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Solving differential equations through matrix

by devoured_elysium
Tags: differential, equations, matrix, solving
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devoured_elysium
#1
May9-07, 11:43 AM
P: 15
Hello

I'd like to know how to solve the following equation with matrix, if possible at all:

d ( x^2 ) / dt^2 + w^2 x = 0

I know how to solve it without having to use a matrix, but I heard it is possible to do it with matrix. How about doing it? How is this method called?

Thanks
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HallsofIvy
#2
May9-07, 01:39 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,323
Is that d ( x^2 ) / dt^2 + w^2 x = 0
or d^2 x/dt^2+ w^2x= 0?

I'm going to assume it is the latter. Define y= dx/dt so dy/dt= d^2x/dt^2 and the equation becomes dy/dt= -w^2x. You now have the two equations dx/dt= y and dy/dt= -w^2x. If you write
[tex]X= \left(\begin{array}{c} x \\ y\end{array}\right)[/tex]
Then the two equations become the single matrix equation
[tex]\frac{dX}{dt}= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)X[/tex]
To solve that, find the eigenvalues of the coefficient array (they are [itex]\pm w i[/itex]). The general solution then can be written as exponentials of those eigenvalues times t or, since they are imaginary, sine and cosine.
devoured_elysium
#3
May9-07, 03:13 PM
P: 15
it's the latter one, as you thought. thanks by the response!

devoured_elysium
#4
May10-07, 06:32 AM
P: 15
Solving differential equations through matrix

Hi again

Sorry but I could not follow the step X=(x, y) and why then it becomes the next equation.
HallsofIvy
#5
May10-07, 07:36 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,323
[tex]\left(\begin{array}{c}\frac{dx}{dt} \\ \frac{dy}{dt}\end{array}\right)= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)[/tex]

Do you see how the matrix multiplication on the right works out?
D H
#6
May10-07, 08:32 AM
Mentor
P: 15,065
Quote Quote by HallsofIvy View Post
[tex]\left(\begin{array}{c}\frac{dx}{dt} \\ \frac{dy}{dt}\end{array}\right)= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)[/tex]
Taking this one step further, define [itex]\mathbf X[/itex] and [itex]\mathbf A[/itex] as
[tex]\begin{array}{rl}
\mathbf X &\equiv \bmatrix x\\y\endbmatrix \\[12pt]
\mathbf A &\equiv \bmatrix 0&&1\\-w^2&&0\endbmatrix
\endarray[/tex]

then

[tex]\frac {d\mathbf X}{dt}= \mathbf A\mathbf X[/tex]

If [itex]\mathbf X[/itex] and [itex]\mathbf A[/itex] were scalars, the solution to the above would be the exponential

[tex]\mathbf X = e^{\mathbf A t}\mathbf X|_{t=0}[/tex]

The series expansion of the exponential function works for matrices as well as scalars (for example, see http://mathworld.wolfram.com/MatrixExponential.html or http://www.sosmath.com/matrix/expo/expo.html).

In this case,

[tex]\mathbf A^2 = -w^2 \mathbf I[/tex]

where [itex]\mathbf I[/itex] is the identity matrix. Thus

[tex]\begin{array}{rl}
(\mathbf A t)^{2n} &= (-1)^n (wt)^{2n} \mathbf I \\
(\mathbf A t)^{2n+1} &= (-1)^n \frac 1 w (wt)^{2n+1} \mathbf A\\
\end{array}[/tex]

The matrix exponential is thus

[tex]\begin{array}{rl}
e^{\mathbf A t} &=\sum_{n=0}^{\infty} \frac {(\mathbf At)^n}{n!}
\\[12pt]
&= \sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n}}{(2n)!} +
\sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n+1}}{(2n+1)!}
\\[12pt]
&= \sum_{n=0}^{\infty} (-1)^n\frac {(wt)^{2n}}{(2n)!} \mathbf I +
\frac 1 w\sum_{n=0}^{\infty}(-1)^n\frac{(wt)^{2n+1}}{(2n+1)!} \mathbf A
\\[12pt]
&= \cos(wt) \mathbf I +\frac 1 w\sin(wt) \mathbf A
\end{array}[/tex]


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