# integro-differential equation with application in quantum mechanics

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.

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.
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,
• 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 $t$. – Robert Israel May 7 '14 at 6:54