Introduction

The majority of the engineering problems are expressed in terms of highly nonlinear differential equations. Obtaining accurate results for these types of engineering problems is relatively unhurt. However, in current years, numerical techniques have significantly been developed for highly nonlinear ODEs. Different numerical and analytical methods are applied to solve some problems. Practically all arithmetical modeling involves linear or nonlinear DEs. To determine those equations impeccably, approximately, or mathematically a few specialists employ various excellent scientific, semi-analytic plans alongside lasting labels of convergence. Various arithmetical techniques to find the solution of such highly nonlinear differential equations; nevertheless, a few critical information, including circular happening purpose, vanishes throughout finding the solution. Hence, analytical methods include emerged to find the solution of highly nonlinear ODEs such as the homotopy perturbation method and Collocation method [1], spectral element method [2], spectral homotopy analysis method [3], optimal homotopy analysis method [4], variational iteration method [5], differential transform method and finite difference method [6].

A growing number of engineering applications are concerned with energy transport by requiring the rapid movement of heat. To increase the heat transfer rate on a surface, fin assembly is commonly used. The heat transfer mechanism of the fin is to conduct heat from the heat source to the fin surface by its thermal conduction and then dissipate heat to the air by the effect of thermal convection. These extended surfaces are extensively used in various industrial applications. Conventional packages of warmth control in porous media consist of sun collectors, reactor cooling, warmness exchangers [7]. The idea of warmth control through permeable fins was delivered via Kiwan and Al-Nimr [8]. The warmth control version system through porous media changed into stepped forward using Darcy’s version [9, 10]. Subsequently the paintings of kiwan and Al-nimr, several research have been accomplished to recognize the idea of warmth switch via porous fins. The thermal evaluation of finite length permeable fins through the insulated tip and established the force of governing physical parameters on nonlinear temperature distribution is studied by Patel and Meher [11]. It was concluded that permeable fins are more competent in dissipating heat to the ambient fluid than solid fins [12, 13]. Some analytical methods are applied to optimize permeable fins with various forecast models available by Kundu and Bhanja [14]. Taklifi et al. [15] studied rectangular permeable fin with magnetohydrodynamic (MHD) effects, explained the MHD effects near the fin tip and also heat-transfer rate in the permeable fin decreases. This type of application has been solved by some scientists applying various analytical and numerical methods [1,2,3,4,5,6]. Recently, various types of MHD thermal boundary layer of a Casson fluid flow problems have been analyzed [16, 17]. Nisar et al. [18] analyzed the steady free convective incompressible electrically conducting Jeffry fluid flow over a stretching surface. Ramzan et al. [19] are examined the role of modified Fourier law in the flow of an MHD Micropolar nano liquid flow with dust particles over a stretched surface. The MHD laminar flow, containing hybrid nanoparticles, with heat transfer phenomenon over a stretching sheet immersed in a porous medium, is investigated by Ali et al. [20]. New soliton solutions for generalized nonlinear differential equations using an effective analytical method explained by Ghanbari et al. [21].

As per our narrative review, we have not seen any research article on this model through wavelets. This impels us to propose the HWM for considered fin flow problem, and proficiency of the current technique is revealed through tables and graph simulation. Wavelet is a function, more precisely, is an arithmetical function applied to split a given function signal into various scale components. In the 1980s, they applied the French word “ondelette,” which means “small wave.” Later, it was transferred to English by translating “onde” into “wave,” charitable “wavelet.” Wavelet theory is developing tremendously due to arithmetical analysis by Stromberg, Grossmann, Morlet, Meyer, and Daubechies. Due to its unique features like compactly supported, orthogonality, and multiresolution analysis. Many mathematicians are applying wavelets in the field of numerical analysis; as a result presently, we have seen many diverse techniques on differential equations such as An investigation with the Hermite wavelets for accurate solutions for heat transfer problems [22, 23], high mass transfer via wavelet frames [24], Laguerre wavelet method [25].

