# A doubt regarding uniqueness of IVP.

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 .

• 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 $\overline{B}(x_0,a) \times \overline{B}(y_0,b)$. Nov 28 '20 at 20:09

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 $$x>0$$.
• @neelkanth: The solution of $(1)$ is unique in the open $x$-interval $(-\infty,0)$. Nov 29 '20 at 9:57