Petrov-Galerkin method for small deflections in fourth-order beam equations in civil engineering

Youssri Hassan Youssri; Ahmed Gamal Atta; Ziad Yousef Abu Waar; Mohamed Orabi Moustafa

doi:10.1515/nleng-2024-0022

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Petrov-Galerkin method for small deflections in fourth-order beam equations in civil engineering

Youssri Hassan Youssri , Ahmed Gamal Atta , Ziad Yousef Abu Waar and Mohamed Orabi Moustafa

Published/Copyright: November 15, 2024

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From the journal Nonlinear Engineering Volume 13 Issue 1

Abstract

This study explores the Petrov–Galerkin method’s application in solving a linear fourth-order ordinary beam equation of the form u ″ ″ + q u = f . The equation entails two distinct boundary conditions: pinned–pinned conditions on u and u ′ , and clamped–clamped conditions on u and u ″ . To satisfy these boundary conditions, we have built two sets of basis functions. The explicit forms of all spectral matrices were reported. The nonhomogeneous boundary conditions were easily handled using perfect transformations, ensuring the numerical solution’s accuracy. Detailed analysis of the method’s convergence was studied. Some numerical examples were presented, accompanied by comparisons with other existing methods in the literature.

Keywords: fourth-order beam equation; Chebyshev polynomials of the third-kind; Petrov–Galerkin method; error analysis

MSC 2010: 65M70; 42C05; 65G99; 35R11

1 Introduction

In this study, our attention is devoted to the following fourth-order boundary value problem [1–4]:

(1) υ ′ ′ ′ ′ ( x ) + q ( x ) υ ( x ) = f ( x ) , x ∈ ( 0 , 1 ) ,

subject to one of the following sets of fixed boundary conditions:

• Pinned–pinned beam

(2) υ ( 0 ) = α 1 , υ ′ ( 0 ) = α 2 , υ ( 1 ) = β 1 , and υ ′ ( 1 ) = β 1 ,

• Clamped–clamped beam

(3) υ ( 0 ) = α 1 , υ ″ ( 0 ) = α 2 , υ ( 1 ) = β 1 , and υ ″ ( 1 ) = β 1 ,

where υ ( x ) represents the lateral displacement of the beam at position x , q ( x ) represents the distribution of the transverse load, and f ( x ) represents any external forces applied to the beam. This mathematical model describes the equilibrium of a beam subjected to both external loads and its weight while accounting for large deformations. The fourth-order derivative term υ ′ ′ ′ ′ ( x ) captures the bending stiffness of the beam and higher-order effects in its deformation. The nonlinearity in q ( x ) allows for considering scenarios where the beam’s response depends on its deflection, making it suitable for modeling nonlinear material behavior or geometric effects. The term u stands for the beam’s displacement, or how much it moves from its original position. The term q represents another factor that affects the beam’s behavior, like its stiffness or the force acting on it. The term f is any external force applied to the beam. The equation tells us how the beam’s displacement changes under the influence of these factors.

Beams, the long, robust structures seen in bridges and buildings, serve an important function in engineering. Engineers frequently need to understand how beams flex and move under varying situations. In this study, we look at a clever approach of solving equations that explain how beams behave. We focus on a specific equation, similar to a mathematical recipe, known as the linear fourth-order ordinary beam equation. This equation explains how beams bend and twist when we apply force on them. For more details, please refer [5,6].

To solve this problem, we employ the Petrov–Galerkin technique [7–11]. Think of it as a unique tool in our arithmetic toolkit. This strategy helps us solve difficult equations by dividing them down into smaller, more manageable chunks. It is similar to completing a large puzzle by focusing on one component at a time. The Petrov–Galerkin approach is particularly useful since it can handle a variety of boundary conditions. These are the rules that specify how the beam acts at its ends, such as whether it is clamped down or free to move.

The structure of this article is as follows: first, we will go over the equation in greater depth and explain why it is crucial. Then, we will look at how the Petrov–Galerkin technique works and why we picked it for our study. Following that, we will provide the findings of our research, including the accuracy of our technique and several examples to demonstrate its efficacy. Finally, we will discuss potential directions for this study and how we conducted our trials utilizing Mathematica. So let us begin our voyage into the realm of beams and equations!

1.1 Account on shifted Chebyshev polynomials of the third kind

The SCP3K υ S , i ( z ) are special ones of the shifted Jacobi polynomials that can be defined as [12,13]

(4) υ S , i ( z ) = 2 2 i 2 i i P i − 1 2 , 1 2 ( 2 z − 1 ) .

