Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 8A / b

Table of contents
- Solution b

Solution

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore give

\frac{d}{d t} e^{tM} = \sum_{n = 1}^{\infty} \frac{t ^{n - 1} M ^{n}}{( n - 1 )!} = M e^{tM}, e^{0 M} = I .

Hence

x (t) = e^{tM} x_{0}

solves the initial value problem. For this particular matrix, direct multiplication gives

M^{2} = - I

. Splitting the matrix exponential into its even and odd powers proves

e^{tM} = I cos t + M sin t = (cos t + 2 sin t - sin t 5 sin t cos t - 2 sin t) .

This is the matrix exponential when the square is minus the identity. If

x_{0} = (x_{0}, y_{0})^{T}

, the general solution is

x (t) = x_{0} cos t + (2 x_{0} + 5 y_{0}) sin t, y (t) = y_{0} cos t - (x_{0} + 2 y_{0}) sin t .

The constants in part (a) are

α = x_{0} /5

and

β = y_{0} + 2 x_{0} /5

, which makes the two forms identical.

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