To the best of our understanding, no study was performed temperature distribution in a rectangular moving porous fin using DTM and HWM. Moreover, we generated the new Hermite wavelet operational matrix method and DTM to re-examine the above model. The highly nonlinear ordinary differential equation was tackled using DTM and HWM. The results are presented through tables and graphical plots for different known physical parameters. Subsequently, the obtained solutions compare with DTM and HWM and the exact solution of the highly nonlinear ODE. Comparison between resulting solutions showed that HWM is more suitable and correct than DTM. Furthermore, the essential physical quantities of real advantage and the control of unique physical known parameters on the temperature distribution's allotment were also incorporated in the present study.

Formulation

The highly nonlinear ODE, which described heat-transfer in a rectangular moving porous fin, can be summarized as introduced by Joneidi et al. [26] and Ndlovu and Moitsheki [5]:

$$ \frac{d}{dx}\left\{ {\left[ {1 + \beta \left( {y - y_{a} } \right)} \right]\frac{dy}{{dx}}} \right\} - Nc\left( {y - y_{a} } \right)^{m + 1} - Np\left( {y - y_{a} } \right)^{2} - Nr\left( {y^{4} - y_{a}^{4} } \right) - Pe\frac{dy}{{dx}} = 0 $$
(2.1)
$$ \,\frac{dy(0)}{{dx}} = 0\,\,\,{\text{and}}\,\,\,y(1) = 1 $$
(2.2)

where \(0 \le x \le 1\). The important dimensionless parameters are introduced in Eq. (2.1).

Parameter

Notation

Temperature

\(y\)

Ambient temperature

\(y_{a}\)

Thermal conductivity gradient

\(\beta\)

Peclet number

\(Pe\)

convection-conduction parameter

\(Nc\)

radiation conduction parameter

\(Nr\)

porosity parameter

\(Np\)

Basic Idea of Hermite Wavelet Method and Differential Transformation Method

In this paper, we apply the Hermite wavelet method and Differential transformation method to the discussed problem.

Hermite Wavelet Method

The Hermite wavelet and function approximation are studied in detail [27,28,29,30,31,32]. To extract the functional matrix, here we use the following procedure. Consider a few Hermite wavelet basis at k = 1 as follows:

$$ \begin{gathered} \phi_{1,0} (x) = \frac{2}{\sqrt \pi } \hfill \\ \phi_{1,1} (x) = \frac{1}{\sqrt \pi }(8x - 4) \hfill \\ \phi_{1,2} (x) = \frac{1}{\sqrt \pi }(32x^{2} - 32x + 4) \hfill \\ \phi_{1,3} (x) = \frac{1}{\sqrt \pi }(128x^{3} - 192x^{2} + 48x + 8) \hfill \\ \phi_{1,4} (x) = \frac{1}{\sqrt \pi }(512x^{4} - 1024x^{3} + 384x^{2} + 128x - 40) \hfill \\ \end{gathered} $$
$$ \phi_{1,5} (x) = \frac{1}{\sqrt \pi }(2048x^{5} - 5120x^{4} + 2560x^{3} + 1280x^{2} - 800x + 16) $$
$$ \phi_{1,6} (x) = \frac{1}{\sqrt \pi }(8192x^{6} - 24576x^{5} + 15360x^{4} + 10240x^{3} - 9600x^{2} + 384x + 368) $$
$$ \phi_{1,7} (x) = \frac{1}{\sqrt \pi }(32768x^{7} - 114688x^{6} + 86016x^{5} + 71680x^{4} - 89600x^{3} + 5376x^{2} + 10304x - 928) $$
$$ \begin{gathered} \phi_{1,8} (x) = \frac{1}{\sqrt \pi }(131072x^{8} - 524288x^{7} + 458752x^{6} + 458752x^{5} - 716800x^{4} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 57344x^{3} + 164864x^{2} - 29696x - 3296) \hfill \\ \end{gathered} $$
$$ \begin{gathered} \phi_{1,9} (x) = \frac{1}{\sqrt \pi }(524288x^{9} - 2359296x^{8} + 2359296x^{7} + 2752512x^{6} - 5160960x^{5} + 516096x^{4} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1978368x^{3} - 534528x^{2} - 118656x + 21440) \hfill \\ \end{gathered} $$
$$ \begin{gathered} \phi_{1,10} (x) = \frac{1}{\sqrt \pi }(2097152x^{10} - 10485760x^{9} + 11796480x^{8} + 15728640x^{7} - 34406400x^{6} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 4128768x^{5} + 19783680x^{4} - 7127040x^{3} - 2373120x^{2} + 857600x + 16448) \hfill \\ \end{gathered} $$
$$ \begin{gathered} \phi_{1,11} (x) = \frac{1}{\sqrt \pi }(8388608x^{11} - 46137344x^{10} + 57671680x^{9} + 86507520x^{8} - 216268800x^{7} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 30277632x^{6} + 174096384x^{5} - 78397440x^{4} - 34805760x^{3} + 18867200x^{2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 723712x - 461696) \hfill \\ \end{gathered} $$
$$ \begin{gathered} \phi_{1,12} (x) = \frac{1}{\sqrt \pi }(33554432x^{12} - 201326592x^{11} + 276824064x^{10} + 461373440x^{9} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1297612800x^{8} + 207618048x^{7} + 1392771072x^{6} - 752615424x^{5} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 417669120x^{4} + 301875200x^{3} + 17369088x^{2} - 22161408x + 561536) \hfill \\ \end{gathered} $$

