Skip to main content
MathOverflow

Return to Question

edited body
Source Link
Bernard
Bernard
  • 427
  • 2
  • 9

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 $f$$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?

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 $f$ 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?

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?

sorry, I got confused when typing my ODE, corrected this.
Source Link
Bernard
Bernard
  • 427
  • 2
  • 9

It is easy to solve the ODE: $\frac {df}{dx} = a - b x^2$$\frac {dx}{dt} = a - b x^2$ with $f(0)=0$$x(0)=0$, $a>0$, and $b>0$, indeed all one has to do is write $df = \frac {dx}{a-bx^2} = \frac 1 {2\sqrt a}(\frac {dx}{\sqrt a-\sqrt bx} + \frac {dx}{\sqrt a+\sqrt bx})$$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 $f$ taking its values in symmetric positive matrices, ie $\frac {df}{dx} = a - x b x$$\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?

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

I am interested in the same equation with $f$ taking its values in symmetric positive matrices, ie $\frac {df}{dx} = a - x b x$, 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?

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 $f$ 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?

Source Link
Bernard
Bernard
  • 427
  • 2
  • 9

ODE in symmetric definite positive matrices

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

I am interested in the same equation with $f$ taking its values in symmetric positive matrices, ie $\frac {df}{dx} = a - x b x$, 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?