This one has completely stumped me. It's from an ODE general exam, I'm given the IVP
$$ y' = \text{min}(y^2, M),$$$$ y(0) = 1.$$
With $M>1$ and I'm asked to give an explicit solution and discuss wether it's unique and global. So, first thing I notice is that since the derivative is bounded, we ought to have Lipschitz continuity everywhere and thus global unique solutions. All right, so now I attempt to solve each part separately and I try to paste them together. Following this idea, we have solutions
$$ y_1(t) = \frac{1}{1-t},$$$$ y_2(t) = Mt + 1.$$
These obviuosly won't paste well together as they are, so my next idea is to try and find the point at which the derivative of $y_1$ reaches $\sqrt{M}$. Since $M>1$, $\sqrt{M} < M$ and we can extend this function for a bit by pasting it with it's differential at $t = 1-1/\sqrt{M}$. Now, this function will continue to grow linearly, so $y^2$ will grow beyond $M$ but the derivative will still be $M$. After some calculations, we have the following candidate:
$$ y(t) = \begin{cases}\frac{1}{1-t} & 0 \leq t < 1 - 1/\sqrt{M}, \\M(t - 1) + 2\sqrt{M} & 1 - 1/\sqrt{M} \leq t.\end{cases}$$
Which seems so work out right, but two things come to mind. First is, I'm guessing that we can't paste some other thing to our straight line because at that point $y^2$ would be greater than $M$. Second is, I'm guessing that the condition $M>1$ precludes starting out with $y_2$ and then pasting another function to extend the line, since around a small enough neighborhood of $0$ we could get $y_2^2$ to be smaller than $M$. I'm not super convinced that this is the way to go with this problem (since I never really saw these kinds of problems in the course that I took) so any feedback would be appreciated. Also, I tried to plot the phase diagram in Mathematica and I did NOT get what I was expecting:(For reference, I used StreamPlot[{{x, Min[x^2, M]}}, {x, -10, 10}, {y, -10, 10})
