We begin this section with the operators \(\mathbf{H}_{X,A,Q}\) and their corresponding quadratic forms in the Hilbert space \(L^{2}(\mathbb{R}_{+}, \mathbb{C}^{m})\). First, we recall some notation for the convenience of the readers. The inequality \(Q(x)\geq(>)0\) means that for any \(Y(x)= (y_{1}(x),y_{2}(x), \ldots,y_{m}(x))^{T}\), \((Q(x)Y(x),Y(x))_{m}\geq(>)0\), and \(Q(x)\) is called bounded from below if \((Q(x)Y(x),Y(x))_{m}\geq-c(Y(x),Y(x))_{m}\), where c is a constant. Then we consider the quadratic forms
$$\begin{aligned} &\mathfrak{q}[Y]:= \int_{0}^{\infty}\bigl(Q(x)Y(x),Y(x)\bigr)_{m}\,dx, \\ &\operatorname{Dom}(\mathfrak{q})=\bigl\{ Y\in L^{2}\bigl( \mathbb{R}_{+},\mathbb{C}^{m}\bigr):\bigl| \mathfrak{q}[Y]\bigr|< \infty \bigr\} . \end{aligned}$$
(3.1)
We denote by \(\mu_{\min}(Q(x))\) and \(\mu_{\max}(Q(x))\) respectively the minimal and maximal eigenvalues of \(Q(x)\), and by \(\mu_{\min}(A_{k})\) and \(\mu_{\max}(A_{k})\) respectively the minimal and maximal eigenvalues of \(A_{k}\). The quadratic form \(\mathfrak{q}[Y]\) is called semibounded from below (above) if and only if so is \(\mu_{\min}(Q(x))(\mu_{\max}(Q(x)))\). We denote by \(\mathbf{t}_{0}[Y]\) the following quadratic form:
$$ \mathbf{t}_{0}[Y]:= \int_{0}^{\infty}\bigl(Y^{\prime}(x),Y^{\prime}(x) \bigr)_{m}\,dx,\quad \operatorname{Dom}\bigl(\mathbf{t}_{0}[Y] \bigr)=W_{0}^{1,2}\bigl(\mathbb{R}_{+}, \mathbb{C}^{m}\bigr). $$
(3.2)
Together with the form \(\mathfrak{q}\), we consider the form
$$\begin{aligned} &\mathbf{t}_{Q}[Y]:=\mathbf{t}_{0}[Y]+ \mathfrak{q}[Y]= \int_{0}^{\infty }\bigl[\bigl(Y^{\prime}(x),Y^{\prime}(x) \bigr)_{m}+\bigl(Q(x)Y(x),Y(x)\bigr)_{m}\bigr]\,dx, \\ &\operatorname{Dom}(\mathbf{t}_{Q}) =W_{0}^{1,2} \bigl(\mathbb{R}_{+},\mathbb {C}^{m}\bigr)\cap \operatorname{Dom}(\mathfrak{q}) \\ &\hphantom{\operatorname{Dom}(\mathbf{t}_{Q})} =\bigl\{ Y\in L^{2}\bigl(\mathbb{R}_{+}, \mathbb{C}^{m}\bigr)\cap AC_{\mathrm{loc}}\bigl(\mathbb{R}_{+}, \mathbb{C}^{m}\bigr),\mathbf{t}_{Q}[Y]< +\infty,Y(0)=\theta \bigr\} . \end{aligned}$$
(3.3)
\(\operatorname{Dom}(\mathbf{t}_{0}[Y])\) is the set of all \(Y\in L^{2}(\mathbb{R}_{+}, \mathbb{C}^{m})\) such that \(Y(x)\) is absolutely continuous, \(Y^{\prime }\in L^{2}(\mathbb{R}_{+},\mathbb{C}^{m})\), and \(Y(0)=\theta\). The form \(\mathbf{t}_{0}[Y]\) is symmetric and closed. \(\operatorname{Dom}(\mathfrak{q})\) is the set of all \(Y\in L^{2}( \mathbb{R}_{+},\mathbb{C}^{m})\) such that \(\int_{0}^{\infty }(Q(x)Y(x),Y(x))_{m}\,dx<+\infty\). The form \(\mathfrak{q}\) is also symmetric and closed. Thus, \(\mathbf{t}_{Q}[Y]=\mathbf{t}_{0}[Y]+\mathfrak{q}[Y]\) is symmetric and closed (see [11], Chapter VI, Theorem 1.31). If the potential \(Q(x)\) is lower semibounded with bound −c, that is, \((Q(x)Y,Y)\geq -c(Y,Y)\), \(\forall Y\in\mathbb{C}^{m}\), then \(\operatorname{Dom}(\mathbf{t}_{Q})\) equipped with the norm \(\| Y\|_{\mathfrak{H}_{Q}}^{2}:=\| Y\| _{W^{1,2}}^{2}+\mathbf{t}_{Q}[Y]+(1+c)\| Y\|_{\mathfrak{H}}^{2}\) is the Hilbert space \(\mathfrak{H}_{Q}:=\mathfrak{H}_{\mathbf{t} _{Q}}:=W_{0}^{1,2}(\mathbb{R}_{+},\mathbb{C}^{m};Q)\). Let
