Volume 2010, Article ID 254928,16pages doi:10.1155/2010/254928
Research Article
Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order m-Point Boundary Value Problem
Xuezhe Lv and Minghe Pei
Department of Mathematics, Beihua University, JiLin City 132013, China
Correspondence should be addressed to Minghe Pei,[email protected] Received 25 November 2009; Accepted 10 March 2010
Academic Editor: Ivan T. Kiguradze
Copyrightq2010 X. Lv and M. Pei. 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.
The existence and uniqueness of positive solution is obtained for the singular second-orderm- point boundary value problemut ft, ut 0 fort∈0,1,u0 0,u1 m−2
i1 αiuηi, wherem≥3,αi>0i1,2, . . . , m−2, 0< η1< η2<· · ·< ηm−2<1 are constants, andft, ucan have singularities fort0 and/ort1 and foru0. The main tool is the perturbation technique and Schauder fixed point theorem.
1. Introduction
In this paper, we investigate the existence and uniqueness of positive solution for the singular second-order differential equation
ut ft, ut 0, t∈0,1 1.1
with them-point boundary conditions
u0 0, u1 m−2
i1
αiu ηi
, 1.2
wherem≥3,αi >0 i1,2, . . . , m−2, 0< η1 < η2<· · ·< ηm−2<1 are constants, andft, u can have singularities fort0 and/ort1 and foru0.
Multipoint boundary value problems for second-order ordinary differential equations arise in many areas of applied mathematics and physics; see1–3 and references therein.
The study of three-point boundary value problems for nonlinear second-order ordinary differential equations was initiated by Lomtatidze 4,5. Since then, the nonlinear second- order multipoint boundary value problems have been studied by many authors; see 1–
3,6–29and references therein. Most of all the works in the above mentioned references are nonsingular multipoint boundary value problems; see1–3,10–17,20–23,25,26,28,29, but the works on the singularities have been quite rarely seen; see4–8,18,19,24,27.
Recently, Du and Zhao7, by constructing lower and upper solutions and together with the maximal principle, proved the existence and uniqueness of positive solutions for the following singular second-orderm-point boundary value problem:
ut ft, ut 0, t∈0,1,
u0 0, u1 m−2
i1
αiu ηi
,
1.3
wherem ≥ 3,0 < αi < 1i 1,2, . . . , m−2, 0 < η1 < η2 < · · · < ηm−2 < 1 are constants, m−2
i1 αi<1,ft, uis singular att0,t1 andu0, under conditions that H1ft, u∈C0,1×0,∞,0,∞,andft, uis decreasing inu;
H2ft, λ/≡0,1
0t1−tft, λt1−tdt <∞,for allλ >0.
The purpose of this paper is to establish existence and uniqueness result of positive solution to SBVP1.1,1.2under conditions that are weaker than conditions in7and hence improve the result in7by using perturbation technique and Schauder fixed point theorem 30.
Throughout this paper, we make the following assumptions:
C0αi>0, i1,2, . . . , m−2 andm−2
i1 αi≤1;
C1f :0,1×0,∞ → 0,∞is continuous and nonincreasing inufor each fixed t∈0,1;
C20<1
0s1−sfs, u0ds <∞for each constantu0∈0,∞.
2. Preliminary
We consider the perturbation problems that are given by ut ft, ut 0, t∈0,1,
u0 h, u1 m−2
i1
αiu ηi
1−m−2
i1
αi
h, 2.1h
wherehis any nonnegative constant.
Definition 2.1. For each fixed constanth≥0,a functionutis said to be a positive solution of BVP2.1hifu∈C0,1∩C20,1withut>0 on0,1such thatut ft, ut 0 holds for allt∈0,1andu0 h,u1 m−2
i1 αiuηi 1−m−2
i1 αih.
Lemma 2.2. Assume that conditionsC1andC2are satisfied. Then, for each fixed constantu0>0,
t→lim0t
η1
t
fs, u0ds0, 2.2
t→lim1−1−t t
ηm−2
fs, u0ds0. 2.3
Proof. We only prove2.2. And2.3can be proved similarly.
