Research Article Numerical Reduced Variable Optimization ...

Research ArticleNumerical Reduced Variable Optimization Methods via ImplicitFunctional Dependence with Applications

Christopher Gunaseelan Jesudason

Department of Chemistry and Center for Theoretical and Computational Physics Faculty of Science University of Malaya50603 Kuala Lumpur Malaysia

Correspondence should be addressed to Christopher Gunaseelan Jesudason jesuumedumy

Received 14 March 2014 Accepted 24 May 2014 Published 27 August 2014

Academic Editor Mariano Torrisi

Copyright copy 2014 Christopher Gunaseelan Jesudason This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

A systematic theoretical basis is developed that optimizes an arbitrary number of variables for (i) modeling data and (ii) thedetermination of stationary points of a function of several variables by the optimization of an auxiliary function of a single variabledeemed the most significant on physical experimental or mathematical grounds from which all the other optimized variables maybe derived Algorithms that focus on a reduced variable set avoid problems associated with multiple minima and maxima thatarise because of the large numbers of parameters For (i) both approximate and exact methods are presented where the singlecontrolling variable k of all the other variables P (119896) passes through the local stationary point of the least squares metric For (ii) anexact theory is developed whereby the solution of the optimized function of an independent variation of all parameters coincideswith that due to single parameter optimization of an auxiliary functionThe implicit function theorem has to be further qualified toarrive at this result A nontrivial real world application of the above implicit methodology to rate constant and final concentrationparameter determination is made to illustrate its utilityThis work is more general than the reduction schemes for conditional linearparameters since it covers the nonconditional case as well and has potentially wide applicability

1 Introduction

The following theory is a systematic development of all func-tions covering properties of constrained and unconstrainedfunctions that are continuous and differentiable to variousspecified degrees [1 2] and the proof of the existence ofimplicit functions [3] for the form of these functions to beoptimized The implicit function theorem is applied in amanner that requires further qualification because the opti-mization problem is of an unconstrained kind without anyredundant variables Methods (i)ab (described in Sections 2and 3 resp) refer to modeling of data [4 Chapter 15 pages773ndash806] where the form of the function 119876MD(P 119896) withindependently varying variables (P 119896) is

119876MD (P 119896) =1198731015840

sum119894=1

(119910119894minus 119891 (P 119905

119894 119896))2 (1)

where 119910119894and 119905119894are datapoints and 119891 is a known function

and optimizations of 119876MD may be termed a least squares

(LS) fit over parameters (P 119896) which are independentlyoptimized for1198731015840 datasets Method (ii) focuses on optimizinga general 119876

119864(P 119896) function not necessarily LS in form

There are many standard and hybrid methods to deal withsuch optimization [4 Chapter 10] such as golden sectionsearches in 1D simplex methods over multidimensions [4pages 499ndash525] steepest descent and conjugate methods[5] and variable metric methods in multidimensions [4pages 521ndash525] Hybrid methods include multidimensional(DFP) secantmethods [6] BFGS secant optimization [7] andRFO rational function optimization [8] which is a Newton-Raphson technique utilizing a rational function rather than aquadratic model for the function close to the solution pointGlobal deterministic optimization schemes combine severalof the above approaches [9 Section 676] Other physicalmethods perhaps less easy to justify analytically includeprobabilistic ldquobasin-hoppingrdquo algorithms [9 Section 674]simulated annealingmethods [10] and genetic algorithms [9page 346] An analytical justification on the other hand is

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 108184 13 pageshttpdxdoiorg1011552014108184

2 Abstract and Applied Analysis

attempted here for these deterministic methods but in real-world applications some of the assumptions (eg C2 conti-nuity compactness of spaces) may not always be obtainedFor what follows the distance metrics used are all Euclideanrepresented by | sdot | or sdot where det sdot represents thedeterminant of the matrix sdot Reduction of the number ofvariables to be optimized is possible in the standard matrixregressionmodel only if conditional linear parameters120573 exist[11] where these 120573 variables do not appear in the final 119878(120579)expression of the least squares function (2) to be optimizedwhereas the 120601 nonconditional linear parameters do and are asubset of the 120579 variables for the existence of each conditionallinear parameter there is a unit reduction in the number ofindependent parameters to be optimizedThese reductions invariable number occur for any ldquoexpectation functionrdquo 119891(x 120579)which is the model or law for which a fitting is requiredwhere there are 119873 different datapoints x

119894 119894 = 1 2 119873

that must be used to determine the 119901 parameter variables 120579[11 page 32 Chapter 2] A conditionally linear parameter 120579

119894

exists if and only if the derivative of the expectation function119891(x 120579) with respect to 120579

119894is independent of 120579

119894 Clearly such a

condition may severely limit the number of parameters thatcan be neglected for the expectation function variables whenthe prescribed matrix regressional techniques are employed[11 Section 355 page 85] where the residual sum of squaresis minimized

119878 (120579) =1003817100381710038171003817y minus 120578(120579)

10038171003817100381710038172 (2)

The119873-vectors 120578(120579) in 119875-dimensional space define the expec-tation surface If the 120579 variables are partitioned into theconditional linear parameters 120573 and the other nonlinearparameters 120601 then the response can be written 120578(120573120601) =A(120601)120573 Golub and Pereyra [12] used a standard Gauss-Newton algorithm to minimise 119878

2(120601) = y minus A(120601)(120601)

2

that depended only on the nonlinear parameters 120601 where(120601) = A+(120601)y with A+ being a defined pseudoinverseof A [11 Section 355 page 85] where A+ and A arematrices The variables must be separable as discussed aboveand the number of variable reduction is only equal tothe number of conditional linear parameters that exists forthe problem In applications the preferred algorithm thatexploits this valuable variable reduction is called variableprojection There are many applications in time resolvedspectroscopy that is heavily dependent on this technique andmany references to the method are given in the review byvan Stokkum et al [13] Recently this method of variableprojection has been extended in a restricted sense [14] in thefield of inverse problems which is not related to our methodof either modeling or optimization nor is the methodologyrelated to the implicit function properties In short much ofthe reported methods developed are 119886119889 ℎ119900119888 meaning thatthey are constructed to face the specific problems at handwith no claim to overall generality and this work too is 119886119889 ℎ119900119888in the sense of suggesting variable reduction with specificclasses of noninverse problems as indicated where the workdevelops a method of reducing the variable number to unityfor all variables in the expectation function space irrespectiveof whether they are conditional or not by approximating their

values by a method of averages (for method (i)a) withoutany form of linear regression being used in determiningtheir approximations during the minimization iterationsand without necessarily using the standard matrix theorythat is valid for a very limited class of functions Methods(i)b and (ii) are on the other hand exact treatments Noldquoeliminationrdquo of conditional linear parameters is involved inthis nonlinear regressionmethodNor is any projection in themathematical sense involved These general methods couldhave useful applications in deterministic systems comprisingmany parameters that are all linked to one variable theprimary one (denoted 119896 here) that is considered on physicalgrounds to be the most significant A generalization of thismethod would be to select a smaller set of variables than thefull parameter list instead of just one variable as illustratedhere Another tool that could be used in conjunction withthe reduced variable method is to employ the various searchalgorithms that has been actively developed to the reductionscheme developed here [15ndash19] As far as we are awarethese optimization methods for multivariable problems allseem to mainly focus on various stochastic or deterministicmethods using discrete algorithms in some type of searchsequence of the domain space of the multivariable domainspace P 119896 119905 as detailed below in more recent publicationsIn other less ambitious works the problem is narrowed tothe specific nature of the system where the object functionis specified and which is amenable to precise treatment forexample [20 21] with a well-defined domain space withoutvariable reduction In another context quite different fromthe current development variable reduction has been appliedto DEA problems [22] Other examples of multiparametercomplex systems include those for multiple-step elementaryreactions each with its own rate constant that gives rise tophotochemical spectra signals that must be resolved unam-biguously [23] but these belong to the class of functions withconditional linear parameters The work here on the otherhand predominantly focuses on accurately determining therange of the function to be optimized by reducing the spaceof the domain Hence this method can be successfully com-binedwith the usual domain searching techniquesmentionedabove to effectively locate stationary points by a two-prongedapproach All these complex and coupled processes in phys-ical theories are related by postulated laws 119884law(P 119896 119905) thatfeature parameters (P 119896) Other examples include quantumchemical calculations with many topological and orientationvariables that need to be optimized with respect to theenergy but in relation to one or a few variables such as themolecular trajectory parameter during a chemical reactionwhere this variable is of primary significance in decidingon the ldquoreasonablenessrdquo of the analysis [9 Section 623page 294] Methods (i)a and (i)b below refer to LS data-fitting algorithms Method (i)a is an approximate methodwhere it is proved under certain conditions it could be amore accurate determination of parameters compared to astandard LS fit using (1) Method (i)b develops a techniquewhere the optimum value for 119876MD with domain values (P 119896)coincides with that of the standard LS method where the(P 119896) variables are varied independently Also discussed arethe relative accuracy of both methods (i)a in Section 22

Abstract and Applied Analysis 3

and (i)b (endnote at end of Section 3) Method (ii) developsa single parameter optimization where the conditions ofan arbitrary 119876OPT(P 119896) function are met simultaneouslynamely

120597119876OPT (119896)

120597119896= 0 997888rarr

120597119876OPT (P 119896)120597P

= 0120597119876OPT (P 119896)

120597119896= 0

(3)

We note that methods (i)a (i)b and (ii) are not relatedto the Adomian decomposition method and its variantsthat expand polynomial coefficients [24] for solutions todifferential equations not connected to estimation theoryindeed here there are no boundary values that determine thesolution of the differential equations

2 Method (i)a Theory

This approximate method utilizes the average of the 119873119888

unique solutions for each value of 119896 defined above where theform of the fitting functionmdasha ldquolawrdquo of nature for instancemdashis specified Deterministic laws of nature are convenientlywritten in the form

119884law = 119884law (P 119896 119905) (4)

linking the variable 119884law to 119905 The components of P 119875119894(119894 =

1 2 119873119901) and 119896 are parameters Verification of a law of

form (4) relies on an experimental dataset (119884exp(119905119894) 119905119894) 119894 =1 2 119873 The 119905 variable could be a vector of variablecomponents of experimentally measured values or a singleparameter as in the kinetics examples below where 119905

119894denotes

values of time 119905 in the domain space The vector form willbe denoted by x Variables (x) are defined as members ofthe ldquodomain spacerdquo of the measurable system and similarly119884law is the defined range or ldquoresponserdquo space of the physicalmeasurement Confirmation or verification of the law is basedon (a) deriving experimentally meaningful values for theparameters (P 119896) and (b) showing a good enough degreeof fit between the experimental set 119884exp(119905119894) and 119884law(119905119894) Inreal world applications to chemical kinetics for instanceseveral methods [25ndash28] and so forth have been devised todetermine the optimal (P 119896) parameters but most if not allthese methods consider the aforementioned parameters asautonomous and independent (eg [26]) A similar scenariobroadly holds for current state-of-the-art applications ofstructural elucidation via energy functions [9 Chapsters4 6] To preserve the viewpoint of the interrelationshipbetween these parameters and the experimental data wedevise schemes that relate P to 119896 for all 119875

