Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2010, Article ID 482467, 20 pages
doi:10.1155/2010/482467
Research Article
On the Solution n-Dimensional of the Product ⊗k
Operator and Diamond Bessel Operator
Wanchak Satsanit and Amnuay Kananthai
Department of Mathematics, Chiangmai University, Chiangmai 50200, Thailand
Correspondence should be addressed to Amnuay Kananthai, malamnaka@science.cmu.ac.th
Received 15 August 2009; Revised 21 November 2009; Accepted 12 January 2010
Academic Editor: Victoria Vampa
Copyright q 2010 W. Satsanit and A. Kananthai. 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.
Firstly, we studied the solution of the equation ⊗k ♦kB ux fx where ux is an unknown
unknown function for x x1 , x2 , . . . , xn ∈ Rn , fx is the generalized function, k is a
positive integer. Finally, we have studied the solution of the nonlinear equation ⊗k ♦kB ux fx, k−1 Lk ΔkB kB ux. It was found that the existence of the solution ux of such an equation
depends on the condition of f and k−1 Lk ΔkB kB ux. Moreover such solution ux is related to the
inhomogeneous wave equation depending on the conditions of p, q, and k.
1. Introduction
The operator ♦k has been first introduced by Kananthai see 1, is named as the Diamond
operator iterated k−times, and is defined by
⎛
⎞2 ⎞k
2 ⎛ pq
p
2
2
∂
∂ ⎠⎟
⎜
−⎝
♦k ⎝
⎠ ,
2
2
∂x
∂x
i1
jp1
i
j
p q n.
1.1
n is the dimension of the space Rn , for x x1 , x2 , . . . , xn ∈ Rn and k is a nonnegative integer.
The operator ♦k can be expressed in the form ♦k Δk k k Δk , where Δk is the Laplacian
operator itrerated k−times defined by
k
Δ ∂2
∂2
∂2
·
·
·
∂x12 ∂x22
∂xn2
k
,
1.2
2
Mathematical Problems in Engineering
and k is the ultrahyperbolic operator iterated k−times defined by
⎛
⎞k
2
2
2
2
2
2
∂
∂
∂
∂
∂
∂
⎝ 2 2 ··· 2 − 2 − 2 − ··· − 2 ⎠ .
∂x1 ∂x2
∂xp ∂xp1 ∂xp2
∂xpq
k
1.3
Kananthai see 1, Theorem 3.1, page 33 has shown that the convolution
x is an elementary solution of the operator ♦k , that is,
−1k Re2k x ∗ RH
2k
♦k −1k Re2k x ∗ RH
2k x δx.
1.4
Next, Kananthai see 2 has studied the linear equation
♦k ux fx.
1.5
This equation is the generalization of the ultrahyperbolic equation and it can be applied to
the wave equation. We obtain ux −1k M2k,2k x ∗ fx as a solution of such an equation
1.5 where
e
M2k,2k RH
2k x ∗ R2k x.
1.6
x is called the ultrahyperbolic kernel defined by 2.2 and Re2k x is called
The function RH
2k
the elliptic kernel defined by 2.8, with α 2k.
Furthermore, Yıldırım et al. see 3 first introduced the ♦kB operator that is named as
Diamond Bessel operator, where ♦kB is defined by
⎡
⎞2 ⎤k
2 ⎛ pq
p
⎥
⎢ Bx i
−⎝
Bx j ⎠ ⎦ ,
♦kB ⎣
i1
1.7
jp1
and Bxi ∂2 /∂xi2 2υi /xi ∂/∂xi , 2υi 2αi 1, αi > −1/2, xi > 0. The operator ♦kB can be
expressed by ♦kB ΔkB kB kB ΔkB , where
ΔkB
n
i1
k
Bx i
.
⎛
⎞k
p
pq
Bx j ⎠ .
kB ⎝ Bxi −
i1
jp1
1.8
1.9
Mathematical Problems in Engineering
3
Next, W. Satsanit has first introduced ⊗k operator and ⊗k is defined by
⎡
⎞3 ⎤k
3 ⎛ pq
p
2
2
∂
∂ ⎠⎥
⎢
⊗k ⎣
−⎝
⎦
2
2
∂x
∂x
i1
jp1
i
j
⎞k ⎡
⎞ ⎛
⎞2 ⎤k
2 p
⎛ pq
p
pq
p
pq
2
2
2
2
2
2
∂
∂ ⎠ ⎢
∂
∂
∂ ⎠ ⎝
∂ ⎠ ⎥
·⎝
⎝
−
⎣
⎦
2
2
2
2
2
2
∂x
∂x
∂x
∂x
∂x
∂x
i1
jp1
i1
i1
jp1
jp1
i
j
i
i
j
j
⎛
k
1
k Δ2 − Δ Δ − 4
k
1
3
♦Δ 3 ,
4
4
1.10
where ♦, Δ, and are defined by 1.1, 1.2, and 1.3 with k 1, respectively.
