Linear ODE Systems in Numerical Methods.

  • Thread starter arizonian
  • Start date
  • #1
18
1
I feel so embarrased asking this question, but this is the place to get answers.

I have a 2nd order ODE with a forcing function that needs to be manipulated and put into a matrix for a numerical method solution, ie Matlab. My question is: Is the matrix composed of a particular solution in the top row and a homogenous solution in the bottom row? Does this satisfy the requirement for two equations? Maybe I should say equation, not solution.

My work:

m d^2x/dt^2 + c dx/dt + kx = 0

d^2x/dt^2 = dx/dt

Substituting y2 for d^2t/dx^2 and y1 for dx/dt, and realizing that y2 is the derivative of y1, I end up with, in matrix form:

(I am using periods to hold the spacing)

[-1/k......-c/mk]..[y2]...=[x2]
[1...............-1]..[y1]...=[x1]

Thank you

Bill

On edit, I realized I forgot the signs in the first equation.
On second edit, I changed the lower equation to simplify what I was after.
 
Last edited:

Answers and Replies

  • #2
18
1
Any body care to comment?

Bill
 
  • #3
6
0
The way you describe the problem is a little confusing. Let me try to paraphrase.
Your original problem is:
[tex] m \frac{d^2 x}{d t^2} + c \frac{dx}{dt} + kx = 0 [/tex]
[tex] m \frac{d^2 x}{d t^2} - \frac{dx}{dt} = 0 [/tex]

And you would like to transfer it to matrix form, right ?
You might do something as follows:

[tex] y_{1}=x [/tex]
[tex] y_{2}= \frac{dx}{dt} [/tex]
And let
[tex] X = \left(\begin{array}{c} y_{1} \\ y_{2} \end{array}\right) [/tex]
[tex] Y = \left(\begin{array}{c}\frac{d y_{1}}{dt} \\ \frac{dy_{2}}{dt} \end{array}\right) [/tex]

And you might turn this problem into

[tex] \left(\begin{array}{cc} c & m \\ -1 & m \end{array}\right) Y=
\left(\begin{array}{cc} -k & 0 \\ 0 & 00 \end{array}\right)X
[/tex]

Is this what you are trying to ask ?

Probably you can just start from here and check the other websites
to solve this question in martrix form. You might as well just take a look
at

http://www.ScienceOxygen.com/math409.html

It might not solve your question directly. But it is with a lot of links
on differential equation. You could start from there to collect some information...
 
  • #4
saltydog
Science Advisor
Homework Helper
1,590
3
I think the original question is a bit confussing. This is how I'd present converting a high-order ODE to a system of first-order ODEs:


[tex]
\frac {d^2y} {dx^2} + \ln{y} = yx
[/tex]

To convert this to a system of ODEs,

let:

[tex]
z[x]=\frac{dy} {dx}
[/tex]

Then:

[tex]
\frac{dy} {dx} = z
[/tex]

[tex]
\frac {dz} {dx} =yx-\ln{y}
[/tex]

You can use the same method for higher-order ODEs, just assigning different variables to each higher-order derivative. I used Mathematica to solve this system with initial conditions:

[tex]
y[0]=1
[/tex]

[tex]
y'[0]=1
[/tex]

I attached a plot of the solution which you can find in "additional options".

SD
 

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  • #5
saltydog
Science Advisor
Homework Helper
1,590
3
Oh yea, I thought the talking dictionary on line was cool, but LaTeX rocks!

SD
 

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