Boundary Value Problems
Volume 2011, Article ID 901796,18pages doi:10.1155/2011/901796
Research Article
Existence Results of Three-Point Boundary Value Problems for Second-order Ordinary Differential Equations
Sheng-Ping Wang
1and Long-Yi Tsai
21Holistic Education Center, Cardinal Tien College of Healthcare and Management, No.11, Zhongxing Road, Sanxing Township, Yilan County 26646, Taiwan
2Department of Mathematical Sciences, National ChengChi University, Taipei 11605, Taiwan
Correspondence should be addressed to Sheng-Ping Wang,[email protected] Received 19 May 2010; Revised 15 September 2010; Accepted 24 September 2010
Academic Editor: Daniel Franco
Copyrightq2011 S.-P. Wang and L.-Y. Tsai. 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 establish existence results of the following three-point boundary value problems: ut ft, ut 0,t ∈ 0,1,BCu0 0 andu1 δuη, where 0 < η < 1 and 0 < δ ≤1. The approach applied in this paper is upper and lower solution method associated with basic degree theory or Schauder’s fixed point theorem. We deal with this problem with the functionfwhich is Carath´eodory or singular on its domain.
1. Introduction
In this paper, we consider three-point boundary value problem
ut ft, ut 0, t∈0,1, 1.1
u0 0, u1 δu
η
, 1.2
where 0< η <1 and 0< δ ≤1.
In the mathematical literature, a number of works have appeared on nonlocal boundary value problems, and one of the first of these was1. Il’in and Moiseev initiated the research of multipoint boundary value problems for second-order linear ordinary differential equations, see2,3, motivated by the study4–6of Bitsadze and Samarskii.
Recently, nonlinear multipoint boundary value problems have been receiving considerable attention, and have been studied extensively by using iteration scheme
e.g., 7, fixed point theorems in cones e.g., 8, and the Leray-Schauder continuation theoreme.g., 9. We refer more detailed treatment to more interesting research 10,11 and the references therein.
The theory of upper and lower solutions is also a powerful tool in studying boundary value problems. For the existence results of two-point boundary value problem, there already are lots of interesting works by applying this essential techniquesee12,13. Recently, it is shown that this method plays an important role in proving the existence of solutions for three-point boundary value problemssee14–16.
Last but not least, as the singular source term appearing in two-point problems, singular three-point boundary value problems also attract more attentione.g.,17.
In this paper, we will discuss the existence of solutions of some general types on three- point boundary value problems by using upper and lower solution method associated with basic degree theory or Schauder’s fixed point theorem.
This paper is organized as follows. InSection 2, we give two lemmas which will be extensively used later. InSection 3, when the source termfis a Carath´eodory function, we consider the Sobolev spaceW2,10,1defined by
W2,10,1:
u∈C10,1|u∈L10,1
, 1.3
and obtain the existence ofW2,1-solution in Theorems3.5and3.11. InSection 4, we discuss the singular case, that is,fmaybe singular at the end pointst0 ort1, or atu0. We will introduce theA-class of functions and another spaceW2,Asee18,19as follows:
A:
h∈L1loc0,1|s1−shs∈L10,1 , W2,A0,1:
u∈W1,10,1|u∈ A ,
1.4
and prove the existence ofW2,A-solution in Theorems4.1and4.4. Some sufficient conditions for constructing upper and lower solutions are given in each section for applications.
2. Preliminaries
DefineG:0,1×0,1 → −∞,∞by Gt, s: 1
1−δηt1−s−Ut, s− δ
1−δηVt, s, 0≤t, s≤1, 2.1 whereδandηare given as1.2and
Ut, s
⎧⎨
⎩
t−s, s≤t, 0, t≤s,
Vt, s
⎧⎨
⎩ t
η−s
, s≤η,
0, η≤s.
2.2
By direct computations, we get the following results.
Lemma 2.1. iThe functionG:0,1×0,1 → −∞,∞defined by2.1, is the Green function corresponding for the problem
ut 0,
u0 0, u1 δu
η
. 2.3
iiThe functionG:0,1×0,1 → −∞,∞defined by2.1, is continuous.
iiiIn the case 0< δη <1, we have
Q−1:max
0≤t≤1 1 0
Gt, sds
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎩ 1 8
1−δη2 1−δη
2 , δη
2−η
≤1, δη
1−η 2
1−δη, δη 2−η
≥1.
