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# A 2-component Camassa-Holm equation, Euler–Bernoulli Beam Problem and Non-
Commutative Continued Fractions

Richard Beals Department of Mathematics, Yale University, New Haven, CT 06520,
USA<EMAIL_ADDRESS>Jacek Szmigielski Department of Mathematics and
Statistics, University of Saskatchewan, 106 Wiggins Road, Saskatoon,
Saskatchewan, S7N 5E6, Canada<EMAIL_ADDRESS>
(May 24, 2021)

###### Abstract

A new approach to the Euler-Bernoulli beam based on an inhomogeneous matrix
string problem is presented. Three ramifications of the approach are
developed:

  1. 1.

motivated by an analogy with the Camassa-Holm equation a class of isospectral
deformations of the beam problem is formulated;

  2. 2.

a reformulation of the matrix string problem in terms of a certain compact
operator is used to obtain basic spectral properties of the inhomogeneous
matrix string problem with Dirichlet boundary conditions;

  3. 3.

the inverse problem is solved for the special case of a discrete Euler-
Bernoulli beam. The solution involves a non-commutative generalization of
Stieltjes’ continued fractions, leading to the inverse formulas expressed in
terms of ratios of Hankel-like determinants.

###### Contents

  1. 1 Introduction
  2. 2 A $2$-Component CH equation
  3. 3 Transfer to an interval
  4. 4 The Dirichlet problem for a beam
  5. 5 Spectral theory
  6. 6 Wronskians and Green’s kernels
  7. 7 The discrete beam
  8. 8 The inverse problem for the discrete beam
  9. 9 Appendix: The consistency conditions; the smooth case

##  1 Introduction

In 1993, Camassa and Holm [5] discovered the shallow water equation

$m_{t}+(um)_{x}+u_{x}m=0,\qquad m=u-u_{xx},$ (1.1)

where subscripts denote partial derivatives. The most attractive novel
property of (1.1) is that it supports non-smooth soliton solutions. These
peaked solitons (peakons) are obtained from the ansatz

$u(x,t)=\sum_{j=1}^{d}m_{j}(t)\,e^{-\left\lvert x-x_{j}(t)\right\rvert},$
(1.2)

involving amplitudes $m_{j}(t)$ and positions $x_{j}(t)$ depending smoothly on
time. It was shown in [5] that $u$ given by (1.2) is a weak solution to the
Camassa–Holm equation (1.1) if and only if the positions $x_{j}(t)$ and
amplitudes $m_{j}(t)$ satisfy the Hamiltonian system

$\dot{x}_{j}=\frac{\partial H}{\partial
m_{j}}=u(x_{j}),\qquad\dot{m}_{j}=-\frac{\partial H}{\partial
x_{j}}=-m_{j}u_{x}(x_{j}),$ (1.3)

with Hamiltonian

$H(x_{1},\dots,x_{d},m_{1},\dots,m_{d})=\frac{1}{2}\sum_{i,j=1}^{d}m_{i}m_{j}e^{-\left\lvert
x_{i}-x_{j}\right\rvert},$

and the convention that
$u_{x}(x_{i})=\bigl{\langle}D_{x}u\bigr{\rangle}(x_{i})$ where
$\bigl{\langle}f\bigr{\rangle}(x_{i})$ means the arithmetic mean of the left
and right hand limits of $f$ at $x_{i}$.

The general solution of (1.3) (for arbitrary $d$) was constructed in [2, 3, 4]
using inverse spectral methods. The main premise of these papers was the
realization that the CH equation can be viewed as an isospectral deformation
of the classical inhomogeneous string problem studied in the $1950$s by M.G.
Krein [15, 14, 13] and, subsequently, by Dym and McKean in [7]; for a review
see [8]. In particular, the peakon solutions (1.2) were shown in [3] to be
directly linked to an isospectral deformation of discrete strings, i.e. those
strings for which the mass density is a linear combination of point masses.
Krein had observed long ago that, in the case of discrete strings, the inverse
string problem can be solved explicitly by using results of Stieltjes on
continued fractions [22]. Stieltjes’ methods were applied in [3] to derive the
determinantal formulas for the amplitudes and positions of peakons and, in
turn, to determine the asymptotic behaviour of peakon solutions.

It has been known at least since the fundamental paper [16] by Lax on the
Korteweg–de Vries equation that the isospectral deformations of boundary value
problems may lead to interesting non-linear equations. The boundary value
problem for the KdV equation is given by the Schrödinger equation on the whole
real axis, in which case the spectrum is a union of continuous and discrete
spectra.

In the late $1970$s, in a series of interesting papers [19, 20, 21], Sabatier
put forward an idea that spectral problems with just discrete spectrum can
also be a source of interesting non-linear problems derived via isospectral
deformations. In particular, the inhomogeneous string boundary value problem
was singled out as a potential source of interesting boundary problems with
the discrete spectrum, to be subjected to isospectral deformations. Sabatier
studied a very limited class of spectral deformations, applicable only to
strictly positive, smooth densities. The situation changed with the discovery
of the CH equation (1.1) and subsequent realization in [2, 3, 4] that the CH
equation is, in disguise, an isospectral deformation of an inhomogeneous
string boundary value problem. This line of research was later generalized to
other equations from the CH family ([17, 6, 11]). In the introduction to [4],
the authors speculated, guided by the work of Barcilon [1] on the Euler-
Bernoulli beam problem, that higher order boundary value problems might
provide an equally rich environment for isospectral deformations. This task
required a new look at the beam boundary value problem that would be naturally
amenable to a Lax type deformation. The present paper aims to offer a new
approach to the beam boundary value problem by rephrasing it as a matrix
inhomogeneous string problem.

In the remainder of this section we outline the main results of this paper and
provide a context for some of the techniques used in our arguments.

In Section 2 we propose a simple derivation of a system of nonlinear equations
that generalizes the CH equation to a new two component equation which is
structurally of the same type as the CH equation. We are keenly aware that
there exist other two-component generalizations of the CH equation, for
example,

$\displaystyle m_{t}=(um)_{x}+u_{x}m+\rho\rho_{x},$ $\displaystyle
m=u-u_{xx},$ $\displaystyle\rho_{t}=(\rho u)_{x},$

first derived using tri-hamiltonian methods by P.J. Olver and Rosenau in [18]
and studied, for example, in [12]. Other generalizations have been proposed as
well [25].

The equation we propose (see (2.5)) takes the form

$\displaystyle n_{t}=(un)_{x}+u_{x}n+vn,$ $\displaystyle
m_{t}=(um)_{x}+u_{x}m-vm,$ $\displaystyle u_{xx}-4u=n+m,$ $\displaystyle
v_{x}=n-m,$

and comes from a matrix valued Lax pair, structurally identical to the
original CH case, but involving two measures $m$ and $n$ rather than one.

In Section 3, using a Liouville transformation, we map the problem to a finite
interval, following a similar procedure used in [2] to study an acoustic
scattering problem. We note that the transformed $x$-member of the Lax pair
(see (3.1)) is a matrix version of an inhomogeneous string boundary value
problem.

In Section 4 we review the pertinent facts about the Euler–Bernoulli beam
problem, which we show can be reformulated as a string problem with a matrix
density – the same system already encountered in Section 3. We then study the
basic spectral properties of that matrix string problem.

In Section 5 we reformulate the Euler–Bernoulli beam problem as a standard
spectral problem for a compact (in fact trace-class) operator $T$ and we study
the properties of that operator on an appropriate Hilbert space. In
particular, we establish basic properties of the resolvent of that operator.

In Section 6, in Theorem 6.2, we derive a closed-form expression for the
resolvent of $T$ and introduce a special element of the resolvent, the Weyl
function, which plays a key role in the formulation of the inverse problem.

In Section 7 we analyze the spectral problem for a discrete Euler-Bernoulli
beam, that is, a beam in which both measures $m$ and $n$ are chosen to be
finite sums of point masses.. This is a beam counterpart of Stietljes’ string.
Similar to what occurs for the string problem, the Weyl function admits a
continued fraction expansion (Section 7), albeit with non-commuting
coefficients. The general concept of continued fractions over non-commutative
rings goes back to Wedderburn [24], but our special non-commutative case is a
direct generalization of continued fractions of Stieltjes’ type, this time
associated to a discrete inverse beam problem, rather than a string problem.
The continued fraction expansion can be rephrased in terms of non-commutative
Padé approximations and we formulate Padé approximation conditions needed for
the inverse problem.

In Section 8 we explicitly solve the inverse problem for the discrete Euler-
Bernoulli beam. To this end we construct a sequence of non-commutative Padé
approximations using the Weyl function, or, to be more precise, the spectral
measure, as an input data and, in the end, we recover the discrete measures
$m$ and $n$. In the process of solving the inverse problem we are prompted to
introduce several variations on the Hankel moment matrix that are germane to
the inverse beam problem. The final formulas bear a remarkable resemblance to
string formulas (see [4]) with the proviso that in the beam problem the usual
Hankel determinants of the moment matrix are replaced by determinants of
suitably reduced Hankel matrices of moments.

The paper concludes with (Appendix) Section 9, in which we provide a detailed
analysis of the Lax pair parametrization giving rise to (2.5).

##  2 A $2$-Component CH equation

The Camassa-Holm (CH) equation (1.1) is the compatibility condition for a pair
of scalar equations on the line, which we take for simplicity to have the form

$\psi_{xx}=(1+\lambda m)\psi,\qquad\psi_{t}=a\psi+b\psi_{x},$ (2.1)

where $-\infty<x<\infty$, and the subscripts in $x$ and $t$ represent
distribution derivatives in $x,t$ respectively.

######  Remark .

In general, the derivatives we consider are distribution derivatives. It will
sometimes be convenient to use $D_{x}$, $D_{t}$, or simply $D$ in the case of
one variable.

The CH flow corresponds to the choice

$b=u+\frac{1}{\lambda},\quad a=-\frac{u_{x}}{2}.$

In this case the compatibility condition reads

$m_{t}=(um)_{x}+u_{x}m,\qquad(u_{xx}-4u)_{x}=2m_{x}.$ (2.2)

This compatibility condition holds even if $m$ is a measure, in which case
$u_{x}$ is of bounded variation, $u$ is continuous, and the term $u_{x}m$
means that on the singular support of $m$ the multiplier is taken to be
$u_{x}(a)=\bigl{\langle}u_{x}\bigr{\rangle}(a)$, where
$\bigl{\langle}f\bigr{\rangle}(a)$ is $\frac{1}{2}[f(a_{-})+f(a_{+})]$.

######  Remark .

Our choice of the coefficients in (2.1) differs slightly from the original Lax
pair in [5]. This change results in a different relation between $u$ and $m$
as one can see by comparing (1.1) and $\eqref{eq:CHZCC}$.

Now, we consider a two-component version

$\displaystyle\Psi_{xx}$ $\displaystyle=(\mathbf{1}+\lambda M)\Psi,\qquad
M=\begin{bmatrix}0&n\\\ m&0\end{bmatrix},$ (2.3a) $\displaystyle\Psi_{t}$
$\displaystyle=a\Psi+b\Psi_{x},$ (2.3b)

where $\Psi,a$ and $b$ are $2\times 2$ matrix functions of $x,t,\lambda$ and
${\bf 1}$ denotes the $2\times 2$ identity matrix. We assume for now that
$a,b$ have only terms of degree $0,-1$ in $\lambda$. As shown in the Appendix,
if we add the assumption that the matrices $a$ and $b$ are bounded as
$x\to\pm\infty$, then, up to a normalization, they have the form

$b=\begin{bmatrix}u&0\\\
0&u\end{bmatrix}+\frac{1}{\lambda}\begin{bmatrix}0&1\\\
1&0\end{bmatrix},\qquad\qquad a=-\frac{1}{2}\begin{bmatrix}u_{x}-v&0\\\
0&u_{x}+v\end{bmatrix}.$ (2.4)

The compatibility conditions split into two constraints

$(u_{xx}-4u)_{x}=(n+m)_{x},\qquad v_{x}=n-m,$

and a system of evolution equations

$n_{t}=(un)_{x}+u_{x}n+vn,\qquad m_{t}=(um)_{x}+u_{x}m-vm.$ (2.5)

Furthermore, if we are interested in bounded $u$ and compactly supported
$m,n$, then we can replace one of the constraints with a more restrictive one,
keeping the other constraint intact,

$u_{xx}-4u=n+m,\qquad v_{x}=n-m.$ (2.6)

We assume that $M$ is compactly supported and take as solutions of (2.6) the
particular choices

$\displaystyle u(x,t)$
$\displaystyle=-\frac{1}{4}\int_{-\infty}^{\infty}e^{-2\left\lvert
x-y\right\rvert}[m(y,t)+n(y,t)]\,dy,$ (2.7) $\displaystyle v(x,t)$
$\displaystyle=\frac{1}{2}\int_{-\infty}^{\infty}\operatorname{sgn}(x-y)[n(y,t)-m(y,t)]\,dy.$
(2.8)

It follows that

$\begin{matrix}\lim_{x\rightarrow\pm\infty}u(x,t)=0=\lim_{x\rightarrow\pm\infty}u_{x}(x,t)\\\
\lim_{x\rightarrow\pm\infty}v(x,t)=\pm\frac{1}{2}\int_{-\infty}^{\infty}[n(y,t)-m(y,t)]\,dy.\end{matrix}$
(2.9)

######  Remark .

The integrals appearing in this paper are all Stieltjes integrals, but we find
it more convenient to write them as $\int f(x)m(x,t)dx$, or, later in the
paper, as $\int f(x)dm(x,t)$ . The former notation is analogous to writing the
point mass at the origin as $\delta(x)dx$.

######  Proposition .

The integral

$\int_{-\infty}^{\infty}[m(y,t)+n(y,t)]\,dy$

is independent of $t$.