The orthogonality relation of υ S , i ( z ) is given by

(5) ∫ 0 1 υ S , i ( z ) υ S , j ( z ) ω ( z ) d z = π 2 δ i , j ,

where ω ( z ) = z 1 − z and δ i , j is the well-known Kronecker delta.

The power form representation of the υ S , k ( z ) can be represented as

(6) υ S , k ( z ) = ∑ i = 0 k ( − 1 ) i ( 2 k + 1 ) 2 2 k − 2 i ( − i + 2 k ) ! i ! ( − 2 i + 2 k + 1 ) ! z k − i ,

(7) υ S , k ( z ) = ∑ i = 0 k 2 2 i ( 2 k + 1 ) ( − 1 ) k − i ( i + k ) ! ( 2 i + 1 ) ! ( − i + k ) ! z i .

Moreover, the inversion formula of υ S , k ( z ) is

(8) z r = Γ ( 2 r + 2 ) 2 2 r ∑ i = 0 r 1 ( − i + r ) ! ( i + r + 1 ) ! υ S , i ( z ) .

Lemma 1

[14] For all nonnegative integers i and j , the following linearization formula is valid:

(9) υ S , i ( z ) υ S , j ( z ) = ∑ p = 0 2 min ( i , j ) ( − 1 ) p υ S , i + j − p ( z ) .

Lemma 2

The following integral formula is valid; for all r > 0 , we have

∫ 0 1 x r υ S , j ( x ) ω ( x ) d x = π Γ r + 3 2 ( r − j + 1 ) j Γ ( j + r + 2 ) .

2 Matrix Petrov–Galerkin approach for pinned–pinned beam fourth-order boundary value problem (40BVP)

In this section, we consider Eq. (12) subject to pinned–pinned beam conditions (2).

To proceed in our proposed Petrov–Galerkin approach, the following transformation

(10) ζ ( x ) ≔ υ ( x ) − a ( x ) ,

where

(11) a ( x ) = x 3 ( υ ′ ( 0 ) + υ ′ ( 1 ) + 2 υ ( 0 ) − 2 υ ( 1 ) ) + x 2 ( − 2 υ ′ ( 0 ) − υ ′ ( 1 ) − 3 υ ( 0 ) + 3 υ ( 1 ) ) + x υ ′ ( 0 ) + υ ( 0 ) ,

is used to convert Eq. (1) governed by (2) into the following modified equation:

(12) ζ ″ ″ ( x ) + q ( x ) ζ ( x ) = g ( x ) , x ∈ ( 0 , 1 ) ,

subject to

(13) ζ ( 0 ) = ζ ( 1 ) = ζ ′ ( 0 ) = ζ ′ ( 1 ) = 0 ,

where g ( x ) = f ( x ) − q ( x ) a ( x ) .

Therefore, instead of solving (1) governed by (2), we can solve the modified equation (12) governed by the homogeneous (13).

2.1 Trial functions

Consider the following basis functions:

(14) A i ( x ) = ( x − x 2 ) 2 υ S , i ( x ) ,

where υ S , i ( x ) is defined in (4).

The orthogonality relation of A i ( x ) is given by

(15) ∫ 0 1 A i ( x ) A j ( x ) ω ˆ ( x ) d x = π 2 δ i , j ,

where ω ˆ ( x ) = ω ( x ) ( x − x 2 ) 4 = ( 1 − x ) − 9 2 x − 7 2 .

Theorem 1

The following two useful integral formulas are given by

(16) ( a ) ∫ 0 1 A i ″ ″ ( x ) V S , r ( x ) ω ( x ) d x = χ i , r , ( b ) ∫ 0 1 υ S , m ( x ) A i ( x ) υ S , r ( x ) ω ( x ) d t = ψ m , i , r ,

where

(17) χ i , r = π 4 2 ( i + 1 ) ( i + 2 ) ( i + 3 ) ( i + 4 ) , i f i = r , 35 i 3 − 15 i 2 r + 45 i 2 − 15 i r 2 − 30 i r + 85 i + 3 r 3 − 3 r 2 − 27 r + 27 , i f i > r , ( i − r ) o d d , 35 i 3 + 15 i 2 r + 60 i 2 − 15 i r 2 + 100 i − 3 r 3 − 12 r 2 + 12 r + 48 , i f i > r , ( i − r ) e v e n , 0 , o t h e r w i s e ,

(18) ψ m , i , r = π 128 ∑ k = ∣ r − m ∣ m + r ( − 1 ) k + m + r × − 1 , i f i + k = 1 , 1 4 , i f i + k = 3 , 0 , o t h e r w i s e , + 3 2 , i f i = k , 1 , i f ∣ i − k ∣ = 2 , 1 4 , i f ∣ i − k ∣ = 4 , 0 , o t h e r w i s e , × ( − 1 ) i − k 2 , i f ( i − k ) e v e n , 1 , o t h e r w i s e .

