Question

Problem 1. (Romeo and Juliet again] The following problems are basically from [Strogatz] 5.3.3 and 5.3.4. (a) Suppose Romeo and Juliet react to each other, but not to themselves: zi = az2, z2 = bz1. What happens? a and b are nonzero constants. (b) Do opposites attract? Analyze ži = azi + bz2, z2 = -bzı – az2. Consider a > 0 and b + 0 in this problem. If a < 0, then we could just switch zı and za to apply our analysis here. Also, no need to discuss the case when a2 = 62. Remember, you can always plot some phase portraits to verify your work.

Answer 1

solution (a): Given that 31 atz Ž 6며 't =) 22 (6):6) *** :) cet are given by The eigenvalues det (A-7 IZ) =0 = ) 2 a b. -2 > =) X=I Jab are real when and equal 2² ab =0 abyo the eigenvalues but opposite in sign. the origin is saddle point, the eigenvalues are purely when ab ao , imaginary - the origin is linear center. a

solution (6): Given that, zi =azi + b Z2 ₂2 = -bzi -azz 21 0 -a 22 b let A- : -a Then by are given the eigenvalues det (A- 010) b a-> 7) -b -a-> 2 => a ca tb => 2 & Ja2b2 Two case may happen: Either ay orash are a case 1: ut à > real and opposite then the eigen values in sign. ... The origin is saddle point The null-clines given by, 음러 (when żi :o) (when 32 = are Z2 Z2 21 S-a

uts drow the phase potrait: 22 11 we see case that in this either Juliet falls for for Romeo while he ignores her Romeo falls in love with Juliet while She ignores him. Case 11 : Then If a <6² eigenvalues are purely imaginary. Origin is liniar cunter. the a The phase potrait is : 72 키 레 care In this Romeo and Juliet switch between states of love and hate.