119894via the set

119884exp(119905119894) 119905119894 and optimize the fit over 119896-space only That isthere is induced a119875

119894(119896) dependency on 119896 via the experimental

set 119884exp(119905119894) 119905119894 The conditions that allow for this will also bestated for the different methods

21 Details of Method (i)a Let 119873119901be the number of com-

ponents of the P parameter 1198731015840 the number of dataset pairs(119884exp(119905119894) 119905119894) and 119873119904 the number of singularities where theuse of a particular dataset (119884exp 119905) leads to a singularity in

the determination of 119875119894(119896) as defined below and which must

be excluded from being used in the determination of 119875119894(119896)

Then (119873119901+ 1) le (119873 minus 119873

119904) for the unique determination

of P 119896 Let 119873119888be the total number of different datasets

that can be chosen which does not lead to singularitiesIf the singularities are not choice dependent that is aparticular dataset pair leads to singularities for all possiblechoices then we have the following definition for 119873

119888where

119873minus119873119904119862119873119901

= 119873119888is the total number of combinations of the

data-sets 119884exp(119905119894) 119905119894 taken119873119901 at a time that does not lead tosingularities in 119875

119894 In general119873

119888is determined by the nature

of the datasets and the way in which the proposed equationsare to be solved Write 119884law in the form

119884law (119905 119896) = 119891 (P 119905 119896) (5)

and for a particular dataset 119884exp(119905119894) 119905119894 write 119891(119894) equiv

119891(P 119905119894 119896) Define the vector function fg with components

119891119892(119894) equiv Yexp(119905119894) minus 119891(119894) = 119891119892(119894)(P 119896) Assume fg isin C1 defined

on an open set 1198700that contains 119896

0

Lemma 1 For any 1198960such that det 120597119891

119892(119894)(P 119896

0)120597119875119895 =

0 exist the unique function P(119896) isin C1(119908119894119905ℎ 119888119900119898119901119900119899119890119899119905119904119875119894(119896) sdot sdot sdot 119875

119873119888

(119896)) defined on 1198700where P(119896

0) = P0 and where

fg(P(119896) 119896) = 0 for every 119896 isin 1198700

Proof The above follows from the implicit function theorem(IFT) [3 Theorem 137 page 374] where 119896 isin 119870

0is the inde-

pendent variable for the existence of the P(119896) function

We seek the solutions for P(119896) subject to the aboveconditions for our defined functions Map 119891 rarr 119884th(P 119905 119896)as follows

119884th (119905 119896) = 119891 (P 119905 119896) (6)

where the term P and its components are defined below andwhere 119896 is a varying parameter For any of the (119894

1 1198942 119894

119873119901

)

combinations denoted by a combination variable 120572 equiv

(1198941 1198942 119894

119873119901

) where 119894119895equiv (119884exp(119905119894

119895

) 119905119894119895

) is a particular datasetpair it is in principle possible to solve for the components ofPin terms of 119896 through the following simultaneous equations

119884exp (1199051198941

) = 119891 (P 1199051198941

119896)

119884exp (1199051198942

) = 119891 (P 1199051198942

119896)

119884exp (119905119894119873119901

) = 119891 (P 119905119894119873119901

119896)

(7)

from Lemma 1 And each 120572 choice yields a unique solution119875119894(119896 120572) (119894 = 1 2 119873

119901) where 119875

119894(119896 120572) isin C1 Hence any

function of 119875119894(119896 120572) involving addition and multiplication is

also in C1 For each 119875119894 there will be 119873

119888different solutions

119875119894(119896 1) 119875

119894(119896 2) 119875

119894(119896119873119888) We can define an arithmetic


mean (there are several possible mean definitions that can beutilized) for the components of P as

119875119894 (119896) =

1

119873119888

119873119888

sum119895=1

119875119894(119896 119895) (8)

In choosing an appropriate functional form for P (8) weassumed equal weightage for each of the dataset combina-tions however the choice is open based on appropriatephysical criteria We verify below that the choice of P(119896)satisfies the constrained variation of the LS method so as toemphasize the connection between the level-surfaces of theunconstrained LS with the line function P(119896)

Each 119875119894(119896 119895) is a function of 119896 whose derivative is known

either analytically or by numerical differentiation To derivean optimized set then for the LS method define

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(9)

Then for an optimized 119896 we have 1198761015840(119896) = 0 Defining

119877 (119896) = 119888 sdot

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119884th (119896 119905119894)) sdot 1198841015840

th (119896 119905119894) (10)

the optimized solution of 119896 corresponds to 119877(119896) = 0

which has been reduced to a one-dimensional problem Thestandard LS variation on the other hand states that thevariables 119875

119879= P 119896 in (5) are independently varied so that

119876119879(119875119879) =

1198731015840

sum119894=1

(119884exp(119905119894) minus 119891(119875119879 119905119894))2

(11)

with solutions for 119876119879in terms of 119875

119879whenever 120597119876

119879120597119875119879=

0 Of interest is the relationship between the single variablevariation in (9) and the total variation in (11) Since P is afunction of 119896 then (11) is a constrained variation where

120575119876 (119896) = 120575119876 (P 119896) = (120597119876120597P

sdot 120575P + 120597119876120597119896) (12)

subjected to 119892119894(P 119896) = 119875

119894minus ℎ119894(119896) = 0 (ie 119875

119894= ℎ119894(119896)

for some function of 119896) and where 119875119894are the components of

P According to the Lagrange multiplier theory [3 Theorem1312 page 381] the function 119891 R119899 rarr R has an optimalvalue at 119909

0subject to the constraints 119892 R119899 rarr R119898 over the

subset 119878 where g = (1198921 1198922sdot sdot sdot 119892119898) vanishes that is 119909

0isin 1198830

where 1198830= 119909 119909 isin 119878 g(119909) = 0 when either of the

following equivalent equations ((13) (14)) are satisfied

119863119903119891 (1199090) +

119898

sum119896=1

120582119896119863119903119892119896(1199090) = 0 (119903 = 1 2 119899) (13)

nabla119891 (1199090) + 1205821nabla1198921(1199090) + sdot sdot sdot + 120582

119898nabla119892119898(1199090) = 0 (14)

where det 119863119895119892119894(1199090) = 0 and the 120582rsquos are invariant real

numbers We refer to 119875119894as any variable that is a function of

119896 constructed on physical or mathematical grounds and notjust to the special case defined in (8) Write

119892119894= 119875119894minus ℎ119894 (119896) = 0 (119894 = 1 2 119873

119901) (15)

where 119863119895119892119894(1199090) = 120575

119894119895since 119863

119895= 120597120597119875

119895and therefore

det 119863119895119892119894(1199090) = 0 We abbreviate the functions 119891(119894) =

119891(P 119905119894 119896) and 119891(119894) = 119891(P 119905

119894 119896) Define

119891119876 (119909) equiv 119876 (P 119896 119905) =

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119891 (119894))2

(16)

where 119884exp(119905119894) are the experimental subspace variables as in(7) with 119909 isin 119883

0defined above We next verify the relation

between 119876(119896) and 119876119879

Verification The solution 1198761015840(119896) = 119877(119896) = 0 of (10) isequivalent to the variation of 119891

119876(119909) defined in (16) subjected

to constraints 119892119894of (15)

Proof Define the Lagrangian to the problem asL = 119891119876(119909) +

sum119873119901

119894=1120582119894119892119894 Then the equations that satisfy the stationary

condition120597L

120597119875119895

= 0 119895 = 1 2 119873119901

120597L

120597119896= 0 (17)

reduce to the (equivalent) simultaneous equations

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119875119895

= 1205821015840

119895 119895 = 1 2 119873

119901 (18)

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896+

119873119901

sum119895=1

1205821015840

119895

120597119901119895

120597119896= 0 (19)

Substituting 1205821015840119895in (18) to (19) leads to

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896

+

119873119901

sum119895=1

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) sdot120597119891 (119894)

120597119875119895

sdot120597119901119895

120597119896= 0

(20)

Since 119889119875119894119889119896 = 120597119901

119894120597119896 then (20) implies 119889119876(P 119896 119905)119889119896 = 0

for the 119876 functions in (11) (12) and (16)

Of interest is the theoretical relationship of the (P 119896)variables of the 119876 functions described by (9) (12) (16)denoted 119876

1and those of the freely varying 119876 function of (1)

denoted 1198762with the variable set which can be written as

1198761= 119876 (P 119905 119896) (21)

1198762= 119876 (P 119905 119896) (22)

which is given by the following theorem where we abbreviate120572119894= (119884exp(119905119894) minus 119891(P 119905119894 119896)) and 120572119894 = (119884exp(119905119894) minus 119891(P 119905119894 119896))

where we note that the 119891 functional form is unique and is ofthe same form for both these 120572 variables


Theorem 2 The unconstrained LS solution to 1198762= 119876(P 119896 119905)

for the independent variables P 119896 is also a solution for theconstrained variation single variable 1198961015840 where P = P(1198961015840) 119896 =1198961015840 Further the two solutions coincide if and only if

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (P 119905119894 119896))120597119891 (P 119905

119894 119896)

120597119875119895

= 0

119895 = 1 2 119873119901

(23)

Proof The 1198762unconstrained solution is derived from the

equations

1205971198762

120597119875119895

= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (24)

1205971198762

120597119896= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (25)

with 119888 being constants If there is a P(119896) dependency then wehave

1198891198761 (119896)

119889119896= 119888 sdot

119873119901

sum119895=1

(

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

)120597119875119895

120597119896+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896

(26)

If the variable set P 119896 satisfies (24) and (25) in uncon-strained variation then the values when substituted into (26)satisfy the equation 119889119876

1(119896)119889119896 = 0 since 119891(119894) and 119891(119894)

are the same functional form This proves the first part ofthe theorem The second part follows from the converseargument where from (26) if 119889119876

1(119896)119889119896 = 0 then setting

one factor to zero in (27) leads to the implication of (28)

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (27)

997904rArr

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (28)

which is the solution set P(1198961015840) 1198961015840 which satisfies1198891198761(119896)119889119896 = 0 and is satisfied by the conditions of

both (27) and (28) Then (27) satisfies (24) and (28) satisfies(25)

The theorem verification and lemma above do notindicate topologically under what conditions a coincidenceof solutions for the constrained and unconstrained mod-els exists Figure 1 depicts the discussion below FromTheorem 2 if set 119860 represents the solution P 119896 for theunconstrained LS method and set 119861 = P 1198961015840 for theconstrained method then 119861 supe 119860 Define 119896 within the range1198961le 119896 le 119896

2 Then 119896 is in a compact space and since P(119896) isin

C1 P(119896) is uniformly continuous [2 Theorem 8 page 79]Then admissible solutions to the above constraint problemwith the inequality 119861 supe 119860 imply119876(P(119896)) ge 119876min where119876min

QT(P k)

QT(Pi P0Pi k)

Q = s1

Q = s2

Q = s3

Pi

(P(k) k)

QT(P0 k0) = Qmin

(P0 k0)

Pi0 + ΔPiPi0 + ΔP

998400i

Q(k)

