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 (D) and replacing on the second one, we arrive at the following equality:

Analogously, differentiating twice on the second equation of ( ) and replacing on the first one, we arrive at

Now, by means of the expression

So, we have that in the expression of the solutions of the two equations on system (D) six real parameters are involved. Now, to fix the value of such parameters, we use the four boundary value conditions imposed on problem(D) together with the fact that

Therefore, we arrive at the following six-dimensional homogeneous linear system:

In consequence, the values of

for which there exist nontrivial solutions of system (D) coincide with the zeroes of the determinant of the matrix

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 one-to-one 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.