Proof. For the first point, it is evidently that $(H^{1}(0,\,1;h),\,(\cdot,\,\cdot ))$ is an inner product space and $\|\cdot \|$ is a norm on $H^{1}(0,\,1;h)$. Besides, the expression (2.5) is deduced from

\[ u\in L^{2}(0,1), hu'\in L^{2}(0,1) \ \Leftrightarrow \ u\in L^{2}(0,1), (hu)'=h'u+hu'\in L^{2}(0,1). \]

For the second point, let $\{u_k\}_{k\in \mathbb {N}}$ be a Cauchy sequence in $H^{1}(0,\,1;h)$. i.e.,

\[ \|u_k-u_n\|_{L^{2}}^{2}+\|hu_k'-hu_n'\|_{L^{2}}^{2}=\|u_k-u_n\|_{H_1(0,1;h)}^{2}\rightarrow 0 \mbox{ as } k, n\rightarrow \infty, \]

which implies that there exists $u,\,v\in L^{2}(0,\,1)$ such that

\[ u_k\rightarrow u \mbox{ and } hu_k'\rightarrow v \mbox{ in } L^{2}(0,1). \]

In addition,

\[ (hu_k)'=h'u_k+hu_k'\rightarrow h'u+v \mbox{ in } L^{2}(0,1), \]

we observe that for each $\phi \in C_c^{\infty }(0,\,1)$:

\[ \int_0^{1}(hu)\phi'{\rm d}\kern0.06em x=\lim_{k\rightarrow \infty}\int_0^{1} (hu_k)\phi'{\rm d}\kern0.06em x={-}\lim_{k\rightarrow \infty}\int_0^{1}(hu_k)'\phi{\rm d}\kern0.06em x={-}\int_0^{1}(h'u+v)\phi{\rm d}\kern0.06em x, \]

one can find that

\[ h'u+hu'=(hu)'=h'u+v \mbox{ in } \mathcal{D}'(0,1), \]

that is $hu'=v \mbox { in } \mathcal {D}'(0,\,1)$, hence $u_k\rightarrow u$ in $H^{1}(0,\,1;h)$ is obtained.

For the third point, as $u\in H^{1}(0,\,1;h)$, we have $(hu)'\in L^{2}(0,\,1)$ by invoking (2.5), and thus

\[ \|hu\|_{H^{1}(0,1)}^{2}=\|hu\|_{L^{2}}^{2}+\|(hu)'\|_{L^{2}}^{2}\leq \sup_{x\in [0,1]}h^{2}(x)\|u\|_{L^{2}}^{2}+\|(hu)'\|_{L^{2}}^{2}<\infty \]

in line with the definition of $h(x)$. Overall, these results indicate that $hu\in H^{1}(0,\,1)$.

Proof. Take $\delta \in (0,\,1)$, thanks to

\[ \left(\inf_{x\in (\delta,1)}h(x)\right)\|u'\|_{L^{2}(\delta,1)}\leq \|hu'\|_{L^{2}(\delta,1)}<\infty, \]

and employing the fact $\inf \limits _{x\in [\delta,1]}h(x)>0$, we readily check $\|u'\|_{L^{2}(\delta,1)}<\infty$, consequently, $u\in H^{1}(\delta,\,1)$. And applying the Sobolev Embedding Theorem to find that $u\in C^{0,\frac {1}{2}}[\delta,\,1]$, due to the arbitrary of $\delta >0$, we deduce that $u\in C^{0,\frac {1}{2}}(0,\,1]$.

Finally, we have $hu\in H^{1}(0,\,1)$ by theorem 2.1, then $hu\in C^{0,\frac {1}{2}}[0,\,1]$ by utilizing the Sobolev Embedding Theorem again.

Proof. Let $\{u_n\}_{n\in \mathbb {N}}\subset H_0^{1}(0,\,1;h)$ be a Cauchy sequence, then $\{u_n\}_{n\in \mathbb {N}}$ is a Cauchy sequence in $H^{1}(0,\,1;h)$, there exists $u\in H^{1}(0,\,1;h)$ such that

(2.7)\begin{equation} u_n\rightarrow u \mbox{ in } H^{1}(0,1;h) \end{equation}

from theorem 2.1. On the other side, since $\{hu_n\}_{n\in \mathbb {N}}\subset H_0^{1}(0,\,1)$ and

\[ \|hu_n-hu_m\|_{H_0^{1}(0,1)}^{2}\leq C\|u_n-u_m\|_{L^{2}}^{2}+\|hu_n'-hu_m'\|_{L^{2}}^{2}\rightarrow 0 \mbox{ as } n,m\rightarrow \infty, \]

there exists $v\in H_0^{1}(0,\,1)$ such that

\[ hu_n\rightarrow v \mbox{ in } H_0^{1}(0,1) \mbox{ as } n\rightarrow \infty. \]

This together with (2.7), that yields $v=hu$ on $(0,\,1)$, and along with the fact that $(hu)(0)=u(1)=0$, so $u\in H_0^{1}(0,\,1;h)$ is derived.

Proof. Assume $h=x^{2}$, for all $n\in \mathbb {N}$, let

\[ u_n(x)= \begin{cases} -n^{\frac{3}{2}}\left(x-\displaystyle\frac{1}{n}\right), & x\in \left[0,\frac{1}{n}\right],\\ 0, & x\in \left(\frac{1}{n}, 1\right]. \end{cases} \]

Then

\begin{align*} \int_0^{1} |u_n(x)|^{2}{\rm d}\kern0.06em x & =\int_0^{\frac{1}{n}} n^{3}\left(x-\frac{1}{n}\right)^{2}{\rm d}\kern0.06em x=\frac{1}{3}, \\ \int_0^{1} |hu_n'(x)|^{2}{\rm d}\kern0.06em x & =\int_0^{\frac{1}{n}}x^{2}n^{3}\left(x-\frac{1}{n}\right)^{2}{\rm d}\kern0.06em x=\frac{1}{5n^{2}}, \end{align*}

and

\[ \lim_{x\rightarrow 0^{+}}h(x)u_n(x)=\lim_{x\rightarrow 0^{+}}x^{2}\left[{-}n^{\frac{3}{2}}\left(x-\frac{1}{n}\right)\right]=0, \quad \lim_{x\rightarrow 1^{-}} u_n(x)=0, \]

which implies that $\{u_n\}_{n\in \mathbb {N}}\subset H_0^{1}(0,\,1;h)$.

Suppose that there exists a subsequence $\{u_{n_k}\}_{k\in \mathbb {N}}$ of $\{u_n\}_{n\in \mathbb {N}}$, and $u\in L^{2}(0,\,1)$, such that

\[ u_{n_k}\rightarrow u \mbox{ strongly in } L^{2}(0,1), \]

then $\|u\|_{L^{2}(0,1)}^{2}=\tfrac {1}{3}$, and

\[ \|hu'\|_{L^{2}(0,1)}=\lim_{n\rightarrow \infty}\|hu_n'\|_{L^{2}(0,1)}=0. \]

Furthermore, there exists a subsequence of $\{u_{n_k}\}_{k\in \mathbb {N}}$, still denoted by itself, such that

\[ u_{n_k}\rightarrow u \mbox{ a.e. in } (0,1). \]

In view of the definition of $u_{n_k} (k\in \mathbb {N})$, the above conclusion implies that $u=0$ a.e. in $(0,\,1)$, but it contradicts to the fact $\|u\|_{L^{2}(0,1)}^{2}=\frac {1}{3}$.

Proof. For all $v\in H_0^{1}(0,\,1;h)$, we have

\[ |(f,v)_{L^{2}(0,1)}|\leq \|f\|_{L^{2}(0,1)}\|v\|_{L^{2}(0,1)}\leq \|f\|_{L^{2}(0,1)}\|v\|_{H_0^{1}(0,1;h)}, \]

and hence $f: H_0^{1}(0,\,1;h)\rightarrow \mathbb {R},\, v\mapsto (f,\,v)_{L^{2}(0,1)}$ is a bounded linear function on $H_0^{1}(0,\,1;h)$.

Next, we shall explain that $B(\cdot,\,\cdot )$ satisfies the Lax–Milgram Theorem. Indeed, through applying Cauchy inequality we obtain

\begin{align*} B(u,v) & =\int_0^{1}h^{2}u'v'{\rm d}\kern0.06em x+\int_0^{1}Vuv{\rm d}\kern0.06em x\leq \|hu'\|_{L^{2}(0,1)}\|hv'\|_{L^{2}(0,1)}+M\|u\|_{L^{2}(0,1)}\|v\|_{L^{2}(0,1)}\\ & \leq \max\{M,1\}\left(\|hu'\|_{L^{2}(0,1)}^2+\|u\|_{L^{2}(0,1)}^2\right)^\frac{1}{2}\left(\|hv'\|_{L^{2}(0,1)}^2+\|v\|_{L^{2}(0,1)}^2\right)^\frac{1}{2}\\ & =\max\{M,1\}\|u\|_{H_0^{1}(0,1;h)}\|v\|_{H_0^{1}(0,1;h)}, \end{align*}

and

\begin{align*} B(u,u)& =\int_0^{1}h^{2}|u'|^{2}{\rm d}\kern0.06em x+\int_0^{1}Vu^{2}{\rm d}\kern0.06em x\geq \|hu'\|_{L^{2}(0,1)}^{2}+m\|u\|_{L^{2}(0,1)}^{2}\\ & \quad \geq \frac{1}{2}\min\{m,1\}\|u\|_{H_0^{1}(0,1;h)}^{2}. \end{align*}

All of these show that $B(\cdot,\,\cdot )$ satisfies the Lax–Milgram Theorem. On the basis of the mentioned analysis and Lax–Milgram Theorem, we see that there exists a unique $u\in H_0^{1}(0,\,1;h)$ such that

\[ B(u,v)=(f,v) \]

for all $v\in H_0^{1}(0,\,1;h)$.

