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The biological interpretation is that the system can

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The biological interpretation is that the system can act like a biochemical switch , but only if the MRNA and protein degrade slowly enough - specifically, their decay rates must satisfy ab < 1 / 2. In this case, there are two stable steady states: one at the origin, meaning the gene is silent and there is no protein around to turn it on; and one where x and y are large, meaning that the gene is active and sustained by the high level of protein. The stable manifold of the saddle acts as a threshold; it determines whether the gene turns on or off, depending on the intial values of x and y . 2. Purpose: To study homoclinic bifurcations. From [Str94], pp 262-263 Notes: This is similar to a heteroclinic bifurcation, in that it involves the unstable manifold of a saddle point. Exercise: Consider the system ˙ x = y ˙ y = μy + x - x 2 + xy (a) Show that the origin is a saddle point for all values of μ . (b) Find any other equilibria and determine their stability. 26
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y x Figure 22: Phase portrait for the a < a c case (c) Numerically plot phase portraits for values of μ between - 1 and - 0 . 5. What happens to the unstable manifold through the origin as μ is varied? (d) Numerically find the critical μ -value at which the stable and unstable manifolds through the origin intersect. (This is called a homoclinic connection or homoclinic orbit .) (e) Plot a few phase portraits for μ -values above and below this critical value and describe what happens to the unstable manifold through the origin. This type of bifurcation is called a homoclinic bifurcation. Solution: (a) The Jacobian of the system at the origin is J = bracketleftbigg 0 1 1 - 2 x + y μ + x bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 0 , 0 = bracketleftbigg 0 1 1 μ bracketrightbigg . The eigenvalues are given by λ 1 , 2 = μ 2 ± μ 2 +4 2 . Now since μ 2 + 4 > μ 2 μ , then it must hold that radicalbig μ 2 + 4 > μ μ . Therefore λ 1 = μ 2 + μ 2 +4 2 > 0 μ and λ 2 = μ 2 - μ 2 +4 2 > 0 μ . Since both eigenvalues are real and are of opposite signs, then the origin is a saddle for all values of μ . (b) The other fixed point, (1 , 0) can be found by solving ˙ x = 0 and ˙ y = 0. The Jacobian evaluated at this point is J = bracketleftbigg 0 1 - 1 μ + 1 bracketrightbigg , and it has eigenvalues λ 1 , 2 = ( μ +1) ± ( μ +1) 2 4 2 . If ( μ + 1) 2 > 4 , λ 1 , 2 R and λ 1 , 2 > 0 if μ > - 1. Therefore, need μ > 1, which then makes this fixed point an unstable node. If ( μ + 1) 2 < 4 , λ 1 , 2 R and λ 1 , 2 are complex conjugates. For μ < - 1, the real parts of λ are negative, therefore the fixed point is a stable spiral. For - 1 < μ < 1, the real parts of λ are positive, therefore the fixed point is an unstable spiral. (c) Figure 23 shows different phase portraits for the system for μ = - 0 . 55 , - 0 . 75 , - 0 . 8 and μ = - 0 . 95. (d) Using Maple, it is possible to numerically determine the critical μ -value, μ c . This critical value, where the stable and unstable manifolds through the origin intersect occurs at μ c = - 0 . 8645.
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