$$ \mathbf{t}_{R}[Y]=\sum_{k=1}^{\infty} \bigl(A_{k}Y(x_{k}),Y(x_{k})\bigr)_{m} $$
(3.4)
be defined on the domain
$$ \operatorname{Dom}(\mathbf{t}_{R})=\bigl\{ Y\in L^{2}\bigl( \mathbb{R}_{+},\mathbb{C}^{m}\bigr), \mathbf{t}_{R}[Y]< + \infty\bigr\} . $$
Denote by \(\mathbf{t}_{R}^{\pm}[Y]\) the sum of k positive (negative) parts of \((A_{k}Y(x_{k}),Y(x_{k}))_{m}\), that is,
$$\begin{aligned} &\mathbf{t}_{R}^{\pm}[Y]=\sum _{k=1}^{\infty }\bigl(A_{k}Y(x_{k}),Y(x_{k}) \bigr)_{m}^{\pm},\\ &\operatorname{Dom}\bigl(\mathbf{t}_{R}^{\pm}[Y]\bigr)=\bigl\{ Y\in W_{0}^{1,2}\bigl(\mathbb{R}_{+}, \mathbb{C}^{m}\bigr), \mathbf{t}_{R}^{\pm}[Y]< + \infty\bigr\} . \end{aligned}$$
(3.5)
Similarly,
$$ \mathfrak{q}^{\pm}[Y]:= \int_{0}^{\infty}\bigl(Q(x)Y(x),Y(x)\bigr)_{m}^{\pm}\,dx, \quad \operatorname{Dom}\bigl(\mathfrak{q}^{\pm}\bigr)=\bigl\{ Y\in AC_{\mathrm{loc}}\bigl(\mathbb{R}_{+},\mathbb {C}^{m}\bigr), \mathfrak{q}^{\pm}[Y]< +\infty\bigr\} . $$
Then we define the form
$$\begin{aligned}& \mathbf{t}_{X,A,Q} =\mathbf{t}_{Q}+\mathbf{t}_{R},\qquad \mathbf{t}_{Q}= \mathbf{t}_{0}+\mathbf{t}_{Q}= \mathbf{t}_{0}+\mathfrak{q}^{+}+\mathfrak{q} ^{-}, \\& \mathbf{t}_{Q}^{+} =\mathbf{t}_{0}+ \mathfrak{q}^{+},\qquad \mathbf{t} _{Q}^{-}= \mathbf{t}_{0}+\mathfrak{q}^{-}, \\& \operatorname{Dom}(\mathbf{t}_{X,A,Q}) = \operatorname{Dom}( \mathbf{t}_{Q})\cap\operatorname{Dom}(\mathbf{t}_{R}), \end{aligned}$$
which is naturally associated with the differential expression, and the form is as follows:
$$\begin{aligned} \mathbf{t}_{X,A,Q}[Y] =& \int_{0}^{\infty}\bigl[\bigl(Y^{\prime}(x),Y^{\prime }(x) \bigr)_{m}+\bigl(Q(x)Y(x),Y(x)\bigr)_{m}\bigr]\,dx \\ &{}+\sum_{k=1}^{\infty}\bigl(A_{k}Y(x_{k}),Y(x_{k}) \bigr)_{m}. \end{aligned}$$
(3.6)
Notation
If the form
\(\mathbf{t}_{X,A,Q}\)
is nonnegative, then
\(\mathfrak{H}_{X,A,Q}:=\mathfrak{H}_{\mathbf{t}_{X,A,Q}}\)
denotes the domain
\(\operatorname{Dom}(\mathbf{t}_{X,A,Q})\)
equipped with the norm
$$\begin{aligned} &\Vert Y\Vert _{\mathfrak{H}_{X,A,Q}}^{2}= \mathbf{t}_{Q}[Y]+ \mathbf{t}_{R}[Y]+\Vert Y\Vert _{L^{2}(\mathbb{R}_{+},\mathbb{C} _{m})}^{2}, \\ &\sup_{x>0} \int_{x}^{x+1}\bigl|\mu_{\max}^{-} \bigl(Q(x)\bigr)\bigr|\,dx< +\infty,\qquad \sup_{x>0}\sum _{x_{k}\in[ x,x+1]}\bigl|\mu_{\max }^{-}(A_{k})\bigr|< + \infty, \end{aligned}$$
(3.7)
where
\(\mu_{\max}^{\pm}(Q(x)):=(\mu_{\max}(Q(x))\pm|\mu_{\max }(Q(x))|)/2\), \(\mu(A)^{\pm}:=\{\mu(A_{k})^{\pm}\}_{1}^{\infty}\), and
\(\mu_{\max}^{\pm}(A_{k}):=(\mu_{\max}(A_{k}) \pm|\mu_{\max }(A_{k})|)/2\).
