Boundary Value Problems
Volume 2011, Article ID 416416,15pages doi:10.1155/2011/416416
Research Article
Existence of Solutions to a Nonlocal Boundary Value Problem with Nonlinear Growth
Xiaojie Lin
School of Mathematical Sciences, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China
Correspondence should be addressed to Xiaojie Lin,[email protected] Received 17 July 2010; Accepted 17 October 2010
Academic Editor: Feliz Manuel Minh ´os
Copyrightq2011 Xiaojie Lin. 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.
This paper deals with the existence of solutions for the following differential equation:xt ft, xt, xt,t ∈0,1, subject to the boundary conditions:x0 αxξ,x1 1
0xsdgs, whereα ≥0, 0 < ξ < 1,f : 0,1×R2 → Ris a continuous function,g : 0,1 → 0,∞is a nondecreasing function withg0 0. Under the resonance conditiong1 1, some existence results are given for the boundary value problems. Our method is based upon the coincidence degree theory of Mawhin. We also give an example to illustrate our results.
1. Introduction
In this paper, we consider the following second-order differential equation:
xt f
t, xt, xt
, t∈0,1, 1.1
subject to the boundary conditions:
x0 αxξ, x1
1
0
xsdgs, 1.2
whereα≥0, 0< ξ <1,f :0,1×R2 → Ris a continuous function,g :0,1 → 0,∞is a nondecreasing function withg0 0. In boundary conditions1.2, the integral is meant in the Riemann-Stieltjes sense.
We say that BVP1.1,1.2is a problem at resonance, if the linear equation
xt 0, t∈0,1, 1.3
with the boundary condition 1.2 has nontrivial solutions. Otherwise, we call them a problem at nonresonance.
Nonlocal boundary value problems were first considered by Bicadze and Samarski˘ı 1 and later by Il’pin and Moiseev2,3. In a recent paper4, Karakostas and Tsamatos studied the following nonlocal boundary value problem:
xt qtf
xt, xt
0, t∈0,1, x0 0, x1
1
0
xsdgs. 1.4
Under the condition 0g0≤g1<1i.e., nonresonance case, they used Krasnosel’skii’s fixed point theorem to show that the operator equationxAxhas at least one fixed point, where operatorAis defined by
Axt t 1−g1
1
0
1
s
qrf
xr, xr
dr dgs t
0
1
s
qrf
xr, xr
dr ds. 1.5
However, ifg1 1i.e., resonance case, then the method in4is not valid.
As special case of nonlocal boundary value problems, multipoint boundary value problems at resonance case have been studied by some authors5–11.
The purpose of this paper is to study the existence of solutions for nonlocal BVP1.1, 1.2at resonance casei.e.,g1 1and establish some existence results under nonlinear growth restriction off. Our method is based upon the coincidence degree theory of Mawhin 12.
2. Main Results
We first recall some notation, and an abstract existence result.
LetY,Z be real Banach spaces, letL : domL ⊂ Y → Z be a linear operator which is Fredholm map of index zeroi.e., ImL, the image of L, KerL, the kernel of L are finite dimensional with the same dimension as the Z/ImL, and let P : Y → Y, Q : Z → Z be continuous projectors such that ImP KerL, KerQImLandY KerL⊕KerP,Z ImL⊕ImQ. It follows thatL|domL∩KerP : domL∩KerP → ImL is invertible; we denote the inverse byKP. LetΩbe an open bounded, subset ofY such that domL∩Ω/∅, the map N:Y → Zis said to beL-compact onΩifQNΩis bounded, andKPI−QN:Ω → Y is compact. LetJ : ImQ → KerLbe a linear isomorphism.
The theorem we use in the following is Theorem IV.13 of12.
Theorem 2.1. LetLbe a Fredholm operator of index zero, and letNbeL-compact onΩ. Assume that the following conditions are satisfied:
iLx /λNxfor everyx, λ∈domL\KerL∩∂Ω×0,1, iiNx /∈ImLfor everyx∈KerL∩∂Ω,
iiidegJQN|KerL,Ω∩KerL,0/0,
whereQ:Z → Zis a projection with ImL KerQ. Then the equationLx Nxhas at least one solution in domL∩Ω.
