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Research article

Uniqueness for a Cauchy problem for the generalized Schrödinger equation

  • Received: 07 November 2022 Revised: 06 December 2022 Accepted: 15 December 2022 Published: 22 December 2022
  • MSC : 35A23, 35A32, 35Q40

  • In this work, we consider a Cauchy problem for the generalized Schrö dinger equation which has important applications in quantum kinetic theory, water wave problems and ferromagnetism. Due to its multidimensionality, it is important from the point of view of modern physics theories such as quantum field theory and string theory. We prove the uniqueness of the solution of the problem in an unbounded domain by using semigeodesic coordinates. The main tool is a pointwise Carleman estimate. To the authors' best knowledge, this is the first study which deals with the solvability of this problem.

    Citation: İsmet Gölgeleyen, Özlem Kaytmaz. Uniqueness for a Cauchy problem for the generalized Schrödinger equation[J]. AIMS Mathematics, 2023, 8(3): 5703-5724. doi: 10.3934/math.2023287

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  • In this work, we consider a Cauchy problem for the generalized Schrö dinger equation which has important applications in quantum kinetic theory, water wave problems and ferromagnetism. Due to its multidimensionality, it is important from the point of view of modern physics theories such as quantum field theory and string theory. We prove the uniqueness of the solution of the problem in an unbounded domain by using semigeodesic coordinates. The main tool is a pointwise Carleman estimate. To the authors' best knowledge, this is the first study which deals with the solvability of this problem.



    The generalized Schrödinger equation appears in many physical models in quantum kinetic theory [1], in the study of water wave problems and in ferromagnetism. Some of these models are Davey-Stewartson [2], Zakharov-Schulman [3] and Ishimori systems [4]. In these integrable systems, we frequently encounter the nonlinear Schrödinger equation and its generalized form, which provide a description of nonlinear waves such as propagation of a laser beam in a medium whose index of refraction is sensitive to the wave amplitude, water waves at the free surface of an ideal fluid and plasma waves, [5]. A detailed analysis of these types of equations was presented in [5]. Moreover, we refer to [6,7,8,9,10] for the new results related to the existence, uniqueness and asymptotic properties of solutions including radial cases for the nonlinear Schrödinger equation. We also refer to [11,12,13,14], where some new integrable systems were introduced based on the non-semisimple Lie algebra.

    In this article, we deal with the generalized Schrödinger equation

    itu(t,x)+ns,j=1asj(x)sju(t,x)n+ms,j=n+1asj(x)sju(t,x)+n+mj=1bj(t,x)ju(t,x)+b0(t,x)u(t,x)=f(t,x) (1.1)

    in the domain Ω={(t,x):tR, xDn+mRn+m}, where the bounded domain Dn+m is defined as

    Dn+m={x=(x1,¯x)Rn+m:x1>0, ¯x=(x2,...,xn+m)Rn+m1}

    and supported by x1=0. Throughout the paper, we use the following notations:

    tu=ut, 21u=2ux21, ju=uxj, sju=2uxsxj (1s, jn+m) and i=1. In (1.1), the coefficients asj (1s, jn+m) are real-valued, and there exist constants M, M1 such that asjC3(¯Dn+m  ) M and feκt2H2(R)M1, where κ>0 is a constant. The coefficients bjLloc(Ω) (j=0,1,...,n+m) are complex-valued, and there exists a constant M2 such that |β0tbj|M2, where 0β02, β0t=β0tβ0. We consider the following problem:

    Problem 1. Find the function u(t,x) in Ω which satisfies equation (1.1) and the following conditions

    u(t,0,¯x)=0, 1u(t,0,¯x)=0. (1.2)

    We set

    U={u(t,x): uC3(Ω), |β0tβ1sβ2ju|Cuexp(κut2)}, (1.3)

    where β0=0,1,2,3; β1, β2=0,1,2; s, j=0,1,2,...,n+m; and Cu, κu>0 are constants depending on u(t,x). In [15,16], the local well-posedness of some initial value problems for the generalized Schrödinger equation was considered in appropriate Sobolev spaces. In [17], a non-standard Cauchy problem for the classical Schrö dinger equation was studied, and solvability of the related inverse problem was investigated. A conditional stability estimate for the ultrahyperbolic Schrödinger equation was obtained in [18], where the ultrahyperbolic part of the equation consists of the term ΔyuΔxu. In [19], a Hölder stability estimate was established for the inverse problem. Problem 1 is a Cauchy problem for the generalized Schrödinger equation because the data are given on the x1=0. We investigate the uniqueness of the solution of the problem, and the main tool is a pointwise Carleman estimate. To the authors' best knowledge, this is the first result devoted to the solvability of Problem 1.

    The main result of this paper is given below:

    Theorem 1. Assume that there exist two positive constants α1,α2 such that

    ns,j=21(asj(x))ζsζjα1|ζ|2, n+ms,j=n+11(asj(x))ηsηjα2|η|2 (1.4)

    for any ζ=(ζ2,...,ζn)Rn1, η=(ηn+1,...,ηn+m)Rm, xDn+m. Then, there exists at most one solution uU satisfying (1.1) and (1.2).

    Theorem 1 is proved in Section 3 by using a Carleman type inequality which we call "Proposition 1". The proof of Proposition 1 will be given in Section 2.

    Here, we suppose that the metric g is given in semigeodesic coordinates, that is, it has the property g11=1, g1j=0, j=2,3,...,n+m; so (1.1) takes the form

    Puitu+21u+ns,j=2asj(x)sjun+ms,j=n+1asj(x)sju+n+mj=1bj(t,x)ju(t,x)+b0(t,x)u=f(t,x). (1.5)

    Since we shall prove the uniqueness, we consider the homogeneous version of Problem 1. Now, we define a new function z=ueκt2, so we can write

    Pz+2iκtz=F(t,x), (1.6)
    z(t,0,¯x)=1z(t,0,¯x)=0, (1.7)

    where F(t,x)=f(t,x)eκt2.

