Integral transformation for nonlinear oscillators

Abstract

This paper introduces a generalized integral transform that encompasses the Laplace, Fourier, and numerous contemporary integral transforms as particular instances, while preserving their defining characteristics. Secondly, we propose the application of the generalized integral transform in the variational iteration method for the straightforward identification of the Lagrange multiplier, thus enabling the resolution of nonlinear oscillator problems. Finally, we present a series of illustrative examples to demonstrate the efficacy of this approach.

Introduction

Laplace and Fourier transforms are the most widely utilized mathematical tools across numerous disciplines. However, they have been employed in isolation, which may result in individuals proficient in the Laplace transform lacking familiarity with the applications of the Fourier transform. As a result of the versatility of these transforms in numerical simulation, new modifications have been introduced on a regular basis to address specific challenges. The integral transformation, as outlined in Table 1, has a multitude of applications in science and engineering, including, but not limited to, mathematics, chemistry, mechanics, and economics.

Table 1.

Relationship between various types of integral transformations and the proposed generalized integral transform.

P (s)	g(s)	Integral formula $P(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t$	Transform name
1	s	${\int }_0^{\infty }{e}^{-st}f(t)\mathrm{d}t$	Laplace transform
1	1/s	${\int }_0^{\infty }{e}^{-t/s}f(t)\mathrm{d}t$	Kamal transform 1
1	s n	${\int }_0^{\infty }{e}^{-s^{n}t}f(t)\mathrm{d}t$	Fareeha transform 2
s	s	$s{\int }_0^{\infty }{e}^{-st}f(t)\mathrm{d}t$	Natural transform 3
s	1/s	$s{\int }_0^{\infty }{e}^{-t/s}f(t)\mathrm{d}t$	Elzaki transform 4
s	s 2	$s{\int }_0^{\infty }{e}^{-s^{2}t}f(t)\mathrm{d}t$	Pourreza transform 5
s 2	s	$s^2{\int }_0^{\infty }{e}^{-st}f(t)\mathrm{d}t$	Mohand transform 6
1/s	s	$1/s{\int }_0^{\infty }{e}^{-st}f(t)\mathrm{d}t$	Aboodh transform 7
1/s	1/s	$1/s{\int }_0^{\infty }{e}^{-t/s}f(t)\mathrm{d}t$	Sumudu transform 8
1/s	s 2	$1/s{\int }_0^{\infty }{e}^{-s^{2}t}f(t)\mathrm{d}t$	Emad–Falih transform 9
1/s2	s	$1/s^2{\int }_0^{\infty }{e}^{-st}f(t)\mathrm{d}t$	Emad–Sara transform 10
1/s2	1/s	$1/s^2{\int }_0^{\infty }{e}^{-t/s}f(t)\mathrm{d}t$	Sawi transform 11
P(s)	s n	$P(s){\int }_0^{\infty }{e}^{-s^{n}t}f(t)\mathrm{d}t$	He transform 12
p(s) is PRF	q(s) is PRF	$p(s){\int }_0^{\infty }{e}^{-q(s)t}f(t)\mathrm{d}t$	Jafari transform (partially) 13

Nadeem et al. ⁶ employed a method combining homotopy perturbation and Mohand transformation to solve the fractional-order Newell–Whitehead–Segel equation. Manimegalai et al. ⁷ investigated the efficacy of the Aboodh transform-based homotopy perturbation method in addressing a generalized oscillatory differential equation. The resulting final coupling results demonstrated a notable improvement in performance compared to those obtained by numerous existing methods. The Sumudu transform ⁸ has been demonstrated to be applicable to certain functions. An additional analogous transform was proposed, based on the Sumudu transform principles, designated as the Elzaki transform. ⁴ An iterative transformation technique that combines the Elzaki transform with iterative approaches can be used to resolve fractional-order linear Klein–Gordon and Hirota–Satsuma-linked KdV equations. Ahmadi et al. ⁵ put forth the Pourreza integral transform, which is a valuable tool for solving both Laguerre and Hermite differential equations. The Emad–Sara transform ¹⁰ and the Emad–Falih transform ⁹ are derived from the Fourier transform. These techniques can be employed for the resolution of ordinary and partial differential equations. Higazy and Aggarwal ¹¹ employed the Sawi transformation to solve a system of ODEs and subsequently calculated the density of chemical reactants in chemical reactions.

In particular, Jafari ¹³ proposed a generalized transform in 2021. In 2023, Khan and Khalid ² put forth the Fareeha transform, which, however, does not encompass a significant number of integral transforms that fall under the Laplace transform category. Additionally, in 2023, He et al. ¹² introduced a more general integral transform that can address issues related to both Fourier and Laplace transforms. While these modifications are effective, engineers are often overwhelmed by the definitions and find it challenging to identify an appropriate transformation for their problems. The objective of this paper is to present a unified approach to integral transformations, with a focus on the selection of appropriate integral transformations based on specific situations to facilitate the simplification of calculations in equation-based problems. In other words, we present a generalized integral transform that encompasses most or even all types of transforms as special cases while retaining their properties. In addition to eliminating the shortcomings of the aforementioned transforms based on the He transform, ¹² we also apply the generalized integral transform to nonlinear oscillator problems. Our principal findings demonstrate that the integration of the generalized integral transform and the variational iteration method ¹⁴ can demonstrate remarkable efficacy even in the context of nonlinear oscillators. Furthermore, we consider to be a promising avenue for future research, and the unification provides a novel perspective for a broader range of potential applications.

Main results

The proposed generalized integral transform

This section is dedicated to the presentation of the generalized integral transform, which encompasses numerous integral transforms that fall under the Laplace transform category, as well as the properties of the Fourier transform, which represents a special case.

Definition 2.1

Let f(t) be a function defined for t ≥ 0, P(s) ≠ 0 for some $s\in \mathbb{C}$ ; we define the generalized integral transform H(s) of f(t) by $H\{f(t)\}=H(s)=P(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t,$ (1)where g(s) = a _n s ⁿ + a_n−1sⁿ⁻¹ + ⋯ + a₁s + a₀ + a₋₁s⁻¹ + a₋₂s⁻² + ⋯ + a_−ms^−m, $n,m\in \mathbb{Z}$ , $a_i\in \mathbb{R}$ , (i = −m, −m + 1,…, 0, 1, 2, …, n), provided that the integral exists.

