Second order brownian motion $\ddot{x}(t) = \xi(t)$

Question

I'd like to solve for the pdf of position $$P(x,t) = \Big\langle \delta\Big(x-\int_0^t dt_1 \int_0^{t_1}dt_2 \xi(t_2)\Big)\Big\rangle $$ for the second order Brownian motion given by a Langevin-type equation $$ \ddot{x}(t) = \xi(t),$$ where $\xi(t)$ is a Gaussian white noise with correlation function $ \langle \xi(t)\xi(s)\rangle = 2D\delta(t-s).$

This does not seem as if it'd be a complex problem, but I have not been able to find references on it. Any advice would be appreciated, including mathematical advice or references. In particular I'd like to understand how to derive the PDE governing $P(x,t)$. I suppose I'd be able to solve it once I had it.

Answer 1

Regarding the joint distribution of position and velocity:

From my previous answer we know the variance $E[(\int_0^tW(u)\,du)^2]=t^3/3\,$ of the position. It is easy to see that the covariance between position $x(t)=\int_0^tW(u)\,du$ and velocity $v(t)=\dot x(t)=W(t)$ is $$ E[\textstyle(\int_0^tW(u)\,du) W(t)]=t^2/2\,. $$ Therefore, the correlation between $x$ and $v$ is

$$ \varrho=\frac{t^2/2}{\sqrt{t^3/3}\sqrt{t}}=\frac{\sqrt{3}}{2}\,. $$ Let's normalize the variables to make them standard normal: $$ \hat{x}(t)=\frac{x(t)}{\sqrt{t^3/3}}\,,~~\hat{v}(t)=\frac{v(t)}{\sqrt{t}}\,. $$ Then, $P(\hat x,\hat v,t)$ is the density of a bivariate normal distribution with correlation $\varrho\,$: $$ P(\hat x,\hat v,t)=\frac{1}{2\pi\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{\hat x^2-2\varrho\,\hat x\,\hat v+\hat v^2}{2(1-\varrho^2)}\Big)\,. $$ Therefore, the joint PDF of position and velocity is \begin{eqnarray}\label{ePDF} P(x,v,t)&=&\frac{\sqrt{3}}{2\pi t^2\sqrt{1-\varrho^2}}\,\exp\Big(-\frac{3x^2/t^3-2\sqrt{3}\varrho\,x\,v/t^2+v^2/t}{2(1-\varrho^2)}\Big)\\ &=&\frac{\sqrt{3}}{\pi t^2}\,\exp\Big(-\frac{6x^2-6\,x\,v\,t+2v^2\,t^2}{t^3}\Big)\,. \end{eqnarray} This function satisfies the Kolmogorov PDE $$ \partial_t P=-v\,\partial_x P+\frac{1}{2}\partial_{vv}P $$ for the diffusion process $$ \left(\begin{array}{c}dx\\dv\end{array}\right)=\left(\begin{array}{c}v\\0\end{array}\right)\,dt+\left(\begin{array}{cc}0&0\\1&0\end{array}\right)\left(\begin{array}{c}dW\\dW_v\end{array}\right) $$ where $W_v$ is a dummy BM that is not driving anything.

The Book of Karatzas and Shreve (Brownian Motion and Stochastic Calculus) is very useful for such problems.

Answer 2

I would suggest using the Fourier transform: $$ \delta(x)=\int\frac{dk}{2\pi}e^{ikx} $$ therefore $$ P(x,t) = \left\langle \int\frac{dk}{2\pi}e^{ik\left[x-\int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2)\right]}\right\rangle= \int\frac{dk}{2\pi}e^{ikx}\left\langle e^{-ik\int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2)}\right\rangle $$ (The order of integration and averaging can be interchanged.)

Now $$ z= \int_0^tdt_1\int_0^{t_1}dt_2\xi(t_2) $$ is a Gaussian random variable, for which we can use the identity $$ \langle e^{\alpha z}\rangle = e^{\frac{\alpha^2\sigma_z^2}{2}} $$ where $\sigma_z$ can be easily calculated. The final Fourier transform is will be a Gaussian integral.

Answer 3

By definition of white noise, the integral (velocity) $$ W(t)=\int_0^t\xi(u)\,du $$ is a standard Brownian motion. The double integral in your $P(x,t)$ formula is then (position) $$ \int_0^tW(u)\,du\,. $$ It is known that this integral is normaly distributed. It has the variance $t^3/3$:

\begin{eqnarray*} \textstyle E\Big[(\int_0^tW(u)\,du)^2\Big]&=&\textstyle E\Big[(\int_0^tW(u)\,du)(\int_0^tW(v)\,dv)\Big]=E\Big[\int_0^t\int_0^tW(u)W(v)\,du\,dv\Big]\\ &=&\textstyle\int_0^t\int_0^t\min(u,v)\,du\,dv =\textstyle\int_0^t\Big(\int_0^vu\,du+\int_v^tv\,du\Big)\,dv\\ &=&\textstyle\int_0^t\frac{v^2}{2}+vt-v^2\,dv =\textstyle\frac{t^3}{6}+\frac{t^3}{2}-\frac{t^3}{3}=\frac{t^3}{3}\,. \end{eqnarray*} This gives us the density of position as the density of a normal distribution with variance $t^3/3\,$: $$ P(x,t)=\frac{\sqrt{3}}{\sqrt{2\pi t^3}}\exp\Big(-\frac{3x^2}{2t^3}\Big)\,. $$ For an arbitrary diffusion coefficient $D$ the formulas are easily modified.