Lagrangean uniqueness versus Eulerian uniqueness

2012-11-29T21:21:23Z

(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 $$ X=\frac{\partial}{\partial t}+\sum_{1\le j\le N}a_j(t,x)\frac{\partial}{\partial x_j},\quad \Sigma\equiv t=0,\quad \vert a_j(t,x)-a_j(t,y)\vert\le C\omega(\vert x-y\vert). $$

Lagrangean uniqueness versus Eulerian uniqueness - MathOverflow

Lagrangean uniqueness versus Eulerian uniqueness