5/3 AN INTRODUCTION TO CYBERNETICS 



not cause the state to change. Algebraically it occurs when T(x) = x. 

 Thus if T is 



. a b c d e f g h 

 dbhaefbe 



then since T(b) = b, the state b is a state of equilibrium under T. 

 So also are e and/. 



If the states are defined by vectors, then, for a vector to be un- 

 changed, each component must be unchanged (by S.3/5). Thus if 

 the state is a vector (x,y), and the transformation is 



^ fx' = 2x-y + 2 

 ■ [y' =x + j; + 3 



then, at a state of equilibrium (x',y') must equal (.v,;), and values 

 for X and ;' must satisfy the equations 



r X = 2.x- - y + 2 

 \y = X + y + 3 

 i.e. f X — y = — 2 



[x = - 3 



So this system has only one state of equilibrium, at (— 3, — 1). 

 Had the equations not been linear there might have been more. 



Exactly the same state, of course, is obtained by using the fact 

 that at a state of equihbrium 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/cit 

 must be zero. So in the system 



dx/dt = 2x - >'2 

 dyfdt = xy — ^ 



the state (|,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 



(i,l)? 

 Ex. 3: Find all the states of equilibrium of the transformation: 



x' = e^^ sin .y, y' = x-. 



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