Lemma 4
If the minimal operator
\(\mathbf{H}_{\min}=\mathbf{H} _{X,A,Q}\)
is lower semibounded, then the operator
\(\mathbf{H}_{X,A,Q}\)
is self-adjoint, \(\mathbf{H}_{X,A,Q}=(\mathbf{H}_{X,A,Q})^{\ast}\).
Proof
Without loss of generality, we assume that \(\mathbf{H}_{X,A,Q}\geqslant I\). It is sufficient to show that \(\ker((\mathbf{H}_{X,A,Q})^{\ast})=\{ \theta \}\), that is, the equation
$$ -Y^{\prime\prime}(x)+Q(x)Y(x)=\theta,\quad x\in\mathbb{R} _{+} \backslash X, Y\in\operatorname{Dom}\bigl((\mathbf{H}_{X,A,Q})^{\ast} \bigr) $$
(3.8)
has only a trivial solution (the derivative is understood in a distribution sense).
Assume the converse, that is, let \(Y(x)=(y_{1}(x),y_{2}(x),\ldots,y_{m}(x))^{T}\) be a solution of equation (3.8). Let \(\chi_{i}\in C_{0}^{\infty}(\mathbb{R}_{+},\mathbb{C} ^{m})\) (\(i=1,2,\ldots,m\)) be such that \(0 \leq \Vert \chi _{i}\Vert \leq1\) and
$$ \chi_{i}(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} e_{i}, & 0\leqslant x\leq1/2, \\ \theta, & x\geqslant1,\end{array}\displaystyle \right .\quad i=1,2,\ldots,m, $$
where \(\Vert \cdot \Vert \) is the Euclidean norm. Let the matrix \(X(x)\) be the combination of \(\chi_{i}(x)\), \(i=1,2,\ldots,m\), that is, \(X(x)=[\chi_{1}(x),\chi_{2}(x),\ldots,\chi_{m}(x)]\). Define
$$ Y_{n}(x):=X_{n}(x)Y(x)=X(x/n)Y(x),\quad n\in\mathbb{N}. $$
Obviously, \(\sup pY_{n}\subset[0,n]\). Since \(Y_{n}(x_{k}-)=Y_{n}(x_{k}+)\) and \(\chi_{i}\in C_{0}^{\infty}(\mathbb{R} _{+},\mathbb{C}^{m})\) (\(i=1,2,\ldots,m\)), we get
$$ Y_{n}^{\prime}(x_{k}+)-Y_{n}^{\prime}(x_{k}-)=X_{n}(x_{k}) \bigl[Y^{\prime }(x_{k}+)-Y^{\prime}(x_{k}-) \bigr]=X_{n}(x_{k})A_{k}Y(x_{k})=A_{k}Y_{n}(x_{k}), $$
and hence \(Y_{n}\in \operatorname{Dom}(\mathbf{H}_{\min})\). Furthermore,
$$\begin{aligned} (\mathbf{H}_{\min}Y_{n},Y_{n}) =& \int_{0}^{\infty}\bigl(\bigl[-Y_{n}^{\prime \prime }(x)+Q(x)Y_{n}(x) \bigr],Y_{n}(x)\bigr)_{m}\,dx \\ =&- \int_{0}^{\infty}\bigl(2X_{n}^{\prime}(x)Y^{\prime}(x)+X_{n}^{\prime \prime}(x)Y(x),X_{n}(x)Y(x) \bigr)_{m}\,dx \\ =&- \int_{0}^{\infty}\bigl[\bigl(2X_{n}^{\prime}(x)Y^{\prime}(x),X_{n}(x)Y(x) \bigr)_{m} +\bigl(X_{n}^{\prime\prime}(x)Y(x),X_{n}(x)Y(x) \bigr)_{m}\bigr]\,dx. \end{aligned}$$
(3.9)
By \(\mathbf{H}_{X,A,Q}\geqslant I\),
