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Hindawi Publishing CorporationISRN Applied MathematicsVolume 2013 Article ID 635263 12 pageshttpdxdoiorg1011552013635263

Research ArticleIterative Scheme for Solving Optimal Transportation ProblemsArising in Reflector Design

Tilmann Glimm1 and Nick Henscheid2

1 Department of Mathematics Western Washington University Bellingham WA 98225 USA2 Program in Applied Mathematics University of Arizona Tucson AZ 85716 USA

Correspondence should be addressed to Tilmann Glimm glimmtwwuedu

Received 29 July 2013 Accepted 5 September 2013

Academic Editors M-H Hsu G Mishuris L Rebollo-Neira and Q Song

Copyright copy 2013 T Glimm and N HenscheidThis is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider the geometric optics problem of finding a system of two reflectors that transform a spherical wavefront into a beamof parallel rays with prescribed intensity distribution Using techniques from optimal transportation theory it has been shownpreviously that this problem is equivalent to an infinite-dimensional linear programming (LP) problem Here we investigatetechniques for constructing the two reflectors numerically by considering the finite-dimensional LP problems which arise asapproximations to the infinite-dimensional problem A straightforward discretization has the disadvantage that the number ofconstraints increases rapidly with the mesh size so only very coarse meshes are practical To address this well-known issue wepropose an iterative solution scheme In each step an LP problem is solved where information from the previous iteration stepis used to reduce the number of necessary constraints As an illustration we apply our proposed scheme to solve a problem withsynthetic data demonstrating that themethod allows formuch finermeshes than a simple discretizationWe also give evidence thatthe scheme converges There exists a growing literature for the application of optimal transportation theory to other beam shapingproblems and our proposed scheme is easy to adapt for these problems as well

1 Introduction

The following beam shaping problem from geometric opticswas described in [1] see Figure 1 Suppose a point sourceemits a spherical wavefront with a given intensity distri-bution The problem we are concerned with consists oftransforming this input beam into an output beam of parallellight rays with a prescribed intensity distributionThis trans-formation is to be achieved with a system of two reflectorsThe problem has some practical importance in engineeringsee further literature cited in [1]

The first rigorous mathematical solution to the problemwas provided in [2] using an approach based on the theoryof optimal transportation [3ndash6] See also the references [7ndash11]which deal with other beam shaping problems using relatedtechniques

The result of [2] is summarized in Section 2 below withTheorem A5 stating the main result The central feature is

that the original reflector design problem is reformulated asan infinite-dimensional constrained optimization problemnamely the problem ofminimizing a certain linear functionalon a function space It is the dual problem for the problem offinding a map from the input aperture to the output aperturewhich minimizes a certain cost functional (see Theorem 3 inSection 2 for the exact statement)

This reformulation of the problem not only is of the-oretical value for questions of existence and uniqueness ofsolutions but it also translates into a practical method forfinding the solution In fact the discretization of the infinite-dimensional constrained optimization problem is a standardlinear programming problem and can be solved numericallyThis was already observed for similar beam shaping problemsby Glimm and Oliker [7 8] and independently by Wang [9]

There has been increased interest in numerical methodsfor optimal transportation problems in the last 10 yearsMost work has concentrated on theMonge-Ampere equation

2 ISRN Applied Mathematics

Table 1 Results of runs with the ldquoSimple Discretizationrdquo Scheme 5 for several edge lengths ℎ Memory was insufficient for the LP solver tohandle the case ℎ = 004 so the maximum number of mesh points that could be handled was about 1750 on each aperture The data usedwas the data from Section 41 Symbols used ℎ edge length119873 number of mesh points in119863119872 number of mesh points in 119879 time (s) timein seconds constr number of constraints

ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877

2Time (s) Constr

009 511 522 368393119864 minus 3 125443119864 minus 3 458391119864 minus 3 191131119864 minus 3 297016 266742008 660 663 296216119864 minus 3 730738119864 minus 4 339579119864 minus 3 108343119864 minus 3 467627 437580007 880 874 233561119864 minus 3 743916119864 minus 4 283982119864 minus 3 115916119864 minus 3 101342 769120006 1214 1214 13051119864 minus 3 712914119864 minus 4 210579119864 minus 3 828341119864 minus 4 174949 1473796005 1752 1757 911319119864 minus 4 266152119864 minus 4 120934119864 minus 3 461313119864 minus 4 459136 3078264

which arises in the special case of optimal transport in R119899

with quadratic cost (also called 1198712 optimal transport) [12ndash

20] although some authors have treated costs proportionalto the distance (1198711 optimal transport) [21] These methodsare based on fluid mechanical approaches [13] or variousfinite difference approaches [17]They are generally faster andallow for much larger mesh sizes than methods based on adiscretization of the linear programming problem but sincethey use the special structure of theMonge-Ampere equationthey are not directly applicable tomore general cost functionsor more general manifolds

A linear programming approach similar to the one weinvestigate here for more general situations for optimalapproachwas investigated byRuschendorf andUckelmann in2000 [22] and more recently in two papers by Canavesi et al[23 24] who also propose a new algorithm inspired by butdistinct from a linear programming approach for the singlereflector problem treated in [8 9]

As noted in [22] and other works an immediate obstaclefor the linear programming approach is that the numberof constraints becomes very large even for relatively coarsemeshesmdashif there are 119872 sample points in the input aperture(denoted by 119863 in Figure 1) and 119873 points in the outputaperture (119879 in Figure 1) then the number of constraints is119872 sdot 119873 For instance Ruschendorf and Uckelmann note in[22] that the LP solver they used SOPLEX was designed tohandle up to 2 million variables This corresponds to meshsizes of approximately 1400 points on the input and outputaperture in our problem if one employs a straightforwarddiscretization schemeWe found that in our implementationthe solver MOSEK (Version 7) could handle about 3 millionconstraints corresponding to approximately 1750 samplepoints on the domains 119879 and 119863 (see Table 1 for details)Ruschendorf and Uckelmann noted in [22] that for betterresults one needs carefully designed programs This is whatwe supply here for our problem we devised a more elaborateiterative method where discretized systems are solved onfiner and finer grids in each step using information fromthe previous solution to choose only a subset of all possibleconstraints This drastically slows the growth of the numberof constraints Details are described in Section 3 With thisstep-wise mesh refinement scheme as an illustration wewere able to obtain solutions on meshes with about 10300points on each aperture using MATLAB and MOSEK 7 acommercially available LP solver The number of constraints

0

Reflector 1

m

Plane

z = z(x)

z

x

S2

Reflector 2

Dz = minusd

120588(m)

T

(x minusd)

Figure 1 Sketch of the reflector problemThe point source is locatedat the origin 0 of the coordinate systemThe coordinate system in thelower left-hand corner explains our use of coordinates The outputbeam propagates in the direction of the negative 119911-axis Points in theplane perpendicular to the 119911-axis are denoted by the vector x isin R2Correspondingly points in three-dimensional space are denoted by(x 119911) See Section 2 for more Reprinted from [2]

needed for this was only a fraction of the size of the ldquofullrdquosystem For instance the run displayed in Table 2 used only033 of all possible constraints

We note that our proposed algorithm does not make anya priori symmetry assumptions on the form the reflectors Itcan also be adapted for the numerical solution of other beamshaping problems for which a formulation using optimaltransportation theory has been found for example those in[7 8 25]

This article is organized as follows in Section 2 we recallthe reformulation of the reflector construction problem as alinear programming problem as given in [2] and fix somenotation We then describe a basic discretization schemefor the numerical solution and propose an improved ldquomeshrefinementrdquo scheme in Section 3 The following Section 4 isdevoted to numerical illustrations and tests of this schemeWe first derive an explicit analytical solution to be used assynthetic data for our numerical work Then we compute thesolution numerically and analyze the error of approximation

ISRN Applied Mathematics 3

Table 2 Results of an iterative run with the ldquoIterative Refinementrdquo Scheme 6 Memory was insufficient to handle the sixth iteration due tomatrix size limitations in MATLAB Note the run times (last two rows) The bulk of the algorithm run times is taken in setting up the linearprogramming problem that is deciding which constraints to include Solving the LP once it is set up is fairly fast (last column)The data usedwas that of Section 41 the number of nearest neighbors searched over 120581nn = 6 and the quasi-activity threshold was 120576 = 001ℎ

2 119896 iterationnumber119872 number of mesh points in119863119873 number of mesh points in 119879 constr number of constraints con dens constraint density thatis used constraints as a percentage of all possible constraints119872sdot119873 time computation time for iterative step in seconds including generatingmeshes building and solving the LP problem LP time time in seconds to solve the LP problem

119896 ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877

2Constr Con dens Time (s) LP time (s)

0 0064 1059 1059 12884119864 minus 3 377819119864 minus 4 188783119864 minus 3 409084119864 minus 4 1121481 100 15962 1231461 00512 1675 1678 940425119864 minus 4 336918119864 minus 4 126944119864 minus 3 363526119864 minus 4 56080 200 208533 09388772 004096 2647 2647 756437119864 minus 4 167174119864 minus 4 753459119864 minus 4 185922119864 minus 4 89724 128 392982 1598313 0032768 4171 4175 364902119864 minus 4 963559119864 minus 5 525941119864 minus 4 121839119864 minus 4 141550 081 969239 2544054 00262144 6582 6567 303932119864 minus 4 88103119864 minus 5 530343119864 minus 4 141908119864 minus 4 224327 052 236379 4325245 002097152 10313 10318 249854119864 minus 4 99045119864 minus 5 39372119864 minus 4 151851119864 minus 4 351514 033 590925 921891

