# proof of Gronwall’s lemma

The inequality

 $\phi(t)\leq K+L\int_{t_{0}}^{t}\psi(s)\phi(s)ds$ (1)

is equivalent to

 $\frac{\phi(t)}{K+L\int_{t_{0}}^{t}\psi(s)\phi(s)ds}\leq 1$

Multiply by $L\psi(t)$ and integrate, giving

 $\int_{t_{0}}^{t}\frac{L\psi(s)\phi(s)ds}{K+L\int_{t_{0}}^{s}\psi(\tau)\phi(% \tau)d\tau}\leq L\int_{t_{0}}^{t}\psi(s)ds$

Thus

 $\ln\left(K+L\int_{t_{0}}^{t}\psi(s)\phi(s)ds\right)-\ln K\leq L\int_{t_{0}}^{t% }\psi(s)ds$

and finally

 $K+L\int_{t_{0}}^{t}\psi(s)\phi(s)ds\leq K\exp\left(L\int_{t_{0}}^{t}\psi(s)ds\right)$

Using (1) in the left hand side of this inequality gives the result.

Title proof of Gronwall’s lemma ProofOfGronwallsLemma 2013-03-22 13:22:23 2013-03-22 13:22:23 jarino (552) jarino (552) 5 jarino (552) Proof msc 26D10