$$ (\mathbf{H}_{\min}Y_{n},Y_{n}) \geq(Y_{n},Y_{n})= \int_{0}^{\infty }\bigl(X_{n}(x)Y(x),X_{n}(x)Y(x) \bigr)_{m}\,dx. $$
(3.10)
On the other hand, transforming the first part of (3.9), integrating by parts, and noting that \(X_{n}(x)\) has a compact support, we get
$$\begin{aligned} \int_{0}^{\infty}\bigl(2X_{n}^{\prime}(x)Y^{\prime}(x),X_{n}(x)Y(x) \bigr)_{m}\,dx =& \int_{0}^{\infty}2Y(x)^{T}X_{n}(x)^{T}X_{n}^{\prime}(x)Y^{\prime}(x)\,dx \\ =&\sum_{i=1}^{m} \int_{0}^{\infty}2y_{i}(x)\chi _{i,i}(x/n)\chi _{i,i}^{\prime}(x/n)y_{i}^{\prime}(x)\,dx \\ =&\sum_{i=1}^{m} \int_{0}^{\infty}\frac{1}{2}\bigl(y_{i}^{2}(x) \bigr)^{ \prime}\bigl(\chi_{i,i}^{2}(x/n) \bigr)^{\prime}\,dx \\ =&-\sum_{i=1}^{m} \int_{0}^{\infty}y_{i}^{2}(x)\bigl[ \chi _{i,i}^{\prime\prime}(x/n)\chi_{i,i}(x/n) \\ &{}+ \bigl( \chi_{i,i}^{\prime}(x/n) \bigr) ^{2} \bigr]\,dx, \end{aligned}$$
(3.11)
where \(\chi_{i,i}(x/n)\) denotes the ith component function of \(\chi _{i}(x/n)\), which is the ith column of the matrix \(X_{n}(x)\) (\(i=1,2,\ldots,m\)). The second part of (3.9) is as follows:
$$\begin{aligned} \int_{0}^{\infty}\bigl(X_{n}^{\prime\prime}(x)Y(x),X_{n}(x)Y(x) \bigr)_{m}\,dx =\sum_{i=1}^{m} \int_{0}^{\infty}y_{i}^{2}(x)\chi _{i,i}^{\prime \prime}(x/n)\chi_{i,i}(x/n)\,dx. \end{aligned}$$
(3.12)
By (3.9), (3.11), and (3.12) we get
$$\begin{aligned} (\mathbf{H}_{\min}Y_{n},Y_{n}) =& \sum_{i=1}^{m} \int _{0}^{\infty }y_{i}^{2}(x) \bigl( \chi_{i,i}^{\prime}(x/n) \bigr) ^{2}\,dx = \int_{0}^{\infty}\bigl(X_{n}^{\prime}(x)Y(x),X_{n}^{\prime}(x)Y(x) \bigr)_{m}\,dx. \end{aligned}$$
(3.13)
Therefore, by (3.10) and (3.13) we obtain
$$\begin{aligned} \int_{0}^{n/2}\bigl(Y(x),Y(x)\bigr)_{m}\,dx =& \int_{0}^{n/2}Y(x)^{T}Y(x)\,dx\leq \int_{0}^{\infty}Y(x)^{T}X_{n}(x)^{T}X_{n}(x)Y(x)\,dx \\ \leq& \int_{0}^{\infty}Y(x)^{T}X_{n}^{\prime}(x)^{T}X_{n}^{\prime }(x)Y(x)\,dx \\ \leq&\frac{C}{n^{2}} \int_{n/2}^{n}Y(x)^{T}Y(x)\,dx= \frac{C}{n^{2}} \int_{n/2}^{n}\bigl(Y(x),Y(x)\bigr)_{m}\,dx, \end{aligned}$$
(3.14)
where \(C:=\sup_{x\in[0,1]}|X^{\prime}(x)|\). Noting that \(Y\in L^{2}( \mathbb{R}_{+},\mathbb{C}^{m})\), inequality (3.14) yields \(Y=\theta\). This contradiction completes the proof. □
Before proceeding further, we need the following fact.
