I have a differential equation in which the equation contains a function. I don't know how to input and call the function to generate the graph I want. can you help me fix it
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function fv=fgliom(T,y)
p1=0.0068;
p2=0.012;
p3=0.002;
i1=4.7*10^-8;
i2=4.7*10^-5;
i3=4.7*10^-8;
c1=510;
c2=510;
a1=510;
a2=510;
a3=510;
s1=1.8*10^-2;
s2=1.8*10^-3;
s3=1.8*10^-3;
v=2;
j=50;
b=0.2;
u=0;
fv=zeros(5,1);
fv(1)= p1*y(1)*(1-(y(1)/c1))-s1*y(1)*(y(2)+y(3))-(i1*y(1)*y(5))/a1+y(1);
fv(2)= p2*y(2)*(1-(y(2)+y(3))/c2)-s2*y(1)*y(2)-u*F[(y(5)]*y(1)-(i2*y(1)*y(5))/a2+y(2);
fv(3)= p3*y(3)*(1-(y(2)+y(3))/c2)-s3*y(1)*y(3)+u*F*(y(5))*y(1);
fv(4)= v*y(1)'*F*(-y(1)'/c1)*y(4)-(i3*y(4)*y(5))/a3+y(4);
fv(5)= j-b*y(5);
function fv=fnaik(T0,y0)
p1=0.0068;
p2=0.012;
p3=0.002;
i1=4.7*10^-8;
i2=4.7*10^-5;
i3=4.7*10^-8;
c1=510;
c2=510;
a1=510;
a2=510;
a3=510;
s1=1.8*10^-2;
s2=1.8*10^-3;
s3=1.8*10^-3;
v=2;
j=50;
b=0.2;
u=10^-3;
fv=zeros(5,1);
fv(1)= p1*y0(1)*(1-(y0(1)/c1))-s1*y0(1)*(y0(2)+y0(3))-(i1*y0(1)*y0(5))/a1+y0(1);
fv(2)= p2*y0(2)*(1-(y0(2)+y0(3))/c2)-s2*y0(1)*y0(2)-u*F*(y0(5))*y0(1)-(i2*y0(1)*y0(5))/a2+y0(2);
fv(3)= p3*y0(3)*(1-(y0(2)+y0(3))/c2)-s3*y0(1)*y0(3)+u*F*(y0(5))*y0(1);
fv(4)= v*y0(1)'*F*(-y0(1)'/c1)*y0(4)-(i3*y0(4)*y0(5))/a3+y0(4);
fv(5)= j-b*y0(5);
function fv=fturun(T1,y1)
p1=0.0068;
p2=0.012;
p3=0.002;
i1=4.7*10^-8;
i2=4.7*10^-5;
i3=4.7*10^-8;
c1=510;
c2=510;
a1=510;
a2=510;
a3=510;
s1=1.8*10^-2;
s2=1.8*10^-3;
s3=1.8*10^-3;
v=2;
j=50;
b=0.2;
u=10^-2;
fv=zeros(5,1);
fv(1)= p1*y1(1)*(1-(y1(1)/c1))-s1*y1(1)*(y1(2)+y1(3))-(i1*y1(1)*y1(5))/a1+y1(1);
fv(2)= p2*y1(2)*(1-(y1(2)+y1(3))/c2)-s2*y1(1)*y1(2)-u*F*(y1(5))*y1(1)-(i2*y1(1)*y1(5))/a2+y1(2);
fv(3)= p3*y1(3)*(1-(y1(2)+y1(3))/c2)-s3*y1(1)*y1(3)+u*F*(y1(5))*y1(1);
fv(4)= v*y1(1)'*F*(-y1(1)'/c1)*y1(4)-(i3*y1(4)*y1(5))/a3+y1(4);
fv(5)= j-b*y1(5);
clc;clear all;format long;
[T y]=ode45('ftuber',0 200,[10000 0 0 2000 0 500 140]',10^-7);
[T0 y0]=ode45('fnaik',0 200,[10000 0 0 2000 0 500 140]',10^-7);
[T1 y1]=ode45('fturun',0 200,[10000 0 0 2000 0 500 140]',10^-7);
figure (1);
plot(T,y(:,1),'.-',T0,y0(:,1),'.-',T1,y1(:,1),'.-');
title('Grafik A')
legend('u=50','u=30','u=25')
xlabel('waktu (hari)');
ylabel('MA(t) (sel)');
figure (2);
plot(T,y(:,1),'.-',T0,y0(:,1),'.-',T1,y1(:,1),'.-');
title('Grafik B')
legend('u=50','u=30','u=25')
xlabel('waktu (hari)');
ylabel('MA(t) (sel)');
figure (3);
plot(T,y(:,1),'.-',T0,y0(:,1),'.-',T1,y1(:,1),'.-');
title('Grafik C')
legend('u=50','u=30','u=25')
xlabel('waktu (hari)');
ylabel('MA(t) (sel)');
figure (4);
plot(T,y(:,1),'.-',T0,y0(:,1),'.-',T1,y1(:,1),'.-');
title('Grafik D')
legend('u=50','u=30','u=25')
xlabel('waktu (hari)');
ylabel('MA(t) (sel)');
figure (5);
plot(T,y(:,1),'.-',T0,y0(:,1),'.-',T1,y1(:,1),'.-');
title('Grafik E')
legend('u=50','u=30','u=25')
xlabel('waktu (hari)');
ylabel('MA(t) (sel)');
here i want
3 Comments
Walter Roberson
on 28 Jul 2022
function [value,isterminal,direction] = eventfun(T1, y1)
value = [y1(5)>0, -y1(1)>0];
isterminal = [true, true];
direction = [0, 0]; %crossing both ways
end
Accepted Answer
Sam Chak
on 28 Jul 2022
Equation 6 is a kind of ON/OFF switch that can be constructed using a scaled version of the signum function.
x = -1:0.001:1;
F = sign(x)/2 + 0.5;
plot(x, F, 'linewidth', 1.5)
ylim([-0.5 1.5]), grid on
If g = y(1), s = y(2), r = y(3), n = y(4), q = y(5), then for
fq = sign(y(5))/2 + 0.5;
For , you can probably create a dummy function for Eq. (7)
gdot = % type out Eq.(7) here
fgdot = sign(-gdot)/2 + 0.5;
8 Comments
Sam Chak
on 28 Jul 2022
Would advise you to check ALL equations again, one by one, term by term:
More Answers (0)
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