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Math Help - Ode

  1. #1
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    Ode

    Hi all, I have an ODE problem.

    Suppose the function f(t,x) is locally Lipschitz on the domain G in R^2, that is, |f(t,x_1)-f(t,x_2)| <= k(t) |x_1 - x_2| for all (t, x_1),(t,x_2) in G. Define I = (a,b) and phi_1(t) and phi_2(t) are 2 continuous functions on I. Assume that, if (t, phi_i(t)) is in G, then the function f(t, phi_i(t)) is an integrable function on I for i = 1, 2. Suppose that for i = 1,2 and t in I,

    phi_i(t) = phi_i(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_i(t)) }ds + E_i(t)

    and (phi_1(t_0) - phi_2(t_0)| <= d

    for some constant d. Show that for all t in (t_0, b) we have

    |phi_1(t) - phi_2(t)| <= d exp(INTEGRATE(from t_0 to t){ k(s) }ds) + E(t) + INTEGRATE(from t_0 to t){ E(s)k(s)exp[INTEGRATE(from t_0 to s) { k(r) } dr] }ds

    where E(t) = |E_1(t)| + |E_2(t)|

    I managed to get d exp(INTEGRATE(from t_0 to t) k(s) ds) using triangle inequality and Gronwall's inequality, but I cannot seem to get the last 2 terms in the inequality.

    Here's what I did:

    |phi_1(t) - phi_2(t)|
    = |phi_1(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds + E_1(t) - phi_2(t_0) - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds - E_2(t)|
    = |phi_1(t_0) - phi_2(t_0) + INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds + E_1(t) - E_2(t)|
    <= |phi_1(t_0) - phi_2(t_0)| + |INTEGRATE (from t_0 to t){ f(s, phi_1(s)) }ds - INTEGRATE (from t_0 to t){ f(s, phi_2(s)) }ds + E_1(t) - E_2(t)|
    <= d + |E_1(t) - E_2(t)| + |INTEGRATE (from t_0 to t){ f(s, phi_1(s)) - f(s, phi_2(s)) }ds|
    <= d + |E_1(t) - E_2(t)| + INTEGRATE (from t_0 to t){ |f(s, phi_1(s)) - f(s, phi_2(s))| }ds
    <= d + |E_1(t) - E_2(t)| + INTEGRATE (from t_0 to t){ k(s)|phi_1(s) - phi_2(s)| } ds
    <= (d + |E_1(t) - E_2(t)|) exp(INTEGRATE (from t_0 to t){ k(s) } ds)
    = d exp(INTEGRATE (from t_0 to t){ k(s) } ds) + (|E_1(t) - E_2(t)|) exp(INTEGRATE (from t_0 to t){ k(s) } ds)

    This is where I got stuck.

    Please help.

    Thank you.

    Regards,
    Rayne
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  2. #2
    Super Member Rebesques's Avatar
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    You already used Gronwall's inequality, so now it's just algebraic manipulations.

    You already showed that |\phi_1(s) - \phi_2(s)|\leq (d+E(s)) {\rm e}^{\int  k(r) dr} (1)
    Keep this in mind. Along your calculations, re-consider the relation

    |\phi_1 - \phi_2|\leq d+E(t)+\int k(s)|\phi_1(s) - \phi_2(s)| ds

    which is valid in this own merit. Substitute (1) into this to get:

    |\phi_1(t) - \phi_2(t)|\leq d+E(t) +\int k(s)\left\{(d+E(s)){\rm e}^{\int k(r)dr}\right\} ds

    or

    |\phi_1(t) - \phi_2(t)|=\left(d+d\int k(s){\rm e}^{\int k(r)dr}ds\right)+E(t)+\int k(s)E(s){\rm e}^{\int k(r)dr} (2)


    Now since k\geq 0, we have d+d\int k(s){\rm e}^{\int k(r)dr}ds\leq d+d\int k(s)ds\leq d{\rm e}^{\int k(s)ds}, and so

    |\phi_1(t) - \phi_2(t)|\leq d{\rm e}^{\int k(s)ds}+E(t)+\int k(s)E(s){\rm e}^{\int k(r)dr} .
    Last edited by Rebesques; September 5th 2007 at 04:58 AM. Reason: msn sucks
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  3. #3
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    Quote Originally Posted by Rebesques View Post
    You already used Gronwall's inequality, so now it's just algebraic manipulations.

    You already showed that |\phi_1(s) - \phi_2(s)|\leq (d+E(s)) {\rm e}^{\int  k(r) dr} (1)
    Keep this in mind. Along your calculations, re-consider the relation

    |\phi_1 - \phi_2|\leq d+E(t)+\int k(s)|\phi_1(s) - \phi_2(s)| ds

    which is valid in this own merit. Substitute (1) into this to get:

    |\phi_1(t) - \phi_2(t)|\leq d+E(t) +\int k(s)\left\{(d+E(s)){\rm e}^{\int k(r)dr}\right\} ds

    or

    |\phi_1(t) - \phi_2(t)|=\left(d+d\int k(s){\rm e}^{\int k(r)dr}ds\right)+E(t)+\int k(s)E(s){\rm e}^{\int k(r)dr} (2)


    Now since k\geq 0, we have d+d\int k(s){\rm e}^{\int k(r)dr}ds\leq d+d\int k(s)ds\leq d{\rm e}^{\int k(s)ds}, and so

    |\phi_1(t) - \phi_2(t)|\leq d{\rm e}^{\int k(s)ds}+E(t)+\int k(s)E(s){\rm e}^{\int k(r)dr} .
    Thank you for replying.

    I don't understand this line:

    Now since k\geq 0, we have d+d\int k(s){\rm e}^{\int k(r)dr}ds\leq d+d\int k(s)ds\leq d{\rm e}^{\int k(s)ds}

    Since k\geq 0, isn't {\rm e}^{\int k(r)dr} \geq 1? So shouldn't d+d\int k(s){\rm e}^{\int k(r)dr}ds\geq d+d\int k(s)ds?
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  4. #4
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    Anyone, please?
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