MATHEMATICS

N. P. KUPTSOV

ON CONDITIONS FOR NON-SELF-ADJOINTNESS OF A LINEAR DIFFERENTIAL OPERATOR OF SECOND ORDER

(Presented by Academician A. N. Kolmogorov on 21 I 1961)

Consider on the half-axis \([0,\infty)\) the operator \(Ly\), defined by the differential expression

\[ -y''+q(x)y \tag{1} \]

and the boundary condition

\[ \cos\alpha\cdot y(0)+\sin\alpha\cdot y'(0)=0. \tag{2} \]

We shall assume the function \(q(x)\) to be real and summable on every finite interval belonging to \([0,\infty)\). It is known ((\(^{1}\), pp. 103—109)) that this operator is either self-adjoint (the defect indices in this case are \((0,0)\)) or has defect indices \((1,1)\). One of the strongest sufficient conditions for self-adjointness of \(Ly\) was indicated by Sears (\(^{3}\)): if there exists a positive nonincreasing function \(Q(x)\) such that

\[ q(x)\geq -Q(x)\quad \text{and}\quad \int_0^\infty \frac{dx}{\sqrt{Q(x)}}=+\infty, \]

then \(Ly\) is a self-adjoint operator.

Some new results in this direction were obtained by Brink (\(^{6}\)). M. A. Naimark (\(^{2}\), p. 270) and V. B. Lidskii (\(^{4}\)) indicated sufficient conditions for non-self-adjointness of \(Ly\):

\(1^\circ\). M. A. Naimark’s criterion. If \(q(x)\) and \(q'(x)\) are absolutely continuous on \([0,\infty)\), \(q'(x)\leq 0\), \(q''(x)\leq 0\), \(\lim_{x\to\infty}q(x)=-\infty\), \(|q'(x)|=\)

\[ =O\bigl(|q(x)|^\alpha\bigr)\quad \left(0<\alpha<\frac{3}{2}\right)\quad \text{and}\quad \int_0^\infty \frac{dx}{\sqrt{|q(x)|}}<\infty, \]

then \(Ly\) has defect indices \((1,1)\).

\(2^\circ\). V. B. Lidskii’s criterion*. The operator \(Ly\) is non-self-adjoint if \(q(x)\) is negative and absolutely continuous on \([0,\infty)\), and

\[ q'(x)\leq \frac{2+\varepsilon}{x}\,q(x) \]

for some \(\varepsilon>0\).

In the present note new sufficient conditions for non-self-adjointness of \(Ly\) are established.

Let \(s(x)>0\) and \(\varphi(x)\) be arbitrary functions absolutely continuous on \([0,\infty)\). Then for any solution of the differential equation

\[ y''-q(x)y=0 \]

* V. B. Lidskii’s criterion was established for systems of the form \(\ddot{x}_i=\sum_{j=1}^n q_{ij}x_j\). Here a particular criterion is formulated for \(n=1\).

the estimate is valid

\[ y^2(x)\leqslant \frac{C}{\sqrt{s(x)}}\exp\left\{\int_0^x \sqrt{\left(\frac{s'}{2s}-2\varphi\right)^2+\frac1s(\varphi'-\varphi^2+q+s)^2}\,d\xi\right\}, \]

where

\[ C=\frac{1}{\sqrt{s(0)}}\left[(\varphi^2(0)+s(0))y^2(0)+2\varphi(0)y(0)y'(0)+y'^2(0)\right]. \]

Hence it follows:

Theorem 1. If there exist functions \(s(x)>0\) and \(\varphi(x)\), absolutely continuous on \([0,\infty)\), such that

\[ \int_0^\infty \frac{1}{\sqrt{s(x)}}\exp\left\{\int_0^x \sqrt{\left(\frac{s'}{2s}-2\varphi\right)^2+\frac1s(\varphi'-\varphi^2+q+s)^2}\,d\xi\right\}\,dx<\infty, \tag{3} \]

then the operator \(Ly\) has deficiency indices \((1,1)\).