Proof

We will prove only part (a), and part (b) is quite similar to part (a); to start the proof, we make use of the analytic form (7), and we can write

(19) A i ( x ) = ∑ k = 0 i 2 2 k ( 2 i + 1 ) ( − 1 ) i − k ( k + i ) ! ( 2 k + 1 ) ! ( − k + i ) ! ( 1 − x ) 2 x k + 2 .

Accordingly

(20) d 4 A i ( x ) d x 4 = ∑ k = 0 i 2 2 k ( 2 i + 1 ) ( − 1 ) i − k ( k + i ) ! ( 2 k + 1 ) ! ( − k + i ) ! ( k + 1 ) 2 × ( ( k + 3 ) ( k + 4 ) x 2 − 2 k ( k + 3 ) x + ( k − 1 ) k ) x k − 2 .

Now, taking the inner product with υ S , j ( x ) and repeated use of Lemma 2 for r = k , k − 1 , k − 2 , we obtain

(21) χ i , r = ∫ 0 1 d 4 A i ( x ) d x 4 υ S , j ( x ) ω ( x ) d x = ∑ k = 0 i π ( k ) 3 ( 4 j 4 + 8 j 3 + 20 j 2 k − 8 j 2 + 20 j k − 12 j + 35 k 2 + 5 k − 12 ) ( 2 k − 2 ) ! 4 k ( k − j ) ! ( k + j + 1 ) ! .

The lateral sum is evaluated via Zeilberger’s algorithm [15], which proves the first part of the theorem.□

2.2 Algorithm of the method

Now, one may set

(22) X = span { A i ( x ) : i = 0 , 1 , … } , Y = { ζ ∈ X : ζ ( 0 ) = ζ ( 1 ) = ζ ′ ( 0 ) = ζ ′ ( 1 ) = 0 } .

Then, any function ζ ( x ) ∈ Y may be approximated as

(23) ζ ( x ) ≈ ζ N ( x ) = ∑ i = 0 N c i A i ( x ) ,

where the residual ℛ ( x ) of Eq. (12) can be written as

(24) ℛ ( x ) = ζ N ″ ″ ( x ) + q ( x ) ζ N ( x ) − g ( x ) .

The application of the Petrov–Galerkin method [16] is used to find ζ N ( x ) ∈ Y such that

(25) ( ℛ ( x ) , υ S , r ( x ) ) ω ( x ) = 0 , 0 ≤ r ≤ N .

Therefore, Eq. (25) can be rewritten alternatively as

(26) ∑ i = 0 N c i ( A i ″ ″ ( x ) , υ S , r ( x ) ) ω ( x ) + ∑ m = 0 N ∑ i = 0 N λ m c i ( υ S , m ( x ) A i ( x ) , υ S , r ( x ) ) ω ( x ) = ( g ( x ) , υ S , r ( x ) ) ω ( x ) ,

(27) ∑ i = 0 N c i χ i , r + ∑ m = 0 N ∑ i = 0 N λ m c i ψ m , i , r = G r ,

where χ i , r = ( A i ″ ″ ( x ) , υ S , r ( x ) ) ω ( x ) , ψ m , i , r = ( υ S , m ( x ) A i ( x ) , υ S , r ( x ) ) ω ( x ) , and G r = ( g ( x ) , υ S , r ( x ) ) ω ( x ) and λ m determined from the following relation:

(28) λ m = 2 π ∫ 0 1 q ( x ) υ S , m ( x ) ω ( x ) d x .

Now, Eq. (27) constitutes a system of algebraic equations of order ( N + 1 ) , which may be solved using the Gauss elimination procedure.

2.3 Error bound

Theorem 2

Assume that ζ N ( x ) is the proposed approximate solution belonging to ζ ( x ) ∈ Y and d i ζ ( x ) d x i ∈ C ( [ 0 , 1 ] ) , i = 0 , 1 , 2 , … , N + 1 , and define

(29) ℳ N = sup x ∈ [ 0 , 1 ] d N + 5 ζ ( x ) d x N + 5 .

