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.
Looking at the equation one can note first that
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.