To show that the system x ˙ = -vx + zy , y ˙ = -vy + ( z - a ) x , z ˙ = 1 - xy is dissipative. To show that the fixed points may be written in parametric form as x * = ±k , y * = ± k - 1 , z * = vk 2 , where v ( k 2 - k - 2 ) = a . To classify the fixed points. Concept Introduction: ➢ The divergence of a vector field is ∇ . f = ∂ ∂ x x ˙ + ∂ ∂ y y ˙ + ∂ ∂ z z ˙ ➢ The fixed points are calculated as x ˙ = 0 , y ˙ = 0 , z ˙ = 0 ➢ The Jacobian matrix is given by A = ( ∂ x ˙ ∂ x ∂ x ˙ ∂ y ∂ x ˙ ∂ z ∂ y ˙ ∂ x ∂ y ˙ ∂ y ∂ y ˙ ∂ z ∂ z ˙ ∂ x ∂ z ˙ ∂ y ∂ z ˙ ∂ z ) ➢ The Eigen value λ can be calculated using the characteristic equation | ( A - λI ) | = 0

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Chapter 9.2, Problem 6E

Interpretation Introduction

Interpretation:

To show that the system $x˙ = -vx + zy$ , $y˙ = -vy + (z - a)x$ , $z˙ = 1 - xy$ is dissipative. To show that the fixed points may be written in parametric form as $x* = ±k$ , $y* = ± k- 1$ , $z* = vk2$ , where $v(k2- k- 2) = a$ . To classify the fixed points.

Concept Introduction:

➢ The divergence of a vector field is

$∇. f = ∂∂xx˙ + ∂∂yy˙ + ∂∂zz˙$

➢ The fixed points are calculated as $x˙ = 0$ , $y˙ = 0$ , $z˙ = 0$

➢ The Jacobian matrix is given by

$A = (∂x˙∂x∂x˙∂y∂x˙∂z∂y˙∂x∂y˙∂y∂y˙∂z∂z˙∂x∂z˙∂y∂z˙∂z)$

➢ The Eigen value $λ$ can be calculated using the characteristic equation

$|(A - λI)| = 0$

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Question 1. Prove that the function f(x) = 2; f: (2,3] → R, is not uniformly continuous on (2,3].

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