3
$\begingroup$

I am looking for results concering the following parabolic PDE $$u\cdot\nabla u + \Delta u = F(x),$$ where $$u\colon\Omega\to\mathbb{R}^2,$$ and $\Omega\subset\mathbb{R}^2$ is a 2D domain (bounded or unbounded), preferably the torus ($\mathbb{T}^2$). I do not specify any boundary condition, as I'm interested in results for any of periodic, Dirichlet, Neumann bd.cond, or unbounded domain.

This equation can be written in scalar form as $$uu_x+vu_y+u_{xx}+u_{yy}=f\\uv_x+vv_y+v_{xx}+v_{yy}=g$$

This equation can be interpreted as time-independent 2D forced viscous Burgers' equation.

I am looking for any known qualitative theoretical results. I've been surprised as little is available in the literature. I searched the net a bit for any existence, uniqueness, multiplicity of solutions result for this equation , but ended up having a bunch of numerical papers, and none theoretical. It looks like this equation is often used as a test-case for numerical methods.

I am aware that the case $F(x)=0$ is simpler, as the equation reduces to a linear one by Hopf-Coles' transform, at least in some specific cases, and there are some results for that.

I am interested in the cases $\|F\|\neq 0$, and especially $\|F\|$ large.

I would also appreciate results for time-dependant version, i.e. $u_t=u\cdot\nabla u + \Delta u + F$.

$\endgroup$
  • $\begingroup$ I am dubious about the Hof-Cole transformation in this vector-valued case. $\endgroup$ – Denis Serre Oct 21 '14 at 9:20
  • $\begingroup$ I was either, but here onlinelibrary.wiley.com/doi/10.1002/fld.1650030302/abstract I found a remark that it can be interpreted as multidim transform $\endgroup$ – jaco Oct 21 '14 at 10:43
  • $\begingroup$ I suspected something like that. This paper considers the special case of irrotational solutions, that is when $u=\nabla \theta$ (incidentally, this assumes that the force is potential). Then the system reduces to a Hamilton-Jacobi equation for the unknown $\theta$. This being scalar, teh Hopf-Cole transformation can be carried out. $\endgroup$ – Denis Serre Oct 21 '14 at 10:47
  • $\begingroup$ good remark, I will modify a bit the question $\endgroup$ – jaco Oct 21 '14 at 11:05
  • $\begingroup$ In my experience, "Burgers' equation" refers to the time-dependent version. I think you should call this "steady state" or "time-independent" to be more clear. $\endgroup$ – David Ketcheson Oct 21 '14 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.