where $$\label{030904-19}
\beta(k)=\frac{{\gamma}_0 R(k)}{{\omega}(k)},\qquad{\omega}_{\epsilon}(k):={\omega}(k) \sqrt{1-({\epsilon}\beta(k))^2}.$$ Note that $
{\lambda}_+^\star={\lambda}_-.
$

Solution of (\[basic:sde:2av\]) {#solution-of-basicsde2av .unnumbered}
-------------------------------

By the Duhamel formula, from we get $$\label{022503-19}
 \hat\Phi(t,k)=e_{{\Omega}_{\epsilon}}(k,t) \hat\Psi(k) - i\gamma_1  \int_0^t e_{{\Omega}_{\epsilon}}(k,t-s){\frak f}{p}_0(s)ds .$$ Here $$\begin{aligned}
\label{011902-19ac}
&e_{{\Omega}_{\epsilon}}(k,t):=\exp\left\{{\Omega}_{\epsilon}(k)t\right\}=\left[
\begin{array}{cc}
e_{{\Omega}_{\epsilon}}^{1,1}(k,t)&e_{{\Omega}_{\epsilon}}^{1,2}(k,t)\\
e_{{\Omega}_{\epsilon}}^{1,2}(k,t)&[e_{{\Omega}_{\epsilon}}^{1,1}]^\star(k,t)
\end{array}\right]\end{aligned}$$ and $$\begin{aligned}
\label{011902-19a}
&
e_{{\Omega}_{\epsilon}}^{1,1}(k,t):=\dfrac14\left\{\left(1+\sqrt{1-({\epsilon}\beta(k))^2}\right)^2e_-(k,t)-({\epsilon}\beta(k))^2e_+(k,t)\right\}, \nonumber\\
&e_{{\Omega}_{\epsilon}}^{1,2}(k,t):=\dfrac{i{\epsilon}\beta(k)}{4} \left(1+\sqrt{1-({\epsilon}\beta(k))^2}\right)(e_-(k,t)-e_+(k,t))
,\\
&
e_\pm(k,t):=e^{{\lambda}_\pm(k)
  t}.\nonumber\end{aligned}$$ Note that $e_\pm^\star(k,t)=e_\mp(k,t)$.

Multiplying scalarly both sides of by ${\frak f}$ and integrating over $k$ we conclude that $$\label{042503-19}
p_0(t)+{\gamma}_1J_{\epsilon}\star p_0(t)=p_0^0(t)$$ Here $$\label{null}
p_0^0(t):=\frac{1}{2i}\int_{{{\mathbb T}}}e_{{\Omega}_{\epsilon}}(k,t)\hat\Psi(k) \cdot {\frak f}dk$$ and $$\label{J-eps}
{J}_{\epsilon}(t):=\frac12\int_{{{\mathbb T}}} \exp\left\{{\Omega}_{\epsilon}(k)t\right\}{\frak
  f}\cdot {\frak f} dk=\int_{{{\mathbb T}}}j_{\epsilon}(t,k)dk,$$ where $$\label{j}
j_{\epsilon}(t,k):=\frac12\exp\left\{{\Omega}_{\epsilon}(k)t\right\}{\frak f}\cdot
   {\frak f}
=e^{-{\epsilon}{\gamma}_0 R(k) t}\left\{\sqrt{1-({\epsilon}\beta(k))^2}\cos\left({\omega}_{\epsilon}(k)t\right)-{\epsilon}\beta(k)\, \sin\left({\omega}_{\epsilon}(k)t\right)\right\}.$$ Taking the Laplace transforms of the both sides of we obtain $$\label{062503-19}
\tilde p_0({\lambda})(1+{\gamma}_1\tilde J_{\epsilon}({\lambda}))=\tilde p_0^0({\lambda}).$$ By a direct calculation one concludes that $$\label{052503-19}
{\rm Re}\, \tilde {J}_{\epsilon}({\lambda})>0,\quad\mbox{for any }{\lambda}\in \mathbb
C_+.$$ Since $J_{\epsilon}(\cdot)$ is real valued, we have $$\tilde {J}_{\epsilon}^\star({\lambda})=\tilde {J}_{\epsilon}({\lambda}^\star)\quad\mbox{for any }{\lambda}\in \mathbb
C_+.$$ From we get $$\label{072503-19}
\tilde p_0({\lambda})=\tilde g_{\epsilon}({\lambda})\tilde p_0^0({\lambda}),$$ with $\tilde g_{\epsilon}({\lambda})$ defined by $$\label{012302-19}
\tilde {g}_{\epsilon}({\lambda}):=\left(1+{\gamma}_1\tilde {J}_{\epsilon}({\lambda})\right)^{-1},\quad {\rm Re}\,{\lambda}>0.$$ Thanks to we obtain $$\label{G}
|\tilde {g}_{\epsilon}({\lambda})|\le 1,\quad {\rm Re}\,{\lambda}>0$$ and, as a result, $$\label{JG}
{\gamma}_1|\tilde {J}_{\epsilon}({\lambda})\tilde {g}_{\epsilon}({\lambda})|\le 2,\quad {\rm Re}\,{\lambda}>0.$$ The following result shows in particular that $\tilde
{g}_{\epsilon}({\epsilon}-i{\omega}(k))$ approximates in some sense $\nu(k)$, 