ON THE GENERALIZED NONLINEAR DIAMOND HEAT KERNEL

In this paper, we study the nonlinear heat equation ∂ ∂t △ k u(x, t) − c 2 ♦ k u(x, t) = f (x, t, u(x, t)), where △ k is the Laplacian operator iterated k− times and is defined by (1.4)and ♦ k is the Diamond operator iterated k− times and is defined by (1.2). We obtain an interesting kernel related to the nonlinear heat equation.


Introduction
Consider the ultra-hyperbolic operator iterated k− times defined by , (1.1) p+q = n.S.E.Trione [8] has shown that the generalized function R 2k (x) defined by (2.4) is the unique elementary solution of the operator k that is k R 2k (x) = δ where x ∈ R n the n− dimensional Euclidean space.Also M. Aguirre Tellez [1, pp.147-149] has proved that R 2k (x) exists only if n is odd with p odd and q even or only if n is even with p odd and q odd.In 1996, A. Kananthai [4] [2] first introduced the Diamond operator defined by (1. 2) The operator ♦ k can be written as the product of the operators in the form where △ k is the Laplacian operator iterated k− times and is defined by and k is the Ultra-hyperbolic operator iterated k− times and defined by(1.1). A. Kananthai [4,p.33] has shown that the convolution (−1 where δ(x) is Dirac-delta distribution and the function R e 2k (x) and R H 2k (υ) are defined by (2.7) and (2.4) respectively with α = β = 2k, k is a nonnegative integer.
It is well known that for the heat equation where △ is the Laplace operator and is defined by as the solution of (1.6).The equation (1.7) can be written in the following form where n and t > 0.Moreover, we obtain E(x, t) → δ as t → 0 where δ is the Dirac delta distribution.
Next, K. Nonlaopon and A. Kananthai [4] have studied the heat equation where is the ultra-hyperbolic operator and is defined by (1.1) with k = 1.They obtain the ultrahyperbolic heat kernel where p + q = n is the dimension of the Euclidean space R n and i = √ −1.Now, the purpose of this article is to study the nonlinear equation where (x, t) = (x 1 , . . ., x n , t) ∈ R n × (0, ∞), t is a time, c is a positive constant, u(x, t) is an unknown function.We consider the equation (1.11) with the following conditions on u and f as follows (1) u(x, t) ∈ C (6k) (R n ) for any t > 0 where C (6k) (R n ) is the space of continuous function with 6k-derivatives.
(2) f satisfies the Lipschitz condition, that is where A is constant with 0 < A < 1. ( Under such conditions of f and u and for the spectrum of as a solution of (1.11), where and Ω ⊂ R n is the spectrum of E(x, t) for any fixed t > 0 and (−1) k R e 2 (x) is defined by (2.4) with α = 2k.The convolution (−1) k R e 2 (x) * E(x, t) is called the Diamond Heat Kernel and all properties it will be studied in detail.Before proceeding the following definitions and concepts of needed.

Preliminaries Definition
The Fourier transform of f (x) is defined by where Also, the inverse of Fourier transform is defined by If f is a distribution with compact supports by the Equation(2.1) can be written as Definition 2.2.The spectrum of the kernel E(x, t) of (1.9) is the bounded support of the Fourier transform E(ξ, t) for any fixed t > 0.
Definition 2.3.Let ξ = (ξ 1 , ξ 2 , . . ., ξ n ) be a point in R n and be denoted by The set of an interior of the forward cone, and Γ + denotes the closure of Γ + .
Let Ω be a spectrum of E(x, t) defined by definition 2.2 for any fixed t > 0 and Ω ⊂ Γ + .Let E(ξ, t) be the Fourier transform of E(x, t) and be defined as (2.3) Definition 2.4.Let x = (x 1 , x 2 , ..., x n ) be a point of the n-dimensional Euclidean space R n and written as where p + q = n is the dimension of the space R n .Let Γ + = {x ∈ R n : x 1 > 0 and υ > 0} be the interior of the forward cone and Γ + denote its closure.For any complex number α, define the function where the constant K n (α) is given by the formula . ( The function R H α (υ) is called the ultra-hyperbolic kernel of Marcel Riesz and was introduced by Y. Nozaki [7].
It is well known that R H α (υ) is an ordinary function if Re(α) ≥ n and it is a distribution of . The elliptic kernel of Marcel Riesz is defined as where with α a complex parameter and n the dimension of R n .It can be shown that R e −2k (x) = (−1) k △ k δ(x) where △ k is defined by (1.4).It follows that R e 0 (x) = δ(x).The function R e 2k (x) is called the elliptic kernel of Marcel Riesz.Lemma 2.1.Let L be the operator and defined by where is the ultra-hyperbolic defined by (1.1) for t ∈ (0, ∞) and c is a positive constant.Then we obtain (2.9) as an elementary solution of (2.8)in the spectrum Ω ⊂ R n for t > 0.
Proof.Let LE(x, t) = δ(x, t), where E(x, t) is an elementary solution of operator L and δ is the Dirac-delta distribution.Thus If we apply the Fourier transform defined by (2.1) to both sides of the equation, we obtain where H(t) is the Heaviside function.Since H(t) = 1 for t > 0. Therefore, which has been already defined by (2.4).Thus where Ω is the spectrum of E(x, t).Thus from (2.4)
(3) lim , for t > 0, where M (t) is a function of t in the spectrum Ω and Γ denotes the Gamma function.Thus E(x, t) is bounded for any fixed t > 0. ( Proof.(1) Since E(x, t) and (−1) k R e 2k (x) are tempered distribution with compact support.Thus (−1) k R e 2k (x) * E(x, t) exists and is a tempered distribution.
(2) We have Since E(x, t) exists, then By the continuity of the convolution, (4) We have By changing to bipolar coordinates exp c 2 t r 2 − s 2 r p−1 s q−1 dr ds dΩ p dΩ q where dξ = r p−1 s q−1 dr ds dΩ p dΩ q , dΩ p and Ω q are the elements of the surface area of the unit sphere in R p and R q respectively.Since Ω ⊂ R n is the spectrum of E(x, t) and we suppose 0 ≤ r ≤ R and 0 ≤ s ≤ T where R and T are constants, we obtain where and Ω q = 2π p/2 Γ q
Theorem 3.2.Given the equation for (x, t) ∈ R n ×(0, ∞), k is a positive number and with the following conditions on u and f as follows (1) u(x, t) is the space of continuous function on R n × (0, ∞).
(2) f satisfies the Lipschitz condition, where A is constant with 0 < A < 1. ( Then we obtain the convolution as a solution of (3.3) for x ∈ Ω 0 where Ω 0 is a compact subset of R n for 0 ≤ t ≤ T with T is constant and (−1)R e 2k (x) is given by (2.4) and E(x, t) is given by (??).
Proof.By (3.1), we have The above can be written in the form or v(x, t) = E(x, t) * f (x, t, u(x, t)).