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Boundary value problem - constrained paramter

  1. Nov 25, 2012 #1
    Let's say I have a set of nonlinear differential equations of the form.

    [tex]x' = f(x,y) \\
    y' = g(x,y)[/tex]

    Where f and g contain some parameter 'a' that is constrained to within certain values.

    Let's say I know x(0), y(0) and x(T), y(T) where T isn't a set value. What methods can I use to solve/integrate this to match the boundary conditions with the parameter 'a' free to change. I suppose if T could be minimized that would be nice but it's not essential, just looking for general methods used to solve these sort of problems.
     
  2. jcsd
  3. Nov 26, 2012 #2

    pasmith

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    Homework Helper

    If I understand correctly, the system you have is this:
    [tex]
    x' = f(x,y,a) \\
    y' = g(x,y,a)
    [/tex]
    where a is an unknown constant lying between certain values. You are given values of x(t) and y(t) at t=0 and t=T, and you need to determine a.

    This is essentially a root-finding problem: given x(0), y(0) and T, the value x(T) is then a function of a. Let's call it X(a). Similarly for Y(a). Thus your problem is to find a such that
    [tex]
    X(a) - x_1 = 0, \\
    Y(a) - y_1 = 0
    [/tex]
    where [itex]x_1[/itex] and [itex]y_1[/itex] are the given values for x(T) and y(T).

    Let's concentrate on the first of those. If we can find a such that [itex]X(a) = x_1[/itex], we can check whether the second equation is satisfied; if it isn't then there is no solution.

    There is no general analytic method for this; it must be done numerically.

    There are a number of root-finding algorithms which may be suitable, but the first step is always to obtain a graph of X(a) for suitable a, and see whether a solution is likely to exist. In general, there is no guarantee that a solution exists, and no guarantee that a solution is unique if it does.

    Given a, you find X(a) (and Y(a)) by solving the ODEs numerically subject to the given conditions at t = 0 and with the given value of a to determine x(T) (and y(T)).
     
  4. Nov 27, 2012 #3
    Hi thanks for the reply.

    That's sort of what I'm after but it seems like that solution will give a fixed choice of a such that the boundary conditions will be met. But what if a can change?

    Let's use a discrete dynamical system as an example to show what I mean.

    Let [tex]x_{n+1} = a x_n[/tex] where a = {2,3}

    We have x(0) = 1, x(T) = 12.

    Clearly having a fixed at 2 or 3 throughout will not give match the boundary conditions (a = 2: 1 -> 2 -> 4 -> 8 -> 16 ..., a = 3: 1 -> 3 -> 9 -> 27 ...)

    But if we change a...

    a = 2

    1 -> 2 -> 4

    a = 3

    4 -> 12.


    So this might be a bad example since I'm literally thinking it out as a type this but going back to the original problem. If a is fixed, we may not be able to find a solution, but the selection of a may still be valid at x(0), it will just change in time.

    Perhaps this should be solved by some sort of multiple shooting method? Such that a can change in each time interval.
     
    Last edited: Nov 27, 2012
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