1
$\begingroup$

Consider the ODE $y'=2\sqrt{|y|}$ where $y\in\Bbb R$. Find all solutions with initial conditions $y(0)=0$.

This is a homework question, so please just small hints...

The most obvious solution is $y(t)=0$. I was thinking that by Picard's Theorem, then this is, locally, the unique solution. But then I realized that $y(t)=t^2$ is another solution, if $t\ge0$ that satisfies the given initial condition. I have two questions:

  1. Why doesn't this contradict Picard's Theorem?
  2. Would I solve this ODE by, just, assuming $t\ge0$ then integrating both sides, then assuming $t\le0$ and integrating both sides? It seems that this technique should work since it looks like any solution which is defined on a neighborhood of 0 must vanish on some neighborhood of 0 (Picard's theorem seems to guarantee this, but please correct me if I am wrong).
$\endgroup$
4
$\begingroup$

1) It does not contradict the theorem because the function $f(y)=2\sqrt{\lvert y \rvert}$ is not locally Lipschitz. Otherwise, you would have $|f(y)-f(0)|=2\sqrt{\lvert y \rvert}\le L|y|$ for $|y|$ sufficiently small. But this is equivalent to $\lvert y \rvert\ge 4/L^2$ for $|y|$ sufficiently small.

2) What you describe is indeed the best approach. You should only check at the end that it gives you a $C^1$ function since solutions are supposed to be $C^1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Quick question. We solving an ODE, do we just wish to solve it on some open set (open in the functions domain) of the point in question (so in a neighborhood of 0 in this case)? For example, in the above example, $f(t) = t^2$ is a solution for $t \geq 0$. So this solution isn't defined on an open neighborhood of 0, but is defined on a relatively open neighborhood containing 0. $\endgroup$ – Jonathan Gafar Mar 3 '16 at 21:42
  • 1
    $\begingroup$ It really depends on what you want, but there is no reason why we shouldn't find a solution it in the largest possible domain (also, the domain only depends on the initial condition, even if the solution is not unique!). $\endgroup$ – John B Mar 3 '16 at 21:44
  • $\begingroup$ So a solution would be any function satisfying the differential equation at the initial condition? What the function does otherwise doesn't matter? For example, in the above example all we care about is that the function satisfies the equation and vanishes at 0? $\endgroup$ – Jonathan Gafar Mar 3 '16 at 21:47
  • 2
    $\begingroup$ Yes, that's the notion of solution, plus it should be $C^1$. $\endgroup$ – John B Mar 3 '16 at 21:49
  • $\begingroup$ Thank you for your patience - ODEs are completely new to me. $\endgroup$ – Jonathan Gafar Mar 3 '16 at 21:51

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.