About the Gronwall inequality

Question

If I have that $$||\eta_u(t)||\leq 1+C_1\int_0^t \frac{1}{||\eta(s)||}||\eta_u(s)||ds$$ and $$\sqrt{1-\frac{2\varepsilon}{C}}||u||\leq ||\eta(s)||\leq 2||u||$$ how to obtain using the Gronwall inequality that $$\displaystyle ||\eta_u(t)||\leq 1+\exp\left(\frac{2C_1}{||u||}t\right)$$

Thank you.

Answer 1

From your norm estimation you have $$ \frac{1}{||\eta(s)||} \leq \sqrt{\frac{C}{C-2\varepsilon}} \frac{1}{||u||} \leq 2\frac{1}{||u||}, $$ for $\varepsilon < \frac{3}{8}C$. Therefore, $$ ||\eta_u(t)||\leq 1+C_1\int_0^t \frac{1}{||\eta(s)||}||\eta_u(s)|| \,ds \leq 1+\frac{2 C_1}{||u||}\int_0^t ||\eta_u(s)|| \,ds $$ Using the Gronwall inequality in integral form (and noting that $1$ is non-decreasing function:) we arrive to $$ ||\eta_u(t)||\leq \exp^{\frac{2C_1}{||u||}t} \leq 1 + \exp^{\frac{2C_1}{||u||}t}. $$