Path Integral Series: 4. Stochastic Differential Equations

路径积分系列：4.随机微分方程 苏剑林 · 2016-06-09 · Mathematics differential equation path integral random

This chapter applies the path integral to stochastic differential equations and arrives at the same result as the asymmetric random walk, thereby proving the equivalence between the two models.

The idea of applying the path integral to stochastic differential equations goes back a long way. Feynman, in his book [5], had already established the relationship between the path integral and linear stochastic differential equations. For the nonlinear case, there has also been considerable research, though the picture is rather muddled—reference [8], for instance, even gives an incorrect result.

Starting from the notion of discretizing the path integral, this article explicitly establishes the Jacobian-determinant relationship between two path-integral measures, thereby constructing the path integral for nonlinear stochastic differential equations as well. The results here agree with those of reference [9].

Concepts

What we study here is only stochastic ordinary differential equations, which differ from ordinary ODEs in the introduction of a Brownian-motion term. A commonly seen class of stochastic differential equations is

$$dx(t)=p(x(t),t)dt + \sqrt{\alpha} dW_t.\tag{48}$$

where $W_t$ denotes a standard Brownian motion. Because of the random term, the solution $x(t)$ is no longer deterministic, but instead follows a certain probability distribution.

In the study of stochastic differential equations, there are many quantities of interest—for instance, the expectation or variance of some quantity related to $x$, or stability, and so on. The field of stochastic differential equations offers all sorts of analytical techniques, but clearly, the most ideal and easiest approach would be to directly obtain the probability distribution of $x(t)$ and then study that distribution. The path integral is precisely a method for obtaining this probability distribution. more

Linear Stochastic Differential Equations

We start from the following linear stochastic differential equation

$$dx(t)=x(t)dt + \sqrt{\alpha} dW_t,\tag{49}$$

We wish to solve this equation, and what we care about is the probability distribution of $x$. So the goal of "solving" it is to find the probability of starting at $x_a$ at time $t_a$ and arriving at $x_b$ at time $t_b$—that is, the propagator (Green's function). To construct the path integral, we need to find the probability functional $x(t)$, starting from $(t_a,x_a)$ and passing along a path to reach $(t_b,x_b)$: $ P [x(t)]$.

We can start from the probability functional of Brownian motion. Since $W(t)$ can uniquely determine $x(t)$, the probability of a given $x(t)$ equals the probability of the corresponding given $W(t)$, i.e.

$$ P [x(t)]\mathscr{D}x(t)= P [W(t)]\mathscr{D}W(t).\tag{50}$$

We already know that for Brownian motion the probability density functional of $W(t)$ is

$$ P [W(t)]=\exp\left(-\int_{t_a}^{t_b} \frac{1}{2}\dot{W}^2(t)dt\right).\tag{51}$$

Note that we're still writing this in the form of a derivative—this is purely formal (the derivative terms appearing in the definition of the path integral are inherently just formal notation anyway), used here as a convenient and memorable symbol. Now we have

$$\begin{aligned} P [x(t)]\mathscr{D}x(t)=& P [W(t)]\mathscr{D}W(t)\\ =&\exp\left(-\int_{t_a}^{t_b} \frac{1}{2}\dot{W}^2(t)dt\right)\mathscr{D}W(t)\\ =&\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}(t)-x(t)\right]^2 dt\right)\mathscr{D}W(t)\end{aligned},\tag{52}$$

So far everything we've done is easy to follow—it's just a change of variables. What's missing is the final step: the relationship between $\mathscr{D}x(t)$ and $\mathscr{D}W(t)$. Both $\mathscr{D}x(t)$ and $\mathscr{D}W(t)$ are defined as limits of $n-1$-fold multiple integrals, and for finite multiple integrals, a change of variables introduces a Jacobian determinant relating the measures. So the same should hold here:

$$\mathscr{D}W(t)=\mathcal{J}[x(t)]\mathscr{D}x(t),\tag{53}$$

Because of the infinite dimensionality of $\mathscr{D}$, the Jacobian determinant here is the determinant of an $\infty\times\infty$-dimensional square matrix.

