Using Row-Echelon Reduction With Differential Operators As Elements

• Jul 12th 2010, 01:06 AM
pi_cubed
Using Row-Echelon Reduction With Differential Operators As Elements
Hello. If this question belongs in another forum, I apologize.

When solving 'n-1' homogeneous equations in 'n' unknowns, is it permissible to use
differential operators in the matrix elements?

For example, when trying to solve :

i = i1 + i2
i = Cv' - Cx'
i1 = x/R
i2 = i3 + i4
i2 = Cx' - Cy'
i3 = y/R
i4 = Cy' - Cu'
i4 = u/R

The variables i, i1, i2, i3 and i4 are instantaneous currents {i.e., i = i(t)}.
The variables x, y, u and v are instantaneous voltages.
The prime marks indicate first derivatives.

I was thinking about putting these equations in row-echelon form, and then
performing reductions. For the primed variables, I was thinking about using
the operator D = d/dt.

Can I do this, being aware that I treat the operator algebraically in the columns,
but use it operationally row-wise?

Thanks!!
• Jul 12th 2010, 02:10 AM
Ackbeet
Sorry, no. That's not allowed. If your matrix starts having operators in it, then finding the eigenvalues (which is ultimately what you're trying to do) becomes very much more complicated. Ph.D. theses have been written on that sort of thing (mine was!).

Your problem looks like a circuit problem. So you have a system of ODE's, correct? How are you expecting to solve it without having n equations? Perhaps you could post the original problem here, or is that the original problem?
• Jul 12th 2010, 02:29 AM
pi_cubed
Problem Definition for Example Given
This problem comes from Paul J. Nahin's book
"An Imaginary Tale (The Story of {radical -1})", Princeton University Press, 1998,
on page 139. This example is a circuit solution for a phase-shift oscillator.

The author says :
"With a total of eight equations in nine variables we can solve for the ratio of any
two, and the particular ration of interest for our circuit is u/v. One way to do this
is to first manipulate the above equations to eliminate all the variables except
for u and v.
This is not difficult to do, but it is rather detailed, and so I will just give you the
answer and encourage you to verify it :
v''' = u''' + [6/RC]u'' + [5/{(RC)^2}]u' + [1/{(RC)^3}]u ."

I was hoping in this instance I could use row-echelon form, reducing to the two
variables u and v. In this situation, will this method work?

If not, would you please suggest a path to follow.

Thanks!
• Jul 12th 2010, 02:44 AM
Ackbeet
Given the nature of what the author is asking you to do, I would plug away with substitutions and differentiation. You're going to have to differentiate in several places, or I'm mistaken. Your initial equations are not so complicated that substitution wouldn't work pretty well, I think.
• Jul 13th 2010, 08:13 AM
pi_cubed
I have solved the problem using the row-echelon method :
1) rewrite all equations so that they are equal to zero;
2) construct the augmented matrix - since the "augmentation column" is all zeroes it can be ignored;
3) perform the normal row operations until reaching reduced row-echelon form, leaving the columns for
u and v with the only non-one entries.

Since the only operations involved in row reduction involving the single differential operator D=d/dt were addition or multiplication, the process went smoothly.
• Jul 13th 2010, 08:19 AM
Ackbeet
You say, "...the only operations involved in row reduction involving the single differential operator D=d/dt were addition or multiplication..."

What kind of multiplications did you come up with? Multiplication can be tricky with differential operators. Operators are slippery, and can give you wrong results unless you use a test function. Can you show me a calculation or two involving multiplication?
• Jul 14th 2010, 03:10 AM
pi_cubed
Clearing up semantic ambiguity

I probably should have written more than that single sentence, which is semantically very vague - sorry about that!

When I was multiplying by D across rows, the elements in each column were simple polynomials in D.

For example : suppose a (zero-augmented) 2x3 echelon matrix,
with left-to-right column headings of 'y', 'v', and 'u', containing two rows
{1, -[(RC)^2](D^2), [4 + 6RCD + [(RC)^2](D^2)]} and
{CD, 0, [-1/R - CD]} .

[D is defined to be "d/dt"; R and C are constants.]

{The two rows represent the equations :
0 = y - [(RC)^2]v'' + 4u + 6RCu' + [(RC)^2]u'' and
0 = Cy' - u/R - Cu' }

Algebraically multiplying the first row by (-CD) and
adding the result to the second row yields :

{0, (R^2)(C^3)(D^3), [-1/R - 5CD - 6R(C^2)(D^2) - (R^2)(C^3)(D^3)]}

which represents :
0 = (R^2)(C^3)v''' - u/R -5Cu' - 6R(C^2)u'' - (R^2)(C^3)u'''
(which gives the answer the author mentioned above wanted).

This is probably so elementary a thing that I overcomplicated it with my lousy ability in English composition.
• Jul 14th 2010, 04:57 AM
Ackbeet
Ah, I see what you're about. I'm going to have to take back some of my words in Post #2. You're not trying to find eigenvalues this way. You're just using these ideas to organize your elimination of variables. Your operator multiplication, I should warn you, while it does work in this case, would not work the same way if anything in your original DE's was nonlinear or if you had non-constant coefficients that have to be differentiated. Operators always look to the right, and operate on whatever is to the right. Because your coefficients are all constants, the derivative operator passes through and ends up on the right of every expression, which is where you want it. If you have non-constant coefficients, or nonlinearities, you'd have to be more careful.

Interesting idea, though. Thanks for posting!
• Jul 14th 2010, 05:06 AM
pi_cubed
Thanks Adrian. You said exactly what I was thinking - you are very eloquent of speech!