MATHEMATICS

M. M. Gekhtman

ON THE SPECTRUM OF A NON-SELF-ADJOINT DIFFERENTIAL OPERATOR OF EVEN ORDER

(Presented by Academician I. M. Vinogradov on January 5, 1963)

In the work of M. A. Naimark \((^1)\) it was proved that the differential operator \(-y''+q(x)y\), considered on the half-line \((0\le x<\infty)\), in the case where \(q(x)\) is a complex-valued function satisfying the condition \(|q(x)|<e^{-\varepsilon_0 x}\), has no positive eigenvalues. This fact, however, does not generalize to the case of differential operators of order \(2n\) \((n\ge 2)\).

Consider in \(\mathcal L_2(-\infty,\infty)\) the differential operator

\[ Ly\equiv (-1)^n\frac{d^{2n}y}{dx^{2n}}+q(x)y, \tag{1} \]

where \(q(x)\) is, generally speaking, a complex-valued function satisfying, for some fixed \(\varepsilon_0>0\), the condition

\[ |q(x)|\le Ce^{-\varepsilon_0|x|}\qquad (-\infty<x<\infty). \tag{2} \]

The present note is devoted to the proof of the following theorem:

Theorem. The eigenvalues of the operator \(L\), under condition (2), can form only a finite set.

Moreover, even in the case where \(q(x)\equiv 0\) outside some finite interval \([a,b]\), the operator \(L\) for \(n\ge 2\) may have positive eigenvalues. In the complex \(\lambda\)-plane we shall indicate the radius of a circle, depending on \(q(x)\), outside which the operator certainly has no eigenvalues.

We pass to the proof. For definiteness we shall assume \(n\) to be an even number.

Let us find the resolvent of the operator \(d^{2n}y/dx^{2n}\) in \(\mathcal L_2(-\infty,\infty)\). To this end one must solve the equation

\[ \frac{d^{2n}y}{dx^{2n}}=s^{2n}y+\delta(x),\qquad 0<\arg s<\frac{\pi}{n}. \]

With the aid of the Fourier transform and the theory of residues we find the resolvent kernel

\[ K_s(x-y)=\frac{\pi i}{ns^{2n-1}}\sum_{k=0}^{n-1}\varepsilon_k e^{is\varepsilon_k|x-y|}; \]

here by \(\varepsilon_k\) are denoted those roots of degree \(2n\) of 1 for which \(\operatorname{Im}s\varepsilon_k>0\).

Therefore the resolvent of the unperturbed operator \(d^{2n}y/dx^{2n}\) is the integral operator

\[ R_s^0 f(x)=\frac{\pi i}{ns^{2n-1}}\int_{-\infty}^{\infty}K_s(x,y)f(y)\,dy,\qquad f\in\mathcal L_2. \tag{3} \]

Let us now consider the equation for determining the resolvent of the operator \(L\)

\[ \frac{d^{2n}y}{dx^{2n}}+q(x)y-s^{2n}y=f,\qquad f\in \mathcal L_2(-\infty,\infty). \]

Applying to this equality the resolvent \(R_s^0\) and denoting by \(T_s\) the operator whose action on \(y(x)\in \mathcal L_2(-\infty,\infty)\) consists in first multiplying \(y(x)\) by \(q(x)\), and then applying \(R_s^0\) to \(q(x)y(x)\in \mathcal L_2(-\infty,\infty)\), by virtue of condition (2), we obtain the integral equation

\[ y(x)+T_s y(x)=R_s^0 f; \tag{4} \]

for \(0<\arg s<\pi/n\), by virtue of \(\operatorname{Im} s\varepsilon_k>0\), the operator \(T_s\) will be a Hilbert–Schmidt operator, and it depends analytically on \(s\); the kernel of this operator is a continuous and bounded function, and therefore Fredholm theory is applicable. The solution of (4) can be written with the aid of Fredholm determinants

\[ y_s(x)=R_s^0 f(x)+\int_{-\infty}^{\infty}\frac{D(x,y,s)}{D(s)}R_s^0 f(y)\,dy, \]

where

\[ D(s)=1+\sum_{k=1}^{\infty}(-1)^k\frac{d_k(s)}{k!}, \tag{5} \]

where

\[ d_k(s)=\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \left| \begin{array}{cccc} K_s(\alpha_1,\alpha_1)&\ldots&K_s(\alpha_1,\alpha_k)\\ K_s(\alpha_2,\alpha_1)&\ldots&K_s(\alpha_2,\alpha_k)\\ \cdot&\cdot&\cdot&\cdot\\ K_s(\alpha_k,\alpha_1)&\ldots&K_s(\alpha_k,\alpha_k) \end{array} \right| \cdot \prod_{j=1}^{k}q(\alpha_j)\,d\alpha_j, \]

\[ K_s(x,y)=\frac{\pi i}{n s^{2n-1}}\sum_{k=0}^{n-1}\varepsilon_k e^{i s\varepsilon_k|x-y|},\qquad 0<\arg s<\frac{\pi}{n}. \tag{6} \]

Lemma 1. \(D(s)\) admits analytic continuation through the positive real semi-axis into the domain \(S\), which is defined by the inequalities
\[ |s|\ge s_0>0,\qquad \operatorname{Im}s>-\varepsilon_0/2,\qquad \operatorname{Re}s>0,\qquad -\pi/n<\arg s<\pi/n. \]
In the domain \(S\), \(K_s(x,y)\) is an analytic function of \(s\) for fixed \(x\) and \(y\), and admits the estimate

\[ |K_s(x,y)|\le \frac{\pi}{|s|^{2n-1}}e^{\frac{\varepsilon_0}{2}|x-y|}. \]

