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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 7 FALL 2007 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 , y ) is a singular point for these equations if P ( x , y ) = Q ( x , y ) = 0 . (7.1) To linearize near this singular point we make the the linear approximations P ( x, y ) ≈ P x ( x , y )( x- x ) + P y ( x , y )( y- y ) , Q ( x, y ) ≈ Q x ( x , y )( x- x ) + Q y ( x , y )( y- y ) (here we have used (7.1)). With X = x- x , Y = y- y , the linearized equations are X ′ = P x ( x , y ) X + P y ( x , y ) Y, Y ′ = Q x ( x , y ) X + Q y ( x , y ) Y. (7.2) 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 (7.2) 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 of5)....
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- Fall '07
- Web page, Eigenvalue, eigenvector and eigenspace, saddle point, phase plane