Using find command to find bifurcation points
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I have code which plots a graph showing the levels of Notch and Delta in a pair of cells. The graph is a bifurcation diagram with 2 bifurcation points. After plotting, I have attempted to find the 2 values of 'a' where the bifurcations happen. I have tried using the find command but only receive empty vectors when I run the code. Function file is at the end of the code.
% Specify initial conditions
dinit=[0 1];
ninit=[1 0];
initialconditions=[dinit; ninit]';
% Set values of a
a=logspace(-8,2,200);
for i=1:numel(a)
% Apply ode45
[t,y]=ode45(@(t,y)twocellfunct(t,y,a(i)),[0 200],initialconditions);
% Calculate maximum value of Notch in cells 1 and 2
mx=max([y(end,3), y(end,4)]);
% Calculate minimum value of Notch in cells 1 and 2
mn=min([y(end,3), y(end,4)]);
% Storing the max/min values of notch for each a
M(:,i)=[mx mn]';
end
% Plot
semilogx(a,M(1,:),'r');
hold on
semilogx(a,M(2,:),'b');
xlabel('a');
ylabel('notch level');
y=ylabel('notch level', 'rot', 90);
set(y, 'Units', 'Normalized', 'Position', [-0.07, 0.5, 0]);
a1=find(abs(M(1,:)-M(2,:))<eps,1,'first');
a2=find(abs(M(1,:)-M(2,:))<eps,1,'last');
function l = twocellfunct(t,y,a)
% Specifying parameters
b=100;
v=1;
k=2;
h=2;
% RHS functions
f=@(x)(x.^k./(a+x.^k));
g=@(x)(1./(1+b.*x.^h));
l = [v.*(g(y(3))-y(1)); v.*(g(y(4))-y(2)); f(y(2))-y(3); f(y(1))-y(4)];
end
3 个评论
the cyclist
2019-11-18
Could you also include the function twocellfunct, so that people can run your code?
the cyclist
2019-11-18
Without being able to run your code, I speculated that eps is too tight a tolerance. Have you tried making that larger?
Ross Mannion
2019-11-18
采纳的回答
the cyclist
2019-11-18
编辑：the cyclist
2019-11-18
Here is a loglog plot of the difference between M(1,:) and M(2,:).
I think what you are actually trying to do is find the first and last points where the difference is larger than some tolerance, not smaller. That's the bifurcation.
It's an inexact science, but the following tolerance will get close. You might be able to refine the estimate with additional rules.
tol = 5.e-3;
a1=find(abs(M(1,:)-M(2,:))>tol,1,'first');
a2=find(abs(M(1,:)-M(2,:))>tol,1,'last');
3 个评论
Ross Mannion
2019-11-18
the cyclist
2019-11-18
a1 and a2 are the indices into a, not the values. You'll get what you want from
a(a1)
a(a2)
Ross Mannion
2019-11-18
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