aa r X i v : . [ m a t h - ph ] A p r Uniqueness Theorems for Point Source Expansions:DIDACKS V ∗ Alan RuftyNovember 28, 2007
Abstract
In the
Principia Mathematica
Sir Isaac Newton proved that concentric mass shells with equivalentmass distributions produce the same external gravitational field and thus that the problem of estimating acontinuous interior mass distribution from external field information alone is ill-posed. What is generallyless well known is that finite collections of point masses contained in some bounded domain producea unique field in the exterior domain, which means that the associated basis functions (often called“fundamental solutions”) are independent. A new proof of this result is given in this paper that can begeneralized to other finite combinations of point source distributions. For example, one result this papershows in R is that a finite combination of (gravitational or electrostatic) point dipole sources containedin some interior region produces a unique field in the corresponding exterior region of interest.Since no direct proofs of uniqueness results of this type are known for the R setting, which is thesetting of primary practical interest, an indirect strategy is necessary. The strategy employed in thepaper is to develop results for analytic functions in the complex plane, C , where logarithmic source basisfunctions correspond to point mass basis functions, and then carry them over to harmonic functions inthe real plane, R , and from there to harmonic functions in R . Although some of the results obtainedcan be generalized from R to R n , for n >
3, far more results are shown for R and for C than areshown for these more general settings. For example, in the complex plane, the paper shows that a finitecombination of higher order poles of any order in the interior of a unit disk always corresponds to aunique analytic function in the exterior of a unit disk. Key words:
Laplace’s equation, inverse problem, potential theory, point sources, Vandermodematrix, fundamental solutions, multipole
AMS subject classification (2000):
Primary 86A20. Secondary 35J05, 30E05, 86A22
Inverse source problems associated with harmonic functions (i.e., ones that satisfy Laplace’s equation) area small area of modern inverse source theory; however, this area encompasses problems in geophysics,geoexploration and electrostatics, as well as many other applied sciences. Moreover this area undoubtedlycontains the oldest substantial mathematical result of any real significance associated with inverse sourcetheory: Sir Isaac Newton showed in his
Principia Mathematica that a continuous spherically symmetricdistribution produces an external field equivalent to that of a point mass located at the sphere’s center, ∗ Approved for public release; distribution is unlimited. R n ( n ≥
2) that shows that a finite set of point masses located inside somebounded region produces a unique field in the exterior region [12]. This paper presents an alternative proofof this R n result, as well as various generalizations of it. One of these generalizations is that a finite set ofpoint dipoles is shown to be independent in R . Other generalizations of this result are shown in the complexplane, C , and the real plane, R . Finally, one open question is analyzed extensively here; namely, are linearcombinations of R combined point mass/point dipole basis functions linearly independent? In the end, theindependence of these basis functions is shown to hinge on the invertibility of a matrix with complex entriesthat is related to the Vandermode matrix [and which is specified by (59)]. Clearly, in general, this generalizedVandermode matrix is invertible, however, currently no proof of this invertibility for all n is known.In what follows, the term point source will be used to denote a general source term [i.e., a point massor (point charge), a point mass dipole (or an electrostatic dipole), a point quadrapole or a higher order(electrostatic) multipole in R n ( n ≥ C ].Because uniqueness results may seem to be somewhat removed from the applications arena, before con-sidering the paper’s basic approach and content in more detail, it is appropriate to briefly consider themotivations for studying point source uniqueness results. The primary motivations can be framed as follows:1. By better understanding and quantifying various point source uniqueness results, the hope is thatgeophysical inverse source theory can be better understood and placed on a firmer physical and math-ematical foundation. (Point source uniqueness results should be viewed as merely a first tentative stepin this goal.)2. Point source uniqueness results show that, in theory, when well formulated algorithms are properlyimplemented, point source determination software always produces reliable results. In particular,by showing that combinations of point sources and point dipoles are linearly independent, one canimmediately infer that the DIDACKS implementation that interpolate for the scalar potential of gravityand the vector components of gravity are mathematically well framed, as discussed in [6, 7, 8] and [9].3. As discussed in Appendix A of [8], this uniqueness result for point sources and point dipoles alsoshows that geophysical collocation procedures for similar data sets (such as geoid height and thevector components of gravity disturbance) are mathematically consistent since the associated error-free covariance matrix can always be inverted.4. Finally, as always, when a different topic is addressed, different mathematical techniques may be calledfor, and thus there is the possibility of uncovering new mathematical theorems, results and techniquesthat may be of interest in other domains. Here, not only are the results dealing with logs and poles in C suggestive, but so are several of the (geometric) theorems that allow for results to be carried over from R to R n . Also, since Vandermode’s matrix in an extended form is used here, it is not unreasonable tothink that there are mathematical links to general interpolation theory, where the Vandermode matrixis often employed, and that this is a two-way street (which means that there may be general theoremsin interpolation theory that allow for more general point source uniqueness theorems than those provenhere).Here item 1. is worth delving into a little deeper, since placing inverse source gravity theory (i.e., R Laplacianinverse source theory) on a firmer mathematical and physical foundation is one goal of the DIDACKSsequence of papers initiated by [7]. In this regard, there are a number of interconnected points that arerelevant, and it is appropriate to point one of them out here. First, consider the point made in [4] that since2ach point source can be replaced by an equivalent sphere of uniform density, one can replace a distributionof point sources with an equivalent set of spheres that produces a more uniform distribution of mass. Thus,consider a collection of point masses that have different strengths from point to point and are located on aregular cubic grid. Since the grid spacing is uniform each point mass can be replaced by a sphere of like sizethat has a corresponding density. From the uniqueness theorems given here one can infer that the estimationproblem for the associated densities of this spherical arrangement is well posed. Moreover, since it is wellknown that as the grid spacing for a set of point masses increases the underlying estimation problem becomesbetter conditioned, one can immediately infer that the same holds for this spherical approximation. Thisresult has other generalizations and interpretations as well. For example, in geoexploration it is commonto use a set of right parallelepipeds of uniform density to approximate continuous mass densities since theyproduce potentials that can be calculated in closed form [5, p. 398] and they yield inverse source densityestimation equations that are (relatively) stable. As a next step, one might consider extending the uniquenessresults here to uniform cubic latices (or one might consider applying DIDACKS theory to the estimation ofsuch cubic distributions). This and other points raised above clearly dovetail with similar discussion in [10].
