A study on different classes of differential equations by semi-analytical and numerical techniques

Sachin K. Narayana; Suguntha Devi Kannadasan; Kumbinarasaiah Srinivasa

doi:10.2478/ijmce-2026-0004

Introduction

Almost every domain relies on the nonlinear ODEs. Their applications arise in the modeling biological, chemical, mechanical, and electrical applications. With the help of calculus, it became apparent that not all ODEs and PDEs could be analytically solved. To tackle this challenge, numerical methods have been devised to approximate solutions for ODEs. Renowned techniques like Heun, Runge-Kutta, Adams-Bashforth, linear multistep, and Euler forward and backward methods emerged to address this need. In addition to analytical and numerical techniques, we have a semi-analytical method to solve such equations [1].

In this paper, we consider one among the semi-analytical methods called HAM to solve the differential equations. In this frame, the initial-value problems (IVPs) of the first order ODEs is taken in the following form

v^{'} (y) = g (z, v (z)), v (z_{0}) = v_{0},

v^{\prime}(y)=g(z,v(z)), v(z_{0})=v_{0},

and also the second-order IVPs of ODEs is considered as

v^{″} (z) = g (z, v (z), v^{'} (z)), v (z_{0}) = v_{0}, v^{'} (z_{0}) = v_{0}^{'} .

v^{\prime\prime}(z)=g(z,v(z),v^{\prime}(z)), v(z_{0})=v_{0}, v^{\prime}(z_{0})=v_{0}^{\prime}.

The n system of ODEs is studied as

\begin{array}{l} \begin{array}{l} \begin{array}{l} v_{1}^{'} (z) & = & g_{1} (z, v_{1} (z), v_{2} (z), \dots, v_{n} (z)), \\ v_{2}^{'} (z) & = & g_{2} (z, v_{1} (z), v_{2} (z), \dots, v_{n} (z)), \\ v_{3}^{'} (z) & = & g_{3} (z, v_{1} (z), v_{2} (z), \dots, v_{n} (z)), \end{array} \\ \begin{array}{l} ⋮ \\ v_{n}^{'} (z) & = & g_{n} (z, v_{1} (z), v_{2} (z), \dots, v_{n} (z)), \end{array} \end{array} \end{array}}

\begin{matrix}v_{1}^{\prime}(z)=g_{1}(z,v_{1}(z),v_{2}(z),\cdot\cdot\cdot,v_{n}(z)),\\ v_{2}^{\prime}(z)=g_{2}(z,v_{1}(z),v_{2}(z),\cdot\cdot\cdot,v_{n}(z))\\ v_{3}^{\prime}(z)=g_{3}(z,v_{1}(z),v_{2}(z),\cdot\cdot\cdot,v_{n}(z)).\\ v_{n}^{\prime}(z)=g_{n}(z,v_{1}(z),v_{2}(z),\cdot\cdot\cdot,v_{n}(z)),\end{matrix}

subject to v_i(0) = vi0 , for i = 0,1,2,· · · ,n.

The time-dependent PDEs being,

v_{z} (z, t) + v_{t} (z, t) = f (z, v, t), v (z, 0) = s (z),

v_{z}(z,t)+v_{t}(z,t)=f(z,v,t), v(z,0)=s(z),

is examined via this work. These problems are frequently utilised to represent real-world problems in engineering, social sciences, physics, and economics; they have garnered a lot of interest from scholars in the field of numerical analysis. It's also crucial to remember that a first-order system may be used to represent chemical kinetic problems, a population model for logistic growth, extremely stiff oscillatory problems, a SIR model, and other problems of a similar nature. Since many of these kinds of problems are thought to lack exact solutions, numerical approximations are obtained. Many experts have used various techniques to provide the numerical solutions, particularly, multi-step methods such as Runge-Kutta methods [2–6], and Hybrid block methods [7, 8]. Since some numerical techniques lead to numerical instability while addressing stiff systems, these issues are regarded as complicated problems. Many experts have worked very hard to develop more effective techniques for addressing stiff systems like A-EBDF: an adaptive method [9], Carrol [10] in a matricial exponentially fitted scheme, modified extended backward differentiation [11]. Haar wavelet and single term Haar wavelet series [12, 13], Matrix free MEBDF method [14], multistep multi derivative hybrid methods [15], stability and accuracy of one-step methods [16], a semi-implicit midpoint rule [17], wavelet methods [18–21], variational iteration method [22], fractional order PDEs [23–25] and so on.

Here, we consider the semi-analytical method for solving DEs called HAM which has the independent parameter, in contrast to perturbation approaches that depend on small or large parameters. The choice of linear operator, auxiliary function, and control convergence parameter is entirely up to the user. This approach often solves high nonlinearity differential equations [26]. Moreover, HAM is an analytical method that generates a sequence of convergent linear equations from a nonlinear one by applying homotopy, or the distortion of one continuous function into another, to solve nonlinear ordinary or partial differential equations. Liao introduced the HAM first in 1992, and it underwent additional changes in 1997 with the addition of the auxiliary parameter h. This non-zero parameter gives the series control over its convergence. We are free to select the auxiliary linear operator, h, the convergence control value, and the initial approximation of the solution because HAM is predicated on the idea of homotopy. This feature distinguishes HAM from other approaches as it allows to select the base functions of the high-order deformation equation's solution as well as its equation type [27]. An analytical approximation technique for highly nonlinear systems is the HAM. In the past, perturbation techniques were often employed. Nevertheless, perturbation techniques heavily rely on the presence of small physical characteristics, and in addition, perturbation approximations frequently diverge as perturbation quantity increases. However, since the HAM is based on homotopy, a fundamental idea in topology, unlike perturbation techniques, it has nothing to do with the presence of small or large physical factors. In particular, the HAM offers a practical means of ensuring that the solution series will converge. The methodology may be used with other approaches for solving nonlinear differential equations, such as Pade approximation and spectral methods [28]. The HAM is an analytic approximation method as opposed to the discrete computing approach of Homotopy continuation. On the other hand, the HAM demonstrates that a nonlinear system may be divided into endless linear systems that can be solved analytically using the Homotopy parameter merely in theory. In the computer era, the HAM was developed as an analytical approximation method to "compute with functions instead of numbers". Combining the HAM with a computer algebra program such as Maple or Mathematica allows to get analytic approximations of a higher-order strongly nonlinear problem easily [29].

The rest of this paper is organized as follows. In Section 2, the preliminaries of the wavelets and the convergence theorems are introduced. In Section 3, HAM for ODE, the systems of ODE and PDE are explained. In Section 4, the application of problems are investigated and results are discussed in terms of tables and graphs. Finally, Section 5 concludes the paper.

