Q:

Why does $ b \ge 0 $ in Gronwall Lemma? 

I'm reading about Gronwall Lemma, formulated this way:

Let $a \in \mathbb{R}$, $b\ge 0$, $T>0$. If $ u: [0, T) \to \mathbb{R} $ continuous and for $ t \in (0, T) $
$$ u(t) \le a + b\int^t_0u(s)ds $$
  then for $ t \in (0,T) $
$$ u(t) \le ae^{bt}.  $$

I can see that $a$ is any real number, so why $b$ has to be greater (or equal) than zero?  Are there some counterexamples for this lemma with $b<0$? 

A:

It is really unclear.  Let's look at a proof and what restrictions result from its construction.  Start with applying an integrating factor
$$
\frac{d}{dt}\left(e^{-bt}\int_0^tu(s)ds\right)=e^{-bt}\left(u(t)-b\int_0^tu(s)ds\right)
\le ae^{-bt}.\tag1
$$
Now integrate on both sides over $[0,t]$ to get
$$
e^{-bt}\int_0^tu(s)ds\le \frac{a}{b}(1-e^{-bt}).\tag2
$$
To finish insert into the original inequality
$$
u(t)\le a+b⋅\int_0^tu(s)ds\le a+b\cdot \frac{a}{b}(e^{bt}-1)=ae^{bt}.\tag3
$$
In this last step it was necessary to multiply the inequality (2) with $b$, which will only work without changing the direction of the inequality if $b\ge 0$.