Forx ∈C10,1, we use the normsx∞ maxt∈0,1|xt|andx max{x∞,x∞} and denote the norm inL10,1by · 1. We will use the Sobolev spaceW2,10,1which may be defined by
W2,10,1
x:0,1−→R|x, xare absolutely continuous on0,1withx∈L10,1 . 2.1
LetY C10,1,ZL10,1.L: domL⊂Y → Zis a linear operator defined by
Lxx, x∈domL, 2.2
where
domL
x∈W2,10,1:x0 αxξ, x1 1
0
xsdgs . 2.3
LetN:Y → Zbe defined as
Nxf
t, xt, xt
, t∈0,1. 2.4
Then BVP1.1,1.2isLxNx.
We will establish existence theorems for BVP1.1,1.2in the following two cases:
casei:α0, g1 1,1
0s dgs/1;
caseii:α1, g1 1,1
0s dgs/1.
Theorem 2.2. Letf :0,1×R2 → Rbe a continuous function and assume that
H1there exist functionsa, b, c, r ∈L10,1and constantθ∈0,1such that for allx, y∈ R2,t∈0,1, it holds that
f
t, x, y≤at|x|btyct
|x|θyθ
rt, 2.5
H2there exists a constantM > 0, such that forx∈domL, if|xt|> M, for allt∈0,1, then
1
0
f
s, xs, xs ds−
1
0
s
0
f
v, xv, xv
dv dgs/0, 2.6
H3there exists a constantM∗>0, such that either
d· 1
0
fs, ds, dds− 1
0
s
0
fv, dv, ddv dgs
<0, for any|d|> M∗, 2.7
or else
d· 1
0
fs, ds, dds− 1
0
s
0
fv, dv, ddv dgs
>0, for any|d|> M∗. 2.8
Then BVP1.1,1.2withα0,g1 1, and1
0s dgs/1 has at least one solution inC10,1 provided that
a1b1< 1
2. 2.9
Theorem 2.3. Letf :0,1×R2 → Rbe a continuous function. Assume that assumption (H1) of Theorem 2.2is satisfied, and
H4there exists a constantM >0, such that forx∈domL, if|xt|> M, for allt ∈0,1, then
1
0
f
s, xs, xs ds−
1
0
s
0
f
v, xv, xv
dv dgs/0, 2.10
H5there exists a constantM∗>0, such that either
e· 1
0
fs, e,0ds− 1
0
s
0
fv, e,0dv dgs
<0, for any|e|> M∗, 2.11
or else
e· 1
0
fs, e,0ds− 1
0
s
0
fv, e,0dv dgs
>0, for any|e|> M∗. 2.12
Then BVP1.1,1.2withα 1, g1 1, and1
0s dgs/1 has at least one solution inC10,1 provided that
a1b1< 1
2. 2.13
3. Proof of Theorems 2.2 and 2.3
We first proveTheorem 2.2via the following Lemmas.
Lemma 3.1. Ifα0,g1 1, and1
0s dgs/1, thenL: domL⊂Y → Zis a Fredholm operator of index zero. Furthermore, the linear continuous projector operatorQ:Z → Zcan be defined by
Qy 1
1−1
0s dgs 1
0
ysds− 1
0
s
0
yvdv dgs
, 3.1
and the linear operatorKP : ImL → domL∩KerPcan be written by
KPy t
0
s
0
yvdv ds. 3.2
Furthermore,
KPy≤y
1, for everyy∈ImL. 3.3
Proof. It is clear that
KerL{x∈domL:xdt, d∈R, t∈0,1}. 3.4
Obviously, the problem
xy 3.5
has a solutionxtsatisfyingx0 0,x1 1
0xsdgs, if and only if 1
0
ysds− 1
0
s
0
yvdv dgs 0, 3.6
which implies that
ImL
y∈Z: 1
0
ysds− 1
0
s
0
yvdv dgs 0 . 3.7
In fact, if3.5has solutionxtsatisfyingx0 0,x1 1
0xsdgs, then from3.5we have
xt x0t t
0
s
0
yvdv ds. 3.8
According tox1 1
0xsdgs,g1 1, we obtain x1 x0
1
0
ysds 1
0
xsdgs
1
0
x0
s
0
yvdv
dgs x0g1
1
0
s
0
yvdv dgs,
3.9
then
1
0
ysds− 1
0
s
0
yvdv dgs 0. 3.10
On the other hand, if3.6holds, setting
xt dt t
0
s
0
yvdv ds, 3.11
where d is an arbitrary constant, then xt is a solution of3.5, andx0 0, and from g1 1 and3.6, we have
x1− 1
0
xsdgs d 1
0
ysds− 1
0
d
s
0
yvdv
dgs d
1−g1
1
0
ysds− 1
0
s
0
yvdv dgs 0.
