Hello everyone I am working on a state dependent time delay - delay differential equation problem where my delay is defined as [x(t) - x(t-tau) = constant] I am trying to sove it with ddesd but am not sure how to code this delay function thanks and regards

question about coding a delay function for ddesd

Torsten 2022-6-11

Include your delay differential equation.

Mohammed Said 2022-6-11

编辑：Mohammed Said 2022-6-12

the full system of equations goes like this

x3(t) - x3(t-tau) = constant

Delay term = (x1(t) - x1(t-tau)

system of equations= x(2)

- (Delay term *x(2) + Delay term * x(4)) - (C1 * x(2)) - (C2 * x(1)) + C3*t

x(3)

-(Delay term *x(2) + Delay term * x(4)) - (C4 * x(4)) - (C5 * x(3)) + C6*t)]

Torsten 2022-6-11

And you say tau depends on the state variables - how is it calculated ?

Mohammed Said 2022-6-11

It is calculated through intepolating with the time index

x3 minus the constant will give you x3 (t-tau) through interpolatin we get t- tau and then we interpolate again with x1 to get x1 (t-tau)

Torsten 2022-6-11

编辑：Torsten 2022-6-11

You don't need to do this interpolation - the delay solver does this.

The only question is: is "tau" a constant during the integration or does it change with x1, x2, x3 or t ?

By the way: The condition x3(t) - x3(t-tau) = constant already fixes x3(t) for all t. So you cannot define it a second time by a differential equation.

Mohammed Said 2022-6-11

Tau changes as x3 changes

So you start with the history of states and based on how far x3 has progressed tau changes to keep the x3(t) - x3 (t-tau) constant

Torsten 2022-6-11

编辑：Torsten 2022-6-11

You prescribe x3 for -tau <= t < 0 as the history of x3.

Then, by the condition x3(t) = x3(t-tau) + constant, x3 is uniquely determined for t >=0. No differential equation for x3 is needed.

Mohammed Said 2022-6-11

编辑：Mohammed Said 2022-6-12

I had a mistake on the equations earlier the proper equations should be

x3(t) - x3(t-tau) = constant

Delay term = (x1(t) - x1(t-tau)

system of equations= x(2)

- (Delay term *x(2) + Delay term * x(4)) - (C1 * x(2)) - (C2 * x(1)) + C3*t

x(4)

-(Delay term *x(2) + Delay term * x(4)) - (C4 * x(4)) - (C5 * x(3)) + C6*t)]

in this system of equations x3 is a position and x4 is a velocity

the delay term couples them to x1 ( also a position) and x2 (a velocity)

sorry for the confusion

Torsten 2022-6-12

I doesn't change my answer.

You can't write a differential equation and an algebraic relation for x3 because using both will lead to a contradiction.

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