Ratio Mathematica Volume 48, 2023

Approximation and moduli of continuity for
a function belonging to Hölder’s class
Hα[0,1) and solving Lane-Emden

differential equation by Boubaker wavelet
technique

Shyam Lal*
Swatantra Yadav†

Abstract

In this paper, Boubaker wavelet is considered. The Boubaker wavelets
are orthonormal. The series of this wavelet is verified for the
function f(t) = t ∀ t ∈[0,1). The convergence analysis of solution
function of Lane-Emden differential equation has been studied. New
Boubaker wavelet estimator E2k,M(f) for the approximation of
solution function f belong to Hölder’s class Hα[0,1) of
order 0 < α ≤ 1, has been developed. Furthermore, the moduli of
continuity of

(
f − S2k,M(f)

)
of solution function f of Lane-Emden

differential equation has been introduced and it has been estimated
for solution function f∈ Hα[0,1) class. These estimator and moduli
of continuity are new and best possible in wavelet analysis. Boubaker
wavelet collocation method has been proposed to solve Lane-Emden
differential equations with unknown Boubaker coefficients. In this
process, Lane-Emden differential equations are reduced into a system
of algebraic equations and these equations are solved by collocation
method. Three Lane-Emden type equations are solved to demonstrate
the applicability of the proposed method. The solutions obtained by
the proposed method are compared with their exact solutions. The

*Department of Mathematics, Institute of Science, Banaras Hindu University, Varanasi -
221005, India; shyam lal@rediffmail.com.

†Department of Mathematics, Institute of Science, Banaras Hindu University, Varanasi -
221005, India; swatantrayadavbhu@gmail.com.



Shyam Lal and Swatantra Yadav

absolute errors are negligible. Thus, this shows that the method de-
scribed in this paper is applicable and accurate.
Keywords: Boubaker wavelet, Boubaker polynomial, Boubaker wavelet
approximation, Moduli of continuity, convergence analysis,
Collocation method, Lane-Emden differential equations.
2020 AMS subject classifications:42C40, 65T60, 45G10, 45B05.1

1 Introduction
Wavelet theory is a newly emerging area of research in a mathematical

sciences. It has applications in engineering disciplines; such as signal
analysis for wave representation and segmentation etc. Wavelets allow
the accurate representation of a variety of signals and operators. Wavelets are
assumed as a basis function {ψn,m(·)} continuously in time domain.
Special feature of wavelets basis is that all functions {ψn,m(·)} are constructed
from a single mother wavelet ψ(·) which is small pulse. Many practical and
physical problems in the field of science and engineering are formulated as intial
and boundry value problems. Approximation of functions by the wavelet method
has been discussed by many researchers like Devorce[2], Morlet[5], Meyer[4],
Debnath[3], Lal and Satish[11]. The wavelet functions have been applied for
finding approximate solutions for some problems arising in numerous branches of
science and engineering.

In this paper, Boubaker wavelet has been studied. This wavelet is defined
by the orthogonal Boubaker polynomials. It has several interesting and useful
properties.
The main aims of present paper are as follows:
(i) To define Boubaker wavelet and to verify Boubaker wavelet series by
examples.
(ii) To study the properties of Boubaker wavelet coefficient in expansion of
characteristic function.
(iii) To estimate the approximation of solution function f of Lane-Emden
differential equations belonging to Hölder’s class Hα[0,1) by the Boubaker wavelet
series.
(iv) To estimate the moduli of continuity of

(
f − S2k,M(f)

)
of solution function

f of Lane-Emden differential equations belonging to Hölder’s class Hα[0,1) by
the Boubaker wavelet series.
(v) To solve Lane-Emden differential equation by Boubaker wavelet series by

1Received on January 28, 2023. Accepted on July 15, 2023. Published on August 1, 2023. DOI:
10.23755/rm.v39i0.1092. ISSN: 1592-7415. eISSN: 2282-8214. ©The Authors. This paper is
published under the CC-BY licence agreement.



Approximation and moduli of continuity...

collocation method.
The paper is organized as follows: In Section-2, Boubaker wavelet,Boubaker
wavelet approximation and moduli of continuity of Function of Hölder’s class
Hα[0,1) are defined. The Boubaker wavelet series is verified by example. In
Section-3 ,theorem concerning the convergence analysis of Boubaker wavelet has
been discussed. In Section-4, theorem concerning Boubaker wavelet
coefficient in the expansion of characteristic function and approximation in Hα[0,1)
have been obtained. In Section-5, theorem concerning the moduli of continu-
ity of

(
f − S2k,M(f)

)
have been determined. In Section-6, Boubaker wavelet

method for solution of differential equation has been discussed . In Section-7,
Lane-Emden differential equations have been solved using Boubaker wavelet
series by collocation method. Finally, the main conclusions are summarized in
Section-8.

2 Definitions and Preliminaries

2.1 Boubaker wavelet

Wavelet functions are constructed from dilation and translation of a definite
function, named mother wavelet ψ . ψb,a may be defined as

ψb,a(t) = |a|−1/2ψ
(
t − b
a

)
a,b ∈ IR,a ̸= 0 (Daubechies[6]) (1)

where a and b are dilation and translation parametres respectively while t is nor-
malized time. By taking a = 1

2k
, b = n

2k
and ψ(t) =

√
(2m + 1)

(2m!)
(m!)2

Bm(t),
where Bm(t) is Boubaker polynomial of degree m, in equation (1), it reduces into
form

ψ(B)n,m(t) =
√
(2m + 1)

(2m!)

(m!)2
2

k
2 Bm(2

kt − n).

In precise,

ψ(B)n,m(t) =

{√
(2m + 1)

(2m!)
(m!)2

2
k
2 Bm(2

kt − n), if n
2k

⩽ t < n+1
2k
,

0, otherwise.

where n = 0,1,2...,2k − 1, k = 0,1,2,3, ..., while m represents the order of
orthogonal Boubaker polynomial (shiralashetti et al.[8]). It has four parameters
m,n,k & t. The orthogonal Boubaker polynomial Bm(t) of order m satisfies the



Shyam Lal and Swatantra Yadav

following conditions:

B0(t) = 1, B1(t) =
1

2
(2t − 1),B2(t) =

1

6
(6t2 − 6t + 1) ;

B3(t) =
1

20
(20t3 − 30t2 + 12t − 1), B4(t) =

1

70
(70t4 − 140t3 + 90t2 − 20t + 1) ;

B5(t) =
1

252
(252t5 − 630t4 + 560t3 − 210t2 + 30t − 1);

B6(t) =
1

924
(924t6 − 2772t5 + 3150t4 − 1680t3 + 420t2 − 42t + 1);

2.2 Boubaker Wavelet Approximation
A function f ∈ L2[0,1) may be expanded in Boubaker wavelet series as

f(t) =
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t), (2)

where the coefficients cn,m are given by

cn,m =< f(t),ψ
(B)
n,m(t) > . (3)

If the series (2) is truncated, then

(S2k,Mf)(t) =
2k−1∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t) = C

T Ψ(t) (4)

where (S2k,Mf) is the (2
k,M)th partial sum of series (2) and C, Ψ(t) are 2kM ×1

matrices given by

C = [c0,0,c0,1, ...,c0,M − 1,c1,0, ...,c1,M−1, ...,c2k−1,0, ...,c2k−1,M−1]T (5)

and

Ψ(t) = [ψ0,0(t),ψ0,1(t), ...,ψ0,M−1(t),ψ1,0(t),ψ1,M−1(t), ...,ψ2k−1,0(t), ...,ψ2k−1,M−1]
T .

(6)
The Boubaker wavelet approximation of f by (S2k,Mf) under norm || ||2, denoted
by E2k,M(f), is defined by

E2k,M(f) = min||f − (S2k,M)||2 (Zygmund)[10]).

If E2k,M(f) → 0 as k → ∞, M→∞,then E2k,M(f) is best approximation of f
order (2k,M) (Zygmund[10]).



Approximation and moduli of continuity...

2.3 Moduli of continuity
The moduli of continuity of a function f ∈ L2[0,1) is defined as

W(f,δ) = sup
0≤h≤δ

||f(· + h) − f(·)||2

= sup
0≤h≤δ

( ∫ 1
0

|f(t + h) − f(t)|2dt
)1

2

It is remarkable to note that W(f,δ) is a non-decreasing function of δ and
W(f,δ) → 0 as δ → 0 , (Chui [1]).

2.4 Function of Hölder’s class Hα[0,1)
A function f ∈ Hα[0,1) if f is continuous and satisfies the inequality

f(x) − f(y) = O(|x − y|α),∀x,y ∈ [0,1) and 0 < α ≤ 1 (Das)[9]).

2.5 Example
The example of this section illustrates the validity of the Boubaker wavelet

series as follows:
Consider the function f : [0,1) → R defined by f(t) = t ∀ t ∈[0,1).
Let

f(t) =
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t). (7)

cn,m = < f(t),ψ
(B)
n,m(t) >=

∫ n+1
2k

n

2k

f(t)ψ(B)n,m(t)dt

=

∫ n+1
2k

n

2k

t
√

(2m + 1)
(2m!)

(m!)2
2

k
2 Bm(2

kt − n) dt

=
√
(2m + 1)

(2m!)

(m!)2
2

k
2

∫ 1
0

v + n

2k
Bm(v)

dv

2k
, 2kt − n = v

=
√

(2m + 1)
(2m!)

(m!)2
1

2
3k
2

∫ 1
0

(v + n)Bm(v)dv.

By above expansion ,taking m = 0 ,

cn,0 =
1

2
3k
2

∫ 1
0

(v + n)B0(v)dv =
1

2
3k
2

∫ 1
0

(v + n)dv =
2n + 1

2
3k+2

2

.

(8)



Shyam Lal and Swatantra Yadav

Next

cn,1 =
2
√
3

2
3k
2

∫ 1
0

(v + n)B1(v)dv =

√
3

2
3k
2

∫ 1
0

(v + n)(2v − 1)dv

=

√
3

3.2
3k+2

2

. (9)

cn,2 =
6
√
5

2
3k
2

∫ 1
0

(v + n)B2(v)dv =

√
5

2
3k
2

∫ 1
0

(v + n)(6v2 − 6v + 1)dv = 0

For, m ≥ 2

cn,m =
√

(2m + 1)
(2m!)

(m!)2
1

2
3k
2

∫ 1
0

(v + n)Bm(v)dv

=
√
(2m + 1)

(2m!)

(m!)2
1

2
3k
2

( ∫ 1
0

vBm(v)dv +

∫ 1
0

nBm(v)dv

)
=

√
(2m + 1)

(2m!)

(m!)2
1

2
3k
2

( ∫ 1
0

1

2
(2v − 1)Bm(v)dv

+

∫ 1
0

1

2
Bm(v)dv +

∫ 1
0

nBm(v)dv

)
=

√
(2m + 1)

(2m!)

(m!)2
1

2
3k
2

( ∫ 1
0

B1(v)Bm(v)dv

+
1

2

∫ 1
0

B0(v)Bm(v)dv + n

∫ 1
0

B0(v)Bm(v)dv

)
=

√
(2m + 1)

(2m!)

(m!)2
1

2
3k
2

(0 + 0 + 0) = 0 ,Bm(v) is orthogonal .

cn,m = 0 ∀n ≥ 2k, by defintion of ψ(B)n,m.