The general method of proof employed is to start with the easiest to handle uniqueness case—which turnsout to be a distribution of simple poles of the form 1 / ( z − z k ) in the complex plane–and then generalize thisresult in various ways: First to uniqueness of logarithmic basis functions in C , then to higher order poles ofthe form 1 / ( z − z k ) n . Next a (well-known) one-to-one correspondence between various types of expansionsin C and R is noted: logarithmic potentials ⇐⇒ point masses; simple poles ⇐⇒ dipoles; poles of order n ⇐⇒ multipoles of higher order for n = 1 , ,
3. It is then shown that uniqueness of an expansion in C implies uniqueness of a like expansion in R —thus proving the R cases. Finally it is shown explicitly thatpoint mass uniqueness results in R imply point mass uniqueness results in R . These correspondences andthe symbols used to represent the associated expansions in the sequel are shown in Table 1. The argumentsemployed can clearly be generalized in two ways: (1) To show that similar results hold for multipoles oforder three or less in R . (2) To show that results obtained in R imply similar results in R N for N ≥ R and R the conservative vector fields of interest can be obtained by taking the gradient of a scalarfield: ~F ( ~X ) = −∇ U ( ~X ) , (1)where, of course, ∇ is the N − dimensional gradient and where ~X ∈ R N . Potentials specified by (1) satisfyLaplace’s equation in N − dimensions: ∇ U = 0. Arguably the most significant cases, and the ones that willbe focused on here, are the electrostatic and gravitational fields in R and R . Because of the issue of thesign associated with the mutual attraction of point sources and the issue of the sign of the energy densityplay no role in what follows, the only relevant difference between the gravitational and electrostatic casesis that point mass strengths ( m k ) are generally assumed to be non-negative, but electric change strengths( q k ) can have either sign. Since point source uniqueness results that hold for sources of either sign alsoclearly hold when the sources are all assumed to be positive and since a historical precedent exists for statinguniqueness results in terms of point masses, in the sequel it will always be assumed that while the scalarpoint source parameters are called a point masses and denoted m k , their assigned values can take on eithersign. Likewise the vector point source distribution that corresponds to a electrostatic dipole or point massdipole will be denoted ~D k and it will be called a point dipole. Higher order point multipoles will also be3 N Description R N Symbol C Analog C SymbolHarmonic Functions U Analytic Functions f Point Masses V Logarithm Terms g Point Dipoles W Simple Poles h Multipoles H ( n ) Higher Order Poles h ( n ) Table 1: Corresponding Types of Potential and Analytic Functionsconsidered in what follows and point masses, point dipoles and point multipoles will be collectively referredto as point sources. In all cases, since either sign is allowed for these point sources and since uniquenessresults pertain to whether the source terms form a linearly dependent set of basis functions in the exteriorregion of interest, the overall choice of sign convention for these basis functions and associated point sourcestrengths does not matter here and the basis sign conventions are chosen for convenience. As previouslynoted, while point source distributions for R N for N > N k denote a finite integer greater than zero and let the subscript k (or k ′ ) be used to index the N k point sources under consideration so that k and k ′ = 1 , , , . . . , N k is alwaysunderstood. To further fix notation in R N for N ≥
2, let the N k fixed distinct point-source locations bespecified by ~X k so that ~X k = ~X k ′ when k ′ = k . It will also be assumed that all of the sources are located insome bounded domain.Introducing a specific symbol for point mass potentials, V , instead of the general symbol U used in (1)results in the following definition V ( ~X ) ≡ N k X k =1 m k ln ( | ~X − ~X k | − ) (2)in R and V ( ~X ) ≡ N k X k =1 m k | ~X − ~X k | (3)in R .Likewise the point dipole potential form is W ( ~X ) ≡ N k X k =1 ~D k · ∇ ln ( | ~X − ~X k | − ) (4)in R and W ( ~X ) ≡ N k X k =1 ~D k · ∇ ( | ~X − ~X k | − ) (5)in R . The relationships of these potentials and their corresponding complex analytic counter parts areshown in Table 1. These correspondences will be addressed below.4ext consider uniqueness results stated in terms of scalar potentials. So long as Laplace’s equation issatisfied in the region of interest the exact shape of the exterior region is immaterial due to the uniquenessof Dirichlet boundary value problems so, without loss of generality, assume that the mass distributions arelocated in some bounding sphere. Further, since the origin of coordinates and the length scale also doesnot change, for the desired final uniqueness results it can be assumed that the harmonic region of interestis | ~X | ≥ < | ~X k | < V ( ~X ) = 0, for all ~X ≥
1, then m k = 0for all k and that the converse also holds. Which is to say that for any finite N k >
0, if m k = 0, for all k ,then V ( ~X ) = 0 for some values of ~X ≥
1. Dipole uniqueness can be stated similarly: If V ( ~X ) = 0, for all | ~X | ≥
1, then ~D k = 0 for all k , or conversely as the condition that for any finite N k > ~D k = 0, for all k , then V ( ~X ) = 0. It is assumed throughout that any nonzero values of m k and ~D k are bounded (while thefixed nature of ~X k rules out the formation of dipoles in the limit m k → ∞ and m k ′ → −∞ with | m k | = | m k ′ | for some pair of points indexed by k and k ′ , the unbounded mass case is still best bypassed).Several observations are relevant. By introducing the R basis functionsΨ k ( ~X ) = ln 1 | ~X − ~X k | − ln 1 | ~X | = ln | ~X || ~X − ~X k | , (6)Equation (2) can be reexpressed as V ( ~X ) = N k X k =1 m k Ψ k ( ~X ) + m ln 1 ~X where m = P N k k =1 m k . The case m = 0 can be easily disposed of since Ψ k → | ~X | → ∞ and it is clearthat | V | → ∞ as | ~X | → ∞ unless m = 0. Consequently, without loss of generality take m = 0, so that theform V ( ~X ) = N k X k =1 m k Ψ k ( ~X ) (7)is always assumed for R in what follows. Likewise all of the harmonic functions considered here will beassumed to vanish at infinity by convention.In R N , uniqueness results for a scalar potential imply uniqueness results for the vector field itself as canbe seen from the following argument. Consider the line integral U ( ~X ) = U ( ~X o ) + Z ~X~X o ∇ U · d ~ℓ (8)where ~ℓ = ~ℓ ( s ) denotes a parameterized path (with arclength s ). The contention is that since ~X and ~X o are arbritrary points in the exterior region and ~ℓ ( s ) is also assumed to lie wholly in this exterior region, ~F = 0 ⇐⇒ V = 0 and conversely ~F = 0 ⇐⇒ U = 0, where ~F and U are related by (1). For example,if ~F ( ~X ′ ) < ~X ′ >
1, then due to the mean value theorem for harmonic functions ∇ U > ~X ′ so that both ~X and ~X o , along with the path connecting them in (8),can be taken to be inside this same neighborhood. This, in turn, means that at least one of the two values U ( ~X ) or U ( ~X o ) must be nonzero. Alternatively, if U ( ~X ) >
