Home Technology Haar wavelet collocation method for existence and numerical solutions of fourth-order integro-differential equations with bounded coefficients
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Haar wavelet collocation method for existence and numerical solutions of fourth-order integro-differential equations with bounded coefficients

  • Rohul Amin EMAIL logo , Muhammad Nawaz , Kamal Shah and Thabet Abdeljawad
Published/Copyright: April 25, 2025
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Abstract

In this article, Haar wavelet collocation method is applied for the solution of fourth-order integro-differential equations. Also, a fixed point approach is used to investigate the existence theory of solution to the considered problem. The fourth-order derivative is approximated using Haar function. In addition, third-, second-, and first-order derivatives together with unknown functions are obtained by the process of successive integrations. On applying the Haar collocation method, the suggested problem of IDEs is transformed to a system of algebraic equations. The Gauss elimination scheme is used for the solution of linear algebraic equations. The precision, effectiveness, and convergence of the Haar approach are checked on some test problems. Different collocation and Gauss points are used to determine the absolute and root mean square errors. To demonstrate the applicability of the proposed method, an experimental rate of convergence is calculated, which is almost equal to 2. The method is accurate, easily applicable, and efficient.

1 Introduction

Functional equations, such as partial differential equations (PDEs), integral and integro-differential equations (IDEs), stochastic equations, and others, are typically the outcome of mathematical modelling of real-world issues. IDEs are found in many mathematical formulations of physical processes and are used in chemical kinetics, fluid dynamics, and biological models (for more information, see [1]). IDEs are frequently encountered in problems involving electrostatics, low-frequency electromagnetism, electromagnetism scattering, and the propagation of elastic and acoustic waves. Some applications in scientific disciplines, including microscopy, seismology, radio astronomy, electron emission, atomic scattering, radar range, plasma diagnostics, X-ray radiography, and optical fibre evaluation, when formulated, we obtain in the form of IDEs [2]. Numerous topics are covered in the said area, for instance, the Volterra population growth model, coexisting biological species, the spread of stocked fish in a new lake, heat radiation, and heat transfer. Usually, it is difficult to obtain the exact or analytical solution for every IDEs; therefore, researchers have always tried to establish sophisticated numerical tools to study such problems for their approximate solutions.

Many researchers worked to compute the approximate solution of fourth-order IDEs. Sweilam [3] has used a variational iteration technique (VIT) for the solution of IDEs of order four. Linear and nonlinear fourth-order IDEs have been solved using VIT with boundary conditions. Sweilam et al. [4] have developed a scheme for the solution of fourth-order IDEs based on the pseudo-spectral procedure. Lakestani and Deghan [5] used Chebyshev functions for approximate solutions of fourth-order IDEs. This technique involves expanding the desired solution as Chebyshev cardinal functions. They convert the IDEs to a system of algebraic equations by using an operational matrix. Ghomanjani et al. [6] used the Bezier curve technique to find an approximate solution of fourth-order IDEs. Based on the Adomian decomposition technique, Singh and Wazwaz [7] developed a numerical scheme to address fourth-order boundary value problems (BVPs) of Volterra IDEs. For a recursive method of solution, they applied the Green functions to transform IDEs into integral equations.

The Haar wavelet collocation (HWC) technique is used for different problems in the literature. Some applications of the HWC technique for Lane–Emden equations [8], nonlinear delay IDEs [9], nonlinear delay integral equations [10], software piracy [11], fractional delay differential equations (DEs) [12,13], fractional IDEs [14,15], third-order boundary-value problems of IDEs [16], system of fractional DEs [17], Burger’s equations [18], Schrodinger’s equations [19], inverse problems [20], and interface problems [21]. Majak et al. [22] introduced a higher order Haar wavelet method. Additionally, this technique is used for DEs [23], vibration analysis of beams [24,25], integral equations [26], IDEs [17], nonlinear PDEs [27], and nonlinear evolution equations [28].

Wu et al. [29] used a homotopy-based stochastic finite-element model updating method to cope with the correlation of static measurement data. Cai et al. [30] studies dynamically controlling terahertz wavefronts with cascaded metasurfaces. Yang and Kai [31] studied the nonlinear coupled Schrödinger equation in fibre Bragg gratings. Zhou et al. [32] developed an iterative threshold algorithm for the sparse optimization problems with a log-sum function. Jiang et al. [33] modeled the uncertainty mechanism using an inclusiveness function for each agent to capture the acceptance degree of opposing opinions. The main advantages of this method are its simplicity and less computation costs: it is due to the sparsity of the transform matrices and the small number of significant wavelet coefficients. The Haar method stands out due to its computational efficiency, direct approximation of solutions, ability to handle discontinuities, and flexibility, especially in solving nonlinear IDEs. Its advantages over traditional methods like Laplace, Sumudu, and others make it a powerful tool in numerical analysis, particularly when solving problems require high accuracy with fewer computational resources [34].

Existence theory is an important consequence of applied analysis. The said analysis is used to investigate whether a particular problem that is under investigation has a solution or not. For this purpose, an important theory was founded by Banach in 1922. After him, many theories and results were established to study the existence theory of solutions for integral, functional, algebraic, and DEs. Previous studies [3537] have valuable contributions to the theory of existence of solutions. This theory tells us whether the problem has a solution or not. If yes, how many solutions are there for a particular problem? Therefore, fixed point theory and nonlinear analysis tools like the degree theories of Mawhin and Schauder have gained much attention in the last few decades. We refer to some work like [3840]. Here, we remark that the existence theory of IDEs has also been considered recently by many researchers in their work. For instance, Shatanawii et al. [41] have studied the existence of solutions to some integral problems. In the same way, Shatanawii et al. [42] and Isik et al. [43] established the existence theory for some IDE problems.

