Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 71 / 3

Let the spatial spacing be $d$, the time step $k$, and $\mu =k/d$. Abbreviate the three new-time coefficients by $a=\mu (1+\mu )/2$, $b=(1+\mu )(2-\mu )$, $c=(1-\mu )(2-\mu )/2$, and the two old-time coefficients by $d_0=2-\mu$, $e_0=1+\mu$. These are the coefficients of the compact third-order advection stencil.

For a smooth solution of the advection equation, $u_t=u_x$ implies $u(x,t+k)=u(x+k,t)$. Substituting the exact solution into the stencil therefore evaluates all five terms on the same characteristic profile, at spatial offsets $(\mu -1)d$, $\mu d$, $(\mu +1)d$, $0$, and $d$. Its Taylor moments areDirect polynomial expansion givesConsequently the unscaled local truncation error isAt a fixed nonzero Courant number, dividing by the time step gives an $O(d^3)$ consistency defect. ThusWhen $K(\mu )=0$, the order is higher because the method simplifies to an exact integer shift: at $\mu =1$ the two sides have the common factor $2+e^{-i\theta }$ and the amplification is $e^{i\theta }$; at $\mu =2$ it is $e^{2i\theta }$. The signed value $\mu =-1$ gives the shift $e^{-i\theta }$, and $\mu =0$ gives the identity. These special cases do not invalidate the third-order statement for the family. Under the stability conditions established next, the usual accumulation of the fourth-order local defect over $O(1/k)$ steps gives third-order convergence for smooth data.

For a Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$, defineThe amplification factor is $G=N/L$. To derive its stability condition, use the coefficient abbreviations from the preceding solution and set $q=\cos \theta$. The squared moduli areHere $a+c={\mu }^2-\mu +1$, $a-c=2\mu -1$, and $4ac=K(\mu )$. Subtraction and factorization giveThus $|G|\le 1$ for every frequency exactly when $K(\mu )\ge 0$, provided the implicit denominator does not vanish. For the physically nonnegative Courant number, this givesFor a signed convention the full real range is $(-\infty ,-1]\cup [0,1]\cup [2,\infty )$.

There is no hidden denominator failure in these ranges. If $K>0$ and $L=0$, the identity forces $N=0$ and $\cos \theta =1$; but $N(0)=3$. If $K=0$, then $|L|=|N|$, and $N$ could vanish only when its two real coefficients have equal magnitudes. Their sum is three, so equality requires $\mu =1/2$, which is not a root of $K$. Hence $L$ is nonzero for every frequency in the allowed ranges. On the compact frequency interval its inverse is bounded for each fixed allowed $\mu$.

When $K<0$, choose a small nonzero frequency. The denominator remains close to three, while the boxed identity gives $|N|>|L|$. Its fixed-frequency numerical mode grows geometrically, proving instability in the usual fixed-Courant refinement limit. Conversely, for $K\ge 0$, the Plancherel theorem converts the mode bound into a discrete L2 norm contraction on the whole-line or periodic grid. The large-Courant branch is possible because the scheme is implicit; stability alone does not promise small error at a large step.