where, \(\phi_{9} (x) = [\phi_{1,0} (x),\phi_{1,1} (x),\phi_{1,2} (x),\,\phi_{1,3} (x),\,\phi_{1,4} (x),\,\phi_{1,5} (x),\,\phi_{1,6} (x),\phi_{1,7} (x),\phi_{1,8} (x)]^{T}\).

At this time, integrate above first 9 basis regarding x limit from 0 to x, subsequently articulate as a linear-combination of Hermite-wavelet basis as,

$$ \int\limits_{0}^{x} {\phi_{1,0} (x)} = \left[ {\begin{array}{*{20}c} \frac{1}{2} & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,1} (x)} = \left[ {\begin{array}{*{20}c} {\frac{ - 1}{4}} & 0 & \frac{1}{8} & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,2} (x)} = \left[ {\begin{array}{*{20}c} {\frac{ - 1}{3}} & 0 & 0 & \frac{1}{12} & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,3} (x)} = \left[ {\begin{array}{*{20}c} \frac{5}{4} & 0 & 0 & 0 & \frac{1}{16} & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,4} (x)} = \left[ {\begin{array}{*{20}c} {\frac{ - 2}{5}} & 0 & 0 & 0 & 0 & \frac{1}{20} & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,5} (x)} = \left[ {\begin{array}{*{20}c} {\frac{ - 23}{3}} & 0 & 0 & 0 & 0 & 0 & \frac{1}{24} & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,6} (x)} = \left[ {\begin{array}{*{20}c} \frac{116}{7} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{28} & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,7} (x)} = \left[ {\begin{array}{*{20}c} \frac{103}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{32} \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\phi_{1,8} (x)} = \left[ {\begin{array}{*{20}c} {\frac{ - 2680}{9}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) + \frac{1}{36}\phi_{1,9} (x). $$

