0
$\begingroup$

I'm not quite positive where to even begin with this problem. Can somebody help me interpret what it is I need to do and/or how to find the correct solution?

By the Picard-Lindelöf existence and uniqueness theorem (also known as Picard's existence theorem or Cauchy–Lipschitz theorem), we can say that $x^\prime = x^{2}$, $x(0) = x_{0}$ has a unique solution on an interval $[-b/M , b/M]$ where $$ \sup_{x \epsilon B_{b} (x_0 )} \parallel f(x) \parallel $$. For $x_0 > 0$, show that $M = (x_0 + b)^2$ and then find a value of $b$ to maximize this interval.

Thanks!

$\endgroup$
2
$\begingroup$

The first thing you want to do is explicitly solve this ODE. I will write $y' = y^2$ and $y(0) = x_0$, using the variable $y$ for the sake of convention. The solution to this is

$$y(x) = \frac{1}{c - x}$$

when $x=0$, you have $1/c$, so set $c = 1/x_0$. You can see that this will not have a solution that exists for all time, since this function blows up at $x=c = 1/x_0$. Thus you will want to assert that the solution exists for $|x| < c$. Now try and rewrite this condition in the form you are asked.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.