On discreteness of spectrum and positivity of the Green’s function for a second order functionaldifferential operator on semiaxis
 Sergey Labovskiy^{1}Email author and
 Mário Frengue Getimane^{2}
https://doi.org/10.1186/168727702014102
© Labovskiy and Getimane; licensee Springer. 2014
Received: 5 February 2014
Accepted: 4 April 2014
Published: 7 May 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.
Keywords
1 Introduction
1.1 Problems and a wellknown result
The second question is existence of positive solutions of the equation $\mathcal{L}u=f$ (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.
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.
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).
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).
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

$T:W\to {L}_{2}({R}_{+},\rho )$ 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
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).
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.
or $u=Gf$, where G is the Green’s operator of ℒ.
2 Discreteness of spectrum
2.1 Operator ${\mathcal{L}}_{0}$
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 $\rho (x)=1/{x}^{2}$, 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 $\lambda >1/4$.
this equation has to be considered in the space ${L}_{2}(R,{e}^{t})$.
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
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.
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.
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 $q(x)$ is bounded.
3 Positive solvability
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.
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 $q(x)\le \mathrm{const}=q$, 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).
Note that $Q(u,v)=(Qu,Tv)$ (see the equality (5.2)).
From Theorem A.5 follows the following.
 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 ${L}_{2}({R}_{+})$. □
□
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 $u(s)={\int}_{0}^{s}{u}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\ge s{u}^{\prime}(s)$, ${B}_{u}\ge {\int}_{{R}_{+}}{u}^{\prime}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds=[u,u]$.
□
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)
the convergence (4.4) would be uniform on the E.
uniformly on the set $\{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}$.
This tends to zero because of (4.1) and (1.3). □
uniformly on the set $\{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}$.
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 $\parallel u\parallel \le 1$.
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.
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).
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.
In this case $X=Y={R}_{+}$, $\mu (e)={\int}_{e}\rho (x)\phantom{\rule{0.2em}{0ex}}dx$, $K(x,dy)=r(x,dy)={d}_{y}r(x,y)$.
Hence, this form is symmetric if the function $\xi (x,y)$ is symmetric: $\xi (x,s)=\xi (s,x)$.
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. $Qu(x)={\int}_{0}^{\mathrm{\infty}}u(t)\phantom{\rule{0.2em}{0ex}}{d}_{t}r(x,t)$. 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.
The latter step can be done in the same manner as in the relation (5.4).
This estimate coincides with (5.3), if ${r}_{1}(t)=\{\begin{array}{cc}0\hfill & \text{if}t\le 0,\hfill \\ 1\hfill & \text{if}t0.\hfill \end{array}$
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
has the unique solution $u={T}^{\ast}f$ for any $f\in H$, where ${T}^{\ast}$ is adjoint operator. Let ${D}_{\mathcal{L}}={T}^{\ast}(H)$.
 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.
Since $T(W)$ is dense in H, the $f=0$. □
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 $u=\lambda {T}^{\ast}Tu$, 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]. □
and $lim{\lambda}_{n}=\mathrm{\infty}$.
Remark A.1 The minimal eigenvalue ${\lambda}_{0}$ satisfies the equality (A.4).
A.2 General case
where ${\mathcal{L}}_{0}={({T}^{\ast})}^{1}$.
The system ${u}_{n}$ can be chosen to form an orthogonal basis in the space W.
is the greatest lower bound of the spectrum [[7], Chapter 6]. Thus we have the following.
A.3 Positive definiteness and spectral radius of $Q{T}^{\ast}$
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 $. □
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.
 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].
 1.the function ν defined on $\mathcal{A}\times \mathcal{B}$ by the equality$\nu (E)={\int}_{X}K(x,{E}_{x})\mu (dx),\phantom{\rule{1em}{0ex}}\mathit{\text{where}}{E}_{x}=\{y:(x,y)\in E\},$
 2.if $f:X\times Y\to [\mathrm{\infty},\mathrm{\infty}]$ is νmeasurable on $X\times Y$, then${\int}_{X\times Y}f(x,y)\phantom{\rule{0.2em}{0ex}}d\nu ={\int}_{X}({\int}_{Y}f(x,y)K(x,dy))\mu (dx).$
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 $A\times B$, $A\in \mathcal{A}$, $B\in \mathcal{B}$.
Declarations
Authors’ Affiliations
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