###### Proof.

According to (2.5), (2.9),

$\begin{split}D_{t}\int_{-\infty}^{\infty}[m(y,t)+n(y,t)]\,dy&=\int_{-\infty}^{\infty}\\{[u(n+m)]_{y}+u_{y}(n+m)+v(n-m)\\}\,dy\\\
&=\int_{-\infty}^{\infty}[u_{y}u_{yy}-4u_{y}u+vv_{y}]\,dy\\\
&=\frac{1}{2}\int_{-\infty}^{\infty}[u_{y}^{2}-4u^{2}+v^{2}]_{y}\,dy=0.\end{split}$

∎

######  Remark .

Note, however, that separately $\int_{-\infty}^{\infty}mdy$ and
$\int_{-\infty}^{\infty}ndy$ are not conserved. Indeed, it follows from (2.6)
that

$D_{t}\int_{-\infty}^{\infty}[n-m]\,dy=-4\int_{-\infty}^{\infty}uv\,dy.$

##  3 Transfer to an interval

As in [3], we transfer the problem on the real axis to the interval $[-1,1]$.
In this section we will refer to the variable on the real axis as $\xi$ and
its counterpart on $[-1,1]$ by $x$. The functions originally defined on the
real axis will carry a tilde. Thus, for example, $M$ from the previous section
will be denoted by $\tilde{M}$. We set

$x=\tanh\xi,\qquad D_{x}=\cosh^{2}\xi D_{\xi}$

Define, for general $f:\mathbf{R}\rightarrow\mathbf{R}$,

$f^{*}(\tanh\xi)=f(\xi).$

Note that $\operatorname{sech}^{2}\xi=1-x^{2}$. Then a straightforward
computation shows that

$(1-x^{2})^{\frac{3}{2}}D_{x}^{2}(f^{*}(1-x^{2})^{-\frac{1}{2}})=[(D_{\xi}^{2}-1)f]^{*}.$

Therefore

$[(D_{\xi}^{2}-\mathbf{1}-\lambda\tilde{M})f]^{*}=(1-x^{2})^{\frac{3}{2}}[D_{x}^{2}-\lambda(1-x^{2})^{-2}\tilde{M}](f^{*}(1-x^{2})^{\frac{1}{2}})$

and solutions to (2.3a), after flipping $x$ with $\xi$ and $M$ with
$\tilde{M}$, correspond to solutions of

$\Phi_{xx}=\lambda M\Phi$ (3.1)

under the map

$\Phi=(1-x^{2})^{\frac{1}{2}}\Psi^{*},\qquad M=(1-x^{2})^{-2}\tilde{M}^{*}.$

The flow equation (2.3b) implies

$\begin{split}D_{t}\Phi=[(\cosh\xi)^{-1}D_{t}\Psi]^{*}&=[(\cosh\xi)^{-1}(\tilde{a}\Psi+\tilde{b}D_{\xi}\Psi]^{*}\\\
&=[(\cosh\xi)^{-1}\tilde{a}\Psi+(\cosh\xi)^{-3}\tilde{b}\cosh^{2}\xi
D_{\xi}\Psi]^{*}\\\
&=\tilde{a}^{*}\Phi+(1-x^{2})^{\frac{3}{2}}\tilde{b}^{*}D_{x}\big{(}(1-x^{2})^{-\frac{1}{2}}\Phi\big{)}\\\
&=\tilde{a}^{*}\Phi+(1-x^{2})\tilde{b}^{*}\big{[}D_{x}+\frac{x}{1-x^{2}}\big{]}\Phi\\\
&=[\tilde{a}^{*}+x\tilde{b}^{*}]\Phi+(1-x^{2})\tilde{b}^{*}D_{x}\Phi.\end{split}$

Thus

$\Phi_{t}=a\Phi+b\Phi_{x},$ (3.2)

where

$a=\tilde{a}^{*}+x\tilde{b}^{*},\qquad b=(1-x^{2})\tilde{b}^{*},$

or, more explicitly,

$\begin{split}a&=\frac{1}{2}\begin{bmatrix}-u_{x}+v&0\\\
0&-u_{x}-v\end{bmatrix}-\frac{1}{2\lambda}\begin{bmatrix}0&\beta_{x}\\\
\beta_{x}&0\end{bmatrix},\\\ b&=\begin{bmatrix}u&0\\\
0&u\end{bmatrix}+\frac{1}{\lambda}\begin{bmatrix}0&\beta\\\
\beta&0\end{bmatrix},\\\ u&=(1-x^{2})\tilde{u}^{*},\qquad
v=\tilde{v}^{*},\qquad\beta=1-x^{2}.\end{split}$ (3.3)

It can be shown that

$u_{xxx}=(\beta(m+n))_{x}+\beta_{x}(m+n),\qquad v_{x}=\beta(n-m),$ (3.4)

and that the flow of $M$ takes the same form as the flow of $\tilde{M}$,
namely,

$n_{t}=(un)_{x}+u_{x}n+vn,\qquad m_{t}=(um)_{x}+u_{x}m-vm.$ (3.5)

In fact (3.4) and (3.5) are consequences of the compatibility conditions for
the Lax pair (3.1) and (3.2). It follows from (2.9) and (3.3) that

$u(\pm 1)=u_{x}(\pm 1)=0,\qquad v(-1)=-v(+1).$ (3.6)

######  Proposition .

The integral

$\int_{-1}^{1}\beta[m+n]\,dx$

is independent of $t$.

###### Proof.

According to (3.5), (3.4) and (3.6),

$\displaystyle D_{t}\int_{-1}^{1}\beta(x)[m(x,t)+n(x,t)]\,dx$
$\displaystyle\qquad\ =\
\int_{-1}^{1}\beta(x)\\{[u(n+m)]_{x}+u_{x}(n+m)+v(n-m)\\}\,dx$
$\displaystyle\qquad\ =\
\int_{-1}^{1}[-\beta_{x}u(m+n)-u(\beta(m+n))_{x}+vv_{x}]\,dx$
$\displaystyle\qquad\ =\ \int_{-1}^{1}(-u_{xxx}u+v_{x}v)\,dx$
$\displaystyle\qquad\ =\
\frac{1}{2}\int_{-1}^{1}[u_{x}^{2}+v^{2}]_{x}\,dx=\frac{1}{2}(u_{x}^{2}+v^{2})\big{|}_{-1}^{1}=0.$

∎

We end this section by stating integral formulas for $u$ and $v$, which, as
can be easily checked, provide the unique solutions of (3.4) subject to
boundary conditions (3.6).

######  Proposition .

$\displaystyle u(x,t)$
$\displaystyle=-\int_{-1}^{1}G_{D}(x,y)^{2}(m(y,t)+n(y,t))\,dy,$
$\displaystyle v(x,t)$
$\displaystyle=\int_{-1}^{1}\operatorname{sgn}(x-y)G_{D}(y,y)(n(y,t)-m(y,t))\,dy.$

Here

$G_{D}(x,y)=\frac{1}{2}\begin{cases}(1+x)(1-y),\quad x<y\\\ (1-x)(1+y),\quad
y<x\end{cases}$ (3.7)

is the Green’s function of the classical Dirichlet string problem

$-D_{x}^{2}f=\lambda\rho f,\quad~{}f(-1)=f(1)=0.$

##  4 The Dirichlet problem for a beam

Vibrations of a beam parametrized by the interval $-1\leq x\leq 1$ are
characterized by the equation

$D^{2}[rD^{2}\phi]\ =\ \lambda^{2}m\phi,\qquad D=D_{x};$ (4.1)

(see [1] or, for a comprehensive view of vibration problems in engineering,
[10]). What we refer to as a beam problem is often referred to as an
Euler–Bernoulli beam problem. The two functions (or positive measures) $r,m$
are the flexural rigidity and mass density. The spectral parameter
$\lambda^{2}$ denotes the square of the frequency.

Setting

$D^{2}\varphi_{1}\ =\ \lambda n\varphi_{2},$ (4.2)

where $n=1/r$, (4.1) becomes

$D^{2}\varphi_{1}\ =\ \lambda n\varphi_{2},\qquad D^{2}\varphi_{2}\ =\ \lambda
m\varphi_{1}.$ (4.3)

The matrix form is

$D^{2}\varphi\ =\ \lambda M\varphi,\qquad\varphi\ =\
\begin{bmatrix}\varphi_{1}\\\ \varphi_{2}\end{bmatrix},\quad M\ =\
\begin{bmatrix}0&n\\\ m&0\end{bmatrix}.$ (4.4)

We require that $\varphi$ be continuous. We want to allow $m$ and $n$ to be
finite positive measures on the interval $-1\leq x\leq 1$. We assume

$\begin{matrix}\hbox{(a)\quad The endpoints $x=\pm 1$ are not atoms for $m$ or
$n$.}\\\ \hbox{(b)\quad The measures $m$ and $n$ have the same
support.}\end{matrix}$ (4.5)

Then $M\varphi$ should be interpreted as

$M\varphi\ =\ \left[\begin{matrix}\varphi_{2}n\\\
\varphi_{1}m\end{matrix}\right].$

######  Remark .

Throughout this section and the next it is convenient to adopt the (more
correct) way of writing integrals with respect to $m$ and $n$, using $dm(x)$,
$dn(x)$ rather than the symbolic $m(x)dx$, $n(x)dx$.

A partial fundamental solution $\Phi(x,\lambda)$ with the property
$\Phi(-1,\lambda)=0,\,$ $D\Phi(-1,\lambda)=~{}~{}{\bf 1}$, the identity
matrix, can be constructed in the form

$\Phi(x,\lambda)\ =\ \sum_{k=0}^{\infty}\lambda^{k}\Phi_{k}(x)$ (4.6)

with

$\Phi_{0}(x)\ =\ (1+x){\bf 1},\qquad\Phi_{k+1}(x)\ =\
\int_{-1}^{x}\left[\int_{-1}^{y}M(z)\Phi_{k}(z)\,dz\right]\,dy.$

To be specific, we take the integrals to run on the intervals $[-1,x)$ and
$[-1,y)$.

######  Proposition .

The functions $\Phi_{k}(x)$ are diagonal for even $k$, off-diagonal for odd
$k$. The non-zero entries $\Phi_{k1}$, $\Phi_{k2}$ are non-negative, positive
at $x=1$, and satisfy the estimate

$0\ \leq\ \Phi_{kj}(x)\ \leq\
2\frac{(1+x)^{k}(\overline{m}+\overline{n})^{k}}{k^{k}\,k\,!}.\qquad\ j=1,2.$
(4.7)

where $\overline{m}$ and $\overline{n}$ denote the total mass of $m$ and $n$.

###### Proof.

The first assertions follow by induction from the construction (4.6) and the
assumption that $m$ and $n$ are positive measures. To prove the estimate, let
$p=m+n$, and define $\psi_{k}$ by

$\psi_{0}(x)\ =\ 1+x,\qquad\psi_{k+1}(x)\ =\
\int_{-1}^{x}\int_{-1}^{y}\psi_{k}(z)\,dp(z)\,dy.$

Each $\Phi_{kj}(x)$ is bounded by $\psi_{k}(x)$. Changing the order of
integration and integrating first with respect to $y$, the result can be
written as

$\displaystyle\psi_{k}(x)\ =\ \int_{-1}^{x}(x-y_{k})\psi_{k-1}dp(y_{k})$
$\displaystyle\ =\
\int_{-1}^{x}\int_{-1}^{y_{k}}\left[(x-y_{k})(y_{k}-y_{k-1})\right]\psi_{k-2}(y_{k-1})\,dp(y_{k-1})\,dp(y_{k})$
$\displaystyle\ =\
\int_{-1}^{x}\cdots\int_{-1}^{y_{2}}\left[(x-y_{k})(y_{k}-y_{k-1})\cdots(y_{2}-y_{1})\right]\psi_{0}(y_{1})\,dp(y_{1})\,dp(y_{2})\cdots\,dp(y_{k}).$

The product in brackets is maximized when the $k$ factors are all the same.
The interval $[y_{1},x]$ has length at most $1+x$, and $\psi_{0}(x)\leq 2$, so

$0\ \leq\ \psi_{k}(x)\leq 2\left[\frac{1+x}{k}\right]^{k}\int_{0\leq
y_{1}<\dots<y_{k}<x}dp^{(k)}(y_{1},y_{2},\dots y_{k}),$ (4.8)

where $p^{(k)}$ denotes the product measure on the $k$-cube $\\{(x_{1},\dots
x_{k}):|x_{j}|\leq 1\\}$ in $\mathbf{R}^{k}$. The domain of integration is one
of $k\,!$ pairwise disjoint domains in the cube that are obtained by permuting
the indices. Each domain has the same measure, so

$0\ \leq\ \int_{0\leq y_{1}<\dots<y_{k}<x}dp^{(k)}(y_{1},y_{2},\dots y_{k})\
\leq\ \frac{\left[\int_{-1}^{x}dp(y)\right]^{k}}{k\,!}.$ (4.9)

Since $\int^{1}_{-1}dp(x)\ =\ \overline{m}+\overline{n}$, the estimates (4.8)
and (4.9) imply (4.7). ∎.

######  Corollary .

(a) The function $\Phi(x,\lambda)$ is continuous in both variables, and entire
as a function of $\lambda$, for fixed $x$.

(b) Each entry of $\Phi(1,\lambda)$ is dominated by

$2\sum_{k=0}^{\infty}\frac{a^{2k}|\lambda|^{k}}{(2k)\,!}\ \leq\
2\exp(a|\lambda|^{1/2}),\qquad a=\sqrt{2(\overline{m}+\overline{n})}.$ (4.10)

.

###### Proof.

(a) The estimates (4.7) imply that the series (4.6) converges uniformly on
bounded sets in $[-1,1]\times\mathbf{C}$.

(b) Since $(2k)\,!\leq(2k)^{k}k\,!$, the estimates (4.7) imply the bound
(4.10). ∎

We are interested in the Dirichlet problem for solutions of (4.4):

$\varphi_{1}(\pm 1,\lambda)\ =\ 0\ =\ \varphi_{2}(\pm 1,\lambda).$ (4.11)

A value $\lambda\in\mathbf{C}$ for which a non-zero solution of (4.4), (4.11)
exists will be referred to as a Dirichlet eigenvalue. Note that zero is not an
eigenvalue.

######  Proposition .

The Dirichlet eigenvalues $\\{\lambda_{\nu}\\}$ are precisely the zeros of
$\Delta$, where

$\Delta(\lambda)\ =\ \det\Phi(1,\lambda).$

They satisfy

$\sum_{\nu}\frac{1}{|\lambda_{\nu}|}\ <\ \infty.$ (4.12)

###### Proof.

Any solution $\varphi$ of the Dirichlet problem with eigenvalue $\lambda$ is a
linear combination of the two columns of $\Phi(\cdot,\lambda)$:

$\varphi(x)\ =\ \Phi(x,\lambda)\,v$

where $v$ is a constant 2-vector. The condition at $x=1$ implies that
$\Phi(1,\lambda)v=0$. Thus the necessary and sufficient condition for the
existence of a non-zero solution (“eigenfunction”) of the Dirichlet problem
(4.4), (4.11) is that $\det\Phi(1,\lambda)=0$.

Corollary 4 implies that $|\Delta(\lambda)|$ is dominated by
$\exp(4a|\lambda|^{1/2})$. Therefore the zeros $\lambda_{\nu}$, if numbered
with $|\lambda_{\nu}|$ non-decreasing, satisfy

$|\lambda_{\nu}|\ \geq\ c\nu^{2}$ (4.13)

for some constant $c>0$. [23, §8.21]. In particular, (4.12) is true. ∎

At the end of this section we will show that the zeros of $\Delta$ are simple.

If $\varphi$ is a solution of the Dirichlet problem (4.4), (4.11), then
$f=D\varphi$ is a solution of

$Df\ =\ \lambda M\varphi.$ (4.14)

It follows that $f$ is a function of bounded variation which is continuous at
any point that is not an atom. In particular, $f$ is continuous at the
endpoints $\pm 1$. This fact, for $f$ and similar functions, justifies the
various integration-by-parts formulas in this and later sections.

The following identity is fundamental to our discussion of the Dirichlet
problem.

######  Lemma .

If $\varphi$ is a solution of the Dirichlet problem with eigenvalue $\lambda$,
and $f=D\varphi$, then

$\int^{1}_{-1}[f_{1}\overline{f}_{2}+f_{2}\overline{f}_{1}]\,dx\ =\
-\lambda\int^{1}_{-1}[|\varphi_{1}|^{2}\,dm+|\varphi_{2}|^{2}\,dn]\ \neq\ 0.$
(4.15)

###### Proof.

It is convenient to write the left side in the form $(D\varphi,\sigma
D\varphi)$, where the inner product $(\ ,\ )$ is the $L^{2}$ inner product for
vector-valued functions:

$(f,g)\ =\ \int^{1}_{-1}f(x)\cdot\overline{g(x)}\,dx\ =\
\int^{1}_{-1}[f_{1}(x)\overline{g_{1}(x)}+f_{2}(x)\overline{g_{2}(x)}]\,dx,$
(4.16)

and $\sigma$ is the matrix

$\sigma\ =\ \begin{bmatrix}0&1\\\ 1&0\end{bmatrix}.$ (4.17)

Then integration by parts gives

$(D\varphi,\sigma D\varphi)\ =\ -(D^{2}\varphi,\sigma\varphi)\ =\
-\lambda(M\varphi,\sigma\varphi),$ (4.18)

which is (4.15).

