A Semi-Analytical Method for Solving Non-Linear Multi-Dimensional Acoustic Wave Equations

The relationship between engineering problems and other sciences requires the integration of knowledge from multiple fields [1]. Engineers and scientists are able to analyze and understand engineering and scientific phenomena using equations and develop effective solutions to the problems facing the environment and society [2]. These equations help us understand the relationships between different variables, leading to new and effective solutions for engineering problems that promote innovation and technological advancement [3].

Many problems in engineering, science, and the environment can be expressed by nonlinear partial differential equations (PDEs) [4]. When the analytical solution is not feasible, solving a nonlinear PDE presents a challenge for engineers and scientists [5]. Consequently, numerous mathematicians and engineers have explored various methods and algorithms to tackle these problems, including, Finite Element Method [6] Complex variable boundary element method (CVBEM) [7], (TAM) [8], homotopy perturbation method (HPM) [9], and some other analytical and numerical methods [10, 11].

The acoustic wave equation describes the transmission of energy through a medium, including air, water, displacement, and acoustic intensity [12]. The speed of acoustic waves relies on the medium's qualities, such as density, elasticity and differing velocities in solids. Examples of acoustic waves include audible sound from speakers, seismic waves creating ground movements, and ultrasound used for medical imaging [13, 14].

Acoustic wave equations have numerous techniques to be used in solving them, for example. An Effective Discontinuous Galerkin Method [15], The Adomian decomposition approach [16], Variation iteration technique [17], Finite difference method [18], locally one method (LOD) [19], Normal mode analysis [20] Method of fundamental solutions [21], Explicit hybridizable discontinuous Galerkin method (HDGM) [22], the Concept of Perturbed Derivative Order [23], adaptive WKB approximation method [24] partial-low-rank method [25].

The TAM proposed by Temimi and Ansari in 2011 [26] to solve non-linear equations. This technique will be employed to address a range of differential equations, it includes second-order nonlinear ordinary differential equations in the field of physics [27], Falkner-Skan equations [28], non-linear multi-dimensional wave equations [29], differential algebraic equations [30], and nonlinear thin film flow issues [31].

This study primarily aims to employ the iterative TAM method for solving the nonlinear acoustic wave equation in one-dimensional (1D), two-dimensional (2D), and three-dimensional (3D) cases. The outcomes are compared with the exact resolution of this issue to show the efficacy and precision of the suggested approach. The following shows how the paper is divided. Section 2 consists of the standard formulation of the problem. Section 3 provides a comprehensive explanation of the core concepts underlying the suggested iterative technique. In Section 4 of the suggested technique, we will analyze the convergent. Section 5 presents a detailed analysis and discussion of the numerical simulation. Section 6 provides the conclusion.

In this section, the basic concepts of TAM will be presented.

Here is the nonlinear partial differential equation that will be presented [33]:

$L\left( u\left( x,t \right) \right)=N\left( u\left( x,t \right) \right)+g\left( x,t \right)=0$   (4)

with the boundary conditions

$B\left( u,\frac{du}{dt} \right)=0$.

The independent variable is denoted as x, while the variable of time is denoted as t. The function u(x, t) represents the unknown function, whereas g(x, t) is a known given function. L denotes a linear operator, N represents a nonlinear operator, and B(.) stands for the boundary operator. Let us start by considering that this is a starting approximation for solving the Eq. (4).

The initial phase of the resolution approach is to solve the following initially value problem$~{{u}_{0}}\left( x,t \right)$:

$L\left( {{u}_{0}}\left( x,t \right) \right)=g\left( x,t \right),~\text{with}~B\left( {{u}_{0}},\frac{d{{u}_{0}}}{dt} \right)=0$    (5)

The next iterative function $u_1(x, t)$ are obtained by solving the following equation:

$L\left(u_1(x, t)\right)=N\left(u_0(x, t)\right)+g(x, t), B\left(u_1, \frac{d u_1}{d t}\right)=0$   (6)

In the same way, we find iterative function $u_2(x, t)$ by solving the subsequent issue

$L\left(u_2(x, t)\right)=N\left(u_1(x, t)\right)+g(x, t), \quad B\left(u_2, \frac{d u_2}{d t}\right)=0$

In general, we can compute the subsequent iterations. $u_{n+1}(x, t)$ by solving the following problem

$\begin{gathered}L\left(u_{n+1}(x, \mathrm{t})\right)=N\left(u_n(x, \mathrm{t})\right)+\mathcal{g}(x, \mathrm{t}), \\ B\left(u_{n+1}, \frac{d u_{n+1}}{d \mathrm{t}}\right)=0\end{gathered}$    (7)

It should be noted that each $u_n$ serves as a solution to Eq. (4). Additionally, increasing the number of iterations enhances the precision of the approximation results.

The solution for Eq. (4) can be obtained by $u=\lim _{n \rightarrow \infty} u_n$.

