Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / a

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- Solution a

Solution

The energy is

E (t) = c_{1}^{2} e^{- 2 λ_{1} t} + c_{2}^{2} e^{- 2 λ_{2} t} + 2 α c_{1} c_{2} e^{- (λ_{1} + λ_{2}) t} .

Thus

E^{'} (0) = - 2 [λ_{1} c_{1}^{2} + λ_{2} c_{2}^{2} + (λ_{1} + λ_{2}) α c_{1} c_{2}] .

This quadratic form can be negative for positive

c_{1}, c_{2}

precisely when

α < - \frac{2 λ _{1} λ _{2}}{λ _{1} + λ _{2}} .

The decaying vectors must therefore be sufficiently nonorthogonal and oppositely directed. This is transient growth from non-normal modes.

With

c_{2} = 1

and

μ = c_{1} / c_{2}

, maximizing

E^{'} (0)

gives

2 λ_{1} μ + (λ_{1} + λ_{2}) α = 0,

so

μ_{opt} = - \frac{( λ _{1} + λ _{2} ) α}{2 λ _{1}} .

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