3.12
Thenx1 1
0xsdgs. Hence3.7is valid.
Fory∈Z, define
Qy 1
1−1
0s dgs 1
0
ysds− 1
0
s
0
yvdv dgs
, 0≤t≤1. 3.13
Lety1y−Qy, and we have
1− 1
0
s dgs
Qy1 1
0
y−Qy sds−
1
0
s
0
y−Qy
vdv dgs
1
0
ysds−Qy− 1
0
s
0
yvdv dgs Qy 1
0
s dgs
1
0
ysds− 1
0
s
0
yvdv dgs−Qy
1− 1
0
s dgs
0,
3.14
thenQy1 0, thusy1 ∈ImL. Hence,ZImLZ1, whereZ1{xt≡d:t∈0,1, d∈R}, also ImL∩Z1{0}. So we haveZImL⊕Z1, and
dim KerLdimZ1co dim ImL1. 3.15
Thus,Lis a Fredholm operator of index zero.
We define a projector P : Y → KerL byP xt x0t. Then we show thatKP
defined in3.2is a generalized inverse ofL: domL∩Y → Z.
In fact, fory∈ImL, we have
LKPyt KPy
t
yt, 3.16 and, forx∈domL∩KerP, we know
KPLxt t
0
s
0
xvdv dsxt−x0−x0t. 3.17
In view ofx∈domL∩KerP,x0 0, andP x0, thus
KPLxt xt. 3.18
This shows thatKP L|domL∩KerP−1. Also we have KPy
∞≤ 1
0
yvdv dsy
1, KPy
∞≤y
1, 3.19
thenKPy ≤ y1. The proof ofLemma 3.1is finished.
Lemma 3.2. Under conditions 2.5 and 2.9, there are nonnegative functions a, b, r ∈ L10,1 satisfying
f
t, x, y≤at|x|btyrt. 3.20
Proof. Without loss of generality, we suppose that c1 1
0|ct|dt β > 0. Take γ ∈ 0,1/2β1/2−a1b1, then there existsM >0 such that
|x|θ≤γ|x|M, yθ≤γyM. 3.21
Let
at at γ ct, bt bt γ ct, rt rt 2Mct. 3.22
Obviously,a, b, r∈L10,1, and
a1≤ a1γc1, b
1≤ b1γc1. 3.23
Then
a1b
1≤ a1b12βγ < 1
2, 3.24
and from2.5and3.21, we have f
t, x, y≤
at γ ct
|x|
bt γ cty2Mct rt
at|x|btyrt. 3.25
Hence we can takea,b, 0, andr to replacea,b,c, andr, respectively, in2.5, and for the convenience omit the bar abovea,b, andr, that is,
f
t, x, y≤at|x|btyrt. 3.26
Lemma 3.3. If assumptions (H1), (H2) andα 0,g1 1, and1
0s dgs/1 hold, then the set Ω1{x∈domL\KerL:LxλNxfor someλ∈0,1}is a bounded subset ofY.
Proof. Suppose thatx∈Ω1andLxλNx. Thusλ /0 andQNx0, so that 1
0
ysds− 1
0
s
0
yvdv dgs 0, 3.27
thus by assumptionH2, there existst0∈0,1, such that|xt0| ≤M. In view of
x0 xt0− t0
0
xtdt, 3.28
then, we have
x0≤Mx
1MLx1 ≤MNx1. 3.29
Again forx ∈ Ω1,x ∈ domL\KerL, thenI −Px ∈ domL∩KerP,LP x 0 thus from Lemma 3.1, we know
I−PxKPLI−Px ≤ LI−Px1Lx1≤ Nx1. 3.30
From3.29and3.30, we have
x ≤ P xI−Pxx0I−Px ≤2Nx1M. 3.31
If2.5holds, from3.31, and3.26, we obtain x ≤2
a1x∞b1x
∞r1 M 2
. 3.32
Thus, fromx∞≤ xand3.32, we have x∞≤ 2
1−2a1
b1x
∞r1 M 2
. 3.33
Fromx∞≤ x,3.32, and3.33, one has x
∞≤ x ≤2
1 2a1 1−2a1
b1x
∞r1M 2
2 1−2a1
b1x
∞r1M 2
,
3.34
that is,
x∞≤ 2
1−2a1b1
r1M 2
:M1. 3.35
From3.35and3.33, there existsM2>0, such that
x∞≤M2. 3.36
Thus
xmax
x∞,x
∞
≤max{M1, M2}. 3.37
Again from2.5,3.35, and3.36, we have x
1Lx1≤ Nx1≤ a1M2b1M1r1. 3.38
Then we show thatΩ1is bounded.
