Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 80 / 4

Put $h=\Delta x$, $k=\Delta t$ and $\mu =k/h$, consistent with $u_t=u_x$ and the exact transport $u(x,t)=F(x+t)$. The PDF's middle implicit value is $u_m^{n+1}$; the converted TeX incorrectly places it at $m+1$.

Write the new-level coefficients at offsets $-1,0,1$ as $a=\mu (1+\mu )/2$, $b=(1+\mu )(2-\mu )$ and $c=(1-\mu )(2-\mu )/2$, and the old-level coefficients as $d=2-\mu$, $e=1+\mu$. Their sums are $a+b+c=d+e=3$. To compute the local truncation error, expand the exact values $F(x_m+t_n+h(\mu +j))$ at the new level and $F(x_m+t_n+jh)$ at the old level. Equivalently the residual symbol isThe coefficients through $z^3$ vanish, whileFor example, the first three moment identities are ${\sum }_j{a}_j(\mu +j{)}^r=e$ for $r=1,2,3$, where $a_{-1}=a,a_0=b,a_1=c$; the fourth-moment difference is $\mu (\mu -2)(\mu -1)(\mu +1)$. Thus the exact residual isNormalizing the equation by $3k$ givesFor fixed nonzero $\mu \notin \{-1,1,2\}$, the method has order three. Its one-step residual is fourth degree, but the differential-equation truncation error is third order because of division by the time step.

There are exact transport exceptions. At $\mu =-1$ the equation reduces to $u_m^{n+1}=u_{m-1}^n$. At $\mu =2$ it gives $u_m^{n+1}=u_{m+2}^n$. At $\mu =1$, the exact shift $u_m^{n+1}=u_{m+1}^n$ satisfies the implicit equation, whose symbol is invertible as shown below. These are exact for every grid profile, not merely fourth-order exceptions. At $\mu =0$ the equation is the zero-time-step identity. These properties are summarized by the rational implicit advection stencil with exact shift exceptions.

For Fourier stability analysis, take spatial modes $e^{im\theta }$ and putA bounded implicit step requires $D$ to stay away from zero, and von Neumann stability analysis requires $|G|\le 1$ for every frequency. Expanding the squared moduli and using $\cos 2\theta =2{\cos }^2\theta -1$ givesConsequently, apart from denominator invertibility, the necessary and sufficient sign condition isIts real solution set is $(-\infty ,-1]\cup [0,1]\cup [2,\infty )$.

To check the denominator rather than assume it, noteIf $\mu \ne 1/2$, a zero can occur only at zero or the Nyquist frequency. The zero-frequency value is nonzero, and the Nyquist zeros occur only at $\mu =(1\pm \sqrt{3})/2$, both outside the proposed stable intervals. At $\mu =1/2$, $D=9/4+(3/4)\cos \theta \ge 3/2$, also nonzero. Continuity on the frequency circle now gives a positive lower bound on $|D|$ for each allowed $\mu$.

The implicit convolution operator is therefore boundedly invertible on the square-summable grid sequences. By the Plancherel theorem, $|G|\le 1$ givesOutside these intervals the modulus difference is negative at every nonzero frequency where $D$ is nonzero. There is an interval on which $|G|>1$, giving arbitrarily large growth of powers of the step operator, so the method is unstable; a zero of $D$ additionally makes the step ill posed.

For the physical choice $k,h>0$, the stable range is $0<\mu \le 1$ or $\mu \ge 2$; zero is the limiting zero-step case. The exact-shift values have unit gain and are included.