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Homework5Sol

# Homework5Sol - Homework 5 AMATH 383 Autumn 2009 Due...

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Homework 5 AMATH 383, Autumn 2009 Due: Wednesday, December 2, after class 1. (4+6+8+6+8 points) Hopf-Bifurcation and Limit Cycle: In class we discussed the Brussela- tor, a chemical reaction that showed an oscillating limit cycle after its equilibrium point changed from stable to unstable. In this problem we study an ODE system similar to the Brusselator equations. It is given by x 0 = α x - y - x ( x 2 + y 2 ) y 0 = x + α y - y ( x 2 + y 2 ) with solution ( x ( t ) , y ( t )) depending on the parameter α and some initial conditions. (a) Find the single equilibrium of this system and compute the eigenvalues of the Jacobian at that point. Show that the equilibrium is a stable focus (damped oscillations) for α < 0 and becomes unstable for α > 0. (b) Using a computer, plot the phase diagram of the system in the range ( x, y ) [ - 2 , 2] 2 , for the parameter choices α = 0 . 5 and α = - 0 . 8. Note the change in behavior of the equilibrium point, which is called Hopf-Bifurcation . For α = - 0 . 8, compute the solution of the ODE system numerically with initial conditions x (0) = 0 . 01 and y (0) = 0 . 01 and plot it in the time range t [0 , 15]. The solution starts to grow exponentially according to an unstable equilibrium, but turns into a stationary oscillation for larger times (a limit cycle). (c) In order to understand this behavior we change the variables ( x ( t ) , y ( t )) to polar coordinates ( r ( t ) , ϕ ( t )) by x ( t ) = r ( t ) cos ϕ ( t ) , y ( t ) = r ( t ) sin ϕ ( t ) with a radius from the origin r ( t ) 0 and an angle ϕ ( t ) [0 , 2 π ]. Show that the derivative of ( x ( t ) , y ( t )) is given by x 0 ( t ) y 0 ( t ) = r 0 ( t ) e 1 + r ( t ) ϕ 0 ( t ) e 2 with vectors e 1 = (cos ϕ, sin ϕ ) and e 2 = ( - sin ϕ, cos ϕ ), and the right hand side can be written ( α - ( x 2 + y 2 ) ) x - y ( α - ( x 2 + y 2 ) ) y + x = ( α r ( t ) - r ( t ) 3 ) e 1 + r ( t ) e 2 . Derive the equations r 0 ( t ) = α r ( t ) - r ( t ) 3 , ϕ 0 ( t ) = 1 by equalizing the derivative and the right hand side and scalar-multiplying the whole equa- tion with e 1 and e 2 . Note, that these two equations are equivalent to the original systems. They only use different coordinates, namely radius and angle. (d) The equation for the radius is similar to a logistic equation. Find its stationary points and their stability. Keep in mind that r 0 and consider the cases α < 0 and α > 0. What is the solution for ϕ ( t )? Use these results to explain the phenomena of a limit cycle for the original system. 1

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Homework, AMATH 383, Autumn 2009 (e) Show that the solution of the equation for r ( t ) with initial conditions r (0) = R is given by r ( t ) = R r α R 2 + ( α - R 2 ) exp( - 2 αt ) . Combine this and the solution for ϕ ( t ) with initial conditions ϕ (0) = 0, to write down the solution for ( x ( t ) , y ( t )). Solution: (a) 0 = α x - y - x ( x 2 + y 2 ) ( x * , y * ) = (0 , 0) 0 = x + α y - y ( x 2 + y 2 ) J = - 3 x 2 - y 2 + α - 1 - 2 xy 1 - 2 xy - x 2 - 3 y 2 + α at (0 , 0) J = α - 1 1 α λ 1 , 2 = α ± i Re( λ 1 , 2 ) = α stable α < 0 unstable α > 0 (b) see plot in mathematica file ( α = - 0 . 5 , 0 . 8) (c) d dt r ( t ) cos ϕ ( t ) r ( t ) sin ϕ ( t ) = r 0 cos ϕ - 0 sinϕ r 0 sin ϕ + 0 cos ϕ = r 0 ( t ) cos ϕ sin ϕ + 0 ( t ) - sin ϕ cos ϕ ( α - ( x 2 + y 2 ) ) x - y ( α - ( x 2 + y 2 ) ) y + x = ( α - r 2 ) r cos ϕ - r sin ϕ ( α - r 2 ) r sin ϕ + r cos ϕ = ( αr - r 3 ) cos ϕ sin ϕ + r - sin ϕ cos ϕ r 0 e 1 + 0 e 2 = ( αr - r 3 ) e 1 + r e 2 since e 1 · e 2 = 0, e 1 , 2
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