In this study, fourth-order linear Volterra–Fredholm IDEs are solved using the HWC technique. Consider the subsequent fourth-order linear IDE:

(1) ξ ( 4 ) ( t ) = a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) + μ 1 a t k 1 ( t , s ) ξ ( s ) d s + μ 2 a b k 2 ( t , s ) ξ ( s ) d s + f ( t ) , t [ 0 , 1 ] ,

subject to

ξ ( 3 ) ( 0 ) = λ 4 , ξ ( 2 ) ( 0 ) = λ 3 , ξ ( 1 ) ( 0 ) = λ 2 and ξ ( 0 ) = λ 1 ,

where a 1 , a 2 , a 3 , a 4 , f are known functions, ξ is an unknown function, k 1 , k 2 : [ 0 , 1 ] × [ 0 , 1 ] R are defined, t and s functions called as the integral’s kernel, μ 1 , μ 2 , λ 1 , λ 2 , λ 3 , and λ 4 are any constant real numbers. In addition, the following fourth-order nonlinear Volterra–Fredholm IDEs are solved using the HWC technique as

(2) ξ ( 4 ) ( t ) = a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) + μ 1 a t k 1 ( t , s ) F ( s , ξ ( s ) ) d s + μ 2 a b k 2 ( t , s ) F ( s , ξ ( s ) ) d s + f ( t ) , t [ 0 , 1 ] ,

subject to

ξ ( 3 ) ( 0 ) = λ 4 , ξ ( 2 ) ( 0 ) = λ 3 , ξ ( 1 ) ( 0 ) = λ 2 and ξ ( 0 ) = λ 1 ,

here F : [ 0 , 1 ] × R R is a nonlinear continuous function. Inspired by the importance of fixed point theory, here we establish a useful result based on Banach fixed point to verify the uniqueness and existence of solutions to the proposed problem.

The structure of this article is as follows. Section 2 addresses Haar functions, and Section 3 presents existence and uniqueness results. The numerical scheme for solving nonlinear fourth-order IDEs is provided in Section 4. In Section 5, convergence is provided. Examples from the literature are provided in Section 6. Section 7 presents the results and comments. Section 8 provides the conclusion.

2 Haar wavelet

The scaling function is [9]

(3) h 1 ( t ) = 1 if t [ 0 , 1 ) , 0 elsewhere .

The Mother wavelet is given by

(4) h 2 ( t ) = 1 if t 0 , 1 2 , 1 if t 1 2 , 1 , 0 elsewhere .

The dilation and translation operations provide h 2 ( t ) , from which all functions in this family are formed. These functions are defined on subintervals of [0, 1). The scaling function can be stated as follows, but all other functions are specified for t [ a , b )

(5) h i ( t ) = 1 if t [ α , β ) , 1 if t [ β , γ ) , 0 otherwise .

where α = k m , β = k + 1 2 m , γ = k + 1 m , for i = 2 , 3 , , 2 M = N , m = 2 j , where k = 0 , 1 , 2 , , m 1 and j = 0 , 1 , 2 , , J , J = 2 M . Translation parameter is K , while the wavelet’s level is indicated by the number j , integer J represents the highest resolution level. 1 + k + m = i is relation between i , m , and k . Expression for every square integrable f ( t ) in [0, 1) is

(6) f ( t ) = i = 1 η i h i ( t ) ,

For purposes of approximation, this is truncated at N terms as

(7) f ( t ) = i = 1 N η i h i ( t ) ,

we use the symbol

(8) P i , 1 ( t ) = 0 t h i ( t ) d t ,

and the definition of h i is used to determine the value of the aforementioned integral, which is provided by

(9) P i , 1 ( t ) = t α if t [ α , β ) , γ t if t [ β , γ ) , 0 elsewhere .

Through integral simplification, we can obtain the values of P i , 2 ( t ) , P i , 3 ( t ) , and P i , 4 ( t ) .

The value of P i , n is

(10) P i , n ( t ) = 0 for t [ 0 , α ) , 1 n ! ( t α ) n for t [ α , β ) , 1 n ! [ ( t α ) n 2 ( α β ) n ] for t [ β , γ ) , 1 n ! [ ( t α ) n 2 ( α β ) n + ( t γ 3 ) n ] , for t [ γ , 1 ) , n = 1 , 2 , 3 ,

If i = 1 , then

P 1 , n ( t ) = t n n ! , n = 1 , 2 , .

The interval I is divided into subintervals as

(11) t j = a + ( b a ) j 0.5 N , j = 1 , 2 , 3 , , N .

The points t j ’s are known as collocation points (CPs). Gauss points (GPs) are defined as

(12) G j = j 1 2 + 3 3 6 h , G j + 1 = j 1 2 + 3 + 3 6 h , j = 1 , 2 , 3 , , N 1 .

3 Existence and uniqueness results

This section is related to existing results. Let denotes X = C 4 [0, 1] Banach space, with norm as

ξ = max t [ 0 , 1 ] { ξ ( t ) , ξ ( 1 ) ( t ) , ξ ( 2 ) ( t ) , ξ ( 3 ) ( t ) } .

We make some assumptions that will be used in the main result:

  1. Since a 1 , a 2 , a 3 , a 4 , and f are continuous, so there exist a * , b * , c * , d * , f * > 0 such that

    a 1 ( t ) a * , a 2 ( t ) b * , a 3 ( t ) c * , a 4 ( t ) d * , f ( t ) f * .

  2. k 1 , k 2 : [ 0 , 1 ] × [ 0 , 1 ] R are continuous and bounded, so k 1 * , k 2 * > 0 such that

    k 1 ( t , s ) k 1 * , k 2 ( t , s ) k 2 * .

  3. Since F is a continuous function and satisfies the Liptchitz condition, such that for any ξ , ν X , one has

    F ( t , ξ ( t ) ) F ( t , ν ( t ) ) L F ξ ν , L F > 0 .