Table 2 displays the generalized integral transform of some basic functions.

Table 2.

New generalized integral transform of some elementary functions.

f(t) = H−1{H(s)}	1	t α , α > 0	e at	sin bt	cos bt	sinh bt	cosh bt	t sin t
H(s)	P(s)/g(s)	P(s)Γ(α + 1)/g(s)α+1	P(s)/[g(s) − a]	bP(s)/[g(s)2 + b2]	p(s)g(s)/[g(s)2 + b2]	bP(s)/[g(s)2 − b2]	P(s)g(s)/[g(s)2 − b2]	2P(s)g(s)/[1 + g(s)2]

Next, we will present the sufficient condition for the existence of integrals.

Theorem 2.2

Let f(t) be a piecewise continuous function of exponential order for all t ≥ 0, that is, f(t) satisfies $\left|f(t)\right|\le M{e}^{kt}$ , where k and M are positive constants. Then the integral exists for all Re[g(s)] > k, where Re[g(s)] denotes the real part of g(s).

Proof

It follows that $\left|H(S)\right|=\left|H\{f(t)\}\right|=\left|P(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t\right|$ $\le \left|P(s)\right|{\int }_0^{\infty }\left|f(t)\right|e^{-\mathit{R}\mathit{e}[g(s)]t}\mathrm{d}t$ $\le \left|P(s)\right|{\int }_0^{\infty }M{e}^{kt}{e}^{-\mathit{R}\mathit{e}[g(s)]t}\mathrm{d}t.$

Moreover, we have $\left|H(s)\right|\le \left|P(s)\right|M{\int }_0^{\infty }{e}^{\left[k-\mathit{R}\mathit{e}[g(s)]\right]t}\mathrm{d}t=\frac{\left|P(s)\right|M}{Re[g(s)]-k},$ where Re[g(s)] > k.

Remark 2.3

1. This theorem ensures the convergence of the generalized integral transform. If f(t) is absolutely integrable function, we can set Re[g(s)] = 0, it has degenerated into the Fourier transform.

2. Jafari transform: $p(s){\int }_0^{\infty }{e}^{-q(s)t}f(t)dt$ , which did not provide definitions and functions for q(s) and p(s), with the only requirement being that they are positive real functions (PRFs). It is important to note that the convergence condition for the Jafari transform necessitates that q(s) > k. In contrast, while the generalized integral transform g(s) cannot fully encompass all positive real functions, we only require that Re[g(s)] > k. This convergence condition is less stringent than that of the Jafari transform. Furthermore, there are no specific requirements for p(s).

In the following theorem, we discuss the linearity of generalized integral transform.

Theorem 2.4

For any constants β and γ and any two functions f₁(t) and f₂(t) whose generalized integral transforms exist, respectively, then $H\{\beta f_1(t)+\gamma f_2(t)\}=\beta H\{f_1(t)\}+\gamma H\{f_2(t)\}.$

Proof

By applying Definition 2.1, we can obtain $\begin{array}{ll}\hfill H\{\beta f_1(t)+\gamma f_2(t)\}& =P(s){\int }_0^{\infty }{e}^{-g(s)t}[\beta f_1(t)+\gamma f_2(t)]\mathrm{d}t\hfill \\ \hfill & =\beta P(s){\int }_0^{\infty }{f}_1(t)e^{-g(s)t}\mathrm{d}t\hfill \\ \hfill & +\gamma P(s){\int }_0^{\infty }{f}_2(t)e^{-g(s)t}\mathrm{d}t\hfill \\ \hfill & =\beta H\{f_1(t)\}+\gamma H\{f_2(t)\}.\hfill \end{array}$

Next, we will utilize the following theorem in the subsequent examples.

Theorem 2.5

Let f(t) has n-th derivative at t ≥ 0, then $\begin{array}{ll}\hfill & \left(i\right)H\{f^{\prime }(t)\}=g(s)H(s)-P(s)f(0);\hfill \\ \hfill & \left(ii\right)H\{f^{\prime \prime }(t)\}=g^2(s)H(s)-g(s)P(s)f(0)-P(s)f^{\prime }(0);\hfill \\ \hfill & \left(iii\right)H\{f^{(n)}(t)\}=g^n(s)H(s)-P(s)\sum _{i=0}^{n-1}{g}^{n-i-1}(s)f^{(i)}(0).\hfill \end{array}$

Proof

(i) It follows from Definition 2.1 that $\begin{array}{ll}\hfill H\{f^{\prime }(t)\}& =P(s){\int }_0^{\infty }{f}^{\prime }(t)e^{-g(s)t}\mathrm{d}t\hfill \\ \hfill & =P(s)\left[e^{-g(s)t}f(t){|}_0^{\infty }+g(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t\right]\hfill \\ \hfill & =P(s)\left[g(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t-f(0)\right].\hfill \end{array}$ Thus, $H\{f^{\prime }(t)\}=g(s)H(s)-P(s)f(0).$

(ii) We assume h(t) = f′(t), then h′(t) = f′′(t). Hence we have $\begin{array}{ll}\hfill H\{h^{\prime }(t)\}& =P(s){\int }_0^{\infty }{h}^{\prime }(t)e^{-g(s)t}\mathrm{d}t\hfill \\ \hfill & =g(s)H\{h(t)\}-P(s)h(0)\hfill \\ \hfill & =g(s)H\{f^{\prime }(t)\}-P(s)f^{\prime }(0)\hfill \\ \hfill & =g(s)[g(s)H(s)-P(s)f(0)]-P(s)f^{\prime }(0)\hfill \\ \hfill & =g^2(s)H(s)-g(s)P(s)f(0)-P(s)f^{\prime }(0).\hfill \end{array}$

(iii) By induction, we can prove (iii).

Subsequently, we will elucidate the convolution theorem for generalized integral transformations.

