L94 - Fall 2003 Math 308/501502 9 Matrix Methods for Linear...

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Fall 2003 Math 308/501–502 9 Matrix Methods for Linear Systems 9.4 Linear Systems in Normal Form Wed, 12/Nov c ± 2003, Art Belmonte Summary Definition of a linear system; matrix notation A linear system is one that may be written in the normal form x 0 ( t ) = Ax ( t ) + f ( t ) or more briefly, x 0 = Ax + f .Here x is an n × 1 column vector function, A is an n × n matrix function, and f is an n × 1 column vector function (the forcing term ). If f = 0 ,the n × 1 zero vector function, then the system is homogeneous ; otherwise it is said to be nonhomogeneous . The j -th row in this vector differential equation is x 0 j = f j ( t ) + n ± k = 1 a jk ( t ) x k ( t ). The sum on the right-hand side consists of j -th element of the forcing term and the dot product of the j -th row of A with the column vector x . Note that the x k appear solely to the first power (hence the phrase “linear”). Moreover, while the a and f k functions may depend on the independent variable t (or be constants), they do not depend on the dependent variables (the x k ). YALO (“Yet Another Linear Operator”) Rewrite x 0 = Ax + f as x 0 - Ax = f and let L [ x ] = x 0 - Ax . Then the system becomes L [ x ] = f . (NOTE: In this form, it’s actually easier to see why the linear system is called homogeneous if f = 0 and nonhomogeneous if f 6= 0 .) That L is a linear operator follows immediately from the linearity of differentiation and matrix muliplication. Initial value problem for a linear system of ODEs This consists of the differential equation x 0 = Ax + f together with the initial condition x ( t 0 ) = x 0 . Existence and Uniqueness Theorem With A ( t ) and f ( t ) defined as above and continuous on an interval I ,let t 0 I and x 0 R n . Then the initial value problem x 0 ( t ) = A ( t ) x ( t ) + f ( t ), x ( t 0 ) = x 0 has a unique solution defined for all t I . Converting higher-order linear ODEs to systems Given the n th order linear equation in standard form y ( n ) ( t ) + n - 1 ± j = 0 p j ( t ) y ( j ) ( t ) = g ( t ), let x 1 = y , x 2 = y 0 , x 3 = y 00 , ... , x n = y ( n - 1 ) . We thus have x 0 k = y ( k ) = x k + 1 , k = 1 , 2 ,... n - 1. Moreover, x 0 n = y ( n ) = g ( t ) - n - 1 ± j = 0 p j ( t ) y ( j ) ( t ). Therefore, we have the linear system x 0 1 = x 2 x 0 2 = x 3 ··· x 0 n - 1 = x n x 0 n = g ( t ) - n ± j = 1 p j ( t ) y ( j ) ( t ).
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This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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L94 - Fall 2003 Math 308/501502 9 Matrix Methods for Linear...

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