If the right side of (4.15) is zero, then $\varphi_{1}=0$ on the support of
$m$ and $\varphi_{2}=0$ on the support of $n$. As a consequence

$(D\varphi,D\varphi)\ =\ -\lambda(M\varphi,\varphi)\ =\
-\lambda\int^{1}_{-1}[\varphi_{1}\overline{\varphi}_{2}\,dm+\varphi_{2}\overline{\varphi}_{1}\,dn]\
=\ 0,$

implying that $\varphi$ is constant, hence zero, a contradiction. ∎

######  Theorem 4.1.

The Dirichlet eigenvalues satisfy the following conditions:

(a) If $\lambda$ is an eigenvalue, so is $-\lambda$.

(b) Each eigenvalue $\lambda$ is real and its eigenspace has dimension one.

###### Proof.

If $\varphi=[\varphi_{1},\varphi_{2}]^{t}$ is an eigenvector with eigenvalue
$\lambda$, it follows immediately from (4.3) that
$[\varphi_{1},-\varphi_{2}]^{t}$ has eigenvalue $-\lambda$. The complex
conjugate $\overline{\varphi}$ has eigenvalue $\overline{\lambda}$.
Integration by parts shows that

$(D^{2}\varphi,\varphi)\ =\ (\varphi,D^{2}\varphi)\ =\
\overline{(D^{2}\varphi,\varphi)},$

so Lemma (4) and (4.18) show that

$0\ \neq\ \lambda\int^{1}_{-1}[|\varphi_{1}|^{2}\,dm+|\varphi_{2}|^{2}\,dn]\
=\
\overline{\lambda}\int^{1}_{-1}[|\varphi_{1}|^{2}\,dm+|\varphi_{2}|^{2}\,dn].$

Therefore $\lambda=\overline{\lambda}$. Then the singular matrix
$\Phi(1,\lambda)$ has positive diagonal entries, so it has two eigenvalues,
namely $0$ and ${\rm tr}\,\Phi(1,\lambda)>0$. Thus the eigenspace for
$\lambda$ has dimension 1. ∎

######  Proposition .

The zeros of $\Delta=\det\Phi(1,\lambda)$ are simple.

###### Proof.

Write

$\Phi(1,\lambda)\ =\ \begin{bmatrix}a(\lambda)&b(\lambda)\\\
c(\lambda)&d(\lambda)\end{bmatrix}\qquad v(\lambda)\ =\
\begin{bmatrix}d(\lambda)\\\ -c(\lambda)\end{bmatrix},$

and let $\varphi(x,\lambda)=\Phi(x,\lambda)v(\lambda)$. The entries of
$\Phi(1,\lambda)$ are non-zero, so $\varphi\neq 0$. We have
$\varphi(-1,\lambda)={\bf 0}$ and

$\varphi(1,\lambda)\ =\ \begin{bmatrix}\varphi_{1}(1,\lambda)\\\
\varphi_{2}(1,\lambda)\end{bmatrix}\ =\ \begin{bmatrix}\Delta(\lambda)\\\
0\end{bmatrix}.$

We want to prove that the derivative $D_{\lambda}\varphi(1,\lambda)$ does not
vanish if $\lambda$ is an eigenvalue. Differentiating with respect to
$\lambda$ shows that at an eigenvalue

$(D_{x}^{2}-\lambda M)D_{\lambda}\varphi\ =\ M\varphi.$

Therefore, by Lemma (4), and because all terms here are real,

$\displaystyle\int^{1}_{-1}[|\varphi_{1}|^{2}\,dm+|\varphi_{2}|^{2}\,dn]\ =\
(M\varphi,\sigma\varphi)$ $\displaystyle\quad\ =\
(D_{x}^{2}D_{\lambda}\varphi,\sigma\varphi)-(\lambda
MD_{\lambda}\varphi,\sigma\varphi)$ $\displaystyle\quad\ =\
-(D_{x}D_{\lambda}\varphi,\sigma D_{x}\varphi)-\lambda(M\varphi,\sigma
D_{\lambda}\varphi)$ $\displaystyle\quad\ =\ -D_{\lambda}\varphi\cdot\sigma
D_{x}\varphi\big{|}^{1}_{-1}+[(D_{\lambda}\varphi,\sigma
D_{x}^{2}\varphi)-(\lambda M\varphi,\sigma D_{\lambda}\varphi)]$
$\displaystyle\quad\ =\
-D_{\lambda}\varphi_{1}(1,\lambda)D_{x}\varphi_{2}(1,\lambda).$

Therefore $D_{\lambda}\Delta(\lambda)=D_{\lambda}\varphi_{1}(1,\lambda)\neq
0$. ∎

##  5 Spectral theory

In this section we rephrase the Dirichlet problem for the Euler-Bernoulli beam
as a standard eigenvalue problem for a compact operator, whose eigenfunctions
are the derivatives $D\varphi$.

Let ${\bf H}$ be the $L^{2}$ space of vector-valued functions on the interval
$[-1,1]$, with the inner product 4.16 and norm

$||f||\ =\ (f,f)^{1/2}.$

We also introduce an indefinite form

$\langle f,g\rangle\ =\
\int^{1}_{-1}[f_{1}(x)\overline{g_{2}(x)}+f_{2}(x)\overline{g_{1}(x)}]\,dx\ =\
(\sigma f,g)\ =\ (f,\sigma g)$ (5.1)

where $\sigma$ is the matrix (4.17).

We know that the Dirichlet eigenvalues $\\{\lambda_{\nu}\\}$ are real, and we
take their eigenvectors $\\{\varphi_{\nu}\\}$ to be real as well. Lemma 4 can
be rephrased as

$\langle f_{\nu},f_{\nu}\rangle\ =\
-\lambda_{\nu}\int^{1}_{-1}[\varphi_{\nu,1}^{2}\,dm+\varphi_{\nu,2}^{2}\,dn]\neq
0.$ (5.2)

Thus $\langle f_{\nu},f_{\nu}\rangle$ and $\lambda_{\nu}$ have opposite signs.
Because of this we index the eigenvalues with

$\dots\ <\ \lambda_{2}\ <\ \lambda_{1}\ <\ 0\ <\lambda_{-1}\ <\ \lambda_{-2}\
<\ \dots\ .$ (5.3)

The calculation that led to (5.2) leads to two formulations for $\langle
f_{\nu},f_{\mu}\rangle$ that show

$\langle f_{\nu},f_{\mu}\rangle\ =\ 0\qquad\hbox{if\ \ $\mu\neq\nu$.}$ (5.4)

Let ${\bf H}_{0}\subset{\bf H}$ consist of the constant functions, and let
${\bf H}_{1}$ be the orthogonal complement:

${\bf H}_{1}\ =\ \left\\{f\in{\bf H}\,:\,\int^{1}_{-1}f=0\right\\}.$

Note that each $f_{\nu}$ belongs to ${\bf H}_{1}$, since
$\int^{1}_{-1}f_{\nu}=\varphi_{\nu}(1)-\varphi_{\nu}(0)=0$.

This discussion leads us to introduce two inverses of $D=D_{x}$ that map ${\bf
H}$ to ${\bf H}$:

$D_{0}^{-1}g(x)\ =\
\begin{cases}\frac{1}{2}\int_{-1}^{x}g(y)\,dy-\frac{1}{2}\int_{x}^{1}g(y)\,dy,&g\in{\bf
H}_{1},\\\ 0,&g\in{\bf H}_{0},\end{cases}$ (5.5)

and

$D_{1}^{-1}g(x)\ =\
\frac{1}{2}\int_{-1}^{x}(y+1)g(y)\,dy+\frac{1}{2}\int_{x}^{1}(y-1)g(y)\,dy.$
(5.6)

######  Lemma .

For any $g\in{\bf H}$,

$D_{0}^{-1}g(-1)\ =\ D_{0}^{-1}g(1)\ =\ 0;\qquad D_{1}^{-1}g\in{\bf H}_{1}.$

###### Proof.

The first pair of identities is clear for $g\in{\bf H}_{1}$ and true by
definition for $g\in{\bf H}_{0}$. The third identity follows from an easy
calculation. ∎

In view of these remarks, the Dirichlet eigenvalue problem can be reformulated
as

$Df\ =\ \lambda MD_{0}^{-1}f,\qquad f\in{\bf H}_{1}.$

or, equivalently,

$f\in{\bf H}_{1},\qquad f\ =\ \lambda Tf,\qquad T\ =\ D_{1}^{-1}MD_{0}^{-1}.$
(5.7)

Thus the problem (4.4), (4.11) is equivalent to a standard eigenvalue problem
for the operator $T$ mapping ${\bf H}_{1}$ to ${\bf H}_{1}$ or ${\bf H}$ to
${\bf H}$. Note that $T=0$ on ${\bf H}_{0}$, since this is true, by definition
of $D_{0}^{-1}$.

######  Lemma .

The operator $T$ is a compact operator in ${\bf H}$. If $f$ is in the image of
$T$, then $f$ has bounded variation and is continuous at the endpoints $x=\pm
1$.

###### Proof.

Any inverse $D^{-1}$ takes $L^{2}$ functions to functions that satisfy a
Hölder condition: if $u=D^{-1}f$ and $x<y$, then $Du=f$ so

$|u(y)-u(x)|\ \leq\ \int_{x}^{y}|f(t)|\,dy\ \leq\
\left[\int_{x}^{y}|f|^{2}\right]^{1/2}\,\left[\int_{x}^{y}dt\right]^{1/2}\
\leq\ ||f||\,(y-x)^{1/2}.$

It follows that the image under $D_{0}^{-1}$ of a bounded sequence in ${\bf
H}$ is uniformly equicontinuous. The same argument applied to the two summands
in (5.5) shows that $|D_{0}^{-1}f(x)|\leq 2||f||$, so the image is also
uniformly bounded. By the theorem of Ascoli–Arzelá there is a uniformly
convergent subsequence, $\\{g_{n}\\}$. Thus $D_{0}^{-1}$ is compact from ${\bf
H}$ to the space of bounded continuous functions. The map $g\to D_{1}^{-1}gM$
from this space to ${\bf H}$ is bounded, so $T$ is compact. Moreover $g\to
D_{1}^{-1}gM$ maps to functions of bounded variation; continuity of $Tf$ at
the endpoints follows from assumption (a) of (4.5). ∎

######  Lemma .

If $K\neq(0)$ is a closed subspace of $\,{\bf H}$ that is invariant under $T$,
then $K$ contains an eigenfunction of $T$.

###### Proof.

Since $T$ is compact, every non-zero point of the spectrum is an eigenvalue.
This applies also to the restriction of $T$ to the invariant subspace $K$.
Compactness implies that the operator $I-\lambda T$ is Fredholm with index
zero. If the restriction to $K$ has no null space, for all
$\lambda\in\mathbf{C}$, then it is invertible, and the inverse is entire and
bounded at $\lambda=\infty$, leading to a contradiction. Therefore $T$ has an
eigenvector in $K$. ∎

######  Lemma .

The operator $T$ is symmetric with respect to the indefinite form $\langle\ ,\
\rangle$.

###### Proof.

We may assume that $f$ and $g$ are in $H_{1}$ and are smooth. Since
$g=D[D_{0}^{-1}g]$ and $D_{0}^{-1}g(\pm 1)=0$ we may integrate by parts to get

$\displaystyle\langle Tf,g\rangle$ $\displaystyle=$ $\displaystyle(Tf,\sigma
DD_{0}^{-1}g\rangle\ =\ -(DTf,\sigma D_{0}^{-1}g)$ (5.8) $\displaystyle=$
$\displaystyle-(MD_{0}^{-1}f,\sigma D_{0}^{-1}g)\ =\ -(D_{0}^{-1}f,M^{t}\sigma
D_{0}^{-1}g).$

Since $M^{t}\sigma$ is diagonal, the last expression is symmetric in $f$ and
$g$. ∎

Let ${\bf H}_{T}$ be the closure in ${\bf H}$ of the range of $T$. Any
solution of (5.7) will belong to ${\bf H}_{T}$.

######  Proposition .

The span of the eigenfunctions $\\{f_{\nu}\\}$ is dense in ${\bf H}_{T}$.

###### Proof.

Let $N_{T}\in{\bf H}_{T}$ be the orthogonal complement of $\\{f_{\nu}\\}$ in
${\bf H}_{T}$ with respect to the standard inner product. Then $\sigma N_{T}$
is orthogonal to ${\bf H}_{T}$ with respect to the indefinite form. Since $T$
is symmetric, $\sigma N_{T}$ is invariant for $T$. By construction, $\sigma
N_{T}$ is orthogonal to every eigenfunction. It follows from Lemma 5 that
$\sigma N_{T}=(0)$. Therefore $N_{T}=(0)$. ∎

######  Proposition .

Let $N$ be the null space of $T$, i.e. $N=\\{f\in{\bf H};Tf=0\\}$. Then
$N=\sigma N$. Moreover, $N$ coincides with each of the subspaces

(a) The orthogonal complement of ${\bf H}_{T}$ with respect to $(\ ,\ )$;

(b) The orthogonal complement of ${\bf H}_{T}$ with respect to $\langle\ ,\
\rangle$.

###### Proof.

Note that $Tf=0$ if and only if either $f\in H_{0}$, in which case $\sigma
H_{0}=H_{0}$, or $f\in H_{1}$ and $M\varphi=0$, where $\varphi=D_{0}^{-1}f$.
This, in turn, is equivalent to the conditions that $\varphi_{1}$ vanish on
the support of $m$ and $\varphi_{2}$ vanish on the support of $n$. In view of
condition (b) of (4.5), this is equivalent to $M\sigma\varphi=0$, so $T\sigma
f=0$ whenever $f\in N$.

The relations between $N$ and the spaces (a), (b) follow from the identities,
valid for every $f,g$ in ${\bf H}$:

$\displaystyle(f,Tg)$ $\displaystyle=$ $\displaystyle\langle\sigma
f,Tg\rangle\ =\ \langle T(\sigma f),g\rangle;$ $\displaystyle\langle
Tf,g\rangle$ $\displaystyle=$ $\displaystyle\langle f,Tg\rangle.$

The first identity shows that $f$ is in space (a) if and only if $T(\sigma
f)=0$, which is equivalent to $Tf=0$. The second shows that $f$ is in space
(b) if and only if $Tf=0$. ∎

There is a natural decomposition of ${\bf H}_{T}$ into two subspaces that are
orthogonal with respect to the indefinite form:

${\bf H}_{T}^{\pm}\ =\ \hbox{closure of the span of
$\\{f_{\nu}\,:\,\pm\nu>0\\}$.}$

We introduce a new inner product in ${\bf H}_{T}$ by setting

$(f,g)_{T}\ =\ \begin{cases}\ \langle f,g\rangle,\quad f,g\in{\bf
H}_{T}^{+};\\\ -\langle f,g\rangle,\quad f,g\in{\bf H}_{T}^{-};\\\ \
0,\quad\qquad f\in{\bf H}_{T}^{+},\ g\in{\bf H}_{T}^{-}.\end{cases}$ (5.9)

Let $\widehat{\bf H}_{T}$ be the completion of ${\bf H}_{T}$ with respect to
the norm $||f||_{T}^{2}\ =\ (f,f)_{T}$. Note that in all cases

$(f,f)_{T}\ =\ |\langle f,f\rangle|\ =\ |(f,\sigma f)|\ \leq||f||\,||\sigma
f||\ =\ ||f||^{2}.$ (5.10)

The $\\{f_{\nu}\\}$ are clearly an orthonormal basis for this space. Since
$Tf_{\nu}=\lambda_{\nu}^{-1}f_{\nu}$, the restriction of $T$ to ${\bf H}_{T}$
extends to a compact self-adjoint operator in $\widehat{\bf H}_{T}$ with
eigenvalues (in the usual sense) $\\{\lambda_{\nu}^{-1}\\}$.

