Re: Boundary value problem.

Peter Spellucci wrote:
In article <f2hui6$gio$1@xxxxxxxxxxxxxxxxxx>,
Amit Bhatia <abhatia@xxxxxxxxxxxxxxxxx> writes:
>
>Hi,
> I am trying to identify the parameters for the following two point >boundary value problem:
>
>y' = f(x,y)=[cos(y(3));sin(y(3));u(y)]
>where -1<=u(y)<=1,
>u(y)= c1*y(2)-c2*y(1)+c3 OR sign(c1*y-c2*x+c3) , c1,c2,c3 are constants >to be determined.
>
>y(a), y(b) are given to me.
>
>It is a given that there exists smooth differentiable solution to the >problem, but there can be more than one such solutions.
>
>
>If I solve this problem using bvp4c in matlab, I get the error
>"Unable to solve the collocation equations -- a singular Jacobian >encountered"
>
>So apparently my initial guess is not good or there is some other >problem. I was wondering if I should be doing something else or using >something different then this Matlab routine?
> The problem is more of parameter estimation probably.
>
>thanks,
>--a.
what does this "OR" mean : two cases , a switch in the definition of y(3)' or what? the sign-case might be harmful if you have a bad initial guess and a sign change occurs, such that a jump occurs in the course of integration
this will led to failure. also, if there exists several (isolated) solutions
then somewhere the jacobian of the system must become singular, and again with a bad choice you will get trouble. did you try "shooting" to get an initial guess?
maybe if you begin with a low precision requirement first and then increase this
step by step, using interpolation of the old solution to get the new guess?
hth
peter

OR means that when the value of u(y) gets saturated we use the maximum/minimum possible, i.e.,

u(y)=c1*y(2)-c2*y(1)+c3, if -1 < c1*y(2)-c2*y(1)+c3 < 1
=1 if c1*y(2)-c2*y(1)+c3 >1
=-1 if c1*y(2)-c2*y(1)+c3 < -1

I am not sure what you meant by initial guess: I have y(0), y(T) already specified. So I need the constants c1, c2, c3 only. I do start with an initial guess for them, but that gives me that Jacobian is singular.
I do not think there are any jumps, however there is non linearity in form of saturation. Also, we can have sines and cosines in solutions, so I am not sure but it may cause problems?
In any case, it is a given that the solution y() is continuous and differentiable.

--a.
.

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