In the 1986 paper "Reaction-Diffusion equations for Interacting Particle Systems" from De Masi Ferrari and Lebowitz, one reads:
$$\frac{\partial}{\partial t}\tilde{c}(q,q';t)=\frac12\left(\frac{\partial^2\tilde{c}}{\partial q'}+\frac{\partial^2\tilde{c}}{\partial q}\right)+[F'(m(q,t))+F'(m(q',t))]\tilde{c}+\delta(q-q')\left[-\left(\frac{\partial m}{\partial r}\right)^2+2F'(m)(1-m^2)+2mF(m)+4f(m)\right]\tag{2.29}$$ where $$F'(m)=2\left[\frac{\gamma}{\gamma_c}-1\right]-6\gamma^2m^2\tag{2.30}$$ $$f(m)=1-\gamma(2-\gamma)m^2\tag{2.31}$$ Here $m(q,t)$ is the solution of $(2.22)$, and $(2.29)$ is to be solved with initial condition $\tilde{c}(q,q';0)=0$.
$\ \ $ Let us now consider the solution of $(2.29)$ around the stationary state $m_0(q)=m(q,t)=0$. It is given by
How do arrive at $2.32a$?
Attempt
Using Duhamel principle
$$\tilde c(q,\tilde q; t ) = \int_0^t P^sf(q,\tilde q; t)\, ds $$
Where $P^sf$ solves
$$\begin{cases} \partial_t u(q,\tilde q, t) = \Delta u(q,\tilde q, t)\\ u(q,\tilde q, s) = 4\left(\frac{\gamma}{\gamma_c} -1\right)\tilde c(q,\tilde q, s) + \delta_{(q-q')}\left[ 4 \left(\frac{\gamma}{\gamma_c} -1\right) + 4\right] \end{cases}$$
This is because we use that $m = 0$.
Now we need to solve this problem, that is, find the value of $P^sf(q,\tilde q, t)$. This is the result of the convolution with the heat kernel after $t-s$ seconds:
$$P^sf(q,\tilde q, t) = K_{t-s}* u(q,\tilde q,s) \int_{R}\int_R \frac{\exp\{-(x-q)^2/2(t-s)\}}{\sqrt{2\pi(t-s)}} \frac{\exp\{(y-\tilde q)^2/2(t-s)\}}{\sqrt{2\pi(t-s)}}\\4\left(\frac{\gamma}{\gamma_c} -1\right)\tilde c(x,y, s) + \delta_{(x-y')}\left[ 4 \left(\frac{\gamma}{\gamma_c} -1\right) + 4\right]\, dx \, dy $$
But should this idea work, one expects to find something similar (once we might change the (t-s) by s) to
$$P^sf(q,\tilde q, t) = (4\pi s)^{-1/2} \exp[-(q-\tilde q)^2/4s]\exp[-4(1 - \gamma/\gamma_c)s] $$
But I don't see how to arrive there.
Any ideas?
