Upper Bound For The Ratios Of Eigenvalues Of Schrodinger Operators With Nonnegative Single-Barrier Potentials

TL;DR

This paper establishes an optimal upper bound for the ratio of eigenvalues of one-dimensional Schrödinger operators with certain nonnegative single-barrier potentials, using a novel approach to analyze the Pr"ufer angle function.

Contribution

It introduces a new method to prove the eigenvalue ratio bound for Schrödinger operators with specific potential conditions, improving understanding of spectral properties.

Findings

Proves the bound $rac{ abla_{n}}{ abla_{m}} extless rac{n^{2}}{m^{2}}$ under given conditions.

Establishes the bound holds for potentials with $ ext{sup}_{x ext{ in }[0,1]} q(x) extless rac{ ext{ extpi}^{2}}{11}$.

Develops a new approach to analyze the monotonicity of the modified Prüfer angle function.

Abstract

In this paper we prove the optimal upper bound $\frac{\lambda_{n}}{\lambda_{m}}\leq\frac{n^{2}}{m^{2}}$ $\Big(\lambda_{n}>\lambda_{m}\geq 11\sup\limits_{x\in[0,1]}q(x)\Big)$ for one-dimensional Schrodinger operators with a nonnegative differentiable and single-barrier potential $q(x)$, such that $\mid q'(x) \mid\leq q^{*},$ where $q^{*}=\frac{2}{15}\min\{q(0) , q(1)\}$. In particular, if $q(x)$ satisfies the additional condition $\sup\limits_{x\in[0,1]}q(x)\leq \frac{\pi^{2}}{11}$, then $\frac{\lambda_{n}}{\lambda_{m}}\leq \frac{n^{2}% }{m^{2}}$ for $n>m\geq 1.$ For this result, we develop a new approach to study the monotonicity of the modified Pr\"{u}fer angle function.

Equations

$$-y^{\prime \prime }+q(x)y=\lambda y,$$

$$y(0)=y(1)=0.$$

$$\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\geq\frac{1}{1+\xi}\Big{(}\frac{n+1}{m+1}\Big{)}^{2}\frac{k}{K},~{}~{}n>m\geq 0,$$

$$\frac{{\lambda }_n}{{\lambda }_m}\le \frac{n^2}{m^2},n>m\ge 1,$$

$$-y^{\prime \prime }+q(x)y=z^{2}y,x\in [0,1],z>0,$$

$$y(0)=0,y^{\prime }(0)=1.$$

$$y(x,z)=r(x,z)\sin \phi (x,z),$$

$$y^{\prime }(x,z)=zr(x,z)\cos \phi (x,z),$$

$$\phi (0,z)=0,$$

$${\phi }^{\prime }=z-\frac{q}{z}{\sin }^2\phi ,$$

$$\frac{r^{\prime }}{r}=\frac{q}{z}\sin \phi \cos \phi .$$

$$\frac{{\lambda }_n}{{\lambda }_m}\le \frac{n^2}{m^2},for{\lambda }_n>{\lambda }_m\ge 11q(x_0).$$

$$\dot{\theta}(x,z)=\frac{2}{z^{2}r^{2}(x)}\int_{0}^{x}r^{2}(t)\frac{q(t)}{z}\Big{(}\sin^{2}\varphi-\varphi\sin\varphi\cos\varphi\Big{)}dt.$$

$$\displaystyle\displaystyle\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}r^{2}(x)\frac{q(x)}{z}\Big{(}\sin^{2}\varphi-\varphi\sin\varphi\cos\varphi\Big{)}dx$$

$$-{\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}{r}^2(x)\frac{q(x)}{z}\phi \sin \phi \cos \phi dx$$

$$=$$

$$-$$

$$g(x,z)=e^{-2{\int }_x^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}\frac{q(s)}{z}\sin \phi \cos \phi ds}$$

$$\ge 1-2{\int }_x^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}\frac{q(s)}{z}\sin \phi \cos \phi ds,$$