    If we apply the Fourier transform to (1.6) and conditions (1.7) with respect to t, we get

    ξˆz+21ˆz+ns,j=2asjsjˆzn+ms,j=n+1asjsjˆz+n+mj=1^bjjz+^b0z2κξˆz=F1+iF2, (1.8)
    ˆz(ξ,0,¯x)=1ˆz(ξ,0,¯x)=0, (1.9)

    where ˆz, ˆF, ^bjjz, ^b0z are the Fourier transforms with respect to t of the functions z, F, bjjz, b0z (j=0,1,...,n+m), respectively. Here, ξ is the parameter of the Fourier transforms and ξz=zξ.

    We write ˆz=w1(ξ,x)+iw2(ξ,x), ^bjjz=b1j+ib2j and ^b0z=b10+ib20 in (1.8), and so we obtain the following system of equations:

    2κξ(wk)21(wk)ns,j=2asjsj(wk)+n+ms,j=n+1asjsj(wk)+ξwk=lk, (1.10)

    where k=1,2;

    l1=F1n+mj=0b1j, l2=F2n+mj=0b2j. (1.11)

    By (1.9), we have

    w1(ξ,0,¯x)=0, 1(w1)(ξ,0,¯x)=0, w2(ξ,0,¯x)=0, 1(w2)(ξ,0,¯x)=0. (1.12)

    Thus, we shall show that this homogeneous problem has only zero solution.

    T. Carleman [20] established the first Carleman estimate in 1939 for proving the unique continuation for a two-dimensional elliptic equation. A Carleman estimate is an L2-weighted estimate with large parameters for a solution to a partial differential equation. These parameters play an essential role and are important how to choose a weight function in order to adjust given geometric configurations, [21]. This estimate is used as a powerful tool to establish the uniqueness and stability results for ill-posed Cauchy problems, [22]. In 1954, C. Müller extended Carleman's result to Rn, [23]. After that, A. P. Calderón [24] and L. Hörmander [25] improved these results based on the concept of pseudo-convexity. In [26,27,28], uniqueness and stability of various problems for the classical Schr ödinger equation were studied using Carleman estimates.

    The Carleman estimate used in this paper asserts a pointwise inequality, while the conventional Carleman estimates are proved in terms of weighted L2-integrals of solutions and boundary data. This type of pointwise Carleman estimate also was used in [22,29,30] for various equations including hyperbolic, parabolic and ultrahyperbolic equations.

    We first reduce (1.10) to a form which is more suitable for a Carleman estimate. For this aim, we define a new variable ˜x1=2x1μ0, μ0>0, and then we define

    wk(ξ,x)=wk(ξ,12(˜x1+μ0)2,¯x)˜wk(ξ,˜x1,¯x).

    Replacing the notations ˜wk,˜asj,˜x1,˜br,˜f with wk,asj,x1,br,f for the sake of simplicity, we have

    P0wk=lk,

    where lk=(x1+μ0)lk, k=1,2, and

    P0wk=(x1+μ0)121(wk)+(x1+μ0)(ns,j=2asjsj(wk)n+ms,j=n+1asjsj(wk)2κξ(wk)ξwk). (2.1)

    Next, we define a subdomain of Dn+m

    Dn+mγ={x:xRn+m, x1>0, α0<ψ(x)<γ+α0<1},

    where 0<γ<1, 2μ0min{α0,γ}, and a Carleman weight function

    φ=eλψν,

    where α0>0, the parameters δ, λ, v are positive numbers, and

    ψ(x)=δx1+12n+mj=2(xjx0j)2+α0. (2.2)

    Proposition 1. There exist δ, ν=ν(δ) and λ=λ(ν) such that the inequality

     ψν+1φ2(P0w)2+2λν(x1+μ0)Sφ2w(P0w)4λ3ν4ψ2ν3φ2w2+2λνφ2(1w)2+λν(x1+μ0)φ2n+mj=2(jw)2+4j=1dj (2.3)

    holds for all δ>δ, ν>ν, λ>λ and wC2(¯Dn+mγ). Here, the constant δ depends on α1,α2,M,ψ,γ, and

    S=nl,m=2m(almlψ)n+ml,m=n+1m(almlψ).

    Lemma 1. Assume that the conditions of Proposition 1 are satisfied. Then, there exist positive constants δ0, ν0=ν0(δ) and λ0=λ0(ν) such that

    ψν+1(P0w)2φ22λνγ3δ(1w)2φ2+λ3ν4δ4ψ2ν3w2φ2+λν(x1+μ0)δ(α1nj=2(jw)2+α2n+mj=n+1(jw)2)φ2+2λνξ(x1+μ0)2(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ))w2φ2+3j=1dj (2.4)

    for all δ>δ0, ν>ν0, λ>λ0 and wC2(¯Dn+mγ). Here, the constant δ0 depends on α1,α2,M,ψ,γ, and dj, 1j3, represent some divergence terms which will be given explicitly in the proof below.

