To illustrate how the symmetries of a problem can

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Unformatted text preview: n Section 2.1 that the eigenspace corresponding to the other eigenvalue is just the orthogonal complement W−3 = span −1 1 . Unit length eigenvectors lying in the two eigenspaces are √ √ 1/√2 −1/ 2 √ b1 = , b2 = 1/ 2 1/ 2 . The theorem of Section 2.1 guarantees that the matrix √ √ 1/√2 −1/ 2 √ B= , 1/ 2 1/ 2 whose columns are b1 and b2 , will diagonalize our system of differential equations. 51 Indeed, if we define new coordinates (y1 , y2 ) by setting √ √ x1 1/√2 −1/ 2 y1 √ = , x2 y2 1/ 2 1/ 2 our system of differential equations transforms to d2 y1 /dt2 d2 y2 /dt2 We set ω1 = 1 and ω2 = √ = −y1 , = −3y2 . 3, so that this system assumes the familiar form 2 d2 y1 /dt2 + ω1 y1 2 2 2 d y2 /dt + ω2 y2 = = 0, 0, a system of two noninteracting harmonic oscillators. The general solution to the transformed system is y1 = a1 cos ω1 t + b1 sin ω1 t, y2 = a2 cos ω2 t + b2 sin ω2 t. In the original coordinates, the general solution to (2.15) is √ √ x1 1/√2 −1/ 2 a1...
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This document was uploaded on 01/12/2014.

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