$$\displaystyle\displaystyle-\frac{1}{2}r^{2}(\varphi^{-1}(i\pi+\frac{\pi}{2}+D)\Big{[}D-\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}\varphi^{\prime}(x)g(x,z)dx\Big{]}$$

$$=$$

$$\displaystyle\displaystyle-(\frac{i\pi+\frac{\pi}{2}}{2})r^{2}(\varphi^{-1}(i\pi+\frac{\pi}{2}+D)\Big{[}1-g(\varphi^{-1}(i\pi+\frac{\pi}{2}),z)\Big{]}$$

$$\ge (\frac{i\pi +\frac{\pi }{2}}{2})r^2({\phi }^{-1}(i\pi +\frac{\pi }{2}+D){\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}\frac{q(x)}{z}\sin \phi \cos \phi dx.$$

$$-{\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}{r}^2(x)\frac{q(x)}{z}\phi \sin \phi \cos \phi dx$$

$$\ge -r^2({\phi }^{-1}(i\pi +\frac{\pi }{2}+D){\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}\frac{q(x)}{z}\phi (x)\sin \phi \cos \phi dx.$$

$${\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}{r}^2(x)\frac{q(x)}{z}{\sin }^2\phi dx$$

$$=r^2({\phi }^{-1}(i\pi +\frac{\pi }{2}+D)){\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}\frac{q(x)}{z}{\sin }^2\phi (x)g(x,z)dx$$

$$\displaystyle\displaystyle\geq r^{2}(\varphi^{-1}(i\pi+\frac{\pi}{2}+D))\Big{[}\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}{\textstyle\frac{q(x)}{z}\sin^{2}\varphi dx}$$

$$\displaystyle\displaystyle-2\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}{\textstyle\frac{q(x)}{z}\sin^{2}\varphi\Big{(}\int_{x}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}\frac{q(s)}{z}\sin\varphi\cos\varphi ds\Big{)}dx}.$$

$$\displaystyle\displaystyle-2\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}{\textstyle\frac{q(x)}{z}\sin^{2}\varphi\Big{(}\int_{x}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}\frac{q(s)}{z}\sin\varphi\cos\varphi ds\Big{)}dx}$$

$$\displaystyle\displaystyle={\textstyle-2\Big{[}\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{x}\frac{q(s)}{z}\sin^{2}\varphi ds\int_{x}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}\frac{q(s)}{z}\sin\varphi\cos\varphi ds\Big{]}}_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}$$

$$\displaystyle\displaystyle-2\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}{\textstyle\frac{q(x)}{z}\sin\varphi\cos\varphi\Big{(}\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{x}\frac{q(s)}{z}\sin^{2}\varphi ds\Big{)}dx}$$

$$\displaystyle\displaystyle=-2\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}{\textstyle\frac{q(x)}{z}\sin\varphi\cos\varphi\Big{(}\int_{\varphi^{-1}(i\pi+\frac{\pi}{2})}^{x}\frac{q(s)}{z}\sin^{2}\varphi ds\Big{)}dx.}$$

$${\int }_{{\phi }^{-1}(i\pi +\frac{\pi }{2})}^{{\phi }^{-1}(i\pi +\frac{\pi }{2}+D)}{r}^2(x)\frac{q(x)}{z}{\sin }^2\phi dx$$

Full text

*2000 Mathematics Subject Classification. Primary 34L15, 34B24.

Key words and phrases. Sturm-Liouville Problems, eigenvalue ratio, single-barrier, Prüfer substitution.*

1 Introduction

Consider the Sturm-Liouville equation acting on $\displaystyle[0,1]$

[TABLE]

with Dirichlet boundary conditions

[TABLE]

Here $\displaystyle q$ is a nonnegative differentiable and single-barrier potential in $\displaystyle[0,1]$.

