On the Solution of the n-dimensional k B Operator

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1 Applied Mathematical Sciences, Vol. 9, 015, no. 10, HIKARI Ltd, On the Solution of the n-dimensional B Operator Sudprathai Bupasiri Faculty of Education Mathematics Program Saon Nahon Rajabhat University Saon Nahon 47000, Thailand Copyright c 014 Sudprathai Bupasiri. 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 wor is properly cited. Abstract In this paper, we consider the solution of the equation Bux) = δ where B is the operator related to the Bessel diamond operator iterated -time and is defind by [ Bx1 ) 4 B = + B x + + B xp Bxp+1 + B x + + B ) ] 4 p+ x p+q, where p + q = n, B xi = + v x i x i i x i, v i = α i + 1, α i > 1 [], x i > 0, i = 1,,..., n, is a nonnegative integer and n is the dimension of R + n. In this wor we study the elementary solution of the operator B. Mathematics Subject Classification: 46F10 Keywords: diamond operator, Bessel diamond operator, O-plus operator 1 Introduction H. Yildirim, M.Z. Sariaya and S. Oztur [5] have first introduced the elementary solution of the n-dimensional Bessel diamond operator and the Fourier- Bessel transform of their convolution and showed that the solution of the convolution form 1) S x) R x) is a unique elementary solution of the B ux) = δ.

2 470 Sudprathai Bupasiri Consider the Bessel ultra-hyperbolic operator iterated -times, [ p B = B xi p+q ] Yildirim, Sariaya and Oztur [5] has shown that the generalized function R x) define by 14) is the unique elementary solution of the operator B, that is B R x) = δ where x R + n = {x : x = x 1,... x n ), x 1 > 0, x > 0,..., x n > 0}. Yildirim, Sariaya and Oztur [5] studied the Bessel diamond operator, iterated -times, ) B = B xi [ p = B xi p+q ] [ p ) B xi + p+q ], 1) Yildirim, Sariaya and Oztur [5] showed that the function ux) = 1) S x) R x) is the unique elementary solution for the operator B, where indicates convolution, and S x), R x) are defined by 14) and 17) respectively, that is, 1) S x) R x) ) = δx). ) B Furthermore, the operator was first studied by Kananthai, Suantai and Longani [1]. The operator can be expressed in the form =. [ p x i x i ) + i p+q x j x j ) ] [ p x i i p+q x j ]. 3) Satsanit [4] has studied the Green function and Fourier transform for o-plus

3 On the solution of the n-dimensional B operator 471 operators, iterated -times, defined by where = x i ) 4 x j ) 4 [ ) ) ] + = + ) = + =. 4) = x i ) + x j The purpose of this wor is to study the operator ) 4 B = B xi ) = B xi Let us denote the operator By 8) and 9) we obtain ) 4 ) ) B = B xi + ) B = B xi + ) p ) B xi + ) ).. ). 5) [ B ) ) ] + B B B = + ) = B + B. 6)

4 47 Sudprathai Bupasiri Thus, 5) can be written as B = B B = B B. 7) For = 1 the operator B can be expressed in the form B = B B = B B where B is the Bessel ultra-hyperbolic operator, B = B x1 + B x + + B xp B xp+1 B xp+ B xp+q, 8) where p + q = n and B is the Laplace Bessel operator, B = B x1 + B x + + B xp + B xp+1 + B xp+ + + B xp+q. 9) From 5) with q = 0 and = 1, we obtain where B = 4 B 10) B = B x1 + B x + + B xp. 11) We can find the elementary solution ux) of the operator B ; that is, Bux) = δ, 1) where δ is the Dirac-delta distribution. Moreover, we found that ux) relates to the elementary solution of the Laplace Bessel operator defined by 9) depending on the conditions of q and of 5) with q = 0 and = 1. In finding the elementary solution of 1), we use the method of convolutions of the generalized function. Preliminary Notes Denoted by Tx y the generalized shift operator acting according to the law [] π π Tx y ϕx) = Cv... ϕ x 1 + y 1 x 1 y 1 cos θ 1,..., ) x n + y n x n y n cos θ n 0 0 Π n sin v i 1 ) dθ 1... dθ n, Γv i +1) where x, y R + n, Cv = Π n. We remar that this shift operator is Γ 1 )Γv i ) closely connected with the Bessel differential operator []. d U dx + v x du dx = d U dy + v du y dy Ux, 0) = fx),

5 On the solution of the n-dimensional B operator 473 U y x, 0) = 0. The convolution operator determined by Tx y is as follow: f ϕ) = fy)tx y ϕx) ) Π n y v i i dy. 13) R + n Convolution 13) is nown as a B-convolution. We note the following properties for the B-convolution and the generalized shift operator: a) T y x 1 = 1. b) T 0 x fx) = fx). c) If fx), gx) CR + n ), gx) is a bounded function, x > 0 and 0 fx) ) Π n x v i i dx <, then R + n Tx y fx)gy) ) Π n y v i i dy = R + n fy)tx y gx) ) Π n y v i i dy. d) From c), we have the following equality for gx) = 1, Tx y fx) ) Π n y v i i dy = fy) ) Π n y v i i dy R + n R + n e) f g)x) = g f)x). Lemma.1 Given the equation B ux) = δx) for x R+ n, where B is the Bessel-ultra hyperbolic operator iterated -times defined by 8). Then ux) = R x) is an elementary solution of the operator B, where for and R x) = V n v K n ) x 1 + x + + x p x p+1 x p+ x n v ) p+q) = K n ) 14) V = x 1 + x + + x p x p+1 x p+ x p+q 15) K n ) = π n+ v 1 ) Γ + n v Γ + p v ) Γ ) 1 Γ) Γ p ). 16)

6 474 Sudprathai Bupasiri Lemma.1 Given the equation B ux) = δx) for x R+ n, where B is the Laplace Bessel operator iterated -times defined by 9). Then ux) = 1) S x) is an elementary solution of the operator B, where S x) = x n v, p + q = n, 17) w n ) and x = x 1 + x + + x n ) 1 18) w n ) = Πn v 1 i Γ v i + ) 1 Γ) ). 19) n+ v 4 Γ n+ v Lemma. The convolution R x) 1) S x) is an elementary solution for the operator B iterated -times and is defined by 1). Lemma.3 R x) and S x) are homogeneous distributions of order n v ). We need to show that R x) and 1) S x) satisfy the Euler equation; that is, n v ) R x) = n x i R x), n v ) S x) = x i n x i S x). x i Lemma.4 The B-convolution of tempered distribution). R x) S x) exists and is a tempered distribution. For the proof of Lemma.1- Lemma.4, see [5], p ). Lemma.5 The function R x) and 1) S x) are the inverse in the convolution algebra of R x) and 1) S x), respectively. That is, R x) R x) = R + x) = R 0 x) = δx), 1) S x) 1) S x) = S + x) = S 0 x) = δx) Lemma.6 The B-convolution of R x) and S x)). Let R x) and S x) defined by 14) and 17) respectively, then we obtain: 1) S x) S m x) = S +m x), where and m are nonnegative integers. ) R x) R m x) = R +m x), where and m are nonnegative integers. For the proof of Lemma.5 and Lemma.6, see [3].

7 On the solution of the n-dimensional B operator Main Results Theorem 3.1 Given the equation BGx) = δx) 0) for x R + n, where B is the operator iterated -times is defined by 6). Then we obtain Gx) is an elementary solution of 0), where Gx) = R 4 x) 1) S 4 x)) C x)) 1 1) where Cx) = 1 R 4x) + 1 1) S 4 x). ) Here C x) denotes the convolution of Cx) itself -times, C x)) 1 denotes the inverse of C x) in the convolution algebra. Moreover Gx) is a tempered distribution. Proof. We have ) BGx) = B + B Gx) = δx) or we can write 1 B + 1 ) 1 B B + 1 ) 1 B Gx) = δx). Convolving both sides of the above equation by R 4 x) 1) S 4 x), 1 B + 1 ) B R 4 x) 1) S 4 x) ) 1 B + 1 ) 1 B Gx) = δx) R 4 x) 1) S 4 x) or 1 B R4 x) 1) S 4 x) ) + 1 B R4 x) 1) S 4 x) )) 1 B + 1 ) 1 B Gx) = δx) R 4 x) 1) S 4 x). By properties of convolutions, 1 B 1) S 4 x) ) R 4 x) B R 4 x)) 1) S 4 x)) B + 1 ) 1 B Gx) = δx) R 4 x) 1) S 4 x).

8 476 Sudprathai Bupasiri By Lemma.1 and., we obtain 1 δ R 4x) + 1 ) 1 δ 1) S 4 x) B + 1 ) 1 B Gx) = δx) R 4 x) 1) S 4 x). or 1 R 4x) + 1 ) 1 1) S 4 x) B + 1 ) 1 B Gx) = R 4 x) 1) S 4 x). Keeping on convolving both sides of the above equation by R 4 x) 1) S 4 x) up to 1 times, we obtain C x) Gx) = R 4 x) 1) S 4 x) ) 3) the symbol denotes the convolution of itself -times. By properties of R x) and S x) in Lemma.7, we have R4 x) 1) S 4 x) ) x) = R4 x) 1) S 4 x). Thus 3) becomes, or C x) Gx) = R 4 x) 1) S 4 x) Gx) = R 4 x) 1) S 4 x) ) C x) ) 1 4) is an elementary solution of 0). We consider the function C x), since R 4 x) 1) S 4 x) is a tempered distribution. Thus Cx) defined by ) is tempered distribution, we obtain C x) is tempered distribution. Now, R 4 x) 1) S 4 x) S, the space of tempered distribution. Choose S D R, where D R is the right-side distribution which is a subspace of D of distribution. Thus R 4 x) 1) S 4 x) D R. It follow that R 4 x) 1) S 4 x) is an element of convolution algebra, since D R is a convolution algebra. Hence Zemanian [6], 1) has a unique solution Gx) = R 4 x) 1) S 4 x) ) C x) ) 1, where C x) ) 1 is an inverse of C x) in the convolution algebra. Gx) is called the Green function of the operator B. Since R 4 x) 1) S 4 x) and C x) ) 1 are lies in S, then by [6], p.15) again, we have R 4 x) 1) S 4 x) ) C x) ) 1 S. Hence Gx) is a tempered distribution.

