Thread: 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

There's the straightforward way: Define the Picard iterations (following the proof of the 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 ,

where means projection of the first coordinate of the vector . What happened here is, that the original problem has been rewritten as .

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 is Lipschitz for . For the usual norm, we have

, where is the Lipschitz constant for .

Now, by applying the Picard theorem with initial condition , we obtain some such that our problem has a unique solution in the interval .


Question: How does come into play here? What does it signify?