We conclude with a brief summary of the advantages anddrawbacks of the method and propose future work

2 Formulation as LinearProgramming Problem

We now briefly review the notation for the problem posedin [1] as well as the result of [2] which reformulates thereflector design problem as a linear programming problemMore details are contained in Appendix A

Consider the configuration shown in Figure 1 A pointsource at the origin 119874 = (0 0 0) generates a sphericalwavefront over a given input aperture 119863 contained in theunit sphere 1198782 We use the bar in the notation for 119863 to stressthat 119863 is a closed set The input beam has a given intensitydistribution By means of two reflectors this wavefront is tobe transformed into a beamof parallel rays propagating in thedirection of the negative 119911-axis This output beam is requiredto have a prescribed intensity distribution The cross sectionof the output beam in a plane perpendicular to the directionof propagation is called the output aperture and is denotedby 119879 Again we use the notation 119879 to stress that 119879 is closedCertain regularity conditions apply to119863 and119879 see [2] for thetechnical details

Denote points in space R3 by pairs (x 119911) where x isin

R2 is the position vector in a plane perpendicular to thedirection of propagation and 119911 isin R is the coordinate inthe (negative) direction of propagation See again Figure 1for our convention on the direction of the 119911-axis Points onthe unit sphere 119878

2 will typically be denoted by m isin 1198782

their components are also written as m = (m119909 119898119911) with

|m119909|2+ 1198982

119911= 1

We fix the output aperture in the plane 119911 = minus119889 We willseek to represent the first reflector as the graph of its polarradius 120588(m) (form isin 119863) and the second reflector as the graphof a function 119911(x) (for 119911 isin 119879) See Figure 1 That is

Reflector 1 1198771= 120588 (m) sdotm | m isin 119863

Reflector 2 1198772= (x 119911 (x)) | x isin 119879

(1)

The geometrical optics approximation is assumed Itfollows from general principles of geometric optics that allrays will have equal length from (0 0 0) to the plane 119911 =

minus119889 this length is called the optical path length and will bedenoted by 119871 We define the reduced optical path length asℓ = 119871 minus 119889

Oliker [1] showed that local energy conservation trans-lates into a complicated partial differential equation ofMonge-Ampere type for 120588(m) As noted in [1] the resultingequation is quite involved and a rigorous analysis of thisequation seems very difficult See equation (59) in [1]

To amend this the problem was reformulated in [2]as an infinite dimensional linear programming problemwhich makes a complete analysis possible both concerningtheoretical results on existence and uniqueness and gives amethod for practical computations

For this the following function 119870(m x) called the costfunction in analogy with the theory of optimal transporta-tion plays an important role

119870 (m x) =ℓ minus ⟨m

119909 x⟩

2ℓ (ℓ2 minus |x|2) (1 + 119898119911)minus

1

4ℓ2

for m = (m119909 119898119911) isin 119863 x isin 119879

(2)

In further preparation the following two transformationsare needed

Definition 1 Let 119911 = 119911(x) be a continuous function definedon 119879 sube R2 Then define the function

(x) = 1

2ℓminus

119911 (x)ℓ2 minus |x|2

for x isin 119879 (3)

Definition 2 Let 120588 = 120588(m) be a continuous function definedon119863 sube 119878

2 with 120588 gt 0 Then define the function

120588 (m) = minus1

2ℓ+

1

2120588 (m) sdot (119898119911+ 1)

for m isin 119863 (4)

The following characterization of the solution (120588 119911) of thereflector problem described above was given in [2]


Theorem 3 If the pair (120588 119911) is a solution of the above reflectordesign problem then the transformed pair (log 120588 log ) mini-mizes the functional

F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (5)

on the function space Adm(119863 119879) = (119903 120577) isin 119862(119863) times 119862(119879) |

119903(m) + 120577(x) ge log119870(m x) for all m isin 119863 x isin 119879

One can solve the reflector construction problem bysolving the infinite-dimensional LP problem in the abovetheorem and then applying the inverse transformations of (3)and (4) to recover 120588(m) and 119911(x) In the remainder of thispaper we will concentrate on numerical solutions for thisproblem

It is also possible to characterize the ray-tracing mapanalytically

Definition 4 Let (120588 119911) isin 119862(119863) times 119862(119879) be a solution of thereflector problem Define its ray-tracing map as a set-valuedmap 120574 119863 rarr subsets of 119879 via

120574 (m) = x isin 119879 | 120588 (m) (x) = 119870 (m x) for m isin 119863

(6)

In [2] it is shown that 120574(m) is in fact single-valuedfor almost all m isin 119863 Therefore 120574 may be regarded as atransformation 120574 119863 rarr 119879 Indeed one can show thata ray emitted from the origin in the direction m isin 119863 willbe reflected to a ray in the negative 119911-direction labeled byx = 120574(m) isin 119879 see [2]

More details and rigorous statements are found inAppendix A

3 Numerical Schemes

In the following we describe two schemes for solving theminimization problem in Theorem 3 numerically First welist the given data on which the solution depends

(i) Domains 119863 sube 1198782 119879 sube R2 (The rigorous result

in [2] required the additional technical assumption(0 0 minus1) (0 0 1) notin 119863 For practical purposes theseassumptions can often be dropped)

(ii) Nonnegative integrable functions 119868(m) m isin 119863 and119871(x) x isin 119879 with int

119863119868119889120590 = int

119879119871119889119909

(iii) A reduced optical length ℓ(iv) The value 120588(m

1) = 1205881for some fixed m

1isin 119863 where

120588 is the radial function of the first reflector (Note thatwithout this constraint the reflectors are not uniquelydeterminedThis can be seen in the expression for thefunctional in Theorem 3 Indeed any transformationof the form 119903 997891rarr 119903 + 119888 120577 997891rarr 119911 minus 119888 for constant 119888 willleave the objective functional unchanged)

The first numerical scheme Scheme 5 is a straightforwarddiscretization of the minimization problem inTheorem 3 viameshes on the domains119863 and 119879

As described in the introduction this scheme ismemory-intensive because every pair of points from 119863 and 119879 givesrise to a constraint To address this issue we propose amore sophisticated scheme Scheme 6 which is an iterativescheme using finer and finermeshes where in each iterationinformation from the previous solution is utilized to reducethe number of constraints

31 ldquoSimple Discretizationrdquo Scheme The characterization ofsolutions to the reflector problem by means of the minimiza-tion of a linear functional inTheorem 3 immediately suggestsa solution algorithm by means of a discretization Essentiallythe same algorithm was used for different problems in [22ndash24]

Scheme 5 (simple discretization) (i) Create a mesh in theinput domain 119863 by choosing sets 119889

(0)

1 119889(0)

2 119889

(0)

119872(0) sube 119863

where the interiors of any 119889(0)119894

and 119889(0)

119895are disjoint (for 119894 = 119895)

and119863 = ⋃119872(0)

119894=1119889(0)

119894 For instance 119889(0)

1 119889(0)

2 119889

119872(0) may be a

triangulation of119863(ii) Similarly create a mesh in the output domain 119879 by

choosing sets 119905(0)1 119905(0)

2 119905

(0)

119873(0) sube 119863 where the interiors of

any 119905(0)119894

and 119905(0)

119895are disjoint (for 119894 = 119895) and 119879 = ⋃

119873(0)

119895=1119905(0)

119895

(iii) Choose sample pointsm(0)119894

isin 119889(0)

119894(for 119894 = 1 119872

(0))and x(0)119895

isin 119905(0)

119895(for 119895 = 1 119873

(0)) Here we may assume thatm(0)1

= m1is the same point as given in the data in (iv) above

(iv) Find the solution (119903(0)119894119872(0)

119894=1 120577(0)

119895119873(0)

119895=1) of the following

LP problem

Minimize119872(0)

sum

119894=1

119903(0)

119894119868 (m(0)119894) area (119889(0)

119894)

+

119873(0)

sum

119895=1

120577(0)

119895119871 (x(0)119895) area (119905(0)

119895)

(7)

subject to 119903(0)

1= log 120588

1

where 1205881is as given in the data in (iv) above and

(8)

119903(0)

119894+ 120577(0)

119895ge log119870(m(0)

119894 x(0)119895)

for 119894 = 1 119872(0) 119895 = 1 119873

(0)

(9)

(v) Find the numbers 120588(0)119894

119894 = 1 119872(0) and 119911

(0)

119895 119895 =

1 119873(0) such that 119903(0)

119894= log 120588(0)

119894and 120577(0)

119895= log (0)

119895

This is straightforward by taking the inverse of thetransformations given in Definition 1 and Scheme 5

Then 120588(0)

119894is an approximation for the true value 120588(m(0)

119894)

of the radial function of the first reflector evaluated at thesample point m(0)

119894for 119894 = 1 119872

(0) Similarly 119911(0)119895

isan approximation for the true value 119911(x(0)

119895) of the function

describing the second reflector evaluated at the sample point


x(0)119895

for 119895 = 1 119873(0) We are using the superscript ldquo(0)rdquo

here to distinguish this solution from additional iterativeapproximations we will obtain with the iterative schemedescribed below