    Proof. If we define a new function ϑ=φw, then we write

    ψν+1(P0w)2φ2=φ2ψν+1((x1+μ0)1φ1(21ϑ+2λνψν11ψ1ϑ+(λ2ν2ψ2ν2(1ψ)2λν(ν+1)ψν2(1ψ)2+λνψν121ψ)ϑ)+(x1+μ0)(ns,j=2asjφ1(sjϑ+2λνψν1sψjϑ+(λ2ν2ψ2ν2sψjψλν(ν+1)ψν2sψjψ+λνψν1sjψ)ϑ)n+ms,j=n+1asjφ1(sjϑ+2λνψν1sψjϑ+(λ2ν2ψ2ν2sψjψλν(ν+1)ψν2sψjψ+λνψν1sjψ)ϑ)2κξϑφ1ξϑφ1))2=ψν+1(y21+(y2+y4+y5)2+y23+2y1y2+2y1y3+2y1(y4+y5)+2y2y3+2y3(y4+y5)), (2.5)

    where

    y1=2κ(x1+μ0)ξϑ,y2=((x1+μ0)121ϑ+(x1+μ0)(ns,j=2asjsjϑn+ms,j=n+1asjsjϑ)),y3=2λνψν1(1ψ(x1+μ0)11ϑ+(x1+μ0)(ns,j=2asjsψjϑn+ms,j=n+1asjsψjϑ)),y4=Kϑ, y5=ξ(x1+μ0)ϑ,

    and

    K=(x1+μ0)1(λ2ν2ψ2ν2(1ψ)2λν(ν+1)ψν2(1ψ)2)+(x1+μ0)ns,j=2asj(λ2ν2ψ2ν2sψjψλν(ν+1)ψν2sψjψ+λνψν1δsj)(x1+μ0)n+ms,j=n+1asj(λ2ν2ψ2ν2sψjψλν(ν+1)ψν2sψjψ+λνψν1δsj).

    Next, we calculate the terms in (2.5). Here, we note that 1ψ=δ, 21ψ=0, sjψ=δsj, where

    δsj={1,s=j0,sj, 2s, jn+m.

    First, we can write

    2ψν+1y1y2=4κ(ν+1)ψν1ψξϑ1ϑ+4κ(ν+1)(x1+μ0)2ψνns,j=2jψasjξϑsϑ+4κ(x1+μ0)2ψν+1ns,j=2j(asj)ξϑsϑ4κ(ν+1)(x1+μ0)2ψνn+ms,j=n+1jψasjξϑsϑ4κ(x1+μ0)2ψν+1n+ms,j=n+1j(asj)ξϑsϑ+d1(ϑ)=2ψν+1y1y6+d1(ϑ),

    where

    y6=(ν+1)ψ1((x1+μ0)11ψ1ϑ+(x1+μ0)(ns,j=2asjsψjϑn+ms,j=n+1asjsψjϑ))+(x1+μ0)(ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ),
    d1=d1(ϑ)=4κ1(ψν+1ξϑ1ϑ)+2κξ(ψν+121ϑ)2κ(x1+μ0)2(ns,j=2(2j(ψν+1asjsϑξϑ)ξ(ψν+1asjsϑjϑ))n+ms,j=n+1(2(ψν+1asjsϑξϑ)ξ(ψν+1asjsϑjϑ))).

    On the other hand, by the inequality 2pqp2q2, we estimate the first, third, fourth and fifth terms in (2.5), so we have

    ψν+1(y21+y23+2y1y3+2y1y2)ψν+1(2y3y6y26)+d1(ϑ).

    Then, we can write

    ψν+1(y21+y23+2y1y3+2y1y2)ψν+1((4λν(ν+1)ψν2(ν+1)2ψ22λντ0ψν1+(ν+1)τ0ψ1)×((x1+μ0)11ψ1ϑ+(x1+μ0)(ns,j=2asjsψjϑn+ms,j=n+1asjsψjϑ))21τ0(2λνψν1(ν+1)ψ1)(x1+μ0)2(ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ)2(x1+μ0)2(ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ)2)+d1(ϑ)2λνψ(x1+μ0)2(ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ)2+d1(ϑ). (2.6)

    In (2.6), we used the inequality 2pqτ0p2q2τ0, for τ0=(ν+1)ψ1>0. Next, we continue to estimate the other terms in (2.5):

    2y1(y4+y5)ψν+1=2(2κ(x1+μ0)ξϑ)(Kϑ+ξϑ(x1+μ0))ψν+1=ξ(2κ(x1+μ0)ψν+1Kϑ2)+ξ(2κ(x1+μ0)2ξψν+1ϑ2)2κ(x1+μ0)2ψν+1ϑ2. (2.7)

    Since 21ψ=0, we have

    2y3y5ψν+1=2λν1(ξ1ψϑ2)2λν(x1+μ0)2ns,j=2j(asjsψξϑ2)+2λν(x1+μ0)2ξϑ2ns,j=2j(asjsψ)+2λν(x1+μ0)2n+ms,j=n+1j(asjsψξϑ2)2λν(x1+μ0)2ξϑ2n+ms,j=n+1j(asjsψ), (2.8)

    and

    2y2y3ψν+1=2λν1ψ1((x1+μ0)2(1ϑ)2)+4λν1ψ(x1+μ0)3(1ϑ)2+2λν1ψns,j=2(2j(asjsϑ1ϑ)1(asjsϑjv)2s(asj)jϑ1ϑ+1(asj)sϑjϑ)2λν1ψn+ms,j=n+1(2j(asjsϑ1ϑ)1(asjsϑjϑ)2s(asj)jϑ1ϑ+1(asj)sϑjϑ)+2λνns,j=2(21(asjsψsϑ1ϑ)21(asj)sψjϑ1ϑj(asjsψ(1ϑ)2)+j(asj)sψ(1ϑ)2)2λνn+ms,j=n+1(21(asjsψsϑ1ϑ)21(asj)sψjϑ1ϑj(asjsψ(1ϑ)2)+j(asj)sψ(1ϑ)2)+2λν(x1+μ0)2(n+ms,j,k,l=2(2j(asjakllϑsϑkψ)l(asjakljϑsϑkψ)2j(asjaklkψ)lϑsϑ+l(asjaklkψ)jϑsϑ))4λν(x1+μ0)2(ns,j=2n+ms,j=n+1(2j(asjakllϑsϑkψ)l(asjakljϑsϑkψ)2j(asjaklkψ)lϑsϑ+l(asjaklkψ)jϑsϑ)). (2.9)

