ODE0.1

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State and prove Gronwall's lemma.

Lemma: Let r(t)\, be continuous for |t-t_0|\le\delta\, and satisfy the inequalities

0\le r(t) \le \epsilon + \delta \Big| \int_{t_0}^t r(s) ds \Big| \,

for some non-negative constants \epsilon\, and \delta\,. Then

0\le r(t)\le \epsilon \exp\left\{ \delta |t-t_0| \right\} \,

Proof:

It is sufficient to consider the case t\ge t_0\,. The proof for the other case is the same. Let

R(t) = \int_{t_0}^t r(s) ds\,

Then R(t)\, is continuous and differentiable, and R'=r\,. Hence the inequality takes the form

0\le R'\le\epsilon+\delta R\,

Now multiply both sides by the non-negative function \exp\left\{ -\delta(t-t_0)\right\}\,, so that

0\le\left[ R \exp\left\{-\delta(t-t_0)\right\}\right]' \le \epsilon \exp\left\{-\delta(t-t_0)\right\}

Now integrate the inequality from t_0\, to t\,:

0\le R\exp\left\{-\delta(t-t_0)\right\} \le \frac{\epsilon}{\delta} \left[1 - \exp\left\{-\delta(t-t_0)\right\}\right]\,

or

0\le R \le \frac{\epsilon}{\delta}\left[\exp\left\{\delta(t-t_0)\right\}-1\right]\,.

But,

0\le r = R' \le \epsilon + \delta R\,

or

0\le r \le \epsilon \exp\left\{\delta(t-t_0)\right\}\,

which completes the proof.


Main Page : Ordinary Differential Equations

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