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I'm currently looking at stochastic differential equations with irregular coefficients such as $W^{1,p}_{loc}$. If I have \begin{align} dX_t=b(X_t)dt+\sigma dW_t, \end{align} where $b\in W^{1,1}_{loc}$ and $\sigma$ is a constant, I can write this SDE as a ODE for every Brownian path $w$ by defining $Y_t=X_t-w_t$ and a new vector field $b^w=b(Y,t)=b(Y_t+w_t)$, so the ODE is \begin{align} dY=b^w(t,Y)dt \end{align} with initial condition $Y_0=y$. Since $b^w$ has Sobolev regularity I can then apply DiPerna-Lions theory (1989), which guarantees the existence and uniqueness of the flow of the ODE.

My question is now what happens if $\sigma$ is not a constant,
\begin{align} dX_t=b(X_t)dt+\sigma(X_t)dW_t. \end{align} Apparently the above argument is NOT correct in general. Is there any ways that the diffusion coefficients $\sigma(X_t)$ can be absorbed into the Brownian motion? or maybe there are some conditions on $\sigma$ under which the above argument still hold?

Many thanks!!

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1 Answer 1

If I understood your question correctly, then the diffusion coefficient can be "absorbed" using a standard trick.

So consider the equation $$dX_t=b(X_t)dt+\sigma(X_t)dW_t.$$ Assume that $\sigma(x)\ge\varepsilon>0$ for all real $x$. Define $Y_t:=\psi(X_t)$, where $$\psi(x):=\int\limits_0^{x} \frac{1}{\sigma(s)} ds.$$

By Ito's lemma,

$$dY_t=\frac{b(X_t)}{\sigma(X_t)}dt-\frac12 \sigma'(X_t)dt+dW_t.$$

Now if we recall that $X_t=\psi^{-1}(Y_t)$, we will finally obtain

$$dY_t=\frac{b(\psi^{-1}(Y_t))}{\sigma(\psi^{-1}(Y_t))}dt-\frac12 \sigma'(\psi^{-1}(Y_t))dt+dW_t.$$ Thus, the diffusion coefficient has "absorbed" and now equals 1. Hope, this helps.

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