Consider the following equation: $$ X_t = 1 + \int_0^t a(u) du$$ where $a(\cdot) : [0,\infty) \to [-1,1]$ is a measurable function and the integral is a Lebesgue integral.
Suppose $a(u) = -1$ for all $u \le 1$. Hence, $X_1 = 0$.
Suppose I want the process to remain at $0$ forever. One way I could achieve this is by setting $a(u) = 0$ for all $u > 1$. Alternatively, I would like to say that the following function should also keep the process at $0$ forever.
\begin{align*} a(u) = \begin{cases} -1 & \text{ if } X_u > 0\\ 1 & \text{ if } X_u \leq 0 \end{cases} \end{align*}
Intuitively, the reason is that by setting $a_u = -1$ whenever $X_u > 0$ I am decreasing $X$. Since I start at $1$, I will hit $0$. Once I am at $0$, I reverse the direction of movement, but only for an instant and the process is pushed back to $0$.
But I have no idea if this is correct. If yes, what is the formal way to argue this?
I don't think you can do that. Indeed we can easily see that this doesn't work because you want $X_t=0$ for all $t\geq 1$. But then we get $a(u) =1$ for all $u\geq 1$. These two things can never happen at the same time..