Applications of the Bernoulli wavelet collocation method in the analysis of MHD boundary layer flow of a viscous fluid

Abstract

This study focuses on the flow of viscous, electrically conducting incompressible fluid over a stretching plate. The Falkner–Skan equation is a nonlinear, third-order boundary value problem. No closed-form solutions are available for this two-point boundary value problem. Here, we developed a new functional matrix of integration using the Bernoulli wavelet and also generated a new technique called Bernoulli wavelet collocation method (BWCM) to solve the nonlinear differential equation that arises in the fluid flow over a stretching plate. The boundary layer model is transformed to a nonlinear ordinary differential equation called the Falkner-Skan type equation using suitable transformation. Using BWCM, we have solved the unbounded governing equations of different types that arise in the MHD boundary-layer flow of a viscous fluid over a stretching plate. Several aspects of this problem are justified using the Haar wavelet and the previously obtained theoretical results. It is observed that the boundary-layer thickness decreases as the pressure gradient and magnetic field parameters increase. The overshoots and undershoots are observed for some particular parameters using BWCM. Furthermore, our research yields dual solutions for some physical parameters, which are investigated for the first time in the literature using the Bernoulli wavelet approach. The nature of the flow problem is discussed through the graphs by varying the physical parameters.

1 Introduction

The boundary layer flow (BLF) of a fluid over a stretching area has many applications in mathematical engineering and other fields, such as condensation of liquid film, aerodynamics, drawing of plastic films, wire drawing, cooling of films or sheets, insulating materials, metallic plates, and conveyor belts, etc., The first work in this field is done by B. C. Sakiadis [1] to examine the BLF on a continuous solid surface. B. K. Datta [2] derived an approximate solution for the Blasius equation using $\delta$-perturbation technique. Abbasbandy [3] used the Adomian decomposition method (ADM) to solve the Blasius equation. Abbasbandy and T. Hayat [4, 5] gave the solution for MHD Falkner Skan equations by employing the Hankel-Pade and homotopy analysis methods (HAM), respectively. M. Khan et al. [6] proposed a new technique known as the homotopy analysis transform method, a combination of HAM and Laplace decomposition approach, to find an approximate solution for Blasius equations. H. Zeb et al. [7] studied the nature of Non-Newtonian Ferrofluid over a stretching plate using the Runge–Kutta method. T. Anusha et al.[8] provided the exact solution for the MHD of nanofluid over stretching or shrinking plate with mass transpiration and Brinkman ratio. S. Liao applied HAM to study incompressible viscous fluid flow over a stretching plate [9, 10] and the MHD flow of non-Newtonian fluid over a stretching sheet[11]. Asaithambi studied the Falkner-Skan problem using the finite difference method [12] and the recursive evaluation of Taylor coefficients [13]. R. B. Kudenatti et al. investigated MHD flow over a stretching plate [14] and with suction and injection [15]. R. B. Kudenatti et al. [16] gave approximate analytical solutions for the equations that arise in the nonlinear stretching surface using the asymptotic function method, method of stretching variables, and Dirichlet series. R. B. Kudenatti [17] proposed the exact solution for BLF over a stretching plate. P. L. Sachdev [18] investigated the Falkner Skan equation by solving exactly. The investigation on heat transfer of incompressible electrically conducting fluid was done by A. Chakrabarti [19] and with a power-law velocity by Chiam [20]. Many researchers have studied MHD stretching sheets in quiescent fluid, micropolar fluids, magneto-convection, nonlinear radiation, and diffusion (M. J. Uddin et al.[21,22,23]). Some other techniques, such as modified ADM, pade approximation (Hayat et al. [24]), and differential transform method (Rashidi [25]), are employed to study the flow over a stretching sheet.

Here, the boundary layer model is transformed into a nonlinear ordinary differential equation called the Falkner-Skan type equation using suitable transformation. The Falkner–Skan equation, originally derived in 1931 by Falkner and Skan, is of central importance to the fluid mechanics of wall-bounded viscous flows. It is derived from the two-dimensional incompressible Navier–Stokes equations for a one-sided bounded flow using similarity analysis. Its solution describes the form of an external laminar boundary layer in the presence of an adverse or favorable streamwise pressure gradient. Despite the apparent simplicity of the Falkner-Skan equation solving it accurately can be fraught with difficulty; these problems mainly stem from its non-linearity and third-degree order. There are some examples of analytical solutions to the Falkner-Skan equations for special cases, but most studies have focused either on demonstrating a solution’s existence and uniqueness or finding a numerical/computational solution for particular boundary-layer conditions.

During the terminating year of the twentieth era, the abundant and philosophical theory of wavelets was created due to the efforts of mathematicians, physicists, and engineers. Wavelet methods have been widely used in image and signal processing, approximation theory, geophysics, and many more. Wavelet-based numerical techniques have become a popular method for solving differential equations. In the latest centuries, much consideration has been dedicated to the newly established approaches for the numerical solution of an equation such approaches include the wavelet methods to solve nonlinear equations arising in fluid problems [26,27,28,29,30,31,32,33,34,35,36,37,38]. Some other techniques, such as Perturbation techniques, are too strongly hooked on ‘‘small parameters’’. Thus, developing some new numerical techniques is advisable, not dependent on small parameters. In this article, we use the Bernoulli wavelet operational method of integration to solve the nonlinear ordinary differential equations. P. Rahimkhani et al. proposed an operational matrix based on Bernoulli wavelets for solving delay differential equations [39] and fractional-order Bernoulli wavelet method to solve pantograph differential equations [40]. Adel and Sabir [41] investigated Lane–Emden pantograph delay differential model using the Bernoulli collocation method. Lal and Kumar [42] numerically investigated Volterra integral equation via the Bernoulli wavelet. A numerical investigation of Volterra Integro-Differential Equations by employing the Bernoulli wavelet method was done by Sahu and Saha Ray [43]. Many researchers have solved nonlinear singular Lane–Emden equations [44], fractional-order differential equations [45], linear and nonlinear problems in the calculus of variations (Hedayati [46]), Fractional Diffusion Wave Equations [47] using the Bernoulli wavelet method. Here, we developed a new operational matrix; using BWCM, we have solved the unbounded governing equations of different types that arise in the MHD BLF of a viscous fluid. Usually, ordinary differential equations (ODEs) describe many physical phenomena in fluid dynamics, mathematical biology, chemical theory, and bio-modeling. Several mathematicians have considered these models in past decades, and many techniques have been developed to describe the above model. But the Falkner Skan type equation is highly nonlinear ODE, so anticipating the exact solution is always impossible. We need to switch to numerical methods to crack such a model. Because of this, we proposed a new novel approach called BWCM to solve the Falkner–Skan type equation. The primary purpose of this study is to present and explain a new BWCM for obtaining the numerical solution to nonlinear ODE that cannot be solved exactly. Here, the magnetohydrodynamic BLF of a viscous fluid is considered to analyze the effects of $M,\,\beta$ and $\varepsilon$ of fluid flow. This work aims to solve a nonlinear differential equation governing the magnetohydrodynamic BLF of a viscous fluid using a novel approach called the BWCM. Nonlinear differential equations are solved through this technique and compared with the exact solution and the Haar wavelet solution. Some plots and tables are presented to show the reliability and simplicity of the method. Here, we computed the solution of the Falkner-Skan equation using a wavelet scheme. There are several previously reported many approaches, such as shooting, Taylor series, Runge–Kutta, and other semi-analytic methods. Interestingly, the methods that directly solve the original non-reduced third-order equation are absent from the literature; to our knowledge, this is the first time a continuous wavelet scheme has been presented to find numerical solutions to the Falkner–Skan equation directly. This approach avoids complicated numerical algorithms and presents valuable information about the numerical behavior of the equation. The accuracy and effectiveness of this approach are established by comparison with published work.

2 Problem formulation

The continuity equation and Navier–Stokes equation for a two-dimensional steady flow of an incompressible viscous fluid in the absence of body force are given by [48]:

$$\nabla \cdot \,\,\vec{q} = 0,$$

(1)

$$\rho (\vec{q}\,.\,\nabla )\,\vec{q} = - \nabla p + \mu \,\nabla^{2} \vec{q},$$

(2)

where $\vec{q}$ is the velocity, $\mu$ is the viscosity, $\rho$ is the density, and $p$ is the pressure.

To find $(\vec{q}\,.\,\nabla )\,\vec{q},\,\nabla^{2} \vec{q}$:

$$(\vec{q}\,.\,\nabla )\,\vec{q} = \left[ {(u\,\hat{i} + v\,\hat{j}).\left( {\hat{i}\,\,\,\frac{\partial }{{\partial x^{{}} }}\, + \,\,\hat{j}\,\,\,\frac{{\partial^{{}} }}{\partial y}} \right)} \right]\,\,(u\,\,\hat{i} + v\,\,\hat{j}) = \left( {u\,\,\frac{\partial u}{{\partial x}} + v\,\,\frac{\partial u}{{\partial y}}} \right)\,\hat{i} + \left( {u\,\,\frac{\partial v}{{\partial x}} + v\,\,\frac{\partial v}{{\partial y}}} \right)\,\,\hat{j}.$$

$$\,\nabla^{2} \vec{q} = \left( {\frac{{\partial^{2} }}{{\partial x^{2} }} + \frac{{\partial^{2} }}{{\partial y^{2} }}} \right)\,(u\,\,\hat{i} + v\,\,\hat{j}) = \left( {\frac{{\partial^{2} u}}{{\partial x^{2} }} + \frac{{\partial^{2} u}}{{\partial y^{2} }}} \right)\,\hat{i} + \left( {\frac{{\partial^{2} v}}{{\partial x^{2} }} + \frac{{\partial^{2} v}}{{\partial y^{2} }}} \right)\,\,\hat{j}.$$

With these (1) and (2) becomes,

$$\frac{\partial u}{{\partial x}}\, + \,\,\frac{\partial v}{{\partial y}}\,\, = \,\,0\,,$$

$$u\,\,\frac{\partial u}{{\partial x}}\,\, + v\,\,\frac{\partial u}{{\partial y}}\,\, = \,\,\frac{ - 1}{\rho }\frac{\partial p}{{\partial x}} + \frac{\mu }{\rho }\left( {\frac{{\partial^{2} u}}{{\partial x^{2} }} + \frac{{\partial^{2} u}}{{\partial y^{2} }}} \right),$$

$$u\,\frac{\partial v}{{\partial x}}\,\, + \,\,v\,\,\frac{\partial v}{{\partial y}} = \,\frac{ - 1}{\rho }\,\,\frac{\partial p}{{\partial y}} + \,\,\left( {\frac{{\partial^{2} v}}{{\partial x^{2} }}\,\, + \,\,\frac{{\partial^{2} v}}{{\partial y^{2} }}} \right).$$

$$\Rightarrow \frac{\partial u}{{\partial x}}\,\, + \,\,\frac{\partial v}{{\partial y}}\,\, = \,\,0,$$

(3)

$$u\,\,\frac{\partial u}{{\partial x}} + v\,\,\frac{\partial u}{{\partial y}} = \frac{ - 1}{\rho }\frac{\partial p}{{\partial x}} + \nu \,\,\left( {\frac{{\partial^{2} u}}{{\partial x^{2} }}\,\, + \,\,\frac{{\partial^{2} u}}{{\partial y^{2} }}} \right),$$

(4)

$$u\,\,\frac{\partial v}{{\partial x}} + v\,\,\frac{\partial v}{{\partial y}} = \frac{ - 1}{\rho }\,\,\frac{\partial p}{{\partial y}} + \nu \,\,\,\left( {\frac{{\partial^{2} v}}{{\partial x^{2} }}\,\, + \,\,\frac{{\partial^{2} v}}{{\partial y^{2} }}} \right),$$

(5)

where $\nu = \frac{\mu }{\rho }$ is kinematic viscosity.

To compare the order of magnitude of each term in Eqs. (3),(4),(5), it is more advantageous to put the Eqs. (3),(4),(5) in dimensionless form by letting,

$$x^{ * } = \,\,\frac{\,\,\,x\,}{L},\,\,\,y^{ * } = \frac{y\,\,}{\delta },\,\,\,u^{ * } = \frac{u\,\,}{U},\,\,\,v^{ * } = \frac{\,\,v}{V},\,\,\,t^{ * } = \frac{\,\,t}{\frac{L}{U}},\,\,\,p^{ * } = \frac{\,p}{{p_{\infty } }},$$

(6)

where $L,\,\,\delta ,\,\,U,\,\,V,$ and $p_{\infty }$ are certain values of the corresponding quantities $x,\,\,y,\,\,u,\,\,v,$ and $p$. All dimensionless quantities are of order unity.

Using (6) in (3), we get,

$$\frac{U}{L}\,\,\frac{{\partial u^{*} }}{{\partial x^{*} }}\,\, + \,\,\frac{V}{\delta }\,\,\frac{{\partial v^{*} }}{{\partial y^{*} }}\,\, = \,\,0.$$

(7)

Integrating Eq. (7), after making use of the condition $\left( {v^{*} } \right)_{{y^{*} = 1}} = 1$,

$$\frac{V}{U} = - \frac{\delta }{L}\int_{\,0}^{\,1} {\,\,\frac{{\partial u^{*} }}{{\partial x^{*} }}} \,\,dy^{*} .$$

(8)

Since the integral in Eq. (8) is of order unity, the order of the velocity ratio $\frac{V}{U}$ is $\frac{\delta }{L}$. Hence $V$ ≪ $U$_, Using (6) in (4),

$$u^{ * } \,\frac{{\partial u^{ * } }}{{\partial x^{ * } }}\,\, + \,\,\frac{V}{U}\,\,\frac{L}{\delta }\,\,\frac{{\partial u^{ * } }}{{\partial y^{ * } }}\, = \,\, - \frac{{p_{\infty } }}{{\rho U^{2} }}\,\,\frac{{\partial p^{ * } }}{{\partial x^{ * } }} + \frac{1}{{\text{Re}}}\,\,\left( {\frac{{\partial^{2} u^{ * } }}{{\partial x^{ * 2} }}\,\, + \,\,\frac{{L^{2} }}{{\delta^{2} }}\,\,\frac{{\partial^{2} u^{ * } }}{{\partial y^{ * 2} }}} \right).$$

(9)

