Differential equation with no unique solution Can someone help me find solutions for the following differential equation: $x'=-t\,\text{sign}(x)\sqrt{|x|}$ with $x(\tau)=\xi$. 

With $x$ a function of $t$ and $\tau$ and $\xi$ constants.
It is possible to compute solutions where $x(t)$ is not changing its sign. Since you wanted to find 'solutions' and not 'all solutions' I will not investigate the case where $x(t)$ changes its sign. (It seems strenuous.)

Now let's take the IVP
$$\begin{cases} x'&amp;=-t\,\text{sign}(x)\sqrt{|x|} \\ x(\tau)&amp;=\xi \end{cases}$$ 

Assuming $x(t)&gt;0$ for all $t$:

We get $x'=-t \sqrt{x}$ which has the solution (by separation of variables) $$x(t)=\frac{1}{16}(t^2-A)^2&gt;0$$ 
with $x'(t)=\frac{1}{4}(t^2-A)t$ where $A&gt;0$ is a constant that has the properties:


$x'(t)&lt;0$ i.e. $\begin{cases} \text{for $t&gt;0$: }~t^2-A&lt;0 \Rightarrow t^2&lt;A \Rightarrow |t|&lt;\sqrt{A} \Rightarrow 0&lt;t&lt;\sqrt{A} \\ \text{for $t&lt;0$: }~t^2-A&gt;0 \Rightarrow t^2&gt;A \Rightarrow |t|&gt;\sqrt{A} \Rightarrow t&lt;-\sqrt{A} \end{cases}$
$x(\tau)=\xi$ i.e. $\begin{cases} \text{for $\tau&gt;0$: }~\tau^2-A=- 4 \sqrt{\xi}&lt;0 \Rightarrow A=\tau^2 + 4\sqrt{\xi} \\ \text{for $\tau&lt;0$: }~\tau^2-A= 4 \sqrt{\xi}&gt;0 \Rightarrow A=\tau^2 - 4\sqrt{\xi} \\ \text{for $\tau=0$: }~ 0-A=-4\sqrt{\xi}&lt;0 \Rightarrow A=4\sqrt{\xi} \end{cases}$


The first condition is problematic for arbitrarily large $t$. Since $A$ is specified by the second condition we get by the first condition that $x(t)=\frac{1}{16}(t^2-A)^2$ is only a solution for $$t \in \begin{cases} \left(-\infty,-\sqrt{\tau^2+4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{\tau^2+4 \sqrt{\xi}} \right) &amp;\text{ if } \tau&gt;0\\ 
\left(-\infty,-\sqrt{\tau^2-4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{\tau^2-4 \sqrt{\xi}} \right) &amp;\text{ if } \tau&lt;0 \wedge  \tau^2&gt;4 \sqrt{\xi} \\ \left(-\infty,-\sqrt{4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{4 \sqrt{\xi}} \right) &amp;\text{ if } \tau=0 \end{cases}$$

Assuming $x(t)&lt;0$ for all $t$: 

We get $x'=t\sqrt{-x}$ i.e. $y'=-t\sqrt{y}$ for $y=-x$. So we are back in our first case just with a minus and the solution has to stay negative.
If $t\mapsto x(t)&gt;0$ is a solution defined in some interval $J$ then $$\left(2\sqrt{x(t)}+{1\over2}t^2\right)'={x'(t)\over\sqrt{x(t)}}+t=0\ ,$$
hence $$2\sqrt{x(t)}+{1\over2}t^2={1\over2}c^2$$
for some $c&gt;0$. This allows to conclude that
$$\sqrt{x(t)}={1\over4}(c^2-t^2)\qquad(-c&lt;t&lt;c)\ ,$$
or
$$x(t)={1\over16}(c^2-t^2)^2\qquad(-c&lt;t&lt;c)\ .\tag{1}$$
It is easy to check that these functions, together with their negatives, indeed solve the given ODE, and at the same time their graphs fill the complete $(t,x)$-plane, minus the line $x=0$, in the desired way.

The function $$x(t):\equiv0\tag{2}$$ is a solution as well, but a special one: The Lipschitz-assumption of the general existence and uniqueness theorem is not fulfilled in the points $(\tau,0)$ with $\tau\ne0$. Therefore it does not come as a surprise that IVPs with such a point as initial point have three solution germs starting there: the solution $(2)$ and two arcs of type $(1)$ going off towards $t=0$. In direction $|t|\to\infty$ there is just the trivial behavior $(2)$. See the following figure: