Boundary Value Problems
Volume 2011, Article ID 827510,15pages doi:10.1155/2011/827510
Research Article
Positive Solutions for Integral Boundary Value
Problem with
φ
-Laplacian Operator
Yonghong Ding
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Yonghong Ding,[email protected]
Received 20 September 2010; Revised 31 December 2010; Accepted 19 January 2011
Academic Editor: Gary Lieberman
Copyrightq2011 Yonghong Ding. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider the existence, multiplicity of positive solutions for the integral boundary value problem with φ-Laplacian φut ft, ut, ut 0, t ∈ 0,1, u0 01urgrdr,
u1 01urhrdr, whereφis an odd, increasing homeomorphism fromRontoR. We show that it has at least one, two, or three positive solutions under some assumptions by applying fixed point theorems. The interesting point is that the nonlinear termfis involved with the first-order derivative explicitly.
1. Introduction
We are interested in the existence of positive solutions for the integral boundary value problem
φutft, ut, ut0, t∈0,1,
u0
1
0
urgrdr, u1
1
0
urhrdr,
1.1
whereφ, f, g, andhsatisfy the following conditions.
H1φ is an odd, increasing homeomorphism from R onto R, and there exist two increasing homeomorphismsψ1andψ2of0,∞onto0,∞such that
ψ1uφv≤φuv≤ψ2uφv ∀u, v >0. 1.2
H2f : 0,1 × 0,∞× −∞,∞ → 0,∞ is continuous. g, h ∈ L10,1 are nonnegative, and 0<01gtdt <1, 0<01htdt <1.
The assumptionH1on the functionφwas first introduced by Wang1,2, it covers two important cases:φu uandφu |u|p−2u, p >1. The existence of positive solutions for two above cases received wide attentionsee3–10. For example, Ji and Ge4studied the multiplicity of positive solutions for the multipoint boundary value problem
φp
utqtft, ut, ut0, t∈0,1,
u0
m
i1
αiuξi, u1 m
i1
βiuξi,
1.3
where φps |s|p−2s, p > 1. They provided sufficient conditions for the existence of at least three positive solutions by using Avery-Peterson fixed point theorem. In5, Feng et al. researched the boundary value problem
φp
utqtft, ut 0, t∈0,1,
u0
m−2
i1
aiuξi, u1 m−2
i1
biuξi,
1.4
where the nonlinear termfdoes not depend on the first-order derivative andφps |s|p−2s,
p >1. They obtained at least one or two positive solutions under some assumptions imposed on the nonlinearity offby applying Krasnoselskii fixed point theorem.
As for integral boundary value problem, whenφu uis linear, the existence of positive solutions has been obtainedsee8–10. In8, the author investigated the positive solutions for the integral boundary value problem
ufu 0,
u0
1
0
uτdατ, u1
1
0
uτdβτ.
1.5
The main tools are the priori estimate method and the Leray-Schauder fixed point theorem. However, there are few papers dealing with the existence of positive solutions when φ
satisfiesH1andf depends on bothuandu. This paper fills this gap in the literature. The aim of this paper is to establish some simple criteria for the existence of positive solutions of BVP1.1. To get rid of the difficulty offdepending onu, we will define a special norm in Banach spaceinSection 2.
2. Preliminaries
The basic space used in this paper is a real Banach spaceC10,1with norm ·
1defined by u1max{uc,uc}, whereucmax0≤t≤1|ut|. Let
K
u∈C10,1|ut≥0, u1
1
0
uthtdt, uis concave on0,1
. 2.1
It is obvious thatKis a cone inC10,1.
Lemma 2.1see7. Letu∈K, η∈0,1/2, thenut≥ηmax0≤t≤1|ut|,t∈η,1−η.
Lemma 2.2. Let u ∈ K, then there exists a constant M > 0 such that max0≤t≤1|ut| ≤
Mmax0≤t≤1|ut|.
Proof. The mean value theorem guarantees that there existsτ ∈0,1, such that
u1 uτ
1
0
htdt. 2.2
Moreover, the mean value theorem of differential guarantees that there existsσ∈τ,1, such
that
1
0
htdt−1
uτ u1−uτ 1−τuσ. 2.3
So we have
|ut| ≤ |uτ|
t
τ
usds
≤
⎛
⎝ 1−τ
1−01htdt 1
⎞ ⎠max
0≤t≤1
ut≤ 2−
1 0htdt 1−01htdtmax0≤t≤1
ut.
2.4
DenoteM 2−01htdt/1−01htdt; then the proof is complete.
Lemma 2.3. Assume that (H1), (H2) hold. Ifuis a solution of BVP1.1, there exists a uniqueδ ∈
0,1, such thatuδ 0andut≥0,t∈0,1.
Proof. From the fact that φu −ft, ut, ut < 0, we know that φut is strictly decreasing. It follows thatutis also strictly decreasing. Thus,utis strictly concave on0, 1. Without loss of generality, we assume thatu0 min{u0, u1}. By the concavity of
u, we know thatut ≥ u0,t ∈ 0,1. So we getu0 01utgtdt ≥ u001gtdt. By 0<01gtdt <1, it is obvious thatu0≥0. Hence,ut≥0,t∈0,1.
Lemma 2.4. Assume that (H1), (H2) hold. Supposeuis a solution of BVP1.1; then
ut 1
1−01grdr
1
0
gr
r
0
φ−1
δ
s
fτ, uτ, uτdτ
dsdr
t
0
φ−1
δ
s
fτ, uτ, uτdτ
ds
2.5
or
ut 1
1−01hrdr
1
0
hr
1
r
φ−1
s
δ
fτ, uτ, uτdτ
dsdr
1
t
φ−1
s
δ
fτ, uτ, uτdτ
ds.
2.6
Proof. First, by integrating1.1on0, t, we have
φutφu0−
t
0
fs, us, usds, 2.7
then
ut φ−1 φu0−
t
0
fs, us, usds
. 2.8
Thus
ut u0
t
0
φ−1
φu0−
s
0
fτ, uτ, uτdτ
ds 2.9
or
ut u1−
1
t
φ−1
φu0−
s
0
fτ, uτ, uτdτ
ds. 2.10
According to the boundary condition, we have
u0 1
1−01grdr
1
0
gr
r
0
φ−1
φu0−
s
0
fτ, uτ, uτdτ
dsdr,
u1 − 1
1−01hrdr
1
0
hr
1
r
φ−1
φu0−
s
0
fτ, uτ, uτdτ
dsdr.
By a similar argument in5,φu0 0δfτ, uτ, uτdτ; then the proof is completed.
Now we define an operatorT by
Tut
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1
1−01grdr
1
0
gr
r
0
φ−1 δ s
fτ, uτ, uτdτ
dsdr
t
0
φ−1 δ s
fτ, uτ, uτdτ
ds, 0≤t≤δ,
1
1−01hrdr
1
0
hr
1
r
φ−1
s
δ
fτ, uτ, uτdτ
dsdr
1
t
φ−1
s
δ
fτ, uτ, uτdτ
ds, δ≤t≤1.
2.12
Lemma 2.5. T:K → Kis completely continuous.
Proof. Letu∈K; then from the definition ofT, we have
Tut
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
φ−1 δ
t
fτ, uτ, uτdτ
≥0, 0≤t≤δ,
−φ−1 t δ
fτ, uτ, uτdτ
≤0, δ≤t≤1.
2.13
So Tut is monotone decreasing continuous and Tuδ 0. Hence, Tut is nonnegative and concave on0, 1. By computation, we can getTu1 01Tuthtdt. This shows that TK ⊂ K. The continuity ofT is obvious since φ−1, f is continuous. Next, we prove thatTis compact onC10,1.
Let D be a bounded subset of K and m > 0 is a constant such that 01fτ, uτ, uτdτ < mforu∈D. From the definition ofT, for anyu∈D, we get
|Tut|<
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
φ−1m
1−01grdr, 0≤t≤δ, φ−1m
1−01hrdr, δ≤t≤1,
Tut< φ−1m, 0≤t≤1.
2.14
Hence,TD is uniformly bounded and equicontinuous. So we have thatTD is compact on
|φTut1−φTut2|< ε. SoφTDis compact onC0,1; it follows thatTDis compact onC0,1. Therefore,TDis compact onC10,1.
Thus,T:K → Kis completely continuous.
It is easy to prove that each fixed point ofTis a solution for BVP1.1.
Lemma 2.6see1. Assume that (H1) holds. Then foru, v∈0,∞,
ψ2−1uv≤φ−1uφv≤ψ1−1uv. 2.15
To obtain positive solution for BVP1.1, the following definitions and fixed point theorems in a cone are very useful.
Definition 2.7. The mapαis said to be a nonnegative continuous concave functional on a cone of a real Banach spaceEprovided thatα:K → 0,∞is continuous and
αtx 1−ty≥tαx 1−tαy 2.16
for allx, y∈Kand 0≤t≤1. Similarly, we say the mapγis a nonnegative continuous convex functional on a cone of a real Banach spaceEprovided thatγ:K → 0,∞is continuous and
γtx 1−ty≤tγx 1−tγy 2.17
for allx, y∈Kand 0≤t≤1.
Letγ and θbe a nonnegative continuous convex functionals onK,αa nonnegative continuous concave functional onK, andψa nonnegative continuous functional onK. Then for positive real numbera, b, c, andd, we define the following convex sets:
Pγ, du∈K|γu< d,
Pγ, α, b, du∈K|αu≥b, γu≤d,
Pγ, θ, α, b, c, du∈K|αu≥b, θu≤c, γu≤d,
Rγ, ψ, a, du∈K|ψu≥a, γu≤d.
2.18
Theorem 2.8see11. LetEbe a real Banach space andK ⊂ Ea cone. Assume thatΩ1 andΩ2 are two bounded open sets inEwith0 ∈Ω1,Ω1 ⊂Ω2. LetT :K∩Ω2\Ω1 → Kbe completely continuous. Suppose that one of following two conditions is satisfied:
1Tu ≤ u,u∈K∩∂Ω1, andTu ≥ u,u∈K∩∂Ω2; 2Tu ≥ u,u∈K∩∂Ω1, andTu ≤ u,u∈K∩∂Ω2.
ThenT has at least one fixed point inΩ2\Ω1.
a nonnegative continuous functional onK satisfyingψλu ≤ λψufor0 ≤ λ ≤ 1, such that for positive numberMandd,
αu≤ψu, u ≤Mγu 2.19
for allu∈Pγ, d. SupposeT :Pγ, d → Pγ, dis completely continuous and there exist positive numbersa, b, andcwitha < bsuch that
S1{u∈Pγ, θ, α, b, c, d|αu> b}/∅andαTu> bforu∈Pγ, θ, α, b, c, d;
S2αTu> bforu∈Pγ, α, b, dwithθTu> c;
S30∈/Rγ, ψ, a, dandψTu< aforu∈Rγ, ψ, a, dwithψu a.
ThenT has at least three fixed pointsu1, u2, u3∈Pγ, d, such that
γui≤dfori1,2,3,
αu1> b,
ψu2> awithαu2< b,
ψu3< a.
3. The Existence of One or Two Positive Solutions
For convenience, we denote
Lmax
⎧ ⎨ ⎩
1
0ψ1−11−sds 1−01gsds ,1
⎫ ⎬
⎭, Nmin 1/2
0
ψ2−1
1 2 −s
ds,
1
1/2
ψ2−1
s−1
2
ds
,
fμ lim sup
ucvc→μ
max t∈0,1
ft, ut, vt
φucvc, fμulim infcvc→μ
min t∈0,1
ft, ut, vt φucvc,
3.1
whereμdenotes 0 or∞.
Theorem 3.1. Assume that (H1) and (H2) hold. In addition, suppose that one of following conditions is satisfied.
iThere exist two constantsr, Rwith0< r <N/LRsuch that
aft, u, v≥φr/Nfort, u, v∈0,1×0, r×−r, rand bft, u, v≤φR/Lfort, u, v∈0,1×0, R×−R, R;
iif∞< ψ11/2L, f0> ψ21/N; iiif0 < ψ11/2L, f
∞> ψ21/N.