Problem (1) can be written in equivalent integral form as

(13) ξ ( t ) = λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + 0 t ( t s ) 3 2 ( a 1 ( s ) ξ ( 3 ) ( s ) + a 2 ( s ) ξ ( 2 ) ( s ) + a 3 ( s ) ξ ( 1 ) ( s ) + a 4 ( s ) ξ ( s ) ) d s + μ 1 0 t ( t s ) 4 2 k 1 ( s , t ) ξ ( s ) d s + μ 2 0 t ( t s ) 3 2 0 1 k 2 ( y , t ) ξ ( y ) d y d s + 0 t ( t s ) 3 2 f ( s ) d s .

Further, we write (13) as

(14) ξ ( t ) = λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + I t 4 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 4 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 4 0 ( I 1 0 k 2 ( t , s ) ξ ( s ) ) + I t 4 0 f ( t ) ,

(15) ξ ( 1 ) ( t ) = λ 2 + λ 3 t + λ 4 t 2 2 + I 3 0 t ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I 3 0 t k 1 ( t , s ) ξ ( t ) + μ 2 I 3 0 t ( I 1 0 k 2 ( t , s ) ξ ( s ) ) + I t 3 0 f ( t ) ,

(16) ξ ( 2 ) ( t ) = λ 3 + λ 4 t + I t 2 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 2 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 2 0 ( I 1 0 k 2 ( t , s ) ξ ( s ) ) + I t 2 0 f ( t ) ,

(17) ξ ( 3 ) ( t ) = λ 4 + I t 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) . + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) = + μ 1 I t 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 0 ( I 1 0 k 2 ( t , s ) ξ ( s ) ) + I t 0 f ( t ) .

Theorem 3.1

Under hypotheses ( A 1 ) and ( A 2 ) , and if the condition Δ < 2 , where Δ is defined

(18) Δ = a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ,

then (1) has a unique solution.

Proof

Let T : X X , by using Eqs. (14), (15), (16), and (17), we have

(19) T ξ ( t ) = λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + I t 4 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 4 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 4 0 ( I 1 0 k 2 ( t , s ) ξ ( t ) ) + I t 4 0 f ( t ) ,

(20) T ξ ( 1 ) ( t ) = λ 2 + λ 3 t + λ 4 t 2 2 + 0 I t 3 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 3 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 3 0 ( I 1 0 k 2 ( t , s ) ξ ( t ) ) + I t 3 0 f ( t ) ,

(21) T ξ ( 2 ) ( t ) = λ 3 + λ 4 t + I t 2 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 2 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 2 0 ( I 1 0 k 2 ( t , s ) ξ ( t ) ) + I t 2 0 f ( t ) ,

(22) T ξ ( 3 ) ( t ) = λ 4 + I t 0 ( a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) ) + μ 1 I t 0 k 1 ( t , s ) ξ ( t ) + μ 2 I t 0 ( I 1 0 k 2 ( t , s ) ξ ( t ) ) + I t 0 f ( t ) .

Since T is continuous and bounded, it is uniformly continuous, which can be described further as follows by taking ξ , v X , then we have from Eq. (19)

T ξ T v I t 4 0 ( a * ξ ( 3 ) ( t ) v ( 3 ) ( t ) + b * ξ ( 2 ) ( t ) v ( 2 ) ( t ) + c * ξ ( 1 ) ( t ) v ( 1 ) ( t ) + d * ξ ( t ) v ( t ) ) + I t 4 0 ( μ 1 k 1 * ξ ( t ) v ( t ) + μ 2 I 1 0 k 2 * ξ ( t ) v ( t ) ) 1 4 ( a * + b * + c * + d * ) ξ v + 1 4 ( μ 1 k 1 * + μ 2 k 2 * ) × ξ v ,

which gives

(23) T ξ T v 1 4 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) .

Now, from Eq. (20), we have for any ξ , v X , as

T ξ ( 1 ) T v ( 1 ) I t 3 0 ( a * ξ ( 3 ) ( t ) v ( 3 ) ( t ) + b * ξ ( 2 ) ( t ) v ( 2 ) ( t ) + c * ξ ( 1 ) ( t ) v ( 1 ) ( t ) + d * ξ ( t ) v ( t ) ) + I t 3 0 ( μ 1 k 1 * ξ ( t ) v ( t ) + μ 2 I 1 0 k 2 * ξ ( t ) v ( t ) ) 1 6 ( a * + b * + c * + d * ) ξ v + 1 6 ( μ 1 k 1 * + μ 2 k 2 * ) ξ v ,

hence, we have

(24) T ξ ( 1 ) T v ( 1 ) 1 6 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) ,

also from Eq. (21), we have

(25) T ξ ( 2 ) T v ( 2 ) ( b a ) 2 3 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) ,

also from Eq. (22), we have

(26) T ξ ( 3 ) T v ( 3 ) 1 2 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) .

Now,

(27) T ξ T v = max t [ 0 , 1 ] { T ξ T v , T ξ ( 1 ) T v ( 1 ) , T ξ ( 2 ) T v ( 2 ) , T ξ ( 3 ) T v ( 3 ) } ,

and

max t [ 0 , 1 ] 1 24 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) , 1 6 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) , 1 3 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) , 1 2 ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) = ( a * + b * + c * + d * + μ 1 k 1 * + μ 2 k 2 * ) × max t [ 0 , 1 ] 1 24 , 1 6 , 1 3 , 1 2 = Δ 2 .

Thus, Eq. (27) implies that

(28) T ξ T v Δ 2 ξ v .

