Inverse problems associated with integrable equations of Camassa-Holm type; explicit formulas on the real axis, I

Keivan Mohajer Department of Mathematics, University of Isfahan, Isfahan, 81746-73441, Iran; k.mohajer@sci.ui.ac.ir Jacek Szmigielski Department of Mathematics and Statistics, University of Saskatchewan, 106 Wiggins Road, Saskatoon, Saskatchewan, S7N 5E6, Canada; szmigiel@math.usask.ca

Abstract

The inverse problem which arises in the Camassa–Holm equation is revisited for the class of discrete densities. The method of solution relies on the use of orthogonal polynomials. The explicit formulas are obtained directly from the analysis on the real axis without any additional transformation to a “string” type boundary value problem known from prior works.

Hongyou Wu in memoriam

Key Words: Camassa-Holm equation, inverse problems, orthogonal polynomials AMS Subject Classification: Primary 37K15.

1 Introduction

The purpose of this paper and its sequel is to present an alternative solution to two inverse problems which appear in the context of Camassa-Holm type equations. We concentrate in this paper on the Camassa-Holm equation (CH)[3]:

u_{t}-u_{xxt}+3uu_{x}=2u_{x}u_{xx}+uu_{xxx},

(1.1)

while in the sequel we study the inverse problem for another equation, discovered by V. Novikov (VN) [7]:

u_{t}-u_{xxt}+4u^{2}u_{x}=3uu_{x}u_{xx}+u^{2}u_{xxx}.

(1.2)

Both these equations admit another, often used, representation

m_{t}+m_{x}u+2mu_{x}=0,\qquad m=u-u_{xx},

(1.3)

for the CH equation, and

m_{t}+(m_{x}u+3mu_{x})\,u=0,\qquad m=u-u_{xx},

(1.4)

for the VN equation. The solution to the inverse problem presented in this paper is obtained directly on the real axis, instead of transforming the problem to the inhomogeneous string problem as was done in [1] and adapting the method used by T. Stieltjes in [8].

We give a self-contained presentation of the inverse problem for Equation (2.6), emphasizing the role of orthogonal polynomials (Proposition 11) and culminating in the elegant inverse formulas stated in Corollary 14. In the sequel, for the case of the VN equation (1.2), we will show that a different class of polynomials is germane to the inverse problem, namely a class of biorthogonal polynomials (Cauchy biorthogonal polynomials) studied in [2].

2 CH inspired inverse problem

The well known Lax pair system for the CH equation can be written as follows (see [3])

\begin{split}&\bigl{(}1-D_{x}^{2}-zm\bigr{)}\psi=0,\\ &\bigl{(}D_{t}+(u+1/z)D_{x}-u_{x}/2-1/z\bigr{)}\psi=0.\end{split}

(2.1)

We would like to point out that this Lax pair yields a slightly different normalization than in (1.3), here, $m=2u-\frac{1}{2}u_{xx}$ . If we set $m(x,t)=\sum_{j=1}^{n}m_{j}(t)\delta_{x_{j}}$ where $m_{j}(t)$ and $x_{j}(t)$ are $C^{\infty}$ functions, we obtain what is called the $n$ -peakon Lax pair for the CH equation. We assume that for every fixed $t\geq 0$ the function $\psi(x,t)$ is continuous on $\mathbb{R}$ and $x_{1}(0)<x_{2}(0)<\dots<x_{n}(0)$ . Now the first equation of the system (2.1) implies that on the interval $(x_{j},x_{j+1})$ , we have

\psi=A_{j}e^{x}+B_{j}e^{-x},

(2.2)

It is well known that requiring $\psi\rightarrow 0$ as $|x|\rightarrow\infty$ is consistent with the system (2.1). Let’s therefore assume that $A_{0}=1$ and $B_{0}=0$ . Then from the continuity of $\psi$ and the jump conditions imposed by the first equation of (2.1) we have

\begin{pmatrix}A_{j}\\ B_{j}\end{pmatrix}=K_{j}(z)\begin{pmatrix}1\\ 0\end{pmatrix},\ \ \ 1\leq j\leq n

(2.3)

where

K_{j}(z)=T_{j}(z)T_{j-1}(z)\dots T_{1}(z),

(2.4)

and

T_{k}(z)=\begin{pmatrix}1-\frac{z}{2}m_{k}&-\frac{z}{2}m_{k}e^{-2x_{k}}\\ \frac{z}{2}m_{k}e^{2x_{k}}&1+\frac{z}{2}m_{k}\end{pmatrix},\ \ \ \ 1\leq k\leq n.

(2.5)

Clearly,

Theorem 1.

The spectrum of the boundary value problem

\bigl{(}1-D_{x}^{2}-zm\bigr{)}\psi=0,\qquad\psi\rightarrow 0,\text{ for $|x|\rightarrow\infty$ }

(2.6)

is given by roots of $A_{n}(z)=0$ .

