Represent I(x) as a Fourier series I(x)≈∑k∈Kbkeikx
The corresponding sum for u is u(x,t)≈∑k∈Kbke−αk2teikx
Such solutions are also accepted by the numerical schemes, but with an amplification factor A different from exp(−αk2t): unq=AneikqΔx=Aneikx