Tutorial work - 2.5 - Deduce stability of equilibrium solutions using taylor series expansions PDF

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Deduce Stability of Equilibrium Solutions using Taylor Series Expansions ...

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2.5: Deduce Stability of Equilibrium Solutions using Taylor Series Expansions Kiam Heong Kwa (Dated: October 5, 2011)

Consider the first order autonomous equation

dy = f (y). dt

point of f , the Taylor series can be reduced slightly to f ′′ (y 0 ) f ′ (y 0 ) (y − y 0 ) + (y − y 0 )2 +··· 1! 2! f (n) (y 0 ) f (n+1) (y 0 ) (y − y 0 )n+1 +··· . + (y − y 0 )n + (n + 1)! n!

f (y) = (1)

(3)

Recall that an equilibrium solution of Eq. (1) is any constant solution y(t ) of the equation; in particular, y(t ) = y 0 for all dy = 0, t ∈ R, where y 0 is the initial value of y (t ). Since dt this is equivalent to the condition that f (y 0 ) = 0. This way one sees that a constant function y(t ) = y 0 is an equilibrium solution of Eq. (1) if and only if f (y 0 ) = 0.

Let n be the smallest positive integer such that f (n) (y 0 ) 6= 0. (Generally, such an integer n depends on the particular critical point y 0 of f .) This reduces the Taylor series further to

Let y (t ) = y 0 be an equilibrium solution of Eq. (1). The (in)stability of y(t ) relies on the local property of f near y 0 . To facilitate our discussion, let us suppose that f is smooth near y 0 in the sense that it can be differentiated as many times as we want in a neighborhood of y 0 . Specifically, there is a Taylor series expansion

Since for y sufficiently close to y 0 , |y −y 0 |N ≪ |y −y 0 |n for all integer N > n, it follows that the local behaviors of f near y 0 f (n) (y 0 ) (y − y 0 )n on the are dominated by the leading term n! right-hand side of Eq. (4). In particular, as a consequence of Taylor’s theorem, it can be shown that the sign of f (y) equals the sign of f (n) (y 0 )(y − y 0 )n for all y sufficiently close to y 0 . As a corollary, one has the following conclusion about the stability of y(t ) = y 0 as an equilibrium solution of Eq. (1):

f ′ (y 0 )

f ′′ (y 0 ) (y − y 0 )2 +··· 2! 1! f (n+1) (y 0 ) f (n) (y 0 ) (y − y 0 )n + (y − y 0 )n+1 +··· + n! (n + 1)!

f (y) = f (y 0 ) +

(y − y 0 ) +

(2)

f (y) =

f (n) (y 0 ) n!

(y − y 0 )n +

f (n+1) (y 0 ) (y − y 0 )n+1 +··· . (4) (n + 1)!

1. If n is odd and f (n) (y 0 ) < 0, then y(t ) is asymptotically stable. 2. If n is odd and f (n) (y 0 ) > 0, then y(t ) is unstable. 3. If n is even, then y(t ) is semi-stable.

valid in a neighborhood of y 0 . Note that since y 0 is a critical...