How to create an array of functions in matlab?

Question

I was trying to solve Lorenz System in matlab by using the method of RK2 and RK4. I had a script for both of the methods, the problem now is how can converge the following

y(1) = @(t,y) 10*(y(2)-y(1));
y(2) = @(t,y) y(1)*(28-y(3))-y(2);
y(3) = @(t,y) y(1)*y(2)-8*y(3)/3;

into simply an column vector of y.

here is what i was hoping for and it never worked:

y = zeros(3,1);
y(1) = @(t,y) 10*(y(2)-y(1));
y(2) = @(t,y) y(1)*(28-y(3))-y(2);
y(3) = @(t,y) y(1)*y(2)-8*y(3)/3;

and following is my RK2 function. my RK4 is similar to this RK2, manybe this can help you understand why I really need an vector of functions.

function y = RK2(fcn,lrange,urange,step,init)
%fcn = vector of functions
%lrange = lower bound
%urange = upper bound
%step = number of steps
%init = initial value

row = size(fcn,1);
stepsize = (urange-lrange)/step;
y = zeros(row,step);
%initializing vector of y
y(:,1) = init;
%initial condition
t = zeros(1,step+1);
%initializing vector of t

if row ~= size(init,1)
    disp('number of functions and number of initial values do not match');
end

for n = 1:step

    t(n) = (n-1)*stepsize;
    t(step+1) = urange;
    y1 = stepsize.*fcn(t(n),y(:,n));
    y2 = stepsize.*fcn(t(n) + stepsize/2, y(:,n) + y1./2);
    y(:,n+1) = y(:,n) + y2;

end

Gabrielle Stewart · Accepted Answer · 2019-04-04 03:29:20Z

3

Alternatively,

f = cell(3,1); % create a cell array

% initialize

    f{1} = @(t) t^2;
    f{2} = @(t) cos(2*t);
    f{3} = @(t) 4*(t^3);

% access properties

    size(f)(1); % access the number of functions
    f{1} % access the first function
    f{2}(17) % evaluate the second function at x = 17

A cell array is a data type with indexed data containers called cells, where each cell can contain any type of data.

Documentation: https://www.mathworks.com/help/matlab/ref/cell.html

Alternatively, In Octave:

f = cell(3,1); # create a cell array

# initialize

    f(1) = @(t) t^2;
    f(2) = @(t) cos(2*t);
    f(3) = @(t) 4*(t^3);

# access properties

    size(f)(1); # access the number of functions
    f{1} # access the first function
    f{2}(17) # evaluate the second function at x = 17

edited Apr 4, 2019 at 3:29

answered Apr 1, 2019 at 16:57

Gabrielle Stewart

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3 Comments

Cris Luengo Over a year ago

You need to do either f(1) = {@(t) t^2}; or f{1} = @(t) t^2;. Also, # is not valid syntax in MATLAB.

Gabrielle Stewart Over a year ago

My apologies, I use Octave which is MATLAB for Linux and I hadn't thought there was a difference between that and MATLAB. In Octave # works for a comment Should this be removed as an answer?

Cris Luengo Over a year ago

No, don't delete it, but do state that it's a solution for Octave. (BTW: MATLAB is supported on Linux, and Octave works just as well on MacOS and Windows and many other OSes, I don't think "MATLAB for Linux" is a correct description of Octave. Octave and MATLAB have a similar syntax, largely compatible, but not identical. There are many differences.)

David · Accepted Answer · 2014-12-08 01:27:20Z

2

This should work, just make the function output a vector:

y = @(t,y) [10*(y(2)-y(1)), y(1)*(28-y(3))-y(2), y(1)*y(2)-8*y(3)/3];
y(1,[1;2;3])
size(y(1,[1;2;3]))

edited Dec 8, 2014 at 1:27

answered Dec 8, 2014 at 0:42

David

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9 Comments

Lechen Yuan Over a year ago

hello, thanks for answering my question. I just tried what you answered, it seems it did make my functions into one vector. the problem is when I try to count the number of rows of this vector it either shoot out 1 or "Dimension argument must be a positive integer scalar within indexing range." do you know how to fix this? or how to count the number of rows in this vector if i made it a row vector. Thank you so much!

David Over a year ago

Try replacing the semicolons in the array with commas to make a row vector not a column vector too.

Lechen Yuan Over a year ago

I tried both ways you provided and neither worked. somehow I just cannot count the number of arguments of that array of functions.

Lechen Yuan Over a year ago

even if I used unmel(y), it always shoot out 1 instead of 3.

David Over a year ago

That's correct. y is a function, not an array. It's only when you give y inputs and evaluate the result that you get an array.

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