# Stability analysis of a non-linear ODE system

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Ron_S on 7 Apr 2023
Edited: Torsten on 23 Apr 2023
I solved the following ODE system using the code:
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta theta alpha CL
F=zeros(13,1);
e1=F(1) == (mu - eta*Sci*C);
e2=F(1) == (theta*Ce - eta*Sci*C);
e3=F(3) == (k1*Sci - k2*Sr);
e4=F(4) == (k1*Sci - k10*Sh);
e5=F(5) == (p1*Sr - k11*R - k14*P*R);
e6=F(6) == (k3*CL*R - k4*Cf);
e7=F(7) == (k4*Cf - k5*Cp + k6*Ce);
e8=F(8) == (k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E);
e9=F(9) == (k7*Ce - k8*E);
e10=F(10) == (p2*Sh - k15*HR*Ce);
e11=F(11) == (alpha - k9*HR*H);
e12=F(12) == (k1*Sci - k13*Sp);
e13=F(13) == (p3*Sp - k16*P);
[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]=solve(e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13,Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P)
Sci =
C =
Sr =
Sh =
R =
Cf =
Cp =
Ce =
E =
HR =
H =
Sp =
P =
with stability points for C_e = mu/theta, C_p = (k6*mu - alpha*theta + k12*mu)/(k5*theta), C_f = -(alpha*theta - k12*mu)/(k4*theta) and E = (k7*mu)/(k8*theta). All other variables are dependent on an input variable labelled CL. I want to do a stability analysis around these points.
I've determined the Jacobian using the code:
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta theta alpha CL
f = [mu - eta*Sci*C,theta*Ce - eta*Sci*C,k1*Sci - k2*Sr,k1*Sci - k10*Sh,p1*Sr - k11*R - k14*P*R,k3*CL*R - k4*Cf,k4*Cf - k5*Cp + k6*Ce,k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E,k7*Ce - k8*E,p2*Sh - k15*HR*Ce,alpha - k9*HR*H,k1*Sci - k13*Sp,p3*Sp - k16*P];
J = jacobian(f, [Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
J = simplify(J)
J =
I now have to determine the eigenvalues by first substituting equilibrium point values for the variables in the Jacobian with the steady state values ( in terms of CL) generated in the code above. I'm not sure how to do this in Matlab?

Torsten on 7 Apr 2023
Edited: Torsten on 7 Apr 2023
You won't succeed in this generality. To determine the eigenvalues, MATLAB had to solve for the roots of a polynomial of degree 13 with symbolic coefficients. This is in general only possible for polynomials up to degree 4. So you have to give values to the parameters of your function, I guess.
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta theta alpha CL
F=zeros(13,1);
e1=F(1) == (mu - eta*Sci*C);
e2=F(1) == (theta*Ce - eta*Sci*C);
e3=F(3) == (k1*Sci - k2*Sr);
e4=F(4) == (k1*Sci - k10*Sh);
e5=F(5) == (p1*Sr - k11*R - k14*P*R);
e6=F(6) == (k3*CL*R - k4*Cf);
e7=F(7) == (k4*Cf - k5*Cp + k6*Ce);
e8=F(8) == (k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E);
e9=F(9) == (k7*Ce - k8*E);
e10=F(10) == (p2*Sh - k15*HR*Ce);
e11=F(11) == (alpha - k9*HR*H);
e12=F(12) == (k1*Sci - k13*Sp);
e13=F(13) == (p3*Sp - k16*P);
[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq]=solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13],[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
f = [rhs(e1),rhs(e2),rhs(e3),rhs(e4),rhs(e5),rhs(e6),rhs(e7),rhs(e8),rhs(e9),rhs(e10),rhs(e11),rhs(e12),rhs(e13)];
%f = [mu - eta*Sci*C,theta*Ce - eta*Sci*C,k1*Sci - k2*Sr,k1*Sci - k10*Sh,p1*Sr - k11*R - k14*P*R,k3*CL*R - k4*Cf,k4*Cf - k5*Cp + k6*Ce,k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E,k7*Ce - k8*E,p2*Sh - k15*HR*Ce,alpha - k9*HR*H,k1*Sci - k13*Sp,p3*Sp - k16*P];
J = jacobian(f, [Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
J = simplify(J)
Jeq = subs(J,[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P],[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq])
eig(Jeq)
Ron_S on 7 Apr 2023
Thank you!
Ron_S on 12 Apr 2023
Further to my initial question, I have now given specific values for my parameters and managed to find the eigen values by finding the roots of the characteristic polynomial generated.
I would now like to run the code for a collection of randomly chosen parameter values (say a hundred or so different parameter sets) within a for loop, and check that the eigenvalues have negative real part in each case. I just don't know how to go about doing that or if it is possible?
Ron
%Using the parameter values, with CL = 0.5
k1 = 3;
k2 = 2;
k3 = 5;
k4 = 4;
k5 = 5;
k6 = 1;
k7 = 4;
k8 = 3;
k9 = 1;
k10 = 1.2;
k11 = 1;
k12 = 9;
k13 = 1;
k14 = 1;
k15 = 1;
k16 = 1;
p1 = 30;
p2 = 4;
p3 = 1.5;
mu = 25;
eta = 10;
alpha = 10;
theta = 10;
CL = 0.5;
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P
F=zeros(13,1);
e1=F(1) == (mu - eta*Sci*C);
e2=F(1) == (theta*Ce - eta*Sci*C);
e3=F(3) == (k1*Sci - k2*Sr);
e4=F(4) == (k1*Sci - k10*Sh);
e5=F(5) == (p1*Sr - k11*R - k14*P*R);
e6=F(6) == (k3*CL*R - k4*Cf);
e7=F(7) == (k4*Cf - k5*Cp + k6*Ce);
e8=F(8) == (k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E);
e9=F(9) == (k7*Ce - k8*E);
e10=F(10) == (p2*Sh - k15*HR*Ce);
e11=F(11) == (alpha - k9*HR*H);
e12=F(12) == (k1*Sci - k13*Sp);
e13=F(13) == (p3*Sp - k16*P);
%Now solve to find the steady state values in terms of (A, B, C, D, Cl and the parameters as listed), then determine the Jacobian J:
[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq]=solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13],[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
f = [rhs(e1),rhs(e2),rhs(e3),rhs(e4),rhs(e5),rhs(e6),rhs(e7),rhs(e8),rhs(e9),rhs(e10),rhs(e11),rhs(e12),rhs(e13)];
J = jacobian(f, [Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
J = simplify(J)
J =
%Substitute the steady state values obtained into the Jacobian:
Jeq = subs(J,[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P],[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq])
Jeq =
%Find the eigenvalue matrix:
eig(Jeq)
ans =
%Matlab does not solve the eigenvalues of symbolic matrices greater that 4 x 4. Instead, it gives it as the characteristic polynomial, in this case, a 13 degree polynomial, (z^13...).
%We can find the roots of this polynomial as follow:
%List the coefficients from the characteristic polynomial sigma_1 as a vector and then find the roots:
p = [1 (6809/45) 7658861/1620 17552999/270 270831481/540 3922302799/1620 1253029309/162 27428625103/1620 21201142273/810 2376322271/81 1924251074/81 347643560/27 11572400/3 400000]
p = 1×14
1.0e+07 * 0.0000 0.0000 0.0005 0.0065 0.0502 0.2421 0.7735 1.6931 2.6174 2.9337 2.3756 1.2876 0.3857 0.0400
r = roots(p)
r =
1.0e+02 * -1.1472 + 0.0000i -0.1465 + 0.0000i -0.0550 + 0.0000i -0.0337 + 0.0182i -0.0337 - 0.0182i -0.0307 + 0.0000i -0.0244 + 0.0000i -0.0022 + 0.0113i -0.0022 - 0.0113i -0.0152 + 0.0000i -0.0113 + 0.0000i -0.0091 + 0.0000i -0.0019 + 0.0000i

Torsten on 13 Apr 2023
Edited: Torsten on 13 Apr 2023
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta alpha theta CL
F=zeros(13,1);
e1=F(1) == (mu - eta*Sci*C);
e2=F(1) == (theta*Ce - eta*Sci*C);
e3=F(3) == (k1*Sci - k2*Sr);
e4=F(4) == (k1*Sci - k10*Sh);
e5=F(5) == (p1*Sr - k11*R - k14*P*R);
e6=F(6) == (k3*CL*R - k4*Cf);
e7=F(7) == (k4*Cf - k5*Cp + k6*Ce);
e8=F(8) == (k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E);
e9=F(9) == (k7*Ce - k8*E);
e10=F(10) == (p2*Sh - k15*HR*Ce);
e11=F(11) == (alpha - k9*HR*H);
e12=F(12) == (k1*Sci - k13*Sp);
e13=F(13) == (p3*Sp - k16*P);