Hence,

$$ \int\limits_{0}^{x} {\,\phi (x)} \,dx = H_{9 \times 9} \,\,\phi_{9} (x) + \,\,\overline{\phi }_{9} (x) $$

where,

$$ H_{9 \times 9} = \left[ {\begin{array}{*{20}c} {\tfrac{1}{2}} & {\tfrac{1}{4}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{ - 1}{4}} & 0 & {\tfrac{1}{8}} & 0 & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{ - 1}{3}} & 0 & 0 & {\tfrac{1}{12}} & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{5}{4}} & 0 & 0 & 0 & {\tfrac{1}{16}} & 0 & 0 & 0 & 0 \\ {\tfrac{ - 2}{5}} & 0 & 0 & 0 & 0 & {\tfrac{1}{20}} & 0 & 0 & 0 \\ {\tfrac{ - 23}{3}} & 0 & 0 & 0 & 0 & 0 & {\tfrac{1}{24}} & 0 & 0 \\ {\tfrac{116}{7}} & 0 & 0 & 0 & 0 & 0 & 0 & {\tfrac{1}{28}} & 0 \\ {\tfrac{103}{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {\tfrac{1}{32}} \\ {\tfrac{ - 2680}{9}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right],\,\,\overline{{\phi_{9} }} (x) = \left[ {\begin{array}{*{20}c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ {\frac{1}{36}\phi_{1,9} (x)} \\ \end{array} } \right]. $$

Subsequently, two times integration of above 9 basis we get

$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,0} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} \frac{3}{16} & \frac{1}{8} & \frac{1}{32} & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,1} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} {\frac{ - 1}{6}} & {\frac{ - 1}{{16}}} & 0 & \frac{1}{96} & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,2} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} {\frac{ - 1}{{16}}} & {\frac{ - 1}{{12}}} & 0 & 0 & \frac{1}{192} & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,3} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} \frac{3}{5} & \frac{5}{16} & 0 & 0 & 0 & \frac{1}{320} & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,4} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} {\frac{ - 7}{{12}}} & {\frac{ - 1}{{10}}} & 0 & 0 & 0 & 0 & \frac{1}{480} & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,5} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} {\frac{ - 22}{7}} & {\frac{ - 23}{{12}}} & 0 & 0 & 0 & 0 & 0 & \frac{1}{672} & 0 \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,6} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} \frac{81}{8} & \frac{29}{7} & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{896} \\ \end{array} } \right]\,\phi_{9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,7} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} \frac{148}{9} & \frac{103}{8} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) + \frac{1}{1152}\phi_{1,9} (x) $$
$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi_{1,8} (x)} \,dxdx} = \left[ {\begin{array}{*{20}c} {\frac{ - 773}{5}} & {\frac{ - 670}{9}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right]\,\phi_{9} (x) + \frac{1}{1440}\phi_{1,10} (x). $$

Hence,

$$ \int\limits_{0}^{x} {\int\limits_{0}^{x} {\phi (x)} \,dx} dx = H_{9 \times 9}^{^{\prime}} \,\,\phi_{9} (x) + \,\,\overline{\phi }_{9}^{^{\prime}} (x) $$

where,

$$ H_{9 \times 9}^{^{\prime}} = \left[ {\begin{array}{*{20}c} {\tfrac{3}{16}} & {\tfrac{1}{8}} & {\tfrac{1}{32}} & 0 & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{ - 1}{6}} & {\tfrac{ - 1}{{16}}} & 0 & {\tfrac{1}{96}} & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{ - 1}{{16}}} & {\tfrac{ - 1}{{12}}} & 0 & 0 & {\tfrac{1}{192}} & 0 & 0 & 0 & 0 \\ {\tfrac{3}{5}} & {\tfrac{5}{16}} & 0 & 0 & 0 & {\tfrac{1}{320}} & 0 & 0 & 0 \\ {\tfrac{ - 7}{{12}}} & {\tfrac{ - 1}{{10}}} & 0 & 0 & 0 & 0 & {\tfrac{1}{480}} & 0 & 0 \\ {\tfrac{ - 22}{7}} & {\tfrac{ - 23}{{12}}} & 0 & 0 & 0 & 0 & 0 & {\tfrac{1}{672}} & 0 \\ {\tfrac{81}{8}} & {\tfrac{29}{7}} & 0 & 0 & 0 & 0 & 0 & 0 & {\tfrac{1}{896}} \\ {\tfrac{148}{9}} & {\tfrac{103}{8}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ {\tfrac{ - 773}{5}} & {\tfrac{ - 670}{9}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} } \right],\,\,\overline{{\phi_{9}^{^{\prime}} }} (x) = \left[ {\begin{array}{*{20}c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ {\frac{1}{1152}\phi_{1,9} (x)} \\ {\frac{1}{1440}\phi_{1,10} (x)} \\ \end{array} } \right]. $$

