In the additional notes (luckily not examinable) for my PDE modelling module I stumbled across the following statement and was wondering how it was derived
$$\partial_t\theta=\alpha\left(\partial^2_x+\partial^2_y+\partial^2_z\right)\theta\tag{1}$$ $$\theta=\theta(x,y,z,t),\,\,\,(x,y,z)\in[0,1]^3\tag{2}$$ subject to: $$\theta(0,y,z,t)=\theta(1,y,z,t)=\theta(x,0,z,t)=\theta(x,1,z,t)=\theta(x,y,0,t)=\theta(x,y,1,t)=0\tag{3}$$ $$\theta(x,y,z,0)=f(x,y,z)\tag{4}$$ then the solution is: $$\theta=\sum_{n=1}^\infty\sum_{m=1}^\infty\sum_{p=1}^\infty B_{n,m,p}e^{-\alpha(n^2+m^2+p^2)\pi^2t}\sin(n\pi x)\sin(m\pi y)\sin(p\pi z)$$ $$B_{n,m,p}=\int_0^1\int_0^1\int_0^1f(x,y,z)\sin(n\pi x)\sin(m\pi y)\sin(p\pi z)\,\text{d}x\text{d}y\text{d}z$$
It is clear to me that the initial conditions are just each of the faces being held at zero for $t\ne0$ and an initial temperature as any function of $f(x,y,z)$ and that the series can be split up apart from the dependent constants, so how is this expression for $B$ derived?
When $t=0$ we have $$f(x,y,z)=\sum_{n,m,p \in \mathbb{N}}B_{n,m,p}\sin(n\pi x) \sin (m \pi y) \sin (p \pi z)$$ The functions $$\phi_{n,m,p}(x,y,z)=\sin(n\pi x) \sin (m \pi y) \sin (p \pi z)$$ form an orthogonal basis of functions in $C^2([0,1]^3)$ (which satisfy the boundary). We have $$\|\phi_{n,m,p}\|^2=\int_{[0,1]}\int_{[0,1]}\int_{[0,1]}\phi_{n,m,p}(x,y,z)^2dxdydz=\frac{1}{2^3}$$
Therefore $$B_{n,m,p}=\color{red}{2^3}\int_{[0,1]}\int_{[0,1]}\int_{[0,1]}f(x,y,z)\sin(n\pi x) \sin (m \pi y) \sin (p \pi z)dxdydz$$ This because any function $f \in C^2([0,1]^3)$ (which satisfies the boundary) can be represented as $$f(x,y,z)=\sum_{n,m,p \in \mathbb{N}}\frac{\langle f,\phi_{n,m,p} \rangle}{\|\phi_{n,m,p}\|^2} \phi_{n,m,p}$$ where $\langle f,\phi_{n,m,p} \rangle/\|\phi_{n,m,p}\|^2=B_{n,m,p}$.