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
i∂tu(t,x)+n∑s,j=1asj(x)∂s∂ju(t,x)−n+m∑s,j=n+1asj(x)∂s∂ju(t,x)+n+m∑j=1bj(t,x)∂ju(t,x)+b0(t,x)u(t,x)=f(t,x) | (1.1) |
in the domain Ω={(t,x):t∈R, x∈Dn+m⊂Rn+m}, where the bounded domain Dn+m is defined as
Dn+m={x=(x1,¯x)∈Rn+m:x1>0, ¯x=(x2,...,xn+m)∈Rn+m−1} |
and supported by x1=0. Throughout the paper, we use the following notations:
∂tu=∂u∂t, ∂21u=∂2u∂x21, ∂ju=∂u∂xj, ∂s∂ju=∂2u∂xs∂xj (1≤s, j≤n+m) and i=√−1. In (1.1), the coefficients asj (1≤s, j≤n+m) are real-valued, and there exist constants M, M1 such that ‖asj‖C3(¯Dn+m ) ≤M and ‖fe−κt2‖H2(R)≤M1, where κ>0 is a constant. The coefficients bj∈L∞loc(Ω) (j=0,1,...,n+m) are complex-valued, and there exists a constant M2 such that |∂β0tbj|≤M2, where 0≤β0≤2, ∂β0t=∂β0∂tβ0. 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): u∈C3(Ω), |∂β0t∂β1s∂β2ju|≤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 Δyu−Δxu. 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
n∑s,j=2∂1(asj(x))ζsζj≥α1|ζ|2, −n+m∑s,j=n+1∂1(asj(x))ηsηj≥α2|η|2 | (1.4) |
for any ζ=(ζ2,...,ζn)∈Rn−1, η=(ηn+1,...,ηn+m)∈Rm, x∈Dn+m. Then, there exists at most one solution u∈U 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
Pu≡i∂tu+∂21u+n∑s,j=2asj(x)∂s∂ju−n+m∑s,j=n+1asj(x)∂s∂ju+n+m∑j=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=ue−κt2, 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)e−κt2.
If we apply the Fourier transform to (1.6) and conditions (1.7) with respect to t, we get
−ξˆz+∂21ˆz+n∑s,j=2asj∂s∂jˆz−n+m∑s,j=n+1asj∂s∂jˆz+n+m∑j=1^bj∂jz+^b0z−2κ∂ξˆz=F1+iF2, | (1.8) |
ˆz(ξ,0,¯x)=∂1ˆz(ξ,0,¯x)=0, | (1.9) |
where ˆz, ˆF, ^bj∂jz, ^b0z are the Fourier transforms with respect to t of the functions z, F, bj∂jz, b0z (j=0,1,...,n+m), respectively. Here, ξ is the parameter of the Fourier transforms and ∂ξz=∂z∂ξ.
We write ˆz=w1(ξ,x)+iw2(ξ,x), ^bj∂jz=b1j+ib2j and ^b0z=b10+ib20 in (1.8), and so we obtain the following system of equations:
2κ∂ξ(wk)−∂21(wk)−n∑s,j=2asj∂s∂j(wk)+n+m∑s,j=n+1asj∂s∂j(wk)+ξwk=lk, | (1.10) |
where k=1,2;
l1=F1−n+m∑j=0b1j, l2=F2−n+m∑j=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=√2x1−μ0, 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=l′k, |
where l′k=(x1+μ0)lk, k=1,2, and
P0wk=(x1+μ0)−1∂21(wk)+(x1+μ0)(n∑s,j=2asj∂s∂j(wk)−n+m∑s,j=n+1asj∂s∂j(wk)−2κ∂ξ(wk)−ξwk). | (2.1) |
