We first introduce some notation and then begin our adventure with a brief introduction to the simplest problem of calculus of variations. We first consider which is assumed to be a twice differentiable with respect to all of its variables, i.e. belongs to . We denote Piecewise Smooth functions on the open interval by . It is understood that if , then

(1)

where is piecewise continuous function,, on and is a constant .

Now let denote the space of all real values piecewise smooth functions defined on . For each function , define the by

.

Assume the points are given and define the set of functions by

.

Note the is a real valued function on .

Our objective is to maximize or equivalently minimize the functional given above on the set . It turns out that many optimization and control problems end up in the form of a functional like the one we defined. So if we can learn how to minimize such a functional on the set of interest, then we can solve a great deal of problems that are very much practical. For example the equations of motion of the Double Pendulum can be found explicitly using Calculus of Variations methods very quickly. However, before we embark on our journey, we first introduce the Holy Grail of Calculus of Variations, a beautiful result , a mathematical jewel:The Lemma of Du Bois Reymond.

Lemma 1:

Part (A)

If is piecewise continuous on and

(2)

, then is a constant on except at a finite number of points. That is a constant and a set of points in such that with The converse is true.

Part (B)

If and are on and

, then a constant such that

The converse holds. Furthermore, is , and at points where is continuous

.

Now we proceed to proof part A .

Proof:

If is to be a constant in the sense stated in the lemma above, then it must satisfy the following equation,

(3)

Since is required to vanish at the and , it follows that

we observe that using the hypothesis (2) implies

(4)

We observe that the function is PWS by defintion (1). Clearly by (3); hence has the properties required in the part A of the lemma. Moreover, except for a possible finite subset of . After substituting in (4) we have that

(5)

Let be a point at which is continuous. If , then must differ from on some positive subinterval of positive length. Integral (5) cannot then vanish as stated and, from this contradiction, we infer that , hence whenever is continuous. This establishes the proof. We leave the part B and converse of part A as an exercise to the reader. The proof should require no more than integration by parts and introductory calculus.

UPCOMING ATTRACTIONS

Euler’s Necessary condition and some problems (already).

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February 7, 2010 at 4:53 am

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