IHJPAS. 36 (3) 2023 

427 

This work is licensed under a Creative Commons Attribution 4.0 International License 

 

   

 

 

 

 

 
 

 

 

 

 

 
 

Abstract 

The wavelets have many applications in engineering and the sciences, especially mathematics. 

Recently, in 2021, the wavelet Boubaker (WB) polynomials were used for the first time to study 

their properties and applications in detail. They were also utilized for solving the Lane-Emden 

equation. The aim of this paper is to show the truncated Wavelet Boubaker polynomials for solving 

variation problems. In this research, the direct method using wavelets Boubaker was presented for 

solving variational problems. The method reduces the problem into a set of linear algebraic 

equations. The fundamental idea of this method for solving variation problems is to convert the 

problem of a function into one that involves a finite number of variables. Different numerical 

examples were given to demonstrate the applicability and validity of this method using the Matlab 

program. Also, the results of this technique were compared with the exact solution, and graphs 

were added to these examples to test the convergence of Wavelet Boubaker polynomials using this 

method. 

Keywords: Boubaker wavelets, Calculus of variation problems, Nonlinear programming, 

Numerical methods. 

1. Introduction 

 

Many problems arising in mathematical physics and geometry are connected with the calculus of 

variations, which is determined by finding the maximal and minimal functional functions. The 

functionals are defined by definite integrals, which include boundary conditions and appear in the 

mathematical formula, see [2]. 

Wavelet theory is an emerging field in mathematical research and is applied in a broad range of 

engineering disciplines. Wavelets are very successful in accurately solving numerical problems. 

[3] Utilized Legendre wavelets to solve variational problems. [4] Used the Haar wavelet to solve 

the same problems. [5] Applied direct restarted Pell to solve the problem of variation, then used 

the spectral method with Chebyshev wavelets in another paper to solve the calculus of variations 

doi.org/10.30526/36.3.3048 

Article history: Received 28 September 2022, Accepted 13 March 2023, Published in July 2023. 

 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

 

Direct Method for Variational Problems Using Boubaker Wavelets 
 

Eman Hassan Ouda 

Department of Applied Science, University of Technology- Iraq. 

 

Eman.H.Ouda@uotechnology.edu.iq 

 

  

 

 

https://creativecommons.org/licenses/by/4.0/
mailto:Eman.H.Ouda@uotechnology.edu.iq
mailto:Eman.H.Ouda@uotechnology.edu.iq


IHJPAS. 36 (3) 2023 

428 
 

[6]. [7-8] using a Spline polynomial with collocation method and [9] used a new technique to find 

the numerical solution for solving isoperimetric problems. 

Many researchers have utilized different procedures to solve calculus of variational problems. [10] 

Developed the new functions for solving the problems of variational, then [11] used a combination 

of many functions with Bernoulli polynomials. [12] Found the approximate solution for boundary 

value problems using the wavelet function. [13] Studied moving or fixed boundary Muntz wavelets 

for solving variation problems. 

This paper is arranged as follows: in Section 2, Orthogonal Boubaker polynomials and their 

properties with recurrence relations In Section 3, Boubaker wavelets and their properties In Section 

4, the application of Boubaker wavelet polynomials for solving variational problems with some 

numerical examples has been presented. In Section 5, the convergence test for the introduced 

method has been studied, and at last the conclusion has been reached. 

   

 

2. Orthogonal Boubaker polynomials and their properties: 

Boubaker polynomials haven't been orthogonal, so by applying the Gram-Schmit process to 

Boubaker polynomials, one can obtain orthogonal Boubaker polynomials, OBm(t). Several papers 

have been applied with different applications in physics, applied sciences, etc. (see [14], [15]). 

Orthogonal Boubaker polynomials (OB) of mth degree were presented on the interval [0, 1] by the 

following equation: [1] 

  

  

  𝑂𝐵𝑚(𝑡) =
(𝑚!)2

(2𝑚)!
∑ (−1)𝑚+𝑘

(𝑚+𝑘)!

(𝑚−𝑘)!(𝑘!)2
𝑡𝑘𝑚𝑘=0            … (1) 

    A recursive relation of the orthogonal Boubaker polynomial on the interval [0, 1] has given as 

follows: 

𝑂𝐵𝑚+1(𝑡) =
((𝑚+1)!)2

(2(𝑚+1))!
[

(2𝑚+1)(2𝑚)!

(𝑚+1)(𝑚!)2
(2𝑡 − 1)𝑂𝐵𝑚(𝑡) −

(𝑚)        2(𝑚−1)!

(𝑚+1)    ((𝑚−1)!)2
𝑂𝐵𝑚−1(𝑡)] , 𝑚 ≥2 

with 𝑂𝐵0(𝑡) = 1 𝑎𝑛𝑑 𝑂𝐵1(𝑡) =
1

2
(2𝑡 − 1). 

 

3. Boubaker wavelet and their properties: 

      The Boubaker wavelets can be defined as below (see [1]) 

    𝑊𝐵𝑛,𝑚(𝑡) = {√2𝑚 + 1   2
𝑘

2   0      𝑜𝑡ℎ𝑒𝑟 𝑤𝑖𝑠𝑒  
(2𝑚)!

(𝑚!)2
𝑂𝐵𝑚(2

𝑘 𝑡 − 𝑛) ,    
𝑛

2𝑘−1
 ≤ 𝑡 <

𝑛+1

2𝑘−1
    … (2) 

where   𝑊𝐵𝑛𝑚(𝑡) = 𝑊𝐵(𝑚, 𝑛, 𝑡)  as Boubaker wavelets so n =0,1, 2, …, 2
k-1 ,k is any positive 

integer , m =0,1,2…,M and t is normalized time . 

The first six 𝑊𝐵𝑛𝑚(𝑡)  with k=1, are given as follows: 



IHJPAS. 36 (3) 2023 

429 
 

𝑊𝐵0(𝑡) = 1,   

𝑊𝐵1(𝑡) =
√3

2
(4𝑡 − 3), 

𝑊𝐵2(𝑡) =
√5

6
(24𝑡2 − 36𝑡 + 13) , 

𝑊𝐵3(𝑡) =
√7

20
(160𝑡3 − 360𝑡2 + 264𝑡 − 63), 

𝑊𝐵4(𝑡) =
√9

70
(1120𝑡4 − 3360𝑡3 + 3720𝑡2 − 1800𝑡 + 321), 

𝑊𝐵5(𝑡) =
√11

252
(8064𝑡5 − 30240𝑡4 + 44800𝑡3 − 32760𝑡2 + 11820𝑡 − 1683), 

𝑊𝐵6(𝑡) =
√13

924
(59136𝑡6 − 266112𝑡5 + 493920𝑡4 − 483840𝑡3 + 263760𝑡2 − 75852𝑡 +

8989) . 

The differential with respect to t of Boubaker wavelet polynomials 𝑊𝐵𝑚̇ (𝑡) is given as follows: 

𝑊𝐵0̇ (𝑡) = 0 , 

𝑊𝐵1̇ (𝑡) = 2√3 , 

𝑊𝐵2̇ (𝑡) =
√5

6
(48𝑡 − 36) , 

𝑊𝐵̇ 3(𝑡) =
√7

20
(480𝑡2 − 720𝑡 + 264), 

𝑊𝐵̇ 4(𝑡) =
√9

70
(4480𝑡3 − 10080𝑡2 + 7440𝑡 − 1800), 

   𝑊𝐵̇ 5(𝑡) =
√11

252
(40320𝑡4 − 120960𝑡3 + 134400𝑡2 − 65520𝑡 + 11820), 

𝑊𝐵̇ 6(𝑡) =
√13

924
(354816𝑡5 − 1330560𝑡4 + 1975680𝑡3 − 1451520𝑡2 + 527520𝑡 − 75852) . 

4. Boubaker wavelet polynomials for solving variational problems: 

         We demonstrate the application of wavelet Boubaker polynomials to solve some variational 

problems. 

A function f (t) is defined over𝐿2[0,1] , the function approximate by wavelet Boubaker 

polynomials is defined as follows: 

𝑓(𝑡) =  ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡) = 𝑐
𝑇 𝑊𝐵(𝑡)

∞

𝑖=0

                                       …  (3)   

where 𝑐𝑖 =⟨𝑓(𝑡), 𝑊𝐵𝑖 (𝑡)⟩ and 〈  , 〉is inner product on 𝐿
2[0,1]. 



