In the last post, we saw that solutions of the non-autonomous Riccati equation
for some functions with .
We also saw that if (1) is autonomous — the polynomial is independent of — then the fixed points of are the roots of , and in particular, the classification of as hyperbolic, parabolic, or elliptic can be deduced from knowledge of the roots of .
Is a similar qualitative analysis available in the non-autonomous case? What if we know that always has two real roots, but those roots may move around? Can we still deduce that is a hyperbolic FLT? This question (along with some more sophisticated variants) arose recently in an example of a dynamical system, which is part of a joint project with Dmitry Dolgopyat and Yakov Pesin, for which we would like to understand the evolution of tangent vectors under a particular flow. This evolution turns out to be governed by a non-autonomous Riccati equation where the roots of may vary, and one would like to nevertheless have information on the map — in particular, to know where its fixed points are.
By differentiating (2) and using (1), one could obtain ODEs for the coefficients . The problem is that these end up being no easier to solve explicitly than (1), and moreover the information we are most interested in — the presence and location of fixed points of — cannot be read directly from these coefficients. The solution is to derive ODEs that deal directly with the fixed points themselves. This is Proposition 2 below — first it’s worth formulating the qualitative result that can eventually be obtained. The second paragraph of this result contains some technical conditions that can be improved with a little more work.
Proposition 1 Let be a time-dependent quadratic polynomial in for , and let be disjoint closed intervals (independent of ) such that has one root in and one root in for each . Suppose that and let be the length of the interval between them. Suppose also that each is decreasing on and increasing on . (There is an analogous result for concave .)
Furthermore, suppose that is such that for every and for every . Let be the minimum and maximum values, respectively, of taken over and . Suppose that , and let . (Note that .)
Under the above conditions, for every , the solution map of (1) is a hyperbolic fractional linear transformation with fixed points and . Moreover, and for every .
Observe that a fractional linear transformation is uniquely specified by its two fixed points and by the derivative at either one of those fixed points. We will keep track of the four quantities , , , . We use the notational convention that denotes either or , and denotes the other one of the two.
whenever the right-hand sides are well-defined.
In the autonomous case, we have for all , and so all that remains is to solve (4) in the form , and we get , which is enough information to compute the FLT .
For Proposition 1, the idea is to observe that (4) together with the technical condition in the second paragraph of the proposition guarantees that , and similarly , as long as . As long as , (3) guarantees that moves in the direction of the corresponding root of , and hence cannot leave .
We end with the proof of Proposition 2.
We will write and for the partial derivatives of . Assuming , the fixed points for can be found by observing that
so that dividing by and taking the limit gives
The first term is given by
In the second term, is given by (3), while for we need to recall that the derivative of a fractional linear transformation is
thanks to the normalisation . Differentiating again gives
and so . Thus we must solve for . Recall that the fixed points are roots of (5), and so their sum and product are given in terms of the coefficients of that equation as
Moreover, we have and . Substituting in the first equation of (7) gives
Eliminating in the equation for the determinant gives
and using the above expressions for , we have
which simplifies to
so that is given in terms of as
and the same equation holds with the indices reversed. We will write this state of affairs as
with the understanding that and are the two indices and , in either order. Now returning to (6), we have
Thus evolves according to the equation (4).