A Derivation of the differential Chapman-Kolmogorov Equation

Summary
I am following the book "The Theory of Open Quantum Systems" by Breuer and Petruccione. I can follow the derivation of the integral CK equation but do not understand their derivation of the differential CK equation.
The integral equation is [tex]T(x_3,t_3|x_1,t_1)=\int \text{d}x_2T(x_3,t_3|x_2,t_2)T(x_2,t_2|x_1,t_1) [/tex] where [itex]T(x_3,t_3|x_1,t_1)[/itex] is the probability density of a Markov process taking the value [itex]x_3[/itex] at time [itex]t_3[/itex] given that it took the value of [itex]x_1[/itex] at time [itex]t_1[/itex]. So far so good. To derive the differential CK equation, they take the time derivative of the integral equation and the result is [tex]\frac{\partial}{\partial t}T(x,t|x',t')=A(t)T(x,t|x',t')[/tex] where [itex]A(t)[/itex] is time translation linear operator defined in terms of a density as [tex]A(t)\rho(x)=\lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x'[T(x,t+\Delta t|x',t)-\delta(x-x')]\rho(x').[/tex] I don't understand how to derive that form of [itex]A(t)[/itex]. What I would think (since they are differentiating w.r.t. the first time parameter) is that
[tex]\begin{align*}
\frac{\partial}{\partial t_3}T(x_3,t_3|x_1,t_1) &= \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x_2[T(x_3,t_3+\Delta t|x_2,t_2)T(x_2,t_2|x_1,t_1)-T(x_3,t_3|x_2,t_2)T(x_2,t_2|x_1,t_1)] \\ &= \lim_{\Delta t\rightarrow 0}\frac{1}{\Delta t}\int \text{d}x_2[T(x_3,t_3+\Delta t|x_2,t_2)-T(x_3,t_3|x_2,t_2)]T(x_2,t_2|x_1,t_1)]
\end{align*}[/tex]
but I cannot get this into the same functional form as what they get. They only have one time parameter in their [itex]A(t)[/itex] but I still have all three time parameters.

What am I missing?

Thanks!
 

Stephen Tashi

Science Advisor
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