Next, we define a subdomain of Dn+m
Dn+mγ={x:x∈Rn+m, x1>0, α0<ψ(x)<γ+α0<1}, |
where 0<γ<1, 2μ0≤min{α0,γ}, and a Carleman weight function
φ=eλψ−ν, |
where α0>0, the parameters δ, λ, v are positive numbers, and
ψ(x)=δx1+12n+m∑j=2(xj−x0j)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)≥4λ3ν4ψ−2ν−3φ2w2+2λνφ2(∂1w)2+λν(x1+μ0)φ2n+m∑j=2(∂jw)2+4∑j=1dj | (2.3) |
holds for all δ>δ∗, ν>ν∗, λ>λ∗ and w∈C2(¯Dn+mγ). Here, the constant δ∗ depends on α1,α2,M,ψ,γ, and
S=n∑l,m=2∂m(alm∂lψ)−n+m∑l,m=n+1∂m(alm∂lψ). |
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)2φ2≥2λνγ−3δ(∂1w)2φ2+λ3ν4δ4ψ−2ν−3w2φ2+λν(x1+μ0)δ(α1n∑j=2(∂jw)2+α2n+m∑j=n+1(∂jw)2)φ2+2λνξ(x1+μ0)2(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ))w2φ2+3∑j=1dj | (2.4) |
for all δ>δ0, ν>ν0, λ>λ0 and w∈C2(¯Dn+mγ). Here, the constant δ0 depends on α1,α2,M,ψ,γ, and dj, 1≤j≤3, 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)−1φ−1(∂21ϑ+2λνψ−ν−1∂1ψ∂1ϑ+(λ2ν2ψ−2ν−2(∂1ψ)2−λν(ν+1)ψ−ν−2(∂1ψ)2+λνψ−ν−1∂21ψ)ϑ)+(x1+μ0)(n∑s,j=2asjφ−1(∂s∂jϑ+2λνψ−ν−1∂sψ∂jϑ+(λ2ν2ψ−2ν−2∂sψ∂jψ−λν(ν+1)ψ−ν−2∂sψ∂jψ+λνψ−ν−1∂s∂jψ)ϑ)−n+m∑s,j=n+1asjφ−1(∂s∂jϑ+2λνψ−ν−1∂sψ∂jϑ+(λ2ν2ψ−2ν−2∂sψ∂jψ−λν(ν+1)ψ−ν−2∂sψ∂jψ+λνψ−ν−1∂s∂jψ)ϑ)−2κ∂ξϑφ−1−ξϑφ−1))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)−1∂21ϑ+(x1+μ0)(n∑s,j=2asj∂s∂jϑ−n+m∑s,j=n+1asj∂s∂jϑ)),y3=−2λνψ−ν−1(∂1ψ(x1+μ0)−1∂1ϑ+(x1+μ0)(n∑s,j=2asj∂sψ∂jϑ−n+m∑s,j=n+1asj∂sψ∂jϑ)),y4=−Kϑ, y5=ξ(x1+μ0)ϑ, |
and
K=(x1+μ0)−1(λ2ν2ψ−2ν−2(∂1ψ)2−λν(ν+1)ψ−ν−2(∂1ψ)2)+(x1+μ0)n∑s,j=2asj(λ2ν2ψ−2ν−2∂sψ∂jψ−λν(ν+1)ψ−ν−2∂sψ∂jψ+λνψ−ν−1δsj)−(x1+μ0)n+m∑s,j=n+1asj(λ2ν2ψ−2ν−2∂sψ∂jψ−λν(ν+1)ψ−ν−2∂sψ∂jψ+λνψ−ν−1δsj). |
Next, we calculate the terms in (2.5). Here, we note that ∂1ψ=δ, ∂21ψ=0, ∂s∂jψ=δsj, where
δsj={1,s=j0,s≠j, 2≤s, j≤n+m. |
First, we can write
2ψν+1y1y2=4κ(ν+1)ψν∂1ψ∂ξϑ∂1ϑ+4κ(ν+1)(x1+μ0)2ψνn∑s,j=2∂jψasj∂ξϑ∂sϑ+4κ(x1+μ0)2ψν+1n∑s,j=2∂j(asj)∂ξϑ∂sϑ−4κ(ν+1)(x1+μ0)2ψνn+m∑s,j=n+1∂jψasj∂ξϑ∂sϑ−4κ(x1+μ0)2ψν+1n+m∑s,j=n+1∂j(asj)∂ξϑ∂sϑ+d1(ϑ)=2ψν+1y1y6+d1(ϑ), |
where
y6=(ν+1)ψ−1((x1+μ0)−1∂1ψ∂1ϑ+(x1+μ0)(n∑s,j=2asj∂sψ∂jϑ−n+m∑s,j=n+1asj∂sψ∂jϑ))+(x1+μ0)(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ), |
d1=d1(ϑ)=−4κ∂1(ψν+1∂ξϑ∂1ϑ)+2κ∂ξ(ψν+1∂21ϑ)−2κ(x1+μ0)2(n∑s,j=2(2∂j(ψν+1asj∂sϑ∂ξϑ)−∂ξ(ψν+1asj∂sϑ∂jϑ))−n+m∑s,j=n+1(2(ψν+1asj∂sϑ∂ξϑ)−∂ξ(ψν+1asj∂sϑ∂jϑ))). |
