Taylor’s Expansion for Composite Functions

Abstract

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

Let 𝑓∈𝐶^𝑁+1([0,1]×ℝ₊×ℝ³×ℝ³₊;ℝ) and 𝑢_𝛼∈𝑊={𝑣∈𝐿^∞(0,𝑇;𝐻²(0, 1))∶𝑣_𝑡∈𝐿^∞(0,𝑇;𝐻¹(0,1))},𝛼∈ℤ^𝑛₊,|𝛼|≤𝑁.

Let two functions ℎ(𝑥,𝑡,𝜀) and𝐹(𝑥,𝑡,𝜀), which depend on (𝑥,𝑡) and 𝜀=(𝜀₁,…,𝜀_𝑛), be defined as follows: ℎ(𝑥,𝑡,𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼(𝑥,𝑡)𝜀^𝛼, (1) where 𝜀^𝛼=𝜀^𝛼1₁⋯𝜀^𝛼𝑛_𝑛,𝜀=𝜀₁,…,𝜀_𝑛∈ℝ^𝑛,𝛼=𝛼₁,…,𝛼_𝑛∈ℤ^𝑛₊,|𝛼|=𝛼₁+⋯+𝛼_𝑛,𝐹(𝑥,𝑡,𝜀)=𝑓𝑥,𝑡,ℎ,ℎ_𝑥,ℎ_𝑡,‖ℎ‖²,ℎ_𝑥²,ℎ_𝑡², (2) 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: 𝐹(𝑥,𝑡,𝜀)=𝑓𝑥,𝑡,𝑔₁(𝜀),𝑔₂(𝜀),𝑔₃(𝜀),𝑔₄(𝜀),𝑔₅(𝜀),𝑔₆(𝜀), (3) where 𝑔₁(𝜀)=ℎ(𝑥,𝑡,𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼(𝑥,𝑡)𝜀^𝛼,𝑔₂(𝜀)=ℎ_𝑥(𝑥,𝑡,𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑥(𝑥,𝑡)𝜀^𝛼,𝑔₃(𝜀)=ℎ_𝑡(𝑥,𝑡,𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑡(𝑥,𝑡)𝜀^𝛼,𝑔₄(𝜀)=‖ℎ(𝑡,𝜀)‖²=_{|𝛼|≤𝑁}𝑢_𝛼(𝑡)𝜀^𝛼2=_{|𝛿|≤2𝑁|𝛼|≤𝑁|𝛽|≤𝑁𝛼+𝛽=𝛿}𝑢_𝛼(𝑡),𝑢_𝛽(𝑡)𝜀^𝛿=_{|𝛼|≤2𝑁}𝑔_𝛼^[4](𝑡)𝜀^𝛼,𝑔₅(𝜀)=ℎ_𝑥(𝑡,𝜀)²=_{|𝛼|≤𝑁}𝑢_𝛼𝑥(𝑡)𝜀^𝛼2=_{|𝛿|≤2𝑁|𝛼|≤𝑁|𝛽|≤𝑁𝛼+𝛽=𝛿}𝑢_𝛼𝑥(𝑡),𝑢_𝛽𝑥(𝑡)𝜀^𝛿=_{|𝛼|≤2𝑁}𝑔_𝛼^[5](𝑡)𝜀^𝛼,𝑔₆(𝜀)=ℎ_𝑡(𝑡,𝜀)²=_{|𝛼|≤𝑁}𝑢_𝛼𝑡(𝑡)𝜀^𝛼2=_{|𝛿|≤2𝑁|𝛼|≤𝑁|𝛽|≤𝑁𝛼+𝛽=𝛿}𝑢_𝛼𝑡(𝑡),𝑢_𝛽𝑡(𝑡)𝜀^𝛿=_{|𝛼|≤2𝑁}𝑔_𝛼^[6](𝑡)𝜀^𝛼, (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.

Problem 1. Establish the functions 𝐹_𝛼=𝐹_𝛼(𝑥,𝑡),|𝛼|≤𝑁 (independent of 𝜀) such that for ‖𝜀‖=𝜀²₁+⋯+𝜀²_𝑛 being small enough, 𝐹(𝑥,𝑡,𝜀)=_{|𝛼|≤𝑁}𝐹_𝛼(𝑥,𝑡)𝜀^𝛼+𝑂‖𝜀‖^𝑁+1; (5) it means that 𝐹−_{|𝛼|≤𝑁}𝐹_𝛼𝜀^𝛼_{𝐿∞(0,𝑇,𝐿2(0,1))}≤𝐶_𝑁‖𝜀‖^𝑁+1, (6) where the constant 𝐶_𝑁 is independent from 𝜀.
By Taylor’s expansion for 𝐹(𝑥,𝑡,𝜀), it implies that 𝐹_𝛼()=1𝑥,𝑡𝐷𝛼!^𝛼𝐹(𝑥,𝑡,0),|𝛼|≤𝑁. (7)

However, for 𝐹(𝑥,𝑡,𝜀) which is given in (3), it is very difficult to calculate 𝐷^𝛼𝐹(𝑥,𝑡,0), and so, the Problem 1 cannot proceed. Whichmethod 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 viewsuggests thatwe need to construct the Taylor-Maclaurin expansion of the composite function. So, first in Section 2 we solve the following problem.

Problem 2. Let Ω be an open subset of ℝ^𝑛and 0∈Ω. Let𝑔=(𝑔₁,…,𝑔_𝑝)∈𝐶^𝑁+1(Ω;ℝ^𝑝)and𝑓∈𝐶^𝑁+1(ℝ^𝑝;ℝ).Seek the representation formula for 𝑓𝑜𝑔, such that for ‖𝑥‖ being small enough, 𝑓𝑜𝑔(𝑥)=𝑓𝑔₁(𝑥),…,𝑔_𝑝(𝑥)=_{|𝛼|≤𝑁}𝑑_𝛼𝑥^𝛼+𝑂‖𝑥‖^𝑁+1, (8) where 𝑑_𝛼,|𝛼|≤𝑁 are calculated from the values of the given functions 𝑓;𝑔₁,…,𝑔_𝑝and 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].

2. Solving the Problem 2

We use the following notations. For a multi-index 𝛼=(𝛼₁,…,𝛼_𝑛)∈ℤ^𝑛₊ and 𝑥=(𝑥₁,…,𝑥_𝑛)∈ℝ^𝑛, we put |𝛼|=𝛼₁+⋯+𝛼_𝑛,𝛼!=𝛼₁!⋯𝛼_𝑛‖𝑥‖=!,𝑥²₁+⋯+𝑥²_𝑛,𝑥^𝛼=𝑥^𝛼1₁⋯𝑥^𝛼𝑛_𝑛,𝛼,𝛽∈ℤ^𝑛₊,𝛼≤𝛽⟺𝛼_𝑖≤𝛽_𝑖∀𝑖=1,…,𝑛. (9)

The following lemma is useful to solve the Problems 1 and 2.

Lemma 3. Let 𝑚,𝑁∈ℕ and 𝑎_𝛼∈ℝ,𝛼∈ℤ^𝑛₊,1≤|𝛼|≤𝑁. Then _{|𝛼|1≤≤𝑁}𝑎_𝛼𝑥^𝛼𝑚=_{|𝛼|𝑚≤≤𝑚𝑁}𝑇_𝑁^(𝑚)[𝑎]_𝛼𝑥^𝛼, (10) where the coefficients 𝑇_𝑁^(𝑚)[𝑎]_𝛼,𝑚≤|𝛼|≤𝑚𝑁 depending on 𝑎=(𝑎_𝛼),𝛼∈ℤ^𝑛₊,1≤|𝛼|≤𝑁 are defined by the recurrent formulas 𝑇_𝑁^(𝑚)[𝑎]_𝛼=𝑢_𝛼|𝛼|∑,1≤≤𝑁,𝑚=1,_{𝛽∈𝐴𝛼(𝑚)(𝑁)}𝑎_𝛼−𝛽𝑇_𝑁^(𝑚−1)[𝑢]_𝛽|𝛼|𝐴,𝑚≤≤𝑚𝑁,𝑚≥2,_𝛼(𝑚)(𝑁)=𝛽∈ℤ^𝑝₊𝛽≤()𝑁.∶𝛽≤𝛼,1≤𝛼−𝛽≤𝑁,𝑚−1≤𝑚−1 (11)

The proof of Lemma 3 can be found in [2].

Now, using Taylor’s expansion of the function 𝑓around the point 𝑔=(𝑔₁,…,𝑔_𝑝)∈ℝ^𝑝, we obtain that 𝑓𝑔₁+𝐻₁,…,𝑔_𝑝+𝐻_𝑝=𝑓𝑔₁,…,𝑔_𝑝+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔₁,…,𝑔_𝑝𝐻^𝛼+𝑅_𝑁,𝐻𝑓,𝐻𝐻=₁,…,𝐻_𝑝∈ℝ^𝑝,‖𝐻‖𝑅issmallenough,_𝑁𝑓,𝐻=(𝑁+1)×_{|𝛼|=𝑁+1}1𝐻𝛼!^𝛼1₀(1−𝜃)^𝑁𝐷^𝛼×𝑓𝑔+𝜃𝐻𝑑𝜃. (12)

Similarly, we use Maclaurin’s expansion of 𝑔_𝑖 as follows: 𝐻_𝑖=𝑔_𝑖(𝑥)−𝑔_𝑖(0)=_{1≤|𝛽|≤𝑁}1𝐷𝛽!^𝛽𝑔_𝑖(0)𝑥^𝛽+𝑅_𝑁𝑔_𝑖,𝑥=𝑃_𝑖+𝑅_𝑖,𝑥∈ℝ^𝑛,‖𝑥‖𝑃issmallenough,_𝑖=_{1≤|𝛽|≤𝑁}1𝐷𝛽!^𝛽𝑔_𝑖(0)𝑥^𝛽,𝑅_𝑖=𝑅_𝑁𝑔_𝑖=(),𝑥𝑁+1_{|𝛽|=𝑁+1}1𝑥𝛽!^𝛽1₀(1−𝜃)^𝑁𝐷^𝛽𝑔_𝑖(𝜃𝑥)𝑑𝜃. (13)

Substituting (13) into (12), we get 𝑓𝑔₁(𝑥),…,𝑔_𝑝(𝑥𝑔)=𝑓₁(0),…,𝑔_𝑝(0)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔₁(0),…,𝑔_𝑝(0)𝐻^𝛼+𝑅_𝑁𝑔(0𝑓,𝐻=𝑓)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)𝑃^𝛼+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)𝐻^𝛼−𝑃^𝛼+𝑅_𝑁𝑔(0𝑓,𝐻=𝑓)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)𝑃^𝛼+‖𝑥‖^𝑁+1𝑅_𝑁⁽¹⁾𝑓,𝑔,𝐻,𝑅, (14) where ‖𝑥‖^𝑁+1𝑅_𝑁⁽¹⁾=𝑓,𝑔,𝐻,𝑅_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)𝐻^𝛼−𝑃^𝛼+𝑅_𝑁,𝑅𝑓,𝐻_𝑁=()𝑓,𝐻𝑁+1_{|𝛼|=𝑁+1}1𝐻𝛼!^𝛼1₀(1−𝜃)^𝑁𝐷^𝛼×𝑓𝑔(0)+𝜃𝐻𝑑𝜃. (15)

Applying Lemma 3, with 𝑎=𝜎^(𝑖)=(𝜎_𝛽^(𝑖)),𝜎_𝛽^(𝑖)=(1/𝛽!)𝐷^𝛽𝑔_𝑖(0),1≤|𝛽|≤𝑁, it implies that 𝑃^𝛼𝑖_𝑖=_{1≤|𝛽|≤𝑁}1𝐷𝛽!^𝛽𝑔_𝑖(0)𝑥^𝛽𝛼𝑖≡_{1≤|𝛽|≤𝑁}𝜎_𝛽^(𝑖)𝑥^𝛽𝛼𝑖=_{𝛼𝑖≤|𝛾𝑖|≤𝛼𝑖𝑁}𝑇^(𝛼1)_𝑁𝜎^(𝑖)_𝛾𝑖𝑥^𝛾𝑖. (16)

Hence, 𝑃^𝛼=𝑃^𝛼1₁⋯𝑃^𝛼𝑝_𝑝=_{1≤|𝛽|≤𝑁}1𝐷𝛽!^𝛽𝑔₁(0)𝑥^𝛽𝛼1⋯_{1≤|𝛽|≤𝑁}1𝐷𝛽!^𝛽𝑔_𝑝(0)𝑥^𝛽𝛼𝑝=_{𝛼1≤|𝛾1|≤𝛼1𝑁}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1𝑥^𝛾1⋯_{𝛼𝑝≤|𝛾𝑝|≤𝛼𝑝𝑁}𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝𝑥^𝛾𝑝=_{𝛼1≤|𝛾1|≤𝛼1⋮𝛼𝑁,𝑝≤|𝛾𝑝|≤𝛼𝑝𝑁}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝𝑥^{𝛾1+⋯+𝛾𝑝}=_{𝛼1≤|𝛾1|≤𝛼1⋮𝛼𝑁,𝑝≤|𝛾𝑝|≤𝛼𝑝𝑁}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝𝑥^𝛿=_{𝛼1≤𝛾1≤𝛼1⋮𝛼𝑁,𝑝≤|𝛾𝑝|≤𝛼𝑝𝑁𝛾1+⋯+𝛾𝑝=𝛿}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝𝑥^𝛿=_{|𝛼|≤|𝛿|≤|𝛼|𝑁𝛼1≤|𝛾1|≤𝛼1⋮𝛼𝑁,𝑝≤|𝛾𝑝|≤𝛼𝑝𝑁𝛾1+⋯+𝛾𝑝=𝛿}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝𝑥^𝛿≡_{|𝛼|≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿=_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+_{𝑁+1≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿,Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)=_{𝛼1≤𝛾1≤𝛼1⋮𝛼𝑁,𝑝≤𝛾𝑝≤𝛼𝑝𝑁𝛾1+⋯+𝛾𝑝=𝛿}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼𝑝)_𝑁𝜎^(𝑝)_𝛾𝑝. (17)

Consequently, 𝑓𝑔(𝑥)=𝑓𝑔(0)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)𝑃^𝛼+‖𝑥‖^𝑁+1𝑅_𝑁⁽¹⁾𝑔(0𝑓,𝑔,𝐻,𝑅=𝑓)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)×_{|𝛼|≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+‖𝑥‖^𝑁+1𝑅_𝑁⁽¹⁾𝑔(0)𝑓,𝑔,𝐻,𝑅=𝑓+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)×_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)×_{𝑁+1≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+‖𝑥‖^𝑁+1𝑅_𝑁⁽¹⁾𝑔(0)𝑓,𝑔,𝐻,𝑅=𝑓+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑔(0)×_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+‖𝑥‖^𝑁+1𝑅_𝑁⁽²⁾𝑔(0)𝑓,𝑔,𝐻,𝑅,𝑥=𝑓+_{|𝛿|1≤≤𝑁1≤|𝛼|≤|𝛿|}1𝐷𝛼!^𝛼𝑓𝑔(0)×Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝)𝑥^𝛿+‖𝑥‖^𝑁+1𝑅_𝑁⁽²⁾𝑔(0)𝑓,𝑔,𝐻,𝑅,𝑥=𝑓+_{|𝛿|1≤≤𝑁}𝑑_𝛿𝑥^𝛿+‖𝑥‖^𝑁+1𝑅_𝑁⁽²⁾𝑓,𝑔,𝐻,𝑅,𝑥. (18)

Clearly, the Problem 2 is solved with 𝑑_𝛿=_{1≤|𝛼|≤|𝛿|}1𝐷𝛼!^𝛼𝑓𝑔(0)Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎^(𝑝). (19)

3. Solving the Problem 1

As an application of the method used in Section 2 for 𝑝=6,Ω=(0,1),𝜀⃗=(𝜀₁,…,𝜀_𝑛)∈ℝ^𝑛, and 𝑓=𝑓(𝑥,𝑡,𝑔₁,…,𝑔₆)∈𝐶^𝑁+1(ℝ³×ℝ³₊;ℝ), for eachfixed (𝑥,𝑡)∈[0,1]×ℝ₊;𝑔_𝑖∈𝐶^𝑁+1(Ω;ℝ), 𝑖=1,2,3, for eachfixed (𝑥,𝑡)∈[0,1]×ℝ₊∶𝑔_𝑖∈𝐶^𝑁+1(Ω;ℝ₊),𝑖=4,5,6, for eachfixed 𝑡∈ℝ₊, we now investigate the Problem 1.

For eachfixed (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 𝑓𝑥,𝑡,𝑔+𝐻=𝑓𝑥,𝑡,𝑔₁+𝐻₁,…,𝑔₆+𝐻₆=𝑓𝑥,𝑡,𝑔₁,…,𝑔₆+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔₁,…,𝑔₆𝐻^𝛼+‖𝐻‖^𝑁+1𝑅_𝑁⁽¹⁾,‖𝐻‖𝑓,𝐻_𝑁+1𝑅_𝑁⁽¹⁾=()𝑓,𝐻𝑁+1_{|𝛼|=𝑁+1}1𝐻𝛼!^𝛼1₀(1−𝜃)^𝑁𝐷^𝛼×𝑓𝑔+𝜃𝐻𝑑𝜃, (20)(21) where 𝐻=(𝐻₁,…,𝐻₆)∈ℝ⁶,‖𝐻‖ is small enough, 𝐷^𝛼𝑖_𝑖𝑓=𝜕𝑓/𝜕𝑔_𝑖,𝑖=1,2,…,6.

Next, the precise structure of the representation formulas for 𝑔₁,…,𝑔₆will be needed below to continue.

For each fixed (𝑥,𝑡)∈[0,1]×ℝ₊∶ we have the following.

(i) The representation formula for 𝑔_𝑖(𝜀)=𝑔_𝑖(𝑥,𝑡,𝜀),𝑖=1,2,3.

We rewrite 𝑔₁(𝜀) as follows 𝑔₁(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼(𝑥,𝑡)𝜀^𝛼≡𝑔₁(0)+_{1≤|𝛽|≤𝑁}𝑔_𝛽^[1](𝑥,𝑡)𝜀^𝛽≡𝑔₁(0)+𝐻₁, (22) in which 𝑔₁(0)=𝑔₁(𝑥,𝑡)=𝑢₀(),𝛽𝑔𝑥,𝑡=0,_𝛽[1](𝑥,𝑡)=𝑢_𝛽(𝑥,𝑡),1≤𝛽≤𝑁. (23)

It is similar to 𝑔₂(𝜀),𝑔₃(𝜀); we write 𝑔₂(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑥(𝑥,𝑡)𝜀^𝛼≡𝑔₂(0)+_{1≤|𝛽|≤𝑁}𝑔_𝛽^[2](𝑥,𝑡)𝜀^𝛽≡𝑔₂(0)+𝐻₂,𝑔₃(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑡(𝑥,𝑡)𝜀^𝛼≡𝑔₃(0)+_{1≤|𝛽|≤𝑁}𝑔_𝛽^[3](𝑥,𝑡)𝜀^𝛽≡𝑔₃(0)+𝐻₃, (24) where 𝑔₂(0)=𝑔₂(𝑥,𝑡)=𝑢_0𝑥(),𝑔𝑥,𝑡₃(0)=𝑔₃(𝑥,𝑡)=𝑢_0𝑡(),𝛽𝑔𝑥,𝑡=0,_𝛽[2](𝑥,𝑡)=𝑢_𝛽𝑥(),𝑔𝑥,𝑡_𝛽[3](𝑥,𝑡)=𝑢_𝛽𝑡(𝑥,𝑡),1≤𝛽≤𝑁. (25)

For eachfixed 𝑡∈ℝ₊∶ we have the following.

(ii) The representation formula for 𝑔_𝑖(𝜀)=𝑔_𝑖(𝑡,𝜀),𝑖=4,5,6.

Write 𝑔₄(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼(𝑡)𝜀^𝛼2=_{|𝛼|≤𝑁|𝛽|≤𝑁}𝑢_𝛼(𝑡),𝑢_𝛽(𝑡)𝜀^𝛼+𝛽=𝑢₀(𝑡)²+_{|𝛿|1≤≤𝑁}𝐺_𝛿[𝑢(𝑡)]𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝑢,𝜀]≡𝑔₄(0)+_{𝛽1≤≤𝑁}𝑔_𝛽^[4](𝑡)𝜀^𝛽+𝑅₄≡𝑔₄(0)+𝐻₄≡𝑔₄(0)+𝑃₄+𝑅₄,𝑔₄(0)=𝑢₀(𝑡)²,𝑔_𝛿^[4](𝑡)=𝐺_𝛿[𝑢(𝑡)]=_{|𝛼|≤𝑁𝛽≤𝑁𝛼+𝛽=𝛿}𝑢_𝛼(𝑡),𝑢_𝛽(𝑡,|𝛿|𝐺)1≤≤𝑁,_𝛿[𝑢(𝑡)]=_{|𝛼|≤𝑁𝛽≤𝑁𝛼+𝛽=𝛿}𝑢_𝛼(𝑡),𝑢_𝛽(𝑡,|𝛿|𝑅)1≤≤2𝑁,₄=‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝑢,𝜀]=_{|𝛿|𝑁+1≤≤2𝑁}𝐺_𝛿[𝑢(𝑡)]𝜀^𝛿. (26)(27)

Similarly, 𝑔₅(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑥(𝑡)𝜀^𝛼2=_{|𝛼|≤𝑁𝛽≤𝑁}𝑢_𝛼𝑥(𝑡),𝑢_𝛽𝑥(𝑡)𝜀^𝛼+𝛽=𝑢_0𝑥(𝑡)²+_{|𝛿|1≤≤𝑁}𝐺_𝛿𝑢_𝑥(𝑡)𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾𝑢_𝑥,𝜀≡𝑔₅(0)+_{1≤|𝛽|≤𝑁}𝑔_𝛽^[5](𝑡)𝜀^𝛽+𝑅₅≡𝑔₅(0)+𝐻₅≡𝑔₅(0)+𝑃₅+𝑅₅, (28) in which 𝑔₅(0)=𝑢_0𝑥(𝑡)²,𝑔_𝛿^[5](𝑡)=𝐺_𝛿𝑢_𝑥(𝑡=_{|𝛼|≤𝑁𝛽≤𝑁})_{𝛼+𝛽=𝛿}𝑢_𝛼𝑥(𝑡),𝑢_𝛽𝑥(𝑡,|𝛿|𝑅)1≤≤𝑁,₅=‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾𝑢_𝑥,𝜀=_{|𝛿|𝑁+1≤≤2𝑁}𝐺_𝛿𝑢_𝑥(𝑡)𝜀^𝛿,𝑔₆(𝜀)=_{|𝛼|≤𝑁}𝑢_𝛼𝑡(𝑡)𝜀^𝛼2=_{|𝛼|≤𝑁𝛽≤𝑁}𝑢_𝛼𝑡(𝑡),𝑢_𝛽𝑡(𝑡)𝜀^𝛼+𝛽=𝑢_0𝑡(𝑡)²+_{|𝛿|1≤≤𝑁}𝐺_𝛿𝑢_𝑡(𝑡)𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾𝑢_𝑡,𝜀≡𝑔₆(0)+_{𝛽1≤≤𝑁}𝑔_𝛽^[6](𝑡)𝜀^𝛽+𝑅₆≡𝑔₆(0)+𝐻₆≡𝑔₆(0)+𝑃₆+𝑅₆, (29)(30) where 𝑔₆(0)=𝑢_0𝑡(𝑡)²,𝑔_𝛿^[6](𝑡)=𝐺_𝛿𝑢_𝑡(𝑡=_{|𝛼|≤𝑁𝛽≤})_{𝑁𝛼+𝛽=𝛿}𝑢_𝛼𝑡(𝑡),𝑢_𝛽𝑡(𝑡,|𝛿|𝑅)1≤≤𝑁,₆=‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾𝑢_𝑡,𝜀=_{|𝛿|𝑁+1≤≤2𝑁}𝐺_𝛿𝑢_𝑡(𝑡)𝜀^𝛿. (31)

Then(22), (24), (26), (28), and (30) imply that 𝑔_𝑖(𝜀)−𝑔_𝑖(0)=𝐻_𝑖=𝑃_𝑖+𝑅_𝑖=_{𝛽1≤≤𝑁}𝑔_𝛽^[𝑖]𝜀^𝛽+𝑅_𝑖,𝑅_𝑖=0,𝑖=1,2,3,𝑅_𝑗=𝑂‖𝜀‖^𝑁+1,𝑗=4,5,6. (32)

We also need the following lemma.

Lemma 4. For all 𝛼=(𝛼₁,…,𝛼_𝑛)∈ℤ^𝑛₊,|𝛼|≥1, then 𝐻^𝛼=𝐻^𝛼1₁𝐻^𝛼2₂𝐻^𝛼3₃𝐻^𝛼4₄𝐻^𝛼5₅𝐻^𝛼6₆=_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+𝑂‖𝜀‖^𝑁+1, (33) where Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾=_{𝛼1≤|𝛾1|≤𝛼1⋮𝛼𝑁,6≤|𝛾6|≤𝛼6𝑁𝛾1+⋯+𝛾6=𝛿}𝑇^(𝛼1)_𝑁𝜎⁽¹⁾_𝛾1⋯𝑇^(𝛼6)_𝑁𝜎⁽⁶⁾_𝛾6,𝜎^(𝑖)=𝜎_𝛽^(𝑖),𝜎_𝛽^(𝑖)=𝑔_𝛽^[𝑖],1≤𝛽≤𝑁,𝑖=1,2,…,6. (34)

Proof of Lemma 4. We have 𝐻^𝛼=𝐻^𝛼1₁𝐻^𝛼2₂𝐻^𝛼3₃𝐻^𝛼4₄𝐻^𝛼5₅𝐻^𝛼6₆=𝑃^𝛼1₁𝑃^𝛼2₂𝑃^𝛼3₃𝑃₄+𝑅₄^𝛼4𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6=𝑃^𝛼1₁𝑃^𝛼2₂𝑃^𝛼3₃𝑃^𝛼4₄𝑃^𝛼5₅𝑃^𝛼6₆+𝑃^𝛼1₁𝑃^𝛼2₂𝑃^𝛼3₃×𝑃₄+𝑅₄^𝛼4𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6−𝑃^𝛼4₄𝑃^𝛼5₅𝑃^𝛼6₆. (35)
Note that 𝑅_𝑗=𝑂(‖𝜀‖^𝑁+1),𝑗=4,5,6, so 𝑃₄+𝑅₄^𝛼4𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6−𝑃^𝛼4₄𝑃^𝛼5₅𝑃^𝛼6₆=𝑃₄+𝑅₄^𝛼4𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6−𝑃^𝛼4₄𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6+𝑃^𝛼4₄𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6−𝑃^𝛼4₄𝑃^𝛼5₅𝑃₆+𝑅₆^𝛼6+𝑃^𝛼4₄𝑃^𝛼5₅𝑃₆+𝑅₆^𝛼6−𝑃^𝛼4₄𝑃^𝛼5₅𝑃^𝛼6₆=𝑃₄+𝑅₄^𝛼4−𝑃^𝛼4₄𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6+𝑃^𝛼4₄𝑃₅+𝑅₅^𝛼5−𝑃^𝛼5₅𝑃₆+𝑅₆^𝛼6+𝑃^𝛼4₄𝑃^𝛼5₅𝑃₆+𝑅₆^𝛼6−𝑃^𝛼6₆=𝑅₄^𝛼4−1_𝑗=0𝑃₄+𝑅₄^𝑗𝑃^𝛼4₄^−1−𝑗×𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6+𝑅₅^𝛼5−1_𝑗=0𝑃₅+𝑅₅^𝑗𝑃^𝛼5₅^−1−𝑗𝑃^𝛼4₄𝑃₆+𝑅₆^𝛼6+𝑅₆^𝛼6−1_𝑗=0𝑃₆+𝑅₆^𝑗𝑃^𝛼6₆^−1−𝑗𝑃^𝛼4₄𝑃^𝛼5₅=‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝐻,𝛼,𝜀]=𝑂‖𝜀‖^𝑁+1. (36)
Combining (35) and (36) yields 𝐻^𝛼=𝐻^𝛼1₁𝐻^𝛼2₂𝐻^𝛼3₃𝐻^𝛼4₄𝐻^𝛼5₅𝐻^𝛼6₆=𝑃^𝛼1₁𝑃^𝛼2₂𝑃^𝛼3₃𝑃₄+𝑅₄^𝛼4𝑃₅+𝑅₅^𝛼5𝑃₆+𝑅₆^𝛼6=𝑃^𝛼1₁𝑃^𝛼2₂𝑃^𝛼3₃𝑃^𝛼4₄𝑃^𝛼5₅𝑃^𝛼6₆+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[]=𝐻,𝛼,𝜀_{𝛽1≤≤𝑁}𝑔_𝛽⁽¹⁾𝜀^𝛽𝛼1⋯_{𝛽1≤≤𝑁}𝑔_𝛽⁽⁶⁾𝜀^𝛽𝛼6+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝐻,𝛼,𝜀]=_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+_{𝑁+1≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝐻,𝛼,𝜀]≡_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽²⁾[𝐻,𝛼,𝜀], (37) where ‖𝜀‖^𝑁+1𝑅_𝑁⁽²⁾[𝐻,𝛼,𝜀]=_{𝑁+1≤|𝛿|≤|𝛼|𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽¹⁾[𝐻,𝛼,𝜀]. (38)
Lemma 4 is proved.

