i have error in this code while using ode45

    %constants
A=2.6e-3;
B=90;
C=0.025;
D=5.0E-3;
E=0.3;
F=0.0041;
G=0.27;
H=83.7;

%Parameter Changes
A1=2.6e-2;
B1=80;
C1=0.025;
D1=6.0E-3;
E1=1;
F1=0.008;
G1=0.27;
H1=83.7;


%hold on


f=@(t,y) [-y(2) y(1) -A(y(1)-B); C*D*(y(3)-E) -C*y(2); -G*y(3) +F*max(0,y(1)-H)];
[t,xa]=ode45(f, [0 180], [250 :0 350]);

Index exceeds the number of array elements. Index must not exceed 1.

Error in solution (line 26)
f=@(t,y) [-y(2) y(1) -A(y(1)-B); C*D*(y(3)-E) -C*y(2); -G*y(3) +F*max(0,y(1)-H)];

Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.

Error in ode45 (line 107)
odearguments(odeIsFuncHandle,odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);

disp(t,xa);
f=@(t,y) [-y(2) y(1) -A1(y(1)-B1); C1*D1*(y(3)-E1) -C1*y(2); -G1*y(3) +F1*max(0,y(1)-H1)];
[t,xb]=ode45(f, [0 180], [250 0  350]);
disp(t,xb);
figure(1)
plot(t,xa(:,1)+50,"LineWidth",2); ylabel("glucose(mm/dl)"); 
xlabel("time(min)");title('Glucose dynamics')
hold on
plot(t,xb(:,1),"LineWidth",2); ylabel("glucose(mm/dl)"); 
xlabel("time(min)");title('Glucose dynamics')
legend ('Orignal','Dexcom Curve')
hold off

%hold on

figure(2)
plot(t,xa(:,2)*33,"LineWidth",2); ylabel("Interstitial insulin(\uU/ml)");
xlabel("time(min)");title('Interstitial insulin dynamics')
hold on
plot(t,xb(:,2),"LineWidth",2); ylabel("Interstitial insulin(\uU/ml)");
xlabel("time(min)");title('Interstitial insulin dynamics')
legend ('Orignal','Dexcom Curve')
hold off 

figure(3)
plot(t,xa(:,3)+5,"LineWidth",2); ylabel("Plasma insulin(\uU/ml)"); 
xlabel("time(min)");title('Plasma insulin dyanmics')
hold on
plot(t,xb(:,3),"LineWidth",2); ylabel("Plasma insulin(\uU/ml)"); 
xlabel("time(min)");title('Plasma insulin dyanmics')
legend ('Orignal','Dexcom Curve')
hold off

Answer 1

[Walter Roberson]

[250 :0 350]

ans = 350

Your vector of initial conditions only has one entry.

Your line of code will not be interpreted as

[250, :0, 350]

and that would be a syntax error if it was.

Likewise it is not interpreted as [250 :0: 350] -- which would give you an empty vector.

The 250 :0 part is being intepreted as (250:0) -- the vector of values starting with 250, incrementing by 1, and ending with the first number that is greater than 0... and since 250 is greater than 0, that would be the empty array.

[kainat rasheed]

@Walter Roberson so how should i actualy right. i am stuck in it

[kainat rasheed]

[kainat rasheed]

@Walter Roberson see these graphs i am writing this code for it. and these are my equations. i think i am doing wrong

[Walter Roberson]

When you call

[t,xa]=ode45(f, [0 180], [250 :0 350]);

the second [] is your initial conditions; with your code, you need three initial conditions.

I recommend that you do not use the : operator in defining your initial conditions. For example,

[t,xa]=ode45(f, [0 180], [250, 30, 350]);

but adjust that for your actual initial conditions. Do not compute them with expressions: just list them.

[Torsten]

Since you chose names for the constants in your code that differ from those above it is impossible for us to correct your function line

f=@(t,y) [-y(2) y(1) -A(y(1)-B); C*D*(y(3)-E) -C*y(2); -G*y(3) +F*max(0,y(1)-H)];

[kainat rasheed]

i know i think i type it wrong as [ 2 : 0 1]

but i am having error in ode45 function . it is saying t is unused here @Walter Roberson thats is my reall problem

[kainat rasheed]

@Walter Roberson now tell me how should i actualy write this ode 45 line in this code

[Walter Roberson]

[~,xa]=ode45(f, [0 180], [2, 0, 1]);

[Walter Roberson]

f=@(t,y) [-y(2) y(1) -A(y(1)-B); C*D*(y(3)-E) -C*y(2); -G*y(3) +F*max(0,y(1)-H)];

your A is a constant not a function. You cannot index A at (y(1)-B)

MATLAB is completely lacking in implied multiplication. There are no cases in MATLAB in which P(Q) means that P is to be multiplied by Q.

[kainat rasheed]

f=@(t,y) [-y(2)y(1)-A(y(1)-B);C*D*(y(3)-E)-C*y(2);-G*y(3)+F*max(0,y(1)-H)];

@Walter Roberson what if it is like this?

[Torsten]

Let's play a game: search the two places where you forgot to include the "*" sign in the function definition.

[kainat rasheed]

@Torsten

[-y(2)*y(1)

A*(y(1)-B)

here?

[kainat rasheed]

@Torsten i think it worked. such small change helped me alot. thankyou so so much !

[kainat rasheed]

@Torsten i enjoyed the game. it solved my problem. thankss

[kainat rasheed]

thanks @Walter Roberson and @Torsten

[Torsten]

@kainat rasheed

You're welcome. I like people having humor.