e-summary

# e-summary - will not tell you whether the solutions go...

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Eigenvalues, Eigenvectors, and Solution Curves Assume that J is the linearization of a two-by-two system at an equilibrium point. Case I: J has distinct, real, non-zero eigenvalues. 1. If λ 1 > λ 2 > 0: λ 1 is the dominant eigenvalue . v 1 is the dominant eigenvector . Solutions near the equilibrium will appear approxi- mately as shown at right. v 1 v 2 2. If λ 1 > 0 λ 2 : We rarely both with the label dominant . Solutions near the equilibrium will appear approxi- mately as shown at right. v 1 v 2 3. If 0 > λ 1 > λ 2 > 0: λ 1 is the dominant eigenvalue . (Note that this is true even though, in absolute value, λ 2 is bigger.) v 1 is the dominant eigenvector . Solutions near the equilibrium will appear approxi- mately as shown at right. v 1 v 2 1

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Case II: J has complex eigenvalues. In this case The eigenvalues are complex conjugates. I.e., λ = a ± bi Near the equilibrium solutions will either circle or spiral around the equilibrium.
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Unformatted text preview: will not tell you whether the solutions go clockwise or counter-clockwise. If | + 1 | < 1, then solutions near the equilibrium must spiral inward. Computing | + 1 | can be a pain. You should always seek to 1. Identify the real and imaginary parts. 2. Compute | + 1 | 2 and simplify if possible. Also, if you can prove this theorem you can save yourself future computation troubles: | + 1 | 2 = ( a + 1) 2 + b 2 , and | + 1 | 2 = tr ( J ) + det ( J ) + 1 (Recall that tr ( J ) is the sum of the diagonal entries of J .) If | + 1 | = 1, then solutions near the equilibrium will closely approximate closed loops, but closed loops are not guaranteed. If | + 1 | > 1, then solutions near the equilibrium will spiral out. 2...
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## This note was uploaded on 10/12/2011 for the course MATH 464 taught by Professor Dougbullock during the Fall '08 term at Boise State.

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