Variant of Grönwall's inequality

Question

I am wondering if there is a Grönwall-type inequality I can apply for an estimate of the form: $$u(t) \leq \alpha(t) + \left(\int_a^t |u(s)|^2ds\right)^{\frac{1}{2}}, \; t \in [a,b],$$ where $u: [a,b] \rightarrow \mathbb{R}$.

Thanks

Answer 1

Start with $$ u(t) \leq \alpha(t) + \left(\int_a^t |u(s)|^2ds\right)^{\frac{1}{2}} $$ as given, you can compute $$ |u(t)|^2 \leq \left[ \alpha(t) + (\int_a^t |u(s)|^2 ds)^\frac12 \right]^2 \leq 2 \alpha(t)^2 + 2 \int_a^t |u(s)|^2 ds $$ where we used the version of AM-GM inequality giving $$ (a + b)^2 = a^2 + 2ab + b^2 \leq 2 (a^2 + b^2). $$

Now you can apply Gronwall's inequality to the function $v(s) = |u(s)|^2$, which gives, for example $$ |u(t)|^2 \leq 2\alpha(t)^2 + \int_0^t 2\alpha(s)^2 e^{2(t-s)} ds $$ in general, or, if $\alpha$ is non-decreasing and non-negative, then $$ |u(t)|^2 \leq 2\alpha(t)^2 e^{2t}. $$