Math 211 Math 211 Lecture #42 The Pendulum Predator-Prey - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #42 The Pendulum Predator-Prey December 6, 2002

2 The Pendulum The Pendulum Return

2 The Pendulum The Pendulum • The angle θ satisfies the nonlinear differential equation mLθ ′′ = − mg sin θ − D θ ′ , Return

2 The Pendulum The Pendulum • The angle θ satisfies the nonlinear differential equation mLθ ′′ = − mg sin θ − D θ ′ , � We will write this as θ ′′ + d θ + b sin θ = 0 . Return

2 The Pendulum The Pendulum • The angle θ satisfies the nonlinear differential equation mLθ ′′ = − mg sin θ − D θ ′ , � We will write this as θ ′′ + d θ + b sin θ = 0 . • Introduce ω = θ ′ to get the system θ ′ = ω ω ′ = − b sin θ − d ω Return

3 Analysis Analysis Return

3 Analysis Analysis • The equilibrium points are ( k π, 0) T where k is any integer. Return

3 Analysis Analysis • The equilibrium points are ( k π, 0) T where k is any integer. � If k is odd the equilibrium point is a saddle. Return

3 Analysis Analysis • The equilibrium points are ( k π, 0) T where k is any integer. � If k is odd the equilibrium point is a saddle. � If k is even the equilibrium point is a center if d = 0 or a sink if d > 0 . Return

4 The Inverted Pendulum The Inverted Pendulum Return Pendulum

4 The Inverted Pendulum The Inverted Pendulum • The angle θ measured from straight up satisfies the nonlinear differential equation mLθ ′′ = mg sin θ − D θ ′ , Return Pendulum

4 The Inverted Pendulum The Inverted Pendulum • The angle θ measured from straight up satisfies the nonlinear differential equation mLθ ′′ = mg sin θ − D θ ′ , or θ ′′ + D mLθ ′ − g L sin θ = 0 . Return Pendulum

4 The Inverted Pendulum The Inverted Pendulum • The angle θ measured from straight up satisfies the nonlinear differential equation mLθ ′′ = mg sin θ − D θ ′ , or θ ′′ + D mLθ ′ − g L sin θ = 0 . � We will write this as θ ′′ + d θ − b sin θ = 0 . Return Pendulum

5 The Inverted Pendulum System The Inverted Pendulum System Return Inverted pendulum Pendulum system

5 The Inverted Pendulum System The Inverted Pendulum System • Introduce ω = θ ′ to get the system θ ′ = ω ω ′ = b sin θ − d ω Return Inverted pendulum Pendulum system

5 The Inverted Pendulum System The Inverted Pendulum System • Introduce ω = θ ′ to get the system θ ′ = ω ω ′ = b sin θ − d ω • The equilibrium point at (0 , 0) T is a saddle point and unstable. Return Inverted pendulum Pendulum system

5 The Inverted Pendulum System The Inverted Pendulum System • Introduce ω = θ ′ to get the system θ ′ = ω ω ′ = b sin θ − d ω • The equilibrium point at (0 , 0) T is a saddle point and unstable. • Can we find an automatic way of sensing the departure of the system from (0 , 0) T and moving the pivot to bring the system back to the unstable point at (0 , 0) T ? Return Inverted pendulum Pendulum system

5 The Inverted Pendulum System The Inverted Pendulum System • Introduce ω = θ ′ to get the system θ ′ = ω ω ′ = b sin θ − d ω • The equilibrium point at (0 , 0) T is a saddle point and unstable. • Can we find an automatic way of sensing the departure of the system from (0 , 0) T and moving the pivot to bring the system back to the unstable point at (0 , 0) T ? � Experimentally the answer is yes. Return Inverted pendulum Pendulum system

6 The Control System The Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies mLθ ′′ = mg sin θ − D θ ′ − v cos θ, Return Inverted pendulum Inverted pendulum system

6 The Control System The Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies mLθ ′′ = mg sin θ − D θ ′ − v cos θ, • The system becomes θ ′ = ω ω ′ = b sin θ − d ω − u cos θ, where u = v/mL. Return Inverted pendulum Inverted pendulum system