Lemma 3.4. If assumption (H2) holds, then the setΩ2 {x∈KerL:Nx∈ImL}is bounded.
Proof. Letx ∈ Ω2, thenx ∈ KerL {x ∈ domL : x dt, d ∈ R, t ∈ 0,1}andQNx 0;
therefore,
1
0
fs, ds, dds− 1
0
s
0
fv, dv, ddv dgs 0, 3.39
From assumptionH2,x∞|d| ≤M, sox|d| ≤M, clearlyΩ2is bounded.
Lemma 3.5. If the first part of condition (H3) ofTheorem 2.2holds, then
d· 1
1−1
0s dgs 1
0
fs, ds, dds− 1
0
s
0
fv, dv, ddv dgs
<0, 3.40
for all|d|> M∗. Let
Ω3{x∈KerL:−λx 1−λJQNx0, λ∈0,1}, 3.41
whereJ: ImQ → KerLis the linear isomorphism given byJd dt, for alld∈R,t∈0,1. Then Ω3is bounded.
Proof. Suppose thatxd0t∈Ω3, then we obtain
λd0t 1−λt 1−1
0s dgs 1
0
fs, d0s, d0ds− 1
0
s
0
fv, d0v, d0dv dgs
, 0≤t≤1, 3.42
or equivalently
λd0 1−λ 1−1
0s dgs 1
0
fs, d0s, d0ds− 1
0
s
0
fv, d0v, d0dv dgs
. 3.43
Ifλ1, thend00. Otherwise, if|d0|> M∗, in view of3.40, one has
λd20 d01−λ 1−1
0s dgs 1
0
fs, d0s, d0ds− 1
0
s
0
fv, d0v, d0dv dgs
<0, 3.44
which contradictsλd20 ≥0. Then|x||d0t| ≤ |d0| ≤ M∗and we obtainx ≤ M∗; therefore, Ω3⊂ {x∈KerL:x ≤M∗}is bounded.
The proof of Theorem 2.2 is now an easy consequence of the above lemmas and Theorem 2.1.
Proof ofTheorem 2.2. LetΩ {x ∈Y :x < δ}such that3
i1Ωi ⊂ Ω. By the Ascoli-Arzela theorem, it can be shown thatKPI−QN :Ω → Y is compact; thusNisL-compact onΩ.
Then by the above Lemmas, we have the following.
iLx /λNxfor everyx, λ∈domL\KerL∩∂Ω×0,1.
iiNx /∈ImLfor everyx∈KerL∩∂Ω.
iiiLetHx, λ −λx 1−λJQNx, withJas inLemma 3.5. We knowHx, λ/0, for x∈KerL∩∂Ω. Thus, by the homotopy property of degree, we get
degJQN|KerL,Ω∩KerL,0 degH·,0,Ω∩KerL,0 degH·,1,Ω∩KerL,0 deg−I,Ω∩KerL,0.
3.45
According to definition of degree on a space which is isomorphic toRn,n <∞, and
Ω∩KerL{dt:|d|< δ}. 3.46
We have
deg−I,Ω∩KerL,0 deg
−J−1IJ, J−1Ω∩KerL, J−1{0}
deg−I,−δ, δ,0 −1/0,
3.47
and then
degJQN|KerL,Ω∩KerL,0/0. 3.48
Then byTheorem 2.1,LxNxhas at least one solution in domL∩Ω, so that the BVP1.1, 1.2has at least one solution inC10,1. The proof is completed.
Remark 3.6. If the second part of conditionH3ofTheorem 2.2holds, that is,
d· 1
1−1
0s dgs 1
0
fs, ds, dds− 1
0
s
0
fv, dv, ddv dgs
>0, 3.49
for all|d|> M∗, then inLemma 3.5, we take
Ω3{x∈KerL:λx 1−λJQNx0, λ∈0,1}, 3.50
and exactly as there, we can prove thatΩ3is bounded. Then in the proof ofTheorem 2.2, we have
degJQN|KerL,Ω∩KerL,0 degI,Ω∩KerL,0 1, 3.51 since 0∈Ω∩KerL. The remainder of the proof is the same.
