# Uniqueness of ODEs

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...

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.

• 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. – Robert Israel Sep 29 '14 at 2:24

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 $x_0/2 + C = 1$.