
Linear Algebra Linear Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  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.  

Tags 
linearization, needed, non linear systems, urgent 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Urgent help needed here pls  Hikimaman  Applied Math  2  June 22nd, 2013 04:03 AM 
Urgent help needed for differentiation  asifrahman1988  Calculus  2  October 27th, 2012 04:06 PM 
Urgent help needed  dilshad123  Algebra  5  February 26th, 2010 03:37 PM 
urgent help needed!!!  adii shamz  Calculus  1  February 27th, 2009 03:31 AM 
Linear Programmingurgent help needed!  jumpman23  Applied Math  3  January 20th, 2009 12:23 PM 