Abstract: In the framework of this work, we used the isogeometric method to solve the least squares problem in

one dimension, on a curve of Rd, d = 1,2, including a semicircle. For this purpose, we presented the isogeometric

method and the tools necessary for the description of this method, namely, the b splines basis, the parameterization

of the Rd, d = 1,2curve. We formulated the least squares problem which is a minimization problem. This

problem was solved by using the Discontinuous Galerkin (DG) and the b splines basis as the approximation basis.

The numerical method was validated by evaluating the error. For this purpose, an inverse inequality was therefore

used.

Key-Words: -Isogeometric method, Least squares problem, B-spline basis, Parameterization, Discontinuous

Galerkin, Inverse inequality.

1 Introduction

The approximation of a function in the sense of least

squares is a tool mathematics important in computer-

aided design (CAD) because the engineer uses it to

approximate a function. The least squares problem is

used in biology for problems of mathematical model-

ing, [1].

In the framework of this work, we use isogeometric

method to solve the least squares problem.

Isogeometric method has been introduced in 2005,

[2], [3], [4]. The objectives of Isogeometric Analy-

sis are to generalize and improve upon Finite Element

Analysis (FEA) in the following ways :

1. To provide more accurate modeling of com-

plex geometries and to exactly represent com-

mon engineering shapes such as circles, cylin-

ders, spheres, ellipsoids, etc.

2. To fix exact geometries at the coarsest level of

discretization and eliminate geometrical errors.

3. To vastly simplify mesh refinement of complex

industrial geometries by eliminating the neces-

sity to communicate with the CAD description

of geometry.

4. To provide refinement procedures, including

classical h- and p- refinements analogues, and

to develop a new refinement procedure called

k-refinement, [5].

The idea of isogeometric method is to build a geo-

metry model and, rather than develop a finite ele-

ment model approximating the geometry, directly use

Isogeometric Method for Least Squares Problem

HAUDIÉ JEAN STÉPHANE INKPÉ1, AGUEMON URIEL2, GOUDJO AURÉLIEN3

1Digital Research and Expertise Unit

Université Virtuelle de Côte d'Ivoire (UVCI)

28 BP 536 Abidjan

CÔTE D'IVOIRE

2Université de KINDIA (UK)

Faculté des Sciences

Département de Mathématiques

GUINÉE

3Université d'Abomey-Calavi (UAC)

Faculté des Sciences et Techniques (FAST)

Département de Mathématiques

BENIN

Received: April 27, 2024. Revised: September 19, 2024. Accepted: October 11, 2024. Published: November 6, 2024.

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the functions describing the geometry in analysis, [6].

These functions are b-splines.

Isogeometric Analysis is approached, using con-

tinuous or discontinuous Galerkin method. In the

framework of the isogeometric method, we use a

parametrization to describe the domain on which, we

solve our problem. This parametrization is used for

numerical integration, [1],[7].

The least squares problem is addressed in one

dimesnsion on a segment in, [1]. The particularity of

our work is to solve our problem in one dimension on

a curve of R2.

2 The least squares problem

formulation

The isogeometric method uses b-splines as basis func-

tions to construct the numerical approximations.

Definition 1. Let x1≤x2≤ ··· ≤ xmbe an increas-

ing sequence of reals, b-splines functions of degree k

are defined by Cox-de Boor-Mansfield recursion for-

mula, [8] :











For 1≤i≤m−1

Ni,0(t) = 1 if t ∈[xi, xi+1[

Ni,0(t) = 0 otherwise

(1)











For k≥1and 1≤i≤m−k−1

Ni,k(t) = t−xi

xi+k−xi

Ni,k−1(t)+

xi+k+1 −t

xi+k+1 −xi+1

Ni+1,k−1(t),

(2)

with the convention x

0:= 0 for all real number x

The set (xi)m

i=1 (1 ≤i≤m)is called knots

vector.

Definition 2. Let (xi)m

i=1 (1 ≤i≤m)be a knots

vector. Let nbe a non-zero natural integer et let

(Pi)n

i=0 be a sequence of points of IRd(d= 1,2).

We call b-spline curve of degree k(d'ordre k+ 1) and

of control points Pi,i= 0, . . . , n, the function Pde-

fined from the interval [x1, xm]into IRdby :

P(t) =

i=0

PiNi,k(t), x1≤t≤xm,0≤k≤n

The set of points (Pi)n

i=0 are the vertices of a polygon

called control polygon of the curve P.

