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Edit: According to the comment of Willie Wong I realize that the previous version was trivial. I thank him for his comment. Now I revise it.

We consider the heat equation $$U_{t}=\Delta U\\U(x,y,0)=F(x,y)$$

Here $U$ is not a real valued function in $x,y$ but for every $t$, $U$ is a vector field on the $x-y$ plane. Actually the unknown $U$ is a map from $\mathbb{R}^{3}$ to $\mathbb{R}^{2}$.(Not to $\mathbb{R}$) In this question we are interested in the following particular initial condition $$F(x,y)=(y-F(x))\partial_{x}-x \partial_{y}$$ where $F$ is an even polynomial. One can replace this initial system with any other polynomial vector field with a band of closed orbits. For example a polynomial hamiltonian vector field $$H_{y}\partial_{x}- H_{x}\partial_{y}$$

What can be said about this PDE? Is it true to say that, for sufficiently small $t$, there is a solution $U$ which is defined for all $(x,y)$? if the answer is yes, how is the dynamic of the vector field $U(x,y,t)$, for each fixed and sufficiently small $t$? Are there some limit cycles? How many limit cycles does $U$ have? What periodic orbits of the initial vector fields can generate "limit cycles' for sufficiently small $t$? The later question can be considered as a heat analogy of abelian integrals described here:

A question on "The weakened Hilbert 16th problem"

What type of criterion can be introduced for counting the number of limit cycles of $U(x,y,t)$, for sufficiently small $t$?

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    $\begingroup$ Isn't your initial data harmonic? (The components have linear coefficients.) So a solution is simply the time-independent one which exists for all $t$. Is there a typo in your question? $\endgroup$
    – Willie Wong
    Commented Oct 15, 2015 at 13:58
  • $\begingroup$ @WillieWong thank you very much for your comment I revise the question. $\endgroup$
    – Ali Taghavi
    Commented Oct 15, 2015 at 14:06
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    $\begingroup$ Wellposedness for parabolic equations with non-decaying (and indeed growing) initial data is a tricky business. But for your new particular initial condition, a solution is $$U(x,y,t) = (y - x^2 - 2t) \partial_x - x \partial_y,$$ which I again note exists for all time. Since the solution is explicit you can analyze the dynamics yourself, I am sure. $\endgroup$
    – Willie Wong
    Commented Oct 15, 2015 at 14:17
  • $\begingroup$ @WillieWong I apologize for triviality of the previous version. Thanks again for your interesting comment. $\endgroup$
    – Ali Taghavi
    Commented Oct 15, 2015 at 14:17
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    $\begingroup$ To forestall further edits: in the case of $F(x)$ being a polynomial you can always find an explicit solution to the PDE. Since $F(x)$ has even degree, fix the degree to be $2k$, and you can postulate the ansatz $$ G(t,x) = F(x) + \sum_{i = 1}^k t P_{2k - 2i}(x) $$ where each of the $P_{2k-2i}$ is a $2k-2i$ degree polynomial in $x$. Then your PDE becomes an entirely algebraic set of equations. Solving PDEs for polynomial (or entire functions, for that matter) data is not really all that interesting, and probably doesn't help you too much in your real goal. $\endgroup$
    – Willie Wong
    Commented Oct 15, 2015 at 14:28

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