Error using vertcat Dimensions of arrays being concatenated are not consistent. PLEASE HELP

I have this code and it show me an error as in the title of the question

function [] = call_func

clear all

close all

tspan = [ 0 15 ]

y0 = 0:0.01:1;

x0 = 0:0.01:1;

theta0 = 0:0.1*pi:2*pi;

l = 1;

options=odeset('RelTol',1e-8,'AbsTol',1e-8)

for k = 1:length(y0)

for s = 1:length(x0)

for w = 1:length(theta0)

[t,x] = ode45(@func,tspan,[x0(s) y0(k) theta0(w)],options);

%legend('b = blue', 'r = red')

if (max(mod(x(:,1),l))<0.5) && (min(mod(x(:,2),l))>0.5) || (max(mod(x(:,1),l))<0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (min(mod(x(:,2),l))>0.5)

c='r.';

else

c='b.';

end

plot(mod(x(end,1),l),mod(x(end,2),l),c);

hold on

end

end

end

end

function dydt = func(t,x)

Vs = 0.0;% true value

alpha =1; % true value

B = 0.12;

w = 6.28;

tau = 0;

L = 1;

dydt = [sin(2*pi*(x(1)-(B/L)*sin(w*tau*t))*cos(2*pi*x(2)) +Vs*cos(x(3)))

cos(2*pi*(x(1)-(B/L)*sin(w*tau*t)))*sin(2*pi*x(2)) +Vs*sin(x(3))

2*pi*alpha*sin(2*x(3))*cos(2*pi*(x(1)-B/L*sin(w*tau*t)))*cos(2*pi*x(2))+2*pi*sin(2*pi*(x(1)-B/L*sin(w*tau*t)))*sin(2*pi*x(2))];

end

Answer 1

[Star Strider]

The problem is in the expression for ‘dydt’, since MATLAB interprets the spaces as delimiters. This will throw the reported error. The easiest way to correct that is to put parentheses around each row, so MATLAB considers each row a distinct expression:

dydt = [(sin(2*pi*(x(1)-(B/L))*sin(w*tau*t))*cos(2*pi*x(2)) +Vs*cos(x(3)))
         (cos(2*pi*(x(1)-(B/L)*sin(w*tau*t)))*sin(2*pi*x(2)) +Vs*sin(x(3)))
         (2*pi*alpha*sin(2*x(3))*cos(2*pi*(x(1)-B/L*sin(w*tau*t)))*cos(2*pi*x(2))+2*pi*sin(2*pi*(x(1)-B/L*sin(w*tau*t)))*sin(2*pi*x(2)))];

There was also a missing parenthesis enclosing the second sin call in the first row. This version of ‘dydt’ runs without error, however it would be advisable to check to be certain that I put the correct closing parenthesis in the correct location.

Beyond that, the code takes forever to execute. It might be better to store the values you want to plot in vectors and then plot them at the end, rather than plotting them in the loop.

EDIT — (7 Apr 2021 at 16:54)

The code reported this run time:

Elapsed time is 918.396917 seconds.

and produced this plot:

.

[Ahmed Sulaiman Ahmed Al Rawahi]

Thanks I understand now what was the problem. The code now is working good, but as you mentioned it take forever to excute. I did not know how to store the values in order to reduce the time taken for excectution since I am still new with MATLAB.

If it possible to help me write the code with the values stored and plot outside the for-loop I would appreciate it in order to reduce the time taken.

Thanks again!

[Star Strider]

As always, my pleasure!

