Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 39B / Solution

Set ${\theta }_k=k\pi /(M+1)$ and extend the proposed vector by $v_0=v_{M+1}=0$. For $v_m=i^m\sin (m{\theta }_k)$, the angle-addition identity givesHence the Toeplitz antisymmetric tridiagonal matrix has eigenvaluesThe real sine vectors are eigenvectors of the real symmetric tridiagonal matrix with both off-diagonals one, with distinct eigenvalues $2\cos {\theta }_k$; thus they are orthogonal and form a basis. Multiplication by the unitary diagonal matrix $\mathrm{diag}(i,i^2,\dots ,i^M)$ preserves orthogonality. Their squared norms are $(M+1)/2$, so multiplying by $\sqrt{2/(M+1)}$ gives a common orthonormal basis for all $a,b$, even when some eigenvalues coincide.

Let $K$ have superdiagonal one, subdiagonal minus one and diagonal zero. The prescribed endpoint values remove the exterior unknowns, and the scheme isIn the common orthonormal eigenbasis, the two eigenvalues are $1\mp i(\mu /2)\cos {\theta }_k$. Every eigenvalue of $B$ is nonzero for real $\mu$, and the amplification matrix $Q=B^{-1}C$ has eigenvaluesThus $Q$ is a unitary matrix and $\Vert u^{n+1}{\Vert }_2=\Vert u^n{\Vert }_2$ exactly. The same holds for the mesh-weighted norm. This matrix argument proves $\boxed{\text{stability for every real }\mu \text{, hence every physical }\mu \ge 0}$, without a Fourier stability calculation.

The printed mesh ratio has a consistency error: advection requires $\mu =\Delta t/\Delta x$, while the PDF prints $\Delta t/(\Delta x{)}^2$. Indeed the centered spatial difference is $2\Delta x{u}_x+O((\Delta x{)}^3)$, so division of the scheme by $\Delta t$ gives leading right-hand coefficient $\mu \Delta x/\Delta t$. The printed choice makes this $1/\Delta x$ rather than one. The stability conclusion is valid for the algebraic scheme with either definition, but does not fix this inconsistency. The continuous first-order advection problem also ordinarily specifies only the inflow endpoint, here $x=1$; zero data at both endpoints restrict admissible initial data. For example the initially boundary-zero profile $\sin \pi x$ develops nonzero outflow $u(0,t)=\sin \pi t$ before inflow changes reach that boundary. The matrix norm calculation concerns the stated two-endpoint discretization and does not establish well-posedness for arbitrary continuous initial data.