# Solutions of first-order matrix differential equations

• Haorong Wu

#### 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!

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.

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.

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.

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.

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