#1




delay differential equation  step size problem
Hello everybody,
I do have a problem with the following onedimensional delay differential equation (dde): (1) x'(t) = f(t,x(t1)) := M * [ x(t)  2.35 * sin(x(t1)) ] , 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 t(n+1):=t(n)+h(n). The collocation method of stage 3 (with Lobattopoints) 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 RungeKutta (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 
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