Economics Dynamics Problems 188

Economics Dynamics Problems 188 - (a) x = c 1 e rt v 1 + c...

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172 Economic Dynamics Figure 4.20. i.e. 4 v s 1 2 v s 2 = 0 2 v s 1 v s 2 = 0 Let v s 1 = 1, then v s 2 = 2. Hence, a second solution is u 2 = e st ± 1 2 ² and v s = ± 1 2 ² The situation is illustrated in fgure 4.20. The solution paths are revealed by the direction feld. The unstable arm oF the saddle is the line through the eigenvector v r , while the stable arm oF the saddle is the line through the eigenvector v s . Case 3 (Real equal roots) In this case λ = r = s . Throughout assume the repeated root is negative. (IF it is positive then the argument is identical but the movement oF the system is reversed.) Therearetwosub-casestoconsiderinlinewithourearlieranalysis:(a)independent eigenvectors, and (b) one independent eigenvector. The two situations were Found to be:
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Unformatted text preview: (a) x = c 1 e rt v 1 + c 2 e rt v 2 (b) x = c 1 e rt v + c 2 [ e rt t v + e rt v 2 ] Consider each case in turn. In example 4.9 we Found For two independent eigen-vectors x ( t ) = c 1 e rt y ( t ) = c 2 e rt Hence, x / y = c 1 / c 2 is independent oF t and depends only on the components oF v r and v s and the arbitrary constants c 1 and c 2 . This is a general result and so all solutions lie on straight lines through the origin, as shown in fgure 4.21. In this case the origin is a proper node that is stable. Had the repeated root been positive, then we would have an unstable proper node. It is this situation we gave an example...
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