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Im looking to find 4 solutions to the ODE :

$\frac{dx}{dt} = |x|^{1/2} , x(0)=0$.

Clearly, $x=0$ is one solution. Using seperation of variables for $x>0$ yields $x= t^2/4$ as another solution, and if we consider $x<0$, I find that $x = -t^2/4$. Could someone give a hint as to where I am missing the last solution?

Thanks!

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    $\begingroup$ There are infinitely many solutions. Can you guess how to obtain them from what you have? $\endgroup$ – Daniel Fischer Oct 16 '13 at 20:09
  • $\begingroup$ my text states there exist 4 different solutions through (0,0)..perhaps a linear combination of x>0 and x<0 is what you mean? $\endgroup$ – still_learning Oct 16 '13 at 20:11
  • $\begingroup$ Does it explicitly say "exactly four"? I doubt it. Not "linear combination", but some form of combination. $\endgroup$ – Daniel Fischer Oct 16 '13 at 20:13
  • $\begingroup$ ahhhhhhhh I finally understand what you mean. the piece-wise functions stay the same- we can just alter the intervals on which theyre defined! $\endgroup$ – still_learning Oct 16 '13 at 20:40
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    $\begingroup$ Yes, you can glue solutions together, since $\lvert x\rvert^{1/2}$ isn't Lipschitz-continuous at $0$. All $$f_{a,b}(t) = \begin{cases} -\frac14(a-t)^2 &, t \leqslant a\\ \qquad 0 &, a \leqslant t \leqslant b\\ \hphantom{-}\frac14(t-b)^2 &, b \leqslant t \end{cases}$$ are solutions ($-\infty \leqslant a \leqslant 0 \leqslant b \leqslant \infty$). $\endgroup$ – Daniel Fischer Oct 16 '13 at 20:45
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I claim that a solution (the 'basic' solution) is

$$ x = \begin{cases} t^2/4 &: t>0 \\ 0&: t \leq 0 \end{cases} $$

Verify that this not only satisfies the DE and IC, but it is continuous everywhere, with a continuous derivative. (you only have to check the continuity and differentiability at $x=0$ since it is obviously smooth everywhere else).

Now think about what happens if you translate this function to the right.

EDIT additional thoughts

Another way to go about this that avoids translation would be to show that you can take $x$ to be either $t^2/4$ or $0$ on $[0,\infty)$ and either $-t^2/4$ or $0$ on $(-\infty,0)$, and that no matter which of these four options you choose, the result satisfies the DE, IC, and is continuous w/ continuous derivative.

I.e. $$ x = \begin{cases} t^2/4 &: t>0 \\ -t^2/4 &: t \leq 0 \end{cases} $$

$$ x = \begin{cases} t^2/4 &: t>0 \\ 0 &: t \leq 0 \end{cases} $$ etc. are all solutions.

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