1. Introduction
In this work, we are concerned with the existence and uniqueness of results to discrete, nonlinear equations governed by the variable Laplacian and considered on a finite connected graph. The methods which we apply pertain to the finite dimensional version of the monotonicity theory and therefore rely on coercivity, monotonicity, and continuity of the relevant operators which may contain non-potential terms. Contrary to what is known about problems with variable exponents in the discrete setting (see for example [
1,
2]), the analysis related to the coercivity of the difference operator must be performed with some care. This is due to the structure of the domain (compare with [
3,
4] where problems with constant exponent are considered). Another problem appears with the proper definition of the weak solution, which must take into account both the graph structure of the domain and also the fact that the exponent is now not constant. Problems arising here require analysis related to some symmetry. Having overcome the problems mentioned, one may resort to checking the existence of solutions to abstract tools known from the applied functional analysis. In this work, we have decided to employ the monotonicity theory, but one expects that other approaches would also be applicable. In addition, we indicate that a variational approach can be employed in case the given problem is potential. However, the growth assumptions would, in such a case, be similar to those used in the monotonicity approach, and this is why we do not directly formulate such results. When checking the solvability of problems on graphs one is also interested in some well-posedness and stability, structural stability including the following.
Let be a simple, connected, undirected, weighted, and finite graph. Set V is a non-empty set of vertices, and is a set consisting of all graph edges, i.e., unordered pairs elements of V. Function is a weight on a graph G, i.e., we assume that
- (i)
,
- (ii)
.
Let
be a continuous function,
,
,
; moreover, we assume that
for all
. Then we consider the following problem:
for
. Our goal is to find sufficient conditions for the existence and uniqueness of the solution to (
1) in the space
Note that
A is a
-dimensional Euclidean space with the norm given by
Now we define two operators, which are necessary in order to properly understand our problem.
Definition 1. Let and
- (i)
The -gradient of the function u in point , , is defined by for all , is called the -directional derivative of the function u in the direction . In the case where , we write
- (ii)
The discrete -Laplacian of the function u at point , , is defined by If we write Δ.
On the right hand side of Formula (
3) there appears a
-dimensional vector that is indexed by
. We would like to underline that to the best of our knowledge, the discrete
-Laplacian on finite graphs have not been considered yet, which means that several auxiliary tools that are known on a graph setting must be worked out in detail. This means that we had to investigate the problem in a detailed manner which involves derivation of some auxiliary tools. The results in the literature cover only the case of the
p-Laplacian and
-Laplacian on graphs, where, however, other methods are applied and monotonicity approaches are not used. Thus, our results are also new in the context of constant p. In case
is a constant function, we may recover the well-known
p-Laplacian on a graph setting for which there are some existence and multiplicity results (see [
5]).
The graph
-Laplacian correspond to anisotropic boundary value problems which on the other hand serve as mathematical models used in elastic mechanics [
6] or image restoration [
7]. Variational continuous anisotropic problems have been started by Fan and Zhang in [
8]. A very extensive study contained in [
9] provides many tools and methods used in the area of anisotropic problems. The general reference for discrete problems is [
10], whereas for the background in graph theory we refer to [
11]. The existence of positive solutions follows the pattern employed for problems with constant
p.
The paper is organized as follows. We start with a summation-by-parts formula and some inequalities that are used further on. The summation-by-parts formula is used in derivation of a notion of a weak solution, whereas relevant inequalities are used in order to apply monotonicity techniques. However, these may useful also when applying other nonlinear analysis tools. Next, existence and uniqueness is concerned. Finally the question of the existence of positive solutions is undertaken.
2. Preliminary and Auxiliary Results
For the weak formulation of problem (
1), we need the following lemma, which is in fact a graph variant of summation by parts
Lemma 1. For any pair of functions , if for all , we havewhere is a Euclidean scalar product in . Proof. Let
. By Formulas (
3)–(
5) we have
We want to emphasize that the symmetry of function
s is obligatory here, because we need equality
. □
Next, we will derive a weak formulation of problem (
1). Multiplying Equation (
1) by some
, and then by taking summation on both sides and using Lemma 1, we get
If the above equation is fulfilled for all
, then it is equivalent to having Equation (
1) for all
. To prove that, for each
, we take
, where
This implies (
1) for all
.
