M294F09WS_LastSol - Math 2940 Worksheet: 7.4, 7.5, 8.1,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 2940 Worksheet: 7.4, 7.5, 8.1, 8.3, 9.1, 9.3 (Solutions) December 3, 2009 1. Let A be an n n matrix and k a scalar. Consider the following two systems: d~x dt = A~x d~ c dt = kA~ c Show that if ~x ( t ) is a solution to the first system, then ~ c ( t ) = ~x ( kt ) is a solution of the second system. Solution : Starting with ~ c ( t ) = ~x ( kt ), we have d~ c dt = k~x ( kt ) = k ( A~x ( kt )) = kA~ c , where the second equality is from the assumption that ~x satisfies the first equation. 2. Consider the interaction of two species of animals in a habitat. We are told that the change of the populations x ( t ) and y ( t ) can be modeled by the equations dx dt = 1 . 4 x- 1 . 2 y dy dt = 0 . 8 x- 1 . 4 y where time t is measured in years. (a) Sketch a phase portrait for this system. From the nature of the problem, we are interested only in the first quadrant. Solution : For the phase portrait, we must find the eigenvalues and eigenvectors of the matrix associated with this system, A = 1 . 4- 1 . 2 . 8- 1 . 4 . Eigenvalues f A ( ) = det( A- I ) = det 1 . 4- - 1 . 2 . 8- 1 . 4- = 2- 1 . 96 + 0 . 96 = 2- 1 = ( - 1)( + 1) = 0 when = 1 ,- 1 Eigenspaces E 1 = Ker( A- I ) = Ker . 4- 1 . 2 . 8- 2 . 4 = Span 2 1 E- 1 = Ker( A + I ) = Ker 2 . 4- 1 . 2 . 8- . 4 = Span 1 3 So our system has the solution ~x ( t ) = c 1 e t 2 1 + c 2 e- t 1 3 . Using this information, we obtain the phase portrait (b) What will happen in the long term? Does the outcome depend on the initial populations? If so, how? Solution : Yes, the outcome does depend on the initial condition. This is obvious from the phase portrait. If ~x is above E- 1 , both populations die out. E- 1 is the line 3 x 1 = x 2 , so this region is 0 < 3 x 1 < x 2 where both populations die out. If ~x is between E- 1 and E 1 , both populations survive. E 1 is the line x 1 = 2 x 2 . So this region is 3 x 1 > x 2 > 1 2 x 1 . If If ~x is below E 1 , again both populations survive. This is the region 0 < x 2 < 1 2 x 1 . 3. Consider the dynamical system d~x dt =- 1- 2 2- 1 ~x (a) Find all real solutions of this system. Solution : The general solution of the system is given by ~x ( t ) = c 1 e 1 t ~v 1 + c 2 e 2 t ~v 2 , where i s are the eigenvalues and ~v i s the eigenvectors of the matrix above, and c i s are scalars. So first we find the eigenvalues: f A ( ) = det( A- I ) = det- 1- - 2 2- 1- = 2 + 2 + 5 = 0 Using the quadratic formula to solve for , =- 2 p 2 2- 4(1)(5) 2 =- 1 2 i Since the eigenvalues are complex, the above formula is not what we want it gives complex solutions to the system. All real solutions are in general given by the formula ~x (...
View Full Document

Page1 / 8

M294F09WS_LastSol - Math 2940 Worksheet: 7.4, 7.5, 8.1,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online