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Economics Dynamics Problems 190

Economics Dynamics Problems 190 - orientation the critical...

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174 Economic Dynamics Figure 4.23. the general solution as x = [ c 1 e rt v + c 2 e rt v 2 + c 2 e rt t v ] = [( c 1 v + c 2 v 2 ) + c 2 t v ] e rt = u e rt Then u = ( c 1 v + c 2 v 2 ) + c 2 t v which is a vector equation of a straight line which passes through the point c 1 v + c 2 v 2 and is parallel to v . Two such points are illustrated in figure 4.23, one at point a ( c 2 > 0) and one at point b ( c 2 < 0). We shall not go further into the mathematics of such a node here. What we can do, however, is highlight the variety of solution paths by means of two numerical examples. The first, in figure 4.24, has the orientation of the trajectories as illus- trated in figure 4.23, while figure 4.25 has the reverse orientation. Whatever the
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Unformatted text preview: orientation, the critical point is again an improper node that is stable. Had r > , then the critical point would be an improper node that is unstable. Case 4 (Complex roots, α ±= and β > ) In this case we assume the roots λ = r and λ = s are complex conjugate and with r = α + β i and s = α − β i , and α ±= 0 and β > 0. Systems having such complex roots can be expressed ˙ x = α x + β y ˙ y = − β x + α y...
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