﻿ 两个微分算子乘积的自伴性

# 两个微分算子乘积的自伴性The Self-Adjointness of the Product of Two Differential Operators

Abstract: In this paper, we study the necessary and sufficient conditions for the self-adjointness of the product of the two first-order differential operators on closed interval, and the self-adjoint vertex conditions of the product of two first-order local differential operators on metric graph are given. Based on the self-adjointness of the product operator on closed interval, the necessary and sufficient conditions which make the product operators be self-adjoint operators are obtained by using the self-adjoint vertex conditions of the higher-order differential operator on metric graph.

1. 引言

2. 闭区间上的两个微分算子乘积的自伴性

2.1. 基础知识

${\left(f,g\right)}_{{L}^{2}\left[a,b\right]}={\int }_{a}^{b}f\left(x\right)\stackrel{¯}{g\left(x\right)}\text{ }\text{d}x$, ${‖f‖}_{{L}^{2}\left[a,b\right]}={\left({\int }_{a}^{b}{|f\left(x\right)|}^{2}\text{d}x\right)}^{\frac{1}{2}}$.

$l\left(f\right)=\underset{j=0}{\overset{n}{\sum }}{c}_{j}\left(x\right){f}^{\left(j\right)}\left(x\right)$,

${\left(l\left(f\right),g\right)}_{{L}^{2}\left[a,b\right]}-{\left(f,l\left(g\right)\right)}_{{L}^{2}\left[a,b\right]}={\left[f,g\right]}_{n}\left(b\right)-{\left[f,g\right]}_{n}\left(a\right)$,

${\left[f,g\right]}_{n}\left(x\right)={\left(\stackrel{¯}{{C}_{g}}\left(x\right)\right)}^{\top }{Q}_{n}\left(x\right){C}_{f}\left(x\right)$,

${\left[f,g\right]}_{n}$ 表示关于 $l\left(f\right)$ 的Lagrange双线性型，

${C}_{g}{\left(x\right)}^{\top }=\left(g\left(x\right),{g}^{\prime }\left(x\right),\cdots ,{g}^{\left(n-1\right)}\left(x\right)\right)$, ${\left({C}_{f}\left(x\right)\right)}^{\top }=\left(f\left(x\right),{f}^{\prime }\left(x\right),\cdots ,{f}^{\left(n-1\right)}\left(x\right)\right)$.

${Q}_{n}\left(x\right)$ 表示关于 ${\left[f,g\right]}_{n}\left(x\right)$ 的Lagrange双线性矩阵。令

${Q}_{n}\left(x\right)={\left({d}_{ij}\left(x\right)\right)}_{n×n}$,

${d}_{ij}\left(x\right)=\left\{\begin{array}{l}\underset{h=i}{\overset{n-j-1}{\sum }}{\left(-1\right)}^{h}\left(\begin{array}{c}h\\ i\end{array}\right)\frac{{\text{d}}^{h-i}{c}_{j+h+1}\left(x\right)}{\text{d}{x}^{h-i}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le j+i\le n-1,\\ 0,\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n-1\le j+i\le \text{2}n-\text{2},\end{array}$ (1)

${Q}_{n}\left(x\right)$ 满足以下性质

${Q}_{n}^{*}\left(x\right)=-{Q}_{n}\left(x\right)$, ${\left({Q}_{n}^{-1}\left(x\right)\right)}^{*}=-{Q}_{n}^{-1}\left(x\right)$.

${L}_{max}\left(f\right)$$\left[a,b\right]$ 上由微分形式 $l\left(f\right)$ 生成的最大算子，其定义域为

${D}_{max}\left(l\right)=\left\{f\in {L}^{2}\left[a,b\right]:{f}^{\left(n-1\right)}\in AC\left[a,b\right],l\left(f\right)\in {L}^{2}\left[a,b\right]\right\}$,

2.2. 闭区间上两个一阶算子乘积的自伴性

$\left\{\begin{array}{l}{L}_{1}\left(f\right)=l\left(f\right)=i{f}^{\prime },\hfill \\ \left({a}_{1}+{a}_{2}i\right)f\left(a\right)+\left({b}_{1}+{b}_{2}i\right)f\left(b\right)=0,\hfill \end{array}$ (2)

