Nonhomogeneous linear systems of DEs Diagonalization, Variation of - PowerPoint PPT Presentation

Diagonalization Variation of parameters Nonhomogeneous linear systems of DE’s Diagonalization, Variation of Parameters ITI 11/04/2020 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Diagonalization 1 Variation of parameters 2 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider a nonhomogeneous system system of first-order DE’s x ′ = P ( t ) x + g ( t ) The general solution can be expressed as x ( t ) = c 1 x (1) ( t ) + · · · + c n x ( n ) ( t ) + v ( t ) Here c 1 x (1) ( t ) + · · · + c n x ( n ) ( t ) is the general solution of the homogeneous system x ′ = P ( t ) x and v ( t ) is a particular solution of the nonhomogeneous system. ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider a nonhomogeneous system system of first-order DE’s x ′ = P ( t ) x + g ( t ) The general solution can be expressed as x ( t ) = c 1 x (1) ( t ) + · · · + c n x ( n ) ( t ) + v ( t ) Here c 1 x (1) ( t ) + · · · + c n x ( n ) ( t ) is the general solution of the homogeneous system x ′ = P ( t ) x and v ( t ) is a particular solution of the nonhomogeneous system. There are several methods for determining v ( t ). ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider a system with constant coefficients x ′ = A x + g ( t ) and let A be a diagonalizable matrix. Consider the matrix of eigenvectors T = ( ξ (1) , . . . , ξ ( n ) ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider a system with constant coefficients x ′ = A x + g ( t ) and let A be a diagonalizable matrix. Consider the matrix of eigenvectors T = ( ξ (1) , . . . , ξ ( n ) ) Define new variables y = T − 1 x , x = T y ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider a system with constant coefficients x ′ = A x + g ( t ) and let A be a diagonalizable matrix. Consider the matrix of eigenvectors T = ( ξ (1) , . . . , ξ ( n ) ) Define new variables y = T − 1 x , x = T y In the new variables the system becomes T y ′ = AT y + g ( t ) ⇒ y ′ = T − 1 AT y + T − 1 g ( t ) y ′ = D y + h ( t ) , h ( t ) = T − 1 g ( t ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Since D is a diagonal matrix, the new system of DE’s is a system of uncoupled equations y ′ i ( t ) = λ i y i ( t ) + h i ( t ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Since D is a diagonal matrix, the new system of DE’s is a system of uncoupled equations y ′ i ( t ) = λ i y i ( t ) + h i ( t ) Using the method of integrating factors we find that the solutions are � y i ( t ) = e λ i t e − λ i s h i ( s ) ds + c i e λ i t ITI Nonhomogeneous systems

Diagonalization Variation of parameters Since D is a diagonal matrix, the new system of DE’s is a system of uncoupled equations y ′ i ( t ) = λ i y i ( t ) + h i ( t ) Using the method of integrating factors we find that the solutions are � y i ( t ) = e λ i t e − λ i s h i ( s ) ds + c i e λ i t The solution in terms of the original variables is obtained by changing the variables back x ( t ) = T y ( t ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � 1 � � λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � 1 � � λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 The change of variables matrix is � 1 � � , T − 1 = 1 � 1 1 − 1 , y = T − 1 x T = − 1 1 1 1 2 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � 1 � � λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 The change of variables matrix is � 1 � � , T − 1 = 1 � 1 1 − 1 , y = T − 1 x T = − 1 1 1 1 2 In the new variables the system reads � 2 e − t − 3 t � − 3 � y + 1 � 0 y ′ = D y + T − 1 g ( t ) = 2 e − t + 3 t 0 − 1 2 ITI Nonhomogeneous systems

Diagonalization Variation of parameters In components the equations are 1 + 3 y 1 = e − t − 3 2 + y 2 = e − t + 3 y ′ y ′ 2 t , 2 t and the solutions are y 1 = e − t − t 2 + 1 y 2 = te − t + 3 t 2 − 3 6 + c 1 e − 3 t , 2 + c 2 e − t 2 ITI Nonhomogeneous systems

