Schrödinger-type identity for Schrödinger free boundary problems

R E S E A R C H

Open Access

Schrödinger-type identity for Schrödinger

free boundary problems

Xingjian Zhang

_{, Duo Liu}

_{, Zining Yan}

_{, Guodong Zhao}

_{and Ye Yuan}

*_{Correspondence:}

zgudo@foxmail.com

1_{Department of Computer Science} and Technology, Harbin

Engineering University, Harbin, China

Abstract

Our aim in this paper is to develop a Schrödinger-type identity for a Schrödinger free boundary problem inRn_{. As an application, we establish necessary and sufficient}

conditions for the product of some distributional functions to satisfy the

Schrödinger-type identity. As a consequence, our results significantly improve and generalize previous work.

Keywords: Schrödinger-type identity; Schrödinger transform; Distribution

1 Introduction and main results

Schrödinger-type identities have been studied extensively in the literature (see [1,12,13, 18] for the Schrödinger equation, [5,14] for Schrödinger systems).

In recent years, many exciting phenomena were found by careful experiments on light waves propagating in nonlinear periodic lattices. These phenomena are governed by the following Schrödinger equation:

Schα(u) = (–)αu+V(x)u–h(x,u) = 0 (1.1)

inRn_{, wheren≥2,α∈(0, 1), (–)}α_{stands for the fractional Laplacian,V is a positive} continuous potential,h∈C(R2_{×R,R). The fractional Laplacian (–)}α_{withα∈(0, 1) of} a functionι∈Sis defined by

G(–)α_{ι(ξ) =|ξ|}2α_{G(ι)(ξ), ∀α∈(0, 1),}

whereSdenotes the Schwartz space of rapidly decreasingC∞functions inRn_and

F(ι)(ξ) = 1 (2π)n/2

RNe

–2πiξ·x_ι(x)dx.

The Schrödinger transformSchαis defined as the following singular integral:

Schα(f)

(x) :=p.v.1

π

R f(y)

x–ydy=lim→0

|y–x|≥

f(y)

x–ydy,

wherex∈R.

The Schrödinger-type identity for Schrödinger free boundary problems

Schα(fg) =fSchα(g)

was first studied in [2–4,6]. It was proved that the above identity holds ifh,g∈L2_(R)

satisfy suppfˆ⊆R+ (R+= [0,∞)) andsuppgˆ ⊆R+ in [20]. In 2015, Wan also obtained more general sufficient conditions by weakening the above condition in [19]. Recently, Lv and Ulker and Huang established the first necessary and sufficient condition in the time domain and a parallel result in the frequency domain for the Poisson inequality in [10,14]. It is natural that there have been attempts to define the complex signal and prove the Schrödinger-type identity in the multidimensional case.

Definition 1.1 The partial Schrödinger transformSchα_joff is given by

(Sch_αjf)(x) :=p.v.1

π

R f(y)

xj–yj

dyj,

wheref ∈Lp(Rn) and 1≤p<∞.

The total Schrödinger transformSchαoff is defined as follows:

Schα(f)(x) :=p.v. 1

πn

Rn

f(y) n

j=1(xj–yj)

dy

= lim

maxj→0

|yj–xj|≥j>0,j=1,2,...,n

f(y) n

j=1(xj–yj)

dy,

wheref ∈Lp_(R

n) and 1≤p<∞. The property

Sch_α(f)_Lp_(Rn₎≤Cpn f Lp_(Rn₎

was proved in [8]. The iterative nature of it inLp(Rn) was shown in [16], wherep> 1. It was shown that

Schα= n

j=1

Schα_j.

The operationsSch_αiandSch_αjcommute with each other, wherei,j= 1, 2, . . . ,n. Now we define the Schrödingerean Fourier transformfˆoff (see [17]) by

ˆ

f(x) =

Rnf(t)e

–ix.t_dt,

wherex∈Rn_{andf ∈L}1_(Rn_). Set

D+=

x:x∈Rn,sgn(–x) = n

j=1

D–=

x:x∈Rn,sgn(–x) = n

j=1

sgn(–xj) = –1 ,

and

D0=

x:x∈Rn,sgn(–x) = n

j=1

sgn(–xj) = 0 .

