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UDC 517.948.35
ON CONDITIONS FOR THE DISCRETENESS OF THE SPECTRUM OF MONOMIAL DIFFERENTIAL OPERATORS
V. A. TKACHENKO
The purpose of the present note is to obtain conditions for the discreteness of the spectrum of the self-adjoint operator \(T\), generated in \(L_2(0,\infty)\) by the differential operation
\[ (-1)^n \frac{d^n}{dx^n}\,p\,\frac{d^n}{dx^n} \]
with a function \(p(x)\) positive on the entire half-axis. At the same time, the question of the asymptotic distribution of the eigenvalues of the operator \(T\) is considered in the case of discreteness of the spectrum.
For \(n=1\) the question of conditions for the discreteness of the spectrum of the operator \(T\), by means of the change of variable
\[ s=\int_0^x \frac{dt}{p(t)} \]
reduces to the same question concerning the frequency spectrum of a semi-infinite string of density \(\rho(s)=p(x)\). In this case, using the theorems of Krein—Kac [1] and Birman [2] on the spectrum of a singular string, it is easy to establish the necessary and sufficient condition for the discreteness of the spectrum of the operator \(T\). However, in the general case, when \(n>1\), the question posed no longer reduces to the corresponding problem on the spectrum of a singular string.
- We begin with the proof of two auxiliary inequalities.
Lemma. For every finite function from the domain of definition \(D_T\) of the operator \(T\), one has
\[ \int_0^\infty \frac{|yy'|\,dx}{\displaystyle \int_x^\infty \frac{dt}{p}} \leq 2\int_0^\infty p\,y'^2\,dx \tag{1} \]
and
\[ \int_0^\infty y^2\,dx \leq C\int_0^\infty |y^{(n-1)}y^{(n)}|\,x^{2n-1}\,dx, \tag{2} \]
where the constant \(C\) depends only on \(n\).
Proof. By the Cauchy—Bunyakovsky inequality, for every finite function from \(D_T\) we have
\[ \int_0^\infty \frac{|yy'|\,dx}{\displaystyle \int_x^\infty \frac{dt}{p}} \leq \left\{ \int_0^\infty \frac{y^2\,dx}{\displaystyle p\left(\int_x^\infty \frac{dt}{p}\right)^2} \right\}^{\frac12} \left\{ \int_0^\infty p\,y'^2\,dx \right\}^{\frac12}. \tag{*} \]
At the same time
\[ \int_0^\infty \frac{y^2 dx}{p\left(\displaystyle\int_x^\infty \frac{dt}{p}\right)^2} =2\left|\int_0^\infty \frac{dx}{p\left(\displaystyle\int_x^\infty \frac{dt}{p}\right)^2} \int_x^\infty yy'\,dt\right|= \]
\[ =2\left|\int_0^\infty yy'\,dt \left( \frac{1}{\displaystyle\int_t^\infty \frac{d\tau}{p}} - \frac{1}{\displaystyle\int_0^\infty \frac{d\tau}{p}} \right)\right| \leqslant 2\int_0^\infty \frac{|yy'|\,dx}{\displaystyle\int_x^\infty \frac{dt}{p}}. \tag{**} \]
Combining \((*)\) and \((**)\), we arrive at (1). To prove (2) we use the inequality
\[ \int_0^\infty x^\alpha y^2 dx \leqslant \frac{4}{(\alpha+1)^2} \int_0^\infty x^{\alpha+2} y'^2 dx \]
which is valid in every case for any finite function from \(D_T\) when \(\alpha \geqslant 0\). Assuming first that \(n>1\), put in the preceding relation
\(\alpha=2k-2\) \((1\leqslant k\leqslant n-1)\) and replace \(y\) by \(y^{(k-1)}\). We then obtain
\[ \int_0^\infty x^{2k-2}\left|y^{(k-1)}\right|^2 dx \leqslant \frac{4}{(2k-1)^2} \int_0^\infty x^{2k}\left|y^{(k)}\right|^2 dx. \]
It follows from this that
\[ \int_0^\infty y^2 dx \leqslant C_1\int_0^\infty x^{2n-2}\left|y^{(n-1)}\right|^2 dx, \]
where
\[ C_1=4^{\,n-1}\prod_{k=0}^{n-1}\frac{1}{(2k-1)^2}. \]
The last inequality, obviously, remains valid also for \(n=1\). But then
\[ \int_0^\infty y^2 dx \leqslant 2C_1\left|\int_0^\infty x^{2n-2}dx \int_x^\infty y^{(n-1)}y^{(n)}dt\right| = \]
\[ =\frac{2C_1}{2n-1} \left|\int_0^\infty y^{(n-1)}y^{(n)}x^{2n-1}dx\right| \leqslant C\int_0^\infty \left|y^{(n-1)}y^{(n)}\right|x^{2n-1}dx, \]
which was to be proved. All changes in the order of integration here are admissible, since under the integral sign there stand finite functions from \(\widetilde D_T\).