Now, firstly, the purpose of this work is to study the equation
⊗k ♦kB ux fx,
1.11
where the operator ⊗k is defined by 1.10 and ♦kB defined by 1.7, fx is a generalized
function and ux is an unknown function. Finally we study the equation
⊗k ♦kB ux f x, k−1 Lk ΔkB kB ux
1.12
with f having a continuous first derivative for all x ∈ Ω ∪ ∂Ω, where Ω is an open subset of
Rn , and ∂Ω denotes the boundary of Ω, f is bounded on Ω, that is, |f| ≤ N, N is constant, as
well as k−1 , Lk , ΔkB , and kB are defined by 1.3, 2.46, 1.8 and 1.9, respectively.
We can find the solution ux of 1.12 that is unique under the boundary condition
k−1 Lk ΔkB kB ux 0 for x ∈ ∂Ω. By 4, page 369 there exists a unique solution Wx of the
equation Wx fx, Wx for all x ∈ Ω with the boundary condition Wx 0 for all
x ∈ ∂Ω where Wx k−1 Lk ΔkB kB ux.
Moreover, if we put p k 1 in k kB Mx Wx, then we found that Mx H
I2 x ∗ I2 x ∗ Wx is a solution of the inhomogeneous equation where I2H x and I2 x are
defined by 2.6 and 2.20 with α 2, γ 2, respectively.
Before going into details, the following definitions and some important concepts are
needed.
2. Preliminaries
Definition 2.1. Let x x1 , x2 , . . . , xn be a point of the n-dimensional Euclidean space Rn ,
denoted by
2
2
2
− xp2
− · · · − xpq
.
υ x12 x22 · · · xp2 − xp1
2.1
4
Mathematical Problems in Engineering
The nondegenerated quadratic form p q n is the dimension of the space Rn . Let Γ {x ∈
Rn : x1 > 0 and u > 0} be the interior of forward cone and let Γ denote its closure. For any
complex number α, define the function
⎧ α−n/2
⎪
⎨υ
,
H
Kn α
Rα υ ⎪
⎩
0,
for x ∈ Γ ,
2.2
for x /
∈ Γ ,
where the constant Kn α is given by the formula
Kn α π n−1/2 Γ2 α − n/2Γ1 − α/2Γα
.
Γ 2 α − p /2 Γ p − α /2
2.3
The function RH
α υ is called the ultrahyperbolic kernel of Marcel Riesz and was introduced
by Nozaki see 5.
It is well known that RH
α υ is an ordinary function if Reα ≥ n and is a distribution of
H
υ
denote the support of RH
α if Reα < n. Let supp RH
α
α υ and suppose that supp Rα υ ⊂
H
Γ , that is, supp Rα υ is compact.
υ is an elementary solution of the operator k ,
From Trione see 6, page 11, RH
2k
that is,
k RH
2k υ δx.
2.4
By putting p 1 in RH
υ and taking into account Legendre’s duplication formula for
2k
Γz, that is
Γ2z 2
2z−1
π
−1/2
1
,
ΓzΓ z 2
2.5
we obtain
IαH υ υα−n/2
,
Hn α
2.6
and υ x12 − x22 − x32 · · · − xn2 where
α 2 − n α
.
Hn α π n−2/2 2α−1 Γ
Γ
2
2
2.7
IαH υ is the hyperbolic kernel of Marcel Riesz.
Definition 2.2. Let x x1 , x2 , . . . , xn be a point of Rn and ω x12 x22 · · · xn2 , then the
function Reα ω denoted the elliptic kernel of Marcel Riesz and is defined by
Reα ω ωα−n/2
,
Wn α
2.8
Mathematical Problems in Engineering
5
where
ω x12 x22 · · · xn2 ,
Wn α 2.9
π n/2 2α Γα/2
Γn − α/2
2.10
where α is a complex parameter and n is the dimension of Rn .
It can be shown that Re−2k x −1k Δk δx where Δk is defined by 1.2. It follows
δx, see 7, page 118.
that
Moreover, we obtain −1k Re2k x is an elementary solution of the operator Δk see 8,
Lemma 2.4, page 31. That is
Re0 x
Δk −1k Re2k x δx.
