delay differential equation - step size problem
I do have a problem with the following one-dimensional delay differential equation (dde):
(1) x'(t) = f(t,x(t-1)) := M * [ -x(t) - 2.35 * sin(x(t-1)) ] , t>0
with the initial condition
(I) x(t) = 2, -1 <= t <= 0.
If M is big, e.g. M = 100, this problem becomes stiff, as you can see in the plot of the attached *.jpg picture.
For this plot I solved (1) with a collocation method of order 4 with constant step size h = 0.001. As you can imagine, this is highly inefficient and I wish to apply a good step size control.
First, I like to explain briefly the mentioned collocation method:
Assume, that we have found an approximative solution s(t) for (1) with the initial condition (I) up to a number t(n)>0. Also, we have chosen a new step size h(n). Let
The collocation method of stage 3 (with Lobatto-points) consists in finding (the uniquely existing) polynomial
p(t) of degree less or equal 3 that satisfies:
(A) p(t(n)) = s(t(n)) [initial condition]
(B) p'(t(n)) = f(t(n),s(t(n)-1))
(C) p'(t(n)+0.5*h(n)) = f(t(n)+0.5*h(n),s(t(n)+0.5*h(n)-1))
(D) p'(t(n+1)) = f(t(n+1),s(t(n+1)-1))
(A)-(D) determines p(t) uniquely and you can find the coefficients of p(t) by simply doing a matrix multiplication.
This method is equivalent to the implicit Runge-Kutta (IRK) method with the coefficients
A = [ 0, 0, 0; 5/24, 1/3, -1/24; 1/6, 2/3, 1/6]
c = [ 0 ; 1/2 ; 1]
b = [1/6 , 2/3 , 1/6].
The big advantage of this collocation method is that you don't have to solve any implicit equations.
Finally, my problem:
I don't know how to find cheaply a method of order 3 (that works with stiff problems!!) to make some error estimation to choose the step size. As I used the collocation method, I do not know the values k(i), i=1,2,3 , of the mentioned IRK method.
So, we cannot just change the vector b to get a formula of order 3.
If anyone could give me any idea for an error estimation, I would be very, very grateful.
I thank you for your time!
^^ Your pilzgericht