Localizability with Range-Difference Measurements
Wu Junfeng, Mu Biqiang, Yi Xinlei, Wei Jieqiang, Johansson Karl Henrik
11 Localizability with Range-DifferenceMeasurements
Junfeng Wu, Biqiang Mu, Xinlei Yi, Jieqiang Wei* and Karl Henrik Johansson
Abstract —Physical position is crucial in location-aware ser-vices or protocols based on geographic information, wherelocalization is performed given a set of sensor measurementsfor acquiring the position of an object with respect to acertain coordinate system. In this paper, we revisit the long-standing localization methods for locating a radiating sourcefrom range-difference measurements, or equivalently, time-difference-of-arrival measurements from the perspective of leastsquares (LS). In particular, we focus on the spherical LSerror model, where the error function is defined as differencebetween the true distance from a signal receiver (sensor) tothe source and its measured value, and the resulting sphericalLS estimation problem. This problem has been known to bechallenging due to the non-convex nature of the hyperbolicmeasurement model. First of all, we prove that the existenceof least-square solutions is universal and that solutions arebounded under some assumption on Jacobian matrix of themeasurement model. Then a necessary and sufficient conditionis presented for the solution characterization based on themethod of Lagrange multipliers. Next, we derive a charac-terization for the uniqueness of the solutions incorporating asecond-order optimality condition. The solution structures forsome spacial cases are also established, contributing to insightson the effects of the Lagrangian multipliers on global solutions.These findings establish a comprehensive understanding of thelocalizability with range-difference measurements, which arealso illustrated with numerical examples.
Keywords: localization; least squares estimation;quadratic function minimization; time difference of arrival(TDoA) I. I
NTRODUCTION
Location-based services and protocols [1] can be seen ina large number of applications, ranging from wireless com-munication [2], [3], [51], internet-of-things [4], transporta-tion [5] to advertising or social networks [6]. It is worthwhileto note that, amid the proliferation of localization-basedapplications in mobile networks in the last two decades, wehave seen constantly surging interests in localization. Thistrend has been boosting a batch of researches on distributed *The corresponding author.Junfeng Wu is with the College of Control Science and Technologyand the State Key Laboratory of Industrial Control Technology, ZhejiangUniversity, Hangzhou, P. R. China. [email protected] .Biqiang Mu is with Key Laboratory of Systems and Control, Institute ofSystems Science, Academy of Mathematics and Systems Science, ChineseAcademy of Sciences, Beijing 100190, China. [email protected] .Xinlei Yi and Karl H. Johansson are with the Division of De-cision and Control Systems, School of Electrical Engineering andComputer Science, KTH Royal Institute of Technology. { xinleiy,kallej } @kth.se .Jieqiang Wei is with Ericsson, Torshamnsgatan 21, 16440, Stockholm,Sweden. [email protected] . localization [7], [8], [9], [49], indoor localization [10], [11],[12], simultaneous localization and mapping [13], [14], toname a few. Localization problems focus on acquiring theposition of an object in a certain coordinate system basedon measurement data from a set of sensors. The methodsfor localization vary, among which localization using fieldgeometry about objects’ relative placement (also known asrange-based localization methods) [15], [16] are practicallyprevalent since the geometric information of an object canbe easily measured by sensors. The localization methodutilizing time difference of arrival (TDoA) [19], [20], [21],[22] is an important example of the lateration approach forpassive localization. It performs with high accuracy since itdoes not rely on time synchronization between the sourceand receivers.In this paper, we consider the problem of locating aradiating source from TDoA measurements. In an idealuniform medium, radiation travels through the medium ata known velocity, and therefore the measurement of rangedifferences from the source is accessible from the TDoAmeasurement. Notice that TDoA measurements are the timedifferences among the arrival times of signal measuredat difference receivers, which can be obtained, corruptedby noise, by using a group of passive sensors. Sourcelocalization from range-difference measurements has beenextensively studied by means of least squares methods. Inparticular, finding the least squares solution with respect tothe spherical error criterion [42] can be formulated as anoptimization problem with quadratic objective function andconstraints, also known as a constrained least square (CLS)range-difference based localization problem. The problemis nonconvex as the Hessian matrix of a quadratic term ofone of the constraints is not positive semi-definite. It resultsin difficulty in finding a global solution. Finding solutionsfor the CLS problem and its variants has been an activeresearch point of the area of localization in the past andrecent literature. A direct method is discarding the quadraticconstraints [43], [27], which gives rise to an unconstrainedleast squares (ULS) problem. An extension of of the ULSmethod is exploring the constraint to reduce the error ofthe ULS solution for generating the final estimate [41],[50]. The spherical-interpolation method is a commonlyused approach proposed that solves (6) approximately [37],[38], [27], by which optimization is done by alternatingrestricted minimizations over two disjoint subsets of thevariables. The subspace minimization method [38], [27] isanother way to solve (6) approximately. In this method, a r X i v : . [ s t a t . A P ] F e b orthogonal projection is used to eliminate the challengingquadratic constraint. As a revisit to the methods reviewedabove, it is pointed by [27] that the spherical-interpolationmethod and the subspace minimization method are identicalto the unconstrained LS method in the sense that thesemethods generate identical solutions. The reference [40]presents a closed-form localization technique, termed thespherical-intersection method, which has similar formulationto the spherical-interpolation one. What is worth noticing isthat the spherical-intersection method gives consideration toquadratic constraint but fails in obtaining a global minimizertoo as it by its nature is alternating optimization. Somesolvers are iterative methods, such as [35], [36]. At eachstep a location estimate is improved by solving a local LSestimation problem using the Taylor-series method. However,proper initialization is needed to avoid false local min-ima [52]. A different category of solvers for the CLS problemis based on the Lagrange multiplier technique. In [29], [50],numerical global solution searching algorithms are developedin virtute of necessary conditions for the global optimalityof the CLS problem, as a consequence of the the analysison generalized trust region problems offered in [28]. Thesolution finding amounts to finding roots of a th orderpolynomial in the two-dimensional or a th order in thethree-dimensional localizations. Such extensions are possiblebecause problem (6) is closed to a generalized trust regionproblem in form. To attack the global solution, the methodsin [29], [50] need an exhaustive search for all the suspected,which is numerically less efficient.There are a few key questions remaining to be answeredfor this range-difference based localizations: (i) How arethe solutions to the CLS problem related to the localiza-tion measurements? (ii) When does the CLS problem forrange-difference based localizations have a global/uniquesolution? (iii) How can the global/unique solution be char-acterized? We establish a series of results attempting toaddress these critical questions for range-difference basedlocalization problems utilizing the persistence of excitationon the localization measurements as well as an extended andstructured Karush-Kuhn-Tucker (KKT) analysis. The maincontributions of our paper are threefold as follows. ( i ) . We show that in the absence of measurement noises,the persistence of excitation on the localization mea-surements describes whether or not the coordinate of aradiating source can be exactly recovered from TDoAmeasurements, which depends on the rank of the Jocobimatrix of the model, and that for a practical CLSproblem where measurement noise is taken into accountthe CLS solutions always exist and under some assump-tion on the Jocobi matrix of the measurement modelthe solution set is bounded. In principle, the conditionsuggests that a larger-sized sensor array inclines to leadto a bounded CLS solution. ( ii ) . we manage to develop a necessary and sufficient con-dition for a global CLS solution in terms of a groupof KKT conditions. Moreover, we also establish a char-acterization for the uniqueness of an optimal solution. The uniqueness result is attached to a stricter conditionon the curvature of a Lagrangian function. ( iii ) . the theoretical results are developed in this paperinspire us to consider some special cases, and uncoverinsightful findings on the position of the Lagrangianmultiplier affecting how the global minimizers can besolved in these cases. They pave a way for us tocompute a global minimizer in a relative easy manner.In this part, we also draw remarks on the multiplicitynature of CLS solutions.Simulation examples are carried out to illustrate ourtheory. We have therefore established a comprehensive un-derstanding of solutions to CLS range-difference based lo-calization problem. The derivation of our results are partlyinspired by [28]. In particular, the reference [28] givescharacterization of the global minimizer of the generalizedtrust region problem and its uniqueness in terms of theLagrange multiplier. However, the additional positivity con-straint, see (6c), leads to challenging difficulty in analysis.The global optimal solution has been investigated in [29],and a sufficient condition and a necessary one are offeredseparatively. The two conditions do not coincide in generaland contain conservativeness, which can be seen from aconcrete example in that work. In contrast, our resultsmanage to close the gap. In addition, we develop a conditionfor the solution uniqueness. To the best of our knowledge,this is the first time that a