The main result of the paper is given by the following theorem.
Assume that \(0< x_{1}< x_{2}\leq1\).
Theorem 3
If
y
is a nontrivial solution of the equation
$$ -y^{\prime\prime}+ \biggl[ \frac{2}{x^{2}}-\frac{2}{x}+q ( x ) \biggr] y=\lambda y $$
(3.1)
with
\(y(x_{1})=y^{\prime}(x_{2})+by(x_{2})=0\), then
$$ \lambda\geq- \biggl[ \frac{2}{x_{1}}+ \biggl( \vert b\vert +2\int _{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) ^{2} \biggr], $$
where
\(b\in \mathbb{R} \)
and
\(q(x)\in L^{2} [ 0,1 ] \).
Proof
Multiplying (3.1) by y and integrating of this equation from \(x_{1}\) to \(x_{2}\) gives the formula
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}-\lambda y^{2}+ \biggl( \frac{2}{x^{2}}- \frac{2}{x} \biggr) y^{2} \biggr] \,dx=0. $$
Since \(\int_{x_{1}}^{x_{2}}\frac{2}{x^{2}}y^{2}\,dx\geq0\) the remaining term will be negative or zero,
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}-\lambda y^{2}-\frac{2}{x}y^{2} \biggr] \,dx\leq0. $$
(3.2)
Integrating the first term by parts gives
$$\begin{aligned} \int_{x_{1}}^{x_{2}}-y^{\prime\prime}y\, dx =&-y^{\prime }(x_{2})y(x_{2})+y^{\prime }(x_{1})y(x_{1})+ \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx \\ =&by^{2}(x_{2})+\int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx. \end{aligned}$$
So (3.2) is equal to
$$ by^{2}(x_{2})+\int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}+q ( x ) y^{2}- \lambda y^{2}-\frac{2}{x}y^{2} \biggr] \,dx\leq0. $$
Moreover, we find \(\int_{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\) and that equals
$$\begin{aligned}& \int_{x_{1}}^{x_{2}}2y(x)y^{\prime }(x)\int _{x_{1}}^{x}q(t)\,dt\,dx \\& \quad = \biggl( 2y^{2}(x)\int_{x_{1}}^{x}q(t) \,dt \biggr) _{x_{1}}^{x_{2}}-\int_{x}^{x_{2}}2y(x)y^{\prime }(x) \int_{x_{1}}^{x}q(t)\,dt\,dx-\int _{x_{1}}^{x_{2}}2y^{2}(x)q(x) \,dx \\& \quad = y^{2}(x_{2})\int_{x_{1}}^{x_{2}}q(t) \,dt-\int _{x_{1}}^{x_{2}}y^{2}(x)q(x)\,dx. \end{aligned}$$
Then we have
$$ \int_{x_{1}}^{x_{2}}q(x)y^{2}(x) \,dx=y^{2}(x_{2})\int _{x_{1}}^{x_{2}}q(t) \,dt-\int_{x_{1}}^{x_{2}}2y(x)y^{\prime }(x)\int _{x_{1}}^{x}q(t)\,dt\,dx. $$
(3.3)
Integrating \(\int_{x_{1}}^{x_{2}}2y(x)y^{\prime}(x)\,dx\) by parts gives
$$\begin{aligned} \int_{x_{1}}^{x_{2}}2y(x)y^{\prime}(x)\,dx =& \bigl[ 2y^{2}(x) \bigr] _{x_{1}}^{x_{2}}-\int _{x_{1}}^{x_{2}}2y(x)y^{\prime}(x)\,dx \\ =&2y^{2}(x_{2})-\int_{x_{1}}^{x_{2}}2y(x)y^{\prime}(x) \,dx. \end{aligned}$$
So we get the formula
$$ y^{2}(x_{2})=\int_{x_{1}}^{x_{2}}2y(x)y^{\prime}(x) \,dx. $$
Adding (3.3) to \(by^{2}(x_{2})\) and using the triangle inequality gives the formula
$$\begin{aligned}& \bigl\vert by^{2}(x_{2})\bigr\vert +\biggl\vert \int _{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\biggr\vert \\& \quad \leq \vert b\vert y^{2}(x_{2})+\biggl\vert y^{2}(x_{2})\int_{x_{1}}^{x_{2}}q(t) \,dt\biggr\vert +\biggl\vert \int_{x_{1}}^{x_{2}}2y(x)y^{\prime }(x) \int_{x_{1}}^{x}q(t)\,dt\,dx\biggr\vert \\& \quad \leq \vert b\vert \int_{x_{1}}^{x_{2}}2yy^{\prime } \,dx+\int_{x_{1}}^{x_{2}}\vert q\vert \,dx\int _{x_{1}}^{x_{2}}2yy^{\prime }\,dx+\int _{x_{1}}^{x_{2}}\bigl\vert 2yy^{\prime}\bigr\vert \int_{x_{1}}^{x}\bigl\vert q(t)\bigr\vert \,dt \,dx. \end{aligned}$$
