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

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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)

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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 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)
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)

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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 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)
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 .

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