Finding equilibrium points for an ODE system

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M
M 2022-9-20
评论： M 2022-9-21
Hi, I have two functions named cdot and ctdot. I want to find the eqilibrium points which means cdot=ctdot=0. Could you please tell me how can I find points (c,ct) which satisfied in cdot=ctdot=0.
c and ct should be positive between [0,2].
Thanks in advance for any help.
vplc=0.16;
delta=0.1;
Ktau=0.045;
Kc=0.1;
K=0.0075;
Kp=0.15;
gamma=5.5;
kb=0.4;
vss=0.044;
alpha0=delta*6.81e-6/(0.002);
alpha1=delta*2.27e-5/(0.002);
Ke=7;
Vs=0.002;
ks=0.1;
Kf=0.18;
kplc=0.055;
ki=2;
A=(-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=(-(0.4.*A.*((Kc.^4).*(Kp.^2))./((p.^2.*c.^2.*gamma.*ct.*Kf))));
jin2=alpha1.*Kce.^4./((gamma.*ct).^4+Kce.^4);
p=(vplc.*c.^2/(c.^2+kplc.^2))./ki;
G1=alpha0+jin2;
G2=((1-h)./tau_max).*c.^4;
Fc=(4.*gamma.*Kf).*((c.^3.*p.^2.*h.*ct)./(Kb.*Kp.^2.*Ktau.^4))-(2.*Vss.*c./Ks.^2);
Fct=((gamma.*Kf.*(c.^4).*(p.^2).*h)./(Kb.*Kp.^2.*Ktau.^4))+((Vs.*K.*gamma.^2)./(Ks.^2))-((4.*gamma.^4.*ct.^3.*alpha1.*Kce.^4)./(Kce.^4+(gamma.*ct).^4).^2);
Fh=(gamma.*Kf.*c.^4.*p.^2.*ct)./(Kb.*Kp.^2.*Ktau.^4);
cdot=(Fct).*(G1)+(Fh).*(G2);
ctdot=-G1.*Fc;
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Star Strider
Star Strider 2022-9-21
‘... find points (c,ct) which satisfied in cdot=ctdot=0’
They don’t intersect, at least in the regions described, as we discussed in how can I find the intersection of two surface. The value of ‘C’ may need to be negative for an intersection to exist.
M
M 2022-9-21
Yeah @Star Strider, it was posted before our discussion and because I wanted to see the figures I preferred to post the question separately.

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回答（1 个）

KSSV
KSSV 2022-9-20
编辑：KSSV 2022-9-20
tol = 10^-5 ; % change the tolerance
idx = abs(cdot)<tol & abs(cddot)<tol ;
[cdot(idx) cddot(idx)]
  2 个评论
M
M 2022-9-20
After I used the function you suggested, the below comment appeared. Does it mean this system does not have an equilibrium point?
ans =
Empty matrix: 1-by-0
However, it is not valid; I am sure it has eq.p
KSSV
KSSV 2022-9-21
编辑：KSSV 2022-9-21
Try changing the tolerance value.

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