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On discreteness of spectrum and positivity of the Green’s function for a second order functionaldifferential operator on semiaxis
Boundary Value Problems volume 2014, Article number: 102 (2014)
Abstract
We study conditions of discreteness of spectrum of the operator defined by \mathcal{L}u=\frac{1}{\rho (x)}{u}^{\u2033}(x){\int}_{0}^{\mathrm{\infty}}u(s)\phantom{\rule{0.2em}{0ex}}{d}_{s}r(x,s), x\in [0,\mathrm{\infty}). The operator has two singularities at the ends of the interval (0,\mathrm{\infty}). The second question is positivity of solutions of the equation \mathcal{L}u=f under boundary conditions u(0)=0, {u}^{\prime}(\mathrm{\infty})=0. The used abstract scheme is close to the wellknown MS Birman’s method in the spectral theory of selfadjoint 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 polardifferential operation. The results may be interesting for researchers in qualitative theory of functionaldifferential equations and spectral theory of selfadjoint operators.
MSC: Primary 34K08; 34K10; secondary 34K12.
1 Introduction
1.1 Problems and a wellknown result
Our first objective is to study the conditions for discreteness of spectrum^{a} of the functionaldifferential operator defined by
with two singularities: at x=0 and at infinity. Note that one particular case of the expression (1.1) is the following operator with one deviation:
The second question is existence of positive solutions of the equation \mathcal{L}u=f (see Definition 1.1).
Note the result of Birman [[1], (Chapter 2, Section 29)] for the spectral problem
where the operation {\mathcal{L}}_{0} is called polardifferential operation. This singular spectral problem is usually considered in the space {L}_{2}({R}_{+},\rho ) of functions that are squareintegrable on {R}_{+} with the positive weight ρ. Birman showed that a necessary and sufficient condition of discreteness of spectrum of the operator {\mathcal{L}}_{0} is
The singularity at the point x=0 is not reflected in this condition. If {\int}_{0}^{1}\rho (x)\phantom{\rule{0.2em}{0ex}}dx=\mathrm{\infty}, condition (1.3) is not sufficient for discreteness. We impose the second condition,
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.
Part of this work is a continuation of the research in the articles [2–4].
1.2 Assumptions, notation
Everywhere below, except for the independent appendix, we use the assumptions and notation introduced in this subsection.
The function ρ is assumed to be measurable and positive almost everywhere on {R}_{+}=[0,\mathrm{\infty}) and satisfying the important condition
Remark 1.1 Comparing (1.3), (1.4), and (1.5) we see that the key role is played by the properties of the function
Conditions (1.3) and (1.4) can be written as {lim}_{s\to \mathrm{\infty}}\mathrm{\Phi}(s)={lim}_{s\to 0}\mathrm{\Phi}(s)=0, and condition (1.5) is boundedness of the \mathrm{\Phi}(s). The latter is sufficient (Lemma 4.3) for the inclusion W\subset {L}_{2}({R}_{+},\rho ) and for boundedness of the operator T (see, below in this subsection). It is close to a necessary condition: if, for example, {lim}_{s\to \mathrm{\infty}}\mathrm{\Phi}(s)=\mathrm{\infty}, then T is unbounded (see Remark 4.3).
Everywhere below the function r(x,s) is assumed to satisfy the following conditions: it is nondecreasing in s\in {R}_{+} for almost all x\in {R}_{+}, it is measurable in x for any s\in {R}_{+}, and
is locally integrable on (0,\mathrm{\infty}). We can assume that r(x,0)=0 for almost all x\in {R}_{+} (as follows from a property of the Stieltjes integral).
Now, let us introduce the function
Finally, let

{L}_{2}({R}_{+},\rho ) be the Hilbert space of all squareintegrable with positive almost everywhere weight ρ on {R}_{+}=[0,\mathrm{\infty}) functions f, i.e.{\int}_{{R}_{+}}f{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx<\mathrm{\infty}, and with scalar product
(f,g)={\int}_{{R}_{+}}f(x)g(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx, 
W be the set of all locally absolutely continuous on {R}_{+} functions u satisfying
{\int}_{{R}_{+}}{u}^{\prime}{(x)}^{2}\phantom{\rule{0.2em}{0ex}}dx<\mathrm{\infty},(1.9)
and the boundary condition u(0)=0; W is a Hilbert space (Lemma 4.1) with scalar product
and

T:W\to {L}_{2}({R}_{+},\rho ) be the operator defined by Tu(x)=u(x), x\in {R}_{+}. Note that W\subset {L}_{2}({R}_{+},\rho ) under condition (1.5) (Lemma 4.3).
