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I am trying to solve for the time dynamics for a simple quantum system (two-site system with sinusoidal coupling and a decay parameter on one site) and the math is looking not so simple.

Here is the integro-differential equation I end up with for the time dynamics on one site:

$$y'(t) = \left(\gamma + \delta\sin{\left(\epsilon t\right)}\right)\exp{\left(-\alpha t\right)}\int_0^tdw\, \left(\gamma + \delta\sin{\left(\epsilon w\right)}\right)\exp{\left(-\beta w\right)}y(w)$$

$$y(0) = 1$$

My question is whether I should expect this equation to have a solution, and what types of methods I should pursue if it is soluble. My higher math background is somewhat limited (I get a little uneasy attempting any complex analysis).

My first thought was Laplace transforms, but looking through Schaum's tables of identities I am not sure if I can massage that expression into something that matches the identities.

Any thoughts would be appreciated.

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Note that the right side is $0$ at $t=0$, so you should have $y'(0) = 0$. Divide both sides by $(\gamma + \delta \sin(\epsilon t)) \exp(-\alpha t)$ and differentiate to get a second-order linear differential equation. Unfortunately, probably not one that has closed-form solutions.

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Given the integro-differential equation \begin{align} y'(t) = \left(\gamma + \delta\sin{\left(\epsilon t\right)}\right)\exp{\left(-\alpha t\right)}\int_0^tdw\, \left(\gamma + \delta\sin{\left(\epsilon w\right)}\right)\exp{\left(-\beta w\right)}y(w) \end{align} with $y(o)=1$ it is readily seen that $y^{'}(0)=0$. Let $f(t) = \gamma + \delta\sin(\epsilon t)$ and differentiate both sides yields, with the help of Leibniz' rule,
\begin{align} y^{''}(t) &= \partial_{t} \left[ f(t) e^{-\alpha t} \int_{0}^{t} f(w) e^{-\beta w} y(w) dw \right] \\ &= f^{2}(t) e^{-(\alpha + \beta)t} y(t) + (- \alpha f + f^{'}) \int_{0}^{t} f e^{-\beta w} y dw \\ &= f^{2}(t) e^{-(\alpha + \beta)t} y(t) + (- \alpha + f^{'}/f) y^{'}. \end{align} Rearranging the terms yields \begin{align} \left[ \partial_{t}^{2} + \left( \alpha - \frac{f^{'}(t)}{f(t)} \right) \partial_{t} - f^{2}(t) e^{-(\alpha + \beta)t} \right] y(t) =0 \end{align} or \begin{align} \left[ \partial_{t}^{2} + \left( \alpha - \frac{\delta \epsilon \cos(\epsilon t)}{\gamma + \delta\sin(\epsilon t)} \right) \partial_{t} - (\gamma + \delta\sin(\epsilon t))^{2} e^{-(\alpha + \beta)t} \right] y(t) =0. \end{align} Taking the small angle approximation reduces this to \begin{align} \left[ \partial_{t}^{2} + \left( \alpha - \frac{\delta \epsilon}{\gamma + \delta\epsilon t} \right) \partial_{t} - (\gamma + \delta\epsilon t)^{2} e^{-(\alpha + \beta)t} \right] y(t) =0. \end{align} This equation can be solved with hard work.

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    $\begingroup$ I don't think "hard work" will get you a closed-form solution here. You could of course get a series solution in powers of $t$. $\endgroup$ – Robert Israel May 7 '14 at 6:54

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