research article optimization of the forcing term for the...

Hindawi Publishing CorporationJournal of MathematicsVolume 2013 Article ID 895876 5 pageshttpdxdoiorg1011552013895876

Research ArticleOptimization of the Forcing Term for the Solution ofTwo-Point Boundary Value Problems

Gianni Arioli

Dipartimento di Matematica ldquoF Brioschirdquo Modellistica e Calcolo Scientifico (MOX) Politecnico di MilanoVia Bonardi 9 20133 Milano Italy

Correspondence should be addressed to Gianni Arioli gianniariolipolimiit

Received 28 August 2012 Accepted 1 January 2013

Academic Editor Alfredo Peris

Copyright copy 2013 Gianni ArioliThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We present a new numerical method for the computation of the forcing term of minimal norm such that a two-point boundaryvalue problem admits a solution The method relies on the following steps The forcing term is written as a (truncated) Chebyshevseries whose coefficients are free parameters A technique derived from automatic differentiation is used to solve the initial valueproblem so that the final value is obtained as a series of polynomials whose coefficients depend explicitly on (the coefficients of)the forcing termThen the minimization problem becomes purely algebraic and can be solved by standard methods of constrainedoptimization for example with Lagrange multipliers We provide an application of this algorithm to the planar restricted threebody problem in order to study the planning of low-thrust transfer orbits

1 Introduction

We consider a two-point boundary value problem for thenonautonomous dynamical system in R119873

1199101015840(119905) = 119891 (119910 (119905)) + 120593 (119905)

119910 (minus1) = 1199100

119910 (1) = 1199101

(1)

where 119891 isin 1198621(R119873R119873) and 120593 isin 119862([minus1 1]R119873) We intro-

duce a new numerical method for the computation of 120593 ofminimal norm among the functions representable with atruncated Chebyshev series such that (1) admits a solutionIn order to evaluate the performance of the algorithm weapply it to a well-known problem that is the study of low-thrust orbits between the Earth and theMoonWe refer to [1]and references there for a discussion of the astrodynamicalproblem

Our new approach is derived from the method intro-duced in [2 3] to consider the dependence of parameters ofthe solution of hyperbolic equations Such method heavilyrelies on a computer representation of algebras of functions

in such a way that the computer can use such functions in astraightforward and transparent way as if they were floatingpoint numbers reducing to a minimum the computationsto be done explicitly by hand The underlying ideas ofthis method stem from automatic differentiation algorithmssee [4ndash8] and recently have been used in the field ofcomputer-assisted proofs see [9 10] for some examples butits potentiality in optimization and control appears to havebeen overlooked In Section 2 we describe the approach in ageneric setting in Section 3 we describe the specific-planar-restricted three body problem that we chose as an examplein Section 4 we present the results that we obtained withthe three body problem in Section 5 we discuss the featuresof the method and compare the Taylor and the Chebyshevrepresentation

2 The Generic Problem

Consider an initial value problem in R119899

1199101015840

119888(119905) = 119891 (119905 119910 (119905)) + 120593119888 (119905)

119910119888 (minus1) = 1199100

(2)

2 Journal of Mathematics

where 119891 is an analytic function 119888 = 119888119894119895 is an 119899 times 119873 real

matrix and the forcing term 120593119888 [minus1 1] rarr R119899 depends on 119888

as follows

(120593119888 (119905))119894 =

119873minus1

sum

119895=0

119888119894119895119879119895 (119905) (3)

where 119879119895(119905) is the 119895th Chebyshev polynomial and (120593

119888(119905))119894

denotes the 119894th component of 120593119888(119905) Our first step is to

compute explicitly the dependence of 119910119888(119905) on 119888 for 119905 isin

(minus1 1]We choose a numeric method to solve the IVP in order

to focus on the issue of the parameter dependence we choosethe simplestmethod for example an explicit first-order Eulerscheme with constant time step and we consider the case 119899 =119873 = 1 We show later that it is straightforward to adapt themethod to other schemes for example a Runge-Kutta schemewith variable time step and to the case 119899119873 gt 1

