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Unformatted text preview: Homework 6, Solutions a) = aL and = 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 = , (1) or 1 x y = 0 and y = , (2) or x = 0 and 1 y x = , (3) or 1 x y = 0 and 1 y x = . (4) (1), (2), and (3) correspond to the three fixed points ( , ) , ( 1 , ) , and ( , 1 ) . The Jacobi matrix in ( , ) is J = bracketleftbigg r 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 1 bracketrightbiggbracketleftbigg x y bracketrightbigg = bracketleftbigg 1 1 bracketrightbigg . (5) The system is non-singular if negationslash = 1. In that case, Cramers Rule tells us that1....
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This note was uploaded on 02/28/2011 for the course MATH 150 taught by Professor Dr.borgers during the Spring '09 term at Tufts.
- Spring '09