Figure 1 Depiction of how the 119896 variation optimizing 119876 leads to asolution on a level surface of the 119876

119879function where 119876

119879le 119876

is the unconstrained minimum The unconstrained 119876 = 119876119879

LS function to be minimized in (11) implies

nabla119876 = 0 (120597119876

120597119896=120597119876

120597119875119894

= 0 for 119894 = 1 2 119873119901) (29)

Defining the constrained function 119876119888(119896) =

sum1198731015840

119894=1(119884exp(119894) minus 119891(P(119896) 119905119894 119896))

2 then 119876119888(119896) = 119876 ∘ 119875

119879where

119875119879

= (1198751(119896) 1198752(119896) 119875

119873119901

(119896) 119896)119879 Because 1198761015840

119888(119896) =

(120597119876120597P 120597119876120597119896) sdot (11987510158401(119896) sdot sdot sdot 119875

119873119901

(119896) 119896)119879 solutions occur

when (i) nabla119876 = 0 corresponding to the coincidence of thelocal minimum of the unconstrained 119876 for the best choicefor the line with coordinates (P(119896) 119896) as it passes throughthe local unconstrained minimum and (ii) 1198751015840

119894(119896) = 0

(119894 = 1 2 119873119901) 120597119876120597119896 = 0 where this solution is a

special case of (iii) when the vector 1198751015840119879is perp to nabla119876 = 0

that is 1198751015840119879is at a tangent to the surface 119876 = 119878

2for some

1198782ge 119876min where this situation is shown in Figure 1 where

the vector is tangent at some point of the surface 119876119879= 1198782

Whilst the above characterizes the topology of a solutiononly the existence of a solution for the line (P(119896) 119896) whichpasses through the point of the unconstrained minimumof 119876119879is proven below under certain conditions where a set

of equations are constructed to allow for this significantapplication that specifies the conditions when the standardLS constrained variation solution implies the same solutionas for the unconstrained variation Also discussed is the casewhen it may be possible for unconstrained solution set 119880


to satisfy the inequality 119876119888(119880) ge 119876

119888 where 119876

119888is a function

designed to accommodate all solutions of (7) as given belowin (30)

22 Discussion of LS Fit for a Function 119876119888with a Possibility

of a Smaller LS Deviation than for P 119896 Parameters Derivedfrom a Free Variation of (11) The LS function metric suchas (11) implied 119876(P 119896) le 119876(P 119896) at a stationary (minimum)point for variables (P 119896) On the other hand the sets ofsolutions of (7) denoted

119894119873119888in number provides for each

set exact solutions P(119896) averaged to P(119896) using (8) If the P119894

119894 = 1 2 119873119888solutions are in a 120575-neighbourhood then we

examine the possibility that the composite functionmetric tobe optimized over all the sets of equations

119894119873119888in number

defined here as

119876119888 (P 119896) = sum

119894isin12sdotsdotsdot119873119888

(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)

where (P119879 119896) is the unconstrained optimized value of (22)

This implies that under these conditions the 119876119888of (30) is

a better measure of fit This will be proven to be the caseunder certain conditions below For what follows the P(119896)

119894

for equation set 119894obtains for all 119896 values of the open set

S 119896 isin S from the IFT including 1198960which minimises (9)

Another possibility that will be discussed briefly later is wherein (30) all P 119896 are free to vary Here we consider the caseof the 119873

119888P values averaged to P for some 119896 We recall the

intermediate-value theorem (IVT) [3 Theorem 438 page87] for real continuous functions 119891 defined over a connecteddomain 119878 which is a subset of some R119899 We assume that the119891 functions immediately below obey the IVT For each P

119894

solution of the 119894set for a specific 119896 = 119896

0we assume that the

function 119876119888119894(P) is a strictly increasing function in the sense

of definition (7) below where

119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)

with 119876119888119894(P)(1198960) = 0 in the following sense

Definition 3 A real function 119891 is (strictly) increasing in aconnected domain 119878 isin R119873 about the origin at 119903

0if relative

to this origin if |1199032| gt |119903

1| (for the boundaries 120597 of ball

119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))

Note A similar definition is obtained for a (strictly) decreas-ing function with the (lt) le inequalities Since the 120597119861boundaries are compact and 119891 is continuous the maximumand minimum values are attained for all ball boundariesWe assume 119891 to be strictly increasing relative to 119903

0for what

follows below

Lemma4 For any region bounded by 120597119861(1199030 1199031) and 120597119861(119903

0 1199032)

with coordinate r (radius 119903 centered about coordinate 1199030)

min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)

Proof Suppose in fact 119891(r) lt min119891(120597119861(1199030 1199031)) then

min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)

which is a contradiction to the definition and a similar proofis obtained for the upper bound

Note Similar conditions apply for the nonstrict inequalitiesle ge

The function that is optimized is

119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)

Define P119894as the solution vector for the equation set

119894

We illustrate the conditions where the solution P119879for a

free variation for the 119876 metric given in (11) can fulfill theinequality where 119876

119888is as defined in (35)

119876119888(P119879) ge 119876

119888(P) (36)

withP given as in (8) A preliminary result is required Definemax P

119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879

Proof Any point P would be located within a sphericalannulus centered at P

119894 with radii chosen so that by Lemma 4

we have the following results

120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575

Define Ann(120575 120575119894P119894) as the space bounded by the boundary

of the balls centered onP119894of radius 120575 and 120575

119894(120575 gt 120575

119894)ThenP isin

Ann(120575 120575119894P119894) by Lemma 5 Since 119876

119888in (30) is not equivalent

to119876119879= 119876 in (11) where wewrite here the free variation vector

solution as P119879 then the above results lead to the following

119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)

where (40) follows from (33) Summing (40) leads to119876119888(P119879 119896) gt 119876

119888(P 119896)


Hence we have demonstrated that it may bemore realisticor accurate to fit parameters based on a function thatrepresents different coupling sets such as119876

119888above rather than

the standard LS method using (30) if P119879lies sufficiently far

away from P We note that if P119879is the solution of the free

variation of the above 119876119888in (30) then from the arguments

presented after the proof of Theorem 2 it follows that

119876119888(P119879) le 119876

119888(P) (41)

which implies that the independent variation of all param-eters in LS optimization of the 119876

119888variation is the most

accurate functional form to use assuming equal weighting ofexperimental measurements than the standard free variationof parameters using the 119876 function of (11)

3 Method (i)b Theory

Whilst it is advantageous in science data analysis to optimizea particular multiparameter function by focusing on a fewkey variables (our 119896 variable of restricted dimensionalitywhich we have applied to a 1-dimensional optimization inthe next section) it has been shown that this method yieldsa solution that is always of higher value for the same 119876function than a full independent parameter optimizationmeaning that it is less accurate The key issue therefore iswhether for any119876 function including those of the119876

119888variety

it is possible to construct a 119896 parameter optimization suchthat the line of parameter variables P(119896) passes through theminimum surface of the 119876 function We develop a theory toconstruct such a function below However method (i)a maystill be advantageous because of the greater simplicity of theequations to be solved and the fact that C1119891(119894) functionswere required whereas here the 119891(119894) functions must be atleastC2 continuous

Theorem 7 For the 119876119879(P 119896) function defined in (11) where

each of the 119891(119894) functions is C2 on an open set R119873119901+1 andwhere 119876

119879is convex the solution at any point 119896 of 120597119876

119879120597119875119895=

0 (119895 = 1 2 119873119901) whenever det120597QT120597Pi120597Pj = 0 at

(119889119876119879119889119896)(1198961015840) = 0 determines uniquely the line equation P(119896)

that passes the minimum of the function 119876119879when 119896 = 1198961015840

Proof As before 119891(119894) = 119891(P 119905119894 119896) so that

119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572

119896= 119884exp(119905119896) minus 119891(119896) and for an

independent variation of the variables (P 119896) at the stationarypoint we have

120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)

The above results for the functions ℎ119895(P 119896) = 0 (119895 =

1 2 119873119901) having a unique implicit function of 119896 denoted

P(119896) by the IFT [3 Theorem 137 page 374] require thatdet 120597ℎ

119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open

set 119878 119896 isin 119878 More formally the expansion of the precedingdeterminant in (45) verifies that a symmetric matrix isobtained for 120597ℎ

119894(P 119896)120597119875

119895due to the commutation of second-

order partial derivatives of 119875119895

120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)

Defining1198761(119896) as a function of 119896 only by expanding119876

119879yields

the total derivative with respect to 119896 as 11987610158401(119896) where

1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)

Then ℎ119894(P 119896) = 0 by construction (43) so that 120597119876

119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0

(for all 119895) and hence

(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)

Substituting (48) derived from (43) and (44) into (47)together with the condition 1198761015840

1(119896) = 0 implies that 119888 sdot

sum1198731015840

119894=1120572119894(120597119891(119894)120597119896) = 0 which satisfies (44) for the free

variation in 119896 Thus 11987610158401(119896) = 0 rArr 120575119876

119879= 0 for independent

variation of (P 119896) So 119876119879fulfills the criteria of a stationary

point at say 119896 = 1198960 since nablaP119896119876119879 = 0 ([2 Proposition 16

page 112]) Suppose that 119876119879is convex where 119875

119879= P0 1198960 is

a minimum point 119875119879isin 119863 a convex subdomain of 119876

119879 Then

at 119875119879 nablaP119896119876119879 = 0 and 119875119879 is also the unique global minimum

over 119863 according to [1 Theorem 32 page 46] Thus 119875119879is

unique whether derived from a free variation of (P 119896) or via119875(119896) dependent parameters with the 119876

1function

Note As before 119876119879

and 1198761may be replaced with the

summation of indexes as for 119876119888in (30) to derive a physically

more accurate fit

4 Method (ii) Theory

Methods (i)a and (i)b which are mutual variants of eachother are applications of the implicit method to modelingproblems to provide a best fit to a function Here anothervariant of the implicit methodology for optimization of atarget or cost function 119876

119864is presented One can for instance

consider 119876119864(P 119896) to be an energy function with coordinates

R = P 119896 where as before the components of P are 119875119895 119895 =

1 2 119873119901 119896 isin R is another coordinate so that R isin R119873119901+1


For bounded systems (such as the molecular coordinates)one can write

119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)

Thus R isin D sub R119873119901+1 is in a compact spaceD Define

120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)

Then the equilibrium conditions become

119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)

Take (50) as the defining equations for 119900119895(P 119896) which is

specified by 119876119864in (50) which casts it in a form compatible

with the IFT where some further qualification is required for(P 119896) Assume 119876

119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876

119864120597119875119895120597119875119894 The matrix of the aforementioned

determinant is symmetric partaking of the properties due tothis fact Then by the IFT [3 Theorem 137 page 374] exist is aunique P(119896) function where for some 119896

0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T

0times R) sube D and 119896

0isin T0such

that 119900119895(P(119896) 119896) = 0 for all 119896 isin T

0 For 119896 an isolated point

119896 = 119896119900 from analysis we find 119896 to be still open Write

1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)

Denote 119896119894as a solution to 1198761015840

1198641(119896) = 0 where 119896

119894isin T0in the

indicated range above in (49)

