## Riccati equations and fractional linear transformations

In a typical course on ordinary differential equations, the Picard–Lindelöf theorem on existence and uniqueness of solutions is followed at some point by an example illustrating that such solutions may not be defined for all time, but may go to infinity in finite time. The simplest example of this phenomenon is the ODE

$\displaystyle \frac{dq}{dt} = q^2, \ \ \ \ \ (1)$

which can be solved by observing that ${\frac{d}{dt}(q^{-1}) = -1}$ and so

$\displaystyle q(t) = (q_0^{-1} - t)^{-1}. \ \ \ \ \ (2)$

Assuming ${q_0>0}$, this is the unique solution on the interval ${[0,q_0^{-1})}$, but the solution is not defined for larger values of ${t}$.

The situation changes slightly if one considers ${q}$ not simply as a real number but rather as a point on the real projective line ${{\mathbb R} P^1}$. The coordinate ${q}$ gives a point on ${{\mathbb R} P^1}$ modeled as ${\hat{\mathbb R} := {\mathbb R}\cup \{\infty\}}$, but we can also consider the model of ${{\mathbb R} P^1}$ as the set of all lines through the origin in ${{\mathbb R}^2}$, and associate to each ${q\in \hat{\mathbb R}}$ the line with slope ${q}$. If ${\theta}$ is the angle from the horizontal axis to this line, then ${q = \tan\theta}$, where we observe that ${\theta}$ is defined modulo ${\pi}$ (not modulo ${2\pi}$, since the line is not oriented), and we have ${\dot{q} = (\sec^2\theta)\dot\theta}$, so that

$\displaystyle \dot\theta = q^2\cos^2\theta = \sin^2\theta. \ \ \ \ \ (3)$

We conclude that the flow induced by (1) on ${{\mathbb R} P^1}$ has a single fixed point ${q=0}$ (which corresponds to ${\theta=k\pi}$), and every other trajectory is homoclinic to this point. The moment at which the trajectory passes through ${\theta=\pi/2}$ corresponds to the finite-time blow-up observed in (2).

In fact, the equation (2) does give the solution of the ODE for all time, including ${t>q_0^{-1}}$, if we consider ${q}$ as a point on ${{\mathbb R} P^1}$. In the coordinates given by ${\theta}$, (2) becomes ${\cot\theta(t) = \cot(\theta_0) - t}$, and so ${\theta(t) = \cot^{-1}(\cot(\theta_0) - t)}$, where the branch of ${\cot^{-1}}$ that we use is determined by the interval ${(k\pi,(k+1)\pi)}$ in which ${\theta_0}$ lies.

Moral: The ODE (1) can be interpreted as an ODE on ${{\mathbb R} P^1}$, and then solutions are defined for all ${t}$.

Now consider a more general ODE than (1):

$\displaystyle \dot{q} = aq^2 + bq + c,\qquad a,b,c\in{\mathbb R}. \ \ \ \ \ (4)$

This is an autonomous Riccati equation in one unknown. It can be shown to have a similar property of finite-time blow-up (for suitable initial conditions), and it turns out that we can profitably play the same game and study (4) as an ODE on ${{\mathbb R} P^1}$.

Write ${p(q) = aq^2 + bq+c}$ for the quadratic polynomial defining the ODE. Then equilibrium points of (4) correspond to roots of ${p}$. In particular, if we let ${\Phi(t)\colon {\mathbb R} P^1\rightarrow {\mathbb R} P^1}$ be the map taking ${q_0}$ to ${q(t)}$ for a solution of (1), then the fixed points of ${\Phi(t)}$ are the roots of ${p}$, so that ${\Phi(t)}$ has either 0, 1, or 2 fixed points.

There is a natural class of transformations of ${{\mathbb R} P^1}$ that each have 0, 1, or 2 fixed points: the fractional linear transformations. These are maps of the form

$\displaystyle q\mapsto \frac{Aq + B}{Cq+D}, \qquad A,B,C,D\in{\mathbb R}. \ \ \ \ \ (5)$

Note that the solution (2) of our first ODE (1) can be written as a fractional linear transformation:

$\displaystyle \Phi(t)(q) = \frac 1{q^{-1} - t} = \frac{q}{-tq + 1}.$

Proposition 1 For any coefficients ${a,b,c\in{\mathbb R}}$, there exists a one-parameter family ${\Phi(t)}$ of fractional linear transformations such that ${\Phi(0)}$ is the identity map, and for every ${q\in\hat{\mathbb R}}$ and every ${t\in{\mathbb R}}$, we have

$\displaystyle \frac{d}{dt} (\Phi(t)(q)) = a (\Phi(t)(q))^2 + b(\Phi(t)(q)) + c. \ \ \ \ \ (6)$

Proof: Compute the derivative of the cross-ratio ${(q_1,q_2;q_3,q_4) = \frac{q_1 - q_3}{q_2-q_3} \frac{q_2-q_4}{q_1-q_4}}$ of four points moving under the evolution given by (4). Show that this derivative is zero and conclude that the cross-ratio is constant, so that ${\Phi(t)}$ preserves cross-ratio for every ${t}$. Fractional linear transformations are precisely those transformations that preserve cross-ratio. $\Box$

Proposition 1 can also be proved by solving (4) explicitly via the change of variables ${q=-\dot{u}/(au)}$, which turns it into a second-order linear equation. However, the proof above does not require any such ansatz, and moreover it immediately yields a stronger result.

