# On the Operator⨁Bk Related to Bessel Heat Equation

Authors: Wanchak Satsanit
Journal: Mathematical Problems in Engineering
 (2010)

---

## Abstract

We study the equation(∂/∂t)u(x,t)=c2⊕Bku(x,t) with the initial condition u(x,0)=f(x) for x∈Rn+. The operator ⊕Bk is the operator iterated k-times and is defined by ⊕Bk=((∑i=1pBxi)4-(∑j=p+1p+qBxi)4)k, where p+q=n is the dimension of the ℝn+, Bxi=∂2/∂xi2+(2vi/xi)(∂/∂xi), 2vi=2αi+1, αi>-1/2, i=1,2,3,…,n, and k is a nonnegative integer, u(x,t) is an unknown function for (x,t)=(x1,x2,…,xn,t)∈Rn+×(0,∞), f(x) is a given generalized function, and c is a positive constant. We obtain the solution of such equation, which is related to the spectrum and the kernel, which is so called Bessel heat kernel. Moreover, such Bessel heat kernel has interesting properties and also related to the kernel of an extension of the heat equation.


---

## Body

## 1. Introduction

It is well known that for the heat equation(1.1)∂∂tu(x,t)=c2Δu(x,t)
with the initial condition
(1.2)u(x,0)=f(x),
where Δ=∑i=1n∂2/∂xi2 is the Laplace operator and (x,t)=(x1,x2,…,xn,t)∈ℝn×(0,∞), we obtain
(1.3)u(x,t)=1(4c2πt)n/2∫ℝnexp(-|x-y|24c2t)f(y)dy
as the solution of (1.1). Equation (1.3) can be written as
(1.4)u(x,t)=E(x,t)*f(x),
where
(1.5)E(x,t)=1(4c2πt)n/2exp(-|x|24c2t).E(x,t) is called the heat kernel, where |x|2=x12+x22+⋯+xn2 and t>0 (see [1, pages 208-209]).In 2004, Yildirim et al. [2, 3] first introduced the Bessel diamond operator ♢Bk iterated k-times, defined by(1.6)♢Bk=((∑i=1pBxi)2-(∑j=p+1p+qBxj)2)k,
where Bxi=∂2/∂xi2+(2υi/xi)(∂/∂xi), 2υi=2αi+1,αi>-1/2,xi>0. The operator ♢Bk can be expressed by ♢Bk=ΔBk□Bk=□BkΔBk, where
(1.7)ΔBk=(∑i=1pBxi)k,(1.8)□Bk=(∑i=1pBxi-∑j=p+1p+qBxj)k.And, Yildirim et al. [2, 3] have shown that the solution of the convolution form
(1.9)u(x)=(-1)kS2k(x)*R2k(x)
is a unique elementary solution of ♢Bk that is
(1.10)♢Bk((-1)kS2k(x)*R2k(x))=δ.Now, the purpose of this work is to study the following equation:(1.11)∂∂tu(x,t)=c2⨁Bku(x,t)
with the initial condition
(1.12)u(x,0)=f(x),forx∈Rn+,
where the operator ⊕Bk is first introduced by Satsanit and Kananthai [4] and is defined by
(1.13)⨁Bk=((∑i=1pBxi)4-(∑j=p+1p+qBxi)4)k=((∑i=1pBxi)2-(∑j=p+1p+qBxi)2)k((∑i=1pBxi)2+(∑j=p+1p+qBxi)2)k.
Let us denote the operator
(1.14)⊚Bk=((∑i=1pBxi)2+(∑j=p+1p+qBxj)2)k.
By (1.7) we obtain
(1.15)⊚Bk=[(∑i=1pBxi2)2+(∑j=p+1p+qBxj2)2]k=[(ΔB+□B2)2+(ΔB-□B2)2]k=(ΔB2+□B22)k.
Thus, (1.13) can be written as
(1.16)⨁Bk=♢Bk⊚Bk,
where ♢Bkand⊚Bk are defined by (1.6) and (1.15), respectively, p+q=n is the dimension of the Rn+=x:x=(x1,x2,…,xn,t),xi>0,i=1,2,3,…,n, u(x,t) is an unknown function, Bxi=∂2/∂xi2+(2υi/xi)(∂/∂xi), 2υi=2αi+1,αi>-1/2,f(x) is the given generalized function, k is a positive integer, and c is a constant.Moreover, Bessel heat kernel has interesting properties and also related to the kernel of an extension of the heat equation. We obtain the solution in the classical convolution form(1.17)u(x,t)=E(x,t)*f(x),
where the symbol * is the B-convolution in (2.3), as a solution of (1.11), which satisfies (1.12), and
(1.18)E(x,t)=Cv∫Ω+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy,
and Ω+⊂Rn+ is the spectrum of E(x,t) for any fixed t>0 and Jvi-1/2(xi,yi) is the normalized Bessel function.Before going into details, the following definitions and some important concepts are needed.