Note that the relationship between $W(t)$ and $x(t)$ is linear, and we already know that the Jacobian determinant of a linear transformation is a constant—no exception here. So we can set this constant aside for now and determine it later by normalization once we have the result. Up to this constant factor, we have

$$\begin{aligned} P [x(t)]\mathscr{D}x(t)=&\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}(t)-x(t)\right]^2 dt\right)\mathscr{D}W(t)\\ =&\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}(t)-x(t)\right]^2 dt\right)\mathscr{D}x(t)\end{aligned},\tag{54}$$

which is to say

$$ P [x(t)]=\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}(t)-x(t)\right]^2 dt\right).\tag{55}$$

With this result, we can now use the path integral to compute the propagator:

$$\begin{aligned}\int P [x(t)]\mathscr{D}x(t)=&\int\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}^2(t)+x^2(t)-2\dot{x}(t) x(t)\right] dt\right)\mathscr{D}x(t)\\ =&\exp\left(\frac{1}{2\alpha}\left(x_b^2-x_a^2\right)\right)\times\\ &\qquad\int\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}^2(t)+x^2(t)\right] dt\right)\mathscr{D}x(t)\end{aligned}.\tag{56}$$

where the path-integral part

$$K(x_b,t_b;x_a,t_a)=\int\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}^2(t)+x^2(t)\right] dt\right)\mathscr{D}x(t)\tag{57}$$

is just a basic Gaussian-type path integral, which we can solve exactly.

Computing the Jacobian Determinant

Let us now derive the path integral for the general form of a stochastic differential equation

$$dx(t)=p(x(t), t)dt + \sqrt{\alpha} dW_t ,\tag{58}$$

As in the previous section, we readily obtain

$$\begin{aligned} P [x(t)]\mathscr{D}x(t)=& P [W(t)]\mathscr{D}W(t)\\ =&\exp\left(-\int_{t_a}^{t_b} \frac{1}{2}\dot{W}^2 dt\right)\mathscr{D}W(t)\\ =&\exp\left(-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}-p(x,t)\right]^2 dt\right)\mathscr{D}W(t)\end{aligned}.\tag{59}$$

The key point is the relationship between $\mathscr{D}x(t)$ and $\mathscr{D}W(t)$. In the linear case we know this is a constant and so can be ignored, to be fixed later during normalization; but in the nonlinear case there's no getting around it—we have to actually compute its determinant.

Discretize the equation as

$$(x_k - x_{k-1}) - \frac{1}{2}\epsilon\left[p(x_k,t_k)+p(x_{k-1},t_{k-1})\right]=\sqrt{\alpha} (W_k-W_{k-1}),\tag{60}$$

where $k=0,1,2,\dots,n$, and $W_0\equiv W(t_a),W_n\equiv W(t_b),x_0\equiv x(t_a),x_k\equiv x(t_b)$ is a given initial condition, $\epsilon=\frac{t_b-t_a}{n}$ is the time step, $x_k\equiv x\left(\frac{k\epsilon}{n}\right)$. Note that we have taken an average over $p(x,t)$. This is necessary—without the averaging we would get an incorrect result. This is because $\frac{x(t)-x(t-\epsilon)}{\epsilon}$, as an approximation to $\dot{x}(t)$, only has zeroth-order accuracy; in fact it is closest to $\dot{x}(t-\epsilon/2)$, at which point it has first-order accuracy $\epsilon$. In other words, when we write $(x_k-x_{k-1})/\epsilon$, what it actually represents is the derivative at the midpoint between $x_k$ and $x_{k-1}$. Hence the discretization of $p(x,t)$ must also be taken at the midpoint; we could use either $\frac{1}{2}\left[p(x_k,t_k)+p(x_{k-1},t_{k-1})\right]$ or $p\left(\frac{x_k+x_{k-1}}{2},\frac{t_k+t_{k-1}}{2}\right)$, both of which have the same accuracy and yield the same result (this is analogous to using the trapezoidal rule or midpoint rectangle rule to approximate the area under a curve—both convey the same idea).

