Existence of solution of first order ordinary differential equation Consider the IVP: $y'(t) = 2\sqrt{y(t)},  \enspace y(0) = a, \enspace t \in \mathbb{R}$.
Then, here $f(t,y)$ is continuous and $\frac{\partial f}{\partial y}$ is discontinuous at $0$. 
So we can't conclude about uniqueness of solution. 
But if $a = 0$, we get an infinite number of solutions.
So, here my confusion is, does the continuity of $\frac{\partial f}{\partial y}$ depend on the value of $a$?
If $a > 0$, what will happen to its continuity? If it is continuous then by Picard's theorem we could conclude the uniqueness.
So 


*

*Is $\frac{\partial f}{\partial y}$ continuous if $a>0$?

*If yes, in Picard's theorem, where is the continuity of $\frac{\partial f}{\partial y}$ defined?


Thanks in advance.
 A: Picard's Existence and Uniqueness Theorem states that:
Given IVP: $y'=f(x,y), \enspace y(x_0)=y_0$ with $f(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$ continuous functions in some open rectangle $R = (a,b)\times(c,d)$ containing $(x_0,y_0)$, then the IVP has a unique solution in the closed interval $I = [x_0-h,x_0+h]$ for some $h>0$. (It also gives a sequence of functions which converge uniformly on I on the solution $y$ of the IVP).
Now, if you meant $y'(t)=2 \sqrt{g(t)}$, for some continuous function $g:J \subset \mathbb{R} \rightarrow \mathbb{R}_0^+,0 \in J$, then the function $f(t,y)$ is continuous on $\mathbb{R}^2$, and since it does not depend on $y$, $\frac{\partial f}{\partial y}(t,y)$ exists and $\frac{\partial f}{\partial y}(t,y)=0$ i.e. continuous on all $\mathbb{R}^2$. Thus the IVP has a unique solution (local) on some closed interval $I$ containing $0$.
If you meant $y'=2 \sqrt{y}$, where $y:J \subset \mathbb{R} \rightarrow \mathbb{R}_0^+$ continuous function, $0 \in J$, $J$ open interval, then, if $a=0$, Picard's theorem does not guarantee you a unique solution (nor does Picard-Lindelof theorem for that matter), because $\frac{\partial f}{\partial y}(t,y)$ does not exist at $(0,0)$. However if $a>0$ then by Picard's theorem with $R=J\times(0,\infty)$ we have a unique solution.
(So, 1. Yes 2. on some open interval containing the point of the intial condition)
