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Homework4Sol - Homework 4 solution AMATH 383 Autumn 2009 1...

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Homework 4 solution AMATH 383, Autumn 2009 November 30, 2009 1. Periodicity of the Volterra System (a) try scaling x = x 0 ˜ x, y = y 0 ˜ y, t = t 0 ˜ t then d x d t = x 0 x d ˜ t d ˜ t d t = x 0 t 0 x d ˜ t d y d t = y 0 t 0 y d ˜ t plug into the equations x d ˜ t = rt 0 ˜ x - at 0 y 0 ˜ x ˜ y y d ˜ t = kt 0 ( - ˜ y + bx 0 k ˜ x ˜ y ) let rt 0 = 1 at 0 y 0 = 1 bx 0 k = 1 we have scales for each variables t 0 = 1 r x 0 = k b y 0 = r a (b) d d t A ( x ( t ) , y ( t )) = ∂A ∂x d x d t + ∂A ∂y d y d t = γ (1 - 1 x ) x (1 - y ) + (1 - 1 y ) γy ( x - 1) = 0 (c) as x 0 , ln x → -∞ , so A ( x, y ) y - ln y + γ (0 - ( -∞ )) + . Similar argument also holds for y 0. D 2 A = γ x 2 0 0 1 y 2 is positive definite, thus is convex Combining with the previous fact that A approaches + at the boundary of the domain x > 0 , y > 0, A must have a global minimum. The minimum point can be found by solving DA = (0 , 0) = ( γ (1 - 1 x ) , 1 - 1 y ) ( x, y ) = (1 , 1) (d) the minimum value of A is A (1 , 1) = 1 + γ , so for any c > A min = 1 + γ , A ( x, y ) = c is a closed curve, corresponding to a periodic solution ( x ( t ) , y ( t )). For any positive initial value of ( x (0) , y (0)) other than (1 , 1), A ( x ( t ) , y ( t )) = A ( x (0) , y (0)) > A min , thus is periodic. see also the plots in mathematica for the shape of A . 1
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Homework, AMATH 383, Autumn 2009 2. Carnivores in Australia (a) try scaling x = x 0 ˜ x, y = y 0 ˜ y, t = t 0 ˜ t then d x d t = x 0 x d ˜ t d ˜ t d t = x 0 t 0 x d ˜ t d y d t = y 0 t 0 y d ˜ t plug into the equations x d ˜ t = rt 0 ˜ x (1 - x 0 K 0 ˜ x ) - at 0 y 0 ˜ x ˜ y y d ˜ t = kt 0 ( - ˜ y + bx 0 k ˜ x ˜ y ) with the scaling given in the problem, we got the dimensionless equation.
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