二次势逆热方程等矢量的两种计算

IF 0.6 Q4 MATHEMATICS, APPLIED
Mohamad Houda, P. Lescot
{"title":"二次势逆热方程等矢量的两种计算","authors":"Mohamad Houda, P. Lescot","doi":"10.4310/maa.2020.v27.n1.a2","DOIUrl":null,"url":null,"abstract":"We determine the isovectors of the backward heat equation with quadratic potential term in the space variable. This generalize the calculations of Lescot-Zambrini (cf.[4]). These results first appeared in the first author’s PhD thesis (Rouen, 2013). 1. Backward Equation The fundamental equation of Euclidean Quantum Mechanics of Zambrini is the backward heat equation with potential V (t, q):  θ ∂η u ∂t = − θ 2 ∂η u ∂q2 + V (t, q)η V u (C (V ) 1 ) η u (0, q) = u(q) where t represents the time variable, q the space variable, and θ is a real parameter, strictly positive (in physics, θ = √ ~). This equation is not well–posed in general, but existence and uniqueness of a solution are insured whenever the initial condition u belongs to the set of analytic vectors for the operator appearing on the right–hand side of the equation (see e.g. [1], Lemma 4, p. 429). We shall denote by ηu(t, q) = η 0 u(t, q) the solution of the backward heat equation with null potential :  θ 2 ∂ηu ∂t = − θ 2 ∂ηu ∂q2 (C (0) 1 ) ηu(0, q) = η 0 u(0, q) = u(q). The authors wish to thank the referees for constructive comments. 1.1. Generalization of results due to Lescot-Zambrini (cf.[4]). Theorem 1.1. Let a(t), b(t) and c(t) be continuous functions and V (t, q) = a(t)q+ b(t)q + c(t). For all initial conditions u, the solution η u (t, q) of: θ ∂η u ∂t = − 4 2 ∂η u ∂q2 + V (t, q)η u (C (V ) 1 ) such that η u (0, q) = u(q) is given by: η u (t, q) = φ1(t, q) ηu (φ2(t, q), φ3(t, q)) , (1.1) Date: 01 November 2019. 1 2 MOHAMAD HOUDA & PAUL LESCOT where: ηu(t, q) = η 0 u(t, q) and φ1(t, q), φ2(t, q), φ3(t, q) only depend on a(t), b(t) and c(t) (via formulas (1.7), (1.8), (1.9), (1.11), (1.12), (1.13) and (1.14)). Proof. We shall make formula 1.1 explicit with the following initial conditions:  φ1(0, q) = 1 φ2(0, q) = 0 φ3(0, q) = q. In this case : η u (0, q) = ηu(0, q) = u(q). We now differentiate formula 1.1 with respect to t and q; ηu and its derivatives will be taken in (φ2(t, q), φ3(t, q)). ∂η u ∂t = ∂φ1 ∂t ηu + φ1( ∂ηu ∂t ∂φ2 ∂t + ∂ηu ∂q ∂φ3 ∂t ) , ∂η u ∂q = ∂φ1 ∂q ηu + φ1( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) , ∂η u ∂q2 = ∂φ1 ∂q2 ηu + 2 ∂φ1 ∂q ( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) + φ1 [ ∂ηu ∂t2 ( ∂φ2 ∂q ) 2 + ∂ηu ∂q2 ( ∂φ3 ∂q ) 2 + 2 ∂ηu ∂t∂q ∂φ2 ∂q ∂φ3 ∂q + ∂ηu ∂t ∂φ2 ∂q2 + ∂ηu ∂q ∂φ3 ∂q2 ] and V (t, q)η u = a(t)q 2 φ1 ηu + b(t)q φ1 ηu + c(t) φ1 ηu. So, it’s enough to have: θφ1 ∂φ2 ∂t − θφ1( ∂φ3 ∂q ) 2 + θ 2 (2 ∂φ1 ∂q ∂φ2 ∂q + ∂φ2 ∂q2 ) = 0 (1.2) θ 2 φ1( ∂φ2 ∂q ) 2 = 0 (1.3) θφ1 ∂φ3 ∂t + θ 2 (2 ∂φ1 ∂q ∂φ3 ∂q + φ1 ∂φ3 ∂q2 ) = 0 (1.4) θ ∂φ2 ∂q ∂φ3 ∂q = 0 (1.5) θ ∂φ1 ∂t + θ 2 ∂φ1 ∂q2 − φ1 ( a(t)q + b(t)q + c(t) ) = 0 . (1.6) Two calculates concerned the isovectors of backward heat equation with potential term 3 As φ1(0, q) = 1, the equation (1.3) gives us: (2 ∂q ) 2 = 0, then φ2 = φ2(t) and (1.5) is then automatically satisfied. So, the equation (1.2) implies: φ1(t, q)( ∂φ2 ∂t − (3 ∂q )) = 0, then for all (t, q) : ∂φ2 ∂t = ( ∂φ3 ∂q ) 2","PeriodicalId":18467,"journal":{"name":"Methods and applications of analysis","volume":"1 1","pages":""},"PeriodicalIF":0.6000,"publicationDate":"2020-01-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"Two computations concerning the isovectors of the backward heat equation with quadratic potential\",\"authors\":\"Mohamad Houda, P. Lescot\",\"doi\":\"10.4310/maa.2020.v27.n1.a2\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"We determine the isovectors of the backward heat equation with quadratic potential term in the space variable. This generalize the calculations of Lescot-Zambrini (cf.[4]). These results first appeared in the first author’s PhD thesis (Rouen, 2013). 1. Backward Equation The fundamental equation of Euclidean Quantum Mechanics of Zambrini is the backward heat equation with potential V (t, q):  θ ∂η u ∂t = − θ 2 ∂η u ∂q2 + V (t, q)η V u (C (V ) 1 ) η u (0, q) = u(q) where t represents the time variable, q the space variable, and θ is a real parameter, strictly positive (in physics, θ = √ ~). This equation is not well–posed in general, but existence and uniqueness of a solution are insured whenever the initial condition u belongs to the set of analytic vectors for the operator appearing on the right–hand side of the equation (see e.g. [1], Lemma 4, p. 429). We shall denote by ηu(t, q) = η 0 u(t, q) the solution of the backward heat equation with null potential :  θ 2 ∂ηu ∂t = − θ 2 ∂ηu ∂q2 (C (0) 1 ) ηu(0, q) = η 0 u(0, q) = u(q). The authors wish to thank the referees for constructive comments. 1.1. Generalization of results due to Lescot-Zambrini (cf.[4]). Theorem 1.1. Let a(t), b(t) and c(t) be continuous functions and V (t, q) = a(t)q+ b(t)q + c(t). For all initial conditions u, the solution η u (t, q) of: θ ∂η u ∂t = − 4 2 ∂η u ∂q2 + V (t, q)η u (C (V ) 1 ) such that η u (0, q) = u(q) is given by: η u (t, q) = φ1(t, q) ηu (φ2(t, q), φ3(t, q)) , (1.1) Date: 01 November 2019. 