Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 38E (Numerical Analysis)

Let all spatial derivatives be evaluated at $x=mh$, $t=nk$. Expand the two adjacent flux contributions symmetrically. The bracket in the scheme isFor clarity, the $h^2$ terms are $a{u}_{xx}+a^{\prime }{u}_x$; odd orders cancel under $h\mapsto -h$. Under four spatial derivatives the next coefficient is $a{u}_{xxxx}/12+a^{\prime }{u}_{xxx}/6+a^{\prime \prime }{u}_{xx}/8+a^{(3)}{u}_x/24$. Time expansion gives $u(x,t+k)=u+k{u}_t+O(k^2)$. Since $u_t=(a{u}_x{)}_x$ and $k=\mu h^2$,This stronger smooth-data residual implies the requested $O(h^3)$ consistency bound. Even a less detailed Taylor expansion with an $O(h^3)$ spatial remainder would suffice for the printed order. Uniform bounds on the needed derivatives over the region under consideration make the estimate uniform in the interior mesh index; the zero boundary values enter the scheme exactly.

The Gershgorin circle theorem says that every eigenvalue of a square matrix lies in at least one disc centred at a diagonal entry with radius the sum of the absolute values of the off-diagonal entries in that row.

For interior unknowns write the update as $u^{n+1}=B{u}^n$. The matrix is real symmetric and tridiagonal, with diagonal $1-\mu (a_{m-1/2}+a_{m+1/2})$ and adjacent entries $\mu a_{m+1/2}$. Each Gershgorin radius is at most $\mu (a_{m-1/2}+a_{m+1/2})$. Its real interval is therefore contained inFor $0<\mu \le 1/(2a_{+})$, every eigenvalue lies in $[-1,1]$. Symmetry gives an orthogonal diagonalization, so $\Vert B^n{\Vert }_2\le 1$ for every $n$: errors in initial data cannot grow. This is stability in the discrete ${\ell }^2$ norm, also in its $h$-weighted version. One can additionally see maximum-norm stability directly: the update weights are nonnegative with row sums at most one once the zero boundary unknowns are removed. Hence $\Vert B{\Vert }_{\infty }\le 1$.