Note that (7) is a discretization of the sum of integrals inthe functionalF(119903 120577) = int

119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909 We

have chosen the simplest discretization here Depending onthe geometry of the problem other discretization schemes[26] may yield better results In practice we have chosenpartitions 119889

119894and 119905119895of approximately uniform size and diam-

eter but depending on the form of the intensity distributionsother partitions may be more appropriate

It is important to point out that this discretization alsoyields a discretized version of the ray-tracing map 120574 (seeDefinition 4) Namely for each index 119894 = 1 119872

(0) thereis at least one corresponding index 119895

lowast 1 le 119895lowastle 119873(0) where

the constraint corresponding to the pair (119894 119895lowast) is active thatis where we have

119903(0)

119894+ 120577(0)

119895lowast = log119870(m(0)

119894 x(0)119895lowast ) (10)

This means that the pointm(0)119895lowast is approximately the image of

the point x(0)119894

under the ray-tracing map 120574

32 ldquoIterative Refinementrdquo Scheme As noted in the introduc-tion one of the drawbacks of Scheme 5 is that the constraintset in the linear programming problem in the penultimatestep becomes large very fast with finer meshes and thecorresponding problem becomes too large to handle forstandard LP solvers We were not able to solve problems withmore than about 1750 mesh points on each aperture on astandard PC with 4GB RAM (119872(0) asymp 1750 119873

(0)asymp 1750)

We addressed this problem by developing an iterativescheme First the problem is solved for a mesh with relativelyfew sample points (say 119872

(0)= 119873(0)

= 1000) Then a finermesh is chosen and the previous solution is used to reducethe number of constraints

Specifically we have the following scheme dependingon a number 120576

(1)gt 0 which we call the ldquoquasi-activity

thresholdrdquo and an integer 120581nn the nearest neighbor searchdepth to be explained in detail below

Scheme 6 (iterative refinement) Use Scheme 5 to find aninitial solution (119903

(0)

119894119872(0)

119894=1 120577(0)

119895119873(0)

119895=1) for119872(0) sample points on

119863 and119873(0) sample points on 119879 respectively

Definition 7 Call a pair (119894 119895) with 119894 isin 1 119872(0) and 119895 isin

1 119873(0) ldquoquasi-activerdquo if

119903 (m(0)119895) + 120577 (x(0)

119895) minus log119870(m(0)

119894 x(0)119895) le 120576(0) (11)

Note that the pair is ldquoactiverdquo in the standard sense of linearprogramming theory if the above term is equal to zero

(i) Create a second finer mesh on 119863 by choosing sets119889(1)

1 119889(1)

2 119889

(1)

119872(1) sube 119863 where the interiors of any

119889(1)

119894and 119889

(1)

119895are disjoint (for 119894 = 119895) and119863 = ⋃

119872(1)

119894=1119889(1)

119894

Here 119872(1) is chosen larger than 119872(0) meaning that

we have a finer mesh(ii) Similarly create a second mesh on 119879 by choosing sets

119905(1)

1 119905(1)

2 119905

(1)

119873(1) sube 119863 where the interiors of any 119905

(1)

119894

and 119905(1)


119873(1)

119895=1119905119895

(1)


isin 119889(1)

119894(for 119894 = 1 119872

(1))and x(1)

119894isin 119905(1)

119895(for 119895 = 1 119873

(1)) Here we mayassume that m(1)

1= m1is the same point as given in

the data in (iv) above(iv) For each pair (119894 119895) with 119894 = 1 119872

(1) and 119895 =

1 119872(1) find the 120581nn nearest neighbors of m(1)

119894

from the previous mesh on 119863 and the 120581nn nearestneighbors of x(1)

119895from the previous mesh on 119879

Definition 8 Say that the pair (119894 119895) is potentially active if atleast one of the 1205812nn pairs consisting of a nearest neighbor ofm(1)119894

and a nearest neighbor of x(1)119895

is quasi-active in the abovesense

(i) Find the solution (119903119894119872(1)

119894=1 120577(1)

119895119873(1)


LP problem

Minimize119872(1)

sum

119894=1

119903(1)

119894119868 (m(1)119894) area (119889(1)

119894)

+

119873(1)

sum

119895=1

120577(1)

119894119871 (x(1)119895) area (119905(1)

119895)


1= log 120588

1

119903(1)

119894+ 120577(1)

119895ge log119870(m(1)

119894 x(1)119895)

for any potentially active pair

(in the above sense)

(12)

(ii) Find the numbers 120588(1)119894

119894 = 1 119872(1) and 119911

(1)

119895 119895 =

1 119873(1) such that 119903(1)

119894= log 120588(1)

119894and 120577(1)

119895= log (1)

119895

The idea of Scheme 6 is to solve the discretized LP on acoarse mesh first and then use this information to reduce thenumber of constraints needed for the LP on a finer meshA key difference between Schemes 5 and 6 is that in thediscretized LP problem in Scheme 6 not all pairs of samplepoints from 119863 and 119879 represented by pairs of indices (119894 119895)are included in the list of constraints In fact we would inprinciple only need to include those where the constraint isactive that is where the corresponding inequality holds withequality Of course this information is not available a prioriInstead we use the following heuristic A constraint shouldonly be included in the LP problem if the correspondingpoint on the second reflector is close to the image of thecorresponding point on the first reflector under the ray-tracing map In terms of the LP problem formulation the


corresponding pair of points is located close to an activepair from the previous iteration We call such a constraintldquopotentially activerdquo Our heuristic is that a constraint shouldbe considered potentially active if there is a pair of itsrespective 120581nn nearest neighbors which corresponds to aquasi-active constraint form the previous iteration Hereldquoquasi-activerdquo constraints include active constraints (wherethe constraint is attained ie the term 119903(m(0)

119895) + 120577(x(0)

119895) minus

log119870(m(0)119894 x(0)119895) is zero) and those which are ldquoalmost activerdquo

in the sense that the term 119903(m(0)119895) + 120577(x(0)

119895) minus log119870(m(0)

119894 x(0)119895)

is positive but small ldquoSmallnessrdquo is encoded in the quasi-activity threshold 120576

(1) Clearly increasing 120576(1) means more

constraints are included but also potentially the solution ofthe LP represents a better approximation of the exact reflectorpair

In our numerical tests we found it advantageous to scalethe functions 119868(m) and 119871(x) so that the approximations forthe integrals int

119863119868119889120590 and int

119879119871119889119909 yield exactly the same result

Note that the algorithm in Scheme 6 can be iterated againWe can use the first iterative solution 120588

(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873

(1) in Scheme 6 to obtain a second iterativesolution 120588

(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon

Note that (7) is a discretization of the sum of integralsin the functionalF(119903 120577) = int

119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909

We note that we have chosen again the simplest discretizationhere for the sum of integrals int

119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909

As in Scheme 6 depending on the geometry of the problemother discretization schemes [26] may yield better results

33 Choice of Thresholds for Iterative Refinement Scheme 6For the iteration of Scheme 6 described above we haveto choose a corresponding sequence of mesh sizes(119872(0) 119873(0)) (119872(1) 119873(1)) (119872(2) 119873(2)) We also need

to choose a corresponding sequence of thresholds 120576(1)

120576(2) 120576(3)

This last question is of a certain practical importanceIndeed if the thresholds 120576

(119896)119896are ldquohighrdquo (eg if they

decreases slowly) then ldquomanyrdquo constraints are included ineach iteration meaning that the size of the LP problemsincreases quickly If the sequence 120576(119896)

119896is ldquolowrdquo then ldquofewrdquo

constraints are included in each iteration This is of course inprinciple desirable but if the thresholds are too low this couldcause that some index 119894 (with 1 le 119894 le 119872

(119896)) may not evenbe included in any of the constraints with any of the indices119895 (with 1 le 119895 le 119873

(119896)) or vice versa This would cause the

LP problem to be unbounded and hence there would be nosolution

We can make a very rough estimate for a good choiceof the sequence 120576

(119896)119896 assuming for simplicity that 119872(119896) asymp

119873(119896) Let us assume that each sample point 119898(119896)

119894in the input

aperture 119863 is paired via the ray-tracing map 120574 with a uniquepoint x(119896)

119895(119894)asymp 120574(119898

(119896)

119894) and vice versa Thus there are 119872

(119896)

such pairs Let 119903(m)m isin 119863 and 120577(x) x isin 119879 denote the

solutions of the reflector problem Then the set of all pointsin (m x)-space close to a given pair (m(119896)

119894 x(119896)119895(119894)

) that satisfy119903(x) minus 119911(x) minus log119870(m x) lt 120576

(119896) is approximately an ellipsoidfor small 120576(119896) This can be seen by using the Taylor expansionaround the pair (m(119896)

119894 x(119896)119895(119894)

) in (m x)-space The volume of

this ellipsoid is proportional to (120576(119896))2

So the proportionof pairs of points that satisfy the above constraint amongall constraints is roughly proportional to (120576

(119896))2

Since eachpair (m(119896)

119894 x(119896)119895) that satisfies this inequality corresponds to a

constraint included in the LP of the 119896th iteration the totalnumber of constraints should be roughly proportional to(119872(119896))3sdot (120576(119896))2 So to keep the number of constraints growing

linearly throughout the iterations based on these heuristicsone may choose

120576(119896)

