Let us consider the problem:

$P(D)u(x)=(c(x)+\lambda )u;\phantom{\rule{1em}{0ex}}x\in {R}^{n},u(x)\in {L}^{2}\left({R}^{n}\right),$

(6)

in which

$c(x)$ is a measurable function,

$c(x)>0$ for all

$x\in \mathrm{\Omega}$,

$|c(x)|\le d<+\mathrm{\infty}$ almost everywhere in Ω,

$c(x)=0$ outside Ω, Ω is measurable and

$P(D)$ is a linear pseudo-differential operator with constant coefficients. Here

$P(iz)\ge $ argument quasi-polynomial

$z\in {R}^{n}$, not depending on

*x* and satisfying the following conditions for all

$z\in {R}^{n}$:

We suppose that

${\int}_{|z|\le \delta}{|P(iz)-\mu |}^{-2}\phantom{\rule{0.2em}{0ex}}dz\to +\mathrm{\infty},\phantom{\rule{1em}{0ex}}\text{if}\mu \to {0}^{-}$

(10)

for each sufficiently small $\delta >0$ and $\mu \to {0}^{-}$.

**Theorem 1** *The problem* (6) *has at least one negative eigenvalue if* Ω *is bounded*.

It is necessary to introduce several lemmas before proving this theorem.

In each case, we consider $\mu <0$. By virtue of (8), there is a function $h(x,\mu )\in {L}^{2}({R}^{n})$ of the Fourier transformation which coincides with ${[P(iz)-\mu ]}^{-1}$. Considering (7), the real and even function $h(x,\mu )$ could be obtained.

**Lemma 1** *Let* $\mu <0$.

*The problem* (6)

*has a nonzero solution if and only if the nonzero solution* $v(x)$ *has the form* $v(x)={\int}_{\mathrm{\Omega}}c(t)v(t)h(x-t,\mu )\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{1em}{0ex}}v\in {L}^{2}(\mu ).$

(11)

*Proof* Applying the Fourier transformation for (6) yields

$(P(iz)-\mu )(Fu(x))(z)=\left[F(c(x)u(x))\right](z)\in {L}^{2}\left({R}^{n}\right).$

Hence, in particular, the integral

$\int \left(F(u(x))\right)(z){e}^{ixz}\phantom{\rule{0.2em}{0ex}}dz$

converges absolutely. From now on,

$ixz=(i{x}_{1}{z}_{1},\dots ,i{x}_{n}{z}_{n})$. It follows from latter relations

$u(x)=\frac{1}{\sqrt{2\pi}}\int \left[F(u(x))\right](z){e}^{ixz}\phantom{\rule{0.2em}{0ex}}dz=\int \left[F(c(t)u(t))\right](z)[{F}_{t}h(x+t,\mu )](z)\phantom{\rule{0.2em}{0ex}}dz,$

where

${F}_{t}$ means that the Fourier transformation has been determined under

*t*. Hence, by virtue of Parseval’s equality, it follows that

$u(x)=\int c(t)u(t)h(x-t,\mu )\phantom{\rule{0.2em}{0ex}}dt.$

Since $c(t)=0$ outside Ω, then $u(x)$; $x\in \mathrm{\Omega}$ is the solution of the problem (11). If $u(t)=0$ where in Ω we obtain $[c(t)u(t)]=0$ for ${R}^{n}$, by virtue of the latter equality $u(x)=0$. The necessity is proved.

Let us prove the sufficiency. Let

$v(x)$ be the nonzero solution of the problem (11). Consider the new problem

$P(D)u=\mu u+c(x)f(x),\phantom{\rule{1em}{0ex}}x\in {R}^{n};u\in {L}^{2}\left({R}^{n}\right),$

(12)

in which

$f(x)=v(x)$ for all

$x\in \mathrm{\Omega}$ and

$f(x)=0$ outside Ω. Since

$c(x)f(x)\in {L}^{2}({R}^{n})$, applying the Fourier transformation for (12), we obtain