    The last term has the following form:

    2y3y4ψν+1=2λν1ψ1((x1+μ0)1Kϑ2)2λν1ψ1((x1+μ0)1K)ϑ2+2λν(x1+μ0)(ns,j=2(j(asjsψKϑ2)j(asjsψK)ϑ2)n+ms,j=n+1(j(asjsψKϑ2)j(asjsψK)ϑ2)). (2.10)

    Then, from (2.6)-(2.10), we can write

    ψν+1(P0w)2φ22λνψ(x1+μ0)2(ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ)2+2λν(1ϑ)2ns=2ass+2λν(x1+μ0)2ξϑ2(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ))2λν(1ϑ)2n+ms=n+1ass+4λν1ψ(x1+μ0)3(1ϑ)22κ(x1+μ0)2ψν+1ϑ22λν(ns,j=2(2sψ1(asj)jϑ1ϑsψj(asj)(1ϑ)2+21ψs(asj)jϑ1ϑ1ψ1(asj)jϑsϑ)n+ms,j=n+1(2sψ1(asj)jϑ1ϑsψj(asj)(1ϑ)2+21ψs(asj)jϑ1ϑ1ψ1(asj)jϑsϑ))+2λν(x1+μ0)2(ns,j,k,l=2(2j(asjaklkψ)lϑsϑ+l(asjaklkψ)jϑsϑ)2ns,j=2n+ms,j=n+1(2j(asjaklkψ)lϑjϑ+l(asjaklkψ)jϑsϑ)+n+ms,j=n+1(2j(asjaklkψ)lϑjϑ+l(asjaklkψ)jϑsϑ))2λν(1ψ1((x1+μ0)1K)+(x1+μ0)ns,j=2(j(asjsψK)n+ms,j=n+1j(asjsψξ))ϑ2+d1(ϑ)+d2(ϑ), (2.11)

    where

    d2=d2(ϑ)=ξ(2κ(x1+μ0)ψν+1Kϑ2)+ξ(2κ(x1+μ0)2ξψν+1ϑ2)2λν1(ξ1ψϑ2)2λν(x1+μ0)2(ns,j=2j(asjsψξϑ2)n+ms,j=n+1j(asjsψξϑ2))+2λν1ψ(1((x1+μ0)2(1ϑ)2)+1((x1+μ0)1Kϑ2))+2λν(ns,j=2(21(asjsψjϑ1ϑ)j(asjsψ(1ϑ)2)+2j(1ψasj1ϑsϑ)1(1ψasjjϑsϑ))n+ms,j=n+1(21(asjsψjϑ1ϑ)j(asjsψ(1ϑ)2)+2j(1ψasj1ϑsϑ)1(1ψasjjϑsϑ)))+2λν(x1+μ0)2(n+ms,j,k,l=2(2j(asjakllϑsϑ)l(asjaklsψjϑsϑ))2ns,j=2n+mk,l=n+1(2j(asjakllϑsϑ)l(asjaklsψjϑsϑ)))+2λν(x1+μ0)(ns,j=2j(asjsψKϑ2)n+ms,j=n+1j(asjsψKϑ2)).

    Here, we note that ||asj||C1(¯Dn+mγ)M, and jψ=(xjx0j),|jψ|=|xjx0j|2γ,(2jn+m) in Dn+mγ. Since 2pq1ε0(x1+μ0)p2ε0(x1+μ0)q2, for all ε0>0 and p,q, we see that

    4λνδn+ms,j=n+1s(asj)1ϑjϑ2λνδ((x1+μ0)mε0n+mj=n+1(jϑ)2+m2M2ε0(x1+μ0)(1ϑ)2), (2.12)
    4λνn+ms,j=n+11(asj)sψ1ϑjϑ2λν(x1+μ0)2γmn+mj=n+1(jϑ)22λνm2M22γ(x1+μ0)(1ϑ)2, (2.13)
    2λνn+ms,j=n+1j(asj)sψ(1ϑ)22λνM2γm2(1ϑ)2, (2.14)
    (ns,j=2j(asj)sϑn+ms,j=n+1j(asj)sϑ)2M2(n1)((n1)2+m)nj=2(jϑ)2M2m((n1)+m2)n+mj=n+1(jϑ)2, (2.15)
    2λν(x1+μ0)2n+ms,j,k,l=2(l(asjaklkψ)jϑsϑ2j(asjaklkψ)lϑsϑ)λν(x1+μ0)26M2(22γ+1)(n+m1)3n+mj=2(jϑ)2, (2.16)
    2λν(x1+μ0)2ns,j=2n+mk,l=n+1(l(asjaklkψ)jϑsϑ2j(asjaklkψ)lϑsϑ)λν(x1+μ0)26M2(22γ+1)((n1)nj=2(jϑ)2+mn+mj=n+1(jϑ)2). (2.17)

    The other terms can be estimated similarly. Hence, we can rewrite (2.11) in the following form:

    ψν+1(P0w)2φ2E1(1ϑ)2+E2nj=2(jϑ)2+E3n+mj=n+1(jϑ)2+E4ϑ2+d1+d2, (2.18)

    where

    E1=4λνδ(x1+μ0)32λνδM2(n1)2ε0(x1+μ0)2λνM2(n1)22γ(x1+μ0)2λν2γ(n1)22λνδM2m2ε0(x1+μ0)2λνM2m22γ(x1+μ0)2λνM2γm22λνM(n1)2λνMm,
    E2=2λν(x1+μ0)(δα1δε0(n1)2γ(n1)3M2(x1+μ0)(22γ+1)(n1)33M2(x1+μ0)(22γ+1)(n1)+M2(x1+μ0)(n1)(m+(n1)2)),
    E3=2λν(x1+μ0)(δα2δε0m2γm3M2(x1+μ0)(22γ+1)m33M2(x1+μ0)(22γ+1)m+M2(x1+μ0)m(m2+(n1))),
    E4=2λν(δ((x1+μ0)1K)x1+(x1+μ0)ns,j=2(j(asjsψK)n+ms,j=n+1j(asjsψK)))2(x1+μ0)2(kψν+1λνξ(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ))).

