Modified Sumudu decomposition method for Lane-Emden-type differential equations

Abstract

In this study, we modify the Sumudu decomposition method and apply the method to study the singular initial value problems which are represented by Lane-Emden-type equations. We also consider both linear and nonlinear cases and provide some examples.

Introduction

The Adomian decomposition method (ADM) is an accurate and efficient method which was proposed initially with the aim of solving frontier physical problems analytically. Since then, ADM has been applied in the solution of nonlinear ordinary or partial differential equations without using linearization or perturbation process. It is also a very powerful method to handle deterministic and stochastic problems that arise in physical, biological and chemical reactions, see Adomain ¹ and Duan and Rach, ² and for the effective convergence region, see Duan et al. ³ The Sumudu Adomian decomposition method (SADM) is a combination of ADM and Sumudu transforms. This method was applied successfully to nonlinear system Volterra integro-differential equations and also used in nonlinear system of partial differential equations, see Eltayeb and colleagues.^4,5 The singular differential equations for Lane–Emden type appear in many areas of mathematical physics and astrophysics in Bert and Zeng. ⁶ In the literature, there are several research works on the similar problems. For example, Momoniat and Harley ⁷ got the solution by reducing the singular equation to the first-order differential equation and applying Lie group analysis, Shawagfeh ⁸ gave the approximate solutions of Lane-Emden-type equations, Cantor-type cylindrical-coordinate method was used by Yang et al. ⁹ to investigate the family of local fractional differential operators on Cantor sets^10–12 studied the variational iteration method for fractional and Zhang et al. ¹³ applied the exponential wavelet iterative shrinkage threshold algorithm for compressed sensing magnetic resonance imaging.

The aim of this article is to apply the modified Sumudu decomposition method to obtain the exact solutions for singular differential equation with exponential and logarithmic nonlinearities. Some examples are given to support the present method. Now let A be a space of functions as follows

$$A=\{f:\exists M,{\tau }_1,{\tau }_2>0,\left|f\right|<M{e}^{t/{\tau }_j},\mathrm{if}x\in {(-1)}^j\times (0,\infty )\}$$

then the Sumudu transform of f is defined as

$$S[f(t);s]=:\underset{0}{\overset{\infty }{\int }}f(\mathrm{sx})e^{-t}\mathrm{dx},s\in (-{\tau }_1,{\tau }_2)$$

Now recall the next theorem which was given in Belgacem et al. ¹⁴

Theorem 1

Let $f\in A$ , then the Sumudu transform integer order derivative of $f^n$ is given as

$$S\left(f^{(n)}\right)=\frac{S(u)}{u^n}-\sum _{k=0}^{n-1}\frac{f^{(k)}(0)}{u^{n-k}}$$

for $n\ge 1$ ; for more details, see Belgacem et al. ¹⁴

In general, to study the Sumudu transform of singular differential equations, we need concepts of Sumudu transform for derivatives.

Theorem 2

Let $S[f(t)]$ be the Sumudu transform of f, then we have

$$S[\mathrm{tf}\prime (t)]=u^2\frac{d}{\mathrm{du}}\left[\frac{S(u)-f(0)}{u}\right]+u\left[\frac{S(u)-f(0)}{u}\right]$$

(1)

$$\begin{array}{c}\multicolumn{1}{c}{S[\mathrm{tf}″(t)]=u^2\frac{d}{\mathrm{du}}\left[\frac{S(u)-f(0)-f\prime (0)}{u^2}\right]}\\ \multicolumn{1}{c}{+u\left[\frac{S(u)-f(0)-f\prime (0)}{u^2}\right]}\end{array}$$