Theorem 2.6

Let H{f₁(t)} = H₁(s), H{f₂(t)} = H₂(s), then $H\{f_1(t)\ast f_2(t)\}=H\{f_1(t)\}H\{f_2(t)\}=\frac{H_1(s)H_2(s)}{P(s)},$ where f₁(t)∗f₂(t) is the convolution of f₁(t) and f₂(t), and is recorded as follows: $f_1(t)\ast f_2(t)={\int }_0^t{f}_1(t-\tau )f_2(\tau )d\tau .$

Proof It follows from Definition 2.1 that $\begin{array}{ll}\hfill H\{f_1(t)\ast f_2(t)\}& =P(s){\int }_0^{\infty }{e}^{-g(s)t}\left({\int }_0^t{f}_1(t-\tau )f_2(\tau )\mathrm{d}\tau \right)\mathrm{d}t\hfill \\ \hfill & =P(s){\int }_0^{\infty }{e}^{-g(s)t}\mathrm{d}t\left({\int }_0^t{f}_1(t-\tau )f_2(\tau )\mathrm{d}\tau \right).\hfill \end{array}$

By altering the order of integration, we have $H\{f_1(t)\ast f_2(t)\}=P(s){\int }_0^{\infty }{f}_2(\tau )\mathrm{d}\tau {\int }_0^{\infty }{e}^{-g(s)t}{f}_1(t-\tau )\mathrm{d}t.$

By the substitution of the variable quantity t − τ = w, the above equation turns into $\begin{array}{ll}\hfill H\{f_1(t)\ast f_2(t)\}& =P(s){\int }_0^{\infty }{f}_2(\tau )\mathrm{d}\tau {\int }_0^{\infty }{e}^{-g(s)t}{f}_1(w)\mathrm{d}w\hfill \\ \hfill & =\frac{H_1(s)H_2(s)}{P(s)}.\hfill \end{array}$

Theorem 2.7

Let P(s) and g(s) be differentiable and g′(s) ≠ 0, and H{f(t)} = H(s) exists. Then $\begin{array}{ll}& \left(i\right)H\left\{tf\left(t\right)\right\}=-\frac{P\left(s\right)}{g^{\prime }\left(s\right)}{\left(\frac{H\left(s\right)}{P\left(s\right)}\right)}^{\prime };\\ & \left(ii\right)H\left\{t^{2}f\left(t\right)\right\}={(-1)}^2\frac{P\left(s\right)}{g^{\prime }\left(s\right)}{\left(\frac{1}{g^{\prime }\left(s\right)}{\left(\frac{H\left(s\right)}{P\left(s\right)}\right)}^{\prime }\right)}^{\prime };\\ & \left(iii\right)H\left\{t^{n}f\left(t\right)\right\}={(-1)}^n\frac{P\left(s\right)}{g^{\prime }\left(s\right)}\underset{n-1}{\underbrace{\left(\frac{1}{g^{\prime }\left(s\right)}\right.\left(\frac{1}{g^{\prime }\left(s\right)}(\right.\dots \left(\frac{1}{g^{\prime }\left(s\right)}\right.}}{\left(\frac{H\left(s\right)}{P\left(s\right)}\right)}^{\prime }\underset{n-1}{\underbrace{{{)}^{\prime }\dots )}^{\prime }}}.\end{array}$

Proof

(i) From Definition 2.1, we can get $H\{f(t)\}=H(s)=P(s){\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t,$ $\frac{H(s)}{P(s)}={\int }_0^{\infty }{e}^{-g(s)t}f(t)\mathrm{d}t.$ Differentiating on both sides of the equation with respect to s, we have $\begin{array}{ll}\hfill {\left(\frac{H(s)}{P(s)}\right)}^{\prime }& =-g^{\prime }(s){\int }_0^{\infty }{e}^{-g(s)t}tf(t)\mathrm{d}t\hfill \\ \hfill & =-g^{\prime }(s)\frac{H\{tf(t)\}}{P(s)}.\hfill \end{array}$ Hence, we obtain $H\{tf(t)\}=-\frac{P(s)}{g^{\prime }(s)}{\left(\frac{H(s)}{P(s)}\right)}^{\prime }.$

(ii) It follows from (i) that

$H\{tf(t)\}=-\frac{P(s)}{g^{\prime }(s)}{\left(\frac{H(s)}{P(s)}\right)}^{\prime }.$

In the same way, it is leads to $H\{t^{2}f(t)\}={(-1)}^2\frac{P(s)}{g^{\prime }(s)}{\left(\frac{1}{g^{\prime }(s)}{\left(\frac{H(s)}{P(s)}\right)}^{\prime }\right)}^{\prime }.$

(iii) By the equivalent process, we can get proof (iii) done.

We then arrive at the following theorem.

Theorem 2.8

Let P(s) and g(s) be differentiable, and the n-order derivative of f(t) is continuous, then $\begin{array}{ll}\hfill & (i)H\{t{f}^{(n)}(t)\}=-\frac{P(s)}{g^{\prime }(s)}\frac{d}{ds}\left[\frac{1}{P(s)}H\{f^{(n)}(t)\}\right];\hfill \\ \hfill & (ii)H\{t^2{f}^{(n)}(t)\}={(-1)}^2\frac{P(s)}{g^{\prime }(s)}\frac{d}{ds}\left[\frac{1}{g^{\prime }(s)}\left(\frac{d}{ds}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right)\right)\right];\hfill \\ \hfill & (iii)H\{t^n{f}^{(m)}(t)\}={(-1)}^n\frac{P(s)}{g^{\prime }(s)}\frac{d}{ds}\left[\underset{n}{\underbrace{\frac{1}{g^{\prime }(s)}\left(\frac{d}{ds}\dots \frac{1}{g^{\prime }(s)}\left(\frac{d}{ds}\right.\right.}}\left.\left.\left(\frac{1}{P(s)}H\{f^{(m)}(t)\}\right)\right)\dots \right)\right].\hfill \end{array}$

Proof

(i) From Definition 2.1, we can get $H\{f^{(n)}(t)\}=P(s){\int }_0^{\infty }{e}^{-g(s)t}{f}^{(n)}(t)\mathrm{d}t.$