characterization of the uniquenessis given in the literature.The remainder of the paper is organized as follows. InSection II, we introduce range-difference localization andhow it is formulated into a CLS localization problem, as wellas the problems of interest in the paper. In Section III, we de-velop characterizations of solutions to the CLS localization.Specifically, the development mainly includes a feasibilitycondition for the problem, a characterization of a globalsolution and a uniqueness characterization. We also givecomparison between our results and the main related onesin literature. In Section IV, we establish a few findings onthe structural properties of global solutions in some specialcases. We also give some numerical examples to illustrateour theory. Finally, some concluding remarks are drawn inSection V. Most of the proofs of the main results can befound in Appendices. Notations . We use σ min ( X ) and σ max ( X ) to denote theeigenvalues of a square real matrix X , which have thesmallest and largest magnitude, respectively. For x, y ∈ R , x ∨ y and x ∧ y stand for the maximum and minimum of x and y , respectively. For a real-valued function h ( x ) : R n → R ,we use ∇ h to denote the gradient of h . For a set A , weuse A to denote the closure of A . Let A be a metricspace, which is a set equipped with a metric d . Supposethat ∅ (cid:54) = B ⊂ A and x ∈ A . The distance of x from A isdefined as d ( x ; A ) = inf { d ( x, y ) : y ∈ A} . a = 0 a a x d d Figure 1. Illustration of range-difference measurements in an array con-sisting of three sensors (i.e., m = 2 ). The array is a line one. The solid dot“ • ” represents the radiating source and the hollow dots “ ◦ ” represent thesensors. II. P
ROBLEM S TATEMENT
A. Range-difference Based Localization
We consider a radiating source and an array consisting of m + 1 sensors that collect signals emitting from the source.We denote the coordinate, with respect to an Euclideancoordinate system, of the source by x ∈ R n and denote thecoordinate of sensor i by a i ∈ R n . In particular we assume a = 0 for sensor , i.e., sensor is set to be at the originof the coordinate system. We use d i to denote the range-difference measurement from sensor i , for i = 1 , . . . , m , tosensor . We adopt the additive measurement error model,in which the measurements of the range differences aremodeled as d i , a i and x can be given as d i = (cid:107) a i − x (cid:107) − (cid:107) x (cid:107) + r i , (1)where (cid:107) · (cid:107) denotes the (cid:96) norm of a vector and r i cap-tures the measurement noise contained in d i , which is alsocalled “equation error” [30]. Therefore, a natural localizationproblem arises, on identifying the unknown source x fromthe measurement data { d i } and the sensor localizations { a i } . We postulate that the additive measurement errorshave mean zero and are independent of the range differenceobservation and the source location, as routinely assumed inthe literature, such as [42].The TDoA techniques, as means of passive localization,has a large number of applications in different positioningsystems [31], [32], [33], [34]. For instance, they have beenwidely used for sound source localization, where the goal isto estimate the coordinates of a sound source using acousticsignals received by an array of microphones. The micro-phones are mounted at fixed and known positions a i , withone of them selected as the reference node. The sound sourcelocates at an unknown position x . The TDOA measurement d i is actually time delay that the sound takes to get receivedby the i th and the reference nodes (microphones), multipliedby the propagation speed of sound in appropriate medium,see Figure 1. Then a corresponding hyperbolic surface canbe derived from a TDoA measurement, for which all pointson the surface will have the same distance difference fromthe said pair of nodes. B. Persistence of Excitation of the Spherical LS Error Model
The spherical LS model built on the distance errorfrom the hypothesized source location to every sensor, atransformed model of (1), has a simpler expression on thelocalization x than (1). Consider the distance from the sourceto the i th sensor. The measured value from the TDoAmeasurement model (1) can be computed as d ix = d i + (cid:107) x (cid:107) and the true value is (cid:107) a i − x (cid:107) . The spherical LS error functionis defined as the difference of the squares of d ix and (cid:107) a i − x (cid:107) : e i = 12 d ix − (cid:107) a i − x (cid:107) . By (1), we obtain the spherical LS model b i = d i (cid:107) x (cid:107) + a (cid:62) i x − e i , for i = 1 , . . . , m, (2)where b i = ( (cid:107) a i (cid:107) − d i ) and e i = (cid:107) a i − x (cid:107) r i + r i .Suppose that the measurement errors were absent (i.e., r i ≡ , accordingly e i ≡ ). Then each of model (1)defines a branch of a hyperbola, of which a i and a arethe two foci and d i is the focal distance. The intersectionof the m equations exist, regarding the true position ofthe source. Meanwhile, the spherical LS error e i of thespherical LS model (2) turns to be zero. Here one basicconcept persistence of excitation (PE) lies in, whether ornot any exact source localization x can be recovered froma given set of localizations { a i : i = 1 , · · · , m } . When theanswer is affirmative, we call the measurement localizations { a i : i = 1 , · · · , m } are persistently exciting (PE). ThisPE is a prerequisite for developing numerical algorithmsto estimate x with noisy data, since otherwise intrinsicambiguity persists: there are at least two different possiblesource locations that correspond to measurements of d i withthe same pattern (distribution from a stochastic perspective).Actually, the PE issue aims to find the condition under whichthe global solution to the spherical LS model (2) is unique.For the models (1) and (2), in the absence of measurementnoises, the measurement localizations { a i , i = 1 , · · · , m } arePE only if their Jocobian matrices at the true location are ofcolumn full rank [48]. The Jocobian matrix J ( x ) of (1) is J ( x ) = − x (cid:62) (cid:107) x (cid:107) + ( a − x ) (cid:62) (cid:107) a − x (cid:107) ... x (cid:62) (cid:107) x (cid:107) + ( a m − x ) (cid:62) (cid:107) a m − x (cid:107) , (3)while the Jacobian matrix J ( x ) of (2) is: J ( x ) = w ... w m x T (cid:107) x (cid:107) + a T ... a Tm (4) = 1 (cid:107) x (cid:107) (cid:107) a − x (cid:107) x (cid:62) + ( a − x ) (cid:62) (cid:107) x (cid:107) ... (cid:107) a m − x (cid:107) x (cid:62) + ( a m − x ) (cid:62) (cid:107) x (cid:107) = − (cid:107) a − x (cid:107) . . . (cid:107) a m − x (cid:107) J ( x ) where w j = (cid:107) a j − x (cid:107) − (cid:107) x (cid:107) for j = 1 , · · · , m are appliedin the second equality. This means that the model (1) isequivalent to the model (2) in the absence of measurementnoises when a i (cid:54) = x for all i = 1 , · · · , m . The measurementlocations { a i , i = 1 , · · · , m } are not PE if they satisfyspecial geometric relations, which is given in the followingproposition. Proposition 1:
Suppose that r i ≡ . The measurementlocations { a i : i = 1 , · · · , m } are not PE if one of thefollowing equivalent conditions hold: ( i ) . for n = 2 all of measurement localizations { a i : i = 1 , · · · , m } are collinear, and for n = 3 all ofmeasurement localizations { a i : i = 1 , · · · , m } arecoplanar; ( ii ) . there exists some x ∈ R n such that J ( x ) is not of offull column rank; ( iii ) . there exists some x ∈ R n such that J ( x ) is not ofof full column rank. C. Practical CLS Solution from the Spherical LS ErrorModel
A fundamental approach to estimate x is to minimize thesummed spherical least squares errors e i ’s, i.e.,( LS ): minimize x ∈ R n m (cid:88) i =1 ( d i (cid:107) x (cid:107) + a (cid:62) i x − b i ) , (5)in which the notation X (cid:62) denotes the transpose of any givenmatrix X ∈ R p × q . The LS problem is nonconvex. Letting y = [ (cid:107) x (cid:107) , x (cid:62) ] (cid:62) , we transform the LS problem equivalentlyinto the following constrained least squares (CLS) problem minimize y ∈ R n +1 f ( y ) (6a) (CLS) : subject to g ( y ) = 0 , (6b) [ y ] ≥ . (6c)In this problem, f ( y ) := (cid:107) Ay − b (cid:107) , where A = d a (cid:62) ... ... d m a (cid:62) m , (7) b = [ b , . . . , b m ] (cid:62) , g ( y ) := y (cid:62) Dy with D = diag(1 , − I n ) , (8)where I n denotes an identity matrix of size n and diag( · , . . . , · ) denotes a diagonal matrix with the argumentsin brackets on the main diagonal, and the notation [ x ] i , i ≤ p ,is used to denote the i th element of x for any given x ∈ R p .The interpretation of each equation in the CLS problem is asfollows: the least square criterion is given by (6b) and (6c)describe the geometric constraints (cid:107) x (cid:107) = [ x ] + · · · + [ x ] n and (cid:107) x (cid:107) ≥ . An optimal value f ∗ of the CLS problem isdefined as f ∗ CLS = inf { f ( y ) : [ y ] ≥ , g ( y ) = 0 , y ∈ R n +1 } . (9) D. Problems of Interest
To find the least squares solution with respect to thespherical error criterion [42], it resorts to solving the CLSproblem (6), which is firstly studied in this reorganizationidea in [37], [38], [39] and has been investigated widely inlater work, such as [40], [41], [42], [43], [27], [44], [29]. Theproblem is nonconvex as the Hessian matrix of a quadraticterm of one of the constraints is not positive semi-definite.To the best of our knowledge, most of the existing methodssolve an constraints is not positive semi-definite. approxi-mate of (6) with some information loss, e.g., one classicalway is to discard the two quadratic constraints (6b) and (6c),and investigate the resulting ULS problem [43], [27], [45],[37], [38]. In [29], one sufficient condition and one necessarycondition are provided respectively for the CLS solution,but there is no explicit and complete characterization for theexact CLS solution.In this paper, we are concerned about solvability ofthe CLS problem, the solution characterization and CLSsolution uniqueness. We are also interested in exploringthe solution structures for some special cases. Answers tothese questions will constitute our understanding of CLSlocalizability and will further facilitate the development ofnumerical algorithms for the exact CLS solution.III. C
HARACTERIZATION OF
CLS L
OCALIZABILITY
In this part, we present conditions on the localizability ofrange-difference based measurements using CLS methods.
A. CLS Solution Existence
First we discuss the existence and boundedness of CLSsolutions. In reality we hope that the solution set is boundedso that when some solving algorithm is implemented it wouldnot output arbitrarily large solution. As will be shown in thefollowing, the boundedness of CLS solutions has connectionswith the Jocobian of the TDoA measurement model (2).
Lemma 1:
The CLS localization problem has a solution.If the sensors are not geometrically collinear in position,or equivalently, if J ( x ) x is nonzero at any nonzero x ∈ R n , or equivalently, if Ay is nonzero for any nonzero y satisfying (6b), the solution set is bounded.The proof is reported in Appendix A. To ensure a boundedCLS solution set, we introduce the following assumption. Assumption 1:
The value of J ( x ) x is nonzero at anynonzero x ∈ R n In the following, we present a necessary and sufficientcondition for a global minimizer of the CLS problem. The condition is developed mainly based on the method ofLagrange multipliers. The proof is reported in Appendix B.
Theorem 1:
Let Assumption 1 hold. A vector y ∈ R n +1 is a global optimal solution of the CLS Problem if and onlyif it falls into one of the following two cases: ( i ) . The vector y satisfies g ( y ) = 0 with [ y ] > and thereexists a λ ∈ R such that ( A (cid:62) A + λD ) y = A (cid:62) b, (10a) v (cid:62) ( A (cid:62) A + λD ) v ≥ , for v satisfies v (cid:62) Dv ≤ (10b) ( ii ) . It holds that y = 0 and there does not exist y (cid:48) ∈ R n +1 and λ ∈ R such that y (cid:48) (cid:54) = 0 , [ y (cid:48) ] > , g ( y (cid:48) ) = 0 and (10) hold.We revisit the ULS problem, an approximate of the CLSproblem, and note that the rank of A (cid:62) A dictates the numberof solutions: when A (cid:62) A is nonsingular there is a uniquesolution of the ULS approximation. However, the situationis not the same for the CLS localization itself. The nonsingu-larity of A (cid:62) A guarantees existence of a global solution of theCLS localization, but not necessarily uniqueness. Our furtherresult uncovers that uniqueness of CLS solution follows froma stricter condition, see Theorem 2. We use the followingexample to illustrate this fact. Example 1:
Consider a line-shaped (see Figure 1) sensorarray consisting of four sensors (i.e., m = 3 ) monitoring aradiating source in a 2-dimension space, i.e., n = 2 . Thecoordinates of the sensors are a = 0 , a = [ √ , √ ] (cid:62) , a = [ − √ , √ ] (cid:62) , a = [0 , − √ ] (cid:62) . We let the range-differences measurements be as follows: d = 1 √ , d = 1 √ , d = 1 √ . Then we compute A = √ √ √ √ − √ √ √ − √ , b = . It can be verified that in this example Assumption 1issatisfied with A (cid:62) A = I . Figure 2 specifies the contour linesto display the value of f ( y ) under the constraints (6b) and(6c) on a plane spanned by [ y ] and [ y ] . As we see inthe figure, all points on a ring (the ring is marked with ared arrow) achieve the minimum of f . This indicates thatsolutions of the CLS problem are not unique.In Theorem 1, the multiplier λ in (10) plays a critical rolein affecting solutions for CLS problem. Define the set of λ ’ssatisfying (10b) as: I = { λ ∈ R : v (cid:62) ( A (cid:62) A + λD ) v ≥ , ∀ v (cid:62) Dv ≤ } . (11)Since A (cid:62) A ≥ , the inequality v (cid:62) ( A (cid:62) A + λD ) v ≥ holdsfor all λ ≤ satisfying v (cid:62) Dv ≤ . Thus, the set I is notempty. The next result gives more refined descriptions of the -3 -2 -1 0 1 2 3 x axis -3-2-10123 x a x i s Figure 2. The contour plot of f ( y ) on a plane spanned by [ y ] (i.e., x axis) and [ y ] (i.e., x axis). The colormap is used to displayed differentlevels of f ( y ) in the plot. All x ’s on a ring, which is marked with a redarrow, achieve the minimum of f . set I . Lemma 2:
Let Assumption 1 hold. The set I can bewritten in the form of I := ( −∞ , λ u ] with a constant λ u ≥ . Moreover, λ u can be computed by solving a linearmatrix inequality (LMI) problem as follows: maximize λ ∈ R λ (12a) subject to A (cid:62) A + λD ≥ . (12b) Proof.
We shall show that I is convex and therefore it isan interval. Since I is not empty, let λ ∈ I . For any λ ∈ ( −∞ , λ ) , we obtain v (cid:62) ( A (cid:62) A + λD ) v > v (cid:62) ( A (cid:62) A + λ D ) v ≥ , ∀ v (cid:62) Dv < . Hence, λ ∈ I . Finally, we conclude λ u ≥ due to ∈ I . (cid:4) The above lemma, as a preliminary result of the valuerange of the multipliers λ ’s that satisfy (10b), depicts aninterval it may locate in. It is succeeded by a concise editionof the CLS solution characterization as follows: Corollary 1:
Let Assumption 1 hold. A vector y ∈ R n +1 is a global optimal solution of the CLS Problem if and onlyif it falls into one of the following two cases: ( i ) . The vector y satisfies g ( y ) = 0 with [ y ] > and thereexists a λ ∈ I such that ( A (cid:62) A + λD ) y = A (cid:62) b . ( ii ) . It holds that y = 0 and there does not exist y (cid:48) ∈ R n +1 and λ ∈ R such that y (cid:48) (cid:54) = 0 , [ y (cid:48) ] > , g ( y (cid:48) ) = 0 and (10a) hold. Proof.