%Now solve to find the steady state values in terms of (A, B, C, D, Cl and the parameters as listed), then determine the Jacobian J:
[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq]=solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13],[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
f = [rhs(e1),rhs(e2),rhs(e3),rhs(e4),rhs(e5),rhs(e6),rhs(e7),rhs(e8),rhs(e9),rhs(e10),rhs(e11),rhs(e12),rhs(e13)];
J = jacobian(f, [Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
J = simplify(J)
J =
%Substitute the steady state values obtained into the Jacobian:
Jeq = subs(J,[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P],[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq])
Jeq =
%Lower (lb) and upper (ub) bounds for the free parameters
lbk1 = 3-0.5*3;
lbk2 = 2-0.5*2;
lbk3 = 5-0.5*5;
lbk4 = 4-0.5*4;
lbk5 = 5-0.5*5;
lbk6 = 1-0.5*1;
lbk7 = 4-0.5*4;
lbk8 = 3-0.5*3;
lbk9 = 1-0.5*1;
lbk10 = 1.2-0.5*1.2;
lbk11 = 1-0.5*1;
lbk12 = 9-0.5*9;
lbk13 = 1-0.5*1;
lbk14 = 1-0.5*1;
lbk15 = 1-0.5*1;
lbk16 = 1-0.5*1;
lbp1 = 30-0.5*30;
lbp2 = 4-0.5*4;
lbp3 = 1.5-0.5*1.5;
lbmu = 25-0.5*25;
lbeta = 10-0.5*10;
lbalpha = 10-0.5*10;
lbtheta = 10-0.5*10;
lbCL = 0.5-0.5*0.5;
ubk1 = 3+0.5*3;
ubk2 = 2+0.5*2;
ubk3 = 5+0.5*5;
ubk4 = 4+0.5*4;
ubk5 = 5+0.5*5;
ubk6 = 1+0.5*1;
ubk7 = 4+0.5*4;
ubk8 = 3+0.5*3;
ubk9 = 1+0.5*1;
ubk10 = 1.2+0.5*1.2;
ubk11 = 1+0.5*1;
ubk12 = 9+0.5*9;
ubk13 = 1+0.5*1;
ubk14 = 1+0.5*1;
ubk15 = 1+0.5*1;
ubk16 = 1+0.5*1;
ubp1 = 30+0.5*30;
ubp2 = 4+0.5*4;
ubp3 = 1.5+0.5*1.5;
ubmu = 25+0.5*25;
ubeta = 10+0.5*10;
ubalpha = 10+0.5*10;
ubtheta = 10+0.5*10;
ubCL = 0.5+0.5*0.5;
%Number of trials
n = 100;
found = zeros(n,1);
% Make a Monte Carlo simulation
for i = 1:n
% Random values for the parameters uniformly distributed between their
% lower and upper bounds
k1num(i) = lbk1 + rand()*(ubk1-lbk1);
k2num(i) = lbk2 + rand()*(ubk2-lbk2);
k3num(i) = lbk3 + rand()*(ubk3-lbk3);
k4num(i) = lbk4 + rand()*(ubk4-lbk4);
k5num(i) = lbk5 + rand()*(ubk5-lbk5);
k6num(i) = lbk6 + rand()*(ubk6-lbk6);
k7num(i) = lbk7 + rand()*(ubk7-lbk7);
k8num(i) = lbk8 + rand()*(ubk8-lbk8);
k9num(i) = lbk9 + rand()*(ubk9-lbk9);
k10num(i) = lbk10 + rand()*(ubk10-lbk10);
k11num(i) = lbk11 + rand()*(ubk11-lbk11);
k12num(i) = lbk12 + rand()*(ubk12-lbk12);
k13num(i) = lbk13 + rand()*(ubk13-lbk13);
k14num(i) = lbk14 + rand()*(ubk14-lbk14);
k15num(i) = lbk15 + rand()*(ubk15-lbk15);
k16num(i) = lbk16 + rand()*(ubk16-lbk16);
p1num(i) = lbp1 + rand()*(ubp1-lbp1);
p2num(i) = lbp2 + rand()*(ubp2-lbp2);
p3num(i) = lbp3 + rand()*(ubp3-lbp3);
munum(i) = lbmu + rand()*(ubmu-lbmu);
etanum(i) = lbeta + rand()*(ubeta-lbeta);
alphanum(i) = lbalpha + rand()*(ubalpha-lbalpha);
thetanum(i) = lbtheta + rand()*(ubtheta-lbtheta);
CLnum(i) = lbCL + rand()*(ubCL-lbCL);
Jeqnum = subs(Jeq,[k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta alpha theta CL],[k1num(i) k2num(i) k3num(i) k4num(i) k5num(i) k6num(i) k7num(i) k8num(i) k9num(i) k10num(i) k11num(i) k12num(i) k13num(i) k14num(i) k15num(i) k16num(i) p1num(i) p2num(i) p3num(i) munum(i) etanum(i) alphanum(i) thetanum(i) CLnum(i)]);
% Compute eigenvalues
v{i} = double(eig(Jeqnum));
% Check if all real parts are <= 0
if any(real(v{i})>0)
found(i) = 1;
end
end
% List the number of trials where eigenvalues with real part > 0 occured
sum(found)
ans = 41
Ron_S on 23 Apr 2023
Dear Torsten,
I'm not 100% sure the code that I now have generates a set of parameters for which at least one eigenvalue of the Jacobian is > 0? I've added the code in lthe last line.
