Register to reply 
Boundary value problem  constrained paramter 
Share this thread: 
#1
Nov2512, 03:41 PM

P: 106

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
Nov2612, 09:50 AM

HW Helper
Thanks
P: 949

[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 rootfinding 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 rootfinding 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
Nov2712, 06:12 AM

P: 106

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. 


Register to reply 
Related Discussions  
Eliminate paramter  Calculus & Beyond Homework  2  
Constrained minimum problem  Calculus  1  
Constrained extreme problem  Calculus & Beyond Homework  6  
Constrained Minimization Problem(HELP ASAP)  Calculus & Beyond Homework  2  
Lagraingian constrained optimization problem  General Math  3 