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It is easy to solve the ODE: $\frac {dx}{dt} = a - b x^2$ with $x(0)=0$, $a>0$, and $b>0$, indeed all one has to do is write $dt = \frac {dx}{a-bx^2} = \frac 1 {2\sqrt a}(\frac {dx}{\sqrt a-\sqrt bx} + \frac {dx}{\sqrt a+\sqrt bx})$ and integrate both parts.

I am interested in the same equation with $x$ taking its values in symmetric positive matrices, ie $\frac {dx}{dt} = a - x b x$, with $a$ and $b$ symmetric positive. In that case, because the involved matrices might not commute, I cannot use the same trick to get a closed form solution.

Would you know of any techniques I can use to solve such a problem?

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Casual guess: some judicious combination of time/path-ordering ( and diagonalization might help. – Steve Huntsman Apr 25 '12 at 20:48
Something is wrong with your scalar version of the ODE. As it is written, it suffices to integrate in $dx$ on both sides to get $f(x)=ax-\frac{b}3x^3+C$. Please correct the text. – Federico Poloni Apr 25 '12 at 20:56
(1) Do you mean df/dx = 1/(a-bx^2)? Or perhaps df/dx = a-bf^2? (2) Is the matrix problem an ODE, i.e. is x a scalar? If so, why write xbx? – Aaron Hoffman Apr 25 '12 at 20:59
sorry, it is indeed (2). To avoid confusion, t is a scalar and x is a positive matrix. When dimension is 1, everything goes well, when dimension is higher I'm not sure what can be done. – Bernard Apr 25 '12 at 21:10
up vote 7 down vote accepted

Consider a solution of the linear system $$ \dot x=by$$ $$\dot y =ax$$ Then $u:=yx^{-1}$ solves your Riccati equation $$\dot u= a- ubu\, .$$

rmk. Note that the solution of the linear system is an exponential, defined for all $t\in\mathbb{R}$; while the solution of the Riccati equation is bounded to the interval where $x$ is invertible (possibly not the whole $\mathbb{R}$) Also, if you have an initial data $u(0)=u_0$, you can take $x(0)=I$ and $y(0)=u_0$ as initial data for the linear system.

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Note that no symmetry or positivity assumption is needed; you may solve your equation this way in any Banach algebra. – Pietro Majer Apr 26 '12 at 6:49

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