Symmetry of a regular
Sturm-Liouville eigenvalue problem
- Notation
-
-
For a matrix $M$ by $M^\top$ we denote the matrix transpose of $M$.
-
In definite integrals instead of $\int_a^b f(x) dx$ we briefly write $\int_a^b f\, dx.$
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By $\mathbb{N}$ we denote the set of positive integers. Let $k \in \mathbb{N}.$ By $0_k$ we denote the square $k\!\times\!k$ matrix whose all entries equal to zero. By $I_k$ we denote the square $k\!\times\!k$ matrix whose diagonal entries equal to $1$ and all nondiagonal entries equal to zero; this matrix is called identity matrix.
- I. Linear algebra preliminaries
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A linear algebra statement. Let $B$ be a $k\!\times\!(2k)$ matrix of full rank $k$ and let $Q$ be a $(2k)\!\times\!(2k)$ invertible antisymmetric matrix (that is $Q^{\top} = -Q$). Assume that $B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k}$. Then ${\mathbf y}^\top\!Q{\mkern+1.5mu}{\mathbf x} = 0$ whenever $B\mkern+1.5mu{\mathbf x} = 0$ and $B{\mkern+1.5mu}{\mathbf y} = 0$.
Proof. Assume that $B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k}.$ This equality implies that the column space $\operatorname{Col}\bigl( Q^{-1}{\mkern-2mu}B^\top \bigr)$ is a subspace of the null space $\operatorname{Nul}(B),$ that is
\begin{equation*}
\operatorname{Col} \bigl( Q^{-1}{\mkern-2mu}B^\top \bigr) \subseteq \operatorname{Nul}(B).
\end{equation*}
By the Rank-Nullity theorem we have
\begin{equation*}
\operatorname{rank}(B) + \dim \operatorname{Nul}(B) = 2k.
\end{equation*}
Hence, since $\operatorname{rank}(B) = k$, we deduce that $\dim \operatorname{Nul}(B) =k.$ Since the $\operatorname{rank}(B) = k$, we have that $\operatorname{rank}\bigl(B^\top\bigr) = k$ as well. Further, since the matrix $Q^{-1}$ is invertible, the rank of $Q^{-1}{\mkern-2mu}B^\top$ is also $k.$ Thus,
\begin{equation*}
\dim \operatorname{Col} \bigl( Q^{-1}{\mkern-2mu}B^\top \bigr) = \dim \operatorname{Nul}(B).
\end{equation*}
The preceding equality of dimensions, together with the inclusion
\begin{equation*}
\operatorname{Col} \bigl( Q^{-1}{\mkern-2mu}B^\top \bigr) \subseteq \operatorname{Nul}(B),
\end{equation*}
yields
\begin{equation*}
\operatorname{Col} \bigl( Q^{-1}{\mkern-2mu}B^\top \bigr) = \operatorname{Nul}(B).
\end{equation*}
Now assume that ${\mathbf x}, {\mathbf y} \in \operatorname{Nul}(B)$. By the preceding displayed set equality, we conclude that there exist vectors ${\mathbf u}, {\mathbf v} \in {\mathbb R}^k$ such that ${\mathbf x} = Q^{-1}{\mkern-2mu}B^\top {\mathbf u}$ and ${\mathbf y} = Q^{-1}{\mkern-2mu}B^\top {\mathbf v}$. Now calculate
\[
{\mathbf y}^\top\!Q{\mkern+1.5mu}{\mathbf x} = \bigl(Q^{-1}{\mkern-2mu}B^\top {\mathbf v} \bigr)^\top Q \bigl(Q^{-1}{\mkern-2mu}B^\top {\mathbf u}\bigr) =
{\mathbf v}^\top\! B \bigl(Q^\top\bigr)^{-1} Q Q^{-1}{\mkern-2mu}B^\top {\mathbf u} = {\mathbf v}^\top\! B{\mkern+1mu}Q^{-1}{\mkern-1mu}B^\top {\mathbf u} = 0.
