I'm trying to prove the theorem general of the Bellman-Gronwall's inequality:
Assume that $u(t)$ be real valued non - negative continuous function, and such that $$u(t)\le u(\tau )+\int_{\tau}^{t}f(t_1)u(t_1)\mathrm{d}t_1, \forall a< \tau <t <b$$ where $f(t)\in C(a,b), f(t)\ge 0$
Then $$u(t_0)\exp \left ( -\int_{t_0}^{t}f(t_1)\mathrm{d}t_1 \right ) \le u(t) \le u(t_0)\exp \left ( \int_{t_0}^{t}f(t_1)\mathrm{d}t_1 \right ), \forall a<t_0 \le t<b$$
- 1/ I'll show that $\bf u(t) \le u(t_0)\exp \left ( \int_{t_0}^{t}f(t_1)\mathrm{d}t_1 \right ) \tag 1$
We use $G-B$:
Let $u(t)$ and $f(t)$ be non - negative continuous functions on $\Bbb R^+$ for which the inequality $$u(t) \le C+ \int_{t_0}^{t}u(s)f(s)\mathrm{d}s$$ holds, where $C$ is a non - negative constant.
Then $$u(t) \le C \cdot \exp \left ( \int_{t_0}^{t}f(s)\mathrm{d}s \right )$$
where $C=u(t_0)$ and $\tau=t_0$. It's obvious. And we're done.
- 2/ I'll show that $$u(t_0)\exp \left ( -\int_{t_0}^{t}f(t_1)\mathrm{d}t_1 \right ) \le u(t) \tag 2$$
But I have stuck when I try to show that $(2)$. Can anyone have a solution?
Any help is always appreciated! Thanks!