By derivation of equation with respect to s, we have $\begin{array}{ll}\hfill \frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right)& =-g^{\prime }(s){\int }_0^{\infty }{e}^{-g(s)t}t{f}^{(n)}(t)\mathrm{d}t\hfill \\ \hfill & =-g^{\prime }(s)\frac{H\{t{f}^{(n)}(t)\}}{P(s)}.\hfill \end{array}$ Hence, $H\{t{f}^{(n)}(t)\}=-\frac{P(s)}{g^{\prime }(s)}\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right).$

(ii) It follows from (i), we can obtain

$H\{t{f}^{(n)}(t)\}=-\frac{P(s)}{g^{\prime }(s)}\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right).$

By derivation of equation with respect to s, we have $\begin{array}{ll}\hfill -\frac{\mathrm{d}}{\mathrm{d}s}\left[\frac{1}{g^{\prime }(s)}\left(\frac{d}{ds}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right)\right)\right]& =-g^{\prime }(s){\int }_0^{\infty }{e}^{-g(s)t}{t}^2{f}^{(n)}(t)\mathrm{d}t\hfill \\ \hfill & =-g^{\prime }(s)\frac{H\{t^2{f}^{(n)}(t)\}}{P(s)}.\hfill \end{array}$ so, $H\{t^2{f}^{(n)}(t)\}={(-1)}^2\frac{P(s)}{g^{\prime }(s)}\frac{\mathrm{d}}{\mathrm{d}s}\left[\frac{1}{g^{\prime }(s)}\left(\frac{d}{ds}\left(\frac{1}{P(s)}H\{f^{(n)}(t)\}\right)\right)\right].$

(iii) By the above process, we can get proof (iii) done.

Generalized integral transform with variational iteration method

The variational iteration method ¹⁴ is a universal tool for solving nonlinear problems. It was first proposed in 1999 and has since been widely used to solve various nonlinear oscillators, particularly MEMS oscillators.^15–18 MEMS oscillators present unique challenges due to their zero initial conditions and singularities, which can make them difficult to solve. However, alternative approaches to identifying their periodic properties have been developed to address these issues.^19–21 The primary challenge in the variational iteration method lies in accurately identifying the Lagrange multiplier. A general approach to this is the variational method. ¹⁴ Although the variational principle is a widely used tool in engineering,^22–24 the identification of the Lagrange multiplier involved in the variational iteration algorithm requires the use of some simplifying assumptions. To address this limitation, numerous modifications have been proposed, including the use of dual Lagrange multipliers. ²⁵ Among the various modifications, the incorporation of the Laplace transform has significantly enhanced the method’s appeal. ²⁶ Subsequent developments include the Aboodh Transform-based Variational Iteration Method ²⁷ and the Elzaki Transform-based Variational Iteration Method, ²⁸ which address specific nonlinear problems. Furthermore, many analytical methods depend upon the initial guess. ²⁹ This paper proposes a generalized integral transform-based Variational Iteration Method.

Consider a general nonlinear oscillator $x^{\prime \prime }(t)+f(x)=0,$ $x(0)=A,x^{\prime }(0)=0.$

We can rewrite the equation in the following form $x^{\prime \prime }+w^{2}x+V(x)=0,$ where $V(x)=f(x)-w^{2}x.$ By the variation iteration method, ¹⁴ the correction functional for equation is given as $x_{m+1}(t)=x_m(t)+{\int }_0^t\lambda (t,\xi )\left[x_m^{\prime \prime }(\xi )+w^2{x}_m(\xi )+\tilde{V}(x_m)\right]\mathrm{d}\xi ,m=\mathrm{0,1,2},\dots$ where λ is the general Lagrange multiplier. And we will suggest an alternative approach for the identification of the multiplier. According to He, ¹⁴ the multiplier can be expressed in the form $\lambda =\lambda (t-\xi ).$

Utilizing generalized integral transform H on both sides of the equation, the correction functional will be transformed into $H\left\{x_{m+1}\left(t\right)\right\}=H\left\{x_m\left(t\right)\right\}+H\left\{{\int }_0^t\lambda (t-\xi )\left[x_m^{\prime \prime }\left(\xi \right)+w^2{x}_m\left(\xi \right)+\tilde{V}\left(x_m\right)\right]\mathrm{d}\xi \right\}.$

Thus $\begin{array}{ll}\hfill H\left\{x_{m+1}(t)\right\}& =H\left\{x_m(t)\right\}+H\left\{\lambda (t)\ast \left(x_m^{\prime \prime }(t)+w^2{x}_m(t)+\tilde{V}(x_m)\right)\right\}\hfill \\ \hfill & =H\left\{x_m(t)\right\}+\frac{1}{P(s)}H\left\{\lambda (t)\right\}H\left\{x_m^{\prime \prime }(t)+w^2{x}_m(t)+\tilde{V}(x_m)\right\}\hfill \\ \hfill & =H\left\{x_m(t)\right\}+\frac{1}{P(s)}H\left\{\lambda (t)\right\}\left[(g^2(s)+w^2)H\left\{x_m(t)\right\}-g(s)P(s)x_m(0)\right.\hfill \\ \hfill & \left.-P(s)x_m^{\prime }(0)+H\left\{\tilde{V}(x_m)\right\}\right].\hfill \end{array}$

Then take the variation with respect to x_m(t). We can obtain $\begin{array}{ll}\hfill \frac{\delta }{\delta x_m}H\left\{x_{m+1}(t)\right\}=\frac{\delta }{\delta x_m}H\left\{x_m(t)\right\}& +\frac{\delta }{\delta x_m}\frac{1}{P(s)}H\left\{\lambda (t)\right\}\left[(g^2(s)+w^2)H\left\{x_m(t)\right\}\right.\hfill \\ \hfill & \left.-g(s)P(s)x_m(0)+H\left\{\tilde{V}(x_m)\right\}\right].\hfill \end{array}$

In the above derivation, we assume that $\frac{\delta H[\tilde{V}(x_m)]}{\delta x_m}=0.$

Upon applying the variation, the equation can be simplified accordingly. $H\left\{\delta x_{m+1}\right\}=H\left\{\delta x_m\right\}+\frac{1}{P(s)}H\left\{\lambda \right\}(g^2(s)+w^2)H\left\{\delta x_m\right\}.$

By applying the extremum condition, we have the follow stationary condition as $H\left\{\delta x_m\right\}+\frac{1}{P(s)}H\left\{\lambda \right\}(g^2(s)+w^2)H\left\{\delta x_m\right\}=0.$ $H\left\{\lambda \right\}=-\frac{P(s)}{g^2(s)+w^2}.$

By applying the generalized integral transform H⁻¹ (see Table 2) on the above equation, we can get the Lagrange multiplier λ $\lambda (t)=-\frac{1}{w}\sin wt.$

So, the iterative formula has the following form $H\left\{x_{m+1}(t)\right\}=H\left\{x_m(t)\right\}-\frac{1}{w}H\left\{{\int }_0^t\sin (t-\xi )\left[x_m^{\prime \prime }(\xi )+w^2{x}_m(\xi )+\tilde{V}(x_m)\right]\mathrm{d}\xi \right\}.$

This approach allows us to derive the fundamental iterative formula for solving this class of nonlinear oscillators. The specific solution process will be illustrated in Example 9.

