- Open Access
On discreteness of spectrum and positivity of the Green’s function for a second order functional-differential operator on semiaxis
© Labovskiy and Getimane; licensee Springer. 2014
- Received: 5 February 2014
- Accepted: 4 April 2014
- Published: 7 May 2014
We study conditions of discreteness of spectrum of the operator defined by , . The operator has two singularities at the ends of the interval . The second question is positivity of solutions of the equation under boundary conditions , . The used abstract scheme is close to the well-known MS Birman’s method in the spectral theory of self-adjoint operators. Conditions for discreteness of spectrum and positivity of the Green’s operator are obtained. The result relates to the MS Birman’s result on the necessary and sufficient condition for discreteness of spectrum of a polar-differential operation. The results may be interesting for researchers in qualitative theory of functional-differential equations and spectral theory of self-adjoint operators.
MSC: Primary 34K08; 34K10; secondary 34K12.
- discreteness of spectrum
- positive solutions
- singular operator
1.1 Problems and a well-known result
The second question is existence of positive solutions of the equation (see Definition 1.1).
The two conditions (1.3) and (1.4) are sufficient for discreteness of the spectrum of (1.2) (Theorem 2.1).b It seems that (1.4) is also necessary one.
1.2 Assumptions, notation
Everywhere below, except for the independent appendix, we use the assumptions and notation introduced in this subsection.
Conditions (1.3) and (1.4) can be written as , and condition (1.5) is boundedness of the . The latter is sufficient (Lemma 4.3) for the inclusion and for boundedness of the operator T (see, below in this subsection). It is close to a necessary condition: if, for example, , then T is unbounded (see Remark 4.3).
is locally integrable on . We can assume that for almost all (as follows from a property of the Stieltjes integral).
be the Hilbert space of all square-integrable with positive almost everywhere weight ρ on functions f, i.e., and with scalar product
W be the set of all locally absolutely continuous on functions u satisfying(1.9)
be the operator defined by , . Note that under condition (1.5) (Lemma 4.3).
1.3 About domain of the operators and ℒ
We impose the boundary conditions , . Both of these conditions are necessary for the implementation of the variational method in the space of W. Then the domain is determined automatically (see Corollary A.1 to Lemma A.1 and Lemma 4.8).
is positively solvable, if it is uniquely solvable for any , and the implication holds.
or , where G is the Green’s operator of ℒ.
has a discrete spectrum.
Proof The discreteness follows from Theorem A.1 and Lemma 4.7. □
Remark 2.1 (Estimate of the greatest lower bound of the spectrum)
This is an accurate estimate. If , then , but is a point of the spectrum, as follows from Example 2.1.
Example 2.1 If , then , and the operator T is bounded, but the two conditions (1.3) and (1.4) are not fulfilled. The spectrum of the operator in , is the interval .
In fact, the value is regular (Remark 2.1). Let .
this equation has to be considered in the space .
If, for example, , , and , (for the may be defined arbitrarily), then . But the corresponding solution . Thus, λ is not a regular value of the operator.
Since the spectrum is a real closed set, is the spectrum.
2.2 General operator ℒ
where is assumed to be nonnegative locally integrable function, and is a measurable function. Note that the notation in (2.5) corresponds to the definition (1.7), if represent expression (2.5) in the form (2.4).
Theorem 2.2 If conditions (1.3) and (1.4) are fulfilled and is bounded, then spectrum of ℒ is discrete. If the function ξ defined by (1.8) is symmetric, i.e. for all , then the spectrum is real and the system of eigenfunctions has the orthogonal basis properties in .
Proof Conditions (1.3) and (1.4) are sufficient conditions of compactness of the operator T (Lemma 4.7). The symmetry condition of allows one to show the identity (see Section 5). Now we can refer to Theorem A.3. □
Using the estimate (5.6) from Theorem A.3 we have the first main result.
has a discrete spectrum. If ,c the spectrum is real, and the system of eigenfunctions has orthogonal basis properties in the spaces W and .
Remark 2.2 The spectrum is not real in general, because of the non-symmetry of the function .
The obtained estimate works well in the case of one deviation, if Q is defined by (2.5). From (2.6) we have the following.
has a discrete spectrum.
then the operator ℒ has discrete spectrum. In this case (2.6) has the form (2.8). It follows also from (5.4). Note that the inequality (2.8) is satisfied, if is bounded.
Sinced and , the operator is positive, and . Thus:
Theorem 3.1 Suppose Q is bounded and . Then the boundary value problem (1.13), (1.14) is uniquely resolvable for any and the Green’s operator G is positive.
Remark 3.1 Since the condition is sufficient for positivity of the Green’s operator G.
