Approximation of Bivariate Functions via Smooth Extensions

Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 102062, 16 pages http://dx.doi.org/10.1155/2014/102062

Research Article Approximation of Bivariate Functions via Smooth Extensions Zhihua Zhang College of Global Change and Earth System Science, Beijing Normal University, Beijing 100875, China Correspondence should be addressed to Zhihua Zhang; [email protected] Received 10 December 2013; Accepted 26 December 2013; Published 10 February 2014 Academic Editors: N. I. Mahmudov and S. Wang Copyright © 2014 Zhihua Zhang. 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. For a smooth bivariate function defined on a general domain with arbitrary shape, it is difficult to do Fourier approximation or wavelet approximation. In order to solve these problems, in this paper, we give an extension of the bivariate function on a general domain with arbitrary shape to a smooth, periodic function in the whole space or to a smooth, compactly supported function in the whole space. These smooth extensions have simple and clear representations which are determined by this bivariate function and some polynomials. After that, we expand the smooth, periodic function into a Fourier series or a periodic wavelet series or we expand the smooth, compactly supported function into a wavelet series. Since our extensions are smooth, the obtained Fourier coefficients or wavelet coefficients decay very fast. Since our extension tools are polynomials, the moment theorem shows that a lot of wavelet coefficients vanish. From this, with the help of well-known approximation theorems, using our extension methods, the Fourier approximation and the wavelet approximation of the bivariate function on the general domain with small error are obtained.

1. Introduction In the recent several decades, various approximation tools have been widely developed [1–14]. For example, a smooth periodic function can be approximated by trigonometric polynomials; a square-integrable smooth function can be expanded into a wavelet series and be approximated by partial sum of the wavelet series; and a smooth function on a cube can be approximated well by polynomials. However, for a smooth function on a general domain with arbitrary shape, even if it is infinitely many time differentiable, it is difficult to do Fourier approximation or wavelet approximation. In this paper, we will extend a function on general domain with arbitrary shape to a smooth, periodic function in the whole space or to a smooth, compactly supported function in the whole space. After that, it will be easy to do Fourier approximation or wavelet approximation. For the higherdimensional case, the method of smooth extensions is similar to that in the two-dimensional case, but the representations of smooth extensions will be too complicated. Therefore, in

this paper, we mainly consider the smooth extension of a bivariate function on a planar domain. By the way, for the one-dimensional case, since the bounded domain is reduced to a closed interval, the smooth extension can be regarded as a corollary of the two-dimensional case. This paper is organized as follows. In Section 2, we state the main theorems on the smooth extension of the function on the general domain and their applications. In Sections 3 and 4, we give a general method of smooth extensions and complete the proofs of the main theorems. In Section 5, we use our extension method to discuss two important special cases of smooth extensions. Throughout this paper, we denote 𝑇 = [0, 1]2 and the interior of 𝑇 by 𝑇𝑜 and always assume that Ω is a simply connected domain. We say that 𝑓 ∈ 𝐶𝑞 (Ω) if the derivatives (𝜕𝑖+𝑗 𝑓/𝜕𝑥𝑖 𝜕𝑦𝑗 ) are continuous on Ω for 0 ≤ 𝑖 + 𝑗 ≤ 𝑞. We say that 𝑓 ∈ 𝐶∞ (Ω) if all derivatives (𝜕𝑖+𝑗 𝑓/𝜕𝑥𝑖 𝜕𝑦𝑗 ) are continuous on Ω for 𝑖, 𝑗 ∈ Z+ . We say that a function ℎ(𝑥, 𝑦) is a 𝛾-periodic function if ℎ(𝑥 + 𝛾𝑘, 𝑦 + 𝛾𝑙) = ℎ(𝑥, 𝑦) ((𝑥, 𝑦) ∈

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𝑇; 𝑘, 𝑙 ∈ Z), where 𝛾 is an integer. We appoint that 0! = 1 and the notation [𝛼] is the integral part of the real number 𝛼.

2. Main Theorems and Applications In this section, we state the main results of smooth extensions and their applications in Fourier analysis and wavelet analysis. 2.1. Main Theorems. Our main theorems are stated as follows.

2.2.2. Fourier Analysis (i) Approximation by Trigonometric Polynomials. Let 𝐹𝑝 be the smooth periodic extension of 𝑓 as in Theorem 2. Then 𝐹𝑝 ∈ 𝐶𝑟 (R2 ) and 𝐹𝑝 = 𝑓 on Ω. By the well-known results [5, 10], we know that the smooth periodic function 𝐹𝑝 can be approximated by bivariate trigonometric polynomials very well. Its approximation error can be estimated by the modulus of continuity of its 𝑟 time derivatives. By Δ∗𝑁, denote the set of all bivariate trigonometric polynomials in the form

Theorem 1. Let 𝑓 ∈ 𝐶∞ (Ω), where Ω ⊂ 𝑇𝑜 and the boundary 𝜕Ω is a piecewise infinitely many time smooth curve. Then for any 𝑟 ∈ Z+ there is a function 𝐹 ∈ 𝐶𝑟 (𝑇) such that (i) 𝐹(𝑥, 𝑦) = 𝑓(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Ω); 𝑖+𝑗

𝑖

𝑗

(ii) (𝜕 𝐹/𝜕𝑥 𝜕𝑦 )(𝑥, 𝑦) = 0 on the boundary 𝜕𝑇 for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟; (iii) on the complement 𝑇\Ω, 𝐹(𝑥, 𝑦) can be expressed locally in the forms 𝐿

∑ 𝜉𝑗 (𝑥) 𝑦𝑗 ,

𝑗=0

𝐿

𝑜𝑟 ∑ 𝜂𝑗 (𝑦) 𝑥𝑗 , 𝑗=0

𝐿

𝑜𝑟 ∑ 𝑐𝑖𝑗 𝑥𝑖 𝑦𝑗 ,

(1)

𝑖,𝑗=0

where 𝐿 is a positive integer and each coefficient 𝑐𝑖𝑗 is constant. ∞

Theorem 2. Let 𝑓 ∈ 𝐶 (Ω), where Ω is stated as in Theorem 1. Then, for any 𝑟 ∈ Z+ , there exists a 1-periodic function 𝐹𝑝 ∈ 𝐶𝑟 (R2 ) such that 𝐹𝑝 (𝑥, 𝑦) = 𝑓(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Ω). Theorem 3. Let 𝑓 ∈ 𝐶∞ (Ω), where Ω is stated as in Theorem 1. Then, for any 𝑟 ∈ Z+ , there exists a function 𝐹𝑐 ∈ 𝐶𝑟 (R2 ) with compact support 𝑇 such that 𝐹𝑐 (𝑥, 𝑦) = 𝑓(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Ω). In Sections 3 and 4, we give constructive proofs of Theorems 1–3. In these three theorems, we assume that 𝑓 ∈ 𝐶∞ (Ω). If 𝑓 ∈ 𝐶𝑞 (Ω) (𝑞 is a nonnegative integer), by using the similar method of arguments of Theorems 1–3, we also can obtain the corresponding results. 2.2. Applications. Here we show some applications of these theorems. 2.2.1. Approximation by Polynomials. Let 𝐹 be the smooth extension of 𝑓 from Ω to 𝑇 which is stated as in Theorem 1. Then 𝐹 ∈ 𝐶𝑟 (𝑇) and 𝐹 = 𝑓 on Ω. By Δ 𝑁, denote the set of all 𝑛1 𝑛2 bivariate polynomials in the form ∑𝑁 𝑛1 ,𝑛2 =−𝑁 𝑐𝑛1 ,𝑛2 𝑥 𝑦 . Then 󵄩 󵄩 inf 󵄩󵄩𝑓 − 𝑃󵄩󵄩󵄩𝐿 𝑝 (Ω) ≤ inf ‖𝐹 − 𝑃‖𝐿 𝑝 (𝑇) ,

𝑃∈Δ 𝑁 󵄩

𝑃∈Δ 𝑁

(2)

where ‖ ⋅ ‖𝐿 𝑝 (𝐷) is the norm of the space 𝐿 𝑝 (𝐷). The righthand side of formula (2) is the best approximation of the extension 𝐹 in Δ 𝑁. By (2), we know that the approximation problem of 𝑓 by polynomials on a domain Ω is reduced to the well-known approximation problem of its smooth extension 𝐹 by polynomials on the square 𝑇 [4, 10].

𝑁

∑ 𝑐𝑛∗1 ,𝑛2 𝑒2𝜋𝑖(𝑛1 𝑥+𝑛2 𝑦) .

(3)

𝑛1 ,𝑛2 =−𝑁

By Theorem 2, we have 󵄩 󵄩 inf 󵄩󵄩𝑓 − 𝑃∗ 󵄩󵄩󵄩 𝑃∗ ∈Δ∗ 󵄩 𝑁

𝐿 𝑝󸀠 (Ω)

󵄩󵄩 󵄩 󵄩󵄩𝐹𝑝 − 𝑃∗ 󵄩󵄩󵄩 ≤ min ∗ ∗ 󵄩 󵄩 𝑃 ∈Δ 𝑁

. 𝐿 𝑝󸀠 (𝑇)

(4)

From this and Theorem 2, we see that the approximation problem of 𝑓 on Ω by trigonometric polynomials is reduced to a well-known approximation problem of smooth periodic functions [5, 7, 10]. (ii) Fourier Series. We expand 𝐹𝑝 into a Fourier series [9] 𝐹𝑝 (𝑥, 𝑦) =

∑ 𝜏𝑛1 ,𝑛2 𝑒2𝜋𝑖(𝑛1 𝑥+𝑛2 𝑦) , (𝑛1 ,𝑛2 )∈Z2

(5)

where 𝜏𝑛1 ,𝑛2 = ∫𝑇 𝐹𝑝 (𝑥, 𝑦)𝑒−2𝜋𝑖(𝑛1 𝑥+𝑛2 𝑦) 𝑑𝑥 𝑑𝑦. By Theorem 2, we obtain that, for (𝑥, 𝑦) ∈ Ω, 𝑓 (𝑥, 𝑦) =

∑ 𝜏𝑛1 ,𝑛2 𝑒2𝜋𝑖(𝑛1 𝑥+𝑛2 𝑦) . (𝑛1 ,𝑛2 )∈Z2

(6)

Denote the partial sum 𝑛1

𝑛2

𝑠𝑛1 ,𝑛2 (𝑥, 𝑦) = ∑ ∑ 𝜏𝑘1 ,𝑘2 𝑒2𝜋𝑖(𝑘1 𝑥+𝑘2 𝑦) .

(7)

𝑘1 =0 𝑘2 =0

Then we have 󵄩󵄩 󵄩 󵄩󵄩𝑓 (𝑥, 𝑦) − 𝑠𝑛1 ,𝑛2 (𝑥, 𝑦)󵄩󵄩󵄩 󵄩 󵄩𝐿 𝑝󸀠 (Ω) 󵄩 󵄩 ≤ 󵄩󵄩󵄩󵄩𝐹𝑝 (𝑥, 𝑦) − 𝑠𝑛1 ,𝑛2 (𝑥, 𝑦)󵄩󵄩󵄩󵄩𝐿

𝑝󸀠

. (𝑇)

(8)

Since the smooth periodic function 𝐹𝑝 can be approximated well by the partial sum of its Fourier series [5, 7, 10], from this inequality, we see that we have constructed a trigonometric polynomial 𝑠𝑛1 ,𝑛2 (𝑥, 𝑦) which can approximate to 𝑓 on Ω very well. (iii) Odd (Even) Periodic Extension. Let 𝐹 be the smooth extension of 𝑓 from Ω to 𝑇 which is stated in Theorem 1. Define 𝐹𝑜 on [−1, 1]2 by 𝐹 (𝑥, 𝑦) , { { { { −𝐹 (−𝑥, 𝑦) , 𝐹𝑜 (𝑥, 𝑦) = { { 𝐹 (−𝑥, −𝑦) , { { {−𝐹 (𝑥, −𝑦) ,

(𝑥, 𝑦) ∈ [0, 1]2 , (𝑥, 𝑦) ∈ [−1, 0] × [0, 1] , (𝑥, 𝑦) ∈ [−1, 0]2 , (𝑥, 𝑦) ∈ [0, 1] × [−1, 0] .

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Then 𝐹𝑜 is an odd function. By Theorem 1, we have 𝐹𝑜 ∈ 𝐶𝑟 ([−1, 1]2 ) and (𝜕𝑖+𝑗 𝐹𝑜 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 on 𝜕([−1, 1]2 ) for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟. Again, doing a 2-periodic extension, we obtain a 2-periodic odd function 𝐹𝑝𝑜 and 𝐹𝑝𝑜 ∈ 𝐶𝑟 (R2 ). By the well-known results [5, 7, 10], 𝐹𝑝𝑜 can be approximated by sine polynomials very well. Moreover, 𝐹𝑝𝑜 can be expanded into the Fourier sine series; that is, ∞

∞

𝐹𝑝𝑜 (𝑥, 𝑦) = ∑ ∑ 𝛼𝑛1 ,𝑛2 sin (𝜋𝑛1 𝑥) sin (𝜋𝑛2 𝑦) , 𝑛1 =1 𝑛2 =1

(10)

where the coefficients 𝛼𝑛1 ,𝑛2 = 4 ∫𝑇 𝐹𝑝𝑜 (𝑥, 𝑦) sin(𝜋𝑛1 𝑥) sin(𝜋𝑛2 𝑦) 𝑑𝑥 𝑑𝑦 [9]. Considering the approximation of 𝐹𝑝𝑜 by the partial sum, the Fejer sum, and the Vallee-Poussin sum [7, 14] of the Fourier sine series of 𝐹𝑝𝑜 , we will obtain the approximation of the original function 𝑓 on Ω by sine polynomials. Define 𝐹𝑒 on [−1, 1]2 as follows: 𝐹 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ [0, 1]2 , { { { {𝐹 (−𝑥, 𝑦) , (𝑥, 𝑦) ∈ [−1, 0] × [0, 1] , 𝐹𝑒 (𝑥, 𝑦) = { {𝐹 (−𝑥, −𝑦) , (𝑥, 𝑦) ∈ [−1, 0]2 , { { (𝑥, 𝑦) ∈ [0, 1] × [−1, 0] . {𝐹 (𝑥, −𝑦) ,

(11)

(i) Periodic Wavelet Series. Let 𝐹𝑝 ∈ 𝐶𝑟 (R2 ) be stated in Theorem 2. Let {𝜓𝜇 }31 be a bivariate smooth wavelet [2]. Then, under a mild condition, the families per , 𝜇 = 1, 2, 3; 𝑚 ∈ Z+ ; Ψper : = {1} ⋃ {𝜓𝜇,𝑚,𝑛

𝑛 = (𝑛1 , 𝑛2 ) , 𝑛1 , 𝑛2 = 0, . . . , 2𝑚 − 1} , = ∑ 𝜓𝜇,𝑚,𝑛 (⋅ + 𝑙) ,

(12)

2𝑚 −1

𝜇=1 𝑚=0 𝑛1 ,𝑛2 =0

𝜓1 (𝑥, 𝑦) = 𝜙 (𝑥) 𝜓 (𝑦) ,

𝜓2 (𝑥, 𝑦) = 𝜓 (𝑥) 𝜙 (𝑦) ,

𝜓3 (𝑥, 𝑦) = 𝜓 (𝑥) 𝜓 (𝑦) ,

(15)

then {𝜓𝜇 (𝑥, 𝑦)}3𝜇=1 is a smooth tensor product wavelet. We expand 𝐹𝑐 into the wavelet series

where 𝜓𝜇,𝑚,𝑛 = 2𝑚 𝜓𝜇 (2𝑚 ⋅ −𝑛) and the wavelet coefficients 𝑐𝜇,𝑚,𝑛 = ∫ 𝐹𝑐 (𝑥, 𝑦) 𝜓𝜇,𝑚,𝑛 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 R2

(17)

= ∫ 𝐹 (𝑥, 𝑦) 𝜓𝜇,𝑚,𝑛 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦. 𝑇

𝑐

Since 𝐹 is a smooth function, the wavelet coefficients 𝑐𝜇,𝑚,𝑛 decay fast. On the other hand, since 𝐹𝑐 (𝑥, 𝑦) = 0, (𝑥, 𝑦) ∈ R2 \𝑇, a lot of wavelet coefficients vanish. In fact, when 𝑚0 ∈ Z and 𝑛0 ∈ Z2 satisfy supp 𝜓𝜇,𝑚0 ,𝑛0 ⊂ (R2 \𝑇), we have 𝑐𝜇,𝑚0 ,𝑛0 = 0. Besides, by condition (iii) in Theorem 1, we know that 𝐹 is univariate or bivariate polynomials on 𝑇 \ Ω. By the moment theorem [2], we know that more wavelet coefficients vanish. For example, let 𝑚∗ ∈ Z and 𝑛∗ = (𝑛1∗ , 𝑛2∗ ) ∈ Z2 satisfy supp 𝜓𝑚∗ ,𝑛2∗ ⊂ [0, 𝛼∗ ], where 𝛼∗ = inf{𝑔(𝑥), 𝑥1 ≤ 𝑥 ≤ 𝑥2 }. Then we have 𝑥1

𝑥2

1

0

𝑥1

𝑥2

𝑐1,𝑚∗ ,𝑛∗ = 2𝑚 (∫ + ∫ + ∫ ) 𝜙𝑚∗ ,𝑛∗ (𝑥) 1

(18)

2

= : 𝐼1 + 𝐼2 + 𝐼3 . By Lemma 8, we know that 𝐿

(13)

(16)

𝜇=1 𝑚∈Z 𝑛∈Z2

0

are a periodic wavelet basis. We expand 𝐹𝑝 into a periodic wavelet series [2] ∞

(ii) Wavelet Approximation. Let 𝐹𝑐 be the smooth function with a compact support as in Theorem 3. Let 𝜓 be a univariate Daubechies wavelet and 𝜙 be the corresponding scaling function [2]. Denoting

× (∫ 𝜓𝑚∗ ,𝑛∗ (𝑦) 𝐹 (𝑥, 𝑦) 𝑑𝑦) 𝑑𝑥

𝜓𝜇,𝑚,𝑛 = 2𝑚 𝜓𝜇 (2𝑚 ⋅ −𝑛)

3

satisfies ‖𝐹𝑝 − 𝑠2𝑀 (𝐹𝑝 )‖𝐿2 (𝑇) = 𝑂(2−2𝑀). From this and 𝐹𝑝 (𝑥, 𝑦) = 𝑓(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Ω), we will obtain an estimate of wavelet approximation for a smooth function 𝑓 on the domain Ω.

𝛼∗

𝑙∈Z2

per 𝐹𝑝 = 𝑑0,0 + ∑ ∑ ∑ 𝑑𝜇,𝑚,𝑛 𝜓𝜇,𝑚,𝑛 .

(14)

𝜇=1 𝑚=0 𝑛1 ,𝑛2 =0

3

2.2.3. Wavelet Analysis

where

3 𝑀−1 2𝑚 −1

per 𝑠2𝑀 (𝐹𝑝 ) = 𝑑0,0 + ∑ ∑ ∑ 𝑑𝜇,𝑚,𝑛 𝜓𝜇,𝑚,𝑛

𝐹𝑐 (𝑥, 𝑦) = ∑ ∑ ∑ 𝑐𝜇,𝑚,𝑛 𝜓𝜇,𝑚,𝑛 (𝑥, 𝑦) ,

Then 𝐹𝑒 is an even function on [−1, 1]2 . By Theorem 1, 𝐹𝑒 ∈ 𝐶𝑟 ([−1, 1]2 ) and (𝜕𝑖+𝑗 𝐹𝑒 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 on 𝜕([−1, 1]2 ) for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟. Again, doing a 2-periodic extension, we obtain a 2-periodic even function 𝐹𝑝𝑒 and 𝐹𝑝𝑒 ∈ 𝐶𝑟 (R2 ). By the well-known result [5, 10], 𝐹𝑝𝑒 can be approximated by cosine polynomials very well. Moreover, 𝐹𝑝𝑒 can be expanded into the Fourier cosine series. Considering the partial sum, the Fejer sum, and the Vallee-Poussin sum [5, 7, 14] of the Fourier cosine series of 𝐹𝑝𝑒 , we will obtain the approximation of the original function 𝑓 on Ω by cosine polynomials.

per 𝜓𝜇,𝑚,𝑛

From this, we can realize the wavelet approximation of 𝑓 on Ω, for example, if 𝑟 = 2, its partial sum

𝐹 (𝑥, 𝑦) = ∑𝜉𝑗 (𝑥) 𝑦𝑗 , 𝑗=0

(𝑥, 𝑦) ∈ 𝐸1 ,

(19)

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where 𝐸1 = {(𝑥, 𝑦) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔(𝑥)} and 𝑔(𝑥) ≥ 𝛼∗ (𝑥1 ≤ 𝑥 ≤ 𝑥2 ). So 𝑥2

𝐼2 = ∫ 𝜙𝑚∗ ,𝑛∗ (𝑥) (∫ 𝑥1

1

𝛼∗

0

H4

𝐿

𝜓𝑚∗ ,𝑛∗ (𝑦) ( ∑ 𝜉𝑗 (𝑥) 𝑦𝑗 ) 𝑑𝑦) 𝑑𝑥. 2

𝑗=0

x3

x4

1

H3

E3

y3

y4

(20) E4

If the Daubechies wavelet 𝜓 chosen by us is 𝐿 time smooth, then, by using the moment theorem and supp 𝜓𝑚∗ ,𝑛2∗ ⊂ [0, 𝛼∗ ], we have 𝛼∗

𝑗

0

2

y2

𝑗

∫ 𝜓𝑚∗ ,𝑛∗ (𝑦) 𝑦 𝑑𝑦 = ∫ 𝜓𝑚∗ ,𝑛∗ (𝑦) 𝑦 𝑑𝑦 = 0, R

2

(21)

(0 ≤ 𝑗 ≤ 2𝑟 + 1) .

3. Proofs of the Main Theorems

3.1. Partition of the Complement of the Domain Ω in 𝑇. Since Ω ⊂ 𝑇𝑜 and 𝜕Ω is a piecewise infinitely many time smooth curve, without loss of generality, we can divide the complement 𝑇\Ω into some rectangles and some trapezoids with a curved side. For convenience of representation, we assume that we can choose four point (𝑥] , 𝑦] ) ∈ 𝜕Ω (] = 1, 2, 3, 4) such that 𝑇\Ω can be divided into the four rectangles

𝐻3 = [𝑥3, 1] × [𝑦3, 1] ,

𝐻4 = [0, 𝑥4 ] × [𝑦4, 1]

0

H2

E1

H1

x2

x1

(22)

1

Figure 1: Partition of the complement of the domain Ω.

From this, we know that 𝑇 can be expressed into a disjoint union as follows: 4

4

1

1

𝑇 = Ω ⋃ (⋃𝐸] ) ⋃ (⋃𝐻] ) ,

We first give a partition of the complement 𝑇 \ Ω.

𝐻2 = [𝑥2, 1] × [0, 𝑦2 ] ,

y1 0

So 𝐼2 = 0. Similarly, since 𝐹(𝑥, 𝑦) is bivariate polynomials on rectangles 𝐻1 and 𝐻3 (see Lemma 11), we have 𝐼1 = 𝐼3 = 0. Furthermore, by (18), we get 𝑐1,𝑚∗ ,𝑛∗ = 0. Therefore, the partial sum of the wavelet series (16) can approximate to 𝐹𝑐 very well and few wavelet coefficients can reconstruct 𝐹𝑐 . Since 𝐹𝑐 = 𝑓 on Ω, the partial sum of the wavelet series (16) can approximate to the original function 𝑓 on the domain Ω very well.

𝐻1 = [0, 𝑥1 ] × [0, 𝑦1 ] ,

E2

Ω

(25)

where each 𝐸] is a trapezoid with a curved side and each 𝐻] is a rectangle (see Figure 1). In Sections 3.2 and 3.3 we will extend 𝑓 to each 𝐸] and continue to extend to each 𝐻] such that the obtained extension 𝐹 satisfies the conditions of Theorem 1. 3.2. Smooth Extension to Each Trapezoid 𝐸] with a Curved Side. By (23), the trapezoid 𝐸1 with a curved side 𝑦 = 𝑔(𝑥) (𝑥1 ≤ 𝑥 ≤ 𝑥2 ) is represented as

and four trapezoids with a curved side 𝐸1 = {(𝑥, 𝑦) ; 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔 (𝑥)} , 𝐸2 = {(𝑥, 𝑦) ; ℎ (𝑦) ≤ 𝑥 ≤ 1, 𝑦2 ≤ 𝑦 ≤ 𝑦3 } , 𝐸3 = {(𝑥, 𝑦) ; 𝑥4 ≤ 𝑥 ≤ 𝑥3 , 𝑔∗ (𝑥) ≤ 𝑦 ≤ 1} ,

𝐸1 = {(𝑥, 𝑦) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔 (𝑥)} . (23)

𝐸4 = {(𝑥, 𝑦) ; 0 ≤ 𝑥 ≤ ℎ∗ (𝑦) , 𝑦1 ≤ 𝑦 ≤ 𝑦4 } , where 𝑔 ∈ 𝐶∞ ([𝑥1 , 𝑥2 ]), ℎ ∈ 𝐶∞ ([𝑦2 , 𝑦3 ]), 𝑔∗ 𝐶∞ ([𝑥4 , 𝑥3 ]), and ℎ∗ ∈ 𝐶∞ ([𝑦1 , 𝑦4 ]) and 0 < 𝑔 (𝑥) < 1

(𝑥1 ≤ 𝑥 ≤ 𝑥2 ) ,

0 < ℎ (𝑦) < 1

(𝑦2 ≤ 𝑦 ≤ 𝑦3 ) ,

0 < 𝑔∗ (𝑥) < 1

(𝑥4 ≤ 𝑥 ≤ 𝑥3 ) ,

0 < ℎ∗ (𝑦) < 1

(𝑦1 ≤ 𝑦 ≤ 𝑦4 ) .

(26)

We define two sequences of functions {𝑎𝑘,1 (𝑥, 𝑦)}∞ 0 and as follows: {𝑏𝑘,1 (𝑥, 𝑦)}∞ 0

∈ 𝑎0,1 (𝑥, 𝑦) =

𝑦 , 𝑔 (𝑥)

𝑏0,1 (𝑥, 𝑦) =

𝑦 − 𝑔 (𝑥) , −𝑔 (𝑥)

𝑘

(24)

(𝑦 − 𝑔 (𝑥)) 𝑦 𝑘+1 𝑎𝑘,1 (𝑥, 𝑦) = ( ) , 𝑘! 𝑔 (𝑥) 𝑏𝑘,1 (𝑥, 𝑦) =

𝑦𝑘 𝑦 − 𝑔 (𝑥) 𝑘+1 ( ) , 𝑘! −𝑔 (𝑥)

𝑘 = 1, 2, . . . .

(27)

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5

By (27), we deduce that for 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝜕𝑙 𝑎𝑘,1 (𝑥, 𝑔 (𝑥)) = 0 𝜕𝑦𝑙 𝜕𝑘 𝑎𝑘,1 (𝑥, 𝑔 (𝑥)) = 1; 𝜕𝑦𝑘

𝜕𝑙 𝑏𝑘,1 (𝑥, 𝑔 (𝑥)) = 0 𝜕𝑦𝑙 𝜕𝑙 𝑏𝑘,1 (𝑥, 0) = 0 𝜕𝑦𝑙

𝑆1(𝑘) (𝑥, 𝑦) = 𝑆1(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,1 (𝑥, 𝑦) (0 ≤ 𝑙 ≤ 𝑘 − 1) ,

(𝜕𝑙 𝑎𝑘,1 ) 𝜕𝑦𝑙

In general, let

×( (𝑥, 0) = 0

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑓 (𝑥, 𝑔 − (𝑥, 𝑔 (𝑥))) (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘

(0 ≤ 𝑙 ≤ 𝑘) , − 𝑏𝑘,1 (𝑥, 𝑦)

(0 ≤ 𝑙 ≤ 𝑘) ;

(0 ≤ 𝑙 ≤ 𝑘 − 1) ,

(𝜕𝑘 𝑏𝑘,1 ) 𝜕𝑦𝑘

𝜕𝑘 𝑆1(𝑘−1) (𝑥, 0) 𝜕𝑦𝑘

(33)

(𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔 (𝑥)) . (𝑥, 0) = 1. (28)

Lemma 4. For any 𝑘 ∈ Z+ , one has 𝑆1(𝑘) ∈ 𝐶∞ (𝐸1 ) and 2𝑘+1

𝑆1(𝑘) (𝑥, 𝑦) = ∑ 𝜁𝑗,1 (𝑥) 𝑦𝑗 , 𝑗=0

On 𝐸1 , we define a sequence of functions {𝑆1(𝑘) (𝑥, 𝑦)}∞ 0 by induction. Let 𝑆1(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝑔 (𝑥)) 𝑎0,1 (𝑥, 𝑦) (𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔 (𝑥)) .

(29)

(𝑥, 𝑦) ∈ 𝐸1 .

(34)

Proof. Since 𝑓 ∈ 𝐶∞ (Ω) and 𝑔 ∈ 𝐶∞ ([𝑥1 , 𝑥2 ]), and 𝑔(𝑥) > 0 (𝑥1 ≤ 𝑥 ≤ 𝑥2 ), by the above construction, we know that 𝑆1(𝑘) ∈ 𝐶∞ (𝐸1 ) for any 𝑘 = 0, 1, . . .. For 𝑘 = 0, since 𝑆1(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝑔 (𝑥)) 𝑎0,1 (𝑥, 𝑦) =

𝑓 (𝑥, 𝑔 (𝑥)) 𝑦, 𝑔 (𝑥)

(35)

(34) holds. We assume that (34) holds for 𝑘 = 𝑙 − 1; that is,

Then, by (27),

2𝑙−1

𝑆1(0)

𝑆1(0)

(𝑥, 0) = 0,

(𝑥, 𝑔 (𝑥)) = 𝑓 (𝑥, 𝑔 (𝑥)) , (𝑥1 ≤ 𝑥 ≤ 𝑥2 ) .

(30)

2𝑙−1 𝜕𝑙 𝑆1(𝑙−1) 𝑗! 𝑗−𝑙 (𝑙−1) (𝑥, 𝑔 = 𝜁𝑗,1 ∑ (𝑥) (𝑔 (𝑥)) , (𝑥)) (𝑗 − 𝑙)! 𝜕𝑦𝑙 𝑗=𝑙

𝑆1(1) (𝑥, 𝑦) = 𝑆1(0) (𝑥, 𝑦) 𝜕𝑆(0) 𝜕𝑓 (𝑥, 𝑔 (𝑥)) − 1 (𝑥, 𝑔 (𝑥))) 𝜕𝑦 𝜕𝑦

𝜕𝑆(0) − 𝑏1,1 (𝑥, 𝑦) 1 (𝑥, 0) 𝜕𝑦 (𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑦 ≤ 𝑔 (𝑥)) . (31) Then, by (27)–(30), we obtain that, for 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝑆1(1)

𝑆1(1) (𝑥, 0) = 0,

𝜕𝑆1(1) (𝑥, 0) = 0. 𝜕𝑦

𝜕𝑙 𝑆1(𝑙−1) 𝜕𝑦𝑙

(37)

(𝑙−1) (𝑥, 0) = 𝑙!𝜁𝑙,1 (𝑥) .

Again, notice that 𝑎𝑙,1 (𝑥, 𝑦) and 𝑏𝑙,1 (𝑥, 𝑦) are polynomials of 𝑦 whose degrees are both 2𝑙 + 1. From this and (33), it follows that (34) holds for 𝑘 = 𝑙. By induction, (34) holds for all 𝑘. Lemma 4 is proved. Below we compute derivatives (𝜕𝑙 𝑆1(𝑘) /𝜕𝑦𝑙 )(𝑥, 𝑦) (0 ≤ 𝑙 ≤ 𝑘) on the curved side Γ1 = {(𝑥, 𝑔(𝑥)) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 } and the bottom side Δ 1 = {(𝑥, 0) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 } of 𝐸1 . Lemma 5. Let 𝑆1(𝑘) (𝑥, 𝑦) be stated as above. For any 𝑘 ∈ Z+ , one has

(𝑥, 𝑔 (𝑥)) = 𝑓 (𝑥, 𝑔 (𝑥)) ,

𝜕𝑆1(1) 𝜕𝑓 (𝑥, 𝑔 (𝑥)) = (𝑥, 𝑔 (𝑥)) , 𝜕𝑦 𝜕𝑦

(36)

𝑗=0

This implies that

Let

+ 𝑎1,1 (𝑥, 𝑦) (

(𝑙−1) 𝑆1(𝑙−1) (𝑥, 𝑦) = ∑ 𝜁𝑗,1 (𝑥) 𝑦𝑗 .

(32)

𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑓 (𝑥, 𝑔 = (𝑥, 𝑔 (𝑥)) , (𝑥)) 𝜕𝑦𝑙 𝜕𝑦𝑙 𝜕𝑙 𝑆1(𝑘) (𝑥, 0) = 0 𝜕𝑦𝑙

(𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑙 ≤ 𝑘) .

(38)

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The Scientific World Journal

Proof. By (30), We have known that, for 𝑘 = 0, (38) holds. Now we assume that (38) holds for 𝑘 − 1. For 𝑥1 ≤ 𝑥 ≤ 𝑥2 , by (33), we have

For 𝑙 = 0, . . . , 𝑘 − 1, by the assumption of induction and (28), we have (𝜕𝑙 𝑆1(𝑘−1) /𝜕𝑦𝑙 )(𝑥, 0) = 0 and (𝜕𝑙 𝑎𝑘,1 /𝜕𝑦𝑙 )(𝑥, 0) = (𝜕𝑙 𝑏𝑘,1 /𝜕𝑦𝑙 )(𝑥, 0) = 0. So

𝜕𝑙 𝑎𝑘,1 𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑆1(𝑘−1) (𝑥, 𝑔 = (𝑥, 𝑔 + (𝑥, 𝑔 (𝑥)) (𝑥)) (𝑥)) 𝜕𝑦𝑙 𝜕𝑦𝑙 𝜕𝑦𝑙 ×(

𝑘

𝜕 𝑓 (𝑥, 𝑔 (𝑥)) − 𝜕𝑦𝑘

𝑙

−

𝜕 𝑏𝑘,1 (𝑥, 𝑔 (𝑥)) 𝜕𝑦𝑙

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑦𝑘

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑦𝑘

𝜕𝑙 𝑆1(𝑘) (𝑥, 0) = 0. 𝜕𝑦𝑙

(𝑥, 𝑔 (𝑥)))

(𝑥, 0) , (0 ≤ 𝑙 ≤ 𝑘) . (39)

For 𝑙 = 0, 1, . . . , 𝑘−1, by the assumption of induction, we have 𝜕𝑙 𝑆1(𝑘−1) 𝜕𝑙 𝑓 (𝑥, 𝑔 (𝑥)) = 𝑙 (𝑥, 𝑔 (𝑥)) . 𝑙 𝜕𝑦 𝜕𝑦 By (28), we have (𝜕𝑙 𝑎𝑘,1 /𝜕𝑦𝑙 )(𝑥, 𝑔(𝑥)) (𝑥, 𝑔(𝑥)) = 0. So we get

=

(41)

For 𝑙 = 𝑘, note that (𝜕𝑘 𝑎𝑘,1 /𝜕𝑦𝑘 )(𝑥, 𝑔(𝑥)) = 1 and (𝜕𝑘 𝑏𝑘,1 /𝜕𝑦𝑘 ) (𝑥, 𝑔(𝑥)) = 0. By (39), we get

=

The second formula of (38) holds. By induction, (38) holds for all 𝑘. From this, we get Lemma 5.

Now we compute the mixed derivatives of 𝑆1(𝑘) (𝑥, 𝑦) on the curved side Γ1 and bottom side Δ 1 of 𝐸1 . Lemma 6. Let Γ1 and Δ 1 be the curved side and the bottom side of 𝐸1 , respectively. Then, for 𝑘 ∈ Z+ , (i) (𝜕𝑖+𝑗 𝑆1(𝑘) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Γ1 ),

=

(𝜕𝑖+𝑗 𝑓/𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦)

Proof. Let 𝑥1 ≤ 𝑥 ≤ 𝑥2 . Then we have

(𝑥, 𝑔 (𝑥)) (42)

𝜕𝑙 𝑓 𝑑 𝜕𝑙−1 𝑓 (𝑥, 𝑔 (𝑥)) ( 𝑙−1 (𝑥, 𝑔 (𝑥))) = 𝑑𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦𝑙−1

𝜕𝑘 𝑓 (𝑥, 𝑔 (𝑥)) . 𝜕𝑦𝑘

+

𝜕𝑙 𝑓 (𝑥, 𝑔 (𝑥)) 𝑔󸀠 (𝑥) , 𝜕𝑦𝑙

𝜕𝑙 𝑎𝑘,1 𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑆1(𝑘−1) 0) = 0) + (𝑥, (𝑥, (𝑥, 0) 𝜕𝑦𝑙 𝜕𝑦𝑙 𝜕𝑦𝑙

−

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑓 (𝑥, 𝑔 − (𝑥, 𝑔 (𝑥))) (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘

𝜕𝑙 𝑏𝑘,1 𝜕𝑘 𝑆1(𝑘−1) 0) (𝑥, (𝑥, 0) , 𝜕𝑦𝑙 𝜕𝑦𝑘 (𝑥1 ≤ 𝑥 ≤ 𝑥2 , 0 ≤ 𝑙 ≤ 𝑘) . (43)

(𝑙 ≥ 1) . (46)

The first formula of (38) holds for 𝑘. By (33), we have

×(

(45)

where 0 ≤ 𝑖 + 𝑗 ≤ 𝑘.

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑓 (𝑥, 𝑔 − (𝑥, 𝑔 (𝑥))) +( (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘 =

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑆1(𝑘) 0) = 0) − (𝑥, (𝑥, (𝑥, 0) = 0. 𝜕𝑦𝑘 𝜕𝑦𝑘 𝜕𝑦𝑘

(ii) (𝜕𝑖+𝑗 𝑆1(𝑘) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 ((𝑥, 𝑦) ∈ Δ 1 ),

𝜕𝑘 𝑆1(𝑘) (𝑥, 𝑔 (𝑥)) 𝜕𝑦𝑘 𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑦𝑘

For 𝑙 = 𝑘, since (𝜕𝑘 𝑎𝑘,1 /𝜕𝑦𝑘 )(𝑥, 0) = 0, (𝜕𝑘 𝑏𝑘,1 /𝜕𝑦𝑘 )(𝑥, 0) = 1, by (43), we have

(40)

0, (𝜕𝑙 𝑏𝑘,1 /𝜕𝑦𝑙 )

𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑓 (𝑥, 𝑔 = (𝑥, 𝑔 (𝑥)) . (𝑥)) 𝜕𝑦𝑙 𝜕𝑦𝑙

(44)

By the Newton-Leibniz formula, we have 𝜕𝑙−1 𝑓 (𝑥, 𝑔 (𝑥)) 𝜕𝑦𝑙−1 =

𝜕𝑙−1 𝑓 (𝑥 , 𝑔 (𝑥1 )) 𝜕𝑦𝑙−1 1 𝑥

+∫ ( 𝑥1

𝜕𝑙 𝑓 𝜕𝑙 𝑓 (𝑡, 𝑔 + (𝑡, 𝑔 (𝑡)) 𝑔󸀠 (𝑡)) 𝑑𝑡. (𝑡)) 𝜕𝑥𝜕𝑦𝑙−1 𝜕𝑦𝑙 (47)

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7

Similarly, replacing 𝑓 by 𝑆1(𝑘) in this formula, we have

Lemma 7. For any positive integer 𝑟, denote 𝑙𝑟 = 𝑟(𝑟 + 1)(𝑟 + 2)(𝑟 + 3). Let

𝜕𝑙−1 𝑆1(𝑘) (𝑥, 𝑔 (𝑥)) 𝜕𝑦𝑙−1 =

(𝑙 )

𝑆 𝑟 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ 𝐸1 , 𝐹 (𝑥, 𝑦) = { 1 𝑓 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ Ω.

𝜕𝑙−1 𝑆1(𝑘) (𝑥1 , 𝑔 (𝑥1 )) 𝜕𝑦𝑙−1 𝑥

+∫ ( 𝑥1

𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑆1(𝑘) (𝑡, 𝑔 + (𝑡, 𝑔 (𝑡)) 𝑔󸀠 (𝑡)) 𝑑𝑡, (𝑡)) 𝜕𝑥𝜕𝑦𝑙−1 𝜕𝑦𝑙 (𝑙 ≥ 1) . (48)

From this and Lemma 5, it follows that, for any 𝑥1 ≤ 𝑥 ≤ 𝑥2 , we have 𝜕𝑙 𝑆1(𝑘) (𝑡, 𝑔 (𝑡)) 𝑑𝑡 ∫ 𝑙−1 𝑥1 𝜕𝑥𝜕𝑦 𝑥

(54)

Then (i) 𝐹 ∈ 𝐶𝑙𝑟 (Ω ⋃ 𝐸1 ) and 𝐹(𝑥, 𝑦) = 𝑓(𝑥, 𝑦) ((𝑥, 𝑦) ∈ Ω); (ii) (𝜕𝑖+𝑗 𝐹/𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0((𝑥, 𝑦) ∈ (𝐸1 ⋂ 𝜕𝑇), 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ). Proof. By the assumption 𝑓 ∈ 𝐶∞ (Ω), Lemma 4: 𝑆1(𝑘) ∈ 𝐶∞ (𝐸1 ), and Lemma 6(i): 𝜕𝑖+𝑗 𝑆1(𝑘) 𝜕𝑖+𝑗 𝑓 (𝑥, 𝑦) = (𝑥, 𝑦) 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗

(55)

((𝑥, 𝑦) ∈ Γ1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝑘) ,

(49)

where Γ1 = Ω ⋂ 𝐸1 , we get (i). By Lemma 6(ii) and 𝐸1 ⋂ 𝜕𝑇 = Δ 1 , we get (ii). Lemma 7 is proved.

Finding derivatives on the both sides of this formula, we get

For ] = 2, 3, 4, by using a similar method, we define 𝑆](𝑘) (𝑥, 𝑦) on the each trapezoid 𝐸] with a curve side. The representations of 𝑆](𝑘) (𝑥, 𝑦) are stated in Section 4.1.

𝜕𝑙 𝑓 =∫ (𝑡, 𝑔 (𝑡)) 𝑑𝑡 𝑙−1 𝑥1 𝜕𝑥𝜕𝑦 𝑥

(1 ≤ 𝑙 ≤ 𝑘) .

𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑓 (𝑥, 𝑔 = (𝑥, 𝑔 (𝑥)) (𝑥)) 𝜕𝑥𝜕𝑦𝑙−1 𝜕𝑥𝜕𝑦𝑙−1

Lemma 8. For any ] = 1, 2, 3, 4, let (1 ≤ 𝑙 ≤ 𝑘) . (50)

Now we start from the equality

(𝑙 ≥ 2) . (51)

Continuing this procedure, we deduce that (i) holds for 0 < 𝑖 + 𝑗 ≤ 𝑘. Letting 𝑙 = 0 in Lemma 5, we have 𝑆1(𝑘) (𝑥, 𝑔(𝑥)) = 𝑓(𝑥, 𝑔(𝑥)); that is, (i) holds for 𝑖 = 𝑗 = 0. So we get (i). By Lemma 5, (𝜕𝑗 𝑆1(𝑘) /𝜕𝑦𝑗 )(𝑥, 0) = 0 (0 ≤ 𝑗 ≤ 𝑘). From this and 𝑆1(𝑘) ∈ 𝐶∞ (𝐸1 ), we have (0 ≤ 𝑖 + 𝑗 ≤ 𝑘) ,

From this, we get the following.

(iii) 𝐹(𝑥, 𝑦) can be expressed in the form: 2𝑙𝑟 +1

𝐹 (𝑥, 𝑦) = ∑ 𝜁𝑗,1 (𝑥) 𝑦𝑗 , 2𝑙𝑟 +1

(2 ≤ 𝑙 ≤ 𝑘) . (52)

so (ii) holds. Lemma 6 is proved.

(ii) (𝜕𝑖+𝑗 𝐹/𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0, (𝑥, 𝑦) ∈ (𝐸] ⋂ 𝜕𝑇) for 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ;

𝑗=0

Similar to the argument from (46) to (50), we get

𝜕𝑖+𝑗 𝑆1(𝑘) (𝑥, 0) = 0 𝜕𝑥𝑖 𝜕𝑦𝑗

(56)

(i) 𝐹 ∈ 𝐶𝑙𝑟 (Ω ⋃ 𝐸] );

𝜕𝑙 𝑓 (𝑥, 𝑔 (𝑥)) 𝑔󸀠 (𝑥) , 𝜕𝑥𝜕𝑦𝑙−1

𝜕𝑙 𝑆1(𝑘) 𝜕𝑙 𝑓 (𝑥, 𝑔 (𝑥)) = 2 𝑙−2 (𝑥, 𝑔 (𝑥)) 2 𝑙−2 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

𝑆](𝑙𝑟 ) (𝑥, 𝑦) , (𝑥, 𝑦) ∈ 𝐸] , 𝑓 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ Ω,

where 𝑙𝑟 = 𝑟(𝑟 + 1)(𝑟 + 2)(𝑟 + 3). Then, for ] = 1, 2, 3, 4, one has the following:

𝜕𝑙 𝑓 𝜕𝑙−1 𝑓 𝑑 (𝑥, 𝑔 (𝑥))) = 2 𝑙−2 (𝑥, 𝑔 (𝑥)) ( 𝑙−2 𝑑𝑥 𝜕𝑥𝜕𝑦 𝜕𝑥 𝜕𝑦 +

𝐹 (𝑥, 𝑦) = {

(53)

𝐹 (𝑥, 𝑦) = ∑ 𝜁𝑗,2 (𝑦) 𝑥𝑗 , 𝑗=0

𝐹 (𝑥, 𝑦) = ∑ 𝜁𝑗,3 (𝑥) 𝑦𝑗 , 𝑗=0

𝐹 (𝑥, 𝑦) = ∑ 𝜁𝑗,4 (𝑦) 𝑥𝑗 , 𝑗=0

(𝑥, 𝑦) ∈ 𝐸2 , (57)

2𝑙𝑟 +1

2𝑙𝑟 +1

(𝑥, 𝑦) ∈ 𝐸1 ,

(𝑥, 𝑦) ∈ 𝐸3 ,

(𝑥, 𝑦) ∈ 𝐸4 .

Proof. By Lemma 7, we have 𝐹 ∈ 𝐶𝑙𝑟 (Ω ⋃ 𝐸1 ) ,

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

((𝑥, 𝑦) ∈ (𝐸1 ⋂ 𝜕𝑇) , 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ) .

(58)

8

The Scientific World Journal

Similar to the argument of Lemma 7, for ] = 2, 3, 4, we have 𝐹 ∈ 𝐶𝑙𝑟 (Ω ⋃ 𝐸] ) ,

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(59)

From this, we get (i) and (ii). The proof of (iii) is similar to the argument of Lemma 4(iii). Lemma 8 is proved. 3.3. Smooth Extension to Each Rectangle 𝐻] . We have completed the smooth extension of 𝑓 to each trapezoid 𝐸] with a curved side. In this subsection we complete the smooth extension of the obtained function 𝐹 to each rectangle 𝐻] . First we consider the smooth extension of 𝐹 to 𝐻1 . We divide this procedure in two steps. 𝑙

Step 1. In Lemma 8, we know that 𝐹(𝑥, 𝑦) = 𝑆4𝑟 (𝑥, 𝑦) on 𝐸4 . (𝑙 ) Now we construct the smooth extension of 𝑆4 𝑟 (𝑥, 𝑦) from (𝑙 ) 𝐸4 to 𝐻1 , where 𝑆4 𝑟 (𝑥, 𝑦) is stated in Section 4.2 and 𝑙𝑟 = 𝑟(𝑟 + 1)(𝑟 + 2)(𝑟 + 3). Let 𝛼𝑘,11 (𝑦) =

𝑘+1

(𝑦 − 𝑦1 ) 𝑦 ( ) 𝑘! 𝑦1

𝑦𝑘 𝑦 − 𝑦1 𝑘+1 ) , 𝛽𝑘,11 (𝑦) = ( 𝑘! −𝑦1

=

(𝑙 )

(𝜕𝑖+𝑗 𝑆4 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦)

(𝜏 )

((𝑥, 𝑦) ∈ (𝐸] ⋂ 𝜕𝑇) , 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ) .

𝑘

(𝜏 )

(ii) (𝜕𝑖+𝑗 𝑀1 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) ((𝑥, 𝑦) ∈ 𝐽1,1 );

(iii) (𝜕𝑖+𝑗 𝑀1 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 ((𝑥, 𝑦) ∈ 𝐽1,2 ); (𝜏 )

(iv) (𝜕𝑖+𝑗 𝑀1 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 ((𝑥, 𝑦) ∈ 𝐽1,3 ), where 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 . Proof. By Lemma 8(iii), we have 2𝑙𝑟 +1

(𝑙 )

𝑆4 𝑟 (𝑥, 𝑦) = ∑ 𝜁𝑗,4 (𝑦) 𝑥𝑗 .

(63)

𝑗=0

(𝑙 )

So (𝜕𝑘 𝑆4 𝑟 /𝜕𝑦𝑘 )(𝑥, 𝑦1 ) is a polynomial of degree 2𝑙𝑟 + 1 with respect to 𝑥. Since 𝛼𝜏𝑟 ,11 (𝑦) and 𝛽𝜏𝑟 ,11 (𝑦) are both polynomials of degree 2𝜏𝑟 + 1, (i) follows from (61). Similar to the argument of Lemma 6, we get (ii) and (iv). Since (0, 𝑦1 ) ∈ (𝐸4 ⋂ 𝜕𝑇), by Lemma 7, we have (𝑙 )

𝜕𝑖+𝑗 𝑆4 𝑟 𝜕𝑖+𝑗 𝐹 (0, 𝑦 ) = (0, 𝑦1 ) = 0, 1 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗

(0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ) . (64)

By the definition of 𝑀1(0) and (64), we have

, (60)

𝑘 = 0, 1, . . . ,

(𝑙 ) 𝑑𝑗 𝛼0,11 𝜕𝑖+𝑗 𝑀1(0) 𝜕𝑖 𝑆4 𝑟 (0, 𝑦) = (0, 𝑦1 ) (𝑦) = 0, 𝑖 𝑗 𝑖 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝑑𝑦𝑗

(65)

(0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 , 𝑦 ∈ R) .

and let (𝑙 )

𝑀1(0) (𝑥, 𝑦) = 𝑆4 𝑟 (𝑥, 𝑦1 ) 𝛼0,11 (𝑦) ,

We assume that

𝑀1(𝑘) (𝑥, 𝑦) = 𝑀1(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,11 (𝑦) (𝑙 )

×(

𝜕𝑘 𝑆4 𝑟 𝜕𝑘 𝑀1(𝑘−1) (𝑥, 𝑦 ) − (𝑥, 𝑦1 )) 1 (61) 𝜕𝑦𝑘 𝜕𝑦𝑘

𝜕𝑖+𝑗 𝑀1(𝑘−1) (0, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗 By (61), we get

𝜕𝑘 𝑀1(𝑘−1) − 𝛽𝑘,11 (𝑦) (𝑥, 0) , 𝜕𝑦𝑘

𝑑𝑗 𝛼𝑘,11 𝜕𝑖+𝑗 𝑀1(𝑘) 𝜕𝑖+𝑗 𝑀1(𝑘−1) (0, 𝑦) = (0, 𝑦) + (𝑦) 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗 𝑑𝑦𝑗

𝑘 = 1, 2, . . . , 𝜏𝑟 ((𝑥, 𝑦) ∈ 𝐻1 ) ,

(𝑙 )

where 𝜏𝑟 = 𝑟(𝑟 + 2).

×(

Lemma 9. Let {𝐽1,𝑙 }41 be four sides of the rectangle 𝐻1 : −

𝐽1,1 = {(𝑥, 𝑦1 ) , 0 ≤ 𝑥 ≤ 𝑥1 } , 𝐽1,2 = {(0, 𝑦) , 0 ≤ 𝑦 ≤ 𝑦1 } , 𝐽1,3 = {(𝑥, 0) , 0 ≤ 𝑥 ≤ 𝑥1 } ,

(62)

Then one has the following 2𝑙 +1

𝜕𝑘+𝑖 𝑀1(𝑘−1) 𝜕𝑘+𝑖 𝑆4 𝑟 (0, 𝑦 ) − (0, 𝑦1 )) 1 𝜕𝑥𝑖 𝜕𝑦𝑘 𝜕𝑥𝑖 𝜕𝑦𝑘

𝑑𝑗 𝛽𝑘,11 𝜕𝑘+𝑖 𝑀1(𝑘−1) (𝑦) (0, 0) . 𝑑𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑘 (67)

𝐽1,4 = {(𝑥1 , 𝑦) , 0 ≤ 𝑦 ≤ 𝑦1 } .

(𝜏 )

1 (0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 − 𝑘 (𝑘 − 1)) . 2 (66)

(1) 𝑖 𝑗 (1) 𝑟 (i) 𝑀1 𝑟 (𝑥, 𝑦) = ∑𝑖,𝑗=0 𝑑𝑖,𝑗 𝑥 𝑦 , where 𝑑𝑖,𝑗 is a constant;

for 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 − (1/2)𝑘(𝑘 + 1), we have 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 − (1/2)𝑘(𝑘 − 1) and 0 ≤ 𝑖 + 𝑘 ≤ 𝑙𝑟 − (1/2)𝑘(𝑘 − 1). Again, by the assumption of induction, we get 𝜕𝑖+𝑗 𝑀1(𝑘−1) (0, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

𝜕𝑘+𝑖 𝑀1(𝑘−1) (0, 𝑦1 ) = 0. 𝜕𝑥𝑖 𝜕𝑦𝑘

(68)

The Scientific World Journal

9

(𝑙 )

By (64), we have (𝜕𝑘+𝑖 𝑆4 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑘 )(0, 𝑦1 ) = 0. From this and (67), we get

2𝑙 +1

(1) 𝑖 𝑗 (1) 𝑟 (v) 𝑁1(𝑟) (𝑥, 𝑦) = ∑𝑖,𝑗=0 𝜏𝑖,𝑗 𝑥 𝑦 , where 𝜏𝑖,𝑗 is a constant.

𝜕𝑖+𝑗 𝑀1(𝑘) (0, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗 (0 ≤ 𝑦 ≤ 𝑦1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 −

𝑘 (𝑘 + 1)) . 2

(69)

Taking 𝑘 = 𝜏𝑟 , we have

𝜏 (0 ≤ 𝑦 ≤ 𝑦1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 − 𝑟 (𝜏𝑟 + 1)) . 2

(70)

(𝜏 )

𝑘

𝛽𝑘,14 (𝑥) =

𝑥 𝑥 − 𝑥1 ) ( 𝑘! −𝑥1

,

(0 ≤ 𝑥 ≤ 𝑥1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 ) . (74) Now we assume that 𝜕𝑖+𝑗 𝑁1(𝑘−1) (𝑥, 𝑦1 ) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗 𝑘 (0 ≤ 𝑥 ≤ 𝑥1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 − (𝑘 − 1)) . 2

and let 𝑁1(0)

(𝜏 )

𝜕𝑗 (𝑆1 𝑟 − 𝑀1 𝑟 ) 𝑑𝑖 𝛼0,14 𝜕𝑖+𝑗 𝑁1(0) (𝑥, 𝑦 ) = (𝑥 , 𝑦 ) (𝑥) = 0, 1 1 1 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑦𝑗 𝑑𝑥𝑖

(71)

𝑘 = 0, 1, . . . ,

(73)

So we have (𝑙 )

𝑙

Step 2. In Lemma 8, we know that 𝐹(𝑥, 𝑦) = 𝑆1𝑟 (𝑥, 𝑦) on 𝐸1 . (𝑙 ) (𝜏 ) We consider the difference 𝑆1 𝑟 (𝑥, 𝑦)−𝑀1 𝑟 (𝑥, 𝑦). Obviously, (𝜏 ) it is infinitely many time differentiable on 𝐸1 since 𝑀1 𝑟 (𝑥, 𝑦) is a polynomial. Now we construct its smooth extension from 𝐸1 to the rectangle 𝐻1 as follows. Let (𝑥 − 𝑥1 ) 𝑥 𝑘+1 𝛼𝑘,14 (𝑥) = ( ) , 𝑘! 𝑥1

(𝑙 )

𝜕𝑖+𝑗 𝑆1 𝑟 𝜕𝑖+𝑗 𝑆4 𝑟 𝜕𝑖+𝑗 𝑓 (𝑥 , 𝑦 ) = (𝑥 , 𝑦 ) = (𝑥 , 𝑦 ) 𝜕𝑥𝑖 𝜕𝑦𝑗 1 1 𝜕𝑥𝑖 𝜕𝑦𝑗 1 1 𝜕𝑥𝑖 𝜕𝑦𝑗 1 1 𝜕𝑖+𝑗 𝑀1 𝑟 (𝑥1 , 𝑦1 ) . = 𝜕𝑥𝑖 𝜕𝑦𝑗

Since 𝑙𝑟 − (𝜏𝑟 /2)(𝜏𝑟 + 1) = (𝑙𝑟 /2) ≥ 𝜏𝑟 , we get (iii). Lemma 9 is proved.

𝑘+1

Proof. The arguments similar to Lemma 9(ii) and (iv) give the conclusions (i) and (ii) of this theorem. Now we prove (iii) and (iv). By Lemma 6(i) and Lemma 9(ii), as well as 𝑙𝑟 ≥ 𝜏𝑟 , we get that, for 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 , (𝑙 )

(𝜏 )

𝜕𝑖+𝑗 𝑀1 𝑟 (0, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

𝑘

(iv) (𝜕𝑖+𝑗 𝑁1(𝑟) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 on 𝐽1,3 , where 0 ≤ 𝑖 + 𝑗 ≤ 𝑟 and {𝐽1,] }41 are stated in (62);

(75)

By (72) and (73), (𝑥, 𝑦) =

(𝑙 ) (𝑆1 𝑟

(𝑥1 , 𝑦) −

(𝜏 ) 𝑀1 𝑟

(𝑥1 , 𝑦)) 𝛼0,14 (𝑥) ,

𝑑𝑖 𝛼𝑘,14 𝜕𝑖+𝑗 𝑁1(𝑘−1) 𝜕𝑖+𝑗 𝑁1(𝑘) (𝑥, 𝑦 ) = (𝑥, 𝑦 ) + (𝑥) 1 1 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗 𝑑𝑥𝑖

𝑁1(𝑘) (𝑥, 𝑦) = 𝑁1(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,14 (𝑥) (𝑙 )

×(

(𝑙 )

(𝜏 )

𝜕𝑘 (𝑆1 𝑟 − 𝑀1 𝑟 ) 𝜕𝑥𝑘

×(

(𝑥1 , 𝑦)

𝜕𝑘 𝑁1(𝑘−1) − (𝑥1 , 𝑦)) 𝜕𝑥𝑘

𝜕𝑥𝑘 𝜕𝑦𝑗

(72) −

𝜕𝑘 𝑁1(𝑘−1) (0, 𝑦) , − 𝛽𝑘,14 (𝑥) 𝜕𝑥𝑘 𝑘 = 0, 1, . . . , 𝑟, ((𝑥, 𝑦) ∈ 𝐻1 ) .

−

(𝑙 )

(i) (𝜕𝑖+𝑗 𝑁1(𝑟) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = (𝜕𝑖+𝑗 𝑆1 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) − (𝜏 ) (𝜕𝑖+𝑗 𝑀1 𝑟 /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) on 𝐽1,4 ; (ii) (𝜕𝑖+𝑗 𝑁1(𝑟) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 on 𝐽1,2 ; (iii) (𝜕𝑖+𝑗 𝑁1(𝑟) /𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 on 𝐽1,1 ;

(𝑥1 , 𝑦1 )

𝜕𝑘+𝑗 𝑁1(𝑘−1) (𝑥1 , 𝑦1 )) 𝜕𝑥𝑘 𝜕𝑦𝑗

𝑑𝑖 𝛽𝑘,14 𝜕𝑘+𝑗 𝑁1(𝑘−1) (0, 𝑦1 ) = 0, (𝑥) 𝑑𝑥𝑖 𝜕𝑥𝑘 𝜕𝑦𝑗

(0 ≤ 𝑥 ≤ 𝑥1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 −

From this, we obtain the following. Lemma 10. 𝑁1(𝑟) (𝑥, 𝑦) possesses the following properties:

(𝜏 )

𝜕𝑘+𝑗 (𝑆1 𝑟 − 𝑀1 𝑟 )

𝑘 (𝑘 + 1) ). 2 (76)

By induction, we get 𝜕𝑖+𝑗 𝑁1(𝑘) (𝑥, 𝑦1 ) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗 𝑘 (𝑘 + 1) (0 ≤ 𝑥 ≤ 𝑥1 , 0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 − ). 2

(77)

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From this and 𝜏𝑟 − (1/2)𝑟(𝑟 + 1) ≥ 𝑟, we get (iii). By Lemma 6(ii) and Lemma 9(iii), we get that (𝑙 ) 𝜕𝑖+𝑗 𝑆1 𝑟 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 0) = 0,

(𝜏 ) 𝜕 𝑀1 𝑟 𝜕𝑥𝑖 𝜕𝑦𝑗 𝑖+𝑗

(0 ≤ 𝑖 + 𝑗 ≤ 𝑙𝑟 ) , (78)

(𝑥, 0) = 0,

(0 ≤ 𝑖 + 𝑗 ≤ 𝜏𝑟 ) .

From this and (72), by using an argument similar to the proof of (iii), we get (iv). (𝑙 ) By Lemma 8(iii) and Lemma 9(i), we deduce that (𝑆1 𝑟 − (𝜏𝑟 ) 𝑀1 )(𝑥1 , 𝑦) is a polynomial of degree 2𝑙𝑟 + 1 with respect to 𝑦. From this and (72), we get (v). Lemma 10 is proved. By Lemmas 9 and 10, we obtain that for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟, (𝜏 )

𝜕𝑖+𝑗 (𝑀1 𝑟 + 𝑁1(𝑟) ) 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) =

(𝑙 ) 𝜕𝑖+𝑗 𝑆4 𝑟 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦)

on 𝐽1,1 ,

(𝜏 )

𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) = 0

(𝜏 )

𝜕𝑖+𝑗 (𝑀1 𝑟 + 𝑁1(𝑟) ) 𝜕𝑥𝑖 𝜕𝑦𝑗

on 𝐽1,2 ⋃ 𝐽1,3 ,

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) ∈ (𝐻1 ⋂ 𝜕𝑇) .

(86)

So we get (ii). From Lemma 9(i), Lemma 10(v), and 𝐹(𝑥, 𝑦) = (𝜏 ) 𝑀1 𝑟 (𝑥, 𝑦) + 𝑁1(𝑟) (𝑥, 𝑦) ((𝑥, 𝑦) ∈ 𝐻1 ), we get (iii). Lemma 11 is proved. For ] = 2, 3, 4, by using a similar method, we define 𝐹(𝑥, 𝑦) = 𝑀](𝜏𝑟 ) (𝑥, 𝑦) + 𝑁](𝑟) (𝑥, 𝑦) ((𝑥, 𝑦) ∈ 𝐻] ), where representations of 𝑀](𝜏𝑟 ) (𝑥, 𝑦) and 𝑁](𝑟) (𝑥, 𝑦) see Section 4.2. 3.4. The Proofs of the Theorems Proof of Theorem 1. Let

(79) 𝜕𝑖+𝑗 (𝑀1 𝑟 + 𝑁1(𝑟) )

Since 𝐻1 ⋂ 𝜕𝑇 = 𝐽1,2 ⋃ 𝐽1,3 , by (80), we deduce that

(80)

𝐹 (𝑥, 𝑦) {𝑓 (𝑥, 𝑦) , = {𝑆](𝑙𝑟 ) (𝑥, 𝑦) , (𝜏 ) (𝑟) {𝑀] 𝑟 (𝑥, 𝑦) + 𝑁] (𝑥, 𝑦) ,

(𝑥, 𝑦) ∈ Ω, (𝑥, 𝑦) ∈ 𝐸] (] = 1, 2, 3, 4) , (𝑥, 𝑦) ∈ 𝐻] (] = 1, 2, 3, 4) .

(87)

(𝑙 )

(𝑥, 𝑦) =

𝜕𝑖+𝑗 𝑆1 𝑟 (𝑥, 𝑦) 𝜕𝑥𝑖 𝜕𝑦𝑗

on 𝐽1,4 . (81)

By (25), 𝐹 has been defined on the unit square 𝑇. The argument similar to Lemma 11(i)-(ii) shows that 𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸1 ⋃ 𝐸4 ⋃ 𝐻1 ) ;

Lemma 11. Let 𝑓 (𝑥, 𝑦) , { { { {𝑆1(𝑙𝑟 ) (𝑥, 𝑦) , 𝐹 (𝑥, 𝑦) = { (𝑙𝑟 ) {𝑆4 (𝑥, 𝑦) , { { (𝜏 ) (𝑟) 𝑟 {𝑀1 (𝑥, 𝑦) + 𝑁1 (𝑥, 𝑦) ,

(𝑥, 𝑦) ∈ Ω, (𝑥, 𝑦) ∈ 𝐸1 , (82) (𝑥, 𝑦) ∈ 𝐸4 , (𝑥, 𝑦) ∈ 𝐻1 .

Then one has (ii) (𝜕𝑖+𝑗 𝐹/𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0, (𝑥, 𝑦) ∈ ((𝐸1 ⋃ 𝐸4 ⋃ 𝐻1 ) ⋂ 𝜕𝑇) for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟; 2𝑙 +1

𝑟 𝑐𝑖𝑗(1) 𝑥𝑖 𝑦𝑗 ((𝑥, 𝑦) ∈ 𝐻1 ), where each 𝑐𝑖𝑗(1) (iii) 𝐹(𝑥, 𝑦) = ∑𝑖,𝑗=0 is constant.

Proof. By Lemma 7, we have 𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸1 ⋃ 𝐸4 ). Since (𝑙 ) 𝑆1 𝑟 ∈ 𝐶𝑟 (𝐸1 ), (𝜏 )

𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸3 ⋃ 𝐸4 ⋃ 𝐻4 ) ;

𝐸1 ⋂ 𝐻1 = 𝐽1,4 ,

(83)

(𝑙 )

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) ∈ ((𝐸1 ⋃ 𝐸4 ⋃ 𝐻1 ) ⋂ 𝜕𝑇) ;

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) ∈ ((𝐸1 ⋃ 𝐸2 ⋃ 𝐻2 ) ⋂ 𝜕𝑇) ;

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) ∈ ((𝐸2 ⋃ 𝐸3 ⋃ 𝐻3 ) ⋂ 𝜕𝑇) ;

𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) ∈ ((𝐸3 ⋃ 𝐸4 ⋃ 𝐻4 ) ⋂ 𝜕𝑇) .

by (81), we deduce that 𝐹 ∈ 𝐶𝑟 (𝐸1 ⋃ 𝐻1 ). Since 𝑆4 𝑟 ∈ 𝐶𝑟 (𝐸4 ), (𝜏 )

𝑀1 𝑟 + 𝑁1(𝑟) ∈ 𝐶𝑟 (𝐻1 ) ,

𝐸4 ⋂ 𝐻1 = 𝐽1,1 ,

(84)

by (79), we deduce that 𝐹 ∈ 𝐶𝑟 (𝐻1 ⋃ 𝐸4 ). So we get (i). By Lemma 8(ii), 𝜕𝑖+𝑗 𝐹 (𝑥, 𝑦) = 0, 𝜕𝑥𝑖 𝜕𝑦𝑗

(88)

𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸2 ⋃ 𝐸3 ⋃ 𝐻3 ) ;

and for 0 ≤ 𝑖 + 𝑗 ≤ 𝑟,

(i) 𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸1 ⋃ 𝐸4 ⋃ 𝐻1 ),

𝑀1 𝑟 + 𝑁1(𝑟) ∈ 𝐶𝑟 (𝐻1 ) ,

𝐹 ∈ 𝐶𝑟 (Ω ⋃ 𝐸1 ⋃ 𝐸2 ⋃ 𝐻2 ) ;

(89) 𝑟

From this and Ω ⋂ 𝜕𝑇 = 0, by (25), we have 𝐹 ∈ 𝐶 (𝑇) and (𝜕𝑖+𝑗 𝐹/𝜕𝑥𝑖 𝜕𝑦𝑗 )(𝑥, 𝑦) = 0 ((𝑥, 𝑦) ∈ 𝜕𝑇, 0 ≤ 𝑖 + 𝑗 ≤ 𝑟). So we get (i) and (ii). Similar to the argument of Lemma 11(iii), we get 2𝑙𝑟 +1

(𝑥, 𝑦) ∈ ((𝐸1 ⋃ 𝐸4 ) ⋂ 𝜕𝑇) .

(85)

𝐹 (𝑥, 𝑦) = ∑ 𝑐𝑖𝑗(]) 𝑥𝑖 𝑦𝑗 , 𝑖,𝑗=0

(𝑥, 𝑦) ∈ 𝐻] (] = 1, 2, 3, 4) , (90)

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where each 𝑐𝑖𝑗(]) is a constant. From this and Lemma 8(iii), we know that, on 𝑇 \ Ω, 𝐹(𝑥, 𝑦) can be expressed locally in the form 2𝑙𝑟 +1

2𝑙𝑟 +1

∑ 𝜉𝑗 (𝑥) 𝑦𝑗

or ∑ 𝜂𝑗 (𝑥) 𝑥𝑗

𝑗=0

𝑗=0

2𝑙𝑟 +1

or ∑ 𝑐𝑖𝑗 𝑥𝑖 𝑦𝑗 ; (91)

4. Representation of the Extension 𝐹 Satisfying Theorem 1 Let 𝑓 and Ω be stated as in Theorem 1 and let Ω be divided as in Section 3.1. The representation of 𝐹 satisfying conditions of Theorem 1 is as follows:

𝑖, 𝑗=0

(iii) holds. We have completed the proof of Theorem 1.

𝐹 (𝑥, 𝑦)

The representation of 𝐹 satisfying the conditions of Theorem 1 is given in Section 4.

{𝑓 (𝑥, 𝑦) , = {𝑆](𝑙𝑟 ) (𝑥, 𝑦) , (𝜏 ) (𝑟) {𝑀] 𝑟 (𝑥, 𝑦) + 𝑁] (𝑥, 𝑦) ,

Proof of Theorem 2. Let 𝐹 be the smooth extension of 𝑓 from Ω to 𝑇 which is stated as in Theorem 1. Define 𝐹𝑝 by 𝐹𝑝 (𝑥 + 𝑘, 𝑦 + 𝑙) = 𝐹 (𝑥, 𝑦)

𝜕𝑖+𝑗 𝐹𝑝 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝑦) = 0

((𝑥, 𝑦) ∈ 𝜕𝑇; 0 ≤ 𝑖 + 𝑗 ≤ 𝑟) .

(93)

Let 𝑇𝑛1 ,𝑛2 = [𝑛1 , 𝑛1 + 1] × [𝑛2 , 𝑛2 + 1](𝑛1 , 𝑛2 ∈ Z). Since 𝐹𝑝 is 1-periodic function, we have 𝐹𝑝 ∈ 𝐶𝑟 (𝑇𝑛1 ,𝑛2 ) and for any 𝑛1 , 𝑛2 ∈ Z, 𝜕𝑖+𝑗 𝐹𝑝 𝜕𝑥𝑖 𝜕𝑦𝑗

(99) where

((𝑥, 𝑦) ∈ 𝑇; 𝑘, 𝑙 ∈ Z) . (92)

Then 𝐹𝑝 is a 1-periodic function of R2 . By Theorem 1, we know that 𝐹𝑝 ∈ 𝐶𝑟 (𝑇) and

((𝑥, 𝑦) ∈ 𝜕𝑇𝑛1 ,𝑛2 ; 0 ≤ 𝑖 + 𝑗 ≤ 𝑟) . (94)

2

𝑟

2

Noticing that R = ⋃𝑛1 ,𝑛2 ∈Z 𝑇𝑛1 ,𝑛2 , we have 𝐹𝑝 ∈ 𝐶 (R ). By (92) and Theorem 1(i), we get 𝐹𝑝 (𝑥, 𝑦) = 𝐹 (𝑥, 𝑦) = 𝑓 (𝑥, 𝑦)

((𝑥, 𝑦) ∈ Ω) .

1

1

(100)

and the rectangles {𝐻] }41 and the trapezoids {𝐸] }41 with a curved side are stated in (22) and (23) and 𝑙𝑟 = 𝑟(𝑟 + 1)(𝑟 + 2)(𝑟 + 3) and 𝜏𝑟 = 𝑟(𝑟 + 2). Below we write out the representations of {𝑆](𝑘) (𝑥, 𝑦)}41 , {𝑀](𝑘) (𝑥, 𝑦)}41 , and {𝑁](𝑘) (𝑥, 𝑦)}41 . 4.1. The Construction of Each 𝑆](𝑘) (𝑥,𝑦). (i) Denote 𝑘

(𝑦 − 𝑔 (𝑥)) 𝑦 𝑘+1 ) , ( 𝑘! 𝑔 (𝑥)

𝑦𝑘 𝑦 − 𝑔 (𝑥) 𝑘+1 𝑏𝑘,1 (𝑥, 𝑦) = ( ) , 𝑘! −𝑔 (𝑥)

(101)

𝑘 = 0, 1, . . . .

Define 𝑆1(𝑘) (𝑥, 𝑦) by induction as follows:

𝑆1(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝑔 (𝑥)) 𝑎0,1 (𝑥, 𝑦) ,

Proof of Theorem 3. Let 𝐹 be the smooth extension of 𝑓 from Ω to 𝑇 which is stated as in Theorem 1. Define 𝐹𝑐 by 𝐹 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ 𝑇, 𝐹𝑐 (𝑥, 𝑦) = { 0, (𝑥, 𝑦) ∈ R2 \ 𝑇.

𝑆1(𝑘) (𝑥, 𝑦) = 𝑆1(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,1 (𝑥, 𝑦) ×(

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑓 (𝑥, 𝑔 − (𝑥, 𝑔 (𝑥))) (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘

(96) − 𝑏𝑘,1 (𝑥, 𝑦)

From Theorem 1(ii), we have

𝜕𝑘 𝑆1(𝑘−1) (𝑥, 0), 𝜕𝑦𝑘 𝑘 = 1, 2, . . . ,

((𝑥, 𝑦) ∈ 𝜕𝑇; 0 ≤ 𝑖 + 𝑗 ≤ 𝑟) .

(97)

((𝑥, 𝑦) ∈ 𝐸1 ) . (102)

(ii) Denote 𝑐

𝑟

2

From this and (96), we get 𝐹 (𝑥, 𝑦) ∈ 𝐶 (R ). By (96) and Theorem 1(i), we get 𝐹𝑐 (𝑥, 𝑦) = 𝐹 (𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) Theorem 3 is proved.

4

(95)

Theorem 2 is proved.

𝜕𝑖+𝑗 𝐹𝑐 (𝑥, 𝑦) = 0 𝜕𝑥𝑖 𝜕𝑦𝑗

4

𝑇 = Ω ⋃ (⋃𝐸] ) ⋃ (⋃𝐻] )

𝑎𝑘,1 (𝑥, 𝑦) = (𝑥, 𝑦) = 0

(𝑥, 𝑦) ∈ Ω, (𝑥, 𝑦) ∈ 𝐸] (] = 1, 2, 3, 4) , (𝑥, 𝑦) ∈ 𝐻] (] = 1, 2, 3, 4) ,

((𝑥, 𝑦) ∈ Ω) .

(98)

𝑘

𝑎𝑘,2 (𝑥, 𝑦) =

𝑘+1

(𝑥 − ℎ (𝑦)) 1−𝑥 ) ( 𝑘! 1 − ℎ (𝑦)

,

𝑘+1

(𝑥 − 1)𝑘 ℎ (𝑦) − 𝑥 𝑏𝑘,2 (𝑥, 𝑦) = ) ( 𝑘! ℎ (𝑦) − 1

,

𝑘 = 0, 1, . . . .

(103)

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Define 𝑆2(𝑘) (𝑥, 𝑦) by induction as follows:

4.2. The Constructions of Each 𝑀](𝑘) (𝑥,𝑦) and 𝑁](𝑘) (𝑥,𝑦). (i) Denote

𝑆2(0) (𝑥, 𝑦) = 𝑓 (ℎ (𝑦) , 𝑦) 𝑎0,2 (𝑥, 𝑦) ,

𝑘

𝑆2(𝑘) (𝑥, 𝑦) = 𝑆2(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,2 (𝑥, 𝑦) ×(

𝜕𝑘 𝑆2(𝑘−1) 𝜕𝑘 𝑓 (ℎ (𝑦) , 𝑦) − (ℎ (𝑦) , 𝑦)) 𝜕𝑥𝑘 𝜕𝑥𝑘

− 𝑏𝑘,2 (𝑥, 𝑦)

𝜕𝑘 𝑆2(𝑘−1) (1, 𝑦) , 𝜕𝑥𝑘

𝛽𝑘,11 (𝑦) =

𝑘+1

𝑘

𝑦 𝑦 − 𝑦1 ) ( 𝑘! −𝑦1

(𝑙 )

𝑀1(0) (𝑥, 𝑦) = 𝑆4 𝑟 (𝑥, 𝑦1 ) 𝛼0,11 (𝑦) ,

𝑘

𝑎𝑘,3 (𝑥, 𝑦) =

𝑘+1 (𝑦 − 𝑔∗ (𝑥)) 1−𝑦 ) , ( 𝑘! 1 − 𝑔∗ (𝑥)

𝑏𝑘,3 (𝑥, 𝑦) =

(𝑦 − 1) 𝑔∗ (𝑥) − 𝑦 𝑘+1 ( ∗ ) , 𝑘! 𝑔 (𝑥) − 1

(𝑙 )

×(

𝜕𝑘 𝑆4 𝑟 𝜕𝑘 𝑀1(𝑘−1) (𝑥, 𝑦 ) − (𝑥, 𝑦1 )) 1 (110) 𝜕𝑦𝑘 𝜕𝑦𝑘

𝑘

𝑘 = 0, 1, . . . .

− 𝛽𝑘,11 (𝑦)

(105) 𝑆3(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝑔∗ (𝑥)) 𝑎0,3 (𝑥, 𝑦) ,

𝑘

𝜕𝑘 𝑆3(𝑘−1) 𝜕𝑘 𝑓 ∗ (𝑥, 𝑔 (𝑥, 𝑔∗ (𝑥))) − (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘

− 𝑏𝑘,3 (𝑥, 𝑦)

𝜕𝑘 𝑆3(𝑘−1) 𝜕𝑦𝑘

(𝑥, 1) ,

𝑘 = 1, 2, . . .

(𝑥 − ℎ∗ (𝑦)) 𝑥 ) ( ∗ 𝑘! ℎ (𝑦) ∗ 𝑥𝑘 𝑥 − ℎ (𝑦) ) ( 𝑘! −ℎ∗ (𝑦)

((𝑥, 𝑦) ∈ 𝐸3 ) . (106)

(𝑥 − 𝑥1 ) 𝑥 𝑘+1 ( ) , 𝑘! 𝑥1

𝑥𝑘 𝑥 − 𝑥1 𝑘+1 𝛽𝑘,14 (𝑥) = ( ) , 𝑘! −𝑥1

(111) 𝑘 = 0, 1, . . . .

(𝑙 )

𝑁1(0) (𝑥, 𝑦) = (𝑆1 𝑟 (𝑥1 , 𝑦) − 𝑀1(2𝑟) (𝑥1 , 𝑦)) 𝛼0,14 (𝑥) 𝑁1(𝑘) (𝑥, 𝑦) = 𝑁1(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,14 (𝑥)

𝑘+1

,

×(

(107)

𝑘+1

,

𝛼𝑘,14 (𝑥) =

Define 𝑁1(𝑘) (𝑥, 𝑦) by induction as follows:

(iv) Denote 𝑘

((𝑥, 𝑦) ∈ 𝐻1 ) .

Denote

𝑆3(𝑘) (𝑥, 𝑦) = 𝑆3(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,3 (𝑥, 𝑦) ×(

𝜕𝑘 𝑀1(𝑘−1) (𝑥, 0) , 𝜕𝑦𝑘

𝑘 = 1, 2, . . . ,

Define 𝑆3(𝑘) (𝑥, 𝑦) by induction as follows:

𝑏𝑘,4 (𝑥, 𝑦) =

𝑘 = 0, 1, . . . .

𝑀1(𝑘) (𝑥, 𝑦) = 𝑀1(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,11 (𝑦)

(iii) Denote

𝑎𝑘,4 (𝑥, 𝑦) =

,

(109)

Define 𝑀1(𝑘) (𝑥, 𝑦) by induction as follows: ((𝑥, 𝑦) ∈ 𝐸2 ) . (104)

𝑘 = 1, 2, . . . ,

(𝑦 − 𝑦1 ) 𝑦 𝑘+1 𝛼𝑘,11 (𝑦) = ( ) , 𝑘! 𝑦1

𝑘 = 0, 1, . . . .

𝜕𝑘 (𝑆1(4𝑟) − 𝑀1(2𝑟) ) 𝜕𝑥𝑘

(𝑥1 , 𝑦)

𝜕𝑘 𝑁1(𝑘−1) (𝑥1 , 𝑦)) − 𝜕𝑥𝑘

Define 𝑆4(𝑘) (𝑥, 𝑦) by induction as follows: 𝑆4(0) (𝑥, 𝑦) = 𝑓 (ℎ∗ (𝑦) , 𝑦) 𝑎0,4 (𝑥, 𝑦) ,

− 𝛽𝑘,14 (𝑥)

𝑆4(𝑘) (𝑥, 𝑦) = 𝑆4(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,4 (𝑥, 𝑦)

𝜕𝑘 𝑁1(𝑘−1) (0, 𝑦) , 𝜕𝑥𝑘

𝑘 = 1, 2, . . .

((𝑥, 𝑦) ∈ 𝐻1 ) .

×(

𝑘

𝜕 𝑓 ∗ (ℎ (𝑦) , 𝑦) − 𝜕𝑥𝑘

𝜕𝑘 𝑆4(𝑘−1) 𝜕𝑥𝑘

(ℎ∗ (𝑦) , 𝑦))

(ii) Denote 𝑘

(𝑥 − 𝑥2 ) 1 − 𝑥 𝑘+1 ) , 𝛼𝑘,21 (𝑥) = ( 𝑘! 1 − 𝑥2

𝜕𝑘 𝑆4(𝑘−1) − 𝑏𝑘,4 (𝑥, 𝑦) (0, 𝑦) , 𝜕𝑥𝑘 𝑘 = 1, 2, . . .

(112)

((𝑥, 𝑦) ∈ 𝐸4 ) . (108)

𝑘+1 (𝑥 − 1)𝑘 𝑥2 − 𝑥 𝛽𝑘,21 (𝑥) = ( ) , 𝑘! 𝑥2 − 1

(113) 𝑘 = 0, 1, . . . .

The Scientific World Journal

13

(𝜏 )

Define 𝑀3(𝑘) (𝑥, 𝑦) by induction as follows:

Define 𝑀2 𝑟 (𝑥, 𝑦) by induction as follows:

(𝑙 )

𝑀3(0) (𝑥, 𝑦) = 𝑆2 𝑟 (𝑥, 𝑦3 ) 𝛼0,31 (𝑦) .

(𝑙 )

𝑀2(0) (𝑥, 𝑦) = 𝑆1 𝑟 (𝑥2 , 𝑦) 𝛼0,21 (𝑥) ,

𝑀3(𝑘) (𝑥, 𝑦) = 𝑀3(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,31 (𝑦)

𝑀2(𝑘) (𝑥, 𝑦) = 𝑀2(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,21 (𝑥)

(𝑙 )

×(

(𝑙 )

𝜕𝑘 𝑆 𝑟 × ( 1𝑘 (𝑥2 , 𝑦) 𝜕𝑥 𝜕𝑘 𝑀2(𝑘−1) (𝑥2 , 𝑦)) − 𝜕𝑥𝑘

− 𝛽𝑘,31 (𝑦)

(114)

𝜕𝑘 𝑀3(𝑘−1) (𝑥, 1) , 𝜕𝑦𝑘

𝑘 = 1, 2, . . .

((𝑥, 𝑦) ∈ 𝐻3 ) .

Denote

𝜕𝑘 𝑀2(𝑘−1) (1, 𝑦) , − 𝛽𝑘,21 (𝑥) 𝜕𝑥𝑘 𝑘 = 1, 2, . . .

𝜕𝑘 𝑀3(𝑘−1) 𝜕𝑘 𝑆2 𝑟 (𝑥, 𝑦 ) − (𝑥, 𝑦3 )) 3 (118) 𝜕𝑦𝑘 𝜕𝑦𝑘

𝑘

𝛼𝑘,32 (𝑥) =

((𝑥, 𝑦) ∈ 𝐻2 ) .

(𝑥 − 𝑥3 ) 1 − 𝑥 𝑘+1 ) , ( 𝑘! 1 − 𝑥3

(119)

𝑘+1 (𝑥 − 1)𝑘 𝑥3 − 𝑥 𝛽𝑘,32 (𝑥) = ( ) , 𝑘! 𝑥3 − 1

Denote

𝑘 = 0, 1, . . . .

Define 𝑁3(𝑘) (𝑥, 𝑦) by induction as follows: (𝑙 )

𝛽𝑘,22 (𝑦) =

𝑦𝑘 𝑦 − 𝑦2 𝑘+1 ) , ( 𝑘! −𝑦2

(115)

𝑁3(𝑘) (𝑥, 𝑦) = 𝑁3(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,32 (𝑥) (𝑙 )

𝑘 = 0, 1, . . . .

×(

Define 𝑁2(𝑟) (𝑥, 𝑦) by induction as follows: (𝑙 )

𝜕𝑦𝑘

(𝜏 )

𝑁2(𝑘) (𝑥, 𝑦) = 𝑁2(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,22 (𝑦) (𝑙 )

(𝜏 )

𝜕𝑘 (𝑆2 𝑟 − 𝑀2 𝑟 ) 𝜕𝑦𝑘 −

𝜕

𝑘

𝑁2(𝑘−1) 𝜕𝑦𝑘

(𝑥, 𝑦2 )

(𝑥3 , 𝑦)

− 𝛽𝑘,32 (𝑥)

𝜕𝑘 𝑁3(𝑘−1) (1, 𝑦) , 𝜕𝑥𝑘

𝑘 = 1, 2, . . .

((𝑥, 𝑦) ∈ 𝐻3 ) .

𝑘

(116) (𝑥, 𝑦2 ))

𝛼𝑘,41 (𝑥) =

(𝑥 − 𝑥4 ) 𝑥 𝑘+1 ( ) , 𝑘! 𝑥4

(121)

𝑘

(𝑥 − 𝑥4 ) 𝑥 𝑘+1 ) , 𝛽𝑘,41 (𝑥) = ( 𝑘! −𝑥4

𝑘 = 0, 1, . . . .

Define 𝑀4(𝑘) (𝑥, 𝑦) by induction as follows:

((𝑥, 𝑦) ∈ 𝐻2 ) .

(𝑙 )

𝑀4(0) (𝑥, 𝑦) = 𝑆3 𝑟 (𝑥4 , 𝑦) 𝛼0,41 (𝑥) , 𝑀4(𝑘) (𝑥, 𝑦) = 𝑀4(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,41 (𝑥)

(iii) Denote

(𝑙 )

×(

𝑘

𝛼𝑘,31 (𝑦) =

(𝑦 − 𝑦3 ) 1 − 𝑦 𝑘+1 ) , ( 𝑘! 1 − 𝑦3

(117)

𝑘

𝛽𝑘,31 (𝑦) =

(120)

(iv) Denote

𝜕𝑘 𝑁2(𝑘−1) − 𝛽𝑘,22 (𝑦) (𝑥, 0) , 𝜕𝑦𝑘 𝑘 = 1, 2, . . .

(𝜏 )

𝜕𝑘 (𝑆3 𝑟 − 𝑀3 𝑟 )

𝜕𝑘 𝑁3(𝑘−1) − (𝑥3 , 𝑦)) 𝜕𝑥𝑘

𝑁2(0) (𝑥, 𝑦) = (𝑆2 𝑟 − 𝑀2 𝑟 ) (𝑥, 𝑦2 ) 𝛼0,22 (𝑦) ,

×(

(𝜏 )

𝑁3(0) (𝑥, 𝑦) = (𝑆3 𝑟 − 𝑀3 𝑟 ) (𝑥3 , 𝑦) 𝛼0,32 (𝑥) ,

𝑘

(𝑦 − 𝑦2 ) 𝑦 𝑘+1 𝛼𝑘,22 (𝑦) = ( ) , 𝑘! 𝑦2

(𝑦 − 1) 𝑦3 − 𝑦 𝑘+1 ( ) , 𝑘! 𝑦3 − 1

𝑘 = 0, 1, . . . .

𝜕𝑘 𝑆3 𝑟 𝜕𝑘 𝑀4(𝑘−1) (𝑥 , 𝑦) − (𝑥4 , 𝑦)) (122) 4 𝜕𝑥𝑘 𝜕𝑥𝑘

− 𝛽𝑘,41 (𝑥, 𝑦)

𝜕𝑘 𝑀4(𝑘−1) (0, 𝑦) , 𝜕𝑥𝑘

𝑘 = 1, 2, . . .

((𝑥, 𝑦) ∈ 𝐻4 ) .

14

The Scientific World Journal We define {𝑆1(𝑘) (𝑥, 𝑦)} on 𝐺1 as follows. Let

Denote

𝑆1(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝜂 (𝑥)) 𝑎1,0 (𝑥, 𝑦)

𝑘

(𝑦 − 𝑦4 ) 1 − 𝑦 𝑘+1 𝛼𝑘,42 (𝑦) = ) , ( 𝑘! 1 − 𝑦4

(123)

𝑘

𝛽𝑘,42 (𝑦) =

(𝑦 − 1) 𝑦4 − 𝑦 𝑘+1 ( ) , 𝑘! 𝑦4 − 1

and let 𝑘0 be the maximal integer satisfying 1+2+⋅ ⋅ ⋅+𝑘0 ≤ 𝑞. For 𝑘 = 1, 2, . . . , 𝑘0 , we define 𝑆1(𝑘) (𝑥, 𝑦) = 𝑆1(𝑘−1) (𝑥, 𝑦) + 𝑎1,𝑘 (𝑥, 𝑦)

𝑘 = 0, 1, . . . .

Define 𝑁4(𝑘) (𝑥, 𝑦) by induction as follows: (𝑙 )

×(

(𝜏 )

𝑁4(0) (𝑥, 𝑦) = (𝑆4 𝑟 − 𝑀4 𝑟 ) (𝑥, 𝑦4 ) 𝛼0,42 (𝑦) ,

(𝑙 )

(𝜏 )

𝜕𝑘 (𝑆4 𝑟 − 𝑀4 𝑟 ) 𝜕𝑦𝑘

𝑘

𝑎𝑘,2 (𝑥, 𝑦) =

(𝑥, 𝑦4 )

𝜕𝑘 𝑁4(𝑘−1) (𝑥, 𝑦4 )) − 𝜕𝑦𝑘

(124)

𝑘 = 1, 2, . . .

By using the extension method given in Section 3, we discuss the two important special cases. 5.1. Smooth Extensions of Functions on a Kind of Domains. Let Ω be a trapezoid with two curved sides: (125)

where 𝐿 1 < 𝜂(𝑥) < 𝜉(𝑥) < 𝐿 2 (𝑥1 ≤ 𝑥 ≤ 𝑥2 ), 𝜂, 𝜉 ∈ 𝐶𝑚 ([𝑥1 , 𝑥2 ]). Denote the rectangle 𝐷 = [𝑥1 , 𝑥2 ] × [𝐿 1 , 𝐿 2 ]. Then 𝐷 = 𝐺1 ⋃ Ω ⋃ 𝐺2 , where 𝐺1 and 𝐺2 are both trapezoids with a curved side:

𝐺2 = {(𝑥, 𝑦) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝜉 (𝑥) ≤ 𝑦 ≤ 𝐿 2 } .

((𝑥, 𝑦) ∈ 𝐺2 ) . (132) Then 𝑆2(𝑘) ∈ 𝐶𝜆 𝑘 (𝐺2 ), where 𝜆 𝑘 is stated as above. An argument similar to Lemmas 5 and 6 shows that, for 0 ≤ 𝑘 ≤ 𝑘0 and 0 ≤ 𝑖 + 𝑗 ≤ min{𝑘, 𝜆 𝑘 }, 𝜕𝑖+𝑗 𝑆1(𝑘) 𝜕𝑖+𝑗 𝑓 (𝑥, 𝜂 = (𝑥, 𝜂 (𝑥)) , (𝑥)) 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑖+𝑗 𝑆2(𝑘) 𝜕𝑖+𝑗 𝑓 𝜉 = (𝑥, (𝑥)) (𝑥, 𝜉 (𝑥)) , 𝜕𝑥𝑖 𝜕𝑦𝑗 𝜕𝑥𝑖 𝜕𝑦𝑗

(133)

(𝑥1 ≤ 𝑥 ≤ 𝑥2 ) . A direct calculation shows that the number

𝑘

𝑘 = 1, 2, . . . (𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝑦 ∈ R) .

𝜕𝑘 𝑆2(𝑘−1) 𝜕𝑘 𝑓 𝜉 − (𝑥, (𝑥)) (𝑥, 𝜉 (𝑥))) , 𝜕𝑦𝑘 𝜕𝑦𝑘

(126)

Suppose that 𝑓 ∈ 𝐶𝑞 (Ω) (𝑞 is a nonnegative integer). We will smoothly extend 𝑓 from Ω to the trapezoids 𝐺1 and 𝐺2 with a curved side, respectively, as in Section 3.2, such that the extension function 𝐹 is smooth on the rectangle 𝐷. Moreover, we will give a precise formula. It shows that the index of smoothness of 𝐹 depends on not only smoothness of 𝑓 but also smoothness of 𝜂, 𝜉. Denote 𝑎0,1 (𝑥, 𝑦) = (𝑦 − 𝐿 1 )/(𝜂(𝑥) − 𝐿 1 ) and (𝑦 − 𝜂 (𝑥)) 𝑦 − 𝐿 1 𝑘+1 ) , ( 𝑘! 𝜂 (𝑥) − 𝐿 1

(131)

𝑆2(𝑘) (𝑥, 𝑦) = 𝑆2(𝑘−1) (𝑥, 𝑦) + 𝑎𝑘,2 (𝑥, 𝑦) ×(

𝐺1 = {(𝑥, 𝑦) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝐿 1 ≤ 𝑦 ≤ 𝜂 (𝑥)} ,

((𝑥, 𝑦) ∈ 𝐺2 ) .

For 𝑘 = 1, 2, . . . , 𝑘0 , define

((𝑥, 𝑦) ∈ 𝐻4 ) .

Ω = {(𝑥, 𝑦) : 𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝜂 (𝑥) ≤ 𝑦 ≤ 𝜉 (𝑥)} ,

(130)

(𝑥1 ≤ 𝑥 ≤ 𝑥2 , 𝑦 ∈ R) .

𝑆2(0) (𝑥, 𝑦) = 𝑓 (𝑥, 𝜉 (𝑥)) 𝑎0,2 (𝑥, 𝑦)

5. Corollaries

𝑎𝑘,1 (𝑥, 𝑦) =

(𝑦 − 𝜉 (𝑥)) 𝐿 − 𝑦 𝑘+1 ) , ( 2 𝑘! 𝐿 2 − 𝜉 (𝑥)

We define {𝑆2(𝑘) (𝑥, 𝑦)} on 𝐺2 as follows. Let

𝜕𝑘 𝑁4(𝑘−1) − 𝛽𝑘,42 (𝑥, 𝑦) (𝑥, 1) , 𝜕𝑥𝑘 𝑘 = 1, 2, . . .

𝜕𝑘 𝑆1(𝑘−1) 𝜕𝑘 𝑓 (𝑥, 𝜂 − (𝑥, 𝜂 (𝑥))) . (𝑥)) 𝜕𝑦𝑘 𝜕𝑦𝑘 (129)

Then 𝑆1(𝑘) ∈ 𝐶𝜆 𝑘 (𝐺1 ), where 𝜆 𝑘 = min{𝑞 − 1 − 2 − ⋅ ⋅ ⋅ − 𝑘, 𝑚}. Denote 𝑎0,2 (𝑥, 𝑦) = (𝐿 2 − 𝑦)/(𝐿 2 − 𝜉(𝑥)) and

𝑁4(𝑘) (𝑥, 𝑦) = 𝑁4(𝑘−1) (𝑥, 𝑦) + 𝛼𝑘,42 (𝑦) ×(

((𝑥, 𝑦) ∈ 𝐺1 ) , (128)

𝜏 (𝑞, 𝑚) = min {[√ 2𝑞 +

(134)

is the maximal value of integers 𝑘 satisfying 𝑘 ≤ 𝜆 𝑘 , where [⋅] expresses the integral part. So 𝜏(𝑞, 𝑚) ≤ 𝜆 𝜏(𝑞,𝑚) . By (133), we get that, for 0 ≤ 𝑖 + 𝑗 ≤ 𝜏(𝑞, 𝑚), (𝜏(𝑞,𝑚))

𝜕𝑖+𝑗 𝑆1 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝜂 (𝑥)) =

𝜕𝑖+𝑗 𝑓 (𝑥, 𝜉 (𝑥)) , 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥, 𝜉 (𝑥)) =

𝜕𝑖+𝑗 𝑓 (𝑥, 𝜉 (𝑥)) , 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝜏(𝑞,𝑚))

(127)

9 3 − ] , 𝑚} 4 2

𝜕𝑖+𝑗 𝑆2 𝜕𝑥𝑖 𝜕𝑦𝑗

(𝑥1 ≤ 𝑥 ≤ 𝑥2 ) .

(135)

The Scientific World Journal

15

Note that (𝜏(𝑞,𝑚))

𝑆1

(𝜏(𝑞,𝑚))

∈ 𝐶𝜆 𝜏(𝑞,𝑚) (𝐺1 ) ,

𝑆2

∈ 𝐶𝜆(𝑞,𝑚) (𝐺2 ) ,

𝜏 (𝑞, 𝑚) ≤ 𝜆 𝜏(𝑞,𝑚) ≤ 𝑞,

𝑆1(0) (𝑥) = 𝑓(𝑥2 )((1 − 𝑥)/(1 − 𝑥2 )) and for 𝑘 =

Define 1, . . . , 𝑞, (136)

(𝑘)

𝑆1(𝑘) (𝑥) = 𝑆1(𝑘−1) (𝑥) − 𝑎1(𝑘) (𝑥) (𝑓(𝑘) (𝑥2 ) − (𝑆1(𝑘−1) )

and the assumption 𝑓 ∈ 𝐶𝑞 (Ω). Now we define a function on 𝐷 by

− 𝑏1(𝑘) (𝑥) (𝑆1(𝑘−1) )

(137)

Theorem 12. Let the domain Ω and the rectangle 𝐷 be stated as above. If 𝑓 ∈ 𝐶𝑞 (Ω), then the function 𝐹𝑞,𝑚 (𝑥, 𝑦), defined in (137), is a smooth extension of 𝑓 from Ω to 𝐷 and 𝐹𝑞,𝑚 ∈ 𝐶𝜏(𝑞,𝑚) (𝐷), where 𝜏(𝑞, 𝑚) is stated in (134).

(𝑞)

(𝑞) (𝑘)

(𝑆1 ) (𝑞) (𝑘) (𝑆1 )

5.2. Smooth Extensions of Univariate Functions on Closed Intervals. Let 𝑓 ∈ 𝐶𝑞 ([𝑥1 , 𝑥2 ]) and [𝑥1 , 𝑥2 ] ⊂ (0, 1). In order to extend smoothly 𝑓 from [𝑥1 , 𝑥2 ] to [0, 𝑥1 ], we construct two polynomials 𝑘

(𝑥 − 𝑥1 ) 𝑥 𝑘+1 𝑎0(𝑘) (𝑥) = ( ) , 𝑘! 𝑥1

(138)

𝑘 = 0, 1, . . . .

Define 𝑆0(0) (𝑥) = 𝑓(𝑥1 )(𝑥/𝑥1 ) and for 𝑘 = 1, . . . , 𝑞,

(𝑘)

(𝑘)

(𝑥1 ))

(0 ≤ 𝑥 ≤ 𝑥1 ) .

(0)

(139) (𝑞)

Then 𝑆0 (𝑥) is a polynomial of degree ≤ 2𝑞 + 1. Similar to the proof of Lemma 5, we get (𝑞) (𝑘)

(𝑆0 )

(𝑞) (𝑘)

(𝑆0 )

(0) = 0,

(𝑥1 ) = 𝑓(𝑘) (𝑥1 ) ,

(140)

𝑘 = 0, 1, . . . , 𝑞. It is also easy to check directly them. Again extend smoothly 𝑓 from [𝑥1 , 𝑥2 ] to [𝑥2 , 1], we construct two polynomials 𝑎1(𝑘)

𝑘

(𝑥 − 𝑥2 ) 1 − 𝑥 𝑘+1 ) , ( (𝑥) = 𝑘! 1 − 𝑥2

𝑏1(𝑘) (𝑥) =

𝑘

𝑘+1

(𝑥 − 1) 𝑥2 − 𝑥 ( ) 𝑘! 𝑥2 − 1

,

(141) 𝑘 = 0, 1, . . . .

(𝑥2 ) = 𝑓(𝑘) (𝑥2 ) ,

(1) = 0

(143)

(𝑘 = 0, 1, . . . , 𝑞) .

Therefore, we obtain the smooth extension 𝐹 from [𝑥1 , 𝑥2 ] to [0, 1] by

Especially, for 𝑞 = 0 and 𝑚 ≥ 0, we have 𝜏(𝑞, 𝑚) = 0, and so 𝐹0,𝑚 ∈ 𝐶(𝐷); for 𝑞 = 2 and 𝑚 ≥ 1, we have 𝜏(𝑞, 𝑚) = 1, and so 𝐹2,1 ∈ 𝐶1 (𝐷); for 𝑞 = 5 and 𝑚 ≥ 2, we have 𝜏(𝑞, 𝑚) = 2, and so 𝐹5,2 ∈ 𝐶2 (𝐷).

− 𝑏0(𝑘) (𝑥) (𝑆0(𝑘−1) )

(𝑥2 ≤ 𝑥 ≤ 1) .

Then 𝑆1 (𝑥) is a polynomial of degree ≤ 2𝑞 + 1. Similar to the proof of Lemma 5, we get

From this and (135), we have 𝐹𝑞,𝑚 ∈ 𝐶𝜏(𝑞,𝑚) (𝐷). This implies the following theorem.

𝑆0(𝑘) (𝑥) = 𝑆0(𝑘−1) (𝑥) − 𝑎0(𝑘) (𝑥) (𝑓(𝑘) (𝑥1 ) − (𝑆0(𝑘−1) )

(1)

(142)

𝑓 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ Ω, { { (𝜏(𝑞,𝑚)) 𝐹𝑞,𝑚 (𝑥, 𝑦) = {𝑆1 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ 𝐺1 , { (𝜏(𝑞,𝑚)) 𝑆 (𝑥, 𝑦) , (𝑥, 𝑦) ∈ 𝐺2 . { 2

𝑥𝑘 𝑥 − 𝑥 𝑘+1 𝑏0(𝑘) (𝑥) = ( 1 ) , 𝑘! 𝑥1

(𝑘)

(𝑥2 ))

𝑓 (𝑥) , { { (𝑞) 𝐹 (𝑥) = {𝑆0 (𝑥) , { (𝑞) {𝑆1 (𝑥) , (𝑞)

𝑥 ∈ [𝑥1 , 𝑥2 ] , 𝑥 ∈ [0, 𝑥1 ] , 𝑥 ∈ [𝑥2 , 1] ,

(144)

(𝑞)

where 𝑆0 (𝑥) and 𝑆1 (𝑥) are polynomials of degree 2𝑞 + 1 defined as above, and 𝐹 ∈ 𝐶𝑞 ([0, 1]) and 𝐹(𝑙) (0) = 𝐹(𝑙) (1) = 0 (𝑙 = 0, 1, ..., 𝑞). From this, we get the following. Theorem 13. Let 𝑓 ∈ 𝐶𝑞 ([𝑥1 , 𝑥2 ]) and [𝑥1 , 𝑥2 ] ⊂ (0, 1). Then there exists a function 𝐹 ∈ 𝐶𝑞 ([0, 1]) satisfying 𝐹(𝑥) = 𝑓(𝑥) (𝑥1 ≤ 𝑥 ≤ 𝑥2 ) and 𝐹(𝑙) (0) = 𝐹(𝑙) (1) = 0 (𝑙 = 0, 1, . . . , 𝑞). Let 𝑓 ∈ 𝐶𝑞 ([𝑥1 , 𝑥2 ]) and [𝑥1 , 𝑥2 ] ⊂ (0, 1), and let 𝐹 be the smooth extension of 𝑓 from [𝑥1 , 𝑥2 ] to [0, 1] which is stated as in Theorem 12. Let 𝐹𝑝 be the 1-periodic extension satisfying 𝐹𝑝 (𝑥 + 𝑛) = 𝐹(𝑥) (0 ≤ 𝑥 ≤ 1, 𝑛 ∈ Z). Then 𝐹𝑝 ∈ 𝐶𝑞 (R) and 𝐹𝑝 (𝑥) = 𝑓(𝑥) (𝑥 ∈ [𝑥1 , 𝑥2 ]). We expand 𝐹(𝑥) into the Fourier series which converges fast. From this, we get trigonometric approximation of 𝑓 ∈ 𝐶𝑞 ([𝑥1 , 𝑥2 ]). We also may do odd extension or even extension of 𝐹 from [0, 1] to [−1, 1], and then doing periodic extension, we get the odd periodic extension 𝐹𝑝𝑜 ∈ 𝐶𝑞 (R) or the even periodic extension 𝐹𝑝𝑒 ∈ 𝐶𝑞 (R). We expand 𝐹𝑝𝑜 or 𝐹𝑝𝑒 into the sine series and the cosine series, respectively. From this, we get the sine polynomial approximation and the cosine polynomial approximation of 𝑓 on [𝑥1 , 𝑥2 ]. For 𝐹 ∈ 𝐶𝑞 (𝑥) (𝑥 ∈ [0, 1]), we pad zero in the outside of [0, 1] and then the obtained function 𝐹𝑐 ∈ 𝐶𝑞 (R). We expand 𝐹𝑐 into a wavelet series which converges fast. By the moment theorem, a lot of wavelet coefficients are equal to zero. From this, we get wavelet approximation of 𝑓 ∈ 𝐶𝑞 ([𝑥1 , 𝑥2 ]).

Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.

16

Acknowledgments This research is partially supported by the National Key Science Program no. 2013CB956604; the Beijing Higher Education Young Elite Teacher Project; the Fundamental Research Funds for the Central Universities (Key Program) no. 105565GK; and the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

References [1] I. Daubechies, “Orthonormal bases of compactly supported wavelets,” Communications on Pure and Applied Mathematics, vol. 41, pp. 909–996, 1998. [2] I. Daubechies, Ten Lectures on Wavelets, vol. 6 of CBMSConference Lecture Notes, SIAM, Philadelphia, Pa, USA, 1992. [3] C. K. Chui, L. L. Schumaker, and J. Stockler, Approximation Theory X: Wavelets, Splines, and Applications, Vanderbilt University Press, 2002. [4] C. K. Chui, W. Schempp, and K. Zeller, Multivariate Approximation Theory, Birkhauser, 1989. [5] C. K. Chui, “Approximation induced by a fourier series,” Rocky Mountain Journal of Mathematics, vol. 4, no. 4, pp. 643–648, 1974. [6] R. Long, Multidimensional Wavelet Analysis, International Publishing, Beijing, China, 1995. [7] G. G. Lorentz, Approximation of Functions, Holt, Rinebart and Winston, 1966. [8] P. Petrushev, “Multivariate n-term rational and piecewise polynomial approximation,” Journal of Approximation Theory, vol. 121, no. 1, pp. 158–197, 2003. [9] E. M. Stein and G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, 1971. [10] A. F. Timan, Theory at Approximation of Functions of a Real Variable, Pergamon, 1963. [11] Z. Zhang, “Convergence and Gibbs phenomenon of periodic wavelet frame series,” Rocky Mountain Journal of Mathematics, vol. 39, no. 4, pp. 1373–1396, 2009. [12] Z. Zhang, “Measures, densities and diameters of frequency bands of scaling functions and wavelets,” Journal of Approximation Theory, vol. 148, no. 2, pp. 128–147, 2007. [13] D. Zhou and K. Jetter, “Approximation with polynomial kernels and SVM classifiers,” Advances in Computational Mathematics, vol. 25, no. 1–3, pp. 323–344, 2006. [14] A. Zygmund, Trigonometric Series, I, II, Cambridge University Press, 2nd edition, 1968.

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