Since \(x\leq x_{2}\) we get
$$ \bigl\vert by^{2}(x_{2})\bigr\vert +\biggl\vert \int _{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\biggr\vert \leq \biggl[ \vert b\vert +2\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr] \int_{x_{1}}^{x_{2}}2yy^{\prime} \,dx. $$
By the Cauchy-Schwarz inequality, we get
$$\begin{aligned}& \bigl\vert by^{2}(x_{2})\bigr\vert +\biggl\vert \int _{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\biggr\vert \\& \quad \leq \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) 2 \biggl[ \int_{x_{1}}^{x_{2}}y^{2} \,dx \biggr] ^{1/2} \bigl[ \int _{x_{1}}^{x_{2}} \bigl(y^{\prime}\bigr)^{2}\,dx \bigr] ^{1/2} \end{aligned}$$
and, from the inequality of \(2A^{1/2}B^{1/2}\leq\varepsilon A+(1/\varepsilon)B\), we get
$$\begin{aligned}& \bigl\vert by^{2}(x_{2})\bigr\vert +\biggl\vert \int _{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\biggr\vert \\& \quad \leq \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) \biggl[ \varepsilon \biggl( \int _{x_{1}}^{x_{2}}y^{2}\,dx \biggr) +1/\varepsilon \biggl( \int_{x_{1}}^{x_{2}}\bigl(y^{\prime} \bigr)^{2}\,dx \biggr) \biggr]. \end{aligned}$$
For any \(\varepsilon>\vert b\vert +2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx\),
$$\begin{aligned}& by^{2}(x_{2})+\int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}+q ( x ) y^{2}- \lambda y^{2}-\frac{2}{x}y^{2} \biggr] \,dx \\& \quad \geq \int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}-\lambda y^{2}- \frac{2}{x}y^{2} \biggr] \,dx-\vert b\vert y^{2}(x_{2})-\biggl\vert \int_{x_{1}}^{x_{2}}qy^{2} \,dx\biggr\vert \\& \quad \geq \int_{x_{1}}^{x_{2}} \bigl[ \bigl( y^{\prime} \bigr) ^{2}-\lambda y^{2} \bigr] \,dx- \frac{2}{x_{1}}\int _{x_{1}}^{x_{2}}y^{2}\,dx \\& \qquad {} - \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) \biggl[ \varepsilon \biggl( \int _{x_{1}}^{x_{2}}y^{2}\,dx \biggr) +1/\varepsilon \biggl( \int_{x_{1}}^{x_{2}}\bigl(y^{\prime} \bigr)^{2}\,dx \biggr) \biggr] \\& \quad \geq \biggl[ 1- \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) \big/\varepsilon \biggr] \int _{x_{1}}^{x_{2}}\bigl(y^{\prime}\bigr)^{2} \,dx \\& \qquad {} + \biggl[ -\lambda-\frac{2}{x_{1}}-\varepsilon \biggl( \vert b \vert +2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) \biggr] \int_{x_{1}}^{x_{2}}y^{2} \,dx \\& \quad = C(\varepsilon)+ \biggl[ -\lambda-\frac{2}{x_{1}}-\varepsilon \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) \biggr] \int_{x_{1}}^{x_{2}}y^{2} \,dx, \end{aligned}$$
where \(C(\varepsilon)>0\). If we assume that
$$ \lambda< - \biggl[ \frac{2}{x_{1}}+ \biggl( \vert b\vert +2\int _{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) ^{2} \biggr], $$
the equation \(\int_{x_{1}}^{x_{2}} [ -y^{\prime\prime }y+q ( x ) y^{2}-\lambda y^{2}-\frac{2}{x}y^{2} ] \,dx\) should be positive. It contradicts our assumption. So we proved the theorem. □
Theorem 4
If
y
is a nontrivial solution of the equation
$$ -y^{\prime\prime}+ \biggl[ \frac{2}{x^{2}}-\frac{2}{x}+q ( x ) \biggr] y=\lambda y $$
with
\(y(x_{1})=y(x_{2})=0\), then
$$ \lambda\geq- \biggl[ \frac{2}{x_{1}}+4 \biggl( \int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) ^{2} \biggr], $$
where
\(q(x)\in L^{2} [ 0,1 ] \).
