We will deal here with the much simplified heat equation ∂u/∂t = ∂2u/∂t2, which omits naming such relevant quantities as heat conductivity and specific heat. We choose units to simplify the equations. All of our mathematical techniques are illustrated in the simpler equation.

Fourier would take as the simplest example a rod of length 2π and an initial temperature u(0, x) = sin(3x) for some integer such as 3, and boundary conditions u(t, 0) = u(t, 2π) = 0. u(t, x) = e–32tsin(3x) solves the equation nicely now. The general solution for this rod with these boundary conditions is ∑ane–n2tsin(nx) with n ranging over positive integers. The values an can be chosen to match any meaningful initial temperature distribution.

The reader might wonder why, with so pretty an analytic solution to these equations, one would resort to difference equations. When the omitted physics is included and these quantities vary with space, and temperature; Fourier’s pretty equations fail. Numerical tricks seem necessary. We learn some ways these numerical techniques may fail by trying them on problems whose solution Fourier knew.