We denote by DD+(Rn),DD–(Rn) andDD0(R

n_{) the set of functions inD(R}n_{) that are}

supported onD+,D–, andD0, respectively.

The Schwartz classS(Rn_{) consists of all functionsϕonR}n_{such that}

sup x∈Rn

xαDβϕ(x)<∞,

whereα,β∈Zn₊.

The Schrödingerean Fourier transformϕˆis a linear homeomorphism fromS(Rn_{) onto} itself. Meanwhile, the following identity holds:

(Schαϕ)∧(x) = (–i)sgn(x)ϕˆ, whereϕ∈S(Rn_).

The Schrödingerean Fourier transformF:S(Rn_)→S(Rn_{) is defined for anyϕ∈S(R}n₎ as follows:

ˆ,ϕ=,ϕˆ,

which is a linear isomorphism fromS(Rn) onto itself. For the detailed properties ofS(Rn) andS(Rn_{), see [3,7,15].}

For∈S(Rn_),λ∈S(Rn_{), it is easy to check that} ˜˘,λ=˜,λˆ=,λ˘˜=ˆ,λ=,λˆ

for anyλ∈S(Rn_{), where} ˜

λ(x) =λ(–x), ˜

is the inverse Schrödingerean Fourier transform defined as follows: ˘,λ=,λ˜.

Therefore in the distributional sense, we obtain

˜˘

=ˆ.

(1) λ∈C∞(Rn_);

(2) Dk_λ∈Lp_(Rn_{),k= 1, 2, . . ., whereC}∞_(Rn_{)consists of infinitely differentiable} functions,

Dkλ(x) = ∂

|k|

∂xk1 1 · · ·∂x

kn n

λ(x).

In the sequel, we denote byD_Lp(Rn) the dual of the corresponding spacesD_Lp(Rn), where

1

p+

1

p = 1.

As a consequence, we have

DRn_⊆SRn_⊆D

LpRn⊆LpRn

and

LpRn⊆D_Lp

Rn_⊆SRn_⊆DRn_.

Definition 1.2 Letf ∈D_Lp(Rn), where 1 <p<∞. Then the Schrödinger transform off is defined as follows:

Schα(f),λ

=f, (–1)nSchαλ

,

whereλ∈D_Lp(Rn).

In [10], Huang proved that the total Schrödinger transform is a linear homeomorphism fromDLp(Rn) onto itself, and that, ifh∈D

Lp(Rn) (1 <p<∞), thenPIh∈D_Lp(Rn) and the Schrödinger transform H defined above is a linear isomorphism fromD_Lp(Rn) onto itself.

Note that, if∈Lp_(Rn_{) (1 <p<∞), then we have}

(H)∧,λ=H,λˆ

= (–1)n,Hλˆ

= (–1)nˇ, (Hλˆ)∧ = (–1)nˇ, (–i)nsgn(·)λˆˆ

=ˇ, (i)nsgn(·)λ˜

=˜ˇ, (i)nsgn(·)λ

=(–i)nsgn(·)ˆ,λ, whereλ∈S(Rn_).

Therefore in the distributional sense

Define

t={tx:x∈},

wheretis a nonzero real number andis a nonempty subset ofR. Hence we have

supp

u

x t

=tsupp(u)

for any nonzero real numbert. For a subsetA⊆R, define

A=

t∈A

t.

2 Schrödinger-type identity forLp_(Rn_)functions

This part is motivated by the need of defining multidimensional complex signals. We de-fine the complex signal of f ∈Lp_(Rn_{) through the total Schrödinger transformSch}

α as

f +iSchα(f).

In this section we investigate the multidimensional Schrödinger-type identitySchα(fg) = fSchα(g) forf ∈S(Rn_{) andg∈L}p_(Rn_{), where 1 <p≤2. In particular, several necessary and} sufficient conditions are obtained.

Theorem 2.1 Suppose that f ∈S(Rn_);g∈Lp_(Rn_{) (1 <p≤2),then the Schrödinger}

trans-form of the function fg satisfies the Schrödinger-type identitySchα(fg) =fSchα(g)if and only

if

Rn

sgn(x) –sgn(t)fˆ(x–t)gˆ(t) dt= 0. (2.1)

Proof According to [10], we use the following equalities:

dx¯=ε2dx,

d¯=εd onSRn, d¯=ε2d onLp_Rn_.