On the other hand, by the inequality 2pq≥−p2−q2, we estimate the first, third, fourth and fifth terms in (2.5), so we have
ψν+1(y21+y23+2y1y3+2y1y2)≥ψν+1(−2y3y6−y26)+d1(ϑ). |
Then, we can write
ψν+1(y21+y23+2y1y3+2y1y2)≥ψν+1((4λν(ν+1)ψ−ν−2−(ν+1)2ψ−2−2λντ0ψ−ν−1+(ν+1)τ0ψ−1)×((x1+μ0)−1∂1ψ∂1ϑ+(x1+μ0)(n∑s,j=2asj∂sψ∂jϑ−n+m∑s,j=n+1asj∂sψ∂jϑ))2−1τ0(2λνψ−ν−1−(ν+1)ψ−1)(x1+μ0)2(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ)2−(x1+μ0)2(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ)2)+d1(ϑ)≥−2λνψ(x1+μ0)2(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ)2+d1(ϑ). | (2.6) |
In (2.6), we used the inequality 2pq≥−τ0p2−q2τ0, for τ0=(ν+1)ψ−1>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=−2λν∂1(ξ∂1ψϑ2)−2λν(x1+μ0)2n∑s,j=2∂j(asj∂sψξϑ2)+2λν(x1+μ0)2ξϑ2n∑s,j=2∂j(asj∂sψ)+2λν(x1+μ0)2n+m∑s,j=n+1∂j(asj∂sψξϑ2)−2λν(x1+μ0)2ξϑ2n+m∑s,j=n+1∂j(asj∂sψ), | (2.8) |
and
2y2y3ψν+1=2λν∂1ψ∂1((x1+μ0)−2(∂1ϑ)2)+4λν∂1ψ(x1+μ0)−3(∂1ϑ)2+2λν∂1ψn∑s,j=2(2∂j(asj∂sϑ∂1ϑ)−∂1(asj∂sϑ∂jv)−2∂s(asj)∂jϑ∂1ϑ+∂1(asj)∂sϑ∂jϑ)−2λν∂1ψn+m∑s,j=n+1(2∂j(asj∂sϑ∂1ϑ)−∂1(asj∂sϑ∂jϑ)−2∂s(asj)∂jϑ∂1ϑ+∂1(asj)∂sϑ∂jϑ)+2λνn∑s,j=2(2∂1(asj∂sψ∂sϑ∂1ϑ)−2∂1(asj)∂sψ∂jϑ∂1ϑ−∂j(asj∂sψ(∂1ϑ)2)+∂j(asj)∂sψ(∂1ϑ)2)−2λνn+m∑s,j=n+1(2∂1(asj∂sψ∂sϑ∂1ϑ)−2∂1(asj)∂sψ∂jϑ∂1ϑ−∂j(asj∂sψ(∂1ϑ)2)+∂j(asj)∂sψ(∂1ϑ)2)+2λν(x1+μ0)2(n+m∑s,j,k,l=2(2∂j(asjakl∂lϑ∂sϑ∂kψ)−∂l(asjakl∂jϑ∂sϑ∂kψ)−2∂j(asjakl∂kψ)∂lϑ∂sϑ+∂l(asjakl∂kψ)∂jϑ∂sϑ))−4λν(x1+μ0)2(n∑s,j=2n+m∑s,j=n+1(2∂j(asjakl∂lϑ∂sϑ∂kψ)−∂l(asjakl∂jϑ∂sϑ∂kψ)−2∂j(asjakl∂kψ)∂lϑ∂sϑ+∂l(asjakl∂kψ)∂jϑ∂sϑ)). | (2.9) |
The last term has the following form:
2y3y4ψν+1=2λν∂1ψ∂1((x1+μ0)−1Kϑ2)−2λν∂1ψ∂1((x1+μ0)−1K)ϑ2+2λν(x1+μ0)(n∑s,j=2(∂j(asj∂sψKϑ2)−∂j(asj∂sψK)ϑ2)−n+m∑s,j=n+1(∂j(asj∂sψKϑ2)−∂j(asj∂sψK)ϑ2)). | (2.10) |
Then, from (2.6)-(2.10), we can write
ψν+1(P0w)2φ2≥−2λνψ(x1+μ0)2(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ)2+2λν(∂1ϑ)2n∑s=2ass+2λν(x1+μ0)2ξϑ2(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ))−2λν(∂1ϑ)2n+m∑s=n+1ass+4λν∂1ψ(x1+μ0)−3(∂1ϑ)2−2κ(x1+μ0)2ψν+1ϑ2−2λν(n∑s,j=2(2∂sψ∂1(asj)∂jϑ∂1ϑ−∂sψ∂j(asj)(∂1ϑ)2+2∂1ψ∂s(asj)∂jϑ∂1ϑ−∂1ψ∂1(asj)∂jϑ∂sϑ)−n+m∑s,j=n+1(2∂sψ∂1(asj)∂jϑ∂1ϑ−∂sψ∂j(asj)(∂1ϑ)2+2∂1ψ∂s(asj)∂jϑ∂1ϑ−∂1ψ∂1(asj)∂jϑ∂sϑ))+2λν(x1+μ0)2(n∑s,j,k,l=2(−2∂j(asjakl∂kψ)∂lϑ∂sϑ+∂l(asjakl∂kψ)∂jϑ∂sϑ)−2n∑s,j=2n+m∑s,j=n+1(−2∂j(asjakl∂kψ)∂lϑ∂jϑ+∂l(asjakl∂kψ)∂jϑ∂sϑ)+n+m∑s,j=n+1(−2∂j(asjakl∂kψ)∂lϑ∂jϑ+∂l(asjakl∂kψ)∂jϑ∂sϑ))−2λν(∂1ψ∂1((x1+μ0)−1K)+(x1+μ0)n∑s,j=2(∂j(asj∂sψK)−n+m∑s,j=n+1∂j(asj∂sψξ))ϑ2+d1(ϑ)+d2(ϑ), | (2.11) |