Preliminaries

In this part of the paper, we present some important definitions and theorems.

Wavelet: A real-valued function Ψ(ξ) satisfies the following conditions [30, 31]: $\int_{- \infty}^{\infty} Ψ (ξ) d ξ = 0 and \int_{- \infty}^{\infty} | Ψ (ξ) |^{2} d ξ = 1 .$ \int_{-\infty}^{\infty}\Psi(\xi)d\xi=0 \text{ and } \int_{-\infty}^{\infty}|\Psi(\xi)|^{2}d\xi=1. This means that Ψ(ξ) is an oscillatory function having unit energy and zero mean.

Theorem 1.

As long as the series $v_{i} (y) = \sum_{m = 0}^{m} v_{i_{m}} (y)$ v_{i}(y)=\sum_{m=0}^{m}v_{i_{m}}(y) converges, it must be the exact solution of equation (14). Where v_im (y) is governed by the m^th order deformation equation [32].

Theorem 2.

Let $ψ_{0}, ψ_{1}, ψ_{2}, \dots$ \psi_0, \psi_1, \psi_2, \dots be the components of solution of equation (14). The series solution $\sum_{k = 0}^{\infty} ψ_{k} (t)$ \sum_{k=0}^{\infty}\psi_{k}(t) converges if ∃ 0 < γ < 1 such that $‖ ψ_{k + 1} ‖ \leq γ ‖ ψ_{k} ‖, \forall k \geq k_{0}$ ||\psi_{k+1}||\le\gamma||\psi_{k}||,for some $k_{0} \in ℕ$ k\ge k_{0} for some k₀ ∈ ℕ [33].

Theorem 3.

Assume that the series solution $\sum_{k = 0}^{\infty} ψ_{k} (t)$ \sum_{k=0}^{\infty}\psi_{k}(t) is convergent to the solution v(y), if the truncation series $Σ_{k = 0}^{m} ψ_{k} (t)$ \Sigma_{k=0}^{m}\psi_{k}(t) is the approximation to the solution v_i(y), then the maximum absolute truncation error is calculated as, $‖ v_{i} (y) - \sum_{k = 0}^{m} ψ_{k} (t) ‖ \leq \frac{1}{1 - γ} γ^{m + 1} ‖ ψ_{0} (t) ‖$ ||v_{i}(y)-\sum_{k=0}^{m}\psi_{k}(t)||\le\frac{1}{1-\gamma}\gamma^{m+1}||\psi_{0}(t)|| [33].

Theorem 4.

Suppose that the functions $D_{*}^{α} v_{i, k} (y)$ D_{*}^{\alpha}v_{i,k}(y) are the approximation of $D_{*}^{α} v_{i} (y)$ D_{*}^{\alpha}v_{i}(y)obtained using Haar wavelets, then we have an exact upper bound as follows:

$‖ D_{*}^{α} v_{i} (y) - D_{*}^{α} v_{i, k} (y) ‖_{E} \leq \frac{M}{Γ (m - α) (m - a)} \frac{1}{{[1 - 2^{2 (α - m)}]}^{\frac{1}{2}}} \frac{1}{k^{m - a}} . where ‖ v_{i} (y) ‖_{E} = {(\int_{0}^{1} {(v_{i})}^{2} (y) d y)}^{\frac{1}{2}}$ ||D_{*}^{\alpha}v_{i}(y)-D_{*}^{\alpha}v_{i,k}(y)||_{E}\le\frac{M}{\Gamma(m-\alpha).(m-a)}\frac{1}{[1-2^{2(\alpha-m)}]^{\frac{1}{2}}\frac{1}{k^{m-a}}. ||v_{i}(y)||_{E}=(\int_{0}^{1}(v_{i})^{2}(y)dy)^{\frac{1}{2}}

[34].

Description of method

3.1

HAM for ODE

Let us consider the ODE with different physical conditions,

M [v (y)] = 0, y \geq 0 .

M[v(y)] = 0, y\ge0.

Where M is the differential operator, and v(y) is the function to be determined.

3.1.1

Zeroth order deformation equation

Let v₀(y) be the initial approximation to the actual solution of equation (5). The zeroth deformation equation is constructed by using the auxiliary function $H (y) (\neq 0)$ H(y)(\ne0) and auxiliary parameter $h (\neq 0)$ h(\ne0) as [31, 35],

(1 - q) S [ψ (y; q) - v_{0} (y)] = q h H (y) M [ψ (y; q)]

(1-q)\mathcal{L}[\psi(y;q)-v_{0}(y)]=qhH(y)M[\psi(y;q)]

Where, ψ(y;q) is an unknown function, 𝒮 is a linear operator. When q=0, equation (6) becomes, ψ(y; 0)= v₀ and at q=1, equation (6) becomes, ψ(y; 1) = v(y). So as the q varies from 0 to 1, the function ψ(y;q) varies from initial approximation v₀ (y) to the actual solution v(y). Defining the m^th order deformation derivatives,

v_{m} (y) = \frac{1}{m!} \frac{\partial^{m} ψ (y; q)}{\partial q^{m}}

v_{m}(y)=\frac{1}{m!}\frac{\partial^{m}\psi(y;q)}{\partial q^{m}}

Expanding ψ(y;q) using the Taylor series with respect to (w.r.t.) q. We get,

ψ (y; q) = v_{0} (y) + \sum_{m = 1}^{\infty} v_{m} (y) q^{m} .

\psi(y;q)=v_{0}(y)+\sum_{m=1}^{\infty}v_{m}(y)q^{m}.

As we know, ψ(y;q) becomes the desired solution at q = 1. At q=1, equation (8) becomes

ψ (y; 1) = v (y) = v_{0} (y) + \sum_{m = 1}^{\infty} v_{m} (y) .

\psi(y;1)=v(y)=v_{0}(y)+\sum_{m=1}^{\infty}v_{m}(y).

Similarly, deformation equation of m^th order is obtained as,

𝒮 [v_{m} (y) - χ_{m} v_{m - 1} (y)] = h H (y) R_{m} (v_{m - 1} (y)) .

\mathcal{L}[v_{m}(y)-\chi_{m}v_{m-1}(y)]=hH(y)R_{m}(v_{m-1}(y)).

Where,

χ_{m} = {\begin{array}{l} 0 if m \leq 1 \\ 1 Otherwise \end{array}

\chi_{m}=\begin{cases}0&if&m\le1\\ 1&Otherwise\end{cases}

R_{m} (v_{m - 1} (y)) = \frac{1}{(m - 1)!} \frac{\partial^{m - 1} [M [ψ (y; q)]]}{\partial q^{m - 1}}

R_{m}(v_{m-1}(y))=\frac{1}{(m-1)!}\frac{\partial^{m-1}[M[\psi(y;q)]]}{\partial q^{m-1}}

Thus v₁(y), v₂(y), v₃(y),· · ·, can be attained on solving equation (10). The m^th order approximation of v (y) [36] is given by

v (y) = \sum_{m = 0}^{m} v_{m} (y) .

v(y)=\sum_{m=0}^{m}v_{m}(y).

Equation (13) is the semi-analytical solution of equation (5).

3.2

HAM for the system of ODE

Let us consider the system of stiff ODEs with different physical conditions [37],

[v_{i}^{'} (y)] = g_{i} (y, v_{1}, v_{2}, \dots, v_{n}), i = 1, 2, 3, \dots, n, y \geq 0 .

[v_{i}^{\prime}(y)]=g_{i}(y,v_{1},v_{2},\cdot\cdot\cdot,v_{n}), i=1,2,3,\cdot\cdot\cdot,n, y\ge0.

subject to the condition:

v_{i} (0) = a_{i}, i = 1, 2, 3, \dots, n .

v_{i}(0)=a_{i}, i=1,2,3,\cdot\cdot\cdot,n.

3.2.1

Zeroth order deformation equation

Let v_i0 (y), i = 1,2,3, · · · , n be the initial approximation to the actual solution of equation (14). The zeroth deformation equations are taking the auxiliary functions ℋ (y) (6≠ 0) and auxiliary parameter h (6≠ 0) as [35, 38],

(1 - q) 𝒮_{i} [ψ_{i} (y; q) - v_{i_{0}} (y)] = q h H_{i} (y) M_{i} [ψ_{i} (y; q)], i = 1, 2, 3, \dots, n

(1-q)\mathcal{L}_{i}[\psi_{i}(y;q)-v_{i_{0}}(y)]=qhH_i(y)M_{i}[\psi_{i}(y;q)], i=1,2,3,\cdot\cdot\cdot,n

subject to the conditions:

ψ_{i} (0; q) = a_{i}, i = 1, 2, 3, \dots, n .

\psi_{i}(0;q)=a_{i}. i=1,2,3,\cdot\cdot\cdot,n.

Where, ψ_i(y;t) are unknown functions, $𝒮_{i}$ \mathcal{L}_{i} are the Linear operators. When q=0 equation (16) becomes, ψ_i(y; 0) = v_i0 (y) and at q=1, equation (16) becomes ψ_i(y; 1) = v_i(y). So as the q varies from 0 to 1, the function ψi(y;q) varies from initial approximation v_i0 (y) to the actual solution v_i(y), i = 1,2,3 · · · n. Defining the m^th order deformation derivatives ,

v_{i_{m}} (y) = \frac{1}{m!} \frac{\partial^{m} ψ_{i} (y; q)}{\partial q^{m}}, i = 1, 2, 3, \dots, n .

v_{im}(y)=\frac{1}{m!}\frac{\partial^{m}\psi_{i}(y;q)}{\partial q^{m}} i=1,2,3,\cdot\cdot\cdot,n.

Expanding ψ_i(y;q) using Taylor series w.r.t. q. i = 1,2,3, · · · ,n. We get,

ψ_{i} (y; q) = v_{i_{0}} (y) + \sum_{m = 1}^{\infty} v_{i_{m}} (y) q^{m}, i = 1, 2, 3, \dots, n .

\psi_{i}(y;q) = v_{i_{0}}(y) + \sum_{m=1}^{\infty}v_{i_m}(y)q^m i=1,2,3,\cdot\cdot\cdot,n

As we know at q = 1ψ_i(y;q) becomes the required solution, equation (19) at q = 1 becomes

ψ_{i} (y; 1) = v_{i} (y) = v_{i_{0}} (y) + \sum_{m = 1}^{\infty} v_{i_{m}} (y), i = 1, 2, 3, \dots, n .

\psi_{i}(y;1)=v_{i}(y)=v_{i_{0}}(y)+\sum_{m=1}^{\infty}v_{i_{m}}(y) i=1,2,3,\cdot\cdot\cdot,n.

Similarly, the m^th order deformation is given by

𝒮_{i} [v_{i_{m}} (y) - χ_{m} v_{i_{m - 1}} (y)] = h H_{i} (y) R_{i, m} (v_{i_{m - 1}} (y)), i = 1, 2, 3, \dots, n .

\mathcal{L}_{i}[v_{i_{m}}(y)-\chi_{m}v_{i_{m-1}}(y)]=hH_{i}(y)R_{i,m}(v_{i_{m-1}}(y)), i=1,2,3,\cdot\cdot\cdot,n.

Where,

χ_{m} = {\begin{array}{l} 0 if m \leq 1, \\ 1 Otherwise . \end{array}

\chi_{m}=\begin{cases}0&if&m\le1,\\ 1&Otherwise.\end{cases}

R_{i, m} (v_{i_{m - 1}} (y)) = \frac{1}{(m - 1)!} \frac{\partial^{m - 1} [M_{i} [ψ_{i} (y; q)]]}{\partial q^{m - 1}}, i = 1, 2, 3, \dots, n .

R_{i,m}(v_{i_{m-1}}(y))=\frac{1}{(m-1)!}\frac{\partial^{m-1}[M_i[\psi_{i}(y;q)]]}{\partial q^{m-1}}, i=1,2,3,\cdot\cdot\cdot,n.

Thus v_i1 (y), v_i2 (y), v_i3 (y),· · · are the first , can be obtained from solving equation (21). The m^th order approximation of v_i (y) [39] is given by

v_{i} (y) = \sum_{m = 0}^{m} v_{i_{m}} (y) .

v_{i}(y)=\sum_{m=0}^{m}v_{i_{m}}(y).

Equation (24) is the semi-analytical solution of equation (14).

3.3

HAM for PDE

Let us consider the PDE with different physical conditions,

M [v (y, t)] = 0, y, t \geq 0 .

\mathcal{M}[v(y,t)]=0, y,t\ge0.

Where ℳ is the differential operator, and v(y,t) is the function to be determined.

3.3.1

Zeroth order deformation equation

Let v₀(y,t) be the initial approximation to the actual solution of equation (25). The zeroth deformation equation is constructed using the auxiliary function ℋ(y,t) (6≠ 0) and auxiliary parameter h (6≠ 0) as [31, 35],

(1 - q) 𝒮 [ψ (y, t; q) - v_{0} (y, t)] = q h H (y, t) M [ψ (y, t; q)] .