2.11
α−p/2
By 2.2 and 2.3 with q 0, then υα−n/2 reduces to ωp
where ωp x12 x22 · · · xp2
and Kn α reduces to Kp α π p−1/2 Γ1 − α/2Γα/Γp − α/2. By using the formula
1
Γ2z 2
,
π
ΓzΓ z 2
1
1
Γ
z Γ
− z πsecπz,
2
2
2z−1
−1/2
2.12
we obtain
Kp α 1 πα sec
Wp α,
2
2
2.13
where Wp α is defined by 2.10 with n p. Thus, for q 0,
RH
α υ πα υα−p/2
πα υα−p/2
Reα ωp ,
2 cos
2 cos
Kp α
2 Wp α
2
2.14
where ωp x12 x22 · · · xp2 . Thus, if α 2k, then
k e RH
2k ωp 2−1 R2k ωp
2.15
for q 0 and ωp x12 x22 · · · xp2 .
Definition 2.3. Let x x1 , x2 , . . . , xn , ν ν1 , ν2 , . . . , νn ∈ Rn . For any complex number α, we
define the function Sα x by
Sα x 2n2|ν|−2α Γn 2|ν| − α/2|x|α−n−2|ν|
.
n ν −1/2
i
Γνi 1/2
i1 2
2.16
6
Mathematical Problems in Engineering
Definition 2.4. Let x x1 , x2 , . . . , xn , ν ν1 , ν2 , . . . , νn ∈ Rn , and V x12 x22 · · · xp2 −
2
2
2
xp1
− xp2
− · · · − xpq
the nondegenerated quadratic form. Denote the interior of the forward
cone by Γ {x ∈ Rn : x1 > 0, x2 > 0, . . . , xn > 0, V > 0}. The function Rγ x is defined by
V γ−n−2|ν|/2
,
Kn γ
Rγ x 2.17
where
π n2|ν|−1/2 Γ 2 γ − n − 2|ν| /2 Γ 1 − γ /2 Γ γ
,
Kn γ Γ 2 γ − p − 2|ν| /2 Γ p − 2|ν| − γ /2
2.18
and γ is a complex number. By putting p 1 in Rγ x and taking into account Legendre’s
duplication formula for Γz, that is,
Γ2z 2
2z−1
π
−1/2
1
ΓzΓ z 2
2.19
we obtain
Iγ x V γ−n−2|ν|/2
,
Nn γ
2.20
and V x12 − x22 − x32 · · · − xn2 where
2 γ − n − 2|ν| γ Γ
.
Nn γ π n2|ν|−1/2 22k−1 Γ
2
2
2.21
Lemma 2.5. Given the equation ΔkB ux δx for x ∈ Rn , where ΔkB is defined by 1.8, then
ux −1k S2k x,
2.22
where S2k x is defined by 2.16, with α 2k.
Proof. See 3, page 379 and 9.
Lemma 2.6. Given the equation kB ux δx for x ∈ Rn , where kB is defined by 1.9, then
ux R2k x,
where R2k x is defined by 2.17, with γ 2k.
Proof. See 3, page 379 and 9.
2.23
Mathematical Problems in Engineering
7
Lemma 2.7. Given that P is a hyperfunction, then
P δk p kδk−1 p 0,
2.24
where δk is the Dirac-delta distribution with k-derivatives.
Proof. See 8, page 233.
Lemma 2.8. Given the equation
k ux 0,
2.25
where k is defined by 1.3 and x x1 , x2 , . . . , xn ∈ Rn , then ux RH
υm is a solution
2k−1
of 2.25 with m n − 4/2, n ≥ 4 and n is even dimension. The function RH
υm is defined
2k−1
by 2.2 with m-derivatives, α 2k − 1, and υ being defined by 2.1.
Proof. We first show the generalized function δm r 2 − s2 where r 2 x12 x22 · · · xp2 and
2
2
2
s2 xp1
xp2
· · · xpq
, p q n, is a solution of the equation
ux 0,
2.26
where is defined by 1.3 with k 1 and x x1 , x2 , . . . , xn ∈ Rn ,
∂ m 2
r − s2 2xi δm1 r 2 − s2 ,
δ
∂xi
∂2 m 2
2
m1
2
2
2 m2
2
2
r
2δ
r
4x
r
,
δ
−
s
−
s
δ
−
s
i
∂xi2
p
∂2 m 2
2
δm r 2 − s2 r
δ
−
s
2
i1 ∂xi
2pδm1 r 2 − s2 4r 2 δm2 r 2 − s2
2pδm1 r 2 − s2 4 r 2 − s2 δm2 r 2 − s2 4s2 δm2 r 2 − s2
2pδm1 r 2 − s2 − 4m 2δm1 r 2 − s2 4s2 δm2 r 2 − s2
2p − 4m 2 δm1 r 2 − s2 4s2 δm2 r 2 − s2 .