\]
In the last step we used the assumption $B{\mkern+1mu}Q^{-1}{\mkern-1mu}B^\top = 0_{k}$.
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The preceding linear algebra statement formulated as an equivalence. (I post this here since it is interesting linear algebra and it is related to the first part. It is not directly related to Sturm-Liouville problem.)
Let $A$ and $B$ be $k\!\times\!(2k)$ matrices of full rank $k$ such that $BA^\top = 0_{k}$. Let $Q$ be a $(2k)\!\times\!(2k)$ invertible antisymmetric matrix (that is $Q^{\top} = -Q$). Then $B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k}$ if and only if $A{\mkern+1mu}Q{\mkern+1mu}A^\top = 0_{k}$.
Proof. Assume that $Q$ is a $(2k)\!\times\!(2k)$ invertible antisymmetric matrix, $A$ and $B$ are $k\!\times\!(2k)$ matrices of full rank $k$ and $BA^\top = 0_{k}$.
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Step 1. By the Rank-Nullity Theorem we have
\begin{equation*}
\operatorname{rank}(B) + \dim \operatorname{Nul}(B) = 2k.
\end{equation*}
As $\operatorname{rank}(B) = k,$ we deduce that $\dim \operatorname{Nul}(B) = k.$ Since $\operatorname{rank}(A) = \operatorname{rank}(A^\top)$ and $\operatorname{rank}(A) = k,$ we have $\operatorname{rank}(A^\top) = k.$ Thus,
\begin{equation*}
\dim \operatorname{Col}(A^\top) = \dim \operatorname{Nul}(B) = k.
\end{equation*}
The equality $BA^\top = 0_{k}$ implies
\begin{equation*}
\operatorname{Col}(A^\top) \subseteq \operatorname{Nul}(B).
\end{equation*}
The preceding two displayed relations yield
\begin{equation} \label{eq-cn1}
\operatorname{Col}(A^\top) = \operatorname{Nul}(B).
\end{equation}
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Step 2. Taking the transpose of the equality $BA^\top = 0_{k}$ we get $AB^\top = 0_{k}.$ Reversing roles of $A$ and $B$ the result proved in Step 1 yields
\begin{equation} \label{eq-cn1a}
\operatorname{Col}(B^\top) = \operatorname{Nul}(A).
\end{equation}
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Step 3. In this step we will prove
\begin{equation*}
B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k} \quad \Rightarrow \quad
A{\mkern+1mu}Q{\mkern+1mu}A^\top = 0_{k}.
\end{equation*}
Assume $B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k}.$ This equality implies
\begin{equation*}
\operatorname{Col}\bigl(Q^{-1}{\mkern-2mu}B^\top\bigr)\subseteq \operatorname{Nul}(B).
\end{equation*}
Since $\operatorname{rank}(B) = k$, we have $\operatorname{rank}(B^\top) = k$, therefore, since $Q^{-1}$ is invertible,
\begin{equation*}
\dim \operatorname{Col}\bigl(Q^{-1}{\mkern-2mu}B^\top\bigr) = k = \dim \operatorname{Nul}(B).
\end{equation*}
Last two displayed relations imply
\begin{equation} \label{eq-cn2}
\operatorname{Col}\bigl(Q^{-1}{\mkern-2mu}B^\top\bigr) = \operatorname{Nul}(B).
\end{equation}
The set equalities \eqref{eq-cn1} and \eqref{eq-cn2} yield
\begin{equation*}
\operatorname{Col}\bigl(Q^{-1}{\mkern-2mu}B^\top\bigr) = \operatorname{Col}(A^\top).
\end{equation*}
The preceding equality, the fact that the columns of $A^\top$ ($k$ of them) are linearly independent and the fact that the columns of $Q^{-1}{\mkern-2mu}B^\top$ ($k$ of them) are linearly independent, imply that there exists an invertible $k\!\times\! k$ matrix $P$ such that
\begin{equation*}
A^\top = Q^{-1}{\mkern-2mu}B^\top{\mkern-2mu}P.