Applications

In this section, the generalized integral transform is utilized to address a range of differential equations problems, including initial value problems (IVPs), ordinary differential equations (ODEs), Volterra integral equations, and fractional integral equations. Finally, we address the issue of nonlinear oscillators using the variational iteration method.

Solving initial value problem by generalized integral transform

Consider IVP as follows: $u^{(m)}(t)+b_1{u}^{(m-1)}(t)+\cdots +b_{m}u(t)=\widehat{v}(t),$ $u(0)=u_0,u^{\prime }(0)=u_1,\dots ,u^{(m-1)}(0)=u_{m-1}.$

Now we use generalized integral transform H. Then we can obtain $\begin{array}{l}\hfill H\left\{u^{(m)}(t)+b_1{u}^{(m-1)}(t)+\cdots +b_{m}u(t)\right\}=H\left\{\widehat{v}(t)\right\},\\ \hfill H\left(\left\{u^{(m)}(t)\right\}+b_{1}H\left\{u^{(m-1)}(t)\right\}\right)+\cdots +H\left\{b_{m}u(t)\right\}=H\left\{\widehat{v}(t)\right\}.\\ \hfill g^m(s)H(s)-P(s)\sum _{i=0}^{m-1}{g}^{m-i-1}(s)u^{(i)}(0)+b_1{g}^{m-1}(s)H(s)\\ \hfill -b_{1}P(s)\sum _{i=0}^{m-2}{g}^{m-i-2}(s)u^{(i)}(0)+\cdots +b_{m}H(s)=V(s),\end{array}$ where $V(s)=H\{\widehat{v}(s)\}$ . After applying the initial condition, we can get $h(s)H(s)=V(s)+\mathrm{\Phi }(s),$ where $\begin{array}{ll}\hfill h(s)& =\left(g^m(s)+b_1{g}^{m-1}(s)+\cdots +b_m\right),\hfill \\ \hfill \mathrm{\Phi }(s)& =P(s)\left(\sum _{i=0}^{m-1}{g}^{m-i-1}(s)u^{(i)}+b_1\sum _{i=0}^{m-2}{g}^{m-i-2}(s)u^{(i)}+\cdots +b_{m}u\right).\hfill \end{array}$ The generalized integral transform H(s) is $H(s)=\frac{V(s)}{h(s)}+\frac{\mathrm{\Phi }(s)}{h(s)}.$

In the end, use generalized integral transform H⁻¹ on both sides of the equation, and then $u(t)=H^{-1}\left\{\frac{V(s)}{h(s)}\right\}+H^{-1}\left\{\frac{\mathrm{\Phi }(s)}{h(s)}\right\}.$

Example 1

Homogenous IVP

Consider the second-order homogenous IVP $\begin{array}{l}\hfill u^{\prime \prime }(t)+u^{\prime }(t)-30u(t)=0,\\ \hfill u(0)=1,u^{\prime }(0)=0.\end{array}$

After applying generalized integral transform H on both sides of the equation, we can obtain $H\{u^{\prime \prime }(t)\}+H\{u^{\prime }(t)\}-30H\{u(t)\}=0,$ $g^2(s)H(s)-P(s)[g(s)u(0)+u^{\prime }(0)+u(0)]+H(s)((a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_{1}s+a_0)-30)=0.$

By using the initial conditions, the equation can be simplified into $[g^2(s)+g(s)-30]H(s)=P(s)\left(g(s)+1\right).$

From above equation, we can get $H(s)=\frac{P(s)\left(g(s)+1\right)}{g^2(s)+g(s)-30}.$ after simplification, $H(s)=\frac{6P(s)}{11\left(g(s)-5\right)}+\frac{5P(s)}{11\left(g(s)+6\right)}.$

Then using generalized integral transform H⁻¹ (see Table 2) on each side of the equation, we have following solution $\begin{array}{ll}\hfill u(t)& =H^{-1}\left\{\frac{6P(s)}{11\left(g(s)-5\right)}+\frac{5P(s)}{11\left(g(s)+6\right)}\right\}\hfill \\ \hfill & =\frac{6}{11}{e}^{5t}+\frac{5}{11}{e}^{-6t}.\hfill \end{array}$

Example 2

Inhomogenous IVP

Consider this third-order inhomogenous IVP $\begin{array}{ll}\hfill u^{\prime }(t)-u(t)& =e^{2t},\hfill \\ \hfill u(0)& =0.\hfill \end{array}$

Using generalized integral transform H (see Table 2) on each side of the equation and applying the initial conditions, we can get $g(s)H(s)-H(s)=\frac{P(s)}{g(s)-2}.$

And then by the simplification of above equation, we have $H(s)=\frac{P(s)}{g(s)-2}-\frac{P(s)}{g(s)-1},$ utilizing generalized integral transform H⁻¹ (see Table 2) on both sides of the equation, the following solution is $u(t)=e^{2t}-e^t.$

