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ISRN Applied Mathematics
Volume 2012 (2012), Article ID 518361, 10 pages
http://dx.doi.org/10.5402/2012/518361
Research Article

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

1Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
2Department of Mathematics, Sinnar University, Singa 107, Sudan

Received 30 January 2012; Accepted 19 February 2012

Academic Editors: C.-H. Lien and S. Prudhomme

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.

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 𝜎𝑢(𝑡)+𝑎(𝑡)𝑢𝑚(𝑡)(𝑚)=𝛽𝑢(𝑡)+𝑚1𝑘=0𝑏𝑘(𝑡)𝑢𝑘𝜎𝑘(𝑡)+𝑓(𝑡),𝑡0,(1.1) under the conditions 𝑚1𝑘=0𝑐𝑖𝑘𝑢(𝑘)(0)=𝜆𝑖,𝑖=0,1,,𝑚1,(1.2) where 𝑎,𝑏𝑘,𝜎𝑘 are analytical functions; 𝛽,𝑐𝑖𝑘 and 𝜆𝑖𝒞. A classical case [1] 𝜎𝑘(𝑡)=𝑡𝜏𝑘,𝑘=0,1,,𝑚,(1.3) where 𝜏𝑘 is positive. Another interesting case [2] is 𝜎𝑘(𝑡)=𝑞𝑘𝑡,𝑘=0,1,,𝑚,(1.4) where 0<𝑞𝑘<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. Consider𝑢(𝑡)=𝛽𝑢(𝑡)𝑎(𝑡)𝑢𝑞𝑡𝑞+𝑓𝑡,𝑢(𝑡),𝑢𝑖𝑡,𝑢𝑞𝑖𝑡,𝑖=1,2,...,1,𝑢(0)=𝑢0.(1.5)

Problem 2. Consider𝑢(𝑚)(𝑡)=𝛽𝑢(𝑡)𝑎(𝑡)𝑢(𝑚)𝑞𝑡𝑞+𝑓𝑡,𝑢(𝑡),𝑢𝑖𝑡,𝑢𝑞𝑖𝑡,𝑢𝑞𝑖𝑡,...,𝑢(𝑚1)𝑞𝑖𝑡,𝑚1𝑘=0𝑐𝑗𝑘𝑢(𝑘)(0)=𝜆𝑗,𝑗=0,1,2,...,𝑚1,(1.6) where 𝑎,𝑓 are analytical functions; 𝛽,𝑐𝑗𝑘 and 𝜆𝑗 are real or complex constants; 0<𝑞𝑖<1,𝑖=1,2,,.

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!𝑘𝑑𝑡𝑘𝑢(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)=𝑁𝑘=0𝑈(𝑘)𝑡𝑡0𝑘,for𝑁.(2.2) And from (2.1) and (2.2): 𝑢(𝑡)=𝑁𝑘=01𝑑𝑘!𝑘𝑑𝑡𝑘𝑢(𝑡)𝑡=𝑡0𝑡𝑡0𝑘,for𝑁,(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]).

tab1
Table 1: Fundamental operations of differential transformation.

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<𝑞,𝑞𝑖<1, for 𝑖=1,2,,𝑚: (I)if 𝑢(𝑡)=𝑓(𝑞𝑡), then 𝑈(𝑘)=𝑞𝑘𝐹(𝑘), (II)if 𝑢(𝑡)=𝑑𝑛𝑓(𝑞𝑡)/𝑑(𝑞𝑡)𝑛, then 𝑈(𝑘)=𝑞𝑘((𝑘+𝑛)!/𝑘!)𝐹(𝑘+𝑛), (III)if 𝑢(𝑡)=𝑔(𝑡)𝑑𝑛𝑓(𝑞𝑡)/𝑑(𝑞𝑡)𝑛, then 𝑈(𝑘)=𝑘=0𝑞𝑘(𝑘+𝑛)(𝑘)!𝐺()𝐹(𝑘+𝑛),(3.1)(IV)if 𝑢(𝑡)=(𝑑𝑛/𝑑(𝑞1𝑡)𝑛)[𝑓(𝑞1𝑡)](𝑑𝑚/𝑑(𝑞2𝑡)𝑚)[𝑔(𝑞2𝑡)], then 𝑈(𝑘)=𝑘=0𝑞1𝑞2𝑘(+𝑛)!(𝑘+𝑚)!(𝑘)!()!𝐹(+𝑛)𝐺(𝑘+𝑚),(3.2)(V)if 𝑢(𝑡)=(𝑑𝑛1/𝑑(𝑞1𝑡)𝑛1)[𝑓1(𝑞1𝑡)](𝑑𝑛2/𝑑(𝑞2𝑡)𝑛2)[𝑓2(𝑞2𝑡)](𝑑𝑛𝑚1/𝑑(𝑞𝑚1𝑡)𝑛𝑚1)[𝑓𝑚1(𝑞𝑚1𝑡)(𝑑𝑛𝑚/𝑑(𝑞𝑚𝑡)𝑛𝑚)[𝑓𝑚(𝑞𝑚𝑡)], then 𝑈(𝑘)=𝑘𝑚1=0𝑚1𝑚2=032=021=0𝑞11𝑞212𝑞𝑚1𝑚2𝑚1𝑞𝑘𝑚1𝑚×1+𝑛1!1!21+𝑛2!21!𝑚1𝑚2+𝑛𝑚1!𝑚1𝑚2!𝑘𝑚1+𝑛𝑚!𝑘𝑚1!×𝐹11+𝑛1𝐹221+𝑛2𝐹𝑚1𝑚1𝑚1+𝑛𝑚1𝐹𝑚𝑘𝑚1+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!𝑘𝑑𝑡𝑘𝑑𝑔(𝑡)𝑛𝑑(𝑞𝑡)𝑛𝑓(𝑞𝑡)𝑡=𝑡0=1𝑘!𝑘=0𝑘𝑑𝑑𝑡𝑑𝑔(𝑡)𝑘𝑑𝑡𝑘𝑑𝑛𝑑(𝑞𝑡)𝑛𝑓(𝑞𝑡)𝑡=𝑡0,(3.4) and from (II), we have 1𝑈(𝑘)=𝑑(𝑘)!𝑘𝑑𝑡𝑘𝑑𝑛𝑑(𝑞𝑡)𝑛𝑓(𝑞𝑡)𝑡=𝑡0=𝑞𝑘(𝑘+𝑛)!(𝑘)!𝐹(𝑘+𝑛).(3.5) By utilizing this value, we get 𝑈(𝑘)=𝑘=0𝑞𝑘(𝑘+𝑛)!(𝑘)!𝐺()𝐹(𝑘+𝑛).(3.6)
(IV) By using the definition of DTM (2.1), we have 1𝑈(𝑘)=𝑑𝑘!𝑘𝑑𝑡𝑘𝑑𝑛𝑑(𝑞1𝑡)𝑛𝑓𝑞1𝑡𝑑𝑚𝑑(𝑞2𝑡)𝑚𝑓𝑞2𝑡𝑡=𝑡0=1𝑘!𝑘=0𝑘𝑑𝑑𝑡𝑑𝑛𝑑(𝑞1𝑡)𝑛𝑓𝑞1𝑡𝑑𝑘𝑑𝑡𝑘𝑑𝑚𝑑(𝑞2𝑡)𝑚𝑔𝑞2𝑡𝑡=𝑡0,(3.7) then from (II), we have 1𝑈()=𝑑!𝑑𝑡𝑑𝑛𝑑𝑞1𝑡𝑛𝑓𝑞1𝑡=𝑞1(+𝑛)!1!𝐹(+𝑛),𝑈(𝑘)=𝑑(𝑘)!𝑘𝑑𝑡𝑘𝑑𝑚𝑑𝑞2𝑡𝑚𝑔𝑞2𝑡=𝑞2𝑘(𝑘+𝑚)!(𝑘)!𝐺(𝑘+𝑚).(3.8) By utilizing these values 𝑈(𝑘)=𝑘=0𝑞1𝑞2𝑘(+𝑛)!(𝑘+𝑚)!(𝑘)!()!𝐹(+𝑛)𝐺(𝑘+𝑚).(3.9)
(V) Let the differential transform of 𝑑𝑛𝑖𝑓𝑖/𝑑(𝑞𝑖𝑡)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)=𝑘𝑚1=0𝑚1𝑚2=032=021=0𝐻11𝐻221𝐻𝑚1𝑚1𝑚1𝐻𝑘𝑚1,(3.10) and from (II), we have 𝐻11=𝑞111+𝑛1!1!𝐹11+𝑛1,𝐻221=𝑞21221+𝑛2!21!𝐹221+𝑛1,𝐻(3.11)𝑚1𝑚1𝑚2=𝑞𝑚1𝑚2𝑚1𝑚1𝑚2+𝑛𝑚1!𝑚1𝑚21!𝐹𝑚1𝑚1𝑚2+𝑛𝑚1,𝐻𝑚𝑘𝑚1=𝑞𝑘𝑚1𝑚𝑘𝑚1+𝑛𝑚!𝑘𝑚1!𝐹𝑚𝑘𝑚1+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: 𝑢1(𝑡)=𝑢(𝑡)+2𝑢𝑡2+12𝑢𝑡2,0<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)+212𝑘1𝑈(𝑘)+212𝑘(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)=1𝑡+2𝑡2!3+𝑡3!4𝑡4!5+𝑡5!6𝑡6!7+𝑡7!8.8!(4.5) The closed form of the above solution, when 𝑁 is 𝑢(𝑡)=𝑒𝑡, 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.

tab2
Table 2: Comparison of the absolute errors for Example 4.1.
tab3
Table 3: Comparison of the absolute errors for Example 4.2.

Example 4.2 (see [6, 9]). Consider the first-order NFDE with proportional delay 𝑢(𝑡)=𝑢(𝑡)+0.1𝑢(0.8𝑡)+0.5𝑢(0.8𝑡)+(0.32𝑡0.5)𝑒0.8𝑡+𝑒𝑡,𝑡0,𝑢(0)=0.(4.6) Let The differential transforms of 𝑡𝑒0.8𝑡, 𝑒0.8𝑡, and 𝑒𝑡 at 𝑡0=0 be 𝛿1(𝑘),𝛿2(𝑘), and 𝛿3(𝑘), respectively: 𝛿1(𝑘)=0,𝑘=0,(1)𝑘1(0.8)𝑘1𝛿𝑘1!,𝑘0,2(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)10.1(0.8)𝑘+0.32𝛿1(𝑘)0.5𝛿2(𝑘)+𝛿3(𝑘)(𝑘+1)10.5(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)=𝑡𝑡2+12𝑡316𝑡4+1𝑡2451𝑡1206+1𝑡72071𝑡50408.(4.10) The closed form of the above solution, when 𝑁, is 𝑢(𝑡)=𝑡𝑒𝑡, 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: 𝑢(𝑡)=𝑢𝑡212𝑡𝑢𝑡2+2,0<𝑡<1,𝑢(0)=1,𝑢(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=012𝑘(𝑘+2)!(𝑘)!𝛿(1)𝑈(𝑘+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: 𝑢3(𝑡)=4𝑡𝑢(𝑡)+𝑢2+𝑢𝑡2+12𝑢𝑡2𝑡2𝑡+1,0<𝑡<1,𝑢(0)=𝑢(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(𝑘2)𝛿(𝑘1)+𝛿(𝑘),(4.15) and can be rewritten as 𝑈(𝑘+2)=3/4+1/2𝑘𝑈(𝑘)+1/2𝑘(𝑘+1)𝑈(𝑘+1)𝛿(𝑘2)𝛿(𝑘1)+𝛿(𝑘)(𝑘+1)(𝑘+2)11/2𝑘+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: 𝑢(𝑡)=𝑢(𝑡)+𝑢𝑡2+𝑢𝑡3+12𝑢𝑡4𝑡4𝑡324𝑡23+21𝑡,𝑢(0)=0,𝑢(0)=0,𝑢(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(𝑘4)(1/2)𝛿(𝑘3)(4/3)𝛿(𝑘2)+21𝛿(𝑘1)(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,𝑘4,1,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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