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 Incoming links: Energy dissipation identity for a linearly damped string

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 14B / Solution Created 2026-09-23 Updated 2026-09-25

Separation

y = T (t) X (x)

and the fixed-end conditions give

X_{n} (x) = sin (nπ x), T_{n}^{''} + T_{n}^{'} + n^{2} π^{2} T_{n} = 0.

Put

ω_{n} = n^{2} π^{2} - \frac{1}{4} .

The initial displacement and zero initial velocity then give the separated solution of the damped string equation

y (t, x) = n = 1 \sum \infty a_{n} e^{- t /2} (cos (ω_{n} t) + \frac{sin ( ω _{n} t )}{2 ω _{n}}) sin (nπ x) .

For the triangular initial displacement,

a_{n} = 2 \int_{0}^{1} y (0, x) sin (nπ x) d x = \frac{4 sin ( nπ /2 )}{n ^{2} π ^{2}} .

Thus the even coefficients vanish and the odd ones alternate in sign.

Let

T_{n} (t) = a_{n} e^{- t /2} (cos (ω_{n} t) + \frac{s i n ( ω _{n} t )}{2 ω _{n}}) .

Then

T_{n}^{'} (t) = - a_{n} e^{- t /2} \frac{n ^{2} π ^{2}}{ω _{n}} sin (ω_{n} t) .

Orthogonality of the sine and cosine modes, equivalently the Parseval identity, gives

E (t) = \frac{1}{4} n = 1 \sum \infty a_{n}^{2} e^{- t} [\frac{n ^{4} π ^{4}}{ω _{n}^{2}} sin^{2} (ω_{n} t) + n^{2} π^{2} (cos (ω_{n} t) + \frac{sin ( ω _{n} t )}{2 ω _{n}})^{2}] .

The sign of its derivative is clearest directly from the equation. Integration by parts, using

y_{t} = 0

at the fixed endpoints, yields

E^{'} (t) = \int_{0}^{1} (y_{t} y_{tt} + y_{x} y_{x t}) d x = \int_{0}^{1} y_{t} (y_{tt} - y_{xx}) d x = - \int_{0}^{1} y_{t}^{2} d x \leq 0.

This is the energy dissipation identity for a linearly damped string.

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