Consequently, this estimate holds

(30) ‖ ζ ( x ) − ζ N ( x ) ‖ L ω ˆ ( x ) 2 ≤ π 1 4 945 ℳ N 32 ( 2 N ) 11 4 ( ( N + 5 ) ! ) .

Proof

Consider the following approximate Taylor expansion of degree ( N + 4 ) for ζ ( x ) about the point x = 0 :

(31) χ N ( x ) = ∑ i = 0 N + 4 d i ζ ( x ) d x i x = 0 x i i ! ,

(32) ζ ( x ) − χ N ( x ) = x N + 5 ( N + 5 ) ! d N + 5 ζ ( x ) d x N + 5 x = c , c ∈ [ 0 , 1 ] .

Because ζ N ( x ) is the best approximation solution of degree ( N + 4 ) for ζ ( x ) , we have, using the idea of best approximation:

(33) ‖ ζ ( x ) − ζ N ( x ) ‖ L ω ˆ ( x ) 2 2 ≤ ‖ ζ ( x ) − χ N ( x ) ‖ L ω ˆ ( x ) 2 2 ≤ ∫ 0 1 ℳ N 2 x 2 ( N + 5 ) ( ( N + 5 ) ! ) 2 ω ˆ ( x ) d x = ℳ N 2 ( ( N + 5 ) ! ) 2 ∫ 0 1 ( 1 − x ) − 9 2 x 2 N + 13 2 d x = 945 π ℳ N 2 Γ 2 N + 15 2 32 Γ ( 2 N + 13 ) ( ( N + 5 ) ! ) 2 .

According to the following inequality [17]:

(34) Γ ( x + a ) Γ ( x + b ) ≤ o x a , b x a − b ,

where x ≥ 1 , x + a > 1 , x + b > 1 , a , b , are any constants and

(35) o x a , b = exp a − b 2 ( x + b − 1 ) + 1 12 ( x + a − 1 ) + ( a − b ) 2 x = 1 + O ( x − 1 ) .

The following estimation may be obtained:

(36) ‖ ζ ( x ) − ζ N ( x ) ‖ L ω ˆ ( x ) 2 2 ≤ 945 π ℳ N 2 32 ( 2 N ) 11 2 ( ( N + 5 ) ! ) 2 .

Therefore, we obtain

(37) ‖ ζ ( x ) − ζ N ( x ) ‖ L ω ˆ ( x ) 2 ≤ π 1 4 945 ℳ N 32 ( 2 N ) 11 4 ( ( N + 5 ) ! ) .□

Theorem 3

Suppose that ζ ( x ) , ζ N ( x ) , and d i ζ ( x ) d x i meet the assumption of Theorem 2. Then, the following estimation holds:

(38) d m d x m [ ζ ( x ) − ζ N ( x ) ] L ω ˆ ( x ) 2 ≤ π 1 4 945 M N 32 ( 2 N − 2 m ) 11 4 ( ( N − m + 5 ) ! ) .

Proof

Assume that d m χ N ( x ) d x m is the Taylor expansion of d m ζ ( x ) d x m about the point x = 0 , and then, the residual between d m ζ ( x ) d x m and d m χ N ( x ) d x m can be written as

(39) d m d x m [ ζ ( x ) − χ N ( x ) ] = x N − m + 5 ( N − m + 5 ) ! d N + 5 ζ ( x ) d x N + 5 x = n , n ∈ [ 0 , 1 ] .

Since d m ζ N ( x ) d x m is the best approximate solution of d m ζ ( x ) d x m , according to the definition of the best approximation, we obtain the desired result by performing steps as in Theorem 2

(40) d m d x m [ ζ ( x ) − ζ N ( x ) ] L ω ˆ ( x ) 2 ≤ π 1 4 945 M N 32 ( 2 N − 2 m ) 11 4 ( ( N − m + 5 ) ! ) .

This completes the proof of this theorem.□

Theorem 4

Assume that ℛ ( x ) is the residual of Eq. (12), and then ‖ ℛ ( x ) ‖ L ω ˆ ( x ) 2 will be sufficiently small for the sufficiently large values of N.

Proof

ℛ ( x ) of Eq. (12) can be written as

(41) R ( x ) = ζ ′ ′ ′ ′ ( x ) + q ( x ) ζ ( x ) − g ( x ) = d 4 d x 4 [ ζ N ( x ) − ζ ( x ) ] + q ( x ) [ ζ N ( x ) − ζ ( x ) ] .