    First, we shall estimate the coefficient E1:

    E1=2λν(x1+μ0)3(δ+E11E12),

    where

    E11=δ(1(x1+μ0)2M2ε0((n1)2+m2)),
    E12=(x1+μ0)22γM2((n1)2+m2)+(x1+μ0)32γM((n1)2+m2)+(x1+μ0)3M(n+m1).

    If we take γ as 0<γ<22ε03M(n1)2+m2, then 1((n1)2+m2)M2ε0(34γ)2>12, and so we obtain

    E11>δ(1((n1)2+m2)M2ε0(34γ)2)>12δ

    in Dn+mγ. Setting

    E12=2γM((n1)2+m2)(M+1)+M(n+m1)

    for δδ1=2E12, and then 12δE120, which implies that

    E1>2λν(x1+μ0)3(δ+12δE12)>2λνγ3δ. (2.19)

    As for the coefficient E2, if we take ε0 such that 0<ε0<α14(n1), then we have

    δ(α1ε0(n1))>34δα1. (2.20)

    Here, we note that

    δ2=4α1(2(n1)+3M2(22+1)(n1)3+6M2(22+1)(n1)+M2(n1)(m+(n1)2)).

    Then, for δδ2, from (2.20) it follows that

    E2=2λν(x1+μ0)(δα1δε0(n1)2γ(n1)3M2(x1+μ0)(22γ+1)(n1)33M2(x1+μ0)(22γ+1)(n1)M2(x1+μ0)(n1)(m+(n1)2))>λνδα1(x1+μ0). (2.21)

    Next, we estimate the coefficient E3. We choose ε0 such that 0<ε0<α24m,

    δ(α2ε0m)>34δα2. (2.22)

    Since μ0<12γ, δ4 and 0<δx1<γ in Dn+mγ, we get (x1+μ0)<34γ. Here, we set

    δ3=4α2(2m+3M2(22+1)m33M2(22+1)m+M2m(m2+(n1))).

    Then, if we take δδ3, from (2.22) it follows that

    E3>2λν(x1+μ0)(34δα214δα2)>λν(x1+μ0)δα2. (2.23)

    Now, let us estimate the coefficient E4. By the definition of the function K, we can write E4 in the following form

    E4=E41+E42+2(x1+μ0)2λνξ(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ)), (2.24)

    where

    E41=2λ3ν3δ31((x1+μ0)2ψ2ν2)2λ3ν3δ(ns,j=21(asjψ2ν2)sψjψn+ms,j=n+11(asjψ2ν2)sψjψ)2λ3ν3δ2(j(ψ2ν2ns,j=2asjsψ)j(ψ2ν2n+ms,j=n+1asjsψ))+4λ3ν3(x1+μ0)ns,j=2n+mk,l=n+1j(asjaklsψψ2ν2)kψlψ2λ3ν3(x1+μ0)2n+ms,j,k,l=2j(asjaklsψkψlψψ2ν2),
    E42=2λ2ν2(δ1((x1+μ0)1E01)+(x1+μ0)(ns,j=2j(asjsψE01)n+ms,j=n+1j(asjsψE01)))2κ(x1+μ0)2ψν+1,
    E01=(x1+μ0)1(ν+1)ψν2δ2+(x1+μ0)(ns,j=2asj((ν+1)ψν2sψjψψν1δsj)n+ms,j=n+1asj((ν+1)ψν2sψjψψν1δsj)).

    We now estimate each of terms in E41, respectively:

    2λ3ν3δ31((x1+μ0)2ψ2ν2)=4λ3ν3δ4(ν+1)(x1+μ0)2ψ2ν3+4λ3ν3δ3(x1+μ0)3ψ2ν2, (2.25)
    2λ3ν3δns,j=21(asjψ2ν2)sψjψ=4λ3ν3δ2(ν+1)ψ2ν3ns,j=2asjsψjψ2λ3ν3δψ2ν2ns,j=21(asj)sψjψ, (2.26)
    2λ3ν3δ2j(ψ2ν2n+ms,j=n+1asjsψ)=4λ3ν3δ2(ν+1)ψ2ν3n+ms,j=n+1asjsψ+2λ3ν3δ2ψ2ν2n+ms,j=n+1(j(asj)sψ+asjδsj), (2.27)
    2λ3ν3(x1+μ0)2ns,j,k,l=2j(asjaklsψkψlψψ2ν2)=4λ3ν3(x1+μ0)2(ν+1)ψ2ν3ns,j,k,l=2asjaklsψkψlψjψ2λ3ν3(x1+μ0)2ψ2ν2ns,j,k,l=2j(asjaklsψkψlψ), (2.28)
    4λ3ν3(x1+μ0)2ns,j=2n+mk,l=n+1j(asjaklsψψ2ν2)kψlψ=8λ3ν3(x1+μ0)2(ν+1)ψ2ν3ns,j=2n+mk,l=n+1asjaklsψjψkψlψ+4λ3ν3(x1+μ0)2ψ2ν2ns,j=2n+mk,l=n+1(j(asj)aklsψ+asjj(akl)sψ+asjaklδsj)kψlψ. (2.29)