Now, we examine the proposed solution to certain linear and nonlinear acoustic wave equations in one, two, and three dimensions.

Example 1. Let's examine the given 1D linear acoustic wave equation by [35]

$\begin{gathered}u_{t t}(x, t)=\frac{1}{V^2} u_{x x}(x, t) \\ \text { with } u(x, 0)=3 \sin \pi \theta, \quad u_t(x, 0)=0\end{gathered}$      (14)

will be solved by using the suggested method, where V=1 [30].

Let's begin by examining the given problem as it is stated

$\begin{gathered}L(u(x, t))=u_{t t}(x, t), \\ N(u(x, t))=v^2 u_{x x}(x, t), \quad g(x, t)=0\end{gathered}$   (15)

The main problem might be expressed as

$\begin{aligned} & L\left(u_0(x, t)\right)=0, \\ & \text { with } u_0(x, 0)=3 \sin \pi \theta, \quad u_{0 t}(x, 0)=0\end{aligned}$     (16)

The subsequent issues can be derived from the overarching general relationship.

$\begin{aligned} L\left(u_{n+1}(x, t)\right) & =g(x, t)+N\left(u_n(x, t)\right)=0, \\ \text { with } u_{n+1}(x, 0) & =3 \sin \pi \theta, \quad u_{(n+1) t}(x, 0)=0\end{aligned}$   (17)

$\begin{gathered}u_{0 t t}(x, t)=0, \\ \text { with } u_0(x, 0)=3 \sin \pi \theta, u_{0 t}(x, 0)=0\end{gathered}$    (18)

To solve the problem set out in Eq. (18) we obtain $u_0=$ $3 \sin \pi \theta$.

The first iteration can be found by assessing the following problem

$\begin{gathered}u_{1 t t}(x, t)=v^2 u_{0 x x}(x, t) \\ \text { with } u_1(x, 0)=3 \sin \pi \theta, u_{1 t}(x, 0)=0 .\end{gathered}$     (19)

Then, the solution of Eq. (19) is

$u_1=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 3 \sin \pi \theta$

The second iteration is

$\begin{gathered}u_{2 t t}(x, t)=v^2 u_{1 x x}(x, t) \\ \text { with } u_2(x, 0)=3 \sin \pi \theta, u_{2 t}(x, 0)=0 .\end{gathered}$     (20)

Then the solution of Eq. (20) is

$\begin{gathered}u_2=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 \sin \pi \theta+\frac{1}{8} t^4 v^4 \sin \pi \theta \\ u_3=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 \sin \pi \theta+\frac{1}{8} t^4 v^4 \sin \pi \theta \\ -\frac{1}{240} t^6 v^6 \sin \pi \theta\end{gathered}$

$\begin{gathered}u_4=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 \sin \pi \theta+\frac{1}{8} t^4 v^4 \sin \pi \theta \\ -\frac{1}{240} t^6 v^6 \sin \pi \theta+\frac{t^8 v^8 \sin \pi \theta}{13440} \\ \begin{array}{c}u_5=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 \sin \pi \theta+\frac{1}{8} t^4 v^4 \sin \pi \theta  \\ -\frac{1}{240} t^6 v^6 \sin \pi \theta+\frac{t^8 v^8 \sin \pi \theta}{13440} \\ -\frac{t^{10} v^{10} \sin \pi \theta}{1209600}\end{array}\end{gathered}$

Finally, by taking the limit

$\begin{gathered}u(x, t)=\lim _{n \rightarrow \infty} u_n \\ u(x, t)=3 \sin \pi \theta-\frac{3}{2} t^2 v^2 \sin \pi \theta+\frac{1}{8} t^4 v^4 \sin \pi \theta \\ -\frac{1}{240} t^6 v^6 \sin \pi \theta+\frac{t^8 v^8 \sin \pi \theta}{13440} \\ -\frac{\mathrm{t}^{10} v^{10} \sin \pi \theta}{1209600}+\cdots\end{gathered}$

Thus, we continue with more iterations until we obtain the exact solution $=3 \sin \pi \theta \operatorname{Cos} \pi v t$.

Example 2. Examine the one-dimensional, nonlinear acoustic wave equation

$\begin{aligned} & u_{t t}-u u_{x x}=2-2\left(x^2+t^2\right) \\ & \text { with } u(x, 0)=x^2, u_t(x, 0)=0.\end{aligned}$        (21)

Eq. (21) will be solved by the TAM.