A function in $\displaystyle[0,1]$ is called single-well (single-barrier) if there is $\displaystyle x_{0}$ point a in $\displaystyle[0,1]$, such that it is monotone decreasing (increasing) in $\displaystyle[0,x_{0}]$ and monotone increasing (decreasing) in $\displaystyle[x_{0},1]$. The point $\displaystyle x_{0}$ is the transition point.

It is known (see   [6]) that the spectrum of Problem $\displaystyle\eqref{1.1}-\eqref{1.2}$ consists of a growing sequence of infinitely point $\displaystyle\lambda_{1}<\lambda_{2}<....<\lambda_{n}...\infty$. Ashbaugh and Benguria [1, 2] proved the optimal bound $\displaystyle\frac{\lambda_{n}}{\lambda_{1}}\leq n^{2},$ for nonnegative potentials. They also established the ratio estimate $\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\leq(\lceil\frac{n}{m}\rceil)^{2},$ for $\displaystyle n>m\geq 1,$ where $\displaystyle\lceil s\rceil$ denotes the smallest integer greater than or equal to $\displaystyle s$. Later, Huang and Law in [5] proved that the eigenvalues of the regular Sturm-Liouville equation $\displaystyle-(p(x)y^{\prime})^{\prime}+q(x)y=\lambda\rho(x)y$ (with Dirichlet boundary conditions) satisfy the lower bound

[TABLE]

for $\displaystyle p\in C^{1}[0,1],$ $\displaystyle q,\rho\in C[0,1]$, $\displaystyle q\geq 0$ and $\displaystyle 0<k\leq p(x)\rho(x)\leq K,$ with $\displaystyle\xi=\frac{K\max{\{pq\}}}{kn^{2}\sigma^{2}\pi^{2}}$ and \displaystyle\sigma=\Big{(}\int_{0}^{1}\frac{1}{p(s)}ds\Big{)}^{-1}. In $\displaystyle 2005$, Horváth and Kiss [4] showed that

[TABLE]

for nonnegative single-well potentials. Their approach is mainly based on the monotonicity of the Prüfer angle as function in $\displaystyle\lambda\geq 0$ (see [4,  Theorem  2.2]). At the end of their paper [4,  Remark 5.1], they gave an example of a single-barrier potential which shows that the associated Prüfer angle is not a monotonous function.

In the present paper, we give additional conditions on the single-barrier potential $\displaystyle q(x)$ for which Theorem 2.2 in [4] and the estimate (1.3) remain valid. Namely, we prove that if $\displaystyle q(x)$ is a nonnegative and single-barrier potential (with transition point $\displaystyle x_{0}\in[0,1]$), such that $\displaystyle|q^{\prime}(x)|\leq q^{*}$ where $\displaystyle q^{*}=\frac{2}{15}\min\{q(0),~{}q(1)\}$, then $\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\leq\frac{n^{2}}{m^{2}}$ for $\displaystyle\lambda_{n}>\lambda_{m}\geq 11q(x_{0})$. In particular, if $\displaystyle q(x)$ satisfies the additional condition $\displaystyle q(x_{0})\leq\frac{\pi^{2}}{11}$, then the last bound estimate holds for $\displaystyle n>m\geq 1.$ Note that, our approach used in this paper can be applied to the case of nonnegative and single-well potentials studied in [4] (without further restrictions on $\displaystyle q(x)$).

2 Preliminaries And The Main Statement

Following [4], we introduce the modified Prüfer transformation.

Denote by $\displaystyle y(x,z)$ the unique solution of the initial value problem

[TABLE]

[TABLE]

The Prüfer variables $\displaystyle r(x,z),$ $\displaystyle\varphi(x,z)$ that we use here are defined by

[TABLE]

[TABLE]

[TABLE]

where $\displaystyle r(x,z)>0$, and then let $\displaystyle\theta(x,z)=\frac{\varphi(x,z)}{z}.$

We denote by prime (resp. dot) the derivative with respect to $\displaystyle x$ (resp. $\displaystyle z$).