(1-q)\mathcal{L}[\psi(<italic>y</italic>,<italic>t</italic>;<italic>q</italic>)-v_{0}(y,t)]=qhH(y,t)M[\psi(<italic>y</italic>,<italic>t</italic>;<italic>q</italic>)].

Where, ψ(y,t;q) is unknown function, 𝒮 is Linear operator. When q=0, equation (26) becomes ψ(y,t;0) = v₀ (y,t). At q=1, equation (26) becomes ψ(y,t;1) = v(y,t;q). So as the q varies from 0 to 1, the function ψ(y,t) varies from initial approximation v₀ (y,t) to the actual solution v(y,t). Defining the m^th order deformation derivatives,

v_{m} (y, t) = \frac{1}{m!} \frac{\partial^{m} ψ (y, t; q)}{\partial q^{m}} .

v_{m}(y,t)=\frac{1}{m!}\frac{\partial^{m}\psi(<italic>y</italic>,<italic>t</italic>;<italic>q</italic>)}{\partial q^{m}}.

Expanding ψ(y,t;q) using the Taylor series w.r.t. q. We get,

ψ (y, t; q) = v_{0} (y, t) + \sum_{m = 1}^{\infty} v_{m} (y, t) q^{m} .

\psi(<italic>y</italic>,<italic>t</italic>;<italic>q</italic>)=v_{0}(y,t)+\sum_{m=1}^{\infty}v_{m}(y,t)q^{m}.

As we know, ψ(y,t;q) becomes the desired solution at q = 1. At q = 1, equation (28) becomes

ψ (y, t; 1) = v (y, t) = v_{0} (y, t) + \sum_{m = 1}^{\infty} v_{m} (y, t)

\psi(y,t;1)=v(y,t)=v_{0}(y,t)+\sum_{m=1}^{\infty}v_{m}(y,t)

Similarly, deformation equation of m^th order is obtained as,

𝒮 [v_{m} (y, t) - χ_{m} v_{m - 1} (y, t)] = h H (y, t) R_{m} (v_{m - 1} (y, t)),

\mathcal{L}[v_{m}(y,t)-\chi_{m}v_{m-1}(y,t)]=hH(y,t)R_{m}(v_{m-1}(y,t)),

where,

χ_{m} = {\begin{array}{l} 0 if m \leq 1, \\ 1 Otherwise . \end{array}

\chi_{m}=\begin{cases}0&if&m\le1,\\ 1&Otherwise.\end{cases}

R_{m} (v_{m - 1} (y, t)) = \frac{1}{(m - 1)!} \frac{\partial^{m - 1} [M [ψ (y, t; q)]]}{\partial q^{m - 1}} .

R_{m}(v_{m-1}(y,t))=\frac{1}{(m-1)!}\frac{\partial^{m-1}[M[\psi(<italic>y</italic>,<italic>t</italic>;<italic>q</italic>)]]}{\partial q^{m-1}}.

Thus v₁(y,t), v₂(y,t), v₃(y,t), · · · can be attained on solving equation (30). The m^th order approximation of v (y) [32, 36, 39] is given by

v (y, t) = \sum_{m = 0}^{m} v_{m} (y, t) .

v(y,t)=\sum_{m=0}^{m}v_{m}(y,t).

Equation (33) is the semi-analytical solution of equation (25).

Applications

We use some package programs to solve the following problems by HAM.

4.1

Application 4.1

Let us consider the following ODE with different physical conditions

x^{2} \frac{u^{''}}{x} + x \frac{u^{'}}{x} + (x^{2} - 0.25) u (x) = 0; u (1) = \sqrt{\frac{2}{π}} \sin (1), u^{'} (1) = \frac{2 \cos (1) - \sin (1)}{\sqrt{2 π}} .

\textstyle x^2\frac{u^''}x+x\frac{u^'}x+(x^2-0.25)u(x)=0;\hspace{1em}u(1)=\sqrt{\frac2\pi}\sin(1),\hspace{1em}u^'(1)=\frac{2\cos(1)-\sin(1)}{\sqrt{2\pi}}.

On applying HAM to equation (34), the m^th order deformation is given by,

D^{2} [u_{m} (x) - χ_{m} u_{m - 1} (x)] = h R_{m} (u_{m - 1} (x)) .

D^{2}[u_{m}(x)-\chi_{m}u_{m-1}(x)]=hR_{m}(u_{m-1}(x)).

Where, $D^{2} = \frac{d^{2}}{d x^{2}}$ D^{2}=\frac{d^{2}}{dx^{2}},

R_{m} (u_{m - 1} (x)) = x^{2} \frac{d^{2} u_{m - 1}}{d x^{2}} + x \frac{d u_{m - 1}}{d x} + (x^{2} - 0.25) u_{m - 1},

R_{m}(u_{m-1}(x))=x^{2}\frac{d^{2}u_{m-1}}{dx^{2}}+x\frac{du_{m-1}}{dx}+(x^{2}-0.25)u_{m-1},

subject to,

\begin{matrix} u_{m} (1) = 0, u_{m}^{'} (1) = 0 \end{matrix}

u(1)=\sqrt{\frac{2}{\pi}}sin(1), u^{\prime}(1)=\frac{2~cos(1)-sin(1)}{\sqrt{2\pi}}, u_{m}(1)=0, u_{m}^{\prime}(1)=0.

Integrating twice on either sides of equation (35), we get

u_{m} (x) = χ_{m} u_{m - 1} (x) + h \int_{0}^{x} \int_{0}^{x} [R_{m} (u_{m - 1} (x))] d x d x + C_{1} + C_{2} x, m \geq 1.

u_{m}(x)=\chi_{m}u_{m-1}(x)+h\int_{0}^{x}\int_{0}^{x}[R_{m}(u_{m-1}(x))]dxdx+C_{1}+C_{2}x,m\ge1.