1 2 MOHAMAD HOUDA & PAUL LESCOT where: ηu(t, q) = η 0 u(t, q) and φ1(t, q), φ2(t, q), φ3(t, q) only depend on a(t), b(t) and c(t) (via formulas (1.7), (1.8), (1.9), (1.11), (1.12), (1.13) and (1.14)). Proof. We shall make formula 1.1 explicit with the following initial conditions:  φ1(0, q) = 1 φ2(0, q) = 0 φ3(0, q) = q. In this case : η u (0, q) = ηu(0, q) = u(q). We now differentiate formula 1.1 with respect to t and q; ηu and its derivatives will be taken in (φ2(t, q), φ3(t, q)). ∂η u ∂t = ∂φ1 ∂t ηu + φ1( ∂ηu ∂t ∂φ2 ∂t + ∂ηu ∂q ∂φ3 ∂t ) , ∂η u ∂q = ∂φ1 ∂q ηu + φ1( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) , ∂η u ∂q2 = ∂φ1 ∂q2 ηu + 2 ∂φ1 ∂q ( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) + φ1 [ ∂ηu ∂t2 ( ∂φ2 ∂q ) 2 + ∂ηu ∂q2 ( ∂φ3 ∂q ) 2 + 2 ∂ηu ∂t∂q ∂φ2 ∂q ∂φ3 ∂q + ∂ηu ∂t ∂φ2 ∂q2 + ∂ηu ∂q ∂φ3 ∂q2 ] and V (t, q)η u = a(t)q 2 φ1 ηu + b(t)q φ1 ηu + c(t) φ1 ηu. So, it’s enough to have: θφ1 ∂φ2 ∂t − θφ1( ∂φ3 ∂q ) 2 + θ 2 (2 ∂φ1 ∂q ∂φ2 ∂q + ∂φ2 ∂q2 ) = 0 (1.2) θ 2 φ1( ∂φ2 ∂q ) 2 = 0 (1.3) θφ1 ∂φ3 ∂t + θ 2 (2 ∂φ1 ∂q ∂φ3 ∂q + φ1 ∂φ3 ∂q2 ) = 0 (1.4) θ ∂φ2 ∂q ∂φ3 ∂q = 0 (1.5) θ ∂φ1 ∂t + θ 2 ∂φ1 ∂q2 − φ1 ( a(t)q + b(t)q + c(t) ) = 0 . (1.6) Two calculates concerned the isovectors of backward heat equation with potential term 3 As φ1(0, q) = 1, the equation (1.3) gives us: (2 ∂q ) 2 = 0, then φ2 = φ2(t) and (1.5) is then automatically satisfied. So, the equation (1.2) implies: φ1(t, q)( ∂φ2 ∂t − (3 ∂q )) = 0, then for all (t, q) : ∂φ2 ∂t = ( ∂φ3 ∂q ) 2\",\"PeriodicalId\":18467,\"journal\":{\"name\":\"Methods and applications of analysis\",\"volume\":\"1 1\",\"pages\":\"\"},\"PeriodicalIF\":0.6000,\"publicationDate\":\"2020-01-01\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Methods and applications of analysis\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.4310/maa.2020.v27.n1.a2\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q4\",\"JCRName\":\"MATHEMATICS, APPLIED\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Methods and applications of analysis","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.4310/maa.2020.v27.n1.a2","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q4","JCRName":"MATHEMATICS, APPLIED","Score":null,"Total":0}
引用次数: 0

摘要