(2)

$$\begin{array}{c}\multicolumn{1}{c}{S[t^{2}f″(t)]=u^4\frac{d^2}{d{u}^2}\left[\frac{S(u)-f(0)-f\prime (0)}{u^2}\right]}\\ \multicolumn{1}{c}{+4u^3\frac{d}{\mathrm{du}}\left[\frac{S(u)-f(0)-f\prime (0)}{u^2}\right]}\\ \multicolumn{1}{c}{+2u^2\left[\frac{S(u)-f(0)-f\prime (0)}{u^2}\right]}\end{array}$$

(3)

$$\begin{array}{c}\multicolumn{1}{c}{S[t^n{f}^{\left(n\right)}(t)]=u^{2n}\frac{d^n}{d{u}^n}(\frac{1}{u^n}S(u)-\frac{1}{u^{n-1}}f(0)-\cdots -\frac{1}{u}{f}^{(n-1)}(0))}\\ \multicolumn{1}{c}{+n^2{u}^{2n-1}\frac{d^{n-1}}{d{u}^{n-1}}(\frac{1}{u^n}S(u)-\frac{1}{u^{n-1}}f(0)-\cdots -\frac{1}{u}{f}^{(n-1)}(0))}\\ \multicolumn{1}{c}{+\cdots +u^{n}n!(\frac{1}{u^n}S(u)-\frac{1}{u^{n-1}}f(0)-\cdots -\frac{1}{u}{f}^{(n-1)}(0))}\end{array}$$

(4)

Proof

To prove equation (1), we use the following formula

$$S[\mathrm{tf}(t)]=u^2\frac{d}{\mathrm{du}}S(u)+\mathrm{uS}(u)$$

given by Asiru. ¹⁵ Then apply

$$\begin{array}{c}\multicolumn{1}{c}{\frac{d}{\mathrm{du}}\left[\frac{S(u)-f(0)}{u}\right]=\frac{d}{\mathrm{du}}\underset{0}{\overset{\infty }{\int }}\frac{1}{u}{e}^{-\frac{t}{u}}f\prime (t)\mathrm{dt}=\underset{0}{\overset{\infty }{\int }}\frac{d}{\mathrm{du}}\left[\frac{1}{u}{e}^{-\frac{t}{u}}\right]f\prime (t)\mathrm{dt}}\\ \multicolumn{1}{c}{=\frac{1}{u^3}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{u}}\mathrm{tf}\prime (t)\mathrm{dt}-\frac{1}{u^2}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{u}}f\prime (t)\mathrm{dt}}\\ \multicolumn{1}{c}{=\frac{1}{u^2}S[\mathrm{tf}\prime (t)]-\frac{1}{u}S[f\prime (t)]}\end{array}$$

and follows that

$$S[\mathrm{tf}\prime (t)]=u^2\frac{d}{\mathrm{du}}\left[\frac{S(u)-f(0)}{u}\right]+u\left[\frac{S(u)-f(0)}{u}\right]$$

The proof of equations (2) and (3) is similar to the proof of equation (1).

Now if we consider solving the following second-order non-homogeneous ordinary differential equations with initial conditions

$$\begin{array}{c}\multicolumn{1}{c}{\frac{d^{2}y}{d{t}^2}+\frac{2n}{t}\frac{\mathrm{dy}}{\mathrm{dt}}+\frac{n(n-1)}{t^2}y+f(t,y)=g(t),n=1,2}\\ \multicolumn{1}{c}{y(0)=A,y\prime (0)=B}\end{array}$$

(5)

where f is a real function, g is a known function and A and B are constants. In particular, if $n=1$ , we obtain

$$\frac{d^{2}y}{d{t}^2}+\frac{2}{t}\frac{\mathrm{dy}}{\mathrm{dt}}+f(t,y)=g(t),y(0)=A,y\prime (0)=B$$

(6)

where $y\prime (0)$ is the value of first derivative at $t=0$ . Multiplying equation (6) with t and applying the transform, we have

$$S[t\frac{d^{2}y}{d{t}^2}+2\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{tf}(t,y)=\mathrm{tg}(t)]$$

