Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 72 / 3 / a

Let $h=\Delta x$, $k=\Delta t$ and $\mu =k/h$. For the exact solution $u(x,t)=F(x+t)$ of the advection equation, the three left-hand samples are translations by $(\mu -1)h$, $\mu h$ and $(\mu +1)h$ relative to $(x_m,t_n)$; the right-hand samples are translations by zero and $h$.

Denote their coefficients by $a=\mu (1+\mu )/6$, $b=(2-\mu )(1+\mu )/3$, $c=(2-\mu )(1-\mu )/6$, $d=(2-\mu )/3$ and $e=(1+\mu )/3$. Taylor expansion compares momentswhere $0^0=1$ in the zeroth moment. Direct calculation givesThus the unscaled exact-solution residual isFor fixed nonzero $\mu$, dividing by the time step yieldsThe scheme is therefore third order under fixed-Courant refinement, with an unscaled one-step defect of order four. Stability then gives third-order accumulated error for smooth data with periodic boundary conditions. Calling the unscaled defect fourth order is a different convention, not fourth-order approximation of the differential equation.

At $\mu =-1,1,2$, the amplification factor in (b) becomes exactly $e^{i\mu \theta }$: the numerical step translates the grid data by that integer number of cells and is exact for this constant-speed advection equation. At $\mu =0$ it is the identity, a zero-time-step limit. These are exact special shifts, rather than merely a cancellation of one term of the Taylor expansion. This is the implicit advection scheme with exact integer shifts.