Theorem 8 The stationary points 119896119894(119894 = 1 2 119873

119896) where

1198761198641(119896) = 0 for 119896 = 119896

119894exist for the range 119896

119894 119896min le

119896 le 119896max of coordinate 119896 if and only if for each of these 119896119894

(i) 119876119864(P 119896) isin C2 and (ii) det120597QE(P ki)120597Pj120597Pi = 0 (for

all 119896119894 119894 = 1 2 119896max) Each of these points 119896

119894isin R1 space

corresponds uniquely in a local sense in the open setT0to some

equilibrium (stationary) point of the target function119876119864(P 119896) in

R119873119901+1 space

Proof If 11987610158401198641(119896) = 0 where 119896 isin T

0 then it also follows from

the IFT that 119900119895= 0 and therefore 120597119876

119864120597119896 = 0 from (53)

which satisfies (50) and (51) for the equilibrium point Theconditions (i) and (ii) of the theorem are a requirement of theIFT Conversely if 119900

119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0

(a stationary or equilibrium point) then by (53) 11987610158401198641(119896) = 0

Hence the coordinates 119896119894 for which 1198761015840

1198641(119896119894) = 0 refer to

the condition where 120575119876119864(P 119896119894) = 0 and uniqueness follows

from the IFT reference to the local uniqueness of the P(119896)function

Note In a bounded system one can choose any of the 119873119901

components 119875119895of P as the 119896 coordinate (denoted 119896

119894) partly

based on the convenience of solving the implicit equationsto determine the 119896

119894minima and thus determine by the

uniqueness criterion the coordinates of the minima inR119873119901+1

space (spanning the119873119901independent variables and 119896)

For nondegenerate coordinate choice meaning that fora particular 119896 coordinate choice there does not exist anequilibrium structure (meaning a set of coordinate values)where for any two structures 119860 and 119861 119896

119860= 119896119861 For such

structures the total number of minima that exists withinthe bounded range in the 119896 coordinate is equal to the totalnumber of minima of the target function119876

119864(P 119896) within the

bounded range Hence a method exists for the very challeng-ing problem of locating and enumerating minima [9 Section51 page 242 ldquoHow many stationary points are thererdquo]From the uniqueness theorem of IFT one could infer pointsin the 119896-axis where nonuniqueness is obtained that iswhenever det 1205972119876

119864120597119875119894120597119875119895 = 0 In such cases for particles

with the same intermolecular potentials permutation of thecoordinates in conjunction with symmetry considerationscould be of use in selecting the appropriate coordinate systemto overcome these systems with degeneracies [9 Section425 page 205 ldquoAppearance and disappearance of symmetryelementsrdquo] Other methods that might address this situationinclude scanning different one dimensional (1D) choices 119896 =119875119894graphs or profiles where if degeneracies exist for choice

119896 = 119875119895 they may not exist for the choice 119896 = 119875

119894in the graph

for a specific point (119875119894

= 119875119895) Thus by scanning through

all or selecting a number of the (1D) 119875119895profiles for 119876

1198641 it

would be possible to make an assignment of the location of aminimum in R119873119901+1 space One is reminded of the methodsthat spectroscopists use in assigning different energy bandsbased on selection rules to uniquely characterize for instancevibrational frequencies A similar analogy is obtained forX-ray reflections where the amplitude variation of the X-ray intensity in reciprocal space can be used to elucidatestructure The minima of the 1D 119896 coordinate scan mustcorrespond to the minima inR119873119901+1 space of the119876

119864function

given that all such minima in 119876119864are locally strict and global

within a small open set about the minima for by continuity1198761198641(119896)minus119876

1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt

1205752 which violates the condition for a maximum

5 Specic Algorithms and Pseudocode forSolution to Optimization Problem UtilizingMethod (i)(a) Method (i)(b) and Method (ii)

We provide suggestions in pseudocode form for the above3 proven methodologies Real world applications of thesemethods are very involved undertakings that are separateresearch topics in their own right Nevertheless we provide adetailed and extended application of method (i)a suitable fora real world chemical kinetics problem where experimentaldata from the published literature are used for method(i)a in Section 531 and the results obtained compared toconventional techniques


51 Pseudocode Algorithm for Method (i)a Of the manyvariations possible the following approach conforms to thetheoretical development

(1) For any physical law for a total of 1198731015840 datapointschoose119873

119901datapoints (set 120572) and solve for parameters

119875119895 119895 = 1 2 119873

119901according to (7) for each of

the sets 120572 119873119888in number for a known value of 119896

The solution set P(119896)120572may be derived analytically

(as in the example below) or by appropriate linearapproximations

(2) Determine from the above set P(119896) either the geo-metric or statistical average (as used here) for the119873

119888

solutions that is P(119896) = sum119873119888120572=1

P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)

(4) Solve the 1D equation at 119896 = 1198960when 1198761015840(119896) = 0

The solution set is P(1198960) 1198960 for the optimization

problem

52 Pseudocode Algorithm for Method (i)b This is an ldquoexactrdquomethod relative to LS variation of all parameters A suitablealgorithm based on the theory could be as follows

(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)

for a particular value of 119896 Since there are 119873119901equa-

tions for ℎ119895 a solution P(119896) exists The solution may

be exact or some linear approximation depending onnature of the problem and convergence criteria

(2) Form the function1198761(119896) = sum

1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0

(4) The solution set to the problem is P(1198960) 1198960 for

optimizing the LS function 119876119879

= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2

53 Pseudocode Algorithm for Method (ii) Here 119876119864(P 119896) is

a general function not necessarily of form 119876119879in (42) Then

for 119873119901variables P define functions 119900

119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)

(1) For a particular 119896 solve [22] for P P(119896) existssince there are119873

119901equations Approximate linearized

solutionsmight also be attempted in the vicinity of thestation point of 119896

(2) Form the function 119876119864119896(119896) = 119876

119864(P 119896)

(3) Solve for 1198960 such that 1198761015840

119864119896(119896) = 0

(4) Solution to the optimization problem of 119876119864(P 119896) by

varying independently all the domain variables isP(1198960) 1198960

531 Application of Method (i)a Algorithm (Section 51) inChemical Kinetics The utility of one of the above triad ofmethods is illustrated in the determination of two parametersin chemical reaction rate studies of 1st and 2nd orderrespectively using data from published literature wheremethod (i)a yields values close within experimental error tothose quoted in the literatureThemethod can directly derivecertain parameters like the final concentration terms (eg120582infin

and 119884infin) if 119896 the rate constant is the single optimizing

variable in this approximation which is not the case in mostconventional methodologies We assume here that the ratelaws and rate constants are not slowly varying functions ofthe reactant or product concentrations which have recentlyfrom simulation been shown to be generally not the case[29] Under this standard assumption the rate equationsbelow are all obtained The first-order reaction studied hereis (i) the methanolysis of ionized phenyl salicylate with dataderived from the literature [30 Table 71 page 381] and thesecond-order reaction analyzed is (ii) the reaction betweenplutonium(VI) and iron(II) according to the data in [31 TableII page 1427] and [32 Tables 2ndash4 page 25]

532 First-Order Results Reaction (i) above corresponds to

PSminus + CH3OH119896119886

997888rarr MSminus + PhOH (56)

where the rate law is pseudo first-order expressed as

rate = 119896119886[PS]minus= 119896119888[CH3OH] [PSminus] (57)

with the concentration of methanol held constant (80 vv)and where the physical and thermodynamical conditions ofthe reaction appear in [30 Table 71 page 381] The change intime 119905 for any material property 120582(119905) which in this case is theabsorbance 119860(119905) (ie 119860(119905) equiv 120582(119905)) is given by

120582 (119905) = 120582infin minus (120582infin minus 1205820) exp (minus119896119886119905) (58)

for a first-order reaction where 1205820refers to the measurable

property value at time 119905 = 0 and 120582infin

is the value at 119905 = infin

which is usually treated as a parameter to yield the best leastsquares fit even if its optimized value is less formonotonicallyincreasing functions (for positive 119889120582119889119905 at all 119905) than anexperimentally determined 120582(119905) at time 119905 In Table 71 of [30]for instance 119860(119905 = 2160 s) = 0897 gt 119860optinfin = 0882 andthis value of119860

infinis used to derive the best estimate of the rate

constant as 165 plusmn 01 times 10minus3 secminus1 in that workFor this reaction the 119875

119894of (5) refers to 120582

infinso that P equiv

120582infin

with 119873119901= 1 and 119896 equiv 119896

119886 To determine the parameter

120582infin

as a function of 119896119886according to (9) based on the entire

experimental (120582exp 119905119894) dataset we invert (58) and write

120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840

(120582exp (119905119894) minus 1205820 expminus119896119905119894)(1 minus expminus119896119905

119894)

(59)


where the summation is for all terms with the i subscriptof the experimental dataset that does not lead to zerosnor singularities such as when 119905

119894= 0 We define the

nonoptimized continuously deformable theoretical curve120582th where 120582th equiv 119884th(119905 119896) in (6) as

120582th (119905 119896) = 120582infin (119896) minus (120582infin (119896) minus 1205820) exp (minus119896119886119905) (60)With such a relationship of the 120582

infinparameter 119875 to 119896 we seek

the least square minimum of 1198761(119896) where 119876

1(119896) equiv 119876 of (9)

for this first-order rate constant 119896 in the form

1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)

where the summation is over all the experimental (120582exp(119905119894) 119905119894)values The solution of the rate constant 119896 corresponds to thezero value of the function which exists for both orders TheP parameters (120582

infinand 119884

infin) are derived by back substitution

into (59) and (65) respectively The Newton-Raphson (NR)numerical procedure [4 page 456] was used to find the rootsto 119875119896 For each dataset there exists a value for 120582

infinand so

the error expressed as a standard deviationmay be computedThe error tolerance for theNRprocedurewas set to 10times10minus10We define the function deviation119891119889 as the standard deviationof the experimental results with the best fit curve where 119891119889 =radic(1119873)sum

119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))

2 Our results are as follows

119896119886= 162 plusmn 09 times 10minus2 sminus1 120582

infin= 088665 plusmn 006 and

119891119889 = 3697 times 10minus3The experimental estimates are 119896

119886= 165 plusmn 01 times 10minus2 sminus1

120582infin= 0882 plusmn 00 and 119891119889 = 8563 times 10minus3The experimental method involves adjusting the 119860

infinequiv

120582infin

to minimize the 119891119889 function and hence no estimate ofthe error in 119860

infincould be made Method (i)a allows direct

calculation of 120582infinand its error without the extraneous fittings

required in the conventional methods It is clear that ourmethod has a lower 119891119889 value and is thus a better fit andthe parameter values can be considered to coincide with theexperimental estimates within experimental error Figure 2shows the close fit between the curve due to our optimizationprocedure and experiment The resulting 119877

119896function (10)

for the first-order reaction based on the published dataset isgiven in Figure 3 The very slight variation between the twocurves could be due to experimental uncertainties as shownin Figure 1

533 Second-Order Results To further test our method wealso analyze the second-order reaction

Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)

whose rate V is given by V = 1198960[PuO2+2][Fe2+] where 119896

0is

relative to the constancy of other ions in solution such asH+ The equations are very different in form to the first-orderexpressions and serves to confirm the viability of the currentmethod