## 2. Preliminaries

The shift operator according to the law remarks that this shift operator connected to the Bessel differential operator (see [2, 3, 5]):
(2.1)Txyφ(x)=Cv*∫0π⋯∫0πφ(x12+y12-2x1y1cosθ1,…,xn2+yn2-2xnyncosθn)×(∏i=1nsin2vi-1θi)dθ1⋯dθn,
where x,y∈Rn+,Cv*=∏i=1nΓ(vi+1)/Γ(1/2)Γ(vi). We remark that this shift operator is closely connected to the Bessel differential operator (see [4]):
(2.2)d2Udx2+2vxdUdx=d2Udy2+2vydUdy,U(x,0)=f(x),Uy(x,0)=0.
The convolution operator determined by the Txy is as follows:
(2.3)(f*φ)(y)=∫Rn+f(y)Txyφ(x)(∏i=1nyi2vi)dy.
Convolution (2.3) is known as a B-convolution. We note the following properties of the B-convolution and the generalized shift operator.  (1)
Txy·1=1.(2)
Tx0·f(x)=f(x).(3)
Iff(x),g(x)∈C(Rn+),g(x) is a bounded function for all x>0, and ∫0∞|f(x)|(∏i=1nxi2vi)dx<∞, then ∫Rn+Txyf(x)g(y)  (∏i=1nyi2vi)dy = ∫Rn+f(y)Txyg(x)  (∏i=1nyi2vi)dy.(4)
From (3),we have the following equality forg(x)=1:∫Rn+Txyf(x)(∏i=1nyi2vi)dy=∫Rn+f(y)(∏i=1nyi2vi)dy.(5)
(f*g)(x)=(g*f)(x).The Fourier-Bessel transformation and its inverse transformation are defined as follows:(2.4)(ℱBf)(x)=Cv∫Rn+f(y)(∏i=1nJvi-1/2(xi,yi)yi2υi)dy,(2.5)(ℱB-1f)(x)=(ℱBf)(-x),Cv=(∏i=1n2υi-1/2Γ(υi+12))-1,
where Jvi-1/2(xi,yi) is the normalized Bessel function which is the eigenfunction of the Bessel differential operator. The following equalities for Fourier-Bessel transformation are true (see [5–7]):
(2.6)ℱBδ(x)=1,(2.7)ℱB(f*g)(x)=ℱBf(x)·ℱBg(x).Definition 2.1.
The spectrum of the kernelE(x,t) of (1.18) is the bounded support of the Fourier Bessel transform ℱBE(y,t) for any fixed t>0.Definition 2.2.
Letx=(x1,x2,…,xn) be a point in ℝn+ and denote by
(2.8)Γ+={x∈ℝn+:x12+x22+⋯+xp2-xp+12-xp+22-⋯-xp+q2>0,ξ1>0}
the set of an interior of the forward cone, and Γ¯+ denotes the closure of Γ+.LetΩ+ be spectrum of E(x,t) defined by (1.18) for any fixed t>0 and Ω⊂Γ¯+. Let FBE(y,t) be the Fourier Bessel transform of E(x,t), which is defined by
(2.9)ℱBE(y,t)={ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]kforξ∈Ω+,0forξ∉Ω+.Lemma 2.3 (Fourier Bessel transform of□Bk operator).
One has(2.10)ℱB□Bku(x)=(-1)kV1k(x)ℱBu(x),
where
(2.11)V1k(x)=(∑i=1pxi2-∑j=p+1p+qxj2)k.Proof.
See [8].Lemma 2.4 (Fourier Bessel transform ofΔBk operator).
One has(2.12)ℱBΔBku(x)=(-1)k|x|2kℱBu(x),
where
(2.13)|x|2k=(x12+x22+⋯+xn2)k.Proof.
See [8].Lemma 2.5 (Fourier Bessel transform of⊕Bk operator).
One has(2.14)ℱB⨁Bku(x)=Vk(x)ℱBu(x),
where
(2.15)Vk(x)=((∑i=1pxi2)4-(∑j=p+1p+qxj2)4)k.Proof.
We can use the mathematical induction method; fork=1, we have
(2.16)ℱB(⨁Bu)(x)=Cv∫Rn+(⨁Bu(y))(∏i=1nJvi-1/2(xi,yi)yi2υi)dy=Cv∫Rn+(♢B⊚Bu(y))(∏i=1nJvi-1/2(xi,yi)yi2υi)dy=Cv∫Rn+(ΔB2+□B2)2g(y)(∏i=1nJvi-1/2(xi,yi)yi2υi)dy,g(y)=♢Bu(y)=ℱB(ΔB2g+□B2g)2(x)=((-1)2(x12+⋯+xn2)2+(-1)2(x12+⋯+xp2-xp+12-⋯-xp+q2)2)2ℱBg(x)=((x12+x22+⋯+xp2)2+(xp+12+⋯+xp+q2)2)ℱB♢Bu(x)=((∑i=1pxi2)2+(∑j=p+1p+qxj2)2)·((∑i=1pxi2)2-(∑j=p+1p+qxj2)2)ℱBu(x)=((x12+x22+⋯+xp2)4-(xp+12+⋯+xp+q2)4)ℱBu(x)=V(x)ℱBu(x),
where V(x)=(x12+x22+⋯+xp2)4-(xp+12+⋯+xp+q2)4. Then, from inverse Fourier transform we obtain
(2.17)⨁Bu(x)=ℱB-1V(x)ℱBu(x).
Assume that the statement is true for k-1, that is,
(2.18)⨁Bk-1u(x)=ℱB-1Vk-1(x)ℱBu(x).
Then, we must prove that is also true for k∈N. So we have
(2.19)⨁Bku(x)=⨁B(⨁Bk-1u(x))=ℱB-1V(x)ℱBℱB-1Vk-1(x)ℱBu(x)=ℱB-1Vk(x)ℱBu(x).
This completes the proof.Lemma 2.6.
Fort,v>0 and x,y∈ℝn, one has
(2.20)∫0∞e-c2x2tx2υdx=Γ(υ)2c2υ+1tυ+1/2,∫0∞e-c2x2tJυ-1/2(xy)x2υdx=Γ(υ+1/2)2(c2t)υ+1/2e-y2/4c2t,
where c is a positive constant.Proof.
See [9].Lemma 2.7.
Let the operatorL be defined by
(2.21)L=∂∂t-c2⨁Bk,
where ⊕Bk is the operator iterated k-times and is given by
(2.22)⨁Bk=((∑i=1pBxi)4-(∑j=p+1p+qBxi)4)k,Bxi=∂2∂xi2+2vixi∂∂xi,p+q=n is the dimension ℝn+, k is a positive integer, (x1,x2,…,xn)∈ℝn+, and c is a positive constant. Then
(2.23)E(x,t)=Cv∫Ω+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy
is the elementary solution of (2.21) in the spectrum Ω+⊂Rn+ for t>0.Proof.
LetLE(x,t)=δ(x,t), where E(x,t) is the elementary solution of L and δ is the Dirac-delta distribution. Thus
(2.24)∂∂tE(x,t)-c2⨁BkE(x,t)=δ(x)δ(t).
Applying the Fourier Bessel transform, which is defined by (2.4) to the both sides of the above equation and using Lemma 2.5 by considering ℱBδ(x)=1, we obtain
(2.25)∂∂tℱBE(x,t)-c2[(x12+x22+⋯+xp2)4-(xp+12+⋯+xp+q2)4]kℱBE(x,t)=δ(t).
Thus, we get
(2.26)ℱBE(x,t)=H(t)ec2t[(x12+x22+⋯+xp2)4-(xp+12+⋯+xp+q2)4]k,
where H(t) is the Heaviside function, because H(t)=1 holds for t≥0.
Therefore,(2.27)ℱBE(x,t)=ec2t[(x12+x22+⋯+xp2)4-(xp+12+⋯+xp+q2)4]k,
which has been already defined by (2.7). Thus from (2.5), we have
(2.28)E(x,t)=Cv∫ℝn+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy,
where Ω+ is the spectrum of E(x,t). Thus, we obtain
(2.29)E(x,t)=Cv∫Ω+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy
as an elementary solution of (2.21) in the spectrum Ω+⊂Rn+ for t>0.