Unfortunately, this criterion is not sufficiently effective, since it is not clear how \(s(x)\) and \(\varphi(x)\) should be chosen. In this respect the following criterion is more convenient:

Theorem 2. Let \(Q(x)\) be an arbitrary positive function on \([0,\infty)\), absolutely continuous together with \(Q'(x)\). If there exists a negative, absolutely continuous function \(\bar q(x)\) on \([0,\infty)\), possessing the properties (\(x_0\) is an arbitrary positive number):

\[ \frac{\bar q'(x)}{\bar q(x)}\geqslant \frac{Q'(x)}{Q(x)}\quad \text{for } x\geqslant x_0; \tag{4} \]

\[ \int_{x_0}^{\infty}\frac{|q(x)-\bar q(x)|}{\sqrt{|\bar q(x)|}}\,dx<\infty; \tag{5} \]

\[ \int_{x_0}^{\infty}\frac{1}{\sqrt{Q(x)}}\exp\left[\frac14\int_{x_0}^{x}\frac{1}{\sqrt{|\bar q(\xi)|}}\left|\frac{Q''(\xi)}{Q(\xi)}-\frac54\frac{Q'^2(\xi)}{Q^2(\xi)}\right|\,d\xi\right]dx<\infty, \tag{6} \]

then the operator \(Ly\) has deficiency indices \((1,1)\).

For the proof it is sufficient to use Theorem 1, putting \(s(x)=-\bar q(x)\) and \(\varphi(x)=Q'(x)/4Q(x)\). Then

\[ \int_{x_0}^{x}\sqrt{\left(\frac{s'}{2s}-2\varphi\right)^2+\frac1s(\varphi'-\varphi^2+q+s)^2}\,d\xi\leqslant \]

\[ \leqslant \int_{x_0}^{x}\left|\frac{\bar q'(\rho)}{2\bar q(\xi)}-\frac{Q'(\xi)}{2Q(\xi)}\right|\,d\xi +\frac14\int_{x_0}^{x}\frac{1}{\sqrt{|\bar q(\xi)|}}\left|\frac{Q''(\xi)}{Q(\xi)}-\frac54\frac{Q'^2(\xi)}{Q^2(\xi)}\right|\,d\xi+ \]

\[ +\int_{x_0}^{x}\frac{1}{\sqrt{|\bar q(\xi)|}}|q(\xi)-\bar q(\xi)|\,d\xi =\frac12\ln\left|\frac{\bar q(x)}{q(0)}\right|-\frac12\ln\frac{Q(x)}{Q(0)}+ \]

\[ +\frac14\int_{x_0}^{x}\frac{1}{\sqrt{|\bar q(\xi)|}}\left|\frac{Q''}{Q(\xi)}-\frac54\frac{Q'^2(\xi)}{Q^2(\xi)}\right|\,d\xi +\int_{x_0}^{x}\frac{|q(\xi)-\bar q(\xi)|}{\sqrt{|\bar q(\xi)|}}\,d\xi. \]

Now, relying on the conditions of Theorem 2, it is easy to establish the validity of relation (3). From Theorem 2 it is not difficult to obtain the above-mentioned criteria of V. B. Lidskii and M. A. Naimark. In the first case it is sufficient to put \(Q(x)=x^{2+\varepsilon}\) and \(\bar q(x)=q(x)\). To derive M. A. Naimark’s criterion, put \(Q(x)=-q(x)\) and \(\bar q(x)=q(x)\). Then conditions (4)

and (5) will be satisfied. Let us proceed to verify condition (6):

\[ \begin{aligned} I(x)&=\int_{x_0}^{x}\frac{1}{\sqrt{|q|}}\left|\,\frac{q''}{q}-\frac{5}{4}\frac{q'^2}{q^2}\right|\,d\xi \leq \int_{x_0}^{x}|q|^{-3/2}|q''|\,d\xi+ \frac{5}{4}\int_{x_0}^{x}|q|^{-5/2}q'^2\,d\xi \\ &=-q'(x)|q(x)|^{-3/2}+q'(x_0)|q(x_0)|^{-3/2} +\frac{11}{4}\int_{x_0}^{x}|q|^{-5/2}q'^2\,d\xi \leq \\ &\leq M|q(x)|^{\alpha-3/2} +\frac{11M}{4}\int_{x_0}^{x}|q|^{\alpha-5/2}|q'|\,d\xi= \\ &=M|q(x)|^{\alpha-3/2} +\frac{11M}{4(3/2-\alpha)} \left[\,|q(x_0)|^{\alpha-3/2}-|q(x)|^{\alpha-3/2}\,\right], \end{aligned} \]

where

\[ M=\sup_{x_0\leq x<\infty}\frac{|q'(x)|}{|q(x)|^\alpha}. \]

Taking into account the relation \(\alpha<3/2\), we easily obtain \(I(x)\leq M_1\), where \(M_1\) does not depend on \(x\). Consequently,

\[ \int_{x_0}^{\infty}\frac{1}{\sqrt{Q(x)}}e^{\frac14 I(x)}\,dx \leq e^{\frac14 M_1}\int_{x_0}^{\infty}\frac{dx}{\sqrt{|q(x)|}}<\infty . \]

Theorem 3. Let \(Q(x)\) be an arbitrary positive, nondecreasing, absolutely continuous function on \([0,\infty)\). Suppose, further, that \(Q'(x)\) is absolutely continuous and that \(Q^{-\beta}(x)\) is convex on \([0,\infty)\) for some \(\beta\in(0,1/2)\).

Finally, suppose

\[ \int_{0}^{\infty}\frac{dx}{\sqrt{Q(x)}}<\infty . \]

If there exists a negative absolutely continuous function \(\overline q(x)\) on \([0,\infty)\), possessing the following properties (\(x_0\) is an arbitrary positive number):

\[ \frac{\overline q'(x)}{\overline q(x)}\geq \frac{Q'(x)}{Q(x)} \quad \text{for } x\geq x_0; \tag{7} \]

\[ \int_{x_0}^{\infty}\frac{|q(x)-\overline q(x)|}{\sqrt{|\overline q(x)|}}\,dx<\infty, \tag{8} \]

then \(Ly\) has deficiency indices \((1,1)\).

Proof. It is enough to verify that condition (6) is fulfilled for \(Q(x)\). From (7) it follows that

\[ |\overline q(x)|\geq C Q(x), \tag{9} \]

where \(C>0\) and does not depend on \(x\). Put \(Q^{-\beta}(x)=\gamma(x)\). From the properties of \(Q(x)\) it follows that \(\gamma'(x)\leq 0\) on \([0,\infty)\) and \(\gamma''(x)\geq 0\) almost everywhere on \([0,\infty)\). Then, taking (9) into account, we may write

\[ K(x)=\int_{0}^{\infty}\frac{1}{\sqrt{|\overline q(x)|}} \left|\frac{Q''}{Q}-\frac{5}{4}\frac{Q'^2}{Q^2}\right|\,dx\leq \]

\[ \leq \frac{1}{\beta\sqrt C}\int_{0}^{\infty}\gamma^{1/2\beta}(x) \left|\frac{\gamma''(x)}{\gamma(x)} +\left(\frac{1}{4\beta}-1\right)\frac{\gamma'^2(x)}{\gamma^2(x)}\right|\,dx\leq \]

\[ \leq C_1\int_{0}^{\infty}\gamma^{1/2\beta-1}\gamma''\,dx + C_2\int_{0}^{\infty}\gamma^{1/2\beta-2}\gamma'^2\,dx, \]

where \(C_1\) and \(C_2\) do not depend on \(x\).

Taking into account the inequality \(\beta < 1/2\), we integrate by parts (\(C_3\) is a constant independent of \(x\)):

\[ K(x) \leqslant C_1\left[\gamma^{1/2\beta-1}\gamma'\bigg|_{0}^{\infty} -\left(\frac{1}{2\beta}-1\right)\int_{0}^{\infty}\gamma^{1/2\beta-2}\gamma'^2\,dx\right]+ \]

\[ +\, C_2\left[\gamma^{1/2\beta-1}\gamma'\bigg|_{0}^{\infty} -\frac{1}{(1/2\beta-1)}\int_{0}^{\infty}\gamma^{1/2\beta-1}\gamma''\,dx\right] \leqslant C_3 . \]

The theorem is proved.

It follows from the work of É. Shnol’ \((^5)\) that in the formulation of Theorem 3 one cannot omit the convexity of the function \(Q^{-\beta}(x)\). From the same work it follows that the differential inequality (7) cannot be replaced by the inequality \(\overline{q}(x)\leqslant -Q(x)\).

Saratov State University
named after N. G. Chernyshevsky

Received
14 I 1961

CITED LITERATURE

\(^{1}\) V. M. Levitan, Expansion in eigenfunctions, Moscow, 1950.
\(^{2}\) M. A. Naimark, Linear differential operators, Moscow, 1954.
\(^{3}\) D. V. Sears, Canad. J. Math., 2, 314 (1950).
\(^{4}\) V. B. Lidskii, DAN, 95, No. 2, 217 (1954).
\(^{5}\) É. Shnol’, UMN, 9, issue 4 (62), 113 (1954).
\(^{6}\) J. Brinck, Math. Scand., 7, No. 1, 219 (1959).