In view of this result, we can claim that Eq. (1) has a unique solution.□

The solution of nonlinear problem Eq. (2) can be written in equivalent integral form as

(29) ξ ( t ) = λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + 0 t ( t s ) 3 2 ( a 1 ( s ) ξ ( 3 ) ( s ) + a 2 ( s ) ξ ( 2 ) ( s ) + a 3 ( s ) ξ ( 1 ) ( s ) + a 4 ( s ) ξ ( s ) ) d s = + μ 1 0 t ( t s ) 4 2 k 1 ( s , t ) F ( s , ξ ( s ) ) d s + μ 2 0 t ( t s ) 3 2 0 1 k 2 ( y , t ) F ( y , ξ ( y ) ) d y d s + 0 t ( t s ) 3 2 f ( s ) d s .

Note: Since we see that max t [ 0 , 1 ] 1 24 , 1 6 , 1 3 , 1 2 = 1 2 , we state the given theorem.

Theorem 3.2

In view of assumptions ( A 1 A 3 ), the nonlinear problem Eq. (2) has a unique solution if the condition

( a * + b * + c * + d * + ( μ 1 k 1 * + μ 2 k 2 * ) L F ) 1 2 < 1

holds.

Proof

The proof is similar and can be followed by the same process as done in Theorem 3.1.□

4 Numerical method

Here, HWC technique is developed for the solution of fourth-order Volterra–Fredholm IDEs. The integration method yields expression for lower derivatives, whereas the Haar function approximates the fourth-order derivative. Additionally, formula (30) is used to obtain the integral involved in Eq. (1)

(30) α β ξ ( t ) d t k = 1 N ξ α + ( β α ) k 0.5 N .

We can obtain the algebraic equations system by placing the CPs and GPs in Eq. (1) and using the HWC technique to it. The unknown Haar coefficients are found using the Gauss elimination method. Ultimately, solution at the CPs is obtained by use of these coefficients. Consider ξ ( 4 ) ( t ) L 2 [ 0 , 1 ) , so ξ ( 4 ) ( t ) is written as

ξ ( 4 ) ( t ) = i = 1 N η i h i ( t ) .

Integrating from 0 to t , we obtain

ξ ( 3 ) ( t ) = ξ ( 3 ) ( 0 ) + i = 1 N η i p i , 1 ( t ) .

Using the initial condition, we have

(31) ξ ( 3 ) ( t ) = λ 4 + i = 1 N η i p i , 1 ( t ) .

Again integrating from 0 to t , we have

ξ ( 2 ) ( t ) = ξ ( 2 ) ( 0 ) + λ 4 t + i = 1 N η i p i , 1 ( t ) .

After establishing an initial condition, we have

(32) ξ ( 2 ) ( t ) = λ 3 + λ 4 t + i = 1 N η i p i , 2 ( t ) .

Again integrating from 0 to t , we have

ξ ( 1 ) ( t ) = ξ ( 1 ) ( 0 ) + λ 3 t + λ 4 t 2 2 + i = 1 N η i p i , 3 ( t ) .

Again using the initial condition, we have

(33) ξ ( 1 ) ( t ) = λ 2 + λ 3 t + λ 4 t 2 2 + i = 1 N η i p i , 3 ( t ) .

Integrating from 0 to t , we obtain

ξ ( t ) ξ ( 0 ) = λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + i = 1 N η i p i , 4 ( t ) .

Using the initial conditions and rearranging, we have

(34) ξ ( t ) = λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + i = 1 N η i p i , 4 ( t ) .

Eq. (34) is known as numerical solution.

4.1 Solution of Volterra–Fredholm fourth-order IDEs

4.1.1 Linear case

Numerical integration of Eq. (1) yields the following equation:

(35) ξ ( 4 ) ( t ) = a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) + μ 1 t N k = 1 N k 1 ( t , s k ) ξ ( s k ) + μ 2 1 N k = 1 N k 2 ( t , s k ) ξ ( s k ) + f ( t ) .

By putting values in Eq. (37), we have

i = 1 N η i h i ( t ) = a 1 ( t ) λ 4 + i = 1 N η i p i , 1 ( t ) + a 2 ( t ) λ 3 + λ 4 t + i = 1 N η i p i , 2 ( t ) + a 3 ( t ) λ 2 + λ 3 t + λ 4 t 2 2 + i = 1 N η i p i , 3 ( t ) + a 4 ( t ) λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + i = 1 N η i p i , 4 ( t ) + μ 1 t N k = 1 N k 1 ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) + μ 2 1 N k = 1 N k 2 ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) + f ( t ) .

After rearranging, we obtain

i = 1 N η i [ h i ( t ) a 1 ( t ) p i , 1 ( t ) a 2 ( t ) p i , 2 ( t ) a 3 ( t ) p i , 3 ( t ) a 4 ( t ) p i , 4 ( t ) μ 1 t N k = 1 N k 1 ( t , s k ) g ( s k ) p i , 4 ( s k ) μ 2 1 N k = 1 N k 2 ( t , s k ) g ( s k ) p i , 4 ( s k ) = a 1 ( t ) λ 4 + a 2 ( t ) ( λ 3 + λ 4 t ) + a 3 ( t ) λ 2 + λ 3 t + λ 4 t 2 2 + a 4 ( t ) λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + μ 1 t N k = 1 N k ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( t ) + μ 2 1 N k = 1 N k ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( t ) + f ( t ) .

By putting CPs t j , where j = 1 , , N , we have

(36) i = 1 N η i [ h i ( t j ) a 1 ( t j ) p i , 1 ( x j ) a 2 ( t j ) p i , 2 ( t j ) a 3 ( t j ) p i , 3 ( t j ) a 4 ( t j ) p i , 4 ( t j ) μ 1 t j a N k = 1 N × t k 1 ( t , s k ) g ( s k ) p i , 4 ( s k ) μ 2 1 N × k = 1 N k 2 ( t j , s k ) g ( s k ) p i , 4 ( s k ) = a 1 ( t j ) λ 4 + a 2 ( t j ) ( λ 3 + λ 4 t j ) + a 3 ( t j ) λ 2 + λ 3 t j + λ 4 t j 2 2 + a 4 ( t j ) λ 1 + λ 2 t j + λ 3 t j 2 2 + λ 4 t j 3 6 + μ 1 t j N k = 1 N k ( t j , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( t j ) + μ 2 1 N k = 1 N k ( t j , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( t j ) + f ( t ) .

