# Existence of strong solution in SDEs and continuity in the time variable

I recently come across some literature in stochastic analysis that uses the following result:

Consider the one-dimensional SDE

$$dX_t= a(t, X_t) \, dt + b(t, X_t) \, dW_t,$$

where $$a, b: \mathbb{R}^2 \rightarrow \mathbb{R}$$ are Borel-measurable functions and $$\{W_t\}_{t \geq 0}$$ is a Wiener process. Suppose that $$a,b$$ also satisfy

$$1. \, \, \, \,\,\,\,\,|a(t,x)- a(t,y)| + |b(t,x) - b(t,y) | \leq K |x-y|, \quad \forall x, y \in \mathbb{R},$$ $$\forall t \geq 0$$, for some constant $$K>0$$, and

$$2. \, \, \, \,\,\,\,\,$$the functions $$t \mapsto a(t,x)$$ and $$t \mapsto b(t,x)$$ are continuous, $$\forall x \in \mathbb{R}$$.

Then, the SDE has a unique strong solution.

The classical result about existence of strong solution in SDEs requires Lipschitzness in the space variable and linear growth in the space variable. I don't see how the latter can be replaced by the requirement of continuity in the time variable. Does anyone know the proof of this result?

Ref: At the bottom of Page 23 of https://spiral.imperial.ac.uk/bitstream/10044/1/28918/3/McMurray-EFV-2015-PhD-Thesis.pdf

• I'm confused because the continuity in time doesn't have anything to do with the behavior in the spacial direction. In particular, what if the coefficients are time-independent? Then this is saying the linear growth in frequency is unnecessary in the classical result you're referring to (which I think is very false but I could be wrong...). Even if the coefficients are time-dependent I think you still need the linear growth in frequency which is not guaranteed (I don't think) with the time-continuity condition... Jun 2, 2016 at 21:42
• @KevinYang spiral.imperial.ac.uk/bitstream/10044/1/28918/3/… Please have a look at the bottom of Page 23..... I am confused too. Jun 2, 2016 at 22:39
• "Lipschitzness in the space variable and linear growth in the space variable": should one of those "space" be "time"? Jun 3, 2016 at 5:30

1 and 2 imply linear growth (locally in $t$). Indeed, for any $T>0$ and $t\in[0,T]$ $$|a(t,x)|\le |a(t,0)| + |a(t,x) - a(t,0)| \le \sup_{s\in[0,T]} |a(s,0)| + K|x|\\\le \big(K\vee ||a(\cdot,0)||_{\infty;[0,T]}\big)(1+|x|).$$ As you see, it is in fact enough that $a(\cdot,0)$ and $b(\cdot,0)$ are locally bounded; with little extra effort one can prove the strong existence if they are only locally $L^1$ and $L^2$, respectively. I'll try to find a reference for the last if you need one.