Volume 2011, Article ID 518431,12pages doi:10.1155/2011/518431
Research Article
Existence of Solutions for Elliptic Systems with Nonlocal Terms in One Dimension
Alberto Cabada,
1J. ´ Angel Cid,
2and Lu´ıs Sanchez
31Departamento de An´alise Matem´atica, Facultade de Matem´aticas, Campus Sur, Universidade de Santiago de Compostela, 15782 Santiago de Compostela, Spain
2Departamento de Matem´aticas, Universidad de Ja´en, Ed. B3, Las Lagunillas, 23009 Ja´en, Spain
3Faculdade de Ciˆencias da Universidade de Lisboa, Centro de Matem´atica e Aplicac¸˜oes Fundamentais, Avenida Professor Gama Pinto 2, 1649-003 Lisboa, Portugal
Correspondence should be addressed to Lu´ıs Sanchez,[email protected] Received 2 June 2010; Revised 11 August 2010; Accepted 26 August 2010 Academic Editor: Gennaro Infante
Copyrightq2011 Alberto Cabada et al. 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 study the solvability of a system of second-order differential equations with Dirichlet boundary conditions and non-local terms depending upon a parameter. The main tools used are a dual variational method and the topological degree.
1. Introduction
In the past decade there has been a lot of interest on boundary value problems for elliptic systems. For general systems of the form
−Δufx, u, v,∇u,∇v, x∈Ω,
−Δvgx, u, v,∇u,∇v, x∈Ω, uv0, in∂Ω,
1.1
whereΩis a domain inRn, a survey was given by De Figueiredo in1 . The specific case of one-dimensional systems, motivated by the problem of finding radial solutions to an elliptic system on an annulus ofRn, has been considered by Dunninger and Wang2 and by Lee 3 , who have obtained conditions under which such a system may possess multiple positive solutions.
On the other hand, systems of two equations that include non-local terms have also been considered recently. These are of importance because they appear in the applied sciences, for example, as models for ignition of a compressible gas, or general physical phenomena where temperature has a central role in triggering a reaction. In fact their interest ranges from physics and engineering to population dynamics. See for instance 4 . The related parabolic problems are also of great interest in reaction-diffusion theory; see5–7 where the approach to existence and blow-up for evolution systems with integral terms may be found.
In this paper we are interested in a simple one-dimensional model: the two-point boundary value problem for the system of second order differential equations with a linear integral term
−ut c 1
0
vsdsgvt 0, for a.e. t∈0,1 ,
−vt c 1
0
usdshut 0, for a.e. t∈0,1 ,
u0 u1 0, v0 v1 0,
1.2
wherec∈R,c /0 andg, h:R → R. First we consider1.2as a perturbation of the nonlocal system and prove that ifg andhgrow linearly, then1.2has a solution provided|c|is not too large. Afterwards, assuming thatg and hare monotone, we will give estimates on the growth of these functions in terms of the parametercto ensure solvability. This will be done on the basis of some spectral analysis for the linear part and a dual variational setting.
2. Preliminaries
Let us introduce some notation: we define L20,1 as the Hilbert space of the Lebesgue measurable functionsfsuch that1
0f2xdx <∞with the usual inner product f, g
L2 1
0
fxgxdx. 2.1
We also define
ACI
f:0,1 −→R:f is absolutely continuous on0,1 , H020,1:
f∈C10,1 :f∈AC0,1 , f∈L20,1, f0 0f1
, 2.2
with the inner product
f, g
H02: f, g
L2 f, g
L2 f, g
L2. 2.3
IfX,·,·XandY,·,·Yare both Hilbert spaces, we will consider the Hilbert product space X×Ywith the inner product
x1, y1 ,
x2, y2 X×Y x1, x2X y1, y2
Y. 2.4
We first study the invertibility of the linear part of1.2.
Lemma 2.1. The linear operatorL:H020,1×H020,1 → L20,1×L20,1, defined by
Lcu, v
−uc 1
0
vsds, −vc 1
0
usds
, 2.5
is invertible if and only ifc / ±12.
Moreover,LcandL−1c are both continuous forc / ±12.
Proof. Letx, y∈L20,1×L20,1. The equationLcu, v x, yis equivalent to
−ut xt−c 1
0
vsds, for a.e. t∈0,1 ,
−vt yt−c 1
0
usds, for a.e. t∈0,1 ,
u0 u1 0, v0 v1 0.
2.6
We denote1
0vsds a,1
0usds b,Xt 1
0Gt, sxsds, andYt 1
0Gt, sysds, where
Gt, s
⎧⎨
⎩
t1−s, if 0≤t < s≤1,
s1−t, if 0≤s≤t≤1 2.7
is the Green’s function associated to−u ht,u0 0 u1. Notice thatX,Y ∈H020,1 are the solutions of−uxt,u0 0u1and−uyt,u0 0u1, respectively.
Now it is easy to see thatu, vis a solution of2.6if and only if ut Xt ca
2 tt−1, vt Yt cb
2 tt−1, 2.8
for somea, b∈Rsuch that
a 1
0
Ysds−cb 12, b
1
0
Xsds−ca 12,
2.9
Clearly this linear system is uniquely solvable for each pair of functionsX,Y ∈ H020,1if and only ifc / ±12.
In order to prove the continuity ofLcit is easy to show that there existsk >0 such that Lcu, vL2×L2≤ku, vH2
0×H20 ∀u, v∈H02×H02. 2.10 By the open mapping theorem we deduce thatL−1c ,c / ±12, is continuous too.
In view of the previous lemma we will assume C0c∈R\ {−12,0,12}.
Lemma 2.2. AssumeC0.Then the operatorKc U◦i◦L−1c :L20,1×L20,1 → L20,1× L20,1 is compact and self-adjoint, where i : H020,1×H020,1 → L20,1×L20,1 is the inclusion andUx, y y, x.
Proof. Since the inclusioniis compactsee8, Theorem VIII.7 andL−1c andUare continuous we obtain the compactness ofKc. On the other hand an easy computation shows that
Kcu, vL2×L2u, KcvL2×L2, 2.11
soKcis a self-adjoint operator.
3. An Existence Result of Perturbative Type
Let us introduce the basic assumption
Hg :R → Randh:R → Rare continuous functions, and set
l:lim sup
|v| → ∞
gv v
, m:lim sup
|u| → ∞
hu u
. 3.1
Theorem 3.1. AssumeH,c /0, and|c| lm/2< π2. Then problem1.2has a solution.
Proof. Consider the homotopy I −λKc ◦ N for all λ ∈ 0,1 , where Nis the Nemitskii operatorN:L20,1×L20,1 → L20,1×L20,1given by
Nu, v
−gu·,−hv· . 3.2
It is easy to check thatI−λKc◦N v, u 0,0if and only ifu, vis a solution of problem
−ut c 1
0
vsdsλgvt 0, for a.e. t∈0,1 ,
−vt c 1
0
usdsλhut 0, for a.e. t∈0,1 , u0 u1 0, v0 v1 0.
3.3
We are going to prove that the possible solutions ofI−λKc◦ N u, v 0,0are bounded independently ofλ∈0,1 . By our assumptions, there existl>0,m>0 andksuch that
|c|lm
2 < π2, gu≤l|u|k ∀u∈R,|hv| ≤m|v|k ∀v∈R. 3.4 Multiplying the first equation of3.3byu, the second one byv, integrating between 0 and 1 and adding both equations we obtain
1
0
us2
vs2ds
≤2|c|
1
0
|us|ds 1
0
|vs|ds λ
1
0
gvs|us||hus||vs| ds
≤ |c|
1 0
u2sds 1
0
v2sds
lm 2
1
0
u2s v2s ds
k 1
0
|us||vs|ds
|c| lm 2
1
0
u2s v2s ds
k
⎛
⎝1
0
u2sds 1/2
1
0
v2sds 1/2⎞
⎠.
3.5
On the other hand, by the Poincar´e inequalitysee9, Chapter 2
π2 1
0
u2s v2sds≤ 1
0
us2vs2ds, 3.6
so we have
π2 1
0
u2s v2s ds≤
|c|lm 2
1 0
u2s v2s ds
k
⎛
⎝1
0
u2sds 1/2
1
0
v2sds 1/2⎞
⎠
3.7
and since|c|lm/2< π2we obtain that1
0u2sds1/2uL2and1
0v2sds1/2vL2
are bounded.
Thus we may invoke the properties of the Leray-Schauder degreesee, e.g.,10 to deduce the existence of a solution for3.3withλ1 which is our problem1.2.
Remark 3.2. Notice that wheng0 h0 0 the solution given byTheorem 3.1may be the trivial one0,0. However, under our assumptions if moreoverg0/0 orh0/0 we obtain a proper solution.
4. Monotone Nonlinearities
In the following lemma we give some estimates for the minimum eigenvalue ofKc.
Lemma 4.1. AssumeC0. If one denotes byμcthe minimum of the eigenvalues ofKc, one has μc −λ20, whereλ0 is the maximum value between 1/2π and the greater positive solution of the equation
−1−cλ22cλ3tan 1
2λ
−1−cλ22cλ3tanh 1
2λ
0. 4.1
More precisely, if one denotes by
c0− π3
π−2 tanhπ/2 ≈ −23.718, c1− 4π3
π−2 tanhπ ≈ −57.811,
4.2
one obtains that
i|μc|1/4π2 ifc∈−∞, c1 ∪12,∞, ii1/4π2 <|μc|<1/π2 ifc∈c1, c0∪−12,0, iii|μc|1/π2 ifcc0,
iv1/π2<|μc|ifc∈c0,−12∪0,12.
Proof. By Lemma 2.2the operatorKc is compact, so its set of eigenvalues is bounded and nonemptysee8, Theorem VI.8 . Moreover we have thatμ−λ2is a negative eigenvalue ofKcif and only if there exists a pairx, y∈H02×H02,x, y/ 0,0, such that
−λ2xt yt −cλ2 1
0
ysds,
−λ2yt xt −cλ2 1
0
xsds, x0 x1 0, y0 y1 0.
D
Differentiating twice on the first equation ofDand replacing on the second one, we arrive at the following equality:
−λ4x4t xt −cλ2 1
0
xtdt. 4.3
In consequence
xt a1cos t
λ
b1sin t
λ
c1et/λd1e−t/λe1. 4.4
Analogously, differentiating twice on the second equation ofDand replacing on the first one, we arrive at
yt a2cos t
λ
b2sin t
λ
c2et/λd2e−t/λe2. 4.5
Now, by means of the expression
yt λ2x4t, 4.6
we deduce that
a2−a1, b2−b1, c2c1, d2d1, 4.7
and thus
yt −a1cos t
λ
−b1sin t
λ
c1et/λd1e−t/λe2. 4.8
So, we have that in the expression of the solutions of the two equations on systemDsix real parameters are involved. Now, to fix the value of such parameters, we use the four boundary value conditions imposed on problemDtogether with the fact that
e1−cλ2 1
0
xtdt, e2−cλ2 1
0
ytdt. 4.9
Therefore, we arrive at the following six-dimensional homogeneous linear system:
a1c1d1e10, a1cos
1 λ
b1sin 1
λ
c1e1/λd1e−1/λe10,
−a1c1d1e20,
−a1cos 1
λ
−b1sin 1
λ
c1e1/λd1e−1/λe20,
a1λsin 1
λ
b1
1−cos 1
λ
λc1
e1/λ−1 λd1
1−e−1/λ λe1
1 1
cλ2
0,
a1λsin 1
λ
b1
cos 1
λ
−1
λc1
e1/λ−1 λd1
1−e−1/λ λe2
1 1
cλ2
0.
4.10 In consequence, the values ofλ > 0 for which there exist nontrivial solutions of systemD coincide with the zeroes of the determinant of the matrix
m
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
1 0 1 1 1 0
cos 1
λ
sin 1
λ
e1/λ e−1/λ 1 0
−1 0 1 1 0 1
−cos 1
λ
−sin 1
λ
e1/λ e−1/λ 0 1
cλ3sin 1
λ
cλ3
1−cos 1
λ
c
−1e1/λ λ3 cλ3
1−e−1/λ cλ21 0
−cλ3sin 1
λ
cλ3
cos 1
λ
−1 c
−1e1/λ λ3 cλ3
1−e−1/λ 0 cλ21
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠ ,
4.11
that is
Detm 4e−1/λ
−1e1/λ
dc, λ 0, 4.12
where
dc, λ
−c2λ1λ2e1/λ
cλ22λ−1−1
−1
×
2c
cos 1
λ
−1
λ3
cλ21 sin
1 λ
.
4.13
We notice that for alln∈Nwe have
d
c, 1 2nπ
0, 4.14
and for allλ /1/nπ, withnodd,
dc, λ
−1−e1/λ sin
1 λ
2cλ3tan 1
2λ
−cλ2−1
2cλ3tanh 1
2λ
−cλ2−1
. 4.15 Hence,λ1 1/2 πis the greatest zero among the sequence 1/2nπ. On the other hand, since c /0,λ1/π is solution of4.15if and only ifcc0and the remaining solutionsλ >1/2π are the zeroes of the last two factors on4.15. A careful study shows that function
pc, λ −1−cλ22cλ3tanh 1
2λ
4.16
is such thatcpc, λis strictly decreasing on0,∞. Moreover
λ→ ∞limpc, λ − 1
12c12, p
c, 1 π
−πc−2ctanhπ/2 π3
π3 . 4.17
In consequence, there is auniquesolution greater than 1/π of the equationpc, λ 0 if and only ifc∈c0,−12. Moreover the greatest zero of functionpc,·belongs to the interval 1/2 π,1/πif and only ifc∈c1, c0.
On the other hand, function
qc, λ −1−cλ22cλ3tan 1
2λ
4.18
satisfies thatcqc, λis strictly decreasing on its domain{λ >0 :λ /1/nπ withnodd}, and
lim
λ→1/π−cqc, λ −∞, lim
λ→1/πcqc, λ ∞, lim
λ→ ∞qc, λ 1
12c−12. 4.19
So, there is auniquesolution greater than 1/πof the equationqc, λ 0 if and only ifc∈0,12. Moreover, since
q
c, 1 2π
− c
4π2 −1, 4.20
it has its greatest zero between 1/2πand 1/πif and only if−4π2< c <0.
