Appendix

There are several approaches to the calculation of the gradients. The one adopted here illustrates techniques that are applicable to other classes of functionals.

Lemma (duBois-Reymond). Suppose  and  for all  such that , then there exists a constant  such that .

 Proof. Given , define

and put

It follows that  as well as  and . See Theorem 7.20 page 148 of Rudin’s Real and Complex Analysis, (third edition, 1987 McGraw-Hill) for this particular direction of the fundamental theorem in the context of absolutely continuous functions. From this conclude that

This in turn implies that

and hence . ÿ

Remark. As usual, the notation  is used in the sense of  (on subintervals of  when needed.)

Unrestricted gradients. First look at the derivation in the case . When , then  leads to

Since this is true for all , it is true for all  such that . Apply the Lemma and conclude that there is a constant  such that . Integrate to get

Now , and

For each  it must be that

Let , and conclude that . Let  and conclude that . This transforms into

Hence,

Next look at the derivation in the case , which is more complicated.

First

When , then  leads to

Since this is true for all , it is true for all  ’s such that . Apply the Lemma and conclude that there is a constant  such that

Next choose  to show that

and then  to see that . Let

so that  and . The transformed equation is     . This is a non-homogeneous linear equation. The solutions have the form  where  is the general solution of the homogeneous problem. To find a particular solution , use the method of variations of parameters. To this end let , so that

Impose the condition , so that

This leads to the system

 with solution

 A particular solution is given by

This particular solution combined with the general homogeneous solution yields

Since  it follows that , and

At the other endpoint

Finally, Leibniz’ rule yields

Hence,

Fixed endpoints. To deal with fixed endpoints it is helpful to have gradient formulas for so-called point functionals. The technique is illustrated by the most difficult of the three cases considered. Suppose the metric is , then

Integrate the last two terms by parts

and

This leads to

The formulas for fixed endpoints are derived s follows, where the technique of the proof is illustrated in the case when the gradient is .

Proof (fixed endpoints). The projected gradient may be written as

It is also true that  and . Together the two equations form a linear system that determines  and . Specifically,

which is equal to

The solution is given by

With

this rewrites as

An impressive simplification produces

yields

ÿ

Projection Lemma.

Proof. Given  there is a unique decomposition  such that , and . Similarly write . With this notation it follows that

It is also true that

Now suppose that  is arbitrary. Since

so it follows that

Now

expresses  as a sum of a vector in , and a vector in . By uniqueness it must be that . It follows that . Obviously, , and hence . ÿ