When I check the specific k1num(i) value in the 1x100 double in the workspace that corresponds with the trial number that has a eigenvalue >0 it does not correspond? (Screenshot attached)
clear
syms Sci C Sr Sh R Cf Cp Ce E HR H Sp P k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta alpha theta CL
F=zeros(13,1);
e1=F(1) == (mu - eta*Sci*C);
e2=F(1) == (theta*Ce - eta*Sci*C);
e3=F(3) == (k1*Sci - k2*Sr);
e4=F(4) == (k1*Sci - k10*Sh);
e5=F(5) == (p1*Sr - k11*R - k14*P*R);
e6=F(6) == (k3*CL*R - k4*Cf);
e7=F(7) == (k4*Cf - k5*Cp + k6*Ce);
e8=F(8) == (k5*Cp - k6*Ce + k9*HR*H - k12*Ce - k7*Ce + k8*E);
e9=F(9) == (k7*Ce - k8*E);
e10=F(10) == (p2*Sh - k15*HR*Ce);
e11=F(11) == (alpha - k9*HR*H);
e12=F(12) == (k1*Sci - k13*Sp);
e13=F(13) == (p3*Sp - k16*P);
%Now solve to find the steady state values in terms of (A, B, C, D, Cl and the parameters as listed), then determine the Jacobian J:
[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq]=solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13],[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
f = [rhs(e1),rhs(e2),rhs(e3),rhs(e4),rhs(e5),rhs(e6),rhs(e7),rhs(e8),rhs(e9),rhs(e10),rhs(e11),rhs(e12),rhs(e13)];
J = jacobian(f, [Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P]);
J = simplify(J)
%Substitute the steady state values obtained into the Jacobian:
Jeq = subs(J,[Sci,C,Sr,Sh,R,Cf,Cp,Ce,E,HR,H,Sp,P],[Scieq,Ceq,Sreq,Sheq,Req,Cfeq,Cpeq,Ceeq,Eeq,HReq,Heq,Speq,Peq])
%Lower (lb) and upper (ub) bounds for the free parameters
lbk1 = 3-0.5*3;
lbk2 = 2-0.5*2;
lbk3 = 5-0.5*5;
lbk4 = 4-0.5*4;
lbk5 = 5-0.5*5;
lbk6 = 1-0.5*1;
lbk7 = 4-0.5*4;
lbk8 = 3-0.5*3;
lbk9 = 1-0.5*1;
lbk10 = 1.2-0.5*1.2;
lbk11 = 1-0.5*1;
lbk12 = 9-0.5*9;
lbk13 = 1-0.5*1;
lbk14 = 1-0.5*1;
lbk15 = 1-0.5*1;
lbk16 = 1-0.5*1;
lbp1 = 30-0.5*30;
lbp2 = 4-0.5*4;
lbp3 = 1.5-0.5*1.5;
lbmu = 25-0.5*25;
lbeta = 10-0.5*10;
lbalpha = 10-0.5*10;
lbtheta = 6-0.5*6;
lbCL = 9-0.5*9;
ubk1 = 3+0.5*3;
ubk2 = 2+0.5*2;
ubk3 = 5+0.5*5;
ubk4 = 4+0.5*4;
ubk5 = 5+0.5*5;
ubk6 = 1+0.5*1;
ubk7 = 4+0.5*4;
ubk8 = 3+0.5*3;
ubk9 = 1+0.5*1;
ubk10 = 1.2+0.5*1.2;
ubk11 = 1+0.5*1;
ubk12 = 9+0.5*9;
ubk13 = 1+0.5*1;
ubk14 = 1+0.5*1;
ubk15 = 1+0.5*1;
ubk16 = 1+0.5*1;
ubp1 = 30+0.5*30;
ubp2 = 4+0.5*4;
ubp3 = 1.5+0.5*1.5;
ubmu = 25+0.5*25;
ubeta = 10+0.5*10;
ubalpha = 10+0.5*10;
ubtheta = 6+0.5*6;
ubCL = 9+0.5*9;
%Number of trials
n = 100;
found = zeros(n,1);
% Make a Monte Carlo simulation