Choose 119870 isin N let ℎ = 2119870 be the time step set 119905119896=

119896ℎ minus 1 and 119884119896(119888) = 119910

119888(119905119896) so that the Euler scheme reads

119884119896+1 (119888) = 119884119896 (119888) + ℎ119891 (119905119896 119884119896 (119888)) + ℎ120593119888 (119905119896) (4)

and 119884119870(119888) = 119910

119888(1) Now write

119884119896 (119888) =

+infin

sum

119895=0

119910119896119895119880119895 (119888) (5)

where 119880119895(119888)119895isinN is a basis of some space of functions119883 of our

choice In this paper we consider both the Taylor expansionfor the set of analytic functions in a disc (ie 119880

119895(119888) = 119888

119895) andthe Chebyshev expansion for the set of continuous functionsin [minus1 1] (ie 119880

119895(119888) = cos(119895 arccos 119888)) Clearly the method

can be generalized to other expansions for example Fourierseries or Legendre polynomials Now we choose an order119872 and we approximate 119884

119896(119888) with the truncated expansion

sum119872

119895=0119910119896119895119880119895(119888) Our aim is to compute the coefficients 119910

119896119895

using an automatic algorithm Since 1199100(119888) = 119910

0 then we have

11991000

= 1199100and 119910

0119895= 0 when 119895 ge 1

The core of the method consists in a generalization ofthe Taylor Models approach for a detailed description ofthe technique with different expansions see [2 3] Herewe only recall the basic ideas First we observe that it isstraightforward to implement on a computer an arithmeticof Taylor or Chebyshev polynomials of one variable Usingobject-oriented programming and operator overloading it ispossible to define a class Fun which represents a Taylor orChebyshev expansion and a set of methods which performthe basic operations and the computation of the elementaryfunctions More precisely an object Fun is represented as thelist of the coefficients It is straightforward to implement aprocedure that given a scalar 120572 and two objects Fun 119879

11198792

computes the object Fun corresponding to 120572(1198791lowast 1198792) where

lowast is either the addition or the multiplication Then givena polynomial 119901(119909) it is possible to compute the object Funcorresponding to 119901(119879

1) and since all analytic functions 119891(119909)

can be approximated with polynomials it is also possibleto compute the object Fun that better approximates 119891(119879

1)

A more algebraic-oriented way of expressing this idea isthe following let 119883 be the space of functions spanned by119880119895(119888)119895isin0119872

and let 119879 119883 rarr R119872+1 be the (invertible)map that extracts the coefficients Our approach consists inlifting the basic arithmetics of119883 toR119872+1 More precisely wecompute operators

oplus otimes R119872+1

timesR119872+1

997888rarr R119872+1

⊙ R timesR119872+1

997888rarr R119872+1

(6)

such that 119879(119891 + 119892) = 119879(119891) oplus 119879(119892) 119879(119891119892) = 119879(119891) otimes

119879(119892) and 119879(120572119892) = 120572 ⊙ 119879(119892) Once this has been done wecan make extensive use of object-oriented programming andoperator overloading in order to make this lifting completelytransparent to the user

The extension of this method to the case 119899 gt 1 isstraightforward The case 119873 gt 1 requires some additionalwork since we have to consider a multivariable Taylor orChebyshev expansion

119884119896 (119888) =

+infin

sum

1198951119895119873=0

1199101198961198951119895119873

1198801198951

(1198881) sdot sdot sdot 119880

119895119873

(119888119873) (7)

The sum of two of such expansions is done componentwiseThe multiplication is less straightforward but a very efficientalgorithm for the case of the Taylor expansion has beenintroduced in [8] and it is not too hard to extend it tothe Chebyshev (or some other) expansion We refer to ourprograms for details

With the class Fun consisting of the objects (7) andthe functions for addition and multiplication in place wecan use a standard implementation of the algorithm (4)observing that given 119884

119896 the methods of the class will take

care of the computation of 119891(119905119896 119884119896) while 120593

119888(119905119896) is explicitly

given by (3) Extending this approach to better integrationschemes say Runge-Kutta or multistep is straightforwardit suffices to implement the chosen algorithm using objectsFun and their methods By implementing and running theintegration scheme of choice with the class Fun we obtainall the coefficients 119910