Taking ‖ . ‖ L ω ˆ ( x ) 2 and using Theorems 2 and 3, we obtain

(42) ‖ ℛ ( x ) ‖ L ω ˆ ( x ) 2 ≤ π 1 4 945 ℳ N 32 ( 2 ( N − 4 ) ) 11 4 ( ( N + 1 ) ! ) + q ( x ) π 1 4 945 ℳ N 32 ( 2 N ) 11 4 ( ( N + 5 ) ! ) .

Finally, it is evident from Eq. (42) that for sufficiently high values of N , ‖ ℛ ( x ) ‖ L ω ˆ ( x ) 2 will be small enough. Thus, the proof of this theorem is complete.□

3 Matrix Petrov–Galerkin approach for clamped–clamped beam 4OBVP

The main idea of this section is to solve Eq. (12) subject to clamped–clamped beam conditions (3).

Using the following transformation:

(43) ζ ( x ) ≔ υ ( x ) − b ( x ) ,

where

(44) b ( x ) = x 3 6 ( υ ″ ( 1 ) − υ ″ ( 0 ) ) + x 2 2 υ ″ ( 0 ) − x 6 ( 2 υ ″ ( 0 ) + υ ″ ( 1 ) ) + υ ( 0 )

under the condition υ ( 0 ) = υ ( 1 ) , is used to convert Eq. (1) governed by (3) into the following modified equation:

(45) ζ ′ ′ ′ ′ ( x ) + q ( x ) ζ ( x ) = h ( x ) , x ∈ ( 0 , 1 ) ,

subject to

(46) ζ ( 0 ) = ζ ( 1 ) = ζ ″ ( 0 ) = ζ ″ ( 1 ) = 0 ,

where h ( x ) = f ( x ) − q ( x ) b ( x ) . Hence, instead of solving (1) governed by (3), we can solve the modified equation (45) governed by the homogeneous (46).

3.1 Trial functions

Consider the following basis functions:

(47) ℬ i ( x ) = x ( 1 − x ) ( ( 4 i 4 + 24 i 3 + 44 i 2 + 24 i + 9 ) υ S , i + 1 ( x ) − ( i + 2 ) ( 2 i 2 + 8 i + 9 ) 2 υ S , i ( x ) + ( i + 1 ) ( 2 i 2 + 4 i + 3 ) 2 υ S , i + 2 ( x ) ) ,

where υ S , i ( x ) is defined in (4).

Theorem 5

The following two useful integral formulas are given by

(48) ( a ) ∫ 0 1 ℬ i ′ ′ ′ ′ ( x ) V S , r ( x ) ω ( x ) d x = ϕ i , r , ( b ) ∫ 0 1 υ S , m ( x ) ℬ i ( x ) υ S , r ( x ) ω ( x ) d t = μ m , i , r ,

where

(49) ϕ i , r = ( i + 1 ) ( 2 i 2 + 4 i + 3 ) 2 ε i + 2 , r − ( i + 2 ) ( 2 i 2 + 8 i + 9 ) 2 ε i , r + ( 4 i 4 + 24 i 3 + 44 i 2 + 24 i + 9 ) ε i + 1 , r

(50) μ m , i , r = ∑ k = ∣ r − m ∣ m + r ( − 1 ) k + m + r ( ( 4 i 4 + 24 i 3 + 44 i 2 + 24 i + 9 ) b ˆ i + 1 , k − ( i + 2 ) ( 2 i 2 + 8 i + 9 ) 2 b ˆ i , k + ( i + 1 ) ( 2 i 2 + 4 i + 3 ) 2 b ˆ i + 2 , k ) ,

and

(51) ε i , r = − π 1 4 ( i − r ) ( i + r ) ( i + r + 2 ) ( 7 i 2 − 2 ( i + 1 ) r + 6 i − r 2 + 8 ) , i f i > r , ( i − r ) e v e n , 1 4 ( i + r + 1 ) ( ( i − r ) 2 − 1 ) ( 7 i 2 + 2 i r + 8 i − r 2 + 9 ) , i f i > r + 1 , ( i − r ) o d d 0 , o t h e r w i s e ,

(52) b ˆ i , k = π 16 1 , i f i = k , − 1 2 , i f ∣ i − k ∣ = 2 , o r i = 0 , k = 1 , o r i = 1 , k = 0 , 0 , o t h e r w i s e .

