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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 $x\in(a,b)$?

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    $\begingroup$ No. Consider $f(x,y) = y^2$. $\endgroup$
    – copper.hat
    Feb 24, 2019 at 23:33

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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 $(-1,2)$.

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Quick answer: No.

The reason is the function $f$ might met the conditions, but be nonlinear.

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  • $\begingroup$ I may not be understanding the theorem correctly but if you know that $f$ and $f_y$ are continuous on $(a,b)$ and you pick some arbitrary point in $(a,b)$ then why does it seem like the theorem doesn't apply to this point? $\endgroup$
    – Jac Frall
    Feb 25, 2019 at 0:05
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    $\begingroup$ @JacFrall: I have given an example that illustrates the issue. A solution exists in $R$ but, loosely, may 'escape out of the top or bottom of the box'. $\endgroup$
    – copper.hat
    Feb 25, 2019 at 0:15
  • $\begingroup$ @copper.hat yes it took some contemplation but I get it now. Thank you $\endgroup$
    – Jac Frall
    Feb 25, 2019 at 0:24

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