4.2.2 Integral Form for the Distribution Function Perturbation

The integral form of this equation is derived using techniques described in [71]. Introducing a quasi-momentum space trajectory $\vec{K}(t^{'})=\vec{k}-\frac{q}{\hbar}\vec{E}_{s}(t-t^{'})$ which is the solution of Newton's equation it is possible to replace the left-hand side of (4.17) by the total derivative:

$$\frac{\partial f_{1}(\vec{K}(t),t)}{\partial t}+\frac{q}{\hbar}\vec{E}_{s}\cdot\nabla f_{1}(\vec{K}(t),t)=\frac{df_{1}(\vec{K}(t),t)}{dt}.$$ (4.16)

Introducing function $g(t)$:

$$g(t)=\int f_{1}(\vec{k}^{'},t)\widetilde{S}(\vec{k}^{'},\vec{K}(t))\,d\vec{k}^{'}-\frac{q}{\hbar}\vec{E}_{1}\cdot\nabla f_{s}(\vec{K}(t)),$$ (4.17)

the first order equation can be written as:

$$\frac{df_{1}(\vec{K}(t),t)}{dt}+\widetilde{\lambda}(\vec{K}(t))f_{1}(\vec{k},t)=g(t).$$ (4.18)

This is an ordinary differential equation which can be solved by multiplying both sides by a function $h(t)$. This function has to fulfill the condition:

$$\frac{dh(t)}{dt}=h(t)\widetilde{\lambda}(t)$$ (4.19)

with the particular solution:

$$h(t)=\exp\biggl[\int_{0}^{t}\widetilde{\lambda}(y)\,dy\biggr].$$ (4.20)

Then the left-hand side of (4.19) is the total derivative of the product $f_{1}(t)h(t)$. Taking into account that $f_{1}(\vec{K}(t_{0}),t_{0})=0$ for $t_{0}<0$ because of $E_{1}(t)=0$ for $t<0$ the solution is obtained:

$$f_{1}(\vec{K}(t),t)=\int_{0}^{t}g(t^{'})\exp\biggl\{-\int_{t^{'}}^{t}\widetilde{\lambda}[\vec{K}(y)]\,dy\biggr\}\,dt^{'}.$$ (4.21)

Substituting (4.18) into (4.22) the following integral form is obtained as:

$$f_{1}(\vec{K}(t),t)= \int_{0}^{t}\,dt^{'}\int \,d\vec{k}^{'}f_{1}(\vec{k}^{'},t^{'})\w... ...t^{'})) \exp\biggl(-\int_{t^{'}}^{t}\widetilde{\lambda}[\vec{K}(y)]\,dy\biggr)-$$ (4.22)

$$-\frac{q}{\hbar}\int_{0}^{t}\vec{E}_{1}(t^{'})\cdot[\nabla f_{s}]... ...\exp\biggl(-\int_{t^{'}}^{t}\widetilde{\lambda}[\vec{K}(y)]\,dy\biggr)\,dt^{'}.$$

S. Smirnov: