as ${\epsilon}\to0+$ (see ).

\[cor010304-19\] Suppose that $K:{{\mathbb T}}\to{{\mathbb R}}_+$ is a uniformly continuous and bounded function satisfying $$\label{lim-is}
\inf_{k\in{{\mathbb T}}}K(k)>0.$$ Then, $$\label{010304-19}
\tilde g_{\epsilon}({\lambda})=\tilde g({\lambda})+{\epsilon}\tilde r_{\epsilon}({\lambda}),\quad
{\lambda}\in\mathbb C_+,$$ where $$\lim_{{\epsilon}\to0+}{\epsilon}^p\int_{{{\mathbb T}}}|\tilde r_{\epsilon}({\epsilon}K(k)-i{\omega}(k))|^pdk=0,$$ for any $p\in[1,+\infty)$.

The proof of Proposition \[cor010304-19\] is shown in Appendix \[appC\].

Define by $g_{\epsilon}(ds)$ the distribution such that $$\label{g-eps}
\tilde {g}_{\epsilon}({\lambda})=\int_0^{+\infty}e^{-{\lambda}t}g_{\epsilon}(ds).$$ From it satisfies $$\label{g-eps1}
g_{\epsilon}(ds)=\delta(ds)-{\gamma}_1 J_{\epsilon}\star g_{\epsilon}(s)ds.$$ The Volterra equation has a unique real-valued solution and ${\gamma}_1 J_{\epsilon}\star g_{\epsilon}(s)$ is a $C^\infty$ smooth function, see e.g. the argument made in Section 3 of [@kors]. The solution of can be then written as follows $$\begin{aligned}
\label{eq:10}
&
\hat\Phi(t,k) =U(t) \hat\Psi(k):=e_{{\Omega}_{\epsilon}}(k,t)\hat\Psi(k) - i\gamma_1  \int_0^t
 e_{{\Omega}_{\epsilon}}(k,t-s){\frak f}p_0^0\star g_{\epsilon}(s) ds \\
&
=e_{{\Omega}_{\epsilon}}(k,t) \hat\Psi(k) - \frac{\gamma_1}{2}  \int_0^tds\int_0^sg_{\epsilon}(ds_1)\int_{{{\mathbb T}}}
  e_{{\Omega}_{\epsilon}}(k,t-s){\bf D}e_{{\Omega}_{\epsilon}}(k,s-s_1) \hat\Psi(\ell) d\ell .\nonumber
 \end{aligned}$$

Dynamics of the energy density when $T=0$ {#sec4}
=========================================

Starting with the present section untill Section \[sec10\] we shall assume that the thermostat temperature $T=0$, see . We maintain this assumption untill Section \[sec10\]. Let $$\label{Psit}
\hat\Psi^{({\epsilon})}(t,k)=\left[
\begin{array}{c}
\hat\psi^{({\epsilon})}_+(t,k)\\
\hat\psi^{({\epsilon})}_-(t,k)
\end{array}\right],$$ where $\hat\psi^{({\epsilon})}_+(t,k):=\hat \psi^{({\epsilon})}(t,k)$ and $\hat\psi^{({\epsilon})}_-(t,k):=\left(\hat
  \psi^{({\epsilon})}\right)^\star(t,-k)$ (cf ). From we get $$\begin{aligned}
 \label{basic:sde:2av1}
&&
 d \hat\Psi^{({\epsilon})}(t,k)=\frac{1}{{\epsilon}}{\Omega}_{\epsilon}(k) \hat\Psi^{({\epsilon})}(t,k) dt
 +i\sqrt{{\gamma}_0}\int_{{{\mathbb T}}}r(k,k'){\bf
   D} \hat\Psi^{({\epsilon})}\left(t,k-k'\right)B(dt,dk')-\frac{i{\gamma}_1}{{\epsilon}}{\frak p}_0(t){\frak
   g}dt,\nonumber\\
 &&
\Psi^{({\epsilon})}(0,k)= \hat\Psi(k).
 \end{aligned}$$

With some abuse of notation 