Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 4 / a

Keep the Courant number $\mu =\Delta t/\Delta x$ fixed and write $h=\Delta x$. An exact solution of $u_t=u_x$ is $u(x,t)=g(x+t)$. Substituting it into the update gives the defectwhere $s=x+t$ and $g^{(3)}$ is evaluated at $s$. The constant, first-derivative and second-derivative terms cancel. Dividing by $\Delta t=\mu h$ gives a second-order consistency error. Hence for fixed $0<\mu <1$, $\mu \ne 1/2$, the scheme has order two.

At the special value $\mu =1/2$, the update reduces to $u_m^{n+1}=u_{m+1}^{n-1}$. This is exactly the transport over two time steps, since $2\Delta t=\Delta x$, so the defect vanishes identically. With exact data at both starting levels it transports their values exactly; errors in those starting levels are still present, and the even and odd time subsequences evolve separately.