Let us also introduce a notation used throughout the paper, namely
For the case of our problem, we need several inequalities known from [
12]. We provide the suitable proofs due to many subtle differences. These differences arise from the fact of how we treat the boundary of the graph. This is in contrast to [
12] where the boundary plays its role, and therefore there appear several restrictions with which we do not need to cope.
Lemma 2. On the space A the following inequalities hold.
- (i)
For every and for every we have - (ii)
For every and for every we have - (iii)
Let for every , and we have - (iv)
For every and for every we have
Proof. To see (i), note that for any fixed
we have
so
Therefore for every
we obtain
Relation (ii) is obtain from the triangle inequality and from inequality (i), namely
At first with inequality (iii), the most important thing is to divide sum with respect to the 1, to effectively use values of
and
Now we can use inequality (ii) to get the following:
Moreover,
Consequently, we finally have
We will show that (iv) holds. First, we will use discrete Hölder’s inequality with
and
for
Then we use simple transformations to get our inequality
and finally
In the end, we notice that for
we get
so it works also for
. □
Lemma 3 (see [
13] p. 3).
Let , and . Then we have the following inequalitywhere ; moreover denotes an inner scalar product in and denotes a Euclidean norm in . The methods which we employ rely on monotonicity notions (see [
14]). By
A we denote the finite dimensional real Banach space, by
its dual,
denotes the duality pairing between
and
A, i.e., the scalar product.
Definition 2. Operator is called
- (i)
bounded if it maps bounded sets into bounded sets,
- (ii)
- (iii)
for all , and (iv) strictly monotone if for any .
Theorem 1. Let be continuous and coercive. Then T is surjective, i.e., for any , there is at least one solution to the equation . If A is additionally strictly monotone, then A is a homeomorphism.
3. Existence and Uniqueness of Solutions
In order to use the monotonicity tools, we introduce the following operator
where
We observe that when if for
it follows that
then
u is a solution to (
1).
Lemma 4. (i) If for all and for all we havewhere , , then operator T is coercive, and (ii) if for all and for all , we havewhere , , and then operator is coercive. Proof. In the first case, we want to find estimate for the operators
,
from the bottom. From the definition of
we have
The second estimate by (iv) from Lemma 2
Hence we have
Moreover, from (
7) we obtain
Finally, we see that
because
operators
and
dominate, and it is easy to see that operator
T is coercive.
In the second case, we want to find an estimate for the operators
,
from the above. First, estimate by (iii) from Lemma 2
Secondly, estimate by (i) from Lemma 2
Now we see that
where
and
Moreover, from (
8) we have
Finally, we see that
Because
operator
dominates, and it is easy to see that oparator
is coercive. □
Now we give an example to illustrate condition (
7).
Example 1. Let and . PutObserve thatandso condition (7) is satisfied with and To illustrate condition (
8) let us take the following example.
Example 2. Let , and . PutObserve thatandso condition (7) is satisfied with and Theorem 2. If assumption (
7)
or (
8)
holds, then problem (
1)
has at least one solution. Proof. Lemma 3 implies that
T or
is coercive. Moreover
T and
are continuous, so the assumptions of Theorem 1 are fulfilled. This means that there exists a solution to problem (
6) that is equivalent to the existence of the solution to problem (
1). □
We can observe that the operators have the following properties.
Lemma 5. (i) Operator is monotone, (ii) Operator is strictly monotone, (iii) if we assume that the continuous function f does not depend on , so and satisfiesfor all and all u, , then operator is monotone. Proof. By direct calculations and by Lemma 3, we get
By direct calculations and by Lemma 3, we also have
By assumption (
9)
so
is monotone.