1.3 About domain of the operators {\mathcal{L}}_{0} and ℒ
If λ is a regular value of {\mathcal{L}}_{0}, then any solution of \frac{1}{\rho}{u}^{\u2033}\lambda u=f satisfies \frac{1}{\rho}{u}^{\u2033}\in {L}_{2}({R}_{+},\rho ). So, we have to assume that the domain D({\mathcal{L}}_{0}) of the operator {\mathcal{L}}_{0} consists of all solutions of the equation
These solutions have locally on (0,\mathrm{\infty}) absolutely continuous derivative {u}^{\prime} and have the form
where {G}_{0}(x,s)=\rho (s)min\{x,s\}. In fact, since {({\int}_{x}^{\mathrm{\infty}}\rho z\phantom{\rule{0.2em}{0ex}}ds)}^{2}\le {\int}_{x}^{\mathrm{\infty}}\rho {z}^{2}\phantom{\rule{0.2em}{0ex}}ds{\int}_{x}^{\mathrm{\infty}}\rho \phantom{\rule{0.2em}{0ex}}ds the integral is convergent, and {u}^{\prime}(x)={\int}_{x}^{\mathrm{\infty}}\rho (t)z(t)\phantom{\rule{0.2em}{0ex}}dt+{u}^{\prime}(\mathrm{\infty}). From this we have
We impose the boundary conditions u(0)=0, {u}^{\prime}(\mathrm{\infty})=0. 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).
Definition 1.1 We say that the boundary value problem
is positively solvable, if it is uniquely solvable for any f\in {L}_{2}({R}_{+},\rho ), and the implication f\ge 0\to u\ge 0 holds.
Note that positive solvability is equivalent to the positivity of the Green’s function G(x,s), which allows one to represent the solution of \mathcal{L}u=f in the form
or u=Gf, where G is the Green’s operator of ℒ.
2 Discreteness of spectrum
2.1 Operator {\mathcal{L}}_{0}
Instead of direct investigation of the equation {\mathcal{L}}_{0}u=f, where f\in {L}_{2}({R}_{+},\rho ), consider the equation
Equation (2.1) is the result of the variational method, in which the primary object is the form [u,v]. Equation (2.1) has the short form
defined on the spaces W and H={L}_{2}({R}_{+},\rho ). Note that W is a Hilbert space (Lemma 4.1), T:W\to {L}_{2}({R}_{+},\rho ) is bounded (Lemma 4.3), and T(W) is dense in {L}_{2}({R}_{+},\rho ). So, we can use the abstract scheme in Appendix 1. According to Corollary A.1 (2.1) is equivalent to an equation {\mathcal{L}}_{0}u=f. By virtue of Lemma 4.8 (2.1) is equivalent to
Theorem 2.1 Suppose conditions (1.3) and (1.4) hold. Then the spectral problem
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)
Let {\lambda}_{0} be the greatest lower bound of spectrum of the operator {\mathcal{L}}_{0}. Thus the problem
for \lambda <{\lambda}_{0} is uniquely resolvable for any f\in {L}_{2}({R}_{+},\rho ), but it is not if \lambda ={\lambda}_{0}. Then in view of (A.4) from inequality (4.3) it follows that
This is an accurate estimate. If \rho (x)=1/{x}^{2}, then {\lambda}_{0}\ge 1/4, but \lambda =1/4 is a point of the spectrum, as follows from Example 2.1.
Example 2.1 If \rho (x)=1/({x}^{2}), then s{\int}_{s}^{\mathrm{\infty}}\rho (s)\phantom{\rule{0.2em}{0ex}}ds\equiv 1, and the operator T is bounded, but the two conditions (1.3) and (1.4) are not fulfilled. The spectrum of the operator {\mathcal{L}}_{0}u={x}^{2}{u}^{\u2033} in {L}_{2}({R}_{+},\rho ), \rho =1/({x}^{2}) is the interval [1/4,\mathrm{\infty}).
In fact, the value \lambda <1/4 is regular (Remark 2.1). Let \lambda >1/4.
