We have been taught the following version of existence and uniqueness theorem in ordinary differential equations:
Theorem: Let $D\subset \mathbb R^2$ be an open connected set and let $f:D\to \mathbb R$. .Suppose $(x_0,y_0)\in D$ and $R=[a,b]\times [c,d]$ is a rectangle in $D$ centred at $(x_0,y_0)$.Let $f$ be continuous on $D$ and $f$ satisfy Lipschitz condition in $y$ on $R$ i.e. $|f(x,y_1)-f(x,y_2)|\leq K|y_1-y_2|$ ,for all $x,y_1,y_2$ such that $(x,y_1),(x,y_2)\in R$.Suppose $M=\max\limits_{(x,y)\in R}{|f(x,y)|}$ and let $h=\min\{a,\frac{b}{M}\}$,then the IVP $\frac{dy}{dx}=f(x,y),y(x_0)=y_0$ has a unique solution on $[x_0-h,x_0+h]$ i.e. there exists a unique differentiable function $\phi:[x_0-h,x_0+h]\to \mathbb R$ such that $\phi'(x)=f(x,,\phi(x))$ for all $x\in [x_0-h,x_0+h]$ and $\phi(x_0)=y_0$.
This is clearly giving us a sufficient condition for existence and uniqueness of solution.But I want to show that this condition is not necessary.Can someone help me with a counterexample?
Also,I found another statement of existence and uniqueness theorem:
Theorem Let $D\subset \mathbb R^2$ be an open connected set with $(x_0,y_0)\in D$.Suppose $f:D\to \mathbb R$ is continuous and $\frac{\partial f}{\partial y}$ is also continuous on $D$,then the IVP $\frac{dy}{dx}=f(x,y),y(x_0)=y_0$ has a unique solution $\phi$ defined on $[x_0-h,x_0+h]$ for some $h>0$.
I want to know which one implies the other and I also want to find counterexample that would show that this is not a necessary condition.Please help me in solving this problem.