$\left\{\begin{array}{l}{L}_{2}\left(f\right)=l\left(f\right)=i{f}^{\prime },\hfill \\ \left({a}_{3}+{a}_{4}i\right)f\left(a\right)+\left({b}_{3}+{b}_{4}i\right)f\left(b\right)=0,\hfill \end{array}$ (3)

$\left\{\begin{array}{l}L\left(f\right)={l}^{2}\left(f\right)=-{f}^{″},\hfill \\ \left({a}_{1}+{a}_{2}i\right)f\left(a\right)+\left({b}_{1}+{b}_{2}i\right)f\left(b\right),\hfill \\ \left({a}_{3}+{a}_{4}i\right)if\left(a\right)+\left({b}_{3}+{b}_{4}i\right)if\left(b\right)=0,\hfill \end{array}$ (4)

$\left\{\begin{array}{l}L\left(f\right)={l}^{2}\left(f\right)=-{f}^{″},\hfill \\ A{C}_{f}\left(a\right)+B{C}_{f}\left(b\right)=0,\hfill \end{array}$ (5)

$A=\left[\begin{array}{cc}{a}_{1}+{a}_{2}i& 0\\ 0& {a}_{3}i-{a}_{4}\end{array}\right]$, $B=\left[\begin{array}{cc}{b}_{1}+{b}_{2}i& 0\\ 0& {b}_{3}i-{b}_{4}\end{array}\right]$

${\left({C}_{f}\left(a\right)\right)}^{\top }=\left(f\left(a\right),{f}^{\prime }\left(a\right)\right)$, ${C}_{f}{\left(b\right)}^{\top }=\left(f\left(b\right),{f}^{\prime }\left( b \right)\right)$

${Q}_{2}\left(x\right)=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$.

$\left\{\begin{array}{l}{a}_{2}{a}_{3}-{a}_{1}{a}_{4}={b}_{2}{b}_{3}-{b}_{1}{b}_{4}\hfill \\ {a}_{1}{a}_{3}+{a}_{2}{a}_{4}={b}_{1}{b}_{3}+{b}_{2}{b}_{4}\hfill \end{array}$

$\text{Rank}\left(A\oplus B\right)=\text{2}$.

$A{Q}_{\text{2}}^{-1}\left(a\right){A}^{*}=B{Q}_{\text{2}}^{-1}\left(b\right){B}^{*}$$\text{Rank}\left(A\oplus B\right)=\text{2}$.

$\begin{array}{c}A{Q}_{2}^{-1}\left(a\right){A}^{*}=\left[\begin{array}{cc}{a}_{1}+{a}_{2}i& 0\\ 0& {a}_{3}i-{a}_{4}\end{array}\right]\cdot \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\cdot \left[\begin{array}{cc}{a}_{1}-{a}_{2}i& 0\\ 0& -{a}_{3}i-{a}_{4}\end{array}\right]\\ =\left[\begin{array}{cc}0& {a}_{2}{a}_{3}-{a}_{1}{a}_{4}-\left({a}_{1}{a}_{3}+{a}_{2}{a}_{4}\right)i\\ {a}_{1}{a}_{4}-{a}_{2}{a}_{3}-\left({a}_{2}{a}_{4}+{a}_{1}{a}_{3}\right)i& 0\end{array}\right]\end{array}$

$\begin{array}{c}B{Q}_{2}^{-1}\left(b\right){B}^{*}=\left[\begin{array}{cc}{b}_{1}+{b}_{2}i& 0\\ 0& {b}_{3}i-{b}_{4}\end{array}\right]\cdot \left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\cdot \left[\begin{array}{cc}{b}_{1}-{b}_{2}i& 0\\ 0& -{b}_{3}i-{b}_{4}\end{array}\right]\\ =\left[\begin{array}{cc}0& {b}_{2}{b}_{3}-{b}_{1}{b}_{4}-\left({b}_{1}{b}_{3}+{b}_{2}{b}_{4}\right)i\\ {b}_{1}{b}_{4}-{b}_{2}{b}_{3}-\left({b}_{2}{b}_{4}+{b}_{1}{b}_{3}\right)i& 0\end{array}\right]\end{array}$,

