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MIT18_03S10_c35

# MIT18_03S10_c35 - 18.03 Class 35 May 3 2010 Linear Phase...

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18.03 Class 35 , May 3, 2010 Linear Phase Portraits: Eigenvalues Rule (usually) 1. Eigenvalues rule! 2. Trace-determinant plane 3. Marginal cases 4. Stability [1] Phase portrait: this means the (x,y) plane (the "phase plane") with trajectories of solutions of u' = Au drawn on it (with direction of time indicated). These homogeneous linear equations exhibit a nice variety of phase portraits, as shown by the Linear Phase Portraits Mathlets. An important fact is this: EIGENVALUES RULE We'll classify the linear phase portraits according to the eigenvalues of the matrix A . Example: those rabbits again: A = [ .3 .1 ; .2 .4 ] p_A(lambda) = lambda^2 - .7 lambda + .1 has roots lambda_1 = .5 , lambda_2 = .2 so we learn that all solutions flee from the origin: the phase portrait is a "source." lambda = .5 : [ -.2 .1 ; .2 -.1 ] ==> v_1 = [ 1 ; 2 ] so one normal mode is u_1 = e^{.5 t} [ 1 ; 2 ] If the number of rabbits in MacGregor's field is twice the number in Jones's, it stays that way forever after. lambda = .2 : [ .1 .1 ; .2 .2 ] ==> v_2 [ 1 ; -1 ] so the other normal mode is u_2 = e^{.2t} [ 1 ; -1 ] . This is not meaningful in itself in our population model, but we can draw it in the phase plane. And the two together provide the general solution. What to the other trajectories look like? For large t , the v_1 component is much bigger than the v_2 component. For small t , the v_1 component is much smaller than the v_2 component. So near the origin the trajectories become asymototic to the eigenline with smaller eigenvalue. This phase portrait is a NODE. The same kind of picture occurs whenever the eigenvalues are real, of the same sign, and distinct.

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[2] When are the eigenvalues non-real? [Slide:] p_A(lambda) = det ( A - lambda I ) If A = [ a b ; c d ] then p_A(lambda) = lambda^2 - (tr A) lambda + (det A) tr(A) = a + d = lambda_1 + lambda_2 det(A) = ad-bc = lambda_1 lambda)2 To find the eigenvalues, complete the square: p_A(lambda) = (lambda - (tr(A)/2))^2 + (det(A) - (tr(A)/2)^2) so lambda1,2 = tr(A)/2 +- sqrt(tr(A)^2/4 - det(A)).
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MIT18_03S10_c35 - 18.03 Class 35 May 3 2010 Linear Phase...

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