where
d2=d2(ϑ)=−∂ξ(2κ(x1+μ0)ψν+1Kϑ2)+∂ξ(2κ(x1+μ0)2ξψν+1ϑ2)−2λν∂1(ξ∂1ψϑ2)−2λν(x1+μ0)2(n∑s,j=2∂j(asj∂sψξϑ2)−n+m∑s,j=n+1∂j(asj∂sψξϑ2))+2λν∂1ψ(∂1((x1+μ0)−2(∂1ϑ)2)+∂1((x1+μ0)−1Kϑ2))+2λν(n∑s,j=2(2∂1(asj∂sψ∂jϑ∂1ϑ)−∂j(asj∂sψ(∂1ϑ)2)+2∂j(∂1ψasj∂1ϑ∂sϑ)−∂1(∂1ψasj∂jϑ∂sϑ))−n+m∑s,j=n+1(2∂1(asj∂sψ∂jϑ∂1ϑ)−∂j(asj∂sψ(∂1ϑ)2)+2∂j(∂1ψasj∂1ϑ∂sϑ)−∂1(∂1ψasj∂jϑ∂sϑ)))+2λν(x1+μ0)2(n+m∑s,j,k,l=2(2∂j(asjakl∂lϑ∂sϑ)−∂l(asjakl∂sψ∂jϑ∂sϑ))−2n∑s,j=2n+m∑k,l=n+1(2∂j(asjakl∂lϑ∂sϑ)−∂l(asjakl∂sψ∂jϑ∂sϑ)))+2λν(x1+μ0)(n∑s,j=2∂j(asj∂sψKϑ2)−n+m∑s,j=n+1∂j(asj∂sψKϑ2)). |
Here, we note that ||asj||C1(¯Dn+mγ)≤M, and ∂jψ=(xj−x0j),|∂jψ|=|xj−x0j|≤√2γ,(2≤j≤n+m) in Dn+mγ. Since −2pq≥−1ε0(x1+μ0)p2−ε0(x1+μ0)q2, for all ε0>0 and p,q, we see that
4λνδn+m∑s,j=n+1∂s(asj)∂1ϑ∂jϑ≥−2λνδ((x1+μ0)mε0n+m∑j=n+1(∂jϑ)2+m2M2ε0(x1+μ0)(∂1ϑ)2), | (2.12) |
4λνn+m∑s,j=n+1∂1(asj)∂sψ∂1ϑ∂jϑ≥−2λν(x1+μ0)√2γmn+m∑j=n+1(∂jϑ)2−2λνm2M2√2γ(x1+μ0)(∂1ϑ)2, | (2.13) |
−2λνn+m∑s,j=n+1∂j(asj)∂sψ(∂1ϑ)2≥−2λνM√2γm2(∂1ϑ)2, | (2.14) |
−(n∑s,j=2∂j(asj)∂sϑ−n+m∑s,j=n+1∂j(asj)∂sϑ)2≥−M2(n−1)((n−1)2+m)n∑j=2(∂jϑ)2−M2m((n−1)+m2)n+m∑j=n+1(∂jϑ)2, | (2.15) |
2λν(x1+μ0)2n+m∑s,j,k,l=2(∂l(asjakl∂kψ)∂jϑ∂sϑ−2∂j(asjakl∂kψ)∂lϑ∂sϑ)≥−λν(x1+μ0)26M2(2√2γ+1)(n+m−1)3n+m∑j=2(∂jϑ)2, | (2.16) |
2λν(x1+μ0)2n∑s,j=2n+m∑k,l=n+1(∂l(asjakl∂kψ)∂jϑ∂sϑ−2∂j(asjakl∂kψ)∂lϑ∂sϑ)≥−λν(x1+μ0)26M2(2√2γ+1)((n−1)n∑j=2(∂jϑ)2+mn+m∑j=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)2φ2≥E1(∂1ϑ)2+E2n∑j=2(∂jϑ)2+E3n+m∑j=n+1(∂jϑ)2+E4ϑ2+d1+d2, | (2.18) |
where
E1=4λνδ(x1+μ0)−3−2λνδM2(n−1)2ε0(x1+μ0)−2λνM2(n−1)2√2γ(x1+μ0)−2λν√2γ(n−1)2−2λνδM2m2ε0(x1+μ0)−2λνM2m2√2γ(x1+μ0)−2λνM√2γm2−2λνM(n−1)−2λνMm, |
E2=2λν(x1+μ0)(δα1−δε0(n−1)−√2γ(n−1)−3M2(x1+μ0)(2√2γ+1)(n−1)3−3M2(x1+μ0)(2√2γ+1)(n−1)+M2(x1+μ0)(n−1)(m+(n−1)2)), |
E3=2λν(x1+μ0)(δα2−δε0m−√2γm−3M2(x1+μ0)(2√2γ+1)m3−3M2(x1+μ0)(2√2γ+1)m+M2(x1+μ0)m(m2+(n−1))), |
E4=−2λν(δ((x1+μ0)−1K)x1+(x1+μ0)n∑s,j=2(∂j(asj∂sψK)−n+m∑s,j=n+1∂j(asj∂sψK)))−2(x1+μ0)2(kψν+1−λνξ(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ))). |
First, we shall estimate the coefficient E1:
E1=2λν(x1+μ0)−3(δ+E11−E12), |
where
E11=δ(1−(x1+μ0)2M2ε0((n−1)2+m2)), |
E12=(x1+μ0)2√2γM2((n−1)2+m2)+(x1+μ0)3√2γM((n−1)2+m2)+(x1+μ0)3M(n+m−1). |
If we take γ as 0<γ<2√2ε03M√(n−1)2+m2, then 1−((n−1)2+m2)M2ε0(34γ)2>12, and so we obtain
E11>δ(1−((n−1)2+m2)M2ε0(34γ)2)>12δ |