Solving system of first-order ODEs by generalized integral transform

Consider the system of first-order ODEs: $\begin{array}{ll}\hfill u_1^{\prime }(t)& =b_{11}{u}_1(t)+b_{12}{u}_2(t)+\cdots +b_{1\alpha }{u}_{\alpha }(t)+v_1(t),\hfill \\ \hfill u_2^{\prime }(t)& =b_{21}{u}_1(t)+b_{22}{u}_2(t)+\cdots +b_{2\alpha }{u}_{\alpha }(t)+v_2(t),\hfill \\ \hfill ⋮\\ \hfill u_{\alpha }^{\prime }(t)& =b_{\alpha 1}{u}_1(t)+b_{\alpha 2}{u}_2(t)+\cdots +b_{\alpha \alpha }{u}_{\alpha }(t)+v_{\alpha }(t),\hfill \end{array}$ and initial conditions are $u_1(0)=u_1^{(0)},u_2(0)=u_2^{(0)},\dots ,u_{\alpha }(0)=u_{\alpha }^{(0)}.$

Now we use generalized integral transform to each side of the equation and then use Theorem 2.2 and Theorem 2.4, now we have $\begin{array}{ll}\hfill H\{u_1^{\prime }(t)\}& =H\left\{b_{11}{u}_1(t)+b_{12}{u}_2(t)+\cdots +b_{1\alpha }{u}_{\alpha }(t)+v_1(t)\right\},\hfill \\ \hfill H\{u_2^{\prime }(t)\}& =H\left\{b_{21}{u}_1(t)+b_{22}{u}_2(t)+\cdots +b_{2\alpha }{u}_{\alpha }(t)+v_2(t)\right\},\hfill \\ \hfill ⋮\\ \hfill H\{u_{\alpha }^{\prime }(t)\}& =H\left\{b_{\alpha 1}{u}_1(t)+b_{\alpha 2}{u}_2(t)+\cdots +b_{\alpha \alpha }{u}_{\alpha }(t)+v_{\alpha }(t)\right\}.\hfill \end{array}$

Assuming that H₁(s), H₂(s), …, H_α(s) are the corresponding generalized transforms of u₁(t), u₂(t), …, u_α(t), with conditions $\begin{array}{ll}g\left(s\right)H_1\left(s\right)-P\left(s\right)u_1\left(0\right)& =b_{11}{H}_1\left(s\right)+b_{12}{H}_2\left(S\right)+\dots +b_{1\alpha }{H}_{\alpha }\left(s\right)+V_1\left(s\right),\\ g\left(s\right)H_2\left(s\right)-P\left(s\right)u_2\left(0\right)& =b_{21}{H}_1\left(s\right)+b_{22}{H}_2\left(S\right)+\dots +b_{2\alpha }{H}_{\alpha }+V_2\left(s\right),\\ ⋮& \\ g\left(s\right)H_{\alpha }\left(s\right)-P\left(s\right)u_{\alpha }\left(0\right)& =b_{\alpha 1}{H}_1\left(s\right)+b_{\alpha 2}{H}_2\left(S\right)+\dots +b_{\alpha \alpha }{H}_{\alpha }+V_{\alpha }\left(s\right).\end{array}$

It can be written as an algebraic system of α linear equations as $g(s)H(s)=P(s)U_0+BH(s)+V,$ and $\begin{array}{c}\hfill H(s)=\left[\begin{array}{c}\hfill H_1(s)\hfill \\ \hfill H_2(s)\hfill \\ \hfill ⋮\hfill \\ \hfill H_{\alpha }(s)\hfill \end{array}\right],U_0=\left[\begin{array}{c}\hfill U_1^{(0)}\hfill \\ \hfill U_2^{(0)}\hfill \\ \hfill ⋮\hfill \\ \hfill U_{\alpha }^{(0)}\hfill \end{array}\right],\hfill \\ \hfill B=\left[\begin{array}{cccc}\hfill b_{11}\hfill & \hfill b_{12}\hfill & \hfill \dots \hfill & \hfill b_{1\alpha }\hfill \\ \hfill b_{21}\hfill & \hfill b_{22}\hfill & \hfill \dots \hfill & \hfill b_{2\alpha }\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill \\ \hfill b_{\alpha 1}\hfill & \hfill b_{\alpha 2}\hfill & \hfill \dots \hfill & \hfill b_{\alpha \alpha }\hfill \end{array}\right],V=\left[\begin{array}{c}\hfill V_1(s)\hfill \\ \hfill V_2(s)\hfill \\ \hfill ⋮\hfill \\ \hfill V_{\alpha }(s)\hfill \end{array}\right]\hfill \end{array}.$

Once g(s)H(s) = P(s)U₀ + BH(s) + V is solved by method of any algebraic means or solved, the generalized integral transform H⁻¹ is used to the values of H₁(s), H₂(s), …, H_α(s) to obtain u₁(t), u₂(t), …, u_α(t).

Example 3

Homogenous system of first-order ODEs

Consider the first-order homogenous system of ODEs: $u_1^{\prime }(t)+7u_2(t)=0,u_1(0)=0,$ $u_2^{\prime }(t)+7u_1(t)=0,u_2(0)=1.$

By employing generalized integral transform H on both sides of the equation, we can get $g(s)H_1(s)-P(s)u_1(0)+7H_2(s)=0,$ $g(s)H_2(s)-P(s)u_2(0)+7H_1(s)=0.$

After introducing the initial conditions, we can obtain $\begin{array}{ll}\hfill g(s)H_1(s)+7H_2(s)& =0,\hfill \\ \hfill g(s)H_2(s)-P(s)+7H_1(s)& =0.\hfill \end{array}$

Solving the above system for H₁(s) and H₂(s), we have $H_1(s)=\frac{7P(s)}{g{(s)}^2+49},$ $H_2(s)=\frac{g(s)P(s)}{g{(s)}^2+49}.$

By utilizing generalized integral transform H⁻¹ (see Table 2), the solution is $u_1(t)=\sin 7t,$ $u_2(t)=\cos 7t.$

Example 4

Inhomogenous system of first-order ODEs

Consider the first-order inhomogenous system of ODEs: $\begin{array}{ll}\hfill u_1^{\prime }(t)& =3u_1(t)+5u_2(t)+e^{-t},\hfill \\ \hfill u_2^{\prime }(t)& =-5u_1(t)+3u_2(t).\hfill \\ \hfill u_1(0)& =0,u_2(0)=1.\hfill \end{array}$

