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Unformatted text preview: 8.1.11 u = a(1 − u) − uv 2 ˙ There is a ﬁxed point at (1, 0) for all a and k . Nullclines curves are given by u= a a + v2 and u= a+k v v = uv 2 − (a + k )v ˙ Figure 1: Nullcline Curves as the parameters are varied These curves intersect when a+k a = 2 a+v v (0.1) solving for v v= a± a2 − 4a(a + k )2 2(a + k ) The discriminant deﬁnes the number of solutions If If If a2 − 4a(a + k )2 > 0 2 solutions a2 − 4a(a + k )2 < 0 No solutions a2 − 4a(a + k )2 = 0 One solution ∂ ∂v Filling this back into equation 0.1 a a + v2 ∂ ∂v a+k v To determine the bifurcation point analytically, use the tangency condition = ⇒ 2av 3 = (a + k )(a + v 2 )2 a = v2 ⇒ k = −a ± √ a 2
√ a 2. So the bifurcation occurs along a bifurcation curve deﬁned by k = −a ± Now to determine the stability of the ﬁxed points 1 Df (u∗ , v ∗ ) = −a − (v ∗ )2 (v ∗ )2 −2u∗ v ∗ 2u v − (a + k )
∗∗ For the ﬁxed point at (1, 0) the eigenvalues are −a and −(a + k ), which corresponds to a stable node. For the ﬁxed points deﬁned by intersection of the nullclines uv = (a + k ) J= Here ∆ = (a + k )(v 2 − a) and τ = (k − v 2 ) So for v 2 < a , ∆ < 0 which corresponds to a saddle, for v 2 > a , ∆ > 0 and so we have a node (it is stable if k ≤ a). The Jacobian for v 2 = a is J= Here ∆ = 0 and τ = −2a +
√ a 2, −a − (v ∗ )2 (v ∗ )2 −2(a + k ) a+k √ −2a − a √ a a 2 which means we have a zero eigenvalue. v2 < a Saddle v 2 = a Nonhyperbolic Fixed Point v2 > a Node, stable if k ≤ a Therefore we have a saddle node bifurcation at k = −a ± √ a 2 (which corresponds to v 2 = a). 2 ...
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This note was uploaded on 02/28/2011 for the course MATH 104 taught by Professor Dr.buzi during the Spring '10 term at Caltech.
 Spring '10
 Dr.Buzi

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