Lemma 5
If
$$\begin{aligned}& C_{0} :=\sup_{n\in\mathbb{N}} \int_{n}^{n+1}\bigl|\mu_{\max }\bigl(Q(t)\bigr)\bigr|\,d t< \infty, \end{aligned}$$
(3.15)
$$\begin{aligned}& C_{1} :=\sup_{n\in\mathbb{N}}\underset{x_{k}\in[ n,n+1]}{\sum} \bigl|\mu_{\max}(A_{k})\bigr|< \infty, \end{aligned}$$
(3.16)
then the forms
\(\mathfrak{q}\)
and
\(\mathbf{t}_{R}:=\mathbf{t}_{R}^{+}+ \mathbf{t}_{R}^{-}\)
are infinitesimally
\(\mathbf{t}_{0}\)-bounded, and hence the form
\(\mathbf{t}_{X,A,Q}\)
is closed and lower semibounded, and
\(\operatorname{Dom}(\mathbf{t}_{X,A,Q})= W^{1,2}(\mathbb{R}_{+},\mathbb{C}^{m})\)
algebraically and topologically.
Proof
By Lemma 3, for any \(\varepsilon>0\), we have the following inequality:
$$\begin{aligned} \Vert Y\Vert _{C(\mathbb{R}_{+},\mathbb{C}^{m})}^{2} \leq &\varepsilon \int_{n}^{n+1}\bigl(Y^{\prime}(t),Y^{\prime }(t) \bigr)_{m}\,dt+C_{\varepsilon} \int_{n}^{n+1}\bigl(Y(t),Y(t)\bigr)_{m}\,dt \\ \leq&\varepsilon\bigl\Vert Y^{\prime}\bigr\Vert _{W^{1,2}([n,n+1], \mathbb{C}^{m})}^{2}+C_{\varepsilon} \Vert Y\Vert _{L^{2}([n,n+1], \mathbb{C}^{m})}^{2}, \end{aligned}$$
(3.17)
where \(x\in[ n,n+1]\), and the constant \(C_{\varepsilon}>0\) does not depend on Y and \(n\in\mathbb{N}\). Combining (3.15) and (3.16) with (3.17), we obtain, for \(Y\in\operatorname{Dom} (\mathbf{t}_{0})=W_{0}^{1,2}(\mathbb{R} _{+},\mathbb{C}^{m})\),
$$\begin{aligned} & \int_{\mathbb{R}_{+}}\bigl(Q(x)Y(x),Y(x)\bigr)_{m}\,dx+\sum _{k=1}^{\infty }\bigl(A_{k}Y(x_{k}),Y(x_{k}) \bigr)_{m} \\ &\quad=\sum_{n=0}^{\infty }\biggl( \int_{n}^{n+1}\bigl(Q(x)Y(x),Y(x) \bigr)_{m}\,dx+\sum_{x_{k}\in[ n,n+1]} \bigl(A_{k}Y(x_{k}),Y(x_{k})\bigr)_{m} \biggr) \\ &\quad\leq\sum_{n=0}^{\infty} \int_{n}^{n+1}\bigl|\mu_{\max }\bigl(Q(x)\bigr)\bigr| \Vert Y\Vert^{2}\,dx+\sum_{x_{k}\in[ n,n+1]}\bigl| \mu_{\max }(A_{k})\bigr|\Vert Y\Vert^{2} \\ &\quad\leq(C_{0}+C_{1})\sum_{n=0}^{\infty} \Vert Y\Vert _{C[n,n+1]}^{2} \\ &\quad\leq(C_{0}+C_{1})\varepsilon \Vert Y\Vert _{W^{1,2}(\mathbb{R} _{+},\mathbb{C}^{m})}^{2}+(C_{0}+C_{1})C_{\varepsilon} \Vert Y\Vert _{L^{2}(\mathbb{R}_{+},\mathbb{C}^{m})}^{2}. \end{aligned}$$
Since \(\varepsilon>0\) is arbitrary, the forms \(\mathfrak{q}\) and \(\mathbf{t} _{R}\) are infinitesimally form bounded with respect to \(\mathbf {t}_{0}\). It remains to apply Lemma 2. □
We refer to \(A:=\{A_{k}\}_{k=1}^{\infty}\subset\mathbb{R}^{m\times m}\) again. We denote by \(l_{m}^{2}(\|A\|)\) the weighted space of m-dimensional vector sequences \(f=\{f_{k}\}_{k=1}^{\infty}\subset\mathbb{C}^{m}\), where \(\|A\|\) is any norm of a matrix A, and every element \(f_{k}\) of the sequence is an m-dimensional column vector. Let \(f_{k}=(f_{k,1},f_{k,2}, \ldots,f_{k,m})^{T}\); then \(\{f_{k,i}\}_{k=1}^{\infty}\in l^{2}\), \(i=1,2,\ldots,m\), and hence \(\|f_{k,i}\|^{2}=\sum_{k=1}^{\infty }|f_{k,i}|^{2}<\infty\) and \(\|f\|^{2}=\sum_{i=1}^{m}\|f_{k,i}\|^{2}\). The space \(l_{m}^{2}(\|A\|)\) equipped with the weighted norm \(\|f\|_{l_{m}^{2}(\|A\|)}=(\sum_{i=1}^{m}\sum_{k=1}^{ \infty}\|A_{k}\||f_{k,i}|^{2})^{\frac{1}{2}}\) is a Banach space.
