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Math Help - inequation

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    461

    inequation

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

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


    u:I \to \mathbb{R}

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

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

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



    Show that:

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

    => 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:  \epsilon > 0

    Define
     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
     u(t) < z_{\epsilon} (t) \forall t \in I

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


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

    I tried:

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

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

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

    multiplied by 1

    = [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]

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

    \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]

    = 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 [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]

    = 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

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

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

    <br />
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]<br />

    Anyone?

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