# First order ordinary differential equation issue

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

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

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?

• 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 $y = \pm 1$. Commented Mar 15 at 16:43
• 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. Commented Mar 15 at 17:02
• This ODE is one of the prime examples that does not have a unique solution. Hint: is $x^{1/3}$ Lipschitz at $x=0\,?$ Commented Mar 15 at 17:06
• It is NOT Lipschitz as its derivative is not bounded. Commented Mar 15 at 17:19
• $x^{1/3}$ 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. Commented Mar 15 at 17:20

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.