# Determining whether an IVP has a unique solution if uniqueness theorem doesn't hold

I have the following initial value problems:

$$\begin{cases} y' = y^{\frac{1}{2}}\cos(x) \\[6pt] y(-2)= 1 \end{cases}$$

$$\begin{cases} y' = y^{\frac{1}{2}}\cos(x) \\[6pt] y(5)=-8 \end{cases}$$

$$\begin{cases} y' = y^{\frac{1}{2}} \cos(x) \\[6pt] y(0)= 0 \end{cases}$$

I want to determine if they have a unique solution.

I'm using the uniqueness theorem that says if $f(x, y)$ and $\partial f/\partial y$ are continuous on a rectangular region enclosing $(x_0, y_0)$, then there is a unique solution on an interval $I$.

$$\frac{\partial f}{\partial y} = \frac{1}{2} \cdot \frac{\cos(x)}{y^{\frac{1}{2}}}$$

So I know that I can only have unique solutions for $y > 0$. For the first IVP, it's easy to find a rectangular region containing $(-2, 1)$, so it must have a unique solution. But the next one lies below the line $y = 0$ and the last one on the line. The textbook says that if the hypothesis of the theorem doesn't hold, then it's inconclusive how many solutions there are.

My question is, how can I determine if there is a unique solution if I can't use the theorem? Should I try to solve the IVPs and just see what happens?

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The second IVP isn't even defined over real numbers, so it can't have a solution. // For the third, "solve and see what happens" is a good approach. – user53153 Feb 20 '13 at 19:59

If you find general solution of your ODE, you'll get that $$y_G = \frac 14 \left ( \sin x + C\right )^2$$ Taking into account IC $$y_0 = \frac 14 \left ( \sin x_0 + C\right )^2 \\ \sin x_0 + C = \pm 2 \sqrt {y_0}$$ So, for any given IC you have at least two values of $C$. But the value of $C$ that makes the whole expression negative ($\sin x_0 + C = -2\sqrt{y_0}$) will also change the sign of the function after taking square root $$y^{\frac 12} = -\frac 12({\sin x + C})$$ so ODE will look like $$y' = -y^{\frac 12} \cos x$$ So $y_0 \ge 0$ guarantees uniqueness of the solution.
Please reconsider your answer. The solution is unique when $y_0>0$. – user53153 Feb 20 '13 at 20:49
Alright. Let's check it. Consider two functions $$y_1 = \frac 14 \left [ \sin x - \sin (-2) + 2\right ]^2 \\ y_2 = \frac 14 \left [ \sin x - \sin (-2) - 2\right ]^2$$ Both of them satisfy IC $$y_1(-2) = \frac 14 \left [ \sin (-2) - \sin (-2) + 2\right ]^2 = 1 \\ y_2(-2) = \frac 14 \left [ \sin (-2) - \sin (-2) - 2\right ]^2 = 1$$ and both of them satisfy ODE $$y_1' = \frac 12 \left [ \sin x - \sin (-2) + 2\right ] \cos x = y_1^{\frac 12} \cos x \\ y_2' = \frac 12 \left [ \sin x - \sin (-2) - 2\right ] \cos x = y_2^{\frac 12} \cos x$$ – Kaster Feb 20 '13 at 20:57
Thanks for checking. As long as square root means positive square root, $\frac 12 \left [ \sin x - \sin (-2) - 2\right ] \cos x$ does not equal $y_2^{\frac 12} \cos x$. // Of course, if we were to consider both equations $y'=\pm y^{1/2}\cos x$, we'd have two solutions - but to two different IVPs. – user53153 Feb 20 '13 at 21:00
You're right about $y_0 = 1$, I forgot about function's behaviour itself. But how about $y_0 = 10^{-6}$. In this case wherever $\sin x \ge \sin x_0 + 2\cdot 10^{-3}$ square root would be taken without sign change. – Kaster Feb 20 '13 at 21:15
Same thing. When you take $C=-\sin x_0-2\sqrt{y_0}$, the expression $\sin x+C$ is negative in a neighborhood of $x_0$. This invalidates the solution $y=\frac14 (\sin x+C)^2$, since $y^{1/2}$ is now $-\frac12(\sin x+C)$. – user53153 Feb 20 '13 at 21:21