problemset9

An eigenvector for 1 a bi is given by v1 such that v1

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Unformatted text preview: nodal source + 1 + 2 3 < a < 3—unstable node = nodal source a = 3—unstable degenerate comb 3 < a—saddle (b)-(c) [18] Here are�pictures for a = 0, 1, 2, −1 + � 2 3, 2.75, 3, 4.�(a = −2 3 omitted.) The picture for some a < −1 − 2 3 would show a nodal sink, and that for � a = −1 − 2 3 would show a defective nodal sink. � � a −b 36. (a) [9] With A = , pA (�) = �2 − 2a� +(a2 + b2 ) = (� − a)2 + b2 , so the eigenvalues ba � � −bi −b are a ± bi. An eigenvector for �1 = a + bi is given by v1 such that v1 = 0, and b −bi � � � � 1 1 we can take v1 = . The corresponding normal mode is e(a+bi)t . Its real and −i −i � � � � cos(bt) sin(bt) at at imaginary parts give linearly independent real solutions, e and e . sin(bt) cos(bt) � � � cos(bt) sin(bt) 1...
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