We study the following Dirichlet boundary value problem:

Definition 3.1.

We say that

is a weak solution to (3.1) if

for all
and a given
.

Theorem 3.2.

Assume that conditions (2.1)–(2.6) are satisfied. Then there exists a unique weak solution
to the Dirichlet boundary value problem (3.1) in the sense of Definition 3.1. In addition, the solution
satisfies the following properties.

we have

where
is a constant independent of
and
;

The idea behind the existence proof is related to [15, 16]. We will first consider the following Obstacle Problem.

Problem 3.

Find a function

in

such that

for all

. Here

Lemma 3.3.

If
is nonempty, then there is a unique solution p to the Problem 3 in
.

Proof of Lemma 3.3.

Our proof will use Proposition 2.2.

Let

and write

It follows from the proof in [15, Proposition 17.2] that
is a closed convex set.

Next we define a mapping

by

Here we used Assumption (2.6), that is,
. Therefore we have
whenever
. Moreover, it follows from inequality (2.7) that
is monotone.

To show that

is coercive on

, fix

. Then

Inequality (2.8) is used to arrive at the last step. This implies that
is coercive on
.

Finally, we show that

is weakly continuous on

. Let

be a sequence that converges to an element

in

. Select a subsequence {

} such that

a.e. in

. Then it follows that

a.e. in

. Moreover,

weakly in

. Since the weak limit is independent of the choice of the subsequence, it follows that

weakly in
. Hence
is weakly continuous on
. We may apply Proposition 2.2 to obtain the existence of
.

Our uniqueness proof is inspired by [15, Lemmas
,
, and Theorem
]. Since
does not satisfy condition (3.4) of
operator in [15], we need to prove the following lemma, which is equivalent to [15, Lemma 3.11]. Then uniqueness can follow immediately from [15, Lemma 3.22].

Lemma 3.4.

If

is a supersolution of (2.16) in

, then

for all nonnegative
.

Proof.

Let

and choose nonnegative sequence

such that

in

. Equation (

2.6) and Hölder inequality imply that

Because

, we obtain

and the lemma follows.

Similar to [15, Corollary 17.3, page 335], one can also obtain the following Corollary.

Corollary 3.5.

Let
be bounded and
. There is a weak solution
to (3.1) in the sense of Definition 3.1.

Proof of Theorem 3.2.

The existence result is given in Corollary 3.5, and we now turn to proof of uniqueness. For a given

, assume that there exists another solution

Then we have that

for all
. If we take
in above equation, from inequality (2.7), we have the following.

(

) when

,

where
is a positive constant;

(

) when

,

Here the Hölder inequality for

, namely,

is applied to the last inequality.

Poincaré's inequality implies that
a.e. We complete the uniqueness proof.

Next we prove (3.3). Taking

in (3.2), we have

From (2.4), and the Hölder inequality, we obtain

Young's inequality with

implies

and (3.3) follows immediately from (2.3) and (2.6).

Finally, we prove (3.4). From weak solution definition (3.2), we know that

Setting

and subtracting

from both sides, we obtain that

Denote the right-hand side by
. Similar to arguments in the uniqueness proof, we arrive at the folloing:

(

) when

,

(

) when

,

Egorov's Theorem implies that for all

, there is a closed subset

of

such that

and

uniformly on

. Application of the absolute continuity of the Lebesgue Integral implies

Theorem 3.2 is proved.