Chemistry Reference
In-Depth Information
x
1
0.3
2
3
4
5
67
8
9
10
11
12
13
14
N
Fig. 2.4 Example 2.3; linear inverse demand and cost functions and identical capacity constrained
firms. Bifurcation diagram of output as the number of firms in the industry increases. Parameter
values are A
D
16, B
D
1, a
D
0:5, c
D
6 and L
D
1
at capacity. In these cases x D L is stable. Note that this holds even if the stability
condition a.N C 1/ < 4 is violated due to N>4=a 1 D 7 (which yields a slope
of the decreasing branch of T.x/less than 1).
The bifurcation diagram reveals that for an increasing number of firms cyclic
and even chaotic behavior of the production sequences can be observed. The first
qualitative change occurs when the number of firms goes from N D 9 up to N D 10
and it involves a particular kind of global bifurcation known as a border collision
bifurcation . This kind of bifurcation is specific to piecewise differentiable dynamical
systems (see Nusse and Yorke (1995) and Zhanybai and Mosekilde (2003)), hence
we describe it in more detail (see the sequence of pictures in Fig. 2.5). In some
of the figures to follow we depict portions of the phase space outside the strategy
space Œ0;L. We do this to illustrate and emphasize that in order to understand global
bifurcations sometimes it is not sufficient to focus on the local properties around the
equilibrium. Let us start with a situation where N D 8 firms are in the market. In
this case we have .A c/=.B.N C 1// > L and the stable equilibrium x D L is
located on the left upward-sloping branch of the graph of T.x/(Fig.2.5a).Now,if
an additional firm enters the market, N D 9,then x D .A c/=.B.N C 1// D L.
Additionally, since ..A c/=B.N 1// 2L.N 1/ D L, the kink of the graph
of the map T also is located exactly on the boundary (see Fig. 2.5b). As the number
of firms is even further increased to N D 10, the steady state of the map enters the
decreasing branch. The derivative T 0 .x/ of the map for this increasing sequence of
N crosses a jump discontinuity, since it assumes the value .1 a/ D 0:5 on the left
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