The orthogonal projection of $\widehat{\bf H}_{T}$ onto the span of the
eigenfunction $f_{\nu}$ is

$E_{\nu}f\ =\ \frac{\langle f,f_{\nu}\rangle}{\langle
f_{\nu},f_{\nu}\rangle}\,f_{\nu}\ =\ \frac{(f,\sigma f_{\nu})}{\langle
f_{\nu},f_{\nu}\rangle}\,f_{\nu}.$ (5.11)

Abusing notation, we write $E_{\nu}$ also for the kernel of the operator
(5.11):

$\displaystyle E_{\nu}f(x)$ $\displaystyle=$
$\displaystyle\int^{1}_{-1}E_{\nu}(x,y)f(y)\,dy,\qquad f\in{\bf H}_{T}$
$\displaystyle E_{\nu}(x,y)$ $\displaystyle=$ $\displaystyle\frac{1}{\langle
f_{\nu},f_{\nu}\rangle}\left[\begin{matrix}f_{\nu,1}(x)f_{\nu,2}(y)&f_{\nu,1}(x)f_{\nu,1}(y)\\\
f_{\nu,2}(x)f_{\nu,2}(y)&f_{\nu,2}(x)f_{\nu,1}(y)\end{matrix}\right].$ (5.12)

We may also view the operator $E_{\nu}$ as a projection of ${\bf H}$ onto the
span of $f_{\nu}$. We view the kernels $E_{\nu}$ as belonging to the Hilbert
space

${\bf H}^{(2)}\ =\ L^{2}(I\times I;M(2,\mathbf{C}))$

of mappings from the square $I\times I=[-1,1]\times[-1,1]$ to the space
$M(2,\mathbf{C})$ of complex $2\times 2$ matrices.

Let us write the integral kernel for the operator $D_{1}^{-1}$ as $\kappa$.
Since $\frac{1}{2}|y\pm 1|\leq 1$ for all $y\in[-1,1]$, we have $|\kappa|\leq
1$.

If $r$ is a bounded positive measure on the interval $[-1,1]$, then

$|D_{1}^{-1}r(x)|\ =\ \left|\int_{-1}^{1}\kappa(x,y)\,dr(y)\right|.$

######  Lemma .

Suppose that $r$ is a bounded positive measure and $\psi$ is a continuous
function on the interval $[-1,1]$. Then

$|[D_{1}^{-1}\psi r](x)|^{2}\ \leq\
\overline{r}\int^{1}_{-1}\psi(y)^{2}\,dr(y),\quad\overline{r}=\int^{1}_{-1}dr(y).$
(5.13)

###### Proof.

This follows by applying the Cauchy–Schwarz inequality to the term on the
right in the inequality

$\left|[D_{1}^{-1}(\psi r)](x)\right|\ =\
\left|\int^{1}_{-1}\kappa(x,y)\psi(y)\,dr(y)\right|\ \leq\
\int^{1}_{-1}|\psi(y)|\,dr(y).\qquad\qed$

######  Proposition .

Each element of the kernel (5.12) has absolute value bounded by
$(\overline{m}+\overline{n})|\lambda_{\nu}|$, where
$\overline{m}=\int^{1}_{-1}dm$ and $\overline{n}=\int^{1}_{-1}dn$.

###### Proof.

Since $f_{\nu}=\lambda_{\nu}Tf_{\nu}$, we have

$f_{\nu,1}\ =\ \lambda_{\nu}D_{1}^{-1}(\varphi_{\nu,2}n),\quad\varphi_{\nu}\
=\ D_{0}^{-1}f_{\nu}.$

By Lemma 5,

$|f_{\nu,1}(x)|^{2}\ \leq\\\
\lambda_{\nu}^{2}\overline{n}\int^{1}_{-1}\varphi_{\nu,2}^{2}\,dn\ \leq\
\overline{n}|\lambda_{\nu}\langle f_{\nu},f_{\nu}\rangle|.$

Similarly,

$|f_{\nu,2}(x)|^{2}\ \leq\ \overline{m}|\lambda_{\nu}\langle
f_{\nu},f_{\nu}\rangle|.$

Therefore each entry of $E_{\nu}$ is bounded by one of
$|\lambda_{\nu}|\overline{m}$, $|\lambda_{\nu}|\overline{n}$, or
$|\lambda_{\nu}|(\overline{m}\overline{n})^{1/2}$. ∎

For later use we introduce the operator $R_{\lambda}$ defined by

$R_{\lambda}\ =\ I-(I-\lambda T)^{-1}\ =\ -\lambda T(1-\lambda T)^{-1}.$
(5.14)

The operator $R_{\lambda}$ maps ${\bf H}$ to ${\bf H}_{T}$ and is compact, as
is the extension to $\widehat{H}_{T}$ of its restriction to ${\bf H}_{T}$.
Since $Tf_{\nu}=\lambda_{\nu}^{-1}f_{\nu}$,

$R_{\lambda}f_{\nu}\ =\
-\frac{\lambda}{\lambda_{\nu}}\left(1-\frac{\lambda}{\lambda_{\nu}}\right)^{-1}f_{\nu}\
=\ \frac{\lambda}{\lambda-\lambda_{\nu}}\,f_{\nu}.$

Therefore, we have a formal expansion

$R_{\lambda}\ =\ \sum_{\nu}\frac{\lambda}{\lambda-\lambda_{\nu}}E_{\nu}$
(5.15)

with a formal kernel

$\displaystyle\widehat{K}_{\lambda}(x,y)$ $\displaystyle=$
$\displaystyle\sum_{\nu}\frac{\lambda}{\lambda-\lambda_{\nu}}E_{\nu}(x,y)$
(5.16) $\displaystyle=$
$\displaystyle\sum_{\nu}\frac{\lambda}{\lambda-\lambda_{\nu}}\frac{1}{\langle
f_{\nu},f_{\nu}\rangle}\left[\begin{matrix}f_{\nu,1}(x)f_{\nu,2}(y)&f_{\nu,1}(x)f_{\nu,1}(y)\\\
f_{\nu,2}(x)f_{\nu,2}(y)&f_{\nu,2}(x)f_{\nu,1}(y)\end{matrix}\right].$

The question is: does this series converge, in some sense, to the kernel of
$R_{\lambda}$?

######  Theorem 5.1.

For each $\lambda$ that is not in the set of eigenvalues
$\\{\lambda_{\nu}\\}$, the partial sums of the series on the right in (5.16)
converge weakly in ${\bf H}^{(2)}$. The weak limit $\widehat{K}_{\lambda}$ is
the kernel for $R_{\lambda}$.

###### Proof.

For $|\lambda_{\nu}|>2|\lambda$|, $|(\lambda-\lambda_{\nu})^{-1}|$ is less
than $2/|\lambda_{\nu}|$. By Proposition 5, the corresponding summand is
$O(1)$ as an element of ${\bf H}^{(2)}$. Linear combinations of matrix
functions $f(x)g(y)^{t}$, $f,g\in{\bf H}$ are dense in ${\bf H}^{(2)}$. If
either $f$ or $g$ is in $N+{\rm span}\\{f_{\nu}\\}$, integration against
(5.16) yields a finite sum which is $(f,R_{\lambda}g)$. This proves the weak
convergence. Let $\widehat{K}_{\lambda}$ be the weak limit. As an element of
${\bf H}^{(2)}$, it induces a bounded operator in ${\bf H}$. This operator
agrees with $R_{\lambda}$ on a dense subspace, so it is $R_{\lambda}$. ∎

In the next section we derive a closed-form expression for the kernel
$\widehat{K}_{\lambda}$.

Theorem 5.1 can be strengthened considerably under a strengthening of the
assumption (4.5) (b), that $m$ and $n$ have the same support: namely that each
is dominated by the other. This can be put in the form

${\rm(b^{\prime})}\hbox{ There is a constant $C$ such that $m+n\leq Cm$ and
$m+n\leq Cn$.}$ (5.17)

######  Lemma .

Under assumption (5.17), for each eigenfunction $f_{\nu}$,

$|\langle f_{\nu},f_{\nu}\rangle|\ \leq\ (f_{\nu},f_{\nu})\ \leq\ C|\langle
f_{\nu},f_{\nu}\rangle|.$ (5.18)

###### Proof.

The first inequality is (5.10). To prove the second inequality, let
$\varphi_{\nu}=D_{0}^{-1}f_{\nu}$. Then

$(f_{\nu},f_{\nu})\ =\ (D\varphi_{\nu},D\varphi_{\nu})\ =\
-(D^{2}\varphi_{\nu},\varphi_{\nu})\ =\
-\lambda_{\nu}(M\varphi_{\nu},\varphi_{\nu})$ (5.19)

and

$\displaystyle(M\varphi_{\nu},\varphi_{\nu})^{2}$ $\displaystyle=$
$\displaystyle\left[\int^{1}_{-1}\varphi_{\nu,1}\varphi_{\nu,2}d(m+n)\right]^{2}$
(5.20) $\displaystyle\leq$
$\displaystyle\int^{1}_{-1}\varphi_{\nu,1}^{2}d(m+n)\int^{1}_{-1}\varphi_{\nu,2}^{2}d(m+n)$
$\displaystyle\leq$ $\displaystyle
C^{2}\int^{1}_{-1}\varphi_{\nu,1}^{2}\,dm\int^{1}_{-1}\varphi_{\nu,2}^{2}\,dn$
$\displaystyle\leq$ $\displaystyle
C^{2}\left\\{\int^{1}_{-1}[\varphi_{\nu,1}^{2}\,dm+\varphi_{\nu,2}^{2}\,dn]\right\\}^{2}$
$\displaystyle=$ $\displaystyle C^{2}\lambda_{\nu}^{-2}\langle
f_{\nu},f_{\nu}\rangle^{2}.$

Together, (5.19) and (5.20) establish the second inequality in (5.18). ∎

This result leads to the following strengthening of the previous convergence
result.

######  Theorem 5.2.

Under assumption (5.17), for each $\lambda$ that is not an eigenvalue, the
series (5.16) converges in $L^{2}$ norm to the kernel of $R_{\lambda}$.

######  Remark .

The preceding arguments can easily be extended to similar series. The formal
series

$\sum_{\nu}\frac{1}{\lambda_{\nu}}E_{\nu}(x,y)$

converges weakly to the kernel of $T$, and under the assumption (5.17) it
converges in $L^{2}$ norm. Also, under assumption (5.17) the formal series

$\sum_{\nu}E_{\nu}(x,y)$

converges weakly to the kernel of the orthogonal projection of ${\bf H}$ onto
${\bf H}_{T}$.

##  6 Wronskians and Green’s kernels

Let $\Phi(x,\lambda)$ be the partial fundamental matrix solution of (4.4),
normalized at $x=-1$, as constructed earlier. Let $\Psi(x,\lambda)$ be the
matrix solution normalized at $x=1$:

$\displaystyle D^{2}\Phi$ $\displaystyle=$ $\displaystyle\lambda
M\Phi,\quad\Phi(-1,\lambda)\ =\ 0,\quad D\Phi(-1,\lambda)\ =\ {\bf 1};$
$\displaystyle D^{2}\Psi$ $\displaystyle=$ $\displaystyle\lambda
M\Psi,\quad\Psi(1,\lambda)\ =\ 0,\quad D\Psi(1,\lambda)\ =\ -{\bf 1}.$

As for $\Phi$, condition (4.5) (a) implies that $\Psi$ and $D_{x}\Psi$ are
continuous at $x=\pm 1$.

If $C$ is a matrix, let

$\widehat{C}\ =\ C^{t}\sigma.$ (6.1)

Differentiating shows that quasi-Wronskians like
$\widehat{\Phi}_{x}\Psi-\widehat{\Phi}\Psi_{x}$ are constant; for example

$\displaystyle D[\widehat{\Phi}_{x}\Psi-\widehat{\Phi}\Psi_{x}]$
$\displaystyle=$
$\displaystyle(\lambda\Phi^{t}M^{t}\sigma\Psi+\Phi_{x}^{t}\sigma\Psi_{x})-(\Phi_{x}^{t}\sigma\Psi_{x}+\lambda\Phi^{t}\sigma
M\Psi)$ $\displaystyle=$ $\displaystyle 0,$

since $M^{t}\sigma=\sigma M$. The value of the constant can be computed by
taking $x=\pm 1$. Considering the various possibilities, we have (taking
$\lambda$ as given, $\lambda\in(\mathbf{C}\setminus\mathbf{R}))$:

$\displaystyle\widehat{\Phi}_{x}\Phi-\widehat{\Phi}\Phi_{x}$ $\displaystyle=$
$\displaystyle 0,\quad{\rm so}\ \ \widehat{\Phi}^{-1}\widehat{\Phi}_{x}\ =\
\Phi_{x}\Phi^{-1};$ (6.2)
$\displaystyle\widehat{\Psi}_{x}\Psi-\widehat{\Psi}\Psi_{x}$ $\displaystyle=$
$\displaystyle 0,\quad{\rm so}\ \ \widehat{\Psi}^{-1}\widehat{\Psi}_{x}\ =\
\Psi_{x}\Psi^{-1};$ (6.3)
$\displaystyle\widehat{\Phi}_{x}\Psi-\widehat{\Phi}\Psi_{x}$ $\displaystyle=$
$\displaystyle-C_{-}\ =\ \sigma\Psi(-1,\lambda)\ =\
\widehat{\Phi}(1,\lambda);$ (6.4)
$\displaystyle\widehat{\Psi}_{x}\Phi-\widehat{\Psi}\Phi_{x}$ $\displaystyle=$
$\displaystyle C_{+}\ =\ -\sigma\Phi(1,\lambda)\ =\
-\widehat{\Psi}(-1,\lambda).$ (6.5)

Combining some of these identities we find that

$\displaystyle C_{-}$ $\displaystyle=$
$\displaystyle-\widehat{\Phi}[\widehat{\Phi}^{-1}\widehat{\Phi}_{x}-\Psi_{x}\Psi^{-1}]\Psi\
=\ \widehat{\Phi}A\Psi;$ (6.6) $\displaystyle C_{+}$ $\displaystyle=$
$\displaystyle\widehat{\Psi}[\widehat{\Psi}^{-1}\widehat{\Psi}_{x}-\Phi_{x}\Phi^{-1}]\Phi\
=\ \widehat{\Psi}A\Phi;$ (6.7)

where

$A(x,\lambda)\ =\ \Psi_{x}\Psi^{-1}-\Phi_{x}\Phi^{-1}.$ (6.8)

The identities (6.4) and (6.7) lead to two additional important identities:

$\Psi C_{-}^{-1}\widehat{\Phi}-\Phi C_{+}^{-1}\widehat{\Psi}\ =\
A^{-1}-A^{-1}\ =\ 0$ (6.9)

and

$\Psi_{x}C_{-}^{-1}\widehat{\Phi}-\Phi_{x}C_{+}^{-1}\widehat{\Psi}\ =\
\Psi_{x}\Psi^{-1}A^{-1}-\Phi_{x}\Phi^{-1}A^{-1}\ =\ {\bf 1}.$ (6.10)

The identities (6.2), (6.3), (6.6), (6.7), (6.8), (6.9) and (6.10) call for
some discussion.

