Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 4

The matrix exponential is the absolutely convergent power seriesIt converges in any finite-dimensional operator norm, since $\Vert L^m\Vert \le \Vert L{\Vert }^m$. Termwise differentiation gives $B^{\prime }=LB$, $B(0)=I$, so the solution is $q(t)=B(t)q_0$. Also $e^{Lt}{e}^{-Lt}=I$, proving invertibility without requiring $L$ itself to be invertible.

The question's expansion in eigenvectors tacitly needs an eigenbasis: an invertible matrix alone need not be a diagonalizable matrix. The exponential solution and the singular-value arguments remain valid without that assumption.

The relevant norm is the operator norm induced by the complex Euclidean norm, also called the matrix 2-norm:The maximum exists because the unit sphere in finite dimensions is compact. It obeys $\Vert Bq{\Vert }_2\le \Vert B{\Vert }_2\Vert q{\Vert }_2$ and is the norm relevant to the specified perturbation energy. Other matrix norms do not in general give this gain.

A normal matrix admits unitary diagonalization of a normal matrix: $L=V\mathrm{diag}({\lambda }_j)V^{†}$ with orthonormal eigenvectors. ThusFor $t\ge 0$, the largest weight belongs to an eigenvalue with maximal real part. The weighted average is bounded by that weight and attains it on the corresponding eigenspace. Therefore the optimal energy amplification of a linear system isIf the largest real part is repeated, any nonzero combination within that maximal eigenspace is optimal. For negative times the ordering reverses; the displayed formula concerns forward evolution.

The singular value decomposition of $B$ is $B=U\Sigma V^{†}$, where $U,V$ are unitary and $\Sigma =\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)$, ${\sigma }_1\ge \cdots \ge {\sigma }_n>0$ since $B$ is invertible. Its right singular vector $v_j$ and left singular vector $u_j$ are the corresponding unit columns of $V$ and $U$, withEquivalently $B^{†}B{v}_j={\sigma }_j^2{v}_j$ and $B{B}^{†}{u}_j={\sigma }_j^2{u}_j$. The positive numbers ${\sigma }_j$ are the singular values. Degenerate singular subspaces allow any orthonormal choice of paired vectors.

For $q(t)=B{q}_0$, the energy ratio is the Rayleigh quotient of the Hermitian positive-definite matrix $B^{†}B$:Expanding $q_0$ in the orthonormal right singular vectors makes this a weighted average of ${\sigma }_j^2$. HenceAny nonzero vector in the largest right-singular subspace is optimal if ${\sigma }_1$ is repeated. Its amplified state is proportional to the paired left singular vector. For a non-normal matrix $L$, this optimal initial direction need not be an eigenvector of $L$, explaining transient growth from non-normal modes.

A normal mode $x\propto e^{-i\omega t}$ has eigenvalue $\lambda =-i\omega$ of $A$. Setting $d_{\omega }=\omega -{\omega }_1$, the characteristic equation isThusOne branch then has positive imaginary frequency and hence positive exponential growth rate. If $4ac<b^2$, both frequencies are real and the distinct-eigenvalue system has bounded oscillatory motion. At equality there is no positive exponential growth rate, but a nontrivial Jordan block can cause algebraic growth, so exponential neutrality must not be confused with Lyapunov stability. The parameter $d$ listed in the PDF does not appear in the printed matrix and has no effect on this calculation.

Put $q=10/Re>0$. The frequency formula becomesFor large positive Reynolds number, $\sqrt{q-q^2/4}=\sqrt{q}[1-q/8+O(q^2)]$. ThereforeMore precisely the real correction is $-5/Re$, and the next imaginary correction is $O(R{e}^{-3/2})$. The growing branch has a rate tending to zero as $Re\to \infty$, despite being unstable at every sufficiently large finite $Re$.

For $q=10/Re$, the eigenvalues areFor physical $Re>0$, this gives the following classification. If $Re>5/2$, then $0<q<4$ and one eigenvalue has positive real part: exponential instability. If $0<Re<5/2$, then $q>4$, both eigenvalues are distinct and imaginary, and solutions are bounded oscillatory combinations. At $Re=5/2$ there is a repeated imaginary eigenvalue ${\lambda }_0=-i{\omega }_1+2i$. The matrix $N=A-{\lambda }_{0}I=\left(\begin{array}{cc}-2i& 1\\ 4& 2i\end{array}\right)$ is nonzero and satisfies $N^2=0$, so $e^{At}=e^{{\lambda }_{0}t}(I+tN)$ has algebraically growing trajectories.

Thus the exponential instability threshold is $Re>5/2$, consistently with $4ac>b^2$, which here is $4q>q^2$. At the threshold the matrix is defective, and the eigenvector-basis assumption from part (a) cannot be used; the matrix-exponential and singular-value formulas still apply.

A normal matrix satisfies $A{A}^{†}=A^{†}A$. Direct multiplication, with all parameters real, gives the normality criterion for a two-mode shear model:For $a=1$, $b=c=q=10/Re$, both $1-q^2=0$ and $q(1+q)=0$ are required. Their only common solution is $q=-1$. HenceThe original PDF really has these signs and $b=c=10/Re$. Thus parts (v) and (vi) have no physical positive-Reynolds-number case as printed. We give their algebraic continuation at $Re=-10$ below, without silently changing the matrix. The tempting value $Re=10$ makes the off-diagonal magnitudes equal but fails the off-diagonal commutator condition.

At the formal value $Re=-10$, $q=-1$ and $A=-i{\omega }_{1}I+\left(\begin{array}{cc}0& 1\\ -1& -i\end{array}\right)$ is a skew-Hermitian matrix. Put $r_{\pm }=(-1\pm \sqrt{5})/2$. Its two eigenvalues and normalized eigenvectors areSubstitution uses $r_{\pm }^2+r_{\pm }-1=0$ and verifies $A{v}_{\pm }={\lambda }_{\pm }{v}_{\pm }$. Their Hermitian inner product issince $r_{+}{r}_{-}=-1$. This provides the requested orthogonality in the formal continuation. There is no corresponding physical $R{e}_N>0$ for the printed system.

At the only algebraic normal value $Re=-10$, the generator is a skew-Hermitian matrix, so its matrix exponential is a unitary matrix. Both eigenvalues have zero real part; henceThere is no distinguished initial amplitude ratio and no perturbation-energy growth. The relations $x_2(0)=i{r}_{\pm }{x}_1(0)$ select the individual orthogonal eigenmodes, but arbitrary nonzero combinations are equally optimal because all singular values are one. If $Re$ is restricted to its physical positive domain, this part has no applicable normal case at all. This resolves the inconsistency in the original printed parameters rather than assuming an unjustified optimal direction.