International Scholarly Research Notices

International Scholarly Research Notices / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 518361 | 10 pages | https://doi.org/10.5402/2012/518361

Extension of Zhou's Method to Neutral Functional-Differential Equation with Proportional Delays

Academic Editor: S. Prudhomme
Received30 Jan 2012
Accepted19 Feb 2012
Published07 Aug 2012

Abstract

The differential transform method (DTM) is a reliable method applied by providing new theorems to develop exact and approximate solutions of neutral functional-differential equation (NFDE) with proportional delays. The results obtained with the proposed methods are in good agreement with one obtained by other methods. The advantages of this technique are illustrated. It is easy to see that the DTM is very accurate and easy to implement in finding analytical solutions of wide classes of linear and nonlinear NFDEs.

1. Introduction

The neutral functional-differential equation (NFDE) is î€·î€·ğœŽğ‘¢(𝑡)+ğ‘Ž(𝑡)𝑢𝑚(𝑡)(𝑚)=𝛽𝑢(𝑡)+𝑚−1𝑘=0𝑏𝑘(𝑡)ğ‘¢ğ‘˜î€·ğœŽğ‘˜î€¸(𝑡)+𝑓(𝑡),𝑡≥0,(1.1) under the conditions 𝑚−1𝑘=0𝑐𝑖𝑘𝑢(𝑘)(0)=𝜆𝑖,𝑖=0,1,…,𝑚−1,(1.2) where ğ‘Ž,𝑏𝑘,ğœŽğ‘˜ are analytical functions; 𝛽,𝑐𝑖𝑘 and ğœ†ğ‘–âˆˆğ’ž. A classical case [1] ğœŽğ‘˜(𝑡)=𝑡−𝜏𝑘,𝑘=0,1,…,𝑚,(1.3) where 𝜏𝑘 is positive. Another interesting case [2] is ğœŽğ‘˜(𝑡)=ğ‘žğ‘˜ğ‘¡,𝑘=0,1,…,𝑚,(1.4) where 0<ğ‘žğ‘˜<1. Both cases are playing an interesting role in many applications (see [1, 2] and references therein). In recent years, there has been a growing interest in the numerical treatment of NFDE, some of which are the Adams method [3], continuous Runge-Kutta methods [4], segmented Tau approximation [5], Homotopy perturbation method [6], one-leg 𝜃-methods [7, 8], and variational iteration method [9].

In this paper we consider the following neutral functional-differential equations with proportional delays.

Problem 1. Considerğ‘¢î…ž(𝑡)=𝛽𝑢(𝑡)âˆ’ğ‘Ž(𝑡)ğ‘¢î…žî€·ğ‘žâ„“ğ‘¡î€¸î€·î€·ğ‘ž+𝑓𝑡,𝑢(𝑡),𝑢𝑖𝑡,ğ‘¢î…žî€·ğ‘žğ‘–ğ‘¡î€¸î€¸,𝑖=1,2,...,ℓ−1,𝑢(0)=𝑢0.(1.5)

Problem 2. Consider𝑢(𝑚)(𝑡)=𝛽𝑢(𝑡)âˆ’ğ‘Ž(𝑡)𝑢(𝑚)î€·ğ‘žâ„“ğ‘¡î€¸î€·î€·ğ‘ž+𝑓𝑡,𝑢(𝑡),𝑢𝑖𝑡,ğ‘¢î…žî€·ğ‘žğ‘–ğ‘¡î€¸,ğ‘¢î…žî…žî€·ğ‘žğ‘–ğ‘¡î€¸,...,𝑢(𝑚−1)î€·ğ‘žğ‘–ğ‘¡,𝑚−1𝑘=0𝑐𝑗𝑘𝑢(𝑘)(0)=𝜆𝑗,𝑗=0,1,2,...,𝑚−1,(1.6) where ğ‘Ž,𝑓 are analytical functions; 𝛽,𝑐𝑗𝑘 and 𝜆𝑗 are real or complex constants; 0<ğ‘žğ‘–<1,𝑖=1,2,…,ℓ.

