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527solns7

# 527solns7 - 642:527 SOLUTIONS ASSIGNMENT 7 FALL 2009 NOTE A...

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642:527 SOLUTIONS: ASSIGNMENT 7 FALL 2009 NOTE: A. There is a Maple worksheet on the “Assignments and solutions” web page which shows the phase plane portraits for problems 2(b) and 2(k). B. Vectors are denoted either by boldface or, for Greek letters, underlining: ζ . Review: Consider x = P ( x, y ), y = Q ( x, y ). A point ( x 0 , y 0 ) is a singular point for these equations if P ( x 0 , y 0 ) = Q ( x 0 , y 0 ) = 0 . (12.1) To linearize near this singular point we make the the linear approximations P ( x, y ) P x ( x 0 , y 0 )( x - x 0 ) + P y ( x 0 , y 0 )( y - y 0 ) , Q ( x, y ) Q x ( x 0 , y 0 )( x - x 0 ) + Q y ( x 0 , y 0 )( y - y 0 ) (here we have used (12.1)). With X = x - x 0 , Y = y - y 0 , the linearized equations are X = P x ( x 0 , y 0 ) X + P y ( x 0 , y 0 ) Y, Y = Q x ( x 0 , y 0 ) X + Q y ( x 0 , y 0 ) Y. (12.2) Section 7.4 2. (a) The equations are x = y , y = 1 - x 4 . To find the singular points we set y = 0 and 1 - x 4 = 0; since 1 - x 4 = (1 - x )(1 + x )(1 + x 2 ) there are two solutions: (1 , 0) and ( - 1 , 0). We analyze the phase plane near each. (i) Near (1 , 0). Here X = x - 1, Y = y . P ( x, y ) = y is already linear and so it is its own linear approximation. Then (12.2) becomes X = Y, Y = - 4 X. From det( A - λI ) = vextendsingle vextendsingle vextendsingle vextendsingle - λ 1 - 4 - λ vextendsingle vextendsingle vextendsingle vextendsingle = λ 2 + 4 = 0 we find the eigenvalues λ = ± 2 i ; this is a center in the linearized system. We can’t tell from this what it is in the true system: it might be a center, a stable focus, or an unstable focus. But notice that there is a conserved quantity, F ( x, y ) = y 2 2 - x + x 5 5 (this is just the total energy if we regard the equations as describing a particle, with position x and velocity y , moving in a potential V ( x ) = - x + x 5 / 5). The level curves of F are then orbits and it is not hard to see that there are closed orbits circling (1 , 0)—so this point stays a center. (ii) Near ( - 1 , 0). Here X = x + 1, Y = y . Here (12.2) becomes X = Y, Y = 4 X. From det( A - λI ) = vextendsingle vextendsingle vextendsingle vextendsingle - λ 1 4 - λ vextendsingle vextendsingle vextendsingle vextendsingle = λ 2 - 4 = 0 we find the eigenvalues λ = ± 2; this is a saddle point in the linearized system. We find the special trajectories: for λ 1 = 2, we solve bracketleftbigg - 2 1 4 - 2 bracketrightbigg bracketleftbigg ξ η bracketrightbigg = 0 to find the eigenvector ζ (1) = bracketleftbigg 1 2 bracketrightbigg ; for λ 2 = - 2, we solve 1

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642:527 SOLUTIONS: ASSIGNMENT 7 FALL 2009 bracketleftbigg 2 1 4 2 bracketrightbigg bracketleftbigg ξ η bracketrightbigg = 0 to find the eigenvector ζ (2) = bracketleftbigg - 1 2 bracketrightbigg . The straight-line trajectories for the saddle point are y = 2 x , which is oriented outward since λ 1 > 0, and y = - 2 x , which is oriented inward since λ 2 < 0. The point ( - 1 , 0) is again a saddle point in the true system.
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527solns7 - 642:527 SOLUTIONS ASSIGNMENT 7 FALL 2009 NOTE A...

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