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April 24th, 2017, 02:52 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, 03:08 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,505 Thanks: 502 Math Focus: Yet to find out. 
No need to shout!

April 24th, 2017, 03:10 AM  #3 
Newbie Joined: Dec 2016 From: England Posts: 7 Thanks: 0 
lol sorry please help

April 25th, 2017, 05:46 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,939 Thanks: 794  Quote:
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 nonlinear 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 nonlinear 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.  

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linearization, needed, non linear systems, urgent 
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