Proposition 2 Let ${a,b,c\colon {\mathbb R}\rightarrow{\mathbb R}}$ be any continuous functions and let ${p_t(q) = a(t)q^2 + b(t)q + c(t)}$. Then there exists a one-parameter family ${\Phi(t)}$ of fractional linear transformations such that ${\Phi(0)}$ is the identity map, and for every ${q\in\hat{\mathbb R}}$ and every ${t\in{\mathbb R}}$, we have

$\displaystyle \frac{d}{dt} (\Phi(t)(q)) = p_t(\Phi(t)(q)) = a (t) (\Phi(t)(q))^2 + b(t) (\Phi(t)(q)) + c(t). \ \ \ \ \ (7)$

Thus we see that even when we have a non-autonomous Riccati equation in which the coefficients in (4) are allowed to vary with ${t}$, the solution map ${\Phi(t)}$ is still a fractional linear transformation for all ${t}$. In the next post I’ll discuss some consequences of this that have turned out to be useful for a problem I’m interested in. In the meantime, let’s recall some basic facts about fractional linear transformations and make some observations about their consequences for autonomous Riccati equations.

Group structure of FLTs. The set of fractional linear transformations is closed under composition and forms a group, which is isomorphic to the group ${PSL(2,{\mathbb R})}$ of ${2\times 2}$ matrices with determinant ${1}$ and ${X,-X}$ identified. The isomorphism is given by associating the FLT ${q\mapsto \frac{Aq+B}{Cq+D}}$ to the matrix ${\left(\begin{smallmatrix} A&B \\ C &D\end{smallmatrix}\right)}$, assuming that the coefficients are normalised so that ${AD-BC=1}$. Often ${PSL(2,{\mathbb R})}$ is used to denote either of the two isomorphic groups.

Classification of FLTs and canonical examples. Every fractional linear transformation of ${{\mathbb R} P^1}$ is either hyperbolic (two fixed points), parabolic (one fixed point), or elliptic (no fixed points). If we consider ${\Phi}$ as a map of the Riemann sphere ${\hat{\mathbb C} = {\mathbb C} \cup\{\infty\}}$, with ${\hat{\mathbb R}}$ embedded as the real line together with the point at infinity, then every ${\Phi}$ has two fixed points (counting multiplicity), and elliptic FLTs are characterised by the fact that their fixed points lie in ${{\mathbb C}\setminus {\mathbb R}}$.

The canonical examples of each type of FLT are:

• ${q\mapsto e^{2\theta} q}$ (hyperbolic), associated to the matrix ${\left(\begin{smallmatrix} e^{\theta} & 0 \\ 0 &e^{-\theta}\end{smallmatrix}\right)}$, where ${\theta\in{\mathbb R}}$;
• ${q\mapsto q+1}$ (hyperbolic), associated to the matrix ${\left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)}$;
• ${q\mapsto \frac{(\cos \theta) q - \sin \theta}{(\sin \theta)q + \cos \theta}}$ (elliptic), associated to ${\left(\begin{smallmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{smallmatrix}\right)}$ for some ${\theta\in{\mathbb R}}$.

Any FLT can be conjugated to the corresponding canonical example: given any ${\Phi\in PSL(2,{\mathbb R})}$ there exists ${\Psi\in PSL(2,{\mathbb R})}$ such that ${\Psi^{-1} \circ \Phi \circ \Psi}$ is in one of the three forms above.

Normal forms for autonomous Riccati equations. Using the conjugacies ${\Psi}$ described above as changes of coordinates in the autonomous Riccati equation (4), every such equation can be brought into one of the following forms:

• ${\dot{q} = \lambda q}$ (if the polynomial ${p}$ has two real roots);
• ${\dot{q} = \lambda}$ (if the polynomial ${p}$ has one real root);
• ${\dot{q} = q^2 + 1}$ (if the polynomial ${p}$ has no real roots).

The solution maps ${\Phi(t)}$ in each of these three cases are hyperbolic, parabolic, and elliptic FLTs, respectively.

I'm an assistant professor of mathematics at the University of Houston. I'm interested in dynamical systems, ergodic theory, thermodynamic formalism, dimension theory, multifractal analysis, non-uniform hyperbolicity, and things along those lines.
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### 4 Responses to Riccati equations and fractional linear transformations

1. Mike says:

Hi Vaughn,

Thanks for posting this stuff on Riccati equations. I’ve been banging my head against the Riccati brick wall for a while, and your posts showed me a different way to attack them.

Cheers.

2. Jim Currie says:

How do you solve quartic equations with linear fractional transformations?

The main result seems to be that via a fractional linear change of coordinates $y=\frac{c_1 x+c_2}{c_3 x+c_4}$ it is possible to turn a quartic (in x) into the form $y^4 + a_2 y^2 + a_0 = 0$, which can then be solved. Moreover, the coefficients $c_j$ can be determined by solving cubics (or lower degree polynomials).