For instance, the modeled equation is of second order. So, we created 9*9 matrices up to the second order-functional matrix of integration.

Basics of Differential Transformation Method

We assume \(y(x)\) to be an analytic function in a domain \(\Omega\) and \(x = x_{i}\) signify some point in \(\Omega\). The solution \(y(x)\) is subsequently assumed by one power-series whose center is positioned at \(x = x_{i}\). The Taylor-series development solution of \(y(x)\) is of the outline [26]

$$ y(x)\,\,\, = \,\,\,\sum\limits_{k = 0}^{\infty } {\frac{{\left( {x - x_{i} } \right)^{4} }}{k!\,}} \left[ {\frac{{d^{k} y(x)}}{{dx^{k} }}} \right]_{{x = x_{i} }} \,\,\forall x \in \Omega $$

In particular, \(x_{i} = 0\) then above equation becomes

$$ y(x) = \sum\limits_{k = 0}^{\infty } {\frac{{x^{4} }}{k!\,}} \left[ {\frac{{d^{k} y(x)}}{{dx^{k} }}} \right]_{x = 0} \,\,\forall x \in \Omega $$

As given in [26], the differential transformation of the solution \(y(x)\) is written as:

$$ Y(k)\,\,\, = \,\,\,\,\sum\limits_{k = 0}^{\infty } {\frac{{{\rm M}^{4} }}{k!\,}} \left[ {\frac{{d^{k} y(x)}}{{dx^{k} }}} \right]_{x = 0} $$

where novel solution \(y(x)\) and transformed solution \(Y(k)\). The differential spectrum of \(Y(k)\) is restricted inside the distance \(x \in \left[ {0,{\rm M}} \right]\) and as well as \({\rm M}\) is some constant. The differential inverse transform of \(Y(k)\) is defined as follows:

$$ y(x) = \sum\limits_{k = 0}^{\infty } {\left( {\frac{x}{\rm M}} \right)^{k} Y\left( k \right)} $$

The solution \(y(x)\) is articulated by a finite power series, and the above equation can also be expressed as:

$$ y(x)\,\, = \,\,\,\sum\limits_{k = 0}^{j} {\left( {\frac{x}{\rm M}} \right)^{k} Y\left( k \right)} . $$

The basic operations of DTM are given in below table.

New function

Transformed function

\(y(x) = ay_{1} (x) \pm by_{2} (x)\)

\(Y(k) = aY_{1} (k) \pm bY_{2} (\,k)\)

\(y(\,x\,) = \frac{{dy_{1} (\,x\,)}}{d\,x}\)

\(Y(k\,)\,\, = \,\,(k + 1)\,\,Y_{1} (\,k + 1\,)\)

\(y\,(x\,)\,\, = \,\,\frac{{d^{2} y_{1} (\,x\,)}}{{dx^{2} }}\)

\(Y\,(\,k\,) = (\,k + 1\,)\,(\,k + 2\,)\,\,Y_{1} (\,k + 2\,)\)

\(y\,(\,x\,) = y_{1} (\,x\,)y_{2} (\,x\,)\)

\(Y\,(\,k\,) = \sum\limits_{l = 0}^{k} {Y_{1} (l)Y_{2} (k - l)}\)

\(y\,(\,x\,) = x^{r}\)

\(Y\,(\,k\,) = \,\,\delta \left( {\,k - r} \right) = \left\{ {\begin{array}{*{20}c} {0\,\,\,\,\,r \ne k} \\ {1\,\,\,\,r = k} \\ \end{array} } \right.\)