\end{equation*}
Consequently, using the fact that $Q$ is antisymmetric, $Q^\top = -Q$,
\begin{equation*}
A = P^\top\!B\bigl(Q^{-1}\bigr)^\top = P^\top\!B\bigl(Q^\top\bigr)^{-1} = - P^\top\!B{\mkern+1mu} Q^{-1}.
\end{equation*}
Now we use $B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k}$ to calculate
\begin{equation*}
A Q A^\top = - P^\top\!B{\mkern+1mu} Q^{-1} Q Q^{-1}{\mkern-2mu}B^\top{\mkern-2mu}P = - P^\top\!B{\mkern+1mu} Q^{-1}{\mkern-2mu}B^\top{\mkern-2mu}P = 0_{k}.
\end{equation*}
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Step 4. In this step we will prove
\begin{equation*}
A{\mkern+1mu}QA^\top = 0_{k}\quad \Rightarrow \quad
B{\mkern+1mu}Q^{-1}{\mkern-2mu}B^\top = 0_{k} .
\end{equation*}
Assume $A{\mkern+1mu}QA^\top = 0_{k}.$ This equality implies
\begin{equation*}
\operatorname{Col}\bigl(QA^\top \bigr)\subseteq \operatorname{Nul}(A).
\end{equation*}
Replacing $Q^{-1}$ with $Q$ and $B$ with $A$, the reasoning that we used in Step 3 to prove \eqref{eq-cn2}, yields
\begin{equation} \label{eq-cn3}
\operatorname{Col}\bigl(QA^\top \bigr) = \operatorname{Nul}(A).
\end{equation}
The set equalities \eqref{eq-cn1a} and \eqref{eq-cn3} yield
\begin{equation*}
\operatorname{Col}\bigl(QA^\top \bigr) = \operatorname{Col}(B^\top).
\end{equation*}
The preceding equality, the fact that the columns of $B^\top$ ($k$ of them) are linearly independent and the fact that the columns of $QA^\top$ ($k$ of them) are linearly independent, imply that there exists an invertible $k\!\times\! k$ matrix $S$ such that
\begin{equation*}
B^\top = QA^\top{\mkern-2mu}S.
\end{equation*}
Consequently, using the fact that $Q$ is antisymmetric, $Q^\top = -Q$,
\begin{equation*}
B = S^\top\!AQ^\top = - S^\top\!A Q.
\end{equation*}
Now we use $A{\mkern+1mu}Q{\mkern+1mu}A^\top = 0_{k}$ to calculate
\begin{equation*}
B Q^{-1}{\mkern-2mu} B^\top = - S^\top\!A QQ^{-1}{\mkern-2mu} QA^\top{\mkern-2mu}S = - S^\top\!A Q A^\top{\mkern-2mu}S = 0_{k}.
\end{equation*}
- II. A Sturm-Liouville differential operator
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A Sturm-Liouville differential operator. Let $a, b$ be real numbers, $a \lt b$. The assumptions about the coefficients:
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$p$ is a continuous piecewise smooth function on the interval $[a,b]$ and $p(x) \gt 0$ for all $x \in [a,b]$,
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$q$ is a piecewise continuous function on $[a,b]$,
We define a Sturm-Liouville differential operator $L$ by
\[
(Ly)(x) = - \bigl(p(x)y'(x)\bigr)' +q(x)y(x), \qquad a \leq x \leq b.
\]
The operator $L$ is defined for functions $y$ which are continuous on $[a,b]$ and such that the function $py'$ is a continuous piecewise smooth function. The set of all such functions $y$ we denote by ${\mathcal D}(L)$.
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Lagrange's identity For all $u, v \in {\mathcal D}(L)$ we have
\[
L(u) v - u L(v) = \bigl(-(pu')v + u (pv') \bigr)'.