For each fixed constantu0>0, let
vt t
η1
t
fs, u0ds fort∈ 0, η1
. 2.4
Then from the conditionsC1andC2, we have
0≤vt≤ η1
t
sfs, u0ds≤ η1
0
sfs, u0ds <∞ fort∈ 0, η1
,
vt η1
t
fs, u0ds−tft, u0 for t∈ 0, η1
.
2.5
Hence from the conditionsC1andC2, we have
η1
0
vtdt≤ η1
0
dt
η1
t
fs, u0ds η1
0
tft, u0dt2
η1
0
tft, u0dt <∞. 2.6
This implies thatvt∈L10, η1, and hence for eacht∈0, η1,
t 0
vτdτ t
0
dτ
η1
τ
fs, u0ds− t
0
τfτ, u0dτ t
η1
t
fs, u0dsvt. 2.7
Thus, it follows from the absolute continuity of integral that limt→0vt 0, that is,
t→lim0t
η1
t
fs, u0ds0. 2.8
This completes the proof of the lemma.
In the following discussionGt, sdenotes Green’s function for Dirichlet problem:
−ut 0, t∈0,1,
u0 u1 0. 2.9
Then Green’s functionGt, scan be expressed as follows:
Gt, s
⎧⎨
⎩
1−ts, 0≤s≤t≤1,
1−st, 0≤t≤s≤1. 2.10
It is easy to see that Green’s functionGt, shas the following simple properties:
i0≤t1−ts1−s≤Gt, s≤s1−sfort, s∈0,1×0,1;
iiGt, s>0 fort, s∈0,1×0,1;
iiiG0, s G1, s 0 fors∈0,1.
By direct calculation, we can easily obtain the following result.
Lemma 2.3. Assume that conditions C0, C1, and C2 are satisfied. Then, ut is a positive solution of BVP2.1h h > 0 if and only if u ∈ C0,1 is a solution of the following integral equation:
ut 1
0
Gt, sfs, usds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usdsh, 2.11h
such thatut> h >0 on0,1.
Lemma 2.4. Assume that conditionsC0,C1,andC2are satisfied. Suppose also thatu∈C0,1 is a solution of the following integral equation:
ut 1
0
Gt, sfs, usds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds, 2.12
such thatut>0 on0,1. Then,utis a positive solution of SBVP1.1,1.2.
Proof. Sinceu∈C0,1is a solution of2.12withut>0 on0,1, then for eacht∈0,1,
t 0
s1−tfs, usds <∞, 1
t
t1−sfs, usds <∞. 2.13
So for eacht∈0,1, we have
t 0
sfs, usds <∞, 1
t
1−sfs, usds <∞. 2.14
For convenience, letc:1/1−m−2
i1 αiηim−2
i1 αi
1
0Gηi, sfs, usds. Taket∈0,1and Δtsuch thatt Δt∈0,1,then from the definition of derivative, the mean value theorem of
integral, and the absolute continuity of integral, we have
Δtlim→0
ut Δt−ut
Δt lim
Δt→0
1 Δt
tΔt 0
s1−t−Δtfs, usds 1
tΔt1−st Δtfs, usds
− t
0
s1−tfs, usds− 1
t
t1−sfs, usds
c lim
Δt→0
1 Δt
− t
0
sΔtfs, usds tΔt
t
s1−t−Δtfs, usds
1
tΔt1−sΔtfs, usds− tΔt
t
t1−sfs, usds
c
− t
0
sfs, usdst1−tft, ut 1
t
1−sfs, usds−t1−tft, ut c
− t
0
sfs, usds 1
t
1−sfs, usdsc.
2.15
Hence
ut − t
0
sfs, usds 1
t
1−sfs, usdsc fort∈0,1. 2.16
Consequentlyu∈C0,1.
Again, from the definition of derivative and the mean value theorem of integrals, we have
Δt→lim0
ut Δt−ut Δt lim
Δt→0
1 Δt
− tΔt
0
sfs, usds 1
tΔt1−sfs, usds t
0
sfs, usds− 1
t
1−sfs, usds
lim
Δt→0
1 Δt
− t1
t
sfs, usds− tΔt
t
1−sfs, usds
lim
Δt→0
1 Δt
− tΔt
t
fs, usds
−ft, ut for t∈0,1.