Then , f(t) =
2k−1∑
n=0

cn,0ψ
(B)
n,0 (t) +

2k−1∑
n=0

cn,1ψ
(B)
n,1 (t) +

∞∑
n=2k

cn,mψ
(B)
n,m(t), (10)

Next,
||f||22 = < f,f >

= <
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t),

∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t) >

=

〈 2k−1∑
n=0

cn,0ψ
(B)
n,0 (t) +

2k−1∑
n=0

cn,1ψ
(B)
n,1 (t) +

∞∑
n=2k

cn,mψ
(B)
n,m(t),

2k−1∑
n=0

cn,0ψ
(B)
n,0 (t) +

2k−1∑
n=0

cn,1ψ
(B)
n,1 (t) +

∞∑
n=2k

cn,mψ
(B)
n,m(t)

〉

=
2k−1∑
n=0

c2n,0||ψ
(B)
n,0 ||

2
2 +

2k−1∑
n=0

c2n,1||ψ
(B)
n,1 ||

2
2 +

∞∑
n=2k

c2n,m||ψ
(B)
n,m||

2
2,



Approximation and moduli of continuity...

=
2k−1∑
n=0

c2n,0 +
2k−1∑
n=0

c2n,1 + 0,{ψn,m}n,m∈Z being orthonormal

=
2k−1∑
n=0

(2n + 1)2

2(3k+2)
+

2k−1∑
n=0

1

3.23k+2
=

1

3
, by eqns (8) and (9).

Also, ||f||22 = < f,f >=
∫ 1
0

|f(t)|2dt =
∫ 1
0

t2dt =
1

3
.

Hence, the Boubaker wavelet expansion (7) is verified for f(t) = t.

3 Convergence analysis

In this section, the convergence analysis of solution of
Lane-Emden differential equation has been studied.

3.1 Theorem

If f be a exact solution of of Lane-Emden differential equation and its Boubaker
wavelet series is

f(·) =
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(·) (11)

then its (2k,M)th partial sums (S2k,Mf)(·) =
∑2k−1

n=0

∑M−1
m=0 cn,mψ

(B)
n,m(·)

converges to f(·) as M→∞, k→ ∞.

Proof of Theorem 3.1

Now, < f,f > =
〈 ∞∑

n=0

∞∑
m=0

cn,mψ
(B)
n,m,

∞∑
n
′
=0

∞∑
m

′
=0

cn′,m′ψ
(B)

n
′
,m

′

〉

=
∞∑
n=0

∞∑
m=0

∞∑
n
′
=0

∞∑
m

′
=0

cn,mcn′,m′ < ψ
(B)
n,m,ψ

(B)

n
′
,m

′ >

=
∞∑
n=0

∞∑
m=0

cn,mcn,m < ψ
(B)
n,m,ψ

(B)
n,m >

=
∞∑
n=0

∞∑
m=0

|cn,m|2||ψ(B)n,m||
2



Shyam Lal and Swatantra Yadav

=
2k−1∑
n=0

∞∑
m=0

|cn,m|2||ψ(B)n,m||
2 +

∞∑
n=2k

∞∑
m=0

|cn,m|2||ψ(B)n,m||
2

=
2k−1∑
n=0

∞∑
m=0

|cn,m|2 + 0, by defintion of {ψn,m}n,m∈Z

2k−1∑
n=0

∞∑
m=0

|cn,m|2 = ||f||22 < ∞,f ∈ L
2[0,1). (12)

For M > N, using eqn (12),

||(S2k,Mf) − (S2k,Nf)||22 =
∣∣∣∣
∣∣∣∣ 2

k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t) −

2k−1∑
n=0

N−1∑
m=0

cn,mψ
(B)
n,m(t)

∣∣∣∣
∣∣∣∣2
2

=

∣∣∣∣
∣∣∣∣ 2

k−1∑
n=0

M−1∑
m=N

cn,mψ
(B)
n,m(t)

∣∣∣∣
∣∣∣∣2
2

=

〈 2k−1∑
n=0

M−1∑
m=N

cn,mψ
(B)
n,m(t),

2k−1∑
n=0

M−1∑
m=N

cn,mψ
(B)
n,m(t)

〉

=
2k−1∑
n=0

M−1∑
m=N

cn,mcn,m < ψ
(B)
n,m(t),ψ

(B)
n,m(t) >

=
2k−1∑
n=0

∞∑
m=0

|cn,m|2||ψ(B)n,m(t)||
2

=
2k−1∑
n=0

M−1∑
m=N

|cn,m|2 → 0 as M → ∞,N → ∞.

Hence,
{
(S2k,Mf)

}
M∈N is a Cauchy sequence in L

2[0,1), L2[0,1) is a Banach
space and hence

{
(S2k,Mf)

}
M∈N converges to a function g(t) ∈ L

2[0,1).
Now we need to prove that g(t) = f(t).
For this

< g(t) − f(t),ψ(B)n0,m0(t) > = < g(t),ψ
(B)
n0,m0

(t) > − < f(t),ψ(B)n0,m0(t) >
= < lim

M→∞
(S2k,Mf)(t), ,ψ

(B)
n0,m0

(t) > −cn0,m0

= lim
M→∞

2k−1∑
n=0

M−1∑
m=0

cn,m < ψ
(B)
n,m(t),ψ

(B)
n0,m0

(t) > −cn0,m0

= cn0,m0 < ψn0,m0(t),ψn0,m0(t) > −cn0,m0
= cn0,m0 − cn0,m0 = 0.



Approximation and moduli of continuity...

Thus < g(t) − f(t),ψ(B)n,m(t) > = 0 ∀ n ⩾ n0,m ⩾ m0.
Then g(t) = f(t).
Hence,

∑2k−1
n=0

∑M−1
m=0 cn,mψ

(B)
n,m(t) converges to f(t) as k → ∞,M → ∞ .

4 Approximation analysis
In this Section, approximation f by (S2k,Mf) is estimated as follows.

4.1 Theorem
Let a function f = χ[n0

2k
,
n0+1

2k
), where n0 is positive integer less than equal to

2k and Boubaker wavelet expansion of f is

f(t) =
2k−1∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t) (13)

then the coefficients cn,m satisfy

cn,m =


O

(
((2m)!

√
(2m+1) 2−k/2)

(m!)2

)
, if n = n0,

0, n ̸= n0,

4.2 Theorem
Let the solution function f of Lane-Emden differential equation be

a uniformly continuous defined in [0,1) such that

|f(t1) − f(t2)| ≤ |t1 − t2|α
1

2m−1m
3
2

, ∀t1, t2 ∈ [0,1),m ≥ 1 (14)

and its Boubaker wavelet series

f(t) =
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t) (15)

having (2k,M)th partial sums

(S2k,Mf)(t) =
2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t) (16)



Shyam Lal and Swatantra Yadav

then Boubaker wavelet approximation E2k,M(f) satisfies

E2k,M(f) = min||f − (S2k,Mf)||2 = O
(

1
2kα

√
M

)
.

Proof of theorem 4.1 For

f = χ[n0
2k

,
n0+1

2k
),

and

cn0,m = < f(t),ψ
(B)
n0,m

(t) >

=

∫ n0+1
2k

n0
2k

f(t)ψ(B)n0,m(t)dt

=

∫ n0+1
2k

n0
2k

χ[n0
2k

,
n0+1

2k
)(t)

√
(2m + 1)

(2m)!

(m!)2
2

k
2 Bm(2

kt − n0) dt

=
√

(2m + 1)
(2m)!

(m!)2
2

k
2

∫ n0+1
2k

n0
2k

Bm(2
kt − n0) dt

=
√

(2m + 1)
(2m)!

(m!)2
2

k
2

∫ 1
0

Bm(v)
dv

2k
, 2kt − n0 = v

=
√

(2m + 1)
(2m)!

(m!)2
1

2
k
2

∫ 1
0

Bm(v) dv

|cn0,m| ≤
√

(2m + 1)
(2m)!

(m!)2
2−

k
2

∫ 1
0

|Bm(v)| dv

≤
√

(2m + 1)
(2m)!

(m!)2
2−

k
2 ,

∫ 1
0

|Bm(v)|dv ≤ 1. (17)

Then

cn,m =


O

(
((2m)!

√
(2m+1) 2−k/2)

(m!)2

)
, if n = n0,

0, n ̸= n0,

Thus, theorem 4.1 is completely established.



Approximation and moduli of continuity...

Proof of theorem 4.2

f(t) − (S2k,Mf)(t) =
2k−1∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t) −

2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t)

=
2k−1∑
n=0

(
M−1∑
m=0

+
∞∑

m=M

)cn,mψ
(B)
n,m(t) −

2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t)

=
2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t) +

2k−1∑
n=0

∞∑
m=M

cn,mψ
(B)
n,m(t)

−
2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t)

=
2k−1∑
n=0

∞∑
m=M

cn,mψ
(B)
n,m(t).

Next

cn,m = < f(t),ψ
(B)
n,m(t) >

=

∫ n+1
2k

n

2k

f(t)ψ(B)n,m(t)dt

=

∫ n+1
2k

n

2k

{f(t) − f(
n

2k
)}ψ(B)n,m(t)dt +

∫ n+1
2k

n

2k

f(
n

2k
)ψ(B)n,m(t)dt

=

∫ n+1
2k

n

2k

{f(t) − f(
n

2k
)}ψ(B)n,m(t)dt + f(

n

2k
)

∫ n+1
2k

n

2k

ψ(B)n,m(t)dt

=

∫ n+1
2k

n

2k

{f(t) − f(
n

2k
)}ψ(B)n,m(t)dt ,

∫ n+1
2k

n

2k

ψ(B)n,m(t)dt = 0,m ≥ 1.

Then

|cn,m| ≤
∫ n+1

2k

n

2k

|f(t) − f(
n

2k
)| |ψ(B)n,m(t)|dt

=
√
(2m + 1)

(2m)!

(m!)2
2

k
2

∫ n+1
2k

n

2k

|f(t) − f(
n

2k
)| |Bm(2kt − n)|dt

=
√
(2m + 1)

(2m)!

(m!)2
1

2
k
2

∫ 1
0

|f(
u + n

2k
) − f(

n

2k
)||Bm(u)|du,2kt − n = u

≤
√

(2m + 1)
(2m)!

(m!)2
1

2
k
2

1

2kα
1

2(m−1)m
3
2

∫ 1
0

|Bm(u)|du



Shyam Lal and Swatantra Yadav

≤
√

(2m + 1)
(2m)!

(m!)2
1

2kα+
k
2

1

2(m−1)m
3
2

,

∫ 1
0

|Bm(u)|du ≤ 1

≤
1

2k(α+
1
2
)m

,
(m!)2

(2m)!
⩽

1

2m−1
. (18)

Next,

||f − (S2k,Mf)||22 =
∫ 1
0

|f(t) − (S2k,Mf)(t)|2dt

=

∫ 1
0

( 2k−1∑
n=0

∞∑
m=M

cn,mψ
(B)
n,m(t)

)2
dt

=

∫ 1
0

( 2k−1∑
n=0

∞∑
m=M

c2n,m(ψ
(B)
n,m(t))

2

+
∑ ∑

0⩽ n̸=n′≤ 2k−1

∑ ∑
M⩽ m ̸=m′≤∞

cn,mcn′,m′ψ
(B)
n,m(t)ψ

(B)
n′,m′(t)

)
dt

=
2k−1∑
n=0

∞∑
m=M

c2n,m

∫ 1
0

(ψ(B)n,m(t))
2 dt

+
∑ ∑

0⩽ n̸=n′≤ 2k−1

∑ ∑
M⩽ m̸=m′≤∞

cn,mcn′,m′

∫ 1
0

ψ(B)n,m(t)ψ
(B)
n′,m′(t)dt

=
2k−1∑
n=0

∞∑
m=M

|cn,m|2||ψ(B)n,m(t)||
2
2

=
2k−1∑
n=0

∞∑
m=M

|cn,m|2 , {ψn,m}n,m∈Z being orthonormal in [0,1)

≤
2k−1∑
n=0

∞∑
m=M

1

2k(2α+1)m2
, by eqn (18).

=
2k

2k(2α+1)

∞∑
m=M

1

m2

≤
2k

2k(2α+1)
(
1

M2
+

∫ ∞
M

dm

m2
), by Cauchy’s intergal test

=
1

22kα
(
1

M2
+

1

M
)

≤
2

22kαM

Then E2k,M(f) = min||f − (S2k,Mf)||2 ⩽
√
2

2kα
√
M

= O(
1

2kα
√
M

)



Approximation and moduli of continuity...

Thus, theorem 4.2 is completely established.

5 Moduli of continuity
The moduli of continuity of

(
f − (S2k,Mf)

)
have been determined in this

section as follows :

5.1 Theorem
If the solution function f of Lane-Emden differential equation satisfies eqns

(14), (15) & (16), then moduli of continuity of
(
f − (S2k,Mf)

)
is given by

W

((
f − (S2k,Mf)

)
,
1

2k

)
= sup

0≤h≤ 1
2k

||
(
f − (S2k,Mf))(· + h) −

(
f − (S2k,Mf))(·)||2

= O

(
1

2kα
√
M

)
Proof of theorem (5.1)
Following the proof of theorem (4.2) ,

||f − (S2k,Mf)||2 = O
(

1
2kα

√
M

)
.

Then

W

((
f − (S2k,Mf)

)
,
1

2k

)
= sup

0≤h≤ 1
2k

||
(
f − (S2k,Mf))(t + h) −

(
f − (S2k,Mf))(t)||2

≤ ||
(
f − (S2k,Mf))||2 + ||

(
f − (S2k,Mf))||2

= 2||
(
f − (S2k,Mf))||2

= 2.O

(
1

2kα
√
M

)
.

W

((
f − (S2k,Mf)

)
,
1

2k

)
= O

(
1

2kα
√
M

)

6 Boubaker wavelet method for solution of
differential equations

In this Section, the solution of Lane-Emden differential equations are obtained
by applying Boubaker wavelet collocation method.



Shyam Lal and Swatantra Yadav

Consider the Lane-Emden differential of the form

f
′′
(t) +

α

t
f

′
(t) + f(t) = h(t), where t ∈[0,1) ( Wazwaz)[7]). (19)

f(0) = a,f
′
(0) = b (20)

The solution of any differential equation can be expanded as Boubaker wavelet
series as follows

f(t) =
∞∑
n=0

∞∑
m=0

cn,mψ
(B)
n,m(t)

Now f(t) is approximated by truncated series

(S2k,Mf)(t) =
2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t) (21)

then the following residual is obtained by substituting (S2k,Mf) from eqn(21) into
eqn(19)

R(t) = t
2k−1∑
n=0

M−1∑
m=0

cn,mψ
′′(B)
n,m (t) + α

2k−1∑
n=0

M−1∑
m=0

cn,mψ
′(B)
n,m (t) + t

2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(t)

− t × h(t).

The collocation method yields

R(ti) = 0, i = 1,2,3, ...,2
kM − 2.

Moreover using the intial conditions eqn (20),

2k−1∑
n=0

M−1∑
m=0

cn,mψ
(B)
n,m(0) = a,

2k−1∑
n=0

M−1∑
m=0

cn,mψ
′(B)
n,m (0) = b. (22)

Hence 2kM system of equations are derived in the unknown coefficients cn,m
which can be computed. This procedure is applied for differential equations of
heigher order.

7 Illustrated Examples
In this Section, three Lane-Emden differential equations have been solved by

using the procedure discussed in previous section-6. Illustrated examples are as
follows:



Approximation and moduli of continuity...

Example (1)

Consider the following Lane-Emden differential equation

f
′′
(t) +

2

t
f

′
(t) + f(t) = 1 + 12t + t3, f(0) = 1, f

′
(0) = 0, 0 ≤ t < 1 (23)

The exact solution of eqn (23) is f(t) = t3 + 1.
Now the differential equation has been solved by applying the procedure described
in Section-6, using Boubaker wavelet method by taking M = 5,k = 0. Consider

f(t) =
4∑

m=0

c0,mψ
(B)
0,m(t)

= c0,0ψ
(B)
0,0 + c0,1ψ

(B)
0,1 + c0,2ψ

(B)
0,2 c0,3ψ

(B)
0,3 + c0,4ψ

(B)
0,4 (24)

f(t) = c0,0 + c0,1
√
3(2t − 1) + c0,2

√
5(6t2 − 6t + 1)

+ c0,3
√
7(20t3 − 30t2 + 12t − 1)

+ c0,43(70t
4 − 140t3 + 90t2 − 20t + 1) (25)

Differentiate eqn (25) with respect to t,

f
′
(t) = c0,1(2

√
3) + c0,2

√
5(12t − 6) + c0,3

√
7(60t2 − 60t + 12)

+ c0,43(280t
3 − 420t3 + 180t − 20) (26)

f
′′
(t) = c0,2(12

√
5) + c0,3

√
7(120t − 60) + c0,43(840t2 − 840t + 180)

Substitute these values of f(t),f
′
(t) and f

′′
(t) in given differential eqn (23)

c0,0 + c0,1
[4√3
t

+
√
3(2t − 1)

]
+ c0,2

[
12
√
5 +

2
√
5

t
(12t − 6)

+
√
5(6t2 − 6t + 1)

]
+ c0,3

[√
7(120t − 60) +

2
√
7

t
(60t2 − 60t + 12)

+
√
7(20t3 − 30t2 + 12t − 1)

]
+ c0,4

[
3(840t2 − 840t + 180)

+
6

t
(280t3 − 420t2 + 180t − 20) + 3(70t4 − 140t3 + 90t2 − 20t) + 1

]
= 1 + 12t + t3 (27)

Using intial condition, f(0) = 1 and f
′
(0) = 0 in eqns (25) and (26)

c0,0 −
√
3c0,1 +

√
5c0,2 −

√
7c0,3 + 3c0,4 = 1 (28)

2
√
3c0,1 − 6

√
5c0,2 + 12

√
7c0,3 − 60c0,4 = 0 (29)



Shyam Lal and Swatantra Yadav

Now collocate the eqn (27) at t1 = 0.5,t2 = 0.7 and t3 = 0.9, which are obtained
by xi =

i− 1
2

2kM
=

i− 1
2

5
,i = 2,4,5 respectively.

