Uniqueness of solution to $x' = t \sqrt{1-x}$ (Proof checking) I am trying to prove that the solution to the problem:
$$ x' = t \sqrt{1-x} $$
with initial condition $x(0) = \tfrac{1}{2}$ is unique.
I have found that the solution to the problem is the function:
$$ x(t) = 1 - \left(\frac{1}{\sqrt{2}} -\frac{t^2}{4} \right)^2$$
It is clear that $f(t,x) = t \sqrt{1-x}$ is not locally Lipschitz at $0$, so we can't invoke Picard-Lindelöf. I am wondering if the next solution is ok:
By doing the change $z = \sqrt{1-x}$ we get the equivalent system:
$$ z' = \frac{t}{2}$$
Since the new system is linear and $g(t,x) = \frac{t}{2}$ is continous, it has an unique solution $z^*$ in a neighborhood of $0$. Therefore, the solution to the original system, given by $x^* = 1 - {z^*}^2$ must also be locally unique.
Is this approach correct? In case it isn't, could you suggest another one? Thank you in advance.

$x(t) = 1 - \left(\frac{1}{\sqrt{2}} -\frac{t^2}{4} \right)^2$ is a solution of the initial value problem, but the solution is not unique.
With $t^* = 2^{3/4}$ we have $x(\pm t^*) = 1$ and $x'(\pm t^*) = 0$, so that, for example,
$$
 x_1(t) = \begin{cases}
 x(t) & \text{ if } x \le t^* \\
 1 & \text{ if } x \ge t^*
\end{cases}
$$
is another solution.
As you noticed, $f(t,x) = t \sqrt{1-x}$ is not locally Lipschitz at points $(t, 1)$. Picard-Lindelöf guarantees a unique solution in a neighborhood of $x(0) = 1/2$, but can not be applied anymore when the solution reaches $x(t^*) = 1$.
The transformed function $z_1 = \sqrt{1-x_1}$ is not differentiable at $t=t^*$, that's why the transformed initial value problem does not show this solution. This means that $z'=-t/2$ is not equivalent to the original problem.