in Dn+mγ. Setting
E′12=√2γM((n−1)2+m2)(M+1)+M(n+m−1) |
for δ≥δ1=2E′12, and then 12δ−E′12≥0, which implies that
E1>2λν(x1+μ0)−3(δ+12δ−E′12)>2λνγ−3δ. | (2.19) |
As for the coefficient E2, if we take ε0 such that 0<ε0<α14(n−1), then we have
δ(α1−ε0(n−1))>34δα1. | (2.20) |
Here, we note that
δ2=4α1(√2(n−1)+3M2(2√2+1)(n−1)3+6M2(2√2+1)(n−1)+M2(n−1)(m+(n−1)2)). |
Then, for δ≥δ2, from (2.20) it follows that
E2=2λν(x1+μ0)(δα1−δε0(n−1)−√2γ(n−1)−3M2(x1+μ0)(2√2γ+1)(n−1)3−3M2(x1+μ0)(2√2γ+1)(n−1)−M2(x1+μ0)(n−1)(m+(n−1)2))>λνδα1(x1+μ0). | (2.21) |
Next, we estimate the coefficient E3. We choose ε0 such that 0<ε0<α24m,
δ(α2−ε0m)>34δα2. | (2.22) |
Since μ0<12γ, δ≥4 and 0<δx1<γ in Dn+mγ, we get (x1+μ0)<34γ. Here, we set
δ3=4α2(√2m+3M2(2√2+1)m3−3M2(2√2+1)m+M2m(m2+(n−1))). |
Then, if we take δ≥δ3, from (2.22) it follows that
E3>2λν(x1+μ0)(34δα2−14δα2)>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(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ)), | (2.24) |
where
E41=−2λ3ν3δ3∂1((x1+μ0)−2ψ−2ν−2)−2λ3ν3δ(n∑s,j=2∂1(asjψ−2ν−2)∂sψ∂jψ−n+m∑s,j=n+1∂1(asjψ−2ν−2)∂sψ∂jψ)−2λ3ν3δ2(∂j(ψ−2ν−2n∑s,j=2asj∂sψ)−∂j(ψ−2ν−2n+m∑s,j=n+1asj∂sψ))+4λ3ν3(x1+μ0)n∑s,j=2n+m∑k,l=n+1∂j(asjakl∂sψψ−2ν−2)∂kψ∂lψ−2λ3ν3(x1+μ0)2n+m∑s,j,k,l=2∂j(asjakl∂sψ∂kψ∂lψψ−2ν−2), |
E42=2λ2ν2(δ∂1((x1+μ0)−1E01)+(x1+μ0)(n∑s,j=2∂j(asj∂sψE01)−n+m∑s,j=n+1∂j(asj∂sψE01)))−2κ(x1+μ0)2ψν+1, |
E01=(x1+μ0)−1(ν+1)ψ−ν−2δ2+(x1+μ0)(n∑s,j=2asj((ν+1)ψ−ν−2∂sψ∂jψ−ψ−ν−1δsj)−n+m∑s,j=n+1asj((ν+1)ψ−ν−2∂sψ∂jψ−ψ−ν−1δsj)). |
We now estimate each of terms in E41, respectively:
−2λ3ν3δ3∂1((x1+μ0)−2ψ−2ν−2)=4λ3ν3δ4(ν+1)(x1+μ0)−2ψ−2ν−3+4λ3ν3δ3(x1+μ0)−3ψ−2ν−2, | (2.25) |
−2λ3ν3δn∑s,j=2∂1(asjψ−2ν−2)∂sψ∂jψ=4λ3ν3δ2(ν+1)ψ−2ν−3n∑s,j=2asj∂sψ∂jψ−2λ3ν3δψ−2ν−2n∑s,j=2∂1(asj)∂sψ∂jψ, | (2.26) |
2λ3ν3δ2∂j(ψ−2ν−2n+m∑s,j=n+1asj∂sψ)=−4λ3ν3δ2(ν+1)ψ−2ν−3n+m∑s,j=n+1asj∂sψ+2λ3ν3δ2ψ−2ν−2n+m∑s,j=n+1(∂j(asj)∂sψ+asjδsj), | (2.27) |
−2λ3ν3(x1+μ0)2n∑s,j,k,l=2∂j(asjakl∂sψ∂kψ∂lψψ−2ν−2)=4λ3ν3(x1+μ0)2(ν+1)ψ−2ν−3n∑s,j,k,l=2asjakl∂sψ∂kψ∂lψ∂jψ−2λ3ν3(x1+μ0)2ψ−2ν−2n∑s,j,k,l=2∂j(asjakl∂sψ∂kψ∂lψ), | (2.28) |
4λ3ν3(x1+μ0)2n∑s,j=2n+m∑k,l=n+1∂j(asjakl∂sψψ−2ν−2)∂kψ∂lψ=−8λ3ν3(x1+μ0)2(ν+1)ψ−2ν−3n∑s,j=2n+m∑k,l=n+1asjakl∂sψ∂jψ∂kψ∂lψ+4λ3ν3(x1+μ0)2ψ−2ν−2n∑s,j=2n+m∑k,l=n+1(∂j(asj)akl∂sψ+asj∂j(akl)∂sψ+asjaklδsj)∂kψ∂lψ. | (2.29) |
Then, from these relations we can write
