Problem in using ddesd for a simple DDE with one dependent variable and two delay terms.

Let's say we have only one independent variable t and one dependent variable x. Then consider the following DDE equation

and with the history

I guess I have to use ddesd as one of my term delays is not a constant delay, i.e. t-cos(t).

Reading the help page of ddesd (https://uk.mathworks.com/help/matlab/ref/ddesd.html) didn't make it clear for me how to do it. Here is my attempt.

sol = ddesd( @dde_equation, @delay, @history, [ 0, 1 ] );
t = sol.x;
x = sol.y;
plot(t, x);
xlabel('t');
ylabel('x(t)');
title('Solution of Delay Differential Equation');
% local functions
function dxdt = ddefun( t, x, Z )
    dxdt = t * Z( 1, 1 ) + ( t ^ 2 ) * Z( 1, 2 ) + t ^ 3;
end
function d = delay( t )
    d = [ delay1( t ); delay2( t ) ];
end
function d1 = delay1( t )
    d1 = t - cos( t );
end
function d2 = delay2( t )
    d2 = t - 2;
end
function v = history( t )
    v = sin( t );
end

But I get the following error message

Error using Matlab_20230517_DDE_1>delay
Too many input arguments.

Error in ddesd>lagvals (line 549)
  d = delays(tnow,ynow);

Error in ddesd (line 146)
Z0 = lagvals(t0,y0,delays,history,t0,y0,[]);

Error in Matlab_20230517_DDE_1 (line 1)
sol = ddesd( @dde_equation, @delay, @history, [ 0, 1 ] );

I would appreciate if someone let me know what is my mistake and how to solve the above DDE using Matlab. I found help pages and examples of DDE in Matlab help confusing and not really well explained.

Answer 1

[Torsten]

sol = ddesd( @ddefun, @delay, @history, [ 0, 1 ] );
t = sol.x;
x = sol.y;
plot(t, x);
xlabel('t');
ylabel('x(t)');
title('Solution of Delay Differential Equation');

% local functions
function dxdt = ddefun( t, x, Z )
dxdt = t * Z( 1, 1 ) + ( t ^ 2 ) * Z( 1, 2 ) + t ^ 3;
end
function d = delay( t, x )
d = [ delay1( t ); delay2( t ) ];
end
function d1 = delay1( t )
d1 = t - cos( t );
end
function d2 = delay2( t )
d2 = t - 2;
end
function v = history( t )
v = sin( t );
end

[Amirhossein Sadeghi Manesh]

@Torsten Thanks. Just to make sure that I understood right. I had one typo (dde_equation --> ddefun) and the other problem is that the delay function MUST have "x" as argument as well for the sake of ddesd. Am I right?

[Torsten]

Yes. Or you have to modify the list of input arguments to "delay":

sol = ddesd( @ddefun, @(t,x)delay(t), @history, [ 0, 1 ] );

and leave the function "delay" as in your posted code.