m2k_dfq_cstcof

m2k_dfq_cstcof - Constant Coefficient Linear Systems...

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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 non-zero 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 non-zero vector such that ( I n- A ) = 0 if and only if is such that det ( I n- A ) p ( ) = 0; here p ( ) is an n-th 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 non-zero 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...
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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.

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m2k_dfq_cstcof - Constant Coefficient Linear Systems...

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