Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 63 / 5

Write $h=\Delta x=1/N$ and collect only the interior values, so the actual vector dimension is $(N-1{)}^2$. The PDF's $N^2$ matrix size and its indices $k,l$ are inconsistent with the declared interior grid $m,k=1,\dots ,N-1$; neither affects the following stability argument when the interior indexing is used consistently.

Use the discrete L2 norm $\Vert u{\Vert }_h^2=h^2{\sum }_{m,k=1}^{N-1}|u_{m,k}{|}^2$, extending $u$ by zero to the boundary. Let $A$ be the horizontal flux operator and $B$ the vertical one. Each face coefficient is shared by its two adjacent grid values, so summation by parts givesAll boundary-face terms are included. Positivity of $a$ makes $A,B$ symmetric negative definite on the interior grid: zero in either energy sum forces each corresponding grid line to be constant, and its zero boundary endpoint forces that constant to vanish. ConsequentlyThis bound is independent of the spatial mesh, proving semidiscrete stability. No time integrator or time-step restriction has yet been introduced.

Put $\delta =\Delta t$ and $L=A+B$, keeping the spatial grid fixed for the temporal error assertion. The matrix exponential expansions giveHence the Lie-Trotter splitting commutator error isThe energy estimate from part 1 makes $e^{tA}$, $e^{tB}$ and $e^{tL}$ contractions, so both $S_{\delta }$ and $E_{\delta }$ have norm at most one. Telescope their $n$th powers:For $n\delta \le T$, this givesThus the Lie-Trotter splitting has global temporal error $O(\Delta t)$. The constant here may depend on the fixed spatial operators; a mesh-uniform joint-limit error estimate needs additional regularity and commutator bounds. If $A$ and $B$ commute, this particular exponential splitting is exact, but variable diffusion coefficients generally remove that commutativity.

The $[1/1]$ Padé approximant of the exponential is $r(z)=(1+z/2)/(1-z/2)$. Replacing each exponential in its original order gives the split Crank-Nicolson methodwith the analogous definition for $B$. Operationally, first solve the $B$ substep and then the $A$ substep. The matrices $I-\delta A/2$, $I-\delta B/2$ are invertible for every $\delta \ge 0$, since $A,B$ are symmetric negative definite.

An orthonormal eigenbasis of $A$ diagonalizes $C_A$. Its eigenvalue for $\lambda \le 0$ isTherefore $\Vert C_A{\Vert }_h\le 1$, and similarly $\Vert C_B{\Vert }_h\le 1$. Submultiplicativity gives the contractivity of split Crank-Nicolson diffusion:No commutativity or simultaneous diagonalization is needed for this product bound. This proves unconditional discrete $L^2$ stability, not positivity of every stencil weight or unconditional maximum-norm monotonicity. Nor do the individually second-order rational substeps make the ordered splitting second order: when $[A,B]\ne 0$, its leading splitting defect remains ${\delta }^2[A,B]/2$.