######  Lemma .

Suppose $\lambda$ is not real. Then the matrix functions $\Phi(x,\lambda)$ and
$\Psi(x,\lambda)$ are invertible for $x\in(-1,1]$, $x\in[-1,1)$, respectively.

###### Proof.

Suppose that $\Phi(x,\lambda)$ is not invertible at some point $x_{0}$ in the
interval $(-1,1)$. Then $x_{0}>-1$ and $\lambda$ is a Dirichlet eigenvalue for
the beam problem restricted to the interval $[-1,x_{0}]$. Therefore $\lambda$
is real. The same argument applies to $\Psi$ for an interval $[x_{1},1]$. ∎∎

It follows that all the expressions above are well-defined when $\lambda$ is
not real and $\left\lvert x\right\rvert<1$. Moreover, (6.4) and (6.5) imply
that $C_{-}$ and $C_{+}$ are invertible if $\lambda$ is not real. In turn,
these imply that $A$ is invertible if $\lambda$ is not real and $\left\lvert
x\right\rvert<1$. In summary,

######  Corollary .

The identities (6.2) – (6.10) are valid for each non-real $\lambda$ and
$\left\lvert x\right\rvert<1$.

######  Theorem 6.1.

The matrix function

$G_{\lambda}(x,y)\ =\
\begin{cases}\Psi(x,\lambda)\,C_{-}^{-1}\,\widehat{\Phi}(y,\lambda),\quad
y<x;\\\ \Phi(x,\lambda)\,C_{+}^{-1}\,\widehat{\Psi}(y,\lambda),\quad
y>x.\end{cases}$ (6.11)

is the Green’s kernel for the equation $D^{2}u-\lambda Mu=f$,
$\lambda\notin\mathbf{R}$.

###### Proof.

Let

$\displaystyle u(x)$ $\displaystyle=$
$\displaystyle\int^{1}_{-1}G_{\lambda}(x,y)f(y)\,dy$ (6.12) $\displaystyle=$
$\displaystyle\Psi(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}(y)f(y)\,dy+\Phi(x)\int_{x}^{1}C_{+}^{-1}\widehat{\Psi}(y)f(y)\,dy.$

Then, using (6.9),

$\displaystyle Du(x)$ $\displaystyle=$
$\displaystyle[\Psi(x)C_{-}^{-1}\widehat{\Phi}(x)-\Phi(x)C_{+}^{-1}\widehat{\Psi}(x)]f(x)$
$\displaystyle\quad+\Psi_{x}(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}(y)f(y)\,dy+\Phi_{x}(x)\int_{x}^{1}C_{+}^{-1}\widehat{\Psi}(y)f(y)\,dy$
$\displaystyle=$ $\displaystyle
0+\Psi_{x}(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}(y)f(y)\,dy+\Phi_{x}(x)\int_{x}^{1}C_{+}^{-1}\widehat{\Psi}(y)f(y)\,dy.$

Therefore, using (6.10),

$\displaystyle D^{2}u(x)$ $\displaystyle=$ $\displaystyle\lambda
M(x)u(x)+[\Psi_{x}C_{-}^{-1}\widehat{\Phi}-\Phi_{x}C_{+}^{-1}\widehat{\Psi}](x)f(x)$
$\displaystyle=$ $\displaystyle\lambda M(x)u(x)+f(x).\qquad\qed$

The kernel $G_{\lambda}$ can be used to calculate the kernel for the operator
$R_{\lambda}$ defined in (5.14). Given $f\in{\bf H}_{1}$ with $g=Df$
integrable, consider the inhomogeneous problem

$(D^{2}-\lambda M)u\ =\ g$ (6.13)

with solution

$u(x)\ =\ \int^{1}_{-1}G_{\lambda}(x,y)g(y)\,dy.$

Let $v=Du$, so (6.13) is equivalent to

$Dv-\lambda MD_{0}^{-1}v\ =\ Df$

or $v-\lambda Tv=f$, which is the same as

$v\ =\ f-R_{\lambda}f.$ (6.14)

Now

$u(x)\ =\ \int^{1}_{-1}G_{\lambda}(x,y)g(y)\,dy.$

Thus the solution to (6.14) is, using (6.9) and (6.10) again,

$\displaystyle v(x)$ $\displaystyle=$ $\displaystyle Du(x)$ $\displaystyle=$
$\displaystyle\Psi_{x}(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}(y)Df(y)\,dy+\Phi_{x}(x)\int_{x}^{1}C_{+}^{-1}\Phi(y)Df(y)\,dy$
$\displaystyle=$ $\displaystyle
f(x)-\Psi_{x}(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}_{y}(y)f(y)\,dy-\Phi_{x}(x)\int_{x}^{1}C_{+}^{-1}\widehat{\Psi}_{y}(y)f(y)\,dy.$

This shows that the kernel

$[G_{\lambda}]_{xy}\ =\
\begin{cases}\Psi_{x}(x)\,C_{-}^{-1}\widehat{\Phi}_{y}(y),\qquad y<x;\\\
\Phi_{x}(x)\,C_{+}^{-1}\,\widehat{\Psi}_{y}(y),\qquad y>x.\end{cases}$ (6.15)

generates $R_{\lambda}$ for functions in ${\bf H}_{1}$. The operator
$R_{\lambda}$ is zero on ${\bf H}_{0}$, while the integral of
$[G_{\lambda}]_{xy}$ against ${\bf 1}$ is

$\int^{1}_{-1}[G(x,y)_{\lambda}]_{xy}\,dy\ =\
\Psi_{x}(x)\int_{-1}^{x}C_{-}^{-1}\widehat{\Phi}_{y}(y)\,dy+\Phi_{x}(x)\int_{x}^{1}C_{+}^{-1}\widehat{\Psi}_{y}(y)\,dy,$

which, by (6.10), is the identity matrix. We can compensate by subtracting
$\frac{1}{2}{\bf 1}$ from $[G_{\lambda}]_{xy}(x,y)$ as kernel.

Summarizing,

######  Theorem 6.2.

The kernel for the operator $R_{\lambda}=-\lambda T(I-\lambda T)^{-1}$ is

$K_{\lambda}(x,y)\ =\ [G_{\lambda}]_{xy}(x,y)-\tfrac{1}{2}{\bf 1}.$ (6.16)

We can now relate this to the kernel $\widehat{K}_{\lambda}$ defined using the
formal series (5.16). We have shown that they each define the same operator
$R_{\lambda}$ in ${\bf H}$. This means that they coincide as elements of the
$L^{2}$ space ${\bf H}^{(2)}$ [*], so we may choose to identify them at each
point $(x,y)$.

######  Theorem 6.3.

The kernels $K_{\lambda}$ and $\widehat{K}_{\lambda}$ are identical.

For later use we define here the Weyl function for the beam Dirichlet problem
to be

$W(\lambda)\ =\ \frac{1}{\lambda}D\Phi(1,\lambda)\Phi(1,\lambda)^{-1},$ (6.17)

The representation (6.16) shows that

$W(\lambda)\ =\ \frac{1}{\lambda}K_{\lambda}(1,1)+\frac{1}{2\lambda}{\bf 1}.$
(6.18)

It follows that $W$ has a pole at the origin with residue

$D\Phi(1,0)\Phi(1,0)^{-1}\ =\ \frac{1}{2}{\bf 1}.$ (6.19)

Then the representation (5.16) shows that, formally at least,

$W(\lambda)\ =\ \frac{1}{2\lambda}{\bf
1}+\sum_{\nu}\frac{1}{\lambda-\lambda_{\nu}}\frac{1}{\langle
f_{\nu},f_{\nu}\rangle}\left[\begin{matrix}f_{\nu,1}(1)f_{\nu,2}(1)&f_{\nu,1}(1)f_{\nu,1}(1)\\\
f_{\nu,2}(1)f_{\nu,2}(1)&f_{\nu,2}(1)f_{\nu,1}(1)\end{matrix}\right].$ (6.20)

(We omit a detailed justification of (6.20) in the general case, since the
only use we shall make is to the case when $\\{\lambda_{\nu}\\}$ is finite.)

We assume here that the $f_{\nu}$ are chosen to be real. It will be useful to
understand the signs of the entries of the summands.

######  Lemma .

If $\nu>0$, then $f_{\nu,1}$ and $f_{\nu,2}$ have the same sign.

###### Proof.

Because of the relationship between eigenfunctions for $\pm\lambda_{\nu}$,
this is equivalent to the statement that if $\nu<0$ then $f_{\nu,1}$ and
$f_{\nu,2}$ have opposite signs. With our choice of indexing, $\nu<0$ means
$\lambda_{\nu}>0$. The corresponding $\phi_{\nu}$ is $\Psi v$ for some fixed
2-vector $v$. Then $\Psi(-1,\lambda_{\nu})v={\bf 0}$. But $\lambda_{\nu}>0$
implies that all entries of $\Psi(-1,\lambda_{\nu})$ are positive (this is the
dual of the argument for Proposition 4) so $v_{1}v_{2}<0$. Then
$D\phi_{\nu}(1)=\Psi_{x}(1,\lambda)v=-v$. Since $f_{\nu}$ is a multiple of
$v$, its entries have opposite signs. ∎

To simplify the notation in (6.20), let

$\alpha_{\nu}\ =\ \frac{f_{\nu,1}(1)}{\sqrt{\langle
f_{\nu},f_{\nu}\rangle}},\quad\beta_{\nu}\ =\
\frac{f_{\nu,2}(1)}{\sqrt{\langle f_{\nu},f_{\nu}\rangle}},\quad\hbox{for\ \
$\nu>0$}.$ (6.21)

By Lemma 6 we may take $\alpha_{\nu}$ and $\beta_{\nu}$ positive. Taking into
account the relation between $f_{\nu}$ and $f_{-\nu}$ and between $\langle
f_{\nu},f_{\nu}\rangle$ and $\langle f_{-\nu},f_{-\nu}\rangle$ it follows that

$\displaystyle W(\lambda)$ $\displaystyle=$
$\displaystyle\frac{1}{2\lambda}{\bf
1}+\sum_{\nu<0}\frac{1}{\lambda-\lambda_{\nu}}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&-\alpha_{\nu}^{2}\\\
-\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right]$ (6.22)
$\displaystyle\quad+\sum_{\nu>0}\frac{1}{\lambda-\lambda_{\nu}}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&\alpha_{\nu}^{2}\\\
\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right],$

where we set $\alpha_{-\nu}=\alpha_{\nu},\,\beta_{-\nu}=\beta_{\nu}$.

##  7 The discrete beam

The discrete beam is characterized by measures $m$ and $n$ that are supported
on discrete points

$-1\ <\ x_{1}\ <\ x_{2}\ <\ \dots\ <\ x_{d-1}\ <\ x_{d}\ <\ 1,$

with masses $m_{j}$, $n_{j}$. For convenience we also define

$x_{0}\ =\ -1,\quad\ \ x_{d+1}\ =\ 1,\quad\ \ l_{j}\ =\ x_{j+1}-x_{j},\quad\ \
M_{0}\ =\ {\bf 0}.$

Here conditions (4.5) and (5.17) both reduce to the assumption that
$m_{j}n_{j}>0$, $j=1,\dots d$.

The partial fundamental solution $\Phi(x,\lambda)$ satisfies $D^{2}\Phi=0$
except at the $x_{j}$, so it is piecewise linear in $x$, and the derivative
$D\Phi$ is piecewise constant. Thus for any given $\lambda$ the function
$\Phi$ is characterized by its values

$\Phi_{j}\ =\ \Phi_{j}(\lambda)\ =\ \Phi(x_{j},\lambda),\quad j=0,\dots d+1.$
(7.1)

Similarly, $D\Phi=\Phi_{j}^{\prime}$ is characterized by its one-sided values

$\Phi_{j}^{\prime}\ =\ D\Phi_{j}(\lambda)\ =\ D\Phi(x_{j}-,\lambda),\quad
j=1,\dots d+1.$ (7.2)

The beam equation $D^{2}\Phi=\lambda M\Phi$, with initial conditions

$\Phi(-1,\lambda)\ =\ {\bf 0},\qquad D\Phi(-1,\lambda)={\bf 1},$

translates to the conditions

$\displaystyle\Phi_{0}$ $\displaystyle=$ $\displaystyle{\bf
0},\qquad\Phi_{j+1}\ =\ \Phi_{j}+l_{j}\Phi^{\prime}_{j+1},$
$\displaystyle\Phi^{\prime}_{1}$ $\displaystyle=$ $\displaystyle{\bf
1},\qquad\Phi^{\prime}_{j+1}\ =\ \Phi^{\prime}_{j}+\lambda M_{j}\Phi_{j}.$

These relations can be put in two forms:

$\left[\begin{matrix}\Phi_{j+1}\\\ \Phi^{\prime}_{j+1}\end{matrix}\right]\ =\
\left[\begin{matrix}{\bf 1}&l_{j}{\bf 1}\\\ {\bf 0}&{\bf
1}\end{matrix}\right]\left[\begin{matrix}\Phi_{j}\\\
\Phi^{\prime}_{j+1}\end{matrix}\right]$ (7.3)

and

$\left[\begin{matrix}\Phi_{j+1}\\\ \Phi^{\prime}_{j+1}\end{matrix}\right]\ =\
\left[\begin{matrix}{\bf 1}+\lambda l_{j}M_{j}&l_{j}{\bf 1}\\\ \lambda
M_{j}&{\bf 1}\end{matrix}\right]\left[\begin{matrix}\Phi_{j}\\\
\Phi^{\prime}_{j}\end{matrix}\right]\ =\ T_{j}\left[\begin{matrix}\Phi_{j}\\\
\Phi^{\prime}_{j}\end{matrix}\right].$ (7.4)

######  Lemma .

Each of $\Phi_{j}$ and $\Phi^{\prime}_{j}$ is a polynomial of degree $j-1$;
the even part is diagonal and the odd part is off-diagonal. The Dirichlet
spectrum has $2d$ elements.

###### Proof.

The first statement follows by induction from the recursion relations (7.4). A
consequence is that the determinant $\Delta(\lambda)=\det\Phi_{d+1}$ is a
polynomial of degree $d$ in $\lambda^{2}$, so the eigenvalues come in $d$
pairs. ∎

As shown in the general case, the eigenvalues are distinct and real.

The Weyl function (6.17) in the discrete case is

$W(\lambda)\ =\ \frac{1}{\lambda}\Phi^{\prime}_{d+1}\Phi_{d+1}^{-1}.$ (7.5)

The recursion relations (7.3) imply that $W(\lambda)$ has a continued fraction
expansion involving non-commuting coefficients [22, 24].