The N × N linear system with unknown η i is represented by Eq. (36). The Gauss elimination method is used to solve this problem and determine the unknown Haar coefficients. By putting these coefficients in Eq. (34), we were able to derive an appropriate approximation solution at CPs.

4.1.2 Nonlinear case

Numerical integration of Eq. (2) yields the following equation:

(37) ξ ( 4 ) ( t ) = a 1 ( t ) ξ ( 3 ) ( t ) + a 2 ( t ) ξ ( 2 ) ( t ) + a 3 ( t ) ξ ( 1 ) ( t ) + a 4 ( t ) ξ ( t ) + μ 1 t N k = 1 N k 1 ( t , s k ) F ( s k , ξ ( s k ) ) + μ 2 1 N k = 1 N k 2 ( t , s k ) F ( s k , ξ ( s k ) ) + f ( t ) .

By putting values in Eq. (2), we have

i = 1 N η i h i ( t ) a 1 ( t ) λ 4 + i = 1 N η i p i , 1 ( t ) a 2 ( t ) λ 3 + λ 4 t + i = 1 N η i p i , 2 ( t ) a 3 ( t ) λ 2 + λ 3 t + λ 4 t 2 2 + i = 1 N η i p i , 3 ( t ) a 4 ( t ) λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 + i = 1 N η i p i , 4 ( t ) μ 1 t N k = 1 N k 1 ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) μ 2 1 N k = 1 N k 2 ( t , s k ) g ( s k ) × λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) f ( t ) = 0 .

After rearranging, we obtain

let G ( t ) = i = 1 N η i [ h i ( t ) a 1 ( t ) p i , 1 ( t ) a 2 ( t ) p i , 2 ( t ) a 3 ( t ) p i , 3 ( t ) a 4 ( t ) p i , 4 ( t ) μ 1 t N k = 1 N k ( t , s k ) g ( s k ) p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t , s k ) g ( s k ) p i , 4 ( s k )

a 1 ( t ) λ 4 a 2 ( t ) ( λ 3 + λ 4 t ) a 3 ( t ) λ 2 + λ 3 t + λ 4 t 2 2 a 4 ( t ) λ 1 + λ 2 t + λ 3 t 2 2 + λ 4 t 3 6 μ 1 t N k = 1 N k ( t , s k ) F × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t , s k ) F × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) f ( t ) .

By putting CPs t j , where j = 1 , , N , we have

(38) G ( t j ) = i = 1 N η i [ h i ( t j ) a 1 ( t j ) p i , 1 ( t j ) a 2 ( t j ) p i , 2 ( t j ) a 3 ( t j ) p i , 3 ( t j ) a 4 ( t j ) p i , 4 ( t j ) μ 1 t j N k = 1 N k ( t j , s k ) g ( s k ) p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t j , s k ) g ( s k ) p i , 4 ( s k ) a 1 ( t j ) λ 4 a 2 ( t j ) ( λ 3 + λ 4 t j ) a 3 ( t j ) λ 2 + λ 3 t j + λ 4 t j 2 2 a 4 ( t j ) × λ 1 + λ 2 t j + λ 3 t j 2 2 + λ 4 t j 3 6 μ 1 t j N k = 1 N k ( t j , s k ) F × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t j , s k ) F × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) f ( t j ) .

This is an N × N nonlinear system with unknown η i . Broyden’s method is used to solve this problem and determine the unknown Haar coefficients. By putting these coefficients in Eq. (34), we were able to derive an appropriate approximation solution at CPs. The Jacobian of (38) is calculated by partial derivatives which is then utilized for Broyden’s method to solve nonlinear equations

G ( t j ) η = h i ( t j ) a 1 ( t j ) p i , 1 ( t j ) a 2 ( t j ) p i , 2 ( t j ) a 3 ( t j ) p i , 3 ( t j ) a 4 ( t j ) p i , 4 ( t j ) μ 1 t j N k = 1 N k ( t j , s k ) × g ( s k ) p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t j , s k ) g ( s k ) p i , 4 ( s k ) μ 1 t j N k = 1 N k ( t j , s k ) F η × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) μ 2 1 N k = 1 N k ( t j , s k ) F η × s k , λ 1 + λ 2 s k + λ 3 s k 2 2 + λ 4 s k 3 6 + i = 1 N η i p i , 4 ( s k ) .

Remark 4.1

Fourth order linear Fredholm IDEs are represented by Eq. (1) if k 1 = 0 , and fourth-order linear Volterra IDEs are represented by Eq. (1) if k 2 = 0 . It is possible to develop equivalent numerical techniques for solving the Volterra and Fredholm IDEs.

5 Convergence

Theorem 5.1

Let w ( x ) = d n u ( x ) d x n L 2 ( R ) be continuous on (0, 1) and its first derivative is bounded. Then, HWC technique will be convergent, i.e. E m vanishes as J goes to infinity, the convergence is of order two [44]:

E m 2 = O 1 2 J + 1 2 ,

where

E m = f ( x ) f M ( x ) , f M ( x ) = i = 0 2 M a i h i ( x ) .

This section uses the HWC technique on certain cases to demonstrate the accuracy as well as efficacy of the suggested methods. The absolute error is represented by E 1 and E 2 , and is defined as: if ξ apc indicates the numerical solution and ξ exc indicates exact solution (the notation exc and apc is used for exact and approximate solution at CPs, while exg and apg are used for exact and approximate solution at GPs) and defined as

E 1 = max ξ exc ξ apc , E 2 = max ξ exg ξ apg .

