Analyzing “Homotopy Perturbation Method for Solving Fourth-Order Boundary Value Problem”

Abstract

We analyze a previous paper by S. T. Mohyud-Din and M. A. Noor (2007) and show the mistakes in it. Then, we demonstrate a more efficient method for solving fourth-order boundary value problems.

1. Problem

Let us consider the fifth-order boundary problem of the type𝑢⁽⁴⁾(𝑥)=𝑓𝑢,𝑢^′,𝑢^′′,𝑢^′′′+𝑔(𝑥) (1.1) with the boundary conditions𝑢(0)=𝛼₁,𝑢^′(0)=𝛼₂,𝑢(1)=𝛽₁,𝑢^′(1)=𝛽₂, (1.2) where 𝑓 and 𝑔 are continuous functions and 𝛼₁, 𝛼₂, 𝛽₁, and 𝛽₂ are real constants.

The homotopy perturbation method (HPM) is employed in [1] for solving such problems. The purpose of this paper is to point out the mistakes in paper [1] and demonstrate more efficient method for solving the problems of type (1.1)-(1.2).

First of all, we show the mistakes.

(1)  In Example 3.2 the approximate solution 𝑢_approx(𝑥)=512+480𝑥+224.00000000226055𝑥²+… (1.3) of the problem 𝑢⁽⁴⁾(𝑥)=𝑢(𝑥)+𝑢^′′(𝑥)+𝑒^𝑥(),𝑢(0)𝑥−3=1,𝑢^′(0)=0,𝑢(1)=0,𝑢^′(1)=−𝑒 (1.4) (see formula (3.19) in [1]) has no relationship with the exact solution 𝑢_exact(𝑥)=(1−𝑥)𝑒^𝑥. And “error” is at least 𝑢_approx(0)−𝑢_exact(0)=512−𝑒=509.28.

It is strange that the authors demonstrated the reliability of the errors in the table (see Table 3.2 in [1]).

(2)  In Example 3.3 the approximate solution𝑢_approx(𝑥)=11.3472−262.827𝑥−3.40768𝑥³+… (1.5) of the problem𝑢⁽⁴⁾=sin𝑥+sin²𝑥−𝑢^′′(𝑥)²,𝑢(0)=0,𝑢^′(0)=1,𝑢(1)=sin1,𝑢^′(1)=cos1 (1.6) (see formula (3.28) in [1]) has no relationship with the exact solution 𝑢=sin𝑥, since 𝑢_approx(0)=11.3472, 𝑢^′_approx(0)=−262.827.

2. Homotopy Perturbation Method

The basic ideas of the standard HPM were given by He [2, 3], and a new interpretation of HPM was given by He [4]. We introduce a new reliable procedure for choosing the initial approximation in HPM. To do so, we consider the following general nonlinear differential equation𝐿𝑢+𝑁𝑢=𝑓(𝑢,𝑥) (2.1) with some initial boundary conditions, where 𝐿 and 𝑁 are, respectively, the linear and nonlinear operators.

According to HPM, we construct a homotopy which satisfies the following relations:𝐻𝑢,𝑝=𝐿𝑢−𝐿𝑣₀+𝑝𝐿𝑣₀+𝑝𝑁𝑢−𝑓(𝑢,𝑥)=0, (2.2) where 𝑝∈[0,1] is an embedding parameter and 𝑣₀ is an initial approximation. When we put 𝑝=0 and 𝑝=1 in (2.2), we obtain𝐻(𝑢,0)=𝐿𝑢−𝐿𝑣₀,𝐻(𝑢,1)=𝐿𝑢+𝑁𝑢−𝑓(𝑢,𝑥), (2.3) respectively. In topology, this is called deformation and 𝐿𝑢−𝐿𝑣₀ and 𝐿𝑢+𝑁𝑢−𝑓(𝑢,𝑥) are called homotopics.