So T

0

Rn

(fε∂tλ+fεgε· ∇ελ) dxdt= 0, T

0

Rn

fεgε·∂t+fεgε⊗gε:ωε() +fεdivε

dxdt

= T

0

Rn

Pωε(gε)ωε(gε) :ωε() –fε·

dxdt

+h(ε)

ε

T

0

S(Rn₎gε·

ddt+q

T

0

Lp_(Rn₎gε·

for any∈C∞₀ (0,T; [C∞(¯)]3_{), which leads to}

T

0

b(fε)∂tλ+b(fε)gε· ∇ελ+

b(fε) –fεb(fε)

divεgε

λdxdt= 0.

Note that (see [19])

ε

¯

ρε(t) |¯gε(t)|2

2 +ρ¯ε(t)ln

¯

ρε(t)

dx¯

+ t

0

ε

P| ¯Dg¯ε| _¯

Dg¯ε:D¯g¯εdx¯ds+h(ε) t

0

1,ε

|¯gε|2d¯ds

+q

t

0

2,ε

|¯gε|2d¯ds

= t 0 ε ¯

ρε¯fε· ¯gεdx¯ds+

ε _|

(ρ¯εg¯ε)0|2

2ρ¯0,ε

+ρ¯0,εln(ρ¯0,ε)

dx¯.

For anyt∈ 0,T, this yields

fε(t)|

gε(t)|2

2 +fε(t)ln

fε(t) dx + t 0

Pωε(gε)ωε(gε)

2

dxds

+h(ε)

ε

t

0

S(Rn₎|

gε|2dds+q t

0

Lp_(Rn₎|

gε|2dds

= t

0

fεgε·vεdxds+

_|

(fεgε)0|2

2f0,ε

+f0,εln(f0,ε)

dx,

√

gε∂α 2λ+ √

gε∂

∗

λα

2

λ

=

|L|=p–1

n

j,k=1

b

g ∂

2_ρ

∂zj∂zk

αjLαkLe–λdS

+

|K|=p

n

k=1

g∂αK

∂zk 2e–λ_dV

+

|L|=p–1

n

j,k=1

g ∂

2_λ

∂zj∂zk – ∂

2_g

∂zj∂zk

αjLαkLe–λdV

+ 2Re

|L|=p–1

n

j=1

αjL

∂g

∂zj

dzL,∂

∗ λα λ and 2Re

|L|=p–1

n

j=1

αjL

∂g

∂zj

dzL,∂

∗ λα λ ≤2

|L|=p–1

1 √

gε

e–λ/2

n

j=1

∂g

∂zj

αjLdzj,√gεe–λ/2∂

∗

λα

≤2

|L|=p–1

1 √

gε n

j=1

∂g

∂zj

αjLdzj λ

√

gε∂

∗

λαλ

≤

|L|=p–1

ς

1 √_g

ε n

j=1

∂g

∂zj

αjL

2

λ +1

ς

√

gε∂

∗

λα

2

λ

for anyt∈ 0,T, where

gε=

f1,ε,ε–1f2,ε,ε–1f3,ε

, vε= (u1,ε,εu2,ε,εu3,ε).

Since the Schrödingerean Fourier transform is injective fromSinto itself,fg,Schα(f)g, fSchα(g)∈Lp(Rn), we have

Schα(fg)

∧

=fSchα(g) ∧

,

which is equivalent to

(–1)nsgn(x)

Rn ˆ

f(x–t)gˆε(t)dt=

Rn(–1)

n_sgn(t)ˆ_f(x–t)gˆ ε(t)dt,

where

sgn(x) = n

j

= 1sgn(xj), x= (x1,x2, . . . ,xn).

So

Rn

sgn(x) –sgn(t)fˆ(x–t)gˆε(t)dt= 0.

Let aj and bj denote nonnegative real numbers in the rest of the paper, where j= 1, 2, . . . ,n.