Proof

The proof of this theorem follows a similar approach to that of Theorem 1. To avoid redundancy, we will omit the detailed proof here.□

3.2 Algorithm of the method

Now, one may set

(53) X ˆ = span { ℬ i ( x ) : i = 0 , 1 , … , N } , Y ˆ = { ζ ∈ X ˆ : ζ ( 0 ) = ζ ( 1 ) = ζ ″ ( 0 ) = ζ ″ ( 1 ) = 0 } .

Then, any function ζ ( x ) ∈ Y ˆ may be written as

(54) ζ ( x ) ≈ ζ N ( x ) = ∑ i = 0 N c i ℬ i ( x ) .

Imitating similar steps as in the previous section, we obtain

(55) ∑ i = 0 N c i ϕ i , r + ∑ m = 0 N ∑ i = 0 N ϖ m c i μ m , i , r = ℋ r ,

where ϕ i , r = ( ℬ i ′ ′ ′ ′ ( x ) , υ S , r ( x ) ) ω ( x ) , μ m , i , r = ( υ S , m ( x ) A i ( x ) , υ S , r ( x ) ) ω ( x ) , and ℋ r = ( h ( x ) , υ S , r ( x ) ) ω ( x ) and ϖ m determined from the following relation:

(56) ϖ m = 2 π ∫ 0 1 q ( x ) υ S , m ( x ) ω ( x ) d x .

Now, Eq. (55) constitutes a system of algebraic equations of order ( N + 1 ) , which may be solved using the Gauss elimination procedure.

3.3 Error bound

Theorem 6

Assume that ζ N ( x ) is the proposed approximate solution belonging to ζ ( x ) ∈ Y ˆ and d i ζ ( x ) d x i ∈ C ( [ 0 , 1 ] ) , i = 0 , 1 , 2 , … , N + 1 . Then, the following estimation holds:

(57) ‖ ζ ( x ) − ζ N ( x ) ‖ 2 ≤ ℳ N ( 2 N + 11 ) ( N + 5 ) ! .

Proof

Based on Eq. (31) and the definition of best approximation, we obtain the following estimation:

(58) ‖ ζ ( x ) − ζ N ( x ) ‖ 2 2 ≤ ‖ ζ ( x ) − χ N ( x ) ‖ 2 2 ≤ ∫ 0 1 ℳ N 2 x 2 ( N + 5 ) ( ( N + 5 ) ! ) 2 d x = ℳ N 2 ( ( N + 5 ) ! ) 2 ∫ 0 1 x 2 ( N + 5 ) d x = ℳ N 2 ( 2 N + 11 ) ( ( N + 5 ) ! ) 2 .

Hence, we obtain

(59) ‖ ζ ( x ) − ζ N ( x ) ‖ 2 ≤ ℳ N ( 2 N + 11 ) ( N + 5 ) ! .□

Theorem 7

Suppose that ζ ( x ) , ζ N ( x ) , and d i ζ ( x ) d x i meet the assumption of Theorem 6. Then, the following estimation holds:

(60) d m d x m [ ζ ( x ) − ζ N ( x ) ] 2 ≤ ℳ N ( 2 ( N − m ) + 11 ) ( N − m + 5 ) ! .

Proof

The proof of this theorem is similar to the proof of Theorem 3 after imitating similar steps as in the previous theorem.□

Theorem 8

Assume that ℛ ( x ) is the residual of Eq. (12), and then, ‖ ℛ ( x ) ‖ 2 will be sufficiently small for the sufficiently large values of N .

Proof

ℛ ( x ) of Eq. (12) can be written as

(61) R ( x ) = ζ ″ ″ ( x ) + q ( x ) ζ ( x ) − g ( x ) = d 4 d x 4 [ ζ N ( x ) − ζ ( x ) ] + q ( x ) [ ζ N ( x ) − ζ ( x ) ] .

Taking ‖ ⋅ ‖ 2 and using Theorems 6 and 7, we obtain

(62) ‖ ℛ ( x ) ‖ 2 ≤ ℳ N ( 2 N + 3 ) ( N + 1 ) ! + q ( x ) ℳ N ( 2 N + 11 ) ( N + 5 ) ! .