=119862

119872(119896) (13)

where 119862 is a constant These heuristics are very rough Inpractice we have found that when using the formula 120576

(119896)=

119862(119872(119896))120572 there is a range of values of 120572 which all produce

similar results See Section 42

4 Numerical Tests

To test the validity of the numerical scheme described abovewe used it on a case where the solution is known in analyticform In the next subsection we first describe this specialanalytic solution and then discuss our results

41 A Data Set We have constructed a particular data setusing an explicit analytical solution described in more detailin Appendix B The data set of this section is obtained fromthemore general solution of Appendix B by choosing 119886 = 119887 =

0 and 119888 = minus04 120572 = 1 and 119877 = 13 We consider the inputaperture

Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)

which is a spherical cap centered at the point (0 0 minus1) Thedata do not represent a physically possible set of reflectorssince there would be blockage However the example isuseful as an illustration and the numerical scheme itselfis completely independent of the shape of the apertures orany a priori symmetries of the problem Using the resultsfrom Appendix B we can now find the output aperture in astraightforward manner as follows

119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)

We choose a constant output intensity

119871 (x) = 1 for x isin 119879 (16)

Using the relation (B7) this gives the input intensity

119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)

Figure 2 Plots of the reflectors given by (18) (a) and (19) (b) in theexplicit data set given in Section 41

The two reflectors are given as follows

radial radius 120588 for 1198771 120588 (m) =

0765

13 + 04 119898119911

(18)

equation for 1198772 119911 = minus025 (119909

2+ 1199102) + 06 (19)

The corresponding reduced optical path length is

ℓ = 119877 + 2120572 = 29 (20)

The two reflectors are shown in Figure 2 1198771is part of a

spheroid and 1198772is part of a paraboloid see Appendix B for

details

42 Results Using the data set from Section 41 we con-ducted a series of numerical tests to establish the validity ofthe numerical Scheme 6 Since there are explicit analyticalexpressions for the two solutions 120588(119898) and 119911(119909) describingthe two reflectors as given in (18) and (19) respectively wewere able to compute the error of approximation directly Seealso Figure 2 for plots of these reflectors

The Schemes 5 and 6 were implemented in MATLABwith the package distmesh [27] for mesh generation ineach iteration step and the commercial linear programming

solver MOSEK (version 7) [28] for solving the resulting LPproblems Computations were performed on an Intel Core i7-2620M270 GHz with 4GB RAM running Windows 7 (64bit)

The mesh generation algorithm is based on a relaxationscheme of forces in a truss structure [27] For this the inputis the desired edge length ℎ not the number of mesh points119872 (for the domain Ω) or119873 (for the domain 119879) respectivelyHowever it can be seen from geometric considerations thatthe relation between the average edge length ℎ and thenumber of mesh points119872 on119863 obeys

ℎ sim1

radic119872 (21)

In the tables below we show both the number of meshpoints 119872119873 and the edge length ℎ as we believe both areinformative

Our primary aim is to demonstrate the practicabilityof the proposed Scheme 6 For comparison we first testedthe performance of the ldquoSimple Discretizationrdquo Scheme 5The results are summarized in Table 1 The maximum meshsizes attainable are about 1750 points corresponding toapproximately 3 million constraints We found that this wasthe maximum size for the LP solver

To test Scheme 6 we chose a sequence of desired edgelengths ℎ(0) ℎ(1) by reducing ℎ by 20 in each step that isℎ(119896)

= 08119896ℎ(0) Then based on the heuristics given above and

in Section 33 we determined the corresponding constraintthresholds with the following formula

120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)

where 119862 and 120583 are constants The heuristic arguments inSection 33 yield 120583 = 2 but we also tested other valuesAs indicated in Section 33 ldquolargerdquo values for 119862 mean thatldquomanyrdquo constraints are included in the LP in each stepwhich potentially means good approximations but also highmemory usage In contrast ldquosmallrdquo values for 119862 improvememory usage but may mean that the LP problem maybecome unboundedThis is the case if constraints that wouldbe active in the LP problem comprising of all constraints areleft out in an iterated LP problem In our tests the only timethis happened was when we chose 119862 = 0 that is if the quasi-activity threshold is set to zero and only active constraintsare considered ldquoquasi-activerdquo We conclude that our resultsare very robust with respect to 119862 and 120583 meaning for a broadrange of values we reached the terminal iteration with ℎ =

002097152 see Table 2 This is because the memory limitingissue in our tests was not the LP solver but the maximum sizeof arrays in MATLAB when computing the cost matrix seeparagraph below We also conducted some informal tests ofthe algorithm dependence on the output and input intensities119868(m) and 119871(x) specifically to know whether the algorithm isstable under small perturbations of 119868(m) and 119871(x) For thiswe added small perturbation 120575 cos(119898

119910) to 119868(m) and 120575 cos(119909)

to 119871(119909) then renormalized 119871(x) so that the two integrals werethe same We found that the algorithm was indeed stablein the sense that the same maximum number of points wasreached for different values of 120575


Table 3 Results of the final iteration of the ldquoIterative Refinementrdquo Scheme 6 with different values of the nearest neighbor search depth 120581nnAll other data used was the same as in Table 2 Note that the nearest neighbor search depth 120581nn has a large influence on the size of the LPproblem but relatively little influence on the errors or run times Column labels are identical to Table 2 except for the first column

120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877


2 002097152 10313 10318 271973119864 minus 4 904017119864 minus 5 438374119864 minus 4 137601119864 minus 4 96256 0090 594776 2824254 002097152 10313 10318 248492119864 minus 4 977635119864 minus 5 392226119864 minus 4 150005119864 minus 4 226006 0212 575217 6098496 002097152 10313 10318 249854119864 minus 4 99045119864 minus 5 39372119864 minus 4 151851119864 minus 4 351514 0330 590925 9218918 002097152 10313 10318 249816119864 minus 4 990729119864 minus 5 394393119864 minus 4 15188119864 minus 4 482058 0453 63224 129115

Results of a typical run are summarized in Table 2 Ourresults show that the proposed scheme is easy to implementand makes it possible to use much finer meshes than asimple discretization scheme The largest number of pointswere about 10000 on each aperture It is notable that whatprevented our tests to include more plot points was not theLP solver but memory problems related to the maximum sizeof matrices in MATLAB during the set-up phase of the LPproblem when computing the matrix of cost coefficients 119870

119894119895

which is of size 119872(119896) times 119873(119896) For the configuration we used

the maximum size of arrays inMATLAB is about 280 millionvariables see [29]

These results give indication that the scheme convergesand that the maximum error for the first reflector decreasesproportional to 119872

120590 or equivalently proportional to ℎ2120590

where 120590 = minus0773 data from Table 2 The same formulationfor scheme 5 yields an exponent of 120590 = minus1181 unsurpris-ingly a slightly more rapid decrease data from Table 1 Thegeometric distribution of errors is indicated in Figure 3

To investigate the role of the nearest neighbor searchdepth 120581nn we performed several tests with different valuesThe results are summarized in Table 3 The value of 120581nnhas very little influence on the error of approximation Thismakes intuitive sense as the crucial issue in each iterationis that those constraints which are active in the ldquofullrdquo LPproblem (ie the one that includes all possible constraints)are included As long as all active constraints are includedone expects to get the same solution Note however thatthe number of constraints changes quite drastically with 120581nnranging from 96256 (009 of all possible constraints) for120581nn = 2 to 482058 (045 of all possible constraints) for120581nn = 8 in this example The time to solve the LP problemsthus also varies from 282 seconds for 120581nn = 2 to 1291

seconds for 120581nn = 8 although those gains are dwarfed by theoverall run times of the algorithm which include assemblingthe LP problem

5 Conclusion and Future Work

We investigated two numerical schemes for solving aninfinite-dimensional optimal transportation problem arisingin reflector design a straightforward discretization and anovel iterative scheme which uses knowledge of the previoussolution in each step to reduce the number of constraintsThe latter scheme is easily adapted to similar transportation

times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08

minus06minus04 minus02

(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)

Figure 3 Error surface plots for reflector 1 (a) and reflector 2 (b)for the data of Table 3 with 120581nn = 2 For better visibility the plotswere obtained by interpolating the error at each mesh point to acontinuous function Note that the error distribution is relativelyuniform although slight spikes of the error tend to occur near theboundary

problems arising in beam shaping problems for example see[7ndash9]

As a proof of concept we implemented both schemesusing MATLAB and the commercial LP solver MOSEK andshowed that the new scheme is easy to implement and makes


it possible to solve the problem on much finer meshes Inparticular we demonstrated that the number of constraintsare greatly reduced by the new scheme to as little as 009of the full number of constraints Our aim was to show thevalidity of the scheme and we contend that performancecould be further improved for instance by using a compiledcomputer language or a more specialized LP solver fortransportation problems

One drawback of the scheme is that it is quite slowSee Table 2 The most time-consuming part of the algorithmwas not the solution of the LP problem but assembling theproblem that is deciding which constraints to be includedWe believe that this step in particular could be improved byusing a compiled computer language instead of MATLAB

There are a number of possible directions for futureresearch We did not rigorously prove that the schemeconverges although we strongly expect that it does Infact the decreasing error shown in Table 2 gives evidencefor this It would be valuable to have a rigorous proof ofconvergence We also expect that PDE-based schemes forsolving the transportation cost with quadratic costs [13 17ndash19] could be adapted to the problem at hand leading to fasteralgorithms

