Past exam of the stochastic modelling of biological processes course of the University of Oxford / 2023 / 3

We use the equilibrium generalized Langevin equation and its generalized fluctuation-dissipation relation, with the factor $5$ in the noise autocorrelation fixing the equilibrium velocity variance. Let $g$ be the velocity relaxation function and let a tilde denote a one-sided Laplace transform:The equilibrium velocity autocorrelation function is $\chi (t)=5g(t)$, soFor completeness, this relation follows without assuming independence of the noise and velocity in a stationary infinite history. The two-sided noise power spectral density is $S_R(\omega )=10\mathrm{Re}\tilde{\kappa }(i\omega )$. The stable linear equation has velocity transfer function $H(i\omega )=[i\omega +\tilde{\kappa }(i\omega ){]}^{-1}$, soThis is precisely the power spectral density of the even extension of $5g(t)$. Stable relaxation removes initial transients with a finite second moment, giving the long-time autocorrelation function defined in the question. All the parameter choices used below have stable relaxation.

Integrating the velocity autocorrelation function twice gives the mean squared displacement from a fixed initial position:A decaying initial transient does not change its asymptotic slope. Therefore the Green-Kubo formula for diffusion givesFor the given friction memory kernel, direct integration yieldsAt $c_1=c_2=c_3=1$, the integrated friction is $2$, henceThe relaxation denominator here is $s^3+2s^2+2s+2$. Its roots have negative real parts by the cubic Routh-Hurwitz criterion, since all coefficients are positive and $2\cdot 2>2$. This justifies the decay and integrability used above. The diffusion coefficient depends on the area under the friction memory kernel, not just its value at zero.

For $c_1=2$, $c_2=4$, $c_3=5$, the friction memory kernel and its Laplace transform areUsing the generalized fluctuation-dissipation relation with equilibrium velocity variance $5$, the velocity autocorrelation function has Laplace transformAll three poles are negative, so the stationary relaxation is stable. A partial fraction decomposition givesTaking the inverse Laplace transform givesThe negative middle coefficient is compatible with an autocorrelation function; individual terms in a partial fraction decomposition need not themselves be separate physical modes with positive weights. Several useful checks follow directly:The Volterra integral equation for $\chi$ predicts exactly ${\chi }^{\prime }(0)=0$ and ${\chi }^{\prime \prime }(0)=-\kappa (0)\chi (0)=-10$. Its integral is also $8-10/9+1/18=125/18=5/\tilde{\kappa }(0)$, agreeing with the Green-Kubo formula for diffusion.

Write $c=c_1=c_3$, so the friction memory kernel is ${\kappa }_c(t)=c{e}^{-ct}$ and has area one for every $c>0$. Its Laplace transform is ${\tilde{\kappa }}_c(s)=c/(s+c)$. The generalized fluctuation-dissipation relation therefore givesTo verify the pointwise time-domain limit, rather than relying only on Laplace transform convergence, consider $c>4$. The two rootsare negative and satisfy ${\lambda }_{+}\to -1$, ${\lambda }_{-}\to -\infty$. Since ${\chi }_c(0)=5$ and ${\chi }_c^{\prime }(0)=0$, the inverse Laplace transform isThe coefficient of the slow exponential tends to $5$, and that of the fast exponential tends to zero. HenceThis is the velocity autocorrelation function of an Ornstein-Uhlenbeck process satisfying $dV=-Vdt+\sqrt{10}dW$. The friction memory kernel becomes local in time. Consistently, the even noise autocorrelation $5c{e}^{-c|\tau |}$ tends to $10\delta (\tau )$ in the sense of distributions, fixing the white noise normalization. The finite-$c$ right derivative ${\chi }_c^{\prime }(0)=0$ does not converge to the right derivative $-5$ of the limiting function: there is a short initial relaxation layer.

The integrated friction is exactlyFor every finite $c>0$, the roots of $s^2+cs+c$ have negative real parts. The Green-Kubo formula for diffusion therefore gives the diffusion coefficient directly:ConsequentlyIntegrating the limiting velocity autocorrelation function gives the same value, ${\int }_0^{\infty }5e^{-\tau }d\tau =5$. Computing $D_c$ before taking the limit also avoids having to justify an interchange of the infinite-time integral and the parameter limit.