Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 3

A matrix is non-normal when it does not commute with its adjoint matrix:Nonorthogonal decaying eigenmodes can interfere constructively and produce transient growth. For example,has two negative eigenvalues but is non-normal. Starting from $x(0)=(0,1{)}^T$ givesAt $t=\log 2$, its squared Euclidean norm is $2.5^2+0.25^2>1=|x(0){|}^2$. The energy grows transiently even though both eigenmodes eventually decay.

Because $L$ is diagonalizable, writewhere every diagonal entry of $\Lambda$ has negative real part. Define the equivalent normThenfor every $t\ge 0$. Equivalently, this norm comes from the positive-definite inner-product matrix $H=V^{-†}{V}^{-1}$. Thus stable eigenvalues always admit a norm with no growth, even though the standard Euclidean norm may show transient amplification.

The first component obeys ${\dot{x}}_1={\lambda }_1{x}_1$, so $x_1(t)=e^{{\lambda }_{1}t}{x}_1(0)$. Variation of constants inthen givesHence the matrix exponential is

WriteThe extremal squared amplification factors are the largest and smallest singular values of $A$ squared, equivalently the eigenvalues $G$ ofSince its trace is $a^2+b^2+c^2$ and its determinant is $a^2{c}^2$, the requested quadratic is

As ${\lambda }_1\to {\lambda }_2=\lambda$, L'Hôpital's rule gives $b\to t{e}^{\lambda t}$, soThe quadratic becomesFor $t\ge 0$, its maximum root is

For $-1/2<\lambda <0$, logarithmic differentiation givesThe unique maximizing time is thereforeWriting $\alpha =-\lambda >0$ and $q=\sqrt{1-4{\alpha }^2}$, the maximum isConsequently, as $\lambda \to 0^{-}$,At $\lambda =0$ exactly there is no finite maximizing time: the Jordan-block shear produces unbounded quadratic growth. The formulas describe how the optimal transient moves to later times and becomes larger as the damping tends to zero.