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Introduction to the Modeling and Analysis of Complex Systems

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132 CHAPTER 8. BIFURCATIONS<strong>the</strong> local bifurcations, as <strong>the</strong>y can be easily analyzed using <strong>the</strong> concepts <strong>of</strong> linear stabilitythat we discussed in <strong>the</strong> previous chapters.Local bifurcations occur when <strong>the</strong> stability <strong>of</strong> an equilibrium point changes betweenstable <strong>and</strong> unstable. Ma<strong>the</strong>matically, this condition can be written down as follows:Local bifurcations occur when <strong>the</strong> eigenvalues λ i <strong>of</strong> <strong>the</strong> Jacobian matrix at an equilibriumpoint satisfy <strong>the</strong> following:For discrete-time models: |λ i | = 1 for some i, while |λ i | < 1 for <strong>the</strong> rest.For continuous-time models: Re(λ i ) = 0 for some i, while Re(λ i ) < 0 for <strong>the</strong> rest.These conditions describe a critical situation when <strong>the</strong> equilibrium point is about <strong>to</strong> changeits stability. We can formulate <strong>the</strong>se conditions in equations <strong>and</strong> <strong>the</strong>n solve <strong>the</strong>m in terms<strong>of</strong> <strong>the</strong> parameters, in order <strong>to</strong> obtain <strong>the</strong>ir critical thresholds. Let’s see how this analysiscan be done through some examples below.8.2 Bifurcations in 1-D Continuous-Time ModelsFor bifurcation analysis, continuous-time models are actually simpler than discrete-timemodels (we will discuss <strong>the</strong> reasons for this later). So let’s begin with <strong>the</strong> simplest example,a continuous-time, first-order, au<strong>to</strong>nomous dynamical system with just one variable:dx= F (x) (8.1)dtIn this case, <strong>the</strong> Jacobian matrix is a 1 × 1 matrix whose eigenvalue is its content itself(because it is a scalar), which is given by dF/dx. Since this is a continuous-time model,<strong>the</strong> critical condition at which a bifurcation occurs in this system is given by( ) ∣ dF ∣∣∣x=xeqRe= dFdx dx ∣ = 0. (8.2)x=xeqLet’s work on <strong>the</strong> following example:dxdt = r − x2 (8.3)The first thing we need <strong>to</strong> do is <strong>to</strong> find <strong>the</strong> equilibrium points, which is easy in this case.Letting dx/dt = 0 immediately givesx eq = ± √ r, (8.4)

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