Proof
Multiplying the equation by y and integrating of this equation from \(x_{1} \) to \(x_{2}\) gives the formula
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}-\lambda y^{2}+ \biggl( \frac{2}{x^{2}}- \frac{2}{x} \biggr) y^{2} \biggr] \,dx=0. $$
Since \(\int_{x_{1}}^{x_{2}}\frac{2}{x^{2}}y^{2}\,dx\geq0\) the remaining term will be negative or zero,
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}-\lambda y^{2}-\frac{2}{x}y^{2} \biggr] \,dx\leq0. $$
(3.4)
Integrating the first term by parts gives
$$ \int_{x_{1}}^{x_{2}}-y^{\prime\prime}y\, dx =-y^{\prime }(x_{2})y(x_{2})+y^{\prime }(x_{1})y(x_{1})+ \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx =\int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx. $$
So (3.4) is equal to
$$ by^{2}(x_{2})+\int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}+q ( x ) y^{2}- \lambda y^{2}-\frac{2}{x}y^{2} \biggr] \,dx\leq0. $$
Moreover, we find \(\int_{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx\) and that equals
$$\begin{aligned}& \int_{x_{1}}^{x_{2}}2y(x)y^{\prime }(x)\int _{x_{1}}^{x}q(t)\,dt\,dx \\& \quad = \biggl( 2y^{2}(x)\int_{x_{1}}^{x}q(t) \,dt \biggr) _{x_{1}}^{x_{2}}-\int_{x}^{x_{2}}2y(x)y^{\prime }(x) \int_{x_{1}}^{x}q(t)\,dt\,dx-\int _{x_{1}}^{x_{2}}2y^{2}(x)q(x) \,dx \\& \quad = -\int_{x_{1}}^{x_{2}}y^{2}(x)q(x)\,dx. \end{aligned}$$
Then we have
$$ \int_{x_{1}}^{x_{2}}q(x)y^{2}(x)\,dx=- \int _{x_{1}}^{x_{2}}2y(x)y^{\prime}(x)\int _{x_{1}}^{x}q(t)\,dt\,dx, $$
since \(x\leq x_{2}\) we get
$$ \biggl\vert \int_{x_{1}}^{x_{2}}q(x)y^{2}(x) \,dx\biggr\vert \leq \int_{x_{1}}^{x_{2}}\vert q \vert \,dx\, \biggl\vert \int_{x_{1}}^{x_{2}}2yy^{\prime} \,dx\biggr\vert . $$
By the Cauchy-Schwarz inequality, we get
$$ \biggl\vert \int_{x_{1}}^{x_{2}}q(x)y^{2}(x) \,dx\biggr\vert \leq 2\int_{x_{1}}^{x_{2}}\vert q \vert \,dx\, \biggl[ \int_{x_{1}}^{x_{2}}y^{2} \,dx \biggr] ^{1/2} \biggl[ \int _{x_{1}}^{x_{2}} \bigl(y^{\prime}\bigr)^{2}\,dx \biggr] ^{1/2}, $$
and from the inequality of \(2A^{1/2}B^{1/2}\leq\varepsilon A+(1/\varepsilon)B\), we get
$$ \biggl\vert \int_{x_{1}}^{x_{2}}q(x)y^{2}(x) \,dx\biggr\vert \leq 2\int_{x_{1}}^{x_{2}}\vert q \vert \,dx\, \biggl[ \varepsilon \biggl( \int_{x_{1}}^{x_{2}}y^{2} \,dx \biggr) +1/\varepsilon \biggl( \int_{x_{1}}^{x_{2}} \bigl(y^{\prime}\bigr)^{2}\,dx \biggr) \biggr]. $$
For any \(\varepsilon>2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx \),
