Possible workaround for non-integer indices

For an assignment for uni, I have to create a RK4 population simulation. This I what I have so far:

close all
clear

alpha = 10^-3;
gamma = 6e-3;
c = 3;
a = 2;

Dt = 0.5;
lambda = sqrt(-6);


y1(1) = 600;
y2(1) = 1000;

t = 1:Dt:6;
k11 = zeros(1,11);
k12 = zeros(1,11);
k21 = zeros(1,11);
k22 = zeros(1,11);
k31 = zeros(1,11);
k32 = zeros(1,11);
k41 = zeros(1,11);
k42 = zeros(1,11);

for i = 1:10
    k11(i) = Dt*dy1(i,y1,y2);
    k12(i) = Dt*dy2(i,y1,y2);
    
    k21(i) = Dt*(dy1(i,y1 + 0.5*k11(i),y2 + 0.5*k12(i)) + 0.5*k11(i));
    k22(i) = Dt*(dy2(i,y1 + 0.5*k11(i),y2 + 0.5*k12(i)) + 0.5*k12(i));
    
    k31(i) = Dt*(dy1(i,y1 + 0.5*k21(i),y2 + 0.5*k22(i)) + 0.5*k21(i));
    k32(i) = Dt*(dy2(i,y1 + 0.5*k21(i),y2 + 0.5*k22(i)) + 0.5*k22(i));
    
    k41(i) = Dt*(dy1(i,y1 + k31(i),y2 + k32(i)) + k31(i));
    k42(i) = Dt*(dy2(i,y1 + k31(i),y2 + k32(i)) + k32(i));    
    
    y1(i+1) = y1(i) + (1/6)*(k11(i) + 2*k21(i) + 2*k31(i) + k41(i));
    y2(i+1) = y2(i) + (1/6)*(k12(i) + 2*k22(i) + 2*k32(i) + k42(i));
end

dy1, dy2, and W are seperate functions.

The only thing that needs to be changed is the input in the dy1 and dy2.

Ideally, I would just fill in i + 0.5*i, which would give the right answer, howerver matlab doesn't allow this. Is there an easy way to achieve the same effect without changing the entire code?

[Bob Thompson]

I'm a little bit confused what you're asking for. You have dy1, dy2, and W as functions so having a non integer input is not automatically disqualified. Your question though asks about non-integer indexing. Are you looking to set your loop index to be a non-integer for the input to dy1, dy2, and W, but worried about indexing all of the k and y matrices? Why not just use i+0.5*i as your function input?

[Tom Oud]

This gives the error:

Subscript indices must either be real positive integers or logicals.

The function dy1 (and dy2) uses the values for i as its input.

function g = dy1(i,y1,y2)

alpha = 10^-3;
gamma = 6e-3;
c = 3;
a = 2;


lambda = sqrt(-6);

%for i = 1:11
g = (a - alpha.*y2(i)).*y1(i) - W(i/2).*y1(i);
%end
end

[Bob Thompson]

You said that W is a function? The error is in dy1, but not in W? Does W or dy2 change the value of i at all? I have been unable to reproduce the error with the codes you have posted, so I suspect that somewhere the value of i is being changed that I am not experiencing.

[Pieter-Bas Draijer]

Hi Bob,

I'm working together with Tom. The W and dy2 are also depended on i (see downbelow). But this doesn't cause the problem. The problem appears when we want to try to use a non real positive integers or logicals in i such as 1.5 and 1.75 instead of 1, 2, 3 etc. Is there a way in which we can work with this kind of numbers in our code above?

w:

function f = W(t)
    f = [];
    if 0 <= t && t < (3/12)
        f = 0;
    end
    if (3/12) <= t && t <= (8/12)
        f = 1;
    end
    if (8/12) < t && t < 1
        f = 0;
    end
     if 1 <= t && t < 1+(3/12)
        f = 0;
    end
    if 1+(3/12) <= t && t <= 1+(8/12)
        f = 1;
    end
    if 1+(8/12) < t && t < 2
        f = 0;
    end
    
       if 2 <= t && t < 2+(3/12)
        f = 0;
    end
    if 2+(3/12) <= t && t <= 2+(8/12)
        f = 1;
    end
    if 2+(8/12) < t && t < 3
        f = 0;
    end
    
       if 3 <= t && t < 3+(3/12)
        f = 0;
    end
    if 3+(3/12) <= t && t <= 3+(8/12)
        f = 1;
    end
    if 3+(8/12) < t && t < 4
        f = 0;
    end
 if 4 <= t && t < 4+(3/12)
        f = 0;
    end
    if 4+(3/12) <= t && t <= 4+(8/12)
        f = 1;
    end
    if 4+(8/12) < t && t < 5
        f = 0;
    end
    