- Using the lemma, let us prove the following theorem.
Theorem 1. If \(p(x)>0\) \((x\geqslant 0)\) and
\[ \lim_{x\to\infty} x^{2n-1}\int_x^\infty \frac{dt}{p(t)}=0, \tag{3} \]
then the operator \(T\) is positive and its spectrum is discrete.
The proof requires only the second part of the theorem. Replace \(y\) by \(y^{(n-1)}\) in (1). We obtain
\[ \int_{0}^{\infty} \frac{\left|y^{(n-1)}y^{(n)}\right|} {\displaystyle\int_{x}^{\infty}\frac{dt}{p}}\,dx \leq 2\int_{0}^{\infty}p\left|y^{(n)}\right|^{2}\,dx . \]
According to the hypothesis of the theorem, for an arbitrary \(\varepsilon>0\) there exists a number \(m=m(\varepsilon)\) such that for all \(x\geq m\) we shall have
\(x^{2n-1}\displaystyle\int_{x}^{\infty}\frac{dt}{p}<\frac{\varepsilon}{2C}\). But then from the preceding inequality and inequality (2), for every \(m\)-finite function from \(\overline{D}_{T}\) it follows that
\[ \int_{m}^{\infty} y^{2}\,dx \leq \frac{\varepsilon}{2} \int_{m}^{\infty} \frac{\left|y^{(n-1)}y^{(n)}\right|} {\displaystyle\int_{x}^{\infty}\frac{dt}{p}}\,dx \leq \varepsilon\int_{m}^{\infty}p\left|y^{(n)}\right|^{2}\,dx . \]
Hence, by virtue of the splitting principle [3], it follows that the spectrum of the operator \(T\) is discrete. The theorem is proved.
For \(n=1\) condition (3) is not only sufficient, but also necessary for the discreteness of the spectrum of the operator \(T\). Condition (3) also proves necessary for any \(n\), if only the function \(p(x)\) is monotone. This circumstance is easily verified by constructing a corresponding system of trial functions. The theorem itself may be formulated in the following form.
Theorem 2. Let \(p(x)>0,\ p'(x)\geq 0\) \((x\geq 0)\). Then, for the spectrum of the operator \(T\) to be discrete, it is necessary and sufficient that
\[ \lim_{x\to\infty}\frac{x^{2n}}{p(x)}=0. \]
- Put, for \(x<z\),
\[ (t^{i},\,t^{r})=\int_{x}^{z}\frac{t^{i+r}}{p(t)}\,dt \qquad (i,\ r=0,1,\ldots,n-1). \]
Let
\[ \Delta_{0}(x,z)=1,\qquad \Delta_{n}(x,z)= \left| \begin{array}{ccc} (t^{\,n-1},t^{\,n-1}) & \cdots & (t^{\,n-1},1)\\ \cdot & \cdot & \cdot\\ (1,t^{\,n-1}) & \cdots & (1,1) \end{array} \right| \qquad (n\geq 1). \]
The following holds.
Theorem 3. For the spectrum of the operator \(T\) to be discrete, it is necessary that
\[ \lim_{x\to\infty} x\,\frac{\Delta_{n}(x,z)}{\Delta_{n-1}(x,z)}=0. \tag{4} \]
Proof. Suppose that the spectrum of \(T\) is discrete, but the theorem is false. Then, for some \(\varepsilon>0\), there exist infinite sequences \(\{x_{k}\}_{k=1}^{\infty}\) and \(\{z_{k}\}_{k=1}^{\infty}\) for which
\[ x_{k}\frac{\Delta_{n}(x_{k},z_{k})}{\Delta_{n-1}(x_{k},z_{k})}>\varepsilon . \]