For Espenson the above stoichiometry is kineticallyequivalent to the reaction scheme [32 equation (2-36)]

PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)

1st-order fitted curve1st-order expt curve

Figure 2 Plot of the experimental and curvewith optimized param-eters showing the very close fit between the twoThe slight differencebetween the two can probably be attributed to experimental errors

which also follows from the work of Newton and Baker [31equations (8) (9) page 1429] whose data [31 Table II page1427] we use and analyze to verify the principles presentedhere Espenson had also used the same data as we have toderive the rate constant and other parameters [32 pages 25-26] which are used to check the accuracy of ourmethodologyThe overall absorbance in this case 119884(119905) is given by [32equation (2-35)]

119884 (119905) =119884infin+ 1198840 (1 minus 120572) minus 119884infin exp (minus119896Δ 0119905)1 minus 120572 exp (minus119896Δ

0119905)

(64)

where 120572 = [A]0[B]0is the ratio of initial concentrations

where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]

and [B]0= 447 times 10minus5M and [A]

0= 382 times 10minus5M A

rearrangement of (64) leads to the equivalent expression [32equation (2-34)]

ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)

According to Espenson one cannot use this equivalent form[32 page 25] ldquobecause an experimental value of 119884

infinwas not

reportedrdquo and he further asserts that if 119884infin

is determinedautonomously then 119896 the rate constant may be determinedThus central to all conventional methods is the autonomousand independent status of both 119896 and 119884

infin We overcome

this interpretation by defining 119884infin

as a function of the totalexperimental spectrum of 119905

119894values and 119896 by inverting (64) to

define 119884infin(119896) as

119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840

119884exp (119905119894) exp (119896Δ 0119905119894) minus 1 + 1198840 (120572 minus 1)(exp (119896Δ

0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6

Figure 3 119877(119896) functions (10) for reactions (i) and (ii) of order oneand two in reaction rate The horizontal bars correspond to the zerovalues of the 119877(119896) functions

where the summation is over all experimental values thatdoes not lead to singularities such as at 119905

119894= 0 In this case the

P parameter is given by119884infin(119896) = 119875

1(119896) 119896119887= 119896 is the varying 119896

parameter of (5)We likewise define a function119884th of 119896 that isalso a function of 119905 but where the 119896 parameter is interpretedas a ldquodistortionrdquo parameter in the following manner

119884(119905 119896)th =119884infin (119896) + 1198840 (1 minus 120572) minus 119884infin (119896) exp (minus119896Δ 0119905)

1 minus 120572 exp (minus119896Δ0119905)

(67)

In order to extract the parameters 119896 and 119884infin we minimize

the square function1198762(119896) for this second-order rate constant

with respect to 119896 given as

1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)

where the summation is over the 119873 experimental 119905119894coor-

dinates Then the solution to the minimization problem iswhen the corresponding 119877(119896) function (10) is zero The NRmethod was used to solve 119877(119896) = 0 with the error toleranceof 10 times 10minus10 With the same notation as in the first-ordercase the second order results are 119896

119887= 9380 plusmn 18 (Ms)minus1

119884infin= 00245 plusmn 0003 and 119891119889 = 9606 times 10minus4The experimental estimates from the conventional meth-

ods are [32 page 25] 119896119887= 9490 plusmn 22 (Ms)minus1 119884

infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)

2nd-order fitted curve2nd-order expt curve

Figure 4 Graph of the experimental and calculated curve based onthe current induced parameter-dependent optimization method

Again the two results are in close agreementThe graph ofthe experimental curve and the one that is derived from ouroptimization method are given in Figure 4

6 Conclusions

The triad of associated implicit function optimization coversboth the topics of modeling of data and the optimizationof arbitrary functions where experimental or theoreticalconsiderations require that a single variable is tagged to aprocess variable that is iteratively relaxing to an equilibriumstationary point Applying method (i)a to chemical kineticsallows for the direct determination of parameters that isnot possible by application of the standard methodologiesThe results presented here show that for linked variablesit is possible to derive all the parameters associated with acurve by considering only one independent variable whichserves as the independent variable for other functions in theoptimization process as illustrated by methods (i)ab Apartfrom possible reduced errors in the computations it mightalso be a more accurate way of deriving parameters that aremore influenced or conditioned (on physical grounds) bythe value of one parameter (such as 119896 here) than others thecurrent methods that give equal weight to all the variablesmight in some cases lead to results that would be consideredldquounphysicalrdquo In complex dynamical systems with multipro-cesses the physical considerations are such that for scientificpurposes it would be advantageous if optimization would beconducted on just one primary coordinate variable such asin attempting to derive the most general stable conformer ina large molecule where there are thousands of local min-ima present if all free coordinate variables are considered


[9 Section 67 page 330] For such systemsmethod (ii)mightbe applicable This generalized potential surface might befound suitable for reaction trajectory calculations [9 Chapter4 page 192 on ldquoFeatures of a landscaperdquo] that require a singlepath variable where the general optimized conformer wouldbe relevant to the study of the potential surfaces and forcefields present

List of Variables

(P 119905 119896) the entire domain space of a function suchas 119891(P 119905 119896) where 119896 is the variableassociated with a sequence ofmeasurements such as along the timecoordinate The vector P with components119875119894is the normal parameter that must be

optimized and 119896 is the specially chosenvariable on experimental grounds that isoptimized whilst constructing functionssuch that P = P(119896) (6)

119891(P 119905 119896) refers to a function that is proposed to be aldquolaw of naturerdquo whose parameters P 119896are to be optimized (7)

119884law(119905 119896) 119884law(119905 119896) = 119891(P 119905 119896) (5)119884th(119905 119896) the theoretical law of nature if P = P(119896) is

determined That is 119884th(119905 119896) = 119891(P 119905 119896)(6)

119884exp(119905119894) an experimentally determined datapointthat ideally represents the range of119891(P 119905 119896) if there was no error that is119884exp(119905119894) = 119891(P1015840 119905 1198961015840) for a perfect fit for all119905119894 for fixed (P1015840 1198961015840) (7)

119875119894(119896) an averaged value for 119875

119894(119896) based on some

specified algorithm (8)119876119883 119876(119896) least squares (LS) function to optimize

119884law(119905 119896) in the case of 119876 (119896) (9) Ingeneral all 119876

119883functions are LS functions

specified by119883 (eg (9) (11) (42) etc)119876119864 General cost or object function to be

optimized not necessarily in LS form (50)119877(119896) = 119876

1015840(119896) (10)

120582119894 Lagrange multipliers associated with the

119876119883optimization (14)

L Lagrangian to the optimization problem(17)

119896119902 chemical kinetics rate constant for

reaction 119902 (eg (56))120582(119905) 120582

infin absorbance measurements for first-orderchemical kinetics reactions at time 119905 andinfinity (eg (58) and (59))

119884(119905) 119884infin absorbance measurements forsecond-order chemical kinetics reactionsat time 119905 and infinity (eg (64) and (66))

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by University of Malaya GrantUMRG(RG07709AFR) and Malaysian Government grantFRGS(FP0842010A)This work was initiated and completedduring a Sabbatical research visit (2012-2013) to the Atom-istic Simulation Centre (ASC) School of Mathematics andPhysics Queenrsquos University Belfast I thank Ruth Lynden-Bell(Chemistry Department Cambridge University) for facili-tating this visit Cordial discussions concerning real worldapplications with faculty at ASC are gratefully acknowledgedI thank my hosts Jorge Kohanoff (ASC) and ChristopherHardacre (Chem Dept QUB) for congenial hospitalityduring this time

References

[1] B D Craven Functions of Several Variables Chapman amp HallLondon UK 1981

[2] J DePree and C Swartz Introduction to Real Analysis JohnWiley amp Sons New York NY USA 1988

[3] T M Apostol Mathematical Analysis Narosa PublishingHouse New Delhi India 2nd edition 2002

[4] WH Press S A TeukolskyW T Vetterling and B P FlanneryNumerical RecipesmdashTheArt of Scientific Computing CambridgeUniversity Press Cambridge UK 3rd edition 2007

[5] J A Snyman Practical Mathematical Optimization An Intro-duction to Basic Optimization Theory and Classical and NewGradient-Based Algorithms Springer New York NY USA2005

[6] W C Davidon ldquoVariable metric method for minimizationrdquoSIAM Journal on Optimization vol 1 no 1 pp 1ndash17 1991

[7] CG Broyden ldquoThe convergence of a class of double-rankmini-mization algorithmsrdquo Journal of the Institute ofMathematics andIts Applications vol 6 pp 76ndash90 1970

[8] A Banerjee N Adams J Simons and R Shepard ldquoSearch forstationary points on surfacesrdquo Journal of Physical Chemistry vol89 no 1 pp 52ndash57 1985

[9] D A Wales Energy Landscapes Cambridge Molecular ScienceCambridgeUniversity Press Cambridge UK 2003 edited by RSaykally A Zewail and D King

[10] S Kirkpatrick C D Gelatt and M P Vecchi ldquoOptimizationby simulated annealingrdquo The American Association for theAdvancement of Science Science vol 220 no 4598 pp 671ndash6801983

[11] D M Bates and D G Watts Nonlinear Regression Analysis andIts Applications Wiley Series in Probability and MathematicalStatistics John Wiley amp Sons New York NY USA 1988

[12] G H Golub and V Pereyra ldquoThe differentiation of pseudo-inverses and nonlinear least squares problems whose variablesseparaterdquo SIAM Journal on Numerical Analysis vol 10 pp 413ndash432 1973

[13] I H M van Stokkum D S Larsen and R van GrondelleldquoGlobal and target analysis of time-resolved spectrardquoBiochimicaet Biophysica ActamdashBioenergetics vol 1657 no 2-3 pp 82ndash1042004

[14] P Shearer and A C Gilbert ldquoA generalization of variable elim-ination for separable inverse problems beyond least squaresrdquoInverse Problems vol 29 no 4 Article ID 045003 27 pages2013


[15] N IMGould Y Loh andD P Robinson ldquoA filtermethodwithunified step computation for nonlinear optimizationrdquo SIAMJournal on Optimization vol 24 no 1 pp 175ndash209 2014

[16] A Milzarek and M Ulbrich ldquoA semismooth Newton methodwith multidimensional filter globalization for 119897

1-optimizationrdquo

SIAM Journal on Optimization vol 24 no 1 pp 298ndash333 2014[17] H M Gomes ldquoTruss optimization with dynamic constraints

using a particle swarm algorithmrdquo Expert Systems with Appli-cations vol 38 no 1 pp 957ndash968 2011

[18] MGendreau and J-Y Potvin EdsHandbook ofMetaheuristicsvol 146 of International Series in Operations Research amp Man-agement Science Springer New York NY USA 2nd edition2010

[19] R Varadhan and P D Gilbert ldquoBB an R package for solving alarge system of nonlinear equations and for optimizing a high-dimensional nonlinear objective functionrdquo Journal of StatisticalSoftware vol 32 no 4 pp 1ndash26 2009

