Variational calculus with bounded derivative constraints

[retrofit81]

[SOLVED] Variational calculus with bounded derivative constraints

After learning about the calculus of variations and optimal control for a bit this semester, I've decided to tackle a "simple" (in the words of my professor) problem meant to illustrate a simplified example of highway construction.

Suppose the cost of constructing a road of length L over uneven terrain is proportional to the difference between the unknown height of the road and the known height of the existing terrain, i.e. COST = k*abs(y(x) - h(x)) for some constant k>0 and non-negative functions y(x) (the height of the road to be built at point x, 0<=x<=L) and h(x) (the known height of the terrain at point x, 0<=x<=L).

I also want to consider the constraints

1. abs(y'(x)) <= M1 > 0,

implying that the grade (slope) of the road is never too excessive, and

2. abs(y''(x)) <= M2 > 0,

implying that the grade itself doesn't change too rapidly.

Finally, the terminal values y(0) and y(L) are free.

Hence the problem is

minimize integral k*abs(y(x) - h(x)) dt on the interval [0,L]

subject to abs(y'(x)) <= M1, abs(y''(x)) <= M2, y(0) = y_0 free, y(L) = y_L free.I'm pretty comfortable with the methods of solving these problems, but this one has me stumped because of the constraints. My textbook doesn't feature anything about incorporating bounded derivative constraints for a CoV problem. Could someone run me through the basics, or maybe point me in a direction?

Would it be easier to formulate this problem via optimal control? If so, I could use a little nudge there too.

Again, I won't need help actually determining the solution, outside of figuring out how to incorporate these constraints. I hope I'm not missing something glaringly obvious. Thanks in advance!

 

[EnumaElish]

I'll have a guess, y(x) = h(x) is an (the?) answer. If so, aren't you jumping thorough many hoops for formality's sake? That is, doesn't your cost function, along with the constraints, imply that it is cheaper to flatten out the entire island of Manhattan before putting roads on it?

 

[retrofit81]

y(x) = h(x) totally seems to be a candidate solution, but to me this doesn't imply that you flatten everything in your path out. Rather, you'd simply follow the contour of the given terrain exactly, neither removing nor adding dirt. It would minimize cost at zero dollars, assuming the constraints were met.

Because there's no guarantee that the terrain h(x) itself allows for the bounds on y' and y'' to be automatically satisfied by simply following the terrain's contour (as a very simple example, the terrain could be h(x) = 5x, but a constraint on y'(x) could potentially be abs(y'(x))<=4), it doesn't seem plausible to infer, in those cases, that y(x) = h(x) is even a candidate solution.

If I'm thinking about it incorrectly, forgive me and maybe we can clarify the issue.

 

[EnumaElish]

I simply misunderstood. One suggestion is to try to express y' and y" in terms of y. For ex., when conjecturing a linear solution, y = a + bx; y' = b; so y = a + y' x which implies y' = (y - a)/x.

Also see "Formal Statement of Necessary Conditions for Minimization Problem" under http://en.wikipedia.org/wiki/Pontryagin's_maximum_principle

 

[retrofit81]

Thanks for your help! :) I've discovered that these kinds of constraints are called "non-holonomic", which has in turn improved my research on the matter. That, coupled with your link on PMP, should be sufficient for now. Thanks again!

 

[EnumaElish]

I was also going to suggest thinking about the following proposition: Let y = f be a solution of the UNconstrained problem. Suppose for every x for which f(x) does not satisfy either constraint, I substitute the value of the constraint for the function in violation. Thus:

y = f except:
y' = M1 where f' > M1
y" = M2 where f" > M2,

and see whether that would solve the constrained problem.