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I am looking at a very basic example and trying to understand uniqueness of ODEs.

The ODE in question is $y'(x)=\sqrt{y}$, which can easily be solved with separation of variables. After this, we have $\sqrt{y} = \dfrac{x}{2}+C$ for some constant $C$.

I am examining the "Fundamental Theorem of Uniqueness for ODEs". I noticed that $\dfrac{\partial}{\partial y} \left( \sqrt{y} \right) = \dfrac{1}{2\sqrt{y}}$, which is not continuous when $y=0$. This leads us to a nonunique solution, in particular, $y \equiv 0$ satisfies the IVP $y'(x) = \sqrt{y}$ where $y(0)=0$.

But other IVPs have nonunique solutions, correct? For example, if $y(0)=1$, then $C=1$ so $\sqrt{y} = \dfrac{x}{2} +1$.

That is, $y(x) = \dfrac{1}{4}(x+2)^2$ and $y(x) = \dfrac{1}{4}(x-2)^2$ are both solutions. Without the Fundamental Theorem, how can I tell that this second IVP $(y(0)=1)$ has a nonunique solution? In this case, directly solving was simple, but I'm certain that is not always the case...

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The fundamental theorem of uniqueness and existence does not have an if and only if condition.

What this means is that if the ODE is continuous at $(x_0,y_0)$ and $\left(\dfrac{\partial F}{\partial y}\right)$ is continuous at $(x_0,y_0)$, then the ODE has a unique solution but if either of those conditions isn't true, which in your case would be the second one, then the ODE may or may not have a solution and that solution may or may not be unique.

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    $\begingroup$ If $F$ and $\dfrac{\partial F}{\partial y}$ are continuous at $(x_0,y_0)$, the IVP with $y(x_0) = y_0$ has a unique solution in some interval around $x_0$. If the solution hits a point where $\dfrac{\partial F}{\partial y}$ is not continuous, it may become nonunique at that point. $\endgroup$ – Robert Israel Sep 29 '14 at 2:24
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First of all, $\sqrt{y} = x/2 + C$ is not a solution on all of $\mathbb R$, because $y$ can't be real when $x/2 + C < 0$. The actual solution is

$$ y = \cases{0 & for $x \le -2C$\cr (x/2+C)^2 & for $x > -2C$\cr}$$

These, together with $y = 0$, form all the solutions. Note that you can't take $y = (x/2 + C)^2$ for all $x$: it wouldn't satisfy the differential equation when $x < -2C$, because on that interval $\sqrt{(x/2+C)^2} = -(x/2 + C)$, not $x/2 + C$.

The IVP with $y(x_0) = 0$ has non-unique solutions, because you could take any of these solutions with $C \le -x_0/2$, or the solution $y = 0$.

In this case an IVP with $y(x_0) = 1$ does have a unique solution, namely the one with $x_0/2 + C = 1$.

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