$\begin{array}{c}[P(iz-\mu )][Fu(x)](z)=\left[F(c(x)f(x))\right]z\in {L}^{2}\left({R}^{n}\right);\hfill \\ u(x)=\frac{1}{\sqrt{2\pi}}\int \left[F(c(x)f(x))\right](z)[{F}_{t}h(x+t,\mu )](z)\phantom{\rule{0.2em}{0ex}}dz=\int c(t)f(t)h(x-t,\mu )\phantom{\rule{0.2em}{0ex}}dt.\hfill \end{array}$

From Parseval’s equality, the solution of the problem (12) exists and it is unique. In particular, when

$x\in \mathrm{\Omega}$, we have

$u(x)={\int}_{\mathrm{\Omega}}c(t)v(t)h(x-t,\mu )\phantom{\rule{0.2em}{0ex}}dt=v(x).$

Considering this inequality and (12), we obtain $c(x)f(x)=c(x)u(x)$, *i.e.*, $u(x)$ is the solution of the problem (6). Thus, the lemma is proved. □

In the case when

$\mu <0$, we consider

$A(\mu )$ as an integral operator, where

$A(\mu )\omega (x)=\sqrt{c(x)}{\int}_{\mathrm{\Omega}}\sqrt{c(t)}h(x-t,\mu )\omega (t)\phantom{\rule{0.2em}{0ex}}dt;\phantom{\rule{1em}{0ex}}\omega \in {L}_{b}^{2}(\mathrm{\Omega}).$

We remember that the operator

$A(\mu )$ is defined only when

$\mu <0$. Since

$P(iz)=P(-iz)$, thus the Fourier transformation for the functions

$h(t,\mu )$,

$h(-t,\mu )$ coincides. That is why

$h(t,\mu )=h(-t,\mu )$. If Ω is bounded, then the kernel

$(\sqrt{c(x)}\sqrt{c(t)}h(x-t,\mu ))$ of the integrated operator

$A(\mu )$ belongs to

${L}_{b}^{2}(\mathrm{\Omega}\times \mathrm{\Omega})$. It follows that the operator

$A(\mu )$ is completely continuous. Its self-adjointness and positiveness are obvious. This enables us to write down the eigenvalues of the operator

$A(\mu )$:

${\lambda}_{1}(\mu )>{\lambda}_{2}(\mu )>\cdots .$

(13)

It is well known that (see [

7])

${\lambda}_{1}(\mu )=\mathrm{Sup}<A(\mu )f,$

(14)

where Sup is determined for all the function $f\in {L}_{b}^{2}(\mathrm{\Omega})$, for which $\parallel f\parallel \le 1$.

From the known results for self-adjoint and quite continuous operators (see [

7]), it follows that

${\lambda}_{k}(\mu )$ continuously depends on

*μ*, where

${|{\lambda}_{1}(\mu )|}^{2}={\parallel A(\mu )\parallel}^{2}={\int}_{\mathrm{\Omega}}{\int}_{\mathrm{\Omega}}c(x)c(t){(h(x-t,\mu ))}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt.$

(15)

**Lemma 2** *Let* Ω

*be bounded when* $x\in \mathrm{\Omega}$.

*Then* - (1)
${\lambda}_{k}(\mu )\to 0$ *at* $\mu \to -\mathrm{\infty}$,

- (2)
${\lambda}_{1}(\mu )\to +\mathrm{\infty}$ *at* $\mu \to {0}^{-}$.

*Proof* Since

${\lambda}_{j}(\mu ){f}_{j}(x,\mu )=A(\mu )f(x,\mu )$,

${\parallel {f}_{j}(x,\mu )\parallel}_{\mathrm{\Omega}}=1$, and

$|c(x)|\le a<+\mathrm{\infty}$, we have

$\begin{array}{rcl}|{\lambda}_{j}(\mu )|& \le & \parallel A(\mu )\parallel ={[{\int}_{\mathrm{\Omega}}{\int}_{\mathrm{\Omega}}c(x)c(t){(h(x-t,\mu ))}^{2}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt]}^{\frac{1}{2}}\\ \le & a{[{\int}_{\mathrm{\Omega}}c(t)\left(h{(x-t,\mu )}^{2}\phantom{\rule{0.2em}{0ex}}dx\right)\phantom{\rule{0.2em}{0ex}}dt]}^{\frac{1}{2}}\\ =& a\parallel h(x,\mu )\parallel \cdot {({\int}_{\mathrm{\Omega}}c(t)\phantom{\rule{0.2em}{0ex}}dt)}^{\frac{1}{2}}\\ \le & a\parallel h(x,\mu )\parallel \cdot {\parallel c(t)\parallel}_{\mathrm{\Omega}}\\ =& {a}_{1}{(\int \frac{dz}{{(P(iz)-\mu )}^{2}})}^{\frac{1}{2}}.\end{array}$

Hence, the first statement follows from (9).

