Let X be the Banach space with the maximum norm . Define a cone by
For the sake of convenience, we give the following conditions:
-
(i)
The nonlinearity satisfies
for any ;
-
(ii)
;
-
(iii)
There exists a positive constant r such that
(3.1)
where , ;
-
(iv)
.
Remark 3.1
-
(1)
If , for all , then the condition (iv) clearly holds.
-
(2)
If , instead of conditions (i) and (ii), we assume that for any , the following inequalities hold:
where are two constants. Also, the condition (3.1) is replaced by
Case I. Singular ϕ-Laplacian operator: ().
Theorem 3.1 If f is continuous, then the problem (1.1) has at least one solution.
Proof Define a set by
From the definition of , (2.2), (2.7), and (2.8), we get, for any ,
and
Thus . Utilizing (2.9) and Arzela-Ascoli theorem, it is easy to verify that the nonlinear operator S is a completely continuous operator. Therefore, the nonlinear operator S has at least one fixed point by Schauder fixed point theorem.
From the definition of S in (2.8), we have
i.e.,
Then we obtain
and
Consequently, we conclude that the fixed point of S is a solution of the problem (1.1). □
Remark 3.2 Theorem 3.1 shows that if ϕ is singular () and f is continuous on , then the problem (1.1) is always solvable.
Case II. Bounded ϕ-Laplacian operator: , .
Theorem 3.2 If the conditions (i)-(iv) hold, then the nonlinear operator B defined by (2.7) has at least one fixed point. Further, the problem (1.1) has at least one positive solution.
Proof Define a set by
From the conditions (ii) and (iv), we obtain, for any ,
Clearly, the nonlinear operator is well defined.
The condition (iii) implies that we find a constant such that
Define a set by
In virtue of the increasing property of , we get, for any ,
Thus, for any , we have
We choose a small positive constant such that and define . Then for any , we find
Thus, for any , we get
In addition, a standard argument involving the Arzela-Ascoli theorem implies that is a completely continuous operator. Therefore, the nonlinear operator B has at least one fixed point by the use of Lemma 2.1. Let be a fixed point of B, then, from (2.6), we obtain
Consequently, we get from Remark 2.1 that the problem (1.1) has a positive solution . □
Remark 3.3 In order to prove the existence of a positive solution of the problem (1.1), we make a change of variable and introduce a first-order differential equation, and investigate the existence of a fixed point of the corresponding nonlinear operator B. The technique can be used for the different domains and ranges of and give an a priori estimate of the solution.
Theorem 3.3
If
and
f
satisfies the condition
for any , then the problem (1.1) has at least one solution.
Proof Adopting a similar technique to (2.2)-(2.6), we define a nonlinear operator by
Then we get, for any ,
Then the operator has at least one fixed point by the use of Schauder fixed point theorem. Applying expression (2.6), we conclude that the problem (1.1) has at least one solution. □
Remark 3.4 Observe that the solution provided by Theorem 3.3 could be trivial or negative.
Theorem 3.4 If the conditions (ii) and (iv) hold and f satisfies the condition
for any , then problem (1.1) has no solution.
Proof Taking arbitrarily , we get from (3.2)
Thus, it implies that there exists a neighborhood such that for any . This implies that the nonlinear operator is not well defined, since the domain of is the interval and thus a solution of (1.1) cannot exist. □
Example 3.1 We consider the following nonhomogeneous boundary value problem with ϕ-Laplacian operator:
(3.3)
where
It is easy to see that the nonlinearity f satisfies the condition (i), that is,
for any . Computation yields
and there exists a positive constant such that the following inequalities hold:
and
Then conditions (ii)-(iv) also hold. Therefore, we find from Theorem 3.2 that the differential equation (3.3) has at least one positive solution.