For second-order equations with nonnegative characteristic form, Keldys [1] and Fichera presented a kind of boundary that is the Keldys-Fichera boundary value problem, with that the associated problem is of well-posedness. However, for higher-order ones, the discussion of well-posed boundary value problem has not been seen. Here we will give a kind of boundary value condition, which is consistent with Dirichlet problem if the equations are elliptic, and coincident with Keldys-Fichera boundary value problem when the equations are of second-order.
We consider the linear partial differential operator
where
is an open set, the coefficients of
are bounded measurable functions, and 
Let
be a series of functions with
. If in certain order we put all multiple indexes
with
into a row
, then
can be made into a symmetric matrix
. By this rule, we get a symmetric leading term matrix of (2.1), as follows:
Suppose that the matrix
is semipositive, that is,
and the odd order part of (2.1) can be written as
where
is the Kronecker symbol. Assume that for all
, we have
We introduce another symmetric matrix
where
is the outward normal at
. Let the following matrices be orthogonal:
satisfying
where
is the transposed matrix of
are the eigenvalues of
and
are the eigenvalues of
. Denote by
For multiple indices
means that
. Now let us consider the following boundary value problem,
for all
and
, where
.
We can see that the item (2.13) of boundary value condition is determined by the leading term matrix (2.2), and (2.12) is defined by the odd term matrix (2.6). Moreover, if the operator
is a not elliptic, then the operator
has to be elliptic.
In order to illustrate the boundary value conditions (2.11)–(2.13), in the following we give an example.
Example 2.1.
Given the differential equation
Here
. Let
and
, then the leading and odd term matrices of (2.15) respectively are
and the orthogonal matrices are
We can see that
, and
as shown in Figure 1.The item (2.12) is
and the item (2.13) is
for all
and
. Since only
, hence we have
however,
and
, therefore,
Thus the associated boundary value condition of (2.15) is as follows:
which implies that
is free on 
Remark 2.2.
In general the matrices
and
arranged are not unique, hence the boundary value conditions relating to the operator
may not be unique.
Remark 2.3.
When all leading terms of
are zero, (2.10) is an odd order one. In this case, only (2.11) and (2.12) remain.
Now we return to discuss the relations between the conditions (2.11)–(2.13) with Dirichlet and Keldys-Fichera boundary value conditions.
It is easy to verify that the problem (2.10)–(2.13) is the Dirichlet problem provided the operator
being elliptic (see [11]). In this case,
for all
. Besides, (2.13) run over all
and
, moreover
is nondegenerate for any
. Solving the system of equations, we get
.
When
, namely,
is of second-order, the condition (2.12) is the form
and (2.13) is
Noticing
thus the condition (2.13) is the form
It shows that when
, (2.12) and (2.13) are coincide with Keldys-Fichera boundary value condition.
Next, we will give the definition of weak solutions of (2.10)–(2.13) (see [12]). Let
where
is defined by
We denote by
the completion of
under the norm
and by
the completion of
with the following norm
Definition 2.4.
is a weak solution of (2.10)–(2.13) if for any
, the following equality holds:
We need to check the reasonableness of the boundary value problem (2.10)–(2.13) under the definition of weak solutions, that is, the solution in the classical sense are necessarily the solutions in weak sense, and conversely when a weak solution satisfies certain regularity conditions, it will surely satisfy the given boundary value conditions. Here, we assume that all coefficients of
are sufficiently smooth.
Let
be a classical solution of (2.10)–(2.13). Denote by
Thanks to integration by part, we have
Since
, we have
Because
satisfies (2.12),
From the three equalities above we obtain (2.30).
Let
be a weak solution of (2.10)–(2.13). Then the boundary value conditions (2.11) and (2.13) can be reflected by the space
. In fact, we can show that if
, then
satisfies
Evidently, when
, we have
If we can verify that for any
, (2.36) holds true, then we get
which means that (2.35) holds true. Since
is dense in
, for
given, let
and
in
. Then
Due to
satisfying (2.36), hence
satisfies (2.36). Thus (2.31) is verified.
Remark 2.5.
When (2.2) is a diagonal matrix, then (2.13) is the form
where
In this case, the corresponding trace embedding theorem can be set, and the boundary value condition (2.13) is naturally satisfied. On the other hand, if the weak solution
of (2.10)–(2.13) belongs to
for some
, then by the trace embedding theorems, the condition (2.13) also holds true.
It remains to verify the condition (2.12). Let
satisfy (2.30). Since
, hence we have
On the other hand, by (2.30), for any
, we get
Because the coefficients of
are sufficiently smooth, and
is dense in
, equality (2.41) also holds for any
. Therefore, due to
, we have
From (2.36), one drives
Furthermore,
From (2.30) and (2.42), one can see that
Noticing
in
, one deduces that
satisfies (2.12) provided
Finally, we discuss the well-posedness of the boundary value problem (2.10)–(2.13).
Let
be a linear space, and
be the completion of
, respectively, with the norm
. Suppose that
is a reflexive Banach space and
is a separable Banach space.
Definition 2.6.
A mapping
is called to be weakly continuous, if for any
in
, one has
Lemma 2.7 (see [3]).
Suppose that
is a weakly continuous, if there exists a bounded open set
, such that
then the equation
has a solution in
.
Theorem 2.8 (existence theorem).
Let
be an arbitrary open set,
and
. If there exist a constant
and
such that
where
is the component of
corresponding to
, then the problem (2.10)–(2.13) has a weak solution in
.
Proof.
Let
be the inner product as in (2.31). It is easy to verify that
defines a bounded linear operator
. Hence
is weakly continuous (see [3]). From (2.42), for
we drive that
Hence we obtain
Thus by Hölder inequality (see [13]), we have
By Lemma 2.7, the theorem is proven.
Theorem 2.9 (uniqueness theorem).
Under the assumptions of Theorem 2.8 with
in (2.48). If the problem (2.10)–(2.13) has a weak solution in
, then such a solution is unique. Moreover, if
in
,
, then the weak solution
of (2.10)–(2.13) is unique.
Proof.
Let
be a weak solution of (2.10)–(2.13). We can see that (2.30) holds for all
. Hence
is well defined. Let
. Then from (2.49) it follows that
, we obtain
, which means that the solution of (2.10)–(2.13) in
is unique. If all the odd terms
of
, then (2.30) holds for all
, in the same fashion we known that the weak solution of (2.10)–(2.13) in
is unique. The proof is complete.
Remark 2.10.
In next subsection, we can see that under certain assumptions, the weak solutions of degenerate elliptic equations are in
.