The following form are provided as,

$\begin{gathered}L(u(x, t))=u_{t t}(x, t), \\ N(u(x, t))=u u_{t t}, \\ g(x, t)=2-2\left(x^2+t^2\right)\end{gathered}$     (22)

The initial problem is

$\begin{gathered}L\left(u_0\right)=2-2\left(x^2+t^2\right), \\ \text { with } u_0(x, 0)=x^2, u_{0 t}(x, 0)=0\end{gathered}$     (23)

The subsequent problems will rely on the generalized iterative formula

$\begin{gathered}L\left(u_{n+1}(x, t)\right)=g(x, t)+N\left(u_n(x, t)\right)=0 \\ \quad \text { with } u_{n+1}(x, 0)=x^2, u_{(n+1) t}(x, 0)=0\end{gathered}$       （24）

will be solving the following initial problem

$\begin{gathered}u_{0 t t}(x, t)=2-2\left(x^2+t^2\right), \\ \text { with } u_0(x, 0)=x^2, u_{0t}(x, 0)=0\end{gathered}$      (25)

we get

$u_0=t^2-\frac{t^4}{6}+x^2-t^2 x^2.$

The initial iteration $u_1(x, t)$ can be found by solving

$\begin{aligned} & u_{1 t t}=u u_{0 x x}+2-2\left(x^2+t^2\right) \\ & \text { with: } u_1(x, 0)=x^2, u_{1 t}(x, 0)=0\end{aligned}$       (26)

The solution of Eq. (26) with initial conditions, will be

$u_1=t^2-\frac{7 t^6}{90}+\frac{t^8}{168}+x^2-\frac{t^4 x^2}{3}+\frac{t^6 x^2}{15}.$

We apply the same process for the iteration $u_2(x, t)$ as follows

$\begin{aligned} & u_{2 t t}=u u_{1 x x}+2-2\left(x^2+t^2\right), \\ & \text { with: } u_2(x, 0)=x^2, u_{2 t}(x, 0)=0.\end{aligned}$   (27)

By solving Eq. (27), we get

$\begin{aligned} u_2=t^2-\frac{37 t^8}{2520} & +\frac{61 t^{10}}{37800}+\frac{7 t^{12}}{17820}-\frac{271 t^{14}}{3439800}+\frac{t^{16}}{302400} \\ & +x^2-\frac{2 t^6 x^2}{45}+\frac{t^8 x^2}{210}+\frac{t^{10} x^2}{405}-\frac{t^{12} x^2}{1485} \\ & +\frac{t^{14} x^2}{20475}, \\ u_5=t^2-\frac{149 t^{10}}{113400} & +\frac{241 t^{12}}{2494800}+\frac{17 t^{14}}{540540}-\frac{11309 t^{16}}{13621608000} \\ & -\frac{110821 t^{18}}{189464184000}-\frac{4019 t^{20}}{16590420000}+\cdots \\ & +x^2-\frac{2 t^{10} x^2}{14175}+\frac{t^{12} x^2}{155925}+\frac{2 t^{14} x^2}{1216215} \\ & +\frac{t^{16} x^2}{8687250}+\frac{2 t^{18} x^2}{65786175} \\ & -\frac{5351 t^{20} x^2}{259602367500} \cdots\end{aligned}$

The exact solution is converged by this series when

$u(x, t)=\lim _{n \rightarrow \infty} u_n(x, t)=x^2+t^2.$

The convergence of the suggested technique is proved using the provided procedure in Eqs. (9)-(12). The iterative scheme for Eq. (21) can be written

$v_0(x, t)=u_0(x, t)=t^2-\frac{t^4}{6}+x^2-t^2 x^2$

Using the technique (TAM), the operator $F\left[v_k\right]$ as represented by Eq. (10) with the term $S_k$ is defined. Thus, the aim of this study is to exercise that proposition by finding a solution to the next challenge

$\begin{aligned} v_{k t t}(x, t)= & \left(\sum_{i=0}^{k-1} v_{i x x}(x, t)\right)\left(\sum_{i=0}^{k-1} v_i(x, t)\right)+2 \\ & -2\left(x^2+t^2\right),\end{aligned}$

with

$\begin{aligned} v_k(x, 0) & =x^2, \quad v_{k t}(x, 0)=0, \quad k \geq 1 \\ v_1=\frac{t^4}{6} & -\frac{7 t^6}{90}+\frac{t^8}{168}+t^2 x^2-\frac{t^4 x^2}{3}+\frac{t^6 x^2}{15}, \\ v_2=\frac{7 t^6}{90}-\frac{13 t^8}{630} & +\frac{61 t^{10}}{37800}+\frac{7 t^{12}}{17820}-\frac{271 t^{14}}{3439800}+\frac{t^{16}}{302400} \\ & +\frac{t^4 x^2}{3}-\frac{t^6 x^2}{9}+\frac{t^8 x^2}{210}+\frac{t^{10} x^2}{405}-\frac{t^{12} x^2}{1485} \\ & +\frac{t^{14} x^2}{20475}, \\ v_3=\frac{37 t^8}{2520} & -\frac{83 t^{10}}{28350}-\frac{739 t^{12}}{2494800}+\frac{4171 t^{14}}{37837800} \\ & -\frac{28177 t^{16}}{6810804000} \\ & -\frac{110821 t^{18}}{189464184000} \ldots \ldots+\frac{2 t^6 x^2}{45}-\frac{t^8 x^2}{126} \\ & -\frac{32 t^{10} x^2}{14175}+\frac{2 t^{12} x^2}{2673}-\frac{17 t^{14} x^2}{405405} \\ & -\frac{t^{16} x^2}{368550}-\frac{703 t^{18} x^2}{546531300}+\cdots\end{aligned}$

