mhh.m

本程序是求解神经电HH方程的较为简单的例子。可以帮助大家了解神经放电的问题

function dy = MHH(t,Y)
%The prgram is used to calculate the MHH equation. 
%Ode function can be used to transfer the program and get the solution.
%
dy = zeros(5,1);
gl = 0.1; g = [1.5 2.0 0.25]; gsr = 0.4; 
T = [2.0 2.0 10.0]; Tsr = 20.0; 
V0 =[-25.0 -25.0 -40.0]; %definiton of V0k.(k=d, r, sd)
s = [-0.25 -0.25 0.09]; %definiton of sk.(k=d, r, sd)
Vl = 60.0; V = [50.0 -90.0 50.0]; Vsr = -90.0;
A1 = 1.3; A2 = 3.0;
Thelta = 0.17; Ita = 0.012;
T = 10; T0 = 25.0;

rou = A1.^((T - T0)/10); fai = A2.^((T - T0)/10);

infa(1) = ( 1 + exp(- s(1) * (Y(1) - V0(1)) ))^(-1);
infa(2) = ( 1 + exp(- s(2) * (Y(1) - V0(2)) ))^(-1);
infa(3) = ( 1 + exp(- s(3) * (Y(1) - V0(3)) ))^(-1);

dy(1) = -gl*(Y(1)-Vl) - rou * (  g(1).*Y(2).*(Y(1)-V(1)) + g(2).*Y(3).*(Y(1)-V(2)) + g(3).*Y(4).*(Y(1)-V(3) + gsr*Y(5)*(Y(1)-Vsr))); 
dy(2) = fai*(infa(1) - Y(2)) / T(1);
dy(3) = fai*(infa(2) - Y(3)) / T(2);
dy(4) = fai*(infa(3) - Y(4)) / T(3);
dy(5) = fai*(-Ita*rou * ( g(3).*Y(3).*(Y(1)-V(3)) )-Y(5))/Tsr;