Using generalized integral transform H (see Table 2) on both sides of the equation simultaneously and applying the initial conditions, we have $\begin{array}{ll}\hfill g(s)H_1(s)& =3H_1(s)+5H_2(s)+\frac{P(s)}{g(s)+1},\hfill \\ \hfill g(s)H_2(s)& =3H_2(s)-5H_1(s).\hfill \end{array}$

By calculating the above equation, we can get $H_1(s)=\frac{1}{41}\left[\frac{4(g(s)-3)}{{(g(s)-3)}^2+5^2}+\frac{230}{{(g(s)-3)}^2+5^2}-\frac{4}{Q(s)+1}\right],$ $H_2(s)=\frac{1}{41}\left[\frac{46(g(s)-3)}{{(g(s)-3)}^2+5^2}-\frac{20}{{(g(s)-3)}^2+5^2}-\frac{5}{g(s)+1}\right].$

By utilizing generalized integral transform H⁻¹ (see Table 2), we can obtain $\begin{array}{ll}\hfill u_1(t)& =\frac{1}{41}{e}^{3t}(4\cos t+46\sin 5t-4e^{-4t}),\hfill \\ \hfill u_2(t)& =\frac{1}{41}{e}^{3t}(46\cos 5t-4\sin 5t-5e^{-4t}).\hfill \end{array}$

Example 5

Consider the following second-order ODE with variable coefficient $a{t}^2{y}^{\prime \prime }(t)+bt{y}^{\prime }(t)+cy(t)=a_0{t}^n,n\in \mathbb{N},$ $y(0)=y^{\prime }(0)=0.$ where a, b, c and a₀ are constants. We apply generalized integral transform H on both sides of the above equation. $aH\{t^2{y}^{\prime \prime }(t)\}+bH\{t{y}^{\prime }(t)\}+cH\{y(t)\}=a_{0}H\{t^n\}.$ (2)

Now in the view of Theorem 2.7 and equation (2), we have $a{(-1)}^2\frac{P(s)}{g(s)}\frac{d}{ds}\left[\frac{1}{g^{\prime }(s)}\frac{d}{ds}\left(\frac{1}{P(s)}H\{y^{\prime \prime }(t)\}\right)\right]-b\frac{P(s)}{g^{\prime }(s)}\frac{d}{ds}\left[\frac{1}{P(s)}H\{y^{\prime }(t)\}\right]+cH\{y(t)\}=a_{0}H\{t^n\},$ $a{(-1)}^2\frac{P(s)}{g(s)}\frac{d}{ds}\left[\frac{1}{g^{\prime }(s)}\frac{d}{ds}\left(\frac{1}{P(s)}{g}^2(s)H(s)\right)\right]-b\frac{P(s)}{g^{\prime }(s)}\frac{d}{ds}\left[\frac{1}{P(s)}g(s)H(s)\right]+cH(s)=a_0\frac{n!P(s)}{g^{n-1}(s)}.$ (3)

In the below, we listed the transformation of equation (3) for different integral transforms:

1. Laplace transform:

$a{s}^2{H}^{\prime \prime }(s)+s(4a-b)H^{\prime }(s)+H(s)(2a-b+c)-a_0\frac{n!}{s^{n+1}}=0.$

• 2. Pourreza transform:

$a{s}^2{H}^{\prime \prime }(s)+s(5a-2b)H^{\prime }(s)+H(s)(3a-2b+4c)-4a_0\frac{n!}{s^{n+1}}=0.$

3. Elzaki transform: $a{s}^2{H}^{\prime \prime }(s)+s(b-4a)H^{\prime }(s)+H(s)(6a-2b+c)-a_{0}n!s^{n+1}=0.$

4. Sawi transform:

$a{s}^2{H}^{\prime \prime }(s)+s(2a+b)H^{\prime }(s)+H(s)(b+c)-a_{0}n!s^{n-1}=0.$

• 5. Sumudu transform:

$a{s}^2{H}^{\prime \prime }(s)+bs{H}^{\prime }(s)+cH(s)-a_{0}n!s^n=0.$

Remark 3.1

It is evident that a second-order ordinary differential equation (ODE) remains. However, the Laplace, Pourreza, and Elzaki transforms may yield a simplified second-order ODE if the coefficients of H′(s) and H(s) are set to zero.

1. If in the equation we have a = 2, b = 5, c = 1, a₀ = 24 and n = 2, then we can apply the Pourreza transform as a best choice. It gives H′′(s) = $\frac{96}{s^7}$ . It can be easily solved. By solving this equation, we find y(t) = $\frac{8}{5}{t}^2$ which is the exact solution. ¹³

2. Let a = 1, b = 4, c = 2, a₀ = 12 and n = 2 in the equation. So the Elzaki transform is a best choice. It gives H′′(s) = 24s². By solving this equation, we find y(t) = t² which is the exact solution. ¹³

3. Let a = 1, b = −2, c = 2, a₀ = 6 and n = 4, then we can apply the Sawi transform as a best choice. It gives H′′(s) = 144s. By solving this equation, we find y(t) = t⁴ which is the exact solution.

Solving integral equations by generalized integral transform

Example 6

Volterra integral equation of the first kind

Volterra integral equation is widely appeared in engineering, ³⁰ and consider the following equation $\cos t-\sin t-1=-{\int }_0^{t}2u(\tau )d\tau .$

By applying generalized integral transform H (see Table 2) and Theorem 2.4, we can obtain $\frac{P(s)}{g{(s)}^2+1}-\frac{g(s)P(s)}{g{(s)}^2+1}+\frac{P(s)}{g(s)}=\frac{2}{P(s)}\frac{P(s)}{g(s)}H(s).$ through simplification, it can be get that $H(s)=\frac{1}{2}\left(\frac{g(s)P(s)}{g{(s)}^2+1}+\frac{g{(s)}^{2}P(s)}{g{(s)}^2+1}\right).$

By using generalized integral transform H⁻¹ (see Table 2), the result is $\begin{array}{ll}\hfill u(t)& =H^{-1}\left\{\frac{1}{2}\left(\frac{g(s)P(s)}{g{(s)}^2+1}+\frac{g{(s)}^{2}P(s)}{g{(s)}^2+1}\right)\right\}\hfill \\ \hfill & =\frac{1}{2}\cos t+\frac{1}{2}\sin t.\hfill \end{array}$