The above type of duplicates is used for the computation of the values of $\beta_i$ for the Eq. (13), in which we can get

$\begin{aligned} & \beta_0=\frac{\left\|v_1\right\|}{\left\|v_0\right\|}=0.0294663<1 \\ & \beta_1=\frac{\left\|v_2\right\|}{\left\|v_1\right\|}=0.0106408<1 \\ & \beta_2=\frac{\left\|v_3\right\|}{\left\|v_2\right\|}=0.00427883<1 \\ & \beta_3=\frac{\left\|v_4\right\|}{\left\|v_3\right\|}=0.00228477<1 \\ & \beta_4=\frac{\left\|v_5\right\|}{\left\|v_4\right\|}=0.00142926<1\end{aligned}$

The $\beta_i$ values for $i \geq 0$ and $\forall(x, t): x \in \mathbb{R}, 0<x, t \leq$ 1 are all smaller than 1, indicating that the suggested iterative approach meets the condition for convergence.

We can conduct an additional investigation to evaluate the accuracy of the approximate solution $\left(u_n\right)$, by employing an absolute error relationship $A b s r_n=N\left[A b s\left[w-u_n\right]\right]$, where $w=x^2+t^2$ is the exact solution. The $3 D$ graph of the $Absr_n$ for the approximate solution found by the recommended iterative procedure is shown in Figure 1. Increasing the number of iterations also cuts down on mistakes and improves the accuracy of the answers. Furthermore, Table 1 presents the absolute error values ($Absr_n$) derived from TAM for the iterations with n={1, 3, 5}. The reduction of error is readily apparent with an increase in the number of repetitions, particularly when t=1.

1.png

Figure 1. Three-dimensional plotted graph for the $Absr_n$ at $n=(1,3,5)$ of example 2, where $t=1$

Table 1. The $Absr_n$ for example 2 computed using the TAM method over three iterations with t=1

Example 3. Consider the 2D linear acoustic wave equation shown below [36]

$\begin{gathered}u_{t t}=\frac{1}{V^2}\left(u_{x x}+u_{y y}\right), \\ \text { with: } u(x, y, 0)=\operatorname{Cos}(-x-y), \\ u_t(x, y, 0)=-\sqrt{2} \operatorname{Sin}(-x-y) .\end{gathered}$     (28)

The three proposed iterative method will be used to solve Eq. (28), V=1 [31]

$u_{0 t t}=0$, with $u_0(x, y, 0)=\operatorname{Cos}(-x-y),$

$u_{0 t}(x, y, 0)=-\sqrt{2} \operatorname{Sin}(-x-y)$

Then

$u_0=\operatorname{Cos}(x+y)+\sqrt{2} t \operatorname{Sin}(x+y)$

The initial iteration $u_1(x, y, t)$ can be found by solving

$\begin{gathered}u_{1 t t}=u_{0 x x}, \text { with } u_1(x, y, 0) \operatorname{Cos}(-x-y), u_{1 t}(x, y, 0)= \\ -\sqrt{2} \operatorname{Sin}(-x-y).\end{gathered}$

We get

$\begin{aligned} & u_1=\operatorname{Cos}(x+y)-t^2 \operatorname{Cos}(x+y)+\sqrt{2} t \operatorname{Sin}(x+y) \\ &-\frac{1}{3} \sqrt{2} t^3 \operatorname{Sin}(x+y)\end{aligned}$

The iteration $u_2(x, y, t)$ can be obtained by solving this problem

$\begin{gathered}u_{2 t t}=u_{1 x x}+u_{1 y y}, \\ \text { with } u_2(x, y, 0) \operatorname{Cos}(-x-y), u_{2 t}(x, y, 0) \\ =-\sqrt{2} \operatorname{Sin}(-x-y),\end{gathered}$

then

$\begin{aligned} & u_2=\operatorname{Cos}(x+y)-t^2 \operatorname{Cos}(x+y)+\frac{1}{6} t^4 \operatorname{Cos}(x+y) \\ &+\sqrt{2} t \operatorname{Sin}(x+y)-\frac{1}{3} \sqrt{2} t^3 \operatorname{Sin}(x+y) \\ &+\frac{t^5 \operatorname{Sin}(x+y)}{15 \sqrt{2}}, \\ & u_5=\operatorname{Cos}(x+y)-t^2 \operatorname{Cos}(x+y)+\frac{1}{6} t^4 \operatorname{Cos}(x+y) \\ &-\frac{1}{90} t^6 \operatorname{Cos}(x+y)+\frac{t^8 \operatorname{Cos}(x+y)}{2520} \\ &-\frac{t^{10} \operatorname{Cos}(x+y)}{113400}+\sqrt{2} t \operatorname{Sin}(x+y) \\ &-\frac{1}{3} \sqrt{2} t^3 \operatorname{Sin}(x+y) \ldots \end{aligned}$

