This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Constant Coefficient Linear Systems Definition A constant coefficient system of linear differential equa tions takes the form Y prime ( t ) = A Y ( t ) + B ( t ) , where A is a constant n n matrix. As usual, the system is homogeneous if B ( t ) 0, inhomogenous otherwise. Example 1 The system y prime 1 ( t ) y prime 2 ( t ) y prime 3 ( t ) = 2 1 1 3 1 1 2 y 1 ( t ) y 2 ( t ) y 3 ( t ) is a constant coefficient, first order, three dimensional, homogeneous system. When we have a constant coefficient scalar system z prime ( t ) = a z ( t ) the general solution takes the form z ( t, c ) = c e at , where c is an arbi trary constant. Since the solutions of a vector system must be vector functions, it makes sense to try for a solution of the form Y ( t, , ) = e t , R n , where is a constant (scalar) and is a constant vector. Substituting this solution form into the system Y prime ( t ) = A Y ( t ) we have e t = A e t ( I n A ) e t = 0 . Since the exponential cannot vanish, we have ( I n A ) = 0 , i . e ., A = ; 1 thus A is just a scalar multiple of , being the scalar. Definition If is a number such that, for some nonzero n dimen sional vector, , ( I n A ) = 0 then is said to be an eigenvalue of the matrix A and is an eigen vector of A corresponding to . Proposition There exists a nonzero vector such that ( I n A ) = 0 if and only if is such that det ( I n A ) p ( ) = 0; here p ( ) is an nth degree polynomial, p ( ) = n + a 1 n 1 + + a n 1 + a n , which we call the characteristic polynomial of the matrix A . The fact that the determinant must be equal to zero for negationslash = 0 to exist as shown is a familiar result from linear algebra. If we let the columns of A be A j and let the columns of the identity matrix be E j , j = 1 , 2 , ..., n, then p ( ) = det ( I n A ) = det [ E 1 A 1 , E 2 A 2 , , E n A n ] = n det I n + + ( 1) n det A , using the linearity of the determinant with respect to the columns of the matrix in question and the form of p ( ) then follows. Example 2 For the 3 3 matrix in the system shown at the beginning of the lecture we have det ( I n A ) = det  2 1 1  3 1 1  2 = (  2)[(  3)(  2) 1] (  2) = 3 7 2 + 14  8 2 = (  1)(  2)(  4); the eigenvalues are 1 , 2 , and 4. It should be noted that eigenvectors are not quite unique; if is an eigenvector of A corresponding to the eigenvalue and if c negationslash = 0 is a constant, then c is also an eigenvector because it is a nonzero vector for which ( I n A )( c ) = 0. Very commonly, in order to fix a particular eigenvector one chooses c = 1 bardbl bardbl , in which case we obtain the eigenvector for which...
View
Full
Document
This note was uploaded on 01/23/2012 for the course MATH 4254 taught by Professor Robinson during the Spring '10 term at Virginia Tech.
 Spring '10
 ROBINSON
 Equations, Linear Systems

Click to edit the document details