Proof. Employing (2.8) and

\[ B(u,u)=(f,u)_{L^{2}(0,1)}\leq \|f\|_{L^{2}(0,1)}\|u\|_{L^{2}(0,1)}\leq \|f\|_{L^{2}(0,1)}\|u\|_{H_0^{1}(0,1;h)}, \]

for which we deduce that $\|u\|_{H_0^{1}(0,1;h)}\leq C \|f\|_{L^{2}(0,1)}$, and thereby

\[ \|Lf\|_{L^{2}(0,1)}=\|u\|_{L^{2}(0,1)}\leq \|u\|_{H_0^{1}(0,1;h)}\leq C\|f\|_{L^{2}(0,1)}. \]

Evidently, the above formula provides evidence that $L: L^{2}(0,\,1)\rightarrow L^{2}(0,\,1)$ is a bounded linear operator. On other side,

\[ (Lf,g)_{L^{2}(0,1)}=(u,g)_{L^{2}(0,1)}=B(u,v)=(f,v)_{L^{2}(0,1)}=(f,Lg)_{L^{2}(0,1)} \]

for all $f,\,g\in L^{2}(0,\,1)$, where $u,\,v$ are the solutions of (2.1) with respect to $f$ and $g$ respectively. Finally, we explicitly note that $L^{*}=L$, i.e., $L$ is a self-adjoint operator.

Proof. As we can see the definition above has states that $\overline {C_c^{\infty }(0,\,1)}^{H^{1}(0,1;h)} \subset H_c^{1}(0,\,1;h)$, it suffices to prove that, $H_c^{1}(0,\,1;h) \subset \overline {C_c^{\infty }(0,\,1)}^{H^{1}(0,1;h)}$. Indeed, for all $u\in H_c^{1}(0,\,1;h)$, there exists a sequence $u_n \in Lip_c(0,\,1)$, where $u_n$ is a Lipschitz continuous function on $(0,\,1)$ and ${\rm supp}u_n$ is compact in $(0,\,1)$, then $u_n'$ exists a.e. on $(0,\,1)$ and $|u_n'|\leq C \mbox { a.e. on } (0,\,1)$, where $C$ is the Lipschitz constant of $u_n$, these facts illustrate that $u_n\in H_0^{1}(0,\,1)$ with compact support on $(0,\,1)$. It is well known that there exists $\{v_n\}_{n\in \mathbb {N}}\subset C_c^{\infty }(0,\,1)$ such that

\[ \|v_n-u_n\|_{H_0^{1}(0,1)}\rightarrow 0, \ n\rightarrow \infty, \]

and we have

\[ \|v_n-u_n\|_{H^{1}(0,1;h)}\leq C\|v_n-u_n\|_{H_0^{1}(0,1)}\rightarrow 0, \ n\rightarrow \infty \]

by the definition of $h(x)$, finally,

\[ \|v_n-u\|_{H^{1}(0,1;h)}\leq \|v_n-u_n\|_{H^{1}(0,1;h)}+\|u_n-u\|_{H^{1}(0,1;h)} \rightarrow 0, \ n\rightarrow \infty, \]

so the desired result is verified.

Proof. 1. Let $\{u_n\}_{n\in \mathbb {N}}\subset H_c^{1}(0,\,1;h)$ be a Cauchy sequence, then there exists $u\in H^{1}(0,\,1;h)$ such that

\[ u_n\rightarrow u \mbox{ in } H^{1}(0,1;h) \mbox{ as } n\rightarrow \infty. \]

Hence, for arbitrary $\epsilon >0$, there exists $N\in \mathbb {N}$, such that for all $n\geq N$ we have $\frac {1}{2^{n}}<\frac {\epsilon }{2}$ and

\[ \|u_n-u\|_{H^{1}(0,1;h)}<\frac{\epsilon}{2}. \]

Note that for each $n\in \mathbb {N}$, there exists $\phi _n\in C_c^{\infty }(0,\,1)$, such that

\[ \|u_n-\phi_n\|_{H^{1}(0,1;h)}<\frac{1}{2^{n}}<\frac{\epsilon}{2}, \]

these evidences suggest that

\[ \|u-\phi_n\|_{H^{1}(0,1;h)}\leq \|u-u_n\|_{H^{1}(0,1;h)}+\|u_n-\phi_n\|_{H^{1}(0,1;h)}<\epsilon \]

for all $n\geq N$. i.e., $u\in H_c^{1}(0,\,1;h)$.

2. For any $u\in H_c^{1}(0,\,1;h)$, there exists $\phi _n\in C_c^{\infty }(0,\,1)$ such that

\[ \|u-\phi_n\|_{H^{1}(0,1;h)} \to 0,\quad n\to \infty, \]

and consider that

\[ \|u-\phi_n\|_{H^{1}(\delta,1)}\leq \left(\min\left\{\inf_{x\in [\delta,1]}h(x), 1\right\}\right)^{{-}1} \|u-\phi_n\|_{H^{1}(0,1;h)}, \]

hence using Sobolev Embedding Theorem, for any $\delta \in (0,\,1)$,

(3.2)\begin{equation} C_1\|u-\phi_n\|_{C^{0,\frac{1}{2}}[\delta,1]}\leq \|u-\phi_n\|_{H^{1}(\delta, 1)} \leq C_2 \|u-\phi_n\|_{H^{1}(0,1;h)}. \end{equation}

Clearly, $\lim \limits _{n\to \infty }\phi _n(1)=u(1)=0$. Besides, it should be noted here that according to the integration absolutely continuous of $h^{-1}$, for arbitrary $\epsilon >0$, there exists $\xi >0$, for all $E\subset (0,\,1),\, |E|<\xi$, we have

(3.3)\begin{equation} \int_E|h^{{-}1}(t)|^{2}{{\rm d}}t<\epsilon^{2}. \end{equation}

Then for all $\delta \in (0,\,1)$, since

\begin{align*} |\phi_n(\delta)-\phi_m(\delta)|& =|\int_\delta^{1}\phi_n'(x)-\phi_m'(x){\rm d}\kern0.06em x|\le \int_\delta^{1}|h^{{-}1}h(\phi_n'(x)-\phi_m'(x)|{\rm d}\kern0.06em x\\ & \le\left(\int_\delta^{1} |h^{{-}1}|^{2}{\rm d}\kern0.06em x\right)^{\frac{1}{2}} \|h(\phi_n'-\phi'_m)\|_{L^{2}(\delta,1)}\\ & \le \epsilon \|h(\phi_n'-\phi'_m)\|_{L^{2}(0,1)} \to 0, n,m\to \infty, \end{align*}

which shows that $\phi _n$ is uniformly convergent, we have $\lim \limits _{n\to \infty }\lim \limits _{\delta \to 0}\phi _n(\delta )=\lim \limits _{\delta \to 0}\lim \limits _{n\to \infty }\phi _n(\delta )=0$, furthermore, we have $u(0)=\lim \limits _{\delta \to 0}u(\delta )=0$ from (3.2) and proposition 2.3.

3. Let $\{u_n\}_{n\in \mathbb {N}}\subset H_c^{1}(0,\,1;h)$ be a bounded set. On one hand, note that

\[ \|u_n\|_{H^{1}(\delta,1)}\leq \left(\min\left\{\inf_{x\in [\delta,1]}h(x), 1\right\}\right)^{{-}1} \|u_n\|_{H^{1}(0,1;h)} \]

for all $\delta \in (0,\,1)$, as a consequence,

(3.4)\begin{equation} C_1\|u_n\|_{C^{0,\frac{1}{2}}[\delta,1]}\leq \|u_n\|_{H^{1}(\delta, 1)}\leq C_2 \|u_n\|_{H^{1}(0,1;h)} \end{equation}

by Sobolev Embedding Theorem.

Due to (3.3), for any $0< x< y<1$, we have

\begin{align*} |u_n(x)-u_n(y)| & = \left|\int_x^{y}u_n'(t){{\rm d}}t\right|\leq \int_x^{y}|u_n'(t)|{{\rm d}}t\leq \int_x^{y}h^{{-}1}h|u_n'(t)|{{\rm d}}t\\ & \leq \left(\int_x^{y}|h^{{-}1}(t)|^{2}{{\rm d}}t\right)^{\frac{1}{2}}\|hu_n'\|_{L^{2}(0,1)}\\ & <\epsilon\|hu_n'\|_{L^{2}(0,1)}\leq C\epsilon, \end{align*}

for all $|x-y|<\xi,\, x,\,y\in (0,\,1)$. This implies that $\{u_n\}_{n\in \mathbb {N}}\subset C[0,\,1]$ is a bounded and equicontinuous sequence according to $u\in C^{0,\frac {1}{2}}(0,\,1]$ and $u(0)=u(1)=0$ for all $u\in H_c^{1}(0,\,1;h)$. The Arzela–Asoli Theorem tells us that there exists a convergent subsequence of $\{u_n\}_{n\in \mathbb {N}}$, still denoted by itself, and $u_0\in C[0,\,1]$, such that

\[ u_n\rightarrow u_0 \mbox{ in } C[0,1] \mbox{ as } n\rightarrow \infty, \]

that is a sure sign that

\[ u_n\rightarrow u_0 \mbox{ strongly in } L^{2}(0,1) \mbox{ as } n\rightarrow\infty. \]

Proof. That is easily verifiable for $H_c^{1}(0,\,1;h)\subsetneq H_0^{1}(0,\,1;h)\subsetneq H^{1}(0,\,1;h)$ based on lemma 3.3 and the definition of $H_0^{1}(0,\,1;h)$.

For $H_0^{1}(0,\,1)\subsetneq H_c^{1}(0,\,1;h)$, we give an example. Let $h(x)=x^{\frac {1}{4}}$, $g_1(x)=\sqrt {x}$, $g_2(x)=(x-1)^{2}$, let $f(x)$ be infinitely differentiable satisfying $f^{(k)}(\frac {1}{2})=g_1^{(k)}(\frac {1}{2})$ and $f^{(k)}(\frac {3}{4})=g_2^{(k)}(\frac {3}{4})$, where $(\cdot )^{(k)}$ is the $k$-th derivative of $(\cdot )$, $k=0,\,1,\,2 \cdots$. Consider

\[ u(x)= \begin{cases} g_1(x), & x\in \left[0,\frac{1}{2}\right], \\ f(x), & x\in \left[\frac{1}{2},\frac{3}{4}\right],\\ g_2(x), & x\in \left[\frac{3}{4},1\right], \end{cases} \]

then

\[ \|u'(x)\|_{L^{2}(0,1)}^{2}=\int_0^{\frac{1}{2}}|\frac{1}{2\sqrt{x}}|^{2}{\rm d}\kern0.06em x+\int_{\frac{1}{2}}^{\frac{3}{4}}|f'(x)|^{2}{\rm d}\kern0.06em x+\int_{\frac{3}{4}}^{1} 4|x-1|^{2}{\rm d}\kern0.06em x \to \infty, \]

clearly, $u\notin H_0^{1}(0,\,1)$. Now let $\xi \in C_c^{\infty }(R)$ satisfy

\[ 0 \le \xi \le 1, \xi|_{[0,1]}\equiv 1,\quad \xi|_{R-[{-}1,2]}\equiv 0, \]

and let

\[ u_m(x)= \begin{cases} u(x)[1-\xi(mx)], & x\in [0,\frac{1}{2}], \\ u(x)[1-\xi(mx+1-m)], & x\in [\frac{1}{2},1], \end{cases} \]

where $m>4$. We observe that $mx>2$ if $x\in (\tfrac {2}{m},\,\tfrac {1}{2}]$ and $mx+1-m< -1$ if $x\in [\tfrac {1}{2},\,1-\tfrac {2}{m})$. According to the definition of $\xi$, we have $\xi =0$ on $(\tfrac {2}{m},\,1-\tfrac {2}{m})$, so that $u_m(x)=u(x)$ on $(\tfrac {2}{m},\,1-\tfrac {2}{m})$. See that

\begin{align*} \int_0^{1}|u_m-u|^{2}{\rm d}\kern0.06em x& =\int_0^{\frac{2}{m}}|u(x)[1-\xi(mx)]-u(x)|^{2}{\rm d}\kern0.06em x\\ & \quad + \int_{1-\frac{2}{m}}^{1} |u(x)[1-\xi(mx+1-m)]-u(x)|^{2}{\rm d}\kern0.06em x\\ & =\int_0^{\frac{2}{m}}|u(x)\xi(mx)|^{2}{\rm d}\kern0.06em x+ \int_{1-\frac{2}{m}}^{1} |u(x)\xi(mx+1-m)|^{2}{\rm d}\kern0.06em x\\ & \le \int_0^{\frac{2}{m}}xdx+ \int_{1-\frac{2}{m}}^{1} |x-1|^{4}{\rm d}\kern0.06em x\le \frac{2}{m^{2}}+\left(\frac{2}{m}\right)^{5} \to 0,m \to \infty, \end{align*}

since $\xi (mx)$, $\xi (mx+1-m)\le 1$. Moreover, since $u_m'=u'(1-\xi )-mu\xi '$ and $\xi =\xi '=0$ on $(\tfrac {2}{m},\,1-\frac {2}{m})$, we have $u_m'(x)=u'(x)$ on $(\frac {2}{m},\,1-\frac {2}{m})$. Then

\begin{align*} \int_0^{1}|hu_m'\!-\!hu'|^{2}{\rm d}\kern0.06em x& =\!\int_0^{1}|hu'(1\!-\xi)-hmu\xi'\!-hu'|^{2}{\rm d}\kern0.06em x \!=\!\int_0^{1}|-hu'\xi-hmu\xi'|^{2}{\rm d}\kern0.06em x\\ & \le 2 \int_0^{1} |hu'\xi|^{2}{\rm d}\kern0.06em x+ 2\int_0^{1} |hmu\xi'|^{2}{\rm d}\kern0.06em x\\ & \le 2 \int_0^{\frac{2}{m}} |\frac{1}{2} x^{\frac{1}{4}} x^{-\frac{1}{2}}|^{2}{\rm d}\kern0.06em x+ 2\int_{1-\frac{2}{m}}^{1} |2x^{\frac{1}{4}}(x-1)|^{2}{\rm d}\kern0.06em x\\ & \quad +C\int_0^{\frac{2}{m}} |mx^{\frac{1}{4}}\sqrt{x}|^{2}{\rm d}\kern0.06em x +C\int_{1-\frac{2}{m}}^{1} |mx^{\frac{1}{4}}(x-1)|^{2}{\rm d}\kern0.06em x\\ & \le \sqrt{\frac{2}{m}}+C\left(\frac{2}{m}\right)^{3}+\frac{C}{\sqrt{m}}+\frac{C}{m^{3}}\to 0,m \to \infty, \end{align*}

therefore $u_m(x) \to u(x)$ in $H^{1}(0,\,1;h)$. Note that $u_m=0$ on $[0,\,\tfrac {1}{m}] \cup [1-\tfrac {1}{m},\,1]$, we can mollify the $u_m$ to produce functions $\omega _m \in C_c^{\infty }(0,\,1)$ such that $\omega _m \to u$ in $H_c^{1}(0,\,1;h)$, thus $u \in H_c^{1}(0,\,1;h)$.

Proof. Define $L_2^{2}(0,\,1)=L^{2}(0,\,1) \times L^{2}(0,\,1)$, where

\[ \|u\|_{L_2^{2}(0,1)}=\left(\int_0^{1} |u_1|^{2}{\rm d}\kern0.06em x+\int_0^{1} |u_2|^{2}{\rm d}\kern0.06em x\right)^{\frac{1}{2}} \]

for $u=(u_1,\,u_2) \in L_2^{2}(0,\,1)$ as the norm of space $L_2^{2}(0,\,1)$. It is no doubt that $L_2^{2}(0,\,1)$ is a separable space in the light of that $L^{2}(0,\,1)$ is a separable. Set

\[ Pu=(u,hu'),u\in H^{1}(0,1;h), \]

evidently, $W=\{Pu|u\in H^{1}(0,\,1;h)\}$ is a subspace of $L_2^{2}(0,\,1)$. From $\|Pu\|_{L_2^{2}(0,1)}=\|u\|_{H^{1}(0,1;h)}$, we know that $P$ is an isometric isomorphism of mapping $H^{1}(0,\,1;h)$ to $W$. In view of the fact that $H^{1}(0,\,1;h)$ is complete, $W$ is a closed subspace of $L_2^{2}(0,\,1)$, furthermore, $W$ is separable. Note that $P$ is an isometric isomorphism, then $H^{1}(0,\,1;h)$ has the same properties. Lemma 3.4 implies that the space $H_c^{1}(0,\,1;h)$ is a separable Hilbert space. Similarly, reflexivity is also obtained.

Proof. Same to the proof of theorem 2.9.

Proof. By the same argument of lemma 2.10 we obtain that $L$ is a self-adjoint bounded linear operator. This together with lemma 3.3 we obtain $L$ is a compact linear operator.

Proof. Set $L=S^{-1}$, then $L$ is a self-adjoint compact linear operator owing to lemma 4.1. Also noteworthy,

\[ (Lf,f)=(u,f)=B(u,u)\ge 0 \]

for any given $f \in L^{2}(0,\,1)$.

Consider $L^{2}(0,1)$ is separable, by invoking theorem D7 (pp. 728) in [Reference Evans13], there exists a countable orthonormal basis of $L^2{(0,1)}$ consisting of eigenvectors of $L$. It should be pointed out that for $\eta \neq 0$, $Lw=\eta w$ if and only if $Sw=\lambda w$ for $\lambda =\frac {1}{\eta }$. Consequently, we prove theorem 4.3.

Proof. We carry out this proof by several steps.

(i) From theorem 4.3, we have

(4.1)\begin{equation} \begin{cases} B(w_k,w_k)=\lambda_k(w_k,w_k)=\lambda_k,\\ B(w_k,w_l)=\lambda_k(w_k,w_l)=0,k,l=1,2\cdots, k\neq l. \end{cases} \end{equation}

Since $\{w_k\}_{k=1}^{\infty }$ is the orthogonal basis of $L^{2}(0,\,1)$, if $u\in H_c^{1}(0,\,1;h)$ and $\|u\|_{L^{2}(0,1)}=1$, then we write

(4.2)\begin{equation} u=\sum_{k=1}^{\infty}d_kw_k, \quad d_k=(u,w_k), \quad \sum_{k=1}^{\infty}d_k^{2}=\|u\|_{L^{2}(0,1)}^2=1, \end{equation}

the series converging in $L^{2}(0,\,1)$. By (4.1), $\frac {w_k}{\sqrt {\lambda _k}}$ is an orthonormal subset of $H_c^{1}(0,\,1;h)$, endowed with the new inner product $B(\cdot,\,\cdot )$. Indeed,

(4.3)\begin{align} c_1\|u\|_{H_c^{1}(0,1;h)}^{2}& \le \int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x+m\int_0^{1}|u|^{2}{\rm d}\kern0.06em x\le B(u,u)\nonumber\\ & \le \int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x+M\int_0^{1}|u|^{2}{\rm d}\kern0.06em x \le c_2\|u\|_{H_c^{1}(0,1;h)}^2. \end{align}

Moreover, $B(w_k,\,u)=\lambda _k(w_k,\,u)=0$ for $k=1,\,2\cdots$, which implies that $u \equiv 0$ due to $\lambda _k>0$ and $\{w_k\}_{k=1}^{\infty }$ is the orthonormal basis of $L^{2}(0,\,1)$. Hence $u=\sum \limits _{k=1}^{\infty }\mu _k \frac {w_k}{\sqrt {\lambda _k}}$ for $\mu _k=B(u,\,\frac {w_k}{\sqrt {\lambda _k}})$, the series converging in $H_c^{1}(0,\,1;h)$. Together with the equality (4.2), we know that $\mu _k= {d_k}{\sqrt {\lambda _k}}$, thus $u=\sum \limits _{k=1}^{\infty }d_kw_k$ is convergent in $H_c^{1}(0,\,1;h)$.

By employing the equalities (4.1) and (4.2), then

\[ B(u,u)=\sum\limits_{k=1}^{\infty}d_k^{2}\lambda_k\ge \lambda_1, \]

and the equality holds for $u=w_1$, clearly, the desired result (i) is obtained.

(ii) We next claim that if $u \in H_c^{1}(0,\,1;h)$ and $\|u\|_{L^{2}(0,1)}=1$, then $u$ is a weak solution of

(4.4)\begin{equation} \begin{cases} Su=\lambda_1u,x\in (0,1),\\ u(0)=u(1)=0 \end{cases} \end{equation}

if and only if

(4.5)\begin{equation} B(u,u)=\lambda_1. \end{equation}

Naturally, (4.4) implies (4.5). On the other side, suppose that (4.5) is established, writing $d_k=(u,\,w_k)$, then

\[ \sum_{k=1}^{\infty}d_k^{2}\lambda_1=\lambda_1=B(u,u)=\sum_{k=1}^{\infty}d_k^{2}B(w_k,w_k)=\sum_{k=1}^{\infty}d_k^{2}\lambda_k. \]

Therefore

\[ \sum_{k=1}^{\infty}d_k^{2}(\lambda_k-\lambda_1)=0. \]

Evidently, we have $d_k=(u,\,w_k)=0$ provided that $\lambda _k>\lambda _1$, then $u=\sum \limits _{k=1}^{n}(u,\,w_k)w_k$ for some $n$ according to the multiplicity finiteness of $\lambda _1$, where $Sw_k=\lambda _1w_k$. Hence

(4.6)\begin{equation} Su=\sum\limits_{k=1}^{n}(u,w_k)Sw_k=\sum\limits_{k=1}^{n}\lambda_1(u,w_k)w_k=\lambda_1u, \end{equation}

the claim is confirmed.

From (i), we see that if $u$ is a eigenfunction of $\lambda _1$, then $|u|$ is one also. By lemma A.1, we must have $|u|$ is positive in $(0,\,1)$ and hence $\lambda _1$ has a positive eigenfunction. This argument indicates that the eigenfunctions of $\lambda _1$ are either positive or negative and thereby it is impossible that two of them are orthogonal, i.e. $\lambda _1$ is simple.

(iii) We will apply the mathematical induction to prove it. Suppose the conclusion is established for $k=2,\,\cdots,\, N-1$. For $u \bot V_{N-1}$, then equality (4.1) must be satisfied, and note that

\[ u=\sum_{k=1}^{\infty}(u,w_k)w_k=\sum_{k=N}^{\infty}(u,w_k)w_k=\sum_{k=N}^{\infty}d_kw_k, \]

therefore

\[ \sum\limits_{k=1}^{\infty}d_k^{2}B(w_k,w_k)=B(u,u)=\sum\limits_{k=N}^{\infty}d_k^{2}B(w_k,w_k), \]

thus we must have $d_1=0,\,\cdots,\, d_{N-1}=0$ by (4.1), this leads to $\sum \limits _{k=N}^{\infty }d_k^{2}=1$. Moreover,

(4.7)\begin{equation} B(u,u)=\sum\limits_{k=N}^{\infty}d_k^{2}B(w_k,w_k)=\sum\limits_{k=N}^{\infty}d_k^{2}\lambda_k\ge \lambda_N\sum\limits_{k=N}^{\infty}d_k^{2}=\lambda_N, \end{equation}

this inequality holds if and only if $u=w_N$.

(iv) Suppose $u_k$ has more than $k$ nodal domains, let $G_1,\,\cdots,\,G_k,\,G_{k+1}$ be the nodal domains of $u_k$ and consider

\[ v=\sum\limits_{n=1}^{k+1}a_nu_k{\chi_{G_n}} \]

where $a_1,\,\cdots,\, a_{k+1}$ are constants to be chosen later on. It is not hard to find that $u_k{\chi _{G_n}} \in H_c^{1}(0,\,1;h)$ for every $n$, and thus $v \in H_c^{1}(0,\,1;h)$. We may choose a nontrivial $(N+1)$-tuple $(a_1,\,\cdots,\, a_{k+1})$ such that $v$ is orthogonal to the eigenfunctions $u_1,\,\cdots,\,u_k$, this is allowed since there are $N$ equations and $N+1$ coefficients. Next, multiplying by an appropriate coefficient such that $\int _{0}^{1}v^{2}{\rm d}\kern0.06em x=1$, we record the new coefficient as $(c_1,\,c_2,\,\cdots,\,c_{k+1})$. According to the assumption that the eigenvalue of $S$ is simple, by the same way as (4.7), we have

(4.8)\begin{equation} B(v,v)=\int_{0}^{1}h^{2}|v'|^{2} {\rm d}\kern0.06em x+\int_{0}^{1}V|v'|^{2} {\rm d}\kern0.06em x\ge \lambda_{k+1}>\lambda_{k}. \end{equation}

Additionally, since $-(h^{2}(u_k{\chi _{G_n}})')'+Vu_k{\chi _{G_n}}=\lambda _{k}u_k{\chi _{G_n}}$ for $n=1,\,2\cdots,\,k+1$, the equality

\[ \int_{G_n}h^{2}|(u_k{\chi_{G_n}})'|^{2} {\rm d}\kern0.06em x+\int_{G_n}V|u_k{\chi_{G_n}}|^{2} {\rm d}\kern0.06em x=\lambda_{k}\int_{G_n}|u_k{\chi_{G_n}}|^{2} {\rm d}\kern0.06em x, \]

leads to

(4.9)\begin{equation} \begin{aligned} B(v,v) & =\int_{0}^{1}h^{2}|v'|^{2}{\rm d}\kern0.06em x+\int_{0}^{1}V|v|^{2}{\rm d}\kern0.06em x\\ & =\sum\limits_{n=1}^{k+1}\int_{G_n}h^{2}|c_n(u_k{\chi_{G_n}})'|^{2}{\rm d}\kern0.06em x+\sum\limits_{n=1}^{k+1}\int_{G_n}V|c_nu_k{\chi_{G_n}}|^{2}{\rm d}\kern0.06em x\\ & =\lambda_k\sum\limits_{n=1}^{k+1}\int_{G_n}c_n^{2}|u_k{\chi_{G_n}}|^{2}{\rm d}\kern0.06em x=\lambda_k, \end{aligned} \end{equation}

this contradicts inequality (4.8).

Proof. Suppose $u$ and $v$ are two nontrivial linear independent solutions corresponding to $\lambda _{k}$, then

(4.10)\begin{equation} \begin{aligned} (h^{2}u')'v-(V-\lambda_{k})uv=0,\\ (h^{2}v')'u-(V-\lambda_{k})uv=0. \end{aligned} \end{equation}

Let $W_h=h^{2}u'v-h^{2}v'u$, then $W_h'=(h^{2}u')'v-(h^{2}v')'u$. By invoking [Reference Evans13] (chapter 6), the interior regularity for $u,\,v \in H_{loc}^{2}(\delta,\,1)$ is obtained for any $\delta \in (0,\,1)$. Then we discover that $u,\,v\in C^{1,\frac {1}{2}}[\delta,\,1]$ take advantage of Sobolev Embedding Theorem for any $\delta \in (0,\,1)$ and $u(1)=v(1)=0$. For all $\phi \in C_c^{\infty }(0,\,1)$, we have $\int _0^{1} W_h'\phi {\rm d}\kern0.06em x= \int _0^{1} ((h^{2}u')'v-(h^{2}v')'u)\phi {\rm d}\kern0.06em x=\int _0^{1} [(V-\lambda _{k})uv-(V-\lambda _{k})uv] \phi {\rm d}\kern0.06em x =0$ by (4.10), therefore $W_h'=0$ a.e. in the sense of distribution for $x\in [0,\,1]$, which implies that $W_h$ is a constant C a.e., combined with the fact that $W_h(1)=0$ and $W_h$ is continuous on $[\delta,\,1]$ for any $\delta \in (0,\,1)$, we immediately obtain that $C=0$ a.e. on $[0,\,1]$.

If $W_h=0$ a.e. on $[0,\,1]$, that is, $u'v-v'u=0$ a.e. on $(0,\,1)$, i.e. $(\frac {u}{v})'v^{2}=0$ a.e. on $(0,\,1)$, also given that there is no positive measure subset in $(0,\,1)$ such that $v= 0$ on it by lemma 4.1.3 on [Reference Bennewitz, Brown and Weikard5], then $(\frac {u}{v})'=0$ a.e. on $(0,\,1)$. This argument shows that $\frac {u}{v}$ is a constant on $(0,\,1)$ a.e., which contradicts that $u$ and $v$ are linear independent, so that there is one and only one linearly independent solution to each $\lambda _{k}$.