When the solution converges to the exact solution, then

$\begin{aligned} & u(x, y, t)=\operatorname{Cos}(\sqrt{2} t-x-y) \\ & \quad=\operatorname{Cos}(x+y)-t^2 \operatorname{Cos}(x+y) \\ & +\frac{1}{6} t^4 \operatorname{Cos}(x+y)-\frac{1}{90} t^6 \operatorname{Cos}(x+y) \\ & +\frac{\mathrm{t} \operatorname{Cos}(x+y)}{2520}-\frac{t^{10} \operatorname{Cos}(x+y)}{113400} \\ & +\sqrt{2} t \operatorname{Sin}(x+y)-\frac{1}{3} \sqrt{2} t^3 \operatorname{Sin}(x+y) \\ & +\frac{t^5 \operatorname{Sin}(x+y)}{15 \sqrt{2}}-\frac{t^7 \operatorname{Sin}(x+y)}{315 \sqrt{2}} \ldots \end{aligned}$

The exact solution can be acquired by $u(x, y, t)=$ $\lim _{n \rightarrow \infty} u_n=\operatorname{Cos}(\sqrt{2} t-x-y)$.

Example 4. Let us examine the one-dimensional nonlinear wave equation.

         ${{u}_{tt}}+{{u}_{yy~=}}u{{u}_{xx}}+2-\left( \frac{{{x}^{2}}+{{y}^{2}}+{{t}^{2}}}{2} \right)$       (29)

with the given initial conditions

$u\left( x,y,0 \right)=\frac{{{x}^{2}}+{{y}^{2}}}{2},~{{u}_{t}}\left( x,y,0 \right)=0.~$

To resolve Eq. (29) utilizing TAM and the given initial conditions, we can express it as follows

$\begin{gathered}L(u)=u_{\mathrm{tt}}(x, y, t), \\ N(u)=u(x, y, t) u_{x x}(x, y, t)-u_{y y}(x, y, t), \\ g(x, y, t)=2-\left(\frac{x^2+y^2+t^2}{2}\right)\end{gathered}$    (30)

The initial problem is

$\begin{gathered}L\left(u_0\right)=2-2\left(x^2+t^2\right), \\ \text { with } u_0(x, y, 0)=\frac{x^2+y^2}{2}, u_{0 t}(x, y, 0)=0\end{gathered}$      (31)

The following issues will be resolved using the generalized iterative formula

$\begin{gathered}L\left(u_{n+1}\right)+N\left(u_n\right)=g(x, y, t), \\ u_{n+1}(x, y, 0)=\frac{x^2+y^2}{2}, u_{n+1}(x, y, 0)=0\end{gathered}$

By solving the Eq. (31), we get

$u_0=t^2-\frac{t^4}{24}+\frac{x^2}{2}-\frac{t^2 x^2}{4}+\frac{y^2}{2}-\frac{t^2 y^2}{4},$

The first iteration $u_1(x, y, t)$ can be obtained by solving

$\begin{gathered}u_{1 t t}=u_0(x, y, t) u_{0 x x}(x, y, t)-u_{0 y y}(x, y, t)+2 -\left(\frac{x^2+y^2+t^2}{2}\right) \\\text { with } u_1(x, y, 0)=\frac{x^2+y^2}{2}, u_{1_{\mathrm{t}}}(x, y, 0)=0\end{gathered}$

The solution will be

$\begin{gathered}u_1=\frac{\mathrm{t}^2}{2}+\frac{\mathrm{t}^4}{12}-\frac{13 \mathrm{t}^6}{720}+\frac{\mathrm{t}^8}{2688}+\frac{x^2}{2}-\frac{\mathrm{t}^4 x^2}{24}+\frac{\mathrm{t}^6 x^2}{240}+\frac{y^2}{2} \\ -\frac{\mathrm{t}^4 y^2}{24}+\frac{\mathrm{t}^6 y^2}{240}.\end{gathered}$

Applying the same process for $u_2$, we have

$\begin{aligned} & u_{2 t t}= u_1(x, y, t) u_{1_{x x}}(x, y, t)-u_{1 y y}(x, y, t)+2 -\left(\frac{x^2+y^2+\mathrm{t}^2}{2}\right) \\ & \text { with } u_2(x, y, 0)=\frac{x^2+y^2}{2}, u_{2 \mathrm{t}}(x, y, 0)=0\end{aligned}$

