52 lotka: discontinuous evolution 



This being so, the mass Z h eliminated per unit of time by A h 

 will, under otherwise similar conditions, also be constant, so that 



Bn = %h = constant (5) 



Again, Z H the rate of elimination of mass per unit of time from 

 A U} stands in a simple relation to the rate of formation Bi of Au 

 If A H lost mass only through A { feeding upon it, while on the 

 other hand Ai gained mass by feeding exclusively on A-^, we 

 should have simply 



Z R = Bi (6) 



In general we can not, however, suppose these conditions to be 

 satisfied, so that we must introduce a coefficient 6, and write 



Z H = dB { = ebMi (7) 



Substituting (7) in (3) we obtain 



^Ms^H-fl&.Mi (8) 



dt 



We are supposing all other conditions constant, and only M n 

 and Mi changing. Under these circumstances b h r, and will 

 be functions of M K and Mi alone 



6, = h (M H , Mi) n = n (M H , M,) d = 9(M H , Mi).... (9) 



Our immediate problem is to find a general solution of the set 

 of differential equations (2) and (8), expressing M H and Mi as 

 functions of t. For this purpose we will first of all simplify our 

 notation by dropping all subscripts, and writing 



M H = X Mi = Y (10) 



Equation (8) and (2) then assume the form 



dX . = B-dbY (11) 



dt 



clY 



^=r = rY (12) 



at 



