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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\}\,


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]\,


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


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


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

which completes the proof.

Main Page : Ordinary Differential Equations