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Unformatted text preview: Chap . 4 Systems of ODEs . Phase Plane . Qualitative Methods Sec . 4 . 1 Systems of ODEs as Models Example 2 . Spend time on Fig. 79 on p. 134 until you feel that you fully understand the difference between (a) (the usual representation in calculus) and (b), because trajectories will play an important role throughout this chapter. Try to understand the reasons for the following. The trajectory starts at the origin. It reaches its highest point where I 2 has a maximum (before t u 1). It has a vertical tangent where I 1 has a maximum, shortly after t u 1. As t increases from there to t u 5, the trajectory goes downward until it almost reaches the I 1axis at 3; this point is a limit as t U u . In terms of t the trajectory goes up faster than it comes down. Problem Set 4 . 1 . Page 135 7 . Electrical network . The problem amounts to the determination of the two arbitrary constants in a general solution of a system of two ODEs in two unknown functions I 1 and I 2 , representing the currents in an electrical network shown in Fig. 78 in Sec. 4.1. You will see that this is quite similar to the corresponding task for a single secondorder ODE. That solution is given by (6), in components I 1 u t U u 2 c 1 e U 2 t ¡ c 2 e U 0.8 t ¡ 3, I 2 u t U u c 1 e U 2 t ¡ 0.8 c 2 e U 0.8 t . Setting t u 0 and using the given initial conditions I 1 u U u 0, I 2 u U u U 3 gives I 1 u U u 2 c 1 ¡ c 2 ¡ 3 u I 2 u U u c 1 ¡ 0.8 c 2 u U 3. (a) (b) From (a) you have c 2 u U 3 U 2 c 1 . From this and (b) you obtain c 1 ¡ 0.8 u U 3 U 2 c 1 U u U 0.6 c 1 U 2.4 u U 3, hence c 1 u 1. Also c 2 u U 3 U 2 c 1 u U 3 U 2 u U 5. This yields the answer I 1 u t U u 2 e U 2 t U 5 e U 0.8 t ¡ 3 I 2 u t U u e U 2 t ¡ 0.8 u U 5 U e U 0.8 t u e U 2 t U 4 e U 0.8 t . You see that the limits are 3 and 0, respectively. Can you see this directly from Fig. 78 for physical reasons? 11 . Conversion of single ODEs to a system is an important process, which always follows the pattern shown in formulas (9) and (10) of Sec. 4.1. The present equation y ¢¢ U 4 y u 0 can be readily solved. A general solution is y u c 1 e 2 t ¡ c 2 e U 2 t . The point of the problem is not to explain a (complicated) solution method for a simple problem, but to explain the relation between systems and single ODEs and their solutions. In the present case the formulas (9) and (10) give y 1 u y , y 2 u y ¢ and y 1 ¢ u y 2 y 2 ¢ u 4 y 1 (because the given equation can be written y ¢¢ u 4 y , hence y 1 ¢¢ u 4 y 1 , but y 1 ¢¢ u y 2 ¢ ). In matrix form (as in Example 3 of the text) this is y ¢ u Ay u 0 1 4 0 y . The characteristic equation is det u A U u I U u U u 1 4 U u u u 2 U 4 u 0. Chap. 4 Systems of ODEs. Phase Plane. Qualitative Methods 31 The eigenvalues are u 1 U 2 and u 2 U u 2. For u 1 you obtain an eigenvector from (13) in Sec. 4.0 with u U u 1 , that is, u A u u 1 I U x U u 2 1 4 u 2 x 1 x 2 U u 2 x 1 ¡ x 2 4 x 1 u 2 x 2 U ....
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
 Spring '12
 M.lee

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