2.17
Henceut −ft, utfort∈0,1. In particular,u∈C0,1.
On the other hand, from2.12, we haveu0 0 and
m−2
i1
αiu ηi
m−2
i1
αi 1 0
G ηi, s
fs, usds ηi
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds
m−2
i1
αi 1 0
G ηi, s
fs, usds
m−2
i1 αiηi
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds
1 1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds u1.
2.18 In summary,utis a positive solution of SBVP1.1,1.2. This completes the proof of the lemma.
Remark 2.5. Assume that all conditions inLemma 2.4hold. Then 1iff ∈C0,1×0,∞,0,∞, we have
u∈C0,1∩C10,1∩C20,1; 2.19
2iff ∈C0,1×0,∞,0,∞, we get
u∈C0,1∩C10,1∩C20,1. 2.20
Lemma 2.6. Assume that conditionsC0,C1,andC2are satisfied. Then, for each constanth >0, BVP2.1hhas a unique solutionut;hwithut;h≥hon0,1.
Proof. We begin by defining an operatorT inDhby
Tut 1
0
Gt, sfs, usds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usdsh, 2.21
whereDh : {u ∈C0,1 :ut ≥hon0,1}is a convex closed set. Then fromLemma 2.2 and the conditionC2, we haveTu∈C0,1andTusatisfies
Tut ft, ut 0, t∈0,1,
Tu0 h, Tu1 m−2
i1
αiTu ηi
1−m−2
i1
αi
h. 2.22
We now apply Schauder fixed point theorem 30to obtain the existence of a fixed point forT. To do this, it suffices to verify thatT is continuous inDhandTDhis a compact set.
Takeu0∈Dh, and let{uk}∞k1⊂Dhsuch that
uk−u0C0,1−→0 ask−→ ∞. 2.23
Then for eacht∈0,1,
ft, ukt−→ft, u0t ask−→ ∞. 2.24
From the definition ofT, we have
Tukt 1
0
Gt, sfs, uksds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, uksdsh. 2.25
Also, from the conditionsC1andC2, we have
ft, u0t ft, ukt≤2ft, h fort∈0,1,
1 0
s1−sfs, hds <∞. 2.26
Thus by Lebesgue-dominated convergence theorem, we have
t∈0,1max|Tukt−Tu0t| ≤ 1
0
Gs, sfs, uks−fs, u0sds
m−2
i1 αi
1−m−2
i1 αiηi 1 0
Gs, sfs, uks−fs, u0sds
1
m−2
i1 αi
1−m−2
i1 αiηi 1 0
s1−sfs, uks−fs, u0sds
−→0 ask−→ ∞.
2.27
Therefore,T :Dh → Dhis continuous.
Next we need to show thatTDhis a relatively compact subset ofC0,1.
1From the definition ofTand the conditionsC1andC2, for eachu∈Dhwe have
0< h≤Tut≤Tht fort∈0,1. 2.28
This implies thatTDhis uniformly bounded.
2For eachu∈Dh, since
Tut − t
0
sfs, usds 1
t
1−sfs, usds 1
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds fort∈0,1,
2.29
then
Tut≤ t
0
sfs, hds 1
t
1−sfs, hds 1
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, hds :Mt fort∈0,1.
2.30
ObviouslyMt≥0 on0,1, and
1 0
Mtdt2
1 0
s1−sfs, hds 1
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, hds
≤2
1 0
s1−sfs, hds 1
1−m−2
i1 αiηi m−2
i1
αi 1 0
s1−sfs, hds
2
m−2
i1 αi
1−m−2
i1 αiηi 1 0
s1−sfs, hds <∞.
