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. =+ + xx t y yx y t 2 γ sin Dependent variables: x , y Parameter: Nonautonomous linear system Nonhomogeneous sin t () 2. ′=− + uu ut 34 2 υ sin Dependent variables: u , Parameters: none Nonautonomous linear system Nonhomogeneous sin t 3. ′ = ′=− 12 21 κ sin Dependent variables: x 1 , x 2 Parameter: Autonomous nonlinear sin x 1 a f system 4. ′ = =− pq qp q t sin Dependent variables: p , q Parameters: none Nonautonomous nonlinear pq system 5. ′ = Sr S I I rSI I RI Dependent variables: R , S , I Parameters: r , 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. ′ = xx , ′ = yy Plugging xye t == into the two differential equations, we get ee tt = = 7. ′ = ′=− xy yx Plugging xt = sin and yt = cos into the two differential equations, we get cos cos sin sin = −= 8. =+ ′=− + + t y 235 Direct substitution 9. ′ = ′ = yz zy 2 2 Direct substitution ! A Habit to Acquire 10. ′ = =− () 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 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 ′ = xy , =− () yx x 1 To determine whether 0 0 , a f is a spiral or center, we draw the trajectory starting at 0 05 ,. af , 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 , and head off to the upper right, or circle around the origin 0 0 , 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 spiraling away from 0 0 , a f ; evidence that both equilibria are unstable 11. ′ = xx y yy x 3cos h -nullcline: = 3cos υ -nullcline: x - and y -axes The equilibrium points are located at the points 03 , a f , ± F H I K π 2 0 , , ± F H I K 3 2 0 , , ± F H I K 5 2 0 , , . –1 –2 1 2 y x 3 –3 2 3 –2 –3 Nullclines of ′ = y , x The diagram that follows indicates that equilibrium at 0 3 , is unstable and that the equilibrium points on the x -axis appear to alternate between saddle points and centers or spirals with the saddle points closest to the origin. . (The x -intercepts at ± 2 , ± 5 2 , ± 9 2 ,

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

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