By means of the change of variable x={e}^{t} the equation
can be transformed to {u}^{\u2033}(t)+{u}^{\prime}(t)\lambda u=\phi, where \phi =f({e}^{t}). Since
this equation has to be considered in the space {L}_{2}(R,{e}^{t}).
The homogeneous equation {u}^{\u2033}+{u}^{\prime}\lambda u=0 has the solution u={e}^{(1/2)t}({c}_{1}cos\delta t+{c}_{2}sin\delta t) for a \delta >0. It is not in {L}_{2}(R,{e}^{t}). Let u={e}^{(1/2)t}sin\delta tv(t)=b(t)v(t). Then
If, for example, v=1/\sqrt{t}, t\ge 1, and v=0, t<0 (for 0\le t\le 1 the v(t) may be defined arbitrarily), then \phi \in {L}_{2}(R,{e}^{t}). But the corresponding solution u\notin {L}_{2}(R,{e}^{t}). Thus, λ is not a regular value of the operator.
Since the spectrum is a real closed set, [1/4,\mathrm{\infty}) is the spectrum.
2.2 General operator ℒ
The operator (1.1) can be represented as \mathcal{L}={\mathcal{L}}_{0}Q, where Q is defined by
The operator Q acts from W to {L}_{2}({R}_{+},\rho ) and is bounded under certain conditions (see, for example, (5.6)). Along with the general case, let us consider one special case (deviating operator):
where q(x) is assumed to be nonnegative locally integrable function, and h(x) is a measurable function. Note that the notation q(x) 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 Q:W\to {L}_{2}({R}_{+},\rho ) is bounded, then spectrum of ℒ is discrete. If the function ξ defined by (1.8) is symmetric, i.e. \xi (x,y)=\xi (y,x) for all x,y\in {R}_{+}, then the spectrum is real and the system of eigenfunctions has the orthogonal basis properties in {L}_{2}({R}_{+},\rho ).
Proof Conditions (1.3) and (1.4) are sufficient conditions of compactness of the operator T (Lemma 4.7). The symmetry condition of \xi (x,y) allows one to show the identity (Qu,Tv)=(Qv,Tu) (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.
Theorem 2.3 Suppose (1.3) and (1.4) hold and
Then the spectral problem
has a discrete spectrum. If \xi (x,y)=\xi (y,x),^{c} the spectrum is real, and the system of eigenfunctions has orthogonal basis properties in the spaces W and {L}_{2}({R}_{+},\rho ).
Remark 2.2 The spectrum is not real in general, because of the nonsymmetry of the function \xi (x,y).
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.
Corollary 2.1 Suppose (1.3) and (1.4) hold and
Then the spectral problem
has a discrete spectrum.
Example 2.2 If Qu(x)=q(x)u(x) the operator ℒ has the representation
If (1.3) and (1.4) are satisfied and
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 q(x) is bounded.
3 Positive solvability
Suppose Q:W\to {L}_{2}({R}_{+},\rho ) is bounded. By the substitution u={T}^{\ast}z the equation \mathcal{L}u=f is reduced to the equation
where K=Q{T}^{\ast} is an integral operator with nonnegative kernel,
Thus, if spectral radius of K is less than unit (\rho (K)<1), then (3.1) is uniquely resolvable and
Since^{d} z={(IK)}^{1}f and u={G}_{0}z, the operator G={G}_{0}{(IK)}^{1} is positive, and G(x,s)\ge 0. Thus:
Theorem 3.1 Suppose Q is bounded and \rho (K)<1. Then the boundary value problem (1.13), (1.14) is uniquely resolvable for any f\in {L}_{2}({R}_{+},\rho ) and the Green’s operator G is positive.
Remark 3.1 Since \rho (K)\le \parallel K\parallel =\parallel {T}^{\ast}Q\parallel \le \parallel {T}^{\ast}\parallel \parallel Q\parallel =\parallel T\parallel \parallel Q\parallel the condition \parallel T\parallel \parallel Q\parallel <1 is sufficient for positivity of the Green’s operator G.
The second main result is presented in the following statement.
Theorem 3.2 If
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.
Corollary 3.1 The equation
(q\ge 0) has a positive solution in W for any f\in {L}_{2}({R}_{+},\rho ), f\ge 0, f\ne 0, if
The following particular case shows that in this estimate the inequality sign < cannot be replaced by ≤.