This estimate can be strengthened by the triangle inequality:

\[ |K_s(x,y)|\le \frac{\pi}{|s_0|^{2n-1}}e^{\frac{\varepsilon_0}{2}(|x|+|y|)}. \tag{7} \]

Using condition (2) and Hadamard’s lemma, we obtain that in the domain \(S\)

\[ |d_k(s)|\le \left(\frac{\pi B!}{n s_0^{2n-1}}\sqrt{k}\right)^k, \tag{8} \]

where

\[ B=\int_{-\infty}^{\infty}|q(x)|e^{\varepsilon_0|x|}\,dx. \]

By virtue of estimate (8), by d’Alembert’s test, the functional series defining \(D(s)\) and consisting of analytic functions converges uniformly in \(S\). Lemma 1 is proved.

It follows from Lemma 1 that in the domain \(0 \leq \arg s < \pi/n\), \(s \ne 0\), the operator \(L\) can have at most a countable set of eigenvalues, with possible accumulation points at infinity and at \(s=0\).

Choose the number \(L_0\) from the condition

\[ 1=\sum_{k=1}^{\infty}\frac{k^{k/2}L_0^k}{k!}, \]

and choose \(s_0>0\) so that

\[ q=\frac{\pi B}{n s_0^{2n-1}}<L_0. \tag{9} \]

Then, for \(|s|\geq s_0\) and \(s\in S\),

\[ |D(s)|\geq 1-\left|\sum_{k=1}^{\infty}\frac{k^{k/2}}{k!}q^k\right|\ne 0, \]

i.e., for sufficiently large \(|s|\), \(D(s)\) has no zeros.

Corollary. The eigenvalues of the operator \(L\) form a bounded set, the only limiting point of which can be the point \(\lambda=0\).

At the same time, we can also give an effective estimate of the disk in the complex plane \(\lambda=s^{2n}\) outside which the operator \(L\) certainly has no eigenvalues. Namely, from formula (9) it follows immediately that, for \(|\lambda|\geq R\), where

\[ R=\left(\frac{\pi\displaystyle\int_{-\infty}^{\infty}|q(x)|e^{\varepsilon_0|x|}\,dx}{nL_0}\right)^{\frac{2n}{2n-1}}, \tag{10} \]

the operator \(L\) has no eigenvalues.

It remains to show that \(s=0\) is not an accumulation point of the eigenvalues of the operator \(L\).

Consider

\[ \frac{d^{2n}y}{dx^{2n}}+q(x)y=s^{2n}y \quad (-\infty<x<\infty),\qquad 0\leq \arg s<\frac{\pi}{n}, \tag{11} \]

where \(q(x)\) satisfies condition (2).

Lemma 2. The differential equation (11) has \(2n\) linearly independent solutions which have the asymptotics

\[ y_i(x,s)=e^{\rho_i sx}\left(1+o(e^{\varepsilon_0 x})\right) \quad (x\to-\infty), \tag{12} \]

and, for each fixed \(x\), \(y_i(x,s)\) is an analytic function of \(s\) in the disk \(2|s|<\varepsilon_0\); here \(\rho_i\) are the roots of degree \(2n\) of \(1\) \((i=1,2,\ldots,2n)\).

We outline the proof. From the differential equation (11), for each fixed \(\rho_i\) we pass to the integral equation

\[ y_i(x,s)=e^{\rho_i xs}-\int_{-\infty}^{x}K(x,z,s)y_i(z,s)\,dz, \tag{13} \]

where

\[ K(x,z,s)=\frac{\displaystyle\sum_{i=1}^{2n}\rho_i e^{\rho_i s(x-z)}q(z)} {2n\,s^{2n-1}}. \]

By expanding in a Taylor series, we verify that \(K(x,z,s)\) is a function analytic in \(s\) over the whole plane for fixed \(x\) and \(z\), and ...

the estimate holds

\[ |K(x,z,s)| \leq \frac{|x-z|^{2n-1}}{(2n-1)!} e^{2|s||x-z|-\varepsilon_0|z|}. \tag{14} \]

We first prove the existence of the required solutions for the integral equations, using estimate (14) by the method of successive approximations.

Lemma 3. The differential equation (11) has \(2n\) linearly independent solutions that have the asymptotics

\[ u_i(x,s)=e^{\rho_i sx}\left(1+o\left(e^{-\varepsilon_0 x}\right)\right)\quad (x\to+\infty),\qquad i=1,2,\ldots,2n, \tag{15} \]

and \(u_i(x,s)\), for each fixed \(x\), are analytic functions of \(s\) in the disk \(2|s|<\varepsilon_0\), for each fixed \(x\in(-\infty,\infty)\).

The proof is analogous to the proof of Lemma 2; one need only consider the integral equation

\[ u_i(x,s)=e^{\rho_i sx}+\int_x^\infty K(x,z,s)u_i(z,s)\,dz. \]

Lemma 4. The point \(\lambda=0\) is not an accumulation point of the eigenvalues of the operator \(L\).

The proof follows from Lemmas 2 and 3, and also from the fact that the determinant \(D(s)\ne0\) in the domain \(S\), \(s\ne0\).

In conclusion, I take this opportunity to express my gratitude to Prof. M. A. Naimark and R. S. Ismagilov for valuable advice and discussion of the results obtained.

Moscow State University
named after M. V. Lomonosov

Received
3 I 1963

References

¹ M. A. Naimark, Tr. Mosk. matem. obshch., 3, 181 (1954).