Appendices

A Details on Theorem 3

Wegive here somemore details onTheorem 3 fromSection 2We use the same notation as in this sectionThis is a summaryof definitions and results from [2]

In [2] the following notion of a reflector pair and itsassociated ray-tracing map is used

Definition A1 Let 119862gt0(119863) denote the set of all continuous

functions on 119863 whose range is contained in the interval(0infin) A pair (120588 119911) isin 119862

gt0(119863) times 119862(119879) is called a reflector

pair if 120588 gt 0 and

120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863

(x) = supmisin119863

(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)

Here notations (4) and (3) were used

Definition A2 Let (120588 119911) isin 119862gt0(119863) times119862(119879) be a reflector pair

Define its reflector map or ray-tracing map as a set-valuedmap 120574 119863 rarr subsets of 119879 via

120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)

In [2] it is shown that 120574(m) is in fact single-valued for almostallm isin 119863 Therefore 120574may be regarded as a transformation120574 119863 rarr 119879

If (120588 119911) is a ldquoreflector pairrdquo in the above sense the follow-ing can be shown [2] justifying the choice of nomenclatureIf physical copies of the corresponding surfaces (1) are madefrom a reflective material then a ray emitted from the originin the directionm isin 119863will be reflected to a ray in the negative119911-direction labeled by x = 120574(m) isin 119879

With this the reflector problem can be formulated rigor-ously

Problem A3 (reflector problem) For given input and outputintensities 119868(m)m isin 119863 and 119871(x) x isin 119879 respectively sat-isfying global energy conservation int

119863119868(m)119889120590 = int

119879119871(x)119889119909

find a pair (120588 119911) isin 119862gt0(119863) times 119862(119879) that satisfies the following

conditions

(i) (120588 119911) is a reflector pair in the sense of Definition A1

(ii) the ray-tracing map 120574 119863 rarr 119879 satisfies

int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)

for any Borel set 120591 sube 119879

Here 119889120590 denotes the standard area element on the sphere1198782 The second condition is local energy conservation

As we indicate below the reflector problem can bereformulated in the following form

Problem A4 (minimize the functional)

F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)

on the set Adm(119863 119879) = (119903 120577) isin 119862(119863) times 119862(119879)119903(m) + 120577(x) gelog119870(m x) for all m isin 119863 x isin 119879

Problem A4 is the dual problem to the problem offinding a measure-preserving map from 119863 sube 119878

2 to 119879 sube

R2 which minimizes the transportation functional 119875 997891rarr

int119863119870(m 119875(m))119868(x)119889120590 Problems A3 and A4 are equivalent

as expressed in the following theorem

Theorem A5 (see [2]) Let (120588 119911) isin 119862gt0(119863) times 119862(119879) be a

reflector pair Then (log 120588 log ) isin Adm(119863 119879) The followingstatements are equivalent

(i) (120588 119911) solves the reflector Problem A3(ii) (log 120588 log ) solves Problem A4

Thus solving the reflector construction problem in Prob-lem A3 is equivalent to solving the infinite-dimensional LPproblem in Problem A4 Indeed it can be shown that asolution as inTheorem A5 exists See Corollary 65 in [2]


B An Explicit Analytic Solution to theReflector Problem

In order to obtain a data set where an explicit analyticsolution is known we consider a special given configura-tion of reflectors and then solve the ldquoforwardrdquo problem ofdetermining the output intensity this system produces fora given input intensity This general family of solutions weobtainedwas used to generate the explicit synthetic data set inSection 41

B1 Construction of a Pair of Reflectors 1198771 1198772 Consider the

following set of two reflectors1198771and119877

2 sketched in Figure 4

Let a = (119886 119887 119888) be a given point in R3 and let 119877 gt 0 and120572 gt 0 be two positive numbers with 119877 gt |a| = radic1198862 + 1198872 + 1198882Now let the first reflector 119877

1be the boundary of a prolate

spheroid (ellipsoid of revolution) with foci at the origin Oand at a and with major diameter 119877 Thus for each pointP on 119877

1 the sum of the distances from each of the two

foci PO + Pa equals 119877 Let the second reflector 1198772be the

boundary of a paraboloid whose main axis is the negative119911-axis and whose focus is at a with focal parameter 2120572 Thusfor each point p = (119909 119910 119911) on 119877

2 the sum of the distance

to the focus and the shifted 119911-component Pa + (119911 minus 119888) equals2120572

Note that by definition of the two reflectors any coneof light rays emitted from the origin 119874 will be transformedinto a beam of parallel rays traveling in the direction of thenegative 119911-axis Indeed a ray emitted from119874will be reflectedoff 1198771toward the focus a This ray will then be reflected

in the direction of the negative 119911-axis by 1198772 See again

Figure 4It is not hard to find explicit expressions for the two

reflectors If we write 1198771= 120588(m) sdot m | m isin 119878

2 and 119877

2=

(x 119911(x)) | x = (119909 119910) isin R2 then we have the followingexpressions

radial function 120588 for 1198771 120588 (m) =

1198772minus |a|2

2 (119877 minusm sdot a) (B1)

equation for 1198772 minus 4120572119902 = |p|2 minus 4120572

2 (B2)

where we used the shifted coordinates (p 119902) = (119909 119910 119911) minus

(119886 119887 119888) = (x 119911) minus a

B2 Ray-Tracing Map 120574 One can now determine the ray-tracing map 120574 corresponding to the reflector pair 119877

1 1198772 See

again Figure 4Letm isin 119878

2 be a given directionThusm = (mx 119898119911) withmx isin R2 and |mx|

2+ 1198982

119911= 1 To determine the image 120574(m)

consider the ray emitted from the origin in the direction mThe ray will encounter the first reflector at the point 120588(m) sdotmand will then be reflected towards the second focus a Thusthe reflected ray can be parameterized by a+120582sdot(aminus120588(m) sdotm)where 120582 is a parameter With (B2) this yields that the point

z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O

Figure 4 Sketch of the reflectors 1198771and 119877

2 1198771is the boundary of

an ellipsoid whose foci are the points119874 and a 1198772is the boundary of

a paraboloid whose focus is a The sketch shows a two-dimensionalcross section A ray given by the direction m isin 119878

2 will be reflectedby 1198771and then by 119877

2to a ray in the direction of the negative 119911-axis

where the reflected ray hits the second reflector is a+120582lowastsdot (aminus

120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2

minus (119888 minus 120588119898119911)2((119877 minus 120588) minus (119888 minus 120588119898

119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)

The projection of this point to the 119909119910-plane is the value of theray-tracingmapThus we have the following explicit formulafor the ray-tracing map

120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886

119887) minus 120588 (m)mx) m = (mx 119898119911) isin 119878

2

(B4)

B3 Solution of the Forward Problem We can now solve theforward problem Given the reflector pair 119877

1 1198772as defined

above an input apertureΩ sube 1198782 and an intensity distribution

119868(m)m isin Ω find the output aperture 119879 sube R2 and the outputintensity 119871(x) x isin 119879

The output aperture is simply the image of Ω under theray-tracing map 120574 119879 = 120574(Ω) To find the induced intensity119871(x) use the defining property

int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)

for all Borel sets 120596 sube Ω This is an energy balance equationThe above integral equation allows us to find an explicit

expression for 119871(x) given 119868(m) or vice versa We consider forsimplicity the case thatΩ is contained in the left hemisphere


1198782

minus= 1198782cap (119909 119910 119911) isin R3 | 119911 lt 0 We can then use

coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus

(119909 119910) 997891997888rarr (119898119909 119898119910 minusradic1 minus 1198982

119909minus 1198982119910)

(B6)

In these coordinates the standard measure on 1198782 is given by

119889120590 = 119889119898119909119889119898119910|119898119911| where 119898

119911= minusradic1 minus 1198982

119909minus 1198982119910 Using

this (B5) is then equivalent to the equation

119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes

the Jacobian of the map 120574 ∘ 120591 With the help of a computeralgebra system like Mathematica this can be evaluatedexplicitly as

119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2

times ( minus 119898119911(2 (119888 + 119877) mx sdot (

119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)

where again119898119911= minusradic1 minus 1198982

119909minus 1198982119910

Acknowledgment

Tilmann Glimm gratefully acknowledges the contributionof V Oliker for the unpublished collaborative work on analgorithm similar to the one described in this paper for adifferent reflector design problemThisworkwas done during2002 to 2004 at Emory University

References

[1] V Oliker ldquoMathematical aspects of design of beam shapingsurfaces in geometrical opticsrdquo in Trends in Nonlinear AnalysisM Kirkilionis S Kromker R Rannacher and F Tomi Eds pp191ndash222 Springer Berlin Germany 2002

[2] T Glimm ldquoA rigorous analysis using optimal transport theoryfor a two-reflector design problem with a point sourcerdquo InverseProblems vol 26 no 4 Article ID 045001 p 16 2010

[3] Y Brenier ldquoPolar factorization andmonotone rearrangement ofvector-valued functionsrdquo Communications on Pure and AppliedMathematics vol 44 no 4 pp 375ndash417 1991

[4] L A Caffarelli ldquoAllocation maps with general cost functionsrdquoin Partial Differential Equations and Applications vol 177 ofLecture Notes in Pure and Applied Mathematics pp 29ndash35Dekker New York NY USA 1996