    Then, from these relations we can write

    E41=4λ3ν3(ν+1)ψ2ν3(δ4(x1+μ0)2+2δ2ns,j=2asjsψjψ2δ2n+ms,j=n+1asjsψjψ+(x1+μ0)2n+ms,j,k,l=2asjaklsψkψlψjψ2(x1+μ0)2ns,j=2n+mk,l=n+1asjaklsψjψkψlψ)+4λ3ν3ψ2ν2(δ3(x1+μ0)3δ2ns,j=21(asj)sψjψ+δ2n+ms,j=n+11(asj)sψjψδ22ns,j=2(j(asj)sψ+asjδsj)+δ22n+ms,j=n+1(jasjsψ+asjδsj)(x1+μ0)22n+ms,j,k,l=2j(asjaklsψkψlψ)+(x1+μ0)2ns,j=2n+mk,l=n+1(j(asj)aklsψ+asjj(akl)sψ+asjaklδsj)kψlψ).

    Remembering that (x1+μ0)<34γ in Dn+mγ, we have δ4(x1+μ0)2>δ4. We set

    δ4=12((42+2)M2(n+m1)2+(22+3)M(n+m1)+4M2((n1)2+m2)),

    then, we see that

    E4>2λ3ν3δ4(ν+1)ψ2ν3+2λνξ(x1+μ0)2(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ))

    for δδ4 and λ>λ0, which yields

    ψν+1(P0w)2φ22λνδγ3(1ϑ)2+λν(x1+μ0)δα1nj=2(jϑ)2+λν(x1+μ0)δα2n+mj=n+1(jϑ)2+2λ3ν4δ4ψ2ν3ϑ2+2λνξ(x1+μ0)2(ns,j=2j(asjsψ)n+ms,j=n+1j(asjsψ))ϑ2+d1(ϑ)+d2(ϑ). (2.30)

    Finally, if we write ϑ=φw in (2.30), then we obtain

    2λνδγ3(1ϑ)2+λνδα1(x1+μ0)nj=2(jϑ)2+λνδα2(x1+μ0)n+mj=n+1(jϑ)2+2λ3ν4δ4ψ2ν3ϑ2=2λνγ3δ(1w)2φ2+λν(x1+μ0)δα1φ2nj=2(jw)2+λν(x1+μ0)δα2φ2n+mj=n+1(jw)2+2λ3ν4δ4ψ2ν3w2φ22λ2ν2γ3δ1(δψν1w2φ2)λ2ν2(x1+μ0)δα1nj=2j(ψν1jψw2φ2)λ2ν2(x1+μ0)δα2n+mj=n+1j(ψν1jψw2φ2)2λ3ν3γ3δ3ψ2ν2w2φ2λ3ν3(x1+μ0)δα1ψ2ν2w2φ2nj=2(jw)2λ3ν3(x1+μ0)δα2ψ2ν2w2φ2n+mj=n+1(jw)2λ2ν2(2γ3δ3(ν+1)ψν2+(x1+μ0)δα1(ν+1)ψν2nj=2(jw)2+(x1+μ0)δα2(ν+1)ψν2n+mj=n+1(jw)2(x1+μ0)δα1ψν1(n1)(x1+μ0)δα2ψν1m)w2φ2. (2.31)

    Choosing νν0=2γ3δ1+32γ2δ3(α1+α2)+1, we have

    w2φ2ψ2ν2(λ3ν4δ4ψ1φ22λ3ν3γ3δ3λ3ν3(x1+μ0)δ(α1nj=2(jw)2+α2n+mj=n+1(jw)2))λ3ν3δ4w2φ2ψ2ν2(ν2γ3δ132γ2δ3(α1+α2))λ3ν3δ4w2φ2ψ2ν2. (2.32)

    On the other hand, if λλ0=2γ3(ν+1)+32γ2(ν+1)(α1+α2), then we get

    λ3ν3δ4ψ2ν2w2φ22λ2ν2γ3δ3(ν+1)ψν2w2φ2λ2ν2(x1+μ0)δ(α1+α2)(ν+1)ψν2w2φ2n+mj=2(jw)2λ2ν3δ4ψ2ν2w2φ2(λ2γ3(ν+1)32γ2(ν+1)(α1+α2))0. (2.33)

    Therefore, for δ>δ0, νν0 and λλ0, from (2.30)-(2.33) we have

    2λνγ3δ(1ϑ)2+λν(x1+μ0)δα1nj=2(jϑ)2+λν(x1+μ0)δα2n+mj=n+1(jϑ)2+2λ3ν4δ4ψ2ν3ϑ22λνγ3δ(1w)2φ2+λν(x1+μ0)δα1φ2nj=2(jw)2+λν(x1+μ0)δα2φ2n+mj=n+1(jw)2+λ3ν4δ4ψ2ν3w2φ2+d3(w), (2.34)

    where δ0=max{4, δ1, δ2, δ3, δ4}, and

    d3=d3(w)=2λ2ν2γ3δ1(ψν1δw2φ2)λ2ν2(x1+μ0)δα1nj=2j(ψν1jψw2φ2)λ2ν2(x1+μ0)δα2n+mj=n+1j(ψν1jψw2φ2).

    From relations (2.30) and (2.34), the proof of Lemma 1 is completed.

