Suppose an $n$-dimensional process $(X\_t)\_{0 \leq t \leq 1}$ satisfies an SDE of the form:


$$dX\_t = u\_t(X\_t) \,dt + dB\_t, ~~X\_0 = 0$$


where $(B\_t)\_{t\geq 0}$ is a Brownian motion with $B\_1 \sim N(0,K)$, and $K$ is positive definite.


Does anyone have a reference for simple conditions on $u\_t$ that will ensure that $\operatorname{law}(X\_t)$ has full support on $\mathbb{R}^n$?


For example, for $t>0$, if there exists a finite constant $C\_t$ such that $|u\_s(x)| \leq C\_t(|x|+1)$ for all $0 \leq s\leq t$, does this ensure $\operatorname{law}(X\_t)$ has full support?


Of course, some regularity on $u\_t$ is needed; consider the standard Brownian bridge, which has $X\_1 = 0$ a.s., but $\operatorname{law}(X\_t)$ has full support for each $0< t<1$.



When does a solution to SDE have full support?
I am writing down a very similar answer to [this one](https://mathoverflow.net/a/443534/129074), that I wrote for the similar question [*SDE with non-degenerate diffusion visits every point*](https://mathoverflow.net/q/438425/129074). I am not sure how closely it answers your question, since it is not a reference for the exact result you quote, but it does follow as a corollary.


The support of the whole trajectory $X$ in $\mathcal C([0,\infty),\mathbb R^n)$ is described by the so-called Stroock-Varadhan support theorem. In [*On the support of diffusion processes with applications to the strong maximum principle*](https://projecteuclid.org/proceedings/berkeley-symposium-on-mathematical-statistics-and-probability/Proceedings-of-the-Sixth-Berkeley-Symposium-on-Mathematical-Statistics-and/Chapter/On-the-support-of-diffusion-processes-with-applications-to-the/bsmsp/1200514345?tab=ChapterArticleLink), their Theorem 3.1 ensures that the (unique) solution to the martingale problem associated with $u\_t+\frac12\Delta\_K$ has full support provided $u:[0,\infty)\times\mathbb R^d\to\mathbb R^d$ is bounded measurable.


A strong solution to the equation $\mathrm dX\_t=u\_t(X\_t)\mathrm dt + \mathrm dB\_t$ will always be a solution to the martingale problem, so this answers your question in the case where (1) you have conditions guaranteeing existence of a strong solution and (2) $u$ is bounded measurable (you can also define the solution as a solution to the martingale problem, in which case (2) is sufficient for existence and uniqueness).


A natural case that doesn't quite fit the above is when $u$ is unbounded but you have a unique strong and weak solution up to some explosion time using some other argument $\tau$. We can still rely on this theorem in more general situations, and to illustrate my point I will consider the classical case where $u$ is continuous in $(t,x)$, locally Lipschitz in $x$ (for instance $u$ is $\mathcal C^1$ in $(t,x)$). Then one can define the solution $X^R>0$ to the equation
$$ \mathrm dX^R\_t = \big(0\vee(2R-|X^R\_t|)\wedge1\big)u\_t(X^R\_t)\mathrm dt + \mathrm dB\_t, $$
which is defined for all times and coincides with $X$ until one of the two exists the ball of radius $R$. This means that
$$\mathbb P(X\_t\in U)\geq\mathbb P(X\_t\in U\text{ and }\forall s\leq t,|X\_s|<R)=\mathbb P(X^R\_t\in U\text{ and }\forall s\leq t,|X^R\_s|<R)$$
for $R$ large enough. By the theorem above, we know that this last probability must be positive, and so is the first one.


A similar argument shows that your condition $|u\_s(x)|\leq C\_t(1+|x|)$ is enough, provided $u$ is measurable, and your notion of solution coincides with the unique martingale problem solution.


I should add that these results rely on applying a [Girsanov argument](https://en.wikipedia.org/wiki/Girsanov_theorem), making a change of probability with density (if I am not mistaken with my use of $K$ and $K^{-1}$)
$$\exp\left(\int\_0^tu\_s(X\_s)K^{-1}\mathrm dB\_s-\frac12\int\_0^tu\_s\mathrm ds\right).$$


