# Lagrangean uniqueness versus Eulerian uniqueness

(1) Lagrangean description. Let us consider a $N\times N$ system of autonomous ODE: $$\dot x=a(x),\quad \mathbb R\ni t\mapsto x(t)\in \mathbb R^N,\quad a:\mathbb R^N\rightarrow \mathbb R^N.$$ Instead of assuming the traditional Cauchy-Lipschitz hypothesis, let me suppose $$\vert a(x_1)-a(x_2)\vert\le \omega(\vert x_1-x_2\vert)$$ where $\omega:(0,+\infty)\rightarrow(0,+\infty)$ continuous increasing, $\omega(0_+)=0$ such that the so-called Osgood condition holds: $$\exists R>0,\quad\text{such that}\quad \int_0^R\frac{dr}{\omega(r)}=+\infty.\tag{Osgood}$$ It is of course satisfied by $\omega(r)=r$ (Lipschitz case) but also by $\omega(r)=r\log(1/r)$ or $r\log(1/r)r\log\log(1/r)$. Existence and uniqueness are not difficult to prove and follow from a Gronwall-type argument. The flow has some regularity properties, is not Lipschitz-continuous in general, but is better than Hölder of any index $<1$.

(2) Eulerian description. We consider the PDE, in fact the autonomous vector field $$X=\sum_{1\le j\le N}a_j(x)\frac{\partial}{\partial x_j}$$ where $a=(a_1,\dots,a_N)$ satisfies the above Osgood assumption. MY QUESTION: How can we prove uniqueness of $L^\infty$ weak solutions $u$ to the IVP $$Xu=f,\quad u_{\vert \Sigma}=0,\quad \text{\Sigma is a C^1 hypersurface transverse to X}?$$

(3) Comment: DiPerna-Lions theory does not apply since derivatives of the function $a$ could even fail to be a measure and may be a distribution of positive order. If you don't like autonomous vector fields, you can formulate the question for $$X=\frac{\partial}{\partial t}+\sum_{1\le j\le N}a_j(t,x)\frac{\partial}{\partial x_j},\quad \Sigma\equiv t=0,\quad \vert a_j(t,x)-a_j(t,y)\vert\le C\omega(\vert x-y\vert).$$

• The uniqueness of the PDE in $L^\infty$ would follow from the solvability backwards of the dual equation in $L^1$. From the solvability of the ODE you can solve the dual equation in the space of measures by pulling back the flow. What I don't quite see is how to show that these measures would stay absolutely continuous so that they are an $L^1$ function. The problem is equivalent to understanding whether the flow of the ODE maps sets of measure zero to sets of measure zero. I don't know if that is true but I think it is equivalent to your question. This paper is related arxiv.org/abs – Luis Silvestre Dec 1 '12 at 5:05
• Sorry, I meant this paper is related, but does not solve your question. Ambrosio, Luigi(I-SNS); Bernard, Patrick(F-PARIS9-A) Uniqueness of signed measures solving the continuity equation for Osgood vector fields. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei Mat. Appl. 19 (2008), no. 3, 237–245. – Luis Silvestre Dec 1 '12 at 5:07
• @Luis Silvestre Thanks for your remarks. I certainly agree with the fact that the quoted paper by Ambrosio \& Bernard does not solve my question. The problem seems to come from the definition of the product of a measure by a measurable function, always possible in measure theory, but not in distribution theory. In particular, the definition of a weak solution for a PDE is indeed using distribution theory, with a formal integration by parts. – Bazin Dec 2 '12 at 19:18