Hindawi Publishing Corporation The Scientific World Journal Volume 2013, Article ID 536280, 7 pages http://dx.doi.org/10.1155/2013/536280

Research Article Taylor’s Expansion for Composite Functions Le Thi Phuong Ngoc1 and Nguyen Anh Triet2 1 2

Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Vietnam Department of Mathematics, University of Architecture of HoChiMinh City, 196 Pasteur Street, District 3, HoChiMinh City, Vietnam

Correspondence should be addressed to Nguyen Anh Triet; [email protected] Received 4 August 2013; Accepted 28 August 2013 Academic Editors: A. Agouzal and J.-S. Chen Copyright © 2013 L. T. P. Ngoc and N. A. Triet. 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. We build a Taylor’s expansion for composite functions. Some applications are introduced, where the proposed technique allows the authors to obtain an asymptotic expansion of high order in many small parameters of solutions.

1. Introduction

where

Let 𝑓 ∈ 𝐶𝑁+1 ([0, 1] × R+ × R3 × R3+ ; R) and 𝑢𝛼 ∈ 𝑊 = {V ∈ 𝐿∞ (0, 𝑇; 𝐻2 (0, 1)) : V𝑡 ∈ 𝐿∞ (0, 𝑇; 𝐻1 (0, 1))}, 𝛼 ∈ Z𝑛+ , |𝛼| ≤ 𝑁. Let two functions ℎ(𝑥, 𝑡, 𝜀) and 𝐹(𝑥, 𝑡, 𝜀), which depend on (𝑥, 𝑡) and 𝜀 = (𝜀1 , . . . , 𝜀𝑛 ), be defined as follows: 𝛼

ℎ (𝑥, 𝑡, 𝜀) = ∑ 𝑢𝛼 (𝑥, 𝑡) 𝜀 , |𝛼|≤𝑁

(1)

𝑔1 (𝜀) = ℎ (𝑥, 𝑡, 𝜀) = ∑ 𝑢𝛼 (𝑥, 𝑡) 𝜀𝛼 , |𝛼|≤𝑁

𝑔2 (𝜀) = ℎ𝑥 (𝑥, 𝑡, 𝜀) = ∑ 𝑢𝛼𝑥 (𝑥, 𝑡) 𝜀𝛼 , |𝛼|≤𝑁

𝑔3 (𝜀) = ℎ𝑡 (𝑥, 𝑡, 𝜀) = ∑ 𝑢𝛼𝑡 (𝑥, 𝑡) 𝜀𝛼 , |𝛼|≤𝑁

󵄩󵄩 󵄩󵄩󵄩2 󵄩󵄩 𝛼󵄩 󵄩 𝑔4 (𝜀) = ‖ℎ (𝑡, 𝜀)‖ = 󵄩󵄩󵄩 ∑ 𝑢𝛼 (𝑡) 𝜀 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩

where

2

𝛼

𝜀𝛼 = 𝜀1 1 ⋅ ⋅ ⋅ 𝜀𝑛𝛼𝑛 ,

𝜀 = (𝜀1 , . . . , 𝜀𝑛 ) ∈ R𝑛 ,

= ∑

𝛼 = (𝛼1 , . . . , 𝛼𝑛 ) ∈ Z𝑛+ , |𝛼| = 𝛼1 + ⋅ ⋅ ⋅ + 𝛼𝑛 ,

(2)

|𝛼|≤2𝑁

where notation ‖ ⋅ ‖ stands for the norm in 𝐿2 (0, 1). The function 𝐹(𝑥, 𝑡, 𝜀) has the form of a composite function as follows:

𝑔3 (𝜀) , 𝑔4 (𝜀) , 𝑔5 (𝜀) , 𝑔6 (𝜀)) ,

|𝛿|≤2𝑁 |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

= ∑ 𝑔𝛼[4] (𝑡) 𝜀𝛼 ,

󵄩 󵄩2 󵄩 󵄩2 𝐹 (𝑥, 𝑡, 𝜀) = 𝑓 (𝑥, 𝑡, ℎ, ℎ𝑥 , ℎ𝑡 , ‖ℎ‖2 , 󵄩󵄩󵄩ℎ𝑥 󵄩󵄩󵄩 , 󵄩󵄩󵄩ℎ𝑡 󵄩󵄩󵄩 ) ,

𝐹 (𝑥, 𝑡, 𝜀) = 𝑓 (𝑥, 𝑡, 𝑔1 (𝜀) , 𝑔2 (𝜀) ,

∑ ∑ ∑ ⟨𝑢𝛼 (𝑡) , 𝑢𝛽 (𝑡)⟩ 𝜀𝛿

󵄩󵄩2 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩2 󵄩󵄩󵄩󵄩 𝛼󵄩 𝑔5 (𝜀) = 󵄩󵄩ℎ𝑥 (𝑡, 𝜀)󵄩󵄩 = 󵄩󵄩 ∑ 𝑢𝛼𝑥 (𝑡) 𝜀 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩 = ∑

∑ ∑ ∑ ⟨𝑢𝛼𝑥 (𝑡) , 𝑢𝛽𝑥 (𝑡)⟩ 𝜀𝛿

|𝛿|≤2𝑁 |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

(3)

= ∑ 𝑔𝛼[5] (𝑡) 𝜀𝛼 , |𝛼|≤2𝑁

2

The Scientific World Journal 󵄩󵄩2 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩2 󵄩󵄩󵄩󵄩 𝛼󵄩 𝑔6 (𝜀) = 󵄩󵄩ℎ𝑡 (𝑡, 𝜀)󵄩󵄩 = 󵄩󵄩 ∑ 𝑢𝛼𝑡 (𝑡) 𝜀 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩 = ∑

2. Solving the Problem 2 We use the following notations. For a multi-index 𝛼 = (𝛼1 , . . . , 𝛼𝑛 ) ∈ Z𝑛+ and 𝑥 = (𝑥1 , . . . , 𝑥𝑛 ) ∈ R𝑛 , we put

∑ ∑ ∑ ⟨𝑢𝛼𝑡 (𝑡) , 𝑢𝛽𝑡 (𝑡)⟩ 𝜀𝛿

|𝛼| = 𝛼1 + ⋅ ⋅ ⋅ + 𝛼𝑛 ,

|𝛿|≤2𝑁 |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

= ∑ 𝑔𝛼[6] (𝑡) 𝜀𝛼 ,

𝛼, 𝛽 ∈ Z𝑛+ ,

(4) where notation ⟨⋅, ⋅⟩ stands for the scalar product in 𝐿2 (0, 1). Under the above assumptions, in order to obtain an asymptotic expansion of high order in many small parameters of solutions in recent papers [1–4], we need to solve the following problem.

it means that 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩𝐹 − ∑ 𝐹 𝜀𝛼 󵄩󵄩󵄩 ≤ 𝐶𝑁‖𝜀‖𝑁+1 , 󵄩󵄩 𝛼 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝐿∞ (0,𝑇,𝐿2 (0,1)) |𝛼|≤𝑁 󵄩

(5)

(6)

where the constant 𝐶𝑁 is independent from 𝜀. By Taylor’s expansion for 𝐹(𝑥, 𝑡, 𝜀), it implies that 1 𝛼 (7) 𝐷 𝐹 (𝑥, 𝑡, 0) , |𝛼| ≤ 𝑁. 𝛼! However, for 𝐹(𝑥, 𝑡, 𝜀) which is given in (3), it is very difficult to calculate 𝐷𝛼 𝐹(𝑥, 𝑡, 0), and so, the Problem 1 cannot proceed. Which method can be used for solving the Problem 1? To answer, let us note that the function 𝐹(𝑥, 𝑡, 𝜀) has the form of a composite function; this view suggests that we need to construct the Taylor-Maclaurin expansion of the composite function. So, first in Section 2 we solve the following problem. 𝐹𝛼 (𝑥, 𝑡) =

Problem 2. Let Ω be an open subset of R𝑛 and 0 ∈ Ω. Let 𝑔 = (𝑔1 , . . . , 𝑔𝑝 ) ∈ 𝐶𝑁+1 (Ω; R𝑝 ) and 𝑓 ∈ 𝐶𝑁+1 (R𝑝 ; R). Seek the representation formula for 𝑓𝑜𝑔, such that for ‖𝑥‖ being small enough,

= ∑ 𝑑𝛼 𝑥𝛼 + 𝑂 (‖𝑥‖𝑁+1 ) ,

(8)

|𝛼|≤𝑁

where 𝑑𝛼 , |𝛼| ≤ 𝑁 are calculated from the values of the given functions 𝑓; 𝑔1 , . . . , 𝑔𝑝 and of their derivatives at a suitable point. Next in Section 3, as an application of the method used, we study the Problem 1. This technique is also a great help for the authors to obtain an asymptotic expansion as they want in recent papers [1–4].

∀𝑖 = 1, . . . , 𝑛.

The following lemma is useful to solve the Problems 1 and Lemma 3. Let 𝑚, 𝑁 ∈ N and 𝑎𝛼 ∈ R, 𝛼 ∈ Z𝑛+ , 1 ≤ |𝛼| ≤ 𝑁. Then 𝑚 𝛼

( ∑ 𝑎𝛼 𝑥 ) = 1≤|𝛼|≤𝑁

∑

(𝑚) 𝑇𝑁 [𝑎]𝛼 𝑥𝛼 ,

(10)

𝑚≤|𝛼|≤𝑚𝑁

(𝑚) where the coefficients 𝑇𝑁 [𝑎]𝛼 , 𝑚 ≤ |𝛼| ≤ 𝑚𝑁 depending on 𝑛 𝑎 = (𝑎𝛼 ), 𝛼 ∈ Z+ , 1 ≤ |𝛼| ≤ 𝑁 are defined by the recurrent formulas (𝑚) 𝑇𝑁 [𝑎]𝛼

1 ≤ |𝛼| ≤ 𝑁, 𝑚 = 1, 𝑢𝛼 , { { { (𝑚−1) = { ∑ 𝑎𝛼−𝛽 𝑇𝑁 [𝑢]𝛽 , 𝑚 ≤ |𝛼| ≤ 𝑚𝑁, 𝑚 ≥ 2, { (𝑚) { (11) 𝛽∈𝐴 (𝑁) { 𝛼 󵄨󵄨 󵄨󵄨 𝑝 𝐴(𝑚) 𝛼 (𝑁) = {𝛽 ∈ Z+ : 𝛽 ≤ 𝛼, 1 ≤ 󵄨󵄨𝛼 − 𝛽󵄨󵄨 ≤ 𝑁, 󵄨 󵄨 𝑚 − 1 ≤ 󵄨󵄨󵄨𝛽󵄨󵄨󵄨 ≤ (𝑚 − 1) 𝑁} .

The proof of Lemma 3 can be found in [2]. Now, using Taylor’s expansion of the function 𝑓 around the point 𝑔 = (𝑔1 , . . . , 𝑔𝑝 ) ∈ R𝑝 , we obtain that 𝑓 (𝑔1 + 𝐻1 , . . . , 𝑔𝑝 + 𝐻𝑝 ) = 𝑓 (𝑔1 , . . . , 𝑔𝑝 ) +

(𝑓𝑜𝑔) (𝑥) = 𝑓 (𝑔1 (𝑥) , . . . , 𝑔𝑝 (𝑥))

𝛼 ≤ 𝛽 ⇐⇒ 𝛼𝑖 ≤ 𝛽𝑖

(9)

2.

Problem 1. Establish the functions 𝐹𝛼 = 𝐹𝛼 (𝑥, 𝑡), |𝛼| ≤ 𝑁 (independent of 𝜀) such that for ‖𝜀‖ = √𝜀12 + ⋅ ⋅ ⋅ + 𝜀𝑛2 being small enough, |𝛼|≤𝑁

𝛼

𝑥𝛼 = 𝑥1 1 ⋅ ⋅ ⋅ 𝑥𝑛𝛼𝑛 ,

‖𝑥‖ = √𝑥12 + ⋅ ⋅ ⋅ + 𝑥𝑛2 ,

|𝛼|≤2𝑁

𝐹 (𝑥, 𝑡, 𝜀) = ∑ 𝐹𝛼 (𝑥, 𝑡) 𝜀𝛼 + 𝑂 (‖𝜀‖𝑁+1 ) ;

𝛼! = 𝛼1 ! ⋅ ⋅ ⋅ 𝛼𝑛 !,

1 𝛼 𝐷 𝑓 (𝑔1 , . . . , 𝑔𝑝 ) 𝐻𝛼 𝛼! 1≤|𝛼|≤𝑁 ∑

+ 𝑅𝑁 [𝑓, 𝐻] , 𝐻 = (𝐻1 , . . . , 𝐻𝑝 ) ∈ R𝑝 , ‖𝐻‖ is small enough, 𝑅𝑁 [𝑓, 𝐻] = (𝑁 + 1 ) ×

1 𝛼 1 𝐻 ∫ (1 − 𝜃)𝑁𝐷𝛼 𝛼! 0 |𝛼|=𝑁+1 ∑

× 𝑓 (𝑔 + 𝜃𝐻) 𝑑𝜃.

(12)

The Scientific World Journal

3

Similarly, we use Maclaurin’s expansion of 𝑔𝑖 as follows:

Applying Lemma 3, with 𝑎 = 𝜎(𝑖) = (𝜎𝛽(𝑖) ), 𝜎𝛽(𝑖) = (1/𝛽!)𝐷𝛽 𝑔𝑖 (0), 1 ≤ |𝛽| ≤ 𝑁, it implies that

𝐻𝑖 = 𝑔𝑖 (𝑥) − 𝑔𝑖 (0) =

𝛼𝑖

1 𝛽 𝐷 𝑔𝑖 (0) 𝑥𝛽 + 𝑅𝑁 [𝑔𝑖 , 𝑥] 𝛽! 1≤|𝛽|≤𝑁 ∑

𝑥 ∈ R𝑛 ,

= 𝑃𝑖 + 𝑅𝑖 ,

‖𝑥‖ is small enough, 𝑃𝑖 =

𝛼 𝑃𝑖 𝑖

1 𝛽 𝐷 𝑔𝑖 (0) 𝑥𝛽 , 𝛽! 1≤|𝛽|≤𝑁

1 𝛽 =( ∑ 𝐷 𝑔𝑖 (0) 𝑥𝛽 ) ≡ ( ∑ 𝜎𝛽(𝑖) 𝑥𝛽 ) 𝛽! 1≤|𝛽|≤𝑁 1≤|𝛽|≤𝑁 =

(13)

∑

(𝛼 )

∑ 𝛼𝑖 ≤|𝛾𝑖 |≤𝛼𝑖 𝑁

𝑇𝑁 1 [𝜎(𝑖) ]𝛾 𝑥𝛾𝑖 . 𝑖

Hence,

𝑅𝑖 = 𝑅𝑁 [𝑔𝑖 , 𝑥]

𝛼

𝛼

𝑃𝛼 = 𝑃1 1 ⋅ ⋅ ⋅ 𝑃𝑝 𝑝

1 𝛽 1 𝑥 ∫ (1 − 𝜃)𝑁𝐷𝛽 𝑔𝑖 (𝜃𝑥) 𝑑𝜃. 𝛽! 0 |𝛽|=𝑁+1

= (𝑁 + 1) ∑

𝛼1

1 𝛽 =( ∑ 𝐷 𝑔1 (0) 𝑥𝛽 ) 𝛽! 1≤|𝛽|≤𝑁

Substituting (13) into (12), we get 1 𝛽 ⋅⋅⋅( ∑ 𝐷 𝑔𝑝 (0) 𝑥𝛽 ) 𝛽! 1≤|𝛽|≤𝑁

𝑓 (𝑔1 (𝑥) , . . . , 𝑔𝑝 (𝑥)) = 𝑓 (𝑔1 (0) , . . . , 𝑔𝑝 (0)) +

=

1 𝛼 𝐷 𝑓 (𝑔1 (0) , . . . , 𝑔𝑝 (0)) 𝐻𝛼 ∑ 𝛼! 1≤|𝛼|≤𝑁

⋅⋅⋅

+

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝑃𝛼 𝛼! 1≤|𝛼|≤𝑁

(14)

=

1

(𝛼 )

∑

(𝛼 )

𝑇𝑁 1 [𝜎(1) ]𝛾 ⋅ ⋅ ⋅ 𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 𝑥𝛾1 +⋅⋅⋅+𝛾𝑝 1

𝑝

𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

=

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝑃𝛼 𝛼! 1≤|𝛼|≤𝑁

(𝛼 )

(𝛼 )

∑ 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁,

𝑇𝑁 1 [𝜎(1) ]𝛾 ⋅ ⋅ ⋅ 𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 𝑥𝛿 1

𝑝

.. .

∑

(1) + ‖𝑥‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝑅] ,

𝑝

.. .

1 𝛼 𝐷 𝑓 (𝑔 (0)) (𝐻𝛼 − 𝑃𝛼 ) 𝛼! 1≤|𝛼|≤𝑁

+ 𝑅𝑁 [𝑓, 𝐻]

𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 𝑥𝛾𝑝

(𝛼 )

∑ 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁,

∑

= 𝑓 (𝑔 (0)) +

(𝛼 )

𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

∑

𝛼𝑝

𝑇𝑁 1 [𝜎(1) ]𝛾 𝑥𝛾1

∑ 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁

+ 𝑅𝑁 [𝑓, 𝐻] = 𝑓 (𝑔 (0)) +

𝛼𝑖

𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

=

where

(𝛼 )

∑ ∑ 𝑇𝑁 1 [𝜎(1) ]𝛾 1 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁, 𝛾1 +⋅⋅⋅+𝛾𝑝 =𝛿 .. . 𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

(1) [𝑓, 𝑔, 𝐻, 𝑅] ‖𝑥‖𝑁+1 𝑅𝑁

(𝛼 )

1 𝛼 = ∑ 𝐷 𝑓 (𝑔 (0)) (𝐻𝛼 − 𝑃𝛼 ) + 𝑅𝑁 [𝑓, 𝐻] , 𝛼! 1≤|𝛼|≤𝑁

⋅ ⋅ ⋅ 𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 𝑥𝛿 𝑝

=

∑

∑

∑

|𝛼|≤|𝛿|≤|𝛼|𝑁 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁, 𝛾1 +⋅⋅⋅+𝛾𝑝 =𝛿

(𝛼 )

𝑇𝑁 1 [𝜎(1) ]𝛾

.. .

