I couldn't understand what's wrong in this.

%Unable to perform assignment because the left and right sides have a different number of elements.

clc
%defining constant
ti = 0; %inital time
tf = 100E-4;% final time
tspan=[ti tf];
o = 2E5;  % detuning frequency
tc = 30E-9;  %photon life time in cavity
tf = 230E-6;  %flouroscence lifetime
a1 = 0.1; %round trip loss
a2 = 0.1;
P1 = 0.2;  %pump strenght
P2 = 0.2;
kc = 3E-3; %critical coupling strength
%s =  0.17;
%k = s.*kc;
I = 1; %saturation intensity
V = 1;
% define function
%y(1) = A1
%y(2) = A2
%y(3) = G1
%y(4) = G2
%y(5) = phase difference
f = @(t,y,k) [  
    (((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;
    
    (((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;
    
    (P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;
    
    (P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;
    
    o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));
    
    ];
k =0:0.01E-3:3.2E-3;
Vf = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));
end



k =[0:0.01E-3 : 3.2E-3] ;
Vp = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;
end

%maybe here is the problem
k =[0:0.01E-3 : 3.2E-3] ;
T = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))
end

Unable to perform assignment because the left and right sides have a different number of elements.

subplot 221
plot((k./3E-3),Vf)

subplot 222
plot((k./3E-3),Vp)

subplot 223
plot((k./3E-3),T)

[Abderrahim. B]

Hi!

You can not assign a matrix of size 3501 x 321 to a matrix (row vector) of size 3501x1.

% The error is in this line:
T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));

I recommend to debug your code.

HTH

Answer 1

[Walter Roberson]

for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    T(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))
end

Your tspan is a vector of length exactly 2. When tspan is a vector with exactly two elements, ode45() will return a vector of T values and a 2D array of Y values, with it internally deciding which times to emit results for. The number of entries returned is relatively unpredictable. The number of rows of Y will be the same as the number of entries in T..

You then index particular columns of the 2D array of Y values -- making calculations for all of the times. The number of results on the right side will be the same as the number of times.

You then try to overwrite one particular time entry with the vector calculated on the right hand side.

Note that before the for loop you pre-allocated T, but your call to ode45() then overwrites all of T. You probably want to use different variables there.

On the other ode45 loops, you were calculating mean() of values, and so only returned a single value rather than a vector. And in the other loops, you did not accidentally use the same variable for two different purposes.

[SAHIL SAHOO]

ti = 0; %inital time
tf = 100E-4;% final time
tspan=[ti tf];
o = 2E5;  % detuning frequency
tc = 30E-9;  %photon life time in cavity
tf = 230E-6;  %flouroscence lifetime
a1 = 0.1; %round trip loss
a2 = 0.1;
P1 = 0.2;  %pump strenght
P2 = 0.2;
kc = 3E-3; %critical coupling strength
%s =  0.17;
%k = s.*kc;
I = 1; %saturation intensity
V = 1;
% define function
%y(1) = A1
%y(2) = A2
%y(3) = G1
%y(4) = G2
%y(5) = phase difference
f = @(t,y,k) [  
    (((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;
    
    (((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;
    
    (P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;
    
    (P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;
    
    o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));
    
    ];
k =0:0.01E-3:3.2E-3;
Vf = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));
end



k =[0:0.01E-3 : 3.2E-3] ;
Vp = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;
end


k =[0:0.01E-3 : 3.2E-3] ;
T = zeros(length(k),1) ; 
for i = 1:length(k)
    [T,Y]= ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
end


k =[0:0.01E-3 : 3.2E-3] ;
M = zeros(length(k),1) ;
for i = 1:length(k)
    M(i) = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));
end


subplot 221
plot((k./3E-3),Vf)

subplot 222
plot((k./3E-3),Vp)

subplot 223
plot((k./3E-3),M)


%is this what you mean?

[Sam Chak]

Hi Sahil,

A brief explanation of what you are trying to simulate will be helpful.

You have a system of ODEs with 5 states and a variable k. Without talking about the looping, is k a constant, or time-varying k(t) that depends on the time t?

