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Show that the problem $$y'(t)=y^{2/3}(t) \text{ with } y(0)=0$$ has infinitely many solutions. Explain why the existence and uniqueness theorem does not apply here

My attempt
By solving the differential equation by the variable separation method, We get:

$\int\frac{1}{y^{2/3}}dy=\int dt$

$\frac{y^{1/3}}{1/3}=t+c$

And by substituting the initial condition $y(0)=0$ we can get $c=0$
Thus $$y(t)=\frac{t^3}{3^3}$$

But from here how should I prove that there are infinitely many solutions?...

And for the second part (Uniqueness theorem) isn't it because for the solution of $y'=f(y)$ to be unique, we need $f$ to have a continuous first derivative. But here in this specific example, $$\frac{d}{dy}f=\frac{2}{3}y^{-1/3}$$ which is not continuous at zero.

Charith
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1 Answers1

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There is only one "separable solution", as you say: it is $y=t^3/3^3$. But to perform separation of variables, you needed to assume that $y \neq 0$ almost everywhere on the domain where you did the integration (so that it is justified to divide by $y^{2/3}$). And this isn't necessarily true, because certainly $y=0$ is a solution.

To get even more solutions, because the equation is autonomous, you can just stay at zero for however long you want, say up to some $t_1>0$, and then switch over to $(t-t_1)^3/3$. You can do the same backward in time. So all of these functions are solutions to the ODE:

$$y(t;t_0,t_1)=\begin{cases} (t-t_0)^3/3^3 & t < t_0 \\ 0 & t_0 \leq t \leq t_1 \\ (t-t_1)^3/3^3 & t>t_1 \end{cases}$$

whenever $-\infty \leq t_0 \leq t_1 \leq \infty$. Of course you must assume that $t_0 \leq 0 \leq t_1$ in order to have a solution to the IVP as well.

Ian
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