######  Proposition .

$W(\lambda)=\cfrac{1}{\lambda
l_{d}\mathbf{1}+\cfrac{1}{M_{d}+\cfrac{1}{\lambda
l_{d-1}\mathbf{1}+\cfrac{1}{M_{d-1}+\cfrac{1}{\ddots+\cfrac{1}{\lambda
l_{0}\mathbf{1}}}}}}}$ (7.6)

###### Proof.

Let $W_{j}=\lambda^{-1}\Phi^{\prime}_{j}\Phi_{j}^{-1}$. Note that, for
$\lambda$ large enough, all $\Phi^{\prime}_{j}$ and $\Phi_{j}$ are invertible.
The relations

$\Phi_{d+1}\ =\
\Phi_{d}+l_{d}\Phi^{\prime}_{d+1},\qquad\Phi^{\prime}_{d+1}=\Phi^{\prime}_{d}+\lambda
M_{d}\Phi_{d},$

imply that

$\Phi_{d+1}(\Phi^{\prime}_{d+1})^{-1}=(\lambda
M_{d}+\Phi^{\prime}_{d}\Phi_{d}^{-1})^{-1}+l_{d}\mathbf{1},$

hence

$W_{d+1}^{-1}=\lambda l_{d}\mathbf{1}+(M_{d}+W_{d})^{-1}.$

Inverting this expression we obtain

$W_{d+1}=[\lambda l_{d}\mathbf{1}+(M_{d}+W_{d})^{-1}]^{-1}.$

Iterating down to $W_{1}=(\lambda l_{0}\mathbf{1})^{-1}$ concludes the proof.
∎

We want to reverse this procedure and recover the data $\\{l_{j}\\}$ and
$\\{M_{j}\\}$ from the function $W$. We follow the procedure of Stieltjes
[22], starting with the determination of certain Padé approximants of $W$. At
step zero, let

$P_{0}\ =\ 0,\quad Q_{0}\ =\ {\bf 1},$

so

$Q_{0}W\ =\ P_{0}+O(\lambda^{-1});\qquad W\ =\
Q_{0}^{-1}P_{0}+O(\lambda^{-1}).$ (7.7)

To proceed, we note that

$T_{j}^{-1}\ =\ \left[\begin{matrix}{\bf 1}&-l_{j}{\bf 1}\\\ -\lambda
M_{j}&{\bf 1}+\lambda l_{j}M_{j}\end{matrix}\right].$ (7.8)

Therefore the identity (7.4) implies that

$T_{d}^{-1}\left[\begin{matrix}\Phi_{d+1}\\\
\Phi^{\prime}_{d+1}\end{matrix}\right]\ =\
\left[\begin{matrix}\Phi_{d+1}-l_{d}\Phi^{\prime}_{d+1}\\\ -\lambda
M_{d}\Phi_{d+1}+({\bf 1}+\lambda
l_{d}M_{d})\Phi^{\prime}_{d+1}\end{matrix}\right]\ =\
\left[\begin{matrix}\Phi_{d}\\\ \Phi^{\prime}_{d}\end{matrix}\right].$

Multiplying each (block) row on the right by $\Phi_{d+1}^{-1}$, we obtain

$\left[\begin{matrix}{\bf 1}-l_{d}(\lambda W)\\\ -\lambda M_{d}+(1+\lambda
l_{d}M_{d})\lambda W\end{matrix}\right]\ =\
\left[\begin{matrix}\Phi_{d}\Phi_{d+1}^{-1}\\\
\Phi^{\prime}_{d}\Phi_{d+1}^{-1}\end{matrix}\right].$

The two equations for $W$ can be rewritten as

$\displaystyle\lambda l_{d}W$ $\displaystyle=$ $\displaystyle{\bf
1}-\Phi_{d}\Phi_{d+1}^{-1}\ =\ {\bf 1}+O(\lambda^{-1});$ (7.9)
$\displaystyle({\bf 1}+\lambda l_{d}M_{d})W$ $\displaystyle=$ $\displaystyle
M_{d}-\lambda^{-1}\Phi^{\prime}_{d}\Phi_{d+1}^{-1}\ =\ M_{d}+O(\lambda^{-2}).$
(7.10)

Set

$P_{1}={\bf 1},\quad Q_{1}\ =\ \lambda l_{d}{\bf 1};\qquad P_{2}\ =\
M_{d},\quad Q_{2}\ =\ \lambda l_{d}M_{d}+{\bf 1}.$ (7.11)

Then $Q_{1}^{-1}P_{1}$ and $Q_{2}^{-1}P_{2}$ are Padé approximants to $W$ on
the left:

$W\ =\ Q_{1}^{-1}P_{1}+O(\lambda^{-2}),\qquad W\ =\
Q_{2}^{-1}P_{2}+O(\lambda^{-3}).$ (7.12)

These two approximates are uniquely determined by the conditions
$P_{1}(0)={\bf 1}$, $Q_{2}(0)={\bf 1}$, respectively; see the next section.

This process can be continued. We have

$[T_{d}T_{d-1}\cdots T_{d-j+1}]^{-1}\left[\begin{matrix}\Phi_{d+1}\\\
\Phi^{\prime}_{d+1}\end{matrix}\right]\ =\ \left[\begin{matrix}\Phi_{d-j+1}\\\
\Phi^{\prime}_{d-j+1}\end{matrix}\right].$ (7.13)

Let us write, in a temporary notation for this section only,

$[T_{d}\dots T_{d-j+1}]^{-1}\ =\
\left[\begin{matrix}a_{j}(\lambda)&-b_{j}(\lambda)\\\
-c_{j}(\lambda)&d_{j}(\lambda)\end{matrix}\right],\quad 1\leq j\leq d.$ (7.14)

######  Lemma .

(a) The polynomials $a_{j}$ and $b_{j}$ have degree $j-1$; the polynomials
$c_{j}$ and $d_{j}$ have degree $j$, $1\leq j\leq d$.

(b) For each $1\leq j\leq d$, $a_{j}(0)=d_{j}(0)\ =\ {\bf 1}$,
$b_{j}(0)=c_{j}(0)={\bf 0}$.

(c) The coefficients of even powers in $a_{j}$,$b_{j}$, $c_{j}$ and $d_{j}$
are diagonal and the coefficients of odd powers are off-diagonal.

###### Proof.

Note that each of these statements is true at $j=1$:

$\left[\begin{matrix}a_{1}&-b_{1}\\\ -c_{1}&d_{1}\end{matrix}\right]\ =\
T_{d}^{-1}\ =\ \left[\begin{matrix}{\bf 1}&-l_{d}{\bf 1}\\\ -\lambda
M_{d}&{\bf 1}+\lambda l_{d}M_{d}.\end{matrix}\right]$

Note that

$\displaystyle\left[\begin{matrix}a_{j+1}&-b_{j+1}\\\
-c_{j+1}&d_{j+1}\end{matrix}\right]\ =\
T_{d-j}^{-1}\left[\begin{matrix}a_{j}&-b_{j}\\\
-c_{j}&d_{j}\end{matrix}\right]$ (7.15)
$\displaystyle=\left[\begin{matrix}a_{j}+l_{d-j}c_{j}&-b_{j}-l_{d-j}d_{j}\\\
-\lambda M_{d-j}a_{j}-({\bf 1}+\lambda l_{d-j}M_{d-j})c_{j}&\lambda
M_{d-j}b_{j}+({\bf 1}+\lambda l_{d-j}M_{d-j})d_{j}\end{matrix}\right].$

The assertion (a) follows by easily by induction. Each $T_{j}(0)\ =\ {\bf 1}$,
which implies (b). Assertion (c) follows from the fact that multiplication by
any entry of $T_{j}^{-1}$ preserves these properties. ∎

In analogy with the computations that led to (7.9) and (7.10), we multiply
each (block) row of the identity

$\left[\begin{matrix}\Phi_{d-j+1}\\\ \Phi^{\prime}_{d-j+1}\end{matrix}\right]\
=\ \left[\begin{matrix}a_{j}&-b_{j}\\\
-c_{j}&d_{j}\end{matrix}\right]\left[\begin{matrix}\Phi_{d+1}\\\
\Phi^{\prime}_{d+1}\end{matrix}\right]$

on the right by $\Phi_{d+1}^{-1}$ and obtain the equations

$\displaystyle\lambda b_{j}W$ $\displaystyle=$ $\displaystyle
a_{j}-\Phi_{d-j+1}\Phi_{d+1}^{-1};$ $\displaystyle d_{j}W$ $\displaystyle=$
$\displaystyle\lambda^{-1}c_{j}+\lambda^{-1}\Phi^{\prime}_{d-j+1}\Phi_{d+1}^{-1}.$

Accordingly, and consistent with previous definitions for $j=1$,

$\displaystyle P_{2j-1}$ $\displaystyle=$ $\displaystyle a_{j},\qquad
Q_{2j-1}=\lambda b_{j},\quad 1\leq j\leq d;$ (7.16) $\displaystyle P_{2j}$
$\displaystyle=$ $\displaystyle\lambda^{-1}c_{j},\qquad Q_{2j}\ =\ d_{j},\quad
0\leq j\leq d.$ (7.17)

Note that since $c_{j}(0)=0$, each of the $P_{k}$, $Q_{k}$ is a polynomial.

In view of (7.7) (7.9), (7.10), and Lemma 7, we have

######  Proposition .

The polynomials $P_{k}$, $Q_{k}$, $1\leq k\leq 2d$, have the properties

(a) $Q_{2j-1}$ and $Q_{2j}$ have degree $j$, $P_{2j-1}$ and $P_{2j}$ have
degree $j-1$;

(b) The coefficient of odd powers of $Q_{2j-1}$ are diagonal, and the
coefficients of even powers are off-diagonal;

(c) The coefficient of even powers of $Q_{2j}$ are diagonal, and the
coefficients of odd powers are off-diagonal.

Moreover

$\displaystyle Q_{2j-1}W$ $\displaystyle=$ $\displaystyle
P_{2j-1}+O(\lambda^{-j});$ (7.18) $\displaystyle Q_{2j}W$ $\displaystyle=$
$\displaystyle P_{2j}+O(\lambda^{-j-1}).$ (7.19)

In the next section we treat the inverse problem: the problem of recovering
the beam data $\\{l_{j}\\}$, $\\{M_{j}\\}$ from $W$. The final step of the
process described there uses the fact that the data can be recovered from the
leading coefficients of the polynomials $\\{Q_{k}\\}$.

######  Proposition .

Let $\langle Q_{k}\rangle$ denote the leading coefficient of $Q_{k}$. Then for
$1\leq j\leq d$,

$\displaystyle\langle Q_{2j-1}\rangle\langle Q_{2j-2}\rangle^{-1}$
$\displaystyle=$ $\displaystyle l_{d-j+1}{\bf 1};$ (7.20)
$\displaystyle\langle Q_{2j}\rangle\langle Q_{2j-1}\rangle^{-1}$
$\displaystyle=$ $\displaystyle M_{d-j+1}.$ (7.21)

###### Proof.

Let $\langle a_{j}\rangle$, $\langle b_{j}\rangle$, $\langle c_{j}\rangle$,
$\langle d_{j}\rangle$ denote the leading coefficients of $a_{j}$, $b_{j}$,
$c_{j}$, $d_{j}$. Because of Lemma 7 (a) and (7.15), it follows that the
recursion for the matrix of principal coefficients is given by

$\left[\begin{matrix}\langle a_{j}\rangle&-\langle b_{j}\rangle\\\ -\langle
c_{j}\rangle&\langle d_{j}\rangle\end{matrix}\right]\ =\
\left[\begin{matrix}l_{d-j+1}\langle c_{j-1}\rangle&-l_{d-j+1}\langle
d_{j-1}\rangle\\\ -l_{d-j+1}M_{d-j+1}\langle
c_{j-1}\rangle&l_{d-j+1}M_{d-j+1}\langle d_{j-1}\rangle\end{matrix}\right].$
(7.22)

At the first step, $Q_{0}={\bf 1}$ and $\langle Q_{1}\rangle=l_{d}{\bf 1}$, so

$\langle Q_{1}\rangle\langle Q_{0}\rangle^{-1}\ =\ \langle Q_{1}\rangle\ =\
l_{d}{\bf 1}.$

At each subsequent step, (7.22) implies that

$\displaystyle\langle Q_{2j-1}\rangle$ $\displaystyle=$ $\displaystyle\langle
b_{j}\rangle\ =\ l_{d-j+1}\langle d_{j-1}\rangle\ =\ l_{d-j+1}\langle
Q_{2j-2}\rangle,$

which proves (7.20). Similarly, at each step (7.22) implies that

$\langle Q_{2j}\rangle\ =\ \langle d_{j}\rangle\ =\ M_{d-j+1}\langle
b_{j}\rangle\ =\ M_{d-j+1}\langle Q_{2j-1}\rangle,$

which proves (7.21). ∎

##  8 The inverse problem for the discrete beam

We shall show that the Weyl function $W$ has an asymptotic expansion

$W(\lambda)\ =\ \frac{1}{\lambda}\,C_{0}+\frac{1}{\lambda^{2}}\,C_{1}+\dots\
\frac{1}{\lambda^{n+1}}\,C_{n}+O\left(\frac{1}{\lambda^{n+2}}\right)\quad\hbox{as\
\ $\lambda\to\infty$.}$ (8.1)

The denominators $Q_{k}$ of the Padé approximants to $W$ can be recovered from
this asymptotic expansion of $W$. For example, subsitute the expansion (8.1)
for $W$ in (7.18) and expand. Since $Q_{2j-1}$ has no constant term and the
constant term of $P_{2j-1}$ is ${\bf 1}$, the term of order $0$ in the
expansion is ${\bf 1}$ and the terms of order $-1,\dots,1-j$ in the expansion
are zero. Writing

$Q_{2j-1}\ =\ \lambda^{j}Q^{(j-)}_{j}+\dots+\lambda^{2}Q^{(j-)}_{2}+\lambda
Q^{(j-)}_{1},$ (8.2)

the resulting system of equations can be written

$\displaystyle\left[\begin{matrix}Q^{(j-)}_{1}&Q^{(j-)}_{2}&\dots&Q^{(j-)}_{j}\end{matrix}\right]\left[\begin{matrix}C_{0}&C_{1}&\dots&C_{j-1}\\\
C_{1}&C_{2}&\dots&C_{j}\\\ &&\dots&\\\
C_{j-1}&C_{j}&\dots&C_{2j-2}\end{matrix}\right]$ $\displaystyle\quad\ =\
\left[\begin{matrix}{\bf 1}&0&\dots&0\end{matrix}\right].$ (8.3)

Write

$Q_{2j}\ =\ \lambda^{j}Q^{(j+)}_{j}+\dots+\lambda Q^{(j+)}_{1}+{\bf 1}.$ (8.4)

Since $Q_{2j}(0)={\bf 1}$, the same argument leads to the system

$\displaystyle\left[\begin{matrix}Q^{(j+)}_{1}&Q^{(j+)}_{2}&\dots&Q^{(j+)}_{j}\end{matrix}\right]\left[\begin{matrix}C_{1}&C_{2}&\dots&C_{j}\\\
C_{2}&C_{3}&\dots&C_{j+1}\\\ &&\dots&\\\
C_{j}&C_{j+1}&\dots&C_{2j-1}\end{matrix}\right]$ $\displaystyle\quad\ =\
-\left[\begin{matrix}C_{0}&C_{1}&\dots&C_{j-1}\end{matrix}\right].$ (8.5)

In principle, the matrix equations (8.3) and (8.5), considered as scalar
equations, consist of $4j$ linear equations in $4j$ unknowns. However we know
that each of the coefficients of $Q^{(j^{\pm})}$ is either a diagonal or an
off-diagonal matrix, so there are only $2d$ unknowns. Moreover, as we shall
show, the same is true of each of the matrices $C_{k}$, so the associated
$2j\times 2j$ matrix for these equations has only $2\cdot j^{2}$ non-zero
entries. As we shall show, each system (8.3) and (8.5) decomposes easily into
two uncoupled systems of $j$ equations in $j$ unknowns, permitting simple
formulas for the leading coefficients.

To understand the $C_{k}$, we return to the formula (6.22) for $W$:

$W(\lambda)\ =\ \frac{1}{2\lambda}{\bf
1}+\sum_{\nu=-d}^{-1}\frac{1}{\lambda-\lambda_{\nu}}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&-\alpha_{\nu}^{2}\\\
-\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right]+\sum_{\nu=1}^{d}\frac{1}{\lambda-\lambda_{\nu}}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&\alpha_{\nu}^{2}\\\
\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right],$

where $\alpha_{\nu}=\alpha_{-\nu}$ and $\beta_{\nu}=\beta_{-\nu}$ are
positive.