A system of three linear equations are derived.

c0,0 + 13.8564c0,1 + 25.7147c0,2 − 31.7490c0,3 − 88.875c0,4 = 7.125 (30)

c0,0 + 10.5902c0,1 + 41.5844c0,2 + 57.79832c0,3 − 21.7675c0,4 = 9.743 (31)

c0,0 + 9.0836c0,1 + 51.7127c0,2 + 166.0120c0,3 + 351.9676c0,4 = 12.529 (32)

Solving these eqns (30),(31) and (32) with (28) and (29)
c0,0 = 1.2499999999, c0,1 = 0.2598076211, c0,2 = 0.1118033988
c0,3 = 0.0188982236, c0,4 = −0.0000000000.
Substitute all these values of c0,0,c0,1,c0,2,c0,3,c0,4 in eqn (25)

f(t) = 1.2499999999 + 0.2598076211
√
3(2t − 1)

+ 0.1118033988
√
5(6t2 − 6t + 1)

+ 0.0188982236
√
7(20t3 − 30t2 + 12t − 1)

− 0.0000000000(70t4 − 140t3 + 90t2 − 20t + 1) (33)

Comparison of exact and Boubaker wavelet solutions are given in table (1)
for k = 0,M = 5 .

Table (1)
t Exact solution Approximate solution Absolute error(×10−15)

0.1 1.001000000000000 1.001000000000000 0
0.2 1.008000000000000 1.0080000000000000 0
0.3 1.027000000000000 1.0270000000000000 0
0.4 1.064000000000000 1.0640000000000000 0
0.5 1.125000000000000 1.1250000000000000 0
0.6 1.216000000000000 1.2160000000000000 0
0.7 1.343000000000000 1.3430000000000000 0
0.8 1.512000000000000 1.5120000000000000 0
0.9 1.729000000000000 1.7290000000000000 0.222044604925031

Table(1):Comparison table of exact and Boubaker wavelet solutions.



Approximation and moduli of continuity...

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

t

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8
f(

t)
Exact solution

Boubaker wavelet solution

Fig.(1):The graphs of Boubaker wavelet and Exact solutions.

Example (2)

Consider the following Lane-Emden differential equation

f
′′
(t)+

2

t
f

′
(t)+f(t) = 6+12t+t2 +t3, y(0) = 0, y

′
(0) = 0, 0 ≤ t < 1 (34)

The exact solution of eqn (34) is f(t) = t3 + t2.
Following the procedure adopted in Example (1);

f(t) =
4∑

m=0

c0,mψ
(B)
0,m(t)

= c0,0ψ
(B)
0,0 + c0,1ψ

(B)
0,1 + c0,2ψ

(B)
0,2 + c0,3ψ

(B)
0,3 + c0,4ψ

(B)
0,4

f(t) = c0,0 + c0,1
√
3(2t − 1) + c0,2

√
5(6t2 − 6t + 1)

+ c0,3
√
7(20t3 − 30t212t − 1)

+ c0,43(70t
4 − 140t3 + 90t2 − 20t + 1) (35)

f
′
(t) = c0,1(2

√
3) + c0,2

√
5(12t − 6) + c0,3

√
7(60t2 − 60t + 12)

+ c0,43(280t
3 − 420t3 + 180t − 20) (36)

f
′′
(t) = c0,2(12

√
5) + c0,3

√
7(120t − 60) + c0,43(840t2 − 840t + 180)



Shyam Lal and Swatantra Yadav

Substitute these values of f(t),f
′
(t) and f

′′
(t) in given differential eqn (34)

c0,0 + c0,1
[4√3
t

+
√
3(2t − 1)

]
+ c0,2

[
12
√
5 +

2
√
5

t
(12t − 6)

√
5(6t2 − 6t + 1)

]
+ c0,3

[√
7(120t − 60) +

2
√
7

t
(60t2 − 60t + 12)

√
7(20t3 − 30t2 + 12t − 1)

]
+ c0,4

[
3(840t2 − 840t + 180)

+
6

t
(280t3 − 420t2 + 180t − 20) + 3(70t4 − 140t3 + 90t2 − 20t) + 1

]
= 6 + 12t + t2 + t3 (37)

Using intial condition, f(0) = 0 and f
′
(0) = 0 in eqns (35) and (36)

c0,0 −
√
3c0,1 +

√
5c0,2 −

√
7c0,3 + 3c0,4 = 0 (38)

2
√
3c0,1 − 6

√
5c0,2 + 12

√
7c0,3 − 60c0,4 = 0 (39)

Now collocating the equations (37) at t1 = 0.5,t2 = 0.7 and t3 = 0.9 ,

c0,0 + 13.856405c0,1 + 25.71478c0,2 − 31.74901c0,3 − 88.875c0,4 = 12.375 (40)

c0,0 + 10.59025c0,1 + 41.58447c0,2 + 57.79832c0,3 − 21.76757c0,4 = 15.233 (41)
c0,0 + 9.08364c0,1 + 51.71279c0,2 + 166.01207c0,3 + 351.96766c0,4 = 18.339 (42)

Solving these eqns (40),(41) and (42) with (38) and (39) ,
c0,0 = 0.5833333333, c0,1 = 0.5484827557, c0,2 = 0.1863389981
c0,3 = 0.0188982236, c0,4 = −0.0000000000.
Substitute all these values of c0,0,c0,1,c0,2,c0,3,c0,4 in eqn (35).

f(t) = 0.5833333333 + 0.5484827557
√
3(2t − 1)

+ 0.1863389981
√
5(6t2 − 6t + 1)

+ 0.0188982236
√
7(20t3 − 30t2 + 12t − 1)

− 0.0000000000(70t4 − 140t3 + 90t2 − 20t + 1) (43)

Comparison of exact and Boubaker wavelet solutions are given in table (2)
for k = 0,M = 5 .

