Finding equilibrium points for an ODE system

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;

[James Tursa]

Can you post an image of the differential equations you are solving?

[Star Strider]

‘... 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]

Yeah @Star Strider, it was posted before our discussion and because I wanted to see the figures I preferred to post the question separately.

Answer 1

[KSSV]

tol = 10^-5 ; % change the tolerance 
idx = abs(cdot)<tol & abs(cddot)<tol ; 
[cdot(idx) cddot(idx)]

[M]

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]

Try changing the tolerance value.