1198961198951119895119873

and therefore the polynomials119884119896(119888) for all 119896 Now consider the polynomial 119884

119870(119888) solving

the boundary value problem (1) consists in computing 119888 isin

R119899119873 such that119884119870(119888) = 119910

1This amounts to solving 119899 algebraic

equations in119873 unknowns If we wish 120593119888 to be small (where

sdot is some norm of our choice) we recall that when119873 gt 1

the system 119884119870(119888) = 119910

1is underdetermined therefore we can

compute an explicit formula Φ(119888) = 120593119888 and exploit the

additional degrees of freedom to set the following constrainedminimization problem find 119888 isin R119899119873 such that

Φ (119888) = min119884119870(119888)=1199101

Φ (119888) (8)

This result can be achieved by standard constrained mini-mization algorithms for example Lagrangemultipliers Notethat this method can be easily extended to solve other kindsof conditions at 119905 = 1 let 119898 le 119899 and 119891 R119899 rarr R119898 some

Journal of Mathematics 3

polynomial functionWe can apply the same approach to find119888 isin R119899119873 such that

Φ (119888) = min119891(119884119870(119888))=0

Φ (119888) (9)

To sum up once the class Fun and some algorithm for solvinginitial value problems for ODErsquos has been implementedboundary value problems with optimization of the forcingterm are reduced to computing the constrained minumumof a polynomial the constraint being represented by apolynomial equation or system of equations We remarkthat in principle such polynomial constrainedminimizationproblems can be solved exactly (up to floating point errors)but this does not imply that the boundary condition is solvedexactly for the original problem since we are truncating theexpansion (7) and we are neglecting the truncation errorOne needs to check a posteriori for example by solving (2)with standard floating point numbers and with the forcingterm provided by (3) with 119888 = 119888 and computing the error120576 = |119910(1) minus 119910

1| or 120576 = |119891(119910(1))|

3 An Application Low-Thrust Orbits inthe Planar RTBP

In order tomodel the dynamics of a spaceship travelling froman orbit around the Earth to an orbit around the Moon weconsider the forced planar RTBP that is the system

+ 2 119910 = Ω119909+ 119891 (119905)

radic2 + 1199102

119910 minus 2 = Ω119910+ 119891 (119905)

119910

radic2 + 1199102

(10)

where

Ω(119909 119910) =1199092

2+1199102

2+

1 minus 120583

radic(119909 + 120583)2+ 1199102

+120583

radic(119909 + 120583 minus 1)2+ 1199102

+ 119862

(11)

We choose 120583 ≃ 00121506683 that is the reduced mass ofthe Earth-Moon system and 119862 = minus1600172454916536 sothat Ω(119871

1) = 0 119871

1being the Lagrangian point between the

primaries The forcing term is given by a scalar function 119891(119905)to be determined times a unit vector directed as the velocityof the spaceship In other words we assume that the thrustis always parallel to the velocity We show how to apply themethod described in the previous section to the problem ofoptimizing the travel from an orbit 167 km above the Earth toan orbit 100 km above the Moon

We choose a connecting orbit passing close to 1198711 Note

that the Jacobi function

119869 (119909 119910 119910) = 2+ 1199102minus 2Ω (119909 119910) (12)

is approximately minus60 on an orbit 167 km above the Earthand approximately minus16 on an orbit 100 km above the Moon

05

04

03

02

01

0

minus01

minus02

minus03

minus04

minus04 minus02 0 02 04 06 08 1 12

Figure 1 A complete orbit

Table 1 Result enforcing 119877 and 119881

119877 119881 120576119879

119891119879 120576

119862119891119862

04 09 103 sdot 10minus7

03012 127 sdot 10minus7

03012

04 085 232 sdot 10minus4

04215 345 sdot 10minus4

04235

045 09 NA NA 766 sdot 10minus5

02104

043 09 658 sdot 10minus10

01540 139 sdot 10minus7

01540

Furthermore in order to ldquocrossrdquo the region around theLagrangian point 119871

1 it is necessary to raise it above 0 We

build the connecting orbit as follows We start from theLyapunov orbit around 119871

