 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/5/c/solution

Let

β = Ω^{2} - κ^{2}

, interpreted as

i κ^{2} - Ω^{2}

in the overdamped case. Factoring the denominator gives

Ω^{2} - ω^{2} + 2 iκω = - (ω - ω_{-}) (ω - ω_{+}), ω_{\pm} = iκ \pm β .

y (t) = \int_{- \infty}^{\infty} G (t - ξ) f (ξ) d ξ,

where

G (t) = - \frac{1}{2 π} \int_{- \infty}^{\infty} \frac{e ^{iω t}}{( ω - ω _{-} ) ( ω - ω _{+} )} d ω .

Thus the requested numerator is

s (ω, t) = - \frac{1}{2 π} e^{iω t} .

For all positive

κ, Ω

κ \pm κ^{2} - Ω^{2}

, both positive. At equality they merge into a double pole.

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