Example 3.1 From Theorem 3.1 and (5.4) it follows that the equation
(q(x)\ge 0) is uniquely resolvable for f\in {L}_{2}({R}_{+},\rho ) in W and has positive solution for f\ge 0, if
In particular, if q(x)\le \mathrm{const}=q, and \rho (x)=1/{x}^{2}, \parallel {T}^{\ast}Q\parallel \le 16{q}^{2}. Thus, if 4q<1, the equation {x}^{2}{u}^{\u2033}q(x)u=f has unique positive solution for any f\in {L}_{2}({R}_{+},\rho ), f\ge 0, f\ne 0.
But if q(x)=\mathrm{const}=1/4, the equation {x}^{2}{u}^{\u2033}(1/4)u=f may not have a solution for some f (this was considered in Example 2.1).
In concluding this section consider one useful assertion. Let
and
Note that Q(u,v)=(Qu,Tv) (see the equality (5.2)).
Let {\lambda}_{0} be the greatest lower bound of spectrum of the operator ℒ. Then by (A.10)
From Theorem A.5 follows the following.
Theorem 3.3 Suppose conditions (1.3) and (1.4) hold. Then {\lambda}_{0} is the smallest eigenvalue of the problem \mathcal{L}u=\lambda u, u(0)={u}^{\prime}(+\mathrm{\infty})=0, and the following statements are equivalent:

1.
\u3008u,u\u3009 is positive definite,

2.
{\lambda}_{0}>0,

3.
\rho (Q{T}^{\ast})<1.
Remark 3.2 We do not suppose that {\lambda}_{0}>\mathrm{\infty} (but if \rho (Q{T}^{\ast})<1 it is so).
4 Auxiliaries propositions
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 u(x)={\int}_{0}^{x}z(s)\phantom{\rule{0.2em}{0ex}}ds establishes a bijection between W and the Hilbert space {L}_{2}({R}_{+}). □
Lemma 4.2 The value {A}_{u}={\int}_{{R}_{+}}\frac{u(s){u}^{\prime}(s)}{s}\phantom{\rule{0.2em}{0ex}}ds satisfies the inequality
Proof This follows from the inequalities
□
Remark 4.1 The estimate (4.1) is accurate, since if u={s}^{1/2\epsilon}, s\ge 1, and u={s}^{1/2+\epsilon}, s\ge 1, then
Remark 4.2 If u(s)\ge 0, {u}^{\prime}(s)\ge 0, {u}^{\u2033}(s)\le 0, then
In fact, denote {B}_{u}={\int}_{{R}_{+}}\frac{{u}^{2}}{{s}^{2}}\phantom{\rule{0.2em}{0ex}}ds. Then {B}_{u}=2{A}_{u}. Since u(s)={\int}_{0}^{s}{u}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\ge s{u}^{\prime}(s), {B}_{u}\ge {\int}_{{R}_{+}}{u}^{\prime}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds=[u,u].
Lemma 4.3 Suppose (1.5) holds. Then T(W)\subset {L}_{2}({R}_{+},\rho ), the operator T:W\to {L}_{2}({R}_{+},\rho ) is bounded, and the norm of T satisfies the estimate
Proof Since (Tu,Tu)={\int}_{{R}_{+}}{u}^{2}\rho \phantom{\rule{0.2em}{0ex}}dx=2{\int}_{{R}_{+}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx{\int}_{0}^{x}u(s){u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds, we can estimate
From this and (4.1)
□
Remark 4.3 If {lim}_{s\to \mathrm{\infty}}\mathrm{\Phi}(s)=\mathrm{\infty} (see (1.6)), then T(W)\not\subset {L}_{2}({R}_{+},\rho ). In fact, if u(s){u}^{\prime}(s)\ge 0, then (Tu,Tu)=2{\int}_{{R}_{+}}\frac{u(s){u}^{\prime}(s)}{s}\mathrm{\Phi}(s)\phantom{\rule{0.2em}{0ex}}ds. It is possible to find a nonincreasing function \psi (s) such that
Now find u such that u(s){u}^{\prime}(s)=\psi (s): u{(s)}^{2}=2{\int}_{0}^{s}\psi (t)\phantom{\rule{0.2em}{0ex}}dt. Since u\ge 0, {u}^{\prime}(s)\ge 0, and {u}^{\prime} is nonincreasing, by Remark 4.2, [u,u]\le 2{A}_{u}={\int}_{{R}_{+}}\frac{\psi (s)}{s}\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty}. But (Tu,Tu)=\mathrm{\infty}.
