# The Unapologetic Mathematician

## Gronwall’s Inequality

We’re going to need another analytic lemma, this one called “Gronwall’s inequality”. If $v:[0,\alpha]\to\mathbb{R}$ is a continuous, nonnegative function, and if $C$ and $K$ are nonnegative constants such that

$\displaystyle v(t)\leq C+\int\limits_0^tKv(s)\,ds$

for all $t\in[0,\alpha]$ then for all $t$ in this interval we have

$\displaystyle v(t)\leq Ce^{Kt}$

That is, we can conclude that $v$ grows no faster than an exponential function. Exponential growth may seem fast, but at least it doesn’t blow up to an infinite singularity in finite time, no matter what Kurzweil seems to think.

Anyway, first let’s deal with strictly positive $C$. If we define

$\displaystyle V(t)=C+\int\limits_0^tKv(s)\,ds>0$

then by assumption we have $v(t)\leq V(t)$. Differentiating, we find $V'(t)=Kv(t)$, and thus

$\displaystyle\frac{d}{dt}\left(\log(V(t))\right)=\frac{V'(t)}{V(t)}=\frac{Kv(t)}{V(t)}\leq K$

Integrating, we find

$\displaystyle\log(V(t))\leq\log(V(0))+Kt=\log(C)+Kt$

Finally we can exponentiate to find

$\displaystyle v(t)\leq V(t)\leq Ce^{Kt}$

proving Gronwall’s inequality.

If $C=0$, in our hypothesis, the hypothesis is true for any $\bar{C}>0$ in its place, and so we see that $v(t)\leq\bar{C}e^{Kt}$ for any positive $\bar{C}$, which means that $v(t)$ must be zero, as required by Gronwall’s inequality in this case.

May 11, 2011 - Posted by | Analysis, Differential Equations

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