Proof. iLetΩ1{u∈K| u1< r},Ω2 {u∈K| u1< R}.
Foru∈∂Ω1, we obtainu∈0, randu ∈−r, r, which impliesft, u, u≥φr/N.
Hence, by2.12andLemma 2.6,
Tucmax 0≤t≤1|Tut|
1 1−01grdr
1
0
gr
r
0
φ−1
δ
s
fτ, uτ, uτdτ
dsdr
δ
0
φ−1
δ
s
fτ, uτ, uτdτ
ds
1 1−01hrdr
1
0
hr
1
r
φ−1
s
δ
fτ, uτ, uτdτ
dsdr
1
δ
φ−1
s
δ
fτ, uτ, uτdτ
ds ≥min ⎧ ⎨ ⎩ 1
1−01grdr
1
0
gr
r
0
φ−1 δ s
fτ, uτ, uτdτ
dsdr
1/2
0
φ−1
1/2
s
fτ, uτ, uτdτ
ds,
1
1−01hrdr
1
0
hr
1
r
φ−1
s
δ
fτ, uτ, uτdτ
dsdr
1
1/2
φ−1
s
1/2
fτ, uτ, uτdτ
ds ≥min 1/2 0
φ−1
1/2
s
fτ, uτ, uτdτ
ds,
1
1/2
φ−1
s
1/2
fτ, uτ, uτdτ
ds ≥min 1/2 0
φ−1
φr N
1
2−s
ds,
1
1/2
φ−1
φ r N
s−1
2 ds ≥ r Nmin 1/2 0
ψ2−1
1 2−s
ds,
1
1/2
ψ2−1
s− 1
2
ds
ru1.
3.2
This implies that
Next, foru∈∂Ω2, we haveft, u, v≤φR/L. Thus, by2.12andLemma 2.6,
Tucmax 0≤t≤1|Tut|
≤ 1
1−01grdr
1
0
gr
1
0
φ−1
1
s
fτ, uτ, uτdτ
dsdr
1
0
φ−1
1
s
fτ, uτ, uτdτ
ds
≤ 1
1−01grdr
1
0
φ−1
1−sφ
R L
ds
≤ R
L
1
0ψ1−11−sds 1−01grdr
≤Ru1.
3.4
From2.13, we have
Tucmax
φ−1 δ 0
fτ, uτ, uτdτ
, φ−1 1 δ
fτ, uτ, uτdτ
≤φ−1
1
0
fτ, uτ, uτdτ
≤φ−1
φ
R L
≤Ru1.
3.5
This implies that
Tu1≤ u1 foru∈∂Ω2. 3.6
Therefore, byTheorem 2.8, it follows thatT has a fixed point inΩ2\Ω1. That is BVP1.1has at least one positive solution such that 0< r ≤ u1≤R.
iiConsideringf∞< ψ11/2L, there existsρ0>0 such that
ft, u, v≤ψ1
1 2L
φucvc fort∈0,1, ucvc≥2ρ0. 3.7
ChoosingM > ρ0such that
maxft, u, v| ucvc≤2ρ0
≤ψ1
1 2L
then for allρ > M, letΩ3{u∈K| u1< ρ}. For everyu∈∂Ω3, we haveucuc≤2ρ. In the following, we consider two cases.
Case 1ucuc≤2ρ0. In this case,
ft, u, u≤ψ1
1 2L
φM≤φ M
2L
≤φρ L
. 3.9
Case 22ρ0≤ ucuc≤2ρ. In this case,
ft, u, u≤ψ1
1 2L
φucuc≤ψ1
1 2L
φ2ρ≤φρ L
. 3.10
Then it is similar to the proof of3.6; we haveTu1≤ u1foru∈∂Ω3.
Next, turning tof0> ψ21/N, there exists 0< ξ < ρsuch that
ft, u, v≥ψ2
1
N
φucvc fort∈0,1, ucvc≤2ξ. 3.11
LetΩ4{u∈K| u1 < ξ}. For everyu∈∂Ω4, we haveucuc≤2ξ. So
ft, u, u≥ψ2
1
N
φucuc≥ψ2
1
N
φu1≥φ
ξ N
. 3.12
Then like in the proof of3.3, we haveTu1≥ u1foru∈∂Ω4. Hence, BVP1.1has at least one positive solution such that 0< ξ≤ u1≤ρ.
iiiThe proof is similar to theiandii; here we omit it.
