#### Abstract

By using the perturbation theories on sums of ranges of nonlinear accretive mappings of Calvert and Gupta (1978), the abstract result on the existence and uniqueness of the solution in of the generalized Capillarity equation with nonlinear Neumann boundary value conditions, where and denotes the dimension of , is studied. The equation discussed in this paper and the methods here are a continuation of and a complement to the previous corresponding results. To obtain the results, some new techniques are used in this paper.

#### 1. Introduction and Preliminary

Since the Laplacian operator with arises from a variety of physical phenomena, such as nonNewtonian fluids, reaction-diffusion problems, and petroleum extraction, it becomes a very popular topic in mathematical fields.

We began our study on this topic in 1995. We used a perturbation result of ranges for *m*-accretive mappings in Calvert and Gupta [1] to obtain a sufficient condition in Wei and He [2] so that the following zero boundary value problem,
has solutions in , where . Later on, a series work of ours has been done from different angles on this kind of equations, cf. [3–7], and so forth.

Especially, in 2008, as a summary of the work done in [2–6], we use some new techniques to work for the following problem with so-called generalized Laplacian operator: where , is a nonnegative constant and denotes the exterior normal derivative of . We showed in Wei and Agarwal [7] that (1.2) has solutions in under some conditions, where , if , and if , for .

Capillarity equation is another important equation appeared in the capillarity phenomenon and we notice that in Chen and Luo [8], the authors studied the eigenvalue problem for the following generalized Capillarity equations:

Their work inspired us and one idea came to our mind. Can we borrow the main ideas dealing with the nonlinear elliptic boundary value problems with the generalized Laplacian operator to study the nonlinear generalized Capillarity equation with Neumann boundary conditions?

We will answer the question in this paper. By using the perturbation results of ranges for *m*-accretive mappings in Calvert and Gupta [1] again, we will study the following one:

More details on (1.4) will be given in Section 2. Our methods and techniques are different from those in Chen and Luo [8].

Now, we list some basic knowledge we need in sequel.

Let be a real Banach space with a strictly convex dual space . We use “” and to denote strong and weak convergence, respectively. For any subset of , we denote by its interior and its closure, respectively. Let denote that space *X* is embedded compactly in space and denote that space is embedded continuously in space . A mapping is said to be hemicontinuous on if , for any . Let denote the * duality mapping* from into defined by
where denotes the generalized duality pairing between and . It is wellknown that is a single-valued mapping since is strictly convex.

Let be a given multivalued mapping. We say that is boundedly-inversely compact if for any pair of bounded subsets and of , the subset is relatively compact in . The mapping is said to be accretive if , for any and , . The accretive mapping is said to be -accretive if , for some .

Let be a given multi-valued mapping. The graph of , , is defined by . Then is said to be monotone if is a monotone subset of in the sense that for any , . The monotone operator is said to be maximal monotone if is maximal among all monotone subsets of in the sense of inclusion. The mapping is said to be strictly monotone if the equality in (1.6) implies that . The mapping is said to be coercive if for all such that .

*Definition 1.1. *The* duality mapping * is said to be satisfying Condition if there exists a function such that

*Definition 1.2. *Let be an accretive mapping and be a duality mapping. We say that satisfies Condition if, for any and , there exists a constant such that

Lemma 1.3 (Li and Guo [9]). *Let be a bounded conical domain in . Then we have the following results.*(a)*If , then ; if and , then ; if and , then for .*(b)*If , then ; if and , then , .*

Lemma 1.4 (Pascali and Sburlan [10]). *If is an everywhere defined, monotone and hemicontinuous operator, then is maximal monotone. If is maximal monotone and coercive, then .*

Lemma 1.5 (Pascali and Sburlan [10]). *If is a proper convex and lower-semicontinuous function, then is maximal monotone from to . *

Lemma 1.6 (Pascali and Sburlan [10]). *If and are two maximal monotone operators in such that , then is maximal monotone. *

Proposition 1.7 (Calvert and Gupta [1]). * Let and be a bounded domain in . For , the duality mapping defined by , for , satisfies Condition ; for and , the duality mapping defined by , for , satisfies Condition , where .*

Lemma 1.8 (see Calvert and Gupta [1]). * Let be a bounded domain in and be a function satisfying Caratheodory's conditions such that *(i)* is monotonically increasing on ; *(ii)*the mapping is well defined, where and .**Let , be the duality mapping defined by
**
for . Then the mapping defined by , for any satisfies Condition . *

Theorem 1.9 (Calvert and Gupta [1]). *Let be a real Banach space with a strictly convex dual . Let be a duality mapping on satisfying Condition . Let be accretive mappings such that *(i)*either both satisfy Condition or and satisfies Condition ,*(ii)* is m-accretive and boundedly inversely compact. **If be a bounded continuous mapping such that, for any , there is a constant satisfying for any . Then. *(a)*.*(b)*.*

#### 2. The Main Results

##### 2.1. Notations and Assumptions of (1.4)

In the next of this paper, we assume , , if , and , if , where . We use , , , and to denote the norms in , , and . Let , , and .

In (1.4), is a bounded conical domain of a Euclidean space with its boundary , (c.f. [4]). We suppose that the Green's Formula is available. Let denote the Euclidean norm in , the Euclidean inner-product and the exterior normal derivative of . is a nonnegative constant.

Let be a given function such that, for each ,(i) is a proper, convex, lower-semicontinuous function with .(ii) (: subdifferential of ) is maximal monotone mapping on with and for each , the function is measurable for .

Let be a given function satisfying Caratheodory's conditions such that for and , the mapping is defined. Further, suppose that there is a function such that , for , .