The integration constants C₁ and C₂ are calculated using equation (37). The HAM series solution when u₀(x) = 0.671397+0.0954005(x² −x) and h = −1is given by

\begin{array}{l} u (x) = 0.410465 + x (0.171623 + x (0.301729 + x (- 0.00759606 + x (- 0.299954 + x (- 0.0113261 + \\ x (0.12512 + x (- 0.00184181 + x (- 0.0160226 + x (0.00206268 + x (- 0.00288598 + x (0.0000767687 + \\ x (- 0.0000536825 + (4.24682 \times 10^{- 7} - 1.9414 \times 10^{- 7} x (x + \dots))))))))))))) \end{array}

u(x)=0.410465+x(0.171623+x(0.301729+x(-0.00759606+x(-0.299954+x(-0.0113261+ x(0.12512 + x(-0.00184181 + x(-0.0160226 + x(0.00206268 + x(-0.00288598 + x(0.0000767687 + x(-0.0000536825+(4.24682\times10^{-7}-1.9414\times10^{-7}x(x+\dots))))))))))))))

Table 1

Comparison of solutions obtained from HAM, HWT, and their absolute errors (AE) with Exact solution for Application 4.1.

x	Exact	HAM	HWT	HAM Error	HWT Error
0.5	0.541049	0.540974	0.54147	7.554 × 10^-5	5.0 × 10^-4
0.6	0.581622	0.581618	0.58234	4.062 × 10^-6	7.2 × 10^-4
0.7	0.614361	0.614361	0.61534	3.496 × 10^-7	9.8 × 10^-4
0.8	0.639926	0.639926	0.64121	2.917 × 10^-7	1.28 × 10^-3
0.9	0.658813	0.658813	0.66043	2.933 × 10^-7	1.62 × 10^-3
1.0	0.671397	0.671397	0.67340	2.929 × 10^-7	2 × 10^-3
1.1	0.677989	0.677989	0.68041	2.904 × 10^-7	2.42 × 10^-3
1.2	0.678866	0.678865	0.68175	2.838 × 10^-7	2.88 × 10^-3
1.3	0.674286	0.674289	0.67767	3.718 × 10^-6	3.38 × 10^-3
1.4	0.663534	0.664524	0.66844	9.9 × 10^-4	3.92 × 10^-3

The HWT [40–42] is also used to solve the problem studied. The exact solution equation (34) is $u (x) = \sqrt{\frac{2}{π x}} \sin (x)$ u(x)=\sqrt{\frac{2}{\pi x}}\sin(x) [43]. Table 1 contains the numerical values of the solutions obtained from HAM and HWT and also their AE with exact solutions. The tables and graphs explain the results. Geometric comparisons of the Exact, HAM, and HWT solutions are shown in Figure 1. Figure 2 presents a graphic representation of the error analysis of HAM and HWT solutions with the exact solution.

4.2

Application 4.2

Let us consider the following ODE with the initial condition [43]

u^{'} (x) = 0.5 (1 - u (x)), u (0) = 0.5

u^{\prime}(x)=0.5(1-u(x)), u(0)=0.5

On applying HAM to equation (38). The m^th order deformation is given by,

D [u_{m} (x) - χ_{m} u_{m - 1} (t)] = h R_{m} (u_{m - 1} (t)) .

D[u_{m}(x)-\chi_{m}u_{m-1}(t)]=hR_{m}(u_{m-1}(t)).

Where, $D = \frac{d}{d x}$ D=\frac{d}{dx}, 40 $R_{m} (u_{m - 1} (t)) = \frac{d u_{m - 1}}{d x} + 0.5 u_{m - 1} - 0.5 (1 - χ_{m}),$ R_{m}(u_{m-1}(t))=\frac{du_{m-1}}{dx}+0.5u_{m-1}-0.5(1-\chi_{m}), subject to, 41 $u_{m} (0) = 0.$ u_{m}(0)=0.

Integrating on the either sides of equation (39), we get

u_{m} (x) = χ_{m} u_{m - 1} (x) + h \int_{0}^{x} [R_{m} (u_{m - 1} (x))] d x + C_{1}, m \geq 1.

u_{m}(x)=\chi_{m}u_{m-1}(x)+h\int_{0}^{x}[R_{m}(u_{m-1}(x))]dx+C_{1}, m\ge1.

C₁ is a integration constant calculated using equation (41). Taking u₀(x) = 0.5 and setting h = −1 successively we obtain,

\begin{matrix} u_{1} (x) & = & 0.25 x, \\ u_{2} (x) & = & - 0.0625 x^{2}, \\ u_{3} (x) & = & 0.0104167 x^{3}, \\ u_{4} (x) & = & 0.00130208 x^{4}, \\ ⋮ . \end{matrix}

u_{1}(x)=0.25x u_{2}(x)=-0.0625x^{2}, u_{3}(x)=0.0104167x^{3} u_{4}(x)=0.00130208x^{4}

Collectively, HAM series solution $u (x) = \sum_{m = 0}^{\infty} u_{m} (x)$ u(x)=\sum_{m=0}^{\infty}u_{m}(x) is given by

u (x) = 0.5 + 0.25 x - 0.0625 x^{2} + 0.0104167 x^{3} - 0.00130208 x^{4} + 0.000130208 x^{5} \dots .

u(x)=0.5+0.25x-0.0625x^{2}+0.0104167x^{3}-0.00130208x^{4}+0.000130208x^{5}\cdot\cdot\cdot

Table 2

Comparison of solutions obtained from HAM, HWT, and their AE with ND Solver solution for Application 4.2.

x	ND Solver	HAM	HWT	HAM Error	HWT Error
0	0.5	0.5	0.5	0.	0
0.1	0.52438	0.52438	0.52439	2.048 × 10^-8	1.120 × 10^-6
0.2	0.54758	0.54758	0.54759	1.718 × 10^-8	4.480 × 10^-6
0.3	0.56964	0.56964	0.56966	1.413 × 10^-8	1.008 × 10^-5
0.4	0.59063	0.59063	0.59065	9.894 × 10^-9	1.792 × 10^-5
0.5	0.61059	0.61059	0.61063	2.164 × 10^-8	2.800 × 10^-5
0.6	0.62959	0.62959	0.62963	1.587 × 10^-8	4.032 × 10^-5
0.7	0.64765	0.64765	0.64771	4.028 × 10^-8	5.488 × 10^-5
0.8	0.66483	0.66483	0.66491	1.566 × 10^-8	7.168 × 10^-5
0.9	0.68118	0.68118	0.68128	1.790 × 10^-8	9.072 × 10^-5
1.0	0.69673	0.69673	0.69685	7.461 × 10^-10	1.120 × 10^-4

This previously stated problem is similarly resolved with the HWT. The numerical values of the solutions derived from HAM and HWT, along with their AE with exact solutions, are shown in Table 2. The tables and graphs provide an explanation of the findings. Figure 3 displays geometric comparisons of the Exact, HAM, and HWT solutions. A graphic illustration of the error analysis of the HWT and HAM answers with the exact solution is shown in Figure 4.

