# Uniqueness of solution for the ODE $y' = 1 + y^{2/3}$, $y(0) = 0$

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?

• 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.
– Alan
Jun 28, 2021 at 15:34
• That is true, and my question was when it doesn't satisfy those conditions, how does one proceed with proving uniqueness of solution. Jun 28, 2021 at 15:37
• 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/…
– Alan
Jun 28, 2021 at 15:38
• 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. Jun 28, 2021 at 15:50
• 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.
– Ian
Jun 28, 2021 at 16:34

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).
• 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 $y = z$. Jun 28, 2021 at 22:12
• You didn’t show a $y$ satisfying it is unique. Jun 28, 2021 at 22:25