9 On the solution of the n-dimensional B operator 477 Theorem 3. Given the equation Bux) = δx), 5) where B is the operator iterated -times defined by 5), δx) is the Diracdelta distribution, x R + n and is a nonnegative integer. Then we obtain or ux) = R x) 1) S x) ) Gx) 6) ux) = R 6 x) 1) 3 S 6 x) ) C x) ) 1 7) is a Green s function or an elementary solution for the operator B iterated -times where B is defined by 5), and Gx) defined by 1). For q = 0, then 5) becomes 4 B ux) = δx), 8) we obtain ux) = S 8 x) is an elementary solution of 8), where 4 B is the Laplace Bessel operator of p-dimension, iterated 4-times and is defined by 11). Moreover, we obtain R 4 x) 1) 3 S 6 x) C x) ) ux) = R x) 9) as an elementary solution of the Bessel ultra-hyperbolic operator iterated - times is defined by 8). Proof. From 7) and 5), we have Bux) = B B) ux) = δx). 30) Convolving both sides of 30) by R x) 1) S x) ) Gx), we obtain R x) 1) S x) ) Gx) ) B B ux) = δx) R x) 1) S x) ) Gx). By properties of convolution B R x) 1) S x) ) B Gx)) ux) = R x) 1) S x) ) Gx). By Lemma.3 and Theorem 3.1, we obtain, δ δ ux) = ux) = R x) 1) S x) ) Gx). By Lemma.7 and 1), we obtain, ux) = R 6 x) 1) 3 S 6 x) ) C x) ) 1 31)

10 478 Sudprathai Bupasiri is an elementary solution or Green s function of B operator. Now, for q = 0 the 5) becomes 4 B ux) = δx), 3) where 4 B is Laplace Bessel operator of p-dimension iterated 4-times. By Lemma., we have ux) = 1) 4 S 8 x) = S 8 x) is an elementary solution of 8). On the other hand, we can also find ux) from 31). Since q = 0, we have R x) reduces to 1) S x). Thus, by 31) for q = 0, we obtain From 31), we have ux) = S 6 x) 1) 6 S 6 x) ) 1) S 4 x) ) 1 = 1) 6 S 6+6 x) 1) S 4 x) ) 1 = S 8 x). ux) = R 6 x) 1) 3 S 6 x) ) C x) ) 1. Convolving the above equation by R 6 x) 1) 3 S 6 x) C x) ). By Lemma.6 and.7, we obtain or R 6 x) 1) 3 S 6 x) C x) ) ux) = R 0 x) S 0 x) δx) R x) R 6 x) 1) 3 S 6 x) C x) ) ux) = δx) δx) δx) R x). It follows that R 6 x) 1) 3 S 6 x) C x) ) ux) = R x) 33) as an elementary solution of the operator B iterated -times defined by 8) Acnowledgements. The author would lie to than the referee for his suggestions which enhanced the presentation of the paper. The author was supported by Saon Nahon Rajabhat University. References [1] A. Kananthai, S. Suantai and V. Longani, On the operator related to the wave equation and Laplacian, Appl. Math. Comput, 1300), pp

11 On the solution of the n-dimensional B operator 479 [] B. M. Levitan, Expansion in Fourier series and integrals with Bessel functions, Uspei Mat., Naua N.S.) 6,4)1951) in Russian). [3] M. Z. Sariaya and H. Yildirim, On the B-convolutions of the Bessel diamond ernel of Riesz, Appl. Math. Comput ), [4] W. Satsanit, Green function and Fourier transform for o-plus operator, Electronic Journal of Differential Equation, ), No. 48, pp [5] H. Yildirim, M. Z. Sariaya and S. Oztur, The solutions of the n- dimensional Bessel diamond operator and the Fourier-Bessel transform of their convolution, Proc. Indian Acad. Sci. Math.Sci.), 1144)004), [6] A. H. Zemanian, Distribution and Transform Analysis, McGraw-Hill, New Yor, Received: October 19, 014; Published: January 7, 015