2.31
ThusM ∈L10,1. From the absolute continuity of integral, we have that for each number ε > 0,there is a positive number δ > 0 such that for allt1, t2 ∈ 0,1,if |t1 −t2| < δ, then
|t2
t1Mtdt|< ε. It follows that for allt1, t2∈0,1with|t1−t2|< δ, we have
|Tut2−Tut1|
t2
t1
Tutdt ≤
t2
t1
Tutdt ≤
t2
t1
Mtdt
< ε. 2.32 ThereforeTDhis equicontinuous on0,1. It follows from Ascoli-Arzela theorem thatTDh is a relatively compact subset ofC0,1. Consequently, by Schauder fixed point theorem30, T has a fixed pointut;h∈Dh. Obviously,ut;h> h >0 on0,1. Hence fromLemma 2.3, ut;his a solution of BVP2.1h.
Next, we will show the uniqueness of solution. Let us suppose thatu1t;h,u2t;hare two different solutions of BVP2.1h. Then there existst0∈0,1such thatu1t0;h/u2t0;h.
Without loss of generality, assume thatu1t0;h> u2t0;h.Letwt:u1t;h−u2t;h,then w0 0, wt0>0,and hence there existst1∈0, t0such that
wt1 0, wt>0 fort∈t1, t0. 2.33
Further we havewt>0 ont1,1. In fact, assume to the contrary that the conclusion is false.
Then there existst2∈t0,1such thatwt2≤0.Thus there existst3 ∈t0, t2such that
wt3 0, wt>0 fort∈t0, t3. 2.34
Sincewt1 0,wt>0 ont1, t0, then
wt −ft, u1t;h ft, u2t;h≥0 for t∈t1, t3. 2.35
It follows fromwt1 wt3 0 thatwt≤0 ont1, t3. This is a contradiction towt>0 ont1, t3.
Now we prove that wt ≥ 0 on 0, t1. In fact, assume to the contrary that the conclusion is false. Then there existst4 ∈0, t1such thatwt4<0. Sincew0 wt1 0, then there existt5, t6with 0≤t5 < t4< t6≤t1such that
wt5 wt6 0, wt<0 fort∈t5, t6. 2.36
Thus,
wt −ft, u1t;h ft, u2t;h≤0 for t∈t5, t6. 2.37
It follows fromwt5 wt6thatwt≥ 0 ont5, t6. This is a contradiction towt< 0 on t5, t6.
In summary, we havewt≥0 on0, t1andwt>0 ont1,1. Thus
wt 1
0
Gt, s
fs, u1s;h−fs, u2s;h ds
t 1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, u1s;h−fs, u2s;h ds
≤0 fort∈0,1.
2.38
This is a contradiction towt>0 ont1,1. This completes the proof of the lemma.
Lemma 2.7. Assume that conditionsC0,C1,and C2are satisfied. Then, the unique solution ut;hof BVP2.1his nondecreasing inh.
Proof. Let 0< h2 < h1, and letut;h1, ut;h2be the solutions of BVP2.1h1 and BVP2.1h2, respectively. We will show
ut;h1≥ut;h2 fort∈0,1. 2.39
Assume to the contrary that the above inequality is false. Then there existst0∈0,1such that ut0;h1< ut0;h2. Sinceu0;h1 h1 > h2 u0;h2, we have that there existst1 ∈0, t0 such that
ut1;h1 ut1;h2, ut;h1< ut;h2 fort∈t1, t0. 2.40
Next we prove ut;h1 < ut;h2 ont0,1.In fact, assume to the contrary that the conclusion is false. Then there existst2∈t0,1such that
ut2;h1 ut2;h2, ut;h1< ut;h2 fort∈t0, t2. 2.41
Hence
ut;h1−ut;h2 −ft, ut;h1 ft, ut;h2≤0 fort∈t1, t2. 2.42
It follows from uti;h1 uti;h2, i 1,2 that ut;h1 ≥ ut;h2 on t1, t2. This is a contradiction tout;h1 < ut;h2ont1, t2. Thusut;h1 < ut;h2ont1,1. This implies that
ut;h1−ut;h2 −ft, ut;h1 ft, ut;h2≤0 fort∈t1,1. 2.43
It follows from ut1;h1−ut1;h2 ≤ 0 thatut;h1−ut;h2 ≤ 0 ont1,1. Hence, from ut;h1< ut;h2ont1,1, we haveu1;h1−u1;h2<0.Thus
u1;h1−u1;h2< u ηm−2;h1
−u ηm−2;h2
. 2.44
There are two cases to consider.
