Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 41C (Numerical Analysis)

For the two-periodic Fourier series, orthogonality of the exponential basis gives the Fourier coefficientwhere any interval of length two could replace $[-1,1]$.

Let $C_{nm}={\hat{c}}_{n-m}$ for $-d\le n,m\le d$. Since $c$ is real,so $C_{mn}=\overline{C_{nm}}$ and $C$ is a Hermitian matrix. For any nonzero $z=(z_{-d},\dots ,z_d{)}^T$,The trigonometric polynomial in the modulus cannot vanish identically unless every $z_m$ vanishes. Since $c(x)>0$, the integral is strictly positive. Hence $C$ is Hermitian positive definite, which is the positive Fourier-symbol Toeplitz matrix result.

Differentiating the truncated Fourier expansion givesThe $n$th Fourier coefficient of the product $c{u}_x$ is the discrete convolutionProjecting the advection equation onto modes $|n|\le d$ therefore givesThusEquivalently, ifthenThis is the Fourier spectral method for variable-coefficient advection.

The matrix $C$ from part (a) is Hermitian positive definite, while the real diagonal matrix $D$ is Hermitian. The stated product theorem, or similarityshows that every eigenvalue of $CD$ is real. Hence every eigenvalue of $B=-CD$ is real; this is the real spectrum of a positive-Hermitian times Hermitian product.

Moreover, $C$ is invertible and $D$ has rank $2d$, so for $d\ge 1$ the matrix $B$ has nonzero real eigenvalues. If $\lambda \ne 0$ is one of them, the corresponding mode of the semidiscrete equation has eigenvalue $i\pi \lambda$. The explicit Euler method has amplification factorfor every $\Delta t>0$. Therefore the explicit Euler discretization is unstable, as in explicit Euler instability on a nonzero imaginary eigenvalue.

Using Euler's formula,ThusOrdering the modes as $(-1,0,1)$, let $a=1/2+i/4$. ThenConsequentlyThis matrix is triangular, so its eigenvalues are its diagonal entries: