I'm completely lost with this problem. I was asked to solve the following equation, limited to first order of $\lambda$, by using Dyson series
$$\ddot{x}+\omega_0 x=\lambda x^3$$
with initial conditions: $x_0=0$ and $\dot{x}=v_0$.
I saw an example where the author transformed the equation to matrix form, something like
$$\dot{Y(t)}=A(t)Y(t)$$
where $Y(t)=\begin{bmatrix}
x(t) \\
v(t)
\end{bmatrix}$, and $A(t)$ can be written as a combined of two other matrices:
$$A(t)=A+\lambda J(t)$$
which then yields a solution as
$$e^{At}Y_0+\lambda e^{At}(\int_0^te^{-At_1}J(t_1)e^{At_1}dt_1)Y_0$$
I don't know if what I wrote makes any sense at all and I have no idea how to start connecting the differential equation with Dyson series. Any tips on how to write the differential equation to matrix form and find the solution based on the example I gave above? I'm accepting links to external material as well.
Best Answer
To put a differential equation into matrix form, usually you just build a vector with enough derivatives of your variable $x$, so that the equation reduces to a first order EDO. Here, we have a second order EDO to start with, so taking the vector $X = ( x \quad \dot x)^T$, $J(X) = (x^3 \quad 0)^3$ and : $$A = \begin{pmatrix} 0 & 1\\-\omega_0^2 &0 \end{pmatrix}$$ we see that the differential equation can be rewritten : $$\dot X = AX+\lambda J(X)$$
Now, from this, we have : $$\frac{\text d}{\text dt}\left(e^{-At}X\right) = \lambda e^{-At}J(X)\hspace{2cm}(*)$$ which gives : $$X(t) = e^{At}X(0)+\lambda\int_0^te^{A(t-t')}J(X(t'))\text dt'$$ We see that $X(t) = e^{At}X(0) +\mathcal O(\lambda)$ and therefore $J(X(t)) = J(e^{At}X(0)) + \mathcal O(\lambda)$. Plugging this in $(*)$, we get : $$X(t) = e^{At}X(0) + \lambda\int_0^te^{A(t-t')}J(e^{At'}X(0))\text dt'+\mathcal O(\lambda^2)$$
Now, to compute $x$ we have to evaluate the exponential of $A$. This is not that difficult a calculation and the result is : $$e^{tA} = \begin{pmatrix} \cos(\omega_0 t) & \frac{1}{\omega_0}\sin(\omega_0t)\\ -\omega_0\sin(\omega_0t) & \cos(\omega_0t)\end{pmatrix}$$ With $X(0) = (0\quad v_0)^T$, we have: $$X(t) = \frac{v_0}{\omega_0}\sin(\omega_0t) + \mathcal O(\lambda)$$ and in the end : $$X(t) = \frac{v_0}{\omega_0}\sin(\omega_0t) + \lambda\int_0^t\cos(\omega_0(t-t'))\bigg(\frac{v_0}{\omega_0} \sin(\omega_0t')\bigg)^3 \text dt' + \mathcal O(\lambda^2)$$
Remark : This is probably not the fastest way to obtain this result, but you asked for the matrix form of the EDO and the Dyson series.