Finally, it is evident from Eq. (62) that for sufficiently high values of N , ‖ ℛ ( x ) ‖ 2 will be small enough. Thus, the proof of this theorem is complete.□

4 Illustrative examples

Example 1

[3] Consider the following equation:

(63) ζ ″ ″ ( x ) + ζ ( x ) = ( x − 1 ) x ( ( x − 1 ) x ( ( x − 1 ) 2 x 2 + 1,680 ) + 480 ) + 24 ; 0 < x < 1 ,

subject to the initial conditions:

(64) ζ ( 0 ) = ζ ( 1 ) = ζ ′ ( 0 ) = ζ ′ ( 1 ) = 0 ,

where the exact solution is ζ ( x ) = x 4 ( 1 − x ) 4 .

Table 1 presents a comparison of the absolute errors between our method and method in the study of Ashyralyev and Ibrahim [3]. Figure 1 illustrates the absolute errors at different values of N . These results prove that the approximate solution is quite near to the analytic one.

Table 1

Comparison of the absolute errors for Example 1

Method in [3]		Our method at N = 4
N	Error
40	1.5237 × 1 0 ‒ 7
80	2.5606 × 1 0 ‒ 9	2.28333 × 1 0 ‒ 16
160	4.0844 × 1 0 ‒ 11

Figure 1

Absolute errors of Example 1 at different values of N .

Example 2

Consider the following equation:

(65) ζ ″ ″ ( x ) + ζ ( x ) = ( 1 + π 4 ) sin ( π x ) ; 0 < x < 1 ,

subject to the initial conditions:

(66) ζ ( 0 ) = 0 , ζ ( 1 ) = 0 , ζ ′ ( 0 ) = π , ζ ′ ( 1 ) = − π ,

where the exact solution is ζ ( x ) = sin ( π x ) .

Table 2 shows the absolute errors at different values of N and x . Figure 2 illustrates the absolute errors at different values of N .

Table 2

Absolute errors of Example 2

x	N = 8	N = 9	N = 10
0.1	1.30596 × 1 0 ‒ 12	6.7269 × 1 0 ‒ 12	1.77081 × 1 0 ‒ 14
0.2	1.39601 × 1 0 ‒ 11	5.74607 × 1 0 ‒ 12	3.10973 × 1 0 ‒ 13
0.3	1.93806 × 1 0 ‒ 11	1.85485 × 1 0 ‒ 12	4.28102 × 1 0 ‒ 13
0.4	4.63873 × 1 0 ‒ 12	2.07313 × 1 0 ‒ 11	1.42775 × 1 0 ‒ 13
0.5	3.8068 × 1 0 ‒ 11	9.73555 × 1 0 ‒ 12	8.83071 × 1 0 ‒ 13
0.6	4.63873 × 1 0 ‒ 12	7.79721 × 1 0 ‒ 12	1.42664 × 1 0 ‒ 13
0.7	1.93807 × 1 0 ‒ 11	1.18509 × 1 0 ‒ 11	4.2788 × 1 0 ‒ 13
0.8	1.39599 × 1 0 ‒ 11	2.8908 × 1 0 ‒ 12	3.11084 × 1 0 ‒ 13
0.9	1.30534 × 1 0 ‒ 12	8.63976 × 1 0 ‒ 13	1.78191 × 1 0 ‒ 14

Figure 2

Absolute errors of Example 2 at different values of N .

Example 3

Consider the following equation:

(67) ζ ″ ″ ( x ) + x ζ ( x ) = ( x + 16 π 4 ) cos ( 2 π x ) ; 0 < x < 1 ,

subject to the initial conditions:

(68) ζ ( 0 ) = 1 , ζ ( 1 ) = 1 , ζ ″ ( 0 ) = − 4 π 2 , ζ ″ ( 1 ) = − 4 π 2 ,

where the exact solution is ζ ( x ) = cos ( 2 π x ) .

Table 3 shows the maximum absolute errors at different values of N . Figure 3 shows the absolute errors (left) and approximate solution (right) when N = 20 .

Table 3

Maximum absolute errors of Example 3

N	4	8	12	16	20
Error	1.02237 × 1 0 ‒ 1	2.87619 × 1 0 ‒ 4	2.04804 × 1 0 ‒ 7	5.23386 × 1 0 ‒ 11	2.09948 × 1 0 ‒ 11

Figure 3

Absolute errors (left) and approximate solution (right) of Example 3 at different values of N .

Example 4

[3] Consider the following equation:

(69) ζ ″ ″ ( x ) + ζ ( x ) = ( x − 1 ) x ( ( x − 1 ) x ( ( x − 1 ) 2 x 2 + 1,680 ) + 480 ) + 24 ; 0 < x < 1 ,

subject to the initial conditions:

(70) ζ ( 0 ) = ζ ( 1 ) = ζ ″ ( 0 ) = ζ ″ ( 1 ) = 0 ,

where the exact solution is ζ ( x ) = x 4 ( 1 − x ) 4 .