\(y(x) = \left[ {y_{1} (x)} \right]^{r}\)

\(Y^{r} (k) = Y^{r - 1} (k) \otimes Y(k) = \sum\limits_{l = 0}^{k} {Y^{r - 1} (l)Y(k - l)}\)

Method of Solutions

Solutions with HWM

Now, assume that

$$ y^{^{\prime\prime}} \left( x \right) = A^{T} \phi \left( x \right) $$
(4.1)

where \(A^{{\text{T}}} = \left[ {C_{1,0} , \ldots C_{1,M - 1} ,C_{2,0} , \ldots C_{2,M - 1} , \ldots C_{{2^{k - 1} ,0}} , \ldots C_{{2^{k - 1} ,M - 1}} } \right]\), \(\phi \left( x \right) = \left[ {\phi_{1,0} , \ldots \phi_{1,M - 1} ,\phi_{2,0} , \ldots \phi_{2,M - 1} , \ldots \phi_{{2^{k - 1} ,0}} , \ldots \phi_{{2^{k - 1} ,M - 1}} } \right]^{{\text{T}}}\).

Integrate Eq. (4.1) integrating \(x\) form \(0\) to \(x\) and using second initial condition \(y^{\prime}\left( 0 \right) = 0\), we get,

$$ y^{^{\prime}} \left( x \right) = A^{{\text{T}}} \left[ {H\phi \left( x \right) + \overline{\phi }\left( x \right)} \right]. $$
(4.2)

Integrate (4.2) with reverence to \(x\) form \(0\) to \(x\),we get

$$ y\left( x \right) = y\left( 0 \right) + A^{{\text{T}}} \left[ {H^{^{\prime}} \phi \left( x \right) + \overline{\phi }^{^{\prime}} \left( x \right)} \right] $$
(4.3)

put \(x = 1\) in (4.3)

$$ y\left( 0 \right) = 1 - \left. {A^{{\text{T}}} \left[ {H^{^{\prime}} \phi \left( x \right) + \overline{\phi }^{^{\prime}} \left( x \right)} \right]} \right|_{x = 1} $$
(4.4)

substitute (4.4) in (4.3)

$$ y\left( x \right) = 1 + A^{{\text{T}}} \left[ {H^{^{\prime}} \phi \left( x \right) + \overline{\phi }^{^{\prime}} \left( x \right)} \right] - \left. {A^{{\text{T}}} \left[ {H^{^{\prime}} \phi \left( x \right) + \overline{\phi }^{^{\prime}} \left( x \right)} \right]} \right|_{x = 1} $$
(4.5)

Fit (4.1), (4.3), and (4.5) in (2.1), and collocate using the following grid points \(x_{i} = \frac{2\,i - 1}{{2N}}\), \(i = 1,2,3, \ldots ,N\). Solve these equations using the Newton's-Raphson method that yields unknown coefficients. Then substitute these coefficients in (4.5) that contribute us the numerical solution of (2.1).

Solution with DTM

Now considering \({\rm M} = 1\), \(m = 0\) and be valid DTM into Eq. (2.1). Now applying the differential transform of Eq. (2.1) with respect to \(x\) we get:

$$ \begin{aligned} {\rm B}_{1} & \left( {k + 1} \right)\left( {k + 2} \right)Y\left( {k + 2} \right) - NcY\left( k \right) + {\rm B}_{2} Y\left( k \right) - Pe\left( {k + 1} \right)Y\left( {k + 1} \right) + {\rm B}_{3} \\ + & \beta \sum\limits_{i\, = \,0}^{k} {\left[ {\left( {k - i + 1} \right)\left( {k - i + 2} \right)Y\left( i \right)Y\left( {k - i + 2} \right) + \left( {i + 1} \right)\left( {k - i + 1} \right)Y\left( {i + 1} \right)Y\left( {k - i + 1} \right)} \right]} \\ - & Np\sum\limits_{i = 0}^{k} {Y\left( i \right)Y\left( {k - i} \right)} - Nr\sum\limits_{i = 0}^{k} {Y^{3} \left( i \right)Y\left( {k - i} \right)} = 0 \\ \end{aligned} $$
(4.6)

with \({\rm B}_{1} = 1 - \beta Y_{a} ,{\rm B}_{2} = 2NpY_{a} ,\,{\rm B}_{3} = NcY_{a} - NpY_{a}^{2} + NrY_{a}^{4}\).