       if 5 <= t && t < 5+(3/12)
        f = 0;
    end
    if 5+(3/12) <= t && t <= 5+(8/12)
        f = 1;
    end
    if 5+(8/12) < t && t < 6
        f = 0;
    end
    
       if 6 <= t && t < 6+(3/12)
        f = 0;
    end
    if 6+(3/12) <= t && t <= 6+(8/12)
        f = 1;
    end
    if 6+(8/12) < t && t < 7
        f = 0;
    end
    
       if 7 <= t && t < 7+(3/12)
        f = 0;
    end
    if 7+(3/12) <= t && t <= 7+(8/12)
        f = 1;
    end
    if 7+(8/12) < t && t < 8
        f = 0;
    end
    
       if 8 <= t && t < 8+(3/12)
        f = 0;
    end
    if 8+(3/12) <= t && t <= 8+(8/12)
        f = 1;
    end
    if 8+(8/12) < t && t < 9
        f = 0;
    end
    
       if 9 <= t && t < 9+(3/12)
        f = 0;
    end
    if 9+(3/12) <= t && t <= 9+(8/12)
        f = 1;
    end
    if 9+(8/12) < t && t < 10
        f = 0;
    end
    
       if 10 <= t && t < 10+(3/12)
        f = 0;
    end
    if 10+(3/12) <= t && t <= 10+(8/12)
        f = 1;
    end
    if 10+(8/12) < t && t < 11
        f = 0;
    end
    
       if 11 <= t && t < 11+(3/12)
        f = 0;
    end
    if 11+(3/12) <= t && t <= 11+(8/12)
        f = 1;
    end
    if 11+(8/12) < t && t < 12
        f = 0;
    end
    
       if 12 <= t && t < 12+(3/12)
        f = 0;
    end
    if 12+(3/12) <= t && t <= 12+(8/12)
        f = 1;
    end
    if 12+(8/12) < t && t < 13
        f = 0;
    end
    end

dy2

function h = dy2(i,y1,y2)

alpha = 10^-3;
gamma = 6e-3;
c = 3;
a = 2;
%y2 = a/alpha;
%y1 = c/gamma;
Dt = 0.5;
lambda = sqrt(-6);
y1(1) = 600;
y2(1) = 1000;

h = (-c + gamma.*y1(i)).*y2(i);
end

[Rik]

You are calling y1 (and y2) not as a function, but as an array. You need to make sure they are functions. The frequent use of the letter d makes me think you want to treat them as differential equations. Can you explain your reasoning of what you're trying to achieve?

I would also suggest rewriting your w function. I would suggest making an nx3 array with the upper bounds, lower bounds and value for f. That way it is harder to make a typo, easier to edit, easier to understand, and likely easier to speed up.

Answer 1

[James Browne]

Greetings,

I thik, at least your problem with using the iterator variable to generate a complex number is that you are using "i" as the iterator and also trying to use "i" in the complex number as the imaginary component.

There are two solutions to this specific problem:

1) Use a different iterator variable such as k and then generate your complex number(s) with something like:

n = k + 0.5*i;

2) You can also use "j" to define the imaginary component of a complex number, so if you have defined "i" as an iterater variable already, you can create a complex number with something like:

n = i + 0.5*j

One last note too, I often write scripts where the iterator variable ("i", in your case) cannot be used directly in part of an equation. Using the iterator to determine a logical value or take on a non-integer number from a sequence are examples of this issue.

The solution is to generate an object in which the desired values are stored and then use the iterator to access them. This opens up the possabilities of using the iterator beyond modifying a number in an equation. The following are some examples of the concept that I am referring to.

animals = {'Cats','Dogs','Hampsters'};
isCool = {true,false,false};
reasons = {'They just are','They smell bad and bark all the time','They smell bad'};
n = length(animals);
for i = 1:n
    if(isCool{i})
        message = strcat(animals{i},' are cool because',reasons{i},'\n');
        fprintf(message)
    end
    if(~isCool{i})
        message = strcat(animals{i},' are not cool because',reasons{i},'\n');
        fprintf(message)
    end
end

The example above shows how you can use the iterator variable to invoke strings and logical values inside of a loop, which you would not otherwise be able to do by using the iterator value directly.

nWaves = 3;
omegas = zeros(1,nWaves);
for i = 1:nWaves
    omegas(i) = i +1.75*i;
end

n = length(omegas);
duration = 4;
dt = 0.01;
t = 0:dt:duration;

for i = 1:n
    y = sin(omegas(i)*t);
    plot(t,y)
    hold on
end
hold off
xlabel('Time (seconds)')
ylabel('Amplitude')
title('A Really Messy Figure')

In the above example, you can see that the iterator is used to invoke non-integer values for angular velocities in the second for loop. Additionally, in the first for loop, you can see that the equation for omegas(i) looks like it should generate a complex number but it does not because the variable "i" was defined in the previous line as a vector containing integer values so it will no longer generate a complex number when used in an equation.

Hope this helps~