}=0,\quad
\psi(\cdot,\cdot,E_{\rm m})=0.\end{gathered}$$ For simplicity we denote ${1\over{S_0(E)}}f$ and ${1\over{S_0(E)}} K$ again by $f$ and $K$. Assume that $I=[0,\infty[{}=:{\mathbb{R}}_+$ and that $K$ is of the *Volterra type operator* (cf. [@engelnagel pp. 447-452]) $$(K\psi)(x,\omega,E)=\int_0^E\int_S
\sigma(x,\omega',\omega,E-E',E)\psi(x,\omega',E') d\omega' dE'.$$ In other words, $K\psi=\int_{S\times I} \tilde{\sigma}(\cdot,\omega',\cdot,E',\cdot)\psi(\cdot,\omega',E') d\omega' dE'$ for a differential cross section $\tilde{\sigma}$ of the form $\tilde{\sigma}(x,\omega',\omega,E',E)=\chi_I(E-E')\sigma(x,\omega',\omega,E-E',E)$, where $\chi_I$ is the characteristic function of $I$. Assume additionally that $\sigma$ has a decomposition $$\sigma(x,\omega',\omega,E-E',E)=\sigma_1(x,\omega',\omega,E)\sigma_2(E-E').$$ Let $K_1(E)$ (for a fixed $E$) be a linear operator $L^2(G\times S)\to L^2(G\times S)$ defined by $$(K_1(E)\phi)(x,\omega)=\int_S\sigma_1(x,\omega',\omega,E)\phi(x,\omega') d\omega'.$$ Then we find that $$(K\psi)(x,\omega,E)=\int_0^E
\sigma_2(E-E')K_1(E)\psi(E') dE'.$$

Define an [*extended space*]{} by ${{{\mathcal{}}}X}:=L^2(G\times S)\times L^1({\mathbb{R}}_+,L^2(G\times S))$ and a linear operator ${{{\mathcal{}}}A}(E):{{{\mathcal{}}}X}\to {{{\mathcal{}}}X}$ for a fixed $E$ by (here the argument of ${\mathbb{R}}_+$ is denoted by $s$) & D([A]{}(E))=D:=W\_[-,0]{}\^2(GS)H\^[1,1]{}(\_+,L\^2(GS)),\
& [A]{}(E):=

(

A(E)&\_0B(E)&[d]{}

)

, where $$H^{1,1}({\mathbb{R}}_+,L^2(G\times S)):= \{F\in L^1({\mathbb{R}}_+,L^2(G\times S))\ |\ F'\in L^1({\mathbb{R}}_+,L^2(G\times S))\},$$ is the domain of the derivative operator $\frac{d}{ds}:L^1({\mathbb{R}}_+,L^2(G\times S))\to L^1({\mathbb{R}}_+,L^2(G\times S))$, the linear operator $A(E):L^2(G\times S)\to L^2(G\times S)$ with domain $\tilde{W}^2_{-,0}(G\times S)$ (independent of $E$) is given by, $$A(E)\phi=-\Big({1\over { S_0(E)}}\omega\cdot\nabla_x\phi+{1\over { S_0(E)}}\Sigma(E)\phi
-{1\over{ S_0(E)}}{{\frac{\partial  S_0}{\partial E}}}(E)\phi\Big),$$ 