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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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<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'' = ''y''(''t'') and ''p'' = ''dy''/''dt''. Consequently, the Hamiltonian ''H''(''p'', ''q'') is a conserved quantity, a constant, equal to ''H'' = {{sfrac|1|2}} + {{sfrac|1|4}} ''ε'' for the given [[initial conditions]]. This implies that both ''y'' and ''dy''/''dt'' have to be bounded:
<math display="block">\left| y(t) \right| \le \sqrt{1 + \tfrac12 \varepsilon} \quad \text{ and } \quad \left| \frac{dy}{dt} \right| \le \sqrt{1 + \tfrac12 \varepsilon} \qquad \text{ for all } t.</math>
===Straightforward perturbation-series solution===
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