Seeking advice about solving an ODE

[BillKet]

Hello! I am trying to solve the time dependent Schrödinger equation for a 2x2 system and I ended up with this ODE:

$$y''=-iA\sin{(\omega t)}y'-B^2y$$

with the initial conditions $y(t=0)=0$ and $y'(t=0)=B$. I can look at it numerically but I was wondering if there is a way to get something analytical out of it. In my case I have $B<<A,\omega$ (not sure if that helps), but I am interested in the way $y$ depends on B, so I can't just drop that term either. Any advice would be greatly appreciated. Thank you!

 

[pasmith]

First set $\tau = \omega t$ so that $$\frac{d^2 y}{d\tau^2} + i\alpha\sin(\tau) \frac{dy}{d\tau} + \epsilon^2 y = 0$$ subject to $y(0) = 0$, $y'(0) = \epsilon$ where $\alpha = A/\omega$ and $\epsilon = B/\omega \ll 1$.

Pose an asymptotic expansion $$y(t) \sim \sum_{k=0}^\infty \epsilon^ky_k(t)$$ in the limit $\epsilon \to 0$. Then the ODE becomes $$(y''_0 + i\alpha\sin(\tau) y'_0) + \epsilon(y''_1 + i\alpha\sin(\tau) y'_1) \\<br /> + \sum_{k=2}^\infty \epsilon^k \left(y''_k + i\alpha\sin(\tau)y'_k + y_{k-2}\right) = 0$$ so that considering coefficients of $\epsilon^k$ we have $$\begin{split}<br /> y''_0 + i\alpha \sin(\tau) y_0' &= 0 \\<br /> y''_1 + i\alpha \sin(\tau) y_1' &= 0 \\<br /> y''_k + i\alpha \sin(\tau) y_k' &= -y_{k-2} \end{split}$$ subject to the initial conditions
$$\begin{gather*}y_k(0) = y'_k(0) = 0, \quad k \neq 1 \\<br /> y_1(0) = 0, \quad y_1'(0) = 1.\end{gather*}$$ The equation for $y_k$ can be solved by an integrating factor once $y_0, \dots, y_{k-1}$ are known. ("Solved" in the sense that the solution can be written in terms of integrals, but the integrals may or may be expressible in terms of elementary functions.) It seems that $y_{2k} \equiv 0$ so only the odd terms contribute.
Note that this expansion is only valid so long as $$\left| \frac{y_{2k+3}}{y_{2k+1}} \right| < \frac 1{\epsilon^2}.$$

 

[anuttarasammyak]

If we omit $-B^2 y$ in RHS, we can solve the simplified ODE,
$$y=B\int_0^t du \ e^{\frac{iA}{\omega}\cos \omega u}$$
We may be able to expect that in a short time the solution of the original ODE does not so much different from it. I would appreciate it If you could check the difference with your numerical solution.

 

[pasmith]

It i spossible to obtain a power series solution. The system is of the form $$\dot x = M(t)x$$ for $x \in \mathbb{R}^2$ where $x = (y, \dot y)^T$ and $$M = \begin{pmatrix} 0 & 1 \\ -B^2 & -iA\sin \omega t \end{pmatrix}.$$ Then setting $$\begin{split}<br /> x(t) &= \sum_{n=0}^\infty a_nt^n \\<br /> M(t) &= \sum_{n=0}^\infty M_nt^n \end{split}$$ we have $$\sum_{n=0}^\infty (n+1)a_{n+1}t^n = \sum_{n=0}^\infty t^n \left[ <br /> \sum_{m=0}^n M_{n-m}a_m<br /> \right]$$ whence $$a_{n+1} = \frac{1}{n+1} \sum_{m=0}^n M_{n-m} a_m, \qquad n \geq 0$$ with $a_0 = (0, B)^T$.