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Question: Find the leading order of the asymptotic expansion for large t:

$\frac{d^2x}{dt}+\varepsilon\beta\frac{dx}{dt}+x+\varepsilon x^3=Fcos(\frac{1}{3}\big(1+\varepsilon\omega)t\big)$

I have tried to solve the problem via regular multiple scale analysis by transforming $t_0=t, t_1=\varepsilon t$ & expressing the solution in the form of an asymptotic expansion $x(t)=y(t_0,t_1)=\sum_{n=0}^\infty \varepsilon^ny_n(t_0,t_1)$.

The leading order problem becomes: $\frac{\partial^2y_0}{\partial t_0}+y_0=Fcos\big(\frac{1}{3}(t_0+\omega t_1)\big)$

solving, i obtain $y_0=R_0(t_1)cos(t_0+\phi_0(t_1))+\frac{9}{8}cos\big(\frac{1}{3}(t_0+\omega t_1)\big)$

After solving the leading order the 1st order problem is as follows

$$\frac{\partial^2 y_1}{\partial t_0}+y_1=sin(t_0+\phi_0)(2R_0'+\beta R_0)+cos(t_0+\phi_0)(2R_0\phi_0')+\frac{\omega}{4}Fcos(\frac{1}{3}(t_0+\omega t_1))+\frac{3}{8}\beta Fsin(\frac{1}{3}(t_0+\omega t_1))-R_0^3\left(\frac{1}{4}cos(3(t_0+\phi_0)))+\frac{3}{4}cos((t_0+\omega t_1))\right)-(\frac{9}{8}F)^3(\frac{1}{4}cos(t_0+\omega t_1)+\frac{3}{4}cos(\frac{1}{3}(t_0+\omega t_1)))-3R_0(\frac{9}{8}F)^2\left(\frac{1}{4}cos\big(\frac{1}{3}t_0+\phi_0-\frac{2}{3}(\omega t_1)\big)+\frac{1}{4}cos(\frac{4}{3}t_0+\phi_i+\frac{2}{3}\omega t_1)+\frac{1}{2}cos(t_0 +\phi_0)\right)-3R_0^2(\frac{9}{8}F)\left(\frac{1}{4}cos(\frac{5}{2}t_0+2\phi_0+\frac{1}{3}\omega t_1)+\frac{1}{4}cos(\frac{3}{2}t_0+2\phi_0-\frac{1}{3}\omega t_1)+\frac{1}{2}cos(\frac{1}{3}(t_0+\omega_1 t_1))\right)$$

demanding the coefficent of the $sin(t_0+\phi_0)$ term to vanish (for large t) i obtain $R_0=0$, thus the equation becomes $$\frac{\partial^2 y_1}{\partial t_0}+y_1=\frac{\omega}{4}Fcos(\frac{1}{3}(t_0+\omega t_1))+\frac{3}{8}\beta Fsin(\frac{1}{3}(t_0+\omega t_1))-(\frac{9}{8}F)^3\left(\frac{1}{4}cos(t_0+\omega t_1)+\frac{3}{4}cos(\frac{1}{3}(t_0+\omega t_1)))\right)$$

the problem is that a secular term remains (i.e. $-(\frac{9}{8}F)^3(\frac{1}{4}cos(t_0+\omega t_1))$)?

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    $\begingroup$ $t$ or $x$? Also the forcing term on the hrs should have the independent variable no? $\endgroup$ – Chinny84 Jul 8 '14 at 10:31
  • $\begingroup$ Hi guys, thanks for the comments, i've corrected any missunderstanding and written my approach for solving, and what problem i have encounter. thanks for the help $\endgroup$ – Arietul Jul 9 '14 at 13:39

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