$$\begin{aligned}& \int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}+q ( x ) y^{2}-\lambda y^{2}- \frac{2}{x}y^{2} \biggr] \,dx \\& \quad \geq \int_{x_{1}}^{x_{2}} \biggl[ \bigl( y^{\prime} \bigr) ^{2}-\lambda y^{2}- \frac{2}{x}y^{2} \biggr] \,dx-\biggl\vert \int _{x_{1}}^{x_{2}}qy^{2}\,dx\biggr\vert \\& \quad \geq \int_{x_{1}}^{x_{2}} \bigl[ \bigl( y^{\prime} \bigr) ^{2}-\lambda y^{2} \bigr] \,dx- \frac{2}{x_{1}}\int _{x_{1}}^{x_{2}}y^{2}\,dx \\& \qquad {} -2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx\, \biggl[ \varepsilon \biggl( \int_{x_{1}}^{x_{2}}y^{2} \,dx \biggr) +1/\varepsilon \biggl( \int_{x_{1}}^{x_{2}} \bigl(y^{\prime}\bigr)^{2}\,dx \biggr) \biggr] \\& \quad \geq \biggl[ 1-\frac{2}{\varepsilon}\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr] \int_{x_{1}}^{x_{2}} \bigl(y^{\prime}\bigr)^{2}\,dx \\& \qquad {} + \biggl[ -\lambda-\frac{2}{x_{1}}-2\varepsilon \int _{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr] \int _{x_{1}}^{x_{2}}y^{2}\,dx \\& \quad = C(\varepsilon)+ \biggl[ -\lambda-\frac{2}{x_{1}}-\varepsilon \biggl( \vert b\vert +2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) \biggr] \int_{x_{1}}^{x_{2}}y^{2} \,dx, \end{aligned}$$
where \(C(\varepsilon)>0\). If we assume that
$$ \lambda< - \biggl[ \frac{2}{x_{1}}+4 \biggl( \int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) ^{2} \biggr], $$
the equation \(\int_{x_{1}}^{x_{2}} [ -y^{\prime\prime }y+q ( x ) y^{2}-\lambda y^{2}-\frac{2}{x}y^{2} ] \,dx\) should be positive. It contradicts our assumption. So we get our result. □
Theorem 5
If
y
is a nontrivial solution of the equation
$$ -y^{\prime\prime}+ \biggl[ \frac{2}{x^{2}}-\frac{2}{x}+q ( x ) \biggr] y=0, $$
(3.5)
where
\(q(x)\in L^{2} [ 0,1 ] \), and if
\(y^{\prime }(x_{1})=y^{\prime}(x_{2})=0\), where
\(0\leq x_{1}\leq x_{2}\leq1\), then
$$ \frac{1}{x_{2}^{2}}+\frac{1}{x_{2}}\leq2 \biggl[ \int_{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr] ^{2}+\frac{1}{2 ( x_{2}-x_{1} ) }\int _{x_{1}}^{x_{2}}\vert q\vert \,dx. $$
Proof
Multiplying (3.5) by y and integrating of this equation from \(x_{1}\) to \(x_{2}\) gives the formula
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}+ \biggl( \frac{2}{x^{2}}- \frac{2}{x} \biggr) y^{2} \biggr] \,dx=0. $$
Since
$$ \int_{x_{1}}^{x_{2}}2y(x)y'(x)\int_{x_{1}}^{x}q(t)\, dt\, dx=y^{2}(x_{2})\int_{x_{1}}^{x_{2}}q(t)\, dt-\int_{x_{1}}^{x_{2}}q(x)y^{2}(x)\, dx $$
we get
$$ \int_{x_{1}}^{x_{2}}q(x)y^{2}(x) \,dx=y^{2}(x_{2})\int _{x_{1}}^{x_{2}}q(t) \,dt-\int_{x_{1}}^{x_{2}}2y(x)y^{\prime }(x)\int _{x_{1}}^{x}q(t)\,dt\,\,dx . $$