Lemma 4.4 The image T(W) is dense in {L}_{2}({R}_{+},\rho ).
Proof The proof is left to the reader. □
The following theorem [[5], p.318] can be used to show compactness.
Theorem 4.1 (Gelfand)
A set E from a separable Banach space X is relatively compact if and only if for any sequence of linear continuous functionals that converge to zero at each point
the convergence (4.4) would be uniform on the E.
Lemma 4.5 Suppose (1.3) holds. If {f}_{n}\in {L}_{2}({R}_{+},\rho ) is bounded, then
uniformly on the set \{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}.
Proof Since
it is sufficient to show that
uniformly on \parallel u\parallel \le 1. We have
The first term tends to zero because of the inequality
and (1.3). The second term is equal to
This tends to zero because of (4.1) and (1.3). □
Lemma 4.6 Suppose (1.4) holds. If {f}_{n}\in {L}_{2}({R}_{+},\rho ) is bounded, then
uniformly on the set \{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}.
Proof Since
it is sufficient to show that
when a\to 0 uniformly on \parallel u\parallel \le 1. We have
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 \mathrm{\Omega}=\{Tu:\parallel u\parallel \le 1\}. We use the criterium of compactness of Gelfand (see Theorem 4.1). Let {f}_{n}\in {L}_{2}({R}_{+},\rho ) be a sequence of functionals such that {f}_{n}(z)\to 0 for any z\in {L}_{2}({R}_{+},\rho ). We have to show that {f}_{n}(Tu)\to 0 uniformly for \parallel u\parallel \le 1.
Let \epsilon >0. Using Lemma 4.5 choose N such that
for all n and for all \parallel u\parallel \le 1. The same we can do with {\int}_{0}^{a}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx for sufficiently small a>0. For this aim we can use Lemma 4.6.
Now we only have to show that {\int}_{a}^{N}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx\to 0 uniformly on the set \parallel u\parallel \le 1. Since
where {\phi}_{n}(s)={\int}_{s}^{N}{f}_{n}(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx, it suffices to show that {\int}_{a}^{N}{\phi}_{n}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\to 0. We show that {\phi}_{n}(s)\to 0 uniformly for s\in [a,N]. Let
Then {\phi}_{n}(s)={f}_{n}({z}_{s}). The set \{{z}_{s}:s\in [a,N]\} is compact. Thus by Theorem 4.1 of Gelfand {f}_{n}({z}_{s}) converges to zero uniformly for these s. So, {\phi}_{n}(s)\to 0 uniformly for s\in [a,N], and {f}_{n}(Tu)\to 0 uniformly for \parallel u\parallel \le 1. □
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).
Proof Denote {h}^{\prime}=f\rho. From (2.1) it follows that
(if we choose v such that the corresponding limits exist). If v=0 for x\notin [a,b]\subset (0,\mathrm{\infty}), then {\int}_{a}^{b}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx={\int}_{a}^{b}h{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx. From this {u}^{\prime}=h+\mathrm{const} on [a,b]. Since the segment [a,b] is arbitrary, the relation {u}^{\prime}=h+\mathrm{const} is fulfilled on the whole semiaxis (0,\mathrm{\infty}). So, {u}^{\u2033}={h}^{\prime}=\rho f. By the first equality in (4.5)
Now, {u}^{\prime}v{}_{x=\mathrm{\infty}}={u}^{\prime}v{}_{x=0}=0. Choosing v such that v(\mathrm{\infty})\ge C>0, we obtain {u}^{\prime}(\mathrm{\infty})=0. □
5 Operator Q. Symmetry and estimates of the norm
Here we consider the operator Q defined by (2.4).
5.1 Symmetry of the form (Qu,Tv)
Under the assumptions imposed on the functions r(x,s) and \xi (x,y) in Section 1.2, from Lemma B.1 (in Appendix 2) we obtain the following statement.
Lemma 5.1
In this case X=Y={R}_{+}, \mu (e)={\int}_{e}\rho (x)\phantom{\rule{0.2em}{0ex}}dx, K(x,dy)=r(x,dy)={d}_{y}r(x,y).
Using Lemma 5.1 the form
can be represented in the symmetrical form
Hence, this form is symmetric if the function \xi (x,y) is symmetric: \xi (x,s)=\xi (s,x).