Let us prove the second statement. By virtue of (13), with

$c(x)=0$ outside Ω and

${({\lambda}_{1}(\mu ))}^{2}=\int \int [c(x)h(x-t,\mu )]\cdot [c(t)h(x-t,\mu )]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dt,$

which is applied to the last integral in Parseval’s inequality, we obtain

${({\lambda}_{1}(\mu ))}^{2}=\int \int \left(F[c(x)h(x-t,\mu )]\right)(z,\xi )\cdot \left(F[c(t)h(x-t,\mu )]\right)(-z,-\xi )\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}d\xi .$

The following equations are correct:

$\begin{array}{rcl}\left(F[c(x)h(x-t,\mu )]\right)(z,\xi )& =& \left({F}_{x}[c(x)\cdot \frac{{e}^{-ix\xi}}{P(i\xi )-\mu}]\right)(z,\xi )\\ =& \frac{1}{P(i\xi )-\mu}\left({F}_{x}[c(x){e}^{-ix\xi}]\right)(z,\xi )\\ =& \frac{1}{\sqrt{2\pi}}\cdot \frac{1}{P(i\xi )-\mu}{\int}_{\mathrm{\Omega}}c(x){e}^{-ix(\xi +z)}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$

In a similar way, we obtain

$\left(F[c(t)h(x-t,\mu )]\right)(-z,-\xi )=\frac{1}{\sqrt{2\pi}}\cdot \frac{1}{P(-i\xi )-\mu}{\int}_{\mathrm{\Omega}}c(t){e}^{it(\xi +z)}\phantom{\rule{0.2em}{0ex}}dt.$

Thus, we have proved the following:

${({\lambda}_{1}(\mu ))}^{2}=\frac{1}{2\pi}\int \int {|{\int}_{\mathrm{\Omega}}c(x){e}^{-ix(\xi +z)}\phantom{\rule{0.2em}{0ex}}dx|}^{2}\cdot \frac{dz\phantom{\rule{0.2em}{0ex}}d\xi}{{[P(i\xi -\mu )]}^{2}}.$

The following estimate is obvious:

${({\lambda}_{1}(\mu ))}^{2}\ge \frac{1}{2\pi}{\int}_{|z|\le \delta}{\int}_{|\xi |\le \delta}{|{\int}_{\mathrm{\Omega}}c(x){e}^{-ix(\xi +z)}\phantom{\rule{0.2em}{0ex}}dx|}^{2}\cdot \frac{dz\phantom{\rule{0.2em}{0ex}}d\xi}{{[P(i\xi )-\mu ]}^{2}},$

(16)

where

${|{\int}_{\mathrm{\Omega}}c(x){e}^{-ix(\xi +z)}\phantom{\rule{0.2em}{0ex}}dx|}^{2}={|{\int}_{\mathrm{\Omega}}c(x)\phantom{\rule{0.2em}{0ex}}dx+{\int}_{\mathrm{\Omega}}c(x)\left[{e}^{-ix(\xi +z)}\right]\phantom{\rule{0.2em}{0ex}}dx|}^{2},$

(17)

*δ* will be chosen in a way such that $|{e}^{-ix(z+\xi )}-1|\le \frac{1}{2}$ for all $x\in \mathrm{\Omega}$ and $|z|\le \delta $, $|\xi |\le \delta $. Since Ω is bounded, we may always obtain the latter.

Considering (16) and (17), we obtain

$\begin{array}{rcl}{({\lambda}_{1}(\mu ))}^{2}& \ge & \frac{1}{4\cdot 2\pi}{({\int}_{\mathrm{\Omega}}c(x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}\cdot {\int}_{|z|\le \delta}\phantom{\rule{0.2em}{0ex}}dz{\int}_{|\xi |\le \delta}\frac{d\xi}{{[P(i\xi -\mu )]}^{2}}\\ =& {b}_{1}(\delta ){\int}_{|z|\le \delta}\frac{d\xi}{{(P(i\xi )-\mu )}^{2}}.\end{array}$

Hence, by virtue of (10), the lemma is proved. □

*Proof of Theorem 1* At the first stage, we suppose that

$c(x)\ge \nu >0$ for all

$x\in \mathrm{\Omega}$. By virtue of Lemmas 1 and 2, where

${\lambda}_{1}({\mu}_{0}(\nu ))=1$ for

${\mu}_{0}(\nu )<0$, if

${f}_{1}(x)$ is the eigenfunction corresponding to the eigenvalue

${\lambda}_{1}({\mu}_{0}(\nu ))$, then

$\varphi (x)={f}_{1}(x)/\sqrt{c(x)}\in {L}^{2}(\mathrm{\Omega}).$

When $\mu ={\mu}_{0}(\nu )$, we have the nonzero solution of the equation (11). It follows from Lemma 1 that ${\mu}_{0}(\nu )$ is the eigenvalue of the problem (6).

For the general case, we put

$c(\nu ,x)=c(x)$ if

$x\in \mathrm{\Omega}$ and

$c(x)\ge \nu $;

$c(\nu ,x)=\nu $ when

$x\in \mathrm{\Omega}$ and

$c(x)\in (0,\nu )$. The nonzero solutions of the equation

$\varphi (\nu ,x)={\int}_{\mathrm{\Omega}}c(\nu ,x)h(x-t,{\mu}_{0}(\nu ))\varphi (\nu ,t)\phantom{\rule{0.2em}{0ex}}dt$

(18)

are chosen in such a way that ${\parallel \varphi (\nu ,x)\parallel}_{\mathrm{\Omega}}$.

The integral operators defined by the right-hand sides of (11) and (18) are defined in

$B(\mu )$,

$B(\nu ,{\mu}_{0}(\nu ))$ respectively. Since Ω is bounded, then both

$\{c(\nu ,x)\}$ and

$\{h(x,{\mu}_{0}(\nu ))\}$ uniformly converge by norm to

$c(x)$ and

$h(x,{\mu}_{0})$ respectively. If

${\mu}_{0}(\nu )\to {\mu}_{0}$, then

$\parallel B({\mu}_{0})-B(\nu ,{\mu}_{0}(\nu ))\parallel \to 0\phantom{\rule{1em}{0ex}}\text{when}{\mu}_{0}(\nu )\to {\mu}_{0}.$

(19)

Considering the choice $\varphi (\nu ,x)$ and the property $\parallel h(x,\mu )\parallel \to 0$, if $\mu \to -\mathrm{\infty}$, we can easily prove the boundedness of ${\mu}_{0}(\nu )$. Noting that when ${\mu}_{0}$ and ${\nu}_{0}\to 0$ for which ${\mu}_{0}({\nu}_{j})\to {\mu}_{0}$, the operator $B({\mu}_{0})$ is completely continuous. In this case, as we know, the set $B({\mu}_{0})$, $\varphi (\nu ,x)$ contains the subsequence $B({\mu}_{0})$, $\varphi ({\nu}_{j},x)$ which converges by norm where $\mu \to {\mu}_{0}$.

From (18) and (19) it follows that $\{\varphi ({\nu}_{j},x)\}$ converges to $\varphi (x)$ by norm where ${\parallel u(x)\parallel}_{\mathrm{\Omega}}=1$. Then $\{B(\nu ,{\mu}_{0}({\nu}_{{j}_{1}}))\varphi ({\nu}_{{j}_{1}},x)\}$ converges to $B({\mu}_{0})\varphi (x)$ by norm and satisfies the equality $u(x)=B({\mu}_{0})\varphi (x)$, *i.e.*, when $\mu ={\mu}_{0}$, the equation (11) has a nonzero solution. Hence, the theorem is proved. □