Preamble

ON THE COMPLEX EIGENVALUES OF A NON-SELF-ADJOINT OPERATOR

M. GEKHTMAN, A. G. KOSTYUCHENKO

DIFFERENTIAL EQUATIONS, 1967, VOL. III, NO. 3

Consider the self-adjoint differential expression
$$l(y) = (-1)^m y^{(2m)} + (p_1(x) y^{(m-1)})^{(m-1)} + \dots + p_m(x)y, \quad \quad \quad (1)$$
whose coefficients are real-valued functions possessing continuous derivatives up to the required order. Furthermore, we shall assume that these coefficients are such that for any compactly supported, sufficiently smooth functions, the following inequality holds:

$$\int_{-\infty}^{\infty} l(f) \overline{f(x)} dx \ge 0. \quad \quad \quad (2)$$

As is well known, the differential expression $l(y)$ generates a symmetric operator in $L_2(-\infty, \infty)$, which we shall denote by $L_0$. This operator admits self-adjoint extensions. Let us fix one such self-adjoint extension of the operator $L_0$ and denote it hereafter by $L$. By virtue of condition (2), the spectrum of the operator $L$ is bounded from below; without loss of generality, we may assume that the spectrum of $L$ is non-negative.

Now, let $q(x)$ be a continuous real-valued function defined on the entire real axis and satisfying the condition $q(x) \to 0$ as $|x| \to \infty$. We further consider the non-self-adjoint differential operator $T$ in $L_2(-\infty, \infty)$ defined by $Tu = Lu + iq(x)u$. It is known that the limit points of the spectrum of the operator under consideration can only lie on the real axis within the spectral domain of the unperturbed self-adjoint operator $L$. This fact was first established by M. A. Naimark \cite{1} for the Sturm-Liouville operator, and subsequently for the Schrödinger operator by I. M. Gelfand \cite{2}. We note that Gelfand's proof is of a general character: in proving that the limit points of the spectrum of the perturbed operator must lie on the real axis, he utilizes only the fact that the kernel of the resolvent of the unperturbed operator is of Carleman type, i.e., $|K(x, y; z)| < C$ when $z$ varies within a bounded region. In the case of self-adjoint differential operators generated by ordinary differential equations, the resolvent kernel is always of Carleman type (see, for example, \cite{3}).

It is known that the entire spectrum of the operator $T$ is located within the strip $|\text{Im } z| \le \max |q(x)|$ of the complex $z$-plane ($z = \sigma + i\tau$). The purpose of this note is to refine this fact. Specifically, we show that as $\sigma \to \infty$, the spectrum of the operator $T$ lies in the region $|\tau| \le C \sigma^{-\alpha}$, where $C$ is a constant and $\alpha > 0$. In the case where $q(x)$ is a non-negative function satisfying certain conditions, a similar assertion for the non-self-adjoint Sturm-Liouville operator was previously proven in \cite{4}. Since the limit spectrum of the operator $T$ is located only on the real axis, as noted above, it is sufficient to show that all eigenvalues of the operator $T$ are situated in the region $|\tau| < K(\sigma)$.

Let us first consider the case where $q(x)$ is a compactly supported function vanishing outside a finite interval. Suppose $z = \sigma + i\tau$ is a complex eigenvalue of the operator $T$, and $u$ is the corresponding eigenfunction, i.e.,

Lu 0 —z 0 u = —iq(x)u 0 . (5)

Let us denote the resolvent of the operator by $R_{\lambda}$. Since the condition $q(x)u \in L_2$ holds, we can apply the resolvent to both sides of the equality. Consequently, instead of the differential equation, we obtain an equivalent integral equation.

and $u_0 = -R_{\lambda} q(x) u_0$. (6)

It is well known that the operator $R_{\lambda}$ is an integral operator with a kernel $K(x, y, \lambda)$, where the kernel is expressed through the spectral function $\theta(x, y; \lambda)$ of the operator by the relation $K(x, y, \lambda) = \int_{-\infty}^{\infty} \frac{d\theta(x, y; \lambda)}{\mu - \lambda}$. It was proven by one of the authors of this note \cite{5} that the spectral function $\theta(x, y; \lambda)$ can be represented as $\theta(x, y; \lambda) = \theta_0(x, y; \lambda) + \dots$

$$ \theta_0(x, y; \lambda) = \begin{cases} \frac{1}{\pi} \frac{\sin \sqrt{\lambda}(x - y)}{x - y}, & \text{if } \lambda > 0, \\ 0, & \text{if } \lambda < 0, \end{cases} $$

[FIGURE:1]

Furthermore, the function $\delta(x, y; \lambda)$ tends to zero uniformly in each finite domain of the variables. If we now denote $v(x) = q(x) u_0(x)$ and take into account the finiteness of the function $q(x)$ on the interval $[a, b]$, we obtain an integral equation for determining the complex eigenvalues and eigenfunctions of the operator.

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In operator form, let us denote for brevity:

On Complex Eigenvalues of a Single Operator

We shall prove that for sufficiently large values of $a$ (with $m$ fixed), the norm of the operator can be made arbitrarily small. From this, it follows that for such $z = a + i\zeta$, the equation in $L_2(-\infty, \infty)$ can only have a trivial solution. Consequently, the specified values are not eigenvalues of the operator.

The norm of the operator can be easily estimated using the following inequality:
$$\|K\| \le C \sqrt{\int \int |K(x, y; z)|^2 \, dx dy}$$
(where $C = \max |q(x)|$).

Furthermore, we will demonstrate that the kernel of the resolvent $R(x, y; z)$ tends to zero uniformly with respect to $x$ and $y$ varying within a bounded domain, provided that $a \to \infty$ (with $m$ fixed). It follows from formula (7) that $Q(x, y; \lambda)$...

The kernel function is defined as:

$$K(x, y; z)$$

We transform the integral on the left side of this equality by performing integration by parts in the second term:

$$\int \Gamma \phi(x, y, \lambda) \, dx - z$$

In performing the transformation (12), we utilized the fact that if $x$ and $y$ vary within a bounded domain and $\lambda \to \infty$, then $\phi(x, y, \lambda)$ behaves uniformly with respect to these variables.

We shall now verify that each term in the formula tends to zero as $z \to \infty$ (with $\tau$ fixed), uniformly with respect to the variables $x$ and $y$ varying within a bounded domain. Let us denote the integral by $\Phi(x, y, z)$. Since $\Phi(x, y, z)$ is the kernel of the resolvent of the operator $L = (-\Delta)$, it is a decaying solution of the equation:

[The text ends here, but typically follows with the corresponding differential equation for the resolvent kernel.]

$b_0 K^{-1} b = \delta(x - y)$

Using the Fourier transform, we obtain $|x-y|$. By calculating the integral using residues, we find $\Phi(x, y; z)$.

M. M. Gekhtman and A. G. Kostyuchenko. The roots of the equation $s$ ($\text{Im } s > 0$). From formula $(*)$, it follows that

$$|\Phi_0(x, y; z)| < \frac{C}{|x-y|}$$

From this estimate, it is clear that $\Phi(x, y; z)$ tends to zero as $z \to \infty$ uniformly with respect to the variables. Let us verify that the last term in formula (12) also tends to zero as $z \to \infty$. Given $\epsilon > 0$, we choose $N(\epsilon) > 0$ such that $|\phi(x, y, \lambda)| < \epsilon$ for $\lambda > N$, uniformly for variables changing within a bounded domain. We transform the integral:
$$\int_0^\infty \dots d\lambda = \dots$$ (13)
We now estimate each term in formula (13).

$|J_1| < C$ as $z \to \infty$,

since it is possible to pass to the limit under the integral sign.

1 2 l < 8 J U = 7 p = " " e -

Due to the arbitrariness of $\delta > 0$, we conclude that $\lim \delta = 0$ as $z \to \infty$. Thus, we have shown that each term in formula (12) tends to zero as $z \to \infty$ (for a fixed $\tau$) uniformly with respect to the variables $x$ and $y$ varying within a bounded domain. This implies that $\lim K(x, y) = 0$ as $z \to \infty$ (for fixed $\tau$) uniformly with respect to the variables in a bounded domain. Therefore, in inequality (11), we can pass to the limit under the integral sign, yielding:

$$\lim \| S_z \| = 0 \quad (z \to \infty).$$

It follows from this that on the line $\tau = \text{const}$, starting from a certain point, all points are of regular type for the operator $L$, and consequently, there are no eigenvalues of the operator on this line beyond that point. We can now prove the following theorem.

Theorem. Let an operator $L$ be defined by the given conditions. Then there exists a non-negative function $K(\sigma)$ such that $\lim K(\sigma) = 0$ as $\sigma \to \infty$, and the entire spectrum of the operator is located within the region defined by $K(\sigma)$.

Proof. If the potential is compactly supported (finite), the theorem is already proven. We shall therefore assume that $q(x)$ satisfies only the general condition. Let $V_0(x) = p(x)$. Since $q(x) \to \infty$, we can represent $p(x)$ as a sum $p(x) = p_1(x) + p_2(x)$, where $p_1(x)$ is compactly supported on $[a, b]$ and $\| p_2 \| < \epsilon$. To prove the theorem, we follow the previous reasoning verbatim, which leads to the necessity of showing that:

On the Complex Eigenvalues of a Certain Operator

In the general case, $\lim \| S_z p \| = 0$ as $z \to \infty$ (for fixed $\tau$). Since $p_1$ is compactly supported, $\lim \| S_z p_1 \| = 0$ as $z \to \infty$. As for the remaining terms, the norm of each can be made arbitrarily small due to the presence of the operator $p_2$ in each term, whose norm satisfies $\| p_2 \| < \epsilon$.

References

References

Naimark, M. A. DAN SSSR, 85, No. 1, 41–44.

Gelfand, I. M. UMN, 7, No. 6, 183–184.

Naimark, M. A. Linear Differential Operators. Moscow, Gostekhizdat, 1954, p. 181.

Gekhtman, M. M. Vestnik Moskovskogo Universiteta, No. 1, 29–35.

Kostyuchenko, A. G. DAN SSSR, 168, No. 1, 1966.

Submitted to the Editorial Board in December 1965.
Lomonosov Moscow State University.