Table (2)
t Exact solution Approximate solution Absolute error (× 10−15)

0.1 0.011000000000000 0.011000000000000 0.017347234759768
0.2 0.048000000000000 0.048000000000000 0.020816681711722
0.3 0.117000000000000 0.117000000000000 0.027755575615629
0.4 0.224000000000000 0.224000000000000 0
0.5 0.375000000000000 0.375000000000000 0
0.6 0.576000000000000 0.576000000000000 0.222044604925031
0.7 0.833000000000000 0.833000000000000 0.111022302462516
0.8 1.152000000000000 1.152000000000000 -0.222044604925031
0.9 1.539000000000000 1.539000000000000 -0.222044604925031



Approximation and moduli of continuity...

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

t

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

f(
t)

Exact solution

Boubaker wavelet solution

Fig.(2):The graphs of Boubaker wavelet and Exact solutions.

Example 3

Consider the following Lane-Emden differential equation

f
′′
(t) +

2

t
f

′
(t) = 2(2t2 + 3)f(t), f(0) = 1, f

′
(0) = 0, 0 ≤ t < 1 (44)

The exact solution of eqn (44) is f(t) = et
2
.

Following the procedure adopted in Example(1);

f(t) =
4∑

m=0

c0,mψ
(B)
0,m(t)

= c0,0ψ
(B)
0,0 + c0,1ψ

(B)
0,1 + c0,2ψ

(B)
0,2 + c0,3ψ

(B)
0,3 + c0,4ψ

(B)
0,4

f(t) = c0,0 + c0,1
√
3(2t − 1) + c0,2

√
5(6t2 − 6t + 1)

+ c0,3
√
7(20t3 − 30t2 + 12t − 1)

+ c0,43(70t
4 − 140t3 + 90t2 − 20t + 1) (45)

f
′
(t) = c0,1(2

√
3) + c0,2

√
5(12t − 6) + c0,3

√
7(60t2 − 60t + 12)

+ c0,43(280t
3 − 420t3 + 180t − 20) (46)

f
′′
(t) = c0,2(12

√
5) + c0,3

√
7(120t − 60) + c0,43(840t2 − 840t + 180)



Shyam Lal and Swatantra Yadav

Substitute these values of f(t),f
′
(t) and f

′′
(t) in given differential eqn (44)

− 2(2t2 + 3)c0,0 + c0,1
[4√3
t

− 2(2t2 + 3)
√
3(2t − 1)

]
+ c0,2

[
12
√
5 +

2
√
5

t
(12t − 6) − 2(2t2 + 3)

√
5(6t2 − 6t + 1)

]
+ c0,3

[√
7(120t − 60) +

2
√
7

t
(60t2 − 60t + 12)

+ 2(2t2 + 3)
√
7(20t3 − 30t2 + 12t − 1)

]
+ c0,4

[
3(840t2 − 840t + 180) +

6

t
(280t3 − 420t2 + 180t − 20)

− 6(2t2 + 3)(70t4 − 140t3 + 90t2 − 20t) + 1
]
= 0 (47)

Using intial condition, f(0) = 1 and f
′
(0) = 0 in eqns (45) and (46)

c0,0 −
√
3c0,1 +

√
5c0,2 −

√
7c0,3 + 3c0,4 = 1 (48)

2
√
3c0,1 − 6

√
5c0,2 + 12

√
7c0,3 − 60c0,4 = 0 (49)

Now collocating the equation (47) at t1 = 0.5,t2 = 0.7 and t3 = 0.9,

−7c0,0 + 13.85640c0,1 + 34.65905c0,2 − 31.74901c0,3 − 97.8750c0,4 = 0 (50)

−7.96c0,0 + 4.38258c0,1 + 46.79361c0,2 + 68.22893c0,3 − 18.73013c0,4 = 0 (51)
−9.24c0,0−5.10531c0,1+41.18002c0,2+163.84467+02c0,3+359.12542c0,4 = 0 (52)

Solving these eqns (50), (51) and (52) with (48) and (49),
c0,0 = 1.38821917722, c0,1 = 0.388372764863, c0,2 = 0.158389887721
c0,3 = 0.02903271870, c0,4 = 0.002368327161
Substitute all these values of c0,0,c0,1,c0,2,c0,3,c0,4 in eqn (45)

f(t) = 1.3882191 + 0.38837276
√
3(2t − 1) + 0.1583898

√
5(6t2 − 6t + 1)

+ 0.02903271870
√
7(20t3 − 30t2 + 12t − 1)

+ 0.002368327161(70t4 − 140t3 + 90t2 − 20t + 1) (53)

Comparison of exact and Boubaker wavelet solutions are given in table (3)
for k = 0,M = 5

Table (3)
t Exact solution Approximate solution Absolute error

0.1 1.010050167084168 1.005192015148452 0.004858151935716
0.2 1.040810774192388 1.023531157694065 0.017279616498324
0.3 1.094174283705210 1.060057300954230 0.034116982750980
0.4 1.173510870991810 1.121003955135684 0.052506915856126
0.5 1.284025416687741 1.213798267334502 0.070227149353240
0.6 1.433329414560340 1.347061021536102 0.086268393024238
0.7 1.632316219955379 1.530606638615246 0.101709581340133
0.8 1.896480879304952 1.775443176336035 0.121037702968916
0.9 2.247907986676472 2.093772329351916 0.154135657324556



Approximation and moduli of continuity...

Table(3):Comparision table of exact and Boubaker wavelet solutions.
Comparison of exact and Boubaker wavelet solutions are given in table (4)
for k = 0,M = 6 .