By solving this equation, we get

$\begin{aligned} u_2=\frac{t^2}{2}+ & \frac{t^6}{180}-\frac{7 t^8}{5760}-\frac{97 t^{10}}{3628800}+\frac{19 t^{12}}{1140480} \\ & -\frac{439 t^{14}}{440294400}+\frac{t^{16}}{77414400}+\frac{x^2}{2}-\frac{t^6 x^2}{360} \\ & +\frac{t^8 x^2}{6720}+\frac{t^{10} x^2}{25920}-\frac{t^{12} x^2}{190080}+\frac{t^{14} x^2}{5241600} \\ & +\frac{y^2}{2}-\frac{t^6 y^2}{360}+\frac{t^8 y^2}{6720}+\frac{t^{10} y^2}{25920}-\frac{t^{12} y^2}{190080} \\ & +\frac{t^{14} y^2}{5241600} .\end{aligned}$

The approximations will continue till $n=5$, for brevity it's not stated.

The convergence analysis of the suggested iterative procedure allows us to determine the values $\beta_i$ for the problem in Eq. (29). Thus, we determine the terms of the series $\sum_{i=0}^{\infty} v_i(x, t)$ in the above Eq. (11), and proceed to deduce them in the following manner:

$\begin{aligned} & \beta_0=\frac{\left\|v_1\right\|}{\left\|v_0\right\|}=0.825234<1 \\ & \beta_1=\frac{\left\|v_2\right\|}{\left\|v_1\right\|}=0.512054<1 \\ & \beta_2=\frac{\left\|v_3\right\|}{\left\|v_2\right\|}=0.211578<1 \\ & \beta_3=\frac{\left\|v_4\right\|}{\left\|v_3\right\|}=0.109301<1 \\ & \beta_4=\frac{\left\|v_5\right\|}{\left\|v_4\right\|}=0.0977504<1\end{aligned}$

where, the$~{{\beta }_{i}}$ values for $~i\ge 0~$and for each $\left( x,y,t \right)$:$~x,y\in R$ , $0<x,y,t\le 1~\text{and}~$are all smaller than 1. Therefore, the suggested iterative approach guarantees convergence.

To evaluate the accuracy of the projected solution for the given example, an absolute error is computed using $Abs{{r}_{n}}$ where $w=\frac{{{x}^{2}}+{{y}^{2}}+{{t}^{2}}}{2}$ is the exact solution. Figure 2 shows a three-dimensional graph showing the absolute error of the approximate solution given by the suggested techniques, showing how well these techniques work. Also, as the number of iterations grows increases up, the mistakes go down, which makes the approximation answer more accurate.

Additionally, Table 2 displays ($Abs{{r}_{n}})$ obtined by TAM for the iterations with n={1, 3, 5}. The reduction of error is readily apparent with an increase in the number of repetitions, particularly when t = 1.

2.png

Figure 2. Three-dimensional plotted graph for the Absr_n at n=(1, 3, 5) of example 4, where t=1

Table 2. The $A b s r_n$ values for example 4 computed using the TAM method over three iterations with t=1

(x, y)	Abs. Errs. for $\boldsymbol{u}_1$	Abs. Errs. for $\boldsymbol{u}_3$	Abs. Errs. for $\boldsymbol{u}_5$
0	0.0656498	0.000151279	4.79946×10-8
0.1	0.048998	0.000149373	4.73387×10-8
0.2	0.0626498	0.000143655	4.53709×10-8
0.3	0.0588998	0.000134126	4.20912×10-8
0.4	0.0536498	0.000120785	3.74996×10-8
0.5	0.0468998	0.000103632	3.15961×10-8
0.6	0.0386498	0.0000826668	2.43807×10-8
0.7	0.0288998	0.0000578903	1.58535×10-8
0.8	0.0176498	0.000029302	6.01436×10-9
0.9	0.0048998	0.00000309807	5.13665×10-9
1	0.0093502	0.0000393099	1.75995×10-8

Example 5. Consider the 3D linear acoustic wave equation, given as [37]

$u_{t t}=V^2\left(u_{x x}+u_{y y}+u_{z z}\right)$      （32）

with the initial condition

$u(x, y, z, 0)=\cos (x) \sin (y) \cos (z), u_t(x, y, z, 0)=0$

Eq. (32) will be solved using the approach of steps that was suggested, V=1 [37]

$\begin{gathered}u_{0 t t}=0, \text { with } u_0(x, y, z, 0)=\cos (x) \sin (y) \cos (z), \\ u_{0 t}(x, y, z, 0)=0\end{gathered}$

Then,

$u_0=\cos (x) \cos (z) \sin (y)$

The first iteration $u_1(x, y, z, t)$ can be get by solving

$\begin{gathered}

u_{1 t t}=V^2\left(u_{0_{x x}}+u_{0_{y y}}+u_{0_{z z}}\right), \text { with } u_1(x, y, z, 0)= \\

\cos (x) \sin (y) \cos (z), u_{1 t}(x, y, z, 0)=0

\end{gathered}$

We get,

$u_1=\cos (x) \cos (z) \sin (y)-\frac{3}{2} t^2 \cos (x) \cos (z) \sin (y),$

The iteration $u_2(x, y, z, t)$ can be obtained by solving this problem:

