Results 1 to 2 of 2

Thread: ODE's help...

  1. #1
    hernik
    Guest

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    There's the straightforward way: Define the Picard iterations (following the proof of the $\displaystyle Y'=F(t,Y), \ Y(t_0)=Y_0$ 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 $\displaystyle y'(t)=\omega(t), \ Y(t)=\left(\begin{array}{c}
    y(t) \\
    \omega(t) \\
    \end{array}\right), \ F(t,Y)=\left(\begin{array}{c}
    \omega(t) \\
    f(t,pr_1Y) \\
    \end{array}\right)$,

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

    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 $\displaystyle F$ is Lipschitz for $\displaystyle Y$. For the usual $\displaystyle \ell^1$ norm, we have

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

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


    Question: How does $\displaystyle \sigma$ come into play here? What does it signify?
    Last edited by Rebesques; Aug 17th 2007 at 03:20 AM. Reason: phone call
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum