Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 68 / 5

Let $h=\Delta x$, $k=\mu h$ and $M_r={\sum }_{j\in \{-2,-1,1,2\}}{a}_j{j}^r$. Along a smooth solution, ${\partial }_t^{r}u={\partial }_x^{r}u$. The exact difference $u(x,t+k)-u(x,t-k)$ contains only odd derivatives. Matching terms through degree four therefore requiresThe first two conditions give $a_{-1}+a_1=a_{-2}+a_2=0$, which also implies $M_4=0$. With $a_{-j}=-a_j$, the remaining equations are $a_1+2a_2=\mu$ and $a_1+8a_2={\mu }^3$. HenceThe fifth spatial moment is $M_5=10{\mu }^3-8\mu$. After dividing the exact-minus-scheme residual by $2k$, its first possible nonzero term isThis is the fourth-order two-step advection stencil. Thus the normalized local truncation error is $O(h^4)$ at fixed positive Courant number, as required. At $\mu =1$ or $2$ the formula translates the physical advection branch exactly, but consistency of a numerical method alone does not decide stability of all two-level perturbations.

Fourier transform on the infinite spatial lattice giveswith polynomial roots $G_{\pm }=iA\pm \sqrt{1-A^2}$. When $|A|$ is uniformly less than one, the polynomial roots have modulus one and a uniform positive separation. Diagonalizing the companion matrix then bounds all its powers independently of $\theta$ and $n$. Parseval's identity proves stability in the discrete $L^2$ norm for arbitrary square-summable starting perturbations at both levels.

At $\mu =1/2$ the coefficients giveConsequently $\mu =1/2$ is stable.

At $\mu =3/2$,The growing polynomial root at this phase has modulusTo turn this into a Cauchy problem linear instability proof, rather than rely on a plane wave of infinite norm, choose a small interval around $\pi /3$ on which the growing polynomial root has modulus at least $1+\delta$. Choose an initial Fourier transform amplitude that is a nonzero square-integrable function supported there, and set the level-one amplitude equal to that polynomial root times the level-zero amplitude. The recurrence then multiplies by the same growing polynomial root at every step on the support. Parseval identity gives $\Vert u^n{\Vert }_2\ge (1+\delta {)}^n\Vert u^0{\Vert }_2$, whereas both initial norms are bounded independently of $h$. Real data can be obtained by adding the conjugate interval around $-\pi /3$. Since $n=T/(\mu h)$ tends to infinity at fixed $T$, $\mu =3/2$ is unstable.