The second main result is presented in the following statement.
then the boundary value problem (1.13), (1.14) is positively solvable (see Definition 1.1).
Proof See Remark 3.1 to Theorem 3.1 and the estimates (4.2) and (5.6). □
Consider the following particular case.
The following particular case shows that in this estimate the inequality sign < cannot be replaced by ≤.
In particular, if , and , . Thus, if , the equation has unique positive solution for any , , .
But if , the equation may not have a solution for some f (this was considered in Example 2.1).
Note that (see the equality (5.2)).
From Theorem A.5 follows the following.
is positive definite,
Remark 3.2 We do not suppose that (but if it is so).
Recall that in all assertions below condition (1.5) is assumed to be fulfilled.
Lemma 4.1 W is a Hilbert space.
Proof The relation establishes a bijection between W and the Hilbert space . □
In fact, denote . Then . Since , .
Now find u such that : . Since , , and is nonincreasing, by Remark 4.2, . But .
Lemma 4.4 The image is dense in .
Proof The proof is left to the reader. □
The following theorem [, p.318] can be used to show compactness.
Theorem 4.1 (Gelfand)
the convergence (4.4) would be uniform on the E.
uniformly on the set .
This tends to zero because of (4.1) and (1.3). □
uniformly on the set .
Now we refer to (1.4) and (4.1). □
Lemma 4.7 If (1.3) and (1.4) hold, then T is compact.
Proof Let . We use the criterium of compactness of Gelfand (see Theorem 4.1). Let be a sequence of functionals such that for any . We have to show that uniformly for .
for all n and for all . The same we can do with for sufficiently small . For this aim we can use Lemma 4.6.
Then . The set is compact. Thus by Theorem 4.1 of Gelfand converges to zero uniformly for these s. So, uniformly for , and uniformly for . □
Remark 4.4 It seems that condition (1.4) is necessary for compactness of T.
Lemma 4.8 Equation (2.1) is equivalent to problem (2.2).
Now, . Choosing v such that , we obtain . □
Here we consider the operator Q defined by (2.4).
5.1 Symmetry of the form
Under the assumptions imposed on the functions and in Section 1.2, from Lemma B.1 (in Appendix 2) we obtain the following statement.
In this case , , .
Hence, this form is symmetric if the function is symmetric: .
5.2 Unique deviation
5.3 General operator Q
Here we consider the general case of the operator Q defined by (2.4), i.e. . Suppose that the function (see (1.8)) is absolutely continuous in y, and . Then does not decrease in x.
The latter step can be done in the same manner as in the relation (5.4).
This estimate coincides with (5.3), if
We do not use general spectral theory (see, for example, [6, 7]). But the scheme below is close to the scheme in [, Chapter 10] except for using a different notation. We find also convenient explicit use of the embedding T from W to H (see below). This scheme was used also in [2–4].
A.1 Positive form
has the unique solution for any , where is adjoint operator. Let .
the image of the operator T is dense in H,
Lemma A.1 If the image of the operator T is dense in H, then is an injection.
Since is dense in H, the . □
Corollary A.1 (Euler equation)
Theorem A.1 The spectrum of ℒ is discrete if and only if T is compact.
Proof Since (A.3) is equivalent to , discreteness of spectrum of ℒ is equivalent to compactness of . But the operators and are compact at the same time [, Chapter 10]. □
Remark A.1 The minimal eigenvalue satisfies the equality (A.4).
A.2 General case
The system can be chosen to form an orthogonal basis in the space W.
is the greatest lower bound of the spectrum [, Chapter 6]. Thus we have the following.
A.3 Positive definiteness and spectral radius of
Lemma A.2 The quadratic form is positive definite if and only if .
Proof From (A.11) it follows that and . If then is positive definite. Conversely from the inequality () it follows that . □
and . If is nonnegative, then spectrum of is in the segment , and ρ is a point of spectrum. In this case from it follows that . So, we have the following.
is positive definite,
We used a change of integration order in an integral, which does not follow from the classic Fubini theorem. The following assertion is taken from the monograph . Note that it was used without proof in [2–4].
- 1.the function ν defined on by the equality
- 2.if is ν-measurable on , then
aThe spectrum of ℒ is discrete if it consists only of eigenvalues of finite multiplicity.
bThe condition is sufficient for (1.4), but it is not necessary (if, for example, for x near zero).
c is defined by (1.8).
d Here I is the identity operator.
eWe call the form associated with the operator Q.
f The measure is a nonnegative, σ-additive function defined on a σ-algebra; the product is the minimal σ-algebra containing the set of all rectangles , , .
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