# Comments on “Homotopy Perturbation Method for Solving Reaction-Diffusion Equations”

Authors: Afgan Aslanov
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

The paper entitled“Homotopy perturbation method for solving reaction diffussion equation” contains some mistakes and misinterpretations along with a false conclusion. Applying the homotopy perturbation method (HPM) in an incorrect manner, the authors have drawn the false conclusion that this approach is efficient for reaction-diffusion type of equation. We show that HPM in the proposed form is not efficient in most cases, and hence, we will introduce the correct form of HPM.


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## Body

## 1. Introduction

In [1] Wang et al. proposed a new approach to apply the HPM for solving the reaction-diffusion problem. We can easily identify in [1] some mistakes, which can mislead the readers, as we show in Section 3 of this paper. The main difficulties in the application of HPM in proposed form are that the corresponding approximate solution is a divergent series on [0,1] and the method of weighted residuals applied in an incorrect form.We consider a reaction-diffusion process governed by the equation(1.1)u′′+un=0,0<x<L,
with boundary conditions(1.2)u(0)=u(L)=0,
where u(x) represents the steady-state temperature for the corresponding reaction-diffusion equation with the reaction term un,n is the power of the reaction term (heat source), and L is the length of the sample (heat conductor). The physical interpretation of (1.1) was given in [2]. For n=3, the problem (1.1)-(1.2) has a unique positive solution (e.g., as can be seen by phase-plane analysis) which is given in terms of elliptic functions. For an arbitrary real number n≠3, the general reaction-diffusion process is more complicated. The problem (1.1)-(1.2) was studied by Lesnic using Adomian decomposition method [2] and by Mo [3] using variational method. In this paper, we analyze the HPM used in [1] and demonstrate the correct form of HPM.


## 2. Homotopy Perturbation Method

In [1] the authors used the HPM in standard form:(2.1)(1-p)(u′′-u0′′)+p(u′′+un)=0,
with initial approximation(2.2)u0=ax(1-x),
where a is an unknown constant. The HPM supposes that the solution of (1.1)-(1.2) has the form(2.3)u=u0+pu1+p2u2+⋯.
And then equating the terms with identical powers of p, the components u1,u2,… of u can be calculated easily. This standard process for n=2 with L=1 gives an approximate solution for u [1]:(2.4)u=ax(1-x)+ax2-a2(130x6-110x5+112x4)-(a-160a2)x.
To find a the authors in [1] then used the method of weighted residuals. Substituting (2.4) into (1.1) results in the following residual:(2.5)R(x,a)=u′′(x)+un(x).
Then they set R(1/3,a)=0 and found a=45.4205 (why 1/3?).It is not clear why the number 1/3 (but not 1/2) plays some special role for determininga. Actually, the problem is that the solution corresponding to 1/2 in fact has no relationship with the exact solution; that is error is large enough comparatively with the case 1/3. But it is not difficult to show that the error (in case of n=2) for 1/3 is also large enough. This solution is not even so good around 1/3. Tables 1 and 2 demonstrate the errors in u′′+un=0 for some values of x. If we use one more term (component) in the HPM in [1] applying the method of weighted residuals (for x=1/3 and n=2), we obtain a=23.949 and an approximate solution which in fact has no relationship with normal solution: for example, the “error” for x=0.9 is 43.780. That is the result received in [1] cannot be improved by taking more terms in approximation.Table 1
Errors in approximate solution for Example3.1.
xuappr=∑k=09ukuappr=∑k=03ukErrorinu′′+u2=0Errorinu′′+u2=0Errorinu′′+u2=0 in [1](u0=b)(u0=asinπx)(u0=b)(u0=asinx)0.000.00.02.5665×10-30.00.13.30093.29760.11452.5942×10-25.00450.26.47126.46916.061×10-39.8435×10-28.41270.39.21239.21095.5852×10-30.136823.10970.411.10811.1074.5057×10-33.5474×10-26.23720.511.7911.7880.005.4003×10-210.760.611.10811.1074.5057×10-33.5474×10-26.23210.79.21239.21095.5852×10-30.136823.11090.93.30093.29760.11452.5942×10-25.0045Table 2
Errors in approximate solution for Example3.2.
xErrorinu′′+u3=0Errorinu′′+u2=0in [1]0.00.00.00.12.7632×10-31.00550.21.7240×10-23.36030.30.136731.79680.40.121864.39070.50.08.0760.60.121864.38450.70.136731.80830.81. 7240 × 10−23.37490.92. 7632 × 10−31.01801.00.00.0The failure of the method related also with the convergence rate of the series solution. It is not difficult to show that the coefficients in the highest degree terms of the second derivative are large enough and therefore go to infinity forx close to 1.To overcome these difficulties we offer the next two versions of the HPM.(1)
Since the solution must be symmetric with respect to the linex=L/2, try to represent the solution as a series of ∑(x-L/2)k. Since the solution reaches the maximum at L/2 take u′(L/2)=0. We construct a homotopy as in (2.1) but with initial approximation:
(2.6)u0=b,
where b is an unknown constant to be further determined. Instead of initial-boundary conditions we consider initial conditions:
(2.7)uk(L2)=0,uk′(L2)=0,
for each component uk,k=1,2,….(2)
To get rapidly convergent series solution we offer to take an initial approximation asu0=asin(πx/L) where a is an unknown constant to be further determined (by taking x=L/2 in u′′+un=0, where u=uapprox).