## 3. Main Results

Theorem 3.1.
Let us consider the equation(3.1)∂∂tu(x,t)-c2⨁Bku(x,t)=0
with the initial condition
(3.2)u(x,0)=f(x),
where ⊕Bk is the operator iterated k-times and is defined by
(3.3)⨁Bk=((∑i=1pBxi)4-(∑j=p+1p+qBxi)4)k,Bxi=∂2∂xi2+2vixi∂∂xi,p+q=n is the dimension ℝn+,k is a positive integer, u(x,t) is an unknown function for (x,t)=(x1,x2,…,xn,t)∈ℝn+×(0,∞),f(x) is the given generalized function, and c is a positive constant. Then
(3.4)u(x,t)=E(x,t)*f(x)
is a solution of (3.1), which  satisfies (3.2), where E(x,t) is given by (2.23). In particular, if one puts k=1 and q=0 in (3.1), then (3.1) reduces to the equation
(3.5)∂∂tu(x,t)-c2ΔB4u(x,t)=0,
which is related to the Bessel heat equation.Proof.
Taking the Fourier Bessel transform, which is defined by (2.4), of both sides of (3.1) for x∈ℝn+ and using Lemma 2.5, we obtain
(3.6)∂∂tℱBu(x,t)=c2((x12+⋯+xp2)4-(xp+12+⋯+xp+q2)4)kℱBu(x,t).
Thus, we consider the initial condition (3.2); then we have the following equality for (3.6):
(3.7)u(x,t)=f(x)*ℱB-1ec2t[(x12+⋯+xp2)4-(xp+12+⋯+xp+q2)4]k.
Here, if we use (2.4) and (2.5), then we have
(3.8)u(x,t)=f(x)*ℱB-1ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k=∫ℝn+ℱB-1ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]kTxyf(x)(∏i=1nyi2υi)dy=∫ℝn+(Cv∫ℝn+ec2tVk(z)∏i=1nJυi-1/2(yi,zi)zi2υidz)Txyf(x)(∏i=1nyi2υi)dy,
where V(z)=(z12+z22+⋯+zp2)4-(zp+12+zp+22+⋯+zp+q2)4. Set
(3.9)E(x,t)=Cv∫ℝn+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy.
Since the integral in (3.9) is divergent, therefore we choose Ω+⊂ℝn+ to be the spectrum of E(x,t), and by (2.21), we have
(3.10)E(x,t)=Cv∫ℝn+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy=Cv∫Ω+ec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-1/2(xi,yi)yi2υidy.
Thus (3.8) can be written in the convolution form
(3.11)u(x,t)=E(x,t)*f(x).
Moreover, since E(x,t) exists, we can see that
(3.12)limt→0E(x,t)=Cv∫Ω+∏i=1nJvi-1/2(xi,yi)yi2vidy=Cv∫ℝn+∏i=1nJvi-1/2(xi,yi)yi2vidy=δ(x),forx∈ℝn+,
hold (see [8]). Thus for the solution u(x,t)=E(x,t)*f(x) of (3.1), then we have
(3.13)limt→0u(x,t)=u(x,0)=δ*f(x)=f(x),
which satisfies (3.2). This completes the proof.Theorem 3.2.
The kernelE(x,t) defined by (3.10) has the following properties.  (1)
E(x,t)∈𝒞∞-the space of continuous function for x∈ℝn, t>0 with infinitely differentiable.(2)
(∂/∂t-c2⊕Bk)E(x,t)=0  for all  x∈ℝn+,t>0.(3)
limt→0E(x,t)=δ  for all x∈ℝn+.Proof.
(1) From (3.10) and
(3.14)∂n∂tnE(x,t)=Cv∫ℝn+∂n∂tnec2t[(y12+⋯+yp2)4-(yp+12+⋯+yp+q2)4]k∏i=1nJvi-(1/2)(xi,yi)yi2υidy.
we have E(x,t)∈C∞ for x∈ℝn+,t>0.
(2) We haveu(x,t)=E(x,t) since u(x,t)=E(x,t)*f(x) holds. Note here that we use the fact f(x)=δ(x) by the Fourier Bessel transformation. Then, we obtain
(3.15)(∂∂t-c2⨁Bk)E(x,t)=0
by direct computation.
(3) This case is obvious by (3.12).
In particular, if we putk=1 and q=0 in (3.1), then (3.1) reduces to the equation
(3.16)∂∂tu(x,t)-c2ΔB4u(x,t)=0,
and we obtain the solution of (3.16) in the convolution form
(3.17)u(x,t)=E(x,t)*f(x),
where E(x,t) is defined by (2.23) with k=1 which is related to Bessel heat equation. This completes the proof.



---

*Source: 497676-2010-08-09.xml*