[Walter Roberson]

N = 50;

ti = 0; %inital time
tf = 100E-4;% final time
tspan = linspace(ti, tf, N);

o = 2E5;  % detuning frequency
tc = 30E-9;  %photon life time in cavity
tf = 230E-6;  %flouroscence lifetime
a1 = 0.1; %round trip loss
a2 = 0.1;
P1 = 0.2;  %pump strenght
P2 = 0.2;
kc = 3E-3; %critical coupling strength
%s =  0.17;
%k = s.*kc;
I = 1; %saturation intensity
V = 1;
% define function
%y(1) = A1
%y(2) = A2
%y(3) = G1
%y(4) = G2
%y(5) = phase difference
f = @(t,y,k) [  
    (((y(3)./V) - a1) * y(1) + k * y(2) * cos(y(5))) ./ tc;
    
    (((y(4)./V) - a2) * y(2) + k * y(1) * cos(y(5))) ./ tc;
    
    (P1 - ((y(3)./V) * (((abs(y(1)))^2 ./I) + 1))) / tf;
    
    (P2 - ((y(4)./V) * (((abs(y(2)))^2 ./I) + 1))) / tf;
    
    o - (k / tc) * (((y(1) ./ y(2)) + (y(2) ./ y(1))) * sin(y(5)));
    
    ];

k = 0 : 0.01E-3 : 3.2E-3;

Vf = zeros(length(k),1) ;
Vp = zeros(length(k),1) ; 
M = zeros(length(k),N) ;

for i = 1:length(k)
    [T,Y] = ode45(@(t,y)f(t,y,k(i)),tspan,[1;1;1;1;0].*10E-2);
    Vf(i) = (2.*(((mean(Y(:,1).*Y(:,2).*cos(Y(:,5)))).^2 + (mean(Y(:,1).*Y(:,2).*sin(Y(:,5)))).^2).^0.5))/(mean((Y(:,1)).^2)+mean((Y(:,2)).^2));
    Vp(i) = ((mean(cos(Y(:,5)))).^2 + (mean(sin(Y(:,5)))).^2).^0.5;
    M(i,:) = 1 - (k(i)./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)));
end

subplot 221
plot((k./3E-3),Vf)

subplot 222
plot((k./3E-3),Vp)

subplot 223
plot((k./3E-3),M)

subplot 224
plot((k./3e-3), mean(M,2));

The third plot pretty much fills an area because for each k value, you have as many M values as you have time points (because you are not doing anything like mean() which you do for the other two calculations.) I added the 4th plot to show you what taking mean() of the M values over time might look like.

Notice that you do not need to keep redoing all those ode45, since you are running them under the same conditions (same k, same time span, same initial conditions.) So for each k value, you can run ode45 once and calculate Vf, Vp, and M values.

[SAHIL SAHOO]

i want to solve Y(:,1) for the different value of k and then by using Y(:,1) i want to calculate T for different value of k

this is the T

T = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))

[Walter Roberson]

Y(:,1) is a vector, with one entry for each output T (time) within the time span. Everything on the right hand side of

T = 1 - (k./kc).*(1/2).*(((Y(:,1) ./ Y(:,2)) + (Y(:,2) ./ Y(:,1))).* sin(Y(:,5)))

is a vector that size, except for the fact that you are using the entire vector k instead of the "current" k value. It seems likely that you want k(i) there instead of k

However.. T is the variable name that you are using to store the time variable output from the ode45 call. Does it make sense to be calculating one time for each timestep (for each k value) ?

Your code does not do anything with those T values that you calculate. Your code does, however, plot M, which you do not seem to be calculating now.

[SAHIL SAHOO]

plot((k./3e-3), mean(M,2));

why you mean by mean(M,2) ?

[Torsten]

From the documentation of "mean":

M = mean(A,dim) returns the mean along dimension dim. For example, if A is a matrix, then mean(A,2) is a column vector containing the mean of each row.

Answer 2

[KSSV]

The error is simple. You are trying to save more number of elements in LHS than it is initialized for.

Demo

A = zeros(1,3) ;   % 1x3 array intialized 
A(1) = rand ;  % no error, you can save one element in one place 
A(2) = rand(1,2) ; % error, you cannot save two elements in a sinlge element place 

Unable to perform assignment because the left and right sides have a different number of elements.

Similarly, check the dimensions in your case, where error occurs.