One option is to save the data to a matrix (along with the colours) in the loop, then plot them at the end. Adding the ‘xmtx’ and ‘cmtx’ matrices and preallocating them:

numrows = numel(y0)*numel(x0)*numel(theta0);
xmtx = zeros(numrows,2);
cmtx = zeros(numrows,3);
k = 0;
for k = 1:length(y0)
    for s = 1:length(x0)
        for w = 1:length(theta0)
            k = k + 1;
            [t,x] = ode15s(@func,tspan,[x0(s) y0(k) theta0(w)],options);
            %legend('b = blue', 'r = red')
            if (max(mod(x(:,1),l))<0.5) && (min(mod(x(:,2),l))>0.5) || (max(mod(x(:,1),l))<0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (min(mod(x(:,2),l))>0.5)
                cv = [1 0 0];
%                 c='r.';
            else
                cv = [0 0 1];
%                 c='b.';
            end
            cmtx(k,:) = cv;
            xmtx(k,:) = [mod(x(end,1),l), mod(x(end,2),l)];
%             hold on
            
        end
    end
end

figure
scatter(xmtx(:,1), xmtx(:,2), 5, cmtx, 'filled')

This is definitely more efficient, with:

Elapsed time is 731.415747 seconds.

(that I consider to be reasonably impressive, representing a greater than 20% increase in speed, and a demonstration of the effectiveness of preallocation) and the figure essentially as before.

I left the code otherwise unchanged.

Note that the three loops together span 214221 iterations!

[Ahmed Sulaiman Ahmed Al Rawahi]

Hello Star Strider,

I hope you are doing well. Sorry to send this late, but I tried to run the code and it show me " Unrecognized variable y0". I want help running the code please.

Here is the code I have,

function [] = call_func

clear all

close all

numrows = numel(y0)*numel(x0)*numel(theta0);

xmtx = zeros(numrows,2);

cmtx = zeros(numrows,3);

k = 0;

for k = 1:length(y0)

for s = 1:length(x0)

for w = 1:length(theta0)

k = k + 1;

[t,x] = ode45(@func,tspan,[x0(s) y0(k) theta0(w)],options);

%legend('b = blue', 'r = red')

if (max(mod(x(:,1),l))<0.5) && (min(mod(x(:,2),l))>0.5) || (max(mod(x(:,1),l))<0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (max(mod(x(:,2),l))<0.5) || (min(mod(x(:,1),l))>0.5) && (min(mod(x(:,2),l))>0.5)

cv = [1 0 0];

% c='r.';

else

cv = [0 0 1];

% c='b.';

end

cmtx(k,:) = cv;

xmtx(k,:) = [mod(x(end,1),l), mod(x(end,2),l)];

% hold on

end

end

end

figure

scatter(xmtx(:,1), xmtx(:,2), 5, cmtx, 'filled')

end

function dydt = func(t,x)

Vs = 0.05;

alpha =1;

B = 0.12;

w = 6.28;

tau = 80;

L = 1;

dydt = [(sin(2*pi*(x(1)-(B/L)*sin(w*tau*t))*cos(2*pi*x(2)) +Vs*cos(x(3))))

( cos(2*pi*(x(1)-(B/L)*sin(w*tau*t)))*sin(2*pi*x(2)) +Vs*sin(x(3)))

(2*pi*alpha*sin(2*x(3))*cos(2*pi*(x(1)-B/L*sin(w*tau*t)))*cos(2*pi*x(2))+2*pi*sin(2*pi*(x(1)-B/L*sin(w*tau*t)))*sin(2*pi*x(2)))];

end

[Star Strider]

The ‘y0’ vector is in the original posted code:

y0 = 0:0.01:1;

and this ode45 call simply selects one element of it as part of the initial conditions vector:

[t,x] = ode45(@func,tspan,[x0(s) y0(k) theta0(w)],options);

exactly as in the code originally posted.

Import relevant parts of the original code into the function (or pass them as arguments to it) and the code should run without error.

The only changes I made were to change ‘dydt’ so it would run without error, introduce the ‘k’ counting variable, and preallocate the matrices that reduced the run time by >20%.

Also, these:

clear all
close all

have no effect inside the function, since the function has its own workspace that is automatically cleared (since there are no persistent variables) whenever it is called.