Proof. Let $u(x)>0$ on $(x_1,\,x_2)$, suppose $v(x)$ has no zero point on $(x_1,\,x_2)$, without loss of generality, let $v(x)>0$ on $(x_1,\,x_2)$. Multiplying equation (4.11) by $u(x)$ and equation (4.12) by $\frac {u^{2}(x)}{v(x)}$ and subtracting, that yields the equality

(4.13)\begin{equation} -(h^{2}(x)u'(x))'u(x)+(h^{2}(x)v'(x))'\frac{u^{2}(x)}{v(x)}+(V_1-V_2)u^{2}(x)=0.\end{equation}

Integrating this equality over $(x_1,\,x_2)$, then yields

(4.14)\begin{equation} \int_{x_1}^{x_2}(V_1-V_2)u^{2}(x){\rm d}\kern0.06em x+\int_{x_1}^{x_2}(h^{2}(x)v'(x))'\frac{u^{2}(x)}{v(x)} -(h^{2}(x)u'(x))'u(x){\rm d}\kern0.06em x=0.\end{equation}

Integrating by parts and utilizing the fact that $u(x_1)=u(x_2)=0$

(4.15)\begin{equation} \begin{aligned} & \int_{x_1}^{x_2}(V_1-V_2)u^{2}(x){\rm d}\kern0.06em x-\int_{x_1}^{x_2}h^{2}(x)v'(x)\left(\frac{u^{2}(x)}{v(x)}\right)'{\rm d}\kern0.06em x+\int_{x_1}^{x_2}h^{2}(x)|u'(x)|^{2}{\rm d}\kern0.06em x\\ & =\int_{x_1}^{x_2}(V_1-V_2)u^{2}(x){\rm d}\kern0.06em x+\int_{x_1}^{x_2}h^{2}(x)\left(\frac{u'(x)v(x)-v'(x)u(x)}{v(x)}\right)^{2}{\rm d}\kern0.06em x\\ & =0, \end{aligned} \end{equation}

however, the left side of the equality is positive, this is a contradiction.

Proof. For $k=1$, theorem 4.4 implies that $u_1$ has no zero point on $(0,\,1)$, so it be verified. For $k=2$, we know that $u_2$ and $u_1$ are orthogonal and $u_1$ is positive on $(0,\,1)$, then $u_2$ has at least one zero. Meanwhile, $u_2$ has at most two node domains according to $\rm (iv)$ of theorem 4.4 and 4.5, hence $u_2$ has exactly one node in $(0,\,1)$.

For $k\ge 3$, consider

\[ \begin{cases} -(h^{2}(x)u'_i(x))'+(V(x)-\lambda_{i})u_i(x)=0,i=2,3,\cdots,\\ - (h^{2}(x)u'_{i+1}(x))'+(V(x)-\lambda_{i+1})u_{{i+1}}(x)=0,i=2,3,\cdots, \end{cases} \]

we observe that $V(x)-\lambda _2>V(x)-\lambda _3$, by invoking lemma 4.6, it can be obtained that $u_3$ has at least two zeros. Combined with $\rm (iv)$ of theorem 4.4, it is apparent that $u_3$ has at most three node domains, therefore $u_3$ must have two zeros in $(0,\,1)$. The rest can be deduced by this method, so that we get $u_k$ of equation (1.1) has exactly $k-1$ zeroes in $(0,\,1)$, $k=1,\,2\cdots$.

Proof. For the sake of simplicity, let us divide the discussion into several parts.

$1^{\circ }$. Let $u_i(t):=u_i(t,\,x)$ $(i=1,\,2)$ denote the $i$-th normalized eigenfunction for problem (4.16), and $u_i^{M}\ (i=1,\,2)$ denote the $i$-th normalized eigenfunction associated with $V(t,\,x)=M$ in (4.16). Taking $t\in (a,\,b)$, by definition in theorem 4.4

\begin{align*} \lambda_1(t)& =\inf_{u\in H_c^{1}(0,1;h)} \frac{{\int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x + \int_0^{1}\,V (t){|u|^{2}}{\rm d}\kern0.06em x}}{{\|u\|_{{L^{2}}(0,1 )}^{2}}} \le \inf_{u\in H_c^{1}(0,1;h)} \frac{{\int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x + \int_0^{1} M{|u|^{2}}{\rm d}\kern0.06em x}}{{\|u\|_{{L^{2}}(0,1)}^{2}}}\\ & = \frac{{\int_0^{1} h^{2}|{(u_1^{M})}'|^{2}{\rm d}\kern0.06em x + \int_0^{1} M{|u_1^{M}|^{2}}{\rm d}\kern0.06em x}}{{\|u_1^{M}\|_{{L^{2}}(0,1 )}^{2}}}\le C, \end{align*}

likewise, we easily check that $\lambda _2$ has the same properties, that is $\lambda _i(t)\leq C,\, \forall t\in (a,\,b),\, i=1,\,2$. Furthermore, we obtain

\[ \int_0^{1} h^{2}|u_i'(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|u_i(t)|^{2}{\rm d}\kern0.06em x=\lambda_i(t){\|u_i(t)\|_{L^{2}(0,1)}^{2}}=\lambda_i(t)\leq C, i=1,2, \]

so that

(4.17)\begin{equation} \|u_i(t)\|_{H_c^{1}(0,1;h)}\leq C,\quad \forall t\in (a,b),\ i=1,2. \end{equation}

$2^{\circ }$. Taking $\{s_n\}_{n\in \mathbb {N}}$ with $s_n\rightarrow t$, there exists a subsequence of $\{s_n\}$ by $1^{\circ }$, still denoted by itself, and $\lambda _i^{*}=\liminf \limits _{n\rightarrow \infty }\lambda _i(s_n)\in \mathbb {R}$ such that $\lambda _i(s_n)\rightarrow \lambda _i^{*},\, i=1,\,2$. In addition, we can further extract a subsequence of $\{s_n\}_{n\in\mathbb{N}}$ from (4.17), still denoted by itself, and $u_i^{*}\in H_c^{1}(0,\,1;h)$ such that

(4.18)\begin{equation} u_i(s_n)\rightarrow u_i^{*} \mbox{ weakly in } H_c^{1}(0,1;h),i=1,2, \end{equation}

by employing lemma 3.3

(4.19)\begin{equation} u_i(s_n)\rightarrow u_i^{*} \mbox{ strongly in } L^{2}(0,1),i=1,2, \end{equation}

at the limit, the constraint

(4.20)\begin{equation} \|u_i^{*}\|_{L^{2}(0,1)}=1,i=1,2\end{equation}

is preserved. Thanks to (4.19), there exists a subsequence of $\{u_i(s_n)\}_{n\in \mathbb{N}}\subset L^{2}(0,\,1)$, still denoted by itself, such that

(4.21)\begin{equation} u_i(s_n)\rightarrow u_i^{*} \mbox{ a.e.}, i=1,2. \end{equation}

Note that $u_i(s_n)$ is the solution of (4.16) with respect to $V(s_n)$, owing to the fact that $V(s_n)\rightarrow V(t)$ strongly in $L^{\infty }(0,\,1)$ as $n\rightarrow \infty$, by applying (4.18) and (4.19), elementary computations lead to

(4.22)\begin{equation} \begin{cases} -(h^{2} {u_i^{*}}')'+V(t)u_i^{*}=\lambda_i^{*}u_i^{*}, & \mbox{in }(0,1),\\ u_i^{*}=0, & \mbox{on }\{0,1\}. \end{cases} \end{equation}

Clearly, $u_i^{*}$ is a nonzero eigenfunction related to $\lambda _i^{*}$ whence (4.20) and (4.22), $i=1,\,2$. Given that $u_1(s_n)$ is the first positive eigenfunction for $V(s_n)$, so $u_1^{*}> 0$ a.e. by (4.21), it is no doubt that $u_1^{*}$ must change sign except $\lambda _1^{*}=\lambda _1(t)$, for this reason we must have

(4.23)\begin{equation} \lambda_1^{*}=\lambda_1(t) \mbox{ and } u_1^{*}=u_1(t). \end{equation}

Besides, since $u_2(s_n)$ is the second eigenfunction so it must change sign once by theorem 4.7, and consider that (4.21) and $(u_2^{*},\,u_1^{*})=0$, the eigenfunction $u_2^{*}$ exactly has one node on $(0,\,1)$, which shows that $u_2^{*}$ is the second eigenfunction for $V^{*}$ according to theorem 4.7, so that $\lambda _2^{*}=\lambda _2(t)$ and $u_2^{*}=u_2(t)$. In view of this, we conclude that $\lambda _i(s)\rightarrow \lambda _i(t)$ as $s\rightarrow t,\, i=1,\,2$.

Now, we shall show $u_i(s)\rightarrow u_i(t)$ in $L^{2}(0,\,1),\, i=1,\,2$. Indeed, if not, there exists a sequence $\{s_n\}_{n\in \mathbb {N}}\subset (a,\,b)$ and $\epsilon _0>0$ such that $s_n\rightarrow t$ and

(4.24)\begin{equation} \|u_1(s_n)-u_1(t)\|_{L^{2}(0,1)}\geq \epsilon_0. \end{equation}

Follow the inequality (4.17), there exists a subsequence of $\{s_n\}_{n\in \mathbb {N}}$, still denoted by itself, and $\hat u_1\in H_c^{1}(0,\,1;h)$ such that $u_1(s_n)$ weakly converges to $\hat u_1$ in $H_c^{1}(0,\,1;h)$ and $u_1(s_n)$ strongly converges to $\hat u_1$ in $L^{2}(0,\,1)$ as $\|\hat u_1\|_{L^{2}(0,1)}=1$. Hence, we can extract a subsequence of $\{u_1(s_n)\}_{n\in \mathbb{N}}\subset L^{2}(0,\,1)$, still denoted by itself, such that $u_1(s_n)\rightarrow \hat u_1 \mbox { a.e.}$, we deduce that $\|\hat u_1-u_1(t)\|_{L^{2}(0,1)}\geq \epsilon _0$ whence (4.24). However, in the same manner as above (see (4.23)), the result $\hat u_1=u_1(t)$ is obtained, which is absurd. The same argument is applied to $u_2(t)$ again, the desired result is then received.

