4.1. Fractional Boundary Conditions
In this part we develop a theory, analogous to the Sturm–Liouville theory, for equations in the form (
26)–(
28) or, equivalently, (
23)–(
25). We recall that the classical Sturm–Liouville theory is recovered by letting
.
The basic idea behind Sturm’s theory shines in the formulation of its oscillation theorem for the eigenfunctions arising from the eigenvalue problem. This is usually referred to as
Sturm’s oscillation theoremThus, when , the convexity is well understood. Things are not so clear if and have sign changes, but there are studies in this direction. Normally, we will assume that and is continuous in but the first few results here are of a general nature and do not necessarily involve the positivity of .
Now, we consider the fractional eigenvalue problem
subject to the boundary conditions
in which
are constants and we are looking for values of
such that there is a non-trivial solution satisfying the boundary conditions (
24) and (
25). Such a solution, when it exists, will be called an eigenfunction corresponding to the eigenvalue
.
Remark 1. We note that (16) has a leading part that is a classical Sturm–Liouville operator (i.e., with ordinary derivatives) while the remaining part consists of fractional derivatives. Thus, whenever y is a solution to (3), the functions themselves satisfy a Sturm-type differential equation. We shall see below that this transformation to the hybrid form (16) has definite advantages for spectral theory. Replacing
q with
in (
3) and applying Lemma 1, the eigenvalue problem (
23), subject to both (
24) and (
25) is now converted to a hybrid eigenvalue problem of the form,
but now, as per Lemma 1, it is subject to the transformed boundary conditions,
Remark 2. The existence and uniqueness theorems for initial value problems proved in [13] for (3) now carry over to the Equation (26) for each λ. So, in particular, the initial value problem for an equation of the formhas a unique -solution in Remark 3. Once we have two solutions, of (16) such that , (so they must be linearly independent) then any other solution φ can be written as a linear combination of these two, as all we need are its initial values and these are given at the outset to ensure its existence and uniqueness. Remark 4. Note that, on account of the earlier existence results, solutions z of (26) are at least absolutely continuous functions with which is also absolutely continuous and (26) is satisfied a.e. In addition, note that the transformation in Lemma 1 is isospectral in the sense that the spectrum is preserved when passing from (23) to (26). We start with the basic theory that will be necessary in the sequel. Once again, as this is an inaugural paper in a new area, we will consider the homogeneous Dirichlet problem for (
26).
Theorem 1. Let not be identically zero. Then, any solution z (real or complex) to the proble, (26) with satisfies Proof. By multiplying (
26) by
and integrating we obtain, after simplification,
By applying Proposition 7 to the second integral, we find that
Since
, Proposition 3 and its conjugation shows us that
with a similar relation to
, i.e.,
from which there follows (
29). □
The next result is analogous to the ordinary Sturm–Liouville case and deals with weight functions w that may possibly change sign in .
Theorem 2. Assume that are continuous in . Let somewhere in , in and in . Then, any eigenvalue of the Dirichlet problem associated with (26) must be real. Proof. From Theorem 1 we know that such a solution
z must satisfy,
Let
, Im
be a possibly complex eigenvalue and
z a corresponding complex eigenfunction. Then,
Since
are real-valued we can take the imaginary part of both sides to find that
However, since
, we must have
Thus, the left hand side of (
30) must be zero. The positivity of
now implies that
identically on
. Since
z is absolutely continuous, this implies that
is a constant, i.e.,
, contrary to the hypothesis that
z is an eigenfunction. Thus,
and all eigenvalues are real. □
Remark 5. The proof above is actually valid for the measurable coefficients subject to the usual conditions for the existence and uniqueness of solutions to the initial value problems, see [13]. Next, we exhibit an orthogonality relationship between the eigenfunctions corresponding to distinct eigenvalues.
Theorem 3. Let w be continuous and non-zero somewhere in , q be continuous, and p be continuous and positive (although this too can be relaxed). Then eigenfunctions and of (26), corresponding to distinct eigenvalues from the Dirichlet problem, are orthogonal in the sense that Proof. Let
and
(
) be the eigenvalue of the eigenfunctions
and
, respectively. Then,
with a similar equation for
in the form
Multiplying (
31) by
and (
32) by
, and then integrating the first term by parts and the second term by parts using Proposition 7, we obtain, after some simplification,
and
Subtracting the last two equations we obtain
and the result follows. □
The next result will be useful later on and deals with the uniqueness of solutions to certain initial value problems associated with (
26).
Theorem 4 ([
13], Theorem 3.4)
. Let be measurable complex-valued functions satisfying andIn addition, let ,and assume that . Then, the only solution to the initial value problem (3) that satisfiesand is continuous on is the trivial solution. Now, as is well known in the ordinary Sturm–Liouville theory, we show that any (non-trivial) solution to (
26) can only have a finite number of zeros in
, if any at all.