Case 1seet1≥ηm−2. In this case, we have
u ηi;h1
−u ηi;h2
≥0, i1,2, . . . , m−2. 2.45
Hence from the boundary conditions of BVP2.1h, we have
u1;h1−u1;h2 m−2
i1
αiu ηi;h1
1−m−2
i1
αi
h1
−m−2
i1
αiu ηi;h2
−
1−m−2
i1
αi
h2
≥m−2
i1
αi
u ηi;h1
−u ηi;h2
≥0.
2.46
This is a contradiction tou1;h1−u1;h2<0.
Case 2seet1< ηm−2. In this case, we have u1;h1−u1;h2< u
ηm−2;h1
−u ηm−2;h2
<0,
u ηm−2;h1
−u ηm−2;h2
≤u ηi;h1
−u ηi;h2
, i1,2, . . . , m−3. 2.47
It follows fromC0that
u1;h1−u1;h2<
m−2
i1
αi
u ηm−2;h1
−u ηm−2;h2
≤m−2
i1
αi
u ηi;h1
−u ηi;h2
. 2.48
This is a contradiction to the boundary conditions of BVP2.1h.
In summary, we have ut;h1 ≥ ut;h2 on 0,1. This completes the proof of the lemma.
3. Main Results
We now state and prove our main results for singular second-orderm-point boundary value problem1.1,1.2.
Theorem 3.1. Assume that conditionsC0,C1,andC2are satisfied. Then, SBVP1.1,1.2has at most one positive solution.
Proof. Suppose thatu1tandu2tare any two positive solutions of SBVP1.1,1.2. We now prove thatu1t≡u2ton0,1. To do this, letvt u1t−u2ton0,1. We will show that vt≡0 on0,1. There are three cases to consider.
Case 1seev1 > 0. In this case, we have that vt ≥ 0 on0,1. In fact, assume to the contrary that the conclusion is false. Then, there existst0 ∈ 0,1such that vt0 < 0. Since v0 0 andv1>0, then there existt1, t2 ∈0,1witht1 < t0< t2such that
vt<0 on t1, t2, vt1 vt2 0. 3.1
Thus
vt u1t−u2t −ft, u1t ft, u2t≤0 fort∈t1, t2. 3.2 Hencevt≥0 ont1, t2, which is a contradiction tovt<0 ont1, t2. Thereforevt≥0 on 0,1. Consequently
vt −ft, u1t ft, u2t≥0 fort∈0,1. 3.3 Thusvtis convex on0,1. Sincev1>0 and
v1 u11−u21 m−2
i1
αiu1
ηi
−m−2
i1
αiu2
ηi
m−2
i1
αiv ηi
, 3.4
then there existsi0 ∈ {1,2, . . . , m−2}such that v
ηi0
max v
ηi
:i1,2, . . . , m−2
>0, 3.5
and hence fromC0and 0< ηi0<1, we have
v1≤m−2
i1
αiv ηi0
≤v ηi0
< 1 ηi0
v ηi0
, 3.6
which is a contradiction to thatvtis convex on0,1.
Case 2seev1 0. In this case, we have that vt ≡ 0 on0,1. In fact, assume to the contrary that the conclusion is false. Then, there existst0 ∈0,1such thatvt0/0. We may assume without loss of generality that vt0 > 0. Then from v0 v1 0, there exist t1, t2∈0,1witht1< t0< t2such that
vt>0 ont1, t2, vt1 vt2 0. 3.7
Thus
vt −ft, u1t ft, u2t≥0 for t∈t1, t2. 3.8
Sincevt1 vt2 0, then
vt≤0 for t∈t1, t2, 3.9
which is a contradiction to thatvt>0 ont1, t2.