$3^{\circ }$. We observe that

\[ \begin{cases} -(h^{2}u'_i(s))'+V(s)u_i(s)=\lambda_i(s)u_i(s),\\ - (h^{2}u'_i(t))'+V(t)u_i(t)=\lambda_i(t)u_i(t), \end{cases} \]

thereby

\[ -(h^{2}u'_i(s)-h^{2}u'_i(t))'+V(s)u_i(s)-V(t)u_i(t)=\lambda_{i}(s)u_i(s)-\lambda_{i}(t)u_i(t),\quad i=1,2, \]

i.e.,

\begin{align*} & -(h^{2}u'_i(s)-h^{2}u'_i(t))'+(V(s)-V(t))u_i(s)+V(t)(u_i(s)-u_i(t))\\ & =(\lambda_i(s)-\lambda_i(t))u_i(s)+\lambda_i(t)(u_i(s)-u_i(t)),i=1,2. \end{align*}

Multiplying both sides by $u_i(s)-u_i(t)$ and then integrating on $(0,\,1)$, we see that

\begin{align*} & \int_0^{1} h^{2}|u'_i(s)-u'_i(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1}(V(s)-V(t))u_i(s)(u_i(s)-u_i(t)){\rm d}\kern0.06em x\\ & \quad +\int_0^{1}V(t)|u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x\\ & =(\lambda_i(s)-\lambda_i(t))\int_0^{1} u_i(s)(u_i(s)-u_i(t)){\rm d}\kern0.06em x +\lambda_i(t)\int_0^{1} |u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x, i=1,2, \end{align*}

utilizing the Hölder inequality, we attain further results

\begin{align*} & \int_0^{1} h^{2}|u'_i(s)-u'_i(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1}V(t)|u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x\\ & \quad=\int_0^{1}(V(t)-V(s))u_i(s)(u_i(s)-u_i(t)){\rm d}\kern0.06em x+(\lambda_i(s)\\ & \qquad -\lambda_i(t))\int_0^{1} u_i(s)(u_i(s)-u_i(t)){\rm d}\kern0.06em x +\lambda_i(t)\int_0^{1} |u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x,\\ & \quad\le C\|u_i(s)\|_{L^{2}(0,1)}\|u_i(s)-u_i(t)\|_{L^{2}(0,1)}+C\|u_i(s)-u_i(t)\|_{L^{2}(0,1)}^{2}\\ & \quad\le C\left(\|u_i(s)-u_i(t)\|_{L^{2}(0,1)}+\|u_i(s)-u_i(t)\|_{L^{2}(0,1)}^{2}\right),\ i=1,2. \end{align*}

This suggests that if $u_i(s)\rightarrow u_i(t)$ in $L^{2}(0,\,1)$, then $u_i(s)\rightarrow u_i(t)$ in $H_c^{1}(0,\,1;h)$, $i=1,\,2$.

$4^{\circ }$. Since

\begin{align*} & \int_0^{1} h^{2} |u'_i(s)-u'_i(t)|^{2}{\rm d}\kern0.06em x +\int_0^{1} V(t)|u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x \\ & \quad=\int_0^{1} h^{2}| u'_i(s)|^{2}{\rm d}\kern0.06em x-2\int_0^{1}h^{2}u'_i(s) u'_i(t){\rm d}\kern0.06em x +\int_0^{1}h^{2} | u'_i(t)|^{2}{\rm d}\kern0.06em x \\ & \qquad+ \int_0^{1} V(t) |u_i(s)|^{2}{\rm d}\kern0.06em x -2\int_0^{1} V(t)u_i(s)u_i(t){\rm d}\kern0.06em x +\int_0^{1} V(t)|u_i(t)|^{2}{\rm d}\kern0.06em x\\ & \quad=\lambda_i(s)+\!\int_0^{1}V(t)|u_i(s)|^{2}{\rm d}\kern0.06em x-\!\!\int_0^{1}V(s)|u_i(s)|^{2}{\rm d}\kern0.06em x-2\lambda_i(t)\int_0^{1} u_i(s)u_i(t){\rm d}\kern0.06em x\!+\!\lambda_i(t)\\ & \quad=(\lambda_i(s)-\lambda_i(t))+\int_0^{1}(V(t)-V(s))|u_i(s)|^{2}{\rm d}\kern0.06em x+\lambda_i(t)\int_0^{1}|u_i(s)-u_i(t)|^{2}{\rm d}\kern0.06em x, \end{align*}

naturally,

\begin{align*} |\lambda_i(s)-\lambda_i(t)| & \leq \int_0^{1}h^{2} |u'_i(s)-u'_i(t)|^{2}{\rm d}\kern0.06em x+C\|u_i(s)-u_i(t)\|_{L^{2}(0,1)}^{2}\\ & \quad+\int_0^{1}|V(s)-V(t)||u_i(s)|^{2}{\rm d}\kern0.06em x, \end{align*}

for $i=1,\,2$. Combine this result with $4^{\circ }$, in addition $V(s)\rightarrow V(t)$ as $s\rightarrow t$, this suggests that when $u_i(s)\rightarrow u_i(t)$ in $L^{2}(0,\,1)$, we have $\lambda _i(s)\rightarrow \lambda _i(t)$, $i=1,\,2$.

$5^{\circ }$. We claim that $\lambda _i(s)\rightarrow \lambda _i(t)$ as $s\rightarrow t$, then $u_i(s)\rightarrow u_i(t)$ in $L^{2}(0,\,1)$ as $s\rightarrow t$, $i=1,\,2$. If not, suppose that there exist $\epsilon >0$ and a sequence $\{u_i(s_n)\}_{n\in \mathbb {N}}$ such that $s_n\rightarrow t$ but $\|u_i(s_n)-u_i(t)\|_{L^{2}(0,1)}\geq \epsilon$. In the same way as $3^{\circ }$, there exists $\hat u_i\in H_c^{1}(0,\,1;h)$ such that $u_i(s_n)$ weakly converges to $\hat u_i$ in $H_c^{1}(0,\,1;h)$, $u_i(s_n)$ strongly converges to $\hat u_i$ in $L^{2}(0,\,1)$ with $\|\hat u_i\|_{L^{2}(0,1)}=1$ and $u_i(s_n)\to \hat {u_i}$ a.e., $i=1,\,2$. Then we obtain $\|\hat {u_i}-u_i(t)\|_{L^{2}(0,1)}\geq \epsilon$.

We see that $u_1(s_n)-u_1(t)\in H_c^{1}(0,\,1;h)$ and

(4.25)\begin{equation} \begin{aligned} \lambda_1(t) & =\inf_{u\in H_c^{1}(0,1;h),u\neq 0}\frac{\int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|u|^{2}{\rm d}\kern0.06em x}{\int_0^{1} |u|^{2} {\rm d}\kern0.06em x}\\ & \le \frac{\int_0^{1}h^{2}|\hat u'_1-u'_1(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|\hat u_1-u_1(t)|^{2}{\rm d}\kern0.06em x}{\int_0^{1} |\hat u_1-u_1(t)|^{2} {\rm d}\kern0.06em x}. \end{aligned} \end{equation}

Similarly, $u_2(s_n)-u_2(t) \in H_c^{1}(0,\,1;h)$ and

\[ \lambda_2(t)=\inf_{\substack{u\in H_c^{1}(0,1;h),u\neq 0,\\ (u,u_1(t))=0 }}\frac{\int_0^{1} h^{2} |u'|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|u|^{2}{\rm d}\kern0.06em x}{\int_0^{1} |u|^{2} {\rm d}\kern0.06em x}. \]

If $(\hat u_2-u_2(t),\,u_1(t)) \neq 0$, we must have

(4.26)\begin{equation} \lambda_2(t)\neq \frac{\int_0^{1}h^{2} | \hat u'_2-u'_2(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|\hat u_2-u_2(t)|^{2}{\rm d}\kern0.06em x}{\int_0^{1} |\hat u_2-u_2(t)|^{2} {\rm d}\kern0.06em x},\end{equation}

and if $(\hat u_2-u_2(t),\,u_1(t)) = 0$, then

(4.27)\begin{equation} \lambda_2(t)\le\frac{\int_0^{1}h^{2} | \hat u'_2-u'_2(t)|^{2}{\rm d}\kern0.06em x+\int_0^{1} V(t)|\hat u_2-u_2(t)|^{2}{\rm d}\kern0.06em x}{\int_0^{1} |\hat u_2-u_2(t)|^{2} {\rm d}\kern0.06em x}. \end{equation}

For $i=1,\,2$, due to $\lambda _i(t)$ is simple, the equality of (4.25) and (4.27) holds if and only if $\hat u_i-u_i(t)=cu_i(t)$, thus $\hat u_1=(c+1)u_1(t)$, the normalization condition guarantees $c=0$ or $c=-2$, since $1=\|\hat u_i\|_{L^{2}(0,1)}=|c+1|\|u_i(t)\|_{L^{2}(0,1)}=|c+1|$. Under the condition $\|\hat u_i-u_i(t)\|_{L^{2}(0,1)}\geq \epsilon$, the possibility of $c= 0$ is excluded, $i=1,\,2$. And simultaneously we also analyse that $c\neq -2$ for $i=1$ due to $\hat u_1\geq 0$ a.e. Without loss of generality, we may choose the sign of the second eigenfunction to be positive to the left of the node and to be negative in the other side. It is easily found that $\hat {u_2}=-u_2(t)$ for $c=-2$, it means that $\hat {u_2}$ and $u_2(t)$ have the same node, so this case can not happen. All these facts have demonstrated that the inequality of (4.25) and (4.27) holds strictly.

