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Chapter 7

# Chapter 7 - CHAPTER 7 Nonlinear Systems of Differential...

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CHAPTER 7 Nonlinear Systems of Differential Equations 7.1 Nonlinear Systems ! Review of Classifications 1. ′ = + ′ = + + x x ty y x y t 2 γ sin Dependent variables: x , y Parameter: γ Nonautonomous linear system Nonhomogeneous γ sin t ( ) 2. ′ = + ′ = − + u u u t 3 4 2 υ υ sin Dependent variables: u , υ Parameters: none Nonautonomous linear system Nonhomogeneous sin t ( ) 3. ′ = ′ = − x x x x 1 2 2 1 κ sin Dependent variables: x 1 , x 2 Parameter: κ Autonomous nonlinear sin x 1 a f system 4. ′ = ′ = p q q pq t sin Dependent variables: p , q Parameters: none Nonautonomous nonlinear pq ( ) system 5. ′ = − ′ = ′ = S rSI I rSI I R I γ γ Dependent variables: R , S , I Parameters: r , γ Autonomous nonlinear SI ( ) system 517

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518 CHAPTER 7 Nonlinear Systems of Differential Equations ! Verification Review Problems 6–9 are all by direct substitution. This substitution is demonstrated in Problem 7. 6. ′ = x x , ′ = y y Plugging x y e t = = into the two differential equations, we get e e e e t t t t = = 7. ′ = ′ = − x y y x Plugging x t = sin and y t = cos into the two differential equations, we get cos cos sin sin t t t t = = − 8. ′ = + ′ = − + + x y t y x y 2 3 5 Direct substitution 9. ′ = ′ = ′ = − x x y z z y 2 2 Direct substitution ! A Habit to Acquire 10. ′ = ′ = ( ) x y y x x 1 h -nullcline: x = 0 and x = 1 υ -nullcline: y = 0 ( x -axis) The equilibrium points are located at the inter- sections of the vertical and horizontal isoclines, which are the points 0 0 , ( ) , 1 0 , ( ) . Arrows are drawn (see figure) to show the direction of the trajectories. It is clear that the point 1 0 , ( ) is unstable, but not clear whether the origin 0 0 , ( ) is a center or a spiral and if it is a spiral whether it spirals in or away from the origin. 2 –1 1 y x x = 0 x = 1 spiral or center Nullclines of ′ = x y , ′ = ( ) y x x 1
SECTION 7.1 Nonlinear Systems 519 To better understand how solutions behave, we draw sample trajectories in the phase plane. Note from the figure, that we still cannot determine whether 0 0 , a f is a center or a spiral point. –2 2 y 2 –2 x Phase portrait of ′ = x y , ′ = ( ) y x x 1 To determine whether 0 0 , a f is a spiral or center, we draw the trajectory starting at 0 05 , . a f , which shows that 0 0 , a f is a stable center. Note that the solution moves around the origin. Examining the previous figures, we con- clude the solutions come in from the 4th quad- rant, either cross the x -axis to the right of 1 0 , a f and head off to the upper right, or circle around the origin 0 0 , a f and then head off to the upper right. Points near the origin 0 0 , a f are circling around the origin. –0.6 0.6 y 1.5 –1.5 x Trajectory starting at 0 05 , . a f spiraling away from 0 0 , a f ; evidence that both equilibria are unstable 11. ′ = ′ = x xy y y x 3cos h -nullcline: y x = 3cos υ -nullcline: x - and y -axes The equilibrium points are located at the points 0 3 , a f , ± F H I K π 2 0 , , ± F H I K 3 2 0 π , , ± F H I K 5 2 0 π , , .

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• Fall '07
• WILLIAMHEUETT
• Linear system, Equilibrium point, Nonlinear system, Stability theory, nonlinear systems, Nonlinear Systems of Differential Equations

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