2.4
Lemma 2.2. Ifh∈ A, then the problem
ut ht 0 2.5
with boundary condition1.2has a unique solutionu∈W2,A0,1such that
ut 1
0
Gt, shsds, 2.6
whereGt, sis defined by2.1.
3. Carath ´eodory Case
In this section we first introduce the Carath´eodory function as follows.
Definition 3.1. A functionft, udefined onE ⊆a, b×Ris called a Carath´eodory function onEif
ifor almost everyt∈a, b, ft,·is continuous onR;
iifor anyu∈R,the functionf·, uis measurable ona, b;
iiifor anyr >0, there existshr ∈La, bsuch that for anyu ∈−r, rand for almost everyt∈a, bwitht, u∈E, we have|ft, u| ≤hrt.
We in this section assume thatfis a Carath´eodory function and discuss the existence ofW2,1-solution by assuming the existence of upper and lower solutions.
3.1. Existence ofW2,1-Solutions
We first introduce the definitions ofW2,1-upper and lower solutions as below.
Definition 3.2. A functionα∈C0,1is called aW2,1-lower solution of problem1.1and1.2 if it satisfies
iα0≤0,α1≤δαη, and
iifor any t0 ∈ 0,1, eitherD−αt0 < Dαt0, or there exists an open intervalI0 ⊆ 0,1containingt0such thatα∈W2,1I0and, for almost everyt∈I0, we have
αt ft, αt≥0. 3.1
Definition 3.3. A functionβ∈C0,1is called aW2,1-upper solution of problem1.1and1.2 if it satisfies
iβ0≥0,β1≥δβη, and
iifor any t0 ∈ 0,1, eitherD−βt0 > Dβt0, or there exists an open intervalI0 ⊆ 0,1containingt0such thatβ∈W2,1I0and, for almost everyt∈I0, we have
βt f t, βt
≤0. 3.2
Before proving our main results, we first consider such a modified problem given as follows:
ut f
t, γt, ut
γt, ut−ut
1|ut| 0, t∈0,1, 3.3
with boundary condition1.2, whereγ:0,1×R → Ris defined by
γt, u
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
αt ifu < αt, u ifαt≤u≤βt, βt ifu > βt.
3.4
Proposition 3.4. Letαtand βtbe respectiveW2,1-lower and upper solutions of problem1.1 and1.2withαt≤βton0,1. Ifu∈W2,10,1is a solution of problem3.3and1.2, then αt≤ut≤βt, for anyt∈0,1.
Proof. Suppose there existst0∈0,1such that
t∈0,1minut−αt ut0−αt0<0. 3.5
Case 1. Ift0 ∈ 0,1, we haveut0−D−αt0 ≤ ut0−Dαt0, which impliesD−αt0 ≥ Dαt0.Hence, byDefinition 3.2and the continuity ofu−αatt0, there exist an open interval I0 ⊆0,1witht0∈I0,α∈W2,1I0and a neighborhoodNoft0contained inI0such that for almost everyt∈I0∩N,
αt ft, αt≥0. 3.6
Furthermore, it follows fromut0−αt0 0 that fort≥t0,t∈N, we have
ut−αt t
t0
us−αs ds
≤ t
t0
−f
s, γs, us
−γs, us−us
1|us| fs, αs
ds t
t0
−fs, αs−αs−us
1|us| fs, αs
ds
<0.
3.7
This implies that the minimum ofu−αcannot occur att0, a contradiction.
Case 2. Ift00, by the definition ofW2,1-lower solutionα0≤0, we then have
0u0≤u0−α0<0. 3.8
And we get a contradiction.
Case 3. Ift01, it follows from the conclusion of Case1that u1−α1≥δ
u η
−α η
> δu1−α1≥u1−α1, 3.9
which is impossible.