(7)

using the properties of the Sumudu transform, we get

$$\frac{\mathrm{dY}}{\mathrm{du}}-\frac{Y\left(u\right)}{u}+2Y\left(u\right)-\frac{y\left(0\right)}{u}+S[\mathrm{tf}(t,y)]=S[\mathrm{tg}(t)]$$

(8)

simplifying equation (8), we have

$$\frac{d}{\mathrm{du}}\left(\mathrm{uY}\left(u\right)\right)-y\left(0\right)+\mathrm{uS}[\mathrm{tf}(t,y)]=\mathrm{uS}[\mathrm{tg}(t)].$$

(9)

Integrating equation (9), we obtain

$$S[y(t)]-\frac{1}{u}\underset{0}{\overset{u}{\int }}y(0)\mathrm{du}+\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[\mathrm{tf}(t,y)]\mathrm{du}=\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[\mathrm{tg}(t)]\mathrm{du}$$

(10)

$f(t,y)$ is defined as follows

$$f(t,y)=K(y(t))+M(y(t))$$

(11)

where $K(y(t))$ and $M(y(t))$ represent the linear and the nonlinear terms, respectively. Doing this way, one can handle equation (9) applying ADM. Furthermore, to address the nonlinear term $M(y(t))$ with modified Sumudu decomposition method (MSDM), we define the solution $y(t)$ and nonlinear function $f(t,y)$ by infinity series

$$y(t)=\sum _{n=0}^{\infty }{y}_n\left(t\right),f(t,y)=\sum _{n=0}^{\infty }{A}_n\left(t\right)$$

(12)

where $A_n$ is denoted by the following formula

$$A_n=\frac{1}{n!}\left[\frac{d^n}{d{\lambda }^n}M{\left(\sum _{n=0}^{\infty }{\lambda }^n{u}_n\right)}_{\lambda =0}\right],n=0,1,2,\dots$$

(13)

Equation (13) is defined as

$$\begin{array}{c}\multicolumn{1}{c}{A_0=M[u_0],A_1={u\prime }_{1}M[u_0]}\\ \multicolumn{1}{c}{A_2=u_{2}M\prime [u_0]+\frac{1}{2!}{u}_{{}_1}^{2}M″[u_0]}\\ \multicolumn{1}{c}{A_3=u_{3}M\prime [u_0]+u_1{u}_{2}M″[u_0]+\frac{1}{3!}{u}_{{}_1}^{3}M″\prime [u_0].}\end{array}$$

(14)

Substituting equation (12) into equation (10), we have

$$\begin{array}{c}\multicolumn{1}{c}{S\left[\sum _{n=0}^{\infty }{y}_n\left(t\right)\right]-\frac{1}{u}\underset{0}{\overset{u}{\int }}y(0)\mathrm{du}}\\ \multicolumn{1}{c}{+\frac{1}{u}\underset{0}{\overset{u}{\int }}u\sum _{n=0}^{\infty }\left(S[\mathrm{tK}(y(t))+\mathrm{tM}(y(t))]\right)\mathrm{du}}\\ \multicolumn{1}{c}{=\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[\mathrm{tg}(t)]\mathrm{du}.}\end{array}$$

(15)

The recursive relation is given by

$$S\left[y_0\left(t\right)\right]=\frac{1}{u}\underset{0}{\overset{u}{\int }}y(0)\mathrm{du}+\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[\mathrm{tg}(t)]\mathrm{du}$$

(16)

$$S\left[y_{n+1}\left(t\right)\right]=-\frac{1}{u}\int u\left(S[\mathrm{tK}(y_n(t))+t{A}_n\left(t\right)]\right)\mathrm{du}$$

(17)

The inverse Sumudu transform of equations (16) and (17) yields

$$y_0\left(t\right)=S^{-1}(\frac{1}{u}\underset{0}{\overset{u}{\int }}y(0)\mathrm{du}+\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[\mathrm{tg}(t)]\mathrm{du})=H(t)$$

(18)

$$y_{n+1}\left(t\right)=S^{-1}(-\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[\mathrm{tK}(y_n(t))+t{A}_n\left(t\right)]\right)\mathrm{du})$$

(19)

where $H(t)$ is considered as the initial solution of equation (18).