[5] W Gangbo and R J McCann ldquoOptimal maps in Mongersquos masstransport problemrdquo Comptes Rendus de lrsquoAcademie des Sciencesvol 321 no 12 pp 1653ndash1658 1995

[6] C Villani Optimal Transport Old and New vol 338 ofGrundlehren derMathematischenWissenschaften [FundamentalPrinciples of Mathematical Sciences] Springer Berlin Germany2009

[7] T Glimm and V Oliker ldquoOptical design of two-reflectorsystems the Monge-Kantorovich mass transfer problem andFermatrsquos principlerdquo IndianaUniversityMathematics Journal vol53 no 5 pp 1255ndash1277 2004

[8] T Glimm and V Oliker ldquoOptical design of single reflectorsystems and the Monge-Kantorovich mass transfer problemrdquoJournal of Mathematical Sciences vol 117 no 3 pp 4096ndash41082003

[9] X-JWang ldquoOn the design of a reflector antenna IIrdquoCalculus ofVariations and Partial Differential Equations vol 20 no 3 pp329ndash341 2004

[10] T Graf On the near-field reector problem and optimal transport[PhD thesis] Emory University 2010

[11] V Oliker ldquoDesigning freeform lenses for intensity and phasecontrol of coherent light with help from geometry and masstransportrdquoArchive for RationalMechanics andAnalysis vol 201no 3 pp 1013ndash1045 2011

[12] E Haber T Rehman and A Tannenbaum ldquoAn efficientnumerical method for the solution of the 119871

2optimal mass

transfer problemrdquo SIAM Journal on Scientific Computing vol32 no 1 pp 197ndash211 2010

[13] J-D Benamou and Y Brenier ldquoA computational fluid mechan-ics solution to the Monge-Kantorovich mass transfer problemrdquoNumerische Mathematik vol 84 no 3 pp 375ndash393 2000

[14] G L Delzanno L Chacon J M Finn Y Chung and GLapenta ldquoAn optimal robust equidistribution method for two-dimensional grid adaptation based on Monge-Kantorovichoptimizationrdquo Journal of Computational Physics vol 227 no 23pp 9841ndash9864 2008

[15] G Awanou ldquoSpline element method for the Monge-Ampereequationrdquo December 2010

[16] L Chacon G L Delzanno and J M Finn ldquoRobust mul-tidimensional mesh-motion based on Monge-Kantorovichequidistributionrdquo Journal of Computational Physics vol 230 no1 pp 87ndash103 2011

[17] J-D Benamou B D Froese and A M Oberman ldquoTwonumerical methods for the elliptic Monge-Ampere equationrdquoMathematical Modelling and Numerical Analysis vol 44 no 4pp 737ndash758 2010

[18] B D Froese and A M Oberman ldquoFast finite difference solversfor singular solutions of the elliptic Monge-Ampere equationrdquoJournal of Computational Physics vol 230 no 3 pp 818ndash8342011

[19] V Zheligovsky O Podvigina and U Frisch ldquoThe Monge-Ampere equation various forms and numerical solutionrdquo Jour-nal of Computational Physics vol 229 no 13 pp 5043ndash50612010

[20] V I Oliker and L D Prussner ldquoOn the numerical solution ofthe equation (120597

21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2

= 119891 and itsdiscretizations Irdquo Numerische Mathematik vol 54 no 3 pp271ndash293 1988

[21] J W Barrett and L Prigozhin ldquoA mixed formulation of theMonge-Kantorovich equationsrdquo Mathematical Modelling andNumerical Analysis vol 41 no 6 pp 1041ndash1060 2007

[22] L Ruschendorf and L Uckelmann ldquoNumerical and analyticalresults for the transportation problem of Monge-KantorovichrdquoMetrika vol 51 no 3 pp 245ndash258 2000


[23] C Canavesi W J Cassarly and J P Rolland ldquoObservationson the linear programming formulation of the single reflectordesign problemrdquo Optics Express vol 20 no 4 pp 4050ndash40552012

[24] C Canavesi W J Cassarly and J P Rolland ldquoImplementationof the linear programming algorithm for freeform reflectordesignrdquo p 84850E-84850E-6 2012

[25] X-J Wang ldquoOn the design of a reflector antennardquo InverseProblem vol 12 no 3 pp 351ndash375 1996

[26] R Cools ldquoAn encyclopaedia of cubature formulasrdquo Journal ofComplexity vol 19 no 3 pp 445ndash453 2003

[27] P-O Persson and G Strang ldquoA simple mesh generator inMatlabrdquo SIAM Review vol 46 no 2 pp 329ndash345 2004

[28] ldquoTheMOSEK optimization softwarerdquo httpwwwmosekcom[29] I The MathWorks Support pages for matlab 2012 httpwww

mathworkscomsupportsolutionsendata1-IHYHFZindex

html

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


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Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


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Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Table 1 Results of runs with the ldquoSimple Discretizationrdquo Scheme 5 for several edge lengths ℎ Memory was insufficient for the LP solver tohandle the case ℎ = 004 so the maximum number of mesh points that could be handled was about 1750 on each aperture The data usedwas the data from Section 41 Symbols used ℎ edge length119873 number of mesh points in119863119872 number of mesh points in 119879 time (s) timein seconds constr number of constraints

ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877

2Time (s) Constr

009 511 522 368393119864 minus 3 125443119864 minus 3 458391119864 minus 3 191131119864 minus 3 297016 266742008 660 663 296216119864 minus 3 730738119864 minus 4 339579119864 minus 3 108343119864 minus 3 467627 437580007 880 874 233561119864 minus 3 743916119864 minus 4 283982119864 minus 3 115916119864 minus 3 101342 769120006 1214 1214 13051119864 minus 3 712914119864 minus 4 210579119864 minus 3 828341119864 minus 4 174949 1473796005 1752 1757 911319119864 minus 4 266152119864 minus 4 120934119864 minus 3 461313119864 minus 4 459136 3078264

which arises in the special case of optimal transport in R119899

with quadratic cost (also called 1198712 optimal transport) [12ndash

20] although some authors have treated costs proportionalto the distance (1198711 optimal transport) [21] These methodsare based on fluid mechanical approaches [13] or variousfinite difference approaches [17]They are generally faster andallow for much larger mesh sizes than methods based on adiscretization of the linear programming problem but sincethey use the special structure of theMonge-Ampere equationthey are not directly applicable tomore general cost functionsor more general manifolds

A linear programming approach similar to the one weinvestigate here for more general situations for optimalapproachwas investigated byRuschendorf andUckelmann in2000 [22] and more recently in two papers by Canavesi et al[23 24] who also propose a new algorithm inspired by butdistinct from a linear programming approach for the singlereflector problem treated in [8 9]

As noted in [22] and other works an immediate obstaclefor the linear programming approach is that the numberof constraints becomes very large even for relatively coarsemeshesmdashif there are 119872 sample points in the input aperture(denoted by 119863 in Figure 1) and 119873 points in the outputaperture (119879 in Figure 1) then the number of constraints is119872 sdot 119873 For instance Ruschendorf and Uckelmann note in[22] that the LP solver they used SOPLEX was designed tohandle up to 2 million variables This corresponds to meshsizes of approximately 1400 points on the input and outputaperture in our problem if one employs a straightforwarddiscretization schemeWe found that in our implementationthe solver MOSEK (Version 7) could handle about 3 millionconstraints corresponding to approximately 1750 samplepoints on the domains 119879 and 119863 (see Table 1 for details)Ruschendorf and Uckelmann noted in [22] that for betterresults one needs carefully designed programs This is whatwe supply here for our problem we devised a more elaborateiterative method where discretized systems are solved onfiner and finer grids in each step using information fromthe previous solution to choose only a subset of all possibleconstraints This drastically slows the growth of the numberof constraints Details are described in Section 3 With thisstep-wise mesh refinement scheme as an illustration wewere able to obtain solutions on meshes with about 10300points on each aperture using MATLAB and MOSEK 7 acommercially available LP solver The number of constraints

0

Reflector 1

m

Plane

z = z(x)

z

x

S2

Reflector 2

Dz = minusd

120588(m)

T

(x minusd)

Figure 1 Sketch of the reflector problemThe point source is locatedat the origin 0 of the coordinate systemThe coordinate system in thelower left-hand corner explains our use of coordinates The outputbeam propagates in the direction of the negative 119911-axis Points in theplane perpendicular to the 119911-axis are denoted by the vector x isin R2Correspondingly points in three-dimensional space are denoted by(x 119911) See Section 2 for more Reprinted from [2]

needed for this was only a fraction of the size of the ldquofullrdquosystem For instance the run displayed in Table 2 used only033 of all possible constraints

We note that our proposed algorithm does not make anya priori symmetry assumptions on the form the reflectors Itcan also be adapted for the numerical solution of other beamshaping problems for which a formulation using optimaltransportation theory has been found for example those in[7 8 25]

This article is organized as follows in Section 2 we recallthe reformulation of the reflector construction problem as alinear programming problem as given in [2] and fix somenotation We then describe a basic discretization schemefor the numerical solution and propose an improved ldquomeshrefinementrdquo scheme in Section 3 The following Section 4 isdevoted to numerical illustrations and tests of this schemeWe first derive an explicit analytical solution to be used assynthetic data for our numerical work Then we compute thesolution numerically and analyze the error of approximation