Using Equation (1.1) one finds the following differential equations for $\displaystyle r(x,z)$ and $\displaystyle\varphi(x,z):$

[TABLE]

and

[TABLE]

It is obvious that $\displaystyle z^{2}$ is an eigenvalue iff $\displaystyle\varphi(\pi,z)$ is a multiple of $\displaystyle\pi$. Denote by $\displaystyle z_{n}$ the square root of $\displaystyle\lambda_{n}$ $\displaystyle(\lambda_{n}=z_{n}^{2})$. Moreover by (2.6), $\displaystyle\varphi^{\prime}(x,z)>0$ for $\displaystyle x\in[0,x_{0}]$ and $\displaystyle z^{2}>q(x_{0}).$ In this case $\displaystyle\varphi^{-1}$ exists and $\displaystyle\varphi^{-1}(k\pi+\frac{\pi}{2})$, $\displaystyle\varphi^{-1}((k+1)\pi)$ ($\displaystyle k\geq 0$ be an integer) are the zeros of $\displaystyle y^{\prime}$ and $\displaystyle y$ in $\displaystyle(0,x_{0}]$, respectively. It is known (e.g., see [6,  chap.1]) that these zeros are decreasing as $\displaystyle z$ increases. We now enunciate the main result of this paper.

Theorem 2.1**.**

For the Sturm-Liouville problem (1.1)-(1.2), if $\displaystyle q(x)$ is a nonnegative differentiable and single-barrier potential with transition point $\displaystyle x_{0}\in[0,1]$ such that $\displaystyle|q^{\prime}(x)|\leq q^{*}$ where $\displaystyle q^{*}=\frac{2}{15}\min\{q(0),~{}q(1)\}$, then

[TABLE]

*In particular, if $\displaystyle q(x)$ satisfies the additional condition $\displaystyle q(x_{0})\leq\frac{\pi^{2}}{11},$ then $\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\leq\frac{n^{2}}{m^{2}},$ for

$\displaystyle n>m\geq 1.$

If for two different $\displaystyle m$ and $\displaystyle n$ equality holds, then $\displaystyle q\equiv 0$ in $\displaystyle[0,1].$*

The proof of Theorem 2.1 will be given in Section 4.

3 Monotonicity Of The Prüfer Angle Function $\displaystyle\theta(x_{0},z).$

In this section, we study the monotonicity of the Prüfer angle function $\displaystyle\theta(x_{0},z).$

Theorem 3.2**.**

Let $\displaystyle q(x)\geq 0$ be monotone increasing and differentiable in $\displaystyle[0,x_{0}]$ such that $\displaystyle q^{\prime}(x)\leq\frac{2}{15}q(0)$. Then $\displaystyle\dot{\theta}(x_{0},z)\geq 0$ for $\displaystyle z\geq\sqrt{11q(x_{0})}.$ If there is a $\displaystyle z\geq\sqrt{11q(x_{0})}$ with $\displaystyle\dot{\theta}(x_{0},z)=0,$ then $\displaystyle q\equiv 0$ in $\displaystyle[0,x_{0}]$.

To prove this theorem we need some preliminary results.

Lemma 3.1**.**

(Corollary  3.3 in [4])

[TABLE]

The following result plays a fundamental role in the sequel.

Lemma 3.2**.**

Let $\displaystyle i\geq 0$ be an integer and assume that $\displaystyle\varphi^{-1}(i\pi+\frac{\pi}{2}+D)\in(0,x_{0}],$ where $\displaystyle 0\leq D\leq\pi.$ Then

[TABLE]

Proof.

By (2.7),

[TABLE]

Since $\displaystyle r^{2}(x)=e^{2\int_{0}^{x}\frac{q(s)}{z}\sin\varphi\cos\varphi ds}$, then

[TABLE]

where $\displaystyle g(x,z)=e^{-2\int_{x}^{\varphi^{-1}(i\pi+\frac{\pi}{2}+D)}\frac{q(s)}{z}\sin\varphi\cos\varphi ds}$.

