Chapter Homogeneous System with Constant Coeffici

Transcription

1 Chapter Homogeneous System with Constant Coefficients

2 Homogeneous System with constant coefficients We consider homogeneous linear systems: x = Ax A is an n n matrix with constant entries () As in 7.4, solutions of () will be denoted by x (t) x () (t) = x 2 (t),,x(k) (t) = x n (t) x k (t) x 2k (t) x nk (t).

3 Principle of superposition Recall from 7.4 the Principle of superposition and the converse: If x (),...,x (n) are solution of (), then, any constant linear combination x = c x () + +c n x (n) (2) is also a solution of the same system (). The converse is also true, if Wronskian W 0.

4 Solutions of () and Eigenvectors Some solutions of () are given by x = ξe rt where Aξ = rξ, (3) which means that r is an eigenvalue of A and ξ is an eigenvector, corresponding to r. Proof. For x = ξe rt, we have x = (rξ)e rt = A(ξe rt ) = Ax

5 n eigenvalues and vectors Suppose A has n eigenvalues r, r 2,...,r n. Pick ξ i, an eigenvector corresponding to r i. So, Aξ i = r i ξ i. A set of n solutions of () is given by So, the Wronskian x () = ξ e r t,...,x (n) = ξ n e rnt (4) W := W(x (),...x (n) ) = x () x (n) = e (r + +r n)t ξ ξ n

6 Real and Distinct Eigenvalue Now assume r, r 2,...,r n are distinct. Then, ξ,...,ξ n are linearly independent. (It needs a proof). Hence the Wronskian W 0. Now assume r, r 2,...,r n are real. By 7.4, x (),...,x (n) form a fundamental set of solutions of (). In other words, any solution of () has the form x = c x () + +c n x (n) = c ξ e r t + +c n ξ n e rnt (5) where c,...,c n are constants, to be determined by the initial value conditions.

7 When n = 2, system of homogeneous linear system, with constant coefficients, looks like ( ) ( )( ) x a b x = c d Then,. dx 2 dx = x x 2 x 2 = ax + bx 2 cx + dx 2 x 2 is independent of t In x x 2 -plane, at each (x, x 2 ) draw a small line segment with slope dx 2 dx. This will be called a direction field. In Matlab, use "pplane8" to draw direction fields.

8 Continued This direction field provides different information than direction field we saw in chapter, which was time dependent. For a direction field analogous to that in chapter, we would require 3-D diagrams. I will deemphasize direction field in this chapter. Motivated students should read everything given in the textbook, on direction fields and/or discuss with to me.

9 Sample III Ex. Sample IV Ex. 5 Computing eigenvalues and eigenvectors Matlab command [V, D]=eig(A) could be used to find eigenvalues and eigenvectors. Advantage with this is that Matlab can handle complex numbers (see 7.6). However, after experimenting with it, I concluded that it does not work very well. It uses the floating numbers, which make things misleading. Throughout, the following would be my strategy: Compute eigenvalues analytically. If an eigenvalue is real, use use TI-84 (rref) to solve and compute eigenvectors. If an eigenvalue is complex (in 7.6), compute eigenvectors analytically.

10 Sample III Ex. Sample IV Ex. 5 Find a general solution of x = ( ) x First, find the eigenvalues: 4 r r = 0 = r2 +2r = 0 = r = 0, 2

11 Eigenvectors fo r = 0 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 0 is given by (use TI-84 "rref"): ( )( ) ( ) 4 3 ξ 0 = = 8 6 ξ 2 0 ( )( ) ( ).75 ξ 0 = 0 0 ξ 2 0 ( ).75 Taking ξ 2 =, ξ = is an eigenvector for r = 0. So, a solution corresponding to r = 0 is: ( ).75 x () = ξe rt =

12 Eigenvectors fo r = 2 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 2 is given by (use TI-84 "rref"): ( )( ) ( ) 4 r 3 ξ 0 = = 8 6 r 0 ( )( ) ( ) 6 3 ξ 0 = = 8 4 ξ 2 0 ( )( ) ( ).5 ξ 0 = 0 0 ξ 2 0 ( ).5 Taking ξ 2 =, ξ = is an eigenvector for r = 2. ξ 2

13 Continued Sample III Ex. Sample IV Ex. 5 So, a solution corresponding to r = 2 is: ( ).5 x (2) = ξe rt = e 2t So, the general solution is: x = c x () + c 2 x (2) = c (.75 ) (.5 + c 2 ) e 2t

14 Sample III Ex. Sample IV Ex. 5 Find a general solution of x = ( i i ) x First, find the eigenvalues: r i i r = 0 = r2 2r = 0 = r = 0, 2

15 Eigenvectors fo r = 0 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 0 is given by : ( )( ) ( i ξ 0 = i 0 ξ 2 ) = { { ξ + iξ 2 = 0 iξ +ξ 2 = 0 = ξ + iξ 2 = 0 0 = 0 ( ) i Taking ξ 2 =, ξ = is an eigenvector for r = 0. So, a solution corresponding to r = 0 is: ( ) i x () = ξe rt =

