Economics Dynamics Problems 182

# Economics Dynamics Problems 182 - x ∗ if Ax = If A is...

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166 Economic Dynamics with the general solution x = c 1 u 1 + c 2 u 2 (4.25) Continuing our example, where λ =− 1 + 2 i , i.e., α =− 1 and β = 2, then v r = ± 2 1 + i ² , i.e. b 1 = ± 2 1 ² and b 2 = ± 0 1 ² Hence u 1 = e t ³± 2 1 ² cos 2 t + ± 0 1 ² sin 2 t ´ u 2 = e t ³± 0 1 ² cos 2 t ± 2 1 ² sin 2 t ´ and x = c 1 u 1 + c 2 u 2 or x ( t ) = c 1 e t 2 cos 2 t 2 c 2 e t sin 2 t = 2 e t ( c 1 cos 2 t c 2 sin 2 t ) y ( t ) = c 1 e t cos 2 t c 1 e t sin 2 t c 2 e t cos 2 t c 2 e t sin 2 t = e t [( c 1 c 2 ) cos 2 t ( c 1 + c 2 ) sin 2 t )] 4.8 Nodes, spirals and saddles Here we shall consider only a two-variable system of the general form ˙x = Ax which to have solutions of the form x = e λ t v must satisfy ( A λ I ) v = 0 and λ must be the eigenvalue and v the eigenvector associated with the matrix A . We shall denote the two eigenvalues as λ = r and λ = s and the two associated eigenvectors v r and v s , respectively. We have already discussed the general solution of the form x = c 1 e rt v r + c 2 e st v s In this section we shall extract some geometric properties from the various possible solutions. First such a system will have a critical point, denoted
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Unformatted text preview: x ∗ , if Ax = . If A is nonsingular, or det( A ) ±= 0, then the only solution is x ∗ = . The only critical point is at the origin. The solution function x = φ ( t ) satis±es the differential equations, and this shows the solution path in the phase plane. In terms of vectors, the situation is illustrated in ±gure 4.15. The ( x , y )-plane denotes the phase plane and the origin is a critical point, ±xed point or equilibrium point. At time t = 0 we have x (0) = x and y (0) = y . At time t there is a vector with coordinates ( x ( t ) , y ( t )) and the movement of the system as time increases is indicated by the arrows along the solution path....
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