119896 ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877











|m119909|2+ 1198982

119911= 1




(1)






119870 (m x) =ℓ minus ⟨m

119909 x⟩

2ℓ (ℓ2 minus |x|2) (1 + 119898119911)minus

1

4ℓ2

for m = (m119909 119898119911) isin 119863 x isin 119879

(2)



(x) = 1

2ℓminus

119911 (x)ℓ2 minus |x|2

for x isin 119879 (3)



120588 (m) = minus1

2ℓ+

1

2120588 (m) sdot (119898119911+ 1)

for m isin 119863 (4)




F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (5)







(6)



3 Numerical Schemes





119863119868119889120590 = int

119879119871119889119909



1isin 119863 where






(0)

1 119889(0)

2 119889

(0)

119872(0) sube 119863


and 119889(0)

119895are disjoint (for 119894 = 119895)

and119863 = ⋃119872(0)

119894=1119889(0)

119894 For instance 119889(0)

1 119889(0)

2 119889

119872(0) may be a


choosing sets 119905(0)1 119905(0)

2 119905

(0)


any 119905(0)119894

and 119905(0)


119873(0)

119895=1119905(0)

119895


isin 119889(0)

119894(for 119894 = 1 119872

(0))and x(0)119895

isin 119905(0)

119895(for 119895 = 1 119873




119894=1 120577(0)

119895119873(0)


LP problem

Minimize119872(0)

sum

119894=1

119903(0)

119894119868 (m(0)119894) area (119889(0)

119894)

+

119873(0)

sum

119895=1

120577(0)

119895119871 (x(0)119895) area (119905(0)

119895)

(7)


1= log 120588

1


(8)

119903(0)

119894+ 120577(0)

119895ge log119870(m(0)

119894 x(0)119895)

for 119894 = 1 119872(0) 119895 = 1 119873

(0)

(9)


119894 = 1 119872(0) and 119911

(0)

119895 119895 =

1 119873(0) such that 119903(0)

119894= log 120588(0)

119894and 120577(0)

119895= log (0)

119895


Then 120588(0)


119894)


119894for 119894 = 1 119872

(0) Similarly 119911(0)119895





x(0)119895




119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909 We








119903(0)

119894+ 120577(0)

119895lowast = log119870(m(0)

119894 x(0)119895lowast ) (10)





(0)asymp 1750)


(0)= 119873(0)






(0)

119894119872(0)

119894=1 120577(0)

119895119873(0)





119903 (m(0)119895) + 120577 (x(0)

119895) minus log119870(m(0)

119894 x(0)119895) le 120576(0) (11)



1 119889(1)

2 119889

(1)


119889(1)

119894and 119889

(1)


119872(1)

119894=1119889(1)

119894



119905(1)

1 119905(1)

2 119905

(1)


(1)

119894

and 119905(1)


119873(1)

119895=1119905119895

(1)


isin 119889(1)

119894(for 119894 = 1 119872

(1))and x(1)

119894isin 119905(1)

119895(for 119895 = 1 119873




(1) and 119895 =


119894







119894=1 120577(1)

119895119873(1)


LP problem

Minimize119872(1)

sum

119894=1

119903(1)

119894119868 (m(1)119894) area (119889(1)

119894)

+

119873(1)

sum

119895=1

120577(1)

119894119871 (x(1)119895) area (119905(1)

119895)


1= log 120588

1

119903(1)

119894+ 120577(1)

119895ge log119870(m(1)

119894 x(1)119895)



(12)


119894 = 1 119872(1) and 119911

(1)

119895 119895 =

1 119873(1) such that 119903(1)

119894= log 120588(1)

119894and 120577(1)

119895= log (1)

119895




119895) + 120577(x(0)

119895) minus



119895) minus log119870(m(0)

119894 x(0)119895)





119863119868119889120590 and int



(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873


(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon


119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909


119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909




120576(2) 120576(3)












119894in the input


119895(119894)asymp 120574(119898

(119896)


(119896)



119894 x(119896)119895(119894)



119894 x(119896)119895(119894)




(119896))2





120576(119896)

=119862

119872(119896) (13)


(119896)=



4 Numerical Tests




Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)


119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)


119871 (x) = 1 for x isin 119879 (16)


119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119896 ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877











|m119909|2+ 1198982

119911= 1




(1)






119870 (m x) =ℓ minus ⟨m

119909 x⟩

2ℓ (ℓ2 minus |x|2) (1 + 119898119911)minus

1

4ℓ2

for m = (m119909 119898119911) isin 119863 x isin 119879

(2)



(x) = 1

2ℓminus

119911 (x)ℓ2 minus |x|2

for x isin 119879 (3)



120588 (m) = minus1

2ℓ+

1

2120588 (m) sdot (119898119911+ 1)

for m isin 119863 (4)




F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (5)







(6)



3 Numerical Schemes





119863119868119889120590 = int

119879119871119889119909



1isin 119863 where






(0)

1 119889(0)

2 119889

(0)

119872(0) sube 119863


and 119889(0)

119895are disjoint (for 119894 = 119895)

and119863 = ⋃119872(0)

119894=1119889(0)

119894 For instance 119889(0)

1 119889(0)

2 119889

119872(0) may be a


choosing sets 119905(0)1 119905(0)

2 119905

(0)


any 119905(0)119894

and 119905(0)


119873(0)

119895=1119905(0)

119895


isin 119889(0)

119894(for 119894 = 1 119872

(0))and x(0)119895

isin 119905(0)

119895(for 119895 = 1 119873




119894=1 120577(0)

119895119873(0)


LP problem

Minimize119872(0)

sum

119894=1

119903(0)

119894119868 (m(0)119894) area (119889(0)

119894)

+

119873(0)

sum

119895=1

120577(0)

119895119871 (x(0)119895) area (119905(0)

119895)

(7)


1= log 120588

1


(8)

119903(0)

119894+ 120577(0)

119895ge log119870(m(0)

119894 x(0)119895)

for 119894 = 1 119872(0) 119895 = 1 119873

(0)

(9)


119894 = 1 119872(0) and 119911

(0)

119895 119895 =

1 119873(0) such that 119903(0)

119894= log 120588(0)

119894and 120577(0)

119895= log (0)

119895


Then 120588(0)


119894)


119894for 119894 = 1 119872

(0) Similarly 119911(0)119895





x(0)119895




119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909 We








119903(0)

119894+ 120577(0)

119895lowast = log119870(m(0)

119894 x(0)119895lowast ) (10)





(0)asymp 1750)


(0)= 119873(0)






(0)

119894119872(0)

119894=1 120577(0)

119895119873(0)





119903 (m(0)119895) + 120577 (x(0)

119895) minus log119870(m(0)

119894 x(0)119895) le 120576(0) (11)



1 119889(1)

2 119889

(1)


119889(1)

119894and 119889

(1)


119872(1)

119894=1119889(1)

119894



119905(1)

1 119905(1)

2 119905

(1)


(1)

119894

and 119905(1)


119873(1)

119895=1119905119895

(1)


isin 119889(1)

119894(for 119894 = 1 119872

(1))and x(1)

119894isin 119905(1)

119895(for 119895 = 1 119873




(1) and 119895 =


119894







119894=1 120577(1)

119895119873(1)


LP problem

Minimize119872(1)

sum

119894=1

119903(1)

119894119868 (m(1)119894) area (119889(1)

119894)

+

119873(1)

sum

119895=1

120577(1)

119894119871 (x(1)119895) area (119905(1)

119895)


1= log 120588

1

119903(1)

119894+ 120577(1)

119895ge log119870(m(1)

119894 x(1)119895)



(12)


119894 = 1 119872(1) and 119911

(1)

119895 119895 =

1 119873(1) such that 119903(1)

119894= log 120588(1)

119894and 120577(1)

119895= log (1)

119895




119895) + 120577(x(0)

119895) minus



119895) minus log119870(m(0)

119894 x(0)119895)





119863119868119889120590 and int



(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873


(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon


119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909


119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909




120576(2) 120576(3)












119894in the input


119895(119894)asymp 120574(119898

(119896)


(119896)



119894 x(119896)119895(119894)



119894 x(119896)119895(119894)




(119896))2





120576(119896)

=119862

119872(119896) (13)


(119896)=



4 Numerical Tests




Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)


119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)


119871 (x) = 1 for x isin 119879 (16)


119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra











F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (5)







(6)



3 Numerical Schemes





119863119868119889120590 = int

119879119871119889119909



1isin 119863 where






(0)

1 119889(0)

2 119889

(0)

119872(0) sube 119863


and 119889(0)

119895are disjoint (for 119894 = 119895)

and119863 = ⋃119872(0)

119894=1119889(0)

119894 For instance 119889(0)

1 119889(0)

2 119889

119872(0) may be a


choosing sets 119905(0)1 119905(0)

2 119905

(0)


any 119905(0)119894

and 119905(0)


119873(0)

119895=1119905(0)

119895


isin 119889(0)

119894(for 119894 = 1 119872

(0))and x(0)119895

isin 119905(0)

119895(for 119895 = 1 119873




119894=1 120577(0)

119895119873(0)


LP problem

Minimize119872(0)

sum

119894=1

119903(0)

119894119868 (m(0)119894) area (119889(0)

119894)

+

119873(0)

sum

119895=1

120577(0)

119895119871 (x(0)119895) area (119905(0)

119895)

(7)


1= log 120588

1


(8)