1with 119869 = 01 and approximate

periodic orbits around the Moon and the Earth Then wecompute approximate connections from the Lyapunov orbitto the orbit around the Earth (backwards in time) and tothe orbit around the Moon (forward in time) Choosing theLyapunov orbit as a starting point is useful if not necessarybecause of its high instability it is much harder to aim atthe Lyapunov orbit than at a stable orbit around a primaryThese orbits are obtained by applying a constant low thrust(ie 119891(119905) = 119888 with small 119888) chosen by simple trial and error

Once these rough orbits are computed we apply thetechnique described above to compute an accurate correctionso that the trajectory satisfies some required boundaryconditions and at the same time the thrust is minimal Moreprecisely we write (10) as a system of 4 first-order equationsby setting 119907 = and 119908 = 119910 we write the modulus of theforcing term as

119891 (119905) = 119888 +

119873minus1

sum

119895=0

119888119895119879119895 (119905) (13)

we choose some initial value on the trajectory at the arbitrarytime 119905 = 0 and we solve the equation with a fourth-orderRunge-Kutta scheme up to some time 119905 = 119879 using objectsFun as scalars the coefficients 119888

119895being the independent

variables of the objects Fun Referring to Section 2 we have119899 = 4 (the dimension of the system) we choose 119873 = 4

and 119872 = 10 We rescale the time in the interval [minus1 1]and apply the method described in Section 2 to obtain(119909(119879) 119910(119879) 119907(119879) 119908(119879)) as a function of 119888

119895 expressed either

as a Taylor series or a Chebyshev series


Table 2 Result enforcing 119877 119881 and tangential trajectory

119877 119881 120576119879

120576perp

119879119891119879 120576

119862120576perp

119862119891119862

042 087 304 sdot 10minus4

848 sdot 10minus5

03627 153 sdot 10minus4

826 sdot 10minus7

03653

042 09 318 sdot 10minus6

600 sdot 10minus7

02540 402 sdot 10minus6

975 sdot 10minus6

02541

044 082 418 sdot 10minus9

337 sdot 10minus10

02044 248 sdot 10minus8

234 sdot 10minus8

02044

046 082 732 sdot 10minus8

143 sdot 10minus7

01439 377 sdot 10minus6

222 sdot 10minus6

01439

Then we choose the condition to be satisfied at119905 = 119879 We tested the following find 119888

119895 such that

(119909(119879) 119910(119879) 119907(119879) 119908(119879)) is the same as in the case 119891(119905) = 119888but 119891 is minimal find 119888

119895 such that the distance of

the spaceship to the Earth or the Moon is assignedtogether with the modulus of the final velocity that is(119909(119879) minus 120583)

2+ 119910(119879) minus 119877

2= 0 and (119907(119879))2 + 119908(119879)2 minus 1198812 = 0

and 119891 is minimal find 119888119895 such that the distance of the

spaceship to the Earth or the Moon is assigned togetherwith the modulus of the final velocity with the additionalrequirement that the final velocity is orthogonal to the lineconnecting the satellite to a primaryAll these problembelongto the class of problem described in the previous section andthe polynomial constrained minimization problems havebeen solved by the Lagrange multipliers method

4 Results

Figure 1 represents an example of an orbit computed byusing the technique Tables 1 and 2 collect some resultsobtained for a trajectory backward in time starting closeto a Lyapunov orbit around 119871

1 more precisely at the

point (119909(0) 119910(0) 119907(0) 119908(0)) = (073356888 minus0017228233

025064405 017220602) and ending close to an orbit aroundthe Earth at time 119879 = minus17 The trajectory with a constantthrust 119888 = 02 ends at 119877 = radic(119909(119879)minus120583)

2+119910(119879) = 0421852

and 119881 = radic(119907(119879))2+119908(119879)

2=0876662

For both tables we computed the thrust of minimal normnecessary to end the trajectory with the values given inthe columns 119877 and 119881 The results of Table 2 represent thesolution with the additional constraint that the final velocityis orthogonal to the segment from the spaceship to theEarth corresponding to an approximate circular orbit Wetested the accuracy of the computation as explained at theend of Section 2 that is we repeated the computation ofthe orbit with standard floating point numbers using theexplicit expression of the forcing term and we used theresult of such computation to verify that the constraints weresatisfied The values 120576