Modified Sumudu decomposition method

In this section, we consider the term $H(t)$ in equation (18) and separate as follows

$$H(t)=H_0(t)+H_1(t)$$

(20)

Instead of the iteration procedure in equations (18) and (19), we suggest the following modification

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=H_0(t)}\\ \multicolumn{1}{c}{y_1\left(t\right)=H_1(t)+S^{-1}(-\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[\mathrm{tK}(y_n(t))+t{A}_n\left(t\right)]\right)\mathrm{du})}\end{array}$$

(21)

$$y_{n+1}\left(t\right)=S^{-1}(-\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[\mathrm{tK}(y_n(t))+t{A}_n\left(t\right)]\right)\mathrm{du}).$$

(22)

The solution through the modified Sumudu decomposition method highly depends on the choice of $H_0(t)$ and $H_1(t)$ .

Numerical examples

In this section, we solve some singular initial value problems (IVP) of Lane–Emden type.

Example 1

Consider the singular homogeneous Lane-Emden-type equation

$$\frac{d^{2}y}{d{t}^2}+\frac{2}{t}\frac{\mathrm{dy}}{\mathrm{dt}}-2(2t^2+3)y=0$$

(23)

with initial conditions

$$y\left(0\right)=1y\prime (0)=0.$$

(24)

According to the MSDM and initial conditions, we get

$$\frac{d}{\mathrm{du}}\left(\mathrm{uY}\left(u\right)\right)-1=\mathrm{uS}\left[2(2t^3+3t)y\right]$$

(25)

If S and $S^{-1}$ represent the transform and inverse transform, respectively, then the recursive relations can be obtained as

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{du}\right)=1}\\ \multicolumn{1}{c}{y_n\left(t\right)=S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[4t^3+6t]y_{n-1}\left(t\right)\right)\mathrm{du}\right).}\end{array}$$

(26)

Consider equation (26) and substitute $y(0)=1$ , then we have

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=1}\\ \multicolumn{1}{c}{y_1\left(t\right)=S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[4t^3+6t]y_0\left(t\right)\right)\mathrm{du}\right)}\\ \multicolumn{1}{c}{=S^{-1}(\frac{24}{5}{u}^5+2u^3)=\frac{t^4}{5}+t^2}\\ \multicolumn{1}{c}{y_2\left(t\right)=\frac{3}{10}{t}^4+\frac{13}{105}{t}^6+\frac{1}{90}{t}^8}\\ \multicolumn{1}{c}{y_3\left(t\right)=\frac{3}{70}{t}^6+\frac{17}{630}{t}^8+\frac{59}{11550}{t}^{10}+\frac{1}{3510}{t}^{12}}\\ \multicolumn{1}{c}{⋮}\end{array}$$

Thus, the series solution is given as

$$\begin{array}{c}\multicolumn{1}{c}{y\left(t\right)=y_0\left(t\right)+y_1\left(t\right)+y_2\left(t\right)+\dots }\\ \multicolumn{1}{c}{=1+t^2+\frac{t^4}{2!}+\frac{1}{3!}{t}^6+\frac{1}{4!}{t}^8+\dots }\end{array}$$

The exact solution is

$$y\left(t\right)=e^{t^2}.$$

Example 2

Consider the singular nonlinear IVP

$$\frac{d^{2}y}{d{t}^2}+\frac{2}{t}\frac{\mathrm{dy}}{\mathrm{dt}}+4(2e^y+e^{\frac{y}{2}})=0$$

(27)

$$y(0)=0y\prime (0)=0.$$

(28)

Applying the MSDM, we have

$$\frac{d}{\mathrm{du}}\left(\mathrm{uY}\left(u\right)\right)-\mathrm{uS}\left[4t(2e^y+e^{\frac{y}{2}})\right]=0$$

(29)

and then, the recursive relations

$$y_0\left(t\right)=S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}0\mathrm{du}\right)=0$$

(30)

$$y_n\left(t\right)=S^{-1}(-\frac{1}{u}\int u\left(S\left[4t{A}_{n-1}\left(t\right)\right]\right)\mathrm{du})$$

(31)

by equation (14), we have

$$\begin{array}{c}\multicolumn{1}{c}{A_0=M[y_0]=2e^{y_0}+e^{\frac{y_0}{2}}}\\ \multicolumn{1}{c}{A_1=y_1(2e^{y_0}+\frac{1}{2}{e}^{\frac{y_0}{2}})}\\ \multicolumn{1}{c}{A_2=y_2(2e^{y_0}+\frac{1}{2}{e}^{\frac{y_0}{2}})+\frac{1}{2!}{y}_1^2(2e^{y_0}+\frac{1}{4}{e}^{\frac{y_0}{2}})}\\ \multicolumn{1}{c}{A_3=y_3(2e^{y_0}+\frac{1}{2}{e}^{\frac{y_0}{2}})+y_1{y}_2(2e^{y_0}+\frac{1}{4}{e}^{\frac{y_0}{2}})}\\ \multicolumn{1}{c}{+\frac{1}{3!}{y}_1^3(2e^{y_0}+\frac{1}{8}{e}^{\frac{y_0}{2}})}\\ \multicolumn{1}{c}{⋮}\end{array}$$

(32)

and then applying equations (31) and (32), we have

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=0}\\ \multicolumn{1}{c}{S\left[y_1\left(t\right)\right]=-4u^2,y_1\left(t\right)=-2t^2}\\ \multicolumn{1}{c}{S\left[y_2\left(t\right)\right]=24u^4,y_2\left(t\right)=t^4}\\ \multicolumn{1}{c}{S\left[y_3\left(t\right)\right]=-4.5!u^6,y_3\left(t\right)=-\frac{2}{3}{t}^6}\\ \multicolumn{1}{c}{S\left[y_4\left(t\right)\right]=4.7!u^8,y_4\left(t\right)=\frac{1}{2}{t}^8}\end{array}$$

The series solution in general is given as follows

$$y(t)=y_0\left(t\right)+y_1\left(t\right)+y_2\left(t\right)+\dots$$

The exact solution given by

$$y(t)=-2\ln (1+t^2).$$

Example 3

Consider the singular nonlinear IVP

$$\begin{array}{c}\multicolumn{1}{c}{\frac{d^{2}y}{d{t}^2}+\frac{2}{t}\frac{\mathrm{dy}}{\mathrm{dt}}-6y=4y\ln \left(y\right)}\\ \multicolumn{1}{c}{y(0)=1y\prime (0)=0.}\end{array}$$

(33)

Multiply equation (33) with t, and taking the Sumudu transform, we obtain

$$\frac{d}{\mathrm{du}}\left(\mathrm{uY}\left(u\right)\right)-1=\mathrm{uS}[6\mathrm{ty}+4\mathrm{yt}\ln \left(y\right)].$$

(34)

Integrate equation (34) and inverse transform yields

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{du}\right)=1}\\ \multicolumn{1}{c}{y_n\left(t\right)=S^{-1}(-\frac{1}{u}\underset{0}{\overset{u}{\int }}u\left(S[6t{y}_{n-1}+4y_{n-1}t\ln \left(y_{n-1}\right)]\right)\mathrm{du})}\end{array}$$