\]
The proof of this identity is by verification. The Fundamental theorem of calculus applied to Lagrange's identity yields Green's formula.
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Green's formula For all $u, v \in {\mathcal D}(L)$ we have
\begin{align*}
\int_a^b (Lu) v dx - \int_a^b u (Lv) dx & = \Bigl.\bigl(-(pu')v + u (pv') \bigr)\Bigr|_a^b \\
& = -(pu')(b)v(b) + u(b) (pv')(b) + (pu')(a)v(a) - u(a) (pv')(a) \\
& =
\left[\!\begin{array}{c} v(a) \\ (pv')(a) \\ v(b) \\ (pv')(b) \end{array} \right]^\top
\left[\!\begin{array}{rrrr} 0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 \end{array} \right]
\left[\!\begin{array}{c} u(a) \\ (pu')(a) \\ u(b) \\ (pu')(b) \end{array} \right].
\end{align*}
We introduce the following notation
\[
{\mathsf b}(y) = \left[\!\begin{array}{c} y(a) \\ (py')(a) \\ y(b) \\ (py')(b) \end{array} \right], \qquad {\mathsf Q} = \left[\!\begin{array}{rrrr} 0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 \end{array} \right].
\]
Here $(py')(a)$ means $p(a)y'(a)$. With this notation, Green's formula reads
\[
\int_a^b (Lu) v dx - \int_a^b u (Lv) dx =
{\mathsf b}(v)^\top\! {\mathsf Q} \, {\mathsf b}(u).
\]
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Notice that the matrix ${\mathsf Q}$ is antisymmetric and orthogonal, ${\mathsf Q}^\top \! {\mathsf Q} = I_4$. Therefore,
\[
{\mathsf Q}^{-1} = {\mathsf Q}^\top = - {\mathsf Q}.
\]
- III. A regular Sturm-Liouville eigenvalue problem
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A regular Sturm-Liouville eigenvalue problem. Let $a, b$ be real numbers, $a \lt b$. The assumptions about the coefficients:
-
$p$ is a continuous piecewise smooth function on the interval $[a,b]$ and $p(x) \gt 0$ for all $x \in [a,b]$,
-
$q$ is a piecewise continuous function on $[a,b]$,
-
$w$ is a piecewise continuous function on $[a,b]$ and $w(x) \gt 0$ for all $x \in (a,b)$.
We consider the following eigenvalue problem: Find all $\lambda$ for which there exists a nonzero $y \in {\mathcal D}(L)$ such that
\begin{equation*}
- \bigl(p(x)y'(x)\bigr)' + q(x)y(x) = \lambda w(x) y(x), \qquad a \leq x \leq b,
\end{equation*}
and $y$ satisfies the boundary conditions
\begin{align*}
\beta_{11} y(a) + \beta_{12} (py')(a) + \beta_{13} y(b) + \beta_{14} (py')(b) & = 0, \\
\beta_{21} y(a) + \beta_{22} (py')(a) + \beta_{23} y(b) + \beta_{24} (py')(b) & = 0.
\end{align*}
We assume that the boundary conditions are linearly independent; that is neither boundary conditions is a scalar multiple of the other.
Setting
\[
{\mathsf B} = \left[\!\begin{array}{rrrr}
\beta_{11} & \beta_{12} & \beta_{13} & \beta_{14} \\
\beta_{21} & \beta_{22} & \beta_{23} & \beta_{24} \end{array} \right]
\]
we can write the boundary conditions simply as
\[
{\mathsf B} \, {\mathsf b}(y) = 0,
\]
where $0$ is a $2\!\times\!1$ matrix. Since we assume that the boundary conditions are linearly independent, the $2\!\times\!4$ matrix ${\mathsf B}$ has rank $2$.
-
To make the above eigenvalue problem resemble an eigenvalue problem for matrices we will introduce a differential operator whose domain includes the boundary conditions. We call this operator $S$ and define it on the domain
\[
{\mathcal D}(S) = \bigl\{ y \in {\mathcal D}(L) : {\mathsf B} \, {\mathsf b}(y) = 0 \bigr\}.