$\begin{aligned} u_{2 t t}= & V^2\left(u_{1 x x}+u_{2 y y}+u_{3_{z z}}\right), \text { with: } u_2(x, y, z, 0)= \\ & \cos (x) \sin (y) \cos (z), u_{2 t}(x, y, z, 0)=0,\end{aligned}$

Then

$\begin{aligned} & u_2=\cos (x) \cos (z) \sin (y)-\frac{3}{2} t^2 \cos (x) \cos (z) \sin (y), \\ & \\ & \begin{aligned} u_5= & \frac{3}{8} t^4 \cos (x) \cos (x) \sin (y) \\ & \cos (z) \sin (y)-\frac{3}{2} t^2 \cos (x) \cos (z) \sin (y) \\ & +\frac{3}{8} t^4 \cos (x) \cos (z) \sin (y) \\ & -\frac{3}{80} t^6 \cos (x) \cos (z) \sin (y) \\ & +\frac{9 t^8 \cos (x) \cos (z) \sin (y)}{4480} \\ & -\frac{3 t^{10} \cos (x) \cos (z) \sin (y)}{44800}\end{aligned}\end{aligned}$

When the solution converges to the exact solution, then

$\begin{aligned} u(x, y, z, & t)=\cos (\sqrt{3} t) \cos (x) \sin (y) \cos (z) \\ & =\cos (x) \cos (z) \sin (y) \\ & -\frac{3}{2} t^2 \cos (x) \cos (z) \sin (y) \\ & +\frac{3}{8} t^4 \cos (x) \cos (z) \sin (y) \\ & -\frac{3}{80} t^6 \cos (x) \cos (z) \sin (y) \\ & +\frac{9 t^8 \cos (x) \cos (z) \sin (y)}{4480} \\ & -\frac{3 t^{10} \cos (x) \cos (z) \sin (y)}{44800}+\cdots\end{aligned}$

The exact solution can be found by $u(x, y, z, t)=$ $\lim _{n \rightarrow \infty} u_n=\cos (\sqrt{3} \mathrm{t}) \cos (x) \sin (y) \cos (z)$.

Example 6. The equation provided is a three-dimensional nonlinear acoustic wave equation.

$\begin{aligned} u_{t t}(x, y, z, t)= & u(x, y, z, t) u_{x x}(x, y, z, t) \\ & +u_{y y}(x, y, z, t) \\ & +2 x u_{z z}(x, y, z, t)-4 x \\ & -2\left(x^2+y^2+z^2+t^2\right)\end{aligned}$      (33)

with the initial conditions

$u(x, y, z, 0)=x^2+y^2+z^2, u_t(x, y, z, 0)=0.$

When we apply the TAM to solve Eq. (33) with the given initial conditions, the following outcomes are obtained

$\begin{gathered}L(u)=u_{t t}(x, y, z, t) \\ N(u)=u(x, y, z, t) u_{x x}(x, y, z, t)+u_{y y}(x, y, z, t) \\ +u_{z z}(x, y, z, t)-2 \\ -2\left(x^2+y^2+z^2+t^2\right)\end{gathered}$

The initial problem is

$\begin{gathered}L\left(u_0\right)=-4 x-2\left(x^2+y^2+z^2+t^2\right) \\ \text { with } u_0(x, y, z, 0)=x^2+y^2+z^2 \\ u_{0_t}(x, y, z, 0)=0\end{gathered}$  (34)

The generalized iterative method can be used to get the next problems

$\begin{gathered}L\left(u_{n+1}\right)=N\left(y_n\right)+g \\ \text { with } u_{n+1}(x, y, z, 0)=x^2+y^2+z^2, \\ u_{n+1}(x, y, z, 0)=0\end{gathered}$

By solving the Eq. (34), one can get

$u_0=-\frac{t^4}{6}-2 t^2 x+x^2-t^2 x^2+y^2-t^2 y^2+z^2-t^2 z^2$

The first iteration $u_1(x, y, z, t)$ can be found by solving

$\begin{aligned} & u_0=u_{1 t t}(x, y, z, t) \\ = & u_0(x, y, z, t) u_{0 x x}(x, y, z, t) \\ + & u_{0 y y}(x, y, z, t)+2 x u_{0_{z z}}(x, y, z, t)-4 x \\ - & 2\left(x^2+y^2+z^2+t^2\right)\end{aligned}$

with $u_1(x, y, z, 0)=x^2+y^2+z^2 \ldots, u_{1_{\mathrm{t}}}(x, y, z, 0)=0$

The solution will be

$\begin{gathered}u_1=t^2-\frac{t^4}{3}-\frac{t^6}{90}+\frac{t^8}{168}-\frac{2 t^4 x}{3}+\frac{2 t^{64} x}{15}+x^2-\frac{t^4 x^2}{3}+\frac{t^6 x^2}{15} \\ +y^2-\frac{t^4 y^2}{3}+\frac{t^6 y^2}{15}+z^2-\frac{t^4 z^2}{3}+\frac{t^6 z^2}{15},\end{gathered}$