5.2 Unique deviation
Consider first the special case when the operator Q is defined by (2.5), i.e. Qu(x)=q(x)u(h(x)). Using (4.1) we have
From this follows the estimate of \parallel Q\parallel:
In particular, when Qu(x)=q(x)u(x),
5.3 General operator Q
Here we consider the general case of the operator Q defined by (2.4), i.e. Qu(x)={\int}_{0}^{\mathrm{\infty}}u(t)\phantom{\rule{0.2em}{0ex}}{d}_{t}r(x,t). Suppose that the function \xi (x,y)=\rho (x){\int}_{0}^{x}r(s,y)\phantom{\rule{0.2em}{0ex}}ds (see (1.8)) is absolutely continuous in y, and \xi (x,y)={\int}_{0}^{y}p(x,t)\phantom{\rule{0.2em}{0ex}}dt. Then p(x,t) does not decrease in x.
In this case
where q(x) is defined by (1.7). Using Lemma 5.1, we obtain
The latter step can be done in the same manner as in the relation (5.4).
From this
Since
it may be presented in the form
This estimate only works well if for all x deviation is concentrated around the point h(x). For example, if r(x,s)=q(x){r}_{1}(sh(x)), {r}_{1}(+\mathrm{\infty})=1, {r}_{1}(\mathrm{\infty})=0 then
This estimate coincides with (5.3), if {r}_{1}(t)=\{\begin{array}{cc}0\hfill & \text{if}t\le 0,\hfill \\ 1\hfill & \text{if}t0.\hfill \end{array}
In this case
Appendix 1: Abstract scheme
We do not use general spectral theory (see, for example, [6, 7]). But the scheme below is close to the scheme in [[7], 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
Let W and H be Hilbert spaces with inner product [u,v] and (f,g), respectively. Let T:W\to H be a linear bounded operator. The equation
has the unique solution u={T}^{\ast}f for any f\in H, where {T}^{\ast} is adjoint operator. Let {D}_{\mathcal{L}}={T}^{\ast}(H).
Assume that

1.
the image T(W) of the operator T is dense in H,

2.
dimkerT=0.
Lemma A.1 If the image T(W) of the operator T is dense in H, then {T}^{\ast} is an injection.
Proof Suppose {T}^{\ast}f=0 for a f\in H. Then for any g\in T(W)
Since T(W) is dense in H, the f=0. □
Corollary A.1 (Euler equation)
The operator {T}^{\ast} has an inverse ℒ defined on the set {D}_{\mathcal{L}}. Equation (A.1) is equivalent to
The spectral problem for the operator ℒ we will write in the form
Let {\lambda}_{0} be the greatest lower bound of the spectrum of ℒ. It is well known (see, for example, [[7], Chapter 6]) that
Since (\mathcal{L}u,Tu)=[{T}^{\ast}\mathcal{L}u,u]=[u,u],
Theorem A.1 The spectrum of ℒ is discrete if and only if T is compact.
Proof Since (A.3) is equivalent to u=\lambda {T}^{\ast}Tu, discreteness of spectrum of ℒ is equivalent to compactness of {T}^{\ast}T. But the operators {T}^{\ast}T and {T}^{\ast} are compact at the same time [[7], Chapter 10]. □
Theorem A.2 Suppose T is compact. Then (A.3) has a nonzero solution {u}_{n} only in the case of \lambda ={\lambda}_{n}, n=0,1,2,\dots , i.e.
The system {u}_{n} forms an orthogonal basis in W. The sequence {\lambda}_{n} forms a nondecreasing sequence of positive numbers,
and lim{\lambda}_{n}=\mathrm{\infty}.
Remark A.1 The minimal eigenvalue {\lambda}_{0} satisfies the equality (A.4).
A.2 General case
Let
be a symmetric bilinear form, u,v\in W. Assume that Q is bounded in both arguments. Moreover, suppose that this form has the representation Q(u,v)=(Qu,Tv), where Q:W\to H is bounded.^{e} Then the equation [u,v](Qu,Tv)=(f,Tv), \mathrm{\forall}v\in W, is equivalent to
and in the set {D}_{{\mathcal{L}}_{0}}={T}^{\ast}(H) to
where {\mathcal{L}}_{0}={({T}^{\ast})}^{1}.
Theorem A.3 Suppose T is compact. Then the equation
has a nonzero solution {u}_{n} only in the case of \lambda ={\lambda}_{n}, n=0,1,2,\dots , i.e.
The system {u}_{n} can be chosen to form an orthogonal basis in the space W.