Diagonalization Variation of parameters In components the equations are 1 + 3 y 1 = e − t − 3 2 + y 2 = e − t + 3 y ′ y ′ 2 t , 2 t and the solutions are y 1 = e − t − t 2 + 1 y 2 = te − t + 3 t 2 − 3 6 + c 1 e − 3 t , 2 + c 2 e − t 2 Changing the variables back we have � � y 1 � � 1 1 x = T y = − 1 1 y 2 � 1 � � � + 1 � � 1 1 x ( t ) = c 1 e − 3 t + c 2 e − t 2 e − t − 1 1 − 1 � 1 � 1 � 4 � � − 1 � + te − t + t 1 2 5 3 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider again the nonhomogeneous system x ′ = P ( t ) x + g ( t ). Assume that a fundamental matrix Ψ( t ) for the homogeneous system x ′ = P ( t ) x has been found. Then the general solution for the homogeneous system is x ( t ) = Ψ( t ) c . ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider again the nonhomogeneous system x ′ = P ( t ) x + g ( t ). Assume that a fundamental matrix Ψ( t ) for the homogeneous system x ′ = P ( t ) x has been found. Then the general solution for the homogeneous system is x ( t ) = Ψ( t ) c . An ansatz for the nonhomogeneous system x ( t ) = Ψ( t ) u ( t ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Consider again the nonhomogeneous system x ′ = P ( t ) x + g ( t ). Assume that a fundamental matrix Ψ( t ) for the homogeneous system x ′ = P ( t ) x has been found. Then the general solution for the homogeneous system is x ( t ) = Ψ( t ) c . An ansatz for the nonhomogeneous system x ( t ) = Ψ( t ) u ( t ) Plugging into the equation we have ✭ Ψ ′ ( t ) u ( t ) + Ψ( t ) u ′ ( t ) = ✭✭✭✭✭✭ ✘ P ( t )Ψ( t ) u ( t ) + g ( t ) ✘✘✘✘ � u ′ ( t ) = Ψ − 1 ( t ) g ( t ) ⇒ u ( t ) = Ψ − 1 ( t ) g ( t ) dt + c � Ψ − 1 ( t ) g ( t ) dt x ( t ) = Ψ( t ) c + Ψ( t ) ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � � � 1 λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � � � 1 λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 A fundamental matrix and its inverse are � e 3 t e − 3 t − e 3 t � e − t � � , Ψ − 1 ( t ) = 1 Ψ( t ) = − e − 3 t e − t e t e t 2 ITI Nonhomogeneous systems

Diagonalization Variation of parameters Let’s consider an example � 2 e − t � − 2 � � 1 x ′ = x + = A x + g ( t ) 1 − 2 3 t The eigenvalue - eigenvector pairs are � 1 � � � 1 λ 1 = − 3 , ξ 1 = λ 2 = − 1 , ξ 2 = − 1 1 A fundamental matrix and its inverse are � e 3 t e − 3 t − e 3 t � e − t � � , Ψ − 1 ( t ) = 1 Ψ( t ) = − e − 3 t e − t e t e t 2 The DE for the coefficient u is � e 2 t − 3 / 2 te 3 t � u ′ ( t ) = Ψ − 1 ( t ) g ( t ) = 1 + 3 / 2 te t ITI Nonhomogeneous systems

Diagonalization Variation of parameters The DE’s for the two components of u and their solutions are 1 = e 2 t − 3 2 te 3 t ⇒ u 1 ( t ) = 1 2 e 2 t − 1 2 te 3 t + 1 6 e 3 t + c 1 u ′ 2 = 1 + 3 2 te t ⇒ u 2 ( t ) = t + 3 2 te t − 3 2 e t + c 2 u ′ ITI Nonhomogeneous systems

Diagonalization Variation of parameters The DE’s for the two components of u and their solutions are 1 = e 2 t − 3 2 te 3 t ⇒ u 1 ( t ) = 1 2 e 2 t − 1 2 te 3 t + 1 6 e 3 t + c 1 u ′ 2 = 1 + 3 2 te t ⇒ u 2 ( t ) = t + 3 2 te t − 3 2 e t + c 2 u ′ Putting everything together we obtain x ( t ) = Ψ( t ) u ( t ) � 1 � � � + 1 � � 1 1 x ( t ) = c 1 e − 3 t + c 2 e − t 2 e − t − 1 1 − 1 � 1 � 1 � 4 � � − 1 � + te − t + t 1 2 5 3 ITI Nonhomogeneous systems