How to solve second-order matrix diffrential equation?

  • Thread starter Thread starter jahandideh
  • Start date Start date
  • Tags Tags
    Diffrential Matrix
jahandideh
Messages
7
Reaction score
0
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 ...
 
Physics news on Phys.org
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?
 
Back
Top