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Steady states and stability (Section 5.2) Steady states [equilibria, ﬁxed points] for the diﬀerential equation of the form dx = f (x) dt are those values of x that satisfy f (x) = 0. Question of interest: what is the stability of such steady states? If x is perturbed from its steady state value x∗, does it return to x∗ or move away from x∗? Stability analysis For equations of the form dx = f (x), there are dt two approaches to determine the stability of ﬁxed points: • Graphical stability analysis • Linear stability analysis 1 Graphical stability analysis for the cool- ing problem with equation dx = k(21 − dt x) Boardwork . . . Graphical stability analysis for the gen- eral problem with equation dx = f (x) dt Boardwork . . . Graphical stability analysis: observa- tions We note the following relationship between the stability of a ﬁxed point x∗ of the diﬀerential equation dx = f (x) and the slope of f (x): dt df • If dx |x=x∗ > 0, then x∗ is unstable df • If dx |x=x∗ < 0, then x∗ is stable 2 Theorem Let x∗ be a ﬁxed point of dx = f (x). dt df • If dx |x=x∗ > 0, then x∗ is unstable. df • If dx |x=x∗ < 0, then x∗ is stable. df • If dx |x=x∗ = 0, then no conclusions can be drawn without further work. df Anything can happen when dx |x=x∗ = 0 = f (x∗). 3 Linear stability analysis We are considering dx = f (x) (1) dt with steady state x∗, that is, f (x∗) = 0. Introduce a small perturbation y from x∗, that is, let x = x∗ + y (2) Substitute (2) into (1), and expand the right- hand-side with a Taylor series to get: d(x∗ + y) = f (x∗ + y) dt dy ∗ ) + df | 2 = f (x x=x∗ y + O(y ) dt dx Since x∗ is a ﬁxed point, we can replace f (x∗) on the right hand side by 0. If, in addition, we can safely neglect all the terms in the Tay- lor series that have been collected in the term O(y 2), then we are left with the following equa- tion for the perturbation: 4 dy df = |x=x∗ y. dt dx df We recognise that dx |x=x∗ is a constant, λ say. The equation for the perturbation thus is the linear equation dy = λy, dt which we studied previously (the world’s sim- plest ODE). The solution for the ODE (with initial condi- tion y(0) = y0) is y(t) = y0eλt. 5 • If λ < 0, then y(t) → 0 as t → ∞. • If λ > 0, then y(t) → ±∞ as t → ∞. That is, the perturbation dies out if λ = f (x∗) < df 0, and grows if λ = dx |x=x∗ > 0. In the spe- df cial case that λ = dx |x=x∗ = 0, the terms col- lected in the term O(y 2) become important, and other techniques of analysis are required. The theorem presented earlier follows. 6 Example The rate of formation of a chemical in a reac- tion is known to be governed by the equation dx = (a − x)(b − x) dt where x(t) is the amount (mass) of the chem- ical of interest at time t, and a and b are amounts of other chemicals present when t = 0, with 0 < a < b. Find the steady states of the reaction and de- termine their stability (with a graphical analysis as well as a linear stability analysis). What does the model predict? Sketch repre- sentative solutions x(t) as a function of t. 7

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posted: | 12/23/2009 |

language: | English |

pages: | 7 |

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