Proof of Gronwall inequality [duplicate]

Hi I need to prove the following Gronwall inequality Let $I := [a, b]$ and let $u, \alpha: I \rightarrow \mathbb{R}$ and $\beta: I \rightarrow [0, \infty)$ continuous functions. Further let

$$u(t) \leq \alpha(t) + \int_a^t \beta(s)u(s){\rm d}s$$

for all $t \in I$. Then the inequality $$u(t) \leq \alpha(t) + \int_a^t\alpha(s)\beta(s)e^{\int_s^t\beta(\sigma){\rm d}\sigma}{\rm d}s$$

holds for all $t\in I$.

marked as duplicate by Hans Lundmark, happymath, Claude Leibovici, naslundx, Joey ZouAug 27 '16 at 16:05

Let $y(t) := e^{-\int_a^t\beta(\sigma){\rm d}\sigma}\cdot\int_a^t\beta(s)u(s){\rm d}s$. Then $$y'(t) = \beta(t)e^{-\int_a^t\beta(\sigma){\rm d}\sigma}[u(t)-\int_a^t\beta(s)u(s){\rm d}s] \leq \alpha(t)\beta(t)e^{-\int_a^t\beta(\sigma){\rm d}\sigma}\ .$$ Now integrate and get $$\int_a^t\alpha(s)\beta(s)e^{-\int_a^s\beta(\sigma){\rm d}\sigma}{\rm d}s \geq y(t)-y(a) = y(t) = e^{-\int_a^t\beta(\sigma){\rm d}\sigma}\int_a^t\beta(s)u(s){\rm d}s\ ,$$ and then $$\int_a^t\beta(s)u(s){\rm d}s \leq e^{\int_a^t\beta(\sigma){\rm d}\sigma}\int_a^t\alpha(s)\beta(s)e^{-\int_a^s\beta(\sigma){\rm d}\sigma}{\rm d}s = \int_a^t\alpha(s)\beta(s)e^{\int_s^t\beta(\sigma){\rm d}\sigma}{\rm d}s\ .$$ At last Gronwall inequality follows from $u(t)-\alpha(t) \leq \int_a^t\beta(s)u(s){\rm d}s$ .