We write

$$\left\{\begin{aligned}&\mathscr{D}x(t)\approx dx_1 dx_2\dots dx_{k-1}\\ &\mathscr{D}W(t)\approx dW_1 dW_2\dots dW_{k-1}\end{aligned}\right. .\tag{61}$$

and then use finite-dimensional calculus to compute the (approximate value of the) Jacobian determinant. Note that the Jacobian matrix from $W_k$ to $x_k$ is triangular, so the determinant is just the product of the diagonal entries, and the diagonal entries are

$$\frac{\partial W_k}{\partial x_k}=\frac{1}{\sqrt{\alpha}}\left(1-\frac{1}{2}\epsilon\frac{\partial p(x_k,t_k)}{\partial x_k}\right)\approx \frac{1}{\sqrt{\alpha}}\exp\left(-\frac{1}{2}\epsilon\frac{\partial p(x_k,t_k)}{\partial x_k}\right),\tag{62}$$

Therefore, up to a constant, we have

$$\mathcal{J}[x(t)]\approx \prod_{k=1}^{n-1} \left(1-\frac{1}{2}\epsilon\frac{\partial p(x_k,t_k)}{\partial x_k}\right)\approx \exp\left[-\frac{1}{2}\left(\sum_{k=1}^{n-1}\frac{\partial p(x_k,t_k)}{\partial x_k}\right)\epsilon\right],\tag{63}$$

Taking the limit $n\to\infty$, we get

$$\mathcal{J}[x(t)]=\exp\left(-\frac{1}{2}\int_{t_a}^{t_b}\frac{\partial p(x)}{\partial x}dt\right),\tag{64}$$

Substituting back gives

$$\begin{aligned} P [x(t)]\mathscr{D}x(t)=&\exp\left\{-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\dot{x}-p(x,t)\right]^2 dt\right\}\\ &\qquad\qquad\exp\left(-\frac{1}{2}\int_{t_a}^{t_b}\frac{\partial p(x,t)}{\partial x}dt\right)\mathscr{D}x(t)\\ =&\exp\left\{-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\left(\dot{x}-p(x,t)\right)^2 + \alpha\frac{\partial p(x,t)}{\partial x} \right]dt\right\}\mathscr{D}x(t) \end{aligned},\tag{65}$$

and hence

$$\begin{aligned} P [x(t)]=&\exp\left\{-\frac{1}{2\alpha}\int_{t_a}^{t_b} \left[\left(\dot{x}-p(x,t)\right)^2 + \alpha\frac{\partial p(x,t)}{\partial x} \right]dt\right\}\\ =&\exp\left\{-\frac{1}{\alpha}\int_{t_a}^{t_b}\left[\frac{1}{2}\dot{x}^2-\dot{x}p(x,t)+\frac{1}{2}p^2(x,t)+\frac{1}{2}\alpha\frac{\partial p(x,t)}{\partial x}\right]dt\right\} \end{aligned}.\tag{66}$$

Noting that

$$\begin{aligned}\frac{d}{dt}\int p(x,t)dx =& \dot{x}\frac{\partial}{\partial x}\int p(x,t)dx+\frac{\partial}{\partial t}\int p(x,t)dx\\ =&\dot{x}p(x,t)+\int \frac{\partial p(x,t)}{\partial t}dx \end{aligned},\tag{67}$$

we can use this identity to rewrite the above as

$$P [x(t)]=\exp\left[-\frac{1}{\alpha}\int_{t_a}^{t_b}\left(\frac{1}{2}\dot{x}^2-V(x,t)\right)dt\right]\exp\left(\frac{1}{\alpha}\int_{x_a}^{x_b}p(x,t)dx\right).\tag{68}$$

where

$$V(x,t)=-\frac{1}{2}\left(\alpha\frac{\partial p}{\partial x}+p^2\right)-\int \frac{\partial p}{\partial t}dx.\tag{69}$$

Comparing this with the result of equation $(16)\sim (19)$, we see that the stochastic differential equation $(48)$ gives exactly the same result as the asymmetric random walk.

The Path Integral Method

We have now proven the equivalence between the stochastic differential equation and the asymmetric random-walk model, and, back in Chapter 1, we had already worked out the partial differential equation satisfied by the asymmetric random-walk model. This means we can immediately write down the partial differential equation corresponding to the stochastic differential equation as well. At this point, we have achieved the mutual translation among all three formulations, with the path integral method serving as the connecting thread throughout. This is a vivid illustration of the power of the path integral method.