I'm stuck trying to prove some generalisations to the Gronwall's inequality.
Firstly, I saw that you can prove it in the case tha $M \in \mathbb{R} $, $u \in C^0[t_0,t_1]$ and $a \geq 0$ and Riemann integrable $$u(t) \leq M + \int_{to}^t a(s)u(s)ds \Rightarrow u(t) \leq M e^{\int_{to}^t a(s) ds}$$ My proof starts with defining $v(t) = \int_{to}^t a(s)u(s)ds$, and then differentiating it, but as $a$ is not continuous, we can't assume that $v$ is differentiable. I don't know if that really is a problem, if i can just say $v'(t) = a(t)u(t)$ without having to worry about continuity. If not, I would like to know how I could start the proof
The second case I saw was a more general case, in which the assumptions are that $$u(t) \leq M + \int_{to}^t \Phi(s,u(s)) ds$$ being $\Phi$ a continuous, non decreasing, positive and with a continuous partial derivative with respect to its second term. I tried as in the previous case, to define $v(t) = \int_{to}^t \Phi(s,u(s)) ds$.
Then, I differentiate, to get: $v'(t) = \Phi(t,u(t)) \leq \Phi(t,M+v(t))$
I thought about differentiating again, to use the chain rule and get $v'(t)$, but I can't do that in an inequality, so I don't know how I would have to continue, maybe taking inverse but I dont know how I would go on from there.