for i = 1:n
% Random values for the parameters uniformly distributed between their
% lower and upper bounds
k1num(i) = lbk1 + rand()*(ubk1-lbk1);
k2num(i) = lbk2 + rand()*(ubk2-lbk2);
k3num(i) = lbk3 + rand()*(ubk3-lbk3);
k4num(i) = lbk4 + rand()*(ubk4-lbk4);
k5num(i) = lbk5 + rand()*(ubk5-lbk5);
k6num(i) = lbk6 + rand()*(ubk6-lbk6);
k7num(i) = lbk7 + rand()*(ubk7-lbk7);
k8num(i) = lbk8 + rand()*(ubk8-lbk8);
k9num(i) = lbk9 + rand()*(ubk9-lbk9);
k10num(i) = lbk10 + rand()*(ubk10-lbk10);
k11num(i) = lbk11 + rand()*(ubk11-lbk11);
k12num(i) = lbk12 + rand()*(ubk12-lbk12);
k13num(i) = lbk13 + rand()*(ubk13-lbk13);
k14num(i) = lbk14 + rand()*(ubk14-lbk14);
k15num(i) = lbk15 + rand()*(ubk15-lbk15);
k16num(i) = lbk16 + rand()*(ubk16-lbk16);
p1num(i) = lbp1 + rand()*(ubp1-lbp1);
p2num(i) = lbp2 + rand()*(ubp2-lbp2);
p3num(i) = lbp3 + rand()*(ubp3-lbp3);
munum(i) = lbmu + rand()*(ubmu-lbmu);
etanum(i) = lbeta + rand()*(ubeta-lbeta);
alphanum(i) = lbalpha + rand()*(ubalpha-lbalpha);
thetanum(i) = lbtheta + rand()*(ubtheta-lbtheta);
CLnum(i) = lbCL + rand()*(ubCL-lbCL);
Jeqnum = subs(Jeq,[k1 k2 k3 k4 k5 k6 k7 k8 k9 k10 k11 k12 k13 k14 k15 k16 p1 p2 p3 mu eta alpha theta CL], ...
[k1num(i) k2num(i) k3num(i) k4num(i) k5num(i) k6num(i) k7num(i) k8num(i) k9num(i) k10num(i) k11num(i) k12num(i) ...
k13num(i) k14num(i) k15num(i) k16num(i) p1num(i) p2num(i) p3num(i) munum(i) etanum(i) alphanum(i) thetanum(i) CLnum(i)]);
% Compute eigenvalues
v{i} = double(eig(Jeqnum));
% Check if all real parts are <= 0
if any(real(v{i})>0)
found(i) = 1;
end
end
% List the number of trials where eigenvalues with real part > 0 occured
sum(found)
% To get a set of parameters for which at least one eigenvalue of the Jacobian is > 0.
k1num(i), k2num(i), k3num(i), k4num(i), k5num(i), k6num(i), k7num(i), k8num(i), k9num(i), k10num(i), k11num(i), k12num(i), k13num(i), k14num(i), k15num(i), k16num(i), p1num(i), p2num(i), p3num(i), munum(i), etanum(i), alphanum(i), thetanum(i), CLnum(i),(found)
Torsten on 23 Apr 2023
Edited: Torsten on 23 Apr 2023
Your output of parameters is after the loop over i, thus for i = n.
If found(n) = 1 by chance, you got a set of parameters for which at least one eigenvalue of the Jacobian is > 0.
But this will usually not be the case.
Try to understand the code and why you have to output the parameters here:
if any(real(v{i})>0)
found(i) = 1;
k1num(i),k2num(i),....
end