Using (6) in (5),

$$\frac{V}{U}\,\,u^{*} \,\,\frac{{\partial v^{*} }}{{\partial x^{*} }}\,\, + \,\frac{{\,V^{2} }}{{U^{2} }}\,\,\frac{L}{\delta }\,\,\frac{{\partial v^{*} }}{{\partial y^{*} }}\,\, = \, - \frac{{L\,p_{\infty } }}{{\rho \,\,\delta \,\,U^{2} }}\,\frac{{\partial p^{*} }}{{\partial x^{*} }} + \frac{1}{{\text{Re}}}\,\,\,\frac{V}{U}\,\left( {\frac{{\partial^{2} v^{*} }}{{\partial x^{*2} }} + \,\frac{{L^{2} }}{{\delta^{2} }}\,\,\frac{{\partial^{2} v^{*} }}{{\partial y^{ * 2} }}} \right).$$

(10)

where ${\text{Re}} \, = \,\frac{UL}{\nu }$ is the Reynolds number $\left( {{\text{Re}} \sim \,\frac{1}{{\delta^{2} }}} \right)$. Neglecting the terms of order of $\delta$ and smaller from Eqs. (9) and (10) and reverting to dimensional variables, we get,

$$u\,\,\frac{\partial u}{{\partial x}}\,\, + \,\,v\,\,\frac{\partial u}{{\partial y}}\,\, = \,\,\frac{ - 1}{\rho }\,\,\frac{\partial p}{{\partial x}}\,\, + \,\,\nu \,\,\frac{{\partial^{2} u}}{{\partial y^{2} }},$$

(11)

$$\frac{\partial p}{{\partial y}}\,\, = \,\,0,$$

(12)

$$\frac{\partial u}{{\partial x}}\,\, + \,\,\frac{\partial v}{{\partial y}}\,\, = \,\,0.$$

(13)

Now we will consider a two-dimensional steady and laminar boundary layer flow of an electrically conducting incompressible viscous fluid due to the stretching surface. A magnetic field $B(x)$ is applied to the flow in the direction normal to the stretching surface. The axes $x$ and $y$ are taken along the boundary-layer flow and to its normal, respectively. Here, the physical properties of the fluid are taken to be constant.

Using the above magnitude analysis along with boundary-layer approximations, we get,

$$\frac{\partial u}{{\partial x}}\,\, + \,\,\frac{\partial v}{{\partial y}}\,\, = \,\,0,$$

(14)

$$u\,\,\frac{\partial u}{{\partial x}}\,\, + \,\,v\,\,\,\frac{\partial u}{{\partial y}}\,\, = \,\,\frac{ - 1}{\rho }\,\,\frac{dp}{{dx}}\,\, + \,\,\nu \,\,\,\frac{{\partial^{2} u}}{{\partial y^{2} }}\, - \frac{{\sigma \,B^{2} u}}{\rho }.$$

(15)

Subjected to the boundary conditions:

$$\begin{gathered} y\,\, = \,\,0\,:\,\,\,u\,\, = \,\,U_{w} (x),\,\,\,\,v\,\, = \,\,0, \hfill \\ \frac{y}{\delta }\,\, \to \,\infty \,:\,\,\,\,\,u\,\, \to \,\,U_{\infty } (x). \hfill \\ \end{gathered}$$

(16)

where the velocity components $u$ and $v$ are the stream-wise and normal velocity components in directions $x$ and $y$, respectively, $U_{\infty } (x)$ is the velocity at the edge of the boundary layer of thickness $\delta$, $U_{w} (x)$ is the velocity of the bounding surface.

At $y = \delta ,\,\,\,\,\frac{\partial u}{{\partial y}} = 0$, then Eq. (15) reduces to

$$U_{\infty } \frac{{dU_{\infty } }}{dx} + \frac{{\sigma \,B^{2} U_{\infty } }}{\rho }\,\, = \,\,\frac{ - 1}{\rho }\frac{dp}{{dx}}.$$

(17)

Using (17) in (15),

$$u\,\,\frac{\partial u}{{\partial x}}\,\, + \,\,v\,\,\,\frac{\partial u}{{\partial y}}\,\, = \,\,U_{\infty } \,\,\frac{{dU_{\infty } }}{dx}\,\, + \,\,\nu \,\,\,\frac{{\partial^{2} u}}{{\partial y^{2} }}\,\, - \,\,\frac{{\sigma \,\,B^{2} (u - U_{\infty } )}}{\rho }.$$

(18)

It is observed that, at the bounding surface, the fluid's velocity will decompose into the mainstream flows either exponentially or algebraically, and that depends on the imposed pressure gradient on the flow. The similarity solutions exist if $U_{w} (x)$, $U_{\infty } (x)$, and $B(x)$ obey the following power-law relations

$$U_{w} (x) = U_{ow} (x)\,x^{n} ,\,\,\,\,\,U_{\infty } (x) = U_{0\infty } (x)\,x^{n} ,\,\,\,\,\,B(x) = B_{0} \,x^{(n - 1)/2} .$$

(19)

where $U_{ow}$, $U_{0\infty }$,$B_{0}$ are non-negative constants and $n$ is associated with either non-uniform stretching speed or pressure gradient. The nondimensional coordinate transformation and stream function are introduced [18, 48],

$$\psi = \sqrt {\frac{2\,\nu \,x\,\,\,U}{{1 + n}}} f(\eta ),\,\,\eta = \,\,y\,\,\sqrt {\frac{(1 + n)\,U}{{2\,\nu \,x}}} .$$

(20)

where the stream function $\psi (x,y)$ satisfying the continuity Eq. (14) and is defined as:

$$u\,\, = \,\, - \,\,\frac{\partial \psi }{{\partial y}},\,\,\,\,\,\,\,\,\,\,\,\,\,v\,\, = \,\,\frac{\partial \psi }{{\partial x}}.$$

(21)

Now, the composite reference velocity $U$ is defined as:

$$U(x)\,\, = \,\,U_{w} (x)\,\, + \,\,U_{\infty } (x).$$

(22)

Using the above transformations, Eqs. (18) and (16) become,

$$f^{\prime\prime\prime}\,(\eta )\,\, + \,\,f(\eta )\,\,\,f^{\prime\prime}(\eta )\,\, + \,\,\beta \,\,\,\left( { \in^{2} - (f^{\prime}(\eta ))^{2} } \right)\,\, + \,\,M^{2\,\,} \,\left( { \in - f^{\prime}(\eta )} \right)\,\, = \,\,0,$$

(23)

$$f\,\,(0)\,\, = \,\,0,\,\,\,\,\,\,\,\,f^{\prime}\,(0)\,\, = \,\,1\,\, - \,\, \in ,\,\,\,\,\,\,\,\,f^{\prime}\,(\infty )\,\, = \,\, \in .$$

(24)

where $\in$ is a composite velocity parameter given by,

$$\in \,\,\, = \,\,\frac{{U_{\infty } \,\,\,\,}}{{U_{w} \,\, + \,\,U_{\infty } }},\,\,\,\,\,\,\,U_{w} + U_{\infty } \,\, \ne \,\,0,$$

(25)

$\beta = \frac{2n}{{n\,\, + \,\,1}}$ measures the stretch rate of the moving boundary and $M = \sqrt {\frac{2\,\sigma }{{\rho (n + 1)U_{0\infty } }}} B_{0}$ is the magnetic parameter (Hartmann number).

