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R E S E A R C H

Open Access

Applications of maximum modulus method

and Phragmén–Lindelöf method for

second-order boundary value problems with

respect to the Schrödinger operator

Zhen Liu

1*

*Correspondence:

[email protected]

1School of Mathematics and

Statistics, Kashi University, Kashi, China

Abstract

In this paper, we present a reliable combination of the maximum modulus method with respect to the Schrödinger operator (Meng in J. Syst. Sci. Complex. 16:446–452,

2003) and Phragmén–Lindelöf method (Shehu in Matematiche 64:57–66,2015) to investigate the solution of a second-order boundary value problem with respect to the Schrödinger operator. We establish the uniqueness of the solution for this problem. The results reveal that this method is effective and simple.

Keywords: Maximum modulus; Schrödinger operator; Uniqueness; Boundary value problem

1 Introduction

Letג= (0,W)×(0,w) be a bounded rectangular domain inR2, which represents a porous

medium with Lipschitz boundaryג=1∪2, where

2=

{0} ×[0,w]∪[0,W]× {w}∪{W} ×[0,w]

is the part in contact with air or covered by fluid, and

1= [0,W]× {0}

is the impervious part ofג. Let P=ג×(0,M), whereM> 0. Let φ be a nonnegative Lipschitz function defined in the closure ofP, which we denote byP.

Define

Σ1=1×(0,M), Σ2=2×(0,M), Σ3=Σ2∩ {φ> 0},

and

Σ4=Σ2∩ {φ= 0}.

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Letχbe a function of the variabletsatisfying

c1≤χ(t)≤c2, a.e.t∈(0,W), (1.1)

for two positive constantsc1andc2, and letω0be a function of the variabletsatisfying

(see [3,4])

0≤ω0(t)≤1, a.e.t∈ג. (1.2)

The governing equation in terms of the deflection functionf(t) is the elliptic boundary value problem with respect to the Schrödinger operator (see [2,5,9,11,16,24]): Find (f,ω)∈L2(0,T;H1(ג))×L(P) such that

f ≥0, 0≤ω≤1, f(1 –ω) = 0 a.e. inP,

f =φ onΣ2,

P

χ(t1)(ft2+ω)ςt2–ωςs

dt ds

גω0(t)ς(t, 0)dt,

ςH1(P), ς= 0 onΣ3,

ς≥0 onΣ4,

ς(t,M) = 0 for a.e.t∈ג.

(1.3)

For the existence of a solution of problem (1.3) in the homogeneous case, we refer to [10] and [19] in the incompressible and compressible cases, respectively. For the heterogeneous case, we refer to [6] in a more general framework under Assumptions (1.1) and (1.2) for both incompressible and compressible cases. For the incompressible case with nonlinear Darcy’s law, we refer to [18] for Dirichlet, Neumann, and generalized boundary condi-tions. Extensions to the quasilinear and incompressible case were obtained in [10,18] in both homogeneous and nonhomogeneous frameworks. Meng [20] was able to extend the above regularity result to a more general framework under weaker assumptions on the data. The uniqueness of a solution for a homogeneous dam with general geometry was established by the method of doubling variables in [8,24], but it is not obvious whether it works in the heterogeneous situation. Extensions to the Schrödinger operator modeling incompressible fluid flow governed by the nonlinear Darcy law with Dirichlet or Neumann boundary conditions were obtained in [15,20], respectively.

On the other hand, the following boundary value problem with respect to the Schrödin-ger operator corresponding to (1.3) is given by [7,14,18,29,33,36]:

f ≥0, 0≤ω≤1, u(1 –ω) = 0 inP, χ(t1)(ft2+ω)t2–ωs= 0 inP,

u=φ onΣ2,

ω(·, 0) =ω0 inג,

χ(t1)(ft2+ων= 0 onΣ1,

χ(t1)(ft2+ων≤0 onΣ4.

