By Ilya Prigogine, Stuart A. Rice

The *Advances in Chemical Physics* sequence offers the chemical physics and actual chemistry fields with a discussion board for serious, authoritative reviews of advances in each region of the self-discipline. full of state-of-the-art study pronounced in a cohesive demeanour no longer discovered in different places within the literature, every one quantity of the *Advances in Chemical Physics* sequence serves because the excellent complement to any complex graduate classification dedicated to the research of chemical physics.

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The form of +KR is given using Hill’s terminology; consult Hill” for clarification of the results we now state. The steady-state concentrations are given by the “diagrammatic” formula . 43) where C = yX and Z is the sum of all directional diagrams. 45) The extreme currents E, of unimolecular networks are “cycles, ” that is, cyclic sequences of reactions such as XI + X, +X, + X, +XI. C. The currents can be normalized so that all elements of E are either 0 or 1. Then the equation vo = Ej says that v, is the s u m of j i over all cycles Ei containing R,.

Therefore e, is an r/2-dimensional simplicial cone. The detailed balance condition is a restriction on e,. 47) Their dimensions are r/2 and r - d, respectively. Also the boundary of eE lies entirely in the coordinate hyperplanes ui= 0 and is thus entirely Since has r/2 - d fewer dimensions contained in the boundary of it lies in the intersection of and an (r - d) - (r/2 - d) = r/2than dimensional linear subspace of So. Similarly l l E is an r/2-dimensional slice through nu. C. Choose the simplicia1 decomposition of e,, so that e, is a facet of and let this facet be given byj,' = 0 for i = r/2 1, .

The sets of entry and exit points may intersect if a = e; then x = e is at least a triple root and the corresponding point in II, is a critical point or tristate point. 8 illustrates how to obtain equations for entry, exit and critical points. e, e,. Nicolis and Prigogine, Ref. 2, p. 8. 53) becomes Xh = -(j, +j 2 ) x 3 + ( j , + j3)x2 - ( j 3+j 4 ) x + ( j 4+jl) n, is a square. 38 B. L. CLARKE The edge of the fold has two roots x = 1 and is an exit point. Dividing the right-hand side by ( x - 1)’ yields the factorization X = - ( X - l)’[(jr + j 2 ) x + 2j, + j , - j31 - ( x - 1)[3j, + j 2 - j 3 + j41 A double root occurs if and only if the second term vanishes, so that the equation of the exit points is 3j1 + j 2 -j 3 +j 4 = 0 which is plotted in Fig.