Here is the theorem
(a) If $f$ is continuous on an open rectangle $$R : \{ a < x < b , c <
y < d \}$$ that contains $(x_0, y_0)$ then the initial value problem
$$ y ^ { \prime } = f ( x , y ) , \quad y \left( x _ { 0 } \right) = y _ { 0 } $$ has at least one solution on some open subinterval of $(a, b)$ that contains $x_0$.(b) If both $f$ and $f_y$ are continuous on $R$ then the equation has
a unique solution on some open subinterval of $(a, b)$ that contains
$x_0$.
My question is
If the conditions of the Existence and Uniqueness theorem are met, does there exists a unique solution for all $x\in(a,b)$?
Best Answer
Take $f(x,y) = y^2$ with $(x_0,y_0) = (0,1)$, then the solution is $y(x) = {1 \over 1-x}$ for $x < 1$.
Note that $f$ is smooth and defined everywhere.
In particular, if we choose the rectangle$(-1,2) \times (0,100)$ then it is impossible to find a solution starting from $(0,1)$ that is defined on $(-1,2)$.