4.3

Application 4.3

Let us consider the following ODE system as follows

\begin{array}{l} u_{1}^{'} (x) = - 1000 u_{1} (x) + 10000 u_{2} {(x)}^{4}, u_{1} (0) = 1, \\ u_{2}^{'} (x) = u_{1}^{'} (x) - u_{2} {(x)}^{4} - u_{2} (x), u_{2} (0) = 1, \end{array}}

u_{1}^{\prime}(x)=-1000u_{1}(x)+10000u_{2}(x)^{4}, \quad u_{1}(0)=1, \\ u_{2}^{\prime}(x)=u_{1}^{\prime}(x)-u_{2}(x)^{4}-u_{2}(x), \quad u_{2}(0)=1,

having the exact solution as u₁(x) = e^−4x and u₂(x) = e^−x [43]. Applying HAM to equation (42), the m^th order deformation are given by,

\begin{array}{l} D [u_{1, m} (x) - χ_{m} u_{1, m - 1} (x)] = h R_{1, m} (u_{1, m - 1} (x)), \\ D [u_{2, m} (x) - χ_{m} u_{2, m - 1} (x)] = h R_{2, m} (u_{2, m - 1} (x)) . \end{array}}

D[u_{1,m}(x)-\chi_{m}u_{1,m-1}(x)]=hR_{1,m}(u_{1,m-1}(x)), D[u_{2,m}(x)-\chi_{m}u_{2,m-1}(x)]=hR_{2,m}(u_{2,m-1}(x)).

Where, $D = \frac{d}{d x}$ D=\frac{d}{dx} and

\begin{aligned} R_{1, m} (u_{1, m - 1} (x)) & = D [u_{1, m - 1} (x)] + 1000 u_{1, m - 1} (x) - 10000 \sum_{j = 0}^{m - 1} u_{2, m - 1 - j} \sum_{i = 0}^{j} u_{2, j - i} \sum_{k = 0}^{i} u_{2, k} u_{2, i - k}, \\ R_{2, m} (u_{2, m - 1} (x)) & = D [u_{1, m - 1} (x)] + \sum_{j = 0}^{m - 1} u_{2, m - 1 - j} \sum_{i = 0}^{j} u_{2, j - i} \sum_{k = 0}^{i} u_{2, k} u_{2, i - k} + u_{2, m - 1} (x), \end{aligned}}

\begin{equation} \begin{rcases} \mathscr{R}_{1,m}(u_{1,m-1}(x)) &= D[u_{1,m-1}(x)]+1000 u_{1,m-1}(x) -10000 \sum_{j=0}^{m-1}u_{2,m-1-j} \sum_{i=0}^{j} u_{2,j-i} \sum_{k=0}^{i} u_{2,k}u_{2,i-k}, \\ \mathscr{R}_{2,m}(u_{2,m-1}(x)) &= D[u_{1,m-1}(x)]+ \sum_{j=0}^{m-1}u_{2,m-1-j} \sum_{i=0}^{j} u_{2,j-i} \sum_{k=0}^{i} u_{2,k}u_{2,i-k} + u_{2,m-1}(x) , \end{rcases} \end{equation}

subject to,

u_{1, m} (0) = 0 u_{2, m} (0) = 0

u_{2,m}(0)=0

Integrating on the either sides of equation (43), we get

\begin{array}{l} u_{1, m} (x) = χ_{m} u_{1, m - 1} (x) + h \int_{0}^{x} [R_{1, m} (u_{1, m - 1} (x))] d x + C_{1}, m \geq 1, \\ u_{2, m} (x) = χ_{m} u_{2, m - 1} (x) + h \int_{0}^{x} [R_{2, m} (u_{2, m - 1} (x))] d x + C_{2}, m \geq 1. \end{array}

u_{1,m}(x)=\chi_{m}u_{1,m-1}(x)+h\int_{0}^{x}[R_{1,m}(u_{1,m-1}(x))]dx+C_{1}, \quad m\ge1, \\ u_{2,m}(x)=\chi_{m}u_{2,m-1}(x)+h\int_{0}^{x}[R_{2,m}(u_{2,m-1}(x))]dx+C_{2}, \quad m\ge1.

The integration constants C₁ and C₂ are calculated using equation (45). Taking u_1,0(x) = 1, u_2,0(x) = 1 and setting h = -1 successively we obtain,

\begin{array}{l} u_{1, 1} (x) = - 4 x, \\ u_{1, 2} (x) = \frac{{(- 4 x)}^{2}}{2}, \\ u_{1, 3} (x) = \frac{{(- 4 x)}^{3}}{6}, \\ u_{1, 4} (x) = \frac{{(- 4 x)}^{4}}{24}, \dots \\ u_{2, 1} (x) = - x, \\ u_{2, 2} (x) = \frac{{(- x)}^{2}}{2}, \\ u_{2, 3} (x) = \frac{{(- x)}^{3}}{6}, \\ u_{2, 4} (x) = \frac{{(- x)}^{4}}{24}, \\ ⋮ . \end{array}

u_{1,1}(x)=-4x, \quad u_{1,2}(x)=\frac{(-4x)^{2}}{2}, \quad u_{1,3}(x)=\frac{(-4x)^{3}}{6}, \quad u_{1,4}(x)=\frac{(-4x)^{4}}{24},\dots \\ u_{2,1}(x)=-x, \quad u_{2,2}(x)=\frac{(-x)^{2}}{2}, \quad u_{2,3}(x)=\frac{(-x)^{3}}{6}, \quad u_{2,4}(x)=\frac{(-x)^{4}}{24},\dots

Collectively, HAM series solution $u_{1} (x) = \sum_{m = 0}^{\infty} u_{1, m} (x)$ u_{1}(x)=\sum_{m=0}^{\infty}u_{1,m}(x) and $u_{2} (x) = \sum_{m = 0}^{\infty} u_{2, m} (x)$ u_{2}(x)=\sum_{m=0}^{\infty}u_{2,m}(x)is given by,

\begin{array}{l} u_{1} (x) = 1 - 4 x + 8 x^{2} - 32 \frac{x^{3}}{3} + 32 \frac{x^{4}}{3} - \dots = e^{- 4 x} . \\ u_{2} (x) = 1 - x + \frac{x^{2}}{2} - \frac{x^{3}}{6} + \frac{x^{4}}{24} - \dots = e^{- x} . \end{array}

u_{1}(x)=1-4x+8x^{2}-32\frac{x^{3}}{3}+32\frac{x^{4}}{3}-\dots=e^{-4x}. \\ u_{2}(x)=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\frac{x^{4}}{24}-\dots=e^{-x}.