E41=4λ3ν3(ν+1)ψ−2ν−3(δ4(x1+μ0)−2+2δ2n∑s,j=2asj∂sψ∂jψ−2δ2n+m∑s,j=n+1asj∂sψ∂jψ+(x1+μ0)2n+m∑s,j,k,l=2asjakl∂sψ∂kψ∂lψ∂jψ−2(x1+μ0)2n∑s,j=2n+m∑k,l=n+1asjakl∂sψ∂jψ∂kψ∂lψ)+4λ3ν3ψ−2ν−2(δ3(x1+μ0)−3−δ2n∑s,j=2∂1(asj)∂sψ∂jψ+δ2n+m∑s,j=n+1∂1(asj)∂sψ∂jψ−δ22n∑s,j=2(∂j(asj)∂sψ+asjδsj)+δ22n+m∑s,j=n+1(∂jasj∂sψ+asjδsj)−(x1+μ0)22n+m∑s,j,k,l=2∂j(asjakl∂sψ∂kψ∂lψ)+(x1+μ0)2n∑s,j=2n+m∑k,l=n+1(∂j(asj)akl∂sψ+asj∂j(akl)∂sψ+asjaklδsj)∂kψ∂lψ). |
Remembering that (x1+μ0)<34γ in Dn+mγ, we have δ4(x1+μ0)−2>δ4. We set
δ4=12((4√2+2)M2(n+m−1)2+(2√2+3)M(n+m−1)+4M2((n−1)2+m2)), |
then, we see that
E4>2λ3ν3δ4(ν+1)ψ−2ν−3+2λνξ(x1+μ0)2(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ)) |
for δ≥δ4 and λ>λ0, which yields
ψν+1(P0w)2φ2≥2λνδγ−3(∂1ϑ)2+λν(x1+μ0)δα1n∑j=2(∂jϑ)2+λν(x1+μ0)δα2n+m∑j=n+1(∂jϑ)2+2λ3ν4δ4ψ−2ν−3ϑ2+2λνξ(x1+μ0)2(n∑s,j=2∂j(asj∂sψ)−n+m∑s,j=n+1∂j(asj∂sψ))ϑ2+d1(ϑ)+d2(ϑ). | (2.30) |
Finally, if we write ϑ=φw in (2.30), then we obtain
2λνδγ−3(∂1ϑ)2+λνδα1(x1+μ0)n∑j=2(∂jϑ)2+λνδα2(x1+μ0)n+m∑j=n+1(∂jϑ)2+2λ3ν4δ4ψ−2ν−3ϑ2=2λνγ−3δ(∂1w)2φ2+λν(x1+μ0)δα1φ2n∑j=2(∂jw)2+λν(x1+μ0)δα2φ2n+m∑j=n+1(∂jw)2+2λ3ν4δ4ψ−2ν−3w2φ2−2λ2ν2γ−3δ∂1(δψ−ν−1w2φ2)−λ2ν2(x1+μ0)δα1n∑j=2∂j(ψ−ν−1∂jψw2φ2)−λ2ν2(x1+μ0)δα2n+m∑j=n+1∂j(ψ−ν−1∂jψw2φ2)−2λ3ν3γ−3δ3ψ−2ν−2w2φ2−λ3ν3(x1+μ0)δα1ψ−2ν−2w2φ2n∑j=2(∂jw)2−λ3ν3(x1+μ0)δα2ψ−2ν−2w2φ2n+m∑j=n+1(∂jw)2−λ2ν2(2γ−3δ3(ν+1)ψ−ν−2+(x1+μ0)δα1(ν+1)ψ−ν−2n∑j=2(∂jw)2+(x1+μ0)δα2(ν+1)ψ−ν−2n+m∑j=n+1(∂jw)2−(x1+μ0)δα1ψ−ν−1(n−1)−(x1+μ0)δα2ψ−ν−1m)w2φ2. | (2.31) |
Choosing ν≥ν0=2γ−3δ−1+32γ2δ−3(α1+α2)+1, we have
w2φ2ψ−2ν−2(λ3ν4δ4ψ−1φ2−2λ3ν3γ−3δ3−λ3ν3(x1+μ0)δ(α1n∑j=2(∂jw)2+α2n+m∑j=n+1(∂jw)2))≥λ3ν3δ4w2φ2ψ−2ν−2(ν−2γ−3δ−1−32γ2δ−3(α1+α2))≥λ3ν3δ4w2φ2ψ−2ν−2. | (2.32) |
On the other hand, if λ≥λ0=2γ−3(ν+1)+32γ2(ν+1)(α1+α2), then we get
λ3ν3δ4ψ−2ν−2w2φ2−2λ2ν2γ−3δ3(ν+1)ψ−ν−2w2φ2−λ2ν2(x1+μ0)δ(α1+α2)(ν+1)ψ−ν−2w2φ2n+m∑j=2(∂jw)2≥λ2ν3δ4ψ−2ν−2w2φ2(λ−2γ−3(ν+1)−32γ2(ν+1)(α1+α2))≥0. | (2.33) |
Therefore, for δ>δ0, ν≥ν0 and λ≥λ0, from (2.30)-(2.33) we have
2λνγ−3δ(∂1ϑ)2+λν(x1+μ0)δα1n∑j=2(∂jϑ)2+λν(x1+μ0)δα2n+m∑j=n+1(∂jϑ)2+2λ3ν4δ4ψ−2ν−3ϑ2≥2λνγ−3δ(∂1w)2φ2+λν(x1+μ0)δα1φ2n∑j=2(∂jw)2+λν(x1+μ0)δα2φ2n+m∑j=n+1(∂jw)2+λ3ν4δ4ψ−2ν−3w2φ2+d3(w), | (2.34) |
where δ0=max{4, δ1, δ2, δ3, δ4}, and
d3=d3(w)=−2λ2ν2γ−3δ∂1(ψ−ν−1δw2φ2)−λ2ν2(x1+μ0)δα1n∑j=2∂j(ψ−ν−1∂jψw2φ2)−λ2ν2(x1+μ0)δα2n+m∑j=n+1∂j(ψ−ν−1∂jψw2φ2). |
From relations (2.30) and (2.34), the proof of Lemma 1 is completed.