Consequently, we obtainαt≤ uton0,1. By the similar arguments as above, we also have
ut≤βt, on0,1. 3.10
Theorem 3.5. LetαtandβtbeW2,1-lower and upper solutions of problem1.1and1.2such thatαt≤βton0,1and letfbe a Carath´eodory function onE, where
E:
t, u∈0,1×R|αt≤u≤βt, t∈0,1
. 3.11
Then problem1.1and1.2has at least one solutionu∈W2,10,1such that, for allt∈0,1,
αt≤ut≤βt. 3.12
Proof. We consider the modified problem3.3and1.2with respect to the givenαtand βt. Consider the Banach space C0,1 with supremum and the operator T : C0,1 → C0,1by
Tut: 1
0
Gt, s
f
s, γs, us
γs, us−us 1|us|
ds, 3.13
foru∈C0,1, whereGt, sis defined as2.1. Sincefis a Carath´eodory function onE, for almost allt∈0,1and for allx∈αt, βt, there exists a functionh∈L0,1, we have
ft, u≤ht. 3.14
Define
K:{u∈C0,1| u ≤M}, 3.15
where
M:max
t∈0,1 1 0
|Gt, s|hs M1ds <∞, 3.16
M1: max
t,u∈0,1×R
γt, u−u
1|u| . 3.17
It is clear thatKis a closed, bounded and convex set inC0,1and one can show thatT :K → Kis a completely continuous mapping by Arzel`a-Ascoli theorem and Lebesgue dominated convergence theorem. By applying Schauder’s fixed point theorem, we obtain thatT has a fixed point inK which is a solution of problem3.3and 1.2. From Proposition 3.4, this fixed point is also a solution of problem1.1and1.2. Hence, we complete the proof.
We further illustrate the use ofTheorem 3.11in the following second-order differential equation:
ut ft, ut ht 0 3.18 with the boundary condition1.2.
Corollary 3.6. Assume that f : 0,1×R → Ris a Carath´eodory function satisfying ft,u/u is essentially bounded for|u| ≥M, whereMis a constant large enough. Assume further thath∈L0,1 and there exists a constant 0<√
A < π/2 such that lim sup
|u| → ∞ max
t∈0,1
ft, u
u ≤A. 3.19
Then, problem3.18and1.2has at least one solution.
Proof. By hypothesis, for any given > 0 small enough such that√
A ≤ π/2 and for almost allt∈0,1, for anyularge enough, we have
ft, u≤Au. 3.20
We now choose an upper solutionβtof the form
βt wt sψt≥0. 3.21
To this end, we compute βf
t, β
ht≤β Aβht w Awht s
ψ Aψ
. 3.22
Clearly, one can choosewsuch that
w Awht 0,
w0 0, w1 δw
η
, 3.23
that is,
wt sin√
At δsin√
Aη
−sin√
A 1
0
sin√
As−1
√A hsds
δsin√ At δsin√
Aη
−sin√
A η
0
sin√ A
η−s
√A hsds
t
0
sin√
As−t
√A hsds,
3.24
and chooseψt lsin√
At, wherel >0, which is a positive solution of ψ Aψ0,
ψ0 0.
3.25
Hence, ifsis large enough, we can show thatβ0 0 andβ1≥δβη, whereδ≤1, which implies thatβtis a positiveW2,1-upper solution. In the same way we construct aW2,1-lower solutionαwt−sψ ≤0 on0,1.
3.2. Nontangency Solution
In this subsection, we afford another stronger lower and upper solutions to get a strict inequality of the solution between them.
Definition 3.7. A functionα∈C0,1is a strictW2,1-lower solution of problem1.1and1.2, if it is not a solution of problem1.1and1.2,α0<0,α1≤δαηand for anyt0 ∈0,1, one of the following is satisfied:
iD−αt0< Dαt0;
iithere exist an intervalI0 ⊆ 0,1and >0 such thatt0 ∈intI0,α ∈W2,1I0and for almost everyt∈I0, for allu∈αt, αt we have
αt ft, u≥0. 3.26
Definition 3.8. A functionβ∈C0,1is a strictW2,1-upper solution of problem1.1and1.2, if it is not a solution of problem1.1and1.2,β0>0,β1≥δβηand for anyt0 ∈0,1, one of the following is satisfied:
iD−βt0> Dβt0,
iithere exist an intervalI0 ⊆ 0,1and >0 such thatt0 ∈intI0,β ∈W2,1I0and for almost everyt∈I0, for allu∈βt−, βtwe have
βt ft, u≤0. 3.27
Remark 3.9. Every strict W2,1-lowerupper solution of problem 1.1 and 1.2 is a W2,1- loweruppersolution.
Now we are going to show that the solution curve of problem1.1and1.2cannot be tangent to upper or lower solutions from below or above.
Proposition 3.10. Letαtandβtbe respective strictW2,1-lower and upper solutions of problem 1.1and1.2withαt≤βton0,1. Ifu∈W2,10,1is a solution of problem1.1and1.2 withα≤u≤βon0,1, thenαt< ut< βt, for anyt∈0,1.
Proof. Asαis not a solution,uis not identical toα. Assume, the conclusion does not hold, then
t0:inf{t∈0,1|ut αt} 3.28
exists. Hence,u−αhas minimum att0, that is,ut0−αt0 0.
