Multiple-scale analysis: Difference between revisions

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Undid revision 1280810603 by Dstrozzi (talk) That cannot be the case: Dropping q^4 from the Hamiltonian would make the problem linear, independent of epsilon
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Differential equation and energy conservation: Added back and improved the comment on the q bound.
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which is a second-order [[ordinary differential equation]] describing a [[nonlinear system|nonlinear]] [[oscillator]]. A solution ''y''(''t'') is sought for small values of the (positive) nonlinearity parameter 0&nbsp;<&nbsp;''ε''&nbsp;≪&nbsp;1. The undamped Duffing equation is known to be a [[Hamiltonian system]]:
<math display="block">\frac{dp}{dt}=-\frac{\partial H}{\partial q}, \qquad \frac{dq}{dt}=+\frac{\partial H}{\partial p}, \quad \text{ with } \quad H = \tfrac12 p^2 + \tfrac12 q^2 + \tfrac14 \varepsilon q^4,</math>
with ''q''&nbsp;=&nbsp;''y''(''t'') and ''p''&nbsp;=&nbsp;''dy''/''dt''. Consequently, the Hamiltonian ''H''(''p'',&nbsp;''q'') is a conserved quantity, a constant, equal to ''H<sub>0</sub>''&nbsp;=&nbsp;{{sfrac|1|2}}&nbsp;+&nbsp;{{sfrac|1|4}}&nbsp;''ε'' for the given [[initial conditions]]. This implies that both ''yq'' and ''dy''/''dtp'' have to be bounded:
<math display="block">\left| y(t)q \right| \le \sqrt{1 + \tfrac12 \varepsilon} \quad \text{ and } \quad \left| \frac{dy}{dt}p \right| \le \sqrt{1 + \tfrac12 \varepsilon} \qquad \text{ for all } t.</math>The bound on q is found by equating H with p = 0 to H<sub>0</sub>: <math>\tfrac12 q^2 + \tfrac14 \varepsilon q^4 = \tfrac12 + \tfrac14 \varepsilon</math>, and then dropping the q<sup>4</sup> term. This is indeed an upper bound on |q|, though keeping the q<sup>4</sup> term gives a smaller bound with a more complicated formula.
 
===Straightforward perturbation-series solution===