我们确定了空间变量中具有二次势项的后向热方程的等向量。这概括了Lescot-Zambrini(参见b[4])的计算。这些结果首次出现在第一作者的博士论文中(Rouen, 2013)。1. 欧几里得量子力学的基本方程是具有势V (t, q)的逆向热方程:θ∂η u∂t =−θ 2∂η u∂q2 + V (t, q)η V u(C (V) 1)η u(0, q) = u(q)其中t表示时间变量,q表示空间变量,θ是实参数,严格正(在物理学中,θ =√~)。一般情况下,这个方程不是适定的,但当初始条件u属于出现在方程右侧的算子的解析向量集时,解的存在唯一性就得到保证(例如,[1],引理4,第429页)。我们将用ηu(t, q) = η 0 u(t, q)表示零势的反向热方程的解:θ 2∂ηu∂t =−θ 2∂ηu∂q2 (C (0) 1) ηu(0, q) = η 0 u(0, q) = u(q)。作者希望感谢审稿人提出的建设性意见。1.1. 由于Lescot-Zambrini(参见[4])结果的概化。定理1.1。设a(t), b(t), c(t)为连续函数,V (t, q) = a(t)q+ b(t)q + c(t)。对于所有初始条件u,解ηu (t, q)为:θ∂η u∂t = - 4 2∂η u∂q2 + V (t, q)η u(C (V) 1),使得ηu (0, q) = u(q)由:ηu (t, q) = φ1(t, q)η u(φ2(t, q), φ3(t, q)), (1.1)1 2穆罕默德HOUDA和保罗LESCOT:ηu (t, q =η0 (t, q)和φ1 (t, q),φ2 (t, q),φ3 (t, q)只取决于(t), b (t)和c (t)(通过公式(1.7)、(1.8)、(1.9),(1.11),(1.12),(1.13)和(1.14))。证明。将公式1.1显式化,初始条件如下:φ1(0, q) = 1 φ2(0, q) = 0 φ3(0, q) = q,则ηu(0, q) = ηu(0, q) = u(q)。现在我们对公式1.1对t和q求导;ηu和它的导数用(φ2(t, q), φ3(t, q))表示。∂uη∂t =∂φ1 tη∂u +φ1(∂ηu∂t∂φ2∂t +∂uη∂问∂φ3∂t),∂ηu∂q =∂φ1 qη∂u +φ1(∂ηu∂t∂φ2∂q +∂uη∂问∂φ3∂q),∂uη∂q2 =∂φ1∂q2η2∂u +φ1∂q(∂ηu∂t∂φ2∂q +∂uη∂问∂φ3∂q) +φ1[∂uη∂t2(∂φ2∂q) 2 +∂uη∂q2(∂φ3∂q) 2 + 2∂ηu∂t∂问∂φ2∂问∂φ3∂q +∂ηu∂t∂φ2∂q2 +∂uη∂问∂φ3∂q2)和V (t, q)ηu = (t)问2φ1ηu + b (t)问φ1ηu + c (t)φ1ηu。它是足够的有:θφ1∂φ2∂t−θφ1(∂φ3∂q) 2 + 2θ(2∂φ1∂q∂φ2∂q +∂φ2∂q2) = 0(1.2)θ2φ1(∂φ2∂q) 2 = 0(1.3)θφ1∂φ3∂t +θ2(2∂φ1∂q∂φ3∂q +φ1∂φ3∂q2) = 0(1.4)θ∂φ2∂问∂φ3∂q = 0(1.5)θ∂φ1∂t +θ2∂φ1∂q2−φ1 ((t) q + b (t) q + c (t)) = 0。(1.6)两个计算涉及势项为3的逆向热方程的等矢量,当φ1(0, q) = 1时,由式(1.3)得到:(2∂q) 2 = 0,则φ2 = φ2(t),则(1.5)自动满足。因此,方程(1.2)暗示:φ1 (t, q)(∂φ2∂t−(3∂q) = 0,然后对所有(t, q):∂φ2∂t =(∂φ3∂q) 2
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Two computations concerning the isovectors of the backward heat equation with quadratic potential
We determine the isovectors of the backward heat equation with quadratic potential term in the space variable. This generalize the calculations of Lescot-Zambrini (cf.[4]). These results first appeared in the first author’s PhD thesis (Rouen, 2013). 1. Backward Equation The fundamental equation of Euclidean Quantum Mechanics of Zambrini is the backward heat equation with potential V (t, q):  θ ∂η u ∂t = − θ 2 ∂η u ∂q2 + V (t, q)η V u (C (V ) 1 ) η u (0, q) = u(q) where t represents the time variable, q the space variable, and θ is a real parameter, strictly positive (in physics, θ = √ ~). This equation is not well–posed in general, but existence and uniqueness of a solution are insured whenever the initial condition u belongs to the set of analytic vectors for the operator appearing on the right–hand side of the equation (see e.g. [1], Lemma 4, p. 429). We shall denote by ηu(t, q) = η 0 u(t, q) the solution of the backward heat equation with null potential :  θ 2 ∂ηu ∂t = − θ 2 ∂ηu ∂q2 (C (0) 1 ) ηu(0, q) = η 0 u(0, q) = u(q). The authors wish to thank the referees for constructive comments. 