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When considering the differential equation $y' = 3y^{\frac{2}{3}}$, I can find solutions $y = (t-a)^3, t > a$ and $y= (t-b)^3, t < b$. Also $y = 0$ is a solution.

I am suppose to show in reference to these two solutions that there are infinite number of solutions to this differential equation satisfying $y(1) = 1$, and why this does not contradict existence and uniqueness.

Problem is, for $t > 0, y > 0$ $f$ and $f_y$ are continuous, so in the domain $y > 0$ the solution should be unique. Could someone explain my flaw in logic>

Arctic Char
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A. Radek Martinez
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1 Answers1

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As $f(t,y)= 3y^{\frac{2}{3}}$ is continuous, the equation $y^\prime= 3y^{\frac{2}{3}}$ has at least one solution according to Peano existence theorem whatever the initial condition is.

However, $f$ is not Lipschitz continuous in $y$ around $0$. Hence we can’t apply the uniqueness part of Picard–Lindelöf theorem around $y=0$. Uniqueness is however true around $t_0$ if $y(t_0) \neq 0$.

And indeed, we can see that for any $b < 0$,

$$y(t)=\begin{cases} (t-b)^3 & t < b\\ 0 & b \le t \le 0\\ t^3 & t > 0 \end{cases}$$ is a solution with the initial condition $y(1)=1$.

mathcounterexamples.net
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