find an approximate solution, up to the order of epsilon

Question

The question is to find an approximate solution, up to the order of epsilon of following problem.

$$y'' + y+\epsilon y^3 = 0$$ $$y(0) = a$$ $$y'(0) = 0$$

I tried to solve the given problem using perturbation theory.

$$y(t) = y_0(t) + \epsilon y_1(t) + \epsilon^2 y_2(t) + \cdots$$ $$t = w(\epsilon)t = (1 + \epsilon w_1 + \epsilon^2 w_2 + \cdots )t$$

However, i failed to find an appropriate approximate solution... help me!!

Answer 1

You don't need a series for $t$, only $y$. So substitute $y=y_0+\epsilon y_1+\epsilon^2 y_2+\ldots$ into the equation:

$$ y_0''+\epsilon y_1''+\epsilon^2 y_2''+y_0+\epsilon y_1+\epsilon^2 y_2+\epsilon\left(y_0+\epsilon y_1+\epsilon^2 y_2\right)^3+O(\epsilon^3)=0. $$

You also have to expand the boundary conditions, which will give $y_0(0)=a$, and all other $y_i(0)=0$, and all $y_i'(0)=0$.

Now look at the equations for each magnitude of $\epsilon$. Firstly, the largest term is $O(1)$, and the equation is $$ y_0''+y_0=0 $$ so $y_0=A\sin(t)+B\cos(t)$, and the boundary condition gives $y_0=a\cos(t)$. Then the $O(\epsilon)$ equation is $$ y_1''+y_1+y_0^3=0,\ y_1(0)=y_1'(0)=0 $$ and the $O(\epsilon^2)$ equation is $$ y_2''+y_2+3y_0^2y_1=0,\ y_2(0)=y_2'(0)=0, $$ at $O(\epsilon^3)$, $$y_3''+y_3+3y_0^2y_2+3y_0y_1^2=0,\ y_3(0)=y_3'(0)=0 $$ and so on. You can solve these (remember $\cos(x)^3=3/4(\cos(3x)+3\cos(x))$) to give a series for $y$.

Answer 2

Following a similar problem in Question about phase shift on multiple-scale analysis, use polar coordinates in phase space $$ (y(t),y'(t))=r(t)·(\,\cos(φ(t)),\,\sin(φ(t))\,) $$ with initial conditions $r(0)=a$ and $φ(0)=0$. The unperturbed equation leads to $(y,y')=a·(\cos(-t),\sin(-t))$. Thus one expects $r(t)=a+O(ϵ)$ and $φ(t)=-t+O(ϵ)$.

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From $r^2=y^2+y'^2$ one obtains $$ rr'=y'(y+y'')=-ϵy^3y'=-ϵr^4\cos^3φ\sinφ $$ or $$ \left(\frac1{r^2}\right)'=-2ϵ\cos^3 t\sin t+O(ϵ^2) $$ which integrates as $$ \frac1{r^2}-\frac1{a^2}=\fracϵ2(\cos^4t-1)+O(ϵ^2) $$ or $$ r(t)=\frac{a}{\sqrt{1-\frac12ϵa^2(1-\cos^4t)}}+O(ϵ^2) $$

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From $\tan(φ)=\frac{y'}{y}$ one gets $$ φ'=\cos^2φ·\frac{y''y-y'^2}{y^2} =\frac{(y''+y)y-(y^2+y'^2)}{r^2} =-ϵr^2\cos^4φ-1\\ =-ϵa^2\cos^4t-1+O(ϵ^2) =-1 -\fracϵ8a^2(\cos4t+4\cos2t+6)+O(ϵ^2) $$ which integrates as $$ φ(t)=-t-\fracϵ{32}a^2(\sin4t+8\sin2t+24t)+O(ϵ^2) $$ which gives the correction to the average angular velocity as $-(1+\frac34ϵa^2)$

Answer 3

This question asks implicitly to use the Lindstedt-Poincaré method of strained parameters. That is to say, a perturbed solution of the Duffing equation $y''(t) + y(t) + \epsilon y(t)^3 = 0$ is written as \begin{aligned} y &= y_0(\tau) + \epsilon y_1(\tau) + \dots \\ \omega &= 1 + \epsilon \omega_1 + \dots \end{aligned} where $\tau = \omega t$. Therefore, the original equation becomes $\omega^2 \ddot y(\tau) + y(\tau) + \epsilon y(\tau)^3 = 0$ with the same initial conditions. A solution of the form $y = y_0(\tau) + \epsilon y_1(\tau) + \dots$ gives at zeroth order $\ddot y_0 + y_0 = 0$ with the same initial conditions, i.e. $$ y_0 = a \cos \tau \, . $$ At the second order, we have $\ddot y_1 + y_1 =-{y_0}^3 +2\omega_1 y_0$ with homogeneous initial conditions, i.e. $$ y_1(\tau) = \tfrac{1}{32}a^3\left(\cos 3\tau - \cos\tau\right) + a\left(\omega_1 - \tfrac{3}{8}a^2\right) \tau\sin\tau \, . $$ To get rid of the secular term, we take $\omega_1 = \frac{3}{8} a^2$, so that the perturbed solution writes $$ y(t) = a\cos\tau + \epsilon\, \tfrac{1}{32} a^3\left(\cos 3\tau - \cos\tau\right) \qquad \text{where}\qquad \tau = \left(1+ \epsilon\,\tfrac{3}{8}a^2\right) t \, . $$