Corollary 2.1 Let f ∈S(Rn_{)and g∈L}p_(Rn_{),where1 <p≤2.If}

suppfˆ⊆

n

j=1

[–aj,bj], suppgˆε⊆ n

j=1

R\(–bj,aj), (2.2)

then the Schrödinger-type identitySchα(fg) =fSchα(g)holds.

Proof We first prove

Rn

sgn(x) –sgn(t)fˆ(x–t)gˆε(t)dt= 0

from Theorem2.1. That is,

D+

sgn(x) –sgn(t)fˆ(x–t)gˆε(t)dt+

D–

Letx∈D+, ift∈D+, the integrand is vanish so (2.1) holds. Ift∈D0, (2.1) holds since the

integration is over a set of measure zero. As for the caset∈D–, assume that there exists t∈D–, such thatt∈suppfˆ(x–·)gˆε(·), thent∈suppgˆε∩D–,x–t∈suppfˆ.

SinceD–∩D+=∅, there existsj∈ {1, 2, . . . ,n}such thatxjtj≤0. We may assume that

xj> 0 andtj≤0. Thanks to (2.2), we havetj≤–bjandxj–tj≤bj, which is impossible. By repeating this argument forx∈D–andx∈D0 (see [6]), we find the same

conclu-sion.

Lemma 2.1 Suppose that f ∈Lp_(Rn_{)and g∈L}q_(Rn_),where

1

p+

1

q=

1

r ≤1 (1 <p,q≤2).

Then

(f ∗g)∧=fˆgˆε

holds.

Proof Letyi= 1, wherei= 1, 2, . . . ,n– 1. Then

f(t, 1, . . . , 1)≥kbf(t,k, . . . ,k),

wherek∈(0, 1). So

ft,k–1, . . . ,k–1≥kbf(t, 1, . . . , 1),

f(t,ky1, . . . ,kyn–1)≤k–bf(t,y1, . . . ,yn–1),

f(t,k, . . . ,k)≤k–bf(t, 1, . . . , 1),

wherek∈(0, 1), which yields

f₀n+–2w(t) > 0, f₀n+–3w(t) > 0, . . . , f₀1+w(t) > 0, w(t) > 0,

gt,f₀n+–2w(t),f₀n+–3w(t), . . . ,f₀1+w(t),w(t)

≤gt,f₀n+–2Ae(t),f₀n+–3Ae(t), . . . ,f₀1+Ae(t),Ae(t)

≤gt,f₀n+–2A,f₀n+–3A, . . . ,f₀1+A,A

=g

t, A

(n– 2)!t

n–2_, A

(n– 3)!t

n–3_{, . . . ,At,A}

≤g(t,A,A, . . . ,A,A) ≤Abg(t, 1, 1, . . . , 1, 1) ≤Abg(1, 1, 1, . . . , 1, 1),

and

ht,f₀n+–2w(t),f₀n+–3w(t), . . . ,f₀1+w(t),w(t)

≤h

t,f₀n+–2

1

Ae(t),f

n–3 0+

1

Ae(t), . . . ,f 1 0+

1

Ae(t),

1

Ae(t)

=h

t,(α–n+ 2)

A(α) t

α–1_,(α–n+ 2) A(α– 1) t

α–2_{, . . . ,}

(α–n+ 2)

A(α–n+ 3)t α–n+2_, 1

At

α–n+1

≤h

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–n+3_,ζ At

α–n+2

≤h

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–n+4_,ζ At

α–n+3

≤. . .

≤h

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–1_,ζ At α–1 ≤ ζ A –b

t–b(α–1)f(t, 1, 1, . . . , 1, 1)

≤

ζ

A

–b

t–b(α–1)_{f(0, 1, 1, . . . , 1, 1).}

So

gt,f₀n+–2w(t),f₀n+–3w(t), . . . ,f₀1+w(t),w(t)

≥g

t,f₀n+–2

1

Ae(t),f

n–3 0+

1

Ae(t), . . . ,f 1 0+

1

Ae(t),

1

Ae(t)

=g

t,(α–n+ 2)

A(α) t

α–1_,(α–n+ 2) A(α– 1) t

α–2_{, . . . ,}

(α–n+ 2)

A(α–n+ 3)t α–n+2_, 1

At

α–n+1

≥g

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–n+3_,ζ At

α–n+2

≥g

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–n+4_,ζ At

α–n+3

≥. . .

≥g

t,ζ

At

α–1_,ζ At

α–1_{, . . . ,}ζ At

α–1_,ζ At α–1 ≥ ζ A –b

tb(α–1)g(t, 1, 1, . . . , 1, 1)

≥

ζ

A

–b

tb(α–1)g(0, 1, 1, . . . , 1, 1)

and

ht,f₀n+–2w(t),f₀n+–3w(t), . . . ,f₀1+w(t),w(t)

≥ht,f₀n+–2Ae(t),f₀n+–3Ae(t), . . . ,f₀1+Ae(t),Ae(t)

≥ht,f₀n+–2A,f₀n+–3A, . . . ,f₀1+A,A

=h

t, A

(n– 2)!t

n–2_, A

(n– 3)!t

n–3_{, . . . ,At,A}

≥f(t,A,A, . . . ,A,A) ≥A–bf(t, 1, 1, . . . , 1, 1) ≥A–bf(1, 1, 1, . . . , 1, 1),

which yields 1

0

1

0

H(s,ς)gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dςds

≤ 1

0

ιq

sβ–1

(β) 1

0

gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς ds ≤ 1 0 ιq

sβ–1_Ab_{g(1, 1, . . . , 1)}

(β)

ds ≤ 1 0 ιq

sβ–1ds

and 1

0

1

0

H(s,ς)fς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dςds

≤ 1

0

ιq

sβ–1

(β) 1

0

fς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dς ds ≤ 1 0 ιq

sβ–1

(β) 1 0 ζ A –b

ς–b(α–1)f(0, 1, 1, . . . , 1)dς

ds

≤tα–n+1(ζ–bAbf(0, 1, 1, . . . , 1))q–1

(α–n+ 1)((β))q–1

1

0

ιq

sβ–1

1

0

ς–b(α–1)dς

ds.

It follows that 1

0

ιq 1

0

gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς ds ≥ 1 ξ ιq 1 ξ

gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς ds ≥ 1 ξ

γ(s)ιq 1

ξ

ρ(ς)gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς ds ≥ 1 ξ

γ(s)ιq 1

ξ

ρ(ς)

ζ

A

b

ςb(α–1)g(0, 1, 1, . . . , 1)dς

ds

=ζbA–bg(0, 1, 1, . . . , 1)q–1 1

ξ

γ(s)ιq 1

ξ

ρ(ς)ςb(α–1)dς

ds

≥tα–n+1ζbA–bg(0, 1, 1, . . . , 1)q–1 ×

1

ξ

γ(s)ιq 1

ξ

ρ(ς)ςb(α–1)dς

and 1

0

1

0

hς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dςds

≥ 1

ξ

ιq 1

ξ

fς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dς

ds

≥ 1

ξ

γ(s)ιq 1

ξ

ρ(ς)fς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dς

ds

≥ 1

ξ

γ(s)ιq 1

ξ

ρ(ς)H(ς,ς)A–bf(1, 1, 1, . . . , 1)dς

ds

≥A–bf(1, 1, 1, . . . , 1)q–1 1

ξ

γ(s)ιq 1

ξ

ρ(ς)H(ς,ς)dς

ds

≥tα–n+1A–bf(1, 1, 1, . . . , 1)q–1 1

ξ

γ(s)ιq 1

ξ

ρ(ς)H(ς,ς)dς

ds,

which yields

T(v,w)(t)≥ 1

At

α–n+1₌ 1 Ae(t),

wheret∈(0, 1).

Then we prove thatT:Qe×Qe→Qeis a mixed monotone operator. We have

1

0

ιq 1

0

gς,f₀n+–2v1(ς), . . . ,f01+v1(ς),v1(ς)

dς

ds

≤ 1

0

ιq 1

0

H(s,ς)gς,f₀n+–2v2(ς), . . . ,f01+v2(ς),v2(ς)

dς

ds.

ThusT(v,w)(t) is nondecreasing invfor anyw∈Qe. Letw1,w2∈Qeandw1≥w2. Then

1

0

P(s,ς)ιq 1

0

Q(s,ς)fς,f₀n+–2w1(ς), . . . ,f01+w1(ς),w1(ς)

dς

ds

≤ 1

0

P(s,ς)ιq 1

0

Q(s,ς)fς,f₀n+–2w2(ς), . . . ,f01+w2(ς),w2(ς)

dς

ds,

i.e.,

T(v,w1)(t)≤T(v,w2)(t), w∈Qe.

ThereforeT(v,w)(t) is nonincreasing inwfor anyv∈Qe. We shall show that the operatorThas a fixed point. It follows that

1

0

P(s,ς)ιq 1

0

Q(s,ς)gς,f₀n+–2tv(ς), . . . ,f₀1+tv(ς),tv(ς)

dς

ds

= 1

0

P(s,ς)ιq 1

0

Q(s,ς)gς,tf₀n+–2v(ς), . . . ,tf₀1+v(ς),tv(ς)

dς

≥ 1

0

P(s,ς)ιq 1

0

Q(s,ς)tbgς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς

ds

≥tb

1

0

P(s,ς)ιq 1

0

Q(s,ς)gς,f₀n+–2v(ς), . . . ,f₀1+v(ς),v(ς)

dς

ds

and 1

0

P(s,ς)ιq 1

0

Q(s,ς)fς,f₀n+–2t–1w(ς), . . . ,f₀1+t–1w(ς),t–1w(ς)

dς

ds

= 1

0

P(s,ς)ιq 1

0

Q(s,ς)fς,t–1f₀n+–2w(ς), . . . ,t–1f₀1+w(ς),t–1w(ς)

dς

ds

≥ 1

0

P(s,ς)ιq 1

0

Q(s,ς)tbfς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dς

ds

≥tb

1

0

P(s,ς)ιq 1

0

Q(s,ς)fς,f₀n+–2w(ς), . . . ,f₀1+w(ς),w(ς)

dς

ds,

we obtain

T

tx,1

ty

≥tbT(x,y), x,y∈Qe,t∈(0, 1),b∈(0, 1).

Therefore

(α–n+ 2)

A(α) t α–1₌ 1

Af

n–2

0+ e(t)≤u(t)

≤Mf₀n+–2e(t) =

A(α–n+ 2)

(α) t

α–1_{, t∈(0, 1).}

Sincef∈Lp_(Rn_{) andg∈L}q_(Rn_{) (see [11]), we have}

f ∗g∈LrRn,

which shows that there exist functionsgn∈S(Rn) such that

g–gn q→0

asn→ ∞,

(f ∗gn)∧=fˆgˆε_n,

and

(fgn)∧=fˆ∗ ˆgεn.

Thus in the distributional sense

lim n→∞(f ∗gn)

On the other hand

fˆ(gˆεn–gˆε,λ)=(gˆεn–gˆε,fˆλ)≤ ˆfλ q ˆgεn–gˆε q→0

asn→ ∞. Hence the result

(f ∗g)∧=fˆgˆε

is obtained.

We define

Sn=

σk:{1, 2, . . . ,n} → {+1, –1}

,

Qσk=

y= (y1,y2, . . . ,yn)∈Rn:yjσk(j) > 0

and

–Qσk=

ξ∈Rn: –ξ∈Qσk

, sgn(ξ) = n

j=1

sgn(ξj),

wherej= 1, 2, . . . ,n.

It follows that ifξ ∈Qσk andη∈–Qσk, thensgn(ξ) =sgn(η) when nis an even, and sgn(ξ) = –sgn(η) whennis an odd.

With these notations we have the following.

Theorem 2.2 Let n be an odd and f ∈S(Rn_{), g ∈L}p_(Rn_{) (1 <p≤2) satisfy suppf}ˆ_, suppgˆε⊆Qσk∪–Qσk with ajσk(j), –bjσk(j)∈suppfˆ(j= 1, 2, . . . ,n)and

suppfˆ⊆ξ∈Qσk:σk(j)ξj≤aj

∪ξ∈–Qσk: –σk(j)ξj≤bj

.