The proof follows from Theorem 1 and the definitionof the set I . B. CLS Localization Uniqueness
In this part, on top of the existence of the CLS solutions,we will present a sufficient and necessary condition for thesolutions’ uniqueness. Then we further present a finding thatrelates the Lagrange multiplier to the multiplicity of the CLSsolutions.The solution characterization in Theorem 1 involves aKKT condition and a second-order condition on the Hessianmatrix A (cid:62) A + λ ∗ D of a Lagrangian function. The nextresult is on uniqueness of the global optimal solutions ofproblem (6). We will find that uniqueness follows a strictersecond-order condition. The proof is reported in Appendix B. Theorem 2:
Let Assumption 1 hold. A vector y ∈ R n +1 is the unique global optimal solution of the CLS Problem ifand only if it falls into one of the following two cases: ( i ) . The vector y satisfies g ( y ) = 0 with [ y ] > and thereexists a λ ∈ R such that ( A (cid:62) A + λD ) y = A (cid:62) b, (13a) v (cid:62) ( A (cid:62) A + λD ) v > , for v satisfies v (cid:62) Dv < and v (cid:62) Dy (cid:54) = 0 . (13b) ( ii ) . y = 0 and there exists no y (cid:48) (cid:54) = 0 , where [ y (cid:48) ] > , and λ ∈ R such that g ( y (cid:48) ) = 0 and (10) hold.The conditions in Lemma 1 are sufficient for the existenceof a bounded global solution of the CLS problem, but notnecessary. The conditions in Theorems 1 and 2 are both exactcharacterizations of some certain properties of the globalsolutions.The following example demonstrates the gap betweenLemma 1 and Theorems 1 and 2. Example 2:
We continue to consider Example 1. In theexample, we have computed that A (cid:62) A = I and illustratedthat the CLS problem has bounded global solutions. Observethat A (cid:62) A + D = diag(2 , , and A (cid:62) b = [ √ , , (cid:62) .We then find that λ ∗ = 1 and any y ∗ , which satisfies [ y ∗ ] = √ and (cid:107) y ∗ (cid:107) = √ , meet the conditions (10). ByTheorem 1, the problem has global solutions, illustrated byFigure 2. On the other hand, for any nonzero vector v ∈ span , , we have v (cid:62) ( A (cid:62) A + λ ∗ D ) v = 0 .The condition (13b) is not met, therefore, by Theorem 2, theproblem’s solution is not unique. Figure 2 is evidence of thisclaim.We next consider another example of an array consistingof four sensors. Assume that the coordinates of them are a = a = 0 , a = [4 , (cid:62) , a = [0 , (cid:62) , and that the range-difference measurements are d = 4 , d = d = 0 . Then we obtain that A = 4 I, b = [ − , , (cid:62) . By Theorem 1, we have y ∗ = −√ [ √ , , (cid:62) and -1 0 1-1-0.8-0.6-0.4-0.200.20.40.60.81 Figure 3. The contour plot of f ( y ) on the -dimensional plane. Thecolormap is used to displayed different levels of f ( y ) in the plot. The point [ −√ , −√ ] , marked by a red circle, is the unique minimizer of theCLS problem in Example 2. λ ∗ = − √ satisfy (10a) and (10b), and therefore −√ [ √ , , (cid:62) is a minimizer of the CLS problem. Nu-merical search suggests that −√ [ √ , , (cid:62) is the uniqueminimizer, see Figure 3. Since λ ∗ = − √ < ,for any v (cid:62) Dv < , we have v (cid:62) ( A (cid:62) A + λ ∗ D ) v > , whichtheoretically supports our numerical result.The following corollary is on the relations between themultiplier λ in (10) or (13) and the multiplicity of the CLSsolutions. It unveils that whether a CLS problem has a singleunique solution is dictated by the position of λ in I . Itdirectly follows from Corollary 1, Theorem 2 and Lemma 2. Corollary 2:
Let Assumption 1 hold. The CLS problemhas multiple global solutions only if there exists a vector y ∈ R n +1 such that g ( y ) = 0 with [ y ] > and ( A (cid:62) A + λ u D ) y = A (cid:62) b . On the other hand, a vector y ∈ R n +1 isa single unique global solution if it falls into one of thefollowing two cases: ( i ) . The vector y satisfies g ( y ) = 0 with [ y ] > and thereexists a λ < λ u such that ( A (cid:62) A + λD ) y = A (cid:62) b . ( ii ) . y = 0 and there exists no y (cid:48) (cid:54) = 0 , where [ y (cid:48) ] > , and λ (cid:48) ∈ I such that ( A (cid:62) A + λ (cid:48) D ) y = A (cid:62) b . Remark 1:
The reference [28] gives characterization ofthe global minimizer of the generalized trust region problemand its uniqueness in terms of the Lagrange multiplier. Inour paper, the existence of global solutions to the CLSlocalization and solution uniqueness is derived partly basedon [28]. On one hand, the problems respectively investigatedin the two papers have similar forms. The additional positiv-ity constraint (6c) require more elaborate analysis to derive conditions for CLS solution and its uniqueness. On the otherhand, the conditions in our paper are less strict than those of[28] for the generalized trust region problem since adding(6c) into the constraints defines a narrowed feasible set.
C. Literature Comparison and Discussions
In this part, we present comparison between our afore-mentioned results and the related ones in literature.First note that a characterization of the global solutions tothe CLS problem was presented in [29], giving a sufficientcondition and a necessary condition respectively. There isa high degree of similarity between the CLS problem andgeneralized trust region problems (GTRS) [28], a classof problems minimizing a quadratic function subject to aquadratic equality constraint that has been reviewed by us inabove. A characterization of the global solution was obtainedin [28]. In the following remarks, we will compare the resultsof this paper with those of [29] and [28], respectively.In [29], a sufficient condition, i.e., [29, Lemma 3.1], and anecessary condition, i.e., [29, Theorem 3.1], for solutions ofthe CLS problem were presented. The gaps in between themtwo are discussed as follows: (i). the sufficient conditionrequires A (cid:62) A + λD to have no negative eigenvalues while thenecessary one is relaxed, allowing A (cid:62) A + λD to have at mostone negative eigenvalue; (ii). the possibility of y ∗ CLS = 0 isexcluded from the sufficient condition while not excluded inthe necessary one. These gaps are not trivial in the sensethat there may exist cases where a global minimizer y ∗ CLS fails to satisfy the sufficient condition as well as caseswhere the necessary condition does not lead to a solution. InExample 3, we demonstrate two cases where the gaps exist.
Example 3 (Example 2 Cont’d):
In this example, wecontinue to study the cases in Example 2. First, we considerthe same scenarios to the first case of Example 2. We haveshown that λ ∗ = 1 and any y ∗ that satisfies [ y ∗ ] = √ and (cid:107) y ∗ (cid:107) = √ meet the conditions (10). Then, we have A (cid:62) A + λ ∗ D = diag(2 , , . The positive semidefiniteness of A (cid:62) A + λ ∗ D indicates the conservativeness of the necessarycondition of [29, Lemma 3.1].We next consider the second scenario of Example 2. It hasbeen computed in Example 2 that y = −√ [ √ , , (cid:62) is an optimal solution in this case(see Figure 3) and λ = − √ satisfies (10a) and (10b). However, sim-ple computation reveals that, having a negative eigenvalue, A (cid:62) A + λD > does not hold. It suggests conservativenessof the sufficient condition of [29, Lemma 3.1].The CLS problem has a close relation with the GTRSproblem as they have a similar form. The form of the GTRSproblem is as follows [28]: (GTRS) : minimize y ∈ R n +1 f ( y ) (14a) subject to g ( y ) = 0 . (14b)Compared to the CLS problem, the GTRS does not containthe constraint [ y ] ≥ . By [28, Theorems 3.2], a vector y ∗ GTRS ∈ R n +1 is a global minimizer if and only if g ( y ∗ GTRS ) = 0 and there exists a λ ∗ GTRS ∈ R such that (10a)and A (cid:62) A + λ ∗ GTRS D ≥ hold. Moreover by [28, Theorems3.3], in case of A (cid:62) A + λ ∗ GTRS
D > , y ∗ GTRS is the uniquesolution.A comparative study to the characterization of solutionsto the GTRS and the CLS problems reveals some closeobservations. First, Theorems 1 and 2 of our paper andTheorems 3.2 and 4.1 of [28] shared similar KKT conditionsbecause the CLS and the GTRS problems have similar formsand it is allowed to have analytic characterization of thesolutions using the same Lagrangian function. On the otherhand, these problems have different second-order conditionsbecause for the CLS problem the constraint (6c) gets ridof the lower napple of the double cone { y : y (cid:62) Dy = 0 } and this allows negative definiteness of the Hessian matrixof (21) in some subspace.IV. S TRUCTURED
CLS L
OCALIZATIONS
The characterization of CLS global minimizers estab-lished in Theorem 1 paves the way for us to locate theminimizers by resorting to seeking for feasible λ ’s. A viableway for doing so is to verify the conditions ( i ) – ( ii ) of The-orem 1 in order: first search multipliers λ ’s satisfying (10b);substituting each λ into (10a), then solve a solution of (10a)and check the feasibility of (6b) and (6c). Besides, we cando better in some special cases, where a global minimizer ofthe CLS problem can be computed easily. This is done onthe basis of the theory developed in the previous section. Inwhat follows, we will proceed to present how it is realized.To guarantee that the CLS problem exists solutions, weassume that Assumption 1 holds in the rest of this section ,and will elaborate how λ ∗ CLS is positioned in differentsituations.
A. Technical Preparations
Define the following set I = { λ ∈ R : A (cid:62) A + λD > } . (15)Obviously I ⊂ I and due to the indefiniteness of D as wellas Assumption 1 the set I is a non-empty bounded interval .This interval can be formally represented by ( λ l , λ u ) , where λ l ∈ R and is the solution of the following LMI problem: minimize λ ∈ R λ subject to A (cid:62) A + λD ≥ . By [28, Theorem 5.3], the closure of I , denoted by I :=[ λ l , λ u ] , has the following property: I = { λ ∈ R : A (cid:62) A + λD ≥ } , and A (cid:62) A + λD ≥ is singular for λ = λ l and λ u . In thepresence of the specific form of D given in (8), we further A double cone is a quadratic surface. Each single cone placing apex toapex is called a napple. reach claims that λ l < and that, as an eigenvalue of A (cid:62) A + λ l D , zero has a geometric multiplicity M = 1 . This claimcan be proved in the following way. If M were greater than , we consider two distinct subspaces U and V , defined as U := { u ∈ R n +1 : ( A (cid:62) A + λ l D ) u = 0 } and V := { v ∈ R n +1 : ( D + I ) v = 0 } . By calculation, the dimensions of U and V satisfy dim( U ) = M and dim( V ) = n . By theGrassmann’s formula [46, 2.18 Corollary], dim( U ) + dim( V ) = dim( U ∪ V ) + dim(
U ∩ V ) . (16)Since dim( U ∪V ) ≤ n +1 , we have that dim( U ∩V ) ≥ M − ≥ . However, for any v ∈ V \ { } , v (cid:62) ( A (cid:62) A + λ l D ) v = v (cid:62) A (cid:62) Av − λ l (cid:107) v (cid:107) > . Hence v (cid:54)∈ U and U ∩ V = { } ,which reaches a contradiction.Letting I := ( −∞ , λ l ) , we have that I = I ∪ I andthat, for any λ ∈ I , the matrix A (cid:62) A + λD is nonsingular,or to be more specific, it has exactly one negative eigenvalueand n positive eigenvalues. We call I the positive-definiteinterval (PDI) for an optimal Lagrangian multiplier and call I the indefinite interval (IDI) , and call λ l and λ u theleft singular point (LSP) and right singular point (RSP) ,respectively. We use y ( λ ) to denote a solution to (10a), wherethe argument λ in y ( λ ) reminds that it is a solution under agiven λ ∈ I ∪ I , and denote h ( λ ) := g ( y ( λ )) . (17)We will show that the function h ( λ ) exhibits some niceproperties and develop analysis in some special cases onthe ground of them. In what follows, we will present thesecases one by one. Let us first recall notions that have beendefined and will be used frequently in the sequel. Notations.
The functions f , g and h are defined in (6)and (17), respectively. The parameter b is defined in (2).The sets I and I are defined in (11) and (15). We haveknowledge of I that it is an open and bounded interval. Theset I is the closure of I , and λ l and λ u are the left and rightend points of I , respectively. The set I is I = I \I . Thevariables y ∗ GTRS ∈ R n +1 and λ ∗ GTRS ∈ R relate to the GTRSproblem (14). They satisfy g ( y ∗ GTRS ) = 0 and (10a) and A (cid:62) A + λ ∗ GTRS D ≥ —in this situation y ∗ GTRS is a globalminimizer of the GTRS problem. Similarly y ∗ CLS ∈ R n +1 and λ ∗ CLS ∈ R are two variables that meet the conditions inTheorem 1. B. The Case of Collinear Target Position ( b = 0 ) We consider the case of b = 0 . This case is special inthe sense that it allows us to exactly obtain the solutionimmediately. When b = 0 , one can, in a straightforwardmanner, derive the following result (The proof is skippedbecause of its simplicity.): Lemma 3: If b = 0 , we have that h ( λ ) = 0 for λ ∈ I ∪I .It can be readily validated that y = 0 and any λ ∈ I ∪ I satisfies condition ( i ) of Theorem 2. Hence, in this case weconclude that the origin is the unique solution to the CLSproblem. Other facts that we would like to point out are that the uniqueness of λ ∗ CLS is not needed for the uniquenessof y ∗ CLS and that when b = 0 the origin is also the uniqueminimizer of f ( y ) .The position of the CLS solution for b = 0 can alsobe inferred geometrically. By (2) we have (cid:107) a i (cid:107) = d i forall sensor i . The geometry in Figure 1 alludes that the bestestimate of the radiating source’s position is collinear with a and any a i . To make sure that Assumption 1 holds, a and a , . . . , a m cannot be in a line. As a consequence, theorigin must be the unique best position estimate.Next we turn to consider the case of b (cid:54) = 0 , which needsto be treated by more refined analysis. C. Searching CLS solution over PDI
In case of b (cid:54) = 0 , h ( λ ) is strictly monotonic on I . Theproperty was originally derived in [28, Theorem 5.2] and isrephrased as follows: Lemma 4: [28, Theorem 5.2] Assume that b (cid:54) = 0 . Thefunction h ( λ ) is strictly decreasing on I .We revisit the GTRS problem (14). Theorem 3.2 of [28]reads that a vector y ∗ GTRS ∈ R n +1 is a global minimizer ofthe GTRS problem if and only if g ( y ∗ GTRS ) = 0 and thereexists a λ ∗ GTRS ∈ I such that (10a) holds. It can be furtherconcluded that, if in addition [ y ∗ GTRS ] ≥ holds, y ∗ GTRS isalso a global minimizer of the CLS problem.Based on the above conclusions, we can first search λ ∗ GTRS over I , if any, and then compute y ∗ GTRS . Lemma 4reveals that h ( λ ) is monotonically decreasing on I . If A (cid:62) A + λ ∗ GTRS
D > , a bisection algorithm can be appliedfor locating λ ∗ GTRS [29] by evaluating the sign of h ( λ ) .Then y ∗ GTRS = y ( λ ∗ GTRS ) = ( A (cid:62) A + λ ∗ GTRS D ) − A (cid:62) b canbe computed naturally. By examining whether [ y ∗ GTRS ] ispositive or not, one can determine whether y ∗ GTRS is a globalminimizer to the CLS problem. To be precise, y ∗ GTRS is aglobal solution to (6) in the presence of [ y ∗ GTRS ] > . Forthe case of A (cid:62) A + λ ∗ GTRS D being positive semidefinite andsingular, one needs other analysis for locating λ ∗ GTRS , whichwill be treated in the following subsection.
D. Searching CLS solution on Singular Points If A (cid:62) A + λ ∗ GTRS D is positive semidefinite and singular,the sign of h ( λ ) does not change on I . In this situation,by [28, Theorem 5.4], the following claims can be estab-lished: ( i ) . The value of λ ∗ GTRS is pushed towards an endpoint,either λ l or λ u , of I . More precisely, λ ∗ GTRS = λ u incase of h ( λ ) > on I and λ ∗ GTRS = λ l otherwise. ( ii ) The limit lim λ → λ ∗ GTRS h ( λ ) exists and lim λ → λ ∗ GTRS y ( λ ) = y ∗ for some y ∗ ∈ R n +1 .It is pointed out by the reference [28] that y ∗ is not necessar-ily a solution to (14). In general, it requires a subtle treatmentto compute a y ∗ GTRS . Besides what has been concluded in [28], in the presence of the specific form of the matrix D ,see (8), we can get access to better properties related to h ( λ ) and y ∗ GTRS , which make it possible to determine the sign of [ y ∗ GTRS ] even prior to solving the exact value. By knowingthe sign of [ y ∗ GTRS ] , we can determine whether y ∗ GTRS isalso a CLS solution. The whole computation procedure willbe detailed for the rest of this subsection.First we focus on how the sign of h ( λ ) affects that of [ y ∗ GTRS ] . The analysis is divided into two parts on theground of either h ( λ ) > or h ( λ ) < .
