"(x)}{f(x)} = -n^2$$
$$\frac{g"(y)}{g(y)} = -m^2$$
$$\frac{h'(t)}{h(t)} = -(m^2 + n^2)$$
This leads to the seaparted solutions:
$$f(x) = A\cos(nx) + B\sin(mx)$$
$$g(y) = C\cos(my) + D\sin(my)$$
$$h(t) = Ee^{-(n^2 + m^2)t}$$
$$u(o,y,t) = u(\pi,y,t) = 0 \space\text{ implies } \space A = 0$$
$$u_y(x,0,t) = u_y(x,\pi,t) = 0 \space\text{ implies } \space D = 0$$
Thus,
$$u(x,y,t) = B\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ To satisfy the initial value, we can exploit the superposition principle:
$$u(x,y,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my)e^{-(n^2 + m^2)t}$$ 
$$u(x,y,0) = \sum_{m=0}^\infty \sum_{n=0}^\infty B_{nm}\sin(nx)\cos(my) = 1$$ 
This was as far as I was able to get. I'm unsure how to satisfy the initial condition given this double sum. Some help would be appreciated!
 A: You can find $m$ and $n$ using boundary conditions. Instead of calling your constant $n$ or $m$, call them $k$ or $\lambda$. $m$ and $n$ are used frequently for natural numbers.
$$u(x=\pi) = 0 \Rightarrow B\sin(\lambda\pi) = 0 $$
And we want a non trivial solution, so $B\ne 0$. Therefore $\sin(\lambda \pi)=0$
$\lambda \pi = \pi n \Rightarrow \lambda = n $
I didn't see you use the BVs so I'm not sure if you did. that step.
Now all that's left is to find the coefficient $B_{nm}$ using the orthogonality properties of your eigenfunctions.
$$B_{nm} = \frac{4}{\pi^2}\int_0^\pi\int_0^\pi f(x,y)\sin(n'x)\sin(m'y)dxdy$$
where $f(x)=1$ in this case.
You don't even have to memorize the integral above to find the coefficient in the future.
You can simply multiply both sides by $\sin(n'x)\sin(m'y)$ and integrate on the domain. Using properties of Kronecker delta, only when $m' = m$ and $n'=n$ will get something that isn't zero. The left hand side will simply always be:
$$ \\B_{nm}\int_0^\pi \sin^2(nx)dx\int_0^\pi \sin^2(my)dy =B_{nm}\frac{\pi}{2}\frac{\pi}{2} = B_{nm}\frac{\pi^2}{4}$$
Now we have:
$$B_{nm}\frac{\pi^2}{4} = \int_0^\pi\int_0^\pi 1\sin(nx)\sin(my)dxdy$$