3. 闭区间上的两个微分算子乘积的自伴性

3.1. 基础知识

${\left(f,g\right)}_{{L}^{2}\left(G\right)}=\underset{k}{\sum }{\int }_{{a}_{k}}^{{b}_{k}}f\left(x\right)\stackrel{¯}{g\left(x\right)}\text{ }\text{d}x$${‖f‖}_{{L}_{2}\left(G\right)}={\underset{k}{\sum }{‖f‖}_{{L}^{{}_{2}}\left[{a}_{k},{b}_{k}\right]}=\underset{k}{\sum }\left({\int }_{{a}_{k}}^{{b}_{k}}{|f\left(x\right)|}^{2}\text{d}x\right)}^{\frac{1}{2}}$.

$\left\{\begin{array}{l}{T}_{max}\left(f\right)=t\left(f\right),f\in {D}_{max}\left(t\right),\\ {D}_{max}\left(t\right)=\left\{f\in {L}^{2}\left(G\right):\forall k\in N,{f}_{k},{{f}^{\prime }}_{k},\cdots ,{f}_{k}^{\left(n-1\right)}\in C\left[{a}_{k},{b}_{k}\right],{f}_{k}^{\left(n-1\right)}\in AC\left[{a}_{k},{b}_{k}\right],t\left(f\right)\in {L}^{2}\left(G\right)\right\}\end{array}$

$M:=\left\{\phi \in {C}_{comp}^{\infty }\left(G\right):\forall v\in V,\text{\hspace{0.17em}}\exists 领域{U}_{v},\text{\hspace{0.17em}}使得\phi 在{U}_{v}中为常值函数\right\}$.

${\left(t\left(f\right),g\right)}_{{L}^{2}\left(G\right)}-{\left(f,t\left(g\right)\right)}_{{L}^{2}\left(G\right)}={\left[f,g\right]}_{v}$,

${\stackrel{^}{f}}_{ni+j-\left(n-1\right)}={f}^{\left(j\right)}\left({\alpha }_{i}\right)$$j=0,\cdots ,n-1,i=1,\cdots ,\delta \left(v\right)$.

${\left[f,g\right]}_{v}=\stackrel{^}{g}*{S}_{v}\stackrel{^}{f}$.

$\underset{j,i}{\sum }{b}_{j,i}{\stackrel{^}{f}}_{ni+j-\left(n-1\right)}=0$, $j=0,\cdots ,n-1,i=1,\cdots ,\delta \left(v\right)$.

${B}_{v}{\left({S}_{v}^{*}\right)}^{-1}\left({B}_{v}^{*}\right)=0$. (6)

3.2. 闭区间上两个一阶算子乘积的自伴性

$t\left(f\right)=i{f}^{\prime }$.

${Q}_{1}$ 表示关于 $t\left(f\right)$ 的Lagrange双线性矩阵， ${S}_{1}$ 表示由 $t\left(f\right)$ 生成的局部算子的边值矩阵。计算可得

${Q}_{1}\left(x\right)=\left[i\right]$，则 ${S}_{1}={\left[\begin{array}{llllll}i\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \ddots \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & i\hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & -i\hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \ddots \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & -i\hfill \end{array}\right]}_{\delta \left(v\right)×\delta \left( v \right)}$

$\left\{\begin{array}{l}{T}_{\text{1}}\left(f\right)=t\left(f\right)=i{f}^{\prime },\hfill \\ {B}_{\text{1}}\stackrel{^}{f}=0,\hfill \end{array}$ (7)

$\left\{\begin{array}{l}{T}_{2}\left(f\right)=t\left(f\right)=i{f}^{\prime },\hfill \\ {B}_{2}\stackrel{^}{f}=0,\hfill \end{array}$ (8)

${t}^{2}\left(f\right)=t\left(t\left(f\right)\right)=-{f}^{″}$.

${Q}_{2}\left(x\right)=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$.

