EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 3, 2021, 881-894 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Operator ⊕km Related to the Wave Equation and Laplacian Sudprathai Bupasiri Faculty of Education, Sakon Nakhon Rajabhat University, Sakon Nakhon, Thailand Abstract. In this article, we study the fundamental solution of the operator ⊕k m, iterated k-times and is defined by ⊕k m = ( p∑ r=1 ∂2 ∂x2r +m2 )4 −  p+q∑ j=p+1 ∂2 ∂x2j 4  k , where m is a nonnegative real number, p + q = n is the dimension of the Euclidean space Rn, x = (x1, x2, . . . , xn) ∈ Rn, k is a nonnegative integer. At first we study the fundamental solution of the operator ⊕k m and after that, we apply such the fundamental solution to solve for the solution of the equation ⊕k mu(x) = f(x), where f(x) is generalized function and u(x) is unknown function for x ∈ Rn. 2020 Mathematics Subject Classifications: 46F10 Key Words and Phrases: Wave equation, Laplace operator, Ultra-hyperbolic operator 1. Introduction We have observed that an operational quantity such as δ(x) becomes meaningful if it is first multiplied by a sufficiently smooth auxiliary function and then integrated over the entire space. This point of view is also taken as the basis for the definition of an arbitrary generalized function. Accordingly, consider the space D consisting of real-valued function φ(x) = φ(x1, x2, . . . , xn), such that the following hold: (1) φ(x) is an infinitely differentiable function defined at every point of Rn. This mean that Dkφ exists for all multi indices k. Such a function is also call a C∞ function. (2) There exists number A such that φ(x) vanishes for r > A. This means that φ(x) has a compact support. Then φ(x) is called a test function. DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.4006 Email address: sudprathai@gmail.com (S. Bupasiri) http://www.ejpam.com 881 © 2021 EJPAM All rights reserved. S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 882 In physical problem, one often encounters idealized concepts such as a force concen- trated at a point ξ or an impulsive force that acts instaneously. These forces are described by the Dirac-delta function δ(x− ξ), which has several significant properties: δ(x− ξ) = 0, x 6= ξ (1)∫ b a δ(x− ξ)dx = { 0 for a, b < ξ or ξ < a, b 1 for a ≤ ξ ≤ b (2) and ∫ ∞ −∞ δ(x− ξ)dx = 1. (3) Equation (3) is a special case of the general formula∫ ∞ −∞ δ(x− ξ)f(x)dx = f(ξ), (4) where f(x) is a sufficiently smooth function. Relation (4) is called the sifting property or the reproducing property of the delta function, and (3) is obtained from it by putting f(x) = 1. Moreover, Kananthai et al. [1] have studied the fundamental solution of the operator ⊕k and the weak solution of the equation ⊕ku(x) = f(x), f(x) is a generalized function where the operator ⊕k is defined by ⊕k = ( p∑ r=1 ∂2 ∂x2r )4 −  p+q∑ j=p+1 ∂2 ∂x2j 4k = ( p∑ r=1 ∂2 ∂x2r )2 −  p+q∑ j=p+1 ∂2 ∂x2j 2k ( p∑ r=1 ∂2 ∂x2r )2 +  p+q∑ j=p+1 ∂2 ∂x2j 2k = �kLk1Lk2 = �kLk (5) where p+q = n is the dimension of the Euclidean space Rn and k is a nonnegative integer. Next, Kananthai et al. [2] have studied the relationship between the operator ⊕k and the wave operator, and the relationship between the operator ⊕k and the Laplacian. Moreover, equation ⊕kK(x) = δ we have K(x) = [RH2k(x) ∗ (−1)kRe2k(x)] ∗ S2k(x) ∗ T2k(x) is the fundamental solution of the operator ⊕k. Later, Kananthai [8] has studied the inversion of the kernel Kα,β,γ,ν related to the operator ⊕k. In 1988, Trione [11] has studied the fundamental solution of the ultra-hyperbolic Klein- Gordon operator iterated k-times such that operator is defined by (� +m2)k = [ ∂2 ∂x21 + ∂2 ∂x22 + · · ·+ ∂2 ∂x2p − ∂2 ∂x2p+1 − ∂2 ∂x2p+2 − · · · − ∂2 ∂x2p+q +m2 ]k . (6) S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 883 Later, Kananthai [7] has studied the fundamental solution for the (♦ + m4)k which related to the Klein-Gordon operator. From equation (5) the operator ⊕km can be expressed in the form ⊕km = ( p∑ r=1 ∂2 ∂x2r +m2 )4 −  p+q∑ j=p+1 ∂2 ∂x2j 4k = ( p∑ r=1 ∂2 ∂x2r +m2 )2 −  p+q∑ j=p+1 ∂2 ∂x2j 2k ( p∑ r=1 ∂2 ∂x2r +m2 )2 +  p+q∑ j=p+1 ∂2 ∂x2j 2k =  p∑ r=1 ∂2 ∂x2r − p+q∑ j=p+1 ∂2 ∂x2j +m2 k  p∑ r=1 ∂2 ∂x2r + p+q∑ j=p+1 ∂2 ∂x2j +m2 k ×  p∑ r=1 ∂2 ∂x2r + i p+q∑ j=p+1 ∂2 ∂x2j +m2 k  p∑ r=1 ∂2 ∂x2r − i p+q∑ j=p+1 ∂2 ∂x2j +m2 k , (7) where i = √ −1, n = p+ q. And the operator( p∑ r=1 ∂2 ∂x2r )2 −  p+q∑ j=p+1 ∂2 ∂x2j 2k is introduced by Kananthai [4] and is named the diamond operator which is defined by ♦k = ( p∑ r=1 ∂2 ∂x2r )2 −  p+q∑ j=p+1 ∂2 ∂x2j 2k . (8) Otherwise, the operator ♦k can also be expressed in the form ♦k = �k4k = 4k�k, where �k is the ultra-hyperbolic operator iterated k-times, is defined by �k = [ ∂2 ∂x21 + ∂2 ∂x22 + · · ·+ ∂2 ∂x2p − ∂2 ∂x2p+1 − ∂2 ∂x2p+2 − · · · − ∂2 ∂x2p+q ]k , (9) and 4k is the Laplace operator iterated k-times, is defined by 4k = [ ∂2 ∂x21 + ∂2 ∂x22 + · · ·+ ∂2 ∂x2n ]k . (10) By putting p = 1 and x1 = t (time) in (9), then we obtain the wave operator � = ∂2 ∂t2 − n−1∑ j=1 ∂2 ∂x2j . (11) S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 884 The operators Lk1 and Lk2 are defined by Lk1 =  p∑ r=1 ∂2 ∂x2r + i p+q∑ j=p+1 ∂2 ∂x2j k (12) and Lk2 =  p∑ r=1 ∂2 ∂x2r − i p+q∑ j=p+1 ∂2 ∂x2j k , (13) following that Lk = Lk1L k 2 = Lk2L k 1 = ( p∑ r=1 ∂2 ∂x2r )2 +  p+q∑ j=p+1 ∂2 ∂x2j 2k . (14) Thus, equation (7) can be written as ⊕km = ( � +m2 )k (4+m2 )k ( L1 +m2 )k ( L2 +m2 )k = ( L2 +m2 )k ( L1 +m2 )k (4+m2 )k ( � +m2 )k (15) and from (7) with q = m = 0 and k = 1, we obtain Laplace operator of p-dimension ⊕0 = 44 p, where 4p = ∂2 ∂x21 + ∂2 ∂x22 + · · ·+ ∂2 ∂x2p . (16) In this article, we further study the fundamental solution of the operator ⊕km, that is ⊕kmH(x,m) = δ, where H(x,m) is the fundamental solution, δ is the Dirac delta distribution, k is a non- negative integer, m is a nonnegative real number and the operator ⊕km is defined by ⊕km = ( p∑ r=1 ∂2 ∂x2r +m2 )4 −  p+q∑ j=p+1 ∂2 ∂x2j 4k . (17) We then also apply such the fundamental solution to solve the solution of the equation ⊕kmu(x) = f(x), where f(x) is a given generalized function and u(x) is an unknown function for x ∈ Rn. S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 885 2. Preliminary Notes In this section, we studied some properties of the ultra-hyperbolic kernel of Marcel Riesz and the fundamental solution of the partial differential operators which will be used as follow. Definition 1. Let x = (x1, x2, . . . , xn) be a point of the n - dimensional space Rn, u = x21 + x22 + · · ·+ x2p − x2p+1 − x2p+2 − · · · − x2p+q, (18) where p+ q = n. Define Γ+ = {x ∈ Rn : x1 > 0 and u > 0} which designates the interior of the forward cone and Γ+ designates its closure and the following functions introduce by Nozaki ([9], p.72) that RHα (x) = { u α−n 2 Kn(α) if x ∈ Γ+ 0 if x 6∈ Γ+, (19) RHα (x) is called the ultra-hyperbolic kernel of Marcel Riesz. Here α is a complex parameter and n the dimension of the space. The constant Kn(α) is defined by Kn(α) = π n−1 2 Γ ( 2+α−n 2 ) Γ ( 1−α 2 ) Γ(α) Γ ( 2+α−p 2 ) Γ (p−α 2 ) (20) and p is the number of positive terms of u = x21 + x22 + · · ·+ x2p − x2p+1 − x2p+2 − · · · − x2p+q, p+ q = n and let supp RHα (x) ⊂ Γ+. Now RHα (x) is an ordinary function if Re α ≥ n and is a distribution of α if Re α < n. Now, if p = 1 then (19) reduces to the function Mα(u) say, and defined by Mα(u) = { u α−n 2 Hn(α) if x ∈ Γ+ 0 if x 6∈ Γ+, (21) where u = x21 − x22 − · · · − x2n and Hn(α) = π (n−1) 2 2α−1Γ(α−n+2 2 ). The function Mα(u) is called the hyperbolic kernel of Marcel Riesz. Lemma 1. Given the equation 4ku(x) = δ for x ∈ Rn, where 4k is the Laplace operator iterated k-times is defined by (10). Then u(x) = (−1)kRe2k(x) is the fundamental solution of the operator 4k where Re2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) |x|2k−n. (22) Proof. See [4]. S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 886 Lemma 2. If �ku(x) = δ for x ∈ Γ+ = {x ∈ Rn : x1 > 0 and u > 0}, where �kis the ultra-hyperbolic operator iterated k-times is defined by (9). Then u(x) = RH2k(x) is the unique fundamental solution of the operator �k where RH2k(x) = u( 2k−n 2 ) Kn(2k) = (x21 + x22 + · · ·+ x2p − x2p+1 − · · · − x2p+q)( 2k−n 2 ) Kn(2k) (23) for Kn(2k) = π n−1 2 Γ ( 2+2k−n 2 ) Γ ( 1−2k 2 ) Γ(2k) Γ ( 2+2k−p 2 ) Γ(p−2k2 ) . (24) Proof. See [11]. Lemma 3. Given the equation ♦ku(x) = δ for x ∈ Rn, then u(x) = (−1)kRe2k(x) ∗ RH2k(x) is the unique fundamental solution of the operator ♦k, where ♦k is the diamond operator iterated k- times is defined by (8), Re2k(x) and RH2k(x) are defined by (22) and (23), respectively. Moreover, (−1)kRe2k(x) ∗RH2k(x) is a tempered distribution. Proof. See [4]. Lemma 4. Given the equation Lk1u(x) = δ for x ∈ Rn, where Lk1 is the operator defined by (12), then u(x) = (−1)k(−i) q 2S2k(x) is the fundamental solution of the operator Lk1, where S2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) [x21 + x22 + · · ·+ x2p − i(x2p+1 + · · ·+ x2p+q)] ( 2k−n 2 ). (25) Lemma 5. Given the equation Lk2u(x) = δ for x ∈ Rn, where Lk2 is the operator defined by (13), then u(x) = (−1)k(i) q 2T2k(x) is the fundamental solution of the operator Lk2, where T2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) [x21 + x22 + · · ·+ x2p + i(x2p+1 + · · ·+ x2p+q)] ( 2k−n 2 ). (26) Lemma 6. Given the equation Lku(x) = δ for x ∈ Rn, then u(x) = S2k(x) ∗ T2k(x) is the fundamental solution of the operator Lk, which is defined by (14), S2k(x) and T2k(x) are defined by (25) and (26), respectively. Proof. The proof of the Lemma 4, Lemma 5 and Lemma 6 are given in [2]. Lemma 7. The function RH−2k(x) and (−1)kRe−2k(x) are the inverse in the convolution algebra of RH2k(x) and (−1)kRe2k(x), respectively. Lemma 8. (1) The function S2k(x) and T2k(x) are the fundamental solution of the operator Lk1 and Lk2 , respectively, where S2k(x) and T2k(x) are defined by (25) and (26) , respectively. S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 887 (2) The function S−2k(x) and T−2k(x) are the inverse in the convolution algebra of S2k(x) and T2k(x) , respectively. Proof. The proof of the Lemma 7 and Lemma 8 are given in [2]. Lemma 9. Given the equation (� + m2)ku(x) = δ for x ∈ Rn, where � is the ultra- hyperbolic operator defined by (9). Then u(x) = W2k(x,m) is the fundamental solution of the operator (�+m2)k. In particular, for m = 0 we have W2k(x,m = 0) = RH2k(x), where W2k(x,m) = +∞∑ r=0 ( −k r ) m2rRH2k+2r(x), (27) RH2k+2r(x) is defined by (23). Proof. Since the operator � defined in equation (9) is a linearly continuous and have 1− 1 mapping, then it has inverse. From Lemma 2 and equation (27) we obtain W2k(x,m) = +∞∑ r=0 ( −k r ) (m2)r�−k−rδ = (� +m2)−kδ. (28) By applying the operator (� +m2)k to both sides of equation (28), we obtain (� +m2)kW2k(x,m) = (� +m2)k · (� +m2)−kδ. Therefore, (� +m2)kW2k(x,m) = δ. Since W2k(x,m) = ( −k 0 ) m2(0)RH2k+2(0)(x) + +∞∑ r=1 ( −k r ) m2rRH2k+2r(x). (29) The second summand of the right-hand member of (29) vanishes for m = 0 and then, we have W2k(x,m = 0) = RH2k(x) which is the fundamental solution of the ultra hyperbolic operator �k . Lemma 10. Given the equation (4+m2)ku(x) = δ for x ∈ Rn, where 4 is Laplace op- erator defined by (10). Then u(x) = Y2k(x,m) is the fundamental solution of the operator (4+m2)k. In particular, for m = 0 we have Y2k(x,m = 0) = (−1)kRe2k(x), where Y2k(x,m) = +∞∑ r=0 ( −k r ) m2r(−1)k+rRe2k+2r(x), (30) Re2k+2r(x) is defined by (22). S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 888 Proof. Since the operator 4 defined by equation (10) is a linearly continuous and have 1− 1 mapping, then it has inverse. From Lemma 1 and equation (30), we obtain Y2k(x,m) = +∞∑ r=0 ( −k r ) (m2)r 4−k−r δ = (4+m2)−kδ. (31) By applying the operator (4+m2)k to both sides of equation (31), we obtain (4+m2)kY2k(x,m) = (4+m2)k · (4+m2)−kδ. Therefore, (4+m2)kY2k(x,m) = δ. Since Y2k(x,m) = ( −k 