\]
That is, the domain of $S$ consists of all the functions $y$ which are continuous on $[a,b]$ and such that the function $py'$ is a continuous piecewise smooth function and which satisfy the boundary conditions.
We define the operator $S$ as follows
\[
S y = \frac{1}{w} L y = \frac{1}{w} \bigl( -(py')' + qy \bigr), \qquad y \in {\mathcal D}(S).
\]
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The above stated eigenvalue problem can now be restated in terms of the operator $S$ as follows: Find all $\lambda$ for which there exists a nonzero $u \in {\mathcal D}(S)$ such that
\[
S u = \lambda \, u.
\]
- IV. Symmetry of a regular Sturm-Liouville eigenvalue problem
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Finally, we can discuss the symmetry of the operator $S$ with respect to the inner product defined by
\[
\langle u,v \rangle = \int_a^b u \, v\, w\, dx
\]
for piecewise continuous functions $u$ and $v$. The operator $S$ is said to be symmetric if
\[
\bigl\langle Su, v \bigr\rangle = \int_a^b (Su)\, v \, w \, dx = \int_a^b u\, (Sv)\, w \, dx \bigl\langle u, Sv \bigr\rangle
\]
for all $u,v \in {\mathcal D}(S)$. Using the definition of $S$ and Green's formula from II.3, we calculate
\begin{align*}
\bigl\langle Su, v \bigr\rangle - \bigl\langle u, Sv \bigr\rangle & = \int_a^b (Su)\, v\, w\, dx - \int_a^b u\, (Sv)\, w\, dx \\
& = \int_a^b (Lu)\, v\, dx - \int_a^b u\, (Lv)\, dx \\
& = {\mathsf b}(v)^\top\! {\mathsf Q} \, {\mathsf b}(u).
\end{align*}
Hence, the operator $S$ is symmetric if and only if
\[
{\mathsf b}(v)^\top\! {\mathsf Q} \, {\mathsf b}(u) = 0 \quad
\text{whenever} \quad {\mathsf B} \, {\mathsf b}(u) = 0 \ \ \text{and} \ \ {\mathsf B} \, {\mathsf b}(v) = 0.
\]
Recall that we discussed this relationship in the Linear algebra section. There we proved: If $B$ is $k\!\times\!(2k)$ matrix of rank $k$ and $Q$ is a $(2k)\!\times\!(2k)$ antisymmetric matrix such that ${B}Q^{-1}\!{B}^\top = 0$, then
\[
{\mathbf y}^\top {Q} {\mathbf x} = 0 \quad \text{whenever} \quad {B}{\mathbf x} = 0 \ \ \text{and} \ \ {B}{\mathbf y} = 0.
\]
-
Thus, if the "boundary" matrix ${\mathsf B}$ is $2\!\times\!4$ matrix of rank $2$ such that
\[
{\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0_2,
\]
then the operator $S$ is symmetric. Why? Since in the Linear algebra section we proved that under the condition
\[
{\mathsf B}{\mathsf Q}^{-1}\!{\mathsf B}^\top = {\mathsf B}{\mathsf Q}^\top\!{\mathsf B}^\top = - {\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0_2
\]
we have
\[
{\mathbf y}^\top {\mathsf Q} {\mathbf x} = 0 \quad \text{whenever} \quad {\mathsf B}{\mathbf x} = 0 \ \ \text{and} \ \ {\mathsf B}{\mathbf y} = 0.
\]
This implies that the operator $S$ is symmetric whenever ${\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0$; that is
\[
\bigl\langle Su, v \bigr\rangle = \int_a^b (Su)\, v\, w\, dx = \int_a^b u\, (Sv)\, w \, dx = \bigl\langle u, S v \bigr\rangle
\]
holds for all $u, v \in {\mathcal D}(S)$.