At CPs and GPs, root mean square error is described as

E 3 = 1 N i = 1 N ξ exc ξ apc 2 , E 4 = 1 N i = 1 N ξ exg ξ apg 2 .

Rate of convergence at CPs and GPs is denoted by R 1 and R 2 respectively and, defined as [45]

R 1 = log [ ξ apc ( N 2 ) ξ apc ( N ) ] log 2 , R 2 = log [ ξ apg ( N 2 ) ξ apg ( N ) ] log 2 .

6 Numerical examples

Here, we enrich our article by providing some examples to elaborate the numerical scheme.

Example 6.1

Consider the fourth-order IDE:

(39) ξ ( 4 ) ( t ) = t ξ ( t ) + e t ξ ( 2 ) ( t ) + 0 1 t s ξ ( s ) d s t e t ( 1 + e t + t ) ,

with ICs

ξ ( 3 ) ( 0 ) = ξ ( 2 ) ( 0 ) = ξ ( 1 ) ( 0 ) = ξ ( 0 ) = 1 .

The exact solution is ξ ( t ) = e t .

Example 6.2

Consider the fourth-order linear FIDE:

(40) ξ ( 4 ) ( t ) = 4 e t ξ ( 1 ) ( t ) + ( t 2 1 ) ξ ( 3 ) ( t ) 0 1 ξ ( s ) d s 2 + e t ( 7 t ( 2 + 4 e t ( 1 + t ) + t ( 3 + t ) ) ) ,

with ICs

ξ ( 3 ) ( 0 ) = 3 , ξ ( 2 ) ( 0 ) = 2 , ξ ( 1 ) ( 0 ) = ξ ( 0 ) = 1 .

The exact solution is ξ ( t ) = 1 + t e t .

Example 6.3

Consider the linear Fredholm-Volterra IDE

(41) ξ ( 4 ) ( t ) = 0 1 t s ξ ( s ) d s + 0 t ( t + s ) ξ ( s ) d s 2 e t ( 1 + t ) + 1 6 ( 6 2 t 5 t 3 ) ,

ICs are ξ ( 3 ) ( 0 ) = ξ ( 2 ) ( 0 ) = ξ ( 0 ) = 1 , ξ ( 1 ) ( 0 ) = 2 , and the exact solution is ξ ( t ) = e t + t .

Example 6.4

Consider the fourth-order BVP IDE [7]:

(42) ξ ( 4 ) ( t ) = t + 5 e t 1 + 0 t ξ ( s ) d s ,

with boundary conditions

ξ ( 0 ) = ξ ( 1 ) ( 0 ) = 1 , ξ ( 1 ) = 1 + e , ξ ( 1 ) ( 1 ) = 2 e .

The exact solution is ξ ( t ) = 1 + t e t .

Example 6.5

Consider the nonlinear Fredholm IDE

(43) ξ ( 4 ) ( t ) ξ ( 3 ) ( t ) + ξ ( 2 ) ( t ) + ξ 2 ( t ) + 0 1 s d s = e 1 + sin ( 1 ) + e t sin ( t ) + ( cos ( t ) + e t ) 2 1 ,

with initial condition ξ ( 0 ) = 2 , ξ ( 1 ) ( 0 ) = 1 , ξ ( 2 ) ( 0 ) = 0 , and ξ ( 3 ) ( 0 ) = 1 . The exact solution is ξ ( t ) = e t + cos t .

Example 6.6

Consider the nonlinear Volterra IDE

(44) ξ ( 4 ) ( t ) ξ ( 3 ) ( t ) + ξ ( 2 ) ( t ) + ( ξ ( 1 ) ( t ) ) 2 + 0 t ξ ( s ) d s = 1 + 3 e t + sin ( t ) ,

with initial condition ξ ( 0 ) = 2 , ξ ( 1 ) ( 0 ) = 1 , ξ ( 2 ) ( 0 ) = 0 , and ξ ( 3 ) ( 0 ) = 1 . The exact solution is ξ ( t ) = e t + cos t .

Example 6.7

Consider the following nonlinear Volterra–Fredholm fourth-order IDE as

(45) ξ ( 4 ) ( t ) = t e t 2 + t 2 e t 2 + t 3 e t 3 + e t 4 + 0 t t s ξ 2 ( s ) d s + 0 1 ( t s ) ξ 2 ( s ) d s t 5 4 t 12 , t [ 0 , 1 ] .

The exact solution is given by ξ ( t ) = t 3 .

7 Results and discussion

The numerical results for linear Volterra, Fredholm, and Volterra–Fredholm fourth-order IDEs are reported in Tables 1, 2, 3, 4, , 6 and 7. The results of our proposed method are compared with [7], which is shown in Table 5. E 1 , E 2 , E 3 , and E 4 are decreased by taking more CPs and GPs. A quasi-Newton optimization technique for finding the roots of a system of nonlinear algebraic equations is the Broyden method. It is a well-liked technique for solving nonlinear problems and works especially well when there are several equations. The rate of convergence also approaches 2. The comparison of the approximate solution with the exact solution for the number of CPs N = 32 is shown in Figures 1, 2, 3, and 4. Comparison of approximate and exact solutions at GPs is also shown in Figures 5 and 6. Comparison of different errors is given in Figure 7. From the figures, it can easily be observed that the approximate solution is more close to the exact solution for the corresponding number of CPs or GPs (Tables 7 and 8).