(35)

where the nonlinear operator $M(y)=y\ln (y)$ is decomposed as in equation (11) in terms of the Adomian polynomials. From equation (13), the first few Adomian polynomials for $M(y)=y\ln (y)$ are computed as follows

$$\begin{array}{c}\multicolumn{1}{c}{A_0=y_0\ln y_0}\\ \multicolumn{1}{c}{A_1=y_1(1+\ln y_0)}\\ \multicolumn{1}{c}{A_2=y_2(1+\ln y_0)+\frac{y_1^2}{2y_0}}\\ \multicolumn{1}{c}{A_3=y_3(1+\ln y_0)+\frac{y_1{y}_2}{y_0}-\frac{y_1^3}{6y_0}}\\ \multicolumn{1}{c}{⋮}\end{array}$$

(36)

Substituting equation (35) into equation (36) gives the following terms

$$\begin{array}{c}\multicolumn{1}{c}{y_0\left(t\right)=S^{-1}\left[\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{du}\right]=1}\\ \multicolumn{1}{c}{y_1\left(t\right)=-S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[6t{y}_0+4y_{0}t\ln \left(y_0\right)]\mathrm{du}\right)}\\ \multicolumn{1}{c}{=S^{-1}\left(2u^2\right)=t^2}\\ \multicolumn{1}{c}{y_2\left(t\right)=-S^{-1}\left(\frac{1}{u}\underset{0}{\overset{u}{\int }}\mathrm{uS}[6t{y}_1+4y_{1}t\ln \left(y_0\right)]\mathrm{du}\right)}\\ \multicolumn{1}{c}{=S^{-1}\left(\frac{10\times 3!}{5}{u}^4\right)=\frac{1}{2}{t}^4}\end{array}$$

similarly, we obtain

$$y_3(t)=\frac{1}{3!}{t}^6,y_4(t)=\frac{1}{4!}{t}^8$$

The series solution is given as

$$y(t)=y_0\left(t\right)+y_1\left(t\right)+y_2\left(t\right)+\cdots$$

The exact solution is

$$y\left(t\right)=e^{t^2}.$$

Now if $n=2$ in equation (5), then we obtain

$$\begin{array}{c}\multicolumn{1}{c}{\frac{d^{2}y}{d{t}^2}+\frac{4}{t}\frac{\mathrm{dy}}{\mathrm{dt}}+\frac{2}{t^2}y+f(t,y)=g(t),}\\ \multicolumn{1}{c}{y(0)=A,y\prime (0)=B.}\end{array}$$

(37)

Multiplying equation (37) by $t^2$ and applying the Sumudu transform, we have

$$\begin{array}{c}\multicolumn{1}{c}{S(t^2\frac{d^{2}y}{d{t}^2}+4t\frac{\mathrm{dy}}{\mathrm{dt}}+2y+t^{2}f(t,y)=t^{2}g(t)),}\\ \multicolumn{1}{c}{y(0)=A,y\prime (0)=B}\end{array}$$

(38)

and applying the properties of the Sumudu transform, we obtain

$$u^2\frac{d^{2}Y}{d{u}^2}+4u\frac{\mathrm{dY}\left(u\right)}{\mathrm{du}}+2Y\left(u\right)+S[t^{2}f(t,y)]=S[t^{2}g(t)]$$

(39)

Simplifying equation (39), we have

$$\frac{d^2}{d{u}^2}(u^{2}Y(u))+S[t^{2}f(t,y)]=S[t^{2}g(t)]$$

(40)

Integrating equation (40), we get

$$\begin{array}{c}\multicolumn{1}{c}{S[y(t)]+\frac{1}{u^2}\underset{0}{\overset{u}{\int }}\left(\underset{0}{\overset{u}{\int }}S[t^{2}f(t,y)]\mathrm{du}\right)\mathrm{du}}\\ \multicolumn{1}{c}{=\frac{1}{u^2}\underset{0}{\overset{u}{\int }}\left(\underset{0}{\overset{u}{\int }}S[t^{2}g(t)]\mathrm{du}\right)\mathrm{du}}\end{array}$$

(41)

Taking the inverse Sumudu transform of equation (41), we obtain

$$y_0(t)=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}\left(\underset{0}{\overset{u}{\int }}S[t^{2}g(t)]\mathrm{du}\right)\mathrm{du}\right)=H(t)$$

(42)

$$y_{n+1}(t)=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}(\underset{0}{\overset{u}{\int }}-S[t^{2}K(y_n(t))+t^2{A}_n(t)]\mathrm{du})\mathrm{du}\right)$$

(43)

Example 4

Consider the nonlinear singular IVP

$$\begin{array}{c}\multicolumn{1}{c}{\frac{d^{2}y}{d{t}^2}+\frac{4}{t}\frac{\mathrm{dy}}{\mathrm{dt}}+\frac{2}{t^2}y+y^2=t^4+12,}\\ \multicolumn{1}{c}{y(0)=0,y\prime (0)=0.}\end{array}$$

(44)

Multiplying equation (44) by $t^2$ and applying the Sumudu transform, we have

$$\begin{array}{c}\multicolumn{1}{c}{S(t^2\frac{d^{2}y}{d{t}^2}+4t\frac{\mathrm{dy}}{\mathrm{dt}}+2y+t^2{y}^2=t^2(t^4+12)),}\\ \multicolumn{1}{c}{y(0)=0,y\prime (0)=0}\end{array}$$

(45)

then we have

$$\frac{d^2}{d{u}^2}(u^{2}Y(u))+S\left[t^2{y}^2\right]=S\left[t^2(t^4+12)\right].$$

(46)

Integrating equation (46) two times with respect to u, and then taking inverse transform on both sides, we get

$$\begin{array}{c}\multicolumn{1}{c}{y_0(t)=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}\left(\underset{0}{\overset{u}{\int }}S[t^6+12t^2]\mathrm{du}\right)\mathrm{du}\right)}\\ \multicolumn{1}{c}{y_{n+1}(t)=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}(\underset{0}{\overset{u}{\int }}-S\left[t^2{y}_n^2\right]\mathrm{du})\mathrm{du}\right)}\end{array}$$

(47)

where the nonlinear term $y_n^2$ is defined by the Adomian polynomials as follows

$$\begin{array}{c}\multicolumn{1}{c}{A_0=y_0^2}\\ \multicolumn{1}{c}{A_1=2y_0{y}_1}\\ \multicolumn{1}{c}{A_2=2y_0{y}_2+y_1^2}\\ \multicolumn{1}{c}{A_3=2y_0{y}_3+2y_1{y}_2}\end{array}$$

(48)

using equation (48) and proceeding as before yield the recursive relationship

$$\begin{array}{c}\multicolumn{1}{c}{y_0(t)=S^{-1}(\frac{6!}{7\times 8}{u}^6+\frac{4!}{4\times 3}{u}^2)}\\ \multicolumn{1}{c}{y_0(t)=\frac{t^6}{56}+t^2}\\ \multicolumn{1}{c}{y_1(t)=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}(\underset{0}{\overset{u}{\int }}-S\left[t^2{y}_0^2\right]\mathrm{du})\mathrm{du}\right)}\\ \multicolumn{1}{c}{=S^{-1}\left(\frac{1}{u^2}\underset{0}{\overset{u}{\int }}(\underset{0}{\overset{u}{\int }}-S\left[t^2{(\frac{t^6}{56}+t^2)}^2\right]\mathrm{du})\mathrm{du}\right)}\\ \multicolumn{1}{c}{=-\frac{t^{14}}{752640}-\frac{t^{10}}{3696}-\frac{t^6}{56}}\end{array}$$

(49)

The series solution in general is given as follows

$$y(t)=y_0\left(t\right)+y_1\left(t\right)+y_2\left(t\right)+\cdots$$

The exact solution is

$$y\left(t\right)=t^2.$$

As a conclusion, we state that the proposed method can be applied to industrial fields such as image processing and fractional calculus.