Lemma 6
Assume that
\(Q(x)\geq0\)
and
\(A_{k}\geq0\)
for
\(k=1,2,\ldots \) . Then the form
\(\mathbf{t}_{X,A,Q}\)
is nonnegative and closed.
Proof
It is obvious that the form \(\mathbf{t}_{X,A,Q}\) is nonegative if \(Q(x)\geq0\) and \(A_{k}\geq0\). Now we prove the closeness of the form \(\mathbf{t}_{X,A,Q}\). Let us equip \(\mathfrak{H}_{X,A,Q}= \operatorname{Dom}(\mathbf{t} _{X,A,Q})\) with the norm
$$ \| Y\|_{\mathfrak{H}_{X,A,Q}}^{2}=\mathbf{t}_{Q}[Y]+\mathbf{t} _{R}[Y]+\| Y\|_{L^{2}(\mathbb{R}_{+},\mathbb{C}^{m})}^{2}. $$
Let \(\{Y_{n}\}_{n=1}^{\infty}\) be a Cauchy sequence in \(\mathfrak{H} _{X,A,Q} \). Since \(W_{0}^{1,2}(\mathbb{R}_{+},\mathbb{C}^{m})\) and \(l_{m}^{2}(\|A\|)\) are Hilbert spaces, there exist \(Y\in W_{0}^{1,2}(\mathbb{ R}_{+},\mathbb{C}^{m})\) and
$$ \{\xi_{k}\}_{k=1}^{\infty}\in l_{m}^{2}\bigl( \|A\|\bigr) $$
such that
$$ \lim_{n\rightarrow\infty}\| Y_{n}-Y\| _{W_{0}^{1,2}(\mathbb{R} _{+},\mathbb{C}^{m})}=0 $$
and
$$ \lim_{n\rightarrow\infty}\sum_{k=1}^{\infty } \|A_{k}\|\bigl|Y_{n}(t_{k})-\xi_{k}\bigr|^{2}=0, $$
where \(\xi_{i}\) and \(Y_{n}\) are m-dimensional vectors. Since the space \(W_{0}^{1,2}(\mathbb{R}_{+},\mathbb{C}^{m})\) is continuously embedded into \(C( \mathbb{R}_{+},\mathbb{C}^{m})\), the Banach space of bounded continuous functions on \(\mathbb{R}_{+}\). Therefore,
$$ \lim_{n\rightarrow\infty}Y_{n}(x)=Y(x), $$
and then \(Y\in\mathfrak{H}_{X,A,Q}\) and
$$ \lim_{n\rightarrow\infty}\| Y_{n}-Y\|_{\mathfrak{H} _{X,A,Q}}=0. $$
In addition, since \(Q(x)\geq0\) and \(\{A_{k}\}_{k=1}^{\infty}\geq0\), \(\mathfrak{H}_{X,A,Q}\) is a Hilbert space with the inner product
$$\begin{aligned} (Y,Z)_{\mathfrak{H}_{X,A,Q}}={}& \int_{0}^{\infty}\bigl(Y^{\prime},Z^{\prime } \bigr)_{m}\,dx+ \int_{0}^{\infty}\bigl(\bigl(Q(x)+E\bigr)Y,Z \bigr)_{m}\,dx\\ &{}+\sum_{k=1}^{\infty } \bigl(A_{k}Y(x_{k}),Z(x_{k})\bigr)_{m}, \end{aligned}$$
where E is the identity matrix. Then the form \(\mathbf{t}_{X,A,Q}\) is closed. □
Lemma 7
If the form
\(\mathbf{t}_{X,A,Q}\)
is lower semibounded, then the set
\(\operatorname{Dom}(H_{X,A,Q}^{0})\)
is a core of the form
\(\mathbf{t}_{X,A,Q}\).