Let H denote the class of strictly increasing homeomorphisms from R onto R. We introduce the following assumption:
Hg ∈ Hand h∈ H.
Let us define the functionalJc:L20,1×L20,1 → Rgiven by
Jc
x, y : 1 2
Kc
xs, ys ,
xs, ys L2×L2 1
0
G∗xs H∗
ys ds, 4.21
whereG∗t:t
0g−1rdrandH∗t:t
0h−1rdrfor allt∈R.
Notice thatG∗andH∗are the Fenchel transform ofgandhsee11 . Theorem 4.2. AssumeH.LetcsatisfyC0and in addition
lim sup
|x| → ∞
gx x
< 1
μc, lim sup
|x| → ∞
hx x
< 1
μc. 4.22 ThenJcattains a minimum at some pointx0, y0.
Moreover,−u0,−v0is a solution of 1.2, where we putv0, u0 Kcx0, y0. Proof.
Claim 1Jcattains a minimum at some pointx0, y0. The spaceL20,1×L20,1is reflexive, and by our assumptionsJcis weakly sequentially lower semicontinuous. In fact,Jcis the sum of a convex continuous functionalcorresponding to the two last summands in the integrand with a weakly sequentially continuous functionalbecause of the compactness ofKc. So, in order to prove thatJchas a minimum, it is enough to show thatJcis coercive. By4.22we takeα >|μc|/2>0 such that
lim sup
|x| → ∞
gx x
< 1
2α, lim sup
|y| → ∞
h y y
< 1
2α. 4.23
So, there existsk >0 such that
gx≤ |x|
2αk, ∀x∈R, h
y ≤ y
2α k, ∀y∈R.
4.24
Thus, for every >0, there existsa >0 such that we have G∗x≥α−x2−a, ∀x∈R, H∗
y ≥α−y2−a, ∀y∈R. 4.25
On the other handsee8, Proposition VI.9 , Kc
xs, ys ,
xs, ys L2×L2 ≥μc
xs, ys ,
xs,ys L2×L2
μc 1
0
x2s y2s ds.
4.26
Takingsuch that 2α−>|μc|, we have Jc
x, y ≥ μc 2
1
0
x2s y2s
ds α−
1
0
x2s y2s
ds−2a, 4.27
and thereforeJcis coercive.
Claim 2. If we denotev0, u0 Kcx0, y0then−u0,−v0is a solution of1.2.
Sincex0, y0is a critical point ofJcthen for allx, y∈L20,1×L20,1we have Jc
x0, y0 x, y Kc
x0s, y0s ,
xs, ys L2×L2
1
0
g−1x0sxs h−1
y0s ys ds 0,
4.28
which implies thatv0s g−1x0s 0 andu0s h−1y0s 0 for a.e.s∈0,1 , where we putv0, u0 Kcx0, y0. Then−u0,−v0is a solution of1.2.
Remark 4.3. Under the more restrictive assumption
sup
v,w∈R
gv−gw
v−w < 1
μc, sup
v,w∈R
hv−hw
v−w < 1
μc, 4.29 it follows thatJcis a strictly monotone operatorsee11 . Hence, when4.29holds,Jchas a unique critical point. The argument ofClaim 2in previous theorem shows that there is a one- to-one correspondence between critical points ofJcand the solutions to1.2. In consequence, the solution of problem1.2is unique.
Remark 4.4. Suppose that under the conditions of the theorem,g0 h0 0. If moreover lim inf
z→0
gz z > 1
μc, lim inf
z→0
hz z > 1
μc, 4.30
we claim that the solution given by the theorem is not the trivial one0,0. In fact letx, y be a normalized eigenvector associated toμc. The properties of eigenvectors imply thatx andyare in fact continuous functions. Since4.30implies 2G∗z≤ kz2and 2H∗z≤ kz2 for somek <−μcand|z|small, an easy computation implies that
Jc
t x, y
<0 4.31
fortsufficiently small. Hence the minimum ofJcis not attained at0,0.
Remark 4.5. Ifg0>0 andh0≥0 org0≥0 andh0>0, we have thatγ 0,0is a lower solution. Moreover if 0< c <12 and
lim sup
x→ ∞
gx x < 2
312−c, lim sup
x→ ∞
hx x < 2
312−c, 4.32
then we can take an upper solution of the formβ at1−t, t1−twitha > 0 and then apply the monotone method.
Acknowledgments
The authors are indebted to the anonymous referees for useful hints to improve the presentation of the paper. The first and the second authors were partially supported by Ministerio de Educaci ´on y Ciencia, Spain, Project MTM2007-61724. The third author was supported by FCT, Financiamento Base 2009.
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