I have come across the following easy-looking problem that is driving me mad. 


I have a family of measures (on the real line $\mathbb R$) $\{\mu\_t\}\_{t>0}$ which is uniformly bounded (the measures being possibly signed). I know that $\mu\_0 = 0$ and that for every $\varphi \in C\_c^\infty([0,+\infty) \times \mathbb R)$ it holds
$$
\iint\_{(0,+\infty) \times \mathbb R} \left[ \partial\_t \varphi(t,x)- \sqrt[3]{x} \partial\_x \varphi(t,x) \right]  d\mu\_t(x)dt = 0. 
$$



> 
> **Question.** Does $\mu\_t \equiv 0$ for every $t>0$? 
> 
> 
> 


At the beginning I thought it was an easy consequence of the method of characteristics, but the point is that the (1d-)vector field $f(x) = -\sqrt[3]{x}$ is *not* Lipschitz, so the standard theory does not apply. Nevertheless, I know that the characteristics of $f$ are unique for every initial data (for $t>0$, am I right?). Do you know any references showing the extension of the method of characteristics to this setting?



Method of characteristics beyond the Lipschitz setting
Let us show that the condition ${\mu\_0}|\_{x>0} = 0$ implies that ${\mu\_T}|\_{x>0} = 0$ (for a.e. $T>0$). Let $\omega \in C\_0^\infty(\mathbb R)$ and $\delta\in (0,1)$ be such that $\omega(\xi) = 0$ for all $\xi\le \delta$. Define
$$
\varphi(t,x):= 
\begin{cases}
0, & \frac{3}{2} x^{\frac{2}{3}} + t - T \le \delta\\
\omega(\frac{2}{3}(\frac{3}{2} x^{\frac{2}{3}} + t - T)^{\frac{3}{2}}),
& \frac{3}{2} x^{\frac{2}{3}} + t - T > \frac{1}{2}\delta
\end{cases}
$$
(clearly this definition is consistent in the intersection of the cases above). Note that $\varphi$ is smooth, compactly supported (in $[0,T]\times \mathbb R$) and $\partial\_t \varphi + f \cdot \partial\_x \varphi = 0$,
so from the equation for $\mu\_t$ we get $0=\int\_{\mathbb R} \varphi(T, x) \, d\mu\_T(x) = \int\_{\mathbb R} \omega(x)\, d\mu\_T(x)$. By arbitrariness of $\omega$ it follows that $\mu\_T|\_{x>0} = 0$.


In a similar fashion one should be able to show that $\mu\_0|\_{x<0}=0$ implies that $\mu\_T|\_{x<0} = 0$.


Therefore $\mu\_t$ can only have the form $\mu\_t = g(t) \, \delta\_0$, and by the equation for $\mu\_t$ we get $g'(t)=0$.


