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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?

Maximilian
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    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 $y = \pm 1$. – Chee Han Mar 15 '24 at 16:43
  • @CheeHan thank you for your response. I understand the explanation regarding the inverse of a square function, but I'm still unclear on how $x=0$ is a solution that can be identified through inspection. Could you provide an example to illustrate this point? This would help me better understand and also allow me to vote for your answer. – Maximilian Mar 15 '24 at 16:57
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    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. – whpowell96 Mar 15 '24 at 17:02
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    This ODE is one of the prime examples that does not have a unique solution. Hint: is $x^{1/3}$ Lipschitz at $x=0,?$ – Kurt G. Mar 15 '24 at 17:06
  • @whpowell96 does this imply that whenever I separate variables and $x$ appears in the denominator, $x=0$ is necessarily a solution? – Maximilian Mar 15 '24 at 17:16
  • @KurtG. how can I prove that $F(t, x)$ is locally Lipschitz continuous at $x=0$? – Maximilian Mar 15 '24 at 17:17
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    It is NOT Lipschitz as its derivative is not bounded. – Stéphane Mottelet Mar 15 '24 at 17:19
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    Because if you differentiate the right side by $x$, that derivative is not continuous at $(0,0)$. See Picard’s Existence and Uniqueness Theorem – Vasili Mar 15 '24 at 17:19
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    $x^{1/3}$ 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. – Kurt G. Mar 15 '24 at 17:20
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    Note that the RHS not being Lipschitz does not necessary imply nonuniqueness. Lipschitz continuity is sufficient, but not necessary for the existence of unique solutions https://math.stackexchange.com/questions/1274119/is-lipschitzs-condition-necessary-for-existence-of-unique-solution-of-an-i-v-p – whpowell96 Mar 15 '24 at 18:14
  • Thank you all for your comments. What I'd like to ask is for an explanation of why $x=0$ is a solution. Based on @whpowell96's comment, I understand that whenever I separate variables and $x$ appears in the denominator, $x=0$ is necessarily a solution. However, I'm not entirely confident about this. – Maximilian Mar 15 '24 at 18:22
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    We can assume $x(t) = k$ for some constant $k$, substitute this into the given differential equation, and solve for possible $k$-values. Usually there is only one $k$ because of the initial condition. – Chee Han Mar 15 '24 at 18:23
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    The function $x(t)=0$ solves the equation. This is true of any ODE of the form $\dot{x} = f(x)$ with $f(0) = 0$. Given any function, you can simply plug it into the ODE and see if it solves it... – whpowell96 Mar 15 '24 at 19:24

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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.

Stéphane Mottelet
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