For large $|\lambda|$,
$(\lambda-\lambda_{\nu})^{-1}=\sum_{n=0}^{\infty}\lambda_{\nu}^{k}/\lambda^{k+1}$,
so

$C_{0}\ =\
\left[\frac{1}{2}+\sum_{\nu=1}^{d}2\alpha_{\nu}\beta_{\nu}\right]{\bf 1},$
(8.6)

and

$C_{k}\ =\
\sum_{\nu=1}^{d}\left\\{\lambda_{-\nu}^{k}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&-\alpha_{\nu}^{2}\\\
-\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right]+\lambda_{\nu}^{k}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&\alpha_{\nu}^{2}\\\
\beta_{\nu}^{2}&\alpha_{\nu}\beta_{\nu}\end{matrix}\right]\right\\},\qquad
k\geq 1.$

Recall that $\nu$ and $\lambda_{\nu}$ have opposite signs, so

$C_{k}\ =\ \begin{cases}\ \
\sum_{\nu=1}^{d}2|\lambda_{\nu}|^{k}\left[\begin{matrix}0&-\alpha_{\nu}^{2}\\\
-\beta_{\nu}^{2}&0\end{matrix}\right],&\ \hbox{$k$\ \ odd};\\\ &\\\ \ \
\sum_{\nu=1}^{d}2|\lambda_{\nu}|^{k}\left[\begin{matrix}\alpha_{\nu}\beta_{\nu}&0\\\
0&\alpha_{\nu}\beta_{\nu}\end{matrix}\right],&\ \hbox{$k$\ \ even,\ \ \ $k\geq
2$.}\end{cases}$

Thus

$C_{k}\ =\ \left[\begin{matrix}a_{k}&0\\\ 0&a_{k}\end{matrix}\right],\quad k\
\ \hbox{even};\qquad C_{k}\ =\ \left[\begin{matrix}0&b_{k}\\\
c_{k}&0\end{matrix}\right],\quad k\ \ \hbox{odd},$ (8.7)

where

$\displaystyle a_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{2}+2\sum_{\nu=1}^{d}\alpha_{\nu}\beta_{\nu};\qquad
a_{k}\ =\ 2\sum_{\nu=1}^{d}|\lambda_{\nu}|^{k}\alpha_{\nu}\beta_{\nu},\quad k\
{\rm even},\,k\geq 2;$ $\displaystyle b_{k}$ $\displaystyle=$
$\displaystyle-\sum_{\nu>0}2|\lambda_{\nu}|^{k}\alpha_{\nu}^{2};\qquad
c_{k}=-\sum_{\nu>0}2|\lambda_{\nu}|^{k}\beta_{\nu}^{2},\quad\ \ \ k\ \ {\rm
odd.}$

Let us consider the systems (8.3) and (8.5) for $j=2$:

$\left[Q^{(2-)}_{1}\ Q^{(2-)}_{2}\right]\left[\begin{matrix}C_{0}&C_{1}\\\
C_{1}&C_{2}\end{matrix}\right]\ =\ [{\bf 1}\ \ {\bf
0}];\qquad\left[Q^{(2+)}_{1}\
Q^{(2+)}_{2}\right]\left[\begin{matrix}C_{1}&C_{2}\\\
C_{2}&C_{3}\end{matrix}\right]\ =\ -[C_{0}\ C_{1}].$ (8.8)

The key structural fact here is that each row or column consists of one
diagonal matrix and one off-diagonal matrix. For larger values of $j$ there is
a similar structure, with diagonal matrices and off-diagonal matrices
alternating. Filling in the entries, the first of the systems (8.8) is

$\left[\begin{matrix}x_{11}&0&0&x_{21}\\\
0&x_{12}&x_{22}&0\end{matrix}\right]\left[\begin{matrix}a_{0}&0&0&b_{1}\\\
0&a_{0}&c_{1}&0\\\ 0&b_{1}&a_{2}&0\\\ c_{1}&0&0&a_{2}\end{matrix}\right]\ =\
\left[\begin{matrix}1&0&0&0\\\ 0&1&0&0\end{matrix}\right],$ (8.9)

where $x_{k1}$, $x_{k2}$ are the non-zero elements in the first and second
rows of the coefficient $Q^{(2-)}_{k}$, respectively.

Because of the way that the positions of zero and non-zero elements in the
rows and columns either match or complement each other, there are
cancellations. For example, the product of the first row of the matrix on the
left with the second or third columns of the matrix on the right is zero.
Therefore the four equations associated to the first row reduce to two, which
can be written as a system

$\left[\begin{matrix}x_{11}&x_{21}\end{matrix}\right]\left[\begin{matrix}a_{0}&b_{1}\\\
c_{1}&a_{2}\end{matrix}\right]\ =\
\left[\begin{matrix}1&0\end{matrix}\right].$ (8.10)

Similarly, the four equations associated with the second row reduce to

$\left[\begin{matrix}x_{12}&x_{22}\end{matrix}\right]\left[\begin{matrix}a_{0}&c_{1}\\\
b_{1}&a_{2}\end{matrix}\right]\ =\
\left[\begin{matrix}1&0\end{matrix}\right].$ (8.11)

A second way to organize this is by a suitable permutation of rows and
columns, so that rows with the same pattern of zero entries are juxtaposed,
and the same for columns. Then the original system of $8$ equations becomes

$\left[\begin{matrix}x_{11}&x_{21}&0&0\\\
0&0&x_{12}&x_{22}\end{matrix}\right]\left[\begin{matrix}a_{0}&b_{1}&0&0\\\
c_{1}&a_{2}&0&0\\\ 0&0&a_{0}&c_{1}\\\ 0&0&b_{1}&a_{2}\end{matrix}\right]\ =\
\left[\begin{matrix}1&0&0&0\\\ 0&0&1&0\end{matrix}\right].$ (8.12)

The same procedure applies in general to the equations for the coefficients
$Q^{(j-)}_{k}$ of $Q_{2j-1}$, yielding an equivalent form in which the
original $2j\times 2j$ matrix is reduced to a diagonal form with two $j\times
j$ matrices, adjoints of each other, on the diagonal. We write this explicitly
below.

A similar analysis of the second of the systems (8.8) yields a different form
of canonical reduction. Here the system has the form

$\left[\begin{matrix}0&x_{11}&x_{21}&0\\\
x_{12}&0&0&x_{22}\end{matrix}\right]\left[\begin{matrix}0&b_{1}&a_{2}&0\\\
c_{1}&0&0&a_{2}\\\ a_{2}&0&0&b_{3}\\\ 0&a_{2}&c_{3}&0\end{matrix}\right]\ =\
-\left[\begin{matrix}a_{0}&0&0&b_{1}\\\ 0&a_{0}&c_{1}&0\end{matrix}\right]$
(8.13)

where $x_{k1}$ and $x_{k2}$ are the non-zero entries of the first and second
rows of the coefficient $Q^{(2+)}_{k}$, respectively. Again the positioning of
the zeros in the rows and columns tells us that these equations reduce to two
uncoupled systems

$\displaystyle\left[\begin{matrix}x_{11}&x_{21}\end{matrix}\right]\left[\begin{matrix}c_{1}&a_{2}\\\
a_{2}&b_{3}\end{matrix}\right]$ $\displaystyle=$
$\displaystyle-\left[\begin{matrix}a_{0}&b_{1}\end{matrix}\right];$ (8.14)
$\displaystyle\left[\begin{matrix}x_{12}&x_{22}\end{matrix}\right]\left[\begin{matrix}b_{1}&a_{2}\\\
a_{2}&c_{3}\end{matrix}\right]$ $\displaystyle=$
$\displaystyle-\left[\begin{matrix}a_{0}&c_{1}\end{matrix}\right].$ (8.15)

As in the case of (8.10), (8.11), the system (8.13) can be rearranged to the
form

$\left[\begin{matrix}x_{11}&x_{21}&0&0\\\
0&0&x_{12}&x_{22}\end{matrix}\right]\left[\begin{matrix}0&0&c_{1}&a_{2}\\\
0&0&a_{2}&b_{3}\\\ b_{1}&a_{2}&0&0\\\ a_{2}&c_{3}&0&0\end{matrix}\right]\ =\
-\left[\begin{matrix}0&0&a_{0}&b_{1}\\\ a_{0}&c_{1}&0&0\end{matrix}\right].$
(8.16)

Let us pass to the general case for the coefficients $Q_{k}^{(2j\pm)}$ of
$Q_{2j-1}$ and $Q_{2j}$. We start with the $(2d+2)\times(2d+2)$ Hankel matrix

$H\ =\ \left[\begin{matrix}C_{0}&C_{1}&C_{2}&\dots&C_{d}\\\
C_{1}&C_{2}&C_{3}&\dots&C_{d+1}\\\ C_{2}&C_{3}&C_{4}&\dots&C_{d+2}\\\
&&&\ddots&\\\ C_{d}&C_{d+1}&C_{d+2}&\dots&C_{2d}\end{matrix}\right].$ (8.17)

Writing out the $2\times 2$ blocks,

$H\ =\ \left[\begin{matrix}a_{0}&0&0&b_{1}&a_{2}&0&0&b_{3}\ \ \dots\ \\\
0&a_{0}&c_{1}&0&0&a_{2}&c_{3}&0\ \ \dots\ \\\ 0&b_{1}&a_{2}&0&0&b_{3}&a_{4}&0\
\ \dots\ \\\ c_{1}&0&0&a_{2}&c_{3}&0&0&a_{4}\ \dots\ \\\
a_{2}&0&0&b_{3}&a_{4}&0&0&b_{5}\ \dots\ \\\ 0&a_{2}&c_{3}&0&0&a_{4}&c_{5}&0\ \
\dots\ \\\ 0&b_{3}&a_{4}&0&0&b_{5}&a_{6}&0\ \dots\ \\\
c_{3}&0&0&a_{4}&c_{5}&0&0&c_{7}\ \dots\ \\\
&&\ddots&&&\ddots&&\end{matrix}\right]$ (8.18)

In a notation that is best explained by (8.12) we introduce two
$(d+1)\times(d+1)$ matrices constructed by reorganizing $H$:

$H^{NW}\ =\ \left[\begin{matrix}a_{0}&b_{1}&a_{2}&b_{3}\ \ \dots\\\
c_{1}&a_{2}&c_{3}&a_{4}\ \ \dots\\\ a_{2}&b_{3}&a_{4}&b_{5}\ \ \dots\\\
c_{3}&a_{4}&c_{5}&a_{6}\ \ \dots\\\ &&\ddots&\end{matrix}\right];\quad\ \
H^{SE}\ =\ \left[\begin{matrix}a_{0}&c_{1}&a_{2}&c_{3}\ \ \ \dots\\\
b_{1}&a_{2}&b_{3}&a_{4}\ \ \dots\\\ a_{2}&c_{3}&a_{4}&c_{5}\ \ \dots\\\
b_{3}&a_{4}&b_{5}&a_{6}\ \ \dots\\\ &&\ddots&\end{matrix}\right].$

We denote the $j\times j$ principal minors of $H^{NW}$ and $H^{SE}$ by
$H^{NW}_{j}$ and $H^{SE}_{j}$, respectively. Note that they are transposes of
each other:

$[H^{NW}_{j}]^{t}\ =\ H^{SE}_{j}.$

Therefore they have the same determinant

$\det H^{NW}_{j}\ =\ \det H^{SE}_{j}\ =\ \Delta_{j}.$ (8.19)

Following the same procedure as for $Q_{3}$, the equations for the
coefficients of $Q_{2j-1}$ are

$\displaystyle\left[\begin{matrix}x_{11}&x_{21}&\dots&x_{j1}\end{matrix}\right]H^{NW}_{j}$
$\displaystyle=$
$\displaystyle\left[\begin{matrix}1&0&\dots&0\end{matrix}\right];$ (8.20)
$\displaystyle\left[\begin{matrix}x_{12}&x_{22}&\dots&x_{j2}\end{matrix}\right]H^{SE}_{j}$
$\displaystyle=$
$\displaystyle\left[\begin{matrix}1&0&\dots&0\end{matrix}\right],$ (8.21)

where $x_{k1}$ and $x_{k2}$ are the non-zero elements in the first and second
rows of the coefficient of $\lambda^{k}$ in $Q_{2j-1}$.

As remarked in (7.20) and (7.21), we can reconstruct the beam data
$\\{l_{j}\\}$, $\\{M_{j}\\}$ from the leading coefficients $\langle
Q_{k}\rangle$ of the polynomials $\\{Q_{k}\\}$. For $Q_{2j-1}$, we want to
compute $x_{j1}$ and $x_{j2}$ in (8.20). By Cramer’s rule, $x_{j1}$ can be
obtained by replacing the last row of the matrix in (8.20) by the right-hand
side of (8.20) and computing the determinant. The same procedure for (8.21)
gives

$x_{j1}\ =\ (-1)^{j-1}\frac{\Delta^{NW}_{j1}}{\Delta_{j}},\qquad x_{j2}\ =\
(-1)^{j-1}\frac{\Delta^{SE}_{j1}}{\Delta_{j}},$ (8.22)

where $\Delta^{NW}_{j1}$ is the determinant of $H^{NW}_{j}$ with the first
column and last row eliminated, and similarly for $\Delta^{SE}_{j1}$. By
Proposition 7, since $Q_{2j-1}$ has degree $j$, the leading coefficient
$\langle Q_{2j-1}\rangle$ is diagonal if $j$ is odd and off-diagonal if $j$ is
even. Thus we have

######  Proposition .

The leading coefficient of $Q_{2j-1}$ is

$\langle Q_{2j-1}\rangle\ =\
\left[\begin{matrix}\frac{\Delta^{NW}_{j1}}{\Delta_{j}}&0\\\
0&\frac{\Delta^{SE}_{j1}}{\Delta_{j}}\end{matrix}\right]$ (8.23)

if $j$ is odd,

$\langle Q_{2j-1}\rangle\ =\
-\left[\begin{matrix}0&\frac{\Delta^{NW}_{j1}}{\Delta_{j}}\\\
\frac{\Delta^{SE}_{j1}}{\Delta_{j}}&0\ \end{matrix}\right]$ (8.24)

if $j$ is even.

