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 Incoming links: Causal Fourier resolvent of a point-driven fixed string

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 5 / b / iii / Solution 2026-10-18

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Expand in the complete Fourier sine series

y (ω, x) = \sum_{n \geq 1} b_{n} (ω) sin (k_{n} x)

and apply Fourier inversion. Substitution of part (ii) gives

y (t, x) = \frac{1}{π L} n = 1 \sum \infty sin (k_{n} x_{0}) sin (k_{n} x) \int_{- \infty}^{\infty} \frac{F ( ω ) e ^{iω t}}{k _{n}^{2} - ω ^{2}} d ω, C = \frac{1}{π L} .

(1)

To check the source normalization, apply

\partial_{t}^{2} - \partial_{x}^{2}

mode by mode. It cancels the denominator; the remaining inverse transform gives

2 π F (t)

, and completeness gives

\frac{2}{L} \sum_{n \geq 1} sin (k_{n} x_{0}) sin (k_{n} x) = δ (x - x_{0}) .

(2)

The endpoints vanish termwise. These identities establish the equation in the distributional sense, with classical validity wherever the series and derivatives converge suitably.

The displayed real-axis integrals in the paper are not ordinary convergent integrals at their poles. One definite choice is the Cauchy principal value at each

ω = \pm k_{n}

. If a causal solution is wanted, use

[k_{n}^{2} - (ω - i 0)^{2}]^{- 1}

instead. Its inverse transform is

H (t) sin (k_{n} t) / k_{n}

, so the causal solution is

y_{ret} (t, x) = \frac{2}{L} \sum_{n \geq 1} \frac{s i n ( k _{n} x _{0} ) s i n ( k _{n} x )}{k _{n}} \int_{- \infty}^{t} F (s) sin [k_{n} (t - s)] d s,

(3)

when the past integral exists. The principal-value choice instead uses the time-symmetric kernel

sin (k_{n} ∣ t ∣) / (2 k_{n})

. No temporal initial data are given, so homogeneous sine and cosine modes remain permissible additions. This is the causal Fourier resolvent of a point-driven fixed string.

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