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Math Help - Proof of Gronwall's Inequality

  1. #1
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    Proof of Gronwall's Inequality

    I am trying to understand the proof to Gronwall's Inequality, but I have one major issue I was hoping someone could get me through. My issue actually makes the proof incorrect.

    First, the statement of Gronwall's Inequality
    Let

    \begin{matrix}u:[a,b]\rightarrow [0,\infty) \\v:[a,b]\rightarrow [0,\infty)\end{matrix}
    be continuous functions and let C be a constant. Then if

    v(t) \le C + \int_a^t\! v(s) u(s)\, ds.
    For t\in [a,b], it follows that

    v(t) \le C \exp\left(\int_a^t\! u(s)\, ds\right)
    for t\in [a,b].

    Proof:
    \dfrac{v(t)}{C + \int_a^t\! v(s) u(s)\, ds} \le 1

    How do we know C + \int_a^t\! v(s) u(s)\, ds > 0?

    \dfrac{v(t) u(t)}{C + \int_a^t\! v(s) u(s)\, ds} \le u(t)
    \dfrac{d}{dt}\ln\left(C + \int_a^t\! v(s) u(s)\, ds\right) \le u(t)
    \ln\left(C + \int_a^t\! v(s) u(s)\, ds\right) -\ln C \le \int_a^t\! u(s)\, ds
    C + \int_a^t\! v(s) u(s)\, ds \le C exp\left(\int_a^t\! u(s)\, ds\right)
    How do we know C > 0?

    EDIT: I guess both my questions boil down to how do we know C > 0? Since, we know \int_a^t\! u(s) v(s)\, ds \ge 0 by \begin{matrix}u:[a,b]\rightarrow [0,\infty) \\v:[a,b]\rightarrow [0,\infty)\end{matrix}

    EDIT2: v(t) \le C + \int_a^t\! v(s) u(s)\, ds \Rightarrow C + \int_a^t\! v(s) u(s)\, ds \ge 0. However, I still don't see how we know C > 0.
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  2. #2
    A Plied Mathematician
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    Plug in u = 0. The result C > 0 pops right out.
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  3. #3
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    I knew it was going to be that simple. Thank you.
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  4. #4
    A Plied Mathematician
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    You're welcome!
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