We turn now to consideration of the coefficients of $Q_{2j}$. In line with
(8.16), we introduce two $d\times d$ matrices that are obtained by
reorganizing $H$ after removing the first two columns and last two rows:

$H^{NE}\ =\ \left[\begin{matrix}c_{1}&a_{2}&c_{3}&a_{4}\ \dots\\\
a_{2}&b_{3}&a_{4}&b_{5}\ \dots\\\ c_{3}&a_{4}&c_{5}&a_{6}\ \dots\\\
a_{4}&b_{5}&a_{6}&b_{7}\ \dots\\\ &&\ddots&\end{matrix}\right];\quad\ \
H^{SW}\ =\ \left[\begin{matrix}b_{1}&a_{2}&b_{3}&a_{4}\ \dots\\\
a_{2}&c_{3}&a_{4}&c_{5}\ \dots\\\ b_{3}&a_{4}&b_{5}&a_{6}\ \dots\\\
a_{4}&c_{5}&a_{6}&c_{7}\ \dots\\\ &&\ddots&\end{matrix}\right].$

Let $H^{NE}_{j}$ and $H^{SW}_{j}$ be the $j\times j$ principal minors of
$H^{NE}$ and $H^{SW}$, respectively. The equations for the coefficients of
$Q_{2j}$ are

$\displaystyle\left[\begin{matrix}x_{11}&x_{21}&\dots&x_{j1}\end{matrix}\right]H^{NE}_{j}$
$\displaystyle=$
$\displaystyle-\left[\begin{matrix}a_{0}&b_{1}&\dots&a_{j-1}\end{matrix}\right];$
(8.25)
$\displaystyle\left[\begin{matrix}x_{12}&x_{22}&\dots&x_{j2}\end{matrix}\right]H^{SW}_{j}$
$\displaystyle=$
$\displaystyle-\left[\begin{matrix}a_{0}&c_{1}&\dots&a_{j-1}\end{matrix}\right].$
(8.26)

Here $x_{k1}$ and $x_{k2}$ are the non-zero entries in the first and second
rows of the coefficient of $\lambda^{k}$ in $Q_{2j}$. Replacing the last row
of $H^{NE}_{J}$ by the negative of the right-hand side of (8.25), then moving
that to be the first row, gives $H^{NW}_{j}$. Applying the same reasoning to
(8.26), we obtain

$x_{j1}\ =\ (-1)^{j}\frac{\Delta_{j}}{\Delta^{NE}_{j}};\qquad x_{j2}\ =\
(-1)^{j}\frac{\Delta_{j}}{\Delta^{SW}_{j}}.$ (8.27)

By Lemma 7, since $Q_{2j}$ has degree $j$, the top coefficient is off-diagonal
if $j$ is odd and diagonal if $j$ is even. Therefore

######  Proposition .

The leading coefficient of $Q_{2j}$ is

$\langle Q_{2j}\rangle\ =\
-\left[\begin{matrix}0&\frac{\Delta_{j}}{\Delta^{NE}_{j}}\\\
\frac{\Delta_{j}}{\Delta^{SW}_{j}}&0\end{matrix}\right]$ (8.28)

if $j$ is odd,

$\langle Q_{2j}\rangle\ =\
\left[\begin{matrix}\frac{\Delta_{j}}{\Delta^{NE}_{j}}&0\\\
0&\frac{\Delta_{j}}{\Delta^{SW}_{j}}\end{matrix}\right]$ (8.29)

if $j$ is even.

We are now in a position to compute the data $\\{l_{k}\\}$, $\\{M_{k}\\}$, via
(7.20), (7.21). Note that

$H^{NW}_{j1}\ =\ H^{SW}_{j-1};\qquad H^{SE}_{j1}\ =\ H^{NE}_{j-1}$ (8.30)

We use these identities to rewrite (8.23) and (8.24).

If $j$ is odd, we have

$\displaystyle l_{d-j+1}{\bf 1}$ $\displaystyle=$ $\displaystyle\langle
Q_{2j-1}\rangle\langle Q_{2j-2}\rangle^{-1}$ $\displaystyle=$
$\displaystyle\left[\begin{matrix}\frac{\Delta^{SW}_{j-1}}{\Delta_{j}}&0\\\
0&\frac{\Delta^{NE}_{j-1}}{\Delta_{j}}\end{matrix}\right]\left[\begin{matrix}\frac{\Delta_{j-1}}{\Delta^{NE}_{j-1}}&0\\\
0&\frac{\Delta_{j-1}}{\Delta^{SW}_{j-1}}\end{matrix}\right]^{-1}\ =\
\frac{\Delta^{SW}_{j-1}\Delta^{NE}_{j-1}}{\Delta_{j}\Delta_{j-1}}\,{\bf 1}.$

If $j$ is even, we have

$l_{d-j+1}{\bf 1}\ =\
\left[\begin{matrix}0&\frac{\Delta^{SW}_{j-1}}{\Delta_{j}}\\\
\frac{\Delta^{NE}_{j-1}}{\Delta_{j}}&0\
\end{matrix}\right]\left[\begin{matrix}0&\frac{\Delta_{j-1}}{\Delta^{NE}_{j-1}}\\\
\frac{\Delta_{j-1}}{\Delta^{SW}_{j-1}}&0\end{matrix}\right]^{-1}\ =\
\frac{\Delta^{SW}_{j-1}\Delta^{NE}_{j-1}}{\Delta_{j}\Delta_{j-1}}\,{\bf 1}.$

If $j$ is odd, we have

$\displaystyle M_{d-j+1}$ $\displaystyle=$ $\displaystyle\langle
Q_{2j}\rangle\langle Q_{2j-1}\rangle^{-1}$ $\displaystyle=$
$\displaystyle-\left[\begin{matrix}0&\frac{\Delta_{j}}{\Delta^{NE}_{j}}\\\
\frac{\Delta_{j}}{\Delta^{SW}_{j}}&0\end{matrix}\right]\left[\begin{matrix}\frac{\Delta^{SW}_{j-1}}{\Delta_{j}}&0\\\
0&\frac{\Delta^{NE}_{j-1}}{\Delta_{j}}\end{matrix}\right]^{-1}\ =\
-\left[\begin{matrix}0&\frac{\Delta_{j}^{2}}{\Delta^{NE}_{j}\Delta^{NE}_{j-1}}\\\
\frac{\Delta_{j}^{2}}{\Delta^{SW}_{j}\Delta^{SW}_{j-1}}&0\end{matrix}\right].$

If $j$ is even, we have

$M_{d-j+1}\ =\ -\left[\begin{matrix}\frac{\Delta_{j}}{\Delta^{NE}_{j}}&0\\\
0&\frac{\Delta_{j}}{\Delta^{SW}_{j}}\end{matrix}\right]\left[\begin{matrix}0&\frac{\Delta^{SW}_{j-1}}{\Delta_{j}}\\\
\frac{\Delta^{NE}_{j-1}}{\Delta_{j}}&0\ \end{matrix}\right]^{-1}\ =\
-\left[\begin{matrix}0&\frac{\Delta_{j}^{2}}{\Delta^{NE}_{j}\Delta^{NE}_{j-1}}\\\
\frac{\Delta_{j}^{2}}{\Delta^{SW}_{j}\Delta^{SW}_{j-1}}&0\end{matrix}\right].$

##  9 Appendix: The consistency conditions; the smooth case

Under an additional assumption of smoothness, the compatibility conditions
that relate

$D_{x}^{2}\Phi\ =\ ({\bf 1}+\lambda M)\Phi,\qquad M\ =\
\left[\begin{matrix}0&n\\\ m&0\end{matrix}\right].$ (A.1)

and

$D_{t}\Phi\ =\ [bD_{x}+a]\Phi,$ (A.2)

namely

$D_{t}D_{x}^{2}\Phi\ =\ D_{x}^{2}D_{t}\Phi\ $

lead to

$\displaystyle\lambda M_{t}\Phi$ $\displaystyle=$
$\displaystyle\left\\{b_{xx}+2a_{x}+\lambda[b,M]\right\\}D_{x}\Phi$ (A.3)
$\displaystyle\ +\left\\{a_{xx}+2b_{x}+\lambda(bM)_{x}+\lambda
b_{x}M+\lambda[a,M]\right\\}\Phi.$

At a given value of $t$, this is a differential equation for $\Phi$ of order
at most one. We are assuming that $\Phi$ is a solution of a nontrivial
second–order equation. We assume that the differential operator in (A.3)
trivializes:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle b_{xx}+2a_{x}+\lambda[b,M],$
(A.4) $\displaystyle\lambda M_{t}$ $\displaystyle=$ $\displaystyle
a_{xx}+2b_{x}+\lambda(bM)_{x}+\lambda b_{x}M+\lambda[a,M],$ (A.5)

since otherwise the system is degenerate.

As in Section 2 we suppose that

$a\ =\ a_{0}+\lambda^{-1}a_{1},\qquad b\ =\ b_{0}+\lambda^{-1}b_{1},$

and that $a_{j}$ and $b_{j}$ are bounded, $j=0,1$. Each equation in (A.4),
(A.5) leads to three equations, for the coefficients of the powers
$\lambda^{k}$, $k=-1,0,1$.

For $k=-1$ the equations are

$(a_{1})_{xx}+2(b_{1})_{x}\ =\ 0\,;\qquad(b_{1})_{xx}+2(a_{1})_{x}\ =0.$ (A.6)

Thus

$(a_{1})_{x}=-\frac{(b_{1})_{xx}}{2},\qquad(b_{1})_{xxx}-4(b_{1})_{x}=0.$
(A.7)

The second equation implies that $b_{1}=C_{1}e^{2x}+C_{2}e^{-2x}+C_{3}$ and
since $b_{1}$ is bounded, $b_{1}$ is a constant matrix, and by the first
equation so is $a_{1}$.

For $k=1$ the equations are

$0\ =\ [b_{0},M]\,;\qquad M_{t}\ =\ (b_{0}M)_{x}+(b_{0})_{x}M+[a_{0},M].$
(A.8)

We assume that $m\neq n$, so first equation in (A.8) implies that the diagonal
part of $b_{0}$ is a multiple of the identity matrix, and the off-diagonal
part is a multiple of $M$:

$b_{0}\ =\ uI+pM.$ (A.9)

Therefore the diagonal part of the second equation in (A.8) gives

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(pmn)_{x}+(pn)_{x}m+(a_{0})_{12}m-(a_{0})_{21}n;$ $\displaystyle
0$ $\displaystyle=$
$\displaystyle(pmn)_{x}+(pm)_{x}n+(a_{0})_{21}n-(a_{0})_{12}m.$ (A.10)

Adding these two equations gives

$0\ =\ 4p_{x}(mn)+3p(mn)_{x}.$

Multiplying by $p^{3}(mn)^{2}$ gives $0=[p^{4}(mn)^{3}]_{x}$, so
$p^{4}(mn)^{3}$ is constant. If $p\neq 0$ then this is a nontrivial a priori
relationship between $m$ and $n$. Therefore we assume $p=0$. With this
assumption, equations (A.10) imply that the off-diagonal part of $a_{0}$ is
proportional to $M$. We can write

$b_{0}\ =\ u{\bf 1},\qquad a_{0}\ =\
\frac{1}{2}\left[\begin{matrix}w(x)+v(x)&0\\\
0&w(x)-v(x)\end{matrix}\right]+qM.$ (A.11)

The remaining information from equations (A.4), (A.5) is contained in the
equations for the $k=0$ part:

$\displaystyle 0$ $\displaystyle=$
$\displaystyle(b_{0})_{xx}+2(a_{0})_{x}+[b_{1},M]\,;$ (A.12) $\displaystyle 0$
$\displaystyle=$
$\displaystyle(a_{0})_{xx}+2(b_{0})_{x}+b_{1}M_{x}+[a_{1},M].,$ (A.13)

since $a_{1}$, $b_{1}$ are constant. Write

$a_{1}\ =\ \left[\begin{matrix}\alpha_{1}&\alpha_{2}\\\
\alpha_{3}&\alpha_{4}\end{matrix}\right]\,;\qquad b_{1}\ =\
\left[\begin{matrix}\beta_{1}&\beta_{2}\\\
\beta_{3}&\beta_{4}\end{matrix}\right].$

Looking at the diagonal terms, then the off-diagonal terms, in (A.12), we find

$\displaystyle u_{xx}+w_{x}$ $\displaystyle=$ $\displaystyle 0;\qquad v_{x}\
=\ \beta_{3}n-\beta_{2}m;$ (A.14) $\displaystyle
2(qn)_{x}+(\beta_{1}-\beta_{4})n$ $\displaystyle=$ $\displaystyle\ 0\ =\
2(qm)_{x}+(\beta_{4}-\beta_{1})m.$ (A.15)

Multiply the left side of (A.15) by $n$, the right side by $m$, and add, to
obtain:

$0\ =\ 2\left[(qn)_{x}m+(qm)_{x}n\right]\ =\
2\left[2q_{x}(mn)+q(mn)_{x}\right].$

As above, unless $q=0$ this gives an a priori relation $q^{2}(mn)=$ constant.
Taking $q=0$, (A.15) implies $\beta_{1}=\beta_{4}$.

Looking at the off-diagonal terms in (A.13), we obtain equations for $n$ and
for $m$ with constant coefficients:

$\beta_{1}n_{x}\ =\ (\alpha_{4}-\alpha_{1})n;\qquad\beta_{1}m_{x}\ =\
(\alpha_{1}-\alpha_{4})m.$

In order to avoid trivial cases, we must assume that $\beta_{1}=0$ and
$\alpha_{1}=\alpha_{4}$. Computing the diagonal part of (A.13), taking into
account (A.14) gives

$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\frac{1}{2}u_{xxx}+2u_{x}+(\beta_{2}m+\beta_{3}n)_{x}]\,;$
(A.16) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\alpha_{2}m-\alpha_{3}n.$ (A.17)

To avoid a trivial linear relation between $m$ and $n$ we need the
off–diagonal terms $\alpha_{2}$, $\alpha_{3}$ of $a_{1}$ to vanish.

Summing up to this point:

$\displaystyle a$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[\begin{matrix}\gamma-u_{x}+v&0\\\ 0&\gamma-
u_{x}-v\end{matrix}\right]+\frac{1}{2\lambda}\left[\begin{matrix}\alpha&0\\\
0&\alpha\end{matrix}\right]\,;$ $\displaystyle b$ $\displaystyle=$
$\displaystyle\left[\begin{matrix}u&0\\\
0&u\end{matrix}\right]+\frac{1}{\lambda}\left[\begin{matrix}0&\beta_{2}\\\
\beta_{3}&0\end{matrix}\right]\,,$

where $\gamma$,$\alpha$, $\beta_{2}$, and $\beta_{3}$ are constant.

Keeping in mind the obvious symmetry between $(\varphi_{1},m)$ on one hand and
$(\varphi_{2},n)$ on the other, we symmetrize by taking
$\beta_{2}=\beta_{3}=\beta$. Moreover, the first Lax equation (A.1) has an
additional gauge symmetry $\Phi\rightarrow\omega(t,\lambda)\Phi$. Under this
gauge transformation

$a\rightarrow\omega_{t}\omega^{-1}+a,$

and thus, by choosing $\omega$ to satisfy
$\omega_{t}+\frac{1}{2}(\gamma+\frac{\alpha}{\lambda})\omega=0$, we can
eliminate both $\alpha$ and $\gamma$ from the parametrization of $a$,
obtaining:

$\displaystyle a$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\begin{matrix}u_{x}-v&0\\\
0&\\-u_{x}+v\end{matrix}\right];$ $\displaystyle b$ $\displaystyle=$
$\displaystyle\left[\begin{matrix}u&0\\\
0&u\end{matrix}\right]+\frac{1}{\lambda}\left[\begin{matrix}0&\beta\\\
\beta&0\end{matrix}\right].$

Finally, we note that this form of $a,b$ implies that the Lax pair has a
scaling symmetry: $\lambda\rightarrow
s\lambda,M\rightarrow\frac{1}{s}M,\beta\rightarrow s\beta$. Choosing the scale
to be $s=\frac{1}{\beta}$ fixes $\beta=1$ and we obtain the final form (see
(2.4))

$\displaystyle a$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\begin{matrix}u_{x}-v&0\\\
0&\\-u_{x}+v\end{matrix}\right];$ $\displaystyle b$ $\displaystyle=$
$\displaystyle\left[\begin{matrix}u&0\\\
0&u\end{matrix}\right]+\frac{1}{\lambda}\left[\begin{matrix}0&1\\\
1&0\end{matrix}\right],$

used in Section 2.

## Acknowledgements

We would like to thank Professors Xiangke Chang and Shihao Li for pointing out
several misprints in the first draft of the paper.

After the current work was completed we learned that Equation 2.5 was also
derived in [9]. We would like to thank Professor Nianhua Li for bringing this
reference to our attention.

Jacek Szmigielski’s research is supported by the Natural Sciences and
Engineering Research Council of Canada (NSERC).

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