The above-mentioned problem is additionally solved using the HWT. Table 3 and 4 present the numerical values of the solutions of u₁ and u₂, respectively, together with their exact solutions and AE. Exact, HAM, and HWT solution geometric comparisons are displayed in Figure 5 and 7 for u₁ and u₂, respectively. An error analysis of the HAM and HWT solutions is shown graphically in Figure 6 and 8 for u₁ and u₂ respectively

Table 3

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for u₁ in Application 4.3.

x	Exact	HAM	HWT	HWT Error
0	1	1	1	0
0.1	0.67032	0.67032	0.67032	1.120 × 10^-7
0.2	0.44932	0.44932	0.44933	8.960 × 10^-7
0.3	0.30119	0.30119	0.30120	3.024 × 10^-6
0.4	0.20189	0.20189	0.20190	7.168 × 10^-6
0.5	0.13533	0.13533	0.13535	1.140 × 10^-5
0.6	0.09071	0.09071	0.09074	2.419 × 10^-5
0.7	0.06081	0.06081	0.06084	3.841 × 10^-5
0.8	0.04076	0.04076	0.04082	5.734 × 10^-5
0.9	0.02732	0.02732	0.02740	8.164 × 10^-5
1.0	0.01831	0.01831	0.01842	1.120 × 10^-4

Table 4

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for u₂ in Application 4.3.

x	ND Solver	HAM	HWT	HAM Error	HWT Error
0	0.5	0.5	0.5	0.	0
0.1	0.52438	0.52438	0.52439	2.048 × 10^-8	1.120 × 10^-6
0.2	0.54758	0.54758	0.54759	1.718 × 10^-8	4.480 × 10^-6
0.3	0.56964	0.56964	0.56966	1.413 × 10^-8	1.008 × 10^-5
0.4	0.59063	0.59063	0.59065	9.894 × 10^-9	1.792 × 10^-5
0.5	0.61059	0.61059	0.61063	2.164 × 10^-8	2.800 × 10^-5
0.6	0.62959	0.62959	0.62963	1.587 × 10^-8	4.032 × 10^-5
0.7	0.64765	0.64765	0.64771	4.028 × 10^-8	5.488 × 10^-5
0.8	0.66483	0.66483	0.66491	1.566 × 10^-8	7.168 × 10^-5
0.9	0.68118	0.68118	0.68128	1.790 × 10^-8	9.072 × 10^-5
1.0	0.69673	0.69673	0.69685	7.461 × 10^-10	1.120 × 10^-4

4.4

Application 4.4

Let us consider the following PDE with the condition

u_{t} + \frac{1}{2} {(u^{2})}_{x} - u (1 - u) = 0; u (x, 0) = e^{- x},

u_{t}+\frac{1}{2}(u^{2})_{x}-u(1-u)=0; \quad u(x,0)=e^{-x},

having the exact solution u(x,t) = e^t−x. By applying HAM to equation (46), the m^th order deformation is given by,

Where, $D = \frac{\partial}{\partial t}$ D=\frac{\partial}{\partial t} and

D [u_{m} (x, t) - χ_{m} u_{m - 1} (x, t)] = h R_{m} (u_{m - 1} (x, t))

D[u_{m}(x,t)-\chi_{m}u_{m-1}(x,t)]=hR_{m}(u_{m-1}(x,t))

R_{m} (u_{m - 1} (x, t)) = D [u_{m - 1} (x, t)] + \frac{1}{2} \frac{\partial}{\partial x} [\sum_{j = 0}^{m - 1} u_{m - 1 - j} (x, t) u_{j} (x, t)] - u_{m - 1} (x, t) + \sum_{j = 0}^{m - 1} u_{m - 1 - j} (x, t) u_{j} (x, t),

R_{m}(u_{m-1}(x,t))=D[u_{m-1}(x,t)]+\frac{1}{2}\frac{\partial}{\partial x}\left[\sum_{j=0}^{m-1}u_{m-1-j}(x,t)u_{j}(x,t)\right]-u_{m-1}(x,t)+\sum_{j=0}^{m-1}u_{m-1-j}(x,t)u_{j}(x,t),

subject to,

u_{m} (x, 0) = 0

u_{m}(x,0)=0

Integrating on the either sides of equation (47), we get

u_{m} (x, t) = χ_{m} u_{m - 1} (x, t) + h \int_{0}^{t} [R_{m} (u_{m - 1} (x, t))] d t + C_{1}, m \geq 1.

u_{m}(x,t)=\chi_{m}u_{m-1}(x,t)+h\int_{0}^{t}[R_{m}(u_{m-1}(x,t))]dt+C_{1}, \quad m\ge1.

The integration constants C₁ and C₂ are calculated using equation (49). Taking u₀(x,t) = e^−x and setting h = −1 successively we obtain,

u_{1} (x, t) = e^{- x} t

u_{1}(x,t)=e^{-x}t. u_{2}(x,t)=e^{-x}\frac{t^{2}}{2} u_{3}(x,t)=e^{-x}\frac{\overline{t^{3}}}{6} u_{4}(x,t)=e^{-x}\frac{t^{4}}{24}

u_{2} (x, t) = e^{- x} \frac{t^{2}}{2},

$u_{2}(x,t)=$ $e^{-x} \frac{t^2}{2}$

u_{3} (x, t) = e^{- x} \frac{t^{3}}{6},

$u_{3}(x,t)=$ $e^{-x} \frac{t^3}{6}$

u_{4} (x, t) = e^{- x} \frac{t^{4}}{24}

$u_{4}(x,t)=$ $e^{-x} \frac{t^4}{24}$

⋮.

Collectively, HAM series solution $u (x, t) = \sum_{m = 0}^{\infty} u_{m} (x, t)$ u(x,t)=\sum_{m=0}^{\infty}u_{m}(x,t) is given by,

u (x, t) = e^{- x} + t e^{- x} + \frac{t^{2}}{2} e^{- x} + \frac{t^{3}}{6} e^{- x} + \frac{t^{4}}{24} e^{- x} + \frac{t^{5}}{120} e^{- x} + \dots = e^{(t - x)}

u(x,t)=e^{-x}+te^{-x}+\frac{t^{2}}{2}e^{-x}+\frac{t^{3}}{6}e^{-x}+\frac{t^{4}}{24}e^{-x}+\frac{t^{5}}{120}e^{-x}+\dots=e^{(t-x)}

This problem is also solved again by HWT, and the results obtained are compared using tables and graphs. Tables 5, 6, 7, 8, 9 and Table 10 contain the numerical values and their errors with exact solutions of the above problem for different values of x. Figures 9, 11, 13, and 15 compare the HWT and HAM solutions with exact solutions, and Figures 10, 12, 14, and 16 represent the graphical interpretation for different values of x and t . Figure 17 depicts the 3-dimensional analysis of the HAM and HWT solutions with Exact solutions.

