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?

  • $\begingroup$ @WillieWong Thanks for your interest. Well, I suspect it is my fault, as I have not written that $\mathbb R^+:=[0,+\infty)$ with zero included (this is a way to recover the "initial datum"). I will edit accordingly right now. Does this clarify? $\endgroup$
    – user111164
    Aug 29, 2018 at 15:35
  • $\begingroup$ Ah I see. So yes, I was missing something. :-) It is clear now. $\endgroup$ Aug 29, 2018 at 17:56
  • $\begingroup$ @WillieWong On the contrary, that was my fault, as I had not specified it. Thanks for your kind comment :-) $\endgroup$
    – user111164
    Aug 29, 2018 at 19:52

1 Answer 1


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 $g'(t)=0$.

  • $\begingroup$ Uh thanks, very interesting! It seems a very good trick to take in mind. I think my problem is solved by your helpful answer, thanks a lot! $\endgroup$
    – user111164
    Aug 30, 2018 at 9:53

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