IHJPAS. 36 (3) 2023 

430 
 

If the series in equation (3) is truncated, then equation (3) can be written as 

𝑓(𝑡) = ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡) = 𝑐
𝑇 𝑊𝐵(𝑡)

𝑁

𝑖=0

                                         …  (4) 

where c =[c0 , c1, …, cN]
T  and WBi (t) = [WB0,WB1, …,WBN]

T 

Differentiating equation (4) with respect t, to obtain 𝑓 =̇  𝑐𝑇 𝑊𝐵̇ (𝑡) 

where c =[c0, c1, …, cN]
T and 𝑊𝐵̇  (t) = [𝑊𝐵0̇ ,𝑊𝐵̇ 1, …,𝑊𝐵̇ N]

T 

The matrix of derivatives D is given as 
𝑑𝑊𝐵(𝑡)

𝑑𝑡
= 𝑊𝐵̇  (𝑡) = 𝐷𝑊𝐵(𝑡) 

Where 𝑊𝐵̇  (𝑡) derivative of wavelet Boubaker functions 

To demonstrate this procedure, we consider this example of finding the minimum of functional 

[14]  

Example1: 

  𝐽(𝑥) = ∫ [�̇� 2(𝑡) + 𝑡�̇�(𝑡) + 𝑥2(𝑡)]𝑑𝑡
1

0
,                                         … (5) 

  with two conditions x(0) = 0, 𝑥(1) =
1

4
                                            … (6) 

and the  exact solution  𝑥(𝑡) =
−𝑒 −𝑡[(−1+𝑒 𝑡)(𝑒−2𝑒 2−2𝑒 𝑡+𝑒 1+𝑡)]

4(−1+𝑒 2)
  

 Now, we use wavelet Boubaker polynomials of M = 4 and M=6 to approximate the function x(t) 

Suppose  𝑥(𝑡) = ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)
5
0   or  𝑥(𝑡) = 𝑐

𝑇 𝑊𝐵(𝑡)                            … (7) 

  And 𝑥2(𝑡) = 𝑐𝑇 𝑊𝐵(𝑊𝐵)𝑇 𝑐                                                                    … (8) 

where c = [c0, c1,… , c5]
T  and  WB (t) = [WB0, WB1, …, WB5], 

 Differentiating equation (7), we get  �̇�(𝑡) = 𝑐𝑇 𝐷 𝑊𝐵(𝑡)                       … (9) 

and    �̇� 2(𝑡) = 𝑐𝑇 𝐷 (𝑊𝐵)(𝐷 (𝑊𝐵))𝑇 𝑐                                                        …(10) 

Substituting equation (7) - (10) into equation (5), we obtain 

𝐽(𝑥) = ∫ [𝑐𝑇 𝐷 (𝑊𝐵)(𝐷 (𝑊𝐵))
𝑇1

0
𝑐 + 𝑡𝑐𝑇 𝐷 (𝑊𝐵) + 𝑐𝑇 𝑊𝐵(𝑊𝐵)𝑇 𝑐]𝑑𝑡    … (11) 

 We can simplify equation (11) to   𝐽(𝑥) =
1

2
𝑐𝑇 𝐻𝑐 + 𝑞𝑇 𝑐                                   … (12) 

where 𝐻 = 2 ∫ [𝐷 (𝑊𝐵)(𝐷 (𝑊𝐵))
𝑇1

0
+ 𝑊𝐵(𝑊𝐵)𝑇 ]𝑑𝑡, and 𝑞𝑇 = ∫ 𝑡 𝐷 (𝑊𝐵)𝑇𝑑𝑡

1

0
, 

Equation (7) with boundary conditions (6), can imply 

𝑥(0) = 𝑐𝑇 𝑊𝐵(0) = 0  and 𝑥(1) = 𝑐𝑇 𝑊𝐵(1) =
1

4
                                           … (13) 



IHJPAS. 36 (3) 2023 

431 
 

 We can rewrite equations (12) & (13), as follows   

Min 𝐽(𝑥) =
1

2
𝑐𝑇 𝐻𝑐 + 𝑞𝑇 𝑐  

subject to Fc-b=0 where 𝐹 = [𝑊𝐵𝑇 (0)    𝑊𝐵𝑇(1) ] , 𝑏 = [0  
1

4
 ] 

we can find the parameter c using Lagrange equation as    

𝑐∗ = −𝐻−1𝑐 + 𝐻−1𝐹𝑇(𝐹𝐻−1𝐹𝑇 )−1(𝐹𝐻−1𝑐 + 𝑏) 

   When M=4 the approximate value of x is    

x(t)= [0.21463728 ,0.04727207 ,-0.01564479, 0.00192278] WB (t) 

    and when M=6 an approximate value of x is    

x(t)= [0.21464181, 0.04746694,-0.01585894, 0.00129175, -0.00023926 ,0.00001933] WB (t) 

Table (1) shows the numerical results for example (1) with M=4 and M=6 compared with exact 

solution, and graphically in figure (1). 

