How to solve second-order matrix diffrential equation?

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    Diffrential Matrix
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Discussion Overview

The discussion revolves around solving a second-order matrix differential equation of the form X'' + AX' + BX = 0, where A, B, and X are 2x2 matrices. The context includes applications in vibrations and the challenges associated with reducing the order of the system and finding solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant introduces the general form of the second-order matrix differential equation and expresses familiarity with first-order systems but seeks help with the second-order case.
  • Another participant asks for clarification on the meaning of the time derivative of the matrix X, suggesting diagonalization and solving for mode shapes as a potential approach.
  • A different participant explains how to reduce the second-order system to a first-order system by introducing a new variable Y = X', leading to a larger system of first-order equations.
  • One participant shares their ability to solve first-order systems and provides an example, but expresses difficulty with the specific second-order equation Y'' = {-5 -2; 2 -2} Y.
  • Another participant suggests that the second-order equation is simply a case with B = 0 and proposes a solution involving the matrix exponential e^{At}.
  • A participant challenges this by clarifying that the equation is indeed a second-order system and expresses concern about the time-consuming nature of reducing the order, while seeking a faster method for calculating e^{At}.
  • They mention various methods for computing e^{At}, including the Cayley-Hamilton theorem, similarity transformations, and Jordan matrices, but favor the Laplace transform for its efficiency.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to solving the second-order matrix differential equation, with no consensus reached on a single method or solution. Some participants propose various techniques while others question the efficiency and practicality of those methods.

Contextual Notes

Participants note the complexity of calculating the matrix exponential and the potential for different methods to yield varying levels of efficiency, but do not resolve the specific challenges associated with each approach.

jahandideh
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hi all
this is the general problem
X\ddot{}+AX\dot{}+BX=0

let A, B,X be 2*2 matrices

its application is in vibrations.

any opinion will be great

I can solve the first-order but ...
 
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Can you define what do you mean by \frac{d}{dt}X. Is it the time derivative of the entries or something else?

You can start diagonalizing the matrices and then solving the mode shapes, the trick as usual in vibtration analysis,
 
A system of differential equations, of any order, can always be reduced to a (larger) system of first order differential equations.
Here, let Y= X.. Then X..= Y. so your equation becomes Y.+ AY+ BX= 0. That, together with X.= Y gives two first order matrix equation. If X is an n by n matrix, then so is Y and letting Z be the matrix n by 2n matrix with the rows of X above the rows of Y, we have
Z^.= \left(\begin{array}{c}X^. \\Y^.\end{array}\right)= \left(\begin{array}{c}Y \\ -AY-AB\end{array}\right)
 
thanx
yes X\dot{} = dX / dt and t is time.

for first-order system of differential equations like:
X\dot{}=AX+BU
the solution is X(t) = e^{At} X(0)+ \int e^{A(t-\tau)} BU(\tau) d\tau
for example I can solve this system : X\dot{} = {0 1 ; 2 3 } X + {0 1} u
but I have problem with this
Y\ddot{} = {-5 -2 ; 2 -2} Y
which Y\ddot{} and Y are 2 by 2 matrices.
 
thanx
yes X\dot{} = dX / dt and t is time.

for first-order system of differential equations like:

X\dot{}=AX+BU

the solution is X(t) = e^{At} X(0)+ \int e^{A(t-\tau)} BU(\tau) d\tau

for example I can solve this system : X\dot{}= {0 1 ; 2 3 } X + {0 1} u
but I have problem with this
Y\ddot{} = {-5 -2 ; 2 -2} Y
which Y\ddot{} and Y are 2 by 2 matrices.
 
Why would you have a problem with that? That's just your general form with B= 0. According to your formu;a, the solution is
Y= e^{\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right)t}Y(0).
 
no! no!
that was not y\dot{}=\left(\begin{array}{cc}-5 &amp; -2 \\ 2 &amp;<br /> -2\end{array}\right) y

this is a second order system of differential equation
y\ddot{}{}=\left(\begin{array}{cc}-5 &amp; -2 \\ 2 &amp;<br /> -2\end{array}\right)y

reducing the order of the system by assuming y\dot{} = p is an idea but it is very time consuming because we have to calculate e^At towice , i am searching for a better way .

by the way to solve e^At (A is a squar matrix) there is several ways like: caley-hamilton theorium , using similarity tansform and Jordan matrix , but I found the Laplace transform a better way so is there any faster way to calculate the e^At?
 

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