Gronwall’s Inequality
We’re going to need another analytic lemma, this one called “Gronwall’s inequality”. If ![v:[0,\alpha]\to\mathbb{R}](https://s0.wp.com/latex.php?latex=v%3A%5B0%2C%5Calpha%5D%5Cto%5Cmathbb%7BR%7D&bg=e6e6e6&fg=333333&s=0 "v:[0,\alpha]\to\mathbb{R}")
is a continuous, nonnegative function, and if 
and 
are nonnegative constants such that
for all ![t\in[0,\alpha]](https://s0.wp.com/latex.php?latex=t%5Cin%5B0%2C%5Calpha%5D&bg=e6e6e6&fg=333333&s=0 "t\in[0,\alpha]")
then for all 
in this interval we have
That is, we can conclude that 
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 . If we define
then by assumption we have 
. Differentiating, we find 
, and thus
Integrating, we find
Finally we can exponentiate to find
proving Gronwall’s inequality.
If 
, in our hypothesis, the hypothesis is true for any 
in its place, and so we see that 
for any positive 
, which means that 
must be zero, as required by Gronwall’s inequality in this case.
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May 11, 2011 -
Posted by John Armstrong |
Analysis, Differential Equations
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