most recent 30 from http://mathoverflow.net 2013-05-25T23:25:23Z http://mathoverflow.net/feeds/question/111476 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111476/a-gronwall-type-inequality Andrew 2012-11-04T16:51:23Z 2012-11-04T19:59:06Z

<p>I want to derive a Gronwall-type inequality from the inequality below. Here all the functions are nonnegative, continuous and if you need some assumptions you may use that. $$ f^2(t) \leqslant g^2(t) + \int_0^t (f(s) +c) f(s) ds \;\;\;\; (t \in [0,T]) $$ So please help!</p>

http://mathoverflow.net/questions/111476/a-gronwall-type-inequality/111478#111478 Michael Renardy 2012-11-04T16:59:28Z 2012-11-04T16:59:28Z

<p>Consider g=0, c=2, f(t)=t.</p> <p>By the way, what did you mean by "if you need some assumptions?" Are there folks out there who prove things without them?</p>

http://mathoverflow.net/questions/111476/a-gronwall-type-inequality/111482#111482 Pietro Majer 2012-11-04T19:59:06Z 2012-11-04T19:59:06Z

<p>Consider the function $$h(t):=\int_0^t f(s)e^{t-s}ds\ ,$$ which solves the ODE $h'=h+f$ with $h(0)=0$, so $$h(t):=\int_0^t \Big(h(s)+f(s)\Big)ds\ .$$ Adding the term $-c\ h(t)$ to both sides, your inequality takes a more familiar form of a Gronwall inequality: $$f(t)^2-c\;h(t)\le g(t)^2 + \int_0^t \Big(f(s)^2 -c\;h(s) \Big)ds $$ relative to the function $f(t)^2-c\; h(t)$.</p>