# ODE0.1

State and prove Gronwall's lemma.

Lemma: Let $r(t)\,$ be continuous for $|t-t_{0}|\leq \delta \,$ and satisfy the inequalities

$0\leq r(t)\leq \epsilon +\delta {\Big |}\int _{{t_{0}}}^{t}r(s)ds{\Big |}\,$

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

$0\leq r(t)\leq \epsilon \exp \left\{\delta |t-t_{0}|\right\}\,$

Proof:

It is sufficient to consider the case $t\geq t_{0}\,$. 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

$0\leq R'\leq \epsilon +\delta R\,$

Now multiply both sides by the non-negative function $\exp \left\{-\delta (t-t_{0})\right\}\,$, so that

$0\leq \left[R\exp \left\{-\delta (t-t_{0})\right\}\right]'\leq \epsilon \exp \left\{-\delta (t-t_{0})\right\}$

Now integrate the inequality from $t_{0}\,$ to $t\,$:

$0\leq R\exp \left\{-\delta (t-t_{0})\right\}\leq {\frac {\epsilon }{\delta }}\left[1-\exp \left\{-\delta (t-t_{0})\right\}\right]\,$

or

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

But,

$0\leq r=R'\leq \epsilon +\delta R\,$

or

$0\leq r\leq \epsilon \exp \left\{\delta (t-t_{0})\right\}\,$

which completes the proof.