The solution of (2.2) is expressed as𝑢(𝑥)=𝑢₀(𝑥)+𝑝𝑢₁(𝑥)+𝑝²𝑢₂(𝑥)+…. (2.4) Hence, the approximate solution of (1.5) can be expressed as𝑢(𝑥,𝑡)=𝑢₀(𝑥)+𝑢₁(𝑥)+𝑢₂(𝑥)+…. (2.5) Mohyud-Din and Noor [1] tried to rewrite the problem as a system of integral equations and then HPM applied for each equation. In the use of HPM, what we are mainly concerned about are the auxiliary operator 𝐿 and the initial guess 𝑣₀. We take 𝐿=(𝑑/𝑑𝑥⁴)(⋅), and𝑣₀=𝛼₁+𝛼₂𝑥+𝐴𝑥³+𝐵𝑥⁴+𝐿⁻¹𝑔(𝑥), (2.6) where 𝐴 and 𝐵 are yet to be determined. Then using the boundary conditions 𝑢(1)=𝛽₁, 𝑢^′(1)=𝛽₂ we determine 𝐴 and 𝐵.

3. Applications

Here we apply the HPM to solve correctly the problems in [1].

Example 3.1 (see [1, Example 3.2]). We have 𝑢⁽⁴⁾(𝑥)=𝑢(𝑥)+𝑢^′′(𝑥)+𝑒^𝑥(𝑥−3) (3.1) with boundary conditions 𝑢(0)=1,𝑢^′(0)=0,𝑢(1)=0,𝑢^′(1)=−𝑒. (3.2) We construct a homotopy which satisfies the relation 𝑢⁽⁴⁾(𝑥)−𝑣₀⁽⁴⁾(𝑥)+𝑝𝑣₀⁽⁴⁾(𝑥)−𝑢(𝑥)−𝑢^′′(𝑥)−𝑒^𝑥(𝑥−3)=0, (3.3) where 𝑣₀=1+𝐴𝑥²+𝐵𝑥³+𝐿⁻¹𝑒^𝑥(𝑥−3). (3.4) Now substituting (3.4) into (3.3), we obtain 𝑢₀⁽⁴⁾+𝑝𝑢₁⁽⁴⁾+𝑝²𝑢₂⁽⁴⁾+⋯−𝑣₀⁽⁴⁾(𝑥)+𝑝𝑣₀⁽⁴⁾(𝑥)−𝑢₀−𝑝𝑢₁−𝑝²𝑢₂−⋯−𝑢₀^′′−𝑝𝑢₁^′′−𝑝²𝑢₂^′′−⋯−𝑒^𝑥(𝑥−3)=0, (3.5) and, equating the coefficients of a like powers of 𝑝, we get a system of equations: 𝑢₀⁽⁴⁾(𝑥)−𝑣₀⁽⁴⁾(𝑥)=0,𝑢₀(0)=1,𝑢^′₀(0)=0,𝑢₀^′′(0)=𝐴,𝑢₀^′′′(0)𝑢=𝐵,₁₍₄₎+𝑣₀⁽⁴⁾(𝑥)−𝑢₀−𝑢₀^′′−𝑒^𝑥(𝑥−3)=0,𝑢₁(0)𝑢=0,_′1(0)=0,𝑢₁^′′(0)=0,𝑢₁^′′′(0)𝑢=0,₂₍₄₎−𝑢₁−𝑢₁^′′=0,𝑢₂(0)=0,𝑢^′₂(0)=0,𝑢₂^′′(0)=0,𝑢₂^′′′(0)𝑢=0,₃₍₄₎−𝑢₂−𝑢₂^′′=0,𝑢₃(0)=0,𝑢^′₃(0)=0,𝑢₃^′′(0)=0,𝑢₃^′′′(0)=0,…. (3.6) Solving (3.6), we get 𝑢₀=1+𝐴𝑥²+𝐵𝑥³+𝐿⁻¹𝑒^𝑥(𝑥−3)=1+𝐴𝑥²+𝐵𝑥³+6𝑥−7𝑒^𝑥+𝑥𝑒^𝑥+52𝑥²+23𝑥³+7=6𝑥−7𝑒^𝑥+𝐴𝑥²+𝐵𝑥³+𝑥𝑒^𝑥+52𝑥²+23𝑥³𝑢+8,₁=20+𝑥⁵1120𝐵+10+𝑥⁴112𝐴+1324+𝑥⁶11360𝐴+144+𝑥⁷11840𝐵++11260614𝑥³+1216𝑥²+𝑒^𝑥()𝑢2𝑥−20+18𝑥,₂=𝐿⁻¹𝑢₁+𝑢₁^′′=𝐿⁻¹32𝑥−36𝑒^𝑥+𝐴𝑥²+𝐴6𝑥⁴+𝐵𝑥³+𝐴𝑥360⁶+𝐵𝑥10⁵+𝐵𝑥840⁷+4𝑥𝑒^𝑥+292𝑥²+133𝑥³+34𝑥⁴+2𝑥15⁵+1𝑥144⁶+1𝑥1260⁷+36=52+𝑥⁶1360𝐴+29720+𝑥⁷1840𝐵+132520+𝑥⁸1110080𝐴+2240+𝑥⁹𝐵+13024022680+𝑥¹⁰𝐴+11814400725760+𝑥¹¹𝐵+16652800+19979200640𝑥³+1244𝑥²+𝑒^𝑥()34𝑥−52+48𝑥+2𝑥⁴+4𝑥15⁵,…. (3.7) Using only three-term approximation, we have 𝑢=𝑢₀+𝑢₁+𝑢₂=6𝑥−7𝑒^𝑥+𝐴𝑥²+𝐵𝑥³+𝑥𝑒^𝑥+52𝑥²+23𝑥³+8+20+𝑥⁵1120𝐵+10+𝑥⁴112𝐴+1324+𝑥⁶11360𝐴+144+𝑥⁷11840𝐵++11260614𝑥³+1216𝑥²+𝑒^𝑥(2𝑥−20)+18𝑥+52+𝑥⁶𝐴+36029720+𝑥⁷𝐵+840132520+𝑥⁸𝐴+1100802240+𝑥⁹𝐵+13024022680+𝑥¹⁰𝐴+11814400725760+𝑥¹¹𝐵+16652800+19979200640𝑥³+1244𝑥²+𝑒^𝑥()34𝑥−52+48𝑥+2𝑥⁴+4𝑥15⁵. (3.8) Now it follows from conditions 𝑢(1)=0, 𝑢^′(1)=−𝑒 that 𝐴=−0.46717 and 𝐵=−0.38354 and, therefore, 𝑢=𝑢₀+𝑢₁+𝑢₂=80+72𝑥−79𝑒^𝑥+7𝑥𝑒^𝑥+32.033𝑥²+9.2832𝑥³+2.0027𝑥⁴+0.34749𝑥⁵+4.4627×10⁻²𝑥⁶+5.0392×10⁻³𝑥⁷+4.0008×10⁻⁴𝑥⁸+3.1409×10⁻⁵𝑥⁹+1.1204×10⁻⁶𝑥¹⁰+4.2558×10⁻⁸𝑥¹¹, (3.9) or, in power series form, 𝑢=1−0.467𝑥²−0.38347𝑥³−0.1223𝑥⁴−0.01918𝑥⁵−6.762×10⁻³𝑥⁶−9.132×10⁻⁴𝑥⁷+4.0008×10⁻⁴𝑥⁸+3.1409×10⁻⁵𝑥⁹. (3.10) Higher accuracy level can be attained by evaluating some more terms of 𝑢(𝑥).

Example 3.2 (see [1, Example 3.3]). We have 𝑢⁽⁴⁾(𝑥)=sin𝑥+sin²𝑥−𝑢^′′(𝑥)² (3.11) with boundary conditions 𝑢(0)=0,𝑢^′(0)=1,𝑢(1)=sin1,𝑢^′(1)=cos1 (3.12) (the exact solution of the problem is 𝑢=sin𝑥).
We construct a homotopy which satisfies the relation𝑢⁽⁴⁾(𝑥)−𝑣₀⁽⁴⁾(𝑥)+𝑝𝑣₀⁽⁴⁾(𝑥)+𝑢^′′(𝑥)²−sin𝑥−sin²𝑥=0, (3.13) where 𝑣₀=𝑥+𝐴𝑥²+𝐵𝑥³+𝐿⁻¹sin𝑥+sin²𝑥. (3.14) Substituting (3.14) into (3.13), we obtain 𝑢₀⁽⁴⁾+𝑝𝑢₁⁽⁴⁾+𝑝²𝑢₂⁽⁴⁾+⋯−𝑣₀⁽⁴⁾(𝑥)+𝑝𝑣₀⁽⁴⁾(𝑥)+𝑢₀^′′+𝑝𝑢₁^′′+𝑝²𝑢₂^′′+⋯²−sin𝑥−sin²𝑥=0, (3.15) and, equating the coefficients of a like powers of 𝑝, we get a system of equations: 𝑢₀⁽⁴⁾(𝑥)−𝑣₀⁽⁴⁾(𝑥)=0,𝑢₀(0)=1,𝑢₀(0)𝑢=0,_′0(0)=1,𝑢₀^′′(0)=𝐴,𝑢₀^′′′(0)𝑢=𝐵,₁₍₄₎+𝑣₀⁽⁴⁾(𝑥)+𝑢₀^′′2−sin𝑥−sin²𝑥=0,𝑢₁(0)𝑢=0,_′1(0)=0,𝑢₁^′′(0)=0,𝑢₁^′′′(0)𝑢=0,₂₍₄₎+2𝑢₀^′′𝑢₁^′′=0,𝑢₂(0)=0,𝑢^′₂(0)=0,𝑢₂^′′(0)=0,𝑢₂^′′′(0)𝑢=0,₃₍₄₎+𝑢₁^′′2+2𝑢₀^′′𝑢₂^′′=0,𝑢₃(0)=0,𝑢^′₃(0)=0,𝑢₃^′′(0)=0,𝑢₃^′′′(0)=0,…. (3.16) Solving (3.16) we get 𝑢₀=𝑥+𝐴𝑥²+𝐵𝑥³+𝐿⁻¹sin𝑥+sin²𝑥=𝑥+𝐴𝑥²+𝐵𝑥³+1132+sin𝑥−132cos2𝑥+6𝑥³−1𝑥16²1−𝑥+𝑥48⁴,𝑢₁1=−1260𝐴𝑥⁷−𝑥⁸115040𝐴+𝐵1680−𝑥⁹116048𝐵−𝐴90720+𝑥¹⁰𝐴+𝐵113400−1100800−11814406𝐴²𝑥⁴−1𝐵40²𝑥⁶−110𝐴𝐵𝑥⁵+𝑂𝑥¹¹,𝑢₂=𝑥¹⁴1𝐴7567560²[phone]𝐴𝐵+123284800𝐴−𝐵1441440²+1𝐵2522520+𝑥¹³1𝐴21621600²−1715405400𝐴𝐵+11853280𝐴−𝐵739200²+43𝐵43243200+𝑥¹¹−23𝐴4989600²+115400𝐴𝐵+13𝐵158400²+𝑥¹⁰17𝐴226800²+14200𝐵𝐴+𝑥¹²−1𝐴249480²−1313326400𝐴𝐵+1907200𝐴+𝐵44352²+𝑥⁹1𝐴3780²+1𝐵336³+1𝐴45³𝑥⁶+4𝐴105²𝐵𝑥⁷+9560𝐴𝐵²𝑥⁸. (3.17) Using only three-term approximation we have 𝑢=𝑥+𝐴𝑥²+𝐵𝑥³+1132+sin𝑥−132cos2𝑥+6𝑥³−1𝑥16²1−𝑥+𝑥48⁴−11260𝐴𝑥⁷−𝑥⁸115040𝐴+𝐵1680−𝑥⁹116048𝐵−𝐴90720+𝑥¹⁰11113400𝐴+1100800𝐵−−11814406𝐴²𝑥⁴−1𝐵40²𝑥⁶−110𝐴𝐵𝑥⁵+𝑥⁹1𝐴3780²+1𝐵336³+1𝐴45³𝑥⁶+4𝐴105²𝐵𝑥⁷+9560𝐴𝐵²𝑥⁸. (3.18) Now by using the conditions 𝑢(1)=sin1, 𝑢^′(1)=cos1, we have a system of equations of degree three. Solving this system numerically (applying some standard computer programs) we have that 𝐴=5.8611×10⁻³, 𝐵=−0.17455 and the series solution 𝑢=𝑥+5.861×10⁻³𝑥²−0.17455𝑥³−6.3333×10⁻⁶𝑥⁴+8.4356×10⁻³𝑥⁵+2.0161×10⁻³𝑥⁶−2.0329×10⁻⁴𝑥⁷−9.2803×10⁻⁵𝑥⁸+𝑂𝑥⁹. (3.19)

4. Conclusion

In this paper we have used the homotopy perturbation method for finding the solution of fourth-order linear and nonlinear boundary value problems. We presented a simple way to choose 𝐿 and 𝑣₀ when we use the homotopy perturbation method. In most cases, our simple choice yields very good approximation of exact solution.