Set
\[ J_k(y)=\int_{x_k}^{z_k} p|y^{(n)}|^2\,dx \]
and let us seek the minimum of the functional \(J_k\) on those functions from its domain of definition which satisfy the boundary conditions
\(y(x_k)=1,\ y'(x_k)=\cdots=y^{(n-1)}(x_k)=y(z_k)=\cdots=y^{(n-1)}(z_k)=0\). The required minimum is attained on the solution \(y_k(x)\) of the equation \((py_k^{(n)})^{(n)}=0\). Integrating this equation, we obtain
\[ y_k^{(n)}=\sum_{i=0}^{n-1} c_i\frac{x^i}{p} \]
and then
\[ y_k^{(n-r-1)}=-\frac{1}{r!}\sum_{i=0}^{n-1} c_i \int_x^{z_k}\frac{t^i(x-t)^r}{p}\,dt \qquad (r=0,1,\ldots,n-1). \]
The unknown constants \(c_i\), by virtue of the boundary conditions, satisfy the system of algebraic equations
\[ \sum_{i=0}^{n-1} c_i(t^i,t^r)=0 \qquad (r=0,1,\ldots,n-2), \]
\[ \sum_{i=0}^{n-1} c_i(t^i,t^{n-1})=(-1)^n(n-1)! . \]
On the other hand, we have
\[ J_k(y_k)=\int_{x_k}^{z_k}p|y_k^{(n)}|^2\,dx =\sum_{i,r=0}^{n-1}c_i c_r(t^i,t^r) \]
and, consequently,
\[ J_k(y_k)=[(n-1)!]^2\frac{\Delta_{n-1}(x_k,z_k)}{\Delta_n(x_k,z_k)}. \]
Finally, let \(\varphi(x)\) be a smooth function on the interval \([0,1]\), with
\(\varphi(0)=\varphi'(0)=\cdots=\varphi^{(n-1)}(0)=\varphi'(1)=\cdots=\varphi^{(n-1)}(1)=0,\ \varphi(1)=1\). Put
\(\varphi_k(x)=\varphi\!\left(\dfrac{x-u_k}{v_k-u_k}\right)\), where \(u_k\le x\le v_k\), and the numbers \(u_k\) and \(v_k\) are chosen in such a way that
\(z_{k-1}<u_k<v_k<x_k<z_k\). Construct a sequence of finite functions with nonintersecting supports by the formulas:
\[ \psi_k(x)= \begin{cases} 0, & x\le u_k,\\ \varphi_k(x), & u_k\le x\le v_k,\\ 1, & v_k\le x\le x_k,\\ y_k(x), & x_k\le x\le z_k,\\ 0, & z_k\le x. \end{cases} \]
Without loss of generality one may assume that \(x_k>2v_k\) and
\[ \int_{u_k}^{v_k} p|\varphi_k^{(n)}|^2\,dx \le \frac{x_k}{\varepsilon}[(n-1)!]^2. \]
But then
\[ \int_{u_k}^{z_k}\psi_k^2\,dx> \int_{v_k}^{x_k}\psi_k^2\,dx =x_k-v_k>\frac{x_k}{2}. \]
Moreover,
\[ \int_{u_k}^{z_k} p\left|\psi_k^{(n)}\right|^2 dx \leq \left\{\frac{x_k}{\varepsilon}+\frac{\Delta_{n-1}(x_k,z_k)}{\Delta_n(x_k,z_k)}\right\}[(n-1)!]^2. \]
Consequently,
\[ \int_0^\infty p\left|\psi_k^{(n)}\right|^2 dx \leq 2\left\{\frac{1}{\varepsilon}+\frac{\Delta_{n-1}(x_k,z_k)} {x_k\Delta_n(x_k,z_k)}\right\}\times \]
\[ \times [(n-1)!]^2\int_0^\infty \psi_k^2 dx \leq \frac{4[(n-1)!]^2}{\varepsilon}\int_0^\infty \psi_k^2 dx . \]
In view of the noncompactness of the system of finite functions \(\psi_k(x)\), it follows from the last inequality that the spectrum of the operator \(T\) has a finite limit point. The contradiction obtained proves the theorem. We note that for \(n=1\) conditions (3) and (4) coincide.
