For second-order equations with nonnegative characteristic form, Keldys  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:
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.
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
In general the matrices and arranged are not unique, hence the boundary value conditions relating to the operator may not be unique.
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 ). 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
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 ). Let
where is defined by
We denote by the completion of under the norm and by the completion of with the following norm
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.
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
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.
A mapping is called to be weakly continuous, if for any in , one has
Lemma 2.7 (see ).
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 .
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 ). From (2.42), for we drive that
Hence we obtain
Thus by Hölder inequality (see ), 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.
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.
In next subsection, we can see that under certain assumptions, the weak solutions of degenerate elliptic equations are in .