3 Preliminaries of Bernoulli wavelet and its properties

The Bernoulli wavelets are defined as [45],

$${\theta }_{n,m}\left(x\right)=\left\{\begin{array}{l}{2}^{\frac{k -1}{2} }\widetilde{{b}_{m}}\left({2}^{k-1}x-\widehat{n}\right), \frac{-\widehat{n}}{{ 2}^{k-1}}\le x<\frac{\widehat{n}+1}{{2}^{k-1}}\\ 0, Otherwise.\end{array},\right.$$

with

$$\widetilde{{b}_{m}}\left(x\right)=\left\{\begin{array}{l} 1 , m=0,\\ \frac{1}{\sqrt{\frac{{\left(- 1 \right)}^{m-1}{\left(m !\right)}^{2}}{\left(2m\right) !}{a}_{2m}}}{b}_{m}\left(x\right), m>0.\end{array}\right.$$

where $m = 0, 1 , 2 , \ldots , M - 1, \hat{n} = n - 1, n = 1, 2, \ldots , 2^{{\left( { k {-} 1} \right)}}$.

The coefficient $\frac{1 }{\sqrt{\frac{{\left( -1 \right)}^{m-1}{\left(m !\right)}^{2}}{\left(2m\right)!}{a}_{ 2m}}}$ is for normality, the dilation parameter is $f={2}^{ - \left(k - 1\right)}$ and the translation parameter$g=\widehat{n}{2}^{- ( k-1 )}$. Here, ${b}_{m}\left(x\right)$ are the familiar Bernoulli polynomials of order $m$ which can be defined by,${b}_{m}\left(x\right)=\sum_{i=0}^{m}\left(\genfrac{}{}{0pt}{}{m}{i}\right){a}_{m-i}{x}^{i}$, where ${a}_{i}, i=0 , 1 ,\dots , m$ are Bernoulli numbers.

A few Bernoulli numbers are:

$${a}_{0}=1, {a}_{1}=\frac{-1}{2}, {a}_{2}=\frac{1}{6}, {a}_{4}=\frac{-1}{30}, {a}_{ 6}=\frac{1}{42}, {a}_{ 8}=\frac{-1}{30}, {a}_{10}=\frac{5}{66}, {a}_{12}=-\frac{691}{2730}, {a}_{14}=\frac{7}{6}, {a}_{16}=-\frac{3617}{510}, {a}_{ 18}=\frac{43867}{798},\dots$$

with ${a}_{2 i+ 1}= 0, i=1, 2, 3, \dots$

Few Bernoulli polynomials are given by

$${b}_{0}= 1, {b}_{1}=- \frac{1}{ 2} + x, {b}_{2}=\frac{1}{6} - x + {x}^{2}, {b}_{3}= \frac{ x}{2} -\frac{3 x}{2}+{x}^{3},$$

$${b}_{4}=-\frac{1}{30}+{x}^{2}-2{x}^{3}+{x}^{4}, {b}_{5}=-\frac{x}{6}+\frac{5{x}^{3}}{3}-\frac{5{x}^{4}}{2}+{x}^{5},$$

$${b}_{6}=\frac{1}{42}-\frac{{x}^{2}}{2}+\frac{5{x}^{4}}{2}-3{x}^{5}+{x}^{6}, {b}_{7}=\frac{x}{6}-\frac{7{x}^{3}}{6}+\frac{7{x}^{5}}{2}-\frac{7{x}^{6}}{2}+{x}^{7},$$

$${b}_{8}=-\frac{1}{30}+\frac{2{x}^{2}}{3}-\frac{7{x}^{4}}{3}+\frac{14{x}^{6}}{3}-4{x}^{7}+{x}^{8}, {b}_{9}=-\frac{3x}{10}+2{x}^{3}-\frac{21{x}^{5}}{5}+6{x}^{7}-\frac{9{x}^{8}}{2}+{x}^{9},$$

$${b}_{10}=\frac{5}{66}-\frac{3{x}^{2}}{2}+5{x}^{4}-7{x}^{6}+\frac{15{x}^{8}}{2}-5{x}^{9}+{x}^{10},\dots ,$$

Theorem 1:

Let $H$ be a Hilbert space and $W$ be a closed subspace of $H$ such that $\dim W < \infty$ and $\{ w_{1} ,w_{2} ,...,w_{n} \}$ is any basis for $W$. Let $g \in$$H$ be arbitrary and $g_{0}$ be the unique best approximation to $g$ out of $W$. Then[45]

$\left\| {g - g_{0} } \right\|_{2} = G_{g}$_, Where $G_{g} = \left( {\frac{{Z(g,w_{1} ,w_{2} ,...w_{n} )}}{{Z(g,w_{1} ,w_{2} ,...w_{n} )}}} \right)^{\frac{1}{2}}$ and $Z$ is introduced in [49] as follows:

$$Z\,(g,\,w_{1} ,\,w_{2} ,...\,w_{n} )\,\, = \,\,\,\left| {\,\,\begin{array}{*{20}c} {\, < g,g > } & { < g,w_{1} > } & {...} & { < g,w_{n} > } \\ { < w_{1} ,g > } & { < w_{1} \,,\,\,w_{1} > } & {...} & { < \,w_{1} \,\,,\,\,w_{n} > } \\ {...} & {...} & {...} & {...} \\ { < \,\,w_{n} ,g > } & { < \,w_{n} ,w_{1} > } & {...} & { < w_{n} ,w_{n} > } \\ \end{array} \,\,} \right|\,.$$

Theorem 2:

Suppose $f:\,[0,1]\,\,\, \to \,\,\,R$$,$ $f \in \,\,L^{2} \,[0, \, 1]$ and $Y = span\,\{ \,\psi_{10} (t),\,\psi_{11} (t),...,\,\psi_{{2^{k - 1} M - 1}} (\,t\,)\}$. If $C\,^{T} \psi \,(t\,)$ is the best approximation of $f$ out of $Y$, then the error bound is given by:

$$\left\| {\,\,\int\limits_{0}^{t} {\,\,f(\,t^{\prime}\,\,)\,\,\,dt^{\prime}\,\, - \,\,C^{T} \,P\psi (\,\,t\,)\,\,} } \right\|_{\,\,\,2} \le \,\,\,\,G_{f} \,\, + \,\,\,\,\gamma_{f}$$

where, $\gamma_{f} = \,\,\,\frac{{M_{1} \,}}{{2^{\,k\, - 1} M}}\,\,\sum\limits_{n\, = 1}^{{2\,^{k - 1} }} {\,\,\left| {c_{n,M - 1} \,} \right|}$, with $M_{1} = \mathop {\max }\limits_{t \in [0,1]} \left| {\psi_{nM - 1} } \right|$, $n\,\, = \,\,1\,\,,...,\,\,2^{k - 1}$.

Theorem 3:

Let $L^{2} [0,1]$ be the Hilbert space generated by the Bernoulli wavelet basis. Let $\eta (x)$ be the continuous bounded function in $L^{2} \,[0,\,1\,]$. Then the Bernoulli wavelet expansion of $\eta (x)$ converges to it.

