# Symmetry of a regular Sturm-Liouville eigenvalue problem

## Branko Ćurgus

Notation
1. For a matrix $M$ by $M^\top$ we denote the matrix transpose of $M$.
2. In definite integrals instead of $\int_a^b f(x) dx$ we briefly write $\int_a^b f\, dx$ assuming that the reader understands which function of $x$ is being integrated.
I. Linear algebra preliminaries
1. A linear algebra statement. Let $B$ be a $k\!\times\!(2k)$ matrix of rank $k$ and let $Q$ be $(2k)\!\times\!(2k)$ such that $Q^{\top}\!Q = I_{2k}$. Assume that $BQB^\top = 0_{k}$. Then ${\mathbf y}^\top Q {\mathbf x} = 0$ whenever $B{\mathbf x} = 0$ and $B{\mathbf y} = 0$.

Proof. Assume that $BQB^\top = 0_{k}$. This equality implies that the column space ${\rm Col}\bigl( QB^\top\!\bigr)$ is a subspace of the null space ${\rm Nul}\,B$. Since the rank of $B$ is $k$, by the Rank-Nullity theorem, the dimension of the null space of $B$ is $k$. Since the rank of $B$ is $k$, the rank of $B^\top$ is $k$ and, since the matrix $Q$ is invertible, the rank of $QB^\top$ is also $k$. Thus, ${\rm Col}\bigl( QB^\top\!\bigr)$ and ${\rm Nul}\,B$ have the same dimension $k$. Together with ${\rm Col}\bigl( QB^\top\!\bigr) \subseteq {\rm Nul}\,B$ the equal dimension implies that ${\rm Col}\bigl( QB^\top\!\bigr) = {\rm Nul}\,B$. Now assume that ${\mathbf x}, {\mathbf y} \in {\rm Nul}\,B$. By the last equality we conclude that there exist vectors ${\mathbf u}, {\mathbf v} \in {\mathbb R}^k$ such that ${\mathbf x} = QB^\top {\mathbf u}$ and ${\mathbf y} = QB^\top {\mathbf v}$. Now calculate ${\mathbf y}^\top\! Q {\mathbf x} = \bigl(QB^\top {\mathbf v} \bigr)^\top Q QB^\top {\mathbf u} = {\mathbf v}^\top\! B Q^\top Q QB^\top {\mathbf u} = {\mathbf v}^\top\! BQB^\top {\mathbf u} = 0.$ In the last step we used the assumption $BQB^\top = 0_{k}$.
2. The same statement formulated as an equivalence. Let $A$ and $B$ be $k\!\times\!(2k)$ matrices of rank $k$ such that $BA^\top = 0_{k}$. Let $Q$ be $(2k)\!\times\!(2k)$ matrix such that $Q^{\top}\!Q = I_{2k}$. Then $BQB^\top = 0_{k}$ if and only if $A Q A^\top = 0_{k}$.

