# On the Use of Elzaki Decomposition Method for Solving Higher-Order Integro-Differential Equations

Authors: Ekhlass S. Al-Rawi; Merna A. Samarchi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

In this article, we implemented the Elzaki decomposition technique (EDM) to solve Volterra–Fredholm integro-differential equations of higher-order. Illustrations are used to test the technique’s accuracy and validity. Comparison among the acquired consequences by EDM and actual solutions have proven the power and accuracy of this technique. This technique is dependable and able to supply analytic remedies for solving such equations.


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## Body

## 1. Introduction

This study is concerned with important problems in the sciences of engineering, which contain both integral and differential factors. In the early 20th century, this kind of equation was first presented by Volterra, where he studied genetic influences on population growth by developing the topic of integral and differential equations [1].Many researchers, engineers, and scientists face problems of an electrical circuit and biological species when solving differential equations resulting from the spread of heat and mass that appear with increasing and decreasing generation rates.In physics, engineering, and biology applications, one can find additional details about the origins of these equations as well as the behavior of advanced integral equations [2]. The Taylor collocation, Adomian decomposition, Chebyshev, Haar wavelet, successive approximation, Bernstein, homotopy perturbation, and other technologies are widely used to work with the IDEs of higher-order [3–13]. In recent years, Elzaki transform (ET) was introduced by Tarig Elzaki where it was used to solve different types of equations with integral, differential, partial integral, integro-differential, and fractional-order equations [14–20].The motivation of the article is to enhance the analysis of the modified Elzaki transformation, known as the Elzaki decomposition method (EDM), to solve the higher-order IDE of Volterra–Fredholm of the form(1)ynz=fz+λ1∫azk1z,τW1yτdτ+λ2∫abk2z,τW2yτdτ,subject to(2)yma=ym,m=0,1,2,…,n−1,where a,b,λ1,λ2,ym are constant values, fz,k1z,τ,k2z,τ are given forms with appropriate interval derivatives a≤z≤τ≤b,W1yτ, and W2yτ are nonlinear functions in yτ.


## 2. Elzaki Transformation Method (ETM) with Their Properties

The Elzaki transform (ET) offτ∈ℜ, ∀τ≥0 as in [21].(3)Efτ=Fω=ω∫0∞fτe−τ/ωdτh1≤ω≤h2.SinceE. is the operator of Elzaki transform and the set ℜ is defined as(4)ℜ=fτ|∃C,h1,h2>0,fτ<Ceτ/hn,ifτ∈−1n×0,∞,h1,h2 can be finite or infinite, and C is a real finite number.IfEfτ=Fω, then fτ is called the inverse transform of Elzaki to Fω and denoted by fτ=E−1Fω, where E−1 is the operator of inverse Elzaki transform. Table 1 shows the transformation of Elzaki for some basic functions [22].Table 1
Special functions by transform of Elzaki.
fτEfτ=Fω1ω2τω3τ22!ω4τn,n∈Nn!ωn+2eaτω2/1−aωsinaτaω3/1+a2ω2cosaτω2/1+a2ω2sinhaτaω3/1−a2ω2coshaτω2/1−a2ω2τeaτω3/1−aω2Special properties of the Elzaki transform are as follows.Theorem 1 (linearity, see [23])..
IfF1ω and F2ω are Elzaki transform (ET) of f1τ and f2τ, respectively, then(5)Eαf1τ+βf2τ=αEf1τ+βEf2τ=αF1ω+βF2ω.
α,β are constants.Theorem 2 (differentiation, see [21])..
If the transform of Elzaki forfτ given by Efτ=Fω, then(6)Ef′τ=Fωω−ωf0,Ef″τ=Fωω2−f0−ωf′0,Efnτ=Fωωn−∑k=0n−1ω2−n+kfk0.Theorem 3 (shift, see [23])..
Iffτ∈ℜ with Elzaki transform Fω, then(7)Eeaτfτ=11−aωFω1−aω.Theorem 4 (convolution theorem, see [23])..
Iff1τ and f2τ are functions with Elzaki transform F1ω and F2ω, then the convolution of f1andf2 is(8)f1∗f2τ=∫0τf1xf2τ−xdx.
Also, Elzaki transform is(9)Ef1∗f2=1ωF1ωF2ω.Theorem 5 (integration, see [23]).
Let Fω denote to the transform of Elzaki for fτ, then the definite integral of fτ,(10)Lτ=∫0τfτdτ,has transform of Elzaki as(11)ELτ=ωFω.


## 3. The Proposed Technique (EDM)

The proposed technique of the method (EDM) is as follows: we suppose that the (1) and (2) have sufficiently differentiable and unique solution yz. We use the Elzaki decomposition method to get an approximation for the integro-differential equations of high order.Applying the ETM of both sides of (1), we get(12)Eynz=Efz+Eλ1∫azk1z,τW1yτdτ+Eλ2∫abk2z,τW2yτdτ.Using Theorem2, we get(13)Eyzωn−∑k=0n−1ω2−n+kyk0=Efz+Eλ1∫azk1z,τW1yτdτ+Eλ2∫abk2z,τW2yτdτ.Solving forEyz, we get(14)Eyz=ωn∑k=0n−1ω2−n+kyk0+ωnEfz+ωnEλ1∫azk1z,τW1yτdτ+ωnEλ2∫abk2z,τW2yτdτ.Now, the EDM represents the solutionyz of our problem as infinite series(15)yz=∑j=0∞yjz.Using the recursive relationship, the componentsE [y (z)] will be obtained, and the polynomials of the infinite series will be substituted for the nonlinear functions W1y and W2y.(16)W1y=∑j=0∞Aj,W2y=∑j=0∞Bj,where Aj and Bj are the Adomian polynomials that are created using Adomian’s specialized algorithms [24].Substituting equations (15) and (16) in (14), we have(17)E∑j=0∞yjz=ωn∑k=0n−1ω2−n+kyk0+ωnEfz+ωnEλ1∫azk1z,τ∑j=0∞Ajdτ+ωnEλ2∫abk2z,τ∑j=0∞Bjdτ.Now,(18)Ey0z=ωn∑k=0n−1ω2−n+kyk0+ωnEfz.The rest components ofEyjz can be found using the previous solutions as(19)Eyj+1z=ωnEλ1∫azk1z,τ∑j=0∞Ajdτ+ωnEλ2∫abk2z,τ∑j=0∞Bjdτ.As a result, the componentsy0,y1,y2,… are identified by applying the inverse Elzaki transform of (18) and (19) to obtain(20)y0z=E−1ωn∑k=0n−1ω2−n+kyk0+E−1ωnEfz,yj+1z=E−1ωnEλ1∫azk1z,τ∑j=0∞Ajdτ+E−1ωnEλ2∫abk2z,τ∑j=0∞Bjdτ,j≥0.


## 4. Numerical Examples

To apply the proposed technique (EDM), various examples of higher orders were taken, and to study the efficiency of the suggested technique, a comparison with the exact solution was made. The calculations of solutions for examples are performed using the Maple program.Example 1.
The third order linear IDE of Volterra–Fredholm is(21)y″′z=−12z2+∫0zyτdτ+∫−ππzyτdτ,0≤z≤1,subject to(22)y0=1,y′0=1,y″0=−1.
Applying to both sides of (21) Elzaki transform, we get(23)Ey″′z=E−12z2+E∫0zyτdτ+E∫−ππzyτdτ.
We use the initial conditions together with Theorems2 and 5 to obtain(24)Eyzω3=1+1ω−ω−ω4+ωEyz+ω3∫−ππyτdτ.
Solving the (15) for Eyz, we get(25)Eyz=ω21−ω4+ω31−ω4−ω41−ω4−ω71−ω4+ω61−ω4∫−ππyτdτ.
Substituting the series assumptions forEys as given in (15) and using the recursive relation, we get(26)Ey0z=ω21−ω4+ω31−ω4−ω41−ω4−ω71−ω4Eyj+1z=ω61−ω4∫−ππyjτdτ,j=0,1,2,….
Taking the inverse transform of (26), we have(27)y0z=z+cosz,y1z=E−1ω61−ω4∫−ππy0τdτ=0,yj+1z=E−1ω61−ω4∫−ππyjτdτ=0,∀j=2,3,….
So, the series solution will be the actual solutionyz=z+cosz.Example 2.
The fourth-order nonlinear IDE of Volterra–Fredholm is(28)y4z=−z630−z46−z22+83z+32+∫0zz−τy2τdτ−2∫01z+τyτdτ,0≤z≤1,subject to(29)y0=1,y′0=0,y″0=2,y″′0=0.
Taking the transformation of Elzaki to (28), we have(30)Ey4z=E−z630−Ez46−Ez22+E83z+E32+E∫0zz−τy2τdτ−E2∫01z+τyτdτ.
Applying the initial conditions together with Theorems2–5, we obtained(31)Eyzω4=1ω2+2−24ω8−4ω6−ω4+83ω3+32ω2+ω2Ey2τ−2E∫01z+τyτdτ.
Now, solving the (31) for Eyz, we get(32)Eyz=ω2+2ω4−24ω12−4ω10−ω8+83ω7+32ω6+ω6Ey2τ−2ω4E∫01z+τyτdτ.
Substituting the series assumptions forEyz as given in (15) and using the recursive relation, we get(33)Ey0z=ω2+2ω4−24ω12−4ω10−ω8+83ω7+32ω6,Eyj+1z=ω6EAjτ−2ω4E∫01z+τyjτdτ,j≥0,where An′s are given by(34)A0τ=y02,A1τ=2y0y1,A2τ=2y0y2+y12,A3τ=2y0y3+2y1y2.⋮.
We apply for both sides of (33) the inverse Elzaki transform; we obtain general solution given by recursive relation as(35)y0z=E−1ω2+2ω4−24ω12−4ω10−ω8+83ω7+32ω6,yj+1z=E−1ω6EAjτ−2E−1ω4E∫01z+τyjτdτ,j≥0,which leads to(36)y0z=1+z2+z416+z545−z6720−z810080−z10151200,y1z=−731152z4−32314400z5+z6720+z810080−z9136080−z10907200+2959875200z11+23119750400z12+37778377600z13+4158118860800z14+1851350500z15+19710461394944000z16−1144353664000z17−4324251415552000z18−14430547072000z19−150634823680000z20−1488663280000z21−4381876509890560000z22+173849401077760000z24+13789640318464000000z26,and so on.
Now, the solution for the (28) will be the following:(37)yEDMz=∑j=02yjz=y0z+y1z+y2z≅1+z2,which is the actual solution.
Table2 indicates the evaluation of the numerical results and the exact solution, where the numerical solution exactly corresponds to the exact values as shown in Figure 1.Table 2
The approximation numericalyEDM and yexact for Example 2.
ZyexactyEDMAbsolute error01100.11.011.0099999111111038.888889690972235e − 080.21.041.0399985444399031.455560097562980e − 060.31.091.0899924623107627.537689237890888e − 060.41.161.1599756417328212.435826717883671e − 050.51.251.2499392145124046.078548759624880e − 050.61.361.3598711815253101.288184746899290e − 040.71.491.4897560634816252.439365183752873e − 040.81.641.6395744204475774.255795524235051e − 040.91.811.8093021312987386.978687012617613e − 04121.9989092722728581.090727727141916e − 03Figure 1
The approximate solutionyEDM with yexact for Example 2.Example 3.
The fifth order nonlinear IDE of Volterra–Fredholm is(38)y5z=−ze3z+∫0ze3z−3τy3τdτ+∫01ez−τyτdτ,0≤z≤1,where(39)y0=y′0=y″0=y″′0=y40=1.
To both sides of (38), we take the Elzaki transform and we get(40)Ey5z=E−ze3z+E∫0ze3z−3τy3τdτ+E∫01ez−τyτdτ.
Applying the initial conditions together with Theorems2–5, we obtained(41)Eyzω5=1ω3+1ω2+1ω+1+ω−ω31−3ω2+ω1−3ωEy3τ+E∫01ez−τyτdτ.
Solving the (41) for Eyz, we get(42)Eyz=ω2+ω3+ω4+ω5+ω6−ω81−3ω2+ω61−3ωEy3τ+ω5E∫01ez−τyτdτ.
Substituting the series assumptions forEyz as given in (15) and using the recursive relation, we get(43)Ey0z=ω2+ω3+ω4+ω5+ω6,Eyj+1z=−ω81−3ω2+ω61−3ωEAjt+ω5E∫01ez−τyjτdτ,j≥0,where An′s are given by(44)A0τ=y03,A1τ=3y1y02,A2τ=3y2y02+3y12y0,A3τ=3y3y02+6y0y1y2+y13.⋮.
We apply for both sides of (43) the inverse Elzaki transform; we obtain the general solution given by recursive relation as(45)y0z=E−1ω2+ω3+ω4+ω5+ω6,yj+1z=−E−1ω81−3ω2+E−1ω61−3ωEAjt+E−1ω5E∫01ez−τyjτdτ,j≥0,which leads to(46)y0z=1+z+z22+z36+z424,y1z=−1243ze3z+87004043046721e3z−3650729−6827z2−137162z3−23108z4−1219243z−878ez−1+878e−1+878e−1z+8716e−1z2+2916e−1z3+2964e−1z4+5ez−21610364543046721−7255552614348907z−246667879565938z2−43322544782969z3−32910491275484z4−1149594251528z5−25787321257640z6−45669799202320z7−2485316533720z8−338599793618560z9−559152907904z10−22123969978240z11−1181277136640z12−16332401850880z13−23254741760z14−10711208637440z15−11358622720z16−130795448320z17.and so on.
Now, the solution for the (38) will be the following:(47)yEDMz=∑j=02yjz=y0z+y1z+y2z,which converges to the exact solution yz=ez, as shown in Figure 2.
Table3 indicates an assessment of the numerical outcomes and the exact solution, and Figure 2 shows that the approximations are very near to the actual values.Figure 2
Comparing the numerical solutionyEDM with yexact for Example 3.Table 3
The numerical approximationyEDM and yexact for Example 3.
ZyexactyEDMAbsolute error01100.11.1051709180756481.1051709180172685.837930139307446e − 110.21.2214027581601701.2214027562600101.900160251366856e − 090.31.3498588075760031.3498587928965311.467947230437972e − 080.41.4918246976412701.4918246346929026.294836829567885e − 080.51.6487212707001281.6487210751316081.955685198229418e − 070.61.8221188003905091.8221183045740174.958164925294994e − 070.72.0137527074704772.0137516138027011.093667775897700e − 060.82.2255409284924682.2255387452292762.183263192012674e − 060.92.4596031111569502.4595990573630464.053793903668890e − 0612.7182818284590462.7182746741420057.154317040125591e − 06


## 5. Conclusion

The purpose of this study is to offer an efficient approach for solving higher-order IDEs. This goal was achieved using the Elzaki decomposition method (EDM). The validity and the accuracy of our approach are checked by solving three examples. To explain the technique, plots of the behavior of the solutions are considered to make the shower that the EDM is a very powerful and efficient method to produce the approximate solution.



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