Lemma 2. Let , there. Consider the solution to the problemwhere , . Then, z can only have a finite number of zeros in . Proof. The solution is clearly non-trivial. We assume, on the contrary, that this solution has an infinite number of zeros, say ; then, they must accumulate somewhere at , say . Since by existence and (as ), it follows that . On the other hand, since , . Since z and are both zero at , this violates Theorem 4 as it implies that z is the trivial solution. The case where is similar and so is omitted. □
Corollary 1. For a fixed λ, every non-trivial solution y to (23) satisfying , , has the property that has a finite number of zeros. Proof. Replace
q in (
37) by
. If, on the contrary, there was a solution
y with an
with an infinite number of zeros, then, according to Rolle’s theorem, so would its derivative, i.e.,
. But this would contradict Lemma 2 in our equations. □
Remark 6. The relationship between the number of zeros of a function and the number of zeros of its left-Riemann–Liouville integral is an old one and goes back to at least Steinig [17]. There the author shows that, generally speaking, there exists functions f having a finite number of zeros whose left-Riemann–Liouville integral has an infinite number of zeros. (Here, we showed that, if f satisfies some specific differential equation, then this is impossible, i.e., the left-Riemann–Liouville integral must have a finite number of zeros.) In order to prove the existence of eigenvalues for the boundary problem (
26) that satisfy
, we can consider the solutions
as a function of
, denoted by
, and show that, in fact,
is, for each
an entire function of
. This will show that the zeros of
are at most countable and we will prove that there is at least one of them (thereby proving the existence of eigenvalues). We will also show that the zeros of
(each one of which gives an eigenvalue) move to the left as
increases so that there can be finite limit as to the number of zeros, such as
.
Lemma 3. Let be non-negative continuous functions on satisfyingwhere is a constant. Ifthen Remark 7. At this point one may conjecture that a general a priori-type inequality similar to the Gronwall inequality is valid for (38); for example, something like the following: Conjecture 5.Let be non-negative continuous functions on satisfying (38); then, is a constant. Then,where . Since the right-hand side of (40) is a solution to (38) with equality for the value of c chosen therein, it appears to be a maximal solution. It is not, however, a maximal solution because the initial condition fails, i.e., the initial condition depends on y itself A counterexample to this general conjecture is given by setting , , , for all It is easy to verify that (38) holds for , but since , (40) fails. In this last set of lemmata we assume that all integrals and derivatives exist and provide a more general integration by parts formula when interior points are involved in the limits of integration.
Lemma 4 (Another integration by parts formula)
. For any there holds Proof. Integrating the left hand-side by parts and applying the definitions of the various fractional terms along with Fubini’s theorem, we find
as required. □
Remark 8. We observe that, by setting in Lemma 4, the first part of Proposition 7 holds.
This idea can be utilized to generate analogous formulae for other fractional integrals, as demonstrated below:
Lemma 5. For any there holds, Lemma 6. For any , there holds Proof. and the result follows. □
Lemma 7. For any , there holds Lemma 8. For any , there holds Remark 9. We observe that, by setting in Lemma 8, the second part of Proposition 7 holds.
4.2. Classical Boundary Conditions
Lemma 9 (A Sturm comparison theorem)
. Let be real-valued continuous functions over such that for all with strict inequality for at least one point in . Let be a nontrivial solution to such that in . Then, every real solution, , to has at least one zero in . Proof. Without a loss of generality, we may take it that
on
. The proof is by contradiction, as usual. We assume, on the contrary, that there is a solution
on
. By multiplying (
46) with
by
and (
52) with
by
and subtracting we obtain
for all
. On the other hand, integrating one of these integrals by parts, say the first, and using the second of Proposition 7 we obtain (suppressing the variables of integration for simplicity of form)
By subtracting the previous two equations and simplifying we obtain
Using the transformed variables
z defined in Lemma 1 and the definitions of the various fractional derivatives, we rewrite the left hand side of (
48) so as to obtain
which, when combined with (
48), yields the identity,
However, since both
and
on
, the right-hand side of (
49) must be strictly negative. On the other hand, since
we also have
, i.e.,
on
. Similarly,
on
. But, the positivity of
p and the fact that
are differentiable in
implies that
. Similarly,
. Similar arguments show that, finally,
and that
. Thus, the left side of (
49) is positive or zero while the right-side is strictly negative. This contradiction proves the lemma. □
Remark 10. Note that when this becomes Sturm’s comparison theorem as the Caputo derivatives introduce a negative sign before the leading coefficient.
Corollary 2. Let be real-valued continuous functions over such that for all . Let be a nontrivial solution tosuch that Then, every real solution tohas the property that has at least one zero in . Proof. This is an immediate consequence of Lemmas 1, 9, and Proposition 4. □
Corollary 3. Let and q both be continuous in and with strict inequality in at least one point of . Then every solution to the equationhas at most one zero in . Proof. The proof is by contradiction. Simply use Lemma 9 with
,
. Note that
has the particular solution
whose only zero is at (since ). Consequently, no nontrivial solution of (52) can exist and have two zeros in . □ Lemma 9, can now be used to guarantee the existence of oscillations in .
Theorem 6. Consider the equationsandwhere is a (nontrivial) solution of (53) satisfying and in . If , then every solution to (54) has at least one zero in . Proof. The proof is a straightforward consequence of Lemma 9 as the condition is equivalent to . □
Remark 11. It will be shown that there exists a value of and a corresponding solution to (23) satisfying the classical Dirichlet boundary conditions at , i.e., , and that it is indeed positive in . Of course, this means that is an eigenvalue of (23) with as an eigenfunction. Theorem 6 then guarantees that any other eigenvalues must have eigenfunctions with zeros in . Conjecture 7. The number of zeros in any non-trivial solution to (3) satisfying on is necessarily finite. Discussion: This is clear if according to Corollary 3. If there is a nontrivial solution, y, with , then the must accumulate somewhere at , say . Since by existence, it follows that . There are two then cases: either or .
Let
. Now, (
3) and definition (
9) force
which, in turn, implies that
must change sign an infinite number of times in
. However, it is conceivable that
remains of one sign throughout a closed interval near
b. We will show that this is impossible. Assume for the moment that
for all
and for some
m (and therefore for infinitely many subsequent
m). Since
, we obtain
for all
. On the other hand,
shows us that
and is therefore bounded by
M, say. Thus,
Of course, boundedness near is sufficient in the preceding argument. A case where is unclear and open.