119879and 120576

119862represent the mean square

of the relative errors on 119877 and 119881 obtained with the Taylorand the Chebyshev computation respectively while 120576perp

119879and

120576perp

119862is |(119909(119879) minus 120583)119907(119879) + 119910(119879)119908(119879)| that is the error on

the orthogonality requirement The columns 119891119879 and 119891

119862

represent the norm of the forcing terms It is clear from thetables that the method can achieve very accurate results andit is also clear that the norm of the forcing term obtained withthe Taylor and the Chebyshev computation is (almost) thesame

5 Conclusions

We discuss and compare the features of the two expansionthat we tested Taylor and Chebyshev Generally speaking(see [2 3]) the main advantages of the Taylor expansionconsist in three features the algorithms are simpler andfaster they provide directly the derivatives with respectto the parameters and furthermore and since one doesnot need to know the radius of convergence a priori onecan apply them and compute a posteriori the interval ofvalidity of the computation The Taylor expansion has alsotwo main disadvantages it does not provide uniform errorsin the interval where the computation is performed andthe radius of convergence may be bounded by poles inthe complex plane The Chebyshev expansion has oppositefeatures the error is uniform in the interval and the regionof convergence is an ellipse as thin as necessary thereforethere are no problems caused by poles with nontrivialimaginary parts On the other hand the algorithm forthe multiplication is much more complicated and muchslower it does not provide any direct computation of thederivatives and finally since the Chebyshev polynomialsare defined in [minus1 1] one needs to choose a priori (via asuitable translationrescaling) the interval where the param-eter ranges finding out only a posteriori if the approxi-mation is acceptable In the application considered in [23] the Chebyshev expansion was a clear winner due tothe fact that when considering the numerical approxi-mation of shock waves and when trying to improve theresolution the problem of poles becomes the main issueHere our judgment is the opposite since the solutions ofthe odersquos that we are studying are analytic functions theissue of poles limiting the radius of convergence of Taylorseries becomes negligible while the disadvantages of theChebyshev expansion in particular the slower algorithmsand the fact that the domain has to be chosen a prioribecome very relevant Additionally we do not have evi-dence of a better performance in terms of accuracy Itis easy to find examples when either expansion performsbetter than the other one but on average the accuracy issimilar

References

[1] G Mingotti F Topputo and F Bernelli-Zazzera ldquoNumericalmethods to design low-energy low-thrust sun-perturbed trans-fers to the moonrdquo in Proceedings of the 49th Israel AnnualConference on Aerospace Sciences Tel Aviv-Haifa Israel March2009


[2] G Arioli and M Gamba ldquoAn algorithm for the studyof parameter dependence for hyperbolic systemsrdquo MOX-Report 052011 httpmoxpolimiititprogettipubblicazioniquaderni05-2011pdf

[3] G Arioli and M Gamba ldquoAutomatic computation of Cheby-shev polynomials for the study of parameter dependencefor hyperbolic systemsrdquo MOX-Report 072011 httpmoxpolimiititprogettipubblicazioniquaderni07-2011pdf

[4] K Makino and M Berz ldquoHigher order verified inclusions ofmultidimensional systems by Taylor modelsrdquo in Proceedings ofthe 3rd World Congress of Nonlinear Analysts part 5 CataniaItaly July 2000

[5] K Makino and M Berz ldquoHigher order verified inclusionsofmultidimensional systems by TaylormodelsrdquoNonlinear Anal-ysis Theory Methods and Applications vol 47 no 5 pp 3503ndash3514 2001

[6] M Berz and K Makino ldquoTaylor models and other validatedfunctional inclusionmethodsrdquo International Journal of Pure andApplied Mathematics vol 6 no 3 pp 239ndash316 2003


[8] M Berz and K Makino ldquoHigher order multivariate automaticdifferentiation validated computation of remainder boundsrdquoTransactions on Mathematics vol 3 no 1 pp 37ndash44 2004

