Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 4

Keep the Courant number $\mu =\Delta t/\Delta x$ fixed and write $h=\Delta x$. An exact solution of $u_t=u_x$ is $u(x,t)=g(x+t)$. Substituting it into the update gives the defectwhere $s=x+t$ and $g^{(3)}$ is evaluated at $s$. The constant, first-derivative and second-derivative terms cancel. Dividing by $\Delta t=\mu h$ gives a second-order consistency error. Hence for fixed $0<\mu <1$, $\mu \ne 1/2$, the scheme has order two.

At the special value $\mu =1/2$, the update reduces to $u_m^{n+1}=u_{m+1}^{n-1}$. This is exactly the transport over two time steps, since $2\Delta t=\Delta x$, so the defect vanishes identically. With exact data at both starting levels it transports their values exactly; errors in those starting levels are still present, and the even and odd time subsequences evolve separately.

For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.