Math 5110/6830
Homework 10.1 Solutions
1. (a) Right hand side vs a
0.6
n=.3
n=.5
n=.9
0.5
0.4
rhs
0.3
0.2
0.1
0
−0.1
−0.2
0
0.2
0.4
0.6
0.8
1
a
(b) Interaction of a and a∞
1
0.8
0.6
0.4
n=.3
n=.5
n=.9
0.2
0
0
0.2
0.4
0.6
a
1
0.8
1
(c) Phase line
0.6
0.5
0.4
rhs
0.3
0.2
0.1
0
−0.1
−0.2
0
0.2
0.4
0.6
0.8
1
a
As n increase, the right hand side shift upward.
2. (a) ao = 0 and no = 1
kn = .05
1
a
n
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
50
100
150
time
200
250
300
Matlab Code
idxmax=3000; %the number of iterations - idxmax*dt is the total time
dt = .1; %delta t
taua=1;
taun=40;
kn=0.05;
ka=0.05;
theta=0.18;
ao=0; %Initial a
no=1; %Initial n
2
a = zeros(1,idxmax+1); %Initialize vectors
n = zeros(1,idxmax+1); %Initialize vectors
a(1) = ao;
n(1) = no;
for i = 1:idxmax
a(i+1) = dt/taua*(-a(i) + 1/(1+exp(-(n(i)*a(i)-theta)/ka))) + a(i);
n(i+1) = dt/taun*(-n(i) + 1/(1+exp((a(i)-.5)/kn))) + n(i);
end
t = 0:dt:idxmax*dt; %time vector
figure(1)
plot(t,a,t,n,’--’)
xlabel(’time’)
legend(’a’,’n’)
title(’kn = .05’)
3. Starting with n high, a will quickly move toward the upper stable steady state. While a is high, n
will slowly decrease. a will follow along the steady state until it reaches the bifurcation point. a
will then drop to the lower steady state curve. With a now low and n will begin to increase until
a reaches the second bifurcation point. a will increase to the upper stable steady state curve and
repeat.
4. When kn = .5 we obtain the following time series plot:
3
kn = .5
1
a
n
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
50
100
150
time
200
250
300
By looking at the nullclines of the system, we can see why:
1
a−nullcline
n−nullcline kn = .05
0.9
n−nullcline kn =.5
0.8
0.7
a
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
n
0.6
0.7
0.8
0.9
1
When kn increases we get 3 fix points. One of which is stable. Therefore, we move towards that
fix point and stay there.
4