More on Riccati equations and fractional linear transformations

In the last post, we saw that solutions of the non-autonomous Riccati equation

$\displaystyle \dot{q} = p_t(q) = a(t) q^2 + b(t) q + c(t) \ \ \ \ \ (1)$

are given by fractional linear transformations — that is, if ${\Phi(t)}$ is the map taking the initial condition ${q=q_0}$ to the solution ${q(t)}$ at time ${t}$, then

$\displaystyle \Phi(t)(q) = \frac{A(t) q + B(t)}{C(t) q + D(t)} \ \ \ \ \ (2)$

for some functions ${A,B,C,D\colon {\mathbb R}\rightarrow {\mathbb R}}$ with ${AD-BC=1}$.

We also saw that if (1) is autonomous — the polynomial ${p=p_t}$ is independent of ${t}$ — then the fixed points of ${\Phi(t)}$ are the roots of ${p}$, and in particular, the classification of ${\Phi(t)}$ as hyperbolic, parabolic, or elliptic can be deduced from knowledge of the roots of ${p}$.

Is a similar qualitative analysis available in the non-autonomous case? What if we know that ${p_t}$ always has two real roots, but those roots may move around? Can we still deduce that ${\Phi(t)}$ 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 ${p_t}$ may vary, and one would like to nevertheless have information on the map ${\Phi(t)}$ — in particular, to know where its fixed points are.

By differentiating (2) and using (1), one could obtain ODEs for the coefficients ${A,B,C,D}$. 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 ${\Phi(t)}$ — 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 ${p_t(q)}$ be a time-dependent quadratic polynomial in ${q}$ for ${t\in[0,T]}$, and let ${I_1,I_2\subset {\mathbb R}}$ be disjoint closed intervals (independent of ${t}$) such that ${p_t}$ has one root in ${I_1}$ and one root in ${I_2}$ for each ${t\in [0,T]}$. Suppose that ${I_1 < I_2}$ and let ${\Delta}$ be the length of the interval between them. Suppose also that each ${p_t}$ is decreasing on ${I_1}$ and increasing on ${I_2}$. (There is an analogous result for concave ${p_t}$.)

Furthermore, suppose that ${J\subset (0,\infty)}$ is such that ${p_t'(q)\in J}$ for every ${q\in I_2}$ and ${p_t'(q)\in -J}$ for every ${q\in I_1}$. Let ${m<0 be the minimum and maximum values, respectively, of ${p_t(q)}$ taken over ${t\in[0,T]}$ and ${q\in I_1 \cup I_2}$. Suppose that ${2m > - \Delta \inf J}$, and let ${J'=J + [2m/\Delta, 2M/\Delta] \supset J}$. (Note that ${J'\subset (0,\infty)}$.)

Under the above conditions, for every ${t\in[0,T]}$, the solution map ${\Phi(t)}$ of (1) is a hyperbolic fractional linear transformation with fixed points ${w_1\in I_1}$ and ${w_2\in I_2}$. Moreover, ${D_q(\Phi(t))(w_2) \in e^{tJ'}}$ and ${D_q(\Phi(t))(w_1) \in e^{-tJ'}}$ for every ${t\in[0,T]}$.

Observe that a fractional linear transformation is uniquely specified by its two fixed points ${w_1,w_2\in \hat{\mathbb C}}$ and by the derivative ${z_j := D_q(\Phi(t))(w_j)}$ at either one of those fixed points. We will keep track of the four quantities ${w_1}$, ${w_2}$, ${z_1}$, ${z_2}$. We use the notational convention that ${w_j}$ denotes either ${w_1}$ or ${w_2}$, and ${w_i}$ denotes the other one of the two.

Proposition 2 The fixed points ${w_j}$ of the FLT ${\Phi(t)}$ evolve according to the ODE

$\displaystyle \dot{w}_j = \frac{p_t(w_j)}{1-z_j}, \ \ \ \ \ (3)$

and the derivatives ${z_j}$ at those fixed points evolve according to

$\displaystyle \dot{z}_j = z_j \left( p_t'(w_j) + \frac{2p_t(w_j)}{w_i-w_j}\right), \ \ \ \ \ (4)$

whenever the right-hand sides are well-defined.

This will be proved momentarily. First we observe that Proposition 2 is enough to solve autonomous Riccati equations, and give a sketch of how it can be used to establish Proposition 1.

In the autonomous case, we have ${p_t(w_j)=0}$ for all ${t}$, and so all that remains is to solve (4) in the form ${\dot{z}_j = z_j p'(w_j)}$, and we get ${z_j(t) = e^{tp'(w_j)}}$, which is enough information to compute the FLT ${\Phi(t)}$.

For Proposition 1, the idea is to observe that (4) together with the technical condition in the second paragraph of the proposition guarantees that ${\frac{d}{dt} (\log z_2)\in J' \subset (0,\infty)}$, and similarly ${\frac d{dt} (\log z_1) \in -J'}$, as long as ${w_j\in I_j}$. As long as ${z_1 < 1 < z_2}$, (3) guarantees that ${w_j}$ moves in the direction of the corresponding root of ${p_t}$, and hence cannot leave ${I_j}$.