Proof. Assume that $Q^{\top}\!Q = I_{2k}$, $BA^\top = 0_{k}$ and $BQB^\top = 0_{k}$. These two assumptions imply ${\rm Col}\bigl( A^\top\!\bigr) \subseteq {\rm Nul}\,B$, and ${\rm Col}\bigl( QB^\top\!\bigr) \subseteq {\rm Nul}\,B$, respectively. As in the proof of the above statement, we conclude that the dimensions of the spaces ${\rm Col}\bigl(A^\top\!\bigr)$, ${\rm Col}\bigl( QB^\top\!\bigr)$ and ${\rm Nul}\,B$ all equal to $k$. Thus, ${\rm Col}\bigl( A^\top\!\bigr) = {\rm Nul}\,B \quad \text{and} \quad {\rm Col}\bigl( QB^\top\!\bigr) = {\rm Nul}\,B,$ and consequently ${\rm Col}\bigl( A^\top\!\bigr) = {\rm Col}\bigl( QB^\top\!\bigr).$ This equality, the fact that the columns of $A^\top$ ($k$ of them) are linearly independent and the fact that the columns of $QB^\top$ ($k$ of them) are linearly independent, imply that there exists an invertible $k\!\times\! k$ matrix $S$ such that $A^\top = QB^\top S.$ Then $A = S^\top\!BQ^\top$. So, we calculate, using the assumptions $Q^\top Q = I_{2k}$ and $B QB^\top = 0_k$, $A Q A^\top = S^\top\!BQ^\top Q QB^\top S = S^\top\!B QB^\top S = 0_{k}.$ Thus we proved $A Q A^\top = 0_k$.
The implication: $Q^{\top}\!Q = I_{2k}$, $BA^\top = 0_{k}$ and $AQA^\top = 0_{k}$ imply $B QB^\top = 0_k$ follows by observing that $AB^\top = 0_{k}$ and applying the already proved implication with with the roles of $A$ and $B$ reversed.
II. A Sturm-Liouville differential operator
1. A Sturm-Liouville differential operator. Let $a, b$ be real numbers, $a \lt b$. The assumptions about the coefficients:
1. $p$ is a continuous piecewise smooth function on the interval $[a,b]$ and $p(x) \gt 0$ for all $x \in [a,b]$,
2. $q$ 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 ${\mathcal D}(L)$.
2. 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.
3. 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 $\int_a^b (Lu) v dx - \int_a^b u (Lv) dx = {\mathsf b}(v)^\top\! {\mathsf Q} \, {\mathsf b}(u).$
4. Notice that the matrix ${\mathsf Q}$ is orthogonal: ${\mathsf Q}^\top \! {\mathsf Q} = {\mathsf Q} {\mathsf Q}^\top = I_4$.
III. A regular Sturm-Liouville eigenvalue problem
1. A regular Sturm-Liouville eigenvalue problem. Let $a, b$ be real numbers, $a \lt b$. The assumptions about the coefficients:
1. $p$ is a continuous piecewise smooth function on the interval $[a,b]$ and $p(x) \gt 0$ for all $x \in [a,b]$,
2. $q$ is a piecewise continuous function on $[a,b]$,
3. $w$ 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 $2$.
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 $S y = \frac{1}{w} L y = \frac{1}{w} \bigl( -(py')' + qy \bigr), \qquad y \in {\mathcal D}(S).$
3. 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
1. Finally, we can discuss the symmetry of the operator $S$ with respect to the inner product defined by $\int_a^b u \, v\, w\, dx$ for piecewise continuous functions $u$ and $v$. The operator $S$ is said to be symmetric if $\int_a^b (Su)\, v \, w \, dx = \int_a^b u\, (Sv)\, w \, dx$ for all $u,v \in {\mathcal D}(S)$. Using the definition of $S$ and Green's formula from II.3, we calculate \begin{align*} \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 an orthogonal $(2k)\!\times\!(2k)$ matrix such that ${B}{Q}{B}^\top = 0$, then ${\mathbf y}^\top {Q} {\mathbf x} = 0 \quad \text{whenever} \quad {B}{\mathbf x} = 0 \ \ \text{and} \ \ {B}{\mathbf y} = 0.$
2. Thus, if the "boundary" matrix ${\mathsf B}$ is $2\!\times\!4$ matrix of rank $2$, ${\mathsf Q}$ is an orthogonal matrix and ${\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0$, we deduce that ${\mathbf y}^\top {\mathsf Q} {\mathbf x} = 0 \quad \text{whenever} \quad {\mathsf B}{\mathbf x} = 0 \ \ \text{and} \ \ {\mathsf B}{\mathbf y} = 0.$ This implies that the operator $S$ is symmetric whenever ${\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0$; that is $\int_a^b (Su)\, v\, w\, dx = \int_a^b u\, (Sv)\, w \, dx$ holds for all $u, v \in {\mathcal D}(S)$.
V. Consequences of the symmetry
1. Here we assume that the matrix ${\mathsf B}$ in the eigenvalue problem III.1 satisfies ${\mathsf B}{\mathsf Q}{\mathsf B}^\top = 0$.
2. 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 $Su = \lambda u \quad \text{and} \quad Sv = \lambda v.$ We calculate $\int_a^b Su \, v \, w \, dx = \lambda \int_a^b u \, v \, w\, dx$ and $\int_a^b u \, Sv \, w \, dx = \mu \int_a^b u \, v \, w \, dx.$ 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 $\lambda \int_a^b u \, v \, w \, dx = \mu \int_a^b u \, v \, w \, dx$ and consequently $\bigl(\lambda - \mu \bigr) \int_a^b u \, v \, w\, dx = 0.$ Since $\lambda \neq \mu$, the last equality yields $\int_a^b u \, v \, w\, dx = 0,$ that is the eigenfunctions are orthogonal with respect to the inner product introduced in IV.1.
3. 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 $Su = \lambda u \quad \text{and} \quad S\overline{u} = \lambda \overline{u}.$ Nest 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.