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

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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)
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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)
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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)
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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)
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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.
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