16 Eigenvectors for r = 2 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 2 is given by (use TI-84 "rref"): ( )( ) ( ) r i ξ 0 = = i r 0 ( )( ) ( ) i ξ 0 = = i ξ 2 0 { { ξ + iξ 2 = 0 iξ ξ 2 = 0 = ξ + iξ 2 = 0 0 = 0 ( ) i Taking ξ 2 =, ξ = is an eigenvector for r = 2. ξ 2

17 Continued Sample III Ex. Sample IV Ex. 5 So, a solution corresponding to λ = 2 is: ( ) i x (2) = ξe rt = e 2t So, the general solution is: x = c x () + c 2 x (2) = c ( i ) ( i + c 2 ) e 2t

18 Sample II Ex. Sample III Ex. Sample IV Ex. 5 Find a general solution of x = x First, find the eigenvalues: r 2 2 r 2 r = 0

19 Continued Sample III Ex. Sample IV Ex. 5 ( r) 2 r r 2 r r 2 = 0 r 3 + 4r 2 + r 4 = 0 = r 3 4r 2 r + 4 = 0 r 2 (r ) 3r(r ) 4(r ) = 0 (r )(r 2 3r 4) = 0 = (r )(r + )(r 4) = 0 r =,, 4

20 Sample III Ex. Sample IV Ex. 5 An eigenvector and solution for r = Eigenvetors for r = is given by (use TI-84 "rref"): r 2 ξ 0 2 r ξ 2 = 0 = 2 r ξ ξ 2 ξ 3 2ξ +ξ 2 + 2ξ 3 = 0 ξ + 3ξ 2 +ξ 3 = 0 2ξ +ξ 2 + 2ξ 3 = 0 ξ 3 = = = (use rref)

21 Sample III Ex. Sample IV Ex. 5 Taking ξ 3 =, ξ = r =. ξ +ξ 3 = 0 ξ 2 = 0 0 = 0 0 is an eigenvector for So, a solution corresponding to r = is: x () = ξe rt = 0 e t

22 An eigenvector and solution for r = Sample III Ex. Sample IV Ex. 5 Similarly, an eigenvector for r = is ξ = So, a solution corresponding to r = is: x (2) = ξe rt = 2 e t 2

23 An eigenvector and solution for r = 4 Sample III Ex. Sample IV Ex. 5 Similarly, an eigenvector for r = 4 is ξ = So, a solution corresponding to r = 4 is: x (2) = ξe rt = e 4t

24 General Solution Sample III Ex. Sample IV Ex. 5 So, the general solution is: = c 0 x = c x () + c 2 x (2) + c 3 x (3) e t + c 2 2 e t + c 3 e 4t

25 Sample IV Ex. 5 Sample III Ex. Sample IV Ex. 5 Solve the initial value problem: ( 5 x = 3 ) x, x(0) = ( 2 ). First, find the eigenvalues: 5 r 3 r = 0 = r2 6r + 8 = r = 2, 4

26 Eigenvectors fo r = 2 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 2 is given by ( )( ) 5 r ξ = 3 r ξ 2 ( 0 0 ) = ( )( ) ( ) 3 ξ 0 = = 3 ξ 2 0 { { 3ξ ξ 2 = 0 3ξ ξ 2 = 0 = 3ξ ξ 2 = 0 0 = 0 ( ) Taking ξ =, ξ = is an eigenvector for r = 2. 3

27 A solution corresponding to r = 2 Sample III Ex. Sample IV Ex. 5 So, a solution corresponding to r = 2 is: ( ) x () = ξe rt = e 2t 3

28 Eigenvectors fo r = 4 Sample III Ex. Sample IV Ex. 5 Eigenvetors for r = 4 is given by ( )( ) 5 r ξ = 3 r ξ 2 ( 0 0 ) = ( )( ) ( ) ξ 0 = = 3 3 ξ 2 0 { { ξ ξ 2 = 0 3ξ 3ξ 2 = 0 = ξ ξ 2 = 0 0 = 0 ( ) Taking ξ =, ξ = is an eigenvector for r = 2.

29 A solution corresponding to r = 4 Sample III Ex. Sample IV Ex. 5 So, a solution corresponding to r = 4 is: ( ) x (2) = ξe rt = e 4t

30 General Solution Sample III Ex. Sample IV Ex. 5 So, the general solution is: x = c x () + c 2 x (2) = c ( 3 ) e 2t + c 2 ( ) e 4t

31 The Particular Solution Sample III Ex. Sample IV Ex. 5 To find the particular solution, use the initial condition: ( ) ( ) ( ) 2 x(0) = c + c 3 2 = ( 3 ( 0 0 )( c ) ( ) 2 = = (use rref) c 2 ) ( ).5 = = c 3.5 =.5, c 2 = 3.5 )( c c 2

32 Sample III Ex. Sample IV Ex. 5 Continued So, the particular solution is: x = c x () + c 2 x (2) = c ( 3 ) e 2t + c 2 ( ) e 4t ( =.5 3 ) ( e 2t ) e 4t

33 Sample III Ex. Sample IV Ex Assignments and Homework Read Example, 2, 3, 4 (They are helpful). Homework: 7.5 (page 405) See the Homework Site!