Studying local uniqueness in the family $ \begin{cases} x' = |x|^{p}\\ x(t_0) = 0\\ \end{cases} $

Question

Consider the IVP:

$ \begin{cases} x' = |x|^{p}\\ x(t_0) = 0\\ \end{cases} $

This is a differential equation that can be handled by separable variable technique. There is a further theorem stating that:

Let $a \in C(J_1),g \in C(J_2),t_0 \in J_1,x_0 \in J_2$ and consider the IVP $\begin{cases} x' = a(t)g(x) \\ x(t_0) = x_0 \\ \end{cases} $ then the IVP

1. Has solutions.
2. If $g(x_0) \neq 0$ then verifies the property of local uniqueness.
3. If $g(x_0) = 0$, $G \in C(J_2)$ such that $G(x_0) = 0$,$\exists \delta > 0.\forall x \in ]x_0,x_0+\delta[.G'(x) = \frac{1}{g(x)}$ and $a(t_0) \neq 0$ then the IVP doesn't verify the property of local uniqueness.

I was told that local uniqueness does not hold for $0 < p < 1$ and that it holds in other cases.

How can I study this family of IVPs?

Answer 1

The best way to study a Cauchy problem for the first order ordinary differential equation \begin{equation} \begin{cases} x^\prime=v(x) \\ x(t_0)\equiv x_0=0 \end{cases}\tag{1}\label{1} \end{equation} is perhaps to use the standard Barrow's formula ([1], §1.5 p. 19) $$ t-t_0=\int\limits_{x_0}^{x(t)}\frac{\mathrm{d}\xi}{v(\xi)}\tag{2}\label{2} $$ and the related existence and uniqueness theorem ([1], §2.2 p. 36), which says that if $v:\mathbb{R}\to\mathbb{R}$ is continuously differentiable, then the solution of $\eqref{1}$ exists, is unique and is given by formula $\eqref{2}$ if $v(x)\neq 0$, or by $$ \quad x(t)=x_0=\mathrm{const.}\:\text{ if }v(x_0)=0.\tag{3}\label{3} $$ (Solutions of type $\eqref{3}$ are commonly called equilibrium points)

In the case under analysis, since $v(x)=\vert x\vert^p\geq 0$ for all $p\in [0,\infty[$ and the chosen initial condition is $x(t_0)=0$, we infer that $x(t)$ is an increasing function of $t$. Therefore we can assume that $x(t)\geq 0$ for all $t\in[t_0,\infty[$, study the following the Cauchy problem, equivalent to the one proposed by Javier: \begin{equation} \begin{cases} x^\prime=x^p \\ x(t_0)=0 \end{cases}\quad \forall p\in[0,\infty[ \tag{4}\label{4} \end{equation} Now let's discuss the various possibilities by solving this Cauchy problem and applying the cited theorem, as suggested by Artem:

1. $p=0$: $x^p\equiv 1$ so $v$ is $C^1$ and non zero, thus there are no equilibrium points and the solution of $\eqref{4}$ exists and is unique. Formula $\eqref{2}$ allows to represent it as follows: $$ t-t_0=\int\limits_0^{x(t)}\mathrm{d}\xi=x(t)\:\iff\:x(t)=t-t_0 $$
2. $0<p<1$: in this case $x^p$ is not $C^1$ in $x=0$ thus the can expect non uniqueness. And as a matter of fact, $$ x(t)\equiv 0\text{ is a solution since }x_0=0\text{ is an equilibrium point,} $$ and formula $\eqref{2}$ gives another solution of $\eqref{4}$: $$ t-t_0=\int\limits_0^{x(t)}\frac{\mathrm{d}\xi}{\xi^p}=\left[\frac{\xi^{1-p}}{1-p}\right]^{x(t)}_0=\frac{x(t)^{1-p}}{1-p}\:\iff\:x(t)=[(1-p)(t-t_0) ]^\frac{1}{1-p} $$
3. $p\geq1$: now $x^p$ is again $C^1$ the solution of $\eqref{4}$ exists and is unique. Since $x_0=0$ is an equilibrium point, it has the form $$ x(t)\equiv 0 $$

[1] Vladimir Igorevic Arnol'd, "Ordinary differential equations", various editions from MIT Press and from Springer-Verlag, MR1162307 Zbl 0744.34001.

Answer 2

If $p < 0$ then the problem doesn't have a solution since $x(t_0) = 0$ would imply that the derivative is not defined at $t_0$, a mandatory condition to be a solution.

When $0 < p < 1$ we can set:

$ G(x) = \begin{cases} 0 & x < 0\\ \frac{x^{1-p}}{1-p} & x > 0\\ \end{cases} $

The idea here is to avoid integrating an absolute value. $G$ is continuous and $G(0) = 0$. Furthermore, $\forall x \in ]0,\infty[.G'(x) = \frac{1}{x^{p}}$ and therefore the IVP doesn't have local uniqueness.

If $p = 0$ then we get the linear equation:

$ \begin{cases} x' = 1\\ x(t_0) = 0\\ \end{cases} $

which has unique solution $x(t) = t-t_0$. Therefore, there is local uniqueness.

If $p > 1$ we should go for part 3. but now the possible primitive to the right is not defined in $x_0 = 0$ so in principle we cannot say anything about it. We can apply Picard-Lindelof:

The domain of the equation is $D = \mathbb{R}^2$ and since $p \ge 1$ $f$ is a continuous function. Then $\frac{df}{dx} = p|x|^{p-1}$ and this is sufficient to have global uniqueness. See the answer to this other question. As Daniele points out in his answer this is clearly, $x \equiv 0$.