Proof:

Let $\eta :[0,1] \to R$ be a continuous function and $\left| {\eta (x)} \right| \le \mu$, where $\mu$ be any real number. Then Bernoulli wavelet expansion of $y(x)$ is as follows,

$$\eta \left( x \right) = \sum\limits_{n = 1}^{{2^\frac{k - 1}{2}}} {\,\sum\limits_{m\,\, = \,\,0}^{M - 1} {a_{n,m} \,} } \theta_{n,\,\,m\,} \left( x \right).$$

$a_{n,\,m} = \left\langle {\eta (x),\,\left. {\theta_{n,\,\,m} \,(x)} \right\rangle } \right.$ denotes inner product.

$$a_{n,\,m} = \int\limits_{0}^{1} {\eta (x){\kern 1pt} \,\,\theta_{n,m} } \left( x \right)\,dx.$$

Since $\theta_{n,m}$ are the orthogonal basis.

$a_{n,m} = \int\limits_{I} {\eta \left( x \right)\frac{{2^\frac{k - 1}{2}}}{{\sqrt {\frac{{( - 1)^{m - 1} (m!)^{2} \alpha_{2m} }}{(2m)!}} }}} \,\,\beta_{m} \left( {2^{k - 1} x\, - \,n\,\, + \,\,1} \right)\,dx,$ where, $I = \,\,\left[ {\frac{n - 1}{{2^{\,k - 1} }}\,} \right.,\,\left. {\,\frac{n}{{2\,^{k - 1} }}} \right).$

The substitute $2^{k - 1} x - n + 1 = y$ then we get,

$$a_{n,\,\,m} \, = \,\,\,\,\frac{{2^\frac{k - 1}{2}}}{{\,\,\sqrt {\frac{{( - \,1)\,^{m - 1} (m\,!)\,^{2} \,\alpha_{2m} }}{(2m)\,\,!}} }}\,\,\,\int\limits_{0}^{1} {\eta \left( {\frac{y\,\, + n\,\, - \,\,1}{{2^{\,k - 1} }}} \right)} \,\,\beta_{m} \,\left( y \right)\,\frac{dy}{{\,\,2^{k - 1} }},$$

$$a_{n,\,m\,} \,\,\, = \,\,\,\frac{{2^\frac{1 - k}{2}}}{{\,\,\sqrt {\frac{{( - 1)^{m - 1} (m\,!)^{2} \alpha_{2m} }}{(2m)!}} }}\,\,\,\int\limits_{0}^{1} {\eta \left( {\frac{y + n - 1}{{2^{k - 1} }}} \right)} \,\,\,\beta_{m} \left( y \right)dy.$$

By generalized mean value theorem,

$a_{n,\,m} \, = \,\,\frac{{2^\frac{ - k + 1}{2}}}{{\sqrt {\frac{{( - 1)^{\,m - 1} (\,m\,!\,)^{2} \alpha_{2m} }}{(2m)!}} }}\,\eta \,\left( {\frac{\,\xi + \,\,n - 1\,}{{\,\,\,\,2\,^{k\, - \,1} }}} \right)\,\,\int\limits_{0}^{1} {\,\beta_{m} \left( y \right)dy} ,\,\,$ for some $\xi \in (0,1),$

Since $\beta_{m} \left( y \right)$ is a bounded continuous function. Put $\int\limits_{0}^{1} {\beta_{m} \left( y \right)dy = h} ,$

$$\left| {a_{n,m} } \right| = \left| {\frac{{2^{{\frac{ - k + 1}{2}}} }}{{\,\sqrt {\frac{{\,\,( - 1)\,^{m\, - 1} \,(m\,!)^{2} \,\alpha_{2m} }}{(2\,m)!}} }}} \right|\,\,\left| {\eta \left( {\frac{\xi + n - 1}{{2^{k - 1} }}} \right)} \right|h.$$

Since, $\eta$ is bounded

Hence, $\left| {a_{n,m} \,} \right|\,\, = \,\left| {\frac{{2^\frac{ - k + 1}{2}\mu h}}{{\sqrt {\frac{{( - 1)\,^{m - 1} (\,m\,!)^{2} \alpha_{2m} }}{(\,2m)!}} }}} \right|.$

Therefore, $\sum\nolimits_{n,m = 0}^{\infty } {a_{n,m} }$ is absolutely convergent. Hence the Bernoulli wavelet series expansion of $\eta \left( x \right)$ converges uniformly to it.

4 Functional matrix of Bernoulli wavelets

Bernoulli wavelet basis at $k=1$ is as follows:

$${\theta }_{\mathrm{1,0}}\left( x\right)=1,$$

$${\theta }_{1, 1}\left( x \right)=\sqrt{3}\left(-1+2x\right),$$

$${\theta }_{\mathrm{1,2}}\left(x\right)=\sqrt{5}\left(1-6x+6{x}^{2}\right),$$

$${\theta }_{\mathrm{1,3}}\left(x\right)=\sqrt{210}\left(x-3{x}^{2}+2{x}^{3}\right),$$

$${\theta }_{\mathrm{1,4}}\left(x\right)=10\sqrt{21}\left(-\frac{1}{30}+{x}^{2}-2{x}^{3}+{x}^{4}\right),$$

$${\theta }_{\mathrm{1,5}}\left(x\right)=\sqrt{\frac{462}{5}} \left(-x+10{x}^{3}-15{x}^{4}+6{x}^{5}\right),$$

$${\theta }_{\mathrm{1,6}}\left(x\right)=\sqrt{\frac{1430}{691}} \left(1-21{x}^{2}+105{x}^{4}-126{x}^{5}+42{x}^{6}\right),$$

$${\theta }_{\mathrm{1,7}}\left(x\right)=2\sqrt{\frac{143}{7}} \left(x-7{x}^{3}+21{x}^{5}-21{x}^{6}+6{x}^{7}\right),$$

$${\theta }_{\mathrm{1,8}}\left(x\right)=\sqrt{\frac{7293}{3617}} \left(-1+20{x}^{2}-70{x}^{4}+140{x}^{6}-120{x}^{7}+30{x}^{8}\right),$$

$${\theta }_{\mathrm{1,9}}\left(x\right)=\sqrt{\frac{1939938}{219335}} \left(-3x+20{x}^{3}-42{x}^{5}+60{x}^{7}-45{x}^{8}+10{x}^{9}\right),$$

$${\theta }_{\mathrm{1,10}}\left(x\right)=22\sqrt{\frac{125970}{174611}} \left(\frac{5}{66}-\frac{3{x}^{2}}{2}+5{x}^{4}-7{x}^{6}+\frac{15{x}^{8}}{2}-5{x}^{9}+{x}^{10}\right),$$

$${\theta }_{\mathrm{1,11}}\left(x\right)=2\sqrt{\frac{676039}{854513}} \left(5x-33{x}^{3}+66{x}^{5}-66{x}^{7}+55{x}^{9}-33{x}^{10}+6{x}^{11}\right).$$

where,

$${\theta }_{10}\left(x\right)={[{\theta }_{1 , 0}\left(x \right),{\theta }_{1, 1}\left( x\right),{\theta }_{1, 2}\left( x\right),{\theta }_{\mathrm{1,3}}\left( x\right),{\theta }_{\mathrm{1,4}}\left( x\right),{\theta }_{\mathrm{1,5}}\left( x\right),{\theta }_{\mathrm{1,6}}\left(x\right),{\theta }_{\mathrm{1,7}}\left(x\right),{\theta }_{\mathrm{1,8}}\left(x\right),{\theta }_{\mathrm{1,9}}\left(x\right)]}^{T}.$$