[20] F M P Raupp L M G Drummond and B F Svaiter ldquoAquadratically convergent Newton method for vector optimiza-tionrdquo Optimization vol 63 no 5 pp 661ndash677 2014

[21] M D Al-Khaleel M J Gander and A E Ruehli ldquoOptimizationof transmission conditions in waveform relaxation techniquesfor RC circuitsrdquo SIAM Journal on Numerical Analysis vol 52no 2 pp 1076ndash1101 2014

[22] A Amirteimoori D K Despotis and S Kordrostami ldquoVari-ables reduction in data envelopment analysisrdquo Optimizationvol 63 no 5 pp 735ndash745 2014

[23] S Solar W Solar and N Getoff ldquoA pulse radiolysis-computersimulation method for resolving of complex kinetics andspectrardquo Radiation Physics and Chemistry vol 21 no 1-2 pp129ndash138 1983

[24] A-M Wazwaz ldquoA note on using Adomian decompositionmethod for solving boundary value problemsrdquo Foundations ofPhysics Letters vol 13 no 5 pp 493ndash498 2000

[25] J J Houser ldquoEstimation ofAinfinin reaction-rate studiesrdquo Journal

of Chemical Education vol 59 no 9 pp 776ndash777 1982[26] P Moore ldquoAnalysis of kinetic data for a first-order reaction

with unknown initial and final readings by the method of non-linear least squaresrdquo Journal of the Chemical Society FaradayTransactions I Physical Chemistry in Condensed Phases vol 68pp 1890ndash1893 1972

[27] W E Wentworth ldquoRigorous least squares adjustment appli-cation to some non-linear equations Irdquo Journal of ChemicalEducation vol 42 no 2 pp 96ndash103 1965

[28] W E Wentworth ldquoRigorous least squares adjustment appli-cation to some non-linear equations IIrdquo Journal of ChemicalEducation vol 42 no 3 pp 162ndash167 1965

[29] C G Jesudason ldquoThe form of the rate constant for elementaryreactions at equilibrium from MD framework and proposalsfor thermokineticsrdquo Journal of Mathematical Chemistry vol 43no 3 pp 976ndash1023 2008

[30] M N Khan Micellar Catalysis vol 133 of Surfactant ScienceSeries Taylor amp Francis Boca Raton Fla USA 2007 edited byA T Hubbard

[31] T W Newton and F B Baker ldquoThe kinetics of the reactionbetween plutonium(VI) and iron(II)rdquo Journal of Physical Chem-istry vol 67 no 7 pp 1425ndash1432 1963

[32] JH EspensonChemical Kinetics andReactionMechanisms vol102 McGraw-Hill Singapore 2nd edition 1995

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


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

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


attempted here for these deterministic methods but in real-world applications some of the assumptions (eg C2 conti-nuity compactness of spaces) may not always be obtainedFor what follows the distance metrics used are all Euclideanrepresented by | sdot | or sdot where det sdot represents thedeterminant of the matrix sdot Reduction of the number ofvariables to be optimized is possible in the standard matrixregressionmodel only if conditional linear parameters120573 exist[11] where these 120573 variables do not appear in the final 119878(120579)expression of the least squares function (2) to be optimizedwhereas the 120601 nonconditional linear parameters do and are asubset of the 120579 variables for the existence of each conditionallinear parameter there is a unit reduction in the number ofindependent parameters to be optimizedThese reductions invariable number occur for any ldquoexpectation functionrdquo 119891(x 120579)which is the model or law for which a fitting is requiredwhere there are 119873 different datapoints x

119894 119894 = 1 2 119873

that must be used to determine the 119901 parameter variables 120579[11 page 32 Chapter 2] A conditionally linear parameter 120579

119894

exists if and only if the derivative of the expectation function119891(x 120579) with respect to 120579

119894is independent of 120579

119894 Clearly such a

condition may severely limit the number of parameters thatcan be neglected for the expectation function variables whenthe prescribed matrix regressional techniques are employed[11 Section 355 page 85] where the residual sum of squaresis minimized

119878 (120579) =1003817100381710038171003817y minus 120578(120579)

10038171003817100381710038172 (2)

The119873-vectors 120578(120579) in 119875-dimensional space define the expec-tation surface If the 120579 variables are partitioned into theconditional linear parameters 120573 and the other nonlinearparameters 120601 then the response can be written 120578(120573120601) =A(120601)120573 Golub and Pereyra [12] used a standard Gauss-Newton algorithm to minimise 119878

2(120601) = y minus A(120601)(120601)

2

that depended only on the nonlinear parameters 120601 where(120601) = A+(120601)y with A+ being a defined pseudoinverseof A [11 Section 355 page 85] where A+ and A arematrices The variables must be separable as discussed aboveand the number of variable reduction is only equal tothe number of conditional linear parameters that exists forthe problem In applications the preferred algorithm thatexploits this valuable variable reduction is called variableprojection There are many applications in time resolvedspectroscopy that is heavily dependent on this technique andmany references to the method are given in the review byvan Stokkum et al [13] Recently this method of variableprojection has been extended in a restricted sense [14] in thefield of inverse problems which is not related to our methodof either modeling or optimization nor is the methodologyrelated to the implicit function properties In short much ofthe reported methods developed are 119886119889 ℎ119900119888 meaning thatthey are constructed to face the specific problems at handwith no claim to overall generality and this work too is 119886119889 ℎ119900119888in the sense of suggesting variable reduction with specificclasses of noninverse problems as indicated where the workdevelops a method of reducing the variable number to unityfor all variables in the expectation function space irrespectiveof whether they are conditional or not by approximating their

values by a method of averages (for method (i)a) withoutany form of linear regression being used in determiningtheir approximations during the minimization iterationsand without necessarily using the standard matrix theorythat is valid for a very limited class of functions Methods(i)b and (ii) are on the other hand exact treatments Noldquoeliminationrdquo of conditional linear parameters is involved inthis nonlinear regressionmethodNor is any projection in themathematical sense involved These general methods couldhave useful applications in deterministic systems comprisingmany parameters that are all linked to one variable theprimary one (denoted 119896 here) that is considered on physicalgrounds to be the most significant A generalization of thismethod would be to select a smaller set of variables than thefull parameter list instead of just one variable as illustratedhere Another tool that could be used in conjunction withthe reduced variable method is to employ the various searchalgorithms that has been actively developed to the reductionscheme developed here [15ndash19] As far as we are awarethese optimization methods for multivariable problems allseem to mainly focus on various stochastic or deterministicmethods using discrete algorithms in some type of searchsequence of the domain space of the multivariable domainspace P 119896 119905 as detailed below in more recent publicationsIn other less ambitious works the problem is narrowed tothe specific nature of the system where the object functionis specified and which is amenable to precise treatment forexample [20 21] with a well-defined domain space withoutvariable reduction In another context quite different fromthe current development variable reduction has been appliedto DEA problems [22] Other examples of multiparametercomplex systems include those for multiple-step elementaryreactions each with its own rate constant that gives rise tophotochemical spectra signals that must be resolved unam-biguously [23] but these belong to the class of functions withconditional linear parameters The work here on the otherhand predominantly focuses on accurately determining therange of the function to be optimized by reducing the spaceof the domain Hence this method can be successfully com-binedwith the usual domain searching techniquesmentionedabove to effectively locate stationary points by a two-prongedapproach All these complex and coupled processes in phys-ical theories are related by postulated laws 119884law(P 119896 119905) thatfeature parameters (P 119896) Other examples include quantumchemical calculations with many topological and orientationvariables that need to be optimized with respect to theenergy but in relation to one or a few variables such as themolecular trajectory parameter during a chemical reactionwhere this variable is of primary significance in decidingon the ldquoreasonablenessrdquo of the analysis [9 Section 623page 294] Methods (i)a and (i)b below refer to LS data-fitting algorithms Method (i)a is an approximate methodwhere it is proved under certain conditions it could be amore accurate determination of parameters compared to astandard LS fit using (1) Method (i)b develops a techniquewhere the optimum value for 119876MD with domain values (P 119896)coincides with that of the standard LS method where the(P 119896) variables are varied independently Also discussed arethe relative accuracy of both methods (i)a in Section 22



120597119876OPT (119896)

120597119896= 0 997888rarr

120597119876OPT (P 119896)120597P

= 0120597119876OPT (P 119896)

120597119896= 0

(3)





119884law = 119884law (P 119896 119905) (4)




119894denotes


119894via the set








Then (119873119901+ 1) le (119873 minus 119873




119888where

119873minus119873119904119862119873119901



119894 In general119873



119884law (119905 119896) = 119891 (P 119905 119896) (5)





0


119892(119894)(P 119896

0)120597119875119895 =


119873119888


0) = P0 and where



0is the inde-



119884th (119905 119896) = 119891 (P 119905 119896) (6)


1 1198942 119894

119873119901

)


(1198941 1198942 119894

119873119901

) where 119894119895equiv (119884exp(119905119894

119895

) 119905119894119895


119884exp (1199051198941

) = 119891 (P 1199051198941

119896)

119884exp (1199051198942

) = 119891 (P 1199051198942

119896)

119884exp (119905119894119873119901

) = 119891 (P 119905119894119873119901

119896)

(7)


119901) where 119875

119894(119896 120572) isin C1 Hence any




119875119894(119896 1) 119875

119894(119896 2) 119875




119875119894 (119896) =

1

119873119888

119873119888

sum119895=1

119875119894(119896 119895) (8)




119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(9)


119877 (119896) = 119888 sdot

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119884th (119896 119905119894)) sdot 1198841015840

th (119896 119905119894) (10)




119876119879(119875119879) =

1198731015840

sum119894=1

(119884exp(119905119894) minus 119891(119875119879 119905119894))2

(11)


119879whenever 120597119876

119879120597119875119879=


120575119876 (119896) = 120575119876 (P 119896) = (120597119876120597P

sdot 120575P + 120597119876120597119896) (12)

subjected to 119892119894(P 119896) = 119875

119894minus ℎ119894(119896) = 0 (ie 119875

119894= ℎ119894(119896)





0isin 1198830



119863119903119891 (1199090) +

119898

sum119896=1

120582119896119863119903119892119896(1199090) = 0 (119903 = 1 2 119899) (13)


119898nabla119892119898(1199090) = 0 (14)




119892119894= 119875119894minus ℎ119894 (119896) = 0 (119894 = 1 2 119873

119901) (15)

where 119863119895119892119894(1199090) = 120575

119894119895since 119863

119895= 120597120597119875

119895and therefore


119891(P 119905119894 119896) and 119891(119894) = 119891(P 119905

119894 119896) Define

119891119876 (119909) equiv 119876 (P 119896 119905) =

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119891 (119894))2

(16)



between 119876(119896) and 119876119879





sum119873119901


condition120597L

120597119875119895

= 0 119895 = 1 2 119873119901

120597L

120597119896= 0 (17)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119875119895

= 1205821015840

119895 119895 = 1 2 119873

119901 (18)

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896+

119873119901

sum119895=1

1205821015840

119895

120597119901119895

120597119896= 0 (19)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896

+

119873119901

sum119895=1

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) sdot120597119891 (119894)

120597119875119895

sdot120597119901119895

120597119896= 0

(20)