Example 7

Volterra integral equation of the second kind

Consider this $u(t)=t+{\int }_0^t(\tau -t)u(\tau )d\tau .$

Utilizing generalized integral transform H (see Table 2) on the both sides of the equation, we can obtain $H(s)=\frac{P(s)}{g{(s)}^2}-\frac{1}{g{(s)}^2}H(s).$

After simplification $H(s)=\frac{P(s)}{g{(s)}^2+1}.$

Employing generalized integral transform H⁻¹ (see Table 2) on both sides of the equation, we have $u(t)=H^{-1}\left\{\frac{P(s)}{g{(s)}^2+1}\right\}=\sin t.$

Example 8

Fractional integral equation

Consider the fractional integral equation $u(t)=v(t)+\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_0^t{(t-\tau )}^{\alpha -1}u(\tau )d\tau ,\alpha \in R^{+}.$

Using generalized integral transform H on both sides of the equation, we have $H\{u(t)\}=H\{v(t)\}+\frac{1}{\mathrm{\Gamma }(\alpha )P(S)}{H}_1(s)H\{u(t)\},$ then $H\{u(t)\}=H\{v(t)\}+\frac{1}{\mathrm{\Gamma }(\alpha )P(S)}\frac{\mathrm{\Gamma }(\alpha )P(s)}{g(s)}H\{u(t)\}.$

By simple calculation, we can obtain $H\{u(t)\}=\frac{g{(s)}^{\alpha }}{g{(s)}^{\alpha }-1}H\{v(t)\},$ and then $u(t)=H^{-1}\left\{\frac{g{(s)}^{\alpha }}{g{(s)}^{\alpha }-1}H\{v(t)\}\right\}.$

Solving nonlinear oscillator equations by generalized integral transform and variation iteration method

Example 9

Consider a general nonlinear oscillator^31,32 $\begin{array}{ll}\hfill (1+\alpha x^2)x^{\prime \prime }+\alpha x{(x^{\prime })}^2-x(1-x^2)& =0,\\ \hfill x(0)& =A,x^{\prime }(0)=0.\end{array}$

To apply the variational iteration method with generalized integral transform H on the equation, we can rewrite equation as $x^{\prime \prime }+w^{2}x+V(x)=0,$ where $V(x)=-(1+w^2)x+\alpha x^2{x}^{\prime \prime }+\alpha x{(x^{\prime })}^2+x^3=0.$

Next, utilizing the iterative formula we previously obtained, the iterative formula is developed as $H\left\{x_{m+1}(t)\right\}=H\left\{x_m(t)\right\}-\frac{1}{w}H\left\{{\int }_0^t\sin w(t-\xi )\left[x_m^{\prime \prime }(\xi )+w^2{x}_m(\xi )+\tilde{V}(x_m)\right]d\xi \right\}.$ $H\left\{x_{m+1}\right\}=H\left\{x_m\right\}-\frac{1}{wP(s)}H\left\{\sin wt\right\}H\left\{x^{\prime \prime }-x_m+\alpha x_m^2{x}_m^{\prime \prime }+\alpha x_m{(x_m^{\prime })}^2+x_m^3\right\}.$

Assuming $x_0(t)=A\cos wt.$ then we have $\begin{array}{ll}\hfill H\left\{x_1(t)\right\}& =H\left\{A\cos wt\right\}-\frac{1}{wP(s)}\left(-A{w}^2-A+\frac{3}{4}{A}^3-\frac{1}{2}\alpha A^3{w}^2\right)H\left\{\sin wt\right\}H\left\{\cos wt\right\}\hfill \\ \hfill & -\frac{1}{wP(s)}\left(\frac{1}{4}{A}^3-\frac{1}{2}\alpha A^3{w}^2\right)H\left\{\sin wt\right\}H\left\{\cos 3wt\right\}.\hfill \end{array}$

The generalized integral transform H⁻¹ (see Table 2) on above equation results in the first-order approximate solution $\begin{array}{ll}\hfill x_1(t)& =A\cos wt-\frac{1}{w}\left(-A{w}^2-A+\frac{3}{4}{A}^3-\frac{1}{2}\alpha A^3{w}^2\right)\left(\frac{1}{2}t\sin wt\right)\hfill \\ \hfill & -\frac{1}{w}\left(\frac{1}{4}{A}^3-\frac{1}{2}\alpha A^3{w}^2\right)\left(\frac{1}{8w}(\cos wt-\cos 3wt)\right).\hfill \end{array}$

No secular-term in x₁ requires that $\left(-A{w}^2-A+\frac{3}{4}{A}^3-\frac{1}{2}\alpha A^3{w}^2\right)=0,$

We can arrive at the following solution $w=\sqrt{\frac{\frac{3}{4}{A}^2-1}{1+\frac{1}{2}\alpha A^2}}.$

It is extremely useful for treating with nonlinear oscillators as illustrated in this example, one single iteration leads to a high accuracy of the solution.

Conclusions

This paper presents a generalized integral transform that encompasses numerous other integral transforms as particular cases, while preserving their intrinsic characteristics. The generalized integral transform can be utilized to address a multitude of differential equations problems, including initial value problems, ordinary differential equations, Volterra integral equations, and fractional integral equations. Concurrently, we have successfully implemented this generalized integral transform in the variational iteration algorithm, thereby facilitating a streamlined identification process for the Lagrange multiplier. By employing the generalized integral transform, we have derived an optimal variational iteration algorithm that converges rapidly, achieving a highly accurate solution in a single iteration. This provides mathematicians and engineers with a novel approach to selecting the most suitable integral transformation for each distinct scenario. In comparison to the Laplace transform, this generalized integral transform presents a more extensive range of potential applications, suggesting a multitude of prospective avenues for future applications in domains such as control systems, circuit analysis, signal processing, and numerous others.

In the future, our objective is to relax the constraints on the conditions for the existence of the integral, thereby enhancing its adaptability in diverse contexts and exploiting its validity across various fields.

Moreover, an effort will be made to integrate the transform with homotopy perturbation methods like that for Aboodh transformation-based homotopy perturbation method ³³ with the objective of enhancing the resolution of nonlinear problems and fractal/fractional differential equations.