Proof
We need to show that \(\operatorname{Dom}(H_{X,A,Q}^{0})\) is dense in \(\operatorname{Dom}(\mathbf{t} _{X,A,Q})\) with respect to the norm \(\| Y\|_{\mathfrak{H} _{X,A,Q}}^{2}=\mathbf{t}_{Q}[Y]+\mathbf{t}_{R}[Y]+\| Y\|_{L^{2}( \mathbb{R}_{+},\mathbb{C}^{m})}^{2}\). Let \(D_{\min}^{\prime}\) be the linear span of \(C^{\infty}\) functions with compact support in a single interval \((x_{i-1},x_{i})\), \(i\in\mathbb{N}\). Each function \(f_{i}\in C_{0}^{\infty}((x_{i-1},x_{i}),\mathbb{C}^{m})\) can be extended to \([0,\infty)\), and the extended function
$$ \tilde{f}_{i}(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} f_{i}(x), & x\in(x_{i-1},x_{i}), \\ \theta, & x\in[0,\infty)\backslash(x_{i-1},x_{i})\end{array}\displaystyle \right . $$
belongs to \(D_{\min}^{\prime}\subset \operatorname{Dom}(H_{X,A,Q}^{0})\).
We need to prove that for \(u\in \operatorname{Dom}(\mathbf{t}_{X,A,Q})\) and for all \(f\in \operatorname{Dom}(H_{X,A,Q}^{0})\),
$$ (u,f)= \int_{0}^{\infty}\bigl(u^{\prime},f^{\prime} \bigr)_{m}\,dx+ \int_{0}^{\infty }\bigl(\bigl(Q(x)+E\bigr)u,f \bigr)_{m}\,dx+\sum_{k=1}^{\infty } \bigl(A_{k}u(x_{k}),f(x_{k})\bigr)_{m}=0 $$
(3.18)
implies that \(u=0\). Equation (3.18) holds for all \(f\in \operatorname{Dom} (H _{X,A,Q}^{0})\), and thus, for each interval \((x_{i-1},x_{i})\), the equation
$$ \int_{x_{i-1}}^{x_{i}}\bigl(u^{\prime},f_{i}^{\prime } \bigr)_{m}+ \int_{x_{i-1}}^{x_{i}}\bigl((Q+E)u,f_{i} \bigr)_{m}=0 $$
holds for all \(f_{i}\in C_{0}^{\infty}((x_{i-1},x_{i}),\mathbb{C}^{m})\). Then \(u^{\prime\prime}=(Q+E)u\) on each interval \((x_{i-1},x_{i})\) in the sense of distributions.
Since equation (3.18) holds for all \(f\in \operatorname{Dom}(H _{X,A,Q}^{0}) \), integrating by parts, we get \(u\in \operatorname{Dom}((H _{X,A,Q}^{0})^{\ast})\). Then by a similar method as in Lemma 4, the only function \(u\in \operatorname{Dom}(\mathbf{t}_{X,A,Q})\) satisfying equation (3.18) is \(u=0\). So we obtain that the set \(\operatorname{Dom} (H_{X,A,Q}^{0})\) is a core of the form \(\mathbf{t}_{X,A,Q}\). □
Lemma 8
If the form
\(\mathbf{t}_{X,A,Q}\)
is closed and
\(\mathbf{H}_{X,A,Q}\)
is self-adjoint, then the operator associated with the form
\(\mathbf{t} _{X,A,Q}\)
coincides with
\(\mathbf{H}_{X,A,Q}=(\mathbf{H}_{X,A,Q})^{\ast}\).
Proof
Integrating by parts, we can get from (1.2) that \(\operatorname{Dom}(\mathbf{H} _{X,A,Q}^{0})\subset \operatorname{Dom}(\mathbf{t}_{X,A,Q})\). For \(u,v\in \operatorname{Dom}(\mathbf{H} _{X,A,Q}^{0})\),
$$\begin{aligned} \mathbf{t}_{X,A,Q}[u,v] =& \int_{0}^{\infty}\bigl[\bigl(u^{\prime},v^{\prime } \bigr)_{m}+\bigl(Q(x)u,v\bigr)_{m}\bigr]\,dx+\sum _{k=1}^{\infty }\bigl(A_{k}u(x_{k}),v(x_{k}) \bigr)_{m} \\ =&\bigl(\mathbf{H}_{X,A,Q}^{0}u,v\bigr). \end{aligned}$$
Assume that T is the self-adjoint operator associated with \(\mathbf {t}_{X,A,Q}\). Since \(\operatorname{Dom}(H_{X,A,Q}^{0})\) is a core of the form \(\mathbf{t} _{X,A,Q}\) by Lemma 7, by the first representation theorem we have \(u\in \operatorname{Dom}(T)\) and \(Tu=\mathbf{H}_{X,A,Q}^{0}u\), and hence \(T=\mathbf{H} _{X,A,Q}\). □
Lemma 9
Assume that
\(Q(x)\geq0\), \(x\in\mathbb{R}_{+}\). Let the forms
\(\mathbf{t}_{Q}\)
and
\(\mathbf{t}_{R}^{+}\)
be defined by (3.3) and (3.5). Then the form
\(\mathbf{t}_{X,A^{+},Q}=\mathbf{t}_{Q}+\mathbf{t}_{R}^{+}\)
is nonnegative and closed.