(3.6)
From the mean value theorem we can write
$$ y^{2}(x_{3})= \bigl[ 1/(x_{2}-x_{1}) \bigr] \int _{x_{1}}^{x_{2}}y^{2}\,dx, $$
(3.7)
where \(x_{3}\in [ x_{1},x_{2} ] \). From (3.7) and integrating \(\int_{x_{1}}^{x_{2}}2y(x)y^{\prime}(x)\,dx\) by parts we have
$$ y^{2}(x_{2})=y^{2}(x_{3})+\int _{x_{1}}^{x_{2}}2yy^{\prime}\,dx=\frac{1}{x_{2}-x_{1}} \int_{x_{1}}^{x_{2}}y^{2}\,dx+ \int _{x_{1}}^{x_{2}}2yy^{\prime}\,dx. $$
Adding (3.6) to \(\int_{x_{1}}^{x_{2}}\frac{2y^{2}}{x}\,dx\) we have
$$\begin{aligned} \int_{x_{1}}^{x_{2}} \biggl( q-\frac{2}{x} \biggr) y^{2}\,dx =& \biggl[ \frac{1}{x_{2}-x_{1}}\int_{x_{1}}^{x_{2}}y^{2} \,dx+\int _{x_{1}}^{x_{2}}2yy^{\prime}\,dx \biggr] \int _{x_{1}}^{x_{2}}q\, dx \\ &{}-\int_{x_{1}}^{x_{2}}2yy^{\prime } \int_{x_{1}}^{x}q(t)\,dt\,\,dx-\int _{x_{1}}^{x_{2}}\frac{2y^{2}}{x}\,dx. \end{aligned}$$
From the triangle inequality we have the formula
$$\begin{aligned} \biggl\vert \int_{x_{1}}^{x_{2}} \biggl( q- \frac{2}{x} \biggr) y^{2}\,dx\biggr\vert \leq& \biggl\vert \biggl[ \frac{1}{x_{2}-x_{1}}\int _{x_{1}}^{x_{2}}y^{2} \,dx+\int_{x_{1}}^{x_{2}}2yy^{\prime}\,dx \biggr] \int_{x_{1}}^{x_{2}}q\, dx\biggr\vert \\ & {}+\biggl\vert \int_{x_{1}}^{x_{2}}2yy^{\prime }\int _{x_{1}}^{x}q\,dt\,\,dx\biggr\vert +\biggl\vert \int_{x_{1}}^{x_{2}}\frac{2y^{2}}{x}\,dx\biggr\vert \\ \leq& \biggl( \frac{1}{x_{2}-x_{1}}\int_{x_{1}}^{x_{2}} \vert q\vert \,dx+\frac{2}{x_{2}} \biggr) \int_{x_{1}}^{x_{2}}y^{2} \,dx+2\int_{x_{1}}^{x_{2}}2\bigl\vert yy^{\prime} \bigr\vert \int_{x_{1}}^{x}q\,dt\,\,dx. \end{aligned}$$
By using the inequality
$$ 2\bigl\vert yy^{\prime}\bigr\vert \leq\varepsilon y^{2}+ ( 1/ \varepsilon ) \bigl( y^{\prime} \bigr) ^{2}, $$
we get
$$\begin{aligned} \biggl\vert \int_{x_{1}}^{x_{2}} \biggl( q- \frac{2}{x} \biggr) y^{2}\,dx\biggr\vert \leq &\biggl[ \frac{1}{x_{2}-x_{1}}\int_{x_{1}}^{x_{2}} \vert q\vert \,dx+\frac{2}{x_{2}} \biggr] \int_{x_{1}}^{x_{2}}y^{2} \,dx \\ &{}+2 \biggl[ \varepsilon\int_{x_{1}}^{x_{2}}y^{2} \,dx+\frac{1}{\varepsilon}\int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx \biggr] \int_{x_{1}}^{x_{2}} \vert q\vert \,dx. \end{aligned}$$
Integrating \(\int_{x_{1}}^{x_{2}}-y^{\prime\prime}y\, dx\) by parts gives
$$\begin{aligned}& \int_{x_{1}}^{x_{2}}-y^{\prime\prime}y\, dx=-y^{\prime }(x_{2})y(x_{2})+y^{\prime }(x_{1})y(x_{1})+ \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx=\int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx, \end{aligned}$$
