non uniqueness for the solution of the following Cauchy problem The following Cauchy Problem:
$$y'=-10 \sqrt y,~~~ y(0) =\dfrac 14$$
doesn't have a unique solution, is that correct? I tried to solve that, and the solution is a parabola, but what I think is that when the parabola reach zero then there is also another solution which is the first part of the parabola and then the solution $y=0$.
The solution for the diff. eq. is $y(x) = \frac14(1-10x)^2$, but when $x=1/10$ the parabola reaches zero and then there is another solution which is $y(x) = \frac14(1-10x)^2$ for $x<1/10$ and $y=0$ for $x\ge1/10$.
Moreover I need to compute the absolute stability interval for Forward Euler applied to the previous ODE. I know how to compute it for the test equation (linear case), but I don't know how to deal with this particular case.
 A: Looking at the equation one can note first that

*

*$y$ has to be always non-negative and

*the slope $y'$ is always negative or zero for $y=0$.

This means that the solution that is initially parabolic gets then absorbed by the $x$-axis, there can be no solution that follows the rising parabolic branch after touching the $x$-axis.

A-stability of the Euler method for an ODE $y'=f(y)$ means roughly that if $\lambda=f'(y_n)<0$, and $y_n$ is close to a fixed point or an asymptote of the equation, then one wants that $λh_n=h_nf'(y_n)\in (-2,0)$, better closer to zero than the other side.
Here $f'(y)=-5y^{-\frac12}$ so that $0<h_n<\frac25\sqrt{y_n}$ as a guideline. With a fixed step size this bound is eventually violated. Let's see what happens in variable step size.
Selecting for instance $h_n=\frac15\sqrt{y_n}$, this gives the iteration
$$
y_{n+1}=y_n+h_nf(y_n)=y_n-2y_n=-y_n
$$
which is obviously not what is desired. So the guideline seems only useful if the local Lipschitz constant can be bounded.
Let's try a smaller $h_n=\frac1{20}\sqrt{y_n}$, then
$$
y_{n+1}=y_n-\frac12y_n=\frac12y_n\implies y_n=2^{-n}y_0.
$$
It follows that $h_n=\frac1{20}2^{-n/2}\sqrt{y_0}$, but
$$
t_n=h_0+h_1+\dots+h_{n-1}
=\frac1{20}\frac{1-2^{-n/2}}{1-2^{-1/2}}\sqrt{y_0}
<\frac{1+2^{-1/2}}{10}\sqrt{y_0},
$$
so that the time steps can not move the time over this finite boundary, which is also not what is expected of a solution.
