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Consider the transport equations $$ (1) \qquad \partial_t u + \operatorname{div}(bu) = 0$$ and $$ (2) \qquad \partial_t u + b \cdot \nabla u= 0$$

Can we define a notion of entropy solutions for (1) and (2) as it is done for the conservation law $$ (3) \qquad \partial_t u + \operatorname{div} f(u)= 0$$ and prove the corresponding well-posedness result?

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If $b$ is divergence free and in $L^1_{loc}$, it makes a lot of sense to define the entropy solution of $u_t + \mathrm{div}(b u) = 0$ as an $L^\infty$ function so that for every convex function $\eta$, we have $$ \partial_t (\eta(u)) + \mathrm{div} (b \eta(u)) \leq 0,$$ in the sense of distributions.

I'm pretty sure there is no uniqueness result in the literature for linear transport equations like this that requires the solution to be "entropy". If the vector field $b$ is further assumed to be of bounded variation, then there is a unique solution in the sense of distributions (lecture notes here: http://cvgmt.sns.it/paper/1149/)

I believe that in order to obtain uniqueness of entropy solutions for rough vector fields (as in $b \in L^1$) we would need a cleverer definition than the one I said above. For example, if we consider the usual example for Bressan's conjecture (See the item 6. in here: https://www.math.psu.edu/bressan/PSPDF/prize1.pdf), we can take initial data $u$ equal to $+1$ and $-1$ in each half of the square and speed up time appropriately so that it is completely mixed by time one. The natural (entropy?) solution after than time would be constant zero regardless of how the vector field $b$ continues. But we can also run the reverse transport equation to unmix it and make it flow back to $+1$ and $-1$ in each half of the square. It would also be an entropy solution according to the attempted definition above (I think ;)). It means that this definition is not preventing the (un)mixing effect appropriately.

I'm not even going to try vector fields with non-zero divergence.

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  • $\begingroup$ Thank you. Your answer prompted a follow up question, which is here: mathoverflow.net/questions/329960/… $\endgroup$ – Riku Apr 25 at 15:09
  • $\begingroup$ Also, why should a definition "prevent the (un)mixing effect appropriately"? And do renormalized solutions do that? $\endgroup$ – Riku Apr 25 at 15:25

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