007 - 5 Equilibrium Polnt Analysis 323 and the eigenvalues...

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Unformatted text preview: 5.] Equilibrium Polnt Analysis 323 and the eigenvalues are - ISO and -200. So (150, 0) is a sink. Finally, the lacobian matrix of (30. 40) is —30 -90 —80 -40 and the eigenvalues are approximately -120 and 50. So (30, 40) is a saddle. (b) . .V ’ 45 4 2 x 35 2 4 )' 5 105 100 0 x 95 x 150 ISS 5 (c) For each of these equilibria we could compute the eigenvalues and eigenvectors of the lin- earization. However. the figures in part (b) provide a great deal of information. It is impor- tant to note that the eigenvalues will be large in absolute value for this system. so solutions tend toward or away from the equilibria very quickly. (a) The equilibrium points are (0. 0). (0, 30). and (10, 0). Remember that we are only looking in the first quadrant. To determine the stability type of each of the equilibrium points we can compute the Jacobian matrix, which is —2x-y+10 —x —2y —2x—2y+30 and evaluate at each of the equilibrium points. At (0, O) the Jacobian becomes 100 030 and the eigenvalues are 10 and 30. The origin is a source. At (0. 30) the Jaeobian matrix is —20 0 —6O —30 and the eigenvalues are —20 and —30. So (0. 30) is a sink. The Jacobian at (10, 0) is -10 -10 0 10 and the eigenvalues are - 10 and 10. So (10, 0) is a saddle. ...
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