6 The Control System The Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies mLθ ′′ = mg sin θ − D θ ′ − v cos θ, • The system becomes θ ′ = ω ω ′ = b sin θ − d ω − u cos θ, where u = v/mL. • Assume the force is a linear response to the detected value of θ , so u = cθ , where c is a constant. Return Inverted pendulum Inverted pendulum system

7 The Controlled System The Controlled System Return Inverted pendulum Inverted pendulum system Controls

7 The Controlled System The Controlled System • The Jacobian at the origin is � 0 1 � J = b − c − d Return Inverted pendulum Inverted pendulum system Controls

7 The Controlled System The Controlled System • The Jacobian at the origin is � 0 1 � J = b − c − d • The origin is asymptotically stable if T = − d < 0 and D = c − b > 0 . Return Inverted pendulum Inverted pendulum system Controls

7 The Controlled System The Controlled System • The Jacobian at the origin is � 0 1 � J = b − c − d • The origin is asymptotically stable if T = − d < 0 and D = c − b > 0 . Therefore require c > b = g L. Return Inverted pendulum Inverted pendulum system Controls

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle, ( x 0 , y 0 ) = ( c/d, a/b ) is a linear center. Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle, ( x 0 , y 0 ) = ( c/d, a/b ) is a linear center. • The axes are invariant. Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle, ( x 0 , y 0 ) = ( c/d, a/b ) is a linear center. • The axes are invariant. • The positive quadrant is invariant. Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle, ( x 0 , y 0 ) = ( c/d, a/b ) is a linear center. • The axes are invariant. • The positive quadrant is invariant. • The solution curves appear to be closed. Return

8 Predator-Prey Predator-Prey Lotka-Volterra system x ′ = ( a − by ) x (prey – fish) y ′ = ( − c + dx ) y (predator – sharks) • Equilbrium points: (0 , 0) is a saddle, ( x 0 , y 0 ) = ( c/d, a/b ) is a linear center. • The axes are invariant. • The positive quadrant is invariant. • The solution curves appear to be closed. Is this actually true? Return

9 Solutions are Periodic Solutions are Periodic Along the solution curve y = y ( x ) we have Return System

9 Solutions are Periodic Solutions are Periodic Along the solution curve y = y ( x ) we have dx = y ( − c + dx ) dy x ( a − by ) . Return System

9 Solutions are Periodic Solutions are Periodic Along the solution curve y = y ( x ) we have dx = y ( − c + dx ) dy x ( a − by ) . The solution is H ( x, y ) = by − a ln y + dx − c ln x = C Return System

9 Solutions are Periodic Solutions are Periodic Along the solution curve y = y ( x ) we have dx = y ( − c + dx ) dy x ( a − by ) . The solution is H ( x, y ) = by − a ln y + dx − c ln x = C • This is an implicit equation for the solution curve. Return System

9 Solutions are Periodic Solutions are Periodic Along the solution curve y = y ( x ) we have dx = y ( − c + dx ) dy x ( a − by ) . The solution is H ( x, y ) = by − a ln y + dx − c ln x = C • This is an implicit equation for the solution curve. ⇒ All solution curves are closed, and represent periodic solutions. Return System

10 Why Fishing Leads to More Fish Why Fishing Leads to More Fish System Return

10 Why Fishing Leads to More Fish Why Fishing Leads to More Fish Compute the average of the fish & shark populations. System Return

10 Why Fishing Leads to More Fish Why Fishing Leads to More Fish Compute the average of the fish & shark populations. dt ln x ( t ) = x ′ d x = System Return

10 Why Fishing Leads to More Fish Why Fishing Leads to More Fish Compute the average of the fish & shark populations. dt ln x ( t ) = x ′ d x = a − by System Return

10 Why Fishing Leads to More Fish Why Fishing Leads to More Fish Compute the average of the fish & shark populations. dt ln x ( t ) = x ′ d x = a − by � T 0 = 1 d dt ln x ( t ) dt T 0 System Return