Solving a delay differential equation with ddesd

Hello,

as the title suggest I am trying to solve a specific type of dde with built-in function ddesd.

The Problem is as follows: The dependent variable is y. The independent variable x is discretized with grid points n=1,…,N. The DDE is of the form dy/dx=f(y(n), y(k), x(n) ,x(k)), where k with k<n is a previous grid point that has to be determined from a side condition at each grid point n (i.e. k varies). The side condition also contains y(n), y(k), x(n) and x(k).

This type of problem can easily be solved by the Euler method. A very reduced (and nonsensical) example for this is below. However, I want to solve it with a higher order method. I tried ddesd but the issue is that I do not know how to obtain y(n), y(k), x(n) and x(k) and how the delay function should be used. How can I get ddesd running for this type of problem? Or is there another preferable solver?

Any help would be greatly appreciated!

Iter=1000;
x=linspace(0,10,Iter);
step=x(2)-x(1);
m=5; %at this grid point integration starts
y=[10.1,10.5,11.6,12.1,13.3]; %specify history of y for grid points n=1:m

for n=m:Iter

    for k=1:n

       if y(n)/y(k)-x(n)/x(k)>0 %If this side condition is fulfilled, we have found k 

           break %break inner loop in this case

       end

    end

    k_n(n)=k; %collect values of k at each iteration

    y_p=y(k)/y(n)+exp(x(k)*x(n)); %derivative

    y(n+1)=y_p*step+y(n); %Euler method

end

subplot(2,1,1)
plot(x,y(1:n))
ylabel('y')
xlabel('x')
subplot(2,1,2)
plot(x(m:n),k_n(m:n))
ylabel('k')
xlabel('x')

Answer 1

[Torsten]

In my opinion, there is no available DDE solver that can cope with your problem.

The problem is that the length d of your delay at X does not only depend on y(X), but on the complete history y(x) for x<=X.

In your example from above,

d(x,y(x)) = sup (d'>0) {y(x)/y(x-d')-x/(x-d') > 0}

As you can see, y(x-d') for all 0<=d'<=x must be available to compute d(x,y(x)).

Best wishes

Torsten.

[More_Qs_than_As]

Ok. I kind of expected that. Thank you for your reply.

Answer 2

[More_Qs_than_As]

anyone?