English: Graphical representation of the solution to the heat equation for an "infinite slab" of width 1 given by:

${\displaystyle \ u_{t}=ku_{xx}}$

where k = .061644 subject to the boundary conditions:

${\displaystyle u_{x}(0,t)=0,\ \ u_{x}(1,t)=0}$

and with the initial heat distribution given by:

${\displaystyle \ u(x,0)=\cos(2x)}$

In this case, the left face (${\displaystyle x=0}$) and the right face (${\displaystyle x=1}$) are perfectly insulated. This image shows how the heat redistributes, moving from the warmer left edge to the cooler right edge, then equalizing to a constant temperature throughout. This temperature happens to be the average value of ${\displaystyle \cos(2x)}$ over [0,1], as one might expect.

The solution:

${\displaystyle u(x,t)={\frac {\sin(2)}{2}}+\sum _{n=1}^{\infty }A_{n}\cos \left(n\pi x\right)\exp \left(-kn^{2}\pi ^{2}t\right)}$

where:

${\displaystyle {\begin{aligned}A_{n}&=2\int _{0}^{1}\cos(2x)\cos(n\pi x)dx\\&=(-1)^{n}{\frac {-4\sin(2)}{n^{2}\pi ^{2}-4}}\\\end{aligned}}}$