Case 1. Sett0∈0,1. Sinceu−αhas minimum att0, we haveD−αt0≥Dαt0. According to theDefinition 3.7, there existI0,0>0 andt1 ∈I0witht1< t0such that, for everyt∈t1, t0, ut≤αt 0,ut1−αt1<0 and for a.e.t∈t1, t0
αt ft, ut≥0. 3.29
Hence, we have the contradiction since 0<
u−α t0−
u−α
t1 − t0
t1
ft, ut αt
dt≤0. 3.30
Case 2. Ift00, by the definition of strictW2,1-lower solution thatα0<0, we then have
0u0−α0>0. 3.31
And we get a contradiction.
Case 3. If t0 1, repeat the same arguments in Case 3 of the proof of Proposition 3.4.
Therefore, we obtainαt< uton0,1. The inequalityut< βton0,1can be proved by the similar arguments as above.
Theorem 3.11. Letαtandβtbe strictW2,1-lower and upper solutions of problem1.1and1.2 such thatαt< βton0,1and letf:E → Rbe a Carath´eodory function, where
E:
t, u∈0,1×R|αt≤ut≤βt, t∈0,1
. 3.32
Then, problem1.1and1.2has at least one solutionu∈W2,10,1such that, for anyt∈0,1,
αt< ut< βt. 3.33
Proof. This is a consequence ofTheorem 3.5andProposition 3.10and hence, we omits this proof.
4. Singular Case
In this section we give a more general existence result thanTheorem 3.11by assuming the existence ofW2,1-lower and upper solutions. This makes us to deal with problem1.1and 1.2, where the functionfis singular at the end pointt0 andt1.
Theorem 4.1. LetαtandβtbeW2,1-lower and upper solutions of problem1.1and1.2such thatαt≤βton0,1and letf:0,1×R → Rsatisfy the following conditions:
ifor almost everyt∈0,1, ft,·is continuous onR;
iifor anyu∈R,the functionf·, uis measurable on0,1;
iiithere exists a functionh∈ Asuch that, for allt, u∈E,
ft, u≤ht, 4.1
where
E:
t, u|t∈0,1, αt≤ut≤βt
⊆0,1×R. 4.2
Then problem1.1and1.2has at least one solutionu∈W2,A0,1such that, for allt∈0,1,
αt≤ut≤βt. 4.3
Proof. Consider the modified problem3.3and1.2with respect to the givenαtandβt and defineT :C0,1 → C0,1by3.13. Note that byLemma 2.2,Tis well defined. Define
P :{u∈C0,1| u ≤N}, 4.4
where
N: max
t∈0,1 1 0
|Gt, s|hs M1ds <∞, 4.5
andM1is defined by3.17. The rest arguments are similar to the proof ofTheorem 3.5.
Remark 4.2. We have similar results of Theorems3.5–4.1, respectively, for1.1equipped with
u0 A, u1 δu
η
, 4.6
whereA∈Ris a constant andδ,ηare given as1.2.
Example 4.3. Consider the problem4.7, for 0< α <1, 0< βi<2−2α,i1,2,
ut 1
tβ11−tβ2utα10, 0< t <1,
u0 0, u
1 2
u1.
4.7
Clearly, 0 is aW2,1-lower solution of4.7and
Gt, s:2t1−s−Ut, s−2Vt, s, 0≤t, s≤1, 4.8
where
Ut, s
⎧⎨
⎩
t−s, s≤t, 0, t≤s,
Vt, s
⎧⎪
⎨
⎪⎩ t
1 2 −s
, s≤ 1 2,
0, 1
2 ≤s.
4.9
From Lemma 2.1, we have max0≤t≤11
0Gt, sds 9/32 and defineh1t : 1/tβ1/1−α1 − tβ2/1−α. Since, for 2−βi/1−α>0,i1,2,
1 0
t1−th1tdt 1
0
t2−β1/1−α−11−t2−β2/1−α−1 <∞, 4.10
that is, h1t ∈ A, we have, from Lemma 2.2, 1
0Gt, s1 − αh1sds ∈ W2,A and max0≤t≤11
0Gt, s1−αh1sdsexists. Let
B: 1
1−9/32αmax
0≤t≤1 1 0
Gt, s1−αh1sds1 4.11
and, byLemma 2.2again, chooseβsuch that
βt αB 1−αh1t 10,
β0 0, β1 β
1 2
.
