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 April 24th, 2017, 01:52 AM #1 Newbie   Joined: Dec 2016 From: England Posts: 7 Thanks: 0 Linearization URGENT HELP NEEDED PLEASE CAN SOMEONE HELP ME FIND THE FIXED POINTS OF THESE DIFFERENTIAL EQUATIONS PLEASE AND THEN HOW I LINEARIZE THEM! dR --- = BR - ARF dt dF --- = -DF + ARF dt A,B, D are constants. April 24th, 2017, 02:08 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. No need to shout! April 24th, 2017, 02:10 AM #3 Newbie   Joined: Dec 2016 From: England Posts: 7 Thanks: 0 lol sorry please help April 25th, 2017, 04:46 AM   #4
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Quote:
 Originally Posted by ryancooperx PLEASE CAN SOMEONE HELP ME FIND THE FIXED POINTS OF THESE DIFFERENTIAL EQUATIONS PLEASE AND THEN HOW I LINEARIZE THEM! dR --- = BR - ARF dt dF --- = -DF + ARF dt A,B, D are constants.
Nor to bark (ARF, ARF)!

Surely you know that a "fixed point" is solution that does not change- a constant solution. And for any constant, the derivative is 0. That is, at a fixed point, dR/dt= BR- ARF= 0 and dF/dt= -DF+ ARF= 0.

We want to solve BR- ARF= R(B- AF)= 0 and -DF+ ARF= F(-D+ AR)= 0.
A product is 0 if and only if at least one of its factors is 0. So for the first equation, we must have either R= 0 or B- AF= 0. If R= 0, the second equation becomes F(-D)= 0 so that D= 0. One fixed point is R= 0, F= 0.

If B- AF= 0, that is if F= B/A, the second equation become (B/A)(-D+ AR)= -BD/A+ BR= 0 so R= D/A. Another fixed point is R= D/A, F= B/A.

Any nonlinear functions could be written as a power series we know that, for the variables close to 0, higher powers are smaller than the variables themselves so can be dropped. About R= 0, F= 0, that is easy. The equations are dR/dt= BR- ARF and dF/dt= -DF+ ARF. The only non-linear terms are the products, ARF. Dropping those gives dR/dt= BR and dF/dt= -DF. Those are easy to solve.

To linearize about R= D/A, F= B/A, I would introduce the new variables, u= R- D/A, v= F- B/A. That way, the fixed point, R= D/A, F= B/A, becomes u= 0, v= 0. Of course since D/A and B/A are constants, du/dt= dR/dt and dv/dt= dF/dt. Also R= u+ D/A and F= v+ B/A.

The equations become du/dt= B(u+ D/A)+ A(u+ D/A)(v+ B/A)= Bu+ BD/A+ Auv+ Bu+ Dv+ BD/A and dv/dt= -D(v+ B/A)+ A(u+ D/A)(v+ B/A)= -Dv- BD/A+ Auv+ Bu+ Dv+ BD/A. The non-linear term is Auv and, since u and v are close to 0, can be dropped. That gives the two linear equations du/dt= Bu+ BD/A+ Bu+ Dv+ BD/A= 2Bu+ DV+ 2BD/A and dv/dt= -Dv- BD/A+ Bu+ Dv+ BD/A= Bu, not quite as easy to solve as the first two but still not very difficult. Tags linearization, needed, non linear systems, urgent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Hikimaman Applied Math 2 June 22nd, 2013 04:03 AM asifrahman1988 Calculus 2 October 27th, 2012 04:06 PM dilshad123 Algebra 5 February 26th, 2010 03:37 PM adii shamz Calculus 1 February 27th, 2009 03:31 AM jumpman23 Applied Math 3 January 20th, 2009 12:23 PM

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