The basic motivation of this work is to extend the differential transform method (DTM) by presenting and proving new theorems to create the exact or approximate solutions to a high degree of accuracy to the Problems 1 and 2. The DTM is a numerical-analytical technique that was first proposed by Zhou (1986) [10], who solved problems in electric circuit analysis. Since then, DTM was successfully applied for a large variety problems. For example, differential-difference equations [11], Volterra integral equation with separable kernels [12], MHD boundary-layer equations [13], linear and nonlinear systems of partial differential equations [14], and nonlinear oscillators with fractional nonlinearities [15]. To the best of our knowledge differential transform method has not be used by any researcher before to solve NFDE. By this method it is possible to obtain highly accurate results when compared with existing results from variational iteration method [9] and homotopy perturbation method [6].

2. Basic Idea of Differential Transform Method

The differential transform of the 𝑘th derivative of a function 𝑢(𝑡) is defined as follows: 1𝑈(𝑘)=𝑑𝑘!𝑘𝑑𝑡𝑘𝑢(𝑡)𝑡=𝑡0.(2.1) The original and transformed functions are denoted throughout this paper by small and capital letters, respectively. The differential inverse transform of 𝑈(𝑘) is defined as 𝑢(𝑡)=𝑁𝑘=0𝑈(𝑘)𝑡−𝑡0𝑘,forğ‘âŸ¶âˆž.(2.2) And from (2.1) and (2.2): 𝑢(𝑡)=𝑁𝑘=01𝑑𝑘!𝑘𝑑𝑡𝑘𝑢(𝑡)𝑡=𝑡0𝑡−𝑡0𝑘,forğ‘âŸ¶âˆž,(2.3) which is actually the Taylor series expansion of 𝑢(𝑡) about the point 𝑡0.

The fundamental mathematical operations performed by one-dimensional differential transform can be obtained from (1.5) and (1.6) in Table 1 (also, see [10]).


Original function Transformed function

𝑢 ( 𝑡 ) = 𝑓 ( 𝑡 ) ± 𝑔 ( 𝑡 ) 𝑈 ( 𝑘 ) = 𝐹 ( 𝑘 ) ± 𝐺 ( 𝑘 )
𝑢 ( 𝑡 ) = 𝑐 𝑓 ( 𝑡 ) 𝑈 ( 𝑘 ) = 𝑐 𝐹 ( 𝑘 ) , 𝑐 ∈ ℛ
𝑢 ( 𝑡 ) = 𝑓 ( 𝑡 ) 𝑔 ( 𝑡 ) ∑ 𝑈 ( 𝑘 ) = 𝑘 ℓ = 0 𝐹 ( ℓ ) 𝐺 ( 𝑘 − ℓ )
𝑢 ( 𝑡 ) = 𝑓 1 ( 𝑡 ) 𝑓 2 ( 𝑡 ) ⋯ 𝑓 𝑚 − 1 ( 𝑡 ) 𝑓 𝑚 ( 𝑡 ) ∑ 𝑈 ( 𝑘 ) = 𝑘 ℓ 𝑚 − 1 = 0 ∑ ℓ ℓ 𝑚 − 1 𝑚 − 2 = 0 ⋯ ∑ ℓ 3 ℓ 2 = 0 × ∑ ℓ 2 ℓ 1 = 0 𝐹 1 ( ℓ 1 ) × 𝐹 2 ( ℓ 2 − ℓ 1 ) ⋯ 𝐹 𝑚 − 1 ( ℓ 𝑚 − 1 − ℓ 𝑚 − 1 ) 𝐹 ( 𝑘 − ℓ 𝑚 − 1 )
𝑢 ( 𝑡 ) = ( 𝑑 𝑛 / 𝑑 𝑡 𝑛 ) 𝑓 ( 𝑡 ) 𝑈 ( 𝑘 ) = ( ( 𝑘 + 𝑛 ) ! / 𝑘 ! ) 𝐹 𝑚 ( 𝑘 )
𝑢 ( 𝑡 ) = 𝑡 𝑚  𝑈 ( 𝑘 ) = 𝛿 ( 𝑘 − 𝑚 ) = 1 , 𝑘 = 𝑚 ; 0 , 𝑘 ≠ 𝑚

3. Main Results

In the following theorems, we find the differential transformations of given functions. These results are very useful in our approach for solving NFDEs.

Theorem 3.1. Suppose that 𝑈(𝑘),𝐹(𝑘), and 𝐺(𝑘) are the differential transformations of the functions 𝑢(𝑡),𝑓(𝑡), respectively, and 𝑔(𝑡) and 0<ğ‘ž,ğ‘žğ‘–<1, for 𝑖=1,2,…,𝑚: (I)if 𝑢(𝑡)=𝑓(ğ‘žğ‘¡), then 𝑈(𝑘)=ğ‘žğ‘˜ğ¹(𝑘), (II)if 𝑢(𝑡)=𝑑𝑛𝑓(ğ‘žğ‘¡)/𝑑(ğ‘žğ‘¡)𝑛, then 𝑈(𝑘)=ğ‘žğ‘˜((𝑘+𝑛)!/𝑘!)𝐹(𝑘+𝑛), (III)if 𝑢(𝑡)=𝑔(𝑡)𝑑𝑛𝑓(ğ‘žğ‘¡)/𝑑(ğ‘žğ‘¡)𝑛, then 𝑈(𝑘)=𝑘ℓ=0ğ‘žğ‘˜âˆ’â„“(𝑘−ℓ+𝑛)(𝑘−ℓ)!𝐺(ℓ)𝐹(𝑘−ℓ+𝑛),(3.1)(IV)if 𝑢(𝑡)=(𝑑𝑛/𝑑(ğ‘ž1𝑡)𝑛)[𝑓(ğ‘ž1𝑡)](𝑑𝑚/𝑑(ğ‘ž2𝑡)𝑚)[𝑔(ğ‘ž2𝑡)], then 𝑈(𝑘)=𝑘ℓ=0ğ‘žâ„“1ğ‘ž2𝑘−ℓ(ℓ+𝑛)!(𝑘−ℓ+𝑚)!(𝑘−ℓ)!(ℓ)!𝐹(ℓ+𝑛)𝐺(𝑘−ℓ+𝑚),(3.2)(V)if 𝑢(𝑡)=(𝑑𝑛1/𝑑(ğ‘ž1𝑡)𝑛1)[𝑓1(ğ‘ž1𝑡)](𝑑𝑛2/𝑑(ğ‘ž2𝑡)𝑛2)[𝑓2(ğ‘ž2𝑡)]⋯(𝑑𝑛𝑚−1/𝑑(ğ‘žğ‘šâˆ’1𝑡)𝑛𝑚−1)[𝑓𝑚−1(ğ‘žğ‘šâˆ’1𝑡)(𝑑𝑛𝑚/𝑑(ğ‘žğ‘šğ‘¡)𝑛𝑚)[𝑓𝑚(ğ‘žğ‘šğ‘¡)], then 𝑈(𝑘)=𝑘ℓ𝑚−1ℓ=0𝑚−1ℓ𝑚−2=0⋯ℓ3ℓ2ℓ=02ℓ1=0ğ‘žâ„“11ğ‘žâ„“2−ℓ12â‹¯ğ‘žâ„“ğ‘šâˆ’1−ℓ𝑚−2𝑚−1ğ‘žâ„“ğ‘˜âˆ’â„“ğ‘šâˆ’1𝑚×ℓ1+𝑛1!ℓ1!ℓ2−ℓ1+𝑛2!ℓ2−ℓ1!⋯ℓ𝑚−1−ℓ𝑚−2+𝑛𝑚−1!ℓ𝑚−1−ℓ𝑚−2!𝑘−ℓ𝑚−1+𝑛𝑚!𝑘−ℓ𝑚−1!×𝐹1ℓ1+𝑛1𝐹2ℓ2−ℓ1+𝑛2⋯𝐹𝑚−1ℓ𝑚−1−ℓ𝑚−1+𝑛𝑚−1𝐹𝑚𝑘−ℓ𝑚−1+𝑛𝑚.(3.3)

Proof. (I), (II) The proof follows immediately by substituting 𝑢(𝑡) into (2.1).
(III) By using the definition of DTM (2.1), we have 1𝑈(𝑘)=𝑑𝑘!𝑘𝑑𝑡𝑘𝑑𝑔(𝑡)𝑛𝑑(ğ‘žğ‘¡)𝑛𝑓(ğ‘žğ‘¡)𝑡=𝑡0=1âŽ¡âŽ¢âŽ¢âŽ£ğ‘˜!𝑘ℓ=0âŽ›âŽœâŽœâŽğ‘˜â„“âŽžâŽŸâŽŸâŽ ğ‘‘â„“ğ‘‘ğ‘¡â„“ğ‘‘ğ‘”(𝑡)𝑘−ℓ𝑑𝑡𝑘−ℓ𝑑𝑛𝑑(ğ‘žğ‘¡)ğ‘›î‚¶âŽ¤âŽ¥âŽ¥âŽ¦ğ‘“(ğ‘žğ‘¡)𝑡=𝑡0,(3.4) and from (II), we have 1𝑈(𝑘−ℓ)=𝑑(𝑘−ℓ)!𝑘−ℓ𝑑𝑡𝑘−ℓ𝑑𝑛𝑑(ğ‘žğ‘¡)𝑛𝑓(ğ‘žğ‘¡)𝑡=𝑡0=ğ‘žğ‘˜âˆ’â„“(𝑘−ℓ+𝑛)!(𝑘−ℓ)!𝐹(𝑘−ℓ+𝑛).(3.5) By utilizing this value, we get 𝑈(𝑘)=𝑘ℓ=0ğ‘žğ‘˜âˆ’â„“(𝑘−ℓ+𝑛)!(𝑘−ℓ)!𝐺(ℓ)𝐹(𝑘−ℓ+𝑛).(3.6)
(IV) By using the definition of DTM (2.1), we have 1𝑈(𝑘)=𝑑𝑘!𝑘𝑑𝑡𝑘𝑑𝑛𝑑(ğ‘ž1𝑡)ğ‘›î€ºğ‘“î€·ğ‘ž1𝑡𝑑𝑚𝑑(ğ‘ž2𝑡)ğ‘šî€ºğ‘“î€·ğ‘ž2𝑡𝑡=𝑡0=1âŽ¡âŽ¢âŽ¢âŽ£ğ‘˜!𝑘ℓ=0âŽ›âŽœâŽœâŽğ‘˜â„“âŽžâŽŸâŽŸâŽ ğ‘‘â„“ğ‘‘ğ‘¡â„“î‚¸ğ‘‘ğ‘›ğ‘‘(ğ‘ž1𝑡)ğ‘›ğ‘“î€·ğ‘ž1𝑡𝑑𝑘−ℓ𝑑𝑡𝑘−ℓ𝑑𝑚𝑑(ğ‘ž2𝑡)ğ‘šğ‘”î€·ğ‘ž2ğ‘¡î€¸î‚¹âŽ¤âŽ¥âŽ¥âŽ¦ğ‘¡=𝑡0,(3.7) then from (II), we have 1𝑈(ℓ)=𝑑ℓ!â„“ğ‘‘ğ‘¡â„“ğ‘‘ğ‘›ğ‘‘î€·ğ‘ž1ğ‘¡î€¸ğ‘›ğ‘“î€·ğ‘ž1𝑡=ğ‘žâ„“1(ℓ+𝑛)!1ℓ!𝐹(ℓ+𝑛),𝑈(𝑘−ℓ)=𝑑(𝑘−ℓ)!ğ‘˜âˆ’â„“ğ‘‘ğ‘¡ğ‘˜âˆ’â„“ğ‘‘ğ‘šğ‘‘î€·ğ‘ž2ğ‘¡î€¸ğ‘šğ‘”î€·ğ‘ž2𝑡=ğ‘ž2𝑘−ℓ(𝑘−ℓ+𝑚)!(𝑘−ℓ)!𝐺(𝑘−ℓ+𝑚).(3.8) By utilizing these values 𝑈(𝑘)=𝑘ℓ=0ğ‘žâ„“1ğ‘ž2𝑘−ℓ(ℓ+𝑛)!(𝑘−ℓ+𝑚)!(𝑘−ℓ)!(ℓ)!𝐹(ℓ+𝑛)𝐺(𝑘−ℓ+𝑚).(3.9)
(V) Let the differential transform of 𝑑𝑛𝑖𝑓𝑖/𝑑(ğ‘žğ‘–ğ‘¡)𝑛𝑖 at 𝑡=𝑡0 for 𝑖=1,2,...,𝑚 be 𝐻𝑖(𝑘), then by using operations of differential transformation given in Table 1, we have 𝑈(𝑘)=𝑘ℓ𝑚−1ℓ=0𝑚−1ℓ𝑚−2=0⋯ℓ3ℓ2ℓ=02ℓ1=0𝐻1ℓ1𝐻2ℓ2−ℓ1⋯𝐻𝑚−1ℓ𝑚−1−ℓ𝑚−1𝐻𝑘−ℓ𝑚−1,(3.10) and from (II), we have 𝐻1ℓ1=ğ‘žâ„“11ℓ1+𝑛1!ℓ1!𝐹1ℓ1+𝑛1,𝐻2ℓ2−ℓ1=ğ‘žâ„“2−ℓ12ℓ2−ℓ1+𝑛2!ℓ2−ℓ1!𝐹2ℓ2−ℓ1+𝑛1,⋮𝐻(3.11)𝑚−1ℓ𝑚−1−ℓ𝑚−2=ğ‘žâ„“ğ‘šâˆ’1−ℓ𝑚−2𝑚−1ℓ𝑚−1−ℓ𝑚−2+𝑛𝑚−1!ℓ𝑚−1−ℓ𝑚−21!𝐹𝑚−1ℓ𝑚−1−ℓ𝑚−2+𝑛𝑚−1,𝐻𝑚𝑘−ℓ𝑚−1=ğ‘žğ‘˜âˆ’â„“ğ‘šâˆ’1𝑚𝑘−ℓ𝑚−1+𝑛𝑚!𝑘−ℓ𝑚−1!𝐹𝑚𝑘−ℓ𝑚−1+𝑛𝑚.(3.12) Substituting those values into (3.10), we obtain (3.3).

4. Illustrative Examples

In this part, we will apply the DTM to solve NFDE with proportional delays.

The numerical solutions of Examples 4.3, 4.4, and 4.5 have been calculated by variational iteration method [9] and homotopy perturbation method [6], which did not yield the exact solutions. However, applying DTM gives the exact solutions of those examples, as we will show later.

Example 4.1 (see [6, 9]). Consider the following first-order NFDE with proportional delay: ğ‘¢î…ž1(𝑡)=−𝑢(𝑡)+2𝑢𝑡2+12ğ‘¢î…žî‚€ğ‘¡2,0<𝑡<1,𝑢(0)=1.(4.1) Taking the differential transform of (4.1) as given in (2.1), we get 1(𝑘+1)𝑈(𝑘+1)=−𝑈(𝑘)+212𝑘1𝑈(𝑘)+212𝑘(𝑘+1)𝑈(𝑘+1),(4.2) which can be rewritten as follows: 𝑈(𝑘+1)=−𝑈(𝑘)(.𝑘+1)(4.3) The differential transform of the initial condition of 𝑢(𝑡) at 𝑡0=0, is 𝑈(0)=1, form (4.3) for 𝑘=0,1,…,8, we can get 1𝑈(1)=−1,𝑈(2)=12!,𝑈(3)=−13!,𝑈(4)=,14!𝑈(5)=−15!,𝑈(6)=16!,𝑈(7)=−17!,𝑈(8)=,8!(4.4) substituting these values into (2.2), to get 𝑡𝑢(𝑡)=1−𝑡+2−𝑡2!3+𝑡3!4−𝑡4!5+𝑡5!6−𝑡6!7+𝑡7!8.8!(4.5) The closed form of the above solution, when ğ‘â†’âˆž is 𝑢(𝑡)=𝑒−𝑡, which is the exact solution. In Table 2 the absolute errors of DTM for 𝑁=7,8, VIM [9] with eight terms and HPM [6] with eight terms (Table 3) are compared.


𝑡 VIM [9] HPM [6]DTM
𝑛 = 8 𝑛 = 8 𝑛 = 7 𝑛 = 8

0.1 3 . 7 2 𝐸 − 4 3 . 3 6 𝐸 − 4 2 . 4 5 𝐸 − 1 3 2 . 8 9 𝐸 − 1 5
0.2 7 . 0 8 𝐸 − 4 5 . 8 0 𝐸 − 4 6 . 2 1 𝐸 − 1 1 1 . 3 8 𝐸 − 1 2
0.3 1 . 0 1 𝐸 − 3 7 . 5 0 𝐸 − 4 1 . 5 7 𝐸 − 9 5 . 2 7 𝐸 − 1 1
0.4 1 . 2 9 𝐸 − 3 8 . 6 4 𝐸 − 4 1 . 5 6 𝐸 − 8 6 . 9 5 𝐸 − 1 0
0.5 1 . 5 4 𝐸 − 3 9 . 3 3 𝐸 − 4 9 . 1 8 𝐸 − 8 5 . 1 2 𝐸 − 9
0.6 1 . 7 6 𝐸 − 3 9 . 6 8 𝐸 − 4 3 . 9 0 𝐸 − 7 2 . 6 2 𝐸 − 8
0.7 1 . 9 7 𝐸 − 3 9 . 7 8 𝐸 − 4 1 . 3 2 𝐸 − 6 1 . 0 4 𝐸 − 7
0.8 2 . 1 5 𝐸 − 3 9 . 6 8 𝐸 − 4 3 . 8 2 𝐸 − 6 3 . 4 2 𝐸 − 7
0.9 2 . 3 2 𝐸 − 3 9 . 4 4 𝐸 − 4 9 . 7 0 𝐸 − 6 9 . 7 9 𝐸 − 7
1.0 2 . 4 7 𝐸 − 3 9 . 1 0 𝐸 − 4 2 . 2 3 𝐸 − 5 2 . 5 0 𝐸 − 6


𝑡 VIM [9] HPM [6] DTM
  𝑛 = 8 𝑛 = 8 𝑛 = 7 𝑛 = 8

0.1 1 . 3 0 𝐸 − 3 1 . 0 6 𝐸 − 3 8 . 2 0 𝐸 − 9 1 . 3 7 𝐸 − 1 0
0.2 2 . 1 4 𝐸 − 3 1 . 3 5 𝐸 − 3 5 . 1 6 𝐸 − 7 1 . 7 3 𝐸 − 8
0.3 2 . 6 3 𝐸 − 3 1 . 1 8 𝐸 − 3 5 . 7 8 𝐸 − 6 2 . 9 1 𝐸 − 7
0.4 2 . 8 4 𝐸 − 3 7 . 6 1 𝐸 − 4 3 . 2 0 𝐸 − 5 2 . 1 5 𝐸 − 6
0.5 2 . 8 3 𝐸 − 3 2 . 3 2 𝐸 − 4 1 . 2 0 𝐸 − 4 1 . 0 1 𝐸 − 5
0.6 2 . 6 7 𝐸 − 3 2 . 9 8 𝐸 − 4 3 . 5 3 𝐸 − 4 3 . 5 8 𝐸 − 5
0.7 2 . 3 9 𝐸 − 3 7 . 6 4 𝐸 − 4 8 . 7 7 𝐸 − 4 1 . 0 4 𝐸 − 4
0.8 2 . 0 4 𝐸 − 3 1 . 1 2 𝐸 − 3 1 . 9 2 𝐸 − 3 2 . 6 1 𝐸 − 4
0.9 1 . 6 4 𝐸 − 3 1 . 3 7 𝐸 − 3 3 . 8 4 𝐸 − 3 5 . 8 8 𝐸 − 4
1.0 1 . 2 2 𝐸 − 3 1 . 5 0 𝐸 − 3 7 . 1 2 𝐸 − 3 1 . 2 1 𝐸 − 3

Example 4.2 (see [6, 9]). Consider the first-order NFDE with proportional delay ğ‘¢î…ž(𝑡)=−𝑢(𝑡)+0.1𝑢(0.8𝑡)+0.5ğ‘¢î…ž(0.8𝑡)+(0.32𝑡−0.5)𝑒−0.8𝑡+𝑒−𝑡,𝑡≥0,𝑢(0)=0.(4.6) Let The differential transforms of 𝑡𝑒−0.8𝑡, 𝑒−0.8𝑡, and 𝑒−𝑡 at 𝑡0=0 be 𝛿1(𝑘),𝛿2(𝑘), and 𝛿3(𝑘), respectively: 𝛿1⎧⎪⎨⎪⎩(𝑘)=0,𝑘=0,(−1)𝑘−1(0.8)𝑘−1𝛿𝑘−1!,𝑘≠0,2(𝑘)=(−1)𝑘(0.8)𝑘,𝛿𝑘!3(𝑘)=(−1)𝑘.𝑘!(4.7) We can obtain the differential transform of (4.6) as 𝑈(𝑘+1)=−𝑈(𝑘)1−0.1(0.8)𝑘+0.32𝛿1(𝑘)−0.5𝛿2(𝑘)+𝛿3(𝑘)(𝑘+1)1−0.5(0.8)𝑘.(4.8) At 𝑡0=0, the initial condition transformed to 𝑈(0)=0, so from (4.8), we have 1𝑈(1)=1,𝑈(2)=−1,𝑈(3)=21,𝑈(4)=−6,1𝑈(5)=124,𝑈(6)=−1120,𝑈(7)=1720,𝑈(8)=−.5040(4.9) Substituting these values into (2.2), we get 𝑢(𝑡)=𝑡−𝑡2+12𝑡3−16𝑡4+1𝑡245−1𝑡1206+1𝑡7207−1𝑡50408.(4.10) The closed form of the above solution, when ğ‘â†’âˆž, is 𝑢(𝑡)=𝑡𝑒−𝑡, which is the exact solution. In Table 2 we compare the absolute errors of DTM for 𝑁=7,8, VIM [9] with eight terms, and HPM [6] with eight terms.

Example 4.3 (see[6, 9]). Consider the following second-order NFDE with proportional delay: ğ‘¢î…žî…ž(𝑡)=ğ‘¢î…žî‚€ğ‘¡2−12ğ‘¡ğ‘¢î…žî…žî‚€ğ‘¡2+2,0<𝑡<1,𝑢(0)=1,ğ‘¢î…ž(0)=0.(4.11) The differential transform for (4.11) is found as =1(𝑘+1)(𝑘+2)𝑈(𝑘+2)2𝑘1(𝑘+1)𝑈(𝑘+1)−2𝑘ℓ=012𝑘−ℓ(𝑘−ℓ+2)!(𝑘−ℓ)!𝛿(ℓ−1)𝑈(𝑘−ℓ+2)+2𝛿(𝑘),(4.12) form the initial condition we can get 𝑈(0)=1 and 𝑈(1)=0. Form (4.12), we get 𝑈(𝑘)=1,𝑘=2,0,𝑘>2.(4.13) Then, by using (2.2), 𝑢(𝑡)=1+𝑡2, which is the exact solution.

Example 4.4 (see [6, 9]). Consider the second-order NFDE with proportional delay: ğ‘¢î…žî…ž3(𝑡)=4𝑡𝑢(𝑡)+𝑢2+ğ‘¢î…žî‚€ğ‘¡2+12ğ‘¢î…žî…žî‚€ğ‘¡2−𝑡2−𝑡+1,0<𝑡<1,𝑢(0)=ğ‘¢î…ž(0)=0.(4.14) The differential transform for (4.14) at 𝑡0=0 is given by 3(𝑘+1)(𝑘+2)𝑈(𝑘+2)=41𝑈(𝑘)+2𝑘1𝑈(𝑘)+2𝑘1(𝑘+1)𝑈(𝑘+1)+212𝑘(𝑘+1)×(𝑘+2)𝑈(𝑘+2)−𝛿(𝑘−2)−𝛿(𝑘−1)+𝛿(𝑘),(4.15) and can be rewritten as 𝑈(𝑘+2)=3/4+1/2𝑘𝑈(𝑘)+1/2𝑘(𝑘+1)𝑈(𝑘+1)−𝛿(𝑘−2)−𝛿(𝑘−1)+𝛿(𝑘)(𝑘+1)(𝑘+2)1−1/2𝑘+1.(4.16)
Form the initial condition we can get 𝑈(0)=𝑈(1)=0, and form (4.16), we get 𝑈(𝑘)=1,𝑘=2,0,𝑘>2.(4.17) Then, by using (2.2), 𝑢(𝑡)=𝑡2, which is the exact solution.

Example 4.5 (see [6, 9]). Consider the following third-order NFDE with proportional delays: ğ‘¢î…žî…žî…ž(𝑡)=𝑢(𝑡)+ğ‘¢î…žî‚€ğ‘¡2+ğ‘¢î…žî…žî‚€ğ‘¡3+12ğ‘¢î…žî…žî…žî‚€ğ‘¡4−𝑡4−𝑡32−4𝑡23+21𝑡,𝑢(0)=0,ğ‘¢î…ž(0)=0,ğ‘¢î…žî…ž(0)=0.(4.18) The differential transform of (4.18) can be written as 1𝑈(𝑘)+2𝑘1(𝑘+1)𝑈(𝑘+1)+3𝑘(𝑘+1)(𝑘+2)𝑈(𝑘+2)𝑈(𝑘+3)=−𝛿(𝑘−4)−(1/2)𝛿(𝑘−3)−(4/3)𝛿(𝑘−2)+21𝛿(𝑘−1)(𝑘+1)(𝑘+2)(𝑘+3)1−(1/2)1/4𝑘.(4.19)
Form the initial condition we can get 𝑈(0)=𝑈(1)=𝑈(2)=0, so from the (4.19), we get 𝑈(𝑘)=0,𝑘≠4,1,𝑘=4.(4.20) Substituting (4.20) in (2.2) gives 𝑢(𝑡)=𝑡4, which is the exact solution.

5. Conclusion

In this study, we extended DTM to the solution of NFDE with proportional delays. New theorems are presented with their proofs. All examples results show that the DTM is more effective than VIM and HPM for solving NFDE with proportional delays. We believe that the ease of implementation and efficiency of the DTM gives it much wider applicability.

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Copyright © 2012 Sabir Widatalla. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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