Case 3seev1 < 0. In this case, similar to the proof of Case1we can easily show that vt≤0 on0,1. Consequently
vt −ft, u1t ft, u2t≤0 fort∈0,1. 3.10
Thusvtis concave on0,1. Sincev1 m−2
i1 αivηi<0, then there existsi1∈ {1,2, . . . , m−
2}such thatvηi1 min{vηi:i1,2, . . . , m−2}<0, and hence from 0< ηi1<1, we have
v1≥m−2
i1
αiv ηi1
≥v ηi1
> 1 ηi1
v ηi1
, 3.11
which is a contradiction to thatvtis concave on0,1.
In summary,vt≡0 on0,1, that is,u1t≡u2ton0,1. This completes the proof of the theorem.
Theorem 3.2. Assume that conditionsC0,C1,andC2are satisfied. Then SBVP1.1,1.2has exactly one positive solution.
Proof. The uniqueness of positive solution to SBVP1.1, 1.2 follows from Theorem 3.1 immediately. Thus we only need to show the existence.
Let {hj}∞j1 be a decreasing sequence that converges to the number 0. Then from Lemma 2.6, BVP2.1hj has a unique solutionut;hj: ujt. FromLemma 2.7and2.11h, we have that for eachj < k,
0≤ujt−ukt≤hj−hk fort∈0,1. 3.12
Thus there existsu∈C0,1such that
jlim→ ∞ujt ut≥0, uniformly on0,1. 3.13
It is easy to see thatutsatisfies boundary conditions1.2.
Now we prove that
ut>0 fort∈0,1. 3.14
At first, we prove that
u ηi0
max u
ηi
:i1,2, . . . , m−2
>0, 3.15
wherei0 ∈ {1,2, . . . , m−2}. In fact, assume to the contrary that the conclusion is false. Then
u1 m−2
i1
αiu ηi
0. 3.16
From the fact that each function in the sequence {uj}∞j1 is concave, we have that ut is concave. It follows fromu0 uηi0 u1 0 thatut ≡ 0 on0,1. Thus whenj is large enough,ujtis small enough such thatujt≤h1on0,1. Hence from conditionC1, we have
uj
ηi0
1
0
G ηi0, s
f
s, ujs ds
ηi0
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
f
s, ujs dshj
>
1 0
G ηi0, s
fs, h1ds >0.
3.17
Letj → ∞, we have
u ηi0
≥ 1
0
G ηi0, s
fs, h1ds >0. 3.18
This is a contradiction touηi0 0. Thusuηi0>0, and henceu1>0. Sinceutis concave, thenut>0 on0,1. Since
ujt 1
0
Gt, sf
s, ujs
ds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
f
s, ujs
dshj, 3.19
then passing to the limit, by Monotone convergence theorem31, we have
ut 1
0
Gt, sfs, usds t
1−m−2
i1 αiηi m−2
i1
αi 1 0
G ηi, s
fs, usds. 3.20
Therefore byLemma 2.4,utis a positive solution of SBVP1.1,1.2. This completes the proof of the theorem.
Finally, we give an example to which our results can be applicable.
Example 3.3. Consider the singular nonlinear second-orderm-point boundary value problem:
u 1
tβ11−tβ2u2−β1 0, t∈0,1,
u0 0, u1 m−2
i1
αiu ηi
,
3.21
wherem ≥ 3, 0 < η1 < η2 < · · · < ηm−2 < 1,αi > 0 i 1,2, . . . , m−2,m−2
i1 αi ≤ 1,and β1, β2∈0,2.
Let
ft, u 1
tβ11−tβ2u2−β1 fort, u∈0,1×0,∞. 3.22
Obviously, the functionft, uis singular att0,1 andu0. It is easy to verify thatft, u satisfies conditionsC1andC2. So fromTheorem 3.2, SBVP3.21has exactly one positive solution. However, we note that Theorem 2 in7cannot guarantee that SBVP3.21 has a unique positive solution, since
1 0
t1−tft, λt1−tdt ∞ forλ >0. 3.23
Acknowledgment
The authors thank the referee for valuable suggestions which led to improvement of the original manuscript.
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