We study the solvability of a system of secondorder differential equations with Dirichlet boundary conditions and nonlocal 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
where is a domain in , a survey was given by De Figueiredo in [1]. The specific case of onedimensional systems, motivated by the problem of finding radial solutions to an elliptic system on an annulus of , has been considered by Dunninger and Wang [2] 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 nonlocal 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 reactiondiffusion theory; see [5–7] where the approach to existence and blowup for evolution systems with integral terms may be found.
In this paper we are interested in a simple onedimensional model: the twopoint boundary value problem for the system of second order differential equations with a linear integral term
where , and . First we consider (1.2) as a perturbation of the nonlocal system and prove that if and grow linearly, then (1.2) has a solution provided is not too large. Afterwards, assuming that and are monotone, we will give estimates on the growth of these functions in terms of the parameter to 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 as the Hilbert space of the Lebesgue measurable functions such that with the usual inner product
We also define
with the inner product
If and are both Hilbert spaces, we will consider the Hilbert product space with the inner product
We first study the invertibility of the linear part of (1.2).
Lemma 2.1.
The linear operator , defined by
is invertible if and only if .
Moreover, and are both continuous for .
Proof.
Let . The equation is equivalent to
We denote , , , and , where
is the Green's function associated to , . Notice that , are the solutions of , and , , respectively.
Now it is easy to see that is a solution of (2.6) if and only if
for some such that
Clearly this linear system is uniquely solvable for each pair of functions , if and only if .
In order to prove the continuity of it is easy to show that there exists such that
By the open mapping theorem we deduce that , , is continuous too.
In view of the previous lemma we will assume
.
Lemma 2.2.
Assume .Then the operator is compact and selfadjoint, where is the inclusion and .
Proof.
Since the inclusion is compact (see [8, Theorem ]) and and are continuous we obtain the compactness of . On the other hand an easy computation shows that
so is a selfadjoint operator.
3. An Existence Result of Perturbative Type
Let us introduce the basic assumption
and are continuous functions,
and set
Theorem 3.1.
Assume , , and .
Then problem (1.2) has a solution.
Proof.
Consider the homotopy for all , where is the Nemitskii operator given by
It is easy to check that if and only if is a solution of problem
We are going to prove that the possible solutions of are bounded independently of . By our assumptions, there exist , and such that
Multiplying the first equation of (3.3) by , the second one by , integrating between and and adding both equations we obtain
On the other hand, by the Poincaré inequality (see [9, Chapter 2])
so we have
and since we obtain that and are bounded.
Thus we may invoke the properties of the LeraySchauder degree (see, e.g., [10]) to deduce the existence of a solution for (3.3) with which is our problem (1.2).
Remark 3.2.
Notice that when the solution given by Theorem 3.1 may be the trivial one . However, under our assumptions if moreover or we obtain a proper solution.
4. Monotone Nonlinearities
In the following lemma we give some estimates for the minimum eigenvalue of .
Lemma 4.1.
Assume . If one denotes by the minimum of the eigenvalues of , one has , where is the maximum value between and the greater positive solution of the equation
More precisely, if one denotes by
one obtains that
(i),
(ii),
(iii),
(iv).
Proof.
By Lemma 2.2 the operator is compact, so its set of eigenvalues is bounded and nonempty (see [8, Theorem ]). Moreover we have that is a negative eigenvalue of if and only if there exists a pair , , such that
Differentiating twice on the first equation of () and replacing on the second one, we arrive at the following equality:
In consequence
Analogously, differentiating twice on the second equation of () and replacing on the first one, we arrive at
Now, by means of the expression
we deduce that
and thus
So, we have that in the expression of the solutions of the two equations on system () six real parameters are involved. Now, to fix the value of such parameters, we use the four boundary value conditions imposed on problem() together with the fact that
Therefore, we arrive at the following sixdimensional homogeneous linear system:
In consequence, the values of for which there exist nontrivial solutions of system () coincide with the zeroes of the determinant of the matrix
that is
where
We notice that for all we have
and for all , with odd,
Hence, is the greatest zero among the sequence . On the other hand, since , is solution of (4.15) if and only if and the remaining solutions are the zeroes of the last two factors on (4.15). A careful study shows that function
is such that is strictly decreasing on . Moreover
In consequence, there is a (unique) solution greater than of the equation if and only if . Moreover the greatest zero of function belongs to the interval if and only if .
On the other hand, function
satisfies that is strictly decreasing on its domain , and
So, there is a (unique) solution greater than of the equation if and only if . Moreover, since
it has its greatest zero between and if and only if .
Let denote the class of strictly increasing homeomorphisms from onto . We introduce the following assumption:
Let us define the functional given by
where and for all .
Notice that and are the Fenchel transform of and (see [11]).
Theorem 4.2.
Assume .Let satisfy and in addition
Then attains a minimum at some point .
Moreover, is a solution of (1.2), where we put .
Proof.
Claim 1 ( attains a minimum at some point ).
The space is reflexive, and by our assumptions is weakly sequentially lower semicontinuous. In fact, is the sum of a convex continuous functional (corresponding to the two last summands in the integrand) with a weakly sequentially continuous functional (because of the compactness of ). So, in order to prove that has a minimum, it is enough to show that is coercive. By (4.22) we take such that
So, there exists such that
Thus, for every , there exists such that we have
On the other hand (see [8, Proposition ]),
Taking such that , we have
and therefore is coercive.
Claim 2.
If we denote then is a solution of (1.2).
Since is a critical point of then for all we have
which implies that and for a.e. , where we put . Then is a solution of (1.2).
Remark 4.3.
Under the more restrictive assumption
it follows that is a strictly monotone operator (see [11]). Hence, when (4.29) holds, has a unique critical point. The argument of Claim 2 in previous theorem shows that there is a onetoone correspondence between critical points of and the solutions to (1.2). In consequence, the solution of problem (1.2) is unique.
Remark 4.4.
Suppose that under the conditions of the theorem, . If moreover
we claim that the solution given by the theorem is not the trivial one . In fact let be a normalized eigenvector associated to . The properties of eigenvectors imply that and are in fact continuous functions. Since (4.30) implies and for some and small, an easy computation implies that
for sufficiently small. Hence the minimum of is not attained at .
Remark 4.5.
If and or and , we have that is a lower solution. Moreover if and
then we can take an upper solution of the form with 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 Educación y Ciencia, Spain, Project MTM200761724. The third author was supported by FCT, Financiamento Base 2009.
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