THE DYNAMICS OF SIMPLE CIRCUITS 



23 



If the outside stimulus is constant, the only case we shall consider, 

 this differential equation can be solved by a quadrature, 



t/ £1 



de 



</>(£ + £> -e 



to obtain t as a function of e : 



t = T(e) 



= at , 



(3) 



(4) 



We must then solve this equation for £ as a function of t . However, 

 certain properties of this solution are obtainable directly from a con- 

 sideration of the form of equation (2). 



We recall that for £ + £ positive, </> and its first derivative are 

 positive, with the derivative decreasing monotonically to zero. For 

 £ + £ negative $ is identically zero. Suppose first that <£'(0) = 1 . 

 Then, since £ is always non-negative, the equation 



£ = <£(! + £) (5) 



has always a single root s which may be zero (Figure 1). For £ > £ 



the right member of equation (2) is negative and £ is decreasing; for 

 £ < £ the right member of equation (2) is positive and £ is increas- 

 ing. Hence £ = £ represents a stable equilibrium. Whenever £ ^ , 

 £ == . Whenever I > , then £ > , and enhancement of a in the 

 amount £„ results, but after withdrawal of the stimulus, when £ = 

 — h , the neuron comes to rest. 



If (j>' (0) > 1 , equation (5) has a single root £ == e > when I > , 

 the single root s = when I < and numerically large, two positive 

 roots besides the root e = when £ < and numerically small (Fig- 

 ure 2), and one positive root besides the root £ = when £ = . If 