Table 1

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.1

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 6.0423 × 1 0 4 7.8699 × 1 0 5 3.2217 × 1 0 4 4.1227 × 1 0 5
2 2 1.8686 × 1 0 4 1.6931 2.3867 × 1 0 5 1.7213 8.4550 × 1 0 5 1.0580 × 1 0 5
2 3 5.1591 × 1 0 5 1.8568 6.5099 × 1 0 6 1.8743 2.1390 × 1 0 5 2.6584 × 1 0 6
2 4 1.3532 × 1 0 5 1.9307 1.6966 × 1 0 6 1.9400 5.3634 × 1 0 6 6.6537 × 1 0 7
2 5 3.4639 × 1 0 6 1.9659 4.3288 × 1 0 7 1.9706 1.3418 × 1 0 6 1.6639 × 1 0 7
2 6 8.7617 × 1 0 7 1.9831 1.0932 × 1 0 7 1.9854 3.3552 × 1 0 7 4.1600 × 1 0 8
2 7 2.2032 × 1 0 7 1.9916 2.7466 × 1 0 8 1.9928 8.3885 × 1 0 8 1.0400 × 1 0 8
2 8 5.5241 × 1 0 8 1.9971 6.8838 × 1 0 9 1.9983 2.0971 × 1 0 8 2.6001 × 1 0 9
Table 2

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.2

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 3.6460 × 1 0 3 4.2994 × 1 0 4 1.9298 × 1 0 3 2.2473 × 1 0 4
2 2 1.1372 × 1 0 3 1.6808 1.2960 × 1 0 4 1.7301 5.0538 × 1 0 4 5.7063 × 1 0 5
2 3 3.1613 × 1 0 4 1.8469 3.5349 × 1 0 5 1.8743 1.2779 × 1 0 4 1.4296 × 1 0 5
2 4 8.3254 × 1 0 5 1.9249 9.2194 × 1 0 6 1.9389 3.2039 × 1 0 5 3.5757 × 1 0 6
2 5 2.1357 × 1 0 5 1.9628 2.3535 × 1 0 6 1.9699 8.0155 × 1 0 6 8.9401 × 1 0 7
2 6 5.4082 × 1 0 6 1.9815 5.9452 × 1 0 7 1.9850 2.0042 × 1 0 6 2.2351 × 1 0 7
2 7 1.3607 × 1 0 6 1.9908 1.4940 × 1 0 7 1.9925 5.0108 × 1 0 7 5.5877 × 1 0 8
2 8 3.4127 × 1 0 7 1.9975 3.7447 × 1 0 8 1.9978 1.2527 × 1 0 7 1.3969 × 1 0 8
Table 3

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.3

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 4.2182 × 1 0 3 2.5768 × 1 0 3 2.8761 × 1 0 3 1.6016 × 1 0 3
2 2 1.1023 × 1 0 3 1.9361 6.5929 × 1 0 4 1.9666 7.2185 × 1 0 4 3.9205 × 1 0 4
2 3 2.8069 × 1 0 4 1.9734 1.6623 × 1 0 4 1.9877 1.8064 × 1 0 4 9.7464 × 1 0 5
2 4 7.0755 × 1 0 5 1.9880 4.1703 × 1 0 5 1.9949 4.5171 × 1 0 5 2.4331 × 1 0 5
2 5 1.7757 × 1 0 5 1.9944 1.0442 × 1 0 5 1.9977 1.1293 × 1 0 5 6.0807 × 1 0 6
2 6 4.4478 × 1 0 6 1.9972 2.6124 × 1 0 6 1.9989 2.8234 × 1 0 6 1.5200 × 1 0 6
2 7 1.1130 × 1 0 6 1.9986 6.5333 × 1 0 7 1.9994 7.0585 × 1 0 7 3.8000 × 1 0 7
2 8 2.7837 × 1 0 7 1.9993 1.6336 × 1 0 7 1.9997 1.7646 × 1 0 7 9.500 × 1 0 8
2 9 6.9610 × 1 0 8 1.9996 4.0844 × 1 0 8 1.9998 4.4116 × 1 0 8 2.3749 × 1 0 8
Table 4

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.4

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 6.6610 × 1 0 4 8.1316 × 1 0 5 3.5331 × 1 0 4 4.2552 × 1 0 5
2 2 2.0704 × 1 0 4 1.6858 2.4517 × 1 0 5 1.7298 9.2425 × 1 0 5 1.0834 × 1 0 5
2 3 5.7458 × 1 0 5 1.8493 6.6820 × 1 0 6 1.8754 2.3364 × 1 0 5 2.7164 × 1 0 6
2 4 1.5120 × 1 0 5 1.9260 1.7418 × 1 0 6 1.9397 5.8573 × 1 0 6 6.7951 × 1 0 7
2 5 3.8774 × 1 0 6 1.9633 4.4450 × 1 0 7 1.9703 1.4653 × 1 0 6 1.6990 × 1 0 7
2 6 9.8169 × 1 0 7 1.9818 1.1227 × 1 0 7 1.9852 3.6640 × 1 0 7 4.2477 × 1 0 8
2 7 2.4698 × 1 0 7 1.9909 2.8210 × 1 0 8 1.9927 9.1603 × 1 0 8 1.0619 × 1 0 8
2 8 6.1939 × 1 0 8 1.9955 7.0704 × 1 0 9 1.9963 2.2901 × 1 0 8 2.6549 × 1 0 9
Table 5

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.7

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 6.5360 × 1 0 3 8.1172 × 1 0 4 3.4721 × 1 0 3 4.2482 × 1 0 4
2 2 2.0243 × 1 0 3 1.6910 2.4478 × 1 0 4 1.7295 9.0777 × 1 0 4 1.0820 × 1 0 4
2 3 5.6038 × 1 0 4 1.7529 6.6710 × 1 0 5 1.8755 2.2944 × 1 0 4 2.7132 × 1 0 5
2 4 1.4725 × 1 0 4 1.7964 1.9281 × 1 0 5 1.7907 5.7517 × 1 0 5 6.7873 × 1 0 6
2 5 3.7729 × 1 0 5 1.8245 4.4373 × 1 0 6 2.1194 1.4389 × 1 0 5 4.6971 × 1 0 7
2 6 9.5482 × 1 0 6 1.8424 1.1207 × 1 0 6 1.9185 3.5979 × 1 0 6 1.2429 × 1 0 7
2 7 5.2266 × 1 0 6 1.8754 3.7298 × 1 0 7 1.9853 4.2298 × 1 0 7 1.1100 × 1 0 7
Table 6

