calculus - Determine the set of all points $(xi, eta)$ for which the initial value problem is not locally uniquely solvable

Determine all solutions of the differential equation $y' = 2 \sqrt{|y|(1-y)}, \quad y \leq 1$. In any case, prepare a sketch and determine the set of all points $(\xi, \eta)$ for which the initial value problem is not locally clearly solvable. (The sketch part is not the problem).

What I already have:

For positive $y$ with $|y|=y$, the solution is:
$y=\frac{1}{2}(\sin(2(t+c_1))+1)$.

For negative $y$ with $|y|=-y$, the solution is:
$y=\frac{1}{4}e^{-c_1-2t}(e^{c_1+2t}+1)^2$.

I hope these are correct.

Now my main problem is to determine the set of all points $(\xi, \eta)$for which the initial value problem is not locally clearly solvable. Can anyone help me with that?

Answer

I do not know if I fully understood your question. But here is a try. You have a differential equation of the form:
\begin{align}
y'= f(y), \ \ y(t_0)=y_0
\end{align}
Where $f(y)=\sqrt[]{|y|(y-1)}$ and $y\leq 1$.

For $0 or $y_0<0$ we have that $f(y)$ is continuously differentiable on $[ y_0-h , y_0+h]$ for some $h>0$ (why?). By the Existence and Uniqueness Theorem there exists a unique solution in some region of $(t_0,y_0)$.

For $y_0=0$ or $y_0=1$ we have that $f(y)$ is not continuously differentiable BUT we cannot say that the solution is not unique since the Existence and Uniqueness Theorem does not say anything about that. So we might say that the candidates for which the solution is not unique is the points $K=\{ (t_0, y_0) | \ (y_0 = 0 \vee y_0=1)\ ,\ t_0\in\mathbb{R}\}$.

One can show with other kind of methods the existence of solutions when the initial value is in $K$, but that might be difficult. You either come up with two different solutions that satisfy the initial value or prove existence and uniqueness with other methods.

Let's construct two solutions that satisfy $f(t_0)=y_0=0$. We can take: $y_1(t)=0$. Let's take the solution that you already have is $\frac 1 2 [\sin(2(t+c))+1]$. Define now:
\begin{align}
y_2(t)= \frac{1}{2}\sin \left(2\left(t-\frac{\pi}{4}-t_0\right)\right) +\frac{1}{2}

\end{align}
for $t\in(-\pi/8+t_0,\pi/8+t_0)$. You can check that this function satisfies $y_2(t_0)=0$ and is not equal to $y_1(t)=0$. So we have constructed two functions that satisfy the ODE for $y(t_0)=0$. Hence a solution exist but is not unique for $y(t_0)=0$.

We know for sure that the set of points $(t_0,y_0)=(t_0,0)$ is not locally clearly solvable. (Is this what you mean not locally clearly solvable?)

I'll leave the rest of the fun for you and let you do something for the points $y(t_0)=y_0=1$.

Let me know if I misunderstood your question.