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Convergence and Inversion Theorems for Generalized Weierstrass Transform A

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Convergence and Inversion Theorems for Generalized Weierstrass Transform A Bulletin of Mathematical Sciences and Applications Online: 2015-10-01 ISSN: 2278-9634, Vol. 13, pp 6-12 doi:10.18052/www.scipress.com/BMSA.13.6 2015 SciPress Ltd, Switzerland Convergence and inversion theorems for generalized Weierstrass transform A. K. Thakur 1, a *, Preeti Tamrakar 2 1* Department of Mathematics, Dr. C. V. Raman University, Bilaspur (C.G.) India-495001 2 Research Scholar, Dr. C. V. Raman University, Bilaspur (C.G.) India-495001 a [email protected] Keywords: Laplace transforms, Stieltje’stransform, Whittaker’s function, Recurrence relation. Abstract. In this paper , we study the Weierstrass Transformation of certain generalized functions and their convolution transformation of certain generalised function and its inversion . We extend such type of results of convergence and inversion theorems for generalized weierstrass transforms (1.1) and (1.2). 1. Introduction The Laplace transform, become of its simple exponential kernel and is dense with its generalizations. The Stieltje’s transform, because of an outcome of iteration of Laplace transform is in now way poorer in respect to generalizations analogous to these, the Weierstrass transforms, ∞ 푓(푠) = ∫ 퐾(푠 − 푡, 1)φ(푡)푑푡, −∞ 2 −1⁄ −푠 ⁄ where (푠, 푡) = (4휋푡) 2푒 4푡,0 < 푡 < ∞ , −∞ ≤ 푠 < ∞, is the familiar source solution of the 휕2푢 휕푢 푏 heat equation = , a particular case of convolution transform, 푓(푠) = ∫ 퐾(푠 − 휕푠2 휕푡 푎 푡)φ(푡)푑푡, 푠 is complex, because of it involving the exponential of the square of the difference of the variable of the transform and that of the integrand, may be thought for generalization with the help of a special function which Can be particularized to exponential function. We have considered the following generalization of Weierstrass transforms by using Whittaker’s function −(푥−푦)2 ∞ 푒 1 8푡 푓(푥) = √ 1 ∫ 푊 1 1 (푥−푦)2 푑훼(푦) (1.1) ⁄2 푣 + , { } 2휋푡 −∞ √푥 − 푦 ⁄2 4 4 4푡 ∞ −3 −1 −1 ⁄4 ⁄2 2 푓(푥) = 2 휋 ∫ (푥 − 푦) 푤 1 1 (푥−푦)2 휑(푦)푑푦 (1.2) 푣+ , { } −∞ 2 4 4 2 2. The convergence theorems on (1.1) Theorem 1. If the integral (1.1) converges to a value A when x = x0, t = 1; then it converges for ∞ < 푥 < −∞, 0 < 푡 < 1 and ∞ −(푥 )² 푙푚 푒 0−푦 /8푡 − ∫ 푤 (푥−푦)² 푑 ∝ (푦) = 퐴 . (2.1) 푣/2+ 1/2,1/4 푡 → 1 √푥0 − 푦 4푡 −∞ Proof: We first quote a known lemma due to Hirschman and Widder [1]. If (x) is of bounded variation in a ≤ 푥 < 푅 for every R > 0, ∝ ( ) exists, 훽(푥) is positive, continuous and non-increasing in a ≤ 푥 < ∞ , then ∞ 훽(푥)푑 ∝ (푥) (2.2) ∫푎 SciPress applies the CC-BY 4.0 license to works we publish: https://creativecommons.org/licenses/by/4.0/ Bulletin of Mathematical Sciences and Applications Vol. 13 7 converges set for fixed 푥, 푥0, 푡 . 2 * 푦 −(푥 −푧)2/8 −1/2 (푥0−푧) (y) = ∫ 푒 0 [x0 – z] 푤 ,1/4 d (z) , (2.3) 0 푣/2+1/4 4 (푥−푦)2 2 2 푥 −푦 1 푤푣/2+1⁄4,1⁄4 −(푥−푦) /8푡 (푥0−푦) /8 0 ⁄2 1⁄2 4푡 훽(푦) = 푒 푒 [ 푡 ] (푥 −푦)2 . (2.4) 푥−푦 푤 +1⁄4,1⁄4 0 푣/2 4 Now (1.1) is convergent for, 푥 = 푥0, 푡 = 1 −(푥 −푦)2/8 2 ∞ 푒 0 (푥0−푦) ⇒∫ 푤푣⁄2 + 1⁄4, 1⁄4 d ∝ (y) is convergent. −∞ √푥0−푦 4 −(푥 −푦)2/8 2 0 ∞ 푒 0 (푥0−푦) ⇒ (∫ + ∫ ) 푤푣⁄2 + 1⁄4, 1⁄4 d ∝ (y) is convergent. −∞ 0 √푥0−푦 4 −(푥 −푦)2/8 2 ∞ 푒 0 (푥0−푦) ⇒∫ 푤푣⁄2 + 1⁄4, 1⁄4 d ∝ (y) is convergent. 0 √푥0−푦 4 ⇒∝∗ ( ) exists. Again since 훽(푦) is continuous and ∝∗ (푦) is of bounded variation is every finite interval, therefore 1 푦 ∫ 1 훽(푦) 푑 ∝∗ (푦) is convergent , (0< 푦 < ∞) . (2.5) (2휋푡)1⁄2 0 1 −푧2 2 2 Also 푤푣⁄2 + 1⁄4, 1⁄4 (푧 /2) o (푧 푒 2 ) for large, |z|. 2 ⇒푤푣⁄2 + 1⁄4, 1⁄4 (푧 /2) is positive, continuous and decreasing. ⇒훽(푦) is positive, continuous and decreasing. ∞ ⇒∫ 훽(푦)d훼∗ (y) is convergent by the lemma quoted. 푦1 1 ∞ ∗ ⇒ ⁄ ∫ 훽(푦) 푑훼 (푦) (2.6) (2휋푡)1 2 푦1 1 ∞ is convergent. (2.5) and (2.6) ⇒ ∫ 훽(푦) 푑 ∝∗ (푦) is convergent. (2휋푡)1⁄2 0 2 (푥−푦) ⁄ 1 ∞ 푒− 8푡 1 1 (푥−푦)2 푣 ( ) ⇒ 1⁄ ∫ 푊 + , 푑 ∝ 푦 (2.7) √(2휋푡 2 푎 √푥−푦 2 4 4 4푡 is convergent. − ∞ < 푥 < ∞ , 0 < 푡 < 1 by the equations (2.3) and (2.4). Consider the integral 2 1 0 푒−(푥−푦) /8푡 (푥−푦)2 ∫ 푤 + 1⁄4, 1⁄4 d ∝ (y) . (2.8) 1⁄ −∞ 푥−푦 푣⁄2 4푡 (2휋√푡) 2 √ 푥−푦 Which on making the substitution = z equals to √2푡 2 −푧 ⁄4 1 푥/√2푡 푒 2 ⁄ ∫ 푤푣⁄2 + 1⁄4, 1⁄4 [푧 /2] dα (x-z√2푡) (2휋√푡)1 2 ∞ √푧√2푡 2 −푧 ⁄4 1 ∞ 푒 2 = - ∫ 푤푣⁄2 + 1⁄4, 1⁄4 [푧 /2] d α (x-z√2푡) √2휋√푡 푥/√2푡 √푧√2푡 which is convergent by the lemma and therefore (2.8) is convergent for −∞ < x < ∞ 0 < 푡 < 1 . (2.7) and (2.8) ⇒ (1.1) is convergent for −∞ < x < ∞ , 0 < 푡 < 1 . To attend the second part of the theorem, we proceed as under −(푥 −푦)2/8푡 2 1 푥0 푒 0 (푥0−푦) = [ ∫ 푤푣⁄2 + 1⁄4, 1⁄4 d∝ (푦) √2휋√푡 −∞ √푥0−푦 4푡 −(푦−푥 )2/8푡 2 ∞ 푒 0 (푦−푥0) + ∫ 푤푣⁄2 + 1⁄4, 1⁄4 d훼(푦) ] (2.9) 푥0 √푦−푥0 4푡 1 0 −푧/8푡 −1⁄4 푧 = [ ∫ 푒 푧 푤푣⁄2 + 1⁄4, 1⁄4 ( ) 푑 ∝ (푥0 − √푧) √2휋√푡 ∞ 4푡 ∞ 푧 + ∫ 푒−푧/8푡 푧−1⁄4푤 + 1⁄4, 1⁄4 ( ) 푑 ∝ (푥 + √푧) ] 