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

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.

Choose a spatial interval with periodic boundary conditions and a uniform grid, so no separate inflow boundary closure of a difference scheme is required. The discrete Fourier transform is a unitary operator in the discrete L2 norm. Every circulant stencil is diagonal in that transform, and a one-step Fourier multiplier is a contraction exactly when its modulus is at most one for every grid frequency. For a mesh-uniform statement, use all $\theta \in [-\pi ,\pi ]$, the limiting frequency range.

A Fourier mode $e^{im\theta }$ giveswith the coefficients from (a). Squaring moduli and putting $v=\cos \theta$ yieldsThis has the required nonnegative sign for every frequency exactly whenOne must also check that the implicit equation is solvable, rather than divide by a vanishing symbol. HereFor $\mu \ne 1/2$, a zero would have to occur at zero or $\pi$; the latter occurs only at $(1\pm \sqrt{3})/2$, both outside the displayed stable intervals. At $\mu =1/2$, $D=3/4+(1/4)\cos \theta \ge 1/2$. Thus there is no denominator zero in the stable range. For fixed such $\mu$, continuity on the compact frequency interval gives a uniformly bounded inverse, and $|G|\le 1$ proves $\Vert U^n{\Vert }_{2,h}\le \Vert U^0{\Vert }_{2,h}$ for all steps.

Outside these intervals the displayed difference is negative at every nonzero frequency where $D\ne 0$, so $|G|>1$. On periodic refining meshes there are frequencies approaching any selected such value. Their amplification is bounded away from one and is repeated $O(h^{-1})$ times over a fixed physical interval when $k=\mu h$, violating a mesh-uniform stability bound. At the two singular parameters the implicit operator itself fails on meshes containing the Nyquist mode. These observations prove necessity as well as sufficiency.

For the physical convention $k>0$, $h>0$, the result is $0<\mu \le 1$ or $\mu \ge 2$, with zero admitted as the identity limit. The negative intervals describe the algebraic contraction result for signed Courant numbers. In particular, the additional large positive interval must not be discarded by imposing the usual explicit-scheme Courant–Friedrichs–Lewy condition: this scheme is implicit. Stability at large fixed Courant number is not a guarantee of a small error constant.