## 3. Applications

Example 3.1.
Consider the problem(3.1)u′′+u2=0,u(0)=u(1)=0
and apply version 1 of HPM. Substituting u=u0+pu1+p2u2+p3u3+⋯ into (2.1) and equating the terms with the identical powers of p, we obtain
(3.2)u0′′+pu1′′+p2u2′′+p3u3′′+⋯-v0′′+p(v0′′+(u0+pu1+p2u2+p3u3+p4u4+p5u5)2)=0,u0=b,u1′′=-u0′′-u02=-(b)2,u1(0.5)=u1′(0.5)=0,u1=-0.5b2(x-0.5)2,u2′′=-2u0u1=-2(b)(-0.5b2(x-0.5)2),u2(0.5)=u2′(0.5)=0,u2=112b3(x-0.5)4,u3′′=-2u0u2-u12=-2(b)(8.3333×10-2b3(x-0.5)4)-(-0.5b2(x-0.5)2)2,u3=-1.3889×10-2b4(x-0.5)6,u4′′=-2u0u3-2u1u2,u4=0.1111156b5(x-0.5)8,u5′′=-2u0u4-2u1u3-u22,u5=-2.7558×10-4b6(x-0.5)10,u6′′=-2u0u5-2u1u4-2u2u3,u6=3.6743×10-5b7(x-0.5)12,u7′′=-2u0u6-2u1u5-2u2u4-u32u7=-4.7948×10-6b8(x-0.5)14,u8′′=-2u0u7-2u1u6-2u2u5-2u3u42,u8=6.1407×10-7b9(x-0.5)16,u9′′=-2u0u8-2u1u7-2u2u6-2u3u5-(u4)2,u9=-7.7577×10-8b10(x-0.5)18.
Let us take uapprox=u0+u1+⋯+u9 and find b from uapprox(0)=0—some numeric method can be used for finding b (we used standard LaTex computation): b=11.79. Thus we get an approximate solution:
(3.3)u=(11.79)-0.5(11.79)2(x-0.5)2+112(11.79)3(x-0.5)4-1.3889×10-2(11.79)4(x-0.5)6+1.9841×10-3(11.79)5(x-0.5)8-2.7558×10-4(11.79)6(x-0.5)10+3.6743×10-5(11.79)7(x-0.5)12-4.7948×10-6(11.79)8(x-0.5)14+6.1407×10-7(11.79)9(x-0.5)16-7.7577×10-8(11.79)10(x-0.5)18.
Now let us apply version 2 of HPM. We take u0=asinπx, and equating the terms with the identical powers of p, we obtain
(3.4)u0=asin⁡(πx),u1′′=-u0′′-u02=π2asin⁡(πx)-a2sin⁡2(πx),u1(0)=u1′(1)=0,u1=-14a2x2+14a2x-18π2a2cos⁡2πx+18π2a2sin⁡2πx+18π2a2-asin⁡πx,u2′′=-2u0u1=-2(asin⁡(πx))(-14a2x2+14a2x-a2cos⁡2πx8π2+a2sin⁡2πx8π2+a28π2-asin⁡πx),u2=-12π2a3x2sin⁡πx-2π3a3xcos⁡πx+12π2a3xsin⁡πx-124π4a3cos⁡2πxsin⁡πx+a3cos⁡πxπ3+a3sin⁡3πx72π4+278π4a3sin⁡πx+a2x22+a2cos⁡2πx4π2-a2sin⁡2πx4π2-a2x2-a2(π+4a)4π3,u3′′=-2u0u2-u12,u3=1160a4x5-1192a4x4-14a2x2-1480a4x6-3π3a3cos⁡πx-818π4a3sin⁡πx-2π5a4sin⁡πx-332π2a4x3+364π2a4x4-435256π4a4x2-18π2a2cos⁡2πx-451288π6a4cos⁡2πx+2518432π6a4cos⁡4πx+18π2a2sin⁡2πx-124π4a3sin⁡3πx+451288π6a4sin⁡2πx+2518432π6a4sin⁡4πx+18π4a3cos⁡2πxsin⁡πx+964π4a4x2cos⁡2πx-964π4a4x2sin⁡2πx+6π3a3xcos⁡πx-32π2a3xsin⁡πx-253072π6a4cos⁡2πxsin⁡2πx+2532π5a4cos⁡πxsin⁡πx-964π4a4xcos⁡2πx+32π2a3x2sin⁡πx+964π4a4xsin⁡2πx-2516π5a4xcos⁡πxsin⁡πx+13840π4a2(180π2a2+4π4a2+960π4+6525a2)x+96136144π6a4+3π3a3+18π2a2.
(We used very easy LaTex tools for finding integrals.) Now we calculate the number a from the relationship
(3.5)u′′+u2=0,
where u≈u0+u1+u2+u3 and x=0.5. We obtain a=11.569. Table 1 shows the errors in the solutions and comparison with the errors in [1].Example 3.2.
Consider the problem(3.6)u′′+u3=0,u(0)=u(1)=0.
Substituting u=u0+pu1+p2u2+p3u3+⋯ into (2.1), where u0=asin(πx), and equating the terms with the identical powers of p, we obtain from
(3.7)u0′′+pu1′′+p2u2′′+p3u3′′+⋯-v0′′+p(v0′′+(u0+pu1+p2u2+p3u3+p4u4+p5u5)3)=0
that
(3.8)u0=asin⁡(πx),u1′′=-u0′′-u03=π2asin⁡(πx)-a3sin⁡3(πx),u1(0)=u1′(1)=0,u1=-112π(13πa3sin⁡3πx-9πa3sin⁡πx+12πasin⁡πx),u2′′=-3u02u1=-3(asin⁡(πx))2(-a3cos⁡2πxsin⁡πx12π2+a3sin⁡3πx36π2+3a3sin⁡πx4π2-asin⁡πx),u2=11440π3(2460πa5sin⁡πx-2903πa5sin⁡3πx+6a5sin⁡5πx5π-3240πa3sin⁡πx+120πa3sin⁡3πx6a5sin⁡5πx5π),u3′′=-3u02u2-3u0u12,u3=-1756000π5(63000π3a3sin⁡3πx-3925250πa7sin⁡πx-1701000π3a3sin⁡πx+4704703πa7sin⁡3πx-126145πa7sin⁡5πx+1307πa7sin⁡7πx+6457500πa5sin⁡πx-253750πa5sin⁡3πx+3150πa5sin⁡5πx3925250π).
To find a we take u=u0+u1+u2+u3 in u′′+u3=0 and replace x=0.5. Thus we obtain a=3.6033. Table 2 demonstrates the errors in the solution and the comparison with the errors in [1].


## 4. Conclusions

We proved that Wang et al. who intended to develop in [1] a new approach to apply the HPM for solving reaction-diffusion equations, included in their paper some mistakes and misinterpretations. They used the method of weighted residuals in an incorrect form, and initial approximation is not so good for the convergence of the series solution. We introduced two versions of the HPM, which are very efficient for initial-boundary type of problems.The main strength of HPM is in fact its fast convergence, but the form of initial approximation and initial-boundary conditions is very important for the efficiency of the HPM.



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*Source: 650472-2012-06-06.xml*