x 1 x 2 t 6 0 for all t in I The two linearly independent solutions of x t Ax t

# X 1 x 2 t 6 0 for all t in i the two linearly

• 54

This preview shows page 9 - 14 out of 54 pages.

x 1 , x 2 ]( t ) 6 = 0 , for all t in I. The two linearly independent solutions of ˙ x ( t ) = Ax ( t ) are often called a fundamental set of solutions Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (9/54)

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 Superposition and Linear Independence Fundamental Solution Eigenvalue Problem Fundamental Solutions Theorem (Fundamental Solutions) Suppose that x 1 ( t ) and x 2 ( t ) are two solutions of ˙ x ( t ) = Ax ( t ) (2) and that their Wronskian is not zero on an interval I . Then x 1 and x 2 form a fundamental set of solutions for (2), and the general solution is given by x ( t ) = c 1 x 1 ( t ) + c 2 x 2 ( t ) , where c 1 and c 2 are arbitrary constants. If there is a given initial condition x ( t 0 ) = x 0 , where x 0 is any constant vector, then this condition determines the constants c 1 and c 2 uniquely. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (10/54)
Introduction Solutions of Two 1 st Order Linear DEs Homogeneous Linear System of Autonomous DEs Case Studies and Bifurcation Superposition and Linear Independence Fundamental Solution Eigenvalue Problem Solving ˙ x = Ax Consider the general problem ˙ x ( t ) = Ax ( t ) , where x = x 1 x 2 , A = a 11 a 12 a 21 a 22 . We attempt a solution of the form x = e λt v , so λe λt v = A e λt v Since e λt is never zero, Av = λ v or ( A - λ I ) v = 0 , where I is the 2 × 2 identity matrix This is the classic eigenvalue problem Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (11/54)

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 Superposition and Linear Independence Fundamental Solution Eigenvalue Problem Eigenvalue Problem Thus, solving the homogeneous DE ˙ x ( t ) = Ax ( t ) is equivalent to solving the eigenvalue problem ( A - λ I ) v = 0 with A = a 11 a 12 a 21 a 22 . From Linear Algebra (Math 254) the eigenvalues are found by solving det | A - λ I | = a 11 - λ a 12 a 21 a 22 - λ = 0 . This gives the characteristic equation λ 2 - ( a 11 + a 22 ) λ + a 11 a 22 - a 12 a 21 = 0 This is a quadratic equation, so easily solved for λ 1 and λ 2 Each λ i is inserted into ( A - λ I ) v = 0 , and the corresponding eigenvectors , v i are found Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (12/54)
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 Real and Different Eigenvalues Consider ˙ x = Ax and assume that the eigenvalue problem ( A - λ I ) v = 0 has real and different eigenvalues , λ 1 and λ 2 The two solutions are x 1 ( t ) = e λ 1 t v 1 and x 2 ( t ) = e λ 2 t v 2 , so the Wronskian is W [ x 1 ( t ) , x 2 ( t )]( t ) = v 11 e λ 1 t v 12 e λ 2 t v 21 e λ 1 t v 22 e λ 2 t = v 11 v 12 v 21 v 22 e ( λ 1 + λ 2 ) t Since e ( λ 1 + λ 2 t ) t is nonzero, the Wronskian is nonzero if and only if det | v 1 , v 2 | = 0.

Subscribe to view the full document.

• Fall '08
• staff

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern