Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2012, Article ID 175292, 13 pages doi:10.1155/2012/175292 Research Article Calculation of the Reproducing Kernel on the Reproducing Kernel Space with Weighted Integral Er Gao,1, 2 Songhe Song,1, 2 and Xinjian Zhang1 1 Department of Mathematics and Systems Science, College of Science, National University of Defense Technology, Changsha 410073, China 2 State key Laboratory of High Performance Computing, National University of Defense Technology, Changsha 410073, China Correspondence should be addressed to Er Gao, gao.nudter@gmail.com Received 28 April 2012; Accepted 19 July 2012 Academic Editor: Livija Cveticanin Copyright q 2012 Er Gao et al. 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 provide a new definition for reproducing kernel space with weighted integral and present a method to construct and calculate the reproducing kernel for the space. The new reproducing kernel space is an enlarged reproducing kernel space, which contains the traditional reproducing kernel space. The proposed method of this paper is a universal method and is suitable for the case of that the weight is variable. Obviously, this new method will generalize a number of applications of reproducing kernel theory to many areas. 1. Introduction A reproducing kernel is a basic tool for studying the spline interpolation of differential operators and is also the base of the reproducing kernel method, which were widely used in numerical analysis, genetic models, pattern analysis, and so forth. The concept of reproducing kernel is derived from the study of the integration equation, and paper 1 studied specially the reproducing kernels and presented its primary theory. From then on, the reproducing kernel theory and the reproducing kernel method have been studied by many authors 2–7. Let W2m 0, T denote the function space on a finite interval 0, T , W2m 0, T {ft, t ∈ 0, T , f m−1 t ∈ L2 0, T }, and this space becomes a reproducing kernel Hilbert space RKHS if we endow it with some inner product. This kind of the reproducing kernel space is the most popular space for solving the boundary value problems using reproducing kernel method. But in paper 8, the author firstly considered the reproducing kernel space with 2 0, T {ut, t ∈ 0, T , u t is an absolute continuous real-valued weighted integral W2,ρ 2 Journal of Applied Mathematics T √ function on 0, T , 0 tu t2 dt < ∞} and used it solving Volterra integral equation with 2 2 0, T and W2,ρ 0, T will be more weakly singular kernel. It is obvious that W22 0, T ⊆ W2,ρ widely applied. In this paper, we are concerned with the reproducing kernel space with weighted m 0, T {ut, t ∈ 0, T , u t, . . . , um−1 are absolute continuous real-valued integral W2,α T functions on 0, T , 0 tα um t2 dt < ∞}, where α is a constant and satisfies 0 ≤ α < 1 m 0, T is W2m 0, T . The method for computing the corresponding when α 0, W2,α reproducing kernel is given. 2. Preliminaries In order to get the main results of the paper, we introduce the method of Zhang for calculating the reproducing kernel of W2m a, b in a nutshell in this section. The method of Zhang has very powerful system modeling capability. The idea is coming from the relationship between the Green function with reproducing kernel. Set L Dm am−1 Dm−1 · · · a1 D a0 t, t ∈ a, b, where aj t ∈ Cj a, b and Ker L {f ∈ W2m a, b : Lf 0}. Definition 2.1. ϕ1 t, . . . , ϕm t are the basis in Ker L. The i-th row of Wronskian matrix i−1 i−1 is ϕ1 t, . . . , ϕm t, and the last line of its inverse matrix is ϕ1 t, . . . , ϕm t. Call ϕ1 t, . . . , ϕm t are the adjunct functions of ϕ1 t, . . . , ϕm t. Lemma 2.2. Assume gt, τ m ei t ei τt − τ0 2.1 i1 and γ1 , . . . , γm is a system of linear independent functions in Ker L and satisfies b γk g·, τuτdτ a b γk g·, τuτdτ, 2.2 a where e1 t, . . . , em t are the dual basis of Ker L relative to γ1 , . . . , γm , and e1 t, . . . , em t are the adjunct functions of e1 t, . . . , em t. Then for any functions f ∈ W2m a, b, they satisfy the form ft m γi f ei t i1 b Gt, τ · Lfτdτ, 2.3 a where Gt, τ is defined below Gt, τ gt, τ − m γi g·, τ ei t i1 and the expression is exclusive. 2.4 Journal of Applied Mathematics 3 Lemma 2.3. L is a linear differential operator. Assume γ1 , . . . , γm are linear independent functions in Ker L, satisfying 2.2. Then W2m a, b is a Hilbert space if the inner product is defined by the following form: m f, h γi f γi h b LftLht dt, a i1 f, h ∈ W2m a, b. 2.5 Lemma 2.4. Under the assumptions of Lemma 2.2 and the inner product 2.5, the Hilbert space W2m a, b is reproducing kernel Hilbert space with the reproducing kernel can be denoted by b m Kt, τ ei tei τ Gt, xGτ, x dx. 2.6 a i1 3. The New Method for Computing the Reproducing Kernel It is known that the reproducing kernel of a reproducing kernel Hilbert space is existence and uniqueness. The reproducing kernel K of a Hilbert space H completely determines the space H. This section discusses the method of calculating reproducing kernels for the following two cases. The first case is when the weight is constant. In the second case we deal with the m 0, T , where 0 ≤ α < 1, and the result of this part is the main result of this general space W2,α paper. 3.1. Case 1: The Weight Is Constant For general space W2m 0, T , let L Dm be the linear differential operator of order m, and let λ1 , λ2 , . . . , λm be the linear independent functions on Ker L, where Ker L is defined by Ker L {f ∈ W2m 0, T , Lf 0}. Let e1 , e2 , . . . , em be the dual basis of Ker L relative to λ1 , λ2 , . . . , λm . That means Lei 0, λi ej δij , i, j 1, 2, . . . , m. 3.1 Let G be the Green’s function of L and satisfy Lt Gt, s δt − s, λi G·, s 0, i 1, . . . , m. 3.2 By the Lemma 2.4, W2m 0, T is a reproducing kernel Hilbert space if the inner product is defined by the following form: m λi f λi h f, h 1 i1 T LftLhtdt, 0 f, h ∈ W2m 0, T 3.3 4 Journal of Applied Mathematics and the reproducing kernel is m K1 t, τ ei tei τ T Gt, xGτ, xdx. 3.4 0 i1 Let m f, h 2 ai λi f λi h i1 T bLftLhtdt, 0 f, h ∈ W2m 0, T , 3.5 where both a1 , a2 , . . . , am and b are positive real numbers. The following proposition holds. Theorem 3.1. Using the above hypothesis, W2m 0, T is a reproducing kernel Hilbert space if it has been endowed with the inner product 3.5 and the reproducing kernel is K2 t, τ m ei tei τ ai i1 T 0 1 Gt, xGτ, x dx. b 3.6 √ √ bL bDm , and λi f √ai λi f. Proof. Let L From It is obvious that λ1 , λ2 , . . . , λm are also the linear independent functions on Ker L. Lemma 2.3, we have that W2m 0, T is a Hilbert space if the inner product is defined by m T f, h L Lft Lhtdt λi f λi h 0 i1 m ai λi f λi h T 0 i1 bLftLhtdt f, h 2 . 3.7 Next, we will proof K2 t, s is the reproducing kernel of the space W2m 0, T with the inner product ·, ·2 . K1 t, s is the reproducing kernel of the space W2m 0, T with the inner product ·, ·1 . In particular, K1 t, s is contained in W2m 0, T . So K2 t, s is also contained in W2m 0, T . For any f ∈ W2m 0, T , fs, K2 t, s 2 m i1 ai λi f λi K2 t, · T 0 bf m s ∂m K2 t, sds. ∂sm 3.8 Journal of Applied Mathematics 5 From 3.1 and 3.2, we have fτ, K2 t, τ m 2 ai λi f λi m ei tei τ i1 ai i1 m λi f λi m i1 T bf 0 ei tei τ T f m 0 i1 m T ∂m τ m ∂τ ∂m τ m ∂τ 0 1 Gt, xGτ, xdx dτ b 3.9 T Gt, xGτ, xdx dτ. 0 Similarly, from 3.1 and 3.2, we obtain ft fτ, K1 t, τ 1 m λi f λi i1 m ei tei τ T 0 i1 f m s ∂m ∂τ m T 3.10 Gt, xGτ, xdx dτ. 0 So ft fτ, K2 t, τ2 holds. The proof is complete. 3.2. Case 2: The Weight Is Variable m In this case, we construct the inner product of the space W2,α 0, T , and calculate the corresponding reproducing kernel. m 0, T , we know that L1 : Define L1 tα/2 Dm . From the definition of the space W2,α m m 2 W2,α 0, T → L 0, T mapping the space W2,α 0, T to the square integrable space on 0, T . Under the hypothesis of Case 1, λ1 , λ2 , . . . , λm are also the linear independent functions m 0, T , L1 f 0}, and e1 , e2 , . . . , em is on Ker L1 , where Ker L1 is defined by Ker L1 {f ∈ W2,α also the dual basis of Ker L1 relative to λ1 , λ2 , . . . , λm . e1 t, . . . , em t are the adjunct functions of e1 t, . . . , em t. Similar to Case 1, define an algorithm ·, ·3 as the following form m λi f λi h f, h 3 i1 T L1 ftL1 htdt, 0 m f, h ∈ W2,α 0, T 3.11 m 0, T . Theorem 3.2. Under the above assumption, ·, ·3 is the inner product of the space W2,α If act in accordance with the four basic rules of the inner product, the proof of this proposition is easy. So one overlaps the proof. m 0, T into two parts Ker L1 and Ker L1 ⊥ , where Ker L1 is the Divide the space W2,α linear space of order m. From the results in 9, one has the following proposition. 6 Journal of Applied Mathematics Proposition 3.3. Under the above assumption, Ker L1 is the reproducing kernel Hilbert space with the inner product below m λi f λi h f, h 4 f, h ∈ Ker L1 3.12 i1 and the corresponding reproducing kernel is K4 t, τ m ei tei τ. 3.13 i1 Let ⎧ m ⎪ m ⎨ 1 ei t ei τ, 1 0 g1 t, τ α/2 ei t ei τt − τ τ α/2 i1 ⎪ τ ⎩ i1 0, G1 t, τ g1 t, τ − m t ≥ τ, t < τ, 3.14 λi g1 ·, τ ei t. i1 It is obvious that G1 t, τ 1/τ α/2 Gt, τ. The following theorem holds. Theorem 3.4. G1 is the Green’s function of L1 and for any f ∈ W2m 0, T , ut L1 ft, satisfies T L1t G1 t, τuτdτ ut, 0 λi G1 ·, τ 0, 3.15 i 1, 2, . . . , m. Proof. For any i 1, 2, ·, m, we have λi G1 ·, τ τ α/2 λi G·, τ. 3.16 From 3.2, λi G1 ·, τ τ α/2 λi G·, τ 0, i 1, 2, ·, m. 3.17 Then from the results in 9, 10, for any f ∈ W2m 0, T , we obtain T Lt 0 Gt, τLfτdτ Lf. 3.18 Journal of Applied Mathematics 7 So T L1t G1 t, τuτdτ tα/2 Lt 0 T 0 t α/2 1 τ α/2 Gt, ττ α/2 Lfτdτ 3.19 Lft L1 ft ut. The proof is complete. Remark 3.5. If acting in accordance with the process of the paper 9, we have T m ei tei τ G1 t, xG1 τ, xdx. K3 t, τ 3.20 0 i1 But K3 t, τ is not the reproducing kernel of W2m 0, T , since K3 t, τ ∈ / W2m 0, T . m Now, we will give an important property of the arbitrary element of W2,α 0, T . m Theorem 3.6. For any f ∈ W2,α 0, T , L1 ft ut. Then there are some real constant c1 , c2 , . . . , cm , satisfying ft m ci ei t T m 0 i1 ei t ei τ uτdτ, 3.21 i1 and the expression is exclusive. m Proof. L1 tα/2 Dm is a linear mapping, and L1 : W2,α 0, T → L2 0, T is a homomorphic 2 mapping. For any ut ∈ L 0, T , we have a function hx satisfies T m ht ei t ei τ uτdτ. 0 3.22 i1 m m Because L1 ht ut ∈ L2 0, T , ht ∈ W2,α 0, T holds. So L1 : W2,α 0, T → L2 0, T is a surjective homomorphism. Ker L1 Span{e1 , e2 , . . . , em } is a linear system and the dimension of the system is m. So from the knowledge of the group homomorphism, we have m L1 : W2,α 0, T / Ker L1 −→ L2 0, T 3.23 is isomorphic. m On the one hand for any ht ∈ W2,α 0, T / Ker L1 , there exists the exclusive ut, satisfying L1 ht ut. On the other hand for the ut ∈ L2 0, T , h0 t 8 Journal of Applied Mathematics T m m i1 ei t ei τuτdτ satisfies h0 t ∈ W2,α 0, T / Ker L1 and L1 h0 t ut. So for 0 m ht ∈ W2,α 0, T / Ker L1 , that holds ht T m 0 3.24 ei t ei τ L1 hτdτ. i1 m 0, T , ft f0 t ht holds, where f0 t ∈ Ker L1 and For any ft ∈ W2,α m ht ∈ W2,α 0, T / Ker L1 . At the same time, the decomposition is exclusive because of the m 0, T / Ker L1 . orthogonality between Ker L1 and W2,α Furthermore, Ker L1 Span{e1 , e2 , . . . , em }, so f0 t m i1 ci ei t, where c1 , c2 , . . . , cm are real numbers, and the expression is exclusive. m 0, T , we have So for any ft ∈ W2,α ft f0 t ht m ci ei t T m 0 i1 ei t ei τ uτdτ, 3.25 i1 where c1 , c2 , . . . , cm are real numbers, and the the expression is exclusive. The proof is complete. Then similar to the Theorem 3.4, we have the following theorem. m Theorem 3.7. G1 is the Green’s function of L1 and for any f ∈ W2,α 0, T , L1 ft ut, satisfies T L1t G1 t, τuτdτ ut. 3.26 0 m 0, T , L1 ft ut, Proof. For any f ∈ W2,α T L1t G1 t, τuτdτ tα/2 Lt 0 T 0 t α/2 1 Gt, ττ α/2 Lfτdτ τ α/2 T Lt 3.27 Gt, τLfτdτ. 0 From Theorem 3.6, we have Lft Lt T Gt, τLfτdτ. 3.28 0 Thus, we know that 3.26 is true. The proof is complete. m Theorem 3.8. Under the above hypothesis and the inner product ·, ·3 , W2,α 0, T is the Hilbert space. Journal of Applied Mathematics 9 m 0, T . Proof. The norm of the space is denoted by u3 u, u3 , where u ∈ W2,α ∞ m Suppose that {fn }n1 is a Cauchy sequence in W2,α 0, T , that is, fnp − fn 2 −→ 0 n −→ ∞. 3 3.29 From Theorems 3.6 and 3.1, we have m 2 fnp − fn 2 λi fnp − λi fn 3 T i1 2 L1 fnp t − fn t dt −→ 0 n −→ ∞. 3.30 0 By the completeness of Ker L1 and L2 0, T , there exist a real number ri , i 1, 2, . . . , m and a real function h ∈ L2 0, T , such that lim λi fn ri , n→∞ T lim n→∞ Set f0 t m i1 ri ei t T 0 i 1, 2, . . . , m, L1 fn t − ht 2 3.31 0. 0 m G1 t, τhτ dτ. It follows that f0 ∈ W2,α 0, T and fn − f0 2 −→ 0 n −→ ∞. 3 3.32 m m So W2,α 0, T is complete. Namely, W2,α 0, T is Hilbert space. The proof is complete. m Theorem 3.9. Under the above hypothesis and the inner product ·, ·3 , W2,α 0, T is the reproducing kernel Hilbert space, and the reproducing kernel is K3 t, τ T m ei tei τ G1 t, xG1 τ, xdx. 3.33 0 i1 Proof. From Theorem 3.8 and Proposition 3.3, we only need to demonstrate that K5 t, τ T G1 t, xG1 τ, xdx. 3.34 0 is the reproducing kernel of Ker L1 ⊥ , where the inner product is defined by f, h 5 T L1 ftL1 htdt, f, h ∈ Ker L1 ⊥ . 0 0, so K5 t, τ ∈ Ker L1 ⊥ . From Theorem 3.7, L1 K5 t, τ / 3.35 10 Journal of Applied Mathematics For any h ∈ Ker L1 ⊥ , hτ, K5 t, τ5 T T L1 hτL1 0 G1 t, xG1 τ, xdx dτ. 3.36 0 From Theorem 3.7, hτ, K5 t, τ5 T L1 hτG1 t, τdτ. 3.37 0 Furthermore, from the definition of the Ker L1 ⊥ , we have T L1 hτG1 t, τdτ 0 m λi hei t T L1 hτG1 t, τdτ. 3.38 0 i1 Finally, from the Theorem 3.6, ht m λi hei t T L1 hτG1 t, τdτ. 3.39 0 i1 So hτ, K5 t, τ5 ht. 3.40 The proof is complete. 4. Example 2 Example 4.1. We consider the space mentioned in the introduction W2,1/2 0, 1 {ut, t ∈ √ 1 0, 1, u t is an absolute continuous real-valued function on 0, 1, 0 tu t2 dt < ∞}. 2 0, 1 is endowed Let L D2 , λ1 u u0, and λ2 u u 0. Using Theorems 3.8 and 3.9, W2,1/2 with the inner product: u, v3 u0v0 u 0v 0 1 √ xu xv xdx and the corresponding reproducing kernel is ⎧ 4xy3/2 4y5/2 ⎪ ⎪ ⎪1 xy − , ⎪ 3 15 ⎨ K x, y ⎪ ⎪ ⎪ 4x3/2 y 4x5/2 ⎪ ⎩1 xy − , 3 15 This result is in accord with Theorem 2.1 of 8. 4.1 0 y ≤ x, 4.2 y > x. Journal of Applied Mathematics 11 2 0, 1 is given by If λ1 u u0 and λ2 u u1 and the inner product of W2,1/2 u, v3 u0v0 u1v1 1 √ xu xv xdx, 4.3 0 using the method of this paper, the reproducing kernel of the this space is 46xy 4 y − 5 xy3/2 4x − 5x3/2 y K x, y 1 − x − y 15 15 15 ⎧ 3/2 y − 5x 4y ⎪ ⎪ ⎪ , y ≤ x, ⎪ ⎨ 15 − ⎪ ⎪ ⎪ 4x3/2 x − 5y ⎪ ⎩ , y > x. 15 4.4 4 Example 4.2. We consider the space W2,α 0, 1 {ut, t ∈ 0, 1, u3 t is an absolute 1 continuous real-valued function on 0, 1, 0 tα u4 t2 dt < ∞}. Let L D4 , λ1 u u0, λ2 u u1, λ3 u u 0 and λ4 u u 1. The inner product is given by u, v3 u0v0 u1v1 u 0v 0 u 1v 1 1 xα u4 xv4 xdx 4.5 0 Similar to Example 4.1, we can compute the reproducing kernel of the reproducing kernel 4 0, 1 is space W2,α 2 K x, y x − 12 x −1 y y x2 −3 2xy2 −3 2y 2 −1 xx2 −1 y y2 −1 x2 1 2x −1 y 1 2y 2−1 xx2 3 3α −1 y − 8y y2 − 720 − 1764α 1624α2 − 735α3 175α4 − 21α5 α6 10x2 −3 2x 1 a −1 y − 3y y2 − −5040 13068α − 13132α2 6769α3 − 1960α4 322α5 − 28α6 α7 R x, y , y ≤ x, − r x, y − r y, x R y, x , y > x, 4.6 12 Journal of Applied Mathematics where r x, y x4−α y2 −3 −42 55α − 14α2 α3 x − 3 −14 23α − 10α2 α3 x2 −2 y 6−7 α−2 α−1 α 360 − 342α 119α2 − 18α3 α4 4−α 2 y −210 107α − 18α2 α3 y −6 11α − 6α2 α3 x3 −3 2y x , 6−7 α−2 α−1 α 360 − 342α 119α2 − 18α3 α4 R x, y y4−α −7 α−6 α−5 αx3 − 3−7 α−6 α−1 αx2 y 6−7 α−6 α−5 α−4 α−3 α−2 α−1 α y4−α 3−7 α−2 α−1 αxy2 − −3 α−2 α−1 αy3 . 6−7 α−6 α−5 α−4 α−3 α−2 α−1 α 4.7 5. Conclusion In this paper, we have proposed a method to compute the reproducing kernel on the reproducing kernel space with weighted integral. Theorems 3.8 and 3.9 are the most important theorems of the paper. To our best knowledge, Theorem 3.6 is the first results about m 0, T . From the example, we know that the reproducing the component of the space W2,α m 0, T , and the proposed method of this paper kernel space of 8 is just one space of the W2,α is a universal method. Acknowledgment The work is supported by NSF of China under Grant no. 10971226. References 1 N. Aronszajn, “Theory of reproducing kernels,” Transactions of the American Mathematical Society, vol. 68, pp. 337–404, 1950. 2 C.-l. Li and M. Cui, “The exact solution for solving a class nonlinear operator equations in the reproducing kernel space,” Applied Mathematics and Computation, vol. 143, no. 2-3, pp. 393–399, 2003. 3 M. Cui and F. 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Jiang, “The exact solution of a class of Volterra integral equation with weakly singular kernel,” Applied Mathematics and Computation, vol. 217, no. 18, pp. 7515–7519, 2011. Journal of Applied Mathematics 13 9 X. J. Zhang and H. Long, “Computing reproducing kernels for W2m a, b I,” Mathematica Numerica Sinica, vol. 30, no. 3, pp. 295–304, 2008. 10 X. J. Zhang and S. R. Lu, “Computing reproducing kernels for W2m a, b II,” Mathematica Numerica Sinica, vol. 30, no. 4, pp. 361–368, 2008. 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