2.27
By Lemma 2.5 with P r 2 − s2 , similarly,
pq
m1 2
∂2 m 2
2
2
2 m2
2
2
δ
r
−2q
4m
2
r
4r
r
.
δ
−
s
−
s
δ
−
s
2
jp1 ∂xj
2.28
8
Mathematical Problems in Engineering
Thus
p
pq
∂2 m 2
∂2 m 2
2
2
r
−
r
δ
−
s
δ
−
s
δm r 2 − s2 2
2
i1 ∂xi
jp1 ∂xi
2 p q − 8m 2 δm1 r 2 − s2 − 4 r 2 − s2 δm2 r 2 − s2
2n − 8m 2δm1 r 2 − s2 4m 2δm1 r 2 − s2
2.29
2n − 4m 2δm1 r 2 − s2 .
If 2n − 4m 2 0, then we have δm r 2 − s2 0. That is, ux δm r 2 − s2 is a solution
of 2.26 with m n − 4/2, n ≥ 4 and n is even dimension. We write
k
ux k−1 ux 0,
2.30
and from the above proof we have k−1 ux δm r 2 − s2 with m n − 4/2, n ≥ 4 and n is
υ, we obtain
even dimension. Convolving the above equation by RH
2k−1
k−1
H
m
2
2
r
ux
R
−
s
RH
∗
∗
δ
υ
υ
2k−1
2k−1
m
H
k−1 RH
,
2k−1 υ ∗ ux R2k−1 υ
where υ r 2 − s2
2.31
m
δ ∗ ux ux RH
2k−1 υ
by 2.2, and υ r 2 − s2 is defined by Definition 2.1.
m
υ
is a solution of 2.25 with m n − 4/2, n ≥ 4 and n is
Thus ux RH
2k−1
even dimension.
Lemma 2.9. Given the equation
⊗k Gx δx,
2.32
∗−1
2k e
∗k
Gx RH
6k x ∗ −1 R4k x ∗ O x
2.33
then
is an elementary solution for the ⊗k operator iterated k−times where ⊗k is defined by 1.10, and
Ox 3 H
1
R4 x −12 Re4 x
4
4
2.34
∗−1
where O∗k x denotes the convolution of Ox itself k−times and O∗k x denotes the inverse of
O∗k x in the convolution algebra. Moreover Gx is a tempered distribution.
Mathematical Problems in Engineering
9
Proof. From 3.1, we have
k
⊗ Gx 3
13
♦Δ 4
4
k
Gx δx,
2.35
or we can write
1
3
♦Δ 3
4
4
1
3
♦Δ 3
4
4
k−1
Gx δx.
2.36
2 e
Convolving both sides of the above equation by RH
6 x ∗ −1 R4 x,
1
3
♦Δ 3
4
4
k−1
3
13
2 e
2 e
H
♦Δ Gx δx ∗ RH
∗ R6 x ∗ −1 R4 x
6 x ∗ −1 R4 x,
4
4
2.37
or
3 H 1 3 H
2 e
R2 x ∗ Δ2 −12 Re4 x ∗ RH
R
R
∗
−1
x
x
x
6
4
4
4
4
1 3 k−1
3
2 e
♦Δ Gx δx ∗ RH
∗
6 x ∗ −1 R4 x.
4
4
2.38
By 2.4 and 2.8, we obtain
3
1 3 k−1
1
3
2 e
2 e
δ ∗ δ ∗ RH
δ
∗
−1
♦Δ
R
Gx δx ∗ RH
∗
x
x
6 x ∗ −1 R4 x.
4
4
4
4
4
4
2.39
Thus
k−1
13
1
3
3 H
2 e
2 e
R x −1 R4 x ∗
♦Δ Gx RH
6 x ∗ −1 R4 x.
4 4
4
4
4
2.40
2 e
Keeping on convolving both sides of the above equation by RH
6 x ∗ −1 R4 x up to k − 1
times, we obtain
∗k
2 e
O∗k x ∗ Gx RH
6 x ∗ −1 R4 x
2.41
where the symbol ∗k denotes the convolution of itself k−times. By properties of Rα x, we
have
2 e
RH
6 x ∗ −1 R4 x
∗k
2k e
RH
6k x ∗ −1 R4k x.
