\sect{Baruch's D31.}
Consider the reaction A$\leftrightarrow$B
for molecules A,B that evolves via a Langevin type equation
\begin{equation}\nonumber
\frac{dN_A(t)}{dt}=k_1N_B(t)-k_2N_A(t) +A(t) \qquad\text{(*)}
\end{equation}
where the total number of molecules $N=N_A(t)+N_B(t)$ is fixed.
$k_1,\,k_2$ are reaction constants and $A(t)$ is random with
averages $\langle A(t)\rangle =0$, $\langle A(t)A(t')\rangle
=C\delta(t-t')$.
\begin{itemize}
\item[(a)] Solve for $\langle N_A(t)\rangle$ with the initial condition
$N_A(0)=0$ and show that its value at long time ${\bar N_A}$
yields $k_1/k_2=f_A/f_B$, where results of question (A24) are used.
\item[(b)] Solve for $\langle [\delta N_A(t)]^2\rangle$ where
$\delta N_A(t)=N_A(t)-{\bar N_A}$ and from its long time form and
the results of question (A24b) show that
$C=\frac{2k_1k_2}{k_1+k_2}N$.
\item[(c)] The chemical potential of B is now modified by a term
$\delta\mu_B(t)=\delta\mu_B(\omega)\eexp{-i\omega t}$. The
response $\alpha(\omega)$ is defined by the long time form
$\langle
N_A(t)\rangle=\alpha(\omega)\delta\mu_B(\omega)\eexp{-i\omega
t}+{\bar N_A}$. Show that $\delta\mu_B(t)=-k_BT\frac{\delta
k_2(t)}{k_2}$ by using the correspondence with question (A24), where
$k_2\rightarrow k_2+\delta k_2(t)$ in Eq. (*). Solve the
resulting equation for $\langle N_A(t)\rangle$ and identify
$\alpha(\omega)$.
\item[(d)] Solve Eq. (*) for the equilibrium fluctuation in the absence
of external forces ($\delta\mu_B(t)=0$)
\begin{equation}\label{fluct}
\phi_{N_A}(\omega)=\int \langle \delta N_A(t)\delta N_A(t+\tau)\rangle
\eexp{i\omega \tau}d\tau
\end{equation}
and check the (classical) fluctuation dissipation theorem. [Use
the $t\rightarrow \infty$ form for the equilibrium correlation].
\end{itemize}
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