Formula from $4^{\circ }$,

\begin{align*} & \int_0^{1}h^{2} |(u'_i(s_n)-u'_i(t))|^{2}{\rm d}\kern0.06em x +\int_0^{1} V(t)|u_i(s_n)-u_i(t)|^{2}{\rm d}\kern0.06em x-\lambda_i(t)\int_0^{1}|u_i(s_n)-u_i(t)|^{2}{\rm d}\kern0.06em x\\ & \quad=(\lambda_i(s_n)-\lambda_i(t))+\int_0^{1}(V(t)-V(s_n))|u_i(s_n)|^{2}{\rm d}\kern0.06em x, i=1,2, \end{align*}

let $s_n \to t$, we check that the right side of above equation converges to $0$, but the limit on the left is nonzero by virtue of the above analysis, which causes a contradiction. So far, the desired result is proved.

$6^{\circ }$. By $3^{\circ }$,

\begin{align*} & -(h^{2}(u'_i(s)-u'_i(t)))'+(V(s)-V(t))u_i(s)+V(t)(u_i(s)-u_i(t))\\ & =(\lambda_i(s)-\lambda_i(t))u_i(s)+\lambda_i(t)(u_i(s)-u_i(t)),i=1,2, \end{align*}

multiplying $u_i(t)$ on the both sides of this equality and integrating on $(0,\,1)$, then yields

\begin{align*} & \int_0^{1}h^{2}(u'_i(s)-u'_i(t)) u'_i(t){\rm d}\kern0.06em x\\ & \quad +\int_0^{1} (V(s)-V(t))u_i(s)u_i(t){\rm d}\kern0.06em x +\int_0^{1} V(t)u_i(t)(u_i(s)-u_i(t)){\rm d}\kern0.06em x \\ & =(\lambda_i(s)-\lambda_i(t))\int_0^{1} u_i(s)u_i(t){\rm d}\kern0.06em x +\lambda_i(t)\int_0^{1} u_i(t)(u_i(s)-u_i(t)) {\rm d}\kern0.06em x, \quad i=1,2, \end{align*}

i.e.

(4.28)\begin{align} & \int_0^{1}h^{2}\frac{u'_i(s)-u'_i(t)}{s-t}u'_i(t){\rm d}\kern0.06em x +\int_0^{1} \frac{V(s)-V(t)}{s-t}u_i(s)u_i(t){\rm d}\kern0.06em x\nonumber\\ & \quad +\int_0^{1} V(t)u_i(t)\frac{u_i(s)-u_i(t)}{s-t} {\rm d}\kern0.06em x \nonumber\\ & =\frac{\lambda_i(s)-\lambda_i(t)}{s-t}\int_0^{1} u_i(s)u_i(t){\rm d}\kern0.06em x +\lambda_i(t)\int_0^{1} u_i(t)\frac{u_i(s)-u_i(t)}{s-t}{\rm d}\kern0.06em x,\quad i=1,2. \end{align}

Also, given that

\[ -(h^{2}u'_i(t))'+V(t)u_i(t)=\lambda_{i}(t)u_i(t), \]

multiplying $\frac {{u_i(s)-u_i(t)}}{s-t}$ on both sides, after integrating on $(0,\,1)$ yields

(4.29)\begin{align} & \int_0^{1} h^{2} u'_i(t)\frac{u'_i(s)-u'_i(t)}{s-t}{\rm d}\kern0.06em x+\int_0^{1} V(t)u_i(t) \frac{u_i(s)-u_i(t)}{s-t}{\rm d}\kern0.06em x\nonumber\\ & \quad =\lambda_i(t)\int_0^{1} u_i(t)\frac{u_i(s)-u_i(t)}{s-t}{\rm d}\kern0.06em x, \end{align}

further,

\[ \frac{\lambda_i(s)-\lambda_i(t)}{s-t}\int_0^{1} u_i(s)u_i(t) {\rm d}\kern0.06em x=\int_0^{1} \frac{V(s)-V(t)}{s-t} u_i(s)u_i(t){\rm d}\kern0.06em x,\quad i=1,2. \]

whence (4.28) and (4.29). Concerning that $u_i(s)\rightarrow u_i(t)$ in $L^{2}(0,\,1)$ as $s\rightarrow t$, from the dominated convergence theorem we have

\begin{align*} \dot{\lambda}_i(t)& =\lim_{s\rightarrow t}\frac{\lambda_i(s)-\lambda_i(t)}{s-t}\int_0^{1} u_i(s)u_i(t) {\rm d}\kern0.06em x=\lim_{s\rightarrow t}\int_0^{1} \frac{V(s)-V(t)}{s-t} u_i(s)u_i(t){\rm d}\kern0.06em x\\ & =\int_0^{1} \dot V(t)|u_i(t)|^{2}{\rm d}\kern0.06em x,\quad i=1,2. \end{align*}

Proof. We shall prove the existence of the minimizer $V^{*}$ such that $\Gamma ^{*}=\Gamma (V^{*})=\inf \limits _{V\in \mathcal {V}} (\lambda _2(V)-\lambda _1(V)).$

Let $\{V^{k}\}_{k\in \mathbb {N}}$ such that

(5.1)\begin{equation} \Gamma(V^{k}) \downarrow \inf\limits_{V\in\mathcal{V}} \Gamma(V)=\Gamma^{*}.\end{equation}

In view of the compactness of the class $\mathcal {V}$, then there exists a subsequence $\{V^{k}\}_{k\in \mathbb {N}}\subset L^{\infty }(0,\,1)$ such that

(5.2)\begin{equation} V^{k}\rightarrow V^{*} \mbox{ weakly star in } L^{\infty}(0,1), \end{equation}

and

(5.3)\begin{equation} \lambda_2(V^{k}) \to \lambda_2^{*},\lambda_1(V^{k}) \to \lambda_1^{*}, \Gamma^{*}=\lambda_2^{*}-\lambda_1^{*}. \end{equation}

Let $\{(\lambda _j^{k},\,u_j^{k})\}_{k\in \mathbb {N}}$ be a sequence concerned with $V^{k}$, $\lambda _j^{k}$ denotes the $j$-th eigenvalue of (1.1), $u_j^{k}$ be the normalized eigenfunction in $H_c^{1}(0,\,1;h)$, $j=1,\,2$. By the definition of eigenvalue, we realize that

\begin{align*} \lambda_1^{k}& =\inf_{u\in H_c^{1}(0,1;h),u\neq 0}\frac{\int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x+\int_0^{1} V^{k}|u|^{2} {\rm d}\kern0.06em x}{\int_0^{1} |u|^{2} {\rm d}\kern0.06em x}\\ & \quad \le \inf_{u\in H_c^{1}(0,1;h),u\neq 0}\frac{\int_0^{1} h^{2}|u'|^{2}{\rm d}\kern0.06em x+\int_0^{1} M|u|^{2} {\rm d}\kern0.06em x}{\int_0^{1} |u|^{2} {\rm d}\kern0.06em x} \le C, \end{align*}

so, in like manner, we obtain that $\lambda _2^{k}\le C$. Further,

(5.4)\begin{equation} \int_0^{1} h^{2}|(u_j^{k})'|^{2}{\rm d}\kern0.06em x+\int_0^{1} V^{k}|u_j^{k}|^{2}{\rm d}\kern0.06em x=\lambda_j^{k}{\|u_j^{k}\|_{L^{2}(0,1)}^{2}}=\lambda_j^{k}\leq C, i=1,2, \end{equation}

we have that $u_j^{k}$ is bounded in $H_c^{1}(0,\,1;h)$, and hence by the compactness of the embedding in lemma 3.3, a subsequence, which we take as $\{u_j^{k}\}_{k\in\mathbb{N}}$ itself, such that

(5.5)\begin{equation} u_j^{k}\rightarrow u_j^{*} \mbox{ weakly in } H_c^{1}(0,1;h),\quad j=1,2, \end{equation}

and

(5.6)\begin{equation} u_j^{k}\rightarrow u_j^{*} \mbox{ strongly in } L^{2}(0,1),j=1,2. \end{equation}

Hence, there exists a subsequence, still take as $\{u_j^{k}\}_{k\in\mathbb{N}}$ itself, such that

(5.7)\begin{equation} u_j^{k}\rightarrow u_j^{*} \mbox{ a.e.}, j=1,2. \end{equation}

For all $v\in H_c^{1}(0,\,1;h)$, seeing that

\[ \int_0^{1}{h^{2}({u_j^{k}})'}v'{\rm d}\kern0.06em x+\int_0^{1} V^{k}u_j^{k}vdx=\lambda_j^{k}\int_0^{1} u_j^{k}vdx,j=1,2, {\forall v \in H_c^{1}(0,1;h)}, \]

let $k\rightarrow \infty$, then by (5.2) (5.3) and (5.5) (5.6)

(5.8)\begin{equation} \int_0^{1}{h^{2}({u_j^{*}})'}v'{\rm d}\kern0.06em x+\int_0^{1} V^{*}u_j^{*}vdx=\lambda_j^{*}\int_0^{1} u_j^{*}vdx,j=1,2, {\forall v \in H_c^{1}(0,1;h)}. \end{equation}

This implies that $\lambda _j^{*}$ be the element of the spectrum for $V^{*}$, $j=1,\,2$. Consider that $u_1^{k}$ is the first eigenfunction involved with $\lambda _1^{k}$, and the ground states are characterized as the positive eigenfunctions on $(0,\,1)$, this fact determines that $u_1^{*}> 0$ on $(0,\,1)$ by (5.7), so that we must have $\lambda _1^{*}=\lambda _1(V^{*})$. In a similar fashion, we see that $u_2^{k}$ change once sign in $(0,\,1)$, so that $u_2^{*}$ change once sign by (5.7) and $(u_2^{*},\,u_1^{*})=0$, utilizing theorem 4.7, we have $u_2^{*}$ is the second eigenfunction for $V^{*}$ and $\lambda _2^{*}=\lambda _2(V^{*})$. All this suggests that $\Gamma (V^{*})=\lambda _2(V^{*})-\lambda _1(V^{*})$.

