Show that $u(t) \leq u(a) e^{\int_a^t f(s) ds}$ Let u(t) be a continuously differentiable function on [a,b] and the following inequality holds $\forall t \in [a,b]$ such that $u'(t) \leq f(t)u(t).$
Show that
$u(t) \leq u(a) e^{\int_a^t f(s) ds}$
I have been just staring at this problem for a while and dont know where to start.
I have done basis rearranging, but cannot make any real progress.
Does anyone have any hints as to where to start, it would be greatly appreciated.
 A: Set $g(t)=u(t)e^{-\int_a^tf(s)ds}$ 
$g'(t)=u'(t)e^{-\int_a^tf(s)ds}-u(t)f(t)e^{-\int_a^tf(s)ds}\leq 0$, hence $g$ is a decreasing function then $g(t)\leq g(a) $.
A: This is the well known Gronwall's inequality . A very rigorous proof can be found from Tao's book, nonlinear dispersive equations, Chapter 1. You should be able to derive it yourself as well. Note a naive separation of variables would not work, as $f$'s sign is undetermined.  
A: This is Gronwall Inequality, but let's use Euler's method:
Maybe we can discretize
$$ u'(t) \approx  \frac{u(t+\Delta t) - u(t) }{\Delta t} \leq  f(t) u(t) $$
then get:
$$ u(t+\Delta t) \leq u(t) \left( 1 +  f(t) \Delta t \right)$$
By induction and using $1 + x \approx e^x$ for $x << 1$:
$$ u(t+ n\Delta t) \leq u(t) \prod_{k=1}^n \left( 1 +  f(t + k \Delta t) \Delta t \right) \approx  u(t)  e^{\sum_{k=1}^n f(t + k \Delta t) \Delta t} $$
And taking the scaling limit, $\Delta t \to 0$.
$$ u(t+a) \leq u(t) e^{\int_0^a f(t) dt}$$

Historical note:  Gronwall's inequality result was proven by Richard Bellman, father of Dynamic Programming and coiner of the "Curse of Dimensionality".
