Does a strong solution to a SDE imply lipschitz condition?

Question

Consider $dX_t=b(X_t,t)dt+\sigma(X_t,t)dB_t$. I know that,

$|b(x,t)-b(y,t)|+|\sigma(x,t)-\sigma(y,t)|\leq D|x-y|$ for some constant D implies the existence and uniqueness of a strong solution. However in Oksendal, to show that $dX_t=sign(X_t)dB_t$ doesn't have a strong solution another proof is presented, even though $sign(x)$ fails the lipschitz condition. I have seen it at a couple of more places for this equation.

My question: Is Lipschitz condition an iff condition for a strong solution?

Answer 1

No, Lipschitz continuity is not a necessary condition for the existence of a strong solution. There is, for instance, the following general result which goes back to Zvonkin:

Let $(B_t)_{t \geq 0}$ be a one-dimensional Bownian motion. If the coefficients of the SDE $$dX_t = b(t,X_t) \, dt + \sigma(t,X_t) \, dB_t, \qquad X_0 = x_0, \tag{1}$$ are bounded, $b$ is measurable, $\sigma$ is continuous and there exist constants $C>0$ and $\epsilon>0$ such that $$|\sigma(t,x)-\sigma(t,y)| \leq C \sqrt{|x-y|} \quad \text{and} \quad |\sigma(t,x)| \geq \epsilon$$ for all $t \geq 0$, $x,y \in \mathbb{R}$, then the SDE $(1)$ has a (unique) strong solution.

Note that the result does not require any regularity assumptions on the drift $b$ (except from measurability) and only Hölder continuity of order $1/2$ for the diffusion coefficient $\sigma$.

A nice overview on known existence and uniqueness results for SDEs can be found in the book Singular Stochastic Differential Equations by Cherny & Engelbert.