[9] G Arioli and H Koch ldquoComputer-assisted methods for thestudy of stationary solutions in dissipative systems appliedto the Kuramoto-Sivashinski equationrdquo Archive for RationalMechanics and Analysis vol 197 no 3 pp 1033ndash1051 2010

[10] G Arioli and H Koch ldquoIntegration of dissipative partialdifferential equations a case studyrdquo SIAM Journal on AppliedDynamical Systems vol 9 no 3 pp 1119ndash1133 2010

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Stochastic AnalysisInternational Journal of


where 119891 is an analytic function 119888 = 119888119894119895 is an 119899 times 119873 real

matrix and the forcing term 120593119888 [minus1 1] rarr R119899 depends on 119888

as follows

(120593119888 (119905))119894 =

119873minus1

sum

119895=0

119888119894119895119879119895 (119905) (3)

where 119879119895(119905) is the 119895th Chebyshev polynomial and (120593

119888(119905))119894

denotes the 119894th component of 120593119888(119905) Our first step is to

compute explicitly the dependence of 119910119888(119905) on 119888 for 119905 isin

(minus1 1]We choose a numeric method to solve the IVP in order

to focus on the issue of the parameter dependence we choosethe simplestmethod for example an explicit first-order Eulerscheme with constant time step and we consider the case 119899 =119873 = 1 We show later that it is straightforward to adapt themethod to other schemes for example a Runge-Kutta schemewith variable time step and to the case 119899119873 gt 1

Choose 119870 isin N let ℎ = 2119870 be the time step set 119905119896=

119896ℎ minus 1 and 119884119896(119888) = 119910

119888(119905119896) so that the Euler scheme reads

119884119896+1 (119888) = 119884119896 (119888) + ℎ119891 (119905119896 119884119896 (119888)) + ℎ120593119888 (119905119896) (4)

and 119884119870(119888) = 119910

119888(1) Now write

119884119896 (119888) =

+infin

sum

119895=0

119910119896119895119880119895 (119888) (5)

where 119880119895(119888)119895isinN is a basis of some space of functions119883 of our

choice In this paper we consider both the Taylor expansionfor the set of analytic functions in a disc (ie 119880

119895(119888) = 119888

119895) andthe Chebyshev expansion for the set of continuous functionsin [minus1 1] (ie 119880

119895(119888) = cos(119895 arccos 119888)) Clearly the method

can be generalized to other expansions for example Fourierseries or Legendre polynomials Now we choose an order119872 and we approximate 119884

119896(119888) with the truncated expansion

sum119872

119895=0119910119896119895119880119895(119888) Our aim is to compute the coefficients 119910

119896119895

using an automatic algorithm Since 1199100(119888) = 119910

0 then we have

11991000

= 1199100and 119910

0119895= 0 when 119895 ge 1

The core of the method consists in a generalization ofthe Taylor Models approach for a detailed description ofthe technique with different expansions see [2 3] Herewe only recall the basic ideas First we observe that it isstraightforward to implement on a computer an arithmeticof Taylor or Chebyshev polynomials of one variable Usingobject-oriented programming and operator overloading it ispossible to define a class Fun which represents a Taylor orChebyshev expansion and a set of methods which performthe basic operations and the computation of the elementaryfunctions More precisely an object Fun is represented as thelist of the coefficients It is straightforward to implement aprocedure that given a scalar 120572 and two objects Fun 119879

11198792

computes the object Fun corresponding to 120572(1198791lowast 1198792) where

lowast is either the addition or the multiplication Then givena polynomial 119901(119909) it is possible to compute the object Funcorresponding to 119901(119879

1) and since all analytic functions 119891(119909)

can be approximated with polynomials it is also possibleto compute the object Fun that better approximates 119891(119879

1)

A more algebraic-oriented way of expressing this idea isthe following let 119883 be the space of functions spanned by119880119895(119888)119895isin0119872

and let 119879 119883 rarr R119872+1 be the (invertible)map that extracts the coefficients Our approach consists inlifting the basic arithmetics of119883 toR119872+1 More precisely wecompute operators

oplus otimes R119872+1

timesR119872+1

997888rarr R119872+1

⊙ R timesR119872+1

997888rarr R119872+1

(6)

such that 119879(119891 + 119892) = 119879(119891) oplus 119879(119892) 119879(119891119892) = 119879(119891) otimes

