Two computations concerning the isovectors of the backward heat equation with quadratic potential

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