Show that the differential equation $\frac{dy}{dx}=\sqrt y,y(0)=0$ has non-unique solution.
To solve this problem I used Lipschitz theorem. My solution goes like this:
Let $f$ be continuous in an unbounded domain $$D:a\leq x\leq b,-\infty\leq y\leq \infty.$$ Let $f$ satisfy a Lipschitz condition in $D$. That is, assume $\exists k\in\Bbb R$such that $|f(x,y_1)-f(x,y_2)|\leq k|y_1-y_2|$ for all $(x,y_1),(x,y_2)\in D.$Then the differential equation $\frac{dy}{dx}=f(x,y)$ and $y(x_0)=y_0$ ( where $(x_0,y_0)\in D$ ) has a unique solution in $[a,b].$( Lipschitz Theorem) Now, here, we have $f(x,y)=\sqrt y$ and $\frac{\partial f}{\partial y}=\frac {1}{2\sqrt y}.$ We see that $f$ is continuous everywhere, but $\frac{\partial f}{\partial y}=\frac {1}{2\sqrt y},$ is not continuous at $y=0$ and also not bounded at $y=0$ and so, Lipschitz condition is not satisfied( since $f$ is said to satisfy a Lipschitz condition with respect to $y$ in a domain $D$ iff $\frac{\partial f}{\partial y}$ is bounded). Thus, we can conclude that the given differential equation $\frac{dy}{dx}=\sqrt y$ and $y(0)=0$ has no unique solution.
But I am confused as I studied Picard's Theorem as well which goes like this:
The differential equation $\frac{dy}{dx}=f(x,y)$ and $y(x_0)=y_0$ has a unique solution on $|x-x_0|\leq h$ , if both $f$ and $\frac{\partial f}{\partial y}$ are continuous in the domain $D=\{(x,y):|x-x_0|\leq h,|y-y_0|\leq k,h,k>0\}$
If we use this theorem to conclude the above assertion in the question is true, then the reasoning would go like this:
Here, $\frac{\partial f}{\partial y}$ is not continuous in the domain $D$ if $(0,y)\in D.$ Thus, Picard's theorem won't be satisfied if $(0,y) \in D.$ But according to Picard's theorem, we have: the differential equation $\frac{dy}{dx}=\sqrt y$ and $y(0)=0$ has a unique solution on $|x|\leq h(,h>0)$, if both $f$ and $\frac{\partial f}{\partial y}$ are continuous in the domain $D=\{(x,y):|x|\leq h,|y|\leq k,h,k>0\}.$ But, $\frac{\partial f}{\partial y}$ is not continuous in $D$ as $(0,y)\in D.$ So, the differential equation $\frac{dy}{dx}=\sqrt y,y(0)=0$ has non-unique solution.
But I am confused whether both of my solutions (one using Lipschitz theorem and the other one using Picard's theorem) is valid or not?
I think, Lipschitz theorem is more appropriate to solve these sort of problems as Picard's theorem, gurantees a unique solution in a particular or (better say,) a bounded domain but the Lipschitz theorem gurantees the existence of a unique solution, in an unbounded domain(Is it?) But, in this case what I did was that:
I showed both the Lipschitz theorem and the Picard's theorem are not satisfied for the set of equations : $\frac{dy}{dx}=\sqrt y,y(0)=0.$
So, in general, if conditions of Lipschitz's theorem are not satisfied, then can we directly conclude that a set of equations, say: $\frac{dy}{dx}=f(x,y)$ and $y(x_0)=y_0$ (where $(x_0,y_0)$ is in the domain of $f$ ) do not have a unique solution in the domain of $f,$ just as the thing was with this case? Then, the usage of Picard's theorem further is unnecessary(here), isn't it?
You cannot use Lipshitz theorem that way - you are making an elementary logic mistake. $A\implies B$ does not imply $\lnot A\implies \lnot B$. In your case $A$: "function satisfies Lipshitz condition", $B$: "differential equation has a unique solution". (To see why, replace $A$ with "$x$ is a fish" and $B$ with "$x$ lives in the water", and consider whales!)
Also, you can notice that the function $y=0$ satisfies the solution, as well as $y=\left(\frac{x-C}{2}\right)^2$ for any $C\in\mathbb R$ and $x\ge C$. This doesn't just mean $C=0$: you can pack up many more functions the following way: choose a constant $C\ge 0$ and say:
$$y=\begin{cases}0&x<C\\\left(\frac{x-C}{2}\right)^2&x\ge C\end{cases}$$
(It is a part of the $x$ axis all the way on the left from $(0,0)$, and on the right all the way up to $(C,0)$, after which point it rises up as a parabola.)