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Regarding the existence of a solution of the problem with respect to the Schrödinger op-erator (1.4), we refer to [15] and [37], respectively, for the evolutionary dam problem with homogeneous coefficients and for a class of free boundary problem with nonlocal bound-ary condition in heterogeneous domain. The regularity of the solution of the problem with nonlocal boundary condition was discussed in [40] (also see, e.g., [13,32]), where it was proved thatωC0([0,T];Lp(ג)) for allp[1, +) in the free boundary problems with respect to the Schrödinger operator of types

divχ(t)∇f+H(t)ωωs

and

divχ(t)∇f+H(t)ω– (f +ω)s,

and thatfC0([0,M];Lp(ג)) for allp[1, 2] in the second-order class.

The uniqueness of a solution of the Schrödinger equation in the homogeneous case for both incompressible and compressible fluids was obtained in [20] by using the method of doubling variables. In the case of a rectangular dam wet at the bottom and dry near to the top, the uniqueness was obtained in [17] and [35] by the fixed point theory (as one of the main tools in this subject; see [21,25]) with respect to the Schrödinger operator, respec-tively, in homogeneous and heterogeneous porous media (see [28]). For the evolution free boundary problem in theory of the Schrödinger equation, we refer to [8,15].

As an application of the Phragmén–Lindelöf method related to a second-order bound-ary value problem with respect to the Schrödinger operator, in this paper, we consider the weak formulation of an evolution dam problem with heterogeneous coefficients (1.3) in ג. We establish the uniqueness of the solution for this problem with respect to the Schrödinger operator. This uniqueness result is new in the general framework of a het-erogeneous and bounded rectangular domain.

2 Main result and its proof

In this section, we obtain our main results that a solution of problem with respect to the Schrödinger operator (1.3) is unique. We assume that

χC1[0,W]. (2.1)

Theorem 2.1 Let(f1,ω1)and(f2,ω2)be two solutions of (1.3).Then

P χ(t1)

f1(t,s) –f2(t,s)

+

t2+

1 –ω2(t,s)

ω{f1>f2}

+1 –ω2(t,s)

+1 –f2t2(t,s)

ω{f1>0}ιςt2dt ds≤0, (2.2)

where

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Proof Note that

(f1,ω1) : (t,s,x,y)→

f1(t,s),ω1(x,y)

,

(f2,ω2) : (t,s,x,y)→

f2(t,s),ω2(x,y)

.

Define

ψ(t,s,x,y) =ς t1+x1 3 ,

t2+x2

3

ι y+s 3

ρ1,σ

t1–x1

3

ρ2,σ

t2–x2

3

ρ3,σ

ys

3

for (t,s,x,y)∈P×P, where

ρ1,σ(r) = 1 σρ1

r

σ

,

ρ2,σ(r) = 1 σρ2

r

σ

,

ρ3,σ(r) = 1 σρ3

r

σ

withρ1,ρ2,ρ3∈D(R),ρ1,ρ2,ρ3≥0,supp(ρ1),supp(ρ2),supp(ρ3)⊂(–1, 1).

So

ψ(·,·,x,y)∈D(P), (2.3) ψ(t,s,·,·)∈D(P), (2.4) where (x,y)∈Pand (t,s)∈P.

Define

ϑ(t,s,x,y) =max (f1(t,s) –f2(t,s))

+

,ψ(t,s,x,y)

, (2.5)

whereis a positive real number (see [22]).

By applying the fixed point theory to (f1,ω1) withk=f2(x,y) andς(t,s) =ϑ(t,s,x,y) for

almost every (x,y)∈Pwe know that

P

χ(t1)(f1t2+ω1)ϑt2dt ds= 0, (2.6)

which, together withf1.(1 –ω1) = 0 a.e. inP, gives that

ω1χ(t1) max

(f1–f2)+

,ψ

t2

=χ(t1) max

(f1–f2)+

,ψ

t2

a.e. inP.

Meanwhile, (2.6) also yields that

P

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and

P×P

χ(t1)(f1t2+ 1)ϑt2d= 0, (2.7)

whered=dt ds dx dy.

Now we apply the fixed point theory to (f2,ω2) with

k=f1(t,s), ς(x,y) =ϑ(t,s,x,y)

to obtain that

P

χ(x1)(f2x2+ω2) ϑ–min f1

,ψ

x2

dy ds= 0.

So

P×P

χ(x1)(f2x2+ω2) ϑ–min f1

,ψ

x2

dy ds dt ds= 0. (2.8)

By subtracting (2.8) from (2.7) we have that

P×P

χ(t1)f1t2ϑt2–χ(x1)f2x2ϑx2

+χ(t1)ϑt2–ω2χ(x1)ϑx2

d

P×P

χ(x1)(f2x2+ω2)min f1

,ψ

x2

d= 0. (2.9)

It follows from (2.3)–(2.5) that

P×P

χ(t1)f1t2ϑx2d= 0, (2.10)

P×P

χ(x1)f2x2ϑt2d= 0, (2.11)

P×P

χ(t1)(∂t2+∂x2)ϑd= 0, (2.12)

P×P

ω2χ(x1)ϑt2d= 0, (2.13)

P×P

χ(x1)(f2x2+ω2)min f1

,ψ

t2

d= 0, (2.14)

P×P

χ(t1)(∂t2+∂x2)min f1

,ψ

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Combining (2.10)–(2.15) and (2.9), we get that

P×P

χ(t1)(∂t2+∂x2)f1–χ(x1)(∂t2+∂x2)f2

(∂t2+∂x2)ϑ

+χ(t1) –ω2χ(x1)

(∂t2+∂x2)ϑ

d

+

P×P

χ(t1) –χ(x1)(∂t2+∂x2)f2

(∂t2+∂x2)min f1 ,ψ d +

P×P

χ(t1) –ω2χ(x1)

(∂t2+∂x2)min f1

,ψ

d= 0. (2.16)

Put

t+x

3 =z,

tx

3 =ξ,

y+s

3 =κ,

ys

3 =τ. (2.17)

Note that (z,κ)∈Pand

(ξ,τ)∈ –W 3 ,

W

3

× –w 3,

w

3

× –M 3 ,

M

3

:=ג1× – M

3,

M

3

:=P1.

We have that

P×P1

χ(z1+ξ1)f1z2(z+ξ,κ+τ)

χ(z1–ξ1)f2z2(zξ,κτ)

ϑz2

+

P×P1

χ(z1+ξ1) –ω2(zξ,κτ)χ(z1–ξ1)

ϑz2

+

P×P1

χ(z1+ξ1) –ω2(zξ,κτ)χ(z1–ξ1)

×min f1 ,ψ z2 +

P×P1

χ(z1+ξ1) –f2z2(zξ,κτ)χ(z1–ξ1)

×min f1 ,ψ

z2

= 0 (2.18)

from (2.16)–(2.17), where=dz dκdξdτ. Put

I,σ=

P×P1

χ(z1+ξ1)f1z2(z+ξ,κ+τ)

χ(z1–ξ1)f2z2(zξ,κτ)

ϑz2dz dκdξdτ,

J,σ=

P×P1

χ(z1+ξ1) –ω2(zξ,κτ)χ(z1–ξ1)

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K1

,σ =

P×P1

χ(z1+ξ1) –ω2(zξ,κτ)χ(z1–ξ1)

×min f1 ,ψ

z2 ,

K2

,σ =

P×P1

χ(z1+ξ1) –f2z2(zξ,κτ)χ(z1–ξ1)

×min f1 ,ψ

z2 .

Lemma 2.2

lim σ→0

lim →0J,σ

=

P

ιω{f1>f2}χ(z1)

1 –ω2(z,κ)

ςz2dz dκ, (2.19)

lim σ→0

lim →0K

1 ,σ = P

ιω{f1>0}χ(z1)

1 –ω2(z,κ)

ςz2dz dκ, (2.20)

lim σ→0

lim →0K

2 ,σ = P

ιω{f1>0}χ(z1)

1 –f2z2(z,κ)

ςz2dz dκ. (2.21)

Proof Set

J,σ=

P×P1

χ(z1+ξ1) –χ(z1–ξ1)

ϑz2

+

P×P1

χ(z1–ξ1)

1 –ω2(zξ,κτ)

ϑz2

:=J1,σ+J2,σ. (2.22)

Combining (2.3)–(2.5) and the fact thatχ(z1+ξ1) –χ(z1–ξ1) does not depend onz2,

we have that

J1,σ = 0.

So

lim σ→0

lim →0J

1

,σ

= 0. (2.23)

Put

lim σ→0

lim →0J

2 ,σ = P ιχ(z1)

1 –ω2(z,κ)

ςz2dz dκ. (2.24)

Set

A=

(f1–f2)+>ψ

and

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We know that

J2

,σ=

(z1–ξ1)

1 –ω2(zξ,κτ)

f1–f2

z2 +

(z1–ξ1)

1 –ω2(zξ,κτ)

ψz2

:=J2,1,σ +J2,2,σ. (2.25)

By applying (2.17) and that (1 –ω2)f2= 0 andf1x2= 0 a.e. inP(see [1,41]), we have that

J2,1

,σ =

B

χ(x1)

1 –ω2(x,y)

f1t2

d.

By applying (2.3) and that the function (x,y)→χ(x1)(1 –ω2(x,y)) does not depend on t2we have that

J2,1

,σ =

P×P χ(x1)

1 –ω2(x,y) min f1 ,ψ t2 d

(x1)

1 –ω2(x,y)

ψt2dt ds dy ds

=

P×P χ(x1)

1 –ω2(x,y) min f1 ,ψ t2 d

P×P χ(x1)

1 –ω2(x,y)

ψt2d

+

(x1)

1 –ω2(x,y)

ψt2d

=

(x1)

1 –ω2(x,y)

ψt2d.

Taking into account the fact thatlim→0|B|= 0 (see [39]), we know that lim

→0J 2,1

,σ = 0,

which yields that

lim σ→0

lim →0J

2,1

,σ

= 0. (2.26)

To estimateJ2,2

,σ, we obtain that

lim σ→0

lim →0J

2,2 ,σ = P

χ(z1)ω{f1>f2}

1 –ω2(z,κ)

ςz2ιdz dκ (2.27)

by passing to the limit as→0 and then asσ→0.

Hence, combining (2.26)–(2.27), we obtain (2.24) by letting→0 andσ→0 in (2.25). Now we pass successively to the limit in (2.22) as→0 and then asσ→0, and using (2.23)–(2.24), we obtain (2.19). Finally, arguing as in the proof (2.19), we obtain (2.20) and

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Lemma 2.3

lim σ→0

lim →0I,σ

P ιχ(z1)

f1(z,κ) –f2(z,κ)

+

z2ςz2dz dκ. (2.28)

Proof Set

I,σ=

A

χ(z1+ξ1)f1z2(z+ξ,κ+τ)

χ(z1–ξ1)f2z2(zξ,κτ)

ψz2

+

B

χ(z1+ξ1)f1z2(z+ξ,κ+τ)

χ(z1–ξ1)f2z2(zξ,κτ)

f1–f2

z2

:=I1,σ+I2,σ. (2.29)

Note thatI2

,σ can be rewritten as

I2

,σ= 1

B

χ(z1+ξ1)f1z2(z+ξ,κ+τ)f1z2(z+ξ,κ+τ)

+χ(z1–ξ1)f2z2(zξ,κτ)f2z2(zξ,κτ)

dz dκdξdτ

(z1–ξ1)f2z2(zξ,κτ)f1z2(z+ξ,κ+τ)

B(χ(z1+ξ1)f1z2(z+ξ,κ+τ)f2z2(zξ,κτ)

:=I2,1,σI2,2,σI2,3,σ.

It follows from (1.1) that

I2

,σ≥–I

2,2

,σI

2,3

,σ. (2.30)

Let us estimateI2,2

,σ. It follows from (2.3) and (2.5) that

I2,2

,σ=

(x1)f2x2(x,y)

f1(t,s) –f2(x,y)

t2

dt ds dy ds

=

P×P

χ(x1)f2x2(x,y)ϑt2d

(x1)f2x2(x,y)ψt2d

=

P×P

χ(x1)f2x2(x,y)ϑt2d

P×P

(10)

+

(x1)f2x2(x,y)ψt2d

=

B

χ(x1)f2x2(x,y)ψt2d.

Taking into account thatlim→0|B|= 0 (see [39]), we obtain that

lim →0

I2,2

,σ

= 0. (2.31)

Similarly,

lim →0

I2,3

,σ

= 0. (2.32)

Combining (2.31)–(2.32), we obtain that

lim →0

I2

,σ

≥0 (2.33)

by passing to the limit as→0 in (2.30). Let us estimateI1

,σ. We obtain that

lim →0

I1

,σ

=

P×P1

ω{f1>f2}

χ(z1+ξ1)f1z2(z+ξ,κ+τ)

χ(z1–ξ1)f2z2(zξ,κτ)

ψz2,

which yields that

lim →0

I1

,σ

=

P×P1

ω{f1>f2}χ(z1+ξ1)

f1(z+ξ,κ+τ)

f2(zξ,κτ)

z2ψz2

+

P×P1

ω{f1>f2}

χ(z1+ξ1)

χ(z1–ξ1)

f2z2(zξ,κτ)ψz2

=1,1+1,2. (2.34)

By applying (2.1) and taking into account the fact thatsupp(ρ1,σ)⊂(–σ,σ) (see [39]), we obtain that

I1,2

σC

P×P1

|ξ1||f2z2||ςz2|ιρ1,σ(ξ1)ρ2,σ(ξ1)ρ3,σ(τ)

σC

P×P1

|f2z2||ςz2|ιρ1,σ(ξ1)ρ2,σ(ξ1)ρ3,σ(τ)

:=σCWσ

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So

lim σ→0I

1,2

σ = 0. (2.35)

ForI1,1

σ , we have that

lim σ→0I

1,1

σ =

P ιχ(z1)

f1(z,κ) –f2(z,κ)

+

z2ςz2dz dκ (2.36)

by passing to the limit asσ→0.

Combining (2.35)–(2.36), we obtain that

lim σ→0

lim →0I

1

,σ

=

P ιχ(z1)

f1(z,κ) –f2(z,κ)

+

z2ςz2dz dκ (2.37)

by lettingσ→0 in (2.34).

Finally, we pass successively to the limit in (2.29) as→0 and then asσ→0, and using (2.33) and (2.37), we obtain (2.28).

Now, using Lemmas2.2and3.1and letting successively→0 andσ→0 in (2.18), we obtain (2.2). This completes the proof of Theorem2.1.

3 An application

In this section, we establish the existence of nontrivial positive solutions of (1.3) withς= ς∗by Theorem2.1. To this end, we first construct a pair of proper super- and subsolutions of (1.3) withς=ς∗(see [23,27]). We define the continuous functions (see [26,30,34,38])

J+(ζ) =J0,

J–(ζ) =

⎧ ⎨ ⎩

J0–peλ3ζ, ζ<ζ1,

eλ4ζ, ζζ

1,

K+(ζ) =

⎧ ⎨ ⎩

ζeσζ, ζ <ζ

2,

(ϕγ – 1)J0, ζζ2,

K–(ζ) =

⎧ ⎨ ⎩

ζeσζL(–ζ)1/2eσζ, ζ <ζ

3,

0, ζζ3,

where

τ1=min

σ∗ 2 ,

α∗ 2d1

, τ2=

ϕ/d1,

andζ1,ζ2,ζ3∈R,p>J0,> 0,> 0,L> 0 will be clarified later.