(3.8)
$$\begin{aligned}& \int_{x_{1}}^{x_{2}}\frac{2}{x^{2}}y^{2}\,dx \geq \int_{x_{1}}^{x_{2}}\frac{2}{x_{2}^{2}}y^{2} \,dx . \end{aligned}$$
(3.9)
From (3.9) and (3.8) and for any \(\varepsilon>\vert b\vert +2\int_{x_{1}}^{x_{2}}\vert q\vert \,dx\), there exists a number \(C(\varepsilon)>0\) such that
$$\begin{aligned}& \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}+ \biggl( \frac{2}{x^{2}}-\frac{2}{x} \biggr) y^{2} \biggr] \,dx \\& \quad = \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx+\int_{x_{1}}^{x_{2}} \biggl( q- \frac{2}{x} \biggr) y^{2}\,dx+\int_{x_{1}}^{x_{2}} \frac{2}{x^{2}}y^{2}\,dx \\& \quad \geq \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx- \biggl( \frac{1}{x_{2}-x_{1}}\int_{x_{1}}^{x_{2}} \vert q\vert \,dx+\frac{2}{x_{2}} \biggr) \int _{x_{1}}^{x_{2}}y^{2}\,dx \\& \qquad {} -2 \biggl( \varepsilon\int_{x_{1}}^{x_{2}}y^{2} \,dx+\frac{1}{\varepsilon}\int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx \biggr) \int_{x_{1}}^{x_{2}} \vert q\vert \,dx+\int_{x_{1}}^{x_{2}} \frac{2}{x_{2}^{2}}y^{2}\,dx \\& \quad = \biggl( 1-\frac{2}{\varepsilon}\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr) \int_{x_{1}}^{x_{2}} \bigl( y^{\prime} \bigr) ^{2}\,dx \\& \qquad {}+ \biggl\{ \frac{2}{x_{2}^{2}}+ \frac{2}{x_{2}}- \biggl( \frac{1}{x_{2}-x_{1}}+2\varepsilon \biggr) \int _{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr\} \int _{x_{1}}^{x_{2}}y^{2}\,dx \\& \quad = C(\varepsilon)+ \biggl\{ \frac{2}{x_{2}^{2}}+\frac{2}{x_{2}}- \biggl( \frac{1}{x_{2}-x_{1}}+2\varepsilon \biggr) \int_{x_{1}}^{x_{2}} \vert q\vert \,dx \biggr\} \int_{x_{1}}^{x_{2}}y^{2} \,dx. \end{aligned}$$
Let us assume that
$$ \frac{2}{x_{2}^{2}}+\frac{2}{x_{2}}- \biggl( \frac{1}{x_{2}-x_{1}}+2 \varepsilon \biggr) \int_{x_{1}}^{x_{2}}\vert q\vert \,dx>0. $$
In this case, we get
$$\begin{aligned} \frac{1}{x_{2}^{2}}+\frac{1}{x_{2}} >&\varepsilon \int_{x_{1}}^{x_{2}} \vert q\vert \,dx+\frac{1}{2 ( x_{2}-x_{1} ) }\int_{x_{1}}^{x_{2}} \vert q\vert \,dx \\ >&2 \biggl( \int_{x_{1}}^{x_{2}}\vert q\vert \,dx \biggr) ^{2}+\frac{1}{2 ( x_{2}-x_{1} ) }\int_{x_{1}}^{x_{2}} \vert q\vert \,dx. \end{aligned}$$
Then we have
$$ \int_{x_{1}}^{x_{2}} \biggl[ -y^{\prime\prime}y+q ( x ) y^{2}+ \biggl( \frac{2}{x^{2}}-\frac{2}{x} \biggr) y^{2} \biggr] \,dx>0. $$
This is a contradiction. So we proved the theorem. □
Conclusion
In the Carlson case, the potentials are in \(L^{2} [ 0,1 ] \), but in our paper, the potentials are not in \(L^{2} [ 0,1 ] \).