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!

  • $\begingroup$ The left inequality is wrong. Take $f=0$. $\endgroup$ – Hans Oct 12 '13 at 4:42
  • $\begingroup$ If you take $f=0$ then $u(t_0) \le u(t) \le u(t_0)$ . Hence, $u(t)=u(t_0)$. And you are incorrect. :( $\endgroup$ – kimtahe6 Oct 12 '13 at 7:02
  • $\begingroup$ Does $u(t) = u(t_0)$ not contradict your assumption, if $u$ is a strictly decreasing function, which also satisfy the first inequality condition? $\endgroup$ – Hans Oct 12 '13 at 15:21
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    $\begingroup$ First, your tone is bit rude while it is you who is seeking help from others. Second, you don't even point out it was Theorem 8 you are referring when asking others to look up a reference. Third, can you not think about whether a statement makes sense or explain why you think it does? Regarding the specific problem, a strictly decreasing function $u$ satisfies the first condition for some $a$, namely $a=0$. Obviously, the first inequality in the conclusion is not satisfied. Either this "theorem" is wrong or there is a missing condition. If you think I am wrong, tell me where. $\endgroup$ – Hans Oct 12 '13 at 17:54
  • 1
    $\begingroup$ Books may have typos too. Books are not bible. Reasoning is not a religion. $\endgroup$ – Hans Oct 12 '13 at 17:58

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