In the following, we present a result for the existence of at least two positive solutions of BVP1.1.
Theorem 3.2. Assume that (H1) and (H2) hold. In addition, suppose that one of following conditions is satisfied.
If0 < ψ11/2L,f∞< ψ11/2L, and there existsm
1>0such that
ft, u, v≥φm1 N
fort∈0,1, m1≤ ucvc≤2m1; 3.13
IIf0> ψ21/N,f∞> ψ21/N, and there existsm2>0such that
ft, u, v≤φm2 L
fort∈0,1,ucvc≤2m2. 3.14
4. The Existence of Three Positive Solutions
In this section, we impose growth conditions onf which allow us to applyTheorem 2.9of BVP1.1.
Let the nonnegative continuous concave functional α, the nonnegative continuous convex functionalsγ,θ, and nonnegative continuous functionalψbe defined on coneKby
γu max
0≤t≤1u
t, ψu θu max
0≤t≤1|ut|, αu η≤mint≤1−η|ut|. 4.1
By Lemmas2.1and2.2, the functionals defined above satisfy
ηθu≤αu≤ψu θu, u1maxγu, θu≤Mγu, 4.2
for allu∈K. Therefore, the condition2.19ofTheorem 2.9is satisfied.
Theorem 4.1. Assume that (H1) and (H2) hold. Let0< a < b≤dη/11−01htdt/01ht1− tdtand suppose thatfsatisfies the following conditions:
P1ft, u, v≤φdfort, u, v∈0,1×0, Md×−d, d;
P2ft, u, v> φb/ηKfort, u, v∈η,1−η×b,b/η1 1−01htdt/01ht1− tdt×−d, d.
P3ft, u, v< φa/Lfort, u, v∈0,1×0, a×−d, d;
Then BVP1.1has at least three positive solutionsu1, u2, andu3satisfying
max 0≤t≤1u
it≤d fori1,2,3, η≤mint≤1−η|u1t|> b,
max0≤t≤1|u2t|> a with minη≤t≤1−η|u2t|< b, max0≤t≤1|u3t|< a,
4.3
whereLdefined as3.1,Kmin{η1/2ψ2−11/2−sds,1/21−ηψ2−1s−1/2ds}.
Proof. We will show that all the conditions ofTheorem 2.9are satisfied.
If u ∈ Pγ, d, then γu max0≤t≤1|ut| ≤ d. With Lemma 2.2 implying max0≤t≤1|ut| ≤Md, so byP1, we haveft, ut, ut≤φdwhen 0≤t≤1. Thus
γTu max
0≤t≤1
Tut
max
φ−1
δ
0
fτ, uτ, uτdτ
, φ−1
1
δ
fτ, uτ, uτdτ
≤φ−1
1
0
fτ, uτ, uτdτ
≤φ−1φdd.
4.4
To check conditionS1ofTheorem 2.9, we choose
u0t b
η
b1−01htdt
η01ht1−tdt1−
t, 0≤t≤1. 4.5
Let
c b η
⎛
⎝1 1−
1 0htdt
1
0ht1−tdt
⎞
⎠. 4.6
Thenu0t ∈Pγ, θ, α, b, c, dandαu0> b, so{u∈ Pγ, θ, α, b, c, d|αu > b}/∅. Hence, foru∈Pγ, θ, α, b, c, d, there isb≤ut≤c,|ut| ≤dwhenη≤t≤1−η. From assumption P2, we have
ft, ut, ut> φ
b ηK
fort∈η,1−η. 4.7
It is similar to the proof of assumptioniofTheorem 3.1; we can easily get that
αTu min
η≤t≤1−η|Tut| ≥ηmax0≤t≤1|Tut|> b foru∈P
γ, θ, α, b, c, d. 4.8
This shows that conditionS1ofTheorem 2.9is satisfied. Secondly, foru∈Pγ, α, b, dwithθTu> c, we have
αTu≥ηθTu≥ηc > b. 4.9
Thus conditionS2ofTheorem 2.9holds.