Substituting (37) into (20), we obtain 𝐹𝑥,𝑡,𝜀^𝛼=𝑓𝑥,𝑡,ℎ,ℎ_𝑥,ℎ_𝑡,‖ℎ‖²,ℎ_𝑥²,ℎ_𝑡²=𝑓𝑥,𝑡,𝑔+𝐻=𝑓𝑥,𝑡,𝑔(0)+𝐻=𝑓𝑥,𝑡,𝑔₁(0)+𝐻₁,…,𝑔₆(0)+𝐻₆=𝑓𝑥,𝑡,𝑔(0)+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)𝐻^𝛼+‖𝐻‖^𝑁+1𝑅_𝑁⁽¹⁾(0)𝑓,𝐻=𝑓𝑥,𝑡,𝑔+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓(0)×𝑥,𝑡,𝑔_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽²⁾[]+‖𝐻‖𝐻,𝛼,𝜀_𝑁+1𝑅_𝑁⁽¹⁾(0)𝑓,𝐻=𝑓𝑥,𝑡,𝑔+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓(0)×𝑥,𝑡,𝑔_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)𝑅_𝑁⁽²⁾[]+‖𝐻‖𝐻,𝛼,𝜀_𝑁+1𝑅_𝑁⁽¹⁾(0)𝑓,𝐻=𝑓𝑥,𝑡,𝑔+_{|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)×_{|𝛼|≤|𝛿|≤𝑁}Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽³⁾(0)𝑓,𝑔,𝐻,𝜀=𝑓𝑥,𝑡,𝑔+_{1≤|𝛿|≤𝑁|𝛼|≤|𝛿|}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)Φ_𝛿×𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽³⁾𝜀⃗(0)𝑓,𝑔,𝐻,=𝑓𝑥,𝑡,𝑔+_{1≤|𝛿|≤𝑁}𝐹_𝛿(𝑥,𝑡)𝜀^𝛿+‖𝜀‖^𝑁+1𝑅_𝑁⁽³⁾𝑓,𝑔,𝐻,𝜀, (39) where ‖𝜀‖^𝑁+1𝑅_𝑁⁽³⁾=‖𝜀‖𝑓,𝑔,𝐻,𝜀_{𝑁+1|𝛼|1≤≤𝑁}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)𝑅_𝑁⁽²⁾[]+‖𝐻‖𝐻,𝛼,𝜀_𝑁+1𝑅_𝑁⁽¹⁾,𝐹𝑓,𝐻_𝛿(𝑥,𝑡)=_{|𝛼|≤|𝛿|}1𝐷𝛼!^𝛼𝑓𝑥,𝑡,𝑔(0)Φ_𝛿𝛼,𝑁,𝜎⁽¹⁾,…,𝜎⁽⁶⁾,𝛿∈ℤ⁶₊,1≤|𝛿|≤𝑁. (40)

Since 𝑢_𝛼∈𝑊={𝑣∈𝐿^∞(0,𝑇,𝐻²(0,1))∶𝑣_𝑡∈𝐿^∞(0,𝑇,𝐻¹(0,1))},𝛼∈ℤ^𝑛₊,|𝛼|≤𝑁, hence 𝑢_𝛼,𝑢_𝛼𝑥,𝑢_𝛼𝑡∈𝐿^∞(0,𝑇,𝐻¹(0,1))⊂𝐿^∞(𝑄_𝑇). Then from (39) and (40)₁, we get 𝐹(𝜀)−_{|𝛿|≤𝑁}𝐹_𝛿𝜀^𝛿_{𝐿∞(0,𝑇,𝐿2(0,1))}=‖𝜀‖^𝑁+1𝑅_𝑁⁽³⁾𝑓,𝑔,𝐻,𝜀_{𝐿∞(0,𝑇,𝐿2(0,1))}≤𝐶_𝑁‖𝜀‖^𝑁+1, (41) for ‖𝜀‖ is small enough; consequently the Problem 1 is solved.

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,‖ℎ_𝑥‖²), where 𝜇∈𝐶^𝑁+1([0,1]×ℝ₊×ℝ×ℝ₊;ℝ),𝑛=𝑝, this result is also obtained in [3].

Acknowledgments

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. B2013-18-05.