= Solution
On $I=g(\mathbb R)$ define $h(y)=g'(g^{-1}(y))$. Differentiation gives
$$
\boxed{h'(y)=\frac{g''(g^{-1}(y))}{g'(g^{-1}(y))}
=-2b(g^{-1}(y)),\qquad |h'(y)|\leq2B.}
$$
The mean value theorem proves <Lipschitz continuity> of $h$ on $I$, with constant $2B$.
If an endpoint of $I$ is finite, the Lipschitz bound gives a finite limiting value of $h$ there. That value must be zero. Otherwise $h$ would be bounded below by a positive number near the endpoint, and
$$
\frac{d}{dy}g^{-1}(y)=\frac1{h(y)}
$$
would make $g^{-1}$ approach a finite limit, contradicting its tending to $+\infty$ or $-\infty$. The <zero extension of a scale diffusion coefficient at finite endpoints> therefore gives a globally Lipschitz function $\bar h$ on $\mathbb R$: set it to zero beyond each finite endpoint and retain $h$ on $I$.
Use the following standard <global existence theorem for stochastic differential equations with Lipschitz coefficients>: globally Lipschitz drift and diffusion coefficients give a nonexplosive, pathwise unique <strong stochastic solution> for every prescribed initial state and driving <Brownian motion>. Global Lipschitz continuity also gives the required linear-growth bound. Apply it to
$$
dY_t=\bar h(Y_t)\,dW_t,\qquad Y_0=g(x).
$$
It remains to check that $Y$ stays in $I$, so the inverse transform is defined.
Before the first boundary time, put $X=g^{-1}(Y)$. Since $(g^{-1})'=1/h$ and $(g^{-1})''=-h'/h^2$, the <Itô formula>, stopped inside compact subintervals of $I$, yields
$$
dX_t=dW_t-\frac12h'(Y_t)\,dt=dW_t+b(X_t)\,dt.
$$
For each finite $T$, up to that boundary time,
$$
|X_t|\leq|x|+\sup_{s\leq T}|W_s|+BT,\qquad t\leq T.
$$
A finite endpoint of $I$ would require $X$ to diverge, which this bound excludes. Therefore $Y$ stays in $I$ at every finite time, and $X=g^{-1}(Y)$ is a global <strong stochastic solution>.
Finally, any two solutions driven by the same $W$ transform into solutions of the globally Lipschitz $Y$ equation. Its <pathwise uniqueness> makes the transforms, and hence their inverses, indistinguishable. Thus
$$
\boxed{X=g^{-1}(Y)\text{ exists globally and is pathwise unique}.}
$$
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