Proof
The proof is similar to that of Lemma 6. □
Lemma 10
Let
\(q_{ij}\in L_{\mathrm{loc}}^{1}( \mathbb{R} _{+})\), and let the forms
\(\mathbf{t}_{Q}\)
and
\(\mathbf{t}_{R}^{\pm}\)
be given by (3.3) and (3.4).
-
(i)
If
\(\mu_{\max}^{-}(Q(x))\)
and
\(\mu_{\max}^{-}(A_{k})\)
satisfy (3.7), then the form
\(\mathbf{t}_{X,A,Q}\)
is closed and lower semibounded. Moreover, \(\operatorname{Dom}(\mathbf{t}_{X,A,Q})=\operatorname{Dom} (\mathbf{t}_{X,A^{+},Q^{+}})\), and the operator associated with
\(\mathbf{t}_{X,A,Q}\)
coincides with
\(\mathbf{H} _{X,A,Q}\).
-
(ii)
If
\(\mu_{\max}(Q(x))=\mu_{\max}^{-}(Q(x))\)
and
\(\mu _{\max }(A_{k})=\mu_{\max}^{-}(A_{k})\), then conditions (3.7) are necessary and sufficient for the form
\(\mathbf{t}_{X,A,Q}\)
to be lower semibounded.
Proof
(i) By the Lemma 9 the form \(\mathbf{t}_{X,A^{+},Q }\)is closed. Moreover, by Lemma 5, \(\mathbf{t}_{Q}^{-}\) and \(\mathbf{t} _{R}^{-} \) are infinitesimally \(\mathbf{t}_{Q}^{+}\)-bounded and hence infinitesimally \(\mathbf{t}_{X,A^{+},Q^{+}}\)-bounded. Using Lemma 2, we complete the proof.
(ii) Sufficiency is implied by (i). Let us prove necessity. Assume the converse, that is, the second condition in (3.7) does not hold. Then there exists \(\{n_{j}\}_{1}^{\infty}\subset\mathbb{N}\) such that
$$ \sum_{x_{k}\in[ n_{j},n_{j}+1]}\mu_{\max}^{-}(A_{k})< -n_{j}, \quad j\in \mathbb{N.} $$
(3.19)
Define \(\varphi_{j}(x):=\varphi(x-n_{j})\), \(j\in\mathbb{N}\), where \(\varphi\in C_{0}^{\infty}(\mathbb{ \mathbb{R}}_{+},\mathbb{C}^{m})\) is such that \(\sup p\varphi\in(-\frac{1}{2},\frac{3}{2})\), \(0\leq \|\varphi \|\leq1\), and \(\varphi(x)=e_{1}\), \(x\in[0,1]\), where \(e_{1}\) denotes the m-dimensional column unit vector whose first component is 1. By (3.19) and the form (3.6) we obtain \(\mathbf{t} _{X,A,Q}[\varphi_{j}]\leq-n_{j}+\|\varphi\|_{W^{1,2}(\mathbb{R}_{+}, \mathbb{C}^{m})}^{2}\), and we note that when \(n_{j}\rightarrow\infty \), \(\mathbf{t}_{X,A,Q}\) is not lower semibounded. Hence, the contradiction finishes the proof. □
Corollary 1
If
$$ \inf_{k\in\mathbb{N}}\frac{\mu_{\max}^{-}(A_{k})}{d_{k}}>-\infty,\qquad d_{k}:=x_{k}-x_{k-1}, $$
(3.20)
then the form
\(\mathbf{t}_{X,A}=\mathbf{t}_{X,A,0}\)
is closed, and lower semibounded, and
\(\operatorname{Dom}(\mathbf{t}_{X,A})=W_{0}^{1,2}(\mathbb {R}_{+},\mathbb{C} ^{m})\). Moreover, the operator associated with the form
\(\mathbf{t}_{X,A}\)
coincides with
\(\mathbf{H}_{X,A}\).
Proof
Clearly, (3.20) yields (3.15) and (3.16). Lemma 10 completes the proof. □