Integrating the above first ten basis concerning $x$ limit from $0$ to $x$, then expressing as a linear combination of Bernoulli wavelet basis as

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,0}}\left( x\right)dx=\left[\begin{array}{cccccccccc}\frac{1}{2}& \frac{1}{2\sqrt{3}}& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left( x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,1} }\left( x\right)dx=\left[\begin{array}{cccccccccc}-\frac{1}{2\sqrt{3}}& 0& \frac{1}{2\sqrt{15}}& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left( x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,2}}\left( x\right)dx=\left[\begin{array}{cccccccccc}0& 0& 0& \frac{1}{\sqrt{42}}& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,3}}\left(x\right)dx=\left[\begin{array}{cccccccccc}\frac{\sqrt{7}}{2\sqrt{30}}& 0& 0& 0& \frac{1}{2\sqrt{10}}& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,4}}\left(x\right)dx=\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& \frac{\sqrt{5}}{3\sqrt{22}}& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,5}}\left( x \right)dx=\left[\begin{array}{cccccccccc}-\sqrt{\frac{11}{210}} & 0& 0& 0& 0& 0& \frac{\sqrt{691}}{10\sqrt{273}}& 0& 0& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,6}}\left(x\right)dx=\left[\begin{array}{cccccccccc}0& 0& 0& 0& 0& 0& 0& \sqrt{\frac{35}{1382}} & 0& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,7}}\left(x\right)dx=\left[\begin{array}{cccccccccc}\frac{\sqrt{143}}{20\sqrt{7}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{3617}}{20\sqrt{357}}& 0\end{array}\right]{\theta }_{10}\left(x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,8}}\left(x \right) dx=\left[\begin{array}{cccccccccc} 0& 0& 0& 00& 0& 0& 00& 0& 0& \frac{\sqrt{219335}}{3\sqrt{962122}}\end{array}\right]{\theta }_{10}\left( x\right),$$

$$\underset{0}{\overset{x}{\int }}{\theta }_{\mathrm{1,9}}\left(x\right)dx=\left[\begin{array}{cccccccccc}-\sqrt{\frac{146965}{2895222}} & 0& 0& 00& 0& 00& 0& 00& 0& 0\end{array}\right]{\theta }_{10}\left( x\right)+\frac{\sqrt{1222277}}{10\sqrt{482537}}{\theta }_{\mathrm{1,10}}\left(x\right).$$

Hence,

$$\underset{0}{\overset{x}{\int }}\theta \left(x\right)dx={B}_{10\times 10}{\theta }_{10}\left(x\right)+\overline{{\theta }_{10}}\left(x\right),$$

where

$${B}_{10\times 10}=\left[\begin{array}{cccccccccc}\frac{1}{2}& \frac{1}{2\sqrt{3}}& 0& 0& 0& 0& 0& 0& 0& 0\\ -\frac{1}{2\sqrt{3}}& 0& \frac{1}{2\sqrt{15}}& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& \frac{1}{\sqrt{42}}& 0& 0& 0& 0& 0& 0\\ \frac{\sqrt{7}}{2\sqrt{30}}& 0& 0& 0& \frac{1}{2\sqrt{10}}& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& \frac{\sqrt{5}}{3\sqrt{22}}& 0& 0& 0& 0\\ -\sqrt{\frac{11}{210}} & 0& 0& 0& 0& 0& \frac{\sqrt{691}}{10\sqrt{273}}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& \sqrt{\frac{35}{1382}} & 0& 0\\ \frac{\sqrt{143}}{20\sqrt{7}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{3617}}{20\sqrt{357}}& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{219335}}{3\sqrt{962122}}\\ -\sqrt{\frac{146965}{2895222}} & 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right],$$

$$\overline{{\theta }_{10}}\left(x\right)=\left[\begin{array}{c}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \frac{\sqrt{1222277}}{10\sqrt{482537}}{\theta }_{\mathrm{1,10}}\left(x\right)\end{array}\end{array}\right]$$

In general, the first integration of the Bernoulli wavelet can be represented as;

$$\underset{0}{\overset{x}{\int }}\theta \left(x\right)dx={B}_{n\times n}{\theta }_{n}\left(x\right)+\overline{{\theta }_{n}}\left(x\right).$$

Next, integrating the above ten basis twice, we get,

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,0}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}\frac{1}{6}& \frac{1}{4\sqrt{3}}& \frac{1}{12\sqrt{5}}& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,1}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}-\frac{1}{4\sqrt{3}}& -\frac{1}{12}& 0& \frac{1}{6\sqrt{70}}& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,2}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}\frac{1}{12\sqrt{5}}& 0& 0& 0& \frac{1}{4\sqrt{105}}& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,3}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}\frac{\sqrt{7}}{4\sqrt{30}}& \frac{\sqrt{7}}{12\sqrt{10}}& 0& 0& 0& \frac{1}{12\sqrt{11}}& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,4}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}-\frac{1}{6\sqrt{21}}& 0& 0& 0& 0& 0& \frac{\sqrt{691}}{6\sqrt{30030}}& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,5}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}-\frac{\sqrt{11}}{2\sqrt{210}}& -\frac{\sqrt{11}}{6\sqrt{70}}& 0& 0& 0& 0& 0& \frac{1}{2\sqrt{390}}& 0& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,6}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}\frac{\sqrt{143}}{4\sqrt{6910}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{3617}}{4\sqrt{352410}}& 0\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,7}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}\frac{\sqrt{143}}{40\sqrt{7}}& \frac{\sqrt{143}}{40\sqrt{21}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{43867}}{84\sqrt{9690}}\end{array}\right]{\theta }_{10}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,8}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}-\frac{5\sqrt{221}}{6\sqrt{119361}}& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right)+\frac{\sqrt{174611}}{6\sqrt{7559530}}{\theta }_{\mathrm{1,10}}\left(t\right),$$

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}{\theta }_{\mathrm{1,9}}\left(t\right)dtdt=\left[\begin{array}{cccccccccc}-\frac{\sqrt{146965}}{2\sqrt{2895222}}& -\frac{\sqrt{146965}}{6\sqrt{965074}}& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]{\theta }_{10}\left(t\right)+\frac{\sqrt{77683}}{2\sqrt{30268230}}{\theta }_{\mathrm{1,11}}\left(t\right).$$

Hence,

$$\underset{0}{\overset{t}{\int }}\underset{0}{\overset{t}{\int }}\theta \left(t\right)dtdt={B}_{10\times 10}^{^{\prime}}{\theta }_{10}\left(t\right)+{\overline{{\theta }_{10}}}^{^{\prime}}\left(t\right),$$