    Lemma 2. If λ exceeds some constant, then for all wC2(Dn+mγ), we have

    2λν(x1+μ0)φ2Sw(P0w)4S0λ3ν3ψ2ν2(δ2+2γM(n1)2+m2)w2φ22λνS0φ2((1w)2+(x1+μ0)2M((n1)nj=2(jw)2+mn+mj=n+1(jw)2))Cλ2ν2ψν1δ2φ2w22λνSξ(x1+μ0)2φ2w2+d4(w), (2.35)

    where C>0 is a constant depending on asj and dimension n+m,

    d4=d4(w)=ξ(2λνκSφ2(x1+μ0)2)+λν1(2φ2Sw1w1(φ2S)w2)λν(x1+μ0)2ns,j=2(j(s(φ2Sasj)w2)2s(φ2Sasjwjw))+λν(x1+μ0)2n+ms,j=n+1(j(s(φ2Sasj)w2)2s(φ2Sasjwjw)),S=nl,m=2m(almlψ)n+ml,m=n+1m(almlψ),S0=(n1)M((n1)2γ+2)+mM(m2γ+2).

    Proof. By using the equality

    2λν(x1+δ0)φ2Sw(P0w)=2λνφ2Sw21w+2λν(x1+δ0)2φ2Sw(ns,j=2asjsjwn+ms,j=n+1asjsjw)4λν(x1+δ0)2φ2Swκξw2λν(x1+δ0)2φ2Sw2ξ

    and taking into account spjq=j((sp)q)j(sp)q, we have

    2λν(x1+μ0)φ2Sw(P0w)=2λνφ2S((1w)2+(x1+μ0)2(ns,j=2asjswjwn+ms,j=n+1asjswjw))+S1w22λν(x1+μ0)2φ2Sξw2+d4(w), (2.36)

    where

    S1=λν(21(φ2S)+(x1+μ0)2(sj(φ2Sns,j=2asj)sj(φ2Sn+ms,j=n+1asj))).

    Remembering that |S|S0 and ||asj||C1(Dn+mγ)M (2s,jn+m), we see that the first term in (2.36) can be written as

    2λνφ2S((1w)2+(x1+μ0)2(ns,j=2asjswjwn+ms,j=n+1asjswjw))2λνS0φ2((1w)2+(x1+μ0)2M((n1)nj=2(jw)2mn+mj=n+1(jw)2)). (2.37)

    Next, as for the coefficient in the second term, we write

    S1=λν(21(φ2)S+21(φ2)1S+φ221S+(x1+μ0)2(ns,j=2(Ss(j(φ2)asj)+s(φ2)(2jSasj+Sj(asj))+φ2(sjSasj+2jSs(asj)+Ssj(asj)))n+ms,j=n+1(Ss(j(φ2)asj)+s(φ2)(2jSasj+Sj(asj))+φ2(sjSasj+2jSs(asj)+Ssj(asj)))). (2.38)

    Since

    21(φ2)=2λνφ2(2λνδ2ψ2ν2+(ν+1)δ2ψν2),j(asjs(φ2))=2λνφ2(asj(2λνψ2ν2sψjψ+(ν+1)ψν2sψjψψν1sjψ)s(asj)ψν1jψ), (2.39)
    |S|S0, |jS|S0,|sjS|(n1)M((n1)2γ+3)+mM(m2γ+3), (2.40)

    we see that

    |S1|4λ3ν3φ2ψ2ν2S0(δ2+(x1+δ0)2(n+ms,j=2|asjsψjψ|)+λ2ν2ψν1δ2φ2S0(2(ν+1)ψ1+4δ+(x1+δ0)2δ2(n+ms,j=2|2(ν+1)φ2asjψ1sψjψ|+|2asjsjψ|+|2s(asj)jψ|+|4sψasj|+|2sψs(asj)|)))+λνφ2(n+ms,j=2|2s(asj)|S0+|sj(asj)|S0)+|21S|+(x1+δ0)2(n+ms,j=2|asj||sjS|)).

    If we choose that

    λλ1=n+ms,j=2(|2s(asj)|+|sj(asj)|)S0+|21S|+(x1+δ0)2(n+ms,j=2|asj||sjS|),

    and v1, then we have

    |S1|4λ3ν3φ2ψ2ν2S0(δ2+(x1+δ0)2(n+ms,j=2|asjsψjψ|))+Cλ2ν2ψν1δ2φ2,

    where

    C=S0(2(ν+1)ψ1+4δ+(x1+δ0)2δ2(n+ms,j=2(|2(ν+1)asjψ1sψjψ|+|2asjsjψ|+|2s(asj)jψ|+|4sψasj|+|2sψs(asj)|)))+ψv+1δ2.

    Here, we note that |asj|M and |jψ|=2γ,(2s,jn+m). We see that

    S14λ3ν3φ2ψ2ν2S0(δ2+2γM(n1)2+2γMm2)Cλ2ν2δ2ψν1φ2. (2.41)

    Then, by relations (2.36) and (2.41), we write

    2λν(x1+μ0)φ2Sw(P0w)4S0λ3ν3ψ2ν2(δ2+2γM(n1)2+m2)w2φ22λνS0φ2((1w)2+(x1+μ0)2M((n1)nj=2(jw)2+mn+mj=n+1(jw)2))Cλ2ν2ψν1δ2φ2w22λνSξ(x1+μ0)2φ2w2+d4(w).

    Proof of Proposition 1.

    We sum inequalities (2.4) and (2.35) and then we obtain

    ψν+1φ2(P0w)2+2λν(x1+μ0)φ2Sw(P0w)4j=1dj+2λν(1w)2φ2(δγ3S0)+λν(x1+μ0)(δα12S0M(n1)(x1+μ0))φ2ns,j=2(jw)2+λν(x1+μ0)(δα22S0Mm(x1+μ0))φ2n+ms,j=n+1(jw)2+4λ3ν3ψ2ν3w2φ2(14νδ4ψS0(δ2+2γM((n1)2+m2))Cδ2ψν+2λν).

