Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 68 / 2

Put $h=\Delta x$, $k=\Delta t$, with fixed positive $\mu =k/h$. Insert a smooth exact solution and divide the residual by $2k$:Taylor expansion givesThe differential equation cancels the leading part, leaving normalized local truncation error $O(k^2+h^2)$. Since $u_{ttt}=({\partial }_x+{\partial }_y{)}^{3}u$ contains mixed derivatives, there is no fixed positive Courant number canceling this leading error for every smooth solution. Thus the method is second order in time and space. Global second-order convergence also requires stability and initial values at both time levels accurate enough to retain this order.

Use a spatial discrete Fourier mode $u_{k,j}^n=a_n{e}^{i(k\xi +j\eta )}$. Its recurrence and amplification polynomial of a multilevel finite difference scheme areThe polynomial roots are $G_{\pm }=is\pm \sqrt{1-s^2}$. For $|s|<1$ both have modulus one and are separated. Since $|\sin \xi +\sin \eta |\le 2$, a fixed $0<\mu <1/2$ bounds their separation below by $2\sqrt{1-4{\mu }^2}$. The two-level companion matrix is therefore diagonalizable with uniformly bounded eigenvector matrix and inverse. To see the uniform bound explicitly, write $V=\left(\begin{array}{cc}{G}_{+}& G_{-}\\ 1& 1\end{array}\right)$. Its determinant is $G_{+}-G_{-}$; all its entries have modulus one, and those of $V^{-1}$ are bounded by the reciprocal of the root separation. The companion matrix powers are $V\mathrm{diag}(G_{+}^n,G_{-}^n)V^{-1}$, bounded independently of frequency and step number. Parseval's identity converts this into a mesh-independent discrete $L^2$ bound for arbitrary perturbations at the two initial levels.

If $\mu >1/2$, choose $\xi =\eta =\pi /2$. Then $s=2\mu >1$, and one polynomial root has modulus $s+\sqrt{s^2-1}>1$. On periodic meshes with the number of points divisible by four, this is an actual grid mode, producing exponential linear instability.

At $\mu =1/2$ the same phases give $(G-i{)}^2$. The companion matrix is not a scalar matrix, so the double polynomial root has a nontrivial Jordan block. The solution $a_n=n{i}^n$ has bounded starting amplitudes but grows like $n$. On a fixed physical time interval $n$ is of order $1/k$, so no mesh-independent stability bound exists. Consequentlyis the stability interval under the standard two-level definition. Testing only polynomial root moduli would misleadingly include the endpoint. A special startup selecting the non-growing branch at that endpoint restricts the perturbations and does not establish stability of the full recurrence. This is the two-dimensional leapfrog stability threshold.