where

$${B}_{10\times 10}^{^{\prime}}=\left[\begin{array}{cccccccccc}\frac{1}{6}& \frac{1}{4\sqrt{3}}& \frac{1}{12\sqrt{5}}& 0& 0& 0& 0& 0& 0& 0\\ -\frac{1}{4\sqrt{3}}& -\frac{1}{12}& 0& \frac{1}{6\sqrt{70}}& 0& 0& 0& 0& 0& 0\\ \frac{1}{12\sqrt{5}}& 0& 0& 0& \frac{1}{4\sqrt{105}}& 0& 0& 0& 0& 0\\ \frac{\sqrt{7}}{4\sqrt{30}}& \frac{\sqrt{7}}{12\sqrt{10}}& 0& 0& 0& \frac{1}{12\sqrt{11}}& 0& 0& 0& 0\\ -\frac{1}{6\sqrt{21}}& 0& 0& 0& 0& 0& \frac{\sqrt{691}}{6\sqrt{30030}}& 0& 0& 0\\ -\frac{\sqrt{11}}{2\sqrt{210}}& -\frac{\sqrt{11}}{6\sqrt{70}}& 0& 0& 0& 0& 0& \frac{1}{2\sqrt{390}}& 0& 0\\ \frac{\sqrt{143}}{4\sqrt{6910}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{3617}}{4\sqrt{352410}}& 0\\ \frac{\sqrt{143}}{40\sqrt{7}}& \frac{\sqrt{143}}{40\sqrt{21}}& 0& 0& 0& 0& 0& 0& 0& \frac{\sqrt{43867}}{84\sqrt{9690}}\\ -\frac{5\sqrt{221}}{6\sqrt{119361}}& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -\frac{\sqrt{146965}}{2\sqrt{2895222}}& -\frac{\sqrt{146965}}{6\sqrt{965074}}& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right],$$

$${\overline{{\theta }_{10}}}^{^{\prime}}\left(t\right)=\left[\begin{array}{c}\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \frac{\sqrt{174611}}{6\sqrt{7559530}}{\theta }_{\mathrm{1,10}}\left(t\right)\\ \frac{\sqrt{77683}}{2\sqrt{30268230}}{\theta }_{\mathrm{1,11}}\left(t\right)\end{array}\end{array}\right]$$

In the same way, we can create matrices of various sizes for our comfort.

5 Method of solution

The semi-infinite domain $[0,\,\infty )$ in (23) has to be transformed to $[0,\,1]$ by using the coordinate transformation $\xi = \frac{\eta }{{\eta_{\infty } }}$ and change of variable $\,F(\xi ) = \frac{f(\eta )}{{\eta_{\infty } }}$. Under this transformation (23) becomes

$$F^{\prime\prime\prime}(\xi ) + \eta_{\infty }^{2} \,F(\xi )\,F^{\prime\prime}(\xi ) + \beta \,\eta_{\infty }^{2} \,\left( { \in^{2} - (F^{\prime}(\xi )^{2} } \right) + M^{2} \,\eta_{\infty }^{2} \,\left( { \in - F^{\prime}(\xi )} \right) = 0.$$

(26)

where $\eta_{\infty }^{{}}$ is an unknown finite boundary which is assumed to satisfy the asymptotic condition $f^{\prime\prime}\,(\,\eta_{\infty } ) = 0$. Also $\eta_{\infty }$ varies with the parameters $\beta ,\,\,M,$ and $\in$. The boundary conditions (24) are transformed to

$$F\,(\,0\,)\,\, = \,\,0\,\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,F^{\prime}\,(\,0\,)\,\, = \,\,\,1\, - \,\, \in \,\,\,,\,\,\,\,\,\,\,\,\,\,F^{\prime}\,(\,1\,)\,\, = \,\, \in .$$

(27)

6 Numerical solution by Bernoulli wavelet collocation method (BWCM)

Consider the model

$$F^{\prime\prime\prime}(\xi ) + \eta_{\infty }^{2} \,F(\xi )\,F^{\prime\prime}(\xi ) + \beta \,\eta_{\infty }^{2} \,\left( { \in^{2} - (F^{\prime}(\xi )^{2} } \right) + M^{2} \,\eta_{\infty }^{2} \,\left( { \in - F^{\prime}(\xi )} \right) = 0,$$

(28)

with boundary conditions

$$F\,(\,0\,)\,\, = \,\,0\,\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,F^{\prime}\,(\,0\,)\,\, = \,\,\,1\, - \,\, \in \,\,\,,\,\,\,\,\,\,\,\,\,\,F^{\prime}\,(\,1\,)\,\, = \,\, \in .$$

(29)

Let

$$F^{\prime\prime\prime}(\xi ) = A^{T} \,G\,(\xi ).$$

(30)

Integrate (30) concerning $\xi$ from $0$ to $\xi$,

$$F^{\prime\prime}(\xi )\, = \,\,F^{\prime\prime}(\,0\,) + A^{T} \left( {P\,\,G(\xi ) + \overline{P}(\xi )} \right).$$

(31)

Integrate (31) concerning $\xi$ from $0$ to $\xi$,

$$F^{\prime}(\xi ) = 1 - \in \, + \,\,\xi \,\,\,F^{\prime\prime}(0) + A^{T} \left( {\,P^{\prime}\,G(\xi ) + \overline{P}^{\prime}(\xi )} \right).$$

(32)

Integrate (32) concerning $\xi$ from $0$ to $\xi$,

$$F\,(\xi )\, = \,\xi \,(1 - \in ) + \frac{{\xi^{2} }}{2}\,\,F^{\prime\prime}(\,\,0\,) + A^{T} \left( { \, P^{\prime\prime}\,G(\xi ) + \overline{P}^{\prime\prime}\,(\xi )} \right).$$

(33)

Put $\xi = 1$ in (32), we get,

$$F^{\prime\prime}(0) = 2 \in - 1 - A^{T} \left( {P^{\prime}\,G(\xi ) + \overline{P}^{\prime}(\xi )} \right)|_{\xi = 1} .$$

(34)

Fit the Eqs. (30) – (34) in Eq. (28) and discretize the resultant equation using the collocation point $\xi_{i} = \frac{2\,i - 1}{{2\,M}}$ where $i = 1,\,2,\,...,\,M$ we get the $M$ system of equations in $M$ unknown coefficients. Solving this system by Newton Raphson method, the value of $A^{T}$ _{is obtained}. Substituting this value in Eq. (33) to get an approximate solution $F(\xi )$.

The wall shear stress or skin-friction coefficient $f^{\prime\prime}(0)$ is given by,

$$f^{\prime\prime}\,(\,0\,) = \frac{1}{{\eta_{\infty } }}\left[ {2 \in - 1 - A^{T} \left( {P^{\prime}\,G\left( \xi \right) + \overline{P}^{\prime}\left( \xi \right)} \right)|_{\xi = 1} } \right].$$

(35)

7 Results and discussion

Case (i): Solution for $\in \, = \,0$ and general $\beta$.

The two-dimensional laminar flow due to stretching sheet without magnetic field is given by [16, 17, 50],

$$f^{\prime\prime\prime}\,\left( \eta \right)\, + \,f\left( \eta \right)\,f^{\prime\prime}\,\left( \eta \right) - \beta \,f^{\prime}\left( \eta \right)^{2} = 0,$$

(36)

$$f\,\left( 0 \right)\, = \,0\,\,,\,\,f^{\prime}\,\left( 0 \right) = \,\,1\,\,,\,f^{\prime}\,\left( \infty \right)\, = 0.$$

Equation (36) has an exact solution for some particular values of $\beta$[26]:

$${\text{For }}\beta { = 1, }f(\eta ) = 1 - e^{ - \eta } .$$

(37)

$${\text{For }}\beta = - 1,f(\eta ) = \sqrt 2 \tanh \left( {\frac{\eta }{\sqrt 2 }} \right).$$

(38)

Figures 1 and 2 show that the solution from BWCM coincides with the exact solution.

