Register to reply

How to solve second-order matrix diffrential equation?

by jahandideh
Tags: diffrential, equation, matrix, secondorder, solve
Share this thread:
jahandideh
#1
Mar28-08, 01:30 AM
P: 7
hi all
this is the general problem
X[tex]\ddot{}[/tex]+AX[tex]\dot{}[/tex]+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 ...
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
trambolin
#2
Mar28-08, 05:45 AM
P: 341
Can you define what do you mean by [itex]\frac{d}{dt}X[/itex]. 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,
HallsofIvy
#3
Mar28-08, 06:44 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
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
[tex]Z^.= \left(\begin{array}{c}X^. \\Y^.\end{array}\right)= \left(\begin{array}{c}Y \\ -AY-AB\end{array}\right)[/tex]

jahandideh
#4
Mar28-08, 07:36 AM
P: 7
How to solve second-order matrix diffrential equation?

thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time.

for first-order system of differential equations like:
[tex]X\dot{}=AX+BU[/tex]
the solution is X(t) = [tex] e^{At} [/tex] X(0)+ [tex]\int e^{A(t-\tau)} BU(\tau) d\tau [/tex]
for example I can solve this system : X\dot{} = {0 1 ; 2 3 } X + {0 1} u
but I have problem with this
[tex]Y\ddot{}[/tex] = {-5 -2 ; 2 -2} Y
which [tex]Y\ddot{}[/tex] and Y are 2 by 2 matrices.
jahandideh
#5
Mar28-08, 07:39 AM
P: 7
thanx
yes [tex]X\dot{}[/tex] = dX / dt and t is time.

for first-order system of differential equations like:

[tex]X\dot{}=AX+BU[/tex]

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

for example I can solve this system : [tex]X\dot{} [/tex]= {0 1 ; 2 3 } X + {0 1} u
but I have problem with this
[tex]Y\ddot{} = {-5 -2 ; 2 -2} Y[/tex]
which [tex]Y\ddot{}[/tex] and [tex]Y[/tex] are 2 by 2 matrices.
HallsofIvy
#6
Mar29-08, 05:58 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
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
[tex]Y= e^{\left(\begin{array}{cc}-5 & -2 \\ 2 & -2\end{array}\right)t}Y(0)[/tex].
jahandideh
#7
Mar30-08, 01:14 AM
P: 7
no! no!
that was not [tex]y\dot{}=\left(\begin{array}{cc}-5 & -2 \\ 2 &
-2\end{array}\right) y[/tex]

this is a second order system of differential equation
[tex]y\ddot{}{}=\left(\begin{array}{cc}-5 & -2 \\ 2 &
-2\end{array}\right)y [/tex]

reducing the order of the system by assuming [tex]y\dot{} = p[/tex] 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?


Register to reply

Related Discussions
How to solve this 2nd order nonlinear differential equation Differential Equations 22
Order of matrix Calculus & Beyond Homework 3
Under damp ing spring, eigen, 2-nd order eqn as 2 1-st order in matrix form. Calculus & Beyond Homework 1
Problem with easy diffrential equation Calculus & Beyond Homework 2
2nd order DE, is there a way to solve this without series? Differential Equations 11