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hw6sol

hw6sol - Homework 6 Solutions a = aL and = bK Eq(3 is...

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Homework 6, Solutions a) α = aL and β = bK . Eq. (3) is obtained from (1) by dividing both sides by K . Eq. (4) is obtained from (2) by dividing both sides by L . b) ( x , y ) is a fixed point of (3), (4) if and only if x = 0 and y = 0 , (1) or 1 x α y = 0 and y = 0 , (2) or x = 0 and 1 y β x = 0 , (3) or 1 x α y = 0 and 1 y β x = 0 . (4) (1), (2), and (3) correspond to the three fixed points ( 0 , 0 ) , ( 1 , 0 ) , and ( 0 , 1 ) . The Jacobi matrix in ( 0 , 0 ) is J = bracketleftbigg r 0 0 s bracketrightbigg . This matrix has two positive eigenvalues, therefor the origin is an unstable node. c) To find a fixed point ( x , y ) in which the two species coexist, we have to solve the linear system (4). In matrix-vector form: bracketleftbigg 1 α β 1 bracketrightbiggbracketleftbigg x y bracketrightbigg = bracketleftbigg 1 1 bracketrightbigg . (5) The system is non-singular if αβ negationslash = 1. In that case, Cramer’s Rule tells us that x = 1 α 1 αβ and y = 1 β 1 αβ .

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