##### 2.2. Existence and Uniqueness of the Solution of (1.4)

*Definition 2.1 (Calvert and Gupta [1]). *Define and .

Further, define a function by
Then for all , is increasing in and . Moreover, satisfies Caratheodory's conditions and the functions are measurable on . And, if then , for , .

Proposition 2.2 (see Calvert and Gupta [1]). *For and , define the mapping by , for all and , then is a bounded, continuous, and m-accretive mapping. *

Moreover, Lemma 1.8 implies that satisfies Condition .

Define by , where , then satisfies the inequality: for any , where is a constant depending on and denotes the duality mapping, where .

Lemma 2.3 (Wei and Agarwal [7]). *The mapping defined by
**
for any , is a proper, convex, and lower-semicontinuous mapping on .*

Moreover, Lemma 1.5 implies that , the subdifferential of , is maximal monotone.

Lemma 2.4 (Wei and He [2]). *Let denote the closed subspace of all constant functions in . Let be the quotient space . For , define the mapping by . Then, there is a constant , such that for all ,
*

Lemma 2.5. * Define the mapping by
**
for any . Then is everywhere defined, strictly monotone, hemicontinuous, and coercive.*

*Proof. **Step *1. is everywhere defined.

From Lemma 1.3, we know that , when . And, , , when . Thus, for all , , , where are positive constants.

For , we have
where and are positive constants. Thus is everywhere defined.*Step *2. is strictly monotone.

For , we have

If, we let , for . Then we know that
since . And, if and only if . Then is strictly monotone. Thus we can easily know that is strictly monotone.*Step *3. is hemicontinuous.

In fact, it suffices to show that, for any and , , as .

By Lebesque's dominated convergence theorem, it follows that
and hence is hemicontinuous.*Step *4. is coercive.

Now, for , Lemma 2.4 implies that is equivalent to and hence we have the following result:
as , which implies that is coercive.

This completes the proof.

*Remark 2.6. * Lemma 2.5 is a key result for later use.

*Definition 2.7. *Define a mapping as follows:
For , let .

Proposition 2.8. * The mapping is m-accretive.*

* Proof. **Step *1. is accretive.*Case *1. If , the duality mapping is defined by for . It then suffices to prove that for any and ,
To this, we are left to prove that both
are available.

Now take for a constant , is defined by . Then is monotone, Lipschitz with and is continuous except at finitely many points on . This gives that
Also,
the last inequality is available since is monotone and .*Case *2. If , the duality mapping is defined by , for . It then suffices to prove that for any and ,
To this, we define the function by
Then is monotone, Lipschitz with and is continuous except at finitely many points on . So .

Then, for , , we have*Step *2. , for every .

First, define the mapping by and for , where denotes the inner product of . Then is maximal monotone [7].

Secondly, for any , define the mapping by , for . Then similar to that in [7], by using Lemmas 1.4, 1.6, 2.3, and 2.5, we know that is maximal monotone and coercive, so that , for any .

Therefore, for any , there exists , such that
From the definition of , it follows that , for all .

This completes the proof.

Lemma 2.9. * The mapping has a compact resolvent for and .*

* Proof. * Since is accretive by Proposition 2.8, it suffices to prove that if and is bounded in , then is relatively compact in . Now define functions by
Noticing that , , and , . We know that for since is monotone, Lipschitz with , and is continuous except at finitely many points on . Then

We now have from that
which gives that
in view of the fact that when for . Again from (2.22), we have . Hence bounded in implies that is bounded in .

We notice that when and when by Lemma 1.3, hence is relatively compact in . This gives that is relatively compact in since the Nemytskii mapping is continuous.

This completes the proof.

*Remark 2.10. *Since , for any and , we have implies that in the sense of distributions.

Proposition 2.11. * For , if there exists such that , then is the unique solution of (1.4). *

* Proof. * First, we show that
is available.

Now implies that . For all , by Remark 2.10, we have
which implies that (2.24) is true.

Secondly, we show that
We will prove (2.26) under the additional condition , where and . Refer to the result of Brezis [11] for the general case.

Now, from (2.24), implies that . By using Green's Formula, we have that for any ,

Then , where is the space of traces of .

Now let the mapping be defined by , for any , where a.e. on . Clearly, where is a proper, convex, and lower-semicontinuous function on . Now define the mapping by for any . Then is maximal monotone since both are continuous. Finally, for any , we have
Hence we get and so . Therefore, we have

Finally, we will show that is unique.

If and , where . Then
since is strictly monotone and is maximal monotone, which implies that .

This completes the proof.

*Remark 2.12. * If for any , then , for all .

Proposition 2.13. * If for any , then .*

* Proof. * We can easily know that in view of Lemmas 1.4 and 2.5. Note that for any with , the linear function is an element of . So there exists a such that
for any . So in view of Remark 2.12.

This completes the proof.

*Definition 2.14 (see [1, 7]). *For , , let be the element with least absolute value if and , where or , respectively, in case . Finally, let (in the extended sense) for . define measurable functions on , in view of our assumptions on .

Proposition 2.15. * Let such that
**
Then . *

* Proof. * Let and satisfy (2.32), by Proposition 2.8, there exists such that, for each , . In the same reason as that in [1], we only need to prove that , for all .

Indeed, suppose to the contrary that , as . Let . Let be defined by , be its subdifferential and for , denote the Yosida-approximation of . Let denote the indefinite integral of with so that . In view of Calvert and Gupta , we have

Now multiplying the equation by , we get that
Since , it follows that . Also, we can know that