0 ) m2(0)(−1)kRe2k+2(0)(x) + +∞∑ r=1 ( −k r ) m2r(−1)kRe2k+2r(x). (32) The second summand of the right-hand member of (32) vanishes for m = 0 and then, we have Y2k(x,m = 0) = (−1)kRe2k(x) which is the fundamental solution of the Laplace operator 4k . Lemma 11. Given the equation (L1 +m2)ku(x) = δ for x ∈ Rn, where L1 is the operator defined by (12). Then u(x) = M2k(x,m) is the fundamental solution of the operator (L1 +m2)k. In particular, for m = 0 we have M2k(x,m = 0) = (−1)k(−i) q 2S2k(x) where M2k(x,m) = +∞∑ r=0 ( −k r ) m2r(−1)k+r(−i) q 2S2k+2r(x), (33) S2k+2r(x) is defined by (25). Proof. Since the operator L1 defined in equation (12) is a linearly continuous and have 1− 1 mapping ,then it has inverse. From Lemma 4 and equation (33), we obtain M2k(x,m) = +∞∑ r=0 ( −k r ) (m2)rL−k−r1 δ = (L1 +m2)−kδ. (34) By applying the operator (L1 +m2)k to both sides of equation (34), we obtain (L1 +m2)kM2k(x,m) = (L1 +m2)k · (L1 +m2)−kδ. Therefore, (L1 +m2)kM2k(x,m) = δ. S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 889 Since M2k(x,m) = ( −k 0 ) m2(0)(−1)k+0(−i) q 2S2k+2(0)(x) + +∞∑ r=1 ( −k r ) m2r(−1)k+r(−i) q 2S2k+2r(x). (35) The second summand of the right-hand member of (35) vanishes for m = 0 and then, we have M2k(x,m = 0) = (−1)k(−i) q 2S2k(x) which is the fundamental solution of the operator Lk1 . Lemma 12. Given the equation (L2 +m2)ku(x) = δ for x ∈ Rn, where L2 is the operator defined by (13). Then u(x) = N2k(x,m) is the fundamental solution of the operator (L2 +m2)k. In particular, for m = 0 we have N2k(x,m = 0) = (−1)k(i) q 2T2k(x), where N2k(x,m) = +∞∑ r=0 ( −k r ) m2r(−1)k+r(i) q 2T2k+2r(x), (36) T2k+2r(x) is defined by (26). Proof. Since the operator L2 defined in equation (13) is a linearly continuous and have 1− 1 mapping, then it has inverse. From Lemma 5 and equation (36), we obtain N2k(x,m) = +∞∑ r=0 ( −k r ) (m2)rL−k−r2 δ = (L2 +m2)−kδ. (37) By applying the operator (L2 +m2)k to both sides of equation (37), we obtain (L2 +m2)kN2k(x,m) = (L2 +m2)k · (L2 +m2)−kδ. Therefore, (L2 +m2)kN2k(x,m) = δ. Since N2k(x,m) = ( −k 0 ) m2(0)(−1)k+0(i) q 2T2k+2(0)(x) + +∞∑ r=1 ( −k r ) m2r(−1)k+r(i) q 2T2k+2r(x). (38) The second summand of the right-hand member of (38) vanishes for m = 0 and then, we have N2k(x,m = 0) = (−1)k(i) q 2T2k(x) which is the fundamental solution of the operator Lk2 . S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 890 Lemma 13. The convolution W2k(x,m) ∗ Y2k(x,m) exists and is a tempered distribution where W2k(x,m) and Y2k(x,m) are defined by (27) and (30), respectively. Proof. See [7]. Lemma 14. The convolution M2k(x,m) ∗N2k(x,m) exists and is a tempered distribution where M2k(x,m) and N2k(x,m) are defined by (33) and (36), respectively. Proof. From (33) and (36), we have M2k(x,m) ∗N2k(x,m) = ( +∞∑ r=0 ( −k r ) m2r(−1)k+r(−i) q 2S2k+2r(x) ) ∗ ( +∞∑ r=0 ( −k r ) m2r(−1)k+r(i) q 2T2k+2r(x) ) = +∞∑ r=0 +∞∑ s=0 ( −k r )( −k s ) m2r+2sS2k+2r(x) ∗ T2k+2r(x). Since the function S2k+2r(x) and T2k+2r(x) are tempered distributions, see( [5],p.34, [2], p.302 and [6], p.97) and the convolution of functions S2k+2r(x) ∗ T2k+2r(x) exists and is also a tempered distribution, see ([3], p.152). Thus, M2k(x,m) ∗ N2k(x,m) exists and also is a tempered distribution. Lemma 15. (The convolution W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m)). The function W2k(x,m) ∗Y2k(x,m) and M2k(x,m) ∗N2k(x,m) are tempered distributions. The convolution W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m) exists and also a tempered distribution. Proof. See [10]. 3. Main Results In this main results, we obtained two theorems and such a solution H(x,m) related to the partial differential operator depends on the condition of p, q, k and m. Theorem 1. Given the equation ⊕kmH(x,m) = δ, (39) where ⊕km is the operator iterated k- times defined by (7), δ is the Dirac delta distribution, x = (x1, x2, . . . , xn) ∈ Rn , k is a nonnegative integer and m is a nonnegative real number. Then we obtain H(x,m) = W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m) (40) S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 891 is the fundamental solution for the operator ⊕km iterated k-times, where ⊕km is defined by (7). In particular, for q = m = 0 then (39) becomes 44k p H(x, 0) = δ, (41) we obtain H(x, 0) = Y8k(x, 0) = Re8k(x) (42) is the fundamental solution of (41), where 44k p is the Laplace operator of p-dimension, iterated 4k-times which is defined by (16). Moreover, from (40) H(x, 0) = [ RH2k(x) ∗ (−1)kRe2k(x) ] ∗ S2k(x) ∗ T2k(x) (43) is the fundamental solution of O-plus operator ⊕k and from (43) we obtain [(−1)kRe−2k(x) ∗ S−2k(x) ∗ T−2k(x)] ∗H(x, 0) = RH2k(x) (44) is the fundamental solution of the ultra-hyperbolic operator �k iterated k-times defined by (9), where Re−2k(x), S−2k(x) and T−2k(x) are inverse of Re2k(x), S2k(x) and T2k(x) respectively. From (43) and (44) with p = 1, q = n− 1, k = 1 and x1 = t, we obtain [(−1)kRe−2(x) ∗ S−2(x) ∗ T−2(x)] ∗H(x, 0) = M2(u) (45) is the fundamental solution of the wave operator is defined by (11) where M2(u) is defined by (21) with α = 2. Proof. From (15) and (39) we have ⊕kmH(x,m) = (( � +m2 )k (4+m2 )k ( L1 +m2 )k ( L2 +m2 )k) H(x,m) = δ. Convolving both sides of the above equation by the convolution W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m) and the properties of convolution with derivatives, we obtain( � +m2 )k W2k(x,m) ∗ ( 4+m2 )k Y2k(x,m) ∗ ( L1 +m2 )k M2k(x,m) ∗ ( L2 +m2 )k N2k(x,m) ∗H(x,m) = Y2k(x,m) ∗W2k(x,m) ∗M2k(x,m) ∗N2k(x,m) ∗ δ. (46) Thus H(x,m) = δ ∗ δ ∗ δ ∗ δ ∗H(x,m) = W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m) (47) by Lemma 9, Lemma 10, Lemma 11 and Lemma 12. Thus we obtain (40) as required. In particular, for q = m = 0 then (39) becomes 44k p H(x, 0) = δ S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 892 where 44k p is the Laplace operator of p-dimension iterated 4k-times. By Lemma 10 , we have H(x, 0) = Y8k(x, 0) = Re8k(x) (48) is the fundamental solution of (41). From Lemma 9, Lemma 10, Lemma 11 and Lemma 12 , we have H(x,m = 0) = W2k(x, 0) ∗ Y2k(x, 0) ∗M2k(x, 0) ∗N2k(x, 0) = [ RH2k(x) ∗ (−1)kRe2k(x) ] ∗ S2k(x) ∗ T2k(x) (49) is the fundamental solution of the O-plus operator ⊕k in [2]. Now we will relate the fundamental solution H(x,m = 0) given by (43) to the fundamental solution of the wave equation is defined by (11). Now from (43) and by Lemma 7 and Lemma 8(2) and the properties of inverses in convolution algebra, we obtain [(−1)kRe−2k(x) ∗ S−2k(x) ∗ T−2k(x)] ∗H(x,m = 0) = δ ∗ δ ∗ δ ∗RH2k(x) = RH2k(x). Actually, by Lemma 2 RH2k(x) is the fundamental solution of the ultra-hyperbolic operator �k iterated k-times is defined by (9). In particular, by putting p = 1, q = n− 1, k = 1 and x1 = t in (43) and (44) then RH2 (x) reduce to M2(u) is defined by (21) with α = 2. Thus we obtain [(−1)kRe−2(x) ∗ S−2(x) ∗ T−2(x)] ∗H(x,m = 0) = M2(u) is the fundamental solution of the wave operator is defined by (11) where u = t2 − x21 − x22 − · · · − x2n−1. Theorem 2. Given the equation ⊕kmu(x) = f(x), (50) where f(x) is a given generalized function and u(x) is an unknown function, we obtain u(x) = H(x,m) ∗ f(x) (51) is a solution of the equation (50), where H(x,m) is the fundamental solution of equation (39). Proof. Convolving both sides of (50) by H(x,m), where H(x,m) is the fundamental solution for ⊕km in Theorem 1, we obtain H(x,m) ∗ ⊕kmu(x) = H(x,m) ∗ f(x) or, ⊕kmH(x,m) ∗ u(x) = H(x,m) ∗ f(x) S. Bupasiri / Eur. J. Pure Appl. Math, 14 (3) (2021), 881-894 893 applying the Theorem 1, we have δ ∗ u(x) = H(x,m) ∗ f(x). Therefore, u(x) = H(x,m) ∗ f(x). Example 1. Consider the equation (m4 +42)k(m4 −42)ku(x) = f(x), (52) where 42 is the biharmonic operator defined by 42 = ( ∂2 ∂x21 + ∂2 ∂x22 + · · ·+ ∂2 ∂x2n )2 , (53) x ∈ Rn, f(x) is a given generalized function and u(x) is an unknown function. For solving the product of biharmonic operators, we can rewrite the equation (52) as (m8 −44)ku(x) = f(x) (54) and we know that the operator in the equation (54) is the operator ⊕km with p = 0 and n = q, we obtain the function H(x,m) = W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m), where W2k(x,m), Y2k(x,m),M2k(x,m) and N2k(x,m) are defined by (27), (30), (33) and (36), respectively , and Re2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) ( x21 + · · ·+ x2n )( 2k−n 2 ) , n = p+ q (55) RH2k(x) = u( 2k−n 2 ) Kn(2k) = (−x21 − x22 − · · · − x2n)( 2k−n 2 ) Kn(2k) (56) for Kn(2k) = π n−1 2 Γ ( 2+2k−n 2 ) Γ ( 1−2k 2 ) Γ(2k) Γ ( 2+2k 2 ) Γ(−2k2 ) , (57) S2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) ( −i ( x21 + · · ·+ x2n ))( 2k−n 2 ) (58) and T2k(x) = Γ ( n−2k 2 ) 22kπ n 2 Γ(k) ( i ( x21 + · · ·+ x2n ))( 2k−n 2 ) . (59) Convolving both sides of (54) by the new fundamental solution H(x,m) = W2k(x,m) ∗ Y2k(x,m) ∗M2k(x,m) ∗N2k(x,m) , we obtain that u(x) = f(x) ∗H(x,m) is the solution of (54). 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