119903(0)

119894+ 120577(0)

119895ge log119870(m(0)

119894 x(0)119895)

for 119894 = 1 119872(0) 119895 = 1 119873

(0)

(9)


119894 = 1 119872(0) and 119911

(0)

119895 119895 =

1 119873(0) such that 119903(0)

119894= log 120588(0)

119894and 120577(0)

119895= log (0)

119895


Then 120588(0)


119894)


119894for 119894 = 1 119872

(0) Similarly 119911(0)119895





x(0)119895




119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909 We








119903(0)

119894+ 120577(0)

119895lowast = log119870(m(0)

119894 x(0)119895lowast ) (10)





(0)asymp 1750)


(0)= 119873(0)






(0)

119894119872(0)

119894=1 120577(0)

119895119873(0)





119903 (m(0)119895) + 120577 (x(0)

119895) minus log119870(m(0)

119894 x(0)119895) le 120576(0) (11)



1 119889(1)

2 119889

(1)


119889(1)

119894and 119889

(1)


119872(1)

119894=1119889(1)

119894



119905(1)

1 119905(1)

2 119905

(1)


(1)

119894

and 119905(1)


119873(1)

119895=1119905119895

(1)


isin 119889(1)

119894(for 119894 = 1 119872

(1))and x(1)

119894isin 119905(1)

119895(for 119895 = 1 119873




(1) and 119895 =


119894







119894=1 120577(1)

119895119873(1)


LP problem

Minimize119872(1)

sum

119894=1

119903(1)

119894119868 (m(1)119894) area (119889(1)

119894)

+

119873(1)

sum

119895=1

120577(1)

119894119871 (x(1)119895) area (119905(1)

119895)


1= log 120588

1

119903(1)

119894+ 120577(1)

119895ge log119870(m(1)

119894 x(1)119895)



(12)


119894 = 1 119872(1) and 119911

(1)

119895 119895 =

1 119873(1) such that 119903(1)

119894= log 120588(1)

119894and 120577(1)

119895= log (1)

119895




119895) + 120577(x(0)

119895) minus



119895) minus log119870(m(0)

119894 x(0)119895)





119863119868119889120590 and int



(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873


(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon


119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909


119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909




120576(2) 120576(3)












119894in the input


119895(119894)asymp 120574(119898

(119896)


(119896)



119894 x(119896)119895(119894)



119894 x(119896)119895(119894)




(119896))2





120576(119896)

=119862

119872(119896) (13)


(119896)=



4 Numerical Tests




Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)


119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)


119871 (x) = 1 for x isin 119879 (16)


119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra










x(0)119895




119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909 We








119903(0)

119894+ 120577(0)

119895lowast = log119870(m(0)

119894 x(0)119895lowast ) (10)





(0)asymp 1750)


(0)= 119873(0)






(0)

119894119872(0)

119894=1 120577(0)

119895119873(0)





119903 (m(0)119895) + 120577 (x(0)

119895) minus log119870(m(0)

119894 x(0)119895) le 120576(0) (11)



1 119889(1)

2 119889

(1)


119889(1)

119894and 119889

(1)


119872(1)

119894=1119889(1)

119894



119905(1)

1 119905(1)

2 119905

(1)


(1)

119894

and 119905(1)


119873(1)

119895=1119905119895

(1)


isin 119889(1)

119894(for 119894 = 1 119872

(1))and x(1)

119894isin 119905(1)

119895(for 119895 = 1 119873




(1) and 119895 =


119894







119894=1 120577(1)

119895119873(1)


LP problem

Minimize119872(1)

sum

119894=1

119903(1)

119894119868 (m(1)119894) area (119889(1)

119894)

+

119873(1)

sum

119895=1

120577(1)

119894119871 (x(1)119895) area (119905(1)

119895)


1= log 120588

1

119903(1)

119894+ 120577(1)

119895ge log119870(m(1)

119894 x(1)119895)



(12)


119894 = 1 119872(1) and 119911

(1)

119895 119895 =

1 119873(1) such that 119903(1)

119894= log 120588(1)

119894and 120577(1)

119895= log (1)

119895




119895) + 120577(x(0)

119895) minus



119895) minus log119870(m(0)

119894 x(0)119895)





119863119868119889120590 and int



(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873


(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon


119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909


119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909




120576(2) 120576(3)












119894in the input


119895(119894)asymp 120574(119898

(119896)


(119896)



119894 x(119896)119895(119894)



119894 x(119896)119895(119894)




(119896))2





120576(119896)

=119862

119872(119896) (13)


(119896)=



4 Numerical Tests




Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)


119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)


119871 (x) = 1 for x isin 119879 (16)


119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119895) + 120577(x(0)

119895) minus



119895) minus log119870(m(0)

119894 x(0)119895)





119863119868119889120590 and int



(1)

119894 119894 = 1 119872

(1) and119911(1)

119895 119895 = 1 119873


(2)

119894 119894 = 1 119872

(2) and 119911(2)

119895 119895 = 1 119873

(2) and soon


119863119903(m)119868(m)119889120590 + int

119879120577(x)119871(x)119889119909


119863119903(m)119868(m)119889120590+int

119879120577(x)119871(x)119889119909




120576(2) 120576(3)












119894in the input


119895(119894)asymp 120574(119898

(119896)


(119896)



119894 x(119896)119895(119894)



119894 x(119896)119895(119894)




(119896))2





120576(119896)

=119862

119872(119896) (13)


(119896)=



4 Numerical Tests




Ω = (119898119909 119898119910 119898119911) sube 1198782| radic1198982119909+ 1198982119910le 08119898

119911lt 0 (14)


119879 = (119909 119910) isin R2| radic1199092 + 1199102 le

17

9 (15)


119871 (x) = 1 for x isin 119879 (16)


119868 (m) =142716049383

(1 minus 119898119911)2

for m = (119898119909 119898119910 119898119911) isin Ω (17)


R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra










R105

0250

minus025minus05

x

05025

0

minus025minus05

y

minus08

minus07

minus06

minus05

z

(a)

01

x

minus10

1

y

06

z 00204

minus02

minus1

R2

(b)




0765

13 + 04 119898119911

(18)


2+ 1199102) + 06 (19)


ℓ = 119877 + 2120572 = 29 (20)



details





ℎ sim1

radic119872 (21)






120576(119896)

= 119862 sdot (ℎ(119896))120583

(119896 = 1 2 3 ) (22)



119910) to 119868(m) and 120575 cos(119909)




120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra











120581nn ℎ 119872 119873 Max error 1198771

1198712error 119877

1Max error 119877

21198712error 119877




119894119895










times10minus4

4

2

006

06

04

04

02

02

0

0

minus02

minus04

minus06

minus08minus08


(a)

times10minus4

5

02

12

10

0minus1

minus1minus2

minus2

(b)








Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Appendices








120588 (m) = supxisin119879

(1

(x)119870 (m x)) forallm isin 119863


(1

120588 (m)119870 (m x)) forall x isin 119879

(A1)




120574 (m) = x isin 119879 | 120588 (m) =1

(x)119870 (m x) form isin 119863

(A2)





119863119868(m)119889120590 = int

119879119871(x)119889119909


conditions



int120574minus1(120591)

119868 (m) 119889120590 = int120591

119871 (x) 119889119909 (A3)





F (119903 120577) = int119863

119903 (m) 119868 (m) 119889120590 + int119879

120577 (x) 119871 (x) 119889119909 (A4)



2 to 119879 sube


























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra



























2 and 119877

2=



1198772minus |a|2



2 (B2)


(119886 119887 119888) = (x 119911) minus a


1 1198772 See



2+ 1198982



z

x

x = 120574(m)

(x z(x))

120588(m) middotm

Reflector R2

(paraboloid)

Reflector R1

(spheroid)

a = (a b c)O








120588 sdotm) where

120582lowast=

2120572

(119877 minus 120588)2


119911))

=2120572

119877 + 119888 minus 120588 (1 + 119898119911)

(B3)


120574 (m) = (119886

119887) +

2120572

119877 + 119888 minus 120588 (m) (1 + 119898119911)

times ((119886


2

(B4)


1 1198772as defined




int120596

119868 (m) 119889120590 = int120574(120596)

119871 (x) 119889x (B5)




1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1198782


coordinates

120591 (119898119909 119898119910) isin R

2| 1198982

119909+ 1198982

119910lt 1 997888rarr 119878

2

minus


119909minus 1198982119910)

(B6)


119889120590 = 119889119898119909119889119898119910|119898119911| where 119898


119909minus 1198982119910 Using


119868 (mx 119898119911)1

10038161003816100381610038161198981199111003816100381610038161003816

= 119871 (120574 (120591 (mx))) 119869 (120574 ∘ 120591) (B7)

Here 119869(120574 ∘ 120591) = | det (120597(120574 ∘ 120591)120597119898119909

120597(120574 ∘ 120591)120597119898119910) | denotes


119869 (120574 ∘ 120591) = 41205722(|a|2 minus 119877

2)2


119886

119887)

minus (1 + 119898119911) |(119886 119887)|

2

minus(119888 + 119877)2(1 minus 119898

119911) )

2

)

minus1

(B8)


119909minus 1198982119910

Acknowledgment


References













2optimal mass










21199111205971199092)(12059721199111205971199102) minus ((120597

2119911120597119909120597119910))

2












html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















html








Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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