Proof. In order to prove (5.9), we demand display that the set $T=\{x\in (0,\,1)\mid m< V^{*}(x)< M\}$ is zero measure. If not, let $T^{k}=\{x\in [0,\,1] \mid m+\frac {1}{k}< V^{*}(x)< M-\frac {1}{k}\}$, then we may write $T=\bigcup \limits _{k = 1}^{\infty } T^{k}$, we assume that at least one of them is a positive measure for $T^{k}$. For any $x^{*}\in T^{k}$ and any measurable sequence of subsets $G_{k,j}\subset T^{k}$ including $x^{*}$, the optimal condition of the fundamental gap extremum problem determines

\[ {\left. {\frac{{{\rm d}\Gamma (V (x,t))}}{{{\rm d}t}}} \right|_{t = 0}} = \int_{G_{k,j}} {\left(|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}\right)} {\rm d}\kern0.06em x = 0, \]

where $V(x,\,t)=V(x)+tP(x)$ and $P(x)=\chi _{G_{k,j}}$ is an admissible perturbation for any small $t\in (-\tfrac {1}{k},\,\tfrac {1}{k})$. We claim that $|u_2^{*}(x)|^{2} = |u_1^{*}(x)|^{2}$ on $T^{k}$ for any $k>1$, so that $|u_2^{*}(x)|^{2} = |u_1^{*}(x)|^{2}$ on the set $T$. Indeed, if there is a subset $G \subset T^{k}$ of positive measure such that $|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}>0$, $P(x) = {\chi _G}$ is an admissible perturbation, then we have

\[ {\left. {\frac{{{\rm d}\Gamma (V (x,t))}}{{{\rm d}t}}} \right|_{t = 0}} = \int_G |u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}{\rm d}\kern0.06em x > 0, \]

however, this is in contradiction with the fact that $V^{*}(x)$ is the optimal function. Likewise same procedure is carried out, we can obtain that there is no positive measure subset on $T$ with $|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}<0$.

Actually, the case $|u_2^{*}(x)|^{2} = |u_1^{*}(x)|^{2}$ only works on the set with zero measure. Without loss of generality, suppose $T^{+} = \{ x \in T \mid {u_2^{*}(x)} > 0\}$ with positive measure, concerning that $u_1^{*}$ has no zero point on $(0,\,1)$, as a result, $T = {T^ + } \cup {T^ - }$, where ${T^ - } = \{ x \in T \mid {u_2^{*}(x)} < 0\}$. On $T^{+}$, seeing that $u_1^{*} -u_2^{*}=0$ a.e., thus $(u_2^{*}-u_1^{*})'=0$ a.e., from equation (1.1) we can infer that

(5.10)\begin{equation} (h^{2}(u_2^{*}-u_1^{*})')'+V(u_1^{*}-u_2^{*})=\lambda_1u_1^{*}-\lambda_2u_2^{*}. \end{equation}

From the proof of theorem 4.5, we know that $u_1^{*},\,u_2^{*}\in H_{loc}^{2}(0,\,1)$ and $u_1^{*},\,u_2^{*}\in C^{1,\frac {1}{2}}(\delta,\,1)$ for any $\delta \in (0,\,1)$. Let $F(x)=(h^{2}(u_2^{*}-u_1^{*})')'+V(u_1^{*}-u_2^{*})-(\lambda _1u_1^{*}-\lambda _2u_2^{*})$, we obtain $F(x)=0$ a.e. on $T^{+}$ in the sense of distribution. The left side of equation (5.10) vanishes on ${T^ + }$ a.e., it directly caused $\lambda _1=\lambda _2$, a contradiction, so that $T$ is a set of zero measure. This ensures that the optimal function is $V^{*} = m{\chi _{\omega }} + M{\chi _{\omega ^{c}}}$ for some $\omega$.

In this regard, we have to study further, there is no doubt that $\chi _{\omega }$ is an admissible perturbation for any small $t\ge 0$ when $V^{*}(x)=m$, theorem 4.8 guarantees

\[ {\left.{\frac{{{\rm d}\Gamma (V (x,t))}}{{{\rm d}t}}} \right|_{t = 0}} = \int_{\omega } |u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}{\rm d}\kern0.06em x \ge 0, \]

for this we may come to a conclusion $|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2}\ge 0$. If this is not true, there must be a set $S \subset \omega _0$ of positive measure such that $|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2} < 0$ on $S$ and

\[ {\left. {\frac{{{\rm d}\Gamma (V (x,t))}}{{{\rm d}t}}} \right|_{t = 0}} = \int_S|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2} {\rm d}\kern0.06em x< 0 \]

for admissible perturbation ${\chi _S}$, $t\ge 0$, this obviously fail to meet the optimality of $V^{*}(x)$. By applying the same argument, the result $|u_2^{*}(x)|^{2} - |u_1^{*}(x)|^{2} \le 0$ as $V^{*}=M$ is straightforward.

Concerned with the normalization condition $\int _0^{1} |u_2^{*}(x)|^{2}-|u_1^{*}(x)|^{2}{\rm d}\kern0.06em x = 0$, which indicates that both the sets $\omega$ and $\omega ^{c}$ are nonempty.

Proof. For any given function $V(x)$, the set $S_1$ and $S_1^{c}$ are nonempty in the light of normalization condition, here $S_1=\{x\in (0,\,1) \mid u_2(x)|^{2} - |u_1(x)|^{2}\ge 0 \}$, thereby $|u_2(x)|^{2} = |u_1(x)|^{2}$ has at least one solution. In reality, the number of solutions for this equation shall not exceed two, the following will be explained in detail.

Consider

(5.11)\begin{equation} \left(\frac{u_2(x)}{u_1(x)}\right)'=\frac{{u_2'(x)}u_1(x)-{u_1'(x)}u_2(x)}{|u_1(x)|^{2}}=\frac{g(x)}{|u_1(x)|^{2}}, \end{equation}

and note that

\begin{align*} (h^{2}(x)g(x))'& =(h^{2}(x)u'_2(x))'u_1(x)-(h^{2}(x)u'_1(x))'u_2(x)\\ & =(V(x)-\lambda_2)u_1(x)u_2(x)-(V(x)-\lambda_1)u_1(x)u_2(x)\\ & =(\lambda_1-\lambda_2)u_1(x)u_2(x), \end{align*}

so that

(5.12)\begin{equation} h^{2}(x)g(x)=\int_0^{x}(h^{2}(t)g(t))'{\rm d}t=\int_0^{x}(\lambda_1-\lambda_2)u_1(t)u_2(t){\rm d}t,\quad x\in (0,\alpha), \end{equation}

and

(5.13)\begin{equation} h^{2}(x)g(x)={-}\int_x^{1}(h^{2}(t)g(t))'{\rm d}t=\int_x^{1}(\lambda_2-\lambda_1)u_1(t)u_2(t){\rm d}t,\quad x\in (\alpha,1), \end{equation}

where $\alpha$ denotes the node of the eigenfunction $u_2(x)$. Without loss of generality, suppose $u_2(x)>0$ in $(0,\,\alpha )$ and $u_2(x)<0$ in $(\alpha,\,1)$, hence $h^{2}(x)g(x)<0$ on $(0,\,1)$ by (5.12) and (5.13), especially, $g(x)<0$ on $(0,\,1)$. Moreover, thanks to (5.11) we have

\[ \left(\frac{|u_2(x)|^{2}}{|u_1(x)|^{2}}\right)'=2\left(\frac{u_2(x)}{u_1(x)}\right) \left(\frac{u_2(x)}{u_1(x)}\right)'<0, \quad x\in (0,\alpha), \]

and

\[ \left(\frac{|u_2(x)|^{2}}{|u_1(x)|^{2}}\right)'=2\left(\frac{u_2(x)}{u_1(x)}\right) \left(\frac{u_2(x)}{u_1(x)}\right)'>0, \quad x\in (\alpha,1), \]

that is, $\frac {|u_2(x)|^{2}}{|u_1(x)|^{2}}$ is strictly monotonic decreasing on $(0,\,\alpha )$ and strictly monotonic increasing on $(\alpha,\,1)$.

Suppose there are two different points $x^{1},\, x^{2}\in (0,\,\alpha )$ such that $|u_2(x)|^{2} = |u_1(x)|^{2}$, which lead to $\frac {|u_2(x^{2})|^{2}}{|u_1(x^{2})|^{2}}=\frac {|u_2(x^{1})|^{2}}{|u_1(x^{1})|^{2}}$, this is opposite to the condition of strictly monotone for $\frac {|u_2(x)|^{2}}{|u_1(x)|^{2}}$ on $(0,\,\alpha )$, hence the equation $|u_1(x)|^{2} =|u_2(x)|^{2}$ has at most one solution on $(0,\,\alpha )$. Likewise, the same conclusion is achieved on $(\alpha,\,1)$. Consequently, there exist two points $x_0$, $x_1$, $0 \le x_0 < \alpha < x_1 \le 1$ with

(5.14)\begin{equation} |u_2(x)|^{2} - |u_1(x)|^{2}\left\{ \begin{array}{l} > 0,x \in (0,{x_0}) \cup (x_1,1),\\ < 0,x \in (x_0,x_1), \end{array} \right. \end{equation}

and both sets are nonempty. Take into consideration that $u_1(x),\,u_2(x)$ are continuous on $(0,\,1)$, we must have $|u_2(x_0)|^{2}=|u_1(x_0)|^{2}$ and $|u_2(x_1)|^{2}=|u_1(x_1)|^{2}$.

Recognizing the facts $|u_2^{*}(x)|^{2} -|u_1^{*}(x)|^{2} \le 0$ when $V^{*}(x)=M$ according to theorem 5.2 and the inequality 5.14, we point out that the set $\omega ^{c}$ only contains one interval.