Lemma 2. If λ exceeds some constant, then for all w∈C2(Dn+mγ), we have
2λν(x1+μ0)φ2Sw(P0w)≥−4S0λ3ν3ψ−2ν−2(δ2+2γM(n−1)2+m2)w2φ2−2λνS0φ2((∂1w)2+(x1+μ0)2M((n−1)n∑j=2(∂jw)2+mn+m∑j=n+1(∂jw)2))−Cλ2ν2ψ−ν−1δ2φ2w2−2λνSξ(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)+λν∂1(2φ2Sw∂1w−∂1(φ2S)w2)−λν(x1+μ0)2n∑s,j=2(∂j(∂s(φ2Sasj)w2)−2∂s(φ2Sasjw∂jw))+λν(x1+μ0)2n+m∑s,j=n+1(∂j(∂s(φ2Sasj)w2)−2∂s(φ2Sasjw∂jw)),S=n∑l,m=2∂m(alm∂lψ)−n+m∑l,m=n+1∂m(alm∂lψ),S0=(n−1)M((n−1)√2γ+2)+mM(m√2γ+2). |
Proof. By using the equality
2λν(x1+δ0)φ2Sw(P0w)=2λνφ2Sw∂21w+2λν(x1+δ0)2φ2Sw(n∑s,j=2asj∂s∂jw−n+m∑s,j=n+1asj∂s∂jw)−4λν(x1+δ0)2φ2Swκ∂ξw−2λν(x1+δ0)2φ2Sw2ξ |
and taking into account ∂sp∂jq=∂j((∂sp)q)−∂j(∂sp)q, we have
2λν(x1+μ0)φ2Sw(P0w)=−2λνφ2S((∂1w)2+(x1+μ0)2(n∑s,j=2asj∂sw∂jw−n+m∑s,j=n+1asj∂sw∂jw))+S1w2−2λν(x1+μ0)2φ2Sξw2+d4(w), | (2.36) |
where
S1=λν(∂21(φ2S)+(x1+μ0)2(∂s∂j(φ2Sn∑s,j=2asj)−∂s∂j(φ2Sn+m∑s,j=n+1asj))). |
Remembering that |S|≤S0 and ||asj||C1(Dn+mγ)≤M (2≤s,j≤n+m), we see that the first term in (2.36) can be written as
−2λνφ2S((∂1w)2+(x1+μ0)2(n∑s,j=2asj∂sw∂jw−n+m∑s,j=n+1asj∂sw∂jw))≥−2λνS0φ2((∂1w)2+(x1+μ0)2M((n−1)n∑j=2(∂jw)2−mn+m∑j=n+1(∂jw)2)). | (2.37) |
Next, as for the coefficient in the second term, we write
S1=λν(∂21(φ2)S+2∂1(φ2)∂1S+φ2∂21S+(x1+μ0)2(n∑s,j=2(S∂s(∂j(φ2)asj)+∂s(φ2)(2∂jSasj+S∂∂j(asj))+φ2(∂s∂jSasj+2∂jS∂s(asj)+S∂s∂j(asj)))−n+m∑s,j=n+1(S∂s(∂j(φ2)asj)+∂s(φ2)(2∂jSasj+S∂∂j(asj))+φ2(∂s∂jSasj+2∂jS∂s(asj)+S∂s∂j(asj)))). | (2.38) |
Since
∂21(φ2)=2λνφ2(2λνδ2ψ−2ν−2+(ν+1)δ2ψ−ν−2),∂j(asj∂s(φ2))=2λνφ2(asj(2λνψ−2ν−2∂sψ∂jψ+(ν+1)ψ−ν−2∂sψ∂jψ−ψ−ν−1∂s∂jψ)−∂s(asj)ψ−ν−1∂jψ), | (2.39) |
|S|≤S0, |∂jS|≤S0,|∂s∂jS|≤(n−1)M((n−1)√2γ+3)+mM(m√2γ+3), | (2.40) |
we see that
|S1|≤4λ3ν3φ2ψ−2ν−2S0(δ2+(x1+δ0)2(n+m∑s,j=2|asj∂sψ∂jψ|)+λ2ν2ψ−ν−1δ2φ2S0(2(ν+1)ψ−1+4δ+(x1+δ0)2δ2(n+m∑s,j=2|2(ν+1)φ2asjψ−1∂sψ∂jψ|+|2asj∂s∂jψ|+|2∂s(asj)∂jψ|+|4∂sψasj|+|2∂sψ∂s(asj)|)))+λνφ2(n+m∑s,j=2|2∂s(asj)|S0+|∂s∂j(asj)|S0)+|∂21S|+(x1+δ0)2(n+m∑s,j=2|asj||∂s∂jS|)). |
If we choose that
λ≥λ1=n+m∑s,j=2(|2∂s(asj)|+|∂s∂j(asj)|)S0+|∂21S|+(x1+δ0)2(n+m∑s,j=2|asj||∂s∂jS|), |
and v≥1, then we have
|S1|≤4λ3ν3φ2ψ−2ν−2S0(δ2+(x1+δ0)2(n+m∑s,j=2|asj∂sψ∂jψ|))+Cλ2ν2ψ−ν−1δ2φ2, |
where
C=S0(2(ν+1)ψ−1+4δ+(x1+δ0)2δ2(n+m∑s,j=2(|2(ν+1)asjψ−1∂sψ∂jψ|+|2asj∂s∂jψ|+|2∂s(asj)∂jψ|+|4∂sψasj|+|2∂sψ∂s(asj)|)))+ψv+1δ2. |
Here, we note that |asj|≤M and |∂jψ|=√2γ,(2≤s,j≤n+m). We see that
S1≥−4λ3ν3φ2ψ−2ν−2S0(δ2+2γM(n−1)2+2γMm2)−Cλ2ν2δ2ψ−ν−1φ2. | (2.41) |
Then, by relations (2.36) and (2.41), we write