119879(119892) and 119879(120572119892) = 120572 ⊙ 119879(119892) Once this has been done wecan make extensive use of object-oriented programming andoperator overloading in order to make this lifting completelytransparent to the user

The extension of this method to the case 119899 gt 1 isstraightforward The case 119873 gt 1 requires some additionalwork since we have to consider a multivariable Taylor orChebyshev expansion

119884119896 (119888) =

+infin

sum

1198951119895119873=0

1199101198961198951119895119873

1198801198951

(1198881) sdot sdot sdot 119880

119895119873

(119888119873) (7)

The sum of two of such expansions is done componentwiseThe multiplication is less straightforward but a very efficientalgorithm for the case of the Taylor expansion has beenintroduced in [8] and it is not too hard to extend it tothe Chebyshev (or some other) expansion We refer to ourprograms for details

With the class Fun consisting of the objects (7) andthe functions for addition and multiplication in place wecan use a standard implementation of the algorithm (4)observing that given 119884

119896 the methods of the class will take

care of the computation of 119891(119905119896 119884119896) while 120593

119888(119905119896) is explicitly

given by (3) Extending this approach to better integrationschemes say Runge-Kutta or multistep is straightforwardit suffices to implement the chosen algorithm using objectsFun and their methods By implementing and running theintegration scheme of choice with the class Fun we obtainall the coefficients 119910

1198961198951119895119873

and therefore the polynomials119884119896(119888) for all 119896 Now consider the polynomial 119884

119870(119888) solving

the boundary value problem (1) consists in computing 119888 isin

R119899119873 such that119884119870(119888) = 119910

1This amounts to solving 119899 algebraic

equations in119873 unknowns If we wish 120593119888 to be small (where

sdot is some norm of our choice) we recall that when119873 gt 1

the system 119884119870(119888) = 119910

1is underdetermined therefore we can

compute an explicit formula Φ(119888) = 120593119888 and exploit the

additional degrees of freedom to set the following constrainedminimization problem find 119888 isin R119899119873 such that

Φ (119888) = min119884119870(119888)=1199101

Φ (119888) (8)

This result can be achieved by standard constrained mini-mization algorithms for example Lagrangemultipliers Notethat this method can be easily extended to solve other kindsof conditions at 119905 = 1 let 119898 le 119899 and 119891 R119899 rarr R119898 some



Φ (119888) = min119891(119884119870(119888))=0

Φ (119888) (9)


1| or 120576 = |119891(119910(1))|



+ 2 119910 = Ω119909+ 119891 (119905)

radic2 + 1199102

119910 minus 2 = Ω119910+ 119891 (119905)

119910

radic2 + 1199102

(10)

where

Ω(119909 119910) =1199092

2+1199102

2+

1 minus 120583

radic(119909 + 120583)2+ 1199102

+120583

radic(119909 + 120583 minus 1)2+ 1199102

+ 119862

(11)


1) = 0 119871





119869 (119909 119910 119910) = 2+ 1199102minus 2Ω (119909 119910) (12)


05

04

03

02

01

0

minus01

minus02

minus03

minus04

minus04 minus02 0 02 04 06 08 1 12



119877 119881 120576119879

119891119879 120576

119862119891119862

04 09 103 sdot 10minus7

03012 127 sdot 10minus7

03012

04 085 232 sdot 10minus4

04215 345 sdot 10minus4

04235


02104

043 09 658 sdot 10minus10

01540 139 sdot 10minus7

01540







119891 (119905) = 119888 +

119873minus1

sum

119895=0

119888119895119879119895 (119905) (13)









