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This question already has an answer here:

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 $.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Wikipedia has a couple. $\endgroup$ – user228113 Aug 27 '16 at 10:16
  • $\begingroup$ Yes i should have looked before sorry guys... $\endgroup$ – mathnoob Aug 27 '16 at 10:20
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You should first look in the forum but anyway:

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

Btw you can find the proof in this forum at least twice...

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