4.12
Note that according to the direct computation, we see thatβis well-defined and is bounded byB. Next, letft, u: 1/tβ11−tβ2uα1. By Young’s inequality, it follows that
βt f t, βt
≤βt αβt 1−α 1
tβ1/1−α1−tβ2/1−α 1 βt αβt 1−αh1t 1.
≤0.
4.13
Hence, suchβtis aW2,1-upper solution of4.7andβt≥0 on0,1. Clearly,fsatisfiesi, iiofTheorem 4.1. By using Young’s inequality again, fort, u∈E:{t, u|t∈0,1,0 ≤ ut≤βt} ⊆0,1×R., we have
ft, u≤αuh1t
≤αBh1t:h2t. 4.14
andh2t∈ A. Therefore,f satisfies the assumptioniiiofTheorem 4.1. Consequently, we conclude that this problem has at least one solutionu∈W2,A0,1such that, for allt∈0,1,
0≤ut≤βt. 4.15
Notice that inTheorem 4.1, one can only deal with the case thatf is singular at end points t 0,t 1. However, when f is singular at u 0, there is no hope to obtain the solutions directly fromTheorem 4.1. We will establish the following theorem to deal with this case by constructing upper and lower solutions to solve this problem.
Theorem 4.4. Assume
H1the functionf:0,1×R → Ris continuous;
H2there existsk > πand for any compact setK⊆0,1, there is >0 such that
ft, u≥k2u, ∀t∈K, u∈0, ; 4.16
H3for someM >0 and 0< γ <
Q, there ish∈ A ∩C0,1such that
ft, u≤γ2uht, ∀t∈0,1, u∈M,∞; 4.17
whereQ−1is defined as inLemma 2.1.
H4for any compact setK⊆0,∞, there ishK∈ Asuch that
ft, u≤hKt, ∀t∈0,1, u∈K. 4.18
Then problem1.1and1.2withδ1 has at least one solution
u∈C0,1,R∪ {0}∩C20,1,R. 4.19
Remark 4.5 see12, Remark 3.1. Assumption H2 is equivalent to the assumption that there existsk > πand a functiona1∈C020,1,Rsuch that:
ia1>0 for allt∈0,1,
iift, u≥k2u, for allt∈0,1, 0< u≤a1t, iiia1t>0, for allt∈0,1/3∪2/3,1, where
C200,1,R:
u∈C20,1,R|u0 u1 0
. 4.20
Proof.
Step 1. Construction of lower solutions. Considerk2 such thatπ < k2 < mink,3πand the function
α2t A2cosk2
t−1
2
, 4.21
whereA2is chosen small enough so that ft, u≥k2u, ∀t∈
1 2 − π
2k2,1 2 π
2k2
, 0< u≤α2t. 4.22
Next, we choosea1from theRemark 4.5, and let
α1t A1a1t, 4.23
whereA1∈0,1is small enough so that for some pointst1∈0,1/3,t2∈2/3,1, we have:
α1t≥α2t, ∀t∈0, t1∪t2,1, 4.24
α2t≥α1t, ∀t∈t1, t2. 4.25
Notice that by4.24and4.25, for anyh:0,1×R → Rsuch that
ht, u≥ft, u, for anyt, u∈0,1×R, 4.26
we have:
α1t ht, α1t≥α1t k2α1t>0, for anyt∈0, t1∪t2,1, 4.27 α2t ht, α2t≥ −k22α2t k2α2t>0, for any t∈t1, t2. 4.28 Step 2. Approximation problems. We define for eachn∈N,n≥1,
ηnt max 1
2n1,min
t,1− 1 2n1
, t∈0,1 4.29
and set
fnt, u max f
ηnt, u
, ft, u
. 4.30
We have that, for each indexn,fn:0,1×R → Ris continuous and
fnt, u ft, u, for any t, u∈Kn×R, 4.31
where
Kn 1
2n1,1− 1 2n1
. 4.32
Hence, the sequence of functions{fn}converges tofuniformly on any setK×R, whereK is an arbitrary compact subset of0,1. Next we define
fnt, u min
f1t, u, . . . ,fnt, u
. 4.33
Each of the functionsfiis a continuous function defined on0,1×R, moreover
f1t, u≥f2t, u≥ · · · ≥fnt, u≥fn1t, u≥ · · · ≥ft, u 4.34 and the sequence{fn}converges tofuniformly on the compact subsets of0,1×Rsince
fnt, u ft, u, ∀t∈Kn, u∈R. 4.35 Define now a decreasing sequence{n} ⊆Rsuch that
nlim→ ∞n0,
ft, u≥k2u, ∀t∈Kn, u∈0, n, 4.36 and consider a sequence of the following approximation problems:
ut fnt, ut 0, BCu0 n, u1 u
η
, Pn
where 0< η <1.