2.42
10
Mathematical Problems in Engineering
Thus,
2k e
O∗k x ∗ Gx RH
6k x ∗ −1 R4k x.
2.43
2 e
Now, consider the function O∗k x, since RH
6 x∗−1 R4 x is a tempered distribution.
Thus Ox defined by 2.34 is a tempered distribution, and we obtain that O∗k x is a
tempered distribution and RH
x ∗ −12k Re4k x ∈ S is the space of tempered distribution.
6k
where DR
is the right-side distribution which is a subspace of D of
Choose S ⊂ DR
distribution.
x ∗ −12k Re4k x ∈ DR
. It follows that RH
x ∗ −12k Re4k x is an element
Thus RH
6k
6k
of convolution algebra, since DR is a convolution algebra. Hence by the method of Zemanian
see 10, 2.33 has a unique solution
∗−1
2k e
∗k
Gx RH
,
6k x ∗ −1 R4k x ∗ O x
2.44
∗−1
where O∗k x is an inverse of O∗k x in the convolution algebra and Gx is called the
Green function of the ⊗k operator.
Lemma 2.10. Given the equation
Lk Kx δx,
2.45
where Lk is the operator defined by
Lk 3 2 1 2
Δ 4
4
k
2.46
and Δ and are defined by 1.2 and 1.3 with k 1, respectively, one obtains that Kx is an
elementary solution of the Lk operator where
∗−1
2k e
∗k
∗
O
Kx RH
R
,
∗
x
−1
x
x
4k
4k
3
1
Ox RH
x −12 Re4 x,
4 4
4
2.47
∗−1
where O∗k x denotes the convolution of Ox itself k−times and O∗k x denotes the inverse of
O∗k x in the convolution algebra. Moreover Kx is a tempered distribution.
Proof. The proof of Lemma 2.10 is similar to the proof of Lemma 2.9.
Lemma 2.11. Given the equation
ux fx, ux,
2.48
where f is defined and has continuous first derivatives for all x ∈ Ω ∪ ∂Ω, where Ω is an open subset
of Rn and ∂Ω is the boundary of Ω, assume that f is bounded, that is, |fx, ux| ≤ N for all x ∈ Ω.
Mathematical Problems in Engineering
11
Then one obtains a continuous function ux as unique solution of 2.48 with the boundary condition
ux 0 for x ∈ ∂Ω.
Proof. We can prove the existence of the solution ux of 2.48 by the method of iterations
and Schuder’s estimates. The details of the proof are given by Courant and Hilbert; see 4,
pages 369–372.
Lemma 2.12. The function RH
x and S−2k x are the inverse of the convolution algebra of RH
and
−2k
2k
S2k , respectively, that is,
H
H
H
RH
−2k x ∗ R2k x R−2k2k x R0 x δ,
2.49
S−2k x ∗ S2k x S−2k2k x S0 x δ.
Proof. See 7, page 158 and 11.
3. Main Results
Theorem 3.1. Given the equation
⊗k ♦kB ux 0,
3.1
where ⊗k is the Otimes operator iterated k−times and ♦kB is Diamond Bessel operator iterated k− times
defined by 1.10 and 1.7, respectively, and ux is an unknown function, one obtains that ux is a
solution of 3.1 where
m
ux Kx ∗ −1k S2k x ∗ R2k x ∗ −1k−1 RH
2k−1 υ
3.2
where Kx is defined by 2.47, as well as S2k x, R2k x, and RH
υm are defined by
2k−1
2.16,2.17, and 2.2 with α 2k, γ 2k and α 2k − 1, respectively.
Proof. Since
k
⊗ 1
3
♦Δ 3
4
4
k
,
♦kB ΔkB kB .
3.3
Consider the homogeneous equation
⊗k ♦kB ux 0.
3.4
The above equation can be written as
or
k
1
3
♦Δ 3
4
4
k
3 2 1 2
Δ 4
4
ΔkB kB ux 0,
k
ΔkB kB ux 0.
3.5
3.6
12
Mathematical Problems in Engineering
That is,
k Lk ΔkB kB ux 0,
3.7
where k , Lk , ΔkB , and kB are defined by 1.3, 2.46, 1.8, and 1.9, respectively. By
Lemma 2.8, we obtain
m
Lk ΔkB kB ux RH
2k−1 υ
3.8
.
Since −1k S2k x, R2k x are the elementary solution of the operators ΔkB and kB ,
respectively, and by Lemma 2.10, we have that Kx is an elementary of the operator Lk
defined by 2.46, that is,
ΔkB −1k S2k x δx,
kB R2k x δx,
3.9
Lk Kx δx.
Convolving both sides of 3.8 by Kx ∗ −1k S2k x ∗ R2k x, we obtain
m
Kx∗−1k S2k x ∗ R2k x ∗ Łk ΔkB kB ux Kx ∗ −1k S2k x∗R2k x∗RH
2k−1 υ
.
3.10
By properties of convolution
m
Łk Kx∗ΔkB −1k S2k x∗kB R2k x ∗ ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−1 υ
.
3.11
By Lemmas 2.10, 2.5, and 2.6, we obtain
m
δx ∗ δx ∗ δx ∗ ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−1 υ
.
3.12
Thus
m
ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−1 υ
3.13
is the solution of 3.1.
Theorem 3.2. Given the equation
⊗k ♦kB ux fx,
3.14
Mathematical Problems in Engineering
13
where ⊗k is the Otimes operator iterated k−times defined by 1.10, and ♦kB is the Diamond Bessel
operator iterated k−times defined by 1.7, fx is the generalized function, ux is an unknown
function, x x1 , x2 , . . . , xn ∈ Rn and n is even,
One obtains that
m
ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
Gx ∗−1k S2k x ∗ R2k x ∗ fx
2k−1 υ
3.15
is a general solution of 3.14 and Gx is defined by 2.33, Kx is defined by 2.47, as well as
S2k x and R2k x are defined by 2.16 and 2.17 with α 2k and γ 2k, respectively.
Proof. Consider the equation
⊗k ♦kB ux fx
3.16
⊗k ΔkB kB ux fx.
3.17
or
Convolving both sides of 3.14 by Gx ∗ −1k S2k x ∗ R2k x, we obtain
Gx ∗ −1k S2k x ∗ R2k x ∗ ⊗k ΔkB kB ux Gx ∗ −1k S2k x ∗ R2k x ∗ fx.
3.18
By properties of convolution,
⊗k Gx ∗ ΔkB −1k S2k x ∗ kB R2k x ∗ ux Gx ∗ −1k S2k x ∗ R2k x ∗ fx.
3.19
By Lemmas 2.9, 2.5, and 2.6, we obtain
δx ∗ δx ∗ δx ∗ ux Gx ∗ −1k S2k x ∗ R2k x ∗ fx.
3.20
ux Gx ∗ −1k S2k x ∗ R2k x ∗ fx.
3.21
Thus
Consider the homogeneous equation
⊗k ♦kB ux 0.
3.22
By Theorem 3.1, we have a homogeneous solution
m
ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−1 υ
.
3.23
14
Mathematical Problems in Engineering
Thus, the general solution of 3.14 is
m
ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
Gx ∗ −1k S2k x∗R2k x∗fx,
υ
2k−1
3.24
as required.
Theorem 3.3. Consider the nonlinear equation
⊗k ♦kB ux f x, k−1 Lk ΔkB kB ux
3.25
where ⊗k , ♦kB , k−1 , Lk , ΔkB , and kB are defined by 1.10, 1.7,1.3,2.44, and 1.9, respectively.
Let f be defined, and having continuous first derivative for all x ∈ Ω ∪ ∂Ω, Ω is an open subset of Rn
and ∂Ω denotes the boundary function, that is,
f x, k−1 Lk ΔkB kB ux ≤ N
3.26
for all x ∈ Ω and the boundary condition
k−1 Lk ΔkB kB ux 0
3.27
k
ux RH
2k−1 x ∗ Gx ∗ −1 S2k x ∗ R2k x ∗ Wx
3.28
for all x ∈ ∂Ω. Then one obtains
as a solution of 3.25 with the boundary condition
m
ux RH
2k−2 υ
∗ Gx ∗ −1k S2k x ∗ R2k x
3.29
for all x ∈ ∂Ω, m n − 4/2, and Wx is a continuous function for x ∈ Ω ∪
∂Ω, while RH
2k−2υ , S2k x, and R2k x are given by 2.2, 2.16, and 2.17 with α 2k − 2, α 2k, and γ 2k, respectively. Moreover, for k 1 one obtains
2 e
k
∗1
Mx RH
−4 x ∗ −1 R−4 x ∗ O x ∗ −1 S−2 x ∗ ux
3.30
as a solution of the inhomogeneous equation
B Mx Wx,
3.31
Mathematical Problems in Engineering
15
where and B are defined by 1.3 and 1.9 with k 1, respectively, and ux is obtained from
3.28. Furthermore, If one puts p k 1, then the operators k and kB reduce to
∂2
∂2
∂2
∂2
−
−
−
·
·
·
−
,
∂x12 ∂x22 ∂x32
∂xn2
B x 1 − Bx 2 − Bx 3 − · · · − Bx n ,
3.32
respectively, and the solution Mx I2H x ∗ I2 x ∗ Wx is the inhomogeneous wave equation
∂2
∂2
∂2
∂2
−
−
−
·
·
·
−
∂x12 ∂x22 ∂x32
∂xn2
· Bx1 − Bx2 − Bx3 − · · · − Bxn Mx Wx,
3.33
where I2H x is defined by 2.6 with α 2 and I2 x is defined by 2.20 with γ 2.
Proof. Since
⊗k ♦kB ux k−1 Lk ΔkB kB ux f x, k−1 Lk ΔkB kB ux ,
3.34
ux has continuous derivative up to order 6k for k 1, 2, 3, . . ., and k−1 Lk ΔkB kB ux exists
as the generalized function. Thus we can assume that
k−1 Lk ΔkB kB ux Wx,
∀x ∈ Ω.
3.35
Then 3.34 can be written in the form
⊗k ♦kB ux Wx fx, Wx.
3.36
By3.26
fx, Wx ≤ N,
x ∈ Ω,
3.37
∀x ∈ ∂Ω.
3.38
and by3.27 Wx 0, x ∈ ∂Ω, or
k−1 Lk ΔkB kB ux 0,
We obtain a unique solution of 3.28 which satisfies 3.27 by Lemma 2.8.
x, −1k S2k x, and R2k x are the elementary solution of the operators
Since RH
2k−1
k−1 , ΔkB , and kB , respectively, and by Lemma 2.10, we have that Kx is an elementary of the
operator Lk where Lk 3/4Δ2 1/42 k , that is,
k−1 RH
2k−1 x δ,
k
B R2k x
δ,
ΔkB −1k S2k x δ,
Lk Kx δ.
3.39
16
Mathematical Problems in Engineering
From 3.35, we have
k−1 Lk ΔkB kB ux Wx.
3.40
Convolving the above equation by
k
RH
2k−1 x ∗ Kx ∗ −1 S2k x ∗ R2k x,
3.41
we obtain
k
k−1 k k k
RH
L ΔB B ux
2k−1 x ∗ Kx ∗ −1 S2k x ∗ R2k x ∗ RH
2k−1 x
∗ Kx ∗ −1 S2k x ∗ R2k x ∗ Wx.
k
3.42
By properties of convolution, we obtain
k
k
k
k
k−1 RH
∗
Δ
∗
Ł
∗
∗ ux
Kx
∗
−1
S
R
x
2k
2k
B
B
2k−1
k
RH
2k−1 x ∗ Kx ∗ −1 S2k x ∗ R2k x ∗ Wx.
3.43
By 3.39 we obtain
k
δ ∗ δ ∗ δ ∗ δ ∗ ux RH
2k−1 x ∗ Kx ∗ −1 S2k x ∗ R2k x ∗ Wx.
3.44
Thus
k
ux RH
2k−1 x ∗ Kx ∗ −1 S2k x ∗ R2k x ∗ Wx,
3.45
as a solution of 3.25.
Next, consider the boundary condition 3.38. From
k−1 Lk ΔkB kB ux 0,
3.46
by Lemma 2.8, we have
m
Lk ΔkB kB ux RH
2k−2 υ
,
3.47
where m n − 4/2, n ≥ 4 and n is even. Convolving both sides of 3.47 by
Kx ∗ −1k S2k x ∗ R2k x,
3.48
Mathematical Problems in Engineering
17
we obtain
Kx∗−1k S2k x ∗R2k x ∗ Lk ΔkB kB ∗ ux
k
Kx ∗−1 S2k x ∗ R2k x ∗
m
RH
2k−2 υ
3.49
.
By the properties of convolution, we obtain
Lk Kx ∗ ΔkB −1k S2k ∗ kB R2k ∗ux
m
Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−2 υ
3.50
.
By 3.39, we obtain
m
.
δ ∗ δ ∗ δ ∗ ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−2 υ
3.51
Thus, for x ∈ ∂Ω and k 2, 3, 4, 5, . . . .,
m
,
ux Kx ∗ −1k S2k x ∗ R2k x ∗ RH
2k−2 υ
3.52
as required.
Now, for k 1 in 3.28, we have
ux δx ∗ Gx ∗ −1S2 x ∗ R2 x ∗ Wx.
3.53
∗−1
2 e
∗1
∗
O
R
x
.
Gx RH
∗
−1
x
x
6
4
3.54
By 2.47, we have
Taking into account 3.53, we obtain
∗−1
2 e
∗1
ux RH
∗ −11 S2 x ∗ R2 x ∗ Wx
6 x ∗ −1 R4 x ∗ O x
3.55
as a solution of 3.25 for k 1.
Convolving both sides of 3.55 by
2 e
∗1
RH
−4 x ∗ −1 R−4 x ∗ O x ∗ −1S−2 x,
3.56
by Lemma 2.12, we obtain
2 e
∗1
H
RH
−4 x ∗ −1 R−4 x ∗ O x ∗ −1S−2 x ∗ ux R2 x ∗ R2 x ∗ Wx.
3.57
18
Mathematical Problems in Engineering
By Lemma 2.6, we obtain
2 e
∗1
Mx RH
−4 x ∗ −1 R−4 x ∗ O x ∗ −1S−2 x ∗ ux
3.58
as a solution of the inhomogeneous equation
B Mx Wx.
3.59
Now, consider the boundary condition for k 1 in 3.27; we have
LΔB B ux 0,
or
B LΔB ux 0
3.60
for x ∈ ∂Ω. Thus by Lemma 2.8, for k 1, we have
LΔB ux δm υ for x ∈ ∂Ω,
3.61
where δm x RH
0 x. Convolving the above equation by Kx ∗ −1S2 x where Kx is
defined by 2.47 with k 1 and S2 x is defined by 2.16 with α 2, we obtain
Kx ∗ −1S2 x ∗ LΔB ux δm υ ∗ Kx ∗ −1S2 x.
3.62
By properties of convolution,
LKx ∗ ΔB −1S2 x ∗ ux δ m υ ∗ Kx ∗ −1S2 x.
3.63
By Lemmas 2.10 and 2.5, we obtain
δx ∗ δx ∗ ux δm υ ∗ Kx ∗ −1S2 x.
3.64
ux δm υ ∗ Kx ∗ −1S2 x.
3.65
It follows that
By 2.47 with k 1, we have
∗−1
2 e
∗1
∗
O
R
x
.
∗
−1
Kx RH
x
x
4
4
3.66
Taking into account 3.65, we obtain
∗−1
2 e
∗1
∗ −1S2 x for x ∈ ∂Ω.
ux δm υ ∗ RH
4 x ∗ −1 R4 x ∗ O x
3.67
Mathematical Problems in Engineering
19
Now consider the case k 1, p 1, and q n − 1, that is, from 3.59, RH
2 x reduced
where I2H x is defined by 2.2 with α 2 and R2 x reduced to I2 x where I2 x
to
is defined by 2.17 with γ 2, and then the operator defined by 1.3 reduces to the wave
operator
I2H x
∗ ∂2
∂2
∂2
∂2
− 2 − 2 − ··· − 2 ,
2
∂x1 ∂x2 ∂x3
∂xn
3.68
B defined by 1.9 reduces to the Bessel wave operator
∗B Bx1 − Bx2 − Bx3 − · · · − Bxn ,
3.69
and then the solution Mx reduced to
Mx I2H x ∗ I2 x ∗ Wx,
3.70
which is the solution of inhomogeneous wave equation
∗ ∗B Mx Wx,
3.71
or
∂2
∂2
∂2
∂2
−
−
−
·
·
·
−
∂x12 ∂x22 ∂x32
∂xn2
· Bx1 − Bx2 − Bx3 − · · · − Bxn Mx Wx.
3.72
With the boundary condition for x ∈ ∂Ω,
L∗ ∗B ΔB ux 0,
3.73
where L∗ 3/4Δ2 1/4∗ 2 and ∗ is defined by 3.68, or for x ∈ ∂Ω and by 3.65, we
obtain
∗−1
ux δ m s ∗ I4H x ∗ −12 Re4 x ∗ D∗1 x
∗ −1S2 x,
3.74
where I4 x is defined by 2.20 with γ 4, s x12 − x22 − x32 − · · · − xn2 , and Dx reduced from
Ox where it is defined by 2.34, that is, Dx 3/4I4H x 1/2−12 Re4 x.
Acknowledgment
The authors would like to thank The Thailand Research Fund and Graduate School, Chiang
Mai University, Thailand, for financial support.
20
Mathematical Problems in Engineering
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