Since 119889119875119894119889119896 = 120597119901

119894120597119896 then (20) implies 119889119876(P 119896 119905)119889119896 = 0





1198761= 119876 (P 119905 119896) (21)

1198762= 119876 (P 119905 119896) (22)






1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (P 119905119894 119896))120597119891 (P 119905

119894 119896)

120597119875119895

= 0

119895 = 1 2 119873119901

(23)


equations

1205971198762

120597119875119895

= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (24)

1205971198762

120597119896= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (25)


1198891198761 (119896)

119889119896= 119888 sdot

119873119901

sum119895=1

(

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

)120597119875119895

120597119896+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896

(26)


1(119896)119889119896 = 0 since 119891(119894) and 119891(119894)


1(119896)119889119896 = 0 then setting


1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (27)

997904rArr

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (28)






QT(P k)

QT(Pi P0Pi k)

Q = s1

Q = s2

Q = s3

Pi

(P(k) k)

QT(P0 k0) = Qmin

(P0 k0)

Pi0 + ΔPiPi0 + ΔP

998400i

Q(k)



119879le 119876



nabla119876 = 0 (120597119876

120597119896=120597119876

120597119875119894

= 0 for 119894 = 1 2 119873119901) (29)


sum1198731015840

119894=1(119884exp(119894) minus 119891(P(119896) 119905119894 119896))

2 then 119876119888(119896) = 119876 ∘ 119875

119879where

119875119879

= (1198751(119896) 1198752(119896) 119875

119873119901

(119896) 119896)119879 Because 1198761015840

119888(119896) =


119873119901

(119896) 119896)119879 solutions occur


119894(119896) = 0




2for some







119888 where 119876

119888is a function








119894119873119888in number

defined here as

119876119888 (P 119896) = sum


(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)




119894






119894


0we assume that the



119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)



0if relative



119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))


0for what

follows below


0 1199032)


min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)


min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)




119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)


119894




119876119888(P119879) ge 119876

119888(P) (36)


119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879




120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575



119894(120575 gt 120575

119894)ThenP isin





119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)


119888(P 119896)








119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











120597119876OPT (119896)

120597119896= 0 997888rarr

120597119876OPT (P 119896)120597P

= 0120597119876OPT (P 119896)

120597119896= 0

(3)





119884law = 119884law (P 119896 119905) (4)




119894denotes


119894via the set








Then (119873119901+ 1) le (119873 minus 119873




119888where

119873minus119873119904119862119873119901



119894 In general119873



119884law (119905 119896) = 119891 (P 119905 119896) (5)





0


119892(119894)(P 119896

0)120597119875119895 =


119873119888


0) = P0 and where



0is the inde-



119884th (119905 119896) = 119891 (P 119905 119896) (6)


1 1198942 119894

119873119901

)


(1198941 1198942 119894

119873119901

) where 119894119895equiv (119884exp(119905119894

119895

) 119905119894119895


119884exp (1199051198941

) = 119891 (P 1199051198941

119896)

119884exp (1199051198942

) = 119891 (P 1199051198942

119896)

119884exp (119905119894119873119901

) = 119891 (P 119905119894119873119901

119896)

(7)


119901) where 119875

119894(119896 120572) isin C1 Hence any




119875119894(119896 1) 119875

119894(119896 2) 119875




119875119894 (119896) =

1

119873119888

119873119888

sum119895=1

119875119894(119896 119895) (8)




119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(9)


119877 (119896) = 119888 sdot

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119884th (119896 119905119894)) sdot 1198841015840

th (119896 119905119894) (10)




119876119879(119875119879) =

1198731015840

sum119894=1

(119884exp(119905119894) minus 119891(119875119879 119905119894))2

(11)


119879whenever 120597119876

119879120597119875119879=


120575119876 (119896) = 120575119876 (P 119896) = (120597119876120597P

sdot 120575P + 120597119876120597119896) (12)

subjected to 119892119894(P 119896) = 119875

119894minus ℎ119894(119896) = 0 (ie 119875

119894= ℎ119894(119896)





0isin 1198830



119863119903119891 (1199090) +

119898

sum119896=1

120582119896119863119903119892119896(1199090) = 0 (119903 = 1 2 119899) (13)


119898nabla119892119898(1199090) = 0 (14)




119892119894= 119875119894minus ℎ119894 (119896) = 0 (119894 = 1 2 119873

119901) (15)

where 119863119895119892119894(1199090) = 120575

119894119895since 119863

119895= 120597120597119875

119895and therefore


119891(P 119905119894 119896) and 119891(119894) = 119891(P 119905

119894 119896) Define

119891119876 (119909) equiv 119876 (P 119896 119905) =

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119891 (119894))2

(16)



between 119876(119896) and 119876119879





sum119873119901


condition120597L

120597119875119895

= 0 119895 = 1 2 119873119901

120597L

120597119896= 0 (17)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119875119895

= 1205821015840

119895 119895 = 1 2 119873

119901 (18)

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896+

119873119901

sum119895=1

1205821015840

119895

120597119901119895

120597119896= 0 (19)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896

+

119873119901

sum119895=1

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) sdot120597119891 (119894)

120597119875119895

sdot120597119901119895

120597119896= 0

(20)

Since 119889119875119894119889119896 = 120597119901

119894120597119896 then (20) implies 119889119876(P 119896 119905)119889119896 = 0





1198761= 119876 (P 119905 119896) (21)

1198762= 119876 (P 119905 119896) (22)






1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (P 119905119894 119896))120597119891 (P 119905

119894 119896)

120597119875119895

= 0

119895 = 1 2 119873119901

(23)


equations

1205971198762

120597119875119895

= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (24)

1205971198762

120597119896= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (25)


1198891198761 (119896)

119889119896= 119888 sdot

119873119901

sum119895=1

(

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

)120597119875119895

120597119896+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896

(26)


1(119896)119889119896 = 0 since 119891(119894) and 119891(119894)


1(119896)119889119896 = 0 then setting


1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (27)

997904rArr

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (28)






QT(P k)

QT(Pi P0Pi k)

Q = s1

Q = s2

Q = s3

Pi

(P(k) k)

QT(P0 k0) = Qmin

(P0 k0)

Pi0 + ΔPiPi0 + ΔP

998400i

Q(k)



119879le 119876



nabla119876 = 0 (120597119876

120597119896=120597119876

120597119875119894

= 0 for 119894 = 1 2 119873119901) (29)


sum1198731015840

119894=1(119884exp(119894) minus 119891(P(119896) 119905119894 119896))

2 then 119876119888(119896) = 119876 ∘ 119875

119879where

119875119879

= (1198751(119896) 1198752(119896) 119875

119873119901

(119896) 119896)119879 Because 1198761015840

119888(119896) =


119873119901

(119896) 119896)119879 solutions occur


119894(119896) = 0




2for some







119888 where 119876

119888is a function








119894119873119888in number

defined here as

119876119888 (P 119896) = sum


(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)




119894






119894


0we assume that the



119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)



0if relative



119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))


0for what

follows below


0 1199032)


min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)


min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)




119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)


119894




119876119888(P119879) ge 119876

119888(P) (36)


119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879




120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575



119894(120575 gt 120575

119894)ThenP isin





119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)


119888(P 119896)








119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119875119894 (119896) =

1

119873119888

119873119888

sum119895=1

119875119894(119896 119895) (8)




119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(9)


119877 (119896) = 119888 sdot

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119884th (119896 119905119894)) sdot 1198841015840

th (119896 119905119894) (10)




119876119879(119875119879) =

1198731015840

sum119894=1

(119884exp(119905119894) minus 119891(119875119879 119905119894))2

(11)


119879whenever 120597119876

119879120597119875119879=


120575119876 (119896) = 120575119876 (P 119896) = (120597119876120597P

sdot 120575P + 120597119876120597119896) (12)

subjected to 119892119894(P 119896) = 119875

119894minus ℎ119894(119896) = 0 (ie 119875

119894= ℎ119894(119896)





0isin 1198830



119863119903119891 (1199090) +

119898

sum119896=1

120582119896119863119903119892119896(1199090) = 0 (119903 = 1 2 119899) (13)


119898nabla119892119898(1199090) = 0 (14)




119892119894= 119875119894minus ℎ119894 (119896) = 0 (119894 = 1 2 119873

119901) (15)

where 119863119895119892119894(1199090) = 120575

119894119895since 119863

119895= 120597120597119875

119895and therefore


119891(P 119905119894 119896) and 119891(119894) = 119891(P 119905

119894 119896) Define

119891119876 (119909) equiv 119876 (P 119896 119905) =

1198731015840

sum

119894=11015840

(119884exp (119905119894) minus 119891 (119894))2

(16)



between 119876(119896) and 119876119879





sum119873119901


condition120597L

120597119875119895

= 0 119895 = 1 2 119873119901

120597L

120597119896= 0 (17)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119875119895

= 1205821015840

119895 119895 = 1 2 119873

119901 (18)

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896+

119873119901

sum119895=1

1205821015840

119895

120597119901119895

120597119896= 0 (19)


1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894))120597119891 (119894)

120597119896

+

119873119901

sum119895=1

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) sdot120597119891 (119894)

120597119875119895

sdot120597119901119895

120597119896= 0

(20)

Since 119889119875119894119889119896 = 120597119901

119894120597119896 then (20) implies 119889119876(P 119896 119905)119889119896 = 0





1198761= 119876 (P 119905 119896) (21)

1198762= 119876 (P 119905 119896) (22)






1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (P 119905119894 119896))120597119891 (P 119905

119894 119896)

120597119875119895

= 0

119895 = 1 2 119873119901

(23)


equations

1205971198762

120597119875119895

= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (24)

1205971198762

120597119896= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (25)


1198891198761 (119896)

119889119896= 119888 sdot

119873119901

sum119895=1

(

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

)120597119875119895

120597119896+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896

(26)


1(119896)119889119896 = 0 since 119891(119894) and 119891(119894)


1(119896)119889119896 = 0 then setting


1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (27)

997904rArr

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (28)






QT(P k)

QT(Pi P0Pi k)

Q = s1

Q = s2

Q = s3

Pi

(P(k) k)

QT(P0 k0) = Qmin

(P0 k0)

Pi0 + ΔPiPi0 + ΔP

998400i

Q(k)



119879le 119876



nabla119876 = 0 (120597119876

120597119896=120597119876

120597119875119894

= 0 for 119894 = 1 2 119873119901) (29)


sum1198731015840

119894=1(119884exp(119894) minus 119891(P(119896) 119905119894 119896))

2 then 119876119888(119896) = 119876 ∘ 119875

119879where

119875119879

= (1198751(119896) 1198752(119896) 119875

119873119901

(119896) 119896)119879 Because 1198761015840

119888(119896) =


119873119901

(119896) 119896)119879 solutions occur


119894(119896) = 0




2for some







119888 where 119876

119888is a function








119894119873119888in number

defined here as

119876119888 (P 119896) = sum


(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)




119894






119894


0we assume that the



119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)



0if relative



119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))


0for what

follows below


0 1199032)


min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)


min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)




119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)


119894




119876119888(P119879) ge 119876

119888(P) (36)


119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879




120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575



119894(120575 gt 120575

119894)ThenP isin





119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)


