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Homework Help: Lipschitz ODE problem

  1. Sep 4, 2007 #1
    1. The problem statement, all variables and given/known data
    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)|

    3. The attempt at a solution

    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
     
  2. jcsd
  3. Sep 6, 2007 #2

    EnumaElish

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    It would be much easier on the eye if you can re-type your post using [ t e x ] ... [ / t e x ] (single spaced).

    If you have not used it before, click on the image to get a very basic starter course: [tex]\TeX[/tex]
     
  4. Sep 6, 2007 #3
    I rewrote the question in tex.

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

    [tex]\phi_i(t) = \phi_i(t_0) + \int^t_{t_0} f(s, \phi_i(s))\,ds + E_i(t)[/tex]

    and [tex]|\phi_1(t_0) - \phi_2(t_0)| \leq \delta[/tex]

    for some constant [tex]\delta[/tex]. Show that for all [tex]t \in (t_0, b)[/tex] we have

    [tex]|\phi_1(t) - \phi_2(t)| \leq \delta e^{\int^t_{t_0} k(s) \,ds} + E(t) + \int^t_{t_0} E(s) k(s) e^{\int^s_{t_0} k(r) \,dr} \,ds[/tex]

    where [tex]E(t) = |E_1(t)| + |E_2(t)|[/tex]

    I managed to get [tex]\delta e^{\int^t_{t_0} k(s) \,ds}[/tex] 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:

    [tex]|\phi_1(t) - \phi_2(t)|[/tex]
    [tex]
    = |\phi_1(t_0) + \int^t_{t_0} f(s, \phi_1(s)) \,ds + E_1(t) - \phi_2(t_0) - \int^t_{t_0} f(s, \phi_2(s)) \,ds - E_2(t)|[/tex]
    [tex]
    = |\phi_1(t_0) - \phi_2(t_0) + \int^t_{t_0} f(s, \phi_1(s)) \,ds - \int^t_{t_0} f(s, \phi_2(s)) \,ds + E_1(t) - E_2(t)|[/tex]
    [tex]
    \leq |\phi_1(t_0) - \phi_2(t_0)| + |\int^t_{t_0} f(s, \phi_1(s)) }ds - \int^t_{t_0} f(s, \phi_2(s)) \,ds + E_1(t) - E_2(t)|[/tex]
    [tex]
    \leq \delta + |E_1(t) - E_2(t)| + |\int^t_{t_0} f(s, \phi_1(s)) - f(s, \phi_2(s)) \,ds|
    [/tex]
    [tex]
    \leq \delta + |E_1(t) + E_2(t)| + \int^t_{t_0} |f(s, \phi_1(s)) - f(s, \phi_2(s))| \,ds
    [/tex]
    [tex]\leq \delta + |E_1(t)| + |E_2(t)| + \int^t_{t_0} k(s)|\phi_1(s) - \phi_2(s)| \, ds[/tex]
    [tex]\leq (\delta + E(t)) e^{\int^t_{t_0} k(s)\, ds}[/tex]
    [tex]
    = \delta e^{\int^t_{t_0} k(s) \, ds} + E(t) e^{\int^t_{t_0} k(s) \, ds[/tex]

    This is where I got stuck.
     
  5. Sep 8, 2007 #4

    EnumaElish

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    How do you go from
    [tex]\delta + |E_1(t)| + |E_2(t)| + \int^t_{t_0} k(s)|\phi_1(s) - \phi_2(s)| \, ds[/tex]
    to the next step?
     
  6. Sep 8, 2007 #5
    [tex]|E_1(t)| + |E_2(t)| = E(t)[/tex]

    Taking Gronwall's inequality, that is,
    [tex]\phi(t) \leq a \int^t_{t_0} \psi(s) \phi(s) \, ds + M,\, \, \, t_0 \leq t \leq t_0 + T[/tex]

    gives for [tex]t_0 \leq t \leq t_0 + T[/tex]

    [tex]\phi(t) \leq M e^{a \int^t_{t_0} \psi(s) \, ds}[/tex]

    Therefore, taking [tex]\delta + E(t)[/tex] as M, [tex]|\phi_1(s) - \phi_2(s)|[/tex] as [tex]\phi(t)[/tex], and [tex]k(s)[/tex] as [tex]\psi(s)[/tex], I get

    [tex](\delta + E(t)) e^{\int^t_{t_0} k(s)\, ds}[/tex]
     
  7. Sep 9, 2007 #6

    EnumaElish

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    Is [itex]a \int^t_{t_0} \psi(s) \phi(s) \, ds + M[/itex] equal to [itex]M e^{a \int^t_{t_0} \psi(s) \, ds[/itex] ? My basics is a little rusty.
     
    Last edited: Sep 9, 2007
  8. Sep 9, 2007 #7
    Yes, in the Gronwall's inequality.
     
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