1) The case h ( λ ) < for λ ∈ I : In this case, λ ∗ GTRS = λ l . We choose z − with (cid:107) z − (cid:107) = 1 such that ( A (cid:62) A + λ l D ) z − = 0 . Since is an eigenvalue of A (cid:62) A + λ l D with geometricmultiplicity , z − is unique up to sign. Then y ∗ GTRS = y ∗ + αz − for some α ∈ R (Recall that lim λ → λ ∗ GTRS y ( λ ) = y ∗ .).Here α is chosen as either of the roots of the followingquadratic equation (see [28, Section 5]): g ( z − ) α + 2 y (cid:62)∗ Dz − α + g ( y ∗ ) = 0 . (18)The roots have the following property. The proof can befound in Appendix C. Lemma 5:
The quadratic equation (18) has two distinctroots.To distinguish the two roots of (18), without loss ofgenerality, we denote them by α − and α − , where α − > α − ,and have the following claim. Lemma 6:
One of [ y ∗ + α − z − ] and [ y ∗ + α − z − ] isnegative and the other is positive, i.e., [ y ∗ + α − z − ] [ y ∗ + α − z − ] < .The proof is given in Appendix C.To summarize, when h ( λ ) < for λ ∈ I , we have λ ∗ CLS = λ l and can solve the CLS problem by resorting toLemma 6. We obtain y ∗ through lim λ → λ l y ( λ ) = y ∗ andchoose a unit eigenvector z − of A (cid:62) A + λ l D with respect toeigenvalue . Then we compute α and α by solving (18).Lemma 6 guarantees that either y ∗ + α z − or y ∗ + α z − isa global minimizer to (6). Example 4:
Consider a sensor array consisting of fivesensors (i.e., m = 4 ) monitoring a radiating source in a3-dimension space, i.e., n = 3 . The coordinates of thesensors are a = 0 , a = [ − , , (cid:62) , a = [1 , − , (cid:62) , a = [1 , , (cid:62) and a = [0 , , (cid:62) . We set the range-differences measurements to be: d = − , d = 1 , d = 0 , d = − . Then we compute A = − − − − , b = . . . It can be verified that A (cid:62) A is nonsingular so the problem has solutions by Lemma 1. We obtain I = ( − , . using cvx in Matlab. The Matlab simulation reads that h ( λ ) < and it is strictly decreasing for λ ∈ I . By the result in thispart, we have λ ∗ GTRS = λ l = − and z − = √ [2 , − , , (cid:62) .Figure 4 plots the values of y ( λ ) for λ ∈ ( − , and readsthat y ( λ ) converges to y ∗ ≈ [ − . , . , , . (cid:62) when λ approaches − . By solving (18), we obtain α − = 2 √ −√ and α − = − √ − √ . Then we find that y ∗ + α − z − =[2 √ − , −√ , √ , (cid:62) and y ∗ + α − z − = [ − √ − , √ , −√ , (cid:62) , which verifies Lemma 6. Finally, weconclude that y ∗ + α − z − is a solution. -1 -0.8 -0.6 -0.4 -0.2 0 -3-2-1012 [y( )] [y( )] [y( )] [y( )] h( ) Figure 4. In Example 4, the Matlab simulation reads that h ( λ ) < is strictly decreasing when λ ∈ ( − , and y ( λ ) converges to y ∗ ≈ [ − , , , (cid:62) when λ approaches − .
2) The case of h ( λ ) > for λ ∈ I : In this case, λ ∗ GTRS = λ u . Similarly, we choose a vector z + with (cid:107) z + (cid:107) =1 such that ( A (cid:62) A + λ u D ) z + = 0 . Then y ∗ GTRS = y ∗ + αz + ,where α is a solution of the quadratic equation: g ( z + ) α + 2 y (cid:62)∗ Dz + α + g ( y ∗ ) = 0 . (19)Without loss of generality, we denote the roots (which is notnecessarily distinct) by α +1 and α +2 , where α +1 ≥ α +2 . Thefollowing result holds. Lemma 7:
The values of [ y ∗ + α +1 z + ] and [ y ∗ + α +2 z + ] are simultaneously negative or positive, i.e., [ y ∗ + α +1 z + ] [ y ∗ + α +2 z + ] > . Moreover, it is true that [ y ( λ )] , [ y ∗ ] , [ y ∗ + α +1 z + ] and [ y ∗ + α +2 z + ] have the same signfor any λ ∈ I .The proof is given in Appendix C.To summarize, when h ( λ ) > for λ ∈ I , λ ∗ GTRS = λ u .We compute y ∗ through lim λ → λ u y ( λ ) = y ∗ and choosea unit eigenvector z + of A (cid:62) A + λ u D with respect to theeigenvalue . Then y ∗ GTRS = y ∗ + α +2 z + or y ∗ + α +2 z + .To decide whether y ∗ GTRS is a global minimizer of theCLS problem, we need to identify the sign of [ y ∗ GTRS ] .Lemma 7 allows us to do this by simply examining thesign of any [ y ( λ )] for λ ∈ I . When ∈ I , a simplemeans is to check the sign of [ y (0)] = [( A (cid:62) A ) − A (cid:62) b ] . If [( A (cid:62) A ) − A (cid:62) b ] > , then y ∗ GTRS is a CLS solution;otherwise it is not.
Example 5 (Example 1 Cont’d):
We continue to considerExample 1. In this example, I = ( − , . Our Matlabsimulation reads that h ( λ ) > is strictly decreasing for λ ∈ I , see Figure 5. By the result in this part, wehave λ ∗ GTRS = λ u = 1 , which is consistent with theresult in Example 2. Moreover, Figure 5 plots the valuesof y ( λ ) for λ ∈ [0 , and reads that y ( λ ) convergesto y ∗ ≈ [0 . , , (cid:62) when λ approaches , and that [ y ∗ ] and [ y ( λ )] ’s are all positive numbers. we choose z + = [0 , , (cid:62) since ( A (cid:62) A + λ u D ) z + = 0 . Solve theequation g ( y ∗ + αz + ) = 0 , and we obtain that α ≈ ± . ,hence y ∗ GTRS = y ∗ + αz + ≈ [0 . , ± . , (cid:62) . Thefirst elements of the two y ∗ GTRS ’s are both positive. So farthe Matlab simulation has validated Lemma 7. The optimalsolution numerically simulated in Matlab rather meets theones computed theoretically in virtue of Theorem 1. -0.1-0.0500.050.10.150.20.250.3 [y( )] [y( )] [y( )] h( ) Figure 5. In Example 5, the Matlab simulation reads that h ( λ ) > isstrictly decreasing when λ ∈ [0 , and y ∗ ≈ [0 . , , (cid:62) . In addition,it can be observed that [ y ∗ ] and [ y ( λ )] ’s are all positive numbers. V. C
ONCLUSIONS
In this work, we were concerned about the problem ofradiating source localization from range-difference measure-ments. We placed the attention to a spherical least squaresapproach that squares the range-difference measurements.By utilizing this model, one can formulate a CLS range-difference based localization problem, which is a nonconvexoptimization problem. Our first result suggested that theresulting localization problem have bounded global solu-tions under some rank condition. A necessary and sufficientcondition for a global CLS solution was derived by meansof the Lagrange multiplier technique. The uniqueness of aglobal solution can be equivalently checked using a strictersecond-order condition. Consequently, by examining thesetwo conditions, we have complete knowledge of the mul-tiplicity nature of CLS solutions. Finally, we studied the structural properties of global CLS solutions for some specialcases. Our results contribute to finding out the locationsof the global solutions in a convenient manners. Numericalalgorithms for computing the CLS solutions by means of ourresearch findings worth future research. In addition, analysison asymptotic properties of the statistical estimator generatedfrom the CLS solution will be an interesting problem.A
PPENDIX
A. P
ROOF . OF L EMMA y such that y (cid:62) A (cid:62) Ay = 0 , where A isgiven in (7), it holds that f ( y ) is constant. In addition,since A (cid:62) A ≥ , for any unbounded sequence ( y k ) k ∈ N , lim inf k →∞ f ( y k ) is unbounded or constant, therefore theCLS problem has at least a global solution in R n +1 .Note that J ( x ) x = Ay holds for y = [ (cid:107) x (cid:107) x (cid:62) ] (cid:62) .When y (cid:62) A (cid:62) Ay > for all nonzero y satisfying (6b),for any unbounded sequence ( y k ) k ∈ N , lim inf k →∞ f ( y k ) isunbounded.To see the equivalence of the conditions, first note thatwhen the sensors are collinear, there exists x (cid:54) = 0 that is x is collinear with a i such that (cid:107) x (cid:107)(cid:107) a i (cid:107) ≥ for all a i . Then x (cid:62) ( a i − x ) = −(cid:107) x (cid:107)(cid:107) a i − x (cid:107) . Since J ( x ) = 1 (cid:107) x (cid:107) (cid:107) a − x (cid:107) x (cid:62) + ( a − x ) (cid:62) (cid:107) x (cid:107) ... (cid:107) a m − x (cid:107) x (cid:62) + ( a m − x ) (cid:62) (cid:107) x (cid:107) , we have J ( x ) x = 0 . On the other hand, for a nonzero x satisfying J ( x ) x = 0 , then ( a i − x ) (cid:62) x = −(cid:107) a i − x (cid:107)(cid:107) x (cid:107) holds for any i = 1 , . . . , m , which further implies that a i − x and x are collinear, that is a , . . . , a m and x are collinear.This completes the proof.A PPENDIX
B. P
ROOFS OF T HEOREMS AND B.1. Proof of Theorem 1
We define the Lagrangian function L ( y, λ ) = f ( y ) + λg ( y ) . (20)Notice that (cid:79) y L ( y, λ ) is well defined. Next we divide theproof into two steps. Necessity.
