Perturbation theory

Take the expansion and calculate the first and second derivatives:

$x=\frac{1}{2}t^2+\varepsilon c_1 t^4+ \varepsilon^2 c_2 t^6+...$

$\dot x=t+4\varepsilon c_1 t^3+6 \varepsilon^2 c_2 t^5+...$

$\ddot x=1+12 \varepsilon c_1 t^2+ 30 \varepsilon^2 c_2 t^4+...$

We will insert this to the original equation of motion $\ddot x+ \varepsilon \dot x^2=1$ , and we will compare the coefficients of the $t^2$ and $t^4$ terms (as long as we are interested only in $c_1$ and $c_2$ , we do not need to go to higher orders). For that we need the $t^2$ and the $t^4$ terms in $\dot x^2$ ,

$\dot x^2=(t+4\varepsilon c_1 t^3+6 \varepsilon^2 c_2 t^5+...)^2= t^2+8\varepsilon c_1 t^4 + ...$

Finally we get

$(1+12 \varepsilon c_1 t^2+ 30 \varepsilon^2 c_2 t^4+...)+ (\varepsilon t^2+8\varepsilon^2 c_1 t^4 + ...) =1$

This is satisfied if $12 c_1+1 =0$ and $30c_2+8c_1=0$ . The second equation yields $c_2/c_1=-4/15$ .

We assume, that we can expand the solution as a power series for the parameter $\varepsilon$ $\begin{aligned} x(t) &= x_0(t) + \varepsilon \cdot x_1(t) + \varepsilon^2 \cdot x_2(t) + \dots \\ &= \sum_{k = 0}^\infty \varepsilon^k \cdot x_k(t) \end{aligned}$ where the $x_k(t)$ are functions, which are independant of $\varepsilon$ . Since $\varepsilon$ is small, we can get a suitable approximations for $x(t)$ , if we evaluate a few partial sums.

Inserting the power series into the differential equation results $\begin{aligned} & & \sum_{k = 0}^\infty \varepsilon^k \cdot \ddot x_k + \varepsilon \left( \sum_{k = 0}^\infty \varepsilon^k \cdot \dot x_k \right)^2 &= 1 \\ \Rightarrow & & \sum_{k = 0}^\infty \varepsilon^k \cdot \ddot x_k + \sum_{k = 1}^\infty \varepsilon^k \cdot \sum_{i + j = k - 1} \dot x_i \dot x_j &= 1 \\ \Rightarrow & & (\ddot x_0 - 1) + \sum_{k = 1}^\infty \varepsilon^k \cdot \left(\ddot x_k + \sum_{i + j = k - 1} \dot x_i \dot x_j \right) &= 0 \\ \Rightarrow & & (\ddot x_0 - 1) + \varepsilon \cdot (\ddot x_1 + \dot x_0^2) + \varepsilon^2 \cdot (\ddot x_2 + 2 \dot x_0 \dot x_1) + \dots + &= 0 \end{aligned}$ Since this must valid for any $\varepsilon$ , all expressions inside parentheses must be zero. Therefore, we get a sequence of differential equations $\begin{aligned} \ddot x_0 - 1 &= 0 \\ \ddot x_1 + \dot x_0^2 &= 0 \\ \ddot x_2 + 2 \dot x_0 \dot x_1 &= 0 \\ \vdots & \end{aligned}$ We get the solutions one by one through integration with initial conditions $x_k(0) = \dot x_k(0) = 0$ $\begin{aligned} x_0 &= \int \left(\int 1 dt \right) dt = \frac{1}{2} t^2 \\ x_1 &= \int \left(\int (-\dot x_0^2) dt \right) dt = - \int \left(\int t^2 dt \right) dt = - \frac{t^4}{12} \\ x_2 &= \int \left(\int (-2 \dot x_0 \dot x_1) dt \right) dt = \int \left(\int \frac{2 t^4}{3} dt \right) dt = \frac{t^6}{45} \end{aligned}$ The final result is of the form $x(t) = \frac{1}{2} t^2 - \varepsilon \frac{t^4}{12} + \varepsilon^2 \frac{t^6}{45} + \dots$ with the coefficients $c_1 = -\dfrac{1}{12}$ and $c_2 = \dfrac{1}{45}$ , so that $\dfrac{c_2}{c_1} = -\dfrac{12}{45} = -\dfrac{4}{15} = -0.2666\dots$ .