In order to get more information about the spectrum we can rewrite equation (2.6) as an integral equation

\psi(x,z)=z\int K(x,y)\psi(y,z)dM(x),

where $M$ is the distribution function of the measure $m(x)$ and $K(x,y)=\frac{1}{2}e^{-|x-y|}$ is the unique Greens function of $1-D^{2}_{x}$ vanishing at $\pm\infty$ . One can easily write the explicit form of this integral equation as:

\psi(x,z)=z\sum_{j=1}^{n}\frac{1}{2}e^{-|x-x_{j}|}\psi(x_{j},z)m_{j},

(2.7)

where, by assumption, all $m_{j}>0$ . Moreover, evaluating at $x_{i}$ , and introducing the matrix notation

\boldsymbol{\psi}=(\psi(x_{1},z),\psi(x_{2},z),\dots,\psi(x_{n},z))^{T},\qquad\mathbf{E}=\big{[}\frac{1}{2}e^{-|x_{i}-x_{j}|}\big{]},

(2.8)

and

\mathbf{P}=\operatorname{diag}(m_{1},m_{2},\dots,m_{n}),

(2.9)

we obtain equation (2.7) in the form of the matrix eigenvalue problem:

\boldsymbol{\psi}=z\mathbf{E}\mathbf{P}\,\boldsymbol{\psi}.

$\mathbf{E}$ is an example of a single-pair matrix (see [5]) which is oscillatory and so is $\mathbf{E}\mathbf{P}$ . Oscillatory matrices form a subset of totally nonnegative matrices, which can be characterized as matrices whose all minors of arbitrary size are nonnegative. Oscilatory matrices are special in that they have simple, strictly positive, spectrum [5]. We therefore have the following description of the spectrum of the boundary value problem (2.6).

Lemma 2.

$A_{n}(z)=\det(I-z\mathbf{E}\mathbf{P})={\displaystyle\prod_{1\leq k\leq n}\big{(}1-\frac{z}{\lambda_{k}}\big{)}}$ where $0<\lambda_{1}<\lambda_{2}<\dots<\lambda_{n}$

We need one more fact.

Theorem 3.

The Weyl function $W(z)=-\frac{B_{n}(z)}{A_{n}(z)}$ is a (shifted) Stieltjes transform of a measure $d\mu(x)=\sum_{j=1}^{n}a_{j}\delta_{\lambda_{j}},\,a_{j}>0$ , that is

W(z)=\gamma+\int\frac{1}{z-x}d\mu(x),

where $\gamma=\int\frac{1}{x}d\mu(x)$ .

Proof.

This theorem follows, after a slight adjustment of notation, from Theorem 4.2 proven in [1] and the observation that $W(0)=0$ which is a consequence of $T_{k}(0)=I$ . ∎

The continued fraction expansion ([8]) turns out to be at the heart of the inverse problem, in a full analogy with the inverse problem for an inhomogeneous string which was studied in the 1950’s by M.G. Krein ([6], [4]).

Theorem 4.

\begin{split}&\frac{B_{n}}{A_{n}}=-e^{2x_{n}}+\\ &\cfrac{1}{-\frac{z}{2}m_{n}e^{-2x_{n}}+\cfrac{1}{e^{2x_{n}}-e^{2x_{n-1}}+\cfrac{1}{-\frac{z}{2}m_{n-1}e^{-2x_{n-1}}+\cfrac{1}{{\ddots}\ \ \ _{+\cfrac{1}{-\frac{z}{2}m_{1}e^{-2x_{1}}+\cfrac{1}{e^{2x_{1}}}}}}}}},\end{split}

(2.10)

Proof.

Since

\begin{pmatrix}A_{n}\\ B_{n}\end{pmatrix}=T_{n}\begin{pmatrix}A_{n-1}\\ B_{n-1}\end{pmatrix},

(2.11)

we have

\frac{B_{n}}{A_{n}}=\frac{\frac{z}{2}m_{n}e^{2x_{n}}A_{n-1}+(1+\frac{z}{2}m_{n})B_{n-1}}{(1-\frac{z}{2}m_{n})A_{n-1}-\frac{z}{2}m_{n}e^{-2x_{n}}B_{n-1}}.

(2.12)

This equation is equivalent to

\frac{B_{n}}{A_{n}}=-e^{2x_{n}}+\cfrac{1}{-\frac{z}{2}m_{n}e^{-2x_{n}}+\cfrac{1}{e^{2x_{n}}+\cfrac{B_{n-1}}{A_{n-1}}}}.

(2.13)

Hence, the proof is complete by induction. ∎

Suppose that $P_{j}/Q_{j}$ $(0\leq j\leq 2n)$ is the $j$ th convergent of the negative of the continued fraction we obtained above so that we are approximating the Weyl function $W(z)$ . Then we observe that

Q_{0}=1,\ \ \ P_{0}=e^{2x_{n}},

\frac{P_{1}}{Q_{1}}=\frac{1+\frac{z}{2}m_{n}}{\frac{z}{2}m_{n}e^{-2x_{n}}},

(2.14)

and

\frac{P_{2}}{Q_{2}}=\frac{e^{2x_{n-1}}-\frac{z}{2}m_{n}(e^{2x_{n}}-e^{2x_{n-1}})}{1-\frac{z}{2}m_{n}(1-e^{2(x_{n-1}-x_{n})})}.

