2D SDEs

We can generalize SDEs to ${\mathbb{R}}^2$. A particle can be represented as

$$\left(X(t),Y(t)\right)\in {\mathbb{R}}^2$$

moving in 2D with random kicks.

$$\begin{array}{rl}dX& =u(X,Y)dt+\sqrt{2D_1(X,Y)}d{B}_1\\ dY& =v(X,Y)dt+\sqrt{2D_2(X,Y)}d{B}_2\end{array}$$

for functions $u,v,D_1,D_2:{\mathbb{R}}^2\to \mathbb{R}$ and independent Brownian Motions $B_1(t)$ and $B_2(t)$.

The 2D nature was actually Brown’s original observation.

2D Ito’s Lemma

We can generalize Ito’s Lemma to find $df$ for $f(X,Y)$. The strategy is to do Taylor expansion to $O(dt)$. The same assumptions hold from above.

$$\begin{array}{rl}df& =f(X+dX,Y+dY)-f(X,Y)\\ & =dX({\partial }_{x}f)+dY({\partial }_{y}f)+\frac{1}{2}\left(\begin{array}{cc}dX& dY\end{array}\right)\left(\begin{array}{cc}{\partial }_{xx}^{2}f& {\partial }_{xy}^{2}f\\ {\partial }_{xy}^{2}f& {\partial }_{yy}^{2}f\end{array}\right)\left(\begin{array}{c}dX\\ dY\end{array}\right)+O(d{t}^{3/2})\\ & =\left(u\frac{\partial f}{\partial x}+v\frac{\partial f}{\partial y}+D_1\frac{{\partial }^{2}f}{\partial x^2}+D_2\frac{{\partial }^{2}f}{\partial y^2}\right)dt+\frac{\partial f}{\partial x}\sqrt{2D_1}d{B}_1+\frac{\partial f}{\partial y}\sqrt{2D_2}d{B}_2\end{array}$$

Example 1 (Active Particle on a Line)

$"\\begin{document}\n\\begin{tikzpicture}[>=stealth, scale=1.2]\n\n$

source code

The diagram depicts $\theta (t)$ providing rotational noise with the particle moving right on the line.

$$\begin{array}{rlrlrl}dX& =v\cos \theta dt& & +0& & ,X(0)=0\\ d\theta & =0& & +\sqrt{2D}dB& & ,\theta (0)=0\end{array}$$

The particle tries to move in the $\theta -$direction but is constrained to the line.

A 2D active particle needs $3$ SDEs to simulate it.

How far does the particle move? We can measure the expectation.

$$\begin{array}{rl}d(\mathbb{E}\left[X\right])& =v\mathbb{E}\left[\cos \theta \right]dt\\ \frac{d}{dt}\mathbb{E}\left[X\right]& =v\mathbb{E}\left[\cos \theta \right]\end{array}$$

How to find $\mathbb{E}\left[\cos \theta \right]$? We can use 2D Ito’s Lemma! For $f(X,\theta )$, we have

$$\begin{array}{rl}d{f}_1& =\left(u\frac{\partial f}{\partial x}+v\frac{\partial f}{\partial y}+D_1\frac{{\partial }^{2}f}{\partial x^2}+D_2\frac{{\partial }^{2}f}{\partial y^2}\right)dt\\ & =\left(0+0+(0)+D\cdot (-\cos \theta )\right)dt\\ & =-D\cos (\theta )dt\end{array}$$

for the deterministic part. For the noise part,

$$\begin{array}{rl}d{f}_2& =0+-\sin (\theta )\sqrt{2D}dB\end{array}$$

Thus,

$$df=d{f}_1+d{f}_2=-D\cos (\theta )dt-\sqrt{2D}\sin (\theta )dB$$

Taking the expectation,

$$\begin{array}{rl}d\mathbb{E}\left[\cos \theta \right]& =-D\mathbb{E}\left[\cos \theta \right]dt-0\\ \frac{d}{dt}\mathbb{E}\left[\cos \theta \right]& =-D\mathbb{E}\left[\cos \theta \right]\\ \mathbb{E}\left[\cos \theta \right]& =\cos (0)e^{-Dt}\\ & =e^{-Dt}\end{array}$$

Going back to our original expectation,

$$\frac{d}{dt}\mathbb{E}\left[X\right]=v{e}^{-Dt}$$

Integrating,

$$\begin{array}{rl}\mathbb{E}\left[X(t)\right]& ={\int }_0^{t}v{e}^{-Ds}ds\\ & ={\left[-\frac{v}{D}{e}^{-Ds}\right]}_0^t\\ & =-\frac{v}{D}\left(e^{-Dt}-1\right)\\ & =\frac{v}{D}\left(1-e^{-Dt}\right)\end{array}$$

as desired.