The boundary-condition in Eq. (2.2), with the intention of we have exerting transformation

$$ Y(1) = 0 $$
(4.7)

The other boundary conditions are considered as follow:

$$ Y(0) = b $$
(4.8)

where \(b\) is the constant, and we will find it with allowing for an additional boundary condition in Eq. (2.2) in point \(x = 1\). Our motivation has:

$$ {\text{Y(2) = }}\frac{{ - {\rm B}_{{1}} + b\left( { - {\rm B}_{3} + Nc + bNp + b^{3} Nr} \right)}}{{\left( {{\rm B}_{2} + b\beta } \right)2!}} $$
$$ {\text{Y(3) = }}\frac{Y(2)Pe}{{\,3\left( {{\rm B}_{2} + b\beta } \right)}} $$
$$ {\text{Y(4) = }}\frac{{{\text{Y(2)}}\left( { - {\rm B}_{{3}} + Nc + 2bNp + b^{3} Nr - 6\beta {\text{Y(2) + bNrY}}^{{2}} {(2)}} \right) + 3Pe{\text{Y(3)}}}}{{12\left( {{\rm B}_{2} + b\beta } \right)}} $$
$$ {\text{Y(5) = }}\frac{{Y(3)\left[ { - {\rm B}_{3} + Nc + 2bNp + b^{3} Nr - 20\beta Y(2) + bNrY^{2} (3)} \right] + 4PeY(4)}}{{\,20\left( {{\rm B}_{2} + b\beta } \right)}} $$
$$ {\text{Y(6) = }}\frac{{\left( { - {\rm B}_{3} + Nc} \right)Y(4) + Np\left( {Y^{2} (2) + 2bY(4)} \right) - 15\beta \left( {Y^{2} (3) + 2Y(2)Y(4)} \right) + Nr\left( {Y^{4} (2) + b^{3} Y(4) + bY^{3} (4)} \right) + 5PeY(5)}}{{30\left( {{\rm B}_{2} + b\beta } \right)}} $$

The above procedure is incessant. Substituting the values of \(Y(0)\,{\text{to}}\,\,Y(n)\) into the most important equation based on DTM, it can be obtained that the series solution in powers of \(x\):

$$ y(x) = Y(0) + Y(1)x + Y(2)x^{2} + Y(3)x^{3} + Y(4)x^{4} + Y(5)x^{5} + Y(6)x^{6} + \cdot \cdot \cdot . $$
(4.9)

To find the numerical value of \(b\), we alternative the boundary-condition from Eq. (2.2) into Eq. (4.9) in point \(x = 1\). Then

$$ y(1) = 1. $$
(4.10)

Solving Eq. (4.10) gives the constant value of \(b\). By substituting the calculated value of \(b\) addicted to Eq. (4.9), we can obtain the polynomial expressions of \(y(x)\).

