Problem 8.3 i) from "An Introduction to Ordinary Differential Equations", Agarwal and Regan asks for the existence and uniqueness of the initial value problem $y' = 1 + y^{2/3}, y(0) = 0$. Applying the Picard-Lindelöf theorem, one gets $$\dfrac{\partial f}{\partial y} = \dfrac{2}{3}\,y^{-1/3},$$ which is obviously not continuous at $y = 0;$ and thus there must be infinite solutions for the above IVP. However, the solution states that there is a unique solution $3\left(y^{1/3} - \tan^{-1}\left(y^{1/3}\right)\right) = x$, and it seems to check out - I cannot find a trivial solution, or any other solution. So in cases where one cannot apply the Picard-Lindelöf theorem, how does one prove that there exists a unique solution for the IVP?

  • $\begingroup$ I'm not familiar with the theorem, but I Just googled it on wiki and it doesn't seem to be a necessary and sufficient criteria, just a sufficient one, so failing to have the requirements doesn't seem to make it fail to have a unique solution. $\endgroup$
    – Alan
    Jun 28, 2021 at 15:34
  • $\begingroup$ That is true, and my question was when it doesn't satisfy those conditions, how does one proceed with proving uniqueness of solution. $\endgroup$
    – Axe Kumar
    Jun 28, 2021 at 15:37
  • $\begingroup$ Well, some wiki-diving and googling found the following as actual necessary and sufficient criteria, but I have no idea how to apply it...been too long since I've done math at this level. ams.org/journals/proc/1967-018-04/S0002-9939-1967-0212240-6/… $\endgroup$
    – Alan
    Jun 28, 2021 at 15:38
  • $\begingroup$ Thanks Alan, it seems to be a useful characterization, but it might not be possible to work out the same procedure for a more complex example. $\endgroup$
    – Axe Kumar
    Jun 28, 2021 at 15:50
  • 5
    $\begingroup$ The idea is that only $y=0$ creates the non-Lipschitz behavior and you cannot stay there for any period of time because $y' \geq 1$ all the time. This wouldn't happen if the $1$ weren't there. $\endgroup$
    – Ian
    Jun 28, 2021 at 16:34

1 Answer 1


The solution to your problem is given by $$ \int_0^y\frac{ds}{1+s^{2/3}}=x. $$ Note that the integral on the left is a regular well defined Riemann integral, which proves that the solution exists (it is given above) and unique (since this expression can be obtained from the original equation by equivalent rearrangements).

  • $\begingroup$ Now to finish it ... Suppose $y$ and $z$ both satisfy this (integral) equation, that is $\int_0^y\,(1 + s^{2/3})^{-1}\, ds = \int_0^z\,(1 + s^{2/3})^{-1}\, ds = x$. Subtracting, $\int_z^y \,(1 + s^{2/3})^{-1}\, ds = 0$. The integrand is positive and continuous, so $y = z$. $\endgroup$ Jun 28, 2021 at 22:12
  • $\begingroup$ @Aruralreader This is absolutely unnecessary. Integral on the left has no issues (the issues appear, which imply non-uniqueness, when this integral is improper). $\endgroup$
    – Artem
    Jun 28, 2021 at 22:15
  • 2
    $\begingroup$ You didn’t show a $y$ satisfying it is unique. $\endgroup$ Jun 28, 2021 at 22:25

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