# Can the unique, stationary solution of a Cauchy problem be prolonged on all $\Bbb R$ even if the differential equation is undefined at one point?

Consider the Cauchy problem$$(*)\begin{cases}y'=\frac{y^2-1}{x^2} \\y(1)=1.\end{cases}$$ We note that $$y'=h(x)k(y)$$ with $$h(x)$$ continuous on $$\Bbb R\setminus\{0\}$$ and $$k(y)$$ continuously differentiable on $$\Bbb R$$, and since $$(x_0,y_0)=(1,1)\in(0,+\infty)\times\Bbb R$$, this is the set we consider.

Cauchy-Lipschitz theorem then guarantees $$y(x)\equiv1$$ is the unique solution of $$(*)$$ on $$(0,+\infty)$$. Since it's differentiable on all $$\Bbb R$$, does it make sense to say it can be prolonged on all $$\Bbb R$$?

• The differential equation does not exist at $x=0$, thus you can not have any solution through that point. That you can continue the function does not matter, you can not continue the solution as solution. – LutzL Jan 15 at 13:59
• @LutzL Many thanks! – Learner Jan 15 at 14:06

## 1 Answer

As per @LutzL's comment: no. That you can continue the function does not matter, you can not continue the solution as solution.