Applying the same process for $u_2$ as stated below

$\begin{gathered}u_{2 t t}(x, y, z, t)=u_1(x, y, z, t) u_{1 x x}(x, y, z, t)+ \\ u_{1_{y y}}(x, y, z, t)+2 x u_{1_{z z}}(x, y, z, t)-4 x-2\left(x^2+y^2+\right. \\ \left.z^2+t^2\right) \text { with } u_2(x, y, z, 0)=x^2+y^2+z^2, \\ u_{2_{\mathrm{f}}}(x, y, z, 0)=0 .\end{gathered}$

We solve this problem Eq. (35) and then we have

$\begin{aligned} u_2=t^2-\frac{2 t^6}{45}- & \frac{5 t^8}{504}+\frac{463 t^{10}}{113400}-\frac{t^{12}}{3564}-\frac{103 t^{14}}{3439800} \\ & +\frac{t^{16}}{302400}-\frac{4 t^6 x}{45}+\frac{t^8 x}{105}+\frac{2 t^{10} x}{405}-\frac{2 t^{12} x}{1485} \\ & +\frac{2 t^{14} x}{20475}+x^2-\frac{2 t^6 x^2}{45}+\frac{t^8 x^2}{210}+\frac{t^{10} x^2}{405} \\ & -\frac{t^{12} x^2}{1485}+\frac{t^{14} x^2}{20475}+y^2-\frac{2 t^6 y^2}{45}+\frac{t^8 y^2}{2.10} \\ & +\frac{\mathfrak{t}^{10} y^2}{405}-\frac{t^{12} y^2}{1485}+\frac{t^{14} y^2}{20475}+z^2-\frac{2 t^6 z^2}{45} \\ & +\frac{t^8 z^2}{210}+\frac{t^{10} z^2}{405}-\frac{t^{12} z^2}{1485}+\frac{t^{14} z^2}{20475}\end{aligned}$

We continue to find approximations until $n=5$, which we will not list for brevity.

To demonstrate convergence, we calculate the values of $\beta_i$ for the issue outlined in Eq. (33). Therefore, by substituting the terms of the series $\sum_{i=0}^{\infty} v_i(x, y, z, t)$ given in Eq. (11), we get

$\begin{gathered}\beta_0=\frac{\left\|v_1\right\|}{\left\|v_0\right\|}=0.475491<1 \\ \beta_1=\frac{\left\|v_2\right\|}{\left\|v_1\right\|}=0.0105437<1 \\ \beta_2=\frac{\left\|v_3\right\|}{\left\|v_2\right\|}=0.0042245<1 \\ \beta_3=\frac{\left\|v_4\right\|}{\left\|v_3\right\|}=0.00226105<1 \\ \beta_4=\frac{\left\|v_5\right\|}{\left\|v_4\right\|}=0.00140942<1\end{gathered}$

3.png

Figure 3. Three-dimensional plotted graph for the $A b s r_n$ at $n=(1,3,5)$ of example 6 , where $t=1$

Table 3. The $A b s r_n$ values for example 6 computed using the TAM method over three iterations with $t=1$

(x, y, z)	Abs. Errs. for $u_1$	Abs. Errs. for $u_3$	Abs. Errs. for $u_5$
0	0.338492	0.00407229	0.00000630417
0.1	0.399825	0.00473579	0.00000724843
0.2	0.477159	0.00557239	0.00000843902
0.3	0.570492	0.00658208	0.00000987593
0.4	0.679825	0.00776486	0.0000115592
0.5	0.805159	0.00912073	0.0000134887
0.6	0.946492	0.0106497	0.0000156646
0.7	1.10383	0.0123517	0.0000180869
0.8	1.27716	0.0142269	0.0000207554
0.9	1.46649	0.0162751	0.0000236703
1	1.67183	0.0184964	0.0000268315

The $\beta_i$ values for $i \geq 0$ and $\forall(x, y, z, t): x, y, z \in R^3, 0<$ $x, y, z, t \leq 1$ are all smaller than 1. Therefore, the suggested iterative method satisfies convergence. To evaluate the estimated solutions accuracy for this example by using $A b s r_n$ of the approximate solution, when the exact solution is $w=$ $x^2+y^2+z^2+t^2$.

Figures 3(a)-(c) display the three-dimensional plot of the absolute value of the approximate solution generated using the TAM, by increasing the iterations, the resolution of the approximate solution will be increased. Furthermore, Table 3 presents the absolute error values derived from TAM for the iterations with $n=\{1,3,5\}$. The diminution of mistakes is clearly evident with an increase in the frequency of repetitions, especially when $t=1$.