119877 119881 120576119879

120576perp

119879119891119879 120576

119862120576perp

119862119891119862

042 087 304 sdot 10minus4

848 sdot 10minus5

03627 153 sdot 10minus4

826 sdot 10minus7

03653

042 09 318 sdot 10minus6

600 sdot 10minus7

02540 402 sdot 10minus6

975 sdot 10minus6

02541

044 082 418 sdot 10minus9

337 sdot 10minus10

02044 248 sdot 10minus8

234 sdot 10minus8

02044

046 082 732 sdot 10minus8

143 sdot 10minus7

01439 377 sdot 10minus6

222 sdot 10minus6

01439


119895 such that




2+ 119910(119879) minus 119877

2= 0 and (119907(119879))2 + 119908(119879)2 minus 1198812 = 0



4 Results



point (119909(0) 119910(0) 119907(0) 119908(0)) = (073356888 minus0017228233


2+119910(119879) = 0421852

and 119881 = radic(119907(119879))2+119908(119879)

2=0876662


119879and 120576



119879and

120576perp



119862


5 Conclusions


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Φ (119888) = min119891(119884119870(119888))=0

Φ (119888) (9)


1| or 120576 = |119891(119910(1))|



+ 2 119910 = Ω119909+ 119891 (119905)

radic2 + 1199102

119910 minus 2 = Ω119910+ 119891 (119905)

119910

radic2 + 1199102

(10)

where

Ω(119909 119910) =1199092

2+1199102

2+

1 minus 120583

radic(119909 + 120583)2+ 1199102

+120583

radic(119909 + 120583 minus 1)2+ 1199102

+ 119862

(11)


1) = 0 119871





119869 (119909 119910 119910) = 2+ 1199102minus 2Ω (119909 119910) (12)


05

04

03

02

01

0

minus01

minus02

minus03

minus04

minus04 minus02 0 02 04 06 08 1 12



119877 119881 120576119879

119891119879 120576

119862119891119862

04 09 103 sdot 10minus7

03012 127 sdot 10minus7

03012

04 085 232 sdot 10minus4

04215 345 sdot 10minus4

04235


02104

043 09 658 sdot 10minus10

01540 139 sdot 10minus7

01540







119891 (119905) = 119888 +

119873minus1

sum

119895=0

119888119895119879119895 (119905) (13)









119877 119881 120576119879

120576perp

119879119891119879 120576

119862120576perp

119862119891119862

042 087 304 sdot 10minus4

848 sdot 10minus5

03627 153 sdot 10minus4

826 sdot 10minus7

03653

042 09 318 sdot 10minus6

600 sdot 10minus7

02540 402 sdot 10minus6

975 sdot 10minus6

02541

044 082 418 sdot 10minus9

337 sdot 10minus10

02044 248 sdot 10minus8

234 sdot 10minus8

02044

046 082 732 sdot 10minus8

143 sdot 10minus7

01439 377 sdot 10minus6

222 sdot 10minus6

01439


119895 such that




2+ 119910(119879) minus 119877

2= 0 and (119907(119879))2 + 119908(119879)2 minus 1198812 = 0



4 Results



point (119909(0) 119910(0) 119907(0) 119908(0)) = (073356888 minus0017228233


2+119910(119879) = 0421852

and 119881 = radic(119907(119879))2+119908(119879)

2=0876662


119879and 120576



119879and

120576perp



119862


5 Conclusions


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119877 119881 120576119879

120576perp

119879119891119879 120576

119862120576perp

119862119891119862

042 087 304 sdot 10minus4

848 sdot 10minus5

03627 153 sdot 10minus4

826 sdot 10minus7

03653

042 09 318 sdot 10minus6

600 sdot 10minus7

02540 402 sdot 10minus6

975 sdot 10minus6

02541

044 082 418 sdot 10minus9

337 sdot 10minus10

02044 248 sdot 10minus8

234 sdot 10minus8

02044

046 082 732 sdot 10minus8

143 sdot 10minus7

01439 377 sdot 10minus6

222 sdot 10minus6

01439


119895 such that




2+ 119910(119879) minus 119877

2= 0 and (119907(119879))2 + 119908(119879)2 minus 1198812 = 0



4 Results



point (119909(0) 119910(0) 119907(0) 119908(0)) = (073356888 minus0017228233


2+119910(119879) = 0421852

and 119881 = radic(119907(119879))2+119908(119879)

2=0876662


119879and 120576



119879and

120576perp



119862


5 Conclusions


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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