204Lecture152006

# 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 diferential equation y 0 =( y y s ) 0 = M ( y y x ) Suppose that the matrix M can be diagonalized over C .L e t the eigenvalues oF M be a 1 + ib 1 ,a 1 ib 1 ,...,a m + ib m m ib m m +1 n m Then For each ±xed i =1 ,...,n , every real solution is oF the Form ( y ( t ) y s ) i = m X 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 X j = m +1 C ij e a j ( t t 0 ) The n 2 parameters { C ij : i ; j m }∪{ D ij : i ; j ,..., 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 X 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 ,wemusthave 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 2 + ib 2 2 ib 2 ,..., a m + ib m m ib m m +1 ,...,a n m every real solution will be of the form ( y ( t ) y s ) i = m X 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 X j = m +1 C ij e a j ( t t 0 ) 2
Since the diferential equation satisFes 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 o± ±reedom in the n 2 coeﬃcients. Remark 6 The constraints among the coeﬃcients C ij ,D ij can be complicated. **One cannot just solve ±or the coeﬃcients o± y 1 ±rom the initial conditions, then derive the coeﬃcients ±or y 2 ,...,y n . ²or example, consider the diferential equation y 1 y 2 0 = 20 01 y 1 y 2 The eigenvalues are 2 and 1. I± 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, ±rom the original diferential 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 ) C 12 =0 C 21 C 22 = y 2

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## This note was uploaded on 03/06/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at Berkeley.

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

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