Project 2(Your Name)April 14, 20101 Part AGiven the system:˙x = 1 − (b + 1)x + ax2y (1)˙y = bx − ax2y, (2)the null-clines of x, are:y =(b + 1)x − 1ax2. (3)The y null-clines are:y =bax. (4)The intersection of the two null-clines will give only one fixed point at (1,ba). The Jacobianabout any point in the xy plane is:J(x,y)= 2axy −(b + 1) ax2b − 2axy −ax2!(5)At the fixed point (1,ba), the Jacobian is:J(1,ba)= b − 1 a−b −a!(6)Now, the fixed point can be characterized by its trace (τ = b − (1 + a)) and determinant(∆ = a). Since a > 0, the fixed point cannot be a saddle. Otherwise, the fixed can be any type.There are three cases:1.0.1 Case 1: τ > 0There are two possibilities, either the fixed point is an unstable node or an unstable spiral. Ifτ2−4∆ < 0, then the fixed point is an unstable spiral. If τ2−4∆ > 0 is positive, then the fixedpoint is a unstable node.11.0.2 Case 2: τ < 0There are two possibilities, either the fixed point is an stable node or an stable spiral. Ifτ2− 4∆ < 0, then the fixed point is an stable spiral. If τ2− 4∆ > 0 is positive, then the fixedpoint is a stable node.1.0.3 Case 3: τ = 0This implies the fixed point is necessarily a center.2 Part B2.0.4 Question aIt is possible that when the fixed point is an unstable spiral that a limit cycle appears. Thishappens when τ2− 4∆ < 0 and τ > 0. First, consider when τ2− 4∆ < 0. This means that:(b − (a + 1))2− 4a < 0 −→ b2− b(2(a + 1)) + (a − 1)2< 0 (7)Note that this is an upward-facing quadratic in b, which means the inequality holds for bbetween the zeros of this quadratic. The zeros are:2(a + 1) ±p−2(a + 1)2− 4(a − 1)2= (1 + a) ± 2√a (8)So, within this region the fixed point is a spiral. In order for it to be unstable, we must have:b − (1 + a) > 0 → b > (1 + a) (9)This means if b ∈ ((a + 1), (a + 1) + 2√a), then the fixed point is necessarily an unstablespiral. The Hopf bifurcation occurs at the lesser end of this interval. When b passes throughthe value bc= (a + 1), the fixed point changes from a stable spiral to an unstable spiral, leadingto limit cycles.2.0.5 Question bFor b < bc, no limit cycle exists but for some b > bclimit cycles do exist.3 Part C2Figure 1: Stable spiralFigure 2: Limit cycle34 Part DWhen b approaches bcfrom the positive side, the limit cycle takes on the period of the nonlinearcenter. We know that the frequency of the nonlinear center is simply√∆. So, the period shouldbe2π√∆or2π√a.5 Part EFigure 5 shows that if b is close enough to bc, then the period of the limit cycle is approximately2π√a= 2π as predicted.Figure 3: b − bc= .5Figure 4: b − bc= .24Figure 5: b − bc=