Spectral Theory in Hilbert Space

62 10. BOUNDARY CONDITIONS
locally absolutely continuous derivative and satisfies −u ′′ + qu = f.
Conversely, if u, f ∈ L 2 (I) and this equation is satisfied, then u ∈
D(T ∗ c ) and T ∗ c u = f.
Proof. Let u1 be a solution of −u ′′
1 + qu1 = f. Assume u0 is in
the domain of Tc and put f0 = Tcu0. Integrating by parts twice we get
(10.5)
I
u0f =
I
u0(−u ′′
1 + qu1) =
I
(−u ′′
0 + qu0)u1 =
I
f0u1.
So, if f is orthogonal to the domain of Tc, then u1 is orthogonal to all
compactly supported elements f0 ∈ L2 (I) for which there is a solution
u0 of −u ′′
0 + qu0 = f0 with compact support. By Corollary 10.5 it
follows that u1 solves −v ′′ + qv = 0 so that f = 0. Thus Tc is densely
defined.
The calculation (10.5) also proves the converse part of the lemma.
Furthermore, if u is in the domain of T ∗ c with T ∗ c u = f we obtain
0 = 〈u0, f〉 − 〈f0, u〉 = 〈f0, u1 − u〉. Just as before it follows that u1 − u
solves the equation −v ′′ + qv = 0. It follows that u solves the equation
−u ′′ + qu = f so that Tc ⊂ T ∗ c . The proof is complete.
Being symmetric and densely defined Tc is closeable, and we define
the minimal operator T0 as the closure of Tc and denote the domain
of T0 (the minimal domain) by D0. Similarly, the maximal operator
T1 is T1 := T ∗ c with domain D1 ⊃ D0. Thus the maximal domain D1
consists of all differentiable functions u ∈ L 2 (I) such that u ′ is locally
absolutely continuous function for which T1u = −u ′′ + qu ∈ L 2 (I).
We can now apply the theory of Chapter 9. The deficiency indices
of T0 are accordingly the number of solutions of −u ′′ + qu = iu and
−u ′′ + qu = −iu respectively which are linearly independent and in
L 2 (I). Since there are only 2 linearly independent solutions for each of
these equations the deficiency indices can be no larger than 2. For the
equation (10.3) the deficiency indices are always equal, since if u solves
−u ′′ + qu = λu, then u solves the equation with λ replaced by λ, and
linear independence is preserved when conjugating functions. Thus, for
our equation there are only three possibilities: The deficiency indices
may both be 2, both may be 1, or both may be 0. We shall see later
that all three cases can occur, depending on the choice of q and I.
We will now take a closer look at how selfadjoint realizations are
determined as restrictions of the maximal operator. Suppose u1 and
u2 ∈ D1. Then the boundary form (cf. Chapter 9) is
(10.6) 〈(u1, T1u1), U(u2, T1u2)〉 = i (u1T1u2 − T1u1u2)
= i
I
(−u1u ′′
2 + u ′′
1u2) = −i
I
I
[u1, u2] ′
= −i lim [u1, u2] ,
K→I
K