Using the inequality

[TABLE]

we obtain

[TABLE]

Similarly,

[TABLE]

Using (3.15) and (3.18),

[TABLE]

On the other hand,

[TABLE]

Integrating by parts, yields

[TABLE]

Therefore,

[TABLE]

From (3.20) and (3.22), we get (3.4). This completes the proof of the lemma. ∎

Using the substitution $\displaystyle t=\varphi(x)-(i+1)\pi$ $\displaystyle(i\geq 0$), we obtain the following result:

Lemma 3.3**.**

Assume that $\displaystyle\varphi^{-1}(i\pi+\frac{\pi}{2}+D)\in(0,x_{0}]$ $\displaystyle(i\geq 0,~{}0\leq D\leq\pi),$ then

[TABLE]

Lemma 3.4**.**

Let $\displaystyle q(x)\geq 0$ be monotone increasing on $\displaystyle[0,x_{0}]$ such that $\displaystyle z^{2}>q(x_{0})$ and $\displaystyle\varphi^{-1}(i\pi+\frac{\pi}{2}+D)\in(0,x_{0}]$ $\displaystyle(i\geq 0)$ with $\displaystyle\frac{\pi}{2}\leq D\leq\pi.$ Then

i)

[TABLE]

ii)

[TABLE]

iii)

there exists $\displaystyle c_{1}\in(i\pi+\frac{\pi}{2},~{}i\pi+\frac{\pi}{2}+D)$ $\displaystyle(i\geq 0)$ such that

[TABLE]

Proof.

i)

By virtue of Lemma 3.3,

[TABLE]

For $\displaystyle z^{2}>q(x_{0}),$ $\displaystyle|\frac{q(\varphi^{-1}(t))}{z^{2}}\sin^{2}(t)|<1$, and hence, the function $\displaystyle\frac{1}{1-\frac{q}{z^{2}}\sin^{2}\varphi}$ is developable in entire series. Thus, since $\displaystyle q(x)\geq 0$ is monotone increasing, then

[TABLE]

By integration by parts, we obtain

[TABLE]

Therefore,

[TABLE]

Then

[TABLE]

Analogously,

[TABLE]

Using (3.40) and (3.44), we find (3.29).

ii)

Clearly, if $\displaystyle|t|\in]0,\frac{\pi}{2}[,$ then $\displaystyle\sin^{2}t-t\sin t\cos t>0,$ and hence,

[TABLE]

Integrating by parts and using (3.38), yields

[TABLE]

Therefore, (3.40) and (3.48) give (3.30).

iii)

As before,

[TABLE]

By the mean value theorem, there exists $\displaystyle c_{1}\in(i\pi+\frac{\pi}{2},i\pi+\frac{\pi}{2}+D)$ such that

[TABLE]

Therefore, from this and (3.53), we obtain (3.33).

∎

Lemma 3.5**.**

*Let $\displaystyle q(x)$ be satisfying the conditions in Lemma 3.4. Then there exists

$\displaystyle c_{2}\in(i\pi+\frac{\pi}{2},i\pi+\frac{\pi}{2}+D)$ $\displaystyle(\frac{\pi}{2}\leq D\leq\pi)$ such that*

[TABLE]

Proof.

For $\displaystyle z^{2}>q(x_{0}),$ we have

[TABLE]

Integrating by parts and taking into account that $\displaystyle q(x)$ be monotone increasing on $\displaystyle[0,x_{0}]$, we find

[TABLE]

By (3.29),

[TABLE]

On the other hand, using (3.38),

[TABLE]

Thus

[TABLE]

Therefore, by (3.62) and (3.64),

[TABLE]

By the mean value theorem, there exists $\displaystyle c_{2}\in(i\pi+\frac{\pi}{2},i\pi+\frac{\pi}{2}+D)$ such that

[TABLE]

From this and (3.66), we get (3.59). ∎

We are now ready to prove Theorem 3.2.