- Let the spectrum of the operator \(T\) be discrete. Denote by \(N(\lambda)\) the number of eigenvalues of \(T\) lying to the left of \(\lambda\). In order to derive an asymptotic formula for \(N(\lambda)\) as \(\lambda\to\infty\), we are forced to assume somewhat more about \(p(x)\) than was done in the condition of Theorem 1. We shall assume that \(p(x)\) is a monotone function and
\[ \lim_{x\to\infty} x^{4n-1}\int_x^\infty \frac{dt}{p}=0 . \]
As in the proof of Theorem 1, for arbitrarily large \(\lambda\) we find a splitting point \(m=m(\lambda)\) such that for every finite function in \(D_T\) one has
\[ \lambda\int_m^\infty y^2 dx \leq \int_m^\infty p\left|y^{(n)}\right|^2 dx . \]
In this case the spectrum of the singular part of the operator split at the point \(m\) lies to the right of \(\lambda\). Following [4], we divide the interval \([0,m]\) into \(k\) parts by points \(x_i\) \((i=0,1,\ldots,k)\) in such a way that \(x_0=0\), \(x_k=m\), and
\[ p^{-\frac{1}{2n}}(x_i)-p^{-\frac{1}{2n}}(x_{i+1}) =\frac{1}{k}p^{-\frac{1}{2n}}(0). \]
Next, on the same interval we construct two step functions \(p^{(1)}(x)\) and \(p^{(2)}(x)\), defining them by the equalities:
\[ \begin{aligned} p^{(1)}(x)&=p_i^{(1)}=p(x_{i-1}),\\ p^{(2)}(x)&=p_i^{(2)}=p(x_i), \end{aligned} \qquad x\in[x_{i-1},x_i]\quad (i=1,2,\ldots,k). \]
From the minimax properties of R. Courant it is not difficult to obtain the two-sided estimate
\[ \frac{1}{\pi}\sqrt[2n]{\lambda}\sum_{i=1}^k \frac{h_i}{\sqrt[2n]{p_i^{(2)}}} -Ak \leq N(\lambda) \leq \frac{1}{\pi}\sqrt[2n]{\lambda}\sum_{i=1}^k \frac{h_i}{\sqrt[2n]{p_i^{(1)}}} +Ak, \]
where \(h_i=x_i-x_{i-1}\) \((i=1,2,\ldots,k)\), and the constant \(A\) depends neither on \(\lambda\) nor on \(h_i\). But
\[ \sum_{i=1}^k \frac{h_i}{\sqrt[2n]{p_i^{(1)}}} = \int_0^m \frac{dt}{\sqrt[2n]{p^{(1)}}} = \int_0^m \frac{dt}{\sqrt[2n]{p}} + \]
\[ +\int_0^m \left\{ \frac{1}{\sqrt[2n]{p^{(1)}}}-\frac{1}{\sqrt[2n]{p}}\right\}dt \leq \int_0^m \frac{dt}{\sqrt[2n]{p}}+\frac{m}{\sqrt[2n]{p(0)}\,k} \]
and, similarly,
\[ \sum_{i=1}^k \frac{h_i}{\sqrt[2n]{p_i^{(2)}}} \geq \int_0^m \frac{dt}{\sqrt[2n]{p}}-\frac{m}{\sqrt[2n]{p(0)}\,k}. \]
Therefore the preceding relation leads to the inequality
\[ \left|N(\lambda)-\frac{1}{\pi}\sqrt[2n]{\lambda}\int_0^m\frac{dt}{\sqrt[2n]{p}}\right| \leq Ak+\frac{m\sqrt[2n]{\lambda}}{\pi\sqrt[2n]{p(0)}\,k}. \]
Minimizing here the right-hand side with respect to \(k\) and taking into account that, under the assumptions made concerning \(p\), the integral
\[ \int_0^\infty \frac{dt}{\sqrt[2n]{p}} \]
exists, we arrive at the relation
\[ \left| \frac{N(\lambda)}{\sqrt[2n]{\lambda}} -\frac{1}{\pi}\int_0^\infty \frac{dt}{\sqrt[2n]{p}} \right| \leq \frac{1}{\pi}\int_m^\infty \frac{dt}{\sqrt[2n]{p}} + B\,\frac{\sqrt m}{\sqrt[4n]{\lambda}}, \]
in which the first term on the right-hand side tends to \(0\) as \(\lambda\) grows, while the constant \(B\) in the second term does not depend on \(\lambda\). But \(m\), in accordance with the preceding, is the largest root of the equation
\[ m^{2n-1}\int_m^\infty \frac{dt}{p} = \frac{1}{2\lambda C}, \]
so that
\[ \frac{\sqrt m}{\sqrt[4n]{\lambda}} = \sqrt[4n]{\,2C m^{4n-1}\int_m^\infty \frac{dt}{p}\,}. \]
Thus, the following has been proved.
Theorem 4. Let \(p(x)>0\), \(p'(x)\geq 0\) \((x\geq 0)\), and
\[ \lim_{x\to\infty} x^{4n-1}\int_x^\infty \frac{dt}{p}=0. \]
Then
\[ \lim_{\lambda\to\infty} \frac{N(\lambda)}{\sqrt[2n]{\lambda}} = \frac{1}{\pi}\int_0^\infty \frac{dx}{\sqrt[2n]{p}}. \]
The author is grateful to his scientific adviser, Prof. I. M. Glazman, for posing the problem and for his attention to the work.
References
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Kats I. S., Krein M. G. Izv. vuzov, MVO SSSR. Matematika, No. 2, (3), 136—153, 1958.
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Birman M. Sh. On the spectrum of singular boundary-value problems. Matem. sb., 55(97): 2, 1961, pp. 125—173.
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Glazman I. M. Direct methods of qualitative spectral analysis of singular differential operators. Fizmatgiz, Moscow, 1964.
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Titchmarsh E. C. and Titchmarsh E. C. On the asymptotic distribution of eigenvalues, Proc. Roy. Soc., 1950, ser. A. v. 200, p. 572—580.
Received by the editors
May 27, 1965
Kharkov Polytechnic Institute
named after V. I. Lenin