Assume the IVP $$ x'=x^2, \quad x(0)=1 $$ By separating the variables and using the initial condition $$ x(t)=(1-t)^{-1} $$ Then, my book states that since $x_0=1>0$ then the solution can only be extended to the left with maximum existence interval $\, I=(-\infty,1)$.

My question is, since the solution $x(t)$ we found is defined for every $t\neq1$, why isn't the maximum interval of its existence $I^*=(-\infty,1)\cup(1,+\infty)$? Why does it depend on the sign of $x_0$?


The solution of an initial value problem is a function which satisfies the differential equation therefore it is continuous and passes through the initial point.

The interval of existence depends on the initial condition and it extends as far as the solution curve is differentiable and continuous as the result.

In your case you have a vertical asymptote at t=1, so the left branches do not pass the vertical asymptote.

With the given initial condition, $(-\infty, 1)$ is as far as you go without losing continuity.

  • $\begingroup$ I understand it at last! Global continuity is what I missed. Thank you very much. $\endgroup$ – Jevaut Nov 7 '18 at 17:42

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