0 푣⁄2 4푡 0 1 ∞ −푧/8푡 −1⁄4 푧 = ∫ 푒 푧 푤푣⁄2 + 1⁄4, 1⁄4 ( ) [푑 ∝ (푥0 + √푧) − 푑 ∝ (푥0 − √푧)] √2휋√푡 푎 4푡 8 BMSA Volume 13 Which is a generalized Laplace transforms converging for 푡 = 1 by the hypothesis. Since it represents a continuous function of t in 0< t <1, therefore its value at t = 1 will be equal to its limit for t =1−. This is Convergence theorem for (1.1). Corollary If we take 푣 = 0, then we get the convergence theorem Hirschman and Widder [1]. 3. The convergence problems on (1.2) Theorem 2. If (1.2) be convergent for some x = 푥0 and 푦 [휑(푧) − 휑(푎)]푑푧 = 0,|푦 − 푎|, y a (3.1) ∫푎 then, ∞ 2 푙푚 (푎 − 푦) 2−3⁄4휋−1⁄2푡−1⁄4 ∫(푎 − 푦)−1⁄2 푤 + 1⁄4, 1⁄4 휑(푦)푑푦 푡 → 0+ 푣⁄2 2푡 −∞ 1 Γ ( ) = 2 휑(푎) (3.2) 1 푣 Γ (2 − 2) 휑(푥)being absolutely integrable in every finite integral and 푣 being a non – positive real no. Proof We first prove the following lemma. If (1.2) be convergent for some 푥 = 푥0 and 푦 [휑(푧) − 휑(푎)] 푑푧 = 푎(푦 − 푎) , y→ 푎+ (3.3) ∫푎 푙푚 ∞ (푎−푦)2 2−3⁄4휋−1⁄2푡−1⁄4 ∫ (푎 − 푦)−1⁄2 푤 + 1⁄4, 1⁄4 휑(푦)푑푦 (3.4) 푡 → 0+ 푎 푣⁄2 2푡 푙푚 푎 (푎−푦)2 2−3⁄4휋−1⁄2푡−1⁄4 ∫ (푎 − 푦)−1⁄2 푤 + 1⁄4, 1⁄4 휑(푦)푑푦 푡 → 0+ −∞ 푣⁄2 2푡 Γ(1/2) 휑(푎) = . Γ(1/2−푣/2) 2 For this we choose 훿 > 0 and write (3.4) as under 푎+훿 ∞ (푎−푦)2 푡−1⁄42−3⁄4휋−1⁄2 [∫ + ∫ .] (푎 − 푦)−1⁄2푤 + 1⁄4, 1⁄4 휑(푦)푑푦 = 퐼 (푡) + 퐼 (푡). 푎 푎+훿 푣⁄2 2푡 1 2 Now set 푦 (푥 −푧)2 ∝ (y) = ∫ (푥 − 푧)−1⁄2 푤 + 1⁄4, 1⁄4 0 휑(푧)푑푧 (3.5) 푎+훿 0 푣⁄2 2 2 푎−푦 −1/2 (푎−푦) ( ) 푤 ⁄ +1⁄4,1⁄4 √푡 푣 2 2푡 and 훽(푦) = (푥 −푦)2 (3.6) (푥 −푦)−1⁄2푤 +1⁄4,1⁄4 0 0 푣⁄2 2 (3.6), in the light of Slater [2] 푘 −푥⁄2 푤푘,푚(푥)~푥 푒 ⇒ 2 (푥 −푦) ⁄4 1 푎−푦 푣 푒 0 훽(푦) = 푣⁄2 ( ) 2⁄ 푡 푥0−푦 푒(푎−푦) 4푡 ⇒훽(푦) is decreasing and 훽(푦) → 0, as y → ∞ , since the denominator is more rapidly growing function as t → 0+. ∞ We have, 퐼 (t) = 2−3⁄4(휋t)−1⁄2 훽 (푦) d ∝ (y) by the equations (3.5) and (3.6) 2 ∫푎+훿 ∞ = 2−3⁄4(휋t)−1⁄2 [{훽(y) (y)}∞ – ∝ (푦) d훽(푦)] 푎+훿 ∫푎+훿 ∞ = - 2−3⁄4(휋t)−1⁄2 훼 (푦) d훽(y), 푠푛푐푒 ∝ (a+훿) = 0 and훽(푦) → 0 as y → ∞ ∫푎+훿 Bulletin of Mathematical Sciences and Applications Vol. 13 9 ∞ ⇒ |퐼 (푡)|≤ 2−3⁄4(휋푡)−1⁄2 M |d {-훽(y)} |. |α(y)| M 2 ∫푎+훿 = 휎(1) as t → 0+ by (3.6). Again 2 −3⁄4 −1⁄2 푎+훿 푎−푦 −1⁄2 (푎−푦) 퐼 (t) = 2 (휋t) ∫ ( ) 푤 ⁄ +1/4, 1/4 휑(푦)dy 2 푎 √푡 푣 2 2푡 2 −3⁄4 −1⁄2 푎+훿 푎−푦 −1⁄2 (푎−푦) (푎−푦)2/4t −(푎−푦)2/4t = 2 (휋t) ∫ ( ) 푤 ⁄ +1/4, 1/4 푒 푒 휑(푦)dy 푎 √푡 푣 2 2푡 푎+훿 푎−푦 2 2 = 2−3⁄4(휋푡)−1/22−푣⁄2−1/4 ∫ 퐷 ( ) 푒(푎−푦) /4푡푒−(푎−푦) /4t 휑(푦)dy 푎 푣 √푡 푎+훿 푎−푦 2 2 = 2−푣⁄2−1(휋푡)−1/2 ∫ 퐷 ( ) 푒(푎−푦) /4푡푒−(푎−푦) /4t 휑(푦)dy.
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