Proof.

If for some $\displaystyle\lambda>0$, $\displaystyle y^{\prime}(x,z)$ has no zeros in $\displaystyle(0,1)$, then $\displaystyle\varphi(x,z)<\frac{\pi}{2}$ for $\displaystyle x\in(0,x_{0})$, and hence

[TABLE]

In the rest of the proof let $\displaystyle\varphi(x_{0},z)=k\pi+\frac{\pi}{2}+D,$ with $\displaystyle k\geq 0$ be an integer and $\displaystyle 0\leq D\leq\pi.$ Then

[TABLE]

where \displaystyle\tau(x)=r^{2}(x)\frac{q(x)}{z}\Big{(}\sin^{2}\varphi-\varphi\sin\varphi\cos\varphi\Big{)}. Since $\displaystyle\varphi\in]0,\frac{\pi}{2}],$ then

[TABLE]

It is easily seen that if $\displaystyle 0\leq D\leq\frac{\pi}{2}$, then

[TABLE]

If $\displaystyle\frac{\pi}{2}\leq D\leq\pi$, then in view of Lemma 3.3, together with (3.30) and (3.33),

[TABLE]

Under the hypotheses $\displaystyle z^{2}\geq 11q(x_{0})$, $\displaystyle\varphi^{\prime}(\varphi^{-1}(c_{1}))\geq\frac{10z}{11}$ and $\displaystyle\frac{1}{1-\frac{q(\varphi^{-1}(c_{1}))}{z^{2}}}\leq\frac{11}{10}$. On the other hand, recall that if $\displaystyle x_{0}=\varphi^{-1}(i\pi+\frac{\pi}{2}+D)$ $\displaystyle(\frac{\pi}{2}\leq D\leq\pi),$ then the solution $\displaystyle y(x,z)$ of (1.1)-(1.2) has exactly $\displaystyle(i+1)$ zeros in $\displaystyle(0,x_{0}].$ In view of Sturm oscillation theorem, we have necessarily $\displaystyle z\geq z_{i+1},$ where $\displaystyle\lambda_{i}=z_{i}^{2}$ are the eigenvalues of Problem (1.1)-(1.2). Since $\displaystyle z_{i+1}\geq(i+1)\pi$, then

[TABLE]

Therefore, by (3.81) and (3.84),

[TABLE]

On the other hand, it can be easily verified that $\displaystyle\frac{\log(1-s)-s}{(1-s)^{2}}\geq\frac{-1}{4}$ for $\displaystyle s\in[0,\frac{1}{11}]$. Using this, together with (3.59) and (3.84) (for $\displaystyle i=0$), we obtain

[TABLE]

Thus, Summing (3.85) and (3.88), one gets

[TABLE]

where \displaystyle{\textstyle G(s)=\frac{s}{4(1-\frac{3}{4}s)}\Big{(}1-\log(1-\frac{3}{4}s)\Big{)}+\frac{1}{4}\frac{(1+s)\log(1-s)}{4(1-s)}-\frac{1}{6}\log(1-\frac{3}{4}s)} and

$\displaystyle q^{\prime}(\varphi^{-1}(c))=\max\{q^{\prime}(\varphi^{-1}(c_{1})),q^{\prime}(\varphi^{-1}(c_{2}))\}.$ It can be shown by straightforward computation that $\displaystyle G(s)\geq\frac{s}{10}$ for $\displaystyle s\in[0,\frac{1}{11}].$ By setting $\displaystyle s=\frac{q(\varphi^{-1}(i\pi+\frac{\pi}{2}))}{z^{2}}\geq\frac{q(0)}{z^{2}}$ and taking into account the condition $\displaystyle q^{\prime}(x)\leq\frac{2}{15}q(0)$, we find out that

[TABLE]

Now, according to Lemma 3.2 together with (3.95), we conclude that

[TABLE]