We end with the proof of Proposition 2.

Proof: The fixed points ${w_j\in\hat{\mathbb C}}$ are the roots of ${q = \frac{Aq+B}{Cq+D}}$, or equivalently of the quadratic equation

$\displaystyle Cq^2 + (D-A)q - B = 0. \ \ \ \ \ (5)$

We will write ${\phi(t,q) = \Phi(t)(q)}$ and ${D_t,D_q}$ for the partial derivatives of ${\phi}$. Assuming ${z_j(t_0) = D_q\phi(t_0,w_j(t_0)) \neq 1}$, the fixed points for ${t\approx t_0}$ can be found by observing that

\displaystyle \begin{aligned} w_j(t) - w_j(t_0)&= \phi(t,w_j(t)) - \phi(t_0,w_j(t_0)) \\ &= (t-t_0)D_t\phi(t_0,w_j(t_0)) + (w_j(t)-w_j(t_0)) D_q\phi(t_0,w_j(t_0)) \\ &\qquad\qquad + o(t-t_0) + o(w_j(t)-w_j(t_0)), \end{aligned}

so that dividing by ${t-t_0}$ and taking the limit gives

$\displaystyle \dot{w}_j = D_t\phi(t_0,w_j(t_0)) + \dot{w}_j \cdot D_q\phi(t_0,w_j(t_0)).$

Recalling the ODE (1) and the definition of ${z_j}$ gives (3).

The derivation of the equation for the evolution of ${z_j}$ is a little more involved. First we differentiate ${z_j = (D_q\phi)(t,w_j(t))}$ to get

$\displaystyle \dot{z}_j = (D_{tq}\phi)(t,w_j(t)) + \dot{w}_j(t) \cdot (D_{qq}\phi)(t,w_j(t)). \ \ \ \ \ (6)$

The first term is given by

$\displaystyle D_{tq}\phi(t,w_j(t)) = D_q D_t\phi(t,w_j(t)) = D_q p_t(\phi(t,w_j(t))) = D_q\phi(t,w_j(t)) \cdot p_t'(\phi(t,w_j(t))) = z_j p_t'(w_j).$

In the second term, ${\dot{w}_j}$ is given by (3), while for ${D_{qq}\phi}$ we need to recall that the derivative of a fractional linear transformation is

$\displaystyle D_q\Phi = D_q\left(\frac{Aq+B}{Cq+D}\right) = \frac 1{(Cq+D)^2},$

thanks to the normalisation ${AD-BC=1}$. Differentiating again gives

$\displaystyle D_{qq}\Phi = -2C(Cq+d)^{-3} = -2C (D_q\Phi)^{3/2},$

and so ${D_{qq}\phi(t,w_j(t)) = -2C(t) z_j^{3/2}}$. Thus we must solve for ${C(t)}$. 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

$\displaystyle A-D = C(w_1 + w_2), \qquad -B = Cw_1 w_2. \ \ \ \ \ (7)$

Moreover, we have ${AD-BC=1}$ and ${z_j^{-1/2} = Cw_j + D}$. Substituting ${D=z_j^{-1/2} - Cw_j}$ in the first equation of (7) gives

$\displaystyle A = z_1^{-1/2} + C w_2 = z_2^{-1/2} + C w_1.$

Eliminating ${B}$ in the equation for the determinant gives

$\displaystyle AD + C^2w_1w_2 = 1,$

and using the above expressions for ${A,D}$, we have

$\displaystyle (z_1^{-1/2} + Cw_2)(z_1^{-1/2} - Cw_1) + C^2 w_1w_2 = 1,$

which simplifies to

$\displaystyle z_1^{-1} + C(w_2 - w_1)z_1^{-1/2} =1,$

so that ${C}$ is given in terms of ${z_j,w_j}$ as

$\displaystyle C = \frac{1-z_1^{-1}}{(w_2-w_1)z_1^{-1/2}},$

and the same equation holds with the indices reversed. We will write this state of affairs as

$\displaystyle C = \frac{1-z_j^{-1}}{(w_i-w_j)z_j^{-1/2}},$

with the understanding that ${i}$ and ${j}$ are the two indices ${1}$ and ${2}$, in either order. Now returning to (6), we have

\displaystyle \begin{aligned} \dot{z}_j &= z_j p_t'(w_j) + \frac{p_t(w_j)}{1-z_j}(-2C(t) z_j^{3/2}) \\ &= z_j p_t'(w_j) -2\frac{p_t(w_j)}{1-z_j} z_j^{3/2}\frac{1-z_j^{-1}}{(w_i-w_j)z_j^{-1/2}} \\ &= z_j p_t'(w_j) + \frac{2p_t(w_j)z_j}{w_i-w_j}. \end{aligned}

Thus ${z_j}$ evolves according to the equation (4). $\Box$