Table 4 presents a comparison of the absolute errors between our method and method in the study of Ashyralyev and Ibrahim [3]. Figure 4 illustrates the absolute errors at different values of N . These results prove that the approximate solution is quite near to the analytic one.

Table 4

Comparison of the absolute errors for Example 4

Method in [3]
N	Error	Our method at N = 4
40	7.8925 × 1 0 ‒ 7
80	1.2332 × 1 0 ‒ 8	1.14925 × 1 0 ‒ 17
160	1.8981 × 1 0 ‒ 10

Figure 4

Absolute errors of Example 4 at different values of N .

Example 5

Consider the following equation:

(71) ζ ″ ″ ( x ) + ( 1 + x ) ζ ( x ) = E α , β ″ ″ ( x ) + ( 1 + x ) E α , β ( x ) ; 0 < x < 1 ,

subject to the initial conditions:

(72) ζ ( 0 ) = E α , β ( 0 ) , ζ ( 1 ) = E α , β ( 1 ) , ζ ′ ( 0 ) = E ′ ( 0 ) , ζ ′ ( 1 ) = E ′ ( 1 ) ,

such that the exact solution of this problem is ζ ( x ) = ∑ k = 0 ∞ x k Γ ( α k + β ) = E α , β ( x ) and E α , β ( x ) is the Mittag–Leffler function [18] of the two parameters α , β > 0 .

Eq. (71) is solved in two cases corresponding to α = β = 1 and α = 1 2 , β = 1 .

Case 1: At α = β = 1 , this exact solution reduces to ζ ( x ) = E 1 , 1 ( x ) = e x . Table 5 shows the maximum absolute errors at different values of N . Figure 5 shows the absolute errors (left) and approximate solution (right) when N = 18 .

Case 2: At α = 1 2 , β = 1 , this exact solution reduces to ζ ( x ) = E 1 2 , 1 ( x ) = e x 2 erfc ( − x ) , where erfc ( − x ) is the complementary error function and defined as

(73) erfc ( x ) = 2 π ∫ x ∞ e − t 2 d t = 1 − erf ( x ) .

Table 5

Maximum absolute errors of Example 4

N	2	6	10	14	18
Error	5.63019 × 1 0 ‒ 6	3.17232 × 1 0 ‒ 11	4.996 × 1 0 ‒ 16	4.996 × 1 0 ‒ 16	2.77556 × 1 0 ‒ 16

Figure 5

Absolute errors (left) and approximate solution (right) of Example 5 at different values of N .

Table 6 shows the maximum absolute errors at different values of N . Figure 6 shows the absolute errors (left) and approximate solution (right) when N = 20 .

Table 6

Maximum absolute errors of Example 4

N	4	8	12	16	20
Error	8.40726 × 1 0 ‒ 4	1.2976 × 1 0 ‒ 6	8.75009 × 1 0 ‒ 10	3.19744 × 1 0 ‒ 13	2.22045 × 1 0 ‒ 14

Figure 6

Absolute errors (left) and approximate solution (right) of Example 5 at different values of N .

5 Concluding remarks

In conclusion, we verified that the Petrov–Galerkin method works very well for solving beam equations. We solved the beam equation subject to different types of boundary conditions that capture real-life situations. Looking ahead, we think it would be interesting to use this method to solve another kind of equation about beams that change over time. We have used Mathematica for writing and running our codes. The results are accurate, the method is efficient, and we aim to extend the method for more complicated models in applied mathematics.

Acknowledgements

The authors would like to thank the anonymous reviewers for carefully reading the article and for their constructive and valuable comments, which have improved the manuscript’s present form.

Funding information: The authors received no financial support for the research.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript, consented to its submission to the journal, reviewed all results, and approved the final version. YHY and AGA designed and conducted the experiments. AGA developed the model code and performed the simulations. ZYA and MOM reviewed the entire text. YHY prepared the manuscript with contributions from all co-authors.
Conflict of interest: The authors declare that they have no conflict of interest.
Data availability statement: The authors did not use any scientific data during this research.

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Received: 2024-05-06

Accepted: 2024-06-12

Published Online: 2024-11-15

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

Articles in the same Issue

https://doi.org/10.1515/nleng-2024-0022

Keywords for this article

fourth-order beam equation; Chebyshev polynomials of the third-kind; Petrov–Galerkin method; error analysis

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