Univariate B-spline basis functions are piecewise

polynomial. They form a partition of unity, have local

support, and are non-negative, [9].

Theorem 1 (existence and uniqueness of solution).

Let Cbe a non-empty closed convex set of a Hilbert

space H. Then for all f∈H, there exists a unique

u∈Csuch that :

kf−uk=min

v∈Ckf−vk(3)

We note u=PC(f), the projection of fon C. u is

therefore the solution of the minimization problem.

The parametrization Fof the physical domain is de-

fined by :

F:b

Ω⊂IR →Ω⊂IRd, d = 1,2

ε7−→ x=F(ε) =

i=0

Cib

Ni,k(ε)

where b

Ωis the parametric domain, the (Ci)n

i=0 are

control points and n+ 1 is the number of basis func-

tions.

We use the standard notation of Sobolev spaces

Hl(Ω).The norm and semi-norm will be denoted res-

pectively by k.kHl(Ω) and |.|Hl(Ω).

When l= 0, Hl(Ω) = L2(Ω).

We use also the inner product on L2(Ω) denoted by

(., .)L2(Ω).

Let us consider the function tF

7−→ F(t)g

7−→ g(F(t)),

Where Fis the parametrization of the domain

Ω⊂Rd, d = 1,2.

X= (Xi)m

i=1 is an open knots vector, with mthe

total number of knots.

Xd= (χi)Nd

i=1 is the set of distinct knots of X.

Let be Ωi={x∈IRd;x=F([χi, χi+1]),},

1≤i≤Nd−1

Thus ∃!q(i)∈ {1, . . . , m}such as χi=Xq(i)where

q(i) =

k=1

nk,

And niis the multiplicity of χiin X.

Let be Ω =

Nd−1

[

i=1

Ωi,with Ωi∩Ωs=∅,∀i6=s.

(Nj,p)j∈Iiis a b-spline basis of degree p,

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Where Ii={k∈ {0, . . . , n};SuppNk,p ∩Ωi6=∅},

SuppNj,p =F([χj;χj+p+1]).

SuppNj,p ∩F([χi;χi+1]) 6=∅ ⇐⇒ j∈

{q(i)−p, . . . , q(i)}.

∀i∈ {1, . . . , Nd−1}, Ni

j,p(x) = (Nj,p(x)if x∈F([χi;χi+1])

0otherwise

We want to approximate a function gby a b-spline

curve ˜gon a domain Ω⊂Rd, d = 1,2in the least

squares sense

Where ˜g(F(t)) =

Nd−1

i=1 X

j∈Ii

jNi

j,p(F(t)),[7].

We are therefore looking for the coefficients (Pi

j),

knowing that :

ZΩi

g(F(t))Ni

k,p(F(t))dF =

ZΩi

Nd−1

i=1 X

j∈Ii

jNi

j,p(F(t))Ni

k,p(F(t))dF

Solving this approximation problem amounts to min-

imizing kg−˜gkL2(Ω).

Before proposing an isogeometric formulation of our

problem, we give a definition and a theorem.

Definition 3. Let πhbe a map which is such that :

∀v∈L2(Ω), πhv∈L2(Ω) with

(πhv, yh)L2(Ω) = (v, yh)L2(Ω),

∀yh∈L2(Ω).(4)

Theorem 2 (isogeometric inverse inequality).Given

the integers land ssuch that 0≤l≤s≤p+ 1 and

a function u∈Hs(Ω),then:

K∈τh

|u−πhu|2

Hl(K)≤Ch2(s−l)kuk2

Hs(Ω),[7],[10]

(5)

where C is independent of h.

Minimizing kg−˜gkL2(Ω) is equivalent to writing

the following equalities :

ZΩi

g(F(t))Ni

k,p(F(t))dF =

ZΩiX

j∈Ii

jNi

j,p(F(t))Ni

k,p(F(t))dF

ZΩiX

j∈Ii

jNi

j,p(F(t))Ni

k,p(F(t))dF =

ZΩi

g(F(t))Ni

k,p(F(t))dF

j∈Ii

jZΩi

j,p(F(t))Ni

k,p(F(t))dF =

ZΩi

g(F(t))Ni

k,p(F(t))dF

j∈Ii

k,j Pi

j=Si

k,∀k∈Ii.(6)

MiPi=Si,∀i∈ {1, . . . , Nd−1},[7].(7)

Where

Mi= (Mi

kj )k,j∈Ii

1≤i≤Nd−1

Pi= (Pi

j)j∈Ii

1≤i≤Nd−1

Si= (Si

k)k∈Ii

1≤i≤Nd−1

With Mi

k,j =ZΩi

j,p(F(t))Ni

k,p(F(t))dF and

k=ZΩi

g(F(t))Ni

k,p(F(t))dF.

To calculate the Pi,we will be interested in studying

the properties of the mass matrix Mi.

Property 1. Miis a square matrix of order (p+ 1)

which is symmetric, [7].

Proof. Knowing that Mi= (Mi

kj )k,j∈Ii

1≤i≤Nd−1

,and that

card(Ii) = p+ 1, Miis a square matrix of order

(p+1).Moreover, Miis a symmetric matrix because

kj =Mi

jk,∀1≤i≤Nd−1,∀k∈Iiand ∀j∈

Ii.

Property 2. Miis an invertible matrix, [7].

We will show that Miis a positive definite matrix.

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Proof. Let Li= (Li

j)j∈Iibe a column vector.

MiLi= (λi

k)k∈Iiwith λi

k=X

j∈Ii

kj Li

So we get :

(Li)tMiLi=X

k∈Ii

λi

kLi

k∈Ii



X

j∈Ii

kj Li

j

Li

k∈IiX

j∈IiZΩiNi

k,p(x)Ni

j,p(x)dxLi

jLi

=ZΩi

X

j∈Ii

j,p(x)Li

j



=



X

j∈Ii

j,p(x)Li

j



L2(Ωi)

So (Li)tMiLi≥0.

(Li)tMiLi= 0 =⇒X

j∈Ii

j,p(x)Li

j= 0 (8)

=⇒Li

j= 0 because the

j,p form a basis of P(p+1)(Nd−1).

=⇒Li

j= 0,∀j∈Ii

Miis therefore positive definite, hence Miis invert-

ible.

Property 3. invM ithe inverse of Mi,is a square

matrix of order (p+ 1) which is symmetric, [7].

Proof. Mibeing a square matrix of order (p+ 1),its

inverse is also a square matrix of order (p+ 1). Mi

being symmetric, we have : (Mi)t=Mi.

(Mi)t=Mi=⇒inv(Mi)t=invMi

=⇒(invMi)t=invMibecause

Miis an invertible square matrix

So, invMiis a symmetric matrix.

The properties of the mass matrix having been enu-

merated, we obtain from the relation 7 that Pi=

(invMi)Si.

Computing an integral over Ωiamounts to com-

puting this integral over b

Ωiby means of the parame-

terization F, [11]. This integral over the parametric

domain is then brought back to the interval [−1; 1],

using a transformation. So, we get :

k,j =ZΩi

j,p(F(t))Ni

k,p(F(t))dF

=Zc

Ωi

j,p(F(ε))Ni

k,p(F(ε))Jac(F(ε))dε

=Zc

Ωib

j,p(ε)b

k,p(ε)d

Jac(ε)dε

because Ni

j,p ◦F=b

j,p and Jac ◦F=d

Jac

k,j =

pos2(i)

r=pos1(i)

j,p(εi

r)b

k,p(εi

r)d

Jac(εi

With li

r=χi+1−χi

2ωi

rthen pos1(i) = (i−1)Npgs+1,

pos2(i) = iNpgs and 1≤i≤Nd−1.

Npgs is the number of Gaussian points per seg-

ment, the ωi

rand the εi

rare respectively the Gaussian

weights and knots.

k=ZΩi

g(F(t))Ni

k,p(F(t))dF (9)

pos2(i)

r=pos1(i)

rg(F(εi

r)) b

k,p(εi

r)d

Jac(εi

r),[7]

3 Numerical solution

In the previous section, we approximated a function

gby a b-spline curve of degree pin the sense of

least squares on a domain Ω.Subsequently, we want

to validate the approximation in the sense of least

squares by verifying the inverse inequality 5, for l= 0

and s= 1.

For l= 0 and s= 1,inverse inequality 5 becomes :

ku−πhukL2(Ω) ≤ChkukH1(Ω),∀u∈H1(Ω),(10)

Where Cis independent of h.

Therefore, we put in an array, the space step h,log(h)

and log kerrorkL2(Ω)

kukH1(Ω) . Then, we determine the

slope of the curve of the log kerrorkL2(Ω)

kukH1(Ω) as a

function of log(h). Then, we construct this curve in

each case.

