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This is a code that my team has been trying to fix but with no luck. There is one line on the graph that looks correct but the rest are all very small and wrong.

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CONNOR
CONNOR 2023 年 11 月 27 日
コメント済み: CONNOR 2023 年 11 月 28 日
clear, clc, close all
load("EnvironmentalForcing.mat")
for i = 1:length(T)
if T(i)>0 && T(i)<35
Tb(i) = (0.000241*(T(i)^2.06737))*((35-T(i))^0.72859);
end
end
Bmax = 1;
uLi = sum(Tb);
uL = 1/uLi;
%uL = 6;
uI = 10;
E = 0.001;
Ap = 5000;
Pi = 930.27249;
Si = Pi/Ap;
Li = 0.01*Si;
Ii = 0;
Ri = uI*Ii;
Pbi = 0;
y0 = [Si,Li,Ii,Ri,Pi,Pbi];
ode = @(n,t,S,L,I,R,P,Pb) SLIRP(n, t, S, L, I, R, P, Pb, Bmax, uL, uI, E, T, Ap);
[t,y] = rk4(ode,tspan, y0);
y = y/Ap;
figure
hold on
plot(t,y(1,:),'k')
plot(t,y(2,:),'b')
plot(t,y(3,:),'g')
plot(t,y(4,:),'y')
plot(t,y(5,:),'m')
legend("S","L","I","R","P")
%% Functions
function [dsdt, dldt, dIdt, drdt, dpdt, dpbdt] = SLIRP(n, t, S, L, I, R, P, Pb, Bmax, uL, uI, E, T, Ap)
T = T(n);
d = t;
if T>0 && T<35
Tb = (0.000241*(T^2.06737))*(35-T)^0.72859;
B = Bmax * Tb;
else
B = 0;
end
TE = (-0.35968 + (0.10789*T) - (0.00214*(T^2)));
dpldt = 1.33*d*TE;
dpbdt = ((172.4*Pb) - (21.2*(Pb^2)))*TE;
dpdt = dpbdt + dpldt;
dsdt = (-B*S*I) + (dpdt*(1/Ap));
dldt = ((S*I) - (uL*L) + E);
dIdt = (uL*L) - (uI*I);
drdt = uI*I;
end
function [tp,yp] = rk4(f, tspan, y0)
q = length(y0);
N = length(tspan);
t0 = tspan(1);
h = tspan(2) - tspan(1);
tp = zeros(N+1,1);
yp = zeros(q, N+1);
tp(1) = t0;
yp(:,1) = y0;
for n=1:N
[k1(1), k1(2), k1(3), k1(4), k1(5), k1(6)] = f(n, tp(n), yp(1,n), yp(2,n), yp(3,n), yp(4,n), yp(5,n), yp(6,n));
[k2(1), k2(2), k2(3), k2(4), k2(5), k2(6)] = f(n, tp(n) + (0.5*h), yp(1,n) + (0.5*h*k1(1)), yp(2,n) + (0.5*h*k1(2)), yp(3,n) + (0.5*h*k1(3)), yp(4,n) + (0.5*h*k1(4)), yp(5,n) + (0.5*h*k1(5)), yp(6,n) + (0.5*h*k1(6)));
[k3(1), k3(2), k3(3), k3(4), k3(5), k3(6)] = f(n, tp(n) + (0.5*h), yp(1,n) + (0.5*h*k2(1)), yp(2,n) + (0.5*h*k2(2)), yp(3,n) + (0.5*h*k2(3)), yp(4,n) + (0.5*h*k2(4)), yp(5,n) + (0.5*h*k2(5)), yp(6,n) + (0.5*h*k2(6)));
[k4(1), k4(2), k4(3), k4(4), k4(5), k4(6)] = f(n, tp(n) + h, yp(1,n) + (h*k3(1)), yp(2,n) + (h*k3(2)), yp(3,n) + (h*k3(3)), yp(4,n) + (h*k3(4)), yp(5,n) + (h*k3(5)), yp(6,n) + (h*k3(6)));
for j = 1:q
fi(j) = (k1(j)/6)+(k2(j)/3)+(k3(j)/3)+(k4(j)/6);
yp(j,n+1) = yp(j,n) + h*fi(j);
end
tp(n+1) = tp(n) + h;
end
end
  2 件のコメント
CONNOR
CONNOR 2023 年 11 月 27 日
This is a second attempt at the same question as I asked before
CONNOR
CONNOR 2023 年 11 月 27 日
This is what the graph is supposed to look like and it is all on the same scale so I dont think that is it

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回答 (1 件)

Image Analyst
Image Analyst 2023 年 11 月 27 日
Yes, because the scale of the different plots is several orders of magnitude different so you can't see most of them if you use the same y scale. I suggest you plot them on separate axes, or use stackedplot
  5 件のコメント
Image Analyst
Image Analyst 2023 年 11 月 28 日
Well we have no idea where your formulas came from so you're going to have to research that yourself. Maybe some of the parameters are off by a few orders of magnitude, or were for different units (e.g. microns instead of meters).
This may help:
CONNOR
CONNOR 2023 年 11 月 28 日
The formulas came from the assignment and same with the parameters. I dont think that those are wrong because I have checked them multiple times. Thanks for the video Ill check it out in hopes it can get me through this wall. Thanks for the help anyway.

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