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

I am trying to solve delay logistic equation with multiple delay terms and after getting solution i am taking derivative of this function but for i am getting error and my error is

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

g = @(t, y, Z, par) par(1) * y * (1 - sum(par(2:end) .* Z));
tau = [1, 1.5,2,2.5,3]; % Array of different delays
par = [1.5, 0.1,0.2,0.3,0.4,0.5];

this is my equation

x_t = deval(sol, t_t);
%% Add noise to state terms
eps = 0; 
x_tn= x_t + eps * randn(size(x_t)) .* x_t;
%% Calculate delayed states for multiple delays
x_d = cell(length(tau), 1);
for k = 1:length(tau)
    x_d{k} = deval(sol, t_t - tau(k));
end
x_dn = cell(size(x_d));
for k = 1:length(x_d)
    x_dn{k} = x_d{k} + eps * randn(size(x_d{k})) .* x_d{k};
end

and I am getting error here in this part

%% Compute derivative
DX = zeros(1, length(t_t)); 
for i = 1:length(t_t)
    Z = cellfun(@(zd) zd(:,i), x_d); % Collect all delayed states at t_t(i)
    DX(i) = g(t_t(i), x_t(i), Z, par);
end
DX=DX';

[Dyuman Joshi]

Please provide the values for variables that you have not defined above suchas - sol, t_t, par etc.

[Muhammad]

sol = dde23(@(t,y,Z) g(t,y,Z,par), tau, phi, tspan);

%% Split into training and testing

tr_r = 0.8;

t_all = sol.x;

max_tau = max(tau); % Find the maximum value in tau

t_all = t_all(t_all >= max_tau);

% t_all = t_all(t_all >= tau);

n_all = length(t_all);

n_train = round(tr_r * n_all);

t_t = t_all(1:n_train);

par = [1.5, 0.1,0.2,0.3,0.4,0.5]; %[growth rate, w1,w2,w3,w4,w5];

[Dyuman Joshi]

phi and tspan are not defined. Please share their values too.

[Muhammad]

phi = @(x) cos(x); and tspan=[0 20];

Answer 1

[Dyuman Joshi]

The output of g(....) is a 5x1 numerical array and DX(i) is a 1x1 numerical element. And you can not assign 5x1 numerical array into 1x1 numerical placeholder.

You can either define DX as a cell array (as you have done for x_d and x_dn) or preallocate a numeric array based on the size of x_d. I've chosen the latter option here -

g = @(t, y, Z, par) par(1) * y * (1 - sum(par(2:end) .* Z));
tau = [1, 1.5,2,2.5,3]; % Array of different delays
par = [1.5, 0.1,0.2,0.3,0.4,0.5];

phi = @(x) cos(x);
tspan=[0 20];

sol = dde23(@(t,y,Z) g(t,y,Z,par), tau, phi, tspan);

%% Split into training and testing
tr_r = 0.8;
t_all = sol.x;
max_tau = max(tau); % Find the maximum value in tau
t_all = t_all(t_all >= max_tau);
% t_all = t_all(t_all >= tau);
n_all = length(t_all);
n_train = round(tr_r * n_all);
t_t = t_all(1:n_train);

x_t = deval(sol, t_t);

%% Add noise to state terms
eps = 0;
x_tn= x_t + eps * randn(size(x_t)) .* x_t;

%% Calculate delayed states for multiple delays
[x_d, x_dn] = deal(cell(length(tau), 1));

%Club the for loops together
for k = 1:length(tau)
    arr = deval(sol, t_t - tau(k));
    x_d{k} = arr;
    x_dn{k} = arr + eps * randn(size(arr)) .* arr;
end


%% Compute derivative
DX = zeros(size(x_d,1), length(t_t)); 

for i = 1:length(t_t)
    Z = cellfun(@(zd) zd(:,i), x_d);
    DX(:,i) =  g(t_t(i), x_t(i), Z, par);
end
DX=DX.'

DX = 86×5

-0.0860 -0.2220 -0.3579 -0.4938 -0.6297 -0.0537 -0.1382 -0.2226 -0.3070 -0.3914 -0.0342 -0.0876 -0.1411 -0.1946 -0.2481 -0.0222 -0.0569 -0.0916 -0.1263 -0.1610 -0.0147 -0.0376 -0.0605 -0.0834 -0.1063 -0.0098 -0.0251 -0.0404 -0.0557 -0.0710 -0.0066 -0.0169 -0.0272 -0.0375 -0.0478 -0.0045 -0.0115 -0.0185 -0.0255 -0.0325 -0.0030 -0.0078 -0.0126 -0.0174 -0.0221 -0.0021 -0.0053 -0.0086 -0.0119 -0.0151

[Muhammad]

once again thank you so much

[Dyuman Joshi]

You're welcome! Glad to have helped :)