# Any idea how to solve the following coupled diffeq?

The coupled differential equations of interest are

\begin{align} i\dot{C}_{+}(t)=\cos(\omega t)e^{2 i \alpha t} C_{-}(t) \end{align}

and

\begin{align} i\dot{C}_{-}(t)=\cos(\omega t)e^{-2 i \alpha t} C_{+}(t) \end{align}

Solving the first equation above for $C_{-}(t)$ differentiating it, plugging it into the second equation above, and simplifying we have

\begin{align} \ddot{C}_{+}(t) = -\cos^2(\omega t)C_{+}(t) + [2i\alpha-\omega\tan(\omega t)]\dot{C}_{+}(t) \end{align}

The initial conditions of interest are $C_{+}(0)=1$ and $C_{-}(0)=0 \implies \dot{C}_{+}(0)=0$ and I wish to have an analytic solution for t>0

For $\alpha \in Reals$ and $\omega \in Reals$

Mathematica can't seem to solve this problem analytically using DSolve and numerically using NDSolve it can only find the solution for a finite time after which it has no solution because "the step size is effectively 0".

I wish to have an analytic solution. I have tried Fourier transforming the equation in order to solve it with little luck but that still may be the way to go.

Thanks for any and all help!

• What makes you suspect there is an "analytic" closed-form solution to this system? – Mark Fischler Jun 19 '18 at 19:17
• No reason... I am just hopeful. It also seems rather trivial looking so I was surprised I was having issues finding one. Also an analytic solution would be significantly more useful for applications. – E. Nerney Jun 19 '18 at 21:30
• I found a solution using a perturbation series expansion for small $\alpha$ but this has limited applications and involves having to use the Jacobi-Anger expansion to do an integral even at first order which makes going to higher order extremely difficult. – E. Nerney Jun 19 '18 at 21:36