Step 3. A lower solution ofPn. It is clear that for anyc∈0, n, fnt, c≥f
ηnt, c
≥k2c >0. 4.37
As the sequence{n}is decreasing, we also have fnt, n min
1≤k≤nfkt, n≥k2n>0. 4.38
Clearly,α3t:nsatisfies
α3t fnt, α3t fnt, n>0. 4.39 It follows from4.25and4.27thatαt:maxα1t, α2t, nis a lower solution ofPn. Step 4. Existence of a solutionu1of4.7such that
maxα1t, α2t, 1≤u1t. 4.40
From assumptionH3, we can findM ≥ maxα1t, α2t, 1andh ∈ A such that, for all t∈0,1,u∈M,∞,
ft, u≤γ2uht. 4.41
Also, one has
f
η1t, u
≤γ2uh η1t
≤γ2uR, 4.42
whereR >0 is a suitable constant. Hence, we obtain, for suchtandu, f1t, u max
f
η1t, u
, ft, u
≤γ2uht R. 4.43 LetCbe a constant such that
C > 1 1−γ2Q−1
Mmax
0≤t≤1 1 0
Gt, shs Rds
. 4.44
Chooseβsuch that
βt γ2Cht R0,
β0 M, β1 β
η
, 4.45
that is,
βt M 1
0
Gt, s
γ2Chs R
ds, 4.46
whereGt, sis defined by2.1. Note thatβis well-defined andM≤βt≤Con0,1since h∈ A. It is easy to see that
βf1
t, β
≤βγ2βht R γ2
β−C
≤0.
4.47
So byRemark 4.2, there is a solutionu1of4.7such that
maxα1t, α2t, 1≤u1t≤βt. 4.48
Step 5. The problemPnhas at least one solutionunsuch that
maxα1t, α2t, n≤unt≤un−1t. 4.49 Notice thatun−1is an upper solution ofPn, since
0un−1t fn−1t, un−1t≥un−1fnt, un−1t,
un−10 n−1 ≥n. 4.50
Step 6. Existence of a solution. Consider the pointwise limit
ut lim
n→ ∞unt, on0,1. 4.51
It is clear that, for anyn≥1,
maxα1t, α2t≤ut ≤unt, for anyt∈0,1 4.52 and thereforeut >0 on0,1. LetK ⊆0,1be a compact interval. There is an indexn∗ n∗Ksuch thatK⊆Knfor alln≥n∗and therefore for thesen≥n∗,
0unt fnt, unt unt ft, unt, ∀t∈K. 4.53
Moreover, we have
supft, u|t∈K, maxα1t, α2t≤u≤un∗t
<∞. 4.54
By Arzel´a-Ascoli theorem it is standard to conclude thatuis a solution of problem1.1and 1.2on the intervalK. SinceKis arbitrary, we find thatu∈C20,1,Rand for allt∈0,1,
ut ft,ut 0. 4.55
Since
u0 lim
n→ ∞n0, 4.56
it remains only to check the continuity ofuatt0. This can be deduced from the continuity ofunand the fact thatun0 n → 0 asn → ∞.
Example 4.6. Consider the following problemP2, forα >0, 0< β1,β2<2,
ut 1
tβ11−tβ2utα 0, 0< t <1,
u0 0, u
1 2
u1.
4.57
Letft, u 1/tβ11−tβ2uα, wheret, u∈0,1×R. Obviously,fsatisfiesH1andH4. Moreover, for any givenk > πand for any compact setK⊆0,1, for >0 small enough, we have
ft, u≥ 1
uα u 1
uα1 ≥k2u, ∀t∈K, u∈0, , 4.58
Hence,H2holds. Furthermore, for M > 0 large enough, 0 < γ < 4/√
3, we have, from Young’s inequality by choosing 1< p <min{1/β1,1/β2}and 1/p1/q1,
ft, u≤ 1
pt−β1p1−t−β2p 1 qu−αq h1t 1
quu−αq−1
≤h1t γ2u, ∀t∈0,1, u∈M,∞,
4.59
whereh1: 1/pt−β1p1−t−β2p∈ A ∩C0,1. Hence,H3holds. ByTheorem 4.4,P2has at least one solution
u∈C0,1,R∪ {0}∩C20,1,R. 4.60
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