AMATH_383_Solution_3 - AMATH 383 Solution #3 Winter 2010...

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AMATH 383 Solution #3 Winter 2010 Problem 9.1 (a). a=0, b=0, then 00 () , ( dx dy rx x ry y dt dt ) .  Setting the right-hand side to zero yields the equilibrium solution: x*=x 0, y*=y 0 . The stability consideration is rather easy because the equations can be solved exactly: 0 0 ( ) ( (0) ) , ( ) ( (0) ) . rt rt x tx x x ey ty y y e  So any perturbation of x from x 0 and y from y 0 will decay exponentially with decay rate r . (b). ( ) (, ) , ( ) dx dy rx x a y f xy y b x gxy dt dt  . The equilibrium solution is obtained by setting f(x*,y*)=0 and g(x*,y*)=0 to yield: 22 / ( 1 ) , / ( 1 aa b b xx y yy x rr r   ) . a b r To determine the stability of ( x*,y* ), we linearize. Let x(t)=x*+u(t) and y(t)=y*+v(t) . Then : 11 12 21 22 ,, du dv au av dt dt  where 11 12 21 22 , (, ) , ff ax y r y xy gg y b y   a r We try u(t)=u 0 e λ t , v(t) =v 0 e λ t , and find λ 2 -p λ +q=0 , where p=a 11 +a 22 =-2r, q=a 11 a 22 -a 12 a 21 =r 2 -ab . Therefore the two roots are: 12 ,. ra b b      Stability if both roots are negative; otherwise instability occurs.
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So stability if , ra b and instability if . b (c) r=1/5 years, a=1/1 year= b . . b The equilibrium is unstable.
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AMATH_383_Solution_3 - AMATH 383 Solution #3 Winter 2010...

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