In certain cases the boundary value problem can be characterized by means of a uniquely determined unbounded self-adjoint operator. In these cases the eigenvalues and eigenfunctions of the boundary value problem are determined by the eigenvalues and eigenvectors of the corresponding operator; these will be called a self-adjoint case of the boundary value problem. In some cases such a characterization is not possible and these will be referred to as ‘symmetric’ cases in general. In classical point of view, our problem cannot be characterized as ‘self-adjoint case’. For ‘self-adjoint characterization’ of the considered problem (1)-(5), we shall define a new Hilbert space as follows.
Denote the determinant of the i th and j th columns of the matrix
by (). Throughout the paper we shall assume that the conditions
hold. Define a new inner-product space ℋ as a direct sum space equipped with the modified inner-product
for . It is easy to see that the relation (6) really defines a new inner product in the direct sum space .
Lemma 1 ℋ is a Hilbert space.
Proof Let , , be any Cauchy sequence in ℋ. Then by (6) the sequences and will be Cauchy sequences in the Hilbert spaces and , respectively. Therefore they are convergent. Let and be limits of these sequences, respectively. Defining we have that and in ℋ. The proof is complete. □
Let us now define the boundary and transmission functionals , , , , , , where , and the linear operator with the domain
and action low
Then problem (1)-(5) can be written in the operator equation form as , in the Hilbert space ℋ.
Theorem 1 The linear operator ℜ is symmetric in the Hilbert space ℋ.
Proof By applying the method of  it is not difficult to prove that the operator ℜ is densely defined in ℋ, i.e., . Now let . By partial integration we have
where, as usual, denotes the Wronskians of the functions u and . From the definitions of boundary functionals we get that
Further, taking in view the definition of ℜ and initial conditions (14)-(19) we can derive that
Finally, substituting (8), (9) and (10) in (7) we obtain that
The proof is complete. □
Theorem 2 The linear operator ℜ is self-adjoint in ℋ.
Proof Since ℜ is symmetric and densely defined on ℋ, it is sufficient to show that if
for all , then and , where and . Writing equality (11) for all by standard Sturm-Liouville theory, we find that and . Then from equality (11) it follows that
On the other hand, by two partial integrations we get
From this equality, by applying the technique of Theorem 2.5 in our previous work , it can be derived easily that , , , and , , . The proof is complete. □
Theorem 3 The operator ℜ has only point spectrum, i.e., .
Proof It suffices to prove that if is not an eigenvalue of ℜ, then is a regular point of ℜ, i.e., . Let not be an eigenvalue of ℜ. The resolvent operator exists and is defined on all of ℋ. By Theorem 2 and the closed graph theorem, we get that is bounded. Thus, . Hence . □