Ordinary Differential Equations 12 | Picard-Lindelöf Theorem

[@brightsideofmaths]

Check the pdf version for the correct definition of the space X: tbsom.de/s/ode12pdf

[@iliTheFallen]

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[@brightsideofmaths]

Thank you very much :) I am always happy to get some support :)

[@StratosFair]

Nice ! Just curious, why did you set the domain X to be functions defined on (-epsilon,epsilon) instead of [0,epsilon) ? We are only looking at what happens for positive times, aren't we ? Also why take the interval to be open instead of closed ? If the interval was chosen closed in the first place, the functions would already be bounded since they are continuous.

[@brightsideofmaths]

We should also look what happens for negative times. Remember that we want to have the orbit around the point. So we want to know how it gets in and out again. The choice of the open domain is just convenience. There are many possibilities for writing down this proof.

[@StratosFair]

@@brightsideofmaths thank you !

[@josetoledo8181]

In the definition of X, I think you should restrict to the curves with image in the neighbourhood where v is lipschitz continuous. Otherwise you can not use the lipschitz condition to prove that phi is a contraction.

[@brightsideofmaths]

Yes, v is locally Lipschitz continuous at every point here.

[@josetoledo8181]

@@brightsideofmaths I'm sorry, but I don´t get it. Let v(x) = x^2, which is locally Lipschitz continuous at every point and set x0 = 0. Now fix any r > 0 and define on (-r, r): f(t) = 3at g(t) = 0 where a is just a constant. Note that both f, g are continuous and bounded real-valued functions on (-r, r) and f(0) = g(0) = x0 = 0. Now we have: (f - g)(t) = 3at [ phi(f) - phi(g) ](t) = integral(0,t) [9a^2*s^2] ds = 3a^2*t^3 so d(f,g) = 3a*r and d(phi(f), phi(g)) = 3a^2*r^3. But if we choose a > 1/r^2 then 3a^2*r^3 > 3a*r. Therefore phi is not a contraction in X, no matter how we choose r. v will be Lipschitz continuous in some neighbourhood of x0 and in the definition of X we must ensure the functions' images are contained in that neighbourhood. Otherwise, we cannot bound the difference in the proof.

[@brightsideofmaths]

I guess you are right that we have to restrict the codomain in the definition of X. What do you recommend for the change?

[@josetoledo8181]

​@@brightsideofmaths I think you could replace the "bounded condition" by the restriction of the codomain to some neighbourhood where v is bounded and satisfies a Lipschitz condition (e.g. a ball centered at x0). Assuming this is a complete metric space, then choose epsilon small enough so phi maps X into X, and the rest of the video remains the same. Thank you for doing all this work, I really appreciate it!

[@brightsideofmaths]

​@@josetoledo8181 Thank you for pointing out the mistake. I really overlooked that before. I will put an appendix into the pdf version and change the video later. You can check it here: tbsom.de/s/ode12pdf for the moment.

[@uysothea6720]

Dear sir help me for calculus differential equation y''=yf(x).