$\left[i{f}^{\prime }\left(x\right)\right]=\left[\begin{array}{ll}0\hfill & i\hfill \end{array}\right]\left[\begin{array}{c}f\left(x\right)\\ {f}^{\prime }\left(x\right)\end{array}\right]$,

${H}_{1}\left(x\right)=\left[0\right]$, ${H}_{2}\left(x\right)=\left[i\right]$.

$i\ast \left(-1\right)=-i$, $1\ast \left(-i\right)=-i$, $0=-1\ast 0+0\ast \left(-1\right)$.

$\left[\begin{array}{c}i{f}^{\prime }\left({\alpha }_{1}\right)\\ i{f}^{\prime }\left({\alpha }_{2}\right)\\ ⋮\\ i{f}^{\prime }\left({\alpha }_{\delta \left(v\right)}\right)\end{array}\right]={\left[\begin{array}{ccccc}H\left({\alpha }_{1}\right)& & & & \\ & \ddots & & & \\ & & H\left({\alpha }_{\delta \left(v\right)-s}\right)& & \\ & & & \ddots & \\ & & & & \\ & & & & H\left({\alpha }_{\delta \left(v\right)}\right)\end{array}\right]}_{\delta \left(v\right)×2\delta \left(v\right)}\left[\begin{array}{c}f\left({\alpha }_{1}\right)\\ {f}^{\prime }\left({\alpha }_{1}\right)\\ f\left({\alpha }_{2}\right)\\ {f}^{\prime }\left({\alpha }_{2}\right)\\ ⋮\\ f\left({\alpha }_{\delta \left(v\right)}\right)\\ {f}^{\prime }\left({\alpha }_{\delta \left(v\right)}\right)\end{array}\right]$

$\left\{\begin{array}{l}{T}_{1}\left(f\right)={t}^{2}\left(f\right)=-{f}^{″},\hfill \\ {B}_{1}{\left(f\left({\alpha }_{1}\right),f\left({\alpha }_{2}\right),\cdots ,f\left({\alpha }_{\delta \left(v\right)}\right)\right)}^{\top }=0,\hfill \\ {B}_{2}{\left(i{f}^{\prime }\left({\alpha }_{1}\right),i{f}^{\prime }\left({\alpha }_{2}\right),\cdots ,i{f}^{\prime }\left({\alpha }_{\delta \left(v\right)}\right)\right)}^{\top }=0,\hfill \end{array}$ (9)

$\left\{\begin{array}{l}T\left(f\right)={t}^{2}\left(f\right)=-{f}^{″},\hfill \\ B{\left(f\left({\alpha }_{1}\right),{f}^{\prime }\left({\alpha }_{1}\right),f\left({\alpha }_{2}\right),{f}^{\prime }\left({\alpha }_{2}\right),\cdots ,f\left({\alpha }_{\delta \left(v\right)}\right),{f}^{\prime }\left({\alpha }_{\delta \left(v\right)}\right)\right)}^{\top }=0,\hfill \end{array}$ (10)

$B={\left[\begin{array}{cccccc}{H}_{3}& 0& {H}_{4}& \cdots & {H}_{\delta \left(v\right)+2}& 0\\ 0& {B}_{2}\stackrel{˜}{{H}_{2}\left({\alpha }_{1}\right)}& 0& \cdots & 0& {B}_{2}\stackrel{˜}{{H}_{2}\left({\alpha }_{\delta \left(v\right)}\right)}\end{array}\right]}_{\left({K}_{1}\left(v\right)+{K}_{2}\left(v\right)\right)×2\delta \left( v \right)}$

${B}_{1}$$\left({K}_{1}\left(v\right)+{K}_{2}\left(v\right)\right)×2\delta \left(v\right)$ 矩阵，其中 ${H}_{3},{H}_{4},\cdots ,{H}_{\delta \left(v\right)+2}$ 都是 ${K}_{1}\left(v\right)×1$ 矩阵，并且 ${B}_{1}=\left[\begin{array}{llll}{H}_{3}\hfill & {H}_{4}\hfill & \cdots \hfill & {H}_{\delta \left(v\right)+2}\hfill \end{array}\right]$$\stackrel{˜}{{H}_{2}\left({\alpha }_{j}\right)}$ 表示 $\delta \left(v\right)×1$ 矩阵。该矩阵中的第j行元素为i，其余元素为零， $j=1,2,\cdots ,\delta \left(v\right)$。矩阵 $B$ 的共轭转置为