Results and Discussions

The HWM and DTM are applied to resolve the highly nonlinear differential equation arising in the heat transfer fin problem, the disadvantages and advantages of HWM are discussed. In the absence of thermal-conductivity gradient (\(\beta = 0\)), solutions of the current problem are tabulated vs. the analytical solution [26] in Tables 1 and 2. An enormously good-looking agreement between the solutions is obtained, which concludes the strength of HWM and DTM. Subsequently, in Figs. 1 and 2, the assessment of the results between exact solution, HWM, and DTM is shown. The presence of thermal-conductivity gradient (\(\beta \ne 0\)), for \(\beta = 0.4\) (Table 3), \(\beta = 0.2\)(Table 4), numerical solution calculated by 4th-order Runge–Kutta of the current problem are given in Tables 3 and 4. In this case, an extremely motivating concurrence connecting the solutions is observed too, which confirms the outstanding strength of the HWM and DTM. After that, in Figs. 3 and 4, the numerical solutions, HWM, and DTM results are compared. Evaluation of the absolute errors of two techniques revealed in Tables 1, 2, 3, 4 exhibit the higher accuracy of HWM than DTM. For this motivation, HWM has been applied to re-examine the unique properties of the governing parameters of the fin problem.

Table 1 The results of the numerical solution, HWM, and DTM of \(y(x)\)
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Table 2 The results of the exact solution, HWM, and DTM for \(y(x)\)
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Fig. 1
figure 1

Comparison of the solutions via exact solution, HWM, and DTM for \(y(x)\) when \(\beta = 0\), \(Nc = 0.25\), \(Np = Nr = Pe = y_{a} = m = 0\)

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Fig. 2
figure 2

Comparison of the results via exact solution, HWM, and DTM for \(y(x)\) when \(\beta = 0,\,\,Nc = 1\), \(Np = Nr = Pe = y_{a} = m = 0\)

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Table 3 The results of the exact solution, HWM, and DTM for \(y(x)\)
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Table 4 The results of the numerical solution, HWM, and DTM for \(y(x)\)
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Fig. 3
figure 3

Comparison of the results via Numerical solution, HWM, and DTM for \(y(x)\) when \(\beta = 0.4\), \(Nc = 1\), \(Np = Nr = Pe = y_{a} = m = 0\)

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Fig. 4
figure 4

Comparison of the results via Numerical solution, HWM, and DTM for \(y(x)\) when \(\beta = 0.2\), \(Nc = 0.25\), \(Np = Nr = Pe = y_{a} = m = 0\)

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The effect of governing parameters \(\beta\), \(y_{a}\), \(Pe\), \(Nc\), \(Nr\), and \(Np\) on temperature distribution as a function of \(x\) is established in Figs. 5, 6, 7, 8, 9, and 10, respectively. The results exhibited in these figures are for \(\beta = y_{a} = Nc = Np = 0.25,\,m = 2,\,Nr = 4\) and \(Pe = 3\) except the varying parameter in any concerned figure. Figures 5 and 6 show the respective effects of \(\beta\) and \(y_{a}\), on the temperature distribution \(y(x)\). It is easy to see that \(y(x)\) increases as \(\beta\) and \(y_{a}\), increases progressively. Figures 7, 8, 9, 10 give the opposite performance of the temperature distribution \(y(x)\) under the influence of \(Pe\), \(Nc\), \(Nr\), and \(Np\), respectively.

Fig. 5
figure 5

The effect of \(\beta\) on \(y(x)\)

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Fig. 6
figure 6

The effect of \(y_{a}\) on \(y(x)\)

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Fig. 7
figure 7

The effect of \(Pe\) on \(y(x)\)

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Fig. 8
figure 8

The effect of \(Nc\) on \(y(x)\)

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Fig. 9
figure 9

The effect of \(Nr\) on \(y(x)\)

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Fig. 10
figure 10

The effect of \(Np\) on \(y(x)\)

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Conclusion

In the present article, the natural convection fin flow problem is resolved using HWM and DTM. The flow contains the nonlinear temperature distribution in a rectangular moving permeable fin. We resolved a highly nonlinear ODE by using DTM and HWM. The outcome process of HWM and DTM show the competence for solving highly nonlinear ODE, which is exposed in the table and figures. Comparison between calculated solutions showed that HWM is more suitable and correct than DTM. Moreover, the HWM is more proficient than numerical methods to solve this type of coupled highly nonlinear ODEs.