Solutions of first-order matrix differential equations

[Haorong Wu]

Homework Statement

How to solve the following matrix differential equation, $[A(t)+B(t) \partial_t]\left | \psi \right >=0$, where $A(t)$ and $B(t)$ are $n\times n$ matrices and $\left | \psi \right >$ is a $n$-vector.

Relevant Equations

None

Hello, there. I am trying to solve the differential equation, $[A(t)+B(t) \partial_t]\left | \psi \right >=0$. However, $A(t)$ and $B(t)$ can not be simultaneous diagonalized. I do not know is there any method that can apprixmately solve the equation.

I suppose I could write the equation as $\partial_t \left | \psi \right >=-B^{-1}(t) A(t)\left | \psi \right >$ with general solutions being $\left | \psi \right >=\exp \left ( \int_0^t -B^{-1}(t') A(t')dt'\right ) \left | c\right >$ and $\left | c\right >$ is a constant vector. Then a first-order approximation may be $\left | \psi \right >=\left (I+ \int_0^t -B^{-1}(t') A(t')dt'\right ) \left | c\right >$.

I am not familiar with matrix differential equations. Does this method have any restrictions or problems? Or there may be other better approximation solutions? Any references would be greatly appreciated.

Thanks!

 

[Mark44]

Homework Statement:: How to solve the following matrix differential equation, $[A(t)+B(t) \partial_t]\left | \psi \right >=0$, where $A(t)$ and $B(t)$ are $n\times n$ matrices and $\left | \psi \right >$ is a $n$-vector.
Relevant Equations:: None

Hello, there. I am trying to solve the differential equation, $[A(t)+B(t) \partial_t]\left | \psi \right >=0$. However, $A(t)$ and $B(t)$ can not be simultaneous diagonalized. I do not know is there any method that can apprixmately solve the equation.

I suppose I could write the equation as $\partial_t \left | \psi \right >=-B^{-1}(t) A(t)\left | \psi \right >$ with general solutions being $\left | \psi \right >=\exp \left ( \int_0^t -B^{-1}(t') A(t')dt'\right ) \left | c\right >$ and $\left | c\right >$ is a constant vector. Then a first-order approximation may be $\left | \psi \right >=\left (I+ \int_0^t -B^{-1}(t') A(t')dt'\right ) \left | c\right >$.

I am not familiar with matrix differential equations. Does this method have any restrictions or problems? Or there may be other better approximation solutions? Any references would be greatly appreciated.

Thanks!

I don't understand what you did. Starting from as $\partial_t \left | \psi \right >=-B^{-1}(t) A(t)\left | \psi \right >$, wouldn't you get $\left | \psi \right >= \left ( \int_0^t -B^{-1}(t') A(t')dt'\right )$? You might be mixing up the concept of an integrating factor with the solution of a differential equation. It's been nearly 25 years since I worked on this stuff, so I could be mistaken.

Nit: Also, you have used the symbol $\partial_t$. Since the matrices and vector are functions of a single variable t, the ordinary derivative would be more suitable, IMO.

 

[Haorong Wu]

Thanks, @Mark44. There is a $\left | \psi \right >$ in the rhs, so it is like the equation as $y'=\alpha y$, which solution is exponential functions. Also, thanks for the suggestions about $\partial_t$, but I use $\partial_t$ for partial and ordinary derivatives if no confusion could occur.

 

[Mark44]

Thanks, @Mark44. There is a $\left | \psi \right >$ in the rhs, so it is like the equation as $y'=\alpha y$, which solution is exponential functions.

OK, I understand. I'm not so familiar with the notation $|\psi>$, as that's probably more of a physics notation rather than one used in mathematics.

 

[Haorong Wu]

OK, I understand. I'm not so familiar with the notation $|\psi>$, as that's probably more of a physics notation rather than one used in mathematics.

Sorry for the confusion. It can be treated as a vector.

 

[pasmith]

Assuming B is invertible, we can rewrite the ODE as <br /> \frac{d}{dt}(B\psi) + \left[ AB^{-1} - \frac{dB}{dt}B^{-1}\right](B\psi) = 0 which is of the standard form <br /> \frac{du}{dt} + C(t)u = 0. But in general there is no closed form solution unless C commutes with \int_0^t C(s)\,ds, when the solution is u(t) = \exp\left(\int_0^t C(s)\,ds\right)u(0).