119888(P 119896)








119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (P 119905119894 119896))120597119891 (P 119905

119894 119896)

120597119875119895

= 0

119895 = 1 2 119873119901

(23)


equations

1205971198762

120597119875119895

= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (24)

1205971198762

120597119896= 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (25)


1198891198761 (119896)

119889119896= 119888 sdot

119873119901

sum119895=1

(

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

)120597119875119895

120597119896+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896

(26)


1(119896)119889119896 = 0 since 119891(119894) and 119891(119894)


1(119896)119889119896 = 0 then setting


1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

= 0 119895 = 1 2 119873119901 (27)

997904rArr

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (28)






QT(P k)

QT(Pi P0Pi k)

Q = s1

Q = s2

Q = s3

Pi

(P(k) k)

QT(P0 k0) = Qmin

(P0 k0)

Pi0 + ΔPiPi0 + ΔP

998400i

Q(k)



119879le 119876



nabla119876 = 0 (120597119876

120597119896=120597119876

120597119875119894

= 0 for 119894 = 1 2 119873119901) (29)


sum1198731015840

119894=1(119884exp(119894) minus 119891(P(119896) 119905119894 119896))

2 then 119876119888(119896) = 119876 ∘ 119875

119879where

119875119879

= (1198751(119896) 1198752(119896) 119875

119873119901

(119896) 119896)119879 Because 1198761015840

119888(119896) =


119873119901

(119896) 119896)119879 solutions occur


119894(119896) = 0




2for some







119888 where 119876

119888is a function








119894119873119888in number

defined here as

119876119888 (P 119896) = sum


(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)




119894






119894


0we assume that the



119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)



0if relative



119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))


0for what

follows below


0 1199032)


min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)


min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)




119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)


119894




119876119888(P119879) ge 119876

119888(P) (36)


119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879




120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575



119894(120575 gt 120575

119894)ThenP isin





119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)


119888(P 119896)








119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119888 where 119876

119888is a function








119894119873119888in number

defined here as

119876119888 (P 119896) = sum


(119884exp minus 119891(119894))2

(30)

could be such that

119876119888(P119879 119896) ge 119876

119888(P 119896) (31)




119894






119894


0we assume that the



119876119888119894 (P) = sum

119895isin119894

(119884exp(119895) minus 119891(119895))2

(32)



0if relative



119861(1199030 1199031) and 119861(119903

0 1199032)) implies both max 119891(120597119861(119903

0 1199032))(gt) ge

max119891(120597119861(1199030 1199031)) and min119891(120597119861(119903

0 1199032))(gt) ge

min119891(120597119861(1199030 1199031))


0for what

follows below


0 1199032)


min119891 (120597119861 (1199030 1199031)) lt 119891 (r) lt max119891 (120597119861 (119903

0 1199032)) (33)


min119891 (120597119861 (1199030 119903)) lt min119891 (120597119861 (119903

0 1199031)) (119903 gt 119903

1) (34)




119876119888 (P) =

119873119888

sum119894=1

119876119888119894 (P) (35)


119894




119876119888(P119879) ge 119876

119888(P) (36)


119894minus P119895 = 120575 for all 119894 119895 = 1 2 119873

119888and 120575P

119894= P minus P

119894

Lemma 5 120575P119894 le 120575

Proof

10038171003817100381710038171003817P minus P119894

10038171003817100381710038171003817=1

119873119888

1003817100381710038171003817100381710038171003817100381710038171003817

sum119902

P119902minus 119873119888P119894

1003817100381710038171003817100381710038171003817100381710038171003817

le1

119873119888

119873119888

sum119895=1

10038171003817100381710038171003817P119895minus P119894

10038171003817100381710038171003817le 120575

(37)

Lemma 6 119876119888(P119879 119896) gt 119876

119888(P 119896) for 120575 lt P

119879minus P119894 lt 120575

119879for

some 120575119879




120598max119894 gt 119876119888119894 (P) gt 120598min119894 (38)

where 119891 = 119876119888119894

in (32) Choose 120575119894so that 120575

119894lt 120575P

119894 lt 120575



119894(120575 gt 120575

119894)ThenP isin





119873119888

sum119894=1

120598min119894 lt 119876119888 (P) =119873119888

sum119894=1

119876119888119894(P) lt

119873119888

sum119894=1

120598max119894 (39)

120598max119894 lt 119876119888119894 (P119879) lt 120598119879119894 (40)


119888(P 119896)








119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















119876119888(P119879) le 119876

119888(P) (41)






119888variety





119879120597119875119895=





119876119879= 119876 (P 119896) =

1198731015840

sum119894=1

(119884exp(119905119894)minus 119891 (119894))

2

(42)

Define 120597119891(119894)120597119875119895= 119891(119894 119895) 120572



120597119876119879

120597119875119895

= ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(43)

120597119876119879

120597119896= 119868 (P 119896) = 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896= 0 (44)




119894(P 119896)120597119875

119895 = det 120597119876

119879120597119875119895120597119875119894 = 0 on an open


119894(P 119896)120597119875



120597ℎ119894 (P 119896)120597119875119895

= 119888 sdot

1198731015840

sum119897

(120572119896

1205972119891 (119897)

120597119875119895120597119875119894

minus120597119891 (119897)

120597119875119895

sdot120597119891 (119897)

120597119875119894

) (45)


119879yields


1198761 (119896) =

1198731015840

sum119894

(120572119894)2 (46)

1198761015840

1(119896) = 119888 sdot

119873119901

sum119895=1

119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

+ 119888 sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119896 (47)


119879120597119875119895=

0 (119895 = 1 2 119873119901) and (43) implies sum119873

1015840

119894=1120572119894(120597119891(119894)120597119875

119895) = 0


(119889119875119895

119889119896sdot

1198731015840

sum119894=1

120572119894

120597119891 (119894)

120597119875119895

) = 0 (48)



sum1198731015840







119879= P0 1198960 is


119879 Then




1function




more accurate fit









119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119897min119894 le 119875119894 le 119897max119894 119894 = 1 2 119873119901

119897min119896 le 119896 le 119897max119896(49)


120597119876119864

120597119875119895

= 119900119895 (P 119896) 119895 = 1 2 119873

119901 (50)

120597119876119864

120597119896= 120581 (P 119896) (51)


119900119895 (P 119896) = 0

120581 (P 119896) = 0(52)




119864is C2 onD and det 120597119900

119894120597119875119895 = 0 where

120597119900119894120597119875119895equiv 1205972119876



0 119900119895(P(1198960) 1198960) = 0

119900119895isin C1 on T

0with (T


0isin T0such




1198761198641(119896) = 119876

119864(P(119896) 119896) and 1198761015840

1198641(119896) = 119889119876

1198641119889119896 so that

1198891198761198641

119889119896=

119873119901

sum119895=1

(120597119876119864

120597119875119895

sdot119889119875119895

119889119896+120597119876119864

120597119896) =

119873119901

sum119895=1

(119900119895sdot119889119875119895

119889119896+120597119876119864

120597119896)

(53)


1198641(119896) = 0 where 119896

119894isin T0in the



119896) where

1198761198641(119896) = 0 for 119896 = 119896


119894 119896min le




119894isin R1 space



R119873119901+1 space




119864120597119896 = 0 from (53)


119895= 0 (119895 = 1 2 119873

119901) and 120597119876

119864120597119896 = 0



1198641(119896119894) = 0 refer to





119894) partly






119860= 119896119861 For such


119864(P 119896) within the





119894in the graph




1198641 it


119864function



1198641(1198960) gt 0 for |119896minus119896

0| lt 120575 and for |119875(119896)minus119875(119896

0)| lt








119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119875119895 119895 = 1 2 119873






119888


P120572(119896)

(3) Determine 119876(119896) (9) as

119876 (119896) =

1198731015840

sum

119894=11015840

(119884exp(119905119894) minus 119884th(119896 119905119894))2

(54)



problem


(1) Solve

ℎ119895 (P 119896) = 119888 sdot

1198731015840

sum119894=1

(119884exp (119905119894) minus 119891 (119894)) 119891 (119894 119895) = 0

119895 = 1 2 119873119901

(55)





1198731015840

119894(120572119894)2 (46) where120572

119894=

119884exp(119905119894) minus 119891(P 119905119894 119896)

(3) Solve 11987610158401(119896) = 0 for some 119896

0 that is 1198761015840

1(1198960) = 0



= 119876(P 119896) =

sum1198731015840

119894=1(119884exp(119905

119894)minus 119891(P 119905

119894 119896))2




119895(P 119896) = 120597119876

119864120597119875119895

119895 = 1 2 119873119901[22] 120581(P 119896) = 120597119876

119864120597119896 as in (51)





119864(P 119896)


119864119896(119896) = 0



















119894of (5) refers to 120582


120582infin

with 119873119901= 1 and 119896 equiv 119896


120582infin



120582infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


119894)

(59)








1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















1(119896) equiv 119876 of (9)


1198761 (119896) =

119873

sum119894=1

(120582exp (119905119894) minus 120582th (119905119894 119896))2

(61)


infinand 119884



infinand so


119873

119894=1(120582exp(119905119894) minus 120582th(119905119894))







infinequiv

120582infin





119896function (10)



Pu (VI) + 2Fe (II)119896119887

997888rarr Pu (IV) + 2Fe (III) (62)


0is



PuO2+2+ Fe2+aq

119896119887

997888rarr PuO+2+ Fe3+aq (63)

0 100 200 30003

04

05

06

07

08

09

Time (s)

A

(t) (

first

orde

r)





0119905)

(64)


where [B]0gt [A]

0and [B] = [Pu(VI)] [A] = [Fe(II)]




ln1 +Δ0(1198840minus 119884infin)

[A]0 (119884119905 minus 119884infin) = ln [

B]0[A]0

+ 119896Δ0119905 (65)


infinwas not



infin We overcome





119884infin (119896) =

1

1198731015840

1198731015840

sum

119894=11015840


0119905119894) minus 1)

(66)


001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










001

4

001

6

001

8

002

002

2

002

4

002

6

002

8

003

minus05

minus04

minus03

minus02

minus01

0

01

02

03

04

05

R(k) fi

rst o

rder

First order

0 500

1000

1500

2000

2500

minus1

0

1

2

3

4

5

R(k) s

econ

d or

der

Second order

First order ka

Second order kb

times10minus6





1(119896) 119896119887= 119896 is the varying 119896



1 minus 120572 exp (minus119896Δ0119905)

(67)




1198762 (119896) =

119873

sum119894=1

(119884exp (119905119894) minus 119884th (119905119894 119896))2

(68)



119887= 9380 plusmn 18 (Ms)minus1



infin= 0025 plusmn

0003

0 10 20 30 40005

01

015

02

025

03

Time (s)

A

(t) (

seco

nd o

rder

)




6 Conclusions




List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











List of Variables








119894(119896) based on some






1015840(119896) (10)





reaction 119902 (eg (56))120582(119905) 120582





Acknowledgments


References


















1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1-optimizationrdquo


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Numerical Reduced Variable Optimization ...

Documents

Transcript of Research Article Numerical Reduced Variable Optimization ...