Exactly the same state, of course, is obtained by using the fact
That at a state of equilibrium each component’s change must be
zero, giving x' – x = 0, y'– y = 0; which leads to the same equations
As before.
If the equations are in differential form, then the statement that
X is to be unchanged with time is equivalent to saying that dx/dt
Must be zero. So in the system
dx/dt = 2x – y2
dy/dt = xy – 1/2
The state (1/2,1) is one of equilibrium, because when x and y have
These values all the derivatives become zero, i.e. the system stops
Moving.
Ex. 1: Verify that U transforms ( – 3, – 1) to ( – 3, – 1).
Ex. 2: Has the system (of the last paragraph) any state of equilibrium other than
(1/2,1)?
Ex. 3: Find all the states of equilibrium of the transformation:
x'= e–y sin x,y' = x2.
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A N I N T R O D UC T I O N T O C Y B E R NE T I C S
STA BI LIT Y
Stable region. If a is a state of equilibrium, T(a) is, as we saw
In S.5/3, simply a. Thus the operation of T on a has generated no
New state.
The same phenomenon may occur with a set of states. Thus,
Suppose T is the (unclosed) transformation
a b c d e f g hT: ↓ p g b f a a b m
It has no state of equilibrium; but the set composed of b and g has
The peculiarity that it transforms thus
b gT: ↓ g b
I. e. the operation of T on this set has generated no new state. Such
A t is stable with respect to T.
Where it was pointed out that only when the set is stable can the
Transformation proceed to all its higher powers unrestrictedly.
Another reason is discussed more fully in S.10/4, where it is
Shown that such stability is intimately related to the idea of some
Entity “surviving” some operation.
Ex. 1: What other sets are stable with respect to T?
Ex. 2: Is the set of states in a basin always stable?
Ex. 3: Is the set of states in a cycle always stable ?
Ex. 4: If a set of states is stable under T, and also under U, is it necessarily stable
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Under UT?
Fig. 5/5/1
This relation between a set of states and a transformation is, of
Course, identical with that described earlier (S.2/4) as “closure”.
(The words “stable set” could have been used from there onwards,
But they might have been confusing before the concept of stability
Was made clear; and this could not be done until other matters had
Been explained first.)
If the transformation is continuous, the set of states may lie in a
Connected region. Thus in Fig. 5/5/1, the region within the bound-
Ary A is stable; but that within B is not, for there are points within
The region, such as P, which are taken outside the region.
The concept of closure, of a stable set of states, is of fundamen-
Tal importance in our studies. Some reasons were given in S.3/2,
76
DI S TUR BAN CE
In the cases considered so far, the equilibrium or stability has
Been examined only at the particular state or states concerned.
Nothing has been said, or implied, about the behaviour at neigh-
Bouring states.
The elementary examples of equilibrium— a cube resting on its
Face, a billiard ball on a table, and a cone exactly balanced on its
Point— all show a state that is one of equilibrium. Yet the cone is
Obviously different, and in an important way, from the cube. The
Difference is shown as soon as the two systems are displaced by
Disturbance from their states of equilibrium to a neighbouring
State. How is this displacement, and its outcome, to be represented
Generally ?
A “disturbance” is simply that which displaces, that which
Moves a system from one state to another. So, if defined accu-
Rately, it will be represented by a transformation having the sys-
Tem’s states as operands. Suppose now that our dynamic system
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Has transformation T, that a is a state of equilibrium under T, and
That D is a given displacement-operator. In plain English we say:
“Displace the system from its state of equilibrium and then let the
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