Suppose y ∗ is a global optimal solution of theoptimization problem (6). We consider two cases.When y ∗ (cid:54) = 0 , the conditions g ( y ∗ ) = 0 and [ y ∗ ] > are readily met. The KKT optimality condition ensuresthat there exists a multiplier λ ∗ such that (cid:79) y L ( y ∗ , λ ∗ ) = 0 ,the condition (10a) follows. It further leads to the followingresult: for any y with [ y ] ≥ , L ( y, λ ∗ ) = L ( y ∗ , λ ∗ ) + ( y − y ∗ ) (cid:62) ( A (cid:62) A + λ ∗ D )( y − y ∗ ) . (21)Since L ( y, λ ∗ ) = f ( y ) whenever g ( y ) = 0 , we have that v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ (22)holds for v ∈ T := { v : g ( y ∗ + v ) = 0 , [ v ] ≥ − [ y ∗ ] } .We will use this fact to show (10b). To this end, denote S := { v : v (cid:62) Dv ≤ } and partition S into a few disjointsubsets, where S = { v : v (cid:62) Dy ∗ > , v (cid:62) Dv < } , (23) S = { v : v (cid:62) Dy ∗ < , v (cid:62) Dv < } , (24) S = { v : v (cid:62) Dy ∗ = 0 , v (cid:62) Dv = 0 } (25)and S = S \ ( S ∪ S ∪ S ) . For any v ∈ S , there always exists a constant ψ > such that g ( y ∗ + ψv ) = g ( y ∗ ) + 2 ψv (cid:62) Dy ∗ + ψ v (cid:62) Dv = g ( y ∗ ) = 0 , Next we will show [ ψv ] > − [ y ∗ ] . Suppose that [ y ∗ ] + [ ψv ] ≤ were true. Then ([ y ∗ ] + [ ψv ] )[ y ∗ ] ≤ ,by which we further have v (cid:62) Dy ∗ = 1 ψ ( y ∗ + ψv ) (cid:62) Dy ∗ = 1 ψ ([ y ∗ ] + [ ψv ] )[ y ∗ ] − ψ [ y ∗ + ψv ] (cid:62) n +1 [ y ∗ ] n +1 ≤ ψ ([ y ∗ ] + [ ψv ] )[ y ∗ ] + 1 ψ (cid:107) [ y ∗ + ψv ] n +1 (cid:107)(cid:107) [ y ∗ ] n +1 (cid:107) =0 . The above inequality contradicts v ∈ S . Therefore, it isclear that ψv ∈ T . Then by (22), we conclude that v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ for any v ∈ S .For any v ∈ S , we obtain that − v ∈ S . Therefore, v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ holds for any v ∈ S . Similarly, forany v ∈ S , we have that g ( y ∗ + φv ) = 0 for any φ ∈ R and [ φv ] +[ y ∗ ] ≥ when φ is sufficiently small. It impliesthat φv ∈ T . Thus, we have v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ for any v ∈ S .Finally, since each element of S is a limit point of S , S or S , the continuity of v (cid:62) ( A (cid:62) A + λ ∗ D ) v in v impliesthat v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ also holds for v ∈ S . So farwe have shown (10b) for the case of y ∗ (cid:54) = 0 .When y ∗ = 0 , we will prove the result by contradiction.Suppose that there exists at least a ˜ y (cid:54) = 0 such that g (˜ y ) = 0 and [˜ y ] > , and there corresponds a λ ∗ satisfying (10).Then by (21) we have f (0) = L (0 , λ ∗ )= L (˜ y, λ ∗ ) + ˜ y (cid:62) ( A (cid:62) A + λ ∗ D )˜ y = f (˜ y ) + ˜ y (cid:62) ( A (cid:62) A + λ ∗ D )˜ y. Since ˜ y (cid:62) D ˜ y = 0 and by Assumption 1, ˜ y (cid:62) ( A (cid:62) A + λD )˜ y > . It thus follows that f (0) > f (˜ y ) , which contradicts thehypothesis. Sufficiency.
Suppose that y ∗ ∈ R n +1 and λ ∗ ∈ R is avector and a multiplier, respectively, which satisfy ( i ) . Thenwe have (cid:79) y L ( y ∗ , λ ∗ ) = 0 , which further yields the resultof (21). Let y (cid:54) = y ∗ be any given vector satisfying g ( y ) = 0 and [ y ] ≥ . Then y and y ∗ have the following geometricrelation (cid:79) g ( y ∗ )( y − y ∗ ) = 2 y ∗(cid:62) D ( y − y ∗ ) (26) = 2 y ∗(cid:62) Dy = 2[ y ∗ ] y − y ∗ ] n ) (cid:62) [ y ] n ≥ y ∗ ] y − (cid:107) [ y ∗ ] n (cid:107)(cid:107) [ y ] n (cid:107) = 0 . This also implies ( y − y ∗ ) (cid:62) (cid:79) g ( y ∗ )( y − y ∗ ) = − y ∗(cid:62) Dy ≤ . (27)In other words, by (27), we can find a vector v = y − y ∗ satisfying v (cid:62) (cid:79) g ( y ∗ ) v ≤ , which together with (21)implies that L ( y, λ ∗ ) = L ( y ∗ , λ ∗ ) + α v (cid:62) ( A (cid:62) A + λ ∗ D ) v ≥ L ( y ∗ , λ ∗ ) , where the inequality follows from (10b). Observing that g ( y ) = g ( y ∗ ) = 0 , then the above inequality yields that f ( y ∗ ) ≤ f ( y ) , which shows that y ∗ is a global optimal solution of prob-lem (6).Next we will show that ( ii ) is a sufficient conditionfor the optimality of y . Suppose that y ∗ = 0 is not anoptimal solution to problem (6). By Lemma 1, we can findat least a vector ˜ y (cid:54) = 0 that is a global minimizer of (5). Bythe argument used for showing the necessity, we concludethat the conditions in (10) hold for ˜ y , which contradicts ( ii ) . (cid:4) B.2. Proof of Theorem 2
The proof resembles that of Theorem 1. Here we onlyprovide with a sketch.
Sufficiency.
We first suppose that y ∗ ∈ R n +1 and λ ∗ ∈ R isa vector and a multiplier, respectively, that satisfy ( i ) . Thenwe have (cid:79) y L ( y ∗ , λ ∗ ) = 0 , further yielding that, for any y with [ y ] ≥ , the equality (21) holds. For any given vector y (cid:54) = y ∗ satisfying g ( y ) = 0 and [ y ] ≥ , by (27), we have ( y − y ∗ ) (cid:62) (cid:79) g ( y ∗ )( y − y ∗ ) ≤ . When ( y − y ∗ ) (cid:62) (cid:79) g ( y ∗ )( y − y ∗ ) < , we have ( y − y ∗ ) (cid:62) Dy ∗ = y (cid:62) Dy ∗ > [ y ∗ ] y − (cid:107) [ y ∗ ] n (cid:107)(cid:107) [ y ] n (cid:107) = 0 . By (21) and (13b), f ( y ) = L ( y, λ ∗ )= L ( y ∗ , λ ∗ ) + ( y − y ∗ ) (cid:62) ( A (cid:62) A + λ ∗ D )( y − y ∗ ) > L ( y ∗ , λ ∗ )= f ( y ∗ ); and when ( y − y ∗ ) (cid:62) (cid:79) g ( y ∗ )( y − y ∗ ) = 0 , due to Assump- tion 1, f ( y ) = L ( y, λ ∗ )= L ( y ∗ , λ ∗ ) + ( y − y ∗ ) (cid:62) ( A (cid:62) A + λ ∗ D )( y − y ∗ )= L ( y ∗ , λ ∗ ) + ( y − y ∗ ) (cid:62) A (cid:62) A ( y − y ∗ ) > L ( y ∗ , λ ∗ ) = f ( y ∗ ) . The above altogether imply that y ∗ is the unique globaloptimal solution of (5). For ( ii ) , notice that y = 0 isa minimizer in this situation and that (10) is a necessarycondition for a y (cid:54) = 0 being an solution to problem (6), bothby Theorem 1. They altogether show that the origin is theunique solution. Necessity.
