# Boundary value problem - constrained paramter

1. Nov 25, 2012

### Deadstar

Let's say I have a set of nonlinear differential equations of the form.

$$x' = f(x,y) \\ y' = g(x,y)$$

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. Nov 26, 2012

### pasmith

If I understand correctly, the system you have is this:
$$x' = f(x,y,a) \\ y' = g(x,y,a)$$
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
$$X(a) - x_1 = 0, \\ Y(a) - y_1 = 0$$
where $x_1$ and $y_1$ 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 $X(a) = x_1$, 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)).

3. Nov 27, 2012

### Deadstar

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 $$x_{n+1} = a x_n$$ 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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