calculus of variations
Imagine a bead of mass on a wire whose endpoints are at and , with lower than the starting position. If gravity acts on the bead with force , what path (arrangement of the wire) minimizes the bead’s travel time from to , assuming no friction?
This is the famed brachistochrone problem, and its solution was one of the first accomplishments of the calculus of variations. Many minimum problems can be solved using the techniques introduced here.
In its general form, the calculus of variations concerns quantities
for which we wish to find a minimum or a maximum.
To make this concrete, let’s consider a much simpler problem than the brachistochrone: what’s the shortest distance between two points and ? Let the variable represent distance along the path, so that . We wish to find the path such that is a minimum. Zooming in on a small portion of the path, we can see that
If we parameterize the path by , then we have
Let’s assume , so that we may simplify (4) to
Now we have
In this case, is particularly simple. Converting to ’s and ’s to make the comparison easier, we have , not the more general covered by the calculus of variations. We’ll see later how to use our ’s simplicity to our advantage. For now, let’s talk more generally.
We wish to find the path described by , passing through a point at and through at , for which the quantity is a minimum, for which in the path produce no first-order change in , which we’ll call a “stationary point.” This is directly analogous to the idea that for a function , the minimum can be found where produce no first-order change in . This is where ; taking a Taylor series expansion of at , we find
with . Of course, since the whole point is to consider , once we neglect terms this is just the point where . This point, call it , could be a minimum or a maximum, so in the usual calculus of a single variable we’d proceed by taking the second derivative, , and seeing if it’s positive or negative to see whether the function has a minimum or a maximum at , respectively.
In the calculus of variations, we’re not considering in —we’re considering in the integral of the relatively complicated function , where . Also, is a functional, and we can think of the minimization problem as the discovery of a minimum in -space as we jiggle the parameters and .
For the shortest-distance problem, it’s clear the maximum time doesn’t exist, since for any finite path length we (intuitively) can always find a curve for which the path’s length is greater than . This is often true, and we’ll assume for this discussion that finding a stationary point means we’ve found a minimum.
Formally, we write the condition that produce no change in as . To make this precise, we simply write
How are we to simplify this mess? We are considering to the path, which suggests a Taylor series expansion of about :
and since we make little error by discarding higher-order terms in and , we have
Keeping in mind that and noting that
a simple application of the product rule which allows us to substitute
we can rewrite the integral, shortening to for convenience, as:
Substituting all of this progressively back into our original expression for , we obtain
Two conditions come to our aid. First, we’re only interested in the neighboring paths that still begin at and end at , which corresponds to the condition at and , which lets us cancel the final term. Second, between those two points, we’re interested in the paths which do vary, for which . This leads us to the condition
The fundamental theorem of the calculus of variations is that for continuous functions with ,
Using this theorem, we obtain
This condition, one of the fundamental equations of the calculus of variations, is called the Euler–Lagrange condition. When presented with a problem in the calculus of variations, the first thing one usually does is to ask why one simply doesn’t plug the problem’s into this equation and solve.
Recall our shortest-path problem, where we had arrived at
Here, takes the place of , takes the place of , and (8) becomes
Even with , this is still ugly. However, because , we can use the Beltrami identity,
(For the derivation of this useful little trick, see the corresponding entry.) Now we must simply solve
which looks just as daunting, but quickly reduces to
That is, the slope of the curve representing the shortest path between two points is a constant, which means the searched curve, i.e. the extremal of this variational problem, must be a straight line. Through this lengthy process, we’ve proved that a straight line is the shortest distance between two points.
To find the actual function given endpoints and , simply integrate with respect to :
and then apply the boundary conditions
Subtracting the first condition from the second, we get , the standard equation for the slope of a line. Solving for , we get
which is the basic equation for a line passing through and .
The solution to the brachistochrone problem, while slightly more complicated, follows along exactly the same lines.
|Title||calculus of variations|
|Date of creation||2013-03-22 12:20:48|
|Last modified on||2013-03-22 12:20:48|
|Last modified by||rspuzio (6075)|