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springps10_soln

springps10_soln - 18.03 PS10 Solutions-Spring 09 1a Char eq...

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18.03 PS10 Solutions—Spring 09 1a. Char. eq. ( λ + 1) 2 = 0 = λ = - 1; A - λI = A + I = - λ 1 - 1 - 2 - λ = 1 1 - 1 - 1 . We see by inspection that ( A + I ) 1 - 1 = 0 , so x 1 = e - t 1 - 1 is a solution. We guess that the other independent solution will have the form x 2 = ( a t + b ) e - t for column vectors a = a 1 a 2 , b = b 1 b 2 . Plugging in x 2 0 = A x 2 , we obtain ( A + I ) a = 0 and ( A + I ) b = a . Take a = 1 - 1 . Given that choice of a , the condition on b is b 1 + b 2 = 1, so b = 1 0 works. x 1 = e - t 1 - 1 and x 2 = 1 - 1 t + 1 0 e - t are two independent solutions. Note that the answer is far from unique. 1b. The 2nd-order equation is x 00 + 2 x 0 + x = 0 . Two solutions are x = e - t , y = te - t . x x 0 = e - t 1 - 1 = x 1 ; y y 0 = te - t 1 - 1 + e - t 0 1 = x 2 - x 1 . 2ac. See Figure 1 for the diagram. Table follows. Region Geom. type Stability Eigenvalues ( T, D ) definition 1 saddle N/A (unstable) λ 1 < 0 < λ 2 D < 0 2 node stable = sink λ 1 < λ 2 < 0 T < 0 , 0 < D < T 2 / 4 3 spiral stable = sink r ± si , r < 0 T < 0 , D > T 2 / 4 4 spiral unstable = source r ± si , r > 0 T > 0 , D > T 2 / 4 5 node unstable = source 0 < λ 1 < λ 2 T > 0 , 0 < D < T 2 / 4 2b. The parabola is given by (i) λ 1 = λ 2 , (ii) s = 0, and (iii) D = T 2 / 4. 2d. The equation for the boundary is T = 0 , D > 0. In terms of the eigenvalues, one can say that they must be purely imaginary or that they are complex with r = 0. As we move across the boundary from T < 0 to T > 0, we begin with spirals pointing inward. The spirals get tighter and tighter until, at the boundary point T = 0, they stop pointing inward at all and become ellipses. Then as T
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