Table (4)
t Exact solution Approximate solution Absolute error

0.1 1.010050167084168 1.000547638755526 0.009502528328642
0.2 1.040810774192388 1.011676778238292 0.029133995954096
0.3 1.094174283705210 1.043773805588921 0.050400478116290
0.4 1.173510870991810 1.103730511655493 0.069780359336317
0.5 1.284025416687741 1.196980758433975 0.087044658253766
0.6 1.433329414560340 1.329537146508637 0.103792268051703
0.7 1.632316219955379 1.510027682492480 0.122288537462899
0.8 1.896480879304952 1.751732446467655 0.144748432837297
0.9 2.247907986676472 2.074620259425891 0.173287727250581

Table(4):Comparison table of exact and Boubaker wavelet solutions.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

t

1

1.2

1.4

1.6

1.8

2

2.2

2.4

f(
t)

Exact solution

Boubaker wavelet solution for k=0, M=5

Boubaker wavelet solution for k=0, M=6

Fig.(3):The graphs of Boubaker wavelet and Exact solutions.

8 Discussion and Conclusion
1. Boubaker wavelets have applications in approximation theory, moduli of

continuity and solution of Lane-Emden differential equations. The Boubaker wavelet
expansion of solution of Lane-Emden Differential equation is verified and its



Shyam Lal and Swatantra Yadav

convergence analysis has been studied. The estimator E2k,M(f) of
(
f − S2k,M(f)

)
has

been developed. Furthermore, moduli of continuity

W
(
(f − (S2k,Mf)),

1

2k
)

≤ 2E2k,M(f)

This shows that moduli of continuity W
((
f − (S2k,Mf)

)
, 1
2k

)
is sharper and better

than the approximation E2k,M(f) of
(
f − S2k,M(f)

)
. Hence the moduli of continuity

W

((
f − (S2k,Mf)

)
, 1
2k

)
has been also estimated in this research paper.

2. A method has been proposed to solve Lane-Emden differential equation by Boubaker
wavelet collocation method. To illustrate the effectiveness and accuracy of the proposed
method, three Lane-Emden differential equations have been solved by proposed method,
It is observed that the exact solutions of considered differential equations are atmost same
to their solutions obtained by proposed method. This is a significant achivement of the
research paper in wavelet analysis.
3. Our results are concerned with Boubaker wavelet estimator E2k,M(f), moduli of

continuity W
((
f − (S2k,Mf)

)
, 1
2k

)
and the solutions of Lane-Emden diffential

equations by this method.
4. (i) By theorem 4.2,

E2k,M(f) = O

(
1

2kα
√
M

)
→ 0 as k → ∞,M → ∞.

.
(ii) As per theorem 5.1,

W

((
f − S2k,M(f)

)
,
1

2k

)
= O

(
1

2kα
√
M

)
→ 0 as k → ∞,M → ∞.

Thus E2k,M(f) and W
((
f − S2k,M(f)

)
, 1
2k

)
are best possible estimation in wavelet

analysis
5. Solution of Lane-Emden differential equation by Boubaker wavelet series by
collocation method is approximately same as exact solution of Lane-Emden differential
equation. Only a few number of Boubaker wavelet basis is needed to achieve the heigh
accuracy. This is significant achivement in wavelet analysis.
6. Limitations and possible future development:
(i) A non-linear Lane-Emden equation can not solved by Boubaker wavelets without
using collocation method
(ii) In general, Boubaker wavelets in one variable are ineffective to solve a problem
expressed in partial differential equations of two or more variables.
(iii) It is known that Hα[0,1) ⊈ Hα2 [0,1).To find the approximate solution of
Lane-Emden differential equation in class Hα2 [0,1).



Approximation and moduli of continuity...

(iv) To define two dimensional Boubaker wavelets and to find the solution of the partial
differential equation by this method.

Acknowledgments Shyam Lal, one of the authors, is thankful to DST - CIMS for
encouragement to this work.
Swatantra Yadav, one of the authors, is grateful to U.G.C (India) for providing financial
assistance in the form of Junior Research Fellowship vide NTA Ref. No:211610150821
Dated:24-03-2022 for his research work.
Authors are greatful to the referee for his valuable comments and suggestions, to improve
the quality of the research paper.

References
[1] C.K.Chui, Wavelets: A mathematical tool for signal analysis, SIAM, Philadel-

phia PA,(1997).

[2] Devore,R.A.: Nonlinear approximation, Acta Numerica, Vol.7, pp. 51-150,
Cambridge University Press, Cambridge (1998).

[3] Debnath, L.: Wavelet Transform and Their Applications, Birkhauser Bostoon,
Massachusetts (2002).

[4] Meyer,Y.: Wavelets; their past and their future, Progress in Wavelet Analaysis
and (Applications) (Toulouse, 1992) (Meyer,Y.and Roques,S. eds) Frontieres,
Gif-sur-Yvette,pp. 9-18 (1998)

[5] Morlet,J.; Arens,G; Fourgeau,E.; and Giard,D.: Wave propagation and sam-
pling theory, part II. Sampling theory and complex waves, Geophysics 47(2),
222-236 (1982)

[6] Daubechies,I.: Ten Lectures on Wavelets, SIAM, Philadelphia, PA (1992).

[7] Wazwaz, A.-M.: A new algorithm for solving differential equations of
Lane–Emden type. Applied Mathematics and Computation, 118(2), 287–310
(2001).

[8] Shiralashetti, Siddu & Lamani, Lata. (2020). Boubaker wavelet based numeri-
cal method for the solution of Abel’S integral equations. 28. 114-124.

[9] Das, G.; Ghosh, T.; Ray, B.K.: Degree of approximation of functions by their
Fourier series in the generalized H¨older metric. Proc. Indian Acad. Sci. Math.
Sci. Vol.106, no.2, pp. 139–153 (1996)

[10] Zygmund A.: Trigonometric Series, vol.I. Cambridge University
Press,Cambridge (1959).



Shyam Lal and Swatantra Yadav

[11] Lal, Shyam, & Satish Kumar. ”CAS wavelet approximation of functions of
Holder class and solutions of Fredholm integral equation.” Ratio Mathematica
[Online], 39 (2020): 187-212.