${B}^{*}={\left[\begin{array}{cc}{H}_{3}^{*}& 0\\ 0& \stackrel{˜}{{H}_{2}^{*}\left({\alpha }_{1}\right)}{B}_{2}^{*}\\ {H}_{4}^{*}& 0\\ ⋮& ⋮\\ {H}_{\delta \left(v\right)+2}^{*}& 0\\ 0& \stackrel{˜}{{H}_{2}^{*}\left({\alpha }_{\delta \left(v\right)}\right)}{B}_{2}^{*}\end{array}\right]}_{2\delta \left(v\right)×\left({K}_{1}\left(v\right)+{K}_{2}\left( v \right)\right)}$

${\left[f,g\right]}_{v}\left(x\right)={\stackrel{^}{f}}^{\top }{S}_{2}\stackrel{^}{g}$,

${S}_{2}={\left[\begin{array}{cccccc}{Q}_{2}\left({\alpha }_{1}\right)& & & & & \\ & \ddots & & & & \\ & & {Q}_{2}\left({\alpha }_{s}\right)& & & \\ & & & -{Q}_{2}\left({\alpha }_{s+1}\right)& & \\ & & & & \ddots & \\ & & & & & -{Q}_{2}\left({\alpha }_{\delta \left(v\right)}\right)\end{array}\right]}_{2\delta \left(v\right)×2\delta \left( v \right)}$

${B}_{1}{\left({S}_{2}^{*}\right)}^{-1}{B}_{2}^{*}=0$${B}_{2}{\left({S}_{2}^{*}\right)}^{-1}{B}_{1}^{*}=0$${K}_{1}\left(v\right)+{K}_{2}\left(v\right)=\delta \left(v\right)$.

$\begin{array}{l}{B}_{1}{\left({S}_{2}^{*}\right)}^{-1}{B}_{1}^{*}\\ =\left[\begin{array}{cccccc}{H}_{3}& 0& {H}_{4}& \cdots & {H}_{\delta \left(v\right)+2}& 0\\ 0& {B}_{2}\stackrel{˜}{{H}_{2}\left({\alpha }_{1}\right)}& 0& \cdots & 0& {B}_{2}\stackrel{˜}{{H}_{2}\left({\alpha }_{\delta \left(v\right)}\right)}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdot \left[\begin{array}{cccccc}{\left({Q}_{2}^{*}\right)}^{-1}\left({\alpha }_{1}\right)& & & & & \\ & \ddots & & & & \\ & & {\left({Q}_{2}^{*}\right)}^{-1}\left({\alpha }_{s}\right)& & & \\ & & & {\left(-{Q}_{2}^{*}\right)}^{-1}\left({\alpha }_{s+1}\right)& & \\ & & & & \ddots & \\ & & & & & {\left(-{Q}_{2}^{*}\right)}^{-1}\left({\alpha }_{\delta \left(v\right)}\right)\end{array}\right]\cdot \left[\begin{array}{cc}{H}_{3}^{*}& 0\\ 0& \stackrel{˜}{{H}_{2}^{*}\left({\alpha }_{1}\right)}{B}_{2}^{*}\\ {H}_{4}^{*}& 0\\ ⋮& ⋮\\ {H}_{\delta \left(v\right)+2}^{*}& 0\\ 0& \stackrel{˜}{{H}_{2}^{*}\left({\alpha }_{\delta \left(v\right)}\right)}{B}_{2}^{*}\end{array}\right]\\ ={\left[\begin{array}{cc}0& {B}_{1}{\left({S}_{1}^{*}\right)}^{-1}{B}_{2}^{*}\\ {B}_{2}{\left({S}_{1}^{*}\right)}^{-1}{B}_{1}^{*}& 0\end{array}\right]}_{\left({K}_{1}\left(v\right)+{K}_{2}\left(v\right)\right)×\left({K}_{1}\left(v\right)+{K}_{2}\left( v \right)}\end{array}$

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