My lecture notes compare two problems, and state that they are equal:
$$\text {maximize } U(s)=\int_0^\infty e^{-\rho t}u(x(t))dt, \text { s.t. } \int_0^\infty x(t)dt\leq s\in \mathbb R$$
The notes state that this can be reformulated as:
$$\text {maximize } U(s)=\int_0^\infty e^{-\rho t}u(x(t))dt, \text { s.t. }$$ $$ \dot u(t)=-x(t),$$ $$u(0)=s,$$ $$u(t)\geq 0 \text { for }t\geq 0$$
I sort of understand why this would capture the same thing, but I don't understand how this conclusion is drawn.
How do we formally derive the second problem from the first one, and vice versa?
Note that
$$ \frac{{\rm d}u}{{\rm d}t} = -x(t) $$
can be written as
$$ u(t) - u(0) = -\int_0^t{\rm d}t'~x(t') ~~~~\Rightarrow~~~~ u(t) = u(0) - \int_0^t{\rm d}t'~x(t') $$
If you take $u(0) = s$, then this is equivalent to
$$ u(t) = s - \int_0^t{\rm d}t'~x(t') \ge 0 $$
This last inequality comes from the constraint in the original problem. It is important to see that this constraint should only be valid for $t\to\infty$, but if you ensure that $u(t)\ge 0, ~\forall t$ then the constraint is automatically satisfied in the limit.