By using the same method as in the proof ofTheorem 2.2and Lemmas3.1–3.5, we can showLemma 3.7andTheorem 2.3.
Lemma 3.7. Ifα1,g1 1, and1
0s dgs/1, thenL: domL⊂Y → Zis a Fredholm operator of index zero. Furthermore, the linear continuous projector operatorQ:Z → Zcan be defined by
Qy 1
1−1
0s dgs 1
0
ysds− 1
0
s
0
yvdv dgs
, 3.52
and the linear operatorKP : ImL → domL∩KerPcan be written by
KPy−t ξ
ξ
0
s
0
yvdv ds t
0
s
0
yvdv ds. 3.53
Furthermore,
KPy≤2y
1, ∀y∈ImL. 3.54
Notice that
KerL{x∈domL:xe, e∈R}, ImL
y∈Z:
1
0
ysds− 1
0
s
0
yvdv dgs 0 . 3.55
Proof ofTheorem 2.3. Let
Ω1{x∈domL\KerL:LxλNxfor someλ∈0,1}. 3.56 Then, forx∈Ω1,LxλNx; thusλ /0,Nx∈ImLKerQ; hence
1
0
ysds− 1
0
s
0
yvdv dgs 0, 3.57
thus, from assumptionH4, there existst0 ∈ 0,1, such that|xt0| < M and in view of x0 xt0−t0
0 xtdt, we obtain
|x0| ≤Mx
∞. 3.58
From x0 xξ, there existst1 ∈ 0, ξ, such thatxt1 0. Thus, fromxt xt1
t
t1xtdt, one has
x
∞≤x
1. 3.59
We letP xx0; hence from3.58and3.59, we have P x|x0| ≤Mx
∞≤Mx
1
MLx1≤MNx1, 3.60 thus, by using the same method as in the proof of Lemmas3.2and3.3, we can prove thatΩ1
is bounded too. Similar to the other proof of Lemmas3.4–3.7andTheorem 2.2, we can verify Theorem 2.3.
Finally, we give two examples to demonstrate our results.
Example 3.8. Consider the following boundary value problem:
xt38sinx31
9t1x, t∈0,1, x0 0, x1
1
0
xsdgs,
3.61
whereα0,
f t, x, y
t38sinx31
9t1y, t∈0,1, 3.62
and gs s2 satisfying g0 0, g1 1, and1
0s dgs 2/3/1, then we can choose at 0,bt 2/9, andrt 10, fort∈0,1; thus
f
t, x, y≤ 2
9y10, a1b1 2
9 < 1 2.
3.63
Since
1
0
f
s, xs, xs ds−
1
0
s
0
f
v, xv, xv
dv dgs
1
0
f
v, xv, xv
dv dgs− 1
0
s
0
f
v, xv, xv
dv dgs
1
0
1
s
f
v, xv, xv
dv dgs,
3.64
andfhas the same sign asxtwhen|xt|>90, we may chooseMM∗90, and then the conditionsH1–H3ofTheorem 2.2are satisfied.Theorem 2.2implies that BVP3.61has at least one solution,x∈C10,1.
Example 3.9. Consider the following boundary value problem:
xt241
7t2xcos x3
, t∈0,1,
x0 x1, x1
1
0
xsdgs,
3.65
whereα1,
f t, x, y
t241
7t2xcos y3
, t∈0,1, 3.66
and gs s2 satisfying g0 0, g1 1, and1
0s dgs 2/3/1, then we can choose at 3/7,bt 0, andrt 6, fort∈0,1; thus
f
t, x, y≤ 3 7|x|6, a1b1 3
7 < 1 2.
3.67
Similar toExample 3.8, we have 1
0
f
s, xs, xs ds−
1
0
s
0
f
v, xv, xv
dv dgs 1
0
1
s
f
v, xv, xv
dv dgs, 3.68 andfhas the same sign asxtwhen|xt|>21, we may chooseMM∗ 21, and then all conditions ofTheorem 2.3are satisfied.Theorem 2.3implies that BVP3.65has at least one solutionx∈C10,1.
Acknowledgment
This work was sponsored by the National Natural Science Foundation of China11071205, the NSF of Jiangsu Province Education Department, NFS of Xuzhou Normal University.
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