Problem 3
Plant's Neuron Interaction Model [9] is given by the
equations


a y_{1}(t)  y^{3}_{1}(t) / 3 + m ( y_{1}(tt)  y_{1,0} ) 
 

r ( y_{1}(t) + a  b y_{2}(t) ) 


When m = 0, these equations have a steady state solution
(y_{1,0},y_{2,0}), i.e., a solution with y_{1}¢(t) = y_{2}¢(t) = 0. Solve the equations on [0,100] with history y_{1}(t) = a y_{1,0}, y_{2}(t) = b y_{2,0} for t £ 0 and
a = 0.8, b = 0.7, r = 0.08 .
The parameters a and b determine how close the
solution starts to the steady state solution. Let us take a = 0.4 and b = 1.8. To determine the steady state solution,
we find from the equation y_{2}¢(t) = 0 that when m = 0,
y_{2,0} = (y_{1,0} + a)/b. Using this in the equation y_{1}¢(t) = 0 for m = 0, we find that y_{1,0} satisfies the algebraic
equation
b y^{3} + (3 b + 1) y + 3 a = 0 . 

After computing all the roots of this cubic equation with
roots, y_{1,0} is the unique real root bigger than 1 r b. For the given a,b this results in y_{1,0} = 2.417960226013935. Using this value, compute y_{2,0} and solve
the problem for t = 20 and various m, say m = 1, 1, 10,10. The figures for two of these values show what you might
find.
Reference
 [9]

N. MacDonald, Biological Delay Systems: Linear Stability
Theory, Cambridge University Press, Cambridge, 1989.