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Completely unstable equilibrium state

Point O is stable the function (2i — SR sin 3(f) is a Lyapunov function for small R. Clearly, the stable separatrices of the saddle points tend to infinity as t —> — 00. Otherwise, they had to tend to a completely unstable periodic trajectory or a completely unstable equilibrium state but there is none. Another possibility is that a stable separatrices of one saddle might coincide with the unstable one of the other saddle thereby forming a separatrix cycle as that shown in Fig. 10.6.2, but with four saddles however this hypothesis contradicts to the negative divergence condition. The unstable separatrices cannot tend to infinity as t -> +cx). To prove this, check that when R is large, V <0 for the function V in (10.6.12). Therefore, all trajectories of the system, as t increases, must get inside some closed curve V == C with C sufficiently large, where they remain forever. The same behavior applies to the separatrices of the saddle. Thus, the only option for the unstable separatrices of the point Oi is that one tends to O and the other to 0 as shown in Fig. 10.6.6. [Pg.157]

When the equilibrium state is topologically saddle, condition (C.2.8) distinguishes between the cases of a simple saddle and a saddle-focus. However, when the equilibrium is stable or completely unstable, the presence of complex characteristic roots does not necessarily imply that it is a focus. Indeed, if the nearest to the imaginary axis (i.e. the leading) characteristic root is real, the stable (or completely imstable) equilibrium state is a node independently of what other characteristic roots are. [Pg.457]

Unstable states defined by negative curvature of A-V isotherms, between the spinodal points. In this region, dp/dV)T,ai Ni > 0. Single-phase equilibrium states of this nature are completely disallowed. Small fluctuations in system properties grow in unrestricted fashion until the system splits into two phases. [Pg.800]

It is clear that the joints were not sufficiently stiff to provide sway restraint. Although the first and third storey heights were to contain partitions the second storey height would have contained only non-load bearing partitions. As it happened none of the partitions were completed at the time of collapse. The structure was in a state of unstable equilibrium and it needed only a small disturbance to precipitate collapse. The failure of one of the joints on the second floor therefore triggered a collapse of the whole building. [Pg.133]

The stability of the equilibrium state is completely determined by the eigenvalues of the matrix A( ). By Liapounov s theorem, a sufficient condition for the local asymptotic stability of the equilibrium state is that all eigenvalues of the matrix A( ) have negative real parts. Conversely, if the matrix A( ) has one or more eigenvalues with positive real parts, the equilibrium point is unstable. Finally, when the matrix A(f ) has points with zero real parts no definitive statement about stability can be made. [Pg.27]

A topological type (m, p) is assigned to each equilibrium state, where m is the number of the characteristic exponents in the open left half-plane and p is that in the open right half-plane. Therefore, m -hp = n. If m = n (m = 0), the equilibrium state is stable (completely unstable). An equilibrimn state is of saddle type when m 0,n. ... [Pg.45]

If I > 0, then the function F is a Lyapunov function for the system obtained from (10.5.30) by inversion of time. Thus, the equilibrium state of system (10.5.29) and hence the fixed point of the map (10.5.27) is completely unstable here. [Pg.145]

Both cases have much in common in the sense that the imstable set of both bifurcating equilibrium states is one-dimensional. If the unstable set of the critical equilibrimn state is of a higher-dimension, then the subsequent picture may be completely different. Figure 14.3.1 depicts such a situation. When the imstable cycle shrinks into the equilibrium state we have a dilemma the representative point may jump either to the stable node 0 or to the stable node 02- Therefore this dangerous boundary must be classified as dynamically indefinite. [Pg.446]

Why must a living system renew itself Evidently this is the price biological systems must pay for their inherent instability, since a totally stable system is a dead system, existing in chemical and thermodynamic equilibrium. In stark contrast, a living system is not completely covalent and hence is unstable, existing in a far-from-equilibrium state which is responsive to fluctuations in the environment. [Pg.31]

Figure 5.21 Bifurcation far from equilibrium, (a) Primary bifurcation is the distance from equilibrium, at which the thermodynamic branching of minimal entropy production becomes unstable. The bifurcation point or critical point corresponds to the concentration (b) Complete diagram of bifurcations. As the non-linear reaction moves away from equilibrium, the number of possible states increases enormously. (Adapted, with permission, from Coveney and Highfield, 1990). Figure 5.21 Bifurcation far from equilibrium, (a) Primary bifurcation is the distance from equilibrium, at which the thermodynamic branching of minimal entropy production becomes unstable. The bifurcation point or critical point corresponds to the concentration (b) Complete diagram of bifurcations. As the non-linear reaction moves away from equilibrium, the number of possible states increases enormously. (Adapted, with permission, from Coveney and Highfield, 1990).
Further examples of reversible processes will be dealt with in subsequent chapters. In most cases we may imagine complete reversibihty to be attained by carrying out the process sufficiently slowly, so that the driving force would be infinitely small and the system would always be in a state of equilibrium. All strictly reversible processes must therefore go through a continuous series of states of equilibrium. From this condition alone it follows that all spontaneous processes in Nature, which lead from unstable states to states of equilibrium, must be irreversible. [Pg.134]

Kinetics. Oxidation reactions, such as those enumerated in Sec. I, are accompanied by the formation of water, carbon oxides, or both, or by the introduction of elemental oxygen in the organic molecule, or by the step-down of an oxidizing compound from an unstable state of high oxidation to a more stable state of lower oxidation. These reactions are exothermic and accompanied by a free-energy decrease. Equilibrium, therefore, is favorable, and in practically all cases no means need be provided to force the completion of the reaction. Indeed, in the majority of cases, steps must be taken to limit the extent of the reaction and prevent complete loss of product through continued oxidation. [Pg.542]


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See also in sourсe #XX -- [ Pg.83 , Pg.102 , Pg.525 ]




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Equilibrium state

Unstability

Unstable

Unstable states

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