Then g∈Lp_(Rn_{)satisfies the Schrödinger-type identitySch}

α(fg) =fSchα(g)if and only if

suppfˆ⊆

ξ∈Qσk: n

j=1

σk(j)ξj

bj ≥

1 ∪

ξ∈

n

j=1

–σk(j)ξj

aj ≥

1 .

Proof Suppose thatQσk is the first octant inR

n_{, that is to say, all theσ}

k(j) = 1, where

j= 1, 2, . . . ,n. Let

ˆ

fs,gn₀+–2sgn(s),g₀n+–3sgn(s), . . . ,g1₀+sgn(s),sgn(s)

= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

f(s,g₀n+–2m(s),g₀n+–3m(s), . . . ,g₀1+m(s),m(s)), sgn(s) <m(s),

f(s,gn–2

0+ sgn(s),g₀n+–3sgn(s), . . . ,g₀1+sgn(s),sgn(s)), m(s)≤sgn(s)≤n(s), f(s,g₀n+–2n(s),g₀n+–3n(s), . . . ,g1₀+n(s),n(s)), sgn(s) >n(s).

Consider the fractional differential equation

Qβ₀+ιp

Qα₀+–n+2n(s)

+fˆs,g₀n+–2sgn(s),g₀n+–3sgn(s), . . . ,g₀1+sgn(s),sgn(s)

= 0,

0 <t< 1,

v(0) = 0, v(1) =av(ξ), Q₀α+–n+2v(0) =Qα₀+–n+2v(1) = 0.

(2.4)

Set 2={v∈E2: v ≤M1ιq(M2L2)}, then2 is a closed, bounded and convex set,

where

L2:= sup

t∈[0,1],v∈2

fˆs,g₀n+–2sgn(s), . . . ,g₀1+sgn(s),sgn(s)+ 1.

The operatorA:2→E2is defined by

Asgn(s) = 1

0

P(s,ς)ιq 1

0

Q(s,ς)fˆς,g₀n+–2v(ς), . . . ,g₀1+v(ς),v(ς)

dς

ds.

Now, we show thatAis a completely continuous operator. It follows that (Av)(s)

= 1

0

P(s,ς)ιq 1

0

Q(s,ς)fˆς,gn₀+–2v(ς), . . . ,g₀1+v(ς),v(ς)

dς

ds

≤ 1

0

P(s,ς)ιq 1

0

Q(s,ς)fˆς,g0n+–2v(ς), . . . ,g₀1+v(ς),v(ς)dς

ds

≤Lq₂–1

1

0

P(s,ς)ιq 1

0

Q(s,ς)dς

ds

≤Lq₂–1

1

0

P(s,s)ιq 1

0

Q(ς,ς)dς

ds

< +∞,

which yields

P(t1,s) –P(t2,s) <

ε

Lq₂–1ιq( 1

0 Q(ς,ς)dς)

.

So

Av(t2) –Av(t1)

≤ 1

0

P(t2,s) –P(t1,s)ιq 1

0

Q(s,ς)fˆς,g₀n+–2v(ς), . . . ,g₀1+v(ς),v(ς)

dς

ds

≤Lq₂–1

1

0

P(t2,s) –P(t1,s)ιq 1

0

Q(ς,ς)dς

ds

≤Lq2–1ιq 1

0

Q(ς,ς)dς

1

0

_{P(t2,s) –P(t1,s)ds}

<ε

We prove that the fractional differential equation has at least one positive solution. Sup-pose thatd(s) is a solution of (2.4) (see [9]), then

d(0) = 0, d(1) =ad(ξ), Q₀α+–n+2d(0) =Qα₀+–n+2d(1) = 0.

So

fs,gn₀+–2n(s),g₀n+–3n(s), . . . ,g₀1+n(s),n(s)

≤ ˆfs,g₀n+–2d(s),g₀n+–3d(s), . . . ,g₀1+d(s),d(s)

≤fs,g₀n+–2m(s),g₀n+–3m(s), . . . ,g₀1+m(s),m(s)

.

