After the sudden parameter change, the <functional derivative> is
$$
\frac{\delta F}{\delta\mathbf p}
=a_F\mathbf p-\kappa\nabla^2\mathbf p.
$$
Each Cartesian component of each <Fourier transform>[Fourier mode] consequently obeys
$$
\dot p_{\mathbf q,i}
=-r(q)p_{\mathbf q,i}+f_{\mathbf q,i},
\qquad
r(q)=\Gamma(a_F+\kappa q^2).
$$
This is an <Ornstein-Uhlenbeck process>. The integrating-factor method gives
$$
p_{\mathbf q,i}(t)
=p_{\mathbf q,i}(t_1)e^{-r(q)(t-t_1)}
+\int_{t_1}^{t}dt'\,
f_{\mathbf q,i}(t')e^{-r(q)(t-t')}.
$$
Because both coefficients are positive, every mode has positive decay rate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/3/c/solution