𝑅𝑁 [𝑓, 𝐻]

𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

1 𝛼 1 = (𝑁 + 1) ∑ 𝐻 ∫ (1 − 𝜃)𝑁𝐷𝛼 𝛼! 0 |𝛼|=𝑁+1 × 𝑓 (𝑔 (0) + 𝜃𝐻) 𝑑𝜃. (15)

(𝛼 )

⋅ ⋅ ⋅ 𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 𝑥𝛿 𝑝

≡

∑ |𝛼|≤|𝛿|≤|𝛼|𝑁

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] 𝑥𝛿

1

(16)

4

The Scientific World Journal =

∑ |𝛼|≤|𝛿|≤𝑁

+

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] 𝑥𝛿 (1)

∑ 𝑁+1≤|𝛿|≤|𝛼|𝑁

= 𝑓 (𝑔 (0)) + ∑ 𝑑𝛿 𝑥𝛿 1≤|𝛿|≤𝑁

(𝑝)

𝛿

Φ𝛿 [𝛼, 𝑁, 𝜎 , . . . , 𝜎 ] 𝑥 ,

+

(2) ‖𝑥‖𝑁+1 𝑅𝑁

[𝑓, 𝑔, 𝐻, 𝑅, 𝑥] . (18)

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] =

(𝛼 )

∑ ∑ 𝑇𝑁 1 [𝜎(1) ]𝛾 1 𝛾 +⋅⋅⋅+𝛾 𝛼1 ≤|𝛾1 |≤𝛼1 𝑁, 1 𝑝 =𝛿 .. .

Clearly, the Problem 2 is solved with 𝑑𝛿 =

𝛼𝑝 ≤|𝛾𝑝 |≤𝛼𝑝 𝑁

1 𝛼 𝐷 𝑓 (𝑔 (0)) Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] . (19) 𝛼! 1≤|𝛼|≤|𝛿| ∑

(𝛼 )

⋅ ⋅ ⋅ 𝑇𝑁 𝑝 [𝜎(𝑝) ]𝛾 .

3. Solving the Problem 1

𝑝

(17) Consequently, 𝑓 (𝑔 (𝑥)) = 𝑓 (𝑔 (0)) +

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝑃𝛼 𝛼! 1≤|𝛼|≤𝑁 ∑

(1) + ‖𝑥‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝑅]

= 𝑓 (𝑔 (0)) + ×

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁 ∑

∑ |𝛼|≤|𝛿|≤|𝛼|𝑁

+

𝑓 (𝑥, 𝑡, 𝑔 + 𝐻)

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] 𝑥𝛿

(1) ‖𝑥‖𝑁+1 𝑅𝑁

= 𝑓 (𝑥, 𝑡, 𝑔1 + 𝐻1 , . . . , 𝑔6 + 𝐻6 )

[𝑓, 𝑔, 𝐻, 𝑅]

= 𝑓 (𝑥, 𝑡, 𝑔1 , . . . , 𝑔6 )

= 𝑓 (𝑔 (0)) + ×

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁 ∑

+ ×

(1)

(𝑝)

Φ𝛿 [𝛼, 𝑁, 𝜎 , . . . , 𝜎 ] 𝑥

(1) + ‖𝐻‖𝑁+1 𝑅𝑁 [𝑓, 𝐻] ,

𝛿

(1) [𝑓, 𝐻] ‖𝐻‖𝑁+1 𝑅𝑁

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁

1 𝛼 1 𝐻 ∫ (1 − 𝜃)𝑁𝐷𝛼 𝛼! 0 |𝛼|=𝑁+1

∑

∑ 𝑁+1≤|𝛿|≤|𝛼|𝑁

= (𝑁 + 1) ∑

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] 𝑥𝛿

×

∑ |𝛼|≤|𝛿|≤𝑁

𝛼

where 𝐻 = (𝐻1 , . . . , 𝐻6 ) ∈ R6 , ‖𝐻‖ is small enough, 𝐷𝑖 𝑖 𝑓 = 𝜕𝑓/𝜕𝑔𝑖 , 𝑖 = 1, 2, . . . , 6. Next, the precise structure of the representation formulas for 𝑔1 , . . . , 𝑔6 will be needed below to continue. For each fixed (𝑥, 𝑡) ∈ [0, 1] × R+ : we have the following. (i) The representation formula for 𝑔𝑖 (𝜀) = 𝑔𝑖 (𝑥, 𝑡, 𝜀), 𝑖 = 1, 2, 3. We rewrite 𝑔1 (𝜀) as follows

1 𝛼 𝐷 𝑓 (𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁 ∑

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(𝑝) ] 𝑥𝛿

(2) + ‖𝑥‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝑅, 𝑥]

= 𝑓 (𝑔 (0))

𝑔1 (𝜀) = ∑ 𝑢𝛼 (𝑥, 𝑡) 𝜀𝛼

1 𝛼 + ∑ 𝐷 𝑓 (𝑔 (0)) ∑ 𝛼! 1≤|𝛿|≤𝑁 1≤|𝛼|≤|𝛿| (1)

|𝛼|≤𝑁

(𝑝)

×Φ𝛿 [𝛼, 𝑁, 𝜎 , . . . , 𝜎 ] 𝑥 (2) + ‖𝑥‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝑅, 𝑥]

(21)

× 𝑓 (𝑔 + 𝜃𝐻) 𝑑𝜃,

(1) + ‖𝑥‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝑅]

= 𝑓 (𝑔 (0)) +

(20)

1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔1 , . . . , 𝑔6 ) 𝐻𝛼 + ∑ 𝛼! 1≤|𝛼|≤𝑁

∑

|𝛼|≤|𝛿|≤𝑁

As an application of the method used in Section 2 for 𝑝 = 6, Ω = (0, 1), 𝜀 ⃗ = (𝜀1 , . . . , 𝜀𝑛 ) ∈ R𝑛 , and 𝑓 = 𝑓(𝑥, 𝑡, 𝑔1 , . . ., 𝑔6 ) ∈ 𝐶𝑁+1 (R3 × R3+ ; R), for each fixed (𝑥, 𝑡) ∈ [0, 1] × R+ ; 𝑔𝑖 ∈ 𝐶𝑁+1 (Ω; R), 𝑖 = 1, 2, 3, for each fixed (𝑥, 𝑡) ∈ [0, 1]×R+ : 𝑔𝑖 ∈ 𝐶𝑁+1 (Ω; R+ ), 𝑖 = 4, 5, 6, for each fixed 𝑡 ∈ R+ , we now investigate the Problem 1. For each fixed (𝑥, 𝑡) ∈ [0, 1] × R+ , using Taylor’s expansion of the function 𝑓 around the point 𝑔 = (𝑔1 , . . . , 𝑔6 ) ∈ R3 × R3+ up to order 𝑁,we obtain that

𝛿

≡ 𝑔1 (0) +

∑ 𝑔𝛽[1] (𝑥, 𝑡) 𝜀𝛽

1≤|𝛽|≤𝑁

≡ 𝑔1 (0) + 𝐻1 ,

(22)

The Scientific World Journal

5

in which

Similarly, 󵄨󵄨 󵄨󵄨 󵄨󵄨𝛽󵄨󵄨 = 0, 󵄨󵄨 󵄨󵄨 1 ≤ 󵄨󵄨𝛽󵄨󵄨 ≤ 𝑁.

𝑔1 (0) = 𝑔1 (𝑥, 𝑡) = 𝑢0 (𝑥, 𝑡) , 𝑔𝛽[1] (𝑥, 𝑡) = 𝑢𝛽 (𝑥, 𝑡) ,

󵄩󵄩2 󵄩󵄩 󵄩 󵄩󵄩 𝛼󵄩 󵄩 𝑔5 (𝜀) = 󵄩󵄩󵄩 ∑ 𝑢𝛼𝑥 (𝑡) 𝜀 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩

(23)

= ∑ ∑ ⟨𝑢𝛼𝑥 (𝑡) , 𝑢𝛽𝑥 (𝑡)⟩ 𝜀𝛼+𝛽 |𝛼|≤𝑁 |𝛽|≤𝑁 󵄩2 󵄩󵄩 = 󵄩󵄩𝑢0𝑥 (𝑡)󵄩󵄩󵄩 + ∑ 𝐺𝛿 [𝑢𝑥 (𝑡)] 𝜀𝛿

It is similar to 𝑔2 (𝜀), 𝑔3 (𝜀); we write 𝑔2 (𝜀) = ∑ 𝑢𝛼𝑥 (𝑥, 𝑡) 𝜀𝛼 |𝛼|≤𝑁

≡ 𝑔2 (0) +

∑ 𝑔𝛽[2] (𝑥, 𝑡) 𝜀𝛽 ≡ 𝑔2 (0) + 𝐻2 ,

1≤|𝛽|≤𝑁

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝑢𝑥 , 𝜀]

≡ 𝑔5 (0) +

(24)

𝑔3 (𝜀) = ∑ 𝑢𝛼𝑡 (𝑥, 𝑡) 𝜀𝛼 |𝛼|≤𝑁

≡ 𝑔3 (0) +

󵄩2 󵄩 𝑔5 (0) = 󵄩󵄩󵄩𝑢0𝑥 (𝑡)󵄩󵄩󵄩 ,

𝑔2 (0) = 𝑔2 (𝑥, 𝑡) = 𝑢0𝑥 (𝑥, 𝑡) , 󵄨 󵄨 𝑔3 (0) = 𝑔3 (𝑥, 𝑡) = 𝑢0𝑡 (𝑥, 𝑡) , 󵄨󵄨󵄨𝛽󵄨󵄨󵄨 = 0, 𝑔𝛽[2]

(𝑥, 𝑡) = 𝑢𝛽𝑥 (𝑥, 𝑡) , 󵄨 󵄨 𝑔𝛽[3] (𝑥, 𝑡) = 𝑢𝛽𝑡 (𝑥, 𝑡) , 1 ≤ 󵄨󵄨󵄨𝛽󵄨󵄨󵄨 ≤ 𝑁.

𝑔𝛿[5] (𝑡) = 𝐺𝛿 [𝑢𝑥 (𝑡)] = ∑ ∑ ∑ ⟨𝑢𝛼𝑥 (𝑡) , 𝑢𝛽𝑥 (𝑡)⟩ , |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

(25) 𝑅5 =

≡ 𝑔6 (0) +

(26)

where

∑ 𝑔𝛽[4] (𝑡) 𝜀𝛽 + 𝑅4 1≤|𝛽|≤𝑁

󵄩2 󵄩 𝑔6 (0) = 󵄩󵄩󵄩𝑢0𝑡 (𝑡)󵄩󵄩󵄩 ,

≡ 𝑔4 (0) + 𝐻4 ≡ 𝑔4 (0) + 𝑃4 + 𝑅4 , 󵄩 󵄩2 𝑔4 (0) = 󵄩󵄩󵄩𝑢0 (𝑡)󵄩󵄩󵄩 ,

𝑔𝛿[6] (𝑡) = 𝐺𝛿 [𝑢𝑡 (𝑡)] = ∑ ∑ ∑ ⟨𝑢𝛼𝑡 (𝑡) , 𝑢𝛽𝑡 (𝑡)⟩ , |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

(𝑡) = 𝐺𝛿 [𝑢 (𝑡)] 𝑅6 = (27)

(1) ‖𝜀‖𝑁+1 𝑅𝑁

[𝑢𝑡 , 𝜀] =

𝑁+1≤|𝛿|≤2𝑁

∑

𝐺𝛿 [𝑢𝑡 (𝑡)] 𝜀𝛿 .

𝑁+1≤|𝛿|≤2𝑁

Then (22), (24), (26), (28), and (30) imply that 𝑔𝑖 (𝜀) − 𝑔𝑖 (0) = 𝐻𝑖 = 𝑃𝑖 + 𝑅𝑖 =

1 ≤ |𝛿| ≤ 2𝑁, 𝐺𝛿 [𝑢 (𝑡)] 𝜀𝛿 .

(31)

1 ≤ |𝛿| ≤ 𝑁,

= ∑ ∑ ∑ ⟨𝑢𝛼 (𝑡) , 𝑢𝛽 (𝑡)⟩ , |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

∑

∑ 𝑔𝛽[6] (𝑡) 𝜀𝛽 + 𝑅6 1≤|𝛽|≤𝑁

≡ 𝑔6 (0) + 𝐻6 ≡ 𝑔6 (0) + 𝑃6 + 𝑅6 ,

1≤|𝛿|≤𝑁

[𝑢, 𝜀] =

(30)

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝑢𝑡 , 𝜀]

(1) + ∑ 𝐺𝛿 [𝑢 (𝑡)] 𝜀𝛿 + ‖𝜀‖𝑁+1 𝑅𝑁 [𝑢, 𝜀]

𝑅4 =

𝑁+1≤|𝛿|≤2𝑁

1≤|𝛿|≤𝑁

|𝛼|≤𝑁 |𝛽|≤𝑁

𝐺𝛿 [𝑢 (𝑡)] = ∑ ∑ ∑ ⟨𝑢𝛼 (𝑡) , 𝑢𝛽 (𝑡)⟩ , |𝛼|≤𝑁 |𝛽|≤𝑁 𝛼+𝛽=𝛿

𝐺𝛿 [𝑢𝑥 (𝑡)] 𝜀𝛿 ,

∑

= ∑ ∑ ⟨𝑢𝛼𝑡 (𝑡) , 𝑢𝛽𝑡 (𝑡)⟩ 𝜀𝛼+𝛽 |𝛼|≤𝑁 |𝛽|≤𝑁 󵄩2 󵄩 = 󵄩󵄩󵄩𝑢0𝑡 (𝑡)󵄩󵄩󵄩 + ∑ 𝐺𝛿 [𝑢𝑡 (𝑡)] 𝜀𝛿

= ∑ ∑ ⟨𝑢𝛼 (𝑡) , 𝑢𝛽 (𝑡)⟩ 𝜀𝛼+𝛽

1 ≤ |𝛿| ≤ 𝑁,

[𝑢𝑥 , 𝜀] =

󵄩󵄩2 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝑔6 (𝜀) = 󵄩󵄩󵄩󵄩 ∑ 𝑢𝛼𝑡 (𝑡) 𝜀𝛼 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩

󵄩󵄩2 󵄩󵄩 󵄩 󵄩󵄩 𝛼󵄩 󵄩 𝑔4 (𝜀) = 󵄩󵄩󵄩 ∑ 𝑢𝛼 (𝑡) 𝜀 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩|𝛼|≤𝑁 󵄩 󵄩

󵄩 󵄩2 = 󵄩󵄩󵄩𝑢0 (𝑡)󵄩󵄩󵄩

(29)

1 ≤ |𝛿| ≤ 𝑁, (1) ‖𝜀‖𝑁+1 𝑅𝑁

For each fixed 𝑡 ∈ R+ : we have the following. (ii) The representation formula for 𝑔𝑖 (𝜀) = 𝑔𝑖 (𝑡, 𝜀), 𝑖 = 4, 5, 6. Write

(1) ‖𝜀‖𝑁+1 𝑅𝑁

1≤|𝛽|≤𝑁

in which

1≤|𝛽|≤𝑁

≡ 𝑔4 (0) +

∑ 𝑔𝛽[5] (𝑡) 𝜀𝛽 + 𝑅5

≡ 𝑔5 (0) + 𝐻5 ≡ 𝑔5 (0) + 𝑃5 + 𝑅5 ,

∑ 𝑔𝛽[3] (𝑥, 𝑡) 𝜀𝛽 ≡ 𝑔3 (0) + 𝐻3 ,

where

𝑔𝛿[4]

(28)

1≤|𝛿|≤𝑁

𝑅𝑖 = 0,

𝑖 = 1, 2, 3,

∑ 𝑔𝛽[𝑖] 𝜀𝛽 + 𝑅𝑖 , 1≤|𝛽|≤𝑁

𝑅𝑗 = 𝑂 (‖𝜀‖𝑁+1 ) ,

𝑗 = 4, 5, 6. (32)

6

The Scientific World Journal We also need the following lemma.

Combining (35) and (36) yields

Lemma 4. For all 𝛼 = (𝛼1 , . . . , 𝛼𝑛 ) ∈ Z𝑛+ , |𝛼| ≥ 1, then 𝛼

𝛼

𝛼

𝛼

𝛼

𝛼

𝛼

𝐻𝛼 = 𝐻1 1 𝐻2 2 𝐻3 3 𝐻4 4 𝐻5 5 𝐻6 6 =

∑ |𝛼|≤|𝛿|≤𝑁

=

Φ𝛿 [𝛼, 𝑁, 𝜎 , . . . , 𝜎 ] 1

𝜎

=

𝜎𝛽(𝑖)

=

𝛼

𝛼

∑

∑

𝛼

𝛼

𝛼

𝛼

𝛼

𝛼

𝛼

𝛼6

𝛼

𝛼5

𝛼6

− 𝑃4 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 )

𝛼

=

+

𝛼

𝛼

𝛼 𝛼 𝛼 + 𝑅6 ) − 𝑃4 4 𝑃5 5 𝑃6 6 𝛼 𝛼 − 𝑃4 4 ] (𝑃5 + 𝑅5 ) 5 (𝑃6

+

𝛼 𝛼 𝑃4 4 𝑃5 5

[(𝑃5 + 𝑅5 ) −

𝛼 𝑃5 5 ] (𝑃6

𝛼6

[(𝑃6 + 𝑅6 ) −

+ 𝑅6 )

+ 𝑅6 )

where (2) ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀]

=

𝑁+1≤|𝛿|≤|𝛼|𝑁

𝑗 𝛼 −1−𝑗 ]

𝑗 𝛼 −1−𝑗

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

(38)

]

𝐹 (𝑥, 𝑡, 𝜀𝛼 )

= 𝑓 (𝑥, 𝑡, 𝑔1 (0) + 𝐻1 , . . . , 𝑔6 (0) + 𝐻6 ) = 𝑓 (𝑥, 𝑡, 𝑔 (0)) 1 𝛼 (1) [𝑓, 𝐻] 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) 𝐻𝛼 + ‖𝐻‖𝑁+1 𝑅𝑁 + ∑ 𝛼! 1≤|𝛼|≤𝑁

𝛼4 −1

𝛼6 −1

∑

= 𝑓 (𝑥, 𝑡, 𝑔 + 𝐻) = 𝑓 (𝑥, 𝑡, 𝑔 (0) + 𝐻)

𝛼6

𝑗 𝛼 −1−𝑗 ] = 𝑅4 [ ∑ (𝑃4 + 𝑅4 ) 𝑃4 4 [ 𝑗=0 ] 𝛼 𝛼 × (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 ) 6 𝛼5 −1

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

󵄩 󵄩2 󵄩 󵄩2 = 𝑓 (𝑥, 𝑡, ℎ, ℎ𝑥 , ℎ𝑡 , ‖ℎ‖2 , 󵄩󵄩󵄩ℎ𝑥 󵄩󵄩󵄩 , 󵄩󵄩󵄩ℎ𝑡 󵄩󵄩󵄩 )

𝛼6

𝛼 𝑃6 6 ]

+ 𝑅5 [ ∑ (𝑃5 + 𝑅5 ) 𝑃5 5 [ 𝑗=0

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

Substituting (37) into (20), we obtain 𝛼6

𝛼6

𝛼5

𝛼 𝑃4 4

(37)

Lemma 4 is proved.

𝛼6

+ 𝑃4 4 (𝑃5 + 𝑅5 ) (𝑃6 + 𝑅6 ) − 𝑃4 4 𝑃5 5 (𝑃6 + 𝑅6 ) 𝛼 𝛼 + 𝑃4 4 𝑃5 5 (𝑃6 𝛼 [(𝑃4 + 𝑅4 ) 4

𝛼6

⋅ ⋅ ⋅ ( ∑ 𝑔𝛽(6) 𝜀𝛽 ) 1≤|𝛽|≤𝑁

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀] .

(𝑃4 + 𝑅4 ) 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 ) 6 − 𝑃4 4 𝑃5 5 𝑃6 6 = (𝑃4 + 𝑅4 ) 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 )

𝛼

(2) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀] ,

Note that 𝑅𝑗 = 𝑂(‖𝜀‖𝑁+1 ), 𝑗 = 4, 5, 6, so 𝛼

𝛼6

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

∑

|𝛼|≤|𝛿|≤𝑁

(35)

𝛼

𝛼

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀]

𝛼

𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 𝑃1 1 𝑃2 2 𝑃3 3 (𝑃4 + 𝑅4 ) 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 ) 6 𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 𝑃1 1 𝑃2 2 𝑃3 3 𝑃4 4 𝑃5 5 𝑃6 6 + 𝑃1 1 𝑃2 2 𝑃3 3 𝛼 𝛼 𝛼 𝛼 𝛼 𝛼 × [(𝑃4 + 𝑅4 ) 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 ) 6 − 𝑃4 4 𝑃5 5 𝑃6 6 ] .

𝛼

𝛼

( ∑ 𝑔𝛽(1) 𝜀𝛽 ) 1≤|𝛽|≤𝑁

≡

𝛼

𝛼

𝛼

𝑁+1≤|𝛿|≤|𝛼|𝑁

𝐻𝛼 = 𝐻1 1 𝐻2 2 𝐻3 3 𝐻4 4 𝐻5 5 𝐻6 6 =

𝛼

+

󵄨 󵄨 1 ≤ 󵄨󵄨󵄨𝛽󵄨󵄨󵄨 ≤ 𝑁, 𝑖 = 1, 2, . . . , 6. (34)

𝑔𝛽[𝑖] ,

Proof of Lemma 4. We have

=

𝛼

|𝛼|≤|𝛿|≤𝑁

(𝜎𝛽(𝑖) ) ,

𝛼

𝛼

=

6

.. .

𝛼6 ≤|𝛾6 |≤𝛼6 𝑁 (𝑖)

𝛼

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀]

(𝛼 )

(𝛼 )

𝑇𝑁 1 [𝜎(1) ]𝛾 ⋅ ⋅ ⋅ 𝑇𝑁 6 [𝜎(6) ]𝛾 ,

𝛼1 ≤|𝛾1 |≤𝛼1 𝑁, 𝛾1 +⋅⋅⋅+𝛾6 =𝛿

𝛼

𝛼1

(6)

∑

𝛼

(1) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀]

where

∑

𝛼

= 𝑃1 1 𝑃2 2 𝑃3 3 𝑃4 4 𝑃5 5 𝑃6 6

(33)

=

𝛼

𝛼

= 𝑃1 1 𝑃2 2 𝑃3 3 (𝑃4 + 𝑅4 ) 4 (𝑃5 + 𝑅5 ) 5 (𝑃6 + 𝑅6 )

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿 + 𝑂 (‖𝜀‖𝑁+1 ) ,

(1)

𝛼

𝛼

𝐻𝛼 = 𝐻1 1 𝐻2 2 𝐻3 3 𝐻4 4 𝐻5 5 𝐻6 6

= 𝑓 (𝑥, 𝑡, 𝑔 (0)) 1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) + ∑ 𝛼! 1≤|𝛼|≤𝑁

𝛼6

𝛼

𝑃4 4 (𝑃6 + 𝑅6 )

× [ ∑ |𝛼|≤|𝛿|≤𝑁

] 𝑃 4𝑃 5 + 𝑅6 [ ∑ (𝑃6 + 𝑅6 ) 𝑃6 6 4 5 [ 𝑗=0 ] (1) = ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀] = 𝑂 (‖𝜀‖𝑁+1 ) . 𝛼

𝛼

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

(2) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝐻, 𝛼, 𝜀] ]

(36)

(1) + ‖𝐻‖𝑁+1 𝑅𝑁 [𝑓, 𝐻]

The Scientific World Journal

7

= 𝑓 (𝑥, 𝑡, 𝑔 (0)) +

Remark 5. (i) This result improves the one in [1], for 𝑛 = 1. (ii) In case of 𝐹(𝑥, 𝑡, 𝜀) = 𝜇(𝑥, 𝑡, ℎ, ‖ℎ𝑥 ‖2 ), where 𝜇 ∈ 𝑁+1 𝐶 ([0, 1]×R+ ×R×R+ ; R), 𝑛 = 𝑝, this result is also obtained in [3].

1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁 ∑

×[ ∑ |𝛼|≤|𝛿|≤𝑁

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿 ]

Acknowledgments

1 𝛼 (2) 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) 𝑅𝑁 [𝐻, 𝛼, 𝜀] 𝛼! 1≤|𝛼|≤𝑁

The authors wish to express their sincere thanks to the referees for their valuable comments and important remarks. This research is funded by Vietnam National University HoChiMinh City (VNU-HCM) under Grant no. B2013-1805.

+ ‖𝜀‖𝑁+1 ∑

(1) + ‖𝐻‖𝑁+1 𝑅𝑁 [𝑓, 𝐻]

= 𝑓 (𝑥, 𝑡, 𝑔 (0)) +

1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) 𝛼! 1≤|𝛼|≤𝑁 ×

∑ |𝛼|≤|𝛿|≤𝑁

+

References

∑

(3) ‖𝜀‖𝑁+1 𝑅𝑁

[1] N. T. Long, “On the nonlinear wave equation 𝑢𝑡𝑡 − 𝐵(𝑡, ‖𝑢‖2 , ‖𝑢𝑥 ‖2 )𝑢𝑥𝑥 = 𝑓(𝑥, 𝑡, 𝑢, 𝑢𝑥 , 𝑢𝑡 , ‖𝑢‖2 , ‖𝑢𝑥 ‖2 ) associated with the mixed homogeneous conditions,” Journal of Mathematical Analysis and Applications, vol. 306, no. 1, pp. 243–268., 2005. [2] N. T. Long and L. X. Truong, “Existence and asymptotic expansion for a viscoelastic problem with a mixed nonhomogeneous condition,” Nonlinear Analysis: Theory, Methods and Applications, vol. 67, no. 3, pp. 842–864, 2007. [3] L. T. P. Ngoc, N. A. Triet, and N. T. Long, “On a nonlinear wave equation involving the term −(𝜕/𝜕𝑥)(𝜇(𝑥, 𝑡, 𝑢, ‖𝑢𝑥 ‖2 )𝑢𝑥 ): linear approximation and asymptotic expansion of solution in many small parameters,” Nonlinear Analysis: Real World Applications, vol. 11, no. 4, pp. 2479–2501, 2010. [4] N. A. Triet, L. T. P. Ngoc, and N. T. Long, “A mixed DirichletRobin problem for a nonlinear Kirchhoff-Carrier wave equation,” Nonlinear Analysis: Real World Applications, vol. 13, no. 2, pp. 817–839, 2012.

Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿

[𝑓, 𝑔, 𝐻, 𝜀]

= 𝑓 (𝑥, 𝑡, 𝑔 (0)) 1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) Φ𝛿 𝛼! 1≤|𝛿|≤𝑁 |𝛼|≤|𝛿|

+ ∑

∑

× [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] 𝜀𝛿 (3) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝜀]⃗

= 𝑓 (𝑥, 𝑡, 𝑔 (0)) + ∑ 𝐹𝛿 (𝑥, 𝑡) 𝜀𝛿 1≤|𝛿|≤𝑁

(3) + ‖𝜀‖𝑁+1 𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝜀] ,

(39) where (3) [𝑓, 𝑔, 𝐻, 𝜀] ‖𝜀‖𝑁+1 𝑅𝑁

1 𝛼 (2) 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) 𝑅𝑁 [𝐻, 𝛼, 𝜀] 𝛼! 1≤|𝛼|≤𝑁

= ‖𝜀‖𝑁+1 ∑

(1) + ‖𝐻‖𝑁+1 𝑅𝑁 [𝑓, 𝐻] ,

1 𝛼 𝐷 𝑓 (𝑥, 𝑡, 𝑔 (0)) Φ𝛿 [𝛼, 𝑁, 𝜎(1) , . . . , 𝜎(6) ] , 𝛼! |𝛼|≤|𝛿|

𝐹𝛿 (𝑥, 𝑡) = ∑

𝛿 ∈ Z6+ , 1 ≤ |𝛿| ≤ 𝑁. (40) Since 𝑢𝛼 ∈ 𝑊 = {V ∈ 𝐿∞ (0, 𝑇, 𝐻2 (0, 1)) : V𝑡 ∈ 𝐿 (0, 𝑇, 𝐻1 (0, 1))}, 𝛼 ∈ Z𝑛+ , |𝛼| ≤ 𝑁, hence 𝑢𝛼 , 𝑢𝛼𝑥 , 𝑢𝛼𝑡 ∈ 𝐿∞ (0, 𝑇, 𝐻1 (0, 1)) ⊂ 𝐿∞ (𝑄𝑇 ). Then from (39) and (40)1 , we get ∞

󵄩󵄩 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩𝐹(𝜀) − ∑ 𝐹 𝜀𝛿 󵄩󵄩󵄩 󵄩󵄩 𝛿 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩𝐿∞ (0,𝑇,𝐿2 (0,1)) |𝛿|≤𝑁 (41) 󵄩 󵄩 󵄩 (3) = ‖𝜀‖𝑁+1 󵄩󵄩󵄩󵄩𝑅𝑁 [𝑓, 𝑔, 𝐻, 𝜀]󵄩󵄩󵄩󵄩𝐿∞ (0,𝑇,𝐿2 (0,1)) ≤ 𝐶𝑁‖𝜀‖𝑁+1 , for ‖𝜀‖ is small enough; consequently the Problem 1 is solved.

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