Consider the following first order ODE:

$y' = f(t,y)$ subject to $y(t_{0}) = y_{0}$.

I would like to show that there exists a unique function $y(\cdot)$ that passes through ($t_{0}, y_{0}$)

The thing is $f$ is neither globally Lipschitz nor continuous. Therefore, I cannot apply standard theorems to prove the global existence. Instead, $f$ is locally Lipschitz, which implies the local existence at each point: For each point ($t_{0}, y_{0}$), there exists a unique function $y(\cdot)$ that satisfies $y' = f(t,y)$ for $t \in [t_{0} - \epsilon, t_{0}+\epsilon$] and the initial condition. My question is by pasting the local solutions, can we get a global solution? What are the necessary and sufficient conditions that make pasting a legitimate action for obtaining the global solution? I would appreciate your comments about this.


2 Answers 2


$y'=y^2$ is a pleasant ODE, but $$ y = \frac{1}{1-t} $$ is a solution with $y(0)=1$

Similar, $y'=1+y^2, \; \; y(0) = 0$ gives $y = \tan t.$


Local Lipschitz for $f$ guarantees global uniqueness (for any given IVP), but not global solutions. There are several condition which guarantee global solutions. For example: $$ \lvert f(t,y)\rvert \le g(y), $$ and $$ \int_{y_0}^\infty \frac{dy}{g(y)}\,=\,\infty. $$

  • $\begingroup$ Thank you for your suggestion Yiorgos. Unfortunately, my $f$ function does not satisfy the condition you mentioned since it can get arbitrarily large. I wonder if there are other conditions under which I can concatenate local solutions to obtain a global solution. $\endgroup$ Commented Jul 28, 2014 at 0:20
  • $\begingroup$ Isn't $y'=y^2$ a counterexample to your claim? $\endgroup$
    – Artem
    Commented Jul 28, 2014 at 0:25
  • $\begingroup$ @Artem : Correct! I made the necessary change. $\endgroup$ Commented Jul 28, 2014 at 7:50
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    $\begingroup$ And this is best possible, because if $g > 0$ with $\int_{y_0}^\infty \frac{dy}{g(y)} = z < \infty$, the solution of $y' = g(y)$ with $y(0) = y_0$ is given implicitly by $\int_{y_0}^{y(t)} \frac{dy}{g(y)} = t$, and this goes to $\infty$ as $t \to z-$. $\endgroup$ Commented Jul 28, 2014 at 8:08
  • $\begingroup$ Why this is sufficient condition? $\endgroup$
    – ends7
    Commented Sep 10, 2023 at 16:12

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