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Consider the following initial value problem for $\dot{x} =F(t,x)$: \begin{equation} \dot{x} = x^{1/3}, \quad x(0)=0 \end{equation} Now, the general solution reads \begin{equation} x(t)= \left[ \frac{2}{3} (t+ C) \right]^{3/2} \end{equation} with $C$ being an arbitrary constant.

If $C=0$ the initial condition is satisfied and the particular solution reads \begin{equation} x(t)= \left[ \frac{2}{3} t \right]^{3/2} \end{equation}

However, the slides mention that two additional solutions exist for $t \ge 0$, namely $x(t)= - \left[ \frac{2}{3} t \right]^{3/2}$ and $x(t)=0$. Why is that? We can invoke the existence and uniqueness theorem, where $\frac{\partial }{\partial x} x^{1/3} = x^{-2/3}$ is not defined at $x=0$, implying that the solution of the IVP is not unique. But where do these other 2 solutions come from?

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    $\begingroup$ That $x = 0$ is a solution comes from inspection (one typically asks if there is a constant solution). The other solution arises because the inverse of square function is not unique, e.g., solving $y^2 = 1$ gives $y = \pm 1$. $\endgroup$
    – Chee Han
    Commented Mar 15 at 16:43
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    $\begingroup$ Note that when you separate variables in this equation, you assume $x\neq 0$. If $x=0$, then $\dot{x}=0$, so the solution is constant. $\endgroup$
    – whpowell96
    Commented Mar 15 at 17:02
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    $\begingroup$ This ODE is one of the prime examples that does not have a unique solution. Hint: is $x^{1/3}$ Lipschitz at $x=0\,?$ $\endgroup$
    – Kurt G.
    Commented Mar 15 at 17:06
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    $\begingroup$ It is NOT Lipschitz as its derivative is not bounded. $\endgroup$ Commented Mar 15 at 17:19
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    $\begingroup$ $x^{1/3}$ is not Lipschitz at $x=0\,.$ If it were we had a unique solution. Draw the graph of the function to see this. Lipschitz means essentially bounded derivative. The formalities are a healthy exercise every ODE student needs to be able to master. $\endgroup$
    – Kurt G.
    Commented Mar 15 at 17:20

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In addition, you have this one parameter family of solutions (and its opposite) for $T\geq 0$ $$ x_T(t)=\begin{cases} 0 & \text{ if } t< T, \\ \left(\frac{2}{3}(t-T)\right)^{3/2} & \text{ if } t\geq T, \end{cases} $$ you can easily check that $t\rightarrow x_T(t)$ is continuously derivable at $t=T$ and verifies the ode.

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