Table 1. Results for Example1  

t xexact xapp.(t) M=4 
M=4 

Absolute error  
xapp.(t) M=6 

M=6 

Absolute error  

0 0.00000000 0.00000000 0.00000000 0.00000000 0.0000000 

0.1 0.04195072 0.04180602 0.00014470 0.04195070 0.00000002 

0.2 0.07931714 0.07922621 0.00009093 0.07931742 0.00000027 

0.3 0.11247322 0.11250475 0.00003152 0.11247334 0.00000011 

0.4 0.14175081 0.14188583 0.00013502 0.14175057 0.00000023 

0.5 0.16744291 0.16761363 0.00017071 0.16744257 0.00000034 

0.6 0.18980668 0.18993234 0.00012566 0.18980660 0.00000007 

0.7 0.20906592 0.20908615 0.00002022 0.20906621 0.00000029 

0.8 0.22541340 0.22531923 0.00009416 0.22541370 0.00000030 

0.9 0.23901272 0.23887579 0.00013693 0.23901258 0.00000014 

1.0 0.2500000000 0.25000000 0.00000000 0.25000000 0.00000000 

 

 

Figure 1. Shows that the Absolute Error when M=6 is very less than M=4 for Example1  

 



IHJPAS. 36 (3) 2023 

432 
 

Example 2: 

Min J= ∫ (�̇�2 + 𝑥2)
1

0
𝑑𝑡       

with boundary conditions 𝑥(0) = 0 ,   𝑥(1) = 1 , 

and exact solution is 𝑥(𝑡) =
𝑒 𝑡−𝑒 −𝑡

𝑒 1−𝑒−1
 

In the same procedure, we can solve example2. Table (2) shows the numerical results for example 

(2) with M=4 and M=6 compared with exact solution, and graphically in figure (2). 

  When M=4 the approximate value of x is    

x(t)= [0.70703400,0.32030428,0.03906027, 0.00769113] WB (t) 

  and when M=6 the approximate value of x is     

x(t)= [0.70703595,0.32001609,0.03929145,0.00870795,0.00060523,0.00007777] WB (t) 

 

Table 2. Results for Example2  

t xexact M=4 xapp.(t) 

M=4 Absolute 

error M=6 xapp.(t) 

M=6 Absolute 

error 

0 0 0.00000000 0.00000000 0.00000000 0.00000000 

0.1 0.08523370 0.08540591 0.00017221 0.08523414 0.00000043 

0.2 0.17132045 0.17145031 0.00012986 0.17131995 0.00000049 

0.3 0.25912183 0.25910993 0.00001190 0.25912108 0.00000075 

0.4 0.34951660 0.34936152 0.00015507 0.34951638 0.00000021 

0.5 0.44340944 0.44318181 0.00022762 0.44340990 0.00000046 

0.6 0.54174007 0.54154756 0.00019250 0.54174071 0.00000063 

0.7 0.64549262 0.64543551 0.00005710 0.64549284 0.00000021 

0.8 0.75570548 0.75582241 0.00011693 0.75570520 0.00000027 

0.9 0.87348169 0.87368498 0.00020329 0.87348147 0.00000021 

1.0 1.00000000 1.00000000 0.00000000 1.00000000 0.00000000 

   

 

 

Figure2. Shows that the Absolute Error when M=6 is very less than M=4 for Example2 . 

 



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433 
 

Example 3:  

Min j= ∫ (�̇�2 −  𝑥2)
1

0
𝑑𝑡       

with boundary conditions 𝑥(0) = 0,  𝑥(1) = 1 , 

and exact solution is 𝑥(𝑡) =  
𝑠𝑖𝑛𝑡

𝑠𝑖𝑛1
 

Table (3) and figure (3) illustrate the results of example (3). 

When M=4 the approximate value of x is  

x(t)= [ 0.80163166, 0.24981351, 0.04537278, 0.00806631] WB (t) 

and when M =6 the approximate value of x is    

x(t)= [0.80164433, 0.24944837, 0.04508324, 0.00682505, 0.00070902, 0.00008006] WB (t) 

 

Table 3. Results for Example3  

t uexact 
M= 4  

uappr (t) 

M=4 

Absolute Error 

M= 6  

uappr.(t) 

M=6 

Absolute Error 

0 0.00000000 0.00000000 0.00000000 0.00000000 0.0000000 

0.1 0.11864154 0.11885365 0.00021211 0.11862523 0.00001631 

0.2 0.23609766 0.23624932 0.00015166 0.23610543 0.00000777 

0.3 0.35119476 0.35116260 0.00003216 0.35121956 0.00002480 

0.4 0.46278285 0.46256910 0.00021374 0.46280277 0.00001992 

0.5 0.56974696 0.56944444 0.00030251 0.56974637 0.00000058 

0.6 0.67101835 0.67076422 0.00025412 0.67099789 0.00002045 

0.7 0.76558514 0.76550406 0.00008108 0.76556103 0.00002411 

0.8 0.85250246 0.85263956 0.00013709 0.85249567 0.00000678 

0.9 0.93090186 0.93114634 0.00024447 0.93091791 0.00001604 

1.0 1.00000000 1.00000000 0.00000000 1.00000000 0.00000000 

 