First consider y ∗ (cid:54) = 0 . Since y ∗ is the uniqueoptimal solution, similar to (22), we have that v (cid:62) ( A (cid:62) A + λ ∗ D ) v > holds for v ∈ T \ { } . Since { v : v (cid:62) Dv < , v (cid:62) Dy ∗ (cid:54) = 0 } = S ∪S , where S and S are given in (23)and (24), following the proof of Theorem 1, we eventuallyobtain ( i ) . For y ∗ = 0 , the proof directly follows from thatof Theorem 1.The above derivations altogether conclude the claim. (cid:4) A PPENDIX
C. P
ROOFS OF LEMMAS IN S ECTION
IVThis appendix contains proofs of some lemmas in Sec-tion IV.
C.1. Proof of Lemma 5
When h ( λ ) < for λ ∈ I , it has been shown in [28]that g ( z − ) > . Since g ( y ( λ )) = h ( λ ) , by continuity wehave g ( y ∗ ) ≤ . (28)We consider the following two cases, respectively.Suppose that g ( y ∗ ) < . It is straightforward to seethat (18) has two different roots: one is positive and theother is negative.Suppose that g ( y ∗ ) = 0 . First we shall show that y (cid:62)∗ Dz − (cid:54) = 0 . Computation suggests that | y (cid:62)∗ Dz − | = | [ y ∗ ] [ z − ] − [ y ∗ ] (cid:62) n +1 [ z − ] n +1 |≥| [ y ∗ ] [ z − ] | − | [ y ∗ ] (cid:62) n +1 [ z − ] n +1 |≥| [ y ∗ ] [ z − ] | − (cid:107) [ y ∗ ] n +1 (cid:107)(cid:107) [ z − ] n +1 (cid:107) Observe that | [ y ∗ ] | = (cid:107) [ y ∗ ] n +1 (cid:107) due to g ( y ∗ ) = 0 and | [ z − ] | > (cid:107) [ z − ] n +1 (cid:107) due to g ( z − ) > . Combining thesefacts, we obtain that | y (cid:62)∗ Dz − | > , implying y (cid:62)∗ Dz − (cid:54) = 0 . (29)Since g ( y ∗ ) = 0 and g ( z − ) > , we finally concludethat (18) has a zero root and a nonzero root.We conclude the result by completing the analysis for theabove two cases. (cid:4) C.2. Proof of Lemma 6
The following technical lemma is functional to the proof.
Lemma 8:
Let y, ¯ y ∈ R n +1 be two vectors satisfying g ( y ) = g (¯ y ) = 0 . Then the following statement are true:(i). If [ y ] [¯ y ] ≥ , it holds that g ( ψy + (1 − ψ )¯ y ) ≥ forany ψ ∈ [0 , .(ii). If [ y ] [¯ y ] ≤ , it holds that g ( ψy + (1 − ψ )¯ y ) ≤ forany ψ ∈ [0 , . Proof.
First we prove (i). We begin with an observation that g ( ψy + (1 − ψ )¯ y )= ψ g ( y ) + (1 − ψ ) g (¯ y ) + 2 ψ (1 − ψ ) y (cid:62) D ¯ y =2 ψ (1 − ψ )[ y ] [¯ y ] − ψ (1 − ψ )[ y ] (cid:62) n +1 [¯ y ] n +1 ≥ ψ (1 − ψ )[ y ] [¯ y ] − ψ (1 − ψ ) (cid:107) [ y ] n +1 (cid:107)(cid:107) [¯ y ] n +1 (cid:107) =0 , where the last inequality follows from the Cauchy-Schwarzinequality [47] and the last equality holds due to the formof D given in (8) and g ( y ) = g (¯ y ) = 0 .To show (ii), similarity, we have g ( ψy + (1 − ψ )¯ y )=2 ψ (1 − ψ )[ y ] [¯ y ] − ψ (1 − ψ )[ y ] (cid:62) n +1 [¯ y ] n +1 ≤ ψ (1 − ψ )[ y ] [¯ y ] + 2 ψ (1 − ψ ) (cid:107) [ y ] n +1 (cid:107)(cid:107) [¯ y ] n +1 (cid:107) =0 , which completes the proof. (cid:4) Proof of Lemma 6.
It is clear that neither y ∗ + α − z − nor y ∗ + α − z − is , since otherwise (10a) is violated giventhat b (cid:54) = 0 . Notice that y ∗ can be expressed as a linearcombination of y ∗ + α − z − and y ∗ + α − z − as follows: y ∗ = α − α − − α − ( y ∗ + α − z − )+ − α − α − − α − ( y ∗ + α − z − ) . (30)In (30), α − and α − cannot happen to be both positive orboth negative. Therefore, − α − α − − α − , α − α − − α − ∈ [0 , and theabove combination is a convex one.In virtue of (28), we will proceed by considering the casein which the equality sign of (28) holds and the case inwhich the strict inequality sign of (28) holds, respectively.Suppose that g ( y ∗ ) < . Then, the form of (18) suggest that α − > and α − < . If [ y ∗ + α − z − ] [ y ∗ + α − z − ] ≥ , thenby Lemma 8 g ( y ∗ ) ≥ , which contradicts the hypothesis.Suppose that g ( y ∗ ) = 0 . By (29), we consider two cases, i.e., y (cid:62)∗ Dz − > and y (cid:62)∗ Dz − < , respectively. First, assumethat y (cid:62)∗ Dz − > . Then α − = 0 and α − < . If [ y ∗ ] [ y ∗ + α − z − ] were larger than , it suggests from computation that y (cid:62)∗ Dz − = α − y (cid:62)∗ Dz − α − = y (cid:62)∗ D ( y ∗ + α − z − ) α − = 1 α − ([ y ∗ ] [ y ∗ + α − z − ] − [ y ∗ ] (cid:62) n +1 [ y ∗ + α − z − ] n +1 ) ≤ α − ([ y ∗ ] [ y ∗ + α − z − ] − (cid:107) [ y ∗ ] n +1 (cid:107)(cid:107) [ y ∗ + α − z − ] n +1 (cid:107) )=0 , where the last equality follows since g ( y ∗ ) = g ( y ∗ + α − z − ) = 0 . It contradicts the hypothesis. On the otherhand, when we assume that y (cid:62)∗ Dz − < , we reach a similarcontradiction, which concludes the result. (cid:4) C.3. Proof of Lemma 7
When h ( λ ) > for λ ∈ I , it has been shown in [28] that g ( z + ) < . In addition, by continuity, we have g ( y ∗ ) ≥ . The first claim can be proved by contradiction. Theprocedure is similar to the proof of Lemma 6 and is omitted.Next we prove the second claim. As α +1 and α +2 are theroots of (19). Therefore, α +1 and α +2 are not both positive orboth negative since g ( y ∗ ) /g ( z + ) ≤ . Then y ∗ is a convexcombination of y ∗ + α +1 z + and y ∗ + α +2 z + . It further impliesthat [ y ∗ ] , [ y ∗ + α +1 z + ] and [ y ∗ + α +2 z + ] have the same sign.In addition, y ( λ ) is a rational function for λ ∈ I [28] and I is a connected set. We then obtain that the image of I underthe function y , denoted by y [ I ] , is connected. If there werea λ (cid:48) ∈ I such that [ y ( λ (cid:48) )] and [ y ∗ ] have distinct signs, wecould find another λ (cid:48)(cid:48) ∈ I satisfying [ y ( λ (cid:48)(cid:48) )] = 0 . Noticethat g ( y ( λ (cid:48)(cid:48) )) = −(cid:107) [ y ( λ (cid:48)(cid:48) )] n +1 (cid:107) ≤ , which contradicts g ( y ( λ (cid:48)(cid:48) )) := h ( λ (cid:48)(cid:48) ) > for all λ ∈ I and completes theproof. (cid:4) R EFERENCES[1] C. Gentile, N. Alsindi, R. Raulefs, and C. Teolis,
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