2λν(x1+μ0)φ2Sw(P0w)≥−4S0λ3ν3ψ−2ν−2(δ2+2γM(n−1)2+m2)w2φ2−2λνS0φ2((∂1w)2+(x1+μ0)2M((n−1)n∑j=2(∂jw)2+mn+m∑j=n+1(∂jw)2))−Cλ2ν2ψ−ν−1δ2φ2w2−2λνSξ(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)≥4∑j=1dj+2λν(∂1w)2φ2(δγ−3−S0)+λν(x1+μ0)(δα1−2S0M(n−1)(x1+μ0))φ2n∑s,j=2(∂jw)2+λν(x1+μ0)(δα2−2S0Mm(x1+μ0))φ2n+m∑s,j=n+1(∂jw)2+4λ3ν3ψ−2ν−3w2φ2(14νδ4−ψS0(δ2+2γM((n−1)2+m2))−Cδ2ψν+2λν). |
By choosing δ∗=max{δ0, δ5}, δ5=max{(1+S0)γ3, (1+2S0γM(n−1))/α1, (1+2S0γMm)/α2}, λ∗≥max{λ0, λ1, C}, ν∗≥max{ν0, 4δ4(1+ψS0(δ2+2Mγ((n−1)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 l′k=(x1+μ0)lk, we write
(l′1)2+(l′2)2≤4(F21+F22)(x1+μ0)2+4(n+m+1)n+m∑j=0(b21j+b22j). | (3.1) |
Using (3.1), we have
2∑k=1(ψν+1(P0wk)2φ2+2λν(x1+μ0)Swk(P0wk)φ2)≤2∑k=1(ψν+1(P0wk)2φ2+(P0wk)2φ2+λ2ν2(x1+μ0)2S2w2kφ2)=2∑k=1((ψν+1+1)(P0wk)2φ2+λ2ν2(x1+μ0)2S2w2kφ2)=(ψν+1+1)((l′1)2+(l′2)2)φ2+λ2ν2(x1+μ0)2S2(w21+w22)φ2≤(ψν+1+1)(x1+μ0)2(4(F21+F22)+4(n+m+1)n+m∑j=0(b21j+b22j))φ2+λ2ν2(x1+μ0)2S2(w21+w22)φ2. | (3.2) |
Next, we apply the Carleman estimate in (3.2):
2∑k=1(ψν+1(P0wk)2φ2+2λν(x1+μ0)Swk(P0wk))φ2≥2λνφ22∑k=1(∂1(wk))2+λν(x1+μ0)φ22∑k=1n+m∑j=2(∂j(wk))2+4λ3ν3ψ−2ν−3φ22∑k=1(wk)2+2∑k=14∑j=1dj(wk). | (3.3) |
By (3.2) and (3.3), we write
2∑k=1(2λνφ2(∂1(wk))2+λ(x1+μ0)φ2n+m∑j=2(∂j(wk))2+4λ3ν3ψ−2ν−3φ2(wk)2)+2∑k=14∑j=1dj(wk)≤(ψν+1+1)(x1+μ0)2(4(F21+F22)+4(n+m+1)n+m∑j=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
2∑k=1(2λν(∂1(wk))2+λ(x1+μ0)n+m∑j=2(∂j(wk))2+4λ3ν3ψ−2ν−3(wk)2)φ2(1+ξ2)2dξ+2∑k=14∑j=1∫∞−∞dj(wk)(1+ξ2)2dξ≤(ψν+1+1)(x1+μ0)2(4M1+8(n+m+1)M32∑k=1n+m∑j=0∫∞−∞(∂j(wk))2(1+ξ2)2dξ)φ2+λ2ν2φ2(x1+μ0)2S22∑k=1∫∞−∞w2kdξ. | (3.5) |
In (3.5), we used the following relation by the Plancherel's theorem:
n+m∑j=0∫∞−∞(1+ξ2)2|^bj∂jz|2dξ≤2n+m∑j=0∫∞−∞∑0≤β0≤2|∂β0t(bj∂jz)|2dt≤2M3n+m∑j=0∫∞−∞(1+ξ2)2|^∂jz|2dξ=2M32∑k=1n+m∑j=0∫∞−∞(1+ξ2)2(∂j(wk))2dξ, |
where M3 is a constant depending on M2. Then, the inequality (3.5) takes the form
E52∑k=1∫∞−∞(∂1(wk))2(1+ξ2)2φ2dξ+E62∑k=1n+m∑j=2∫∞−∞(∂j(wk))2(1+ξ2)2φ2dξ+E72∑k=1∫∞−∞(wk)2(1+ξ2)2φ2dξ≤−2∑k=14∑j=1∫∞−∞dj(wk)(1+ξ2)2dξ, | (3.6) |
where
E5=2λν−8(n+m+1)(ψν+1+1)(x1+μ0)2M3,E6=λν(x1+μ0)−8(n+m+1)(ψν+1+1)(x1+μ0)2M3,E7=4λ3ν3ψ−2ν−3−(λ2ν2S2−8(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=2λν−16(n+m+1)M3>λ | (3.7) |
for ν≥ν1=12+8(n+m+1)M3, and
E6=(x1+μ0)2(λν−16(n+m+1))M3>(x1+μ0)2λ | (3.8) |
for ν≥ν2=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
E7≥4λ3ν3ψ−2ν−3−λ2ν2(S0)2−16(n+m+1)M3−8M1≥2λ3ν3ψ−2ν−3. | (3.9) |
Hence, for λ>1 and v>max{v1,v2}, (3.6)-(3.9) yield
2∑k=1∫∞−∞(λ(∂1(wk))2+λ(x1+μ0)2n+m∑j=2(∂j(wk))2+2λ3ν3ψ−2ν−3(wk)2)(1+ξ2)2φ2dξ≤−2∑k=14∑j=1∫∞−∞dj(wk)(1+ξ2)2dξ. | (3.10) |
Since 2(1+ξ2)2>1 and φ2>1 on Dn+mγ, we write
2∑k=1∫∞−∞(wk)2dξ≤−1λ3ν32∑k=14∑j=1∫∞−∞dj(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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