□
Theorem 3. If assumption (
7)
or (
8)
holds, and moreover assumption (
9)
is fulfilled, then problem (
1)
has a unique solution. Proof. By direct calculations and by Lemma 3,
By direct calculations and by Lemma 3,
By assumption (
9),
operator
is monotone. □
Now we give an example to illustrate the above theorem.
Example 3. Let and . PutObserve thatandso condition (
7)
is satisfied for all , and all with and Note also thatfor all and all , , so condition (
9)
is satisfied. The assumptions of Theorem 3 are fulfilled, so problem (
1)
has a unique solution with any function As far as we know, Theorem 2 cannot be obtained by a variational method because the function f depends on .
On the other hand, the assertion from Theorem 3 can be obtained by variational methods, with small changes in assumptions.
4. Positive Solutions
In this section, we will look for positive solutions to problem (
1). By a positive solution to problem (
1), we mean the solution, which has only positive values on
V. Positive solutions to (
1) are again investigated in the space
A, considered with the norm (
2).
We introduce the following notion:
It is easy to see that for all
we have
If we assume that
then we can define a new problem,
The solution to this problem fulfilled (
11) for all
. The only difference between (
11) and (
6) is the right hand side, i.e., the second argument of
f in (
11) is
not
.
Let us formulate an auxiliary result which plays an important role in proving all the existence results in this section. This result shows that any solution to (
6) is in fact a positive solution and simultaneously it is the positive solution to (
1). It may be viewed as a kind of a discrete maximum principle.
Lemma 6. Assume that (
10)
holds, and assume that is a solution to problem (
11).
Then is a positive solution to (
1)
. Proof. A straightforward computation shows that for every
and
, we have the following inequality:
Moreover,
holds, because
Assume that
is a solution to (
11). We set
, and we get
Because
f and
q are functions with positive values only and because (
12) holds, the terms on the right hand side of (
14) are together non-negative.
Due to (
13), the term on the left hand side of (
14) is non-positive.
Therefore, Equation (
14) holds only if its both sides are equal to zero, which leads to
for all
Thus,
for all
, which means that for
problems (
6) and (
11) are equivalent, so in fact
is solution of (
1).
Moreover,
for all
. Indeed, assume that there exists
such that
. Then, by (
1) we have
Because the term on the left is non-positive and the term on the right is positive, we have a contradiction. Thus
for all
, it follows that
is a positive solution to (
1). □
To illustrate assumption (
10), take the following example.
Example 4. Let and putIt is easy to see thatfor all , for all and for all Remark 1. We can easily get analogical results for negativity of solutions, by a change of signs in assumption (
10),
in exactly the same way. Remark 2. Theorems 2 and 3 now hold when we add condition (
10)
to its assumptions. 5. Final Comments
With reference to the graph domain setting, some questions and comments arise. We underline here that due to our approach we do not need to take the boundary into account, contrary to what is known from the literature. Should the graph became disjointed, we would obtain two different settings, and in fact two different problems would come under consideration. Toward the multiplicity of solutions, one may employ variational tools and critical point theory (see for example [
15] for some tools available).
Now we turn the question of the Hadamard well-posedness of the given problem. To be more precise, let us formulate some background. Let
Z be a metric space and let
. We can consider the following family of problems for
with assumptions (independent on the parameter) leading to the existence and uniqueness of solutions employed in
Section 3. By using the continuity of the solution operator in the uniqueness case as follows from the approach suggested in [
16], we may obtain the Hadamard well-posedness, which says, roughly speaking, that small deviations from the unique solution return to such solution in the limit. Nevertheless, in case the solution is nonunique, we suggest some continuous dependence on parameters result as well. However, in this case we can approximate some solution with a subsequence only. Moreover, no procedure for how to obtain such subsequence is to be recovered, as suggested by the case of the Galerkin-type approximations in the non-unique case.
Another question concerns the structural stability, i.e., some changes introduced in the graph on which the problem is defined. Such a question is related to the above in case the weights are continuous, and we may proceed as sketched in the above remarks.