Write h=Δx, k=Δt and δxxui=(ui+1−2ui+ui−1)/h2. Insert a smooth exact heat equation solution into the Crank-Nicolson method. Expanding about the midpoint (xi,tn+k/2) givesThus the residual, defined as left side minus right side, isSince ut=Duxx implies uttt=Duxxtt,This proves second-order numerical consistency in both space and time. About the old time level, uncombined Taylor remainders can include O(k3); the midpoint expression displays the symmetric leading errors most cleanly. Multiplying by k gives the one-step defect O(k3+kh2). Stable evolution with suitable initial and boundary data then gives global error O(k2+h2); consistency alone does not prove global convergence.
ku(t+k/2)−u(t−k/2)=ut+24k2uttt+O(k4),
2D[δxxu(t−k/2)+δxxu(t+k/2)]=D[uxx+12h2uxxxx+8k2uxxtt]+O(h4+k4+h2k2).
T=24k2uttt−8Dk2uxxtt−12Dh2uxxxx+O(h4+k4+h2k2).
T=−12k2uttt−12Dh2uxxxx+O(h4+k4+h2k2).
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