Need help to solve 5 simultaneous first order differential equations with Initial Condition.

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I have to solve following first order ordinary differential equations and plot the values of Concentrations (Cmn, Cecm, Crec, Ccirc, Ccells) against time (t).
where value of r(t) is described as following.
Below is my script, but I am getting constant errors. or something i dont understand.
clc; close; clear;
%Constant Parameters
Cmn0 = 6.2E-6; %Initial conc at the MN (
tr = 1805; %release period (s)
ka = 5.01E6; %Association rate (in 1/s)
kd = 5E-4; %Dissociation rate (in 1/s)
ki = 5.05E-3; %Internalization rate (in 1/s)
kc = 5E-3; %Circulation uptake rate (in 1/s)
Rtot = 1.85E-6 ; %initial receptor concentration (in umol/mm^3)
syms r(t) C_mn(t) C_ecm(t) C_rec(t) C_circ(t) C_cells(t) %creating symbolic variable
ode1 = diff(r) == Cmn0/tr;
ode2 = diff(C_mn) == -r;
ode3 = diff(C_ecm) == r- (ka * C_ecm * (Rtot - C_rec - C_cells)) + kd * C_rec - kc * C_ecm;
ode4 = diff(C_rec) == (ka * C_ecm * (Rtot - C_rec - C_cells)) - ((kd + ki)* C_rec);
ode5 = diff(C_circ) == kc * C_ecm;
ode6 = diff(C_cells) == ki * C_rec;
odes = [ode1; ode2; ode3; ode4; ode5; ode6]
cond1 = r(0) == Cmn0/tr;
cond2 = C_mn(0) == 6.2E-6;
cond3 = C_ecm(0) == 0;
cond4 = C_rec(0) == 0;
cond5 = C_circ(0) == 0;
cond6 = C_cells(0) == 0;
conds = [cond1; cond2; cond3; cond4; cond5; cond6];
[VF,Sbs] = odeToVectorField(odes);
odsefcn = matlabFunction(VF,'File', 'Consolvefun')
[t, C] = ode45(@Consolvefun, [0 5000], conds);

Respuesta aceptada

Alan Stevens
Alan Stevens el 14 de Abr. de 2021
You have a stiff system (ka~10^6, kd~10^-4), so use ode15s rather than ode45. The following works. I'll leave you to decide if the results make sense!
C0 = [6.2E-6, 0, 0, 0, 0];
tspan = 0:10:5000;
[t, C] = ode15s(@fn, tspan, C0);
C_mn = C(:,1);
C_ecm = C(:,2);
C_rec = C(:,3);
C_circ = C(:,4);
C_cells = C(:,5);
plot(t,C_mn,t,C_ecm,t,C_rec,t,C_circ,t,C_cells),grid
xlabel('t'),ylabel('C')
legend('Cmn','Cecm','Crec','Ccirc','Ccells')
function dCdt = fn(t, C)
%Constant Parameters
Cmn0 = 6.2E-6; %Initial conc at the MN (
tr = 1805; %release period (s)
ka = 5.01E6; %Association rate (in 1/s)
kd = 5E-4; %Dissociation rate (in 1/s)
ki = 5.05E-3; %Internalization rate (in 1/s)
kc = 5E-3; %Circulation uptake rate (in 1/s)
Rtot = 1.85E-6 ; %initial receptor concentration (in umol/mm^3)
r = Cmn0/tr*(t<=tr);
C_mn = C(1);
C_ecm = C(2);
C_rec = C(3);
C_circ = C(4);
C_cells = C(5);
dCdt = [ -r;
r- ka * C_ecm * (Rtot - C_rec - C_cells) + kd * C_rec - kc * C_ecm;
ka * C_ecm * (Rtot - C_rec - C_cells) - (kd + ki)* C_rec;
kc * C_ecm;
ki * C_rec];
end
  3 comentarios
HARSH ZALAVADIYA
HARSH ZALAVADIYA el 14 de Abr. de 2021
And also the condition for r(t) says that when t>tr, it is 0, where as in the graph its not saturating at 1. Thats the part I am confused that how do i put that in code?

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