Numerical tests are performed using Fortran and

Gnuplot.

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Experience 1. g(x) = ex, x ∈Ω1=] −1; 1[

p= 2,kukH1(Ω) =√e2−e−2.

The parametrization of Ω1is given by :

1. The knots vector :

X= [0000.1250.250.3750.50.6250.750.875

111]

2. The control points : A1(−1; −0.75),

A2(−0.5; −0.25),A3(0; 0),A4(0.25; 0.5)

and A5(0.75; 1).

Step h log(h)log kerreurkL2(Ω)

kukH1(Ω) Rate

h= 0.125 −2.079 −7.556 -

2−2.773 −9.270 2.47

4−3.466 −10.996 2.48

8−4.159 −12.727 2.49

Table 1.Results of experiment 1

Fig.1: line of experiment 1

Experience 2. g(x) = x3, x ∈Ω1=] −1; 1[

p= 2,kukH1(Ω) =q136

35 .

Step h log(h)log kerreurkL2(Ω)

kukH1(Ω) Rate

h= 0.125 −2.079 −6.518 -

2−2.773 −8.249 2.498

4−3.466 −9.982 2.498

8−4.159 −11.713 2.498

Table 2.Results of experiment 2

Fig.2: line of experiment 2

Experience 3. g(x, y) = x2+y2,(x, y)∈Ω2,

the half-circle with center (0,0) and radius 1.

p= 2,kukH1(Ω) =qπ+8

The parameterization of Ω2is given by :

-The knots vector :

X=[0 0 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1

1 1]

-The control points :

A1(1; 0), A2(1; 0.414), A3(0.707; 0.707), A4(0.414; 1),

A5(0; 1), A6(0; 1), A7(−0.414; 1), A8(−0.707; 0.707),

A9(−1; 0.414),A10(−1; 0),[7].

Step h log(h)log kerreurkL2(Ω)

kukH1(Ω) Rate

h= 0.5−0.693 −3.497 -

2−1.386 −5.020 2.19

4−2.079 −6.555 2.21

8−2.772 −8.386 2.64

Table 3.Results of experiment 3

Fig.3: line of experiment 3

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Fig.4: Half-circle with center (0; 0) and radius 1

Three numerical tests were carried out on domains

of IRd,d= 1,2.The tables 1,2and 3permit us to

determine the slope of log kerrorkL2(Ω)

kukH1(Ω) as func-

tion of log(h).In each case, the slope is quadratic.

This slope is observed thanks to)igures 1,)LJXUH2

and)LJXUH.

The inverse inequality 5was verified for each of the

three experiments, after performing a mesh refine-

ment. Moreover, with regard to the experiment 3, it

should be noted that the domain Ω2is a domain used

for isogeometric method for 1Dproblems and not in

finite elements. The)igure 4Zas represented thanks

to domain Ω2.To get Ω2, we have built a parametriza-

tion of our domain. This parametrization is used only

in the framework of the isogeometric method. This

shows that isogeometric method allows us to correct

the shortcomings of the finite element method.

4 Conclusion

The isogeometric method approximates a function by

a b-spline curve on a domain IRd, d = 1,2.Nume-

rical tests have been done to validate numerically an

isogeometric inverse inequality and show the neces-

sity of using the isogeometric method, to the detri-

ment of the finite element method, in one dimension.

In perspective, we can use the isogeometric method

with NURBS as the basis of approximation, to solve

the least squares problems in two and three dimen-

sions. We can use this approach for modeling prob-

lems. This project is currently ongoing.

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perbolic partial differential equations, Doctoral

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Volume 23, 2024

[10] John Evans, Thomas Hughes, Explicit trace

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Contribution of Individual Authors to the Cre-

ation of a Scientific Article:

Goudjon was in charge of the least squares problem

formulation, while Haudié worked on the isogeome-

try method. Aguemon wrote the paper and carried out

the numerical tests.

Sources of Funding for Research Presented in a

Scientific Article or Scientific Article Itself:

No funding was received for conducting this study.

Conflicts of interest:

The authors have no conflicts of interest to declare

that are relevant to the content of this article.

Creative Commons Attribution License 4.0

(Attribution 4.0 International , CC BY 4.0)

This article is published under the terms of the

Creative Commons Attribution License 4.0

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