Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 127-136 , 2010 Applied Mathematics
On The Weak Solutions of The Compound Ultra-Hyperbolic Bessel Equation
Piladda Srisombat1, Kamsing Nonlaopon1;2
1Department of Mathematics, Khon Kaen University, Khon Kaen 40002, Thailand.
e-mail:5150201704@ stdm ail.kku.ac.th
2Center of Excellence in Mathematics, CHE, Si Ayuthaya Rd., Bangkok 10400,
Thai-land.
e-mail:nkam si@ kku.ac.th
Received Date: August 14, 2009 Accepted Date: December 28, 2009
Abstract. In this article, we have studied the compound ultra-hyperbolic Bessel equation of the form
m X r=0
Cr rB;cu(x) = f (x);
where r
B;c is the ultra-hyperbolic Bessel operator iterated r-times, f is a given generalized function, u is an unknown function, x 2 R+
n and Cr is a constant. In this work, we study the weak solution u(x) of above the equation which is of the form ultra-hyperbolic Bessel operator and moreover, such a solution is unique.
Key words: Dirac-delta distribution; Ultra-hyperbolic Bessel operator; Tem-pered distribution.
2000 Mathematics Subject Classi…cation: 46F10. 1. Introduction
I. M. Gelfand and G. E. Shilov [3] have …rst introduced the elementary solution of the n-dimensional classical ultra-hyperbolic operator. S. E. Trione [9] has shown that the n-dimensional ultra-hyperbolic equation has u(x) = R2k(x) as a unique elementary solution. Later, M. A. Tellez [8] has proved that R2k(x) exists only for case p is odd with p + q = n.
H. Yildirim et al. [10] have introduced the Bessel ultra-hyperbolic type operator iterated k-times with x 2 R+n = fx : x = (x1; x2; : : : ; xn); x1> 0; : : : ; xn> 0g; (1.1) kB = (Bx1+ Bx2+ + Bxp Bxp+1 Bxp+q) k where p + q = n; Bxi = @2 @x2 i + 2vi xi @ @xi where 2vi = 2 i+ 1; i > 1 2 [5], k is nonnegative integer and n is the dimension of R+n and studied the elementary solution of this operator. Moreover, they have introduced the Bessel diamond operator and have studied the elementary solution of this operator and also the Fourier-Bessel transform of the elementary solution.
A. Kananthai and K. Nonlaopon [4] have studied the weak solution of the com-pound ultra-hyperbolic equation. Next, M. Z. Sarikaya and H. Yildirim [7] have studied the weak solution of the compound Bessel ultra-hyperbolic equation. Later, S. Bupasiri and K. Nonlaopon [1] have studied the weak solution of the compound equation related to the ultra-hyperbolic operator of the form
(1.2)
m X r=0
Cr rcu(x) = f (x);
where rc is the operator which related to the ultra-hyperbolic type operator iterated r-times, de…ned by
(1.3) rc= 0 @1 c2 p X i=1 @2 @x2 i p+q X j=p+1 @2 @x2 j 1 A r :
A. Saglam et al. [6] have developed the operator of (1.1), de…ned by
(1.4) kB;c= 1
c2(Bx1+ Bx2+ + Bxp) (Bxp+1+ + Bxp+q)
k
and is called the ultra-hyperbolic Bessel operator iterated k-times. Moreover, they studied the product of the ultra-hyperbolic Bessel operator related to elas-tic waves.
In this article, we will consider the equation
(1.5) kB;cu(x) = f (x)
where u(x) and f (x) are some generalized function. We will develop the equation (1.5) to the form
(1.6)
m X k=0
Ck kB;cu(x) = f (x);
which is called the compound ultra-hyperbolic Bessel equation and by conven-tion 0B;cu(x) = u(x). In …nding the solutions of (1.6), we use the properties of convolutions for the generalized functions.
2. Preliminaries
Denoted by Ty the generalized shift operator acting according to the law [5]: Txy'(x) = Cv Z 0 Z 0 ' q x2 1+ y21 2x1y1cos 1; : : : ; p x2 n+ yn2 2xnyncos n n Y i=1 sin2vi 1 i ! d 1 d n; where x; y 2 R+ n; Cv = Qn i=1 (vi+1) (1
2) (vi). We remark that this shift operator is
closely connected with the Bessel di¤erential operator [5]: d2U dx2 + 2v x dU dx = d2U dy2 + 2v y dU dy U (x; 0) = f (x); Uy(x; 0) = 0:
The convolution operator determined by the Ty is as follows:
(2.1) (f ')(y) = Z R+n f (y)Txy'(x) n Y i=1 y2vi i ! dy:
Convolution (2.1) known as a B-convolution. We note the following properties of the B-convolution and the generalized shift operator.