Therefore, by (3.76), (3.79), (3.80) and (3.98) we have $\displaystyle\dot{\theta}(x_{0},z)\geq 0$ for $\displaystyle z^{2}\geq 11q(x_{0})$ and $\displaystyle q^{\prime}(x)\leq\frac{2}{15}q(0).$ Obviously, if there is a $\displaystyle z\geq\sqrt{11q(x_{0})}$ with $\displaystyle\dot{\theta}(x_{0},z)=0,$ then in view of (3.76),

[TABLE]

which is not possible unless $\displaystyle q\equiv 0$ in $\displaystyle[0,x_{0}]$. The theorem is proved. ∎

4 Proof of Theorem 2.1

Let the potential $\displaystyle q(x)$ be monotone increasing in $\displaystyle[0,x_{0}]$ and monotone decreasing in $\displaystyle[x_{0},1].$ We denote by $\displaystyle\tilde{q}(x)$ the reverse of the potential, i.e., $\displaystyle\tilde{q}(x)=q(1-x)$. Then $\displaystyle\tilde{y}(x,z)$ is the solution of the initial value problem

[TABLE]

As in section 2, we define the associated Prüfer transformation

[TABLE]

As in [4], we have the following relations:

[TABLE]

where $\displaystyle\lambda_{n}=z_{n}^{2}$ is an eigenvalue of Problem (1.1)-(1.2).

Proof.

of Theorem 2.1. As $\displaystyle\tilde{q}(x)=q(1-x),$ then $\displaystyle\tilde{q}(x)$ be monotone increasing in $\displaystyle[0,1-x_{0}]$ and monotone decreasing in $\displaystyle[1-x_{0},1]$. We have $\displaystyle\tilde{q}^{\prime}(x)=-q^{\prime}(1-x)\leq\frac{2}{15}\tilde{q}(0)=\frac{2}{15}q(1)$, thus for $\displaystyle x\in[0,1]$, $\displaystyle|q^{\prime}(x)|\leq q^{*}$ where $\displaystyle q^{*}=\frac{2}{15}\min\{q(0),~{}q(1)\}$. Therefore by Theorem 3.2, $\displaystyle\tilde{\theta}(1-x_{0},z)$ is increasing for $\displaystyle z\geq\sqrt{11\tilde{q}(1-x_{0})}=\sqrt{11q(x_{0})}.$ Consequently, the function $\displaystyle\Psi(z)=\theta(x_{0},z)+\tilde{\theta}(1-x_{0},z)$ is increasing for $\displaystyle z\geq\sqrt{11q(x_{0})}$. By (4.11), $\displaystyle z_{n}\Psi(z_{n})=n\pi.$ Let $\displaystyle m$ be an integer such that $\displaystyle m<n$ and $\displaystyle\lambda_{m}\geq 11q(x_{0})$. Then

[TABLE]

which implies that $\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\leq\frac{n^{2}}{m^{2}}.$ On the other hand, if $\displaystyle q(x_{0})\leq\frac{\pi^{2}}{11},$ then

[TABLE]

where $\displaystyle q^{-}=\min_{x\in[0,1]}q(x).$ Thus, in this case $\displaystyle\frac{\lambda_{n}}{\lambda_{m}}\leq\frac{n^{2}}{m^{2}}$ for all $\displaystyle n>m\geq 1.$ If equality holds, then $\displaystyle\Psi(z_{n})=\Psi(z_{m}),$ so that $\displaystyle\dot{\Psi}(z)=0$ for some $\displaystyle z>0.$ Hence $\displaystyle\dot{\theta}(x_{0},z)=\dot{\tilde{\theta}}(1-x_{0},z)=0,$ and in view of Theorem 3.2, $\displaystyle q\equiv 0$ in $\displaystyle[0,x_{0}]$ and $\displaystyle\tilde{q}\equiv 0$ in $\displaystyle[0,1-x_{0}],$ i.e., $\displaystyle q\equiv 0$ in $\displaystyle[0,1].$ ∎