- V. Consequences of the symmetry
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Here we assume that the matrix ${\mathsf B}$ in the eigenvalue problem III.1 satisfies ${\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0$.
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Orthogonality of eigenfunctions corresponding to the distinct eigenvalues. Let $\lambda$ and $\mu$ be distinct eigenvalues of the eigenvalue problem introduced in III.1 and let $u$ and $v$, respectively, be corresponding eigenfunctions. Then $u, v \in {\mathcal D}(S)$ and
\[
Su = \lambda u \quad \text{and} \quad Sv = \mu v.
\]
We calculate
\[
\bigl\langle Su, v \bigr\rangle = \int_a^b Su \, v \, w \, dx = \lambda \int_a^b u \, v \, w\, dx \langle u, v \rangle
\]
and
\[
\bigl\langle u, S v \bigr\rangle = \int_a^b u \, Sv \, w \, dx = \mu \int_a^b u \, v \, w \, dx = \langle u, v \rangle.
\]
By IV.2 we have
\[
\int_a^b (Su)\, v\, w\, dx = \int_a^b u\, (Sv)\, w \, dx.
\]
Therefore the last three equalities imply
\[
\lambda \langle u, v \rangle = \mu \langle u, v \rangle.
\]
Consequently
\[
\bigl(\lambda - \mu \bigr) \langle u, v \rangle = 0.
\]
Since $\lambda \neq \mu$, the last equality yields
\[
\int_a^b\! u v w\, dx = \langle u, v \rangle = 0,
\]
that is the eigenfunctions are orthogonal with respect to the inner product introduced in IV.1.
-
All eigenvalues are real. Let $\lambda$ be an eigenvalue of the eigenvalue problem introduced in III.1 and let $u$ be a corresponding eigenfunction. Here we allow that $\lambda$ is possibly a complex number and that $u$ is a complex function. Then $u \in {\mathcal D}(S)$ and
\[
-(pu')' + qu = \lambda w u.
\]
Since we work only with real functions, we did not emphasize that the functions $p$, $q$ and $w$ are real and that all entries of the matrix ${\mathsf B}$ are real. Taking the complex conjugate of the last displayed equality we get
\[
-(p\overline{u}')' + q\overline{u} = \overline{\lambda} w \overline{u}.
\]
Here the bar above a symbol denotes its complex conjugate. Taking the complex conjugate in the boundary conditions we see that the function $\overline{u}$ also satisfies the boundary conditions;that is ${\mathsf B} {\mathsf b}(\overline{u}) = 0$. Therefore $\overline{u} \in {\mathcal D}(S)$. Using the operator $S$, the last two displayed equalities can be written as
\[
Su = \lambda u \quad \text{and} \quad S\overline{u} = \lambda \overline{u}.
\]
Next we calculate
\[
\int_a^b Su \, \overline{u} \, w\, dx = \lambda \int_a^b u \, \overline{u} \, w \,dx
\]
and
\[
\int_a^b u \, S\overline{u} \, w\, dx = \overline{\lambda} \int_a^b u \, \overline{u} \, w\, dx.
\]
Since by IV.2 we have
\[
\int_a^b (Su)\, v\, w\, dx = \int_a^b u\, (Sv)\, w \, dx,
\]
the last three equalities imply
\[
\lambda \int_a^b u \, \overline{u} \, w \,dx
= \overline{\lambda} \int_a^b u \, \overline{u} \, w \, dx
\]
and consequently
\[
\bigl(\lambda - \overline{\lambda} \bigr) \int_a^b u \, \overline{u} \, w\, dx = 0.
\]
Since $u \, \overline{u} = |u|^2 \geq 0$, and for at least one $x_0 \in [a,b]$ we have $|u|^2(x_0) \gt 0$, and $w(x_0) \gt 0$ we have $\int_a^b |u|^2 w\, dx > 0$. Therefore the last displayed equality implies
\[
\lambda = \overline{\lambda}
\]
that is $\lambda$ is a real number.