So

fs,gn₀+–2q(s),g₀n+–3q(s), . . . ,g₀1+q(s),q(s)

≤ ˆfs,g₀n+–2d(s),g₀n+–3d(s), . . . ,g₀1+d(s),d(s)

≤fs,g₀n+–2p(s),g₀n+–3p(s), . . . ,g₀1+p(s),p(s)

,

which yields

Qβ₀+ιp

Qα–n+2 0+ n(s)

=Qβ₀+ιp

Qα–n+2 0+ (Fp)(s)

=fs,g₀n+–2p(s),g₀n+–3p(s), . . . ,g₀1+p(s),p(s)

.

From the above discussions, we have

Qβ₀+ιp

Qα₀+–n+2n(s)

–Qβ₀+ιp

Qα₀+–n+2d(s)

=fs,g₀n+–2p(s),g₀n+–3p(s), . . . ,g₀1+p(s),p(s)

–fˆs,g₀n+–2d(s),g₀n+–3d(s), . . . ,g₀1+d(s),d(s)

≥0,

wheret∈[0, 1].

If we letz(s) =ιp(Q0α+–n+2n(s)) –ιp(Qα0+–n+2d(s)), then z(0) =z(1) = 0. By Lemma2.1, we

havez(s)≤0. Hence,

ιp

Qα₀+–n+2n(s)

≤ιp

Qα₀+–n+2d(s)

,

wheres∈[0, 1].

Sinceιpis monotone increasing,

Qα–n+2

0+ n(s)≤Q₀α+–n+2d(s),

that is,

By the assumption thatsuppfˆ,suppgˆε⊆σk(j)∪–σk(j), we obtain

Qσk ˆ

f(x–t)gˆε(s)ds= 0,

wherex∈–Qσk, and

–Qσk ˆ

f(x–t)gˆε(s)ds= 0,

wherex∈Qσ_k. So

suppgˆεχQσk⊆

ξ∈Qσk: n

j=1

ξj

bj≥

1 (2.5)

as the other case can be obtained in a similar way. Letλ=ˆfχ–Qσ_k and=gˆεχQσk. We decomposeinto

=1+2

withsupp1⊆

n

j=1[0,bj] andsupp2⊆Qσk\ n

j=1(0,bj). By (2.5) we obtain

(1∗λ)(x) = –(2∗λ)(x),

wherex∈–Qσk. Meanwhile

supp(2∗λ)⊆supp2+suppλ⊆Qσk\ n

j=1

(0,bj) + n

j=1

[–bj, 0]⊆Rn\(–Qσk) (2.6)

and

supp(1∗λ)⊆supp1+suppλ⊆

n

j=1

[0,bj] + n

j=1

[–bj, 0]⊆ n

j=1

[–bj,bj]. (2.7)

By (2.6) it is clear that

supp(1∗λ)⊆Rn\(–)Qσk.

This together with (2.7) implies that

conv supp(1∗λ)⊆

ξ∈Rn: –bj≤ξj≤bj, n

j=1

ξj

aj ≥

We claim that, for anyξ∈supp1,

n

j=1

ξj

aj ≥1

holds.

If it is invalid, then there isξ1_{∈conv supp}

1satisfying

n

j=1

ξj

bj < 1.

Note thatξ2_{=b∈suppλsatisfies}

n

j=1

ξj

bj = –n.

Since

conv supp(1∗λ) =conv supp1+conv suppλ,

there exists some pointξ∈conv supp(1∗λ) such that

n

j=1

ξj

aj < 1 –n.

This contradicts (2.8). We conclude that

supp1=suppgˆεχQσ_k ⊆

ξ∈Qσ_k: n

j=1

ξj

bj ≥1 .

This completes the proof.

3 Conclusions

This paper was mainly devoted to developing the Schrödinger-type identity for a Schrö-dinger free boundary problem in Rn_{. As an application, we established necessary and} sufficient conditions for the product of some distributional functions to satisfy the Schrödinger-type identity. As a consequence, our results significantly improved and gen-eralized previous work.

Acknowledgements

The authors are thankful to the honorable reviewers for their valuable suggestions and comments, which improved the paper.

Funding Not applicable.

Availability of data and materials Not applicable.

Ethics approval and consent to participate Not applicable.

Competing interests

The authors declare that they have no competing interests.

Consent for publication Not applicable.

Authors’ contributions

All authors read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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