(a) Ty
x 1 = 1: (b) T0
x f (x) = f (x): (c) If f (x); g(x) 2 C(R+
n); g(x) is a bounded function all x > 0 and Z R+ n jf(x)j n Y i=1 x2vi i ! dx < 1; then Z R+n Txyf (x)g(y) n Y i=1 y2vi i ! dy = Z R+n f (y)Txyg(x) n Y i=1 y2vi i ! dy:
(d) From (c), we have the following equality for g(x) = 1. Z R+n Txyf (x) n Y i=1 y2vi i ! dy = Z R+n f (y) n Y i=1 y2vi i ! dy: (e) (f g)(x) = (g f )(x):
De…nition 2.1. Let x = (x1; x2; : : : ; xn) be a point of the n-dimensional space R+n,
(2.2) V = c2 x21+ x22+ : : : + xp2 x2p+1 x2p+2 x2p+q;
where p+q = n, the interior of forward cone de…ned by += fx 2 R+n : x1> 0; x2> 0; : : : ; xn> 0 and V > 0g. For any complex number , de…ne
(2.3) RH;c(x) = 8 < : V n22jvj Kn;v(x) ; for x 2 +; 0; for x =2 +; where (2.4) Kn;v( ) = n+2jvj 1 2 2+ 2jvj 2 1 2 ( ) 2+ p 2jvj 2 p+2jvj 2 : Lemma 2.1. RH
2k;c(x) is a homogeneous distribution of order ( n 2 jvj). In particular, it is a tempered distribution.
Proof. We need to show that RH
;c(x) satis…es the Euler equation n X i=1 xi @ @xi RH;c(x) = ( n 2jvj)RH;c(x): Now n X i=1 xi @ @xi RH;c(x) = 1 Kn;v( ) n X i=1 xi @ @xi c2(x21+ + x2p) x2p+1 x2p+q n 2jvj 2 = n 2jvj Kn;v( ) c2(x21+ + x2p) x2p+1 x2p+q n 2jvj 2 2 c2(x21+ + x2p) x2p+1 x2p+q = 1 Kn;v( ) ( n 2jvj) c2(x21+ + x2p) x2p+1 x2p+q n 2jvj 2 = ( n 2jvj)V n 2jvj 2 Kn;v( ) = ( n 2jvj)RH;c(x): Hence RH
;c(x) is a homogeneous distribution of order ( n 2jvj). W. F. Donoghue [2] prove that every homogeneous distribution is a tempered distrib-ution. So RH;c(x) is a tempered distribution. This is complete of proof.
Lemma 2.2. Given the equations
(2.5) kB;cu (x) = (x) ;
where k
B;c is de…ned by (1.4) and x 2 R+n, then we obtain u (x) = RH2k;c(x) as an unique elementary solution of (2.5), where RH2k;c(x) is de…ned by (2.3). The proof of this lemma is given in [6].
Lemma 2.3.(The B-convolutions of tempered distributions) (a) k
B;c (x) u (x) = kB;cu (x), where u is any tempered distribution. (b)Let R2k;cH (x) and RH2m;c(x) be de…ned by (2.3), then RH2k;c(x) RH2m;c(x) exists and is a tempered distribution.
(c)Let RH
2k;c(x) and RH2m;c(x) be de…ned by (2.3), then RH2k;c(x) RH2m;c(x) = RH2k+2m;c(x), where k and m are nonnegative integer.
(d)Let RH
2k;c(x) and RH2m;c(x) be de…ned by (2.3) and if RH2k;c(x) RH2m;c(x) = (x) then RH
2k;c(x) is an inverse of RH2m;c(x) in the B convolution algebra, de-noted by RH 2k;c(x) = R H 1 2m;c (x), moreover R H 1 2m;c (x) is unique. Proof.
(a)First, we consider the case k = 1, now
B;c (x) = 1 c2 p X i=1 @2 (x) @x2 i +2vi xi @ (x) @xi ! 0 @ p+q X j=p+1 @2 (x) @x2 j +2vj xj @ (x) @xj 1 A ; p+q = n
and let '(x) be a testing function in the Schwartz space S. By the de…nition of B-convolution, we have
h B;c (x) u(x); '(x)i = hu(x); h B;c (x); '(x + y)ii = * u(x); * 1 c2 p X i=1 @2 (y) @x2 i +2vi xi @ (y) @xi ! 0 @ p+q X j=p+1 @2 (y) @x2 j +2vj xj @ (y) @xj 1 A ; '(x + y) ++
= * u(x); * (y); 1 c2 p X i=1 @2'(x + y) @x2 i +2vi xi @'(x + y) @xi ! 0 @ p+q X j=p+1 @2'(x + y) @x2 j +2vj xj @'(x + y) @xj 1 A ++ = * u(x); 1 c2 p X i=1 @2'(x) @x2 i +2vi xi @'(x) @xi ! 0 @ p+q X j=p+1 @2'(x) @x2 j +2vj xj @'(x) @xj 1 A + = * 1 c2 p X i=1 @2u(x) @x2 i +2vi xi @u(x) @xi ! 0 @ p+q X j=p+1 @2u(x) @x2 j +2vj xj @u(x) @xj 1 A ; '(x) + = h B;cu(x); '(x)i : (2.6) It follow that B;c (x) u(x) = B;cu(x): Similarly for any k, we can show that
k
B;c (x) u(x) = kB;cu(x): (b)Since RH
2k;c(x) and RH2m;c(x) are tempered distribution by Lemma 2.1. Now chosen supp RH
2k;c(x) = K +, where K is a compact set and + designates of +closure. Then R2k;cH (x) is a tempered distribution with compact support. By [2], RH
2k;c(x) RH2m;c(x) exists and is a tempered distribution. (c) From equation k+mB;c u(x) = (x) we obtain u(x) = RH
2k+2m;c(x) by Lemma 2.2. For any m is a nonnegative integer, we write
k+m B;c u(x) =
k
B;c mB;cu(x) = (x); then by Lemma 2.2, we have the following equality
m
B;cu(x) = RH2k;c(x): B-convolving both sides by RH2m;c(x) we obtain
RH2m;c(x) mB;cu(x) = RH2k;c(x) R2m;cH (x) or
m
B;cRH2m;c(x) u(x) = RH2k;c(x) R2m;cH (x): Then from Lemma 2.2, we have the following equality
(x) u(x) = RH2k;c(x) RH2m;c(x): It follows that
From the fact that u(x) = RH2k+2m;c(x) we obtain RH2k;c(x) RH2m;c(x) = RH2k+2m;c(x): (d)Since RH
2k;c(x) and RH2m;c(x) are tempered distributions with compact supports, thus RH
2k;c(x) and RH2m;c(x) are the elements of space of B-convolution algebra u0 of distribution. Now RH2k;c(x) RH2m;c(x) = (x) then by A. H. Zemanian [11] show that RH
2k;c(x) = RH2m;c1(x) is a unique inverse. Lemma 2.4. Let RH
2k;c(x) and Kn;v(2k), be de…ned by (2.3) and (2.4). Then (a)Kn;v(2k + 2) = 2k (2k + 2 n 2 jvj) Kn;v(2k),
(b) k
B;cRH2m;c(x) = RH2m 2k;c(x), where k and m are nonnegative integer, (c)RH
2k;c(x) = kB;c (x), where k is a nonnegative integer. Proof.
(a)From (2.4), we have
Kn;v(2k + 2) = n+2jvj 1 2 2+2k+2 n 2jvj 2 1 2k 2 2 (2k + 2) 2+2k+2 p 2jvj 2 p+2jvj 2k 2 2 = n+2jvj 1 2 (2k+2 n 2jvj) 2 2k+2 n 2jvj 2 ( 2 1+2k) 1 2k 2 (2k)(2k + 1) (2k) 2k+2 p 2jvj 2 2k+2 p 2jvj 2 2 p+2jvj 2k 2 p+2jvj 2k 2 = 2k(2k + 2 n 2jvj)Kn;v(2k): (b)By Lemma 2.3.(c), we have (x) R2m;cH (x) = RH2k;c(x) RH2m 2k;c(x) k B;cRH2k;c(x) R2m;cH (x) = RH2k;c(x) RH2m 2k;c(x) RH2k;c(x) kB;cR2m;cH (x) = RH2k;c(x) RH2m 2k;c(x); and k B;cR2m;cH (x) = RH2m 2k;c(x): (c)For m = k; by Lemma 2.4.(b) we have
m
B;cRH2m;c(x) = RH0;c(x); RH0;c(x) = (x): For m = 0, by Lemma 2.4.(b) we have
k
B;cRH0;c(x) = RH2k;c(x) or kB;c (x) = RH2k;c(x):
3. Main Results
Theorem 3.1. Given the compound equation related to the ultra-hyperbolic Bessel operator of the form
(3.1) m X r=0 Cr rB;cu (x) = f (x) ; where r
B;c is the ultra-hyperbolic Bessel operator iterated r-times de…ned by (1.4), f is a given generalized function, u is an unknown function, x 2 R+n, n is odd and Cris a constant. Then (3.1) has a unique weak solution
(3.2) u (x) = f (x) RH2m;c(x) CmRH0;c+ w (x) RH2;c 1 ; where (3.3) w (x) = Cm 1+ Cm 2 V 2(4 n 2jvj) + Cm 3 V2 2 4(4 n 2jvj)(6 n 2jvj)+ : : : +C0 V m 1 2 4 6:::2(m 1)(4 n 2jvj)(6 n 2jvj):::(2m n 2jvj) and V de…ned by (2.2) and CmRH0;c(x) + w (x) RH2;c(x)
1
is an inverse of CmRH0;c(x) + w (x) RH2;c(x) :
Proof. By Lemma 2.3.(a), equation (3.1) can be written as
(Cm mB;c (x) + Cm 1 m 1B;c (x) + + C1 B;c (x) + C0 (x)) u(x) = f (x): B-convolving both sides by RH
2m;c(x) de…ned by (2.3), we obtain h Cm mB;cRH2m;c(x) + Cm 1 m 1B;c R H 2m;c(x) + + C1 B;cRH2m;c(x) + C0RH2m;c(x) i u(x) = f (x) RH2m;c(x):
By Lemma 2.2 and Lemma 2.4.(b), we obtain h Cm (x) + Cm 1RH2;c(x) + Cm 2RH4;c(x) + + C1RH2(m 1);c(x) + C0RH2m;c(x) i u(x) = f (x) RH2m;c(x): (3.4)
By Lemma 2.4.(a), we obtain RH4;c(x) = V4 n22jvj Kn(4) = RH2;c(x) V 2(4 n 2jvj)Kn(2) : Similarly, RH6;c(x) = RH2;c(x) V 2 2 4(4 n 2jvj)(6 n 2jvj) R8;c(x) = RH2;c(x) V3 2 4 6(4 n 2jvj)(6 n 2jvj)(8 n 2jvj) .. . RH2m;c(x) = RH2;c(x) V m 1 2 4 6 2(m 1)(4 n 2jvj)(6 n 2jvj) (2m n 2jvj):
Thus we obtain the function w(x) of (3.3). Now w(x) is continuous and in…nitely di¤erentiable in classical sense for n is odd. Since RH2;c(x) is a tempered distri-bution with compact support, hence w(x)RH
2;c(x) also is tempered distribution with compact support and so CmRH0;c(x) + w(x)R2;cH(x). By Lemma 2.3.(d), CmRH0;c(x) + w(x)RH2;c(x) has a unique inverse denote by
CmRH0;c(x) + w(x)RH2;c(x) 1
: Now (3.4) can be written as
(CmRH0;c(x) + w(x)RH2;c(x)) u(x) = f (x) RH2m;c(x); RH0;c(x) = (x): B-convolving both sides by CmRH0;c(x) + w(x)RH2;c(x)
1
, we have u(x) = f (x) RH2m;c(x) CmRH0;c(x) + w(x)RH2;c(x)
1 : Since RH2m;c(x) is a unique by Lemma 2.2 and CmRH0;c(x) + w(x)R2;cH(x)
1
also is a unique by Lemma 2.3.(d), it follows that u(x) is a unique weak solution of (3.1) with odd-dimensional n: This completes the proof.
4. Acknowledgements
This work is supported by the Thailand Research Fund, the Commission on Higher Education (contract number MRG5180058), and the Centre of Excel-lence in Mathematics, Thailand.
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