0
$\begingroup$

Is there an "interesting" class of analytically-integrable, divergence-free vector fields over $\mathbb{R}^2$ and/or $\mathbb{R}^3$?

That is, I'm looking for a large class of vector fields given by $V(x;\eta)$ for $x\in\mathbb{R}^k$, $k\in\{2,3\}$, parameterized by a fixed set of values $\eta\in\mathbb{R}^n$ such that:

  • $\nabla_x \cdot V(x;\eta)=0\ \forall x\in\mathbb{R}^k$
  • There exists an easily-evaluated (closed-form?) function $\Phi(t;x_0,\eta)$ such that the ODE $p'(t)=V(p(t);\eta)$ is solved by taking $p(t)=\Phi(t;p(0),\eta).$

For example, the vector fields $V(x;A)=Ax$ admit $\Phi(t;p_0,\eta)=e^{At}p_0$, but the subset of $A$'s giving divergence-free vector fields is not too rich/interesting.

$\endgroup$
  • $\begingroup$ Have you tried the 'canonical' 2D fields, with Hamiltonian $H(x,y)=sin(k\pi x)cos(m\pi y)$ $\endgroup$ – Piyush Grover Jul 15 '15 at 12:28
  • $\begingroup$ I'm mainly interested in the 3D case, although I guess a similar separable trick can be applied. My intuition (perhaps incorrect) was that it is easy to show that integral curves stay on level sets, but that a closed form $\Phi$ is harder to come by. I'll try some calculations on paper! $\endgroup$ – Justin Jul 15 '15 at 15:32
1
$\begingroup$

You probably need to specify more conditions, as the ones you give are too loose to be interesting. For example, consider the divergence-free vector field in the $xy$-plane given by $$ V = f(x)\,\frac{\partial\ }{\partial y}\,. $$ It is integrated by $$ \Phi(t,x_0,y_0) = \bigl(x_0, y_0 + t\,f(x_0)\bigr). $$ Now let the arbitrary function $f$ depend on as many $\eta$-parameters as you want. (For example, you could let $f$ range over the polynomials in $x$ of degree $n{-}1$ or less.)

Now, conjugating this example by any area-preserving diffeomorphism of the plane (and there are very many of these, even polynomial ones), and you can easily construct infinitely many such families that can be analytically integrated, even though that won't be apparent to the eye. The situation in $\mathbb{R}^3$ is even more flexible.

Maybe you need to specify what you mean by 'interesting'.

$\endgroup$
  • $\begingroup$ Thanks for your help! I'm looking for a class of analytically integrable, div-free vector fields that reasonably approximates a large part of $\{V:\nabla\cdot V=0\}$ -- e.g. a "Taylor series"-like expansion in terms of divergence free, analytically integrable vector fields. I couldn't find many that circulate around more than one point. This would be quite useful for a numerical integration problem I'm studying -- any pointers would be quite useful! $\endgroup$ – Justin Jul 15 '15 at 10:53
0
$\begingroup$

Any Hamiltonian vectorfield is divergence free. In two dimensions, the converse also holds. Therefore, you can obtain explicit solutions by finding the level curves of the Hamiltonian.

$\endgroup$
  • $\begingroup$ Well, no. Actually, that will only give you one 'first integral'; it won't actually tell you how to integrate the flow on each level set of the Hamiltonian. The OP wanted to find the explicit formula for the flow $\Phi$ of the vector field $V$, and finding a first integral is not enough to do that. $\endgroup$ – Robert Bryant Jul 15 '15 at 11:56
  • $\begingroup$ @Robert Bryant: It depends on what you mean by "explicit." On each level curve of the Hamiltonian, you have to solve an autonomous ODE in one dimension. This can always be done in terms of integrations and inverse functions. Of course, whether you can do those "explicitly" is another question, but one that is usually deemed at a "lower level." $\endgroup$ – Michael Renardy Jul 15 '15 at 12:35
  • $\begingroup$ I know the question I posed is a bit ill-posed :-) - still working on formulation. In this case, as Robert suggests, I'm looking to incorporate these formulas into a larger technique for numerical integration, so an explicit or nearly-explicit form for $\Phi$ rather than just the level set plus a 1D problem would be preferable. If I have to do algorithmic work to find $\Phi(t;p_0,\eta)$ after extracting the level set, it may be less clear of an advantage algorithmically beyond simpler methods for numerical integration. Thanks all for your help! Please keep the input coming! $\endgroup$ – Justin Jul 15 '15 at 15:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.