0, then ~X o can be set to a point at infinity andfrom (8) it is clear that ~F = 0 must occur at someplace along the line integral. Without loss of generality,uniqueness results will thus be stated in terms of scalar potentials for convenience.5niqueness in the complex setting is addressed first since this is the easiest route to the desired R results,which can be readily generalized to R . It is useful to have a common (and commonly used) symbolism foraddressing uniqueness issues in R and C . Let x and y denote standard Cartesian coordinates in eithersetting: In R , ~X ≡ ( x, y ) T and ~X k ≡ ( x k , y k ) T (where T denotes the transpose); while in C , z = x + i y and z k = x k + i y k . In both settings p x + y ≥ p x k + y k <
1. Recall from elementary treatments ofanalytic functions that there is a general mapping between harmonic functions defined over some subregionof R and analytic functions defined over the corresponding subregion of the complex plane as indicated byTable 1. This mapping can be done uniquely when certain reasonable conditions are met with regards tobranch-cuts and the nature of the region under consideration and it is assumed that the reader is familiarwith them. Specifically, if U is harmonic in R let f ( z ) denote the unique analytic function in C whose realcomponent corresponds to U ( x, y ), in which case f ( z ) will be called the standard completion of U ( x, y ).The standard completion of the sum of real functions obeys the principle of linear superposition, so that, forexample, the standard completion of V is a linear superposition of terms that are the standard completionsof the Ψ k ( ~X ). Specifically, let standard completions of Ψ k ( ~X ) be denoted ψ k ( z ). Since the real part of ln z is ln | z | , it is obvious from (6) that the standard completion of Ψ k is given by ψ k ( z ) = ln z ( z − z k ) := ln 1 z − z k − ln 1 z (9)and thus (using the notation indicated in Table 1 for the corresponding complex logarithmic case) g ( z ) canbe written as g ( z ) ≡ N k X k =1 µ k ψ k ( z ) , (10)where µ k ∈ C .As before, a general series based on ln [1 /z − z k ] can be contemplated here since it corresponds to addinga term involving 1 /z to (10); however, it is readily proved that when this term is present it always causes g ( z ) = 0 for large z and so its presence need not be considered further. This might all seem straightforward,but even here some care is called for. Moreover, since the basic strategy used in the sequel will be to proveuniqueness (i.e., linear independence) in one setting and then to obtain uniqueness results in all other settingsby applying a linear uniqueness preserving mapping, the general theorems that are required are best statedexplicity (for clarity and uniformity of exposition). Even though the underlying concepts are well known,these linear independence preserving theorems do not necessarily take conventional forms. This section discusses uniqueness preserving mappings that are used in the sequel. As noted at the end ofthe last section, the first of these mappings is associated with the act of standard analytic completion andentails a unique correspondence between analytic functions in the complex pane and harmonic functions in R . Since uniqueness results will first be shown in the complex setting and then mapped into the R setting,first consider what uniqueness means in the complex setting: Definition 3.1
A set of basis functions { f k } N k k =1 is said to produce a unique expansion in the complexsetting when they meet the following criteria. First each basis function must be a bounded analytic functionfor | z | ≥
1. Second each basis function must vanish at infinity. Third, a sum of the form f ( z ) = N k X k =1 µ k f k ( z ) with µ k ∈ C (11)6ust be linearly independent for | z | ≥
1, where linear independence means that f ( z ) = 0 holds if and onlyif µ k = 0 for all k .Let Re { f } denote the real part of f and Im { f } the imaginary part, then if u k ( x, y ) ≡ Re { f k ( z ) } , v k ( x, y ) ≡ Im { f k ( z ) } , α k ≡ Re { µ k } and β k ≡ Im { µ k } ; from f k ( z ) = u k ( x, y )+ i v k ( x, y ) and µ k = α k + i β k it follows that Re { f } = N k X k =1 Re { µ k f k ( z ) } = N k X k =1 [ α k u k ( x, y ) − β k v k ( x, y )] . (12)Since µ k = 0 for all k implies that both α k = 0 and β k = 0 hold for all k , uniqueness of the analytic setof basis functions { f k } N k , implies simultaneous uniqueness of the pair of conjugate harmonic basis functionsets { u k ( x, y ) } N k k =1 and { v k ( x, y ) } N k k =1 . Uniqueness in the R setting is defined analogously to Definition 3.1.This result can be summarized as a theorem: Theorem 1.
If the set of basis functions { f k } N k k =1 are unique in the complex setting and if u k ( x, y ) ≡ Re { f k ( z ) } and v k ( x, y ) ≡ Im { f k ( z ) } , then the combined set of basis functions { u k ( x, y ) } N k k =1 ∪{ v k ( x, y ) } N k k =1 are linearly independent or unique in the R setting. Notice that since the act of standard harmonic completion is unique, if uniqueness can be shown in R fora sequence of harmonic conjugate pairs then uniqueness for basis functions of the form (11) follows.For the complex setting, in what follows only two general types of basis functions will need to be consid-ered: (1) Logarithmic potentials discussed above of the form ψ k ( z ) (which, as discussed latter, are analyticfrom the branch-cut considerations in the first part of Appendix A). (2) Poles of the form µ k / ( z − z k ) n , forfinite n >
0. For f ( z ) = P N k k =1 µ k ψ k ( z ), the restriction β k = 0 can be made since the argument dependentparts of the logarithmic term occuring in (9) [i.e., Im { ψ k ( z ) } ] are not of general interest in the R harmonicsetting. Uniqueness results in the complex plane for a series of logarithmic basis functions can thus be usedto show point mass uniqueness in R .Next consider a series of poles of fixed order n : h ( n ) ( z ) = N k X k =1 µ k ( z − z k ) n with µ k ∈ C . (13)The case n = 1, h ( z ) ≡ h (1) ( z ), is of special interest and results in the well-known linear combination ofsimple poles: h ( z ) = N k X k =1 µ k ( z − z k ) . (14)Uniqueness results will first be shown for an expansion of this form. Since µ k z − z k = µ k ( z ∗ − z ∗ k ) | z − z k | = α k ( x − x k ) + β k ( y − y k )( x − x k ) + ( y − y k ) + i β k ( x − x k ) − α k ( y − y k )( x − x k ) + ( y − y k ) (15)the real part of h ( z ) corresponds to an R dipole expansion W ( ~X ) given by (4), as will now be explicitlyshown.The various point dipole terms contained in (4) are also known as first order multipoles. For futurereference it is useful to have a consistent notation for delineating multipole basis functions of various orders.The order of a multipole corresponds to the number of subscripts it has, so a first order multipole has asingle subscript that can take on the values 1 , , , . . . , N in R N . Thus the subscripts of a dipole (or first7rder multipole) basis function are associated with the various directions in R N . In R this subscript takeson two values so that the two first order R multipole basis functions associated with the source position ~X k can be written as M [ k ] j ≡ M j ( ~X, ~X k ) for j = 1 ,
2. The R dipole or first order multipole basis functionoriented along the x -axis in is given by M [ k ] ≡ ∂∂x ln 1 p ( x − x k ) + ( y − y k ) = − x − x k ( x − x k ) + ( y − y k ) . (16)Likewise a unit dipole or multiple aligned along the y -axis in R is given by M [ k ] ≡ ∂∂y ln 1 p ( x − x k ) + ( y − y k ) = − y − y k ( x − x k ) + ( y − y k ) . (17)Thus uniqueness results in the complex plane for a series of simple poles can be used to show dipole uniquenessin the R setting. Theorem 1 (and its reverse) clearly involves a change of setting form C to R (or R to C ) This section considers relatively straightforward results about uniqueness of expansions of poles and unique-ness of expansions of logarithmic basis functions. In particular, the theorem that shows that an expansionin terms of simple poles, h ( z ) as given by (14), is unique is readily stated and proved. The analogous resultis also easy to prove for expansions of higher-order poles of a given type (i.e., where all the poles are of somespecified order), as well logarithmic basis functions. A consideration of the more difficult uniqueness resultsfor mixed types of expansions are postponed until Section 7 and, in the end, concrete proofs of these mixeduniqueness results prove to be elusive.Before proceeding to a statement of the desired theorem, several observations are in order. First, if z k = 0 for any k , then it is simply necessary to translate and rescale, when required, so that z k = 0 canbe assumed. Second, as in R for concentric spheres, for uniform circular distributions of continuous simplepoles in C and R immediate counter examples can be constructed. Third, in an attempt to derive thewanted uniqueness results, it is tempting to try to directly apply the standard theory of poles and residuesassociated with analytic function theory. For example, applying the residue theorem by taking a closed lineintegral around the unit disk immediately shows that P N k k =1 µ k = 0; however, additional progress quicklybecomes difficult since, in order to make further progress, it is necessary to consider paths that extend intothe interior region, where some form of analytic continuation inside the unit disk must be used, but this is,at best, questionable. These issues are addressed further in Appendix A.For simple poles the desired uniqueness theorem is: Theorem 2. If h ( z ) has the form specified by (14) where N k is finite, < | z k | < and z k ′ = z k for k ′ = k ,then h ( z ) = 0 for all | z | ≥ if and only if µ k = 0 for all k .Proof. Here µ k = 0 for all k trivially implies h ( z ) = 0 for all z , so only the converse needs to be considered.Throughout the proof assume that 0 < | z k | < z k ′ = z k for k ′ = k and that | z | ≥
1. The proof will be bycontradiction, so assume to the contrary that a non-unique expansion exists where µ k = 0, but h ( z ) = 0 forall | z | ≥
1. If µ k = 0 occurs for any k in this expansion, then drop out these terms, reindex the µ k ’s andreduce the value of N k , so that µ k = 0 can be assumed to hold for all k without loss of generality. Using the8eometric series, each of the pole terms appearing in (14) can be reexpressed as1 z − z k = ∞ X n =0 z nk z n +1 , (18)which has the same overall form as the power series f ( z ) = ∞ X n =1 a n z n , where a n ∈ C . (19)As pointed out in many elementary treatments of complex variables, f ( z ) = 0 for all z if and only if a n = 0for all n ≥
1, which is a useful condition here. Substituting (18) into (14) allows h to be rewritten in theform: h ( z ) = ∞ X j =1 b j z − j , where b j = N k X k =1 z j − k µ k . (20)By assumption h ( z ) ≡ z k ’s and µ k ’s with z k = z k ′ and µ k = 0 for k = 1 2 , , · · · N k with N k >
0. Also as noted above, the factors b j occuring here are unique so b j ≡ j = 0 , , , , · · · .From (20) this condition on the b j ’s can be rewritten in matrix form as e G µ = 0 where e G := · · · z z z · · · z N k z z z · · · z N k z z z · · · z N k ... ... ... ... (21)and where µ = ( µ , µ , µ , · · · , µ N k ) T . Consider the square matrix formed by the first N k rows and N k columns of e G . This square matrix is the Vandermode matrix, which has a determinant that is well knownto be nonzero when the z k are distinct [3]. Thus the only solution to (21) is µ = 0, contrary to our originalassumption, proving the uniqueness of the form given by (14).An analogous result can easily be shown for the logarithmic form given by (10) g ( z ) := N k X k =1 ρ k ψ k ( z ) , (22)where ρ ∈ C is the source parameter and, as before, ψ k ( z ) = ln z ( z − z k ) . (23)Here, since z k = 0 implies ψ k ( z ) = 0, so z = 0 and | z | ≥ > | z k | > ψ k ( z ) has no branch cuts for | z | ≥
1, which means that it is analyticfor | z | ≥ ψ k ( z ) = ∞ X n =1 z nk nz n . (24)9lthough (24) can be obtained in various ways, it is easy to see that it is the correct series by simplycomparing the derivative of the RHS of (24) with the series expansion of the derivative of the RHS of (23).Substituting the series expansion (24) into (22) gives g ( z ) = N k X k =1 ∞ X n =1 z nk ρ k nz n , (25)and setting the various powers of z n to zero yields an equation set similar to (21) that must hold if g ( z ) = 0is to hold: e G ′ ρ = 0 where e G ′ := z z z · · · z N k z / z / z / · · · z N k / z / z / z / · · · z N k / (26)and where ρ = ( ρ , ρ , ρ , · · · , ρ N k ) T . As before, it is necessary to consider only the first N k rows of e G ′ ,which can be reexpressed as a simple matrix product of the Vandermode matrix G and two other N k × N k matrices. If particular, if G ′ denotes this N k × N k matrix, then G ′ := N − GX , where N := · · ·
00 2 0 · · ·
00 0 3 · · · · · · N k and X := z k · · · z · · ·
00 0 z · · · · · · z N k . (27)The condition e G ′ ρ = 0 thus becomes N − GX ρ = 0, which immediately implies ρ = 0 since all three matricesinvolved are invertible. This result can be restated as the desired uniqueness of logarithmic expansions givenby the form (10): Theorem 3. If g ( z ) has the form specified by (10), where N k is finite, < | z k | < and z k ′ = z k for k ′ = k ,then g ( z ) = 0 for all | z | ≥ if and only if µ k = 0 for all k . Next consider the issue of the uniqueness of higher-order pole expansions of the form h ( m ) ( z ) := N k X k =1 ν k ( z − z k ) m (28)where m is a positive integer, ν k ∈ C and the usual restrictions apply to z and z k . A series expansion for( z − z k ) − m = z − m (1 − z k /z ) − m can be found directly from the binomial series1(1 − x ) m = 1 + mx + m ( m + 1)2! x + m ( m + 1)( m + 2)3! x + · · · + m ( m + 1) · · · ( m + n − n ! x n + · · · . (29)The general coefficients here are related to the usual binomial coefficients. Introducing the special (andnon-standard) symbol S m, n for these signed coefficients and replacing x with z k /z thus produces1( z − z k ) m = 1 z m ∞ X n =0 S m n (cid:16) z k z (cid:17) n where S m, := 1 and S m, n := ( m + n − n !( m − n > . (30)10he expression of interest is thus h ( m ) ( z ) = 1 z m ∞ X n =1 S m n z n N k X k =1 z nk ν k = 0 , (31)which leads to the following matrix condition for the coefficients: GX ν = 0 (32)with ν := ( ν , ν , ν , · · · , ν N k ) T . The solvability of this matrix equation directly yields a theorem expressingthe uniqueness of higher-order pole representations: Theorem 4.
For finite m > , if h ( m ) ( z ) has the form specified by (28), where N k is finite, < | z k | < and z k ′ = z k for k ′ = k , then h ( m ) ( z ) = 0 for all | z | ≥ if and only if ν k = 0 for all k . Here as m −→ ∞ the theorem no longer holds since ( z − z k ) − m −→ | z | ≥
1. Also Theorem 4pertains only to poles of the same order. For very small N k it is easy to show that expansions of mixedorders of poles are linearly independent by direct algebraic means. R { h ( z ) } = N k X k =1 ~D k · ∇ ln ( | ~X − ~X k | − )where ~D k = ( − α k , − β k ) T , which immediately shows that dipole expansions in R are unique by Theorems 1and 2. In a like fashion, it is clear that Theorem 4 in conjunction with Theorem 1 (with a passive transforma-tion coefficient used as needed) also implies that higher order R multipole expansions of order n are uniqueif they can be fully represented by two basis functions at each location. As noted in Section 2, in additionto the n = 1 case, this program can be carried out for n = 2 and 3, but for n > ~X k and so this correspondence cannot be one-to-one.First, it is useful to build on the multipole notation introduced in conjunction with (16) and (17) byletting an i , i ′ , j or j ′ subscript preceeded by a comma denote ∂/∂x when the subscript in question takeson the value 1 and to denote ∂/∂y when it takes on the value 2 (in what follows it will be assumed that i , i ′ , j and j ′ always take on the values 1 to N in R N and that similar partials are implied for R N , but onlythe R case will be considered in this section). Multipole basis functions of any order can then be definedby expressions of the form M [ k ] i i ′ j ... j ′ ≡ M [ k ] i ,i ′ ,j ,..., j ′ where (as noted before) the number of subscripts corresponds to the order of the multipole. For n = 2 thereare only two independent basis functions, say M [ k ] and M [ k ] since obviously M [ k ] = M [ k ] and from Laplace’sequation M [ k ] = − M [ k ] . Likewise for n = 3, if M [ k ] and M [ k ] are selected as the two independent basisfunctions, then M [ k ] = − M [ k ] , M [ k ] = − M [ k ] and analogous relationships hold for permuted indices.For n = 4 it is easy to check that all of the other basis functions can be obtained from M [ k ] , M [ k ] and M [ k ] . This shows the set of quadrupole basis functions for some given source point cannot be largerthan the indicated set, but it does not show that the indicated sets are indeed independent. Showing the11ctual independence of the indicated basis functions at a common source point follows from direct algebraicmanipulation and involves first calculating them explicitly and then multiplying through by the commondenominator, but this straightforward algebraic manipulation will not be done here.For n <
4, since there are only two independent basis multipole basis functions of order n at each sourcelocation, it is useful to introduce a more compact notation for these cases so that the resulting potential canbe written more efficiently in terms of the above independent basis multipole functions H ( n ) ( ~X ) ≡ N k X k =1 (cid:16) a ( n ) k A ( n ) k + b ( n ) k B ( n ) k (cid:17) (33)where, for n = 1 A (1) k ≡ M [ k ] and B (1) k ≡ M [ k ] , (34)for n = 2 A (2) k ≡ M [ k ] and B (2) k ≡ M [ k ] , (35)and for n = 3 A (3) k ≡ M [ k ] and B (3) k ≡ M [ k ] . (36)Also in (33) the coefficients a ( n ) k and b ( n ) k ∈ R . Subsequently the superscript indicating the order is oftenomitted from a ( n ) k and b ( n ) k .Explicit expressions for A ( n ) k and B (1) k can be easily written: A (1) k = ∂∂x ln ( | ~X − ~X k | − ) = − x − x k | ~X − ~X k | , B (1) k = − x − x k | ~X − ~X k | (37a) A (2) k = ( x − x k ) − ( y − y k ) | ~X − ~X k | , B (2) k = 2 ( x − x k )( y − y k ) | ~X − ~X k | (37b) A (3) k = 2( x − x k ) | ~X − ~X k | [3( y − y k ) − ( x − x k ) ] and B (3) k = 2( y − y k ) | ~X − ~X k | [3( x − x k ) − ( y − y k ) ] . (37c)The higher order complex poles corresponding to the expressions in (37) for n = 2 and n = 3 can easily befound, as in (15), by multiplying the numerator and denominator of µ k / ( z − z k ) n by ( z ∗ − z ∗ k ) n :Re n µ k ( z − z k ) o = α k [( x − x k ) − ( y − y k ) ] | z − z k | + 2 β k ( x − x k )( y − y k ) | z − z k | (38a)Re n µ k ( z − z k ) o = α k ( x − x k )[( x − x k ) − y − y k ) ] | z − z k | + β k ( y − y k )[3( x − x k ) − ( y − y k ) ] | z − z k | (38b)Since | ~X − ~X k | = | z − z k | , from (37) it thus follows that (38) can be immediately be rewritten asRe n µ k ( z − z k ) o = α k A (2) k + β k B (2) k (39a)Re n µ k ( z − z k ) o = − α k A (3) k + β k B (3) k (39b)Clearly equations (39) imply an invertible passive coefficient mapping, so that Re { h n ) ( z ) } ⇐⇒ H ( n ) ( ~X )holds. Thus Theorem 4 implies that multipole expansions of order three or less in R are unique and theseresults can be summarized in the following formally as:12 heorem 5. In R , for finite < n < , if H ( n ) ( ~X ) has the form specified by (33) where N k is finite, < | ~X k | < and ~X k ′ = ~X k for k ′ = k , then H ( n ) ( ~X ) = 0 for all | ~X | ≥ if and only if a k = 0 and b k = 0 for all k . or less formally as Theorem 6.
A finite expansion of R points masses, point dipoles or point quadrupoles is unique. Since there are more than two multipole basis functions for each source location of an R multipole oforder four or higher, no simple correspondence exists between a given single higher order complex pole andthe corresponding multipoles at the same location. Given that all of the uniqueness results considered so farrest on such a correspondence, it is clear that most of the readily obtainable results in R have been foundand multipoles of order n > R results to R or even R N , for N > R R has direct consequences in R . Towards that end the following simple lemma will proveto be useful: Lemma 1. In R N , for N > , an array of N k distinct points forms at most N k × ( N k − unique directionswhen all possible lines containing two or more array points are considered; moreover, it is possible to rotatecoordinates so that all of the points are distinct when projected into the N − dimensional hyperplaneorthogonal to some preferred coordinate axis.Proof. Since two points determine a straight line, the maximin number of independent lines occurs when nothree or more points are collinear; i.e., chose any of the N k points and a second distinct point. Factoring inthe fact that any line determines two possible directions yields 2 N k × ( N k −
1) for a maximum number ofpossible directions. (Besides the fact that many points may be collinear, many of the lines formed may beparallel, so the actual number might be much smaller than 2 N k × ( N k − N ’th axis) choose a directionthat does not coincide with any of the finite possible directions along which two or more points line up. Sinceno two points line up, their projection into the orthogonal hyperplane of the preferred direction is distinctand the lemma follows.This lemma leads to the following lemma: Lemma 2.
If a point mass uniqueness counter-example exists in R , then one exists in R . roof. If a counter example exists in R then from (3) it can be assumed, without loss of generality, that an N k exists with m k = 0 for all N k ≥ k ≥ V ( ~X ) = N k X k =1 m k | ~X − ~X k | = 0 (40)for all | ~X | ≥ ~X k ′ = ~X k for all k ′ = k ). As before, without loss of generality, it is assumed that P N k k =1 m k = 0 and ~X k = 0 for all k . Thus, for later convenience a set of point masses P N k k =1 m k = 0 at theorigin can be explicitly added to (40).Let ~X ≡ ( x, y, z ) T and ~X k ≡ ( x k , y k , z k ) T where there is no danger here of confusing z and z k withthe complex variables introduced earlier. Assume that the coordinates have been chosen in accord withLemma 1 where the preferred direction has been taken to be along the z − axis so that in the x − y plane the N k points are all distinct, so that ( x k , y k , T = ( x k ′ , y k ′ , T for all k ′ = k . Clearly a linear superpositionsof potentials along the z − axis also obey the criteria of (40) so that L Z z = − L N k X k =1 m k | ~X − ~X k | − m k | ~X | ! d z = 0 (41)Observe thatlim L →∞ L Z z = − L d z p ( x − x k ) + ( y − y k ) + ( z − z k ) = lim L →∞ L Z z =0 d z p ( x − x k ) + ( y − y k ) + z and that since L Z z =0 d z √ a + z d z = ln (cid:16) L + p a + L (cid:17) − ln a it follows that L Z z = − L (cid:18) √ a + z − √ b + z (cid:19) d z = 2 ln p ( a/L ) + 11 + p ( b/L ) + 1 ! − (cid:16) ab (cid:17) , where a ≡ p ( x − x k ) + ( y − y k ) and b ≡ p x + y . It thus follows that ∞ Z z = −∞ | ~X − ~X k | − | ~X | ! d z = 2Ψ k ( ~X ) (42)and thus that N k X k =1 m k Ψ k ( ~X ) = 0 with m k = 0 for all k. (43)14 comparison of (43) and (7) allows one to one to conclude from Lemma 2 and Theorem 6 that Theorem 7.
A finite expansion of R points masses is unique, where the usual conditions are assumed toapply. Next consider the question of uniqueness for the R dipole expansion given by (5): W ( ~X ) ≡ N k X k =1 ~D k · ∇ ( | ~X − ~X k | − ) . (44)It is possible to state and prove a dipole analog of Lemma 2: Lemma 3.
If a point dipole uniqueness counter-example exists in R then one exists in R .Proof. Suppose, as in Lemma 2, to the contrary that an R counterexample exists and thus that for some N k distinct ~X k , that for some W given by the right hand side of (44) W ≡ | ~D k | 6 = 0, for all k . FromLemma 1 there are only a finite number of directions parallel to the lines determined by two or more of the ~X k ’s. To this finite collection of directions add the vectors { ~D k } N k k =1 and then select a preferred z coordinatedirection that is different from all of these directions and denote this direction by ˆ k . Since | ˆ k · ~D k | < | ~D k | forall k and, by construction, not only does ( x k , y k , T = ( x k ′ , y k ′ , T for all k ′ = k hold, as in the proof ofLemma 2, but the projection of ~D k on the x − y plane is nonzero. Let ~D { } k denote this projection. Further,as before, the case P N k k =1 ~D k = 0 presents no real difficulties and ( x k , y k , T = 0 can also be assumed. Thentaking the integral along the z − axis as in (42) gives [after adding an analogous term at the origin to the oneadded to (41)] ∞ Z z = −∞ ∂∂ x | ~X − ~X k | − | ~X | ! d z = 2 ∂∂ x Ψ k ( ~X ) (45a) ∞ Z z = −∞ ∂∂ y | ~X − ~X k | − | ~X | ! d z = 2 ∂∂ y Ψ k ( ~X ) (45b) ∞ Z z = −∞ ∂∂ z | ~X − ~X k | − | ~X | ! d z = 0 (45c)Using (45) in (44) allows the R counterexample to be restated as ∞ Z z = −∞ W ( ~X ) d z = 2 N k X k =1 ~D { } k ·∇ Ψ k ( x, y ) = 0 (46)where | ~D { } k | 6 = 0 for all k .Since Theorem 6 shows that a counterexample of the form specified by (46) cannot exist, the followingtheorem is an immediate consequence of Lemma 3: 15 heorem 8. A finite expansion of R point dipoles is unique. Clearly, one would expect the results of Lemma 2 and thus Theorems 7 and 8 to generalize to R N for N >
3, with only the technical difficulty of explicitly obtaining the N dimensional integral analogs of (42)and (45) to stand in the way; however, two other remaining issues of far greater practical importance areless clear: (1) Proving quadrupole and other multipole uniqueness results in R . (2) Proving expansionsof mixed types of point sources are unique. It is unclear how the first issue should be approached, but anapproach to the second issue can be based on the theorems of Section 3. As before, results in the complexplane will be the starting point. This section deals with uniqueness results for expansions of mixed type. Only the complex setting will beconsidered here even though results for mixed types in C can be directly extended to R and R mixed typeresults. There is a pragmatic reason for considering only the complex setting: Uniqueness results for mixedtypes of expansions are very difficult to prove and, in the end, given that there is only a limited amount ofsuccess here in studying the complex case, consideration of the real case is irrelevant.Thus two particular kinds of mixed type analytic expansions will be of primary interest here: (1) Expan-sions consisting of both simple pole terms and second order pole terms. (2) Expansions consisting of simplepoles and logarithmic basis functions.First, observe that as far as uniqueness results for expansions of mixed type are concerned, it onlynecessary to consider the general case where there are (possibly) a like number of either simple poles andlogarithmic basis functions that are located at the same point or, in a like manner, to consider only the caseconsisting of simple poles and second order poles that are located at the same point. Thus, without lossof generality, for the simple pole and second order pole case consider the following expression for poles ofmixed type: h (1 , ( z ) ≡ N k X k =1 µ k ( z − z k ) + N k X k =1 ν k ( z − z k ) (47)For m = 2, (30) gives 1( z − z k ) = 1 z ∞ X n =0 ( n + 1) z nk z n , (48)which can be combined with (18) to yield h (1 , ( z ) = 1 z N k X k =1 µ k ∞ X n =0 z nk z n + 1 z N k X k =1 ν k ∞ X n =0 ( n + 1) z nk z n . (49)The RHS of (49) can be rewritten as z · h (1 , ( z ) = N k X k =1 µ k + N k X k =1 µ k ∞ X n =1 z nk z n + N k X k =1 ν k ∞ X n =1 nz n − k z n . (50)Setting the successive powers of z − n individually to zero on the RHS of (50) (and simply ignoring the16ondition P N k k =1 µ k = 0) gives the following set of equations that must hold when z · h (1 , ( z ) = 0: z z z · · · z N k · · · z z z · · · z N k z z z · · · z N k z z z · · · z N k z z z · · · z N k ... ... ... ... ... ... ... ... z Nk z Nk z Nk · · · z Nk N k N k z Nk − N k z Nk − N k z Nk − · · · N k z Nk − N k z Nk +1 z Nk +1 z Nk +1 · · · z Nk +1 N k ( N k +1) z Nk ( N k +1) z Nk ( N k +1) z Nk · · · ( N k +1) z Nk N k ... ... ... ... ... ... ... ... µ µ µ ... µ Nk ν ν ν ... ν Nk = 0 . (51)The first 2 N k rows of this matrix equation set can immediately be rewritten in the following block matrixform: GX NGGX Nk +1 ( N + N k I ) GX Nk ! µν ! = 0 . (52)Here I is the N k × N k identity matrix and the other matrices in (52) have been previously defined. (52) isequivalent to the two coupled equations sets GX µ + NG ν = 0 (53) GX N k +1 µ + ( N + N k I ) GX Nk ν = 0 (54)Here the first of these equations uniquely determines µ in terms of ν : µ = − X − G − NG ν . (55)When this is substituted into the second matrix equation and the result is rearranged slightly the followingmatrix equation for ν results [ GX N k ( N k I − G − NG ) + NGX Nk ] ν = 0 , (56)which can be multiplied from the left by X − N k G − to give[( N k I − G − NG ) + X − Nk G − NGX Nk ] ν = 0 . (57)Here (56) immediately implies that if the matrix [( N k I − G − NG ) + X − Nk G − NGX Nk ] is invertible then ν = 0 must hold. In turn, this matrix is invertible if the following matrix is invertible: C := [ N k I − N + GX − N k G − NGX Nk G − ] , (58)which can be rewritten as C := N k I − N + U − NU , where U := GX Nk G − . (59)Showing that C is invertible is not as easy at it might first appear. Thus, for example, first observe that C can be rewritten as C = A + B , where A := N k I − N and B := U − NU . Obviously A is positive definite;furthermore, it is clear that the sum of two positive definite matrices is positive definite, so if it can be shown17hat B is positive definite, then C will be positive definite and thus invertible. Next observe that it is trivialto find the eigenvectors of B and that the corresponding eigenvalues are given by the diagonal elements of N . As one can quickly convince him or herself, this, however, does not imply that B is positive definite,since, among other things, the eigenvectors are not orthogonal to each other (clearly B is not normal since BB T = B T B and normality is assumed in relevant theorems of interest). Given that the author’s attemptsat proving that C is invertible have not met with success, the issue is open.Here the question naturally arises as to what the analog of the C matrix would have been if the set of n rows starting at row mn + 1 had been taken instead of at row n + 1. In this case the system of equationsthat result are C m ν = 0 , (60)where: C m := mN k I − G − NG − + X − mNk G − NGX mNk . (61)Here it is also natural to also raise the question about what the analog of the C is when logarithmic basisfunctions are included. Thus consider the conditions that must hold if ϕ ( z ) = h (1 , , ( z ) = 0, where h (1 , , ( z ) := N k X k =1 (cid:24)(cid:22) ρ k ψ k ( z ) + µ k ( z − z k ) + ν k ( z − z k ) (cid:21) . (62)Since d ψ k ( z ) d z = 1 z − z − z k = − ∞ X n =1 z nk z n +1 = − ∞ X n =2 z n − k z n (63)it is obvious that the correct power series expansion for ψ k ( z ) is ψ k ( z ) = ∞ X n =1 n z nk z n (64)Substituting this expansion along with1 z − z k = ∞ X n =0 z nk z n +1 = ∞ X n =1 z nk − z n (65a)and 1( z − z k ) = ∞ X n =1 nz n − k z n +1 = ∞ X n =2 ( n − z n − k z n (65b)yields h (1 , , ( z ) = N k X k =1 (cid:24)(cid:22) ∞ X n =1 ρ k n z nk z n + ∞ X n =1 z n − k z n µ k + ∞ X n =2 ( n − z n − k z n ν k (cid:21) (66)Breaking out the n = 1 terms separately and reindexing yields: h (1 , , ( z ) = 1 z N k X k =1 ( ρ k z k + µ k ) + 1 z ∞ X n =1 z n N k X k =1 (cid:24)(cid:22) ρ k z n +1 k ( n + 1) + z nk µ k + nz n − k ν k (cid:21) (67)18ntroducing ρ ′ k = z k ρ k and ν ′ k = µ k /z k yields h (1 , , ( z ) = 1 z N k X k =1 ( ρ k z k + µ k ) + 1 z ∞ X n =1 z n N k X k =1 (cid:24)(cid:22) ρ ′ k ( n + 1) + µ k + nν ′ k (cid:21) z nk . (68)Ignoring the first term on the RHS of (68) and setting the other powers of 1 /z to zero yields an equation setwhose first 3 n rows can be written in block matrix form as: ( N + I ) − G GX NG [ N + ( N k + 1)] − I ) GX Nk GX Nk ( N + N k I ) GX Nk [ N + (2 N k + 1)] − I ) GX Nk GX Nk ( N + 2 N k I ) GX Nk ρ ′ µν ′ = 0 . (69)When there are no second order pole terms, (69) can be rewritten as G ( N + I ) GXGX Nk [ N + ( N k + 1) I ] GX Nk ! ρ ′ µ ! = 0 , (70)which, by following analogous steps used to obtain (57), can be rewritten as[( N k I − G − NG ) + X − Nk G − NGX Nk ] µ = 0 . (71)Hence combined logarithmic and simple pole basis functions are independent if the C matrix introducedearlier is nonsingular. 19 ppendix ABranch Cut and Analytic Continuation Side Issues in C This appendix addresses two distinct questions (or misperceptions) that some readers may have: (1) Sincethe logarithmic point source basis functions ψ k [as given by (9)] are used, and logarithmic functions generallyhave brach cuts in C , does ψ k have problematic branch-cuts and, if not, why not? (2) When only simple polesmay be present in the interior region, since it seems natural to assume that the analytic continuation of a zerofunction is always a zero function, can one extend analytic continuation into the interior of the unit disk alongvarious paths while assuming that f ( z ) = 0 still holds until there are areas of overlap, so that one can simplyintegrate around each pole separately and then directly prove the desired complex plane uniqueness resultsfor simple poles from a direct application of the residue theorem? This analytic continuation procedure maybe tempting, because for a finite collection of potential simple poles in the interior, Theorem 2 guaranteesthat if f ( z ) = 0 holds in the exterior region, then µ k = 0, so this analytic continuation procedure seems towork for this case. As far as counter examples go, one need only consider a simple circle with a uniformdensity constant simple pole strength and a compensating interior pole, but what about a denumerable setof separated simple poles, which is the case of prime interest here? The reader who is not bothered by thislast sort of question may simply skip the second part of this appendix that deals with this issue. Branch-cut Issues
First consider branch-cut related issues in the exterior of a unit disk, when all the source points resideinside the unit disk. Although simple and higher order poles obviously do not have such branch cuts, at firstglance it might appear that ψ k ( z ) given by (9) does have branch cuts: ψ k ( z ) := ln 1 z − z k − ln 1 z (A-1)since, for example, ln z = ln r + iθ . (A-2)The truth of the matter, however, is not so straightforward and it turns out that there are no branch cutsin the exterior of the unit complex disk for ψ k ( z ). To see this geometrically, first let ℓ k := | z − z k | andlet φ k equal to the angle between the positive x -axis direction and the vector parallel to the line segmentconnecting z k to z [which is to say, the angle between the R line segment connecting the points ( x k , y k ) T and ( x, y k ) T and the line segment connecting ( x k , y k ) T and ( x, y ) T ]. Then ψ k ( z ) := ln rℓ k + i ( θ − φ k ) (A-3)and when one draws a plane figure displaying the various relevant lines and angles, it is perfectly obviousthat, for any choice of z k , θ and φ k are equal in value at two places as z traces out a closed loop aroundthe unit disk (keeping in mind that | z | ≥ π/ ≥ | θ − φ k | must always hold and, hence ψ k ( z ), for each k . is uniquely defined for z | ≥ ψ k ( z ) = ln z ( z − z k ) := ln 1(1 − z k /z ) (A-4)20nd noting that since Re { − z k /z } > > | z k /z | ≥ | Re { z k /z }| ), theabsolute value of the argument of 1 / (1 − z k /z ) is less than π/ ψ k for | z | ≥
1, for | z | < z k and the origin. An expansion of the form (10) thus has a collection of branch cuts thatform a star-like pattern in the interior of the unit disk.Finally, since a linear superposition of analytic function is analytic the use of { ψ k } N k k =1 as a set of basisfunctions, as in (10) entails no branch-cut interpretational issues at all. It is clear, however, that one cannotdecide to treat the RHS of a ψ k fit, or expansion such as (10), as a single logarithmic function using thestandard properties for combining logarithms–when a linear combination of ψ k ’s are combined into a singlecomposite logarithmic function brach-cut issues are arbitrarily introduced. (Since the argument of the logof some term is assumed to be between 0 and 2 π , even absorbing the constant µ k into ψ k by itself causesproblems and introduces unwanted restrictions, because the complex part of µ k ψ k is not similarly restricted.) Analytic Continuation Issues
Next, consider the analytic continuation issues, which are perhaps best addressed by consideration ofcounter-examples. It may seem reasonable to argue that continuation of the zero function is a special case,especially when it done on either side of a neighborhood where it is known that, at most, one simple poleresides and it would seem that what happens outside the greater region under consideration does not matter.Thus, suppose that an expansion of the form (14) is being considered, and that the poles are ordered suchthat z is the location of the simple pole that is closest to the analytic region ( | z | > f ( z ) = 0. Then, if one analytically continues the zero function into the unit disk close to z along a path to one side of this pole, it would seem obvious that f ( z ) = 0 over this entire region. On theother hand, if one starts from the same region and does the same thing on the other side of the pole thenin the region of overlap between these two analytic continuations, it is obvious that f ( z ) = 0. From this,one can conclude that the simple pole necessarily has a weight of zero: µ k = 0. Proceeding in a like fashionto each succeeding pole, one could thus argue that f ( z ) = 0 for | z | > µ k = 0 for all k , whichis the desired result. Although, at each step, since there are two analytic regions of continuation that arenot simply connected, and it is known that analytic continuation into regions that are not simply connectedare problematic; the simple pole in question produces no branch cuts in either of these two regions and,furthermore, the f ( z ) always agrees in the region of overlap. No function is simpler than the zero functionand one might argue that in this special case of analytic continuation there is no problem, at least in thisparticular instance, and, when used in this way, it leads to the right result.To see what is wrong with this argument, consider the following. It will only be necessary to showthat one counter example exists (here some liberty will be taken and it will merely be shown that somecounter-example is very likely to exist, without explicitly constructing the explicit counter-example itself).Let { s j } ∞ j =1 be a sequence of points in the interior of the unit circle (excluding the origin). Consider the setof corresponding points in the exterior of the unit circle given by t k = 1 /s ∗ j . Then it is well known that thesequence of values { f ( t j ) } ∞ j =1 completely characterizes the analytic function f ( z ). Consider a DIDACKS fitof the form ϕ ( z ) = N X k =1 a k z − s k , (A-5)which always exist for N < ∞ since the associated linear system was always shown to be solvable in [7]. Itdoes not seem unreasonable to assume that for some sequence of points { t k } and some choice of analyticfunction f ( z ) that this system remains solvable as N → ∞ . [For example, it is clearly perfectly acceptableto choose an f ( z ) that has the form (A-5) itself, with some appropriate choice of { a j } ∞ j =1 and { s j } ∞ j =1 , so21ong as | a j | 6 = 0 (in the argument that follows, it is important that the sum in (A-5) contains an infinitenumber of simple poles).]Now, f ( z ) for | z | > { f ( p k ) } ∞ k =1 , where p k = t j for all j and k . Here let z k = 1 /p ∗ k and assume that a fit of the form (14)with n = ∞ also exists. Then, for | z | ≥ f ( z ) = 0 = ∞ X k =1 a k z − s k − ∞ X k =1 µ k z − z k . (A-6)Since by construction, a j = 0 for all j and µ k = 0 for all k , it is clear that the strategy of analyticallycontinuing the zero function so as to encompass an isolated simple pole must fail if any representation of f ( z ) = 0 exists that has the form (A-6); moreover, for this to occur it is only necessary that for some a j = 0for all j and some µ k = 0 for all k , that s j and p k exist such that ∞ X k =1 a k z − s k = ∞ X k =1 µ k z − z k , (A-7)where (for all j and k ) 0 < | s k | <
1, 0 < | p k | < s j = p k . It seems most probable that two suchbounded sequences of simple poles exist. 22 eferences [1] Dennis S. Bernstein, Matrix Mathematics: Theory, Facts, and Formulas With Application to LinearSystems Theory , Princeton University Press, Princeton, N.J., 2005.[2] Martin D. Buhmann,
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