204Lecture152006 - Economics 204 Lecture 15Friday Revised...

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Economics 204 Lecture 15–Friday, August 18, 2006 Revised 8/18/06, Revisions indicated by ** Getting Real Solutions from Complex Solutions If M is a real matrix and λ is a complex eigenvalue, corre- sponding eigenvector(s) must be complex. If y is a real solution of the undiagonalized equation, the solu- tion z of the diagonalized equation, which is the representation of y in terms of the basis V of eigenvectors, will be complex. However, we can determine the form of the real solutions once we know the eigenvalues. Theorem 5 Consider the differential equation y = ( y y s ) = M ( y y x ) Suppose that the matrix M can be diagonalized over C . Let the eigenvalues of M be a 1 + ib 1 , a 1 ib 1 , . . . , a m + ib m , a m ib m , a m +1 , . . . , a n m Then for each fixed i = 1 , . . . , n , every real solution is of the form ( y ( t ) y s ) i = m j =1 e a j ( t t 0 ) ( C ij cos b j ( t t 0 ) + D ij sin b j ( t t 0 )) + n m j = m +1 C ij e a j ( t t 0 ) The n 2 parameters { C ij : i = 1 , . . . , n ; j = 1 , . . . , n m }∪{ D ij : i = 1 , . . . , n ; j = 1 , . . . , 1
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have n real degrees of freedom. The parameters are uniquely determined from the n real initial conditions of an Initial Value Problem. Proof: Rewrite the expression for the solution y as ( y ( t ) y s ) i = n j =1 γ ij e λ j ( t t 0 ) Recall that the non-real eigenvalues occur in conjugate pairs, so suppose that λ j = a + ib, λ k = a ib so the expression for ( y ( t ) y s ) contains the pair of terms γ ij e λ j ( t t 0 ) + γ ik e λ k ( t t 0 ) = γ ij e a ( t t 0 ) (cos b ( t t 0 ) + i sin b ( t t 0 )) + γ ik e a ( t t 0 ) (cos b ( t t 0 ) i sin b ( t t 0 )) = e a ( t t 0 ) (( γ ij + γ ik ) cos b ( t t 0 ) + i ( γ ij γ ik ) sin b ( t t 0 )) = e a ( t t 0 ) ( C ij cos b ( t t 0 ) + D ij sin b ( t t 0 )) Since this must be real for all t , we must have C ij = γ ij + γ ik R and D ij = i ( γ ij γ ik ) R so γ ij and γ ik are complex conjugates; this can also be shown directly from the matrix formula for y in terms of z . Thus, if the eigenvalues λ 1 , . . . , λ n are a 1 + ib 1 , a 1 ib 1 , a 2 + ib 2 , a 2 ib 2 , . . . , a m + ib m , a m ib m , a m +1 , . . . , a n m every real solution will be of the form ( y ( t ) y s ) i = m j =1 e a j ( t t 0 ) ( C ij cos b j ( t t 0 ) + D ij sin b j ( t t 0 )) + n m j = m +1 C ij e a j ( t t 0 ) 2
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Since the differential equation satisfies a Lipschitz condition, the Initial Value Problem has a unique solution determined by the n real initial conditions. Thus, the general solution has exactly n real degrees of freedom in the n 2 coefficients. Remark 6 The constraints among the coefficients C ij , D ij can be complicated. **One cannot just solve for the coefficients of y 1 from the initial conditions, then derive the coefficients for y 2 , . . . , y n . For example, consider the differential equation y 1 y 2 = 2 0 0 1 y 1 y 2 The eigenvalues are 2 and 1. If we set y 1 ( t ) = C 11 e 2( t t 0 ) + C 12 e t t 0 y 2 ( t ) = C 21 e 2( t t 0 ) + C 22 e t t 0 we get y 1 ( t 0 ) = C 11 + C 12 y 2 ( t 0 ) = C 21 + C 22 which doesn’t have a unique solution. However, from the original differential equation, we have y 1 ( t ) = y 1 ( t 0 ) e 2( t t 0 ) , y 2 ( t ) = y 2 ( t 0 ) e t t 0 so C 11 = y 1 ( t 0
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