Investigate uniqueness of the following differential equation

Question

This is Problem 1.13 from Teschl's book (ODEs and Dynamical Systems). The equation is as follows:

$$ x' = \begin{cases} -t\sqrt{|x|}, &\text{ if } x \ge 0\\ t\sqrt{|x|}, &\text{ if } x \le 0. \end{cases} $$

It asks to show that the initial value problem $x(0) = x_0$ has unique global solution for every $x_0 \in \mathbb{R}$. There is a hint saying that if $x(t)$ is a solution so is $-x(t)$ and $x(-t)$, so it suffices to consider $x_0 \ge 0$ and $t \ge 0$. For $x_0 > 0$, I managed to find the solution

$$ x(t) = \begin{cases} (\sqrt{x_0} - \frac{1}{4}t^2)^2, &\text{ if } |t| \le 2x_0^{1/4}\\ 0, &\text{ otherwise. } \end{cases} $$

It is easy to see that this is a solution by using the Mean Value Theorem when calculating $x'(t)$ for $t = \pm 2x_0^{1/4}$, because $x$ is continuous at these points and $x'(\theta) \to 0$ as $\theta \to t$. For $x_0 = 0$ the obvious solution is $x(t) = 0$ for all $t$. The problem is that I do not know how to show these solutions are unique.

If $x$ is a is solution and $x_0 > 0$, then by continuity there is an open interval $(t_1, t_2)$ containing $0$ where $x(t)$ is positive, and so it makes sense to write

$$ -s = \frac{x'(s)}{\sqrt{x(s)}}, $$

for $s \in (t_1, t_2)$, and to integrate both sides

\begin{align} -\frac{1}{2}t^2 = \int_{0}^t -s\ ds = \int_0^t \frac{x'(s)}{\sqrt{x(s)}}\ ds = 2(\sqrt{x(t)} - \sqrt{x(0)}). \end{align}

This yields $x(t) = (\sqrt{x_0} - \frac{1}{4}t^2)^2$ for $t \in (t_1, t_2)$. Assuming that $(t_1, t_2)$ is a maximal interval where $x(t)$ is positive, and since $(t_1, t_2) \subseteq (-2x_0^{1/4}, 2x_0^{1/4})$, we see by continuity of $x$ that $x(t_1) = x(t_2) = 0$ and thus $(t_1, t_2) = (-2x_0^{1/4}, 2x_0^{1/4})$. How do I show that $x(t) = 0$ when $t$ is not in this interval? I suspect the same technique in this last step can be used to show that $x(t) = 0$ when $x_0 = 0$.

Answer 1

Alright, I've done it. Suppose first $x_0 = 0$ and let $x(t)$ be a solution. Assume by contradiction that $x(t) > 0$ for some $t > 0$ and set $$ d = \inf\{a < t: x(s) > 0 \text{ for every } s \in (a, t) \}. $$ By continuity $d$ is well defined, $0 \le d < t$ and $x(d) = 0$. By the Mean Value Theorem, we see that $$ x(t) = x(t) - x(d) = x'(\theta)(t - d), $$ for some $\theta \in (d, t)$, so $$ x(t) = -\theta\sqrt{|x(\theta)|}(t - d) \le 0 $$ which is a contradiction. If $x(t) < 0$ the proof should be analogous and we can do the same thing to finish what I wrote above for the case $x_0 > 0$.

Answer 2

To prove uniqueness, let's assume two solutions $x(t)$ and $y(t)$, $x(0)=y(0)=x_0>0$. Let $u=x-y$ with $u(0)=u_0=0$. We subtract the equations for $x$ and $y$ and obtain $$ u'(t)=-t\frac{u}{\sqrt{x}+\sqrt{y}}, $$ which yields $$ u(t)=u_0 \exp\left(-\int_0^t \frac{s}{\sqrt{x(s)}+\sqrt{y(s)}}ds\right). $$ Since $u_0=0$, then $u\equiv0$.