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Thread: ODE's help...

  1. #1

    ODE's help...

    Hello, can i get some help in the next ODE?
    Consider the oblong O={(t,y):abs(t-t0)<=a,abs(y-y0)<=b} and if f is....(f satisfies the picard theorem anyway..) then show that the initial value problem: y''=f(t,y), y(t0)=y0, y'(t0)=y'0 has oly one solution in the space [t0-σ,t0+σ]...thanx
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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    There's the straightforward way: Define the Picard iterations (following the proof of the Y'=F(t,Y), \ Y(t_0)=Y_0 initial value problem), show a fixed point exists, and use the Gronwall lemma for uniqueness.

    However, who can be bothered to do it? So bypass all of that with a trick.

    Define y'(t)=\omega(t), \ Y(t)=\left(\begin{array}{c}<br />
y(t) \\<br />
  \omega(t) \\<br />
        \end{array}\right), \ F(t,Y)=\left(\begin{array}{c}<br />
\omega(t) \\<br />
  f(t,pr_1Y) \\<br />

    where pr_1 means projection of the first coordinate of the vector Y. What happened here is, that the original problem y'(t)=f(t,y) has been rewritten as Y'=F(t,Y).

    Big deal? I mean, we have increased the dimension, is it good? In general, no. But here, we have gained the right to use the (original) Picard theorem!

    So, we need to check whether F is Lipschitz for Y. For the usual \ell^1 norm, we have

    |F(t,Y_1)-F(t,Y_2)|=|f(t,pr_1Y_1)-f(t,pr_1Y_2)|=|f(t,y_1)-f(t,y_2)|\leq L|y_1-y_2|, where L is the Lipschitz constant for f.

    Now, by applying the Picard theorem with initial condition Y(t_0)=\left(\begin{array}{c}<br />
y(t_0) \\<br />
  \omega(t_0) \\<br />
        \end{array}\right)=\left(\begin{array}{c}<br />
y_0 \\<br />
  y'_0 \\<br />
        \end{array}\right), we obtain some \sigma>0 such that our problem has a unique solution Y(t)=\left(\begin{array}{c}<br />
y(t) \\<br />
  y'(t) \\<br />
        \end{array}\right) in the interval [t_0-\sigma,t_0+\sigma].

    Question: How does \sigma come into play here? What does it signify?
    Last edited by Rebesques; Aug 17th 2007 at 04:20 AM. Reason: phone call
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