The desire is to show that u t and w t are real valued solutions forming a

The desire is to show that u t and w t are real

This preview shows page 31 - 36 out of 54 pages.

The desire is to show that u ( t ) and w ( t ) are real-valued solutions forming a fundamental set for ˙ x = Ax Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (31/54)
Image of page 31

Subscribe to view the full document.

Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Complex Eigenvalues 4 Since x 1 ( t ) = u ( t ) + i w ( t ) is a solution to the DE ˙ x 1 = Ax 1 , we have 0 = ˙ x 1 - Ax 1 = ( ˙ u + i ˙ w ) - A ( u + i w ) = ( ˙ u - Au ) + i ( ˙ w - Aw ) This vector is zero if and only if the real and imaginary parts are zero, so ˙ u - Au = 0 and ˙ w - Aw = 0 or u ( t ) and w ( t ) are real-valued solutions of ˙ x = Ax It remains to show u ( t ) and w ( t ) form a fundamental set of solutions , which is done with the Wronskian Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (32/54)
Image of page 32
Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Complex Eigenvalues 5 The two solutions are u ( t ) = e μt ( a cos( νt ) - b sin( νt )) and w ( t ) = e μt ( a sin( νt )+ b cos( νt )) , so let a = a 1 a 2 and b = b 1 b 2 , then the Wronskian satisfies W [ u , w ]( t ) = e μt ( a 1 cos( νt ) - b 1 sin( νt )) e μt ( a 1 sin( νt ) + b 1 cos( νt )) e μt ( a 2 cos( νt ) - b 2 sin( νt )) e μt ( a 2 sin( νt ) + b 2 cos( νt )) = ( a 1 b 2 - a 2 b 1 ) e 2 μt Assume ν 6 = 0 and the eigenvectors are v 1 = a + i b and v 2 = a - i b , a 1 + ib 1 a 1 - ib 1 a 2 + ib 2 a 2 - ib 2 = - 2 i ( a 1 b 2 - a 2 b 1 ) 6 = 0 by our Theorem from Linear Algebra Thus, the Wronskian shows u ( t ) and w ( t ) form a fundamental set of solutions to our problem Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (33/54)
Image of page 33

Subscribe to view the full document.

Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Complex Eigenvalues 6 Example 5: Consider the example: ˙ x 1 ˙ x 2 = 3 - 2 4 - 1 x 1 x 2 Find the general solution to this problem and create a phase portrait. From above we need to find the eigenvalues and eigenvectors, so solve det 3 - λ - 2 4 - 1 - λ = λ 2 - 2 λ + 5 = 0 , which is the characteristic equation with solutions λ = 1 ± 2 i (complex eigenvalues) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (34/54)
Image of page 34
Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Real and Different Eigenvalues Complex Eigenvalues Repeated Eigenvalues Bifurcation Example and Stability Diagram Complex Eigenvalues 7 Example 5 (cont): For λ 1 = 1 + 2 i we have: 3 - λ 1 - 2 4 - 1 - λ 1 ξ 1 ξ 2 = 2 - 2 i - 2 4 - 2 - 2 i ξ 1 ξ 2 = 0 0 This results in the eigenvector ξ (1) = 1 1 - i .
Image of page 35

Subscribe to view the full document.

Image of page 36
  • Fall '08
  • staff

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes