11.5: Exercises

Exercise $\PageIndex{1}$

Find the time-domain Green’s function of the critically-damped harmonic oscillator ($\gamma = \omega_0$).

Exercise $\PageIndex{2}$

Consider an overdamped harmonic oscillator ($\gamma > \omega_0$) subjected to a random driving force $f(t)$, which fluctuates between random values, which can be either positive or negative, at each time $t$. The random force satisfies $$\left\langle f(t)\right\rangle = 0 \quad\mathrm{and}\;\;\;\left\langle f(t) f(t')\right\rangle = A \, \delta(t-t'),$$ where $\left\langle\cdots\right\rangle$ denotes an average taken over many realizations of the random force and $A$ is some constant. Using the causal Green’s function, find the correlation function $\left\langle x(t_1)\, x(t_2) \right\rangle$ and the mean squared deviation $\left\langle [x(t+\Delta t) - x(t)]^2 \right\rangle.$

Answer

For the over-damped oscillator, the Green’s function is $$G(t,t') = \Theta(t-t')\, \frac{e^{-\gamma(t-t')}}{\Gamma} \sinh\big[\Gamma(t-t')\big], \quad\mathrm{where}\;\,\Gamma = \sqrt{\gamma^2 - \omega_0^2}.$$ Hence, the response to the force $f$ is $$x(t) = \frac{1}{m\Gamma} \int^t_{-\infty} dt'\; e^{-\gamma(t-t')} \sinh\big[\Gamma(t-t')\big] f(t').$$ From this, we get the following expression for the desired correlation function: $$\begin{align} \nonumber\langle x(t_1)\, x(t_2)\rangle = \frac{1}{m^2\Gamma^2}& \int^{t_1}_{-\infty} dt' \int^{t_2}_{-\infty} dt'' \; e^{-\gamma(t_1-t')}\, e^{-\gamma(t_2-t'')} \\ &\times \sinh\big[\Gamma(t_1-t')\big] \sinh\big[\Gamma(t_2-t'')\big] \; \langle f(t') f(t'')\rangle.\end{align}$$ Note that the $\langle\cdots\rangle$ can be shifted inside the integrals, because it represents taking the mean over independent sample trajectories. Now, without loss of generality, let us take $$t_1 \ge t_2.$$ Since $\langle f(t') f(t'')\rangle = A \delta(t'-t'')$ which vanishes for $t' \ne t''$, the double integral only receives contributions from values of $t'$ not exceeding $t_2$ (which is the upper limit of the range for $t''$). Thus, we revise $\int^{t_1} dt'$ into $\int^{t_2} dt'$. The delta function then reduces the double integral into a single integral, which can be solved and simplified with a bit of tedious algebra: $$\begin{align} \langle x(t_1)\, x(t_2)\rangle &= \frac{A}{m^2\Gamma^2} e^{-\gamma(t_1+t_2)} \int^{t_2}_{-\infty} dt' e^{2\gamma t'} \sinh\big[\Gamma(t'-t_1)\big] \, \sinh\big[\Gamma(t'-t_2)\big] \\ &= \frac{A}{8m^2\Gamma^2} e^{-\gamma(t_1+t_2)}\Bigg[\frac{e^{-\Gamma t_1} e^{(2\gamma+\Gamma)t_2}}{\gamma+\Gamma} + \frac{e^{\Gamma t_1}e^{(2\gamma-\Gamma)t_2}}{\gamma-\Gamma} \nonumber \\ &\qquad\qquad\qquad\qquad\qquad - \frac{e^{-\Gamma t_1} e^{(\Gamma+2\gamma)t_2} + e^{\Gamma t_1} e^{(-\Gamma+2\gamma)t_2}}{\gamma}\Bigg] \\ &= \frac{A}{8m^2\Gamma\gamma} \left[\frac{e^{-(\gamma-\Gamma)(t_1-t_2)}}{\gamma-\Gamma} - \frac{e^{-(\gamma+\Gamma)(t_1-t_2)}}{\gamma+\Gamma} \right].\end{align}$$ Hence, $$\begin{align} \left\langle [x(t+\Delta t) - x(t)]^2\right\rangle &= 2\Big[\left\langle x(t)^2\right\rangle - \left\langle x(t+\Delta t) x(t)\right\rangle\Big] \\ &= \frac{A}{4m^2\Gamma\gamma} \left[\frac{1-e^{-(\gamma-\Gamma)\Delta t}}{\gamma-\Gamma} - \frac{1-e^{-(\gamma+\Gamma)\Delta t}}{\gamma+\Gamma} \right].\end{align}$$