# Existence and uniqueness of the solutions of $\dot{x}(t)= - (x(t))^{1/3}$

For the IVP: $\left\{ \begin{array}{c} \dot{x}(t) = -x(t)^{1/3} \\ x(0)=x_0 \end{array}\right.$ I am trying to prove existence and uniqueness and the maximal range of existence.

As $\dot{x}(t) = F(t,x(t)) = -(x^{1/3})$ is well-defined and continuous for all real $x$, except negative numbers since then the value of $F(t,x(t))$ would have to be complex, there exists a solution for $x\geq 0$.

As $\frac{\partial\,F(t,x(t))}{x} = \frac{-1}{3\,x^{2/3}}$ is well-defined and continuous only for positive values of $x$, then there is a unique solution for $x>0$.

I know of 3 solutions for $x=0$, and I see that the Picard-Lindel\"of theorem says that, assuming a Lipschitz condition is held, that a unique solution should hold for the $x>0$. Is solving the ode then the only method left to determine the maximal time of existence? (Because I have seen when this theorem gave an answer that was too narrow).

Since any solution is such that $x(t)\geqslant0$ and $(x^{2/3})'(t)=-2/3$, the solutions when $x_0\gt0$ are exactly the functions such that $x(t)^{2/3}=x_0^{2/3}-2t/3$ for every $t\geqslant0$ small enough, that is, $x(t)=(x_0^{2/3}-2t/3)^{3/2}$ on the maximal interval $(-\infty,3x_0^{2/3}/2)$. If $x_0=0$, the unique solution is $x(t)=0$ and the maximal interval is $(-\infty,+\infty)$.

• I guess I don't understand why, of the 3 solutions I found for $(t_0,x_0)=(0,0)$, $x(t)=0$ wouldn't be one of the solutions. $F(t,x(t))$ is continuous there, but $dF/dx$ is undefined so, assuming the Lipschitz condition holds, wouldn't there just be a loss of uniqueness there? I haven't calculated for a Lipschitz constant - is that why there are no solutions for $x_0=0$ (I am doing it now)?
– nate
Sep 20, 2013 at 16:49
• The previous version was not correct, see revised version.
– Did
Sep 20, 2013 at 16:58
• Oh, I cannot even apply $|f(x)-f(y)| \leq K\,|x-y|,\;\forall x,y \in 0 \ni E$ because $f(x<0)$ is not defined for real numbers. Hence it cannot be Lipschitz.
– nate
Sep 20, 2013 at 16:59
• Ummm, one last thing. How do you know the solution to the IVP for $x(0)=0$ is unique, since the above stated theorem can't be used?
– nate
Sep 20, 2013 at 19:44
• @nate For a separable equation the uniqueness can be determined by the fact that after the separation of variables the corresponding integrals are divergent. Sep 22, 2013 at 2:25