    By choosing δ=max{δ0, δ5}, δ5=max{(1+S0)γ3, (1+2S0γM(n1))/α1, (1+2S0γMm)/α2}, λmax{λ0, λ1, C}, νmax{ν0, 4δ4(1+ψS0(δ2+2Mγ((n1)2+m2)+δ2)}, we have (2.3) for δ>δ, ν>ν, λ>λ. This completes the proof.

    In order to prove Theorem 1, by (1.11) and the equality lk=(x1+μ0)lk, we write

    (l1)2+(l2)24(F21+F22)(x1+μ0)2+4(n+m+1)n+mj=0(b21j+b22j). (3.1)

    Using (3.1), we have

    2k=1(ψν+1(P0wk)2φ2+2λν(x1+μ0)Swk(P0wk)φ2)2k=1(ψν+1(P0wk)2φ2+(P0wk)2φ2+λ2ν2(x1+μ0)2S2w2kφ2)=2k=1((ψν+1+1)(P0wk)2φ2+λ2ν2(x1+μ0)2S2w2kφ2)=(ψν+1+1)((l1)2+(l2)2)φ2+λ2ν2(x1+μ0)2S2(w21+w22)φ2(ψν+1+1)(x1+μ0)2(4(F21+F22)+4(n+m+1)n+mj=0(b21j+b22j))φ2+λ2ν2(x1+μ0)2S2(w21+w22)φ2. (3.2)

    Next, we apply the Carleman estimate in (3.2):

    2k=1(ψν+1(P0wk)2φ2+2λν(x1+μ0)Swk(P0wk))φ22λνφ22k=1(1(wk))2+λν(x1+μ0)φ22k=1n+mj=2(j(wk))2+4λ3ν3ψ2ν3φ22k=1(wk)2+2k=14j=1dj(wk). (3.3)

    By (3.2) and (3.3), we write

    2k=1(2λνφ2(1(wk))2+λ(x1+μ0)φ2n+mj=2(j(wk))2+4λ3ν3ψ2ν3φ2(wk)2)+2k=14j=1dj(wk)(ψν+1+1)(x1+μ0)2(4(F21+F22)+4(n+m+1)n+mj=0(b21j+b22j))φ2+λ2ν2(x1+μ0)2S2(w21+w22)φ2. (3.4)

    If we multiply (3.4) by (1+ξ2)2 and integrate with respect to ξ over the interval (,), we have

    2k=1(2λν(1(wk))2+λ(x1+μ0)n+mj=2(j(wk))2+4λ3ν3ψ2ν3(wk)2)φ2(1+ξ2)2dξ+2k=14j=1dj(wk)(1+ξ2)2dξ(ψν+1+1)(x1+μ0)2(4M1+8(n+m+1)M32k=1n+mj=0(j(wk))2(1+ξ2)2dξ)φ2+λ2ν2φ2(x1+μ0)2S22k=1w2kdξ. (3.5)

    In (3.5), we used the following relation by the Plancherel's theorem:

    n+mj=0(1+ξ2)2|^bjjz|2dξ2n+mj=00β02|β0t(bjjz)|2dt2M3n+mj=0(1+ξ2)2|^jz|2dξ=2M32k=1n+mj=0(1+ξ2)2(j(wk))2dξ,

    where M3 is a constant depending on M2. Then, the inequality (3.5) takes the form

    E52k=1(1(wk))2(1+ξ2)2φ2dξ+E62k=1n+mj=2(j(wk))2(1+ξ2)2φ2dξ+E72k=1(wk)2(1+ξ2)2φ2dξ2k=14j=1dj(wk)(1+ξ2)2dξ, (3.6)

    where

    E5=2λν8(n+m+1)(ψν+1+1)(x1+μ0)2M3,E6=λν(x1+μ0)8(n+m+1)(ψν+1+1)(x1+μ0)2M3,E7=4λ3ν3ψ2ν3(λ2ν2S28(n+m+1)(ψν+1+1)M3+4(ψν+1+1)M1)(x1+μ0)2.

    We estimate the coefficients E5,E6,E7: For 0<ψ<η<1, and λ>1, we have

    E5=2λν16(n+m+1)M3>λ (3.7)

    for νν1=12+8(n+m+1)M3, and

    E6=(x1+μ0)2(λν16(n+m+1))M3>(x1+μ0)2λ (3.8)

    for νν2=1+16(n+m+1)M3+C. If λ>max{λ2,λ3}, λ2=(S0)2, λ3=16(n+m+1)M3+8M1, and then we have

    E74λ3ν3ψ2ν3λ2ν2(S0)216(n+m+1)M38M12λ3ν3ψ2ν3. (3.9)

    Hence, for λ>1 and v>max{v1,v2}, (3.6)-(3.9) yield

    2k=1(λ(1(wk))2+λ(x1+μ0)2n+mj=2(j(wk))2+2λ3ν3ψ2ν3(wk)2)(1+ξ2)2φ2dξ2k=14j=1dj(wk)(1+ξ2)2dξ. (3.10)

    Since 2(1+ξ2)2>1 and φ2>1 on Dn+mγ, we write

    2k=1(wk)2dξ1λ3ν32k=14j=1dj(wk)(1+ξ2)2dξ. (3.11)

    Integrating inequality (3.11) over the domain Dn+mγ and passing to the limit as λ, we get w1=w2=0,, that is, ˆz=0 which implies w(x)=0 in Dn+mγ. Thus, Theorem 1 is proved.

    In this study, we consider a Cauchy problem for the generalized Schrö dinger equation. We prove the uniqueness of the solution of the problem by using the Carleman estimate. By similar arguments, a stability estimate can be obtained. Due to its multidimensional structure, the equation has an important place in some modern physics theories such as M-theory and twistor theory. In the future, we are planning to investigate the existence of the solution of the problem.

    This research received no external funding.

    The authors declare no conflict of interest.



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