 

 
Figure 3. Shows that the Absolute Error when M=6 is very less than M=4 for Example3  

 

 

 



IHJPAS. 36 (3) 2023 

434 
 

5. The Convergence Test of wavelet Boubaker polynomials: 

       By (theorem (1), see [1]) if x(t) is continually defined on [0,1] and α(t) is the approximate of  

𝛼∗(t) by applying Boubaker wavelet. Also, suppose that x (t) is bounded by a positive constant that 

is |𝑥(𝑡)| < 𝜖. Then, the coefficients of x (t) are bounded. 

     The state can be expanded using Wavelet Boubaker polynomials, as: 

𝑥𝑖𝑁 (𝑡) = ∑ 𝑐𝑖𝑘 𝑊𝐵𝑘 (𝑡)
𝑁
𝑘=1         

 𝑥𝑖 (𝑡) = 𝑥𝑖𝑁 (𝑡) + ∑ 𝑐𝑖𝑘 𝑊𝐵𝑘 (𝑡)
∞
𝑘=𝑁+1  

or  𝑥𝑖 (𝑡) = 𝑥𝑖𝑁 (𝑡) + 𝑒𝑖 (𝑡)                                                                     …(14) 

The coefficients in equation (14) is limited by [1] 

Such that the residual ‖𝑒(𝑡)‖ is less than some 𝜀 

Where  𝑒(𝑡) = 𝑚𝑎𝑥 {𝑒1(𝑡), 𝑒2(𝑡), … , 𝑒𝑁(𝑡)}, 

Then, using the convergence test for the technique to the variable x in terms of N proposed 𝐿2norm 

of   𝑥𝑖  , i=1,2,…,n 

[∫ (𝑥(𝑡) − 𝑥𝑖𝑁 (𝑡))
2𝑑𝑡

1

0

]

1
2

< 𝜀𝑖 ,     𝑖=1,2,…,𝑛 

  𝜀 = 𝑚𝑎𝑥{𝜀1(𝑡), 𝜀2(𝑡), … , 𝜀𝑛(𝑡) } ,  

[∫ (𝑥(𝑡) − 𝑥𝑁 (𝑡))
2𝑑𝑡

1

0
]

1

2
< 𝜀, 

 

Using the Boubaker wavelet polynomials for approximating x variables, we get 

[∫ ( ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)

𝑁+𝑀

𝑖=0

− ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)

𝑁

𝑖=0

)2𝑑𝑡
1

0

]

1
2

< 𝜀 

  = [∫ ( ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)

𝑁+𝑀

𝑖=𝑁+1

)2𝑑𝑡
1

0

]

1
2

< 𝜀 

= [∫ ( ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)

𝑁+𝑀

𝑖=𝑁+1

)( ∑ 𝑐𝑖 𝑊𝐵𝑖 (𝑡)

𝑁+𝑀

𝑖=𝑁+1

)𝑑𝑡
1

0

]

1
2

<  𝜀 

= ∑ ∑ 𝑐𝑖 𝑐𝑗

𝑁+𝑀

𝑗=𝑁+1

∫ 𝑊𝐵𝑖 (𝑡)
1

0

𝑊𝐵𝑗 (𝑡)  𝑑𝑡 <  𝜀

𝑁+𝑀

𝑖=𝑁+1

 

 

Hence, the Wavelet Boubaker polynomials can be reduced to the form 

 



IHJPAS. 36 (3) 2023 

435 
 

∑ 𝑐𝑖
2

𝑁+𝑀

𝑖=𝑁+1

< 𝜀 

when the squares of the remaining coefficients becomes neglected, a favorable approximation to 

the solution is achieved. 

 

6. Conclusion 

In this paper, a direct method with Wavelet Boubaker polynomials was achieved as an 

evaluation solution for reducing the variational problems into quadratic programming 

problems. Only a few terms of Wavelet Boubaker polynomials were needed to obtain an 

accurate solution. The numerical results were compared with the exact solution, which proved 

the capability and validity of such problems with easy steps. Matlab plotting was also used to 

demonstrate the results. 

 

 

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3. Razzaghi, M.; Yousefi, S., Legendre Wavelets Direct Method for Variational Problems, 

Elsevier, Mathematics and Computers in Simulation, 2000, 53, 185-192. 

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