Fig. 1

Comparison of $f(\eta )$ with analytic solution (37) for $\beta = 1.$

Fig. 2

Comparison of $f(\eta )$ with analytic solution (38) for $\beta = - 1.$

Case(ii): Solution for $\in \, = \,0$ and general $\beta$ and $M$.

The two-dimensional boundary layer flow due to the stretching plate with the magnetic field is given by [24, 25]:

$$\left. \begin{gathered} f^{\prime\prime\prime}\,(\,\eta \,)\,\, + \,\,f\,(\,\eta \,)\,\,f^{\prime\prime}\,(\,\eta \,)\,\, - \,\,\beta \,\,f^{\prime}(\eta )^{2} - M^{2} \,f^{\prime}(\,\eta \,)\,\, = \,\,0, \hfill \\ f\,(\,0\,)\,\, = \,\,0\,\,,\,\,\,\,\,\,\,\,f^{\prime}\,(\,0\,)\, = \,\,1\,\,,\,\,\,\,\,\,\,\,f^{\prime}\,(\,\infty \,)\,\, = \,\,0. \hfill \\ \end{gathered} \right\}$$

(39)

The exact solution for $\beta = 1$ and general $M$ is given by [11]:

$$f(\eta ) = \frac{{1 - e^{{ - \,\eta \,\sqrt {1 + M^{2} \,} }} }}{{\sqrt {1 + M^{2} \,} }}.$$

(40)

From Table 1, we can say that BWCM yields a better solution and consumes less time when compared to the Haar wavelet method.

Table 1 Comparison of values of $f^{\prime\prime}(0)$ for $\beta = 1$ and different values of $M$

Figures 3 and 4 show that the basic solution and velocity profiles satisfy the solution in (40).

Fig. 3

Comparison of solution $f(\eta )$ with analytic solution (40) for different magnetic parameters

Fig. 4

Comparison of solution $f^{\prime}(\eta )$ with analytic solution (40) for different magnetic parameters

Case (iii): Solution for $\in \, = \,1$ and general $\beta$ and $M$.

This case is described by the MHD Falkner-Skan equation [4]:

$$\left. \begin{gathered} f^{\prime\prime\prime}\,\,(\,\eta \,)\,\, + \,\,f\,(\,\eta \,)\,\,f^{\prime\prime}\,(\,\eta \,)\,\, + \,\beta \,\,\left( {\,1 - \,\,f^{\prime}(\eta )^{2} } \right)\,\, + \,\,M^{2\,} \,\left( {\,1 - f^{\prime}(\eta )} \right)\,\, = \,\,0, \hfill \\ f\,(\,0\,)\,\, = \,\,0\,\,,\,\,\,\,\,\,\,\,f^{\prime}\,(\,0\,)\,\, = \,\,0\,\,,\,\,\,\,\,\,\,\,f^{\prime}\,(\,\infty \,)\,\, = \,\,1. \hfill \\ \end{gathered} \right\}$$

(41)

The velocity profile $f^\prime\,(\,\eta \,)$ and shear flow $f^{\prime\prime}(\eta )$ for the various values of pressure gradient $\beta$ with and without magnetic field are shown in Figs. 5 and 6. It is noted that as the pressure gradient $\beta$ increases from zero, the boundary-layer thickness becomes smaller and smaller. The boundary-layer thickness further becomes smaller in the presence of the magnetic field because more energy is released into the flow by the applied magnetic which accelerates the motion of the fluid particles, hence fluid particles move faster. Further increase in $M$ makes the flow more stable. From the Fig. 6, we can also observe that in both cases ${M}^{2}=0$ and ${M}^{2}=10$ $f^{\prime\prime}\left( \eta \right)$ asymptotically tends to zero as$\eta \to \infty$.

Fig. 5

Variation of velocity profiles for various values of $\beta$ and $M$ in (41)

Fig. 6

Shear stress profiles for various values of $\beta$ and $M$ in (41)

Case (iv): Solution for $0 < \in \, < 1$ and general $\beta$ and $M$.

The basic solution $f(\eta )$ is calculated for two sets of values of $M$ (without magnetic field and with magnetic field) and different values of $\in$ which are shown in Figs. 7 and 8. Figures 9 and 10 are obtained by taking $\beta =-1$ and $\beta =-0.4$ are the same as those produced in [17]. Also, when${M}^{2}=0$, the velocity profiles experience both undershoots and overshoots near the wedge, and when $M^{2} \, > \,0$, the flow becomes more stable. As the magnetic field increases, the point of intersection of these solutions is very much closer to the wedge surface. It is seen that when ${M}^{2}=0$ the thickness of the boundary layer is a little large when compared to ${M}^{2}=10$ because the magnetic field releases the energy to flow, thereby making the fluid particles move fast; as a result, the boundary-layer thickness becomes small and makes the flow more stable and mild. It is also seen that as $\beta$ increases from negative ($\beta =-1, -0.4$) to positive ($\beta =0.5, 1.5$), the undershoots $\left({f}^{^{\prime}}\left(\eta \right)<\in \right)$ and overshoots $\left({f}^{^{\prime}}\left(\eta \right)>\in \right)$ vanishes and the flow become mild. This is seen in Figs. 11 and 12 for two different values of$M$. Also, the point of intersection is still nearer to the wedge surface because the flow is accelerated, and the boundary layer thickness becomes smaller and smaller. In addition to this, we can notice that when $\in >\frac{1}{2}$ and$\in <\frac{1}{2}$, all the velocity profiles move toward the endpoints. From Fig. 13, we can say BWCM successfully predicts the double solution for some set of physical parameters when ${\eta }_{\infty }$ is taken to be as large as possible. The double solution is observed only in the absence of a magnetic field, and in the presence of a magnetic field, we found only a single solution.

Fig. 7

Graph of solution $f(\eta )$ for the class of the Falkner-Skan problem at $\beta =1.5, {M}^{2}=0$ and different values of $\in$

Fig. 8

Graph of solution $f(\eta )$ for the class of the Falkner-Skan problem at $\beta =1.5, {M}^{2}=10$ and different values of $\in$

Fig. 9

Variation of a velocity profile for $\beta =-1$ in the absence (first set of solution) and presence of the magnetic field

Fig. 10

Variation of a velocity profile for $\beta =-0.4$ in the absence and presence of the magnetic field

Fig. 11

Velocity profiles with and without magnetic field at $\beta =0.5$

Fig. 12

Velocity profiles with and without magnetic field at $\beta =1.5$

Fig. 13

Variation of velocity profiles for $\beta =-1$ in the absence of magnetic field (second set solutions)

8 Conclusion

In this study, we developed a new operational matrix of integration by the Bernoulli wavelet and the new technique called BWCM. This proposed method studied the two-dimensional boundary layer flow of viscous fluid in the presence of a magnetic field. As we know, many semi-analytical techniques are needed for small parameters, but such difficulties don’t arise in the proposed technique. Here, we found the wavelet solution, including the double solution, by varying the parameters such as, $\in \,,\,\beta ,\,and\,M.$ The following are the important findings from this study:

• As the pressure gradient $\beta$ increases from zero, the boundary-layer thickness becomes smaller and smaller. The boundary-layer thickness further becomes smaller in the presence of the magnetic field $M$.

• The velocity profile experiences both overshoots and undershoots and vanishes when $\beta$ increases from negative to positive.

• Dual solutions are observed only in the absence of the magnetic field. In the presence of a magnetic field, we found only one solution.