Table 5

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for x = 0 in Application 4.4.

t	Exact	HAM	HWT
0	1	1	1
0.1	1.10517	1.10517	1.10517
0.2	1.22140	1.22140	1.22140
0.3	1.34986	1.34986	1.34986
0.4	1.49182	1.49182	1.49182
0.5	1.64872	1.64872	1.64872
0.6	1.82212	1.82212	1.82212
0.7	2.01375	2.01375	2.01375
0.8	2.22554	2.22554	2.22554
0.9	2.45960	2.45960	2.45960
1.0	2.71828	2.71828	2.71828

Table 6

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for x = 1 in Application 4.4.

t	Exact	HAM	HWT Error	HWT
0	0.36787	0.36787	0.36787	0
0.1	0.40657	0.40657	0.41224	5.672 × 10^-3
0.2	0.44932	0.44932	0.47202	2.268 × 10^-2
0.3	0.49658	0.49658	0.54763	5.104 × 10^-2
0.4	0.54881	0.54881	0.63956	9.075 × 10^-2
0.5	0.60653	0.60653	0.74833	1.418 × 10^-1
0.6	0.67032	0.67032	0.87451	2.042 × 10^-1
0.7	0.74081	0.74081	1.01870	2.779 × 10^-1
0.8	0.81873	0.81873	1.18175	3.630 × 10^-1
0.9	0.90483	0.90483	1.36432	4.594 × 10^-1
1.0	1	1	1.56721	5.672 × 10^-1

Table 7

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for t = 0 in Application 4.4.

x	Exact	HAM	HWT Error
0	1	1	1
0.1	0.90483	0.90483	0.90483
0.2	0.81873	0.81873	0.81873
0.3	0.74081	0.74081	0.74081
0.4	0.67032	0.67032	0.67032
0.5	0.60653	0.60653	0.60653
0.6	0.54881	0.54881	0.54881
0.7	0.49658	0.49658	0.49658
0.8	0.44932	0.44932	0.44932
0.9	0.40656	0.40656	0.40656
1.0	0.36787	0.36787	0.36787

Table 8

Comparison of solutions obtained from HAM, HWT, and their AE with Exact solution for t = 1 in Application 4.4.

x	Exact	HAM	HWT Error	HWT
0	2.71828	2.71828	2.71828	0
0.1	2.45960	2.45960	2.46021	5.672 × 10^-4
0.2	2.22554	2.22554	2.23012	4.357 × 10^-3
0.3	2.01375	2.01375	2.0291	1.531 × 10^-2
0.4	1.82212	1.82212	1.8584	3.630 × 10^-2
0.5	1.64872	1.64872	1.7196	7.090 × 10^-2
0.6	1.49182	1.49182	1.6143	1.225 × 10^-1
0.7	1.34986	1.34986	1.5444	1.945 × 10^-1
0.8	1.22140	1.22140	1.5118	2.904 × 10^-1
0.9	1.10517	1.10517	1.5187	4.134 × 10^-1
1.0	1	1	1.5672	5.672 × 10^-1

Table 9

Comparison of absolute error AE of the HAM and HWT solutions with Exact solution for t = 0.1, t = 0.01, and t = 0.001 in Application 4.4.

x	Error at t = 0.1		Error at t = 0.01		Error at t = 0.001
x	HAM	HWT	HAM	HWT	HAM	HWT
0	0	0	0	0	0	0
0.1	0	5.672 × 10^-6	0	5.672 × 10^-8	0	5.672 × 10^-10
0.2	0	4.537 × 10^-5	0	4.537 × 10^-7	0	4.537 × 10^-9
0.3	0	1.531 × 10^-4	0	1.531 × 10^-6	0	1.531 × 10^-8
0.4	0	3.630 × 10^-4	0	3.630 × 10^-6	0	3.630 × 10^-8
0.5	0	7.090 × 10^-4	0	7.090 × 10^-6	0	7.090 × 10^-8
0.6	0	1.225 × 10^-3	0	1.225 × 10^-5	0	1.225 × 10^-7
0.7	0	1.945 × 10^-3	0	1.945 × 10^-5	0	1.945 × 10^-7
0.8	0	2.904 × 10^-3	0	2.904 × 10^-5	0	2.904 × 10^-7
0.9	0	4.134 × 10^-3	0	4.134 × 10^-5	0	4.134 × 10^-7
1.0	0	5.672 × 10^-3	0	5.672 × 10^-5	0	5.672 × 10^-7

Table 10

Comparison of absolute error AE of the HAM and HWT solutions with Exact solution for x = 0.1, x = 0.01, and x = 0.001 in Application 4.4.

t	Error at x = 0.1		Error at x = 0.01		Error at x = 0.001
t	HAM	HWT	HAM	HWT	HAM	HWT
0	0	0	0	0	0	0
0.1	0	5.671 × 10^-6	0	5.672 × 10^-9	0	5.671 × 10^-12
0.2	0	2.268 × 10^-5	0	2.268 × 10^-8	0	2.268 × 10^-11
0.3	0	5.104 × 10^-5	0	5.104 × 10^-8	0	5.104 × 10^-11
0.4	0	9.075 × 10^-5	0	9.075 × 10^-8	0	9.075 × 10^-11
0.5	0	1.418 × 10^-4	0	1.418 × 10^-7	0	1.418 × 10^-10
0.6	0	2.041 × 10^-4	0	2.041 × 10^-7	0	2.041 × 10^-10
0.7	0	2.779 × 10^-4	0	2.779 × 10^-7	0	2.779 × 10^-10
0.8	0	3.630 × 10^-4	0	3.630 × 10^-7	0	3.630 × 10^-10
0.9	0	4.594 × 10^-4	0	4.594 × 10^-7	0	4.594 × 10^-10
1.0	0	5.672 × 10^-4	0	5.672 × 10^-7	0	5.672 × 10^-10

Conclusions

In this research, we suggested a scheme for solving the ODEs, system of ODEs, and PDEs using the homotopy analysis and a Haar wavelet methodology. Theorems were used to discuss convergence analysis. To exhibit the efficiency and efficacy of the suggested scheme, tables and figures describing the results were provided, along with numerical examples illustrating the method's effectiveness. The error analysis reveals that the HAM solutions are more accurate than HWT and all other techniques found in literature, and the obtained results agree with the exact or ND solver solutions. Results show that the suggested technique can potentially solve complex ODE and PDE problems efficiently.

A study on different classes of differential equations by semi-analytical and numerical techniques

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