Problem 3

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

y1¢(t) = a  y1(t) - y31(t) / 3 + m  ( y1(t-t) - y1,0 ) y2¢(t) = r  ( y1(t) + a - b  y2(t) )

When m = 0, these equations have a steady state solution (y_1,0,y_2,0), i.e., a solution with y₁¢(t) = y₂¢(t) = 0. Solve the equations on [0,100] with history y₁(t) = a y_1,0,  y₂(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 y₂¢(t) = 0 that when m = 0, y_2,0 = (y_1,0 + a)/b. Using this in the equation y₁¢(t) = 0 for m = 0, we find that y_1,0 satisfies the algebraic equation

-b y3 + (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.