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Consider IVP$$y’=f(x,y), y(x_0)=y_0$$ where $f$ being continuous and Lipschitz on a rectangle $$|x-x_0|\leq a, |y-y_0|\leq b.$$ Then by existence and uniqueness theorem, the above differential equation has a unique solution in some neighbourhood of $x_0$ (say) $ (x_0-h, x_0+h)$ where $h$ is as given in uniqueness theorem. Now let if I solved the problem as usual method( like variable separable etc.) and got a solution (say) $y$ with domain as $D_y$ . Clearly ( as I think ) $$(x_0-h, x_0+h)\subseteq D_f$$

Now my question is that can I say that $y$ is unique solution of above ordinary differential equation in the domain $D_f ?$ or not guarantee outside $(x_0-h, x_0+h)$ even if differential equation has solution outside it . I am trying by taking example like $$y’=y^{2},y(0)=1$$ by solving as I got solution $y=\frac{1}{1-x}$ and I am not seeing any other solution in $(-\infty, 1)$. Please suggest . Thanks .

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    $\begingroup$ Well, without extra conditions, uniqueness will only follow on the intersection $(x_0-h,x_0+h) \cap D_f = (x_0-h,x_0+h)$. You would need the sufficient conditions to apply where the solution 'exits' the rectangle $\overline{B}(x_0,a) \times \overline{B}(y_0,b)$. $\endgroup$
    – copper.hat
    Nov 28 '20 at 20:09
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It's not clear what you mean by $D_f$. Given the right side $f$ of an ODE and an initial point $(x_0,y_0)\in D_f$ one can ask for solutions to such an IVP. When the assumptions to the existence and uniqueness theorem are fulfilled at $(x_0,y_0)$ are fulfilled then there is indeed an $h>0$ such that there is a unique solution $x\mapsto y(x)$ of the IVP, defined in an interval $(x_0-h,x_0+h)$. One then wants to "continue this solution analytically onto a maximal $x$-interval".

To do this we need the assumptions of the basic theorem in a global setting. Otherwise it may happen that further away from $x_0$ we reach bad zones where the uniqueness is no longer guaranteed. Consider the following example: $$y'= 3|y|^{2/3},\quad y(-1)=-1\ .\tag{1}$$ This IVP has in the open $x$-interval $(-2,0)$ the unique solution $x\mapsto y(x)=x^3$. When this solution arrives at $(0,0)$ it can be continued by $y(x)=x^3$ for $x>0$, but also by $y(x)\equiv0$ for $x>0$.

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  • $\begingroup$ Clear sir .... thank you .... $\endgroup$
    – neelkanth
    Nov 29 '20 at 2:27
  • $\begingroup$ sir can I say it will be unique as as long it does not cross zero ? $\endgroup$
    – neelkanth
    Nov 29 '20 at 4:15
  • $\begingroup$ @neelkanth: The solution of $(1)$ is unique in the open $x$-interval $(-\infty,0)$. $\endgroup$ Nov 29 '20 at 9:57
  • $\begingroup$ ok sir I will post different questions related to this ... thank you . $\endgroup$
    – neelkanth
    Nov 29 '20 at 12:25
  • $\begingroup$ sir please can you look at my this post math.stackexchange.com/questions/3927828/… $\endgroup$
    – neelkanth
    Nov 29 '20 at 18:20

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