E 1 error comparison of [7] for Example 6.4

t E 1 errors of [7] HWC E 1 errors
0.1 3.54 × 1 0 8 2.19 × 1 0 9
0.3 2.23 × 1 0 7 7.49 × 1 0 9
0.5 3.63 × 1 0 7 3.59 × 1 0 8
0.7 2.92 × 1 0 7 1.02 × 1 0 7
0.9 1.02 × 1 0 8 2.26 × 1 0 7
Figure 1 
               Comparisons of numerical and the exact solutions for 32 CPs of Example 6.1.
Figure 1

Comparisons of numerical and the exact solutions for 32 CPs of Example 6.1.

Figure 2 
               Comparisons of numerical and the exact solutions for 32 CPs of Example 6.2.
Figure 2

Comparisons of numerical and the exact solutions for 32 CPs of Example 6.2.

Figure 3 
               Comparison of numerical and exact solutions for 32 CPs of Example 6.3.
Figure 3

Comparison of numerical and exact solutions for 32 CPs of Example 6.3.

Figure 4 
               Comparison of numerical and exact solutions for 32 CPs of Example 6.7.
Figure 4

Comparison of numerical and exact solutions for 32 CPs of Example 6.7.

Figure 5 
               Comparison of numerical and exact solutions for 32 CPs of Example 6.5.
Figure 5

Comparison of numerical and exact solutions for 32 CPs of Example 6.5.

Figure 6 
               Comparison of numerical and exact solutions for 32 CPs of Example 6.6.
Figure 6

Comparison of numerical and exact solutions for 32 CPs of Example 6.6.

Figure 7 
               Comparison of different errors for Example 6.7.
Figure 7

Comparison of different errors for Example 6.7.

Table 7

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.5

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 8.9282 × 1 0 4 4.6941 × 1 0 4 9.4328 × 1 0 5 4.9107 × 1 0 5
2 2 2.8234 × 1 0 4 1.5011 1.2358 × 1 0 4 1.4255 2.8778 × 1 0 5 1.2565 × 1 0 5
2 3 7.9046 × 1 0 5 1.5921 3.1292 × 1 0 5 1.5528 7.8841 × 1 0 6 3.1539 × 1 0 6
2 4 2.0893 × 1 0 5 1.6423 7.8477 × 1 0 6 1.7123 2.0601 × 1 0 6 7.8919 × 1 0 7
2 5 5.3697 × 1 0 6 1.6931 1.9635 × 1 0 6 1.8084 5.2633 × 1 0 7 1.9734 × 1 0 7
2 6 1.3610 × 1 0 6 1.7940 4.9097 × 1 0 7 1.9001 1.3301 × 1 0 7 4.9338 × 1 0 8
2 7 3.4261 × 1 0 7 1.8623 1.2275 × 1 0 7 1.9397 3.3432 × 1 0 8 1.2335 × 1 0 8
Table 8

E 1 , R 1 , E 2 , R 2 , E 3 , and E 4 for Example 6.6

N = 2 J + 1 E 1 R 1 E 2 R 2 E 3 E 4
2 5.6647 × 1 0 4 3.0213 × 1 0 4 7.4263 × 1 0 5 3.8892 × 1 0 5
2 2 1.7584 × 1 0 4 1.5012 7.9631 × 1 0 5 1.6201 2.2744 × 1 0 5 1.0102 × 1 0 5
2 3 4.8595 × 1 0 5 1.6102 2.0167 × 1 0 5 1.7778 6.2151 × 1 0 6 2.5462 × 1 0 6
2 4 1.2750 × 1 0 5 1.6301 5.0579 × 1 0 6 1.8406 1.6202 × 1 0 6 6.3779 × 1 0 7
2 5 3.2641 × 1 0 6 1.7290 1.2655 × 1 0 6 1.9407 4.1336 × 1 0 7 1.5952 × 1 0 7
2 6 8.2567 × 1 0 7 1.8244 3.1644 × 1 0 7 1.9991 1.0438 × 1 0 7 3.9886 × 1 0 8
2 7 2.0763 × 1 0 7 1.9901 7.9113 × 1 0 8 2.0215 2.6225 × 1 0 8 9.9718 × 1 0 9

8 Conclusion

In this study, the solution to fourth-order IDEs has been studied by using the HWC technique. We have established few prerequisites for the presented problem’s like existence and uniqueness of solution by the application of fixed point analysis. We have established an algorithm for the computation via using HWC tools. Then, several examples were investigated to demonstrate our results. For the validation of our procedure, maximum absolute errors for various CPs have been recorded in several tables. A comparison of the exact solution has been made with the numerical solution for the considered examples. Figures have been used to compare the exact and approximate solutions for each example. The suggested scheme showed better accuracy and effectiveness which can be further improved by increasing the number of CPs. The MATLAB 2016 software has been used for all computational tasks. In future, we plan to extend the HWC technique for solution of higher order IDEs, fractional order IDEs, system of IDEs, IDEs with two point and integral boundary conditions, and 2D and 3D problems of higher order IDEs or statements.

Acknowledgment

K. Sha and T. Abdeljawad are thankful to Prince Sultan University for APC and support through the TAS research lab.

  1. Funding information: The authors state no funding involved.

  2. Author contributions: All authors have approved the final version of the manuscript. R. A., M. N., K. S., and T. A. wrote the main manuscript, and R. A. and M. N. prepared the tables figures and reviewed the manuscript.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: The data used in this research are included within the article.

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Received: 2024-12-09
Revised: 2025-02-22
Accepted: 2025-03-19
Published Online: 2025-04-25

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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