If the form \u3008u,v\u3009=[u,v](Qu,Tv) is lower semibounded, i.e.
then spectrum of ℒ is semibounded, and
is the greatest lower bound of the spectrum [[7], Chapter 6]. Thus we have the following.
Theorem A.4 If (A.9) holds and T is compact then the eigenvalues {\lambda}_{n} have a minimum and can be put in increasing order
A.3 Positive definiteness and spectral radius of Q{T}^{\ast}
Let \rho =\rho (Q{T}^{\ast}) be the spectral radius of the operator Q{T}^{\ast}. Note that the two operators Q{T}^{\ast} and {T}^{\ast}Q have the same spectral radius. So
Lemma A.2 The quadratic form \u3008u,u\u3009 is positive definite if and only if \rho <1.
Proof From (A.11) it follows that \rho [u,u](Qu,Tu)\ge 0 and \u3008u,u\u3009=[u,u](Qu,Tu)\ge (1\rho )[u,u]. If \rho <1 then \u3008u,u\u3009 is positive definite. Conversely from the inequality [u,u](Qu,Tu)\ge \epsilon [u,u] (\epsilon >0) it follows that \rho \le 1\epsilon. □
So if \rho <1, then by (A.10)
Conversely, suppose {\lambda}_{0}>0. Then
and \rho \le 1. If (Qu,Tu) is nonnegative, then spectrum of {T}^{\ast}Q is in the segment [0,\rho ], and ρ is a point of spectrum. In this case from {\lambda}_{0}>0 it follows that \rho <1. So, we have the following.
Theorem A.5 Suppose (Qu,Tu) is nonnegative. The following assertions are equivalent:

1.
\u3008u,u\u3009 is positive definite,

2.
{\lambda}_{0}>0,

3.
\rho (Q{T}^{\ast})<1.
Appendix 2: A generalization of the Fubini theorem
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 [8]. Note that it was used without proof in [2–4].
Lemma B.1 Let (X,\mathcal{A}) and (Y,\mathcal{B}) be measurable spaces, μ be a measure^{f} on (X,\mathcal{A}), K:X\times \mathcal{B}\to [0,\mathrm{\infty}] be kernel (i.e. for μalmost all x\in X, K(x,\cdot ) is a measure on (Y,\mathcal{B}), \mathrm{\forall}B\in \mathcal{B}, K(\cdot ,B) is μmeasurable on X). Then

1.
the function ν defined on \mathcal{A}\times \mathcal{B} by the equality
\nu (E)={\int}_{X}K(x,{E}_{x})\mu (dx),\phantom{\rule{1em}{0ex}}\mathit{\text{where}}{E}_{x}=\{y:(x,y)\in E\},
is measure;

2.
if f:X\times Y\to [\mathrm{\infty},\mathrm{\infty}] is νmeasurable on X\times Y, then
{\int}_{X\times Y}f(x,y)\phantom{\rule{0.2em}{0ex}}d\nu ={\int}_{X}({\int}_{Y}f(x,y)K(x,dy))\mu (dx).
Remark B.1 The function ν is the Lebesgue extension from the set of rectangles
Endnotes
^{a}The spectrum of ℒ is discrete if it consists only of eigenvalues of finite multiplicity.
^{b}The condition {\int}_{0}^{1}x\rho (x)\phantom{\rule{0.2em}{0ex}}dx<\mathrm{\infty} is sufficient for (1.4), but it is not necessary (if, for example, \rho (x)=1/({x}^{2}lnx) for x near zero).
^{c}\xi (x,y) is defined by (1.8).
^{d} Here I is the identity operator.
^{e}We call Q(u,v) the form associated with the operator Q.
^{f} The measure is a nonnegative, σadditive function defined on a σalgebra; the product \mathcal{A}\times \mathcal{B} is the minimal σalgebra containing the set of all rectangles A\times B, A\in \mathcal{A}, B\in \mathcal{B}.
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The main idea of this paper was proposed by SL. He prepared Sections 13 and the Appendix. MFG prepared Sections 45. Both authors read and approved the final manuscript
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Labovskiy, S., Getimane, M.F. On discreteness of spectrum and positivity of the Green’s function for a second order functionaldifferential operator on semiaxis. Bound Value Probl 2014, 102 (2014). https://doi.org/10.1186/168727702014102
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DOI: https://doi.org/10.1186/168727702014102