1.1. Generalization of results due to Lescot-Zambrini (cf.[4]). Theorem 1.1. Let a(t), b(t) and c(t) be continuous functions and V (t, q) = a(t)q+ b(t)q + c(t). For all initial conditions u, the solution η u (t, q) of: θ ∂η u ∂t = − 4 2 ∂η u ∂q2 + V (t, q)η u (C (V ) 1 ) such that η u (0, q) = u(q) is given by: η u (t, q) = φ1(t, q) ηu (φ2(t, q), φ3(t, q)) , (1.1) Date: 01 November 2019. 1 2 MOHAMAD HOUDA & PAUL LESCOT where: ηu(t, q) = η 0 u(t, q) and φ1(t, q), φ2(t, q), φ3(t, q) only depend on a(t), b(t) and c(t) (via formulas (1.7), (1.8), (1.9), (1.11), (1.12), (1.13) and (1.14)). Proof. We shall make formula 1.1 explicit with the following initial conditions:  φ1(0, q) = 1 φ2(0, q) = 0 φ3(0, q) = q. In this case : η u (0, q) = ηu(0, q) = u(q). We now differentiate formula 1.1 with respect to t and q; ηu and its derivatives will be taken in (φ2(t, q), φ3(t, q)). ∂η u ∂t = ∂φ1 ∂t ηu + φ1( ∂ηu ∂t ∂φ2 ∂t + ∂ηu ∂q ∂φ3 ∂t ) , ∂η u ∂q = ∂φ1 ∂q ηu + φ1( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) , ∂η u ∂q2 = ∂φ1 ∂q2 ηu + 2 ∂φ1 ∂q ( ∂ηu ∂t ∂φ2 ∂q + ∂ηu ∂q ∂φ3 ∂q ) + φ1 [ ∂ηu ∂t2 ( ∂φ2 ∂q ) 2 + ∂ηu ∂q2 ( ∂φ3 ∂q ) 2 + 2 ∂ηu ∂t∂q ∂φ2 ∂q ∂φ3 ∂q + ∂ηu ∂t ∂φ2 ∂q2 + ∂ηu ∂q ∂φ3 ∂q2 ] and V (t, q)η u = a(t)q 2 φ1 ηu + b(t)q φ1 ηu + c(t) φ1 ηu. So, it’s enough to have: θφ1 ∂φ2 ∂t − θφ1( ∂φ3 ∂q ) 2 + θ 2 (2 ∂φ1 ∂q ∂φ2 ∂q + ∂φ2 ∂q2 ) = 0 (1.2) θ 2 φ1( ∂φ2 ∂q ) 2 = 0 (1.3) θφ1 ∂φ3 ∂t + θ 2 (2 ∂φ1 ∂q ∂φ3 ∂q + φ1 ∂φ3 ∂q2 ) = 0 (1.4) θ ∂φ2 ∂q ∂φ3 ∂q = 0 (1.5) θ ∂φ1 ∂t + θ 2 ∂φ1 ∂q2 − φ1 ( a(t)q + b(t)q + c(t) ) = 0 . (1.6) Two calculates concerned the isovectors of backward heat equation with potential term 3 As φ1(0, q) = 1, the equation (1.3) gives us: (2 ∂q ) 2 = 0, then φ2 = φ2(t) and (1.5) is then automatically satisfied. So, the equation (1.2) implies: φ1(t, q)( ∂φ2 ∂t − (3 ∂q )) = 0, then for all (t, q) : ∂φ2 ∂t = ( ∂φ3 ∂q ) 2
求助全文
通过发布文献求助,成功后即可免费获取论文全文。 去求助
来源期刊
Methods and applications of analysis
Methods and applications of analysis MATHEMATICS, APPLIED-
自引率
33.30%
发文量
3
×
引用
GB/T 7714-2015
复制
MLA
复制
APA
复制
导出至
BibTeX EndNote RefMan NoteFirst NoteExpress
×
提示
您的信息不完整,为了账户安全,请先补充。
现在去补充
×
提示
您因"违规操作"
具体请查看互助需知
我知道了
×
提示
确定
请完成安全验证×
copy
已复制链接
快去分享给好友吧!
我知道了
右上角分享
点击右上角分享
0
联系我们:info@booksci.cn Book学术提供免费学术资源搜索服务,方便国内外学者检索中英文文献。致力于提供最便捷和优质的服务体验。 Copyright © 2023 布克学术 All rights reserved.
京ICP备2023020795号-1
ghs 京公网安备 11010802042870号
Book学术文献互助
Book学术文献互助群
群 号:481959085
Book学术官方微信