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Thread: inequation

  1. #1
    Senior Member
    Joined
    Nov 2008
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    461

    inequation

    Hi!
    -------------

    $\displaystyle

    I = (t_0, t_0 + T) \subset \mathbb{R}, T > 0

    $


    $\displaystyle u:I \to \mathbb{R} $

    $\displaystyle u \in C^0(\overline{I}) $

    $\displaystyle f,g:I \to \mathbb{R} $

    $\displaystyle f,g \in C^0(\overline{I}) , f(t) \ge 0 \ \forall t \in I $



    Show that:

    if (1)::: $\displaystyle u(t) - u_0 \le \int^t_{t_0} ((f(s)-u(s)+g(s))ds \forall t \in I$ and $\displaystyle u (t_0) = u_0 $

    => $\displaystyle u(t) \le exp(\int^t_{t_0} f(x) dx) [u_0 + \int_{t_0}^t g(s)*exp(- \int^s_{t_0} f(x) dx)ds]$



    Hint: $\displaystyle \epsilon > 0$

    Define
    $\displaystyle z_{\epsilon} (t) = exp(\int^t_{t_0} f(x) dx) [u_0 + \epsilon + \int^t_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds] $

    Show that
    $\displaystyle u(t) < z_{\epsilon} (t) \forall t \in I$

    by assuming that $\displaystyle \exists t_1 \in I : t_1 > t_0$ and $\displaystyle u(t_1) = z_{eps}(t_1)$
    Therefor use (1) and
    $\displaystyle \frac{d}{dt} z_{\epsilon}(t), z_{\epsilon} (t_0)$ and
    $\displaystyle u(t) < z_{\epsilon}(t) \forall t \in [t_0,t_1)$


    -------------------------

    I tried:

    $\displaystyle z_{\epsilon}(t_0) = u_0 + \epsilon $

    $\displaystyle \exists t_1 \in I : t_1 > t_0$ and $\displaystyle u(t_1) = z_{eps}(t_1)$

    $\displaystyle u(t_1) \le \int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0$

    multiplied by 1

    $\displaystyle = [exp(\int^{t_1}_{t_0} f(x) dx)][-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0]$

    $\displaystyle u(t_1) = z_{\epsilon}(t_1)$

    $\displaystyle \Rightarrow [exp(\int^{t_1}_{t_0} f(x) dx)][-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0]$

    $\displaystyle = exp(\int^{t_1}_{t_0} f(x) dx) [u_0 + \epsilon + \int^{t_1}_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds] $

    Divide by $\displaystyle [exp(\int^{t_1}_{t_0} f(x) dx)]$

    $\displaystyle [-exp(\int^{t_1}_{t_0} f(x) dx)] * [\int^{t_1}_{t_0} [ f(s)u(s) + g(s) ] ds + u_0] $

    $\displaystyle = u_0 + \epsilon + \int^{t_1}_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx) ds]$

    I dont have any clue what to do now.

    Im not sure about this, too

    $\displaystyle \frac{d}{dt} z_{\epsilon}(t) $

    $\displaystyle = exp(f(t))*(u_0 + \epsilon + \int^t_{t_0} g(s) exp(-\int^s_{t_0} f(x) dx)ds) $
    $\displaystyle + exp (\int^t_{t_0} f(x) dx) [g(t) *exp(-\int^s_{t_0}f(x) dx)] $

    any help would be really apprectiated

    thanks!
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  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    461

    I want to show this

    $\displaystyle
    u(t) \le exp(\int^t_{t_0} f(x) dx) [u_0 + \int_{t_0}^t g(s)*exp(- \int^s_{t_0} f(x) dx)ds]
    $

    Anyone?

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