Gronwall inequality - Knowlpedia

Statement

Let $u\ge0$ be continuous on $[a,b]$ , let $k\ge0$ be integrable there, and let $A\ge0$ . If

u(t)\le A+\int_a^t k(s)u(s)\,ds,

then Gronwall's inequality gives

u(t)\le A\exp\left(\int_a^t k(s)\,ds\right).

Proof

For $A>0$ , let $v(t)=A+\int_a^tku$ . Then $v'\le kv$ almost everywhere. The derivative of $v(t)e^{-\int_a^tk}$ is nonpositive, so $v(t)\le Ae^{\int_a^tk}$ . If $A=0$ , apply the same argument with any positive $A$ and let it decrease to zero.

Source term

If an absolutely continuous $u$ satisfies $u'\le ku+g$ almost everywhere, with integrable $k,g$ , an integrating factor gives

u(t)\le u(a)e^{\int_a^t k}+\int_a^t e^{\int_s^t k}g(s)\,ds.

This version does not require $k\ge0$ . The coefficient must be integrable on the interval where the estimate is used.