Finally, asψ0 0< a, there holds 0∈/Rγ, ψ, a, d. Suppose thatu∈Rγ, ψ, a, dwith
ψu a; then by the assumptionP3,
ft, ut, ut< φa L
fort∈0,1. 4.10
So like in the proof of assumptioniofTheorem 3.1, we can get
ψTu max
0≤t≤1|Tut|< a. 4.11
Thus BVP1.1has at least three positive solutionsu1, u2, andu3satisfying
max 0≤t≤1
uit≤d fori1,2,3, min
η≤t≤1−η|u1t|> b,
max
0≤t≤1|u2t|> a with minη≤t≤1−η|u2t|< b, max0≤t≤1|u3t|< a.
4.12
5. Examples
In this section, we give three examples as applications.
Example 5.1. Letφu |u|u,gt ht 1/2. Now we consider the BVP
φuft, ut, ut0, t∈0,1,
u0 1
2
1
0
utdt, u1 1
2
1
0
utdt,
5.1
whereft, u, v 1t18u4cosvfort, u, v∈0,1×0,∞×−∞,∞.
Letψ1u ψ2u u2,u >0. Choosingr1, R100. By calculations we obtain
L 4
3, N
2 3
1 2
3/2
, φ r
N
18, φ
R L
752. 5.2
Fort, u, v∈0,1×0,1×−1,1,
ft, u, v 1t18u4cosv
≥18×4>18, 5.3
fort, u, v∈0,1×0,100×−100,100,
ft, u, v 1t18u4cosv
≤2×118×5<752.
5.4
Hence, byTheorem 3.1, BVP5.1has at least one positive solution.
Example 5.2. Letφu u,gt ht 1/2. Consider the BVP
φuft, ut, ut0, t∈0,1,
u0 1
2
1
0
utdt, u1 1
2
1
0
utdt,
5.5
whereft, u, v 1t1/10u1/100v21 u
Letψ1u ψ2u u,u >0. ThenL1, N 1/8. It easy to see
f0f∞∞> ψ2
1
N
8. 5.6
Choosingm21/10, fort∈0,1,ucvc≤2m2.
ft, u, v 1t
1 10u
1 100 v
2 1 u
cvc2
!
≤2
1 10
1 5
1 100
1 25
1 1 25
.
39 1250 <
1 10 φ
m
2
L
.
5.7
Hence, byTheorem 3.2, BVP5.5has at least two positive solutions.
Example 5.3. Letφu |u|u,gt ht 1/2; consider the boundary value problem
uuft, ut, ut0, t∈0,1,
u0 u1 1
2
1
0
utdt,
5.8
where
ft, u, v
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
sint
104 2500u 6 1
104
v
105
3
, u≤12,
sint
104 2500·12 6 1
104
v
105
3
, u >12.
5.9
Choosinga1/10,b1,η1/4,d105, then by calculations we obtain that
L 4
3, K
2 3
1 4
3/2
, φ
b ηK
2304, φa L
9
1600. 5.10
It is easy to check that
ft, u, v< φd 1010 for 0≤t≤1, 0≤u≤3·105, −105≤v≤105,
ft, u, v>2304 for 1 4 ≤t≤
3
4, 1≤u≤12, −10
5≤v≤105,
ft, u, v< 9
1600 for 0≤t≤1, 0≤u≤ 1 10, −10
5≤v≤105.
Thus, according toTheorem 4.1, BVP5.8has at least three positive solutionsu1, u2, andu3 satisfying
max 0≤t≤1u
it≤105 fori1,2,3, 1/4min≤t≤3/4|u1t|>1,
max
0≤t≤1|u2t|> 1
10 with1/4min≤t≤3/4|u2t|<1, max0≤t≤1|u3t|< 1 10.
5.12
Acknowledgments
The research was supported by NNSF of China 10871160, the NSF of Gansu Province 0710RJZA103, and Project of NWNU-KJCXGC-3-47.
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