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Astron. Astrophys. 348, 38-42 (1999)

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Appendix A: completeness of [FORMULA]

In this appendix we will verify explicitly that the set of functions defined in Eq. (15) is complete, in the sense that Eqs. (3) and (4) can be recovered. For the purpose, we will apply the Fourier theorem (Brezis 1987). In the following, we will assume that [FORMULA] is a smooth vector field (we stress that this condition is needed only for the proof provided below; the method remains applicable to more general cases).

Let us consider a solution of the form (6). Then, because of the orthonormality condition (13), we have [FORMULA], so that

[EQUATION]

Now we observe that the previous equation holds for any [FORMULA]: then it holds also for any linear combination [FORMULA] of [FORMULA]. Thus

[EQUATION]

In the last step we have integrated by parts ([FORMULA] is the unit vector orthogonal to the boundary [FORMULA] of [FORMULA]).

We now use this equation to show that the chosen set of functions, described by Eq. (15), is complete , while, e.g., a similar set made of sine functions would not be complete. By the nature of the chosen set of functions we already know that we can properly represent any smooth function f. Using this property, we want to show that the two terms [FORMULA] and [FORMULA] entering in the r.h.s. of Eq. (A2) vanish on [FORMULA] and on [FORMULA] respectively.

For the purpose, we observe that if cosines are used as set of functions, we can "build" any function f provided that the function has periodic derivatives on the boundary. In particular, if [FORMULA] is an arbitrary open subset of [FORMULA], there is a function f that is positive on A and vanishes on [FORMULA]. Now suppose per absurdum that the solution obtained from the direct method does not satisfy Eq. (3), so that, e.g., [FORMULA] on a point [FORMULA]. Then, for the sign persistence theorem, this quantity must be strictly positive in a neighborhood A of [FORMULA]. However, if we take a function f which is positive on A and vanishes elsewhere, the rhs of Eq. (A2) will be positive, while the lhs vanishes, which is contradictory. This proves that Eq. (3) is verified by cosines.

In a similar manner, now that we have "disposed of" the first term, we observe that using cosines we can build a function f that vanishes everywhere on the boundary of [FORMULA] except for a neighborhood. In other words, given an open subset [FORMULA] of the boundary [FORMULA], there is a function f that is positive on B and vanishes on [FORMULA]. Note that this property would not be satisfied if the set of functions [FORMULA] were based on sines. Using a proof similar to the one given above, we obtain that [FORMULA] must vanish, thus leading to Eq. (4).

One might worry that, on the boundary, the chosen set of functions of Eq. (15) has zero derivative in the direction of [FORMULA], i.e. [FORMULA]. This might suggest that the boundary condition (4) cannot be reproduced. In reality, although this is true pointwise , this does not affect the convergence in [FORMULA] that is relevant for our problem (see Eqs. (7) and (12)). This important point is best illustrated by the following example.

Suppose that we measure a constant field [FORMULA] on a square field [FORMULA] of side [FORMULA]. The obvious solution for [FORMULA] in this case is [FORMULA]. Suppose now that we try to use a set of functions made of sines. Then the corresponding coefficients [FORMULA] would be proportional to the integrals [FORMULA]. Hence, every coefficient [FORMULA] vanishes. This proves that a set based on sines is not complete (condition (12) is not satisfied or, equivalently, a curl-free vector field [FORMULA] leads to a vanishing mass map). On the other hand, if we use the set (15), the coefficients [FORMULA] do not vanish and the corresponding mass distribution is given by

[EQUATION]

This function can be shown to reduce to [FORMULA].

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© European Southern Observatory (ESO) 1999

Online publication: July 16, 1999
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