Lemma 3.1 Assume thatϕ>γ.Then(1.3)withς=ςadmits a solution satisfying

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Proof Define

m:=

0

RI(t,s)e

λ∗(tcs)dt ds,

n:=

0

RI(t,s)

tαseλ∗(tcs)dt ds.

It is easy to see thatm,nare bounded and

ζm+n≤(ζ+K)mα

0

RI(t,s)se

λ∗(tcs)dt ds,

and soζm+n< 0 for allζ< –K. Moreover, Lemma2.3indicates that

d2λ∗2–αλ∗+ϕmγ = 0, 2d2λ∗–c∗+ϕn= 0.

Note that ifh1≥0,h2≥0 withh1+h2> 0, then h1h2

h1+h2 ≤

min{u,v}.

Ifζ<ζ2, thenI+(ζ) = – ζeσζ

, and it suffices to prove that

αK+(ζ)≥d2K+(ζ) +ϕ(IK+)(ζ) –γK+(ζ). (3.1)

Note that

(IK+)(ζ) =

0

RI(t,s)K+

ζtcsdt ds

0

+∞

ζζ∗–αs

I(t,s)I+

ζtαsdt ds

= –

0

RI(t,s)

ζtcseλ∗(ζtcs)dt ds

= –

0

RI(t,s)

ζ+tcseλ∗(ζ+tcs)dt ds

= – ζeλζ

0

RI(t,s)e

λ∗(tcs)dt ds

ζ

0

RI(t,s)

tcseλ∗(tcs)dt ds

= – ζeλζmeλζn.

It is obvious that (3.1) holds in the case where the following inequality holds.

αK+(ζ)≥d2K+(ζ) –ϕ ζe

σζ

mϕeσζnγK+(ζ).

It follows that

K

+(ζ) = –e

σζ

1 +σζ,

K

+(ζ) = –e

σζ

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for anyζ <ζ2by Lemma2.2and the definition ofI+(ζ).

So

αK+(ζ) –d2K+(ζ) +ϕ ζeσζ

m+ϕeσζn+γK

+(ζ)

=Λλ

σ∗,αζ+Λσ,α ζeσζ = 0,

which yields (3.1). Putζ>ζ2. Then

ϕJ+(ζ)(IK+)(ζ) J+(ζ) + (IK+)(ζ)

ϕJ0(

ϕ γ – 1)J0

J0+ (ϕγ – 1)J0 =γ ϕ

γ – 1

J0.

Denote

p=J01(Kζ)

+sup ζ<0

ϕe(σ∗–τ1)ζ(ζm+n) ατ1–d1τ12

and

J0–peλ3ζ=eλ4ζ.

Ifζ<ζ1, then

J–(ζ) =J0–peτ1ζ> 0

and

αJ(ζ)≤d1J–(ζ) –ϕ(II+)(ζ),

whereζ<ζ1.

It follows that

(IK+)(ζ) =

0

RI(t,s)I+

ζtαsdt ds

0

+∞

ζζ∗–αs

I(t,s)I+

ζtαsdt ds

= – ζeσζmeσζn

sinceζ<ζ1≤ζ∗–K.

To verify that

αJ(ζ)≤d1J–(ζ) +ϕ ζeσζ

m+ϕeσζn,

by simple calculations, we have

ατ1peλ3ζ ≤–d1λ23peλ3ζ +ϕ ζeσζ

m+ϕeσζn, ζ<ζ

1,

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It suffices to confirm that

αJ(ζ)≤d1J–(ζ) –ϕJ–(ζ),

which is equivalent to

αλ4eλ4ζ ≤d1λ24e

λ4ζ ϕeλ4ζ,

and this is also evident by the definition ofλ4.

LetLM1≥√–ζ2be such that J–(ζ)≥J0/2,

ζ <ζ3:= – L

2

ζ2

for> 0 andζ2< 0.

It is obvious that the definition ofζ3implies that

K–(ζ)≤K+(ζ)

for allζ ∈R. Ifζ<ζ3, then

K–(ζ) = – ζeλζ

L(–ζ)1/2ζ≤– ζeλζ=K+(ζ),

which yields that

(IK–)(ζ) =

0

RI(t,s)K

ζtcsdt ds

0

RI(t,s)K+

ζtcsdt ds

= – ζeλζmζn.

So

ϕJ–(ζ)(IK–)(ζ) J–(ζ) + (IK–)(ζ)

ϕ(IK–)(ζ)

ϕJ20(IK–)(ζ)

J0

2 + (IK–)(ζ)

ϕ(IK–)(ζ)

≥–2ϕ

J0

(IK–)(ζ)

2

≥–2ϕ

2

J0

e2λζ(ζm+n)2,

which gives that

cK(ζ)≤d2K–(ζ) +ϕ(IK–)(ζ) –γK–(ζ) –

2ϕ2

J0

(15)

By a simple calculation we have that

K

–(ζ) =K+(ζ) +Le

σζ

1 2(–ζ)

–1/2σ(–ζ)1/2

,

K

–(ζ) =K+(ζ) +Leσζ1

4(–ζ)

–3/2+σ(–ζ)–1/2σ∗2

(–ζ)1/2

.

Since –ζ+y+αs≥0 for anyζ <ζ3,|y|<K, ands≥0, there existsθ∈(0, 1) such that

ζ+t+αs1/2

= (–ζ)1/2+1 2(–ζ)

–1/2t+αs1

8

ζ+θt+αs–3/2t+αs2

≤(–ζ)1/2+1 2(–ζ)

–1/2t+αs

by applying the Taylor theorem (see [12,31]), which yields that

(IK–)(ζ) =

0

RI(t,s)I

ζtαsdt ds

=

0

K

K

I(t,s)I

ζtαsdt ds

0

K

K

I(t,s)–ζtαseσ∗(ζtαs)

Lζtαs1/2∗(ζtαs)dt ds

≥–

0

K

K

I(t,s)ζtαseσ∗(ζtαs)dt ds

L

0

K

K

I(t,s)

(–ζ)1/2+1 2(–ζ)

–1/2t+αseσ∗(ζtαs)dt ds

= – ζeσζmζnL(–ζ)1/2ζm+1 2L(–ζ)

–1/2eσζ

n.

So

αK+(ζ) +ζ

1 2(–ζ)

–1/2σ(–ζ)1/2

≤d2K+(ζ) +d2Leσζ1

4(–ζ)

–3/2+σ(–ζ)–1/2σ∗2(–ζ)1/2

ϕ ζeσζmϕeσζnϕL(–ζ)1/2ζm+ϕ 2L(–ζ)

–1/2eσζ

n

γK+(ζ) +γL(–ζ)1/2ζ

–2ϕ

2

J0

e2σζ(ζm+n)2,

which is true provided that

d2Leσζ1

4(–ζ)

–3/22ϕ2 J0

(16)

Put

M2:=sup

ζ<0

8ϕ2(ζm+n)2(–ζ)3/2eσζ d2J0

+ 1

for anyζ <ζ3, whereL:=M1+M2. Whenζ>ζ3, it is straightforward to show the required

result. The proof is complete.

4 Conclusions

In this paper, we further studied the Phragmén–Lindelöf method related to a second-order order boundary value problem with respect to the Schrödinger operator. We also pre-sented some mathematical consequences of the method including a stability result. The main technical tools used to develop the mathematical analysis are local and global bifur-cation, monotonicity techniques, the augmented Phragmén–Lindelöf method, blow-up arguments, and some techniques used in the previous works. As an application, we proved the uniqueness of a solution for the definite problem of a parabolic variational inequality.

Acknowledgements

The work was supported by the Natural Science Foundation of Xinjiang Uygur Autonomous Region of China (No. 2019D01A04).

Funding Not applicable.

Abbreviations Not applicable.

Availability of data and materials Not applicable.

Competing interests

The author declares that there are no competing interests.

Authors’ contributions

The author completed the paper and approved the final manuscript.

Publisher’s Note

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

Received: 16 July 2019 Accepted: 24 September 2019

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