(2.15)

Therefore, it can be verified that

\begin{split}&Q_{2}=-(e^{2x_{n}}-e^{2x_{n-1}})Q_{1}+Q_{0},\\ &P_{2}=-(e^{2x_{n}}-e^{2x_{n-1}})P_{1}+P_{0},\end{split}

(2.16)

\begin{split}&Q_{3}=\frac{z}{2}m_{n-1}e^{-2x_{n-1}}Q_{2}+Q_{1},\\ &P_{3}=\frac{z}{2}m_{n-1}e^{-2x_{n-1}}P_{2}+P_{1}.\end{split}

(2.17)

These equations guide us to the following theorem

Theorem 5.

\begin{split}&Q_{2k}=-(e^{2x_{n-k+1}}-e^{2x_{n-k}})Q_{2k-1}+Q_{2k-2},\\ &P_{2k}=-(e^{2x_{n-k+1}}-e^{2x_{n-k}})P_{2k-1}+P_{2k-2},\end{split}

(2.18)

\begin{split}&Q_{2k+1}=\frac{z}{2}m_{n-k}e^{-2x_{n-k}}Q_{2k}+Q_{2k-1},\\ &P_{2k+1}=\frac{z}{2}m_{n-k}e^{-2x_{n-k}}P_{2k}+P_{2k-1}.\end{split}

(2.19)

Proof.

By induction. ∎

Corollary 6.

P_{j}Q_{j-1}-P_{j-1}Q_{j}=(-1)^{j-1},\ \ \ \ 1\leq j\leq 2n.

(2.20)

Proof.

Use induction and the previous theorem. ∎

Corollary 7.

\begin{split}&\deg(P_{2k+1})=\deg(Q_{2k+1})=k+1,\\ &\deg(P_{2k})=\deg(Q_{2k})=k.\end{split}

(2.21)

In preparation for the inverse problem we introduce the following notation.

Notation: Given a polynomial $p(z)$ we denote by $p[j]$ the coefficient at the $j$ th power of $z$ . Moreover, $n-k$ is abbreviated to $k^{\prime}$ .

With this notation in place we have

Theorem 8.

	$\displaystyle Q_{2k}[0]=1,\quad P_{2k}[0]=e^{2x_{k^{\prime}}},$
	$\displaystyle Q_{2k+1}[0]=0,\quad P_{2k+1}[0]=1,\qquad P_{2k+1}[1]=\frac{1}{2}\sum_{j=0}^{k}m_{j^{\prime}}$

Proof.

By induction. ∎

The Weyl function of the $n$ -peakon problem reads:

Proposition 9.

W(z)=\frac{P_{2n}(z)}{Q_{2n}(z)}.

The inverse problem hinges on the following approximation formulas:

Theorem 10.

\begin{split}&W(z)-\frac{P_{2k+1}(z)}{Q_{2k+1}(z)}=\mathcal{O}(\frac{1}{z^{2k+1}}),\ \ \ \ k=0,\dots,n-1,\ \ \ z\to\infty,\\ &W(z)-\frac{P_{2k}(z)}{Q_{2k}(z)}=\mathcal{O}(\frac{1}{z^{2k+1}}),\ \ \ \ k=0,\dots,n-1,\ \ \ z\to\infty.\end{split}

(2.22)

Proof.

By the previous theorem we have

\frac{P_{2k+1}}{Q_{2k+1}}-\frac{P_{2k}}{Q_{2k}}=\frac{1}{Q_{2k+1}Q_{2k}}=\mathcal{O}(\frac{1}{z^{2k+1}}),\ \ \ \ z\to\infty.

Also,

\frac{P_{2k}}{Q_{2k}}-\frac{P_{2k-1}}{Q_{2k-1}}=\frac{-1}{Q_{2k}Q_{2k-1}}=\mathcal{O}(\frac{1}{z^{2k-1}}),\ \ \ \ z\to\infty.

From these and the definition of the Weyl function we get

W(z)-\frac{P_{2k-1}}{Q_{2k-1}}=\frac{P_{2n}(z)}{Q_{2n}(z)}-\frac{P_{2k-1}}{Q_{2k-1}}=\mathcal{O}(\frac{1}{z^{2k-1}}),\ \ \ \ z\to\infty.

Similarly, we see that

W(z)-\frac{P_{2k}}{Q_{2k}}=\mathcal{O}(\frac{1}{z^{2k+1}}),\ \ \ z\to\infty.

∎

Since $Q_{2k+1}(0)=0$ we set ${Q_{2k+1}(z)=z\,q_{2k+1}(z),\,\deg\,q_{2k+1}=k.}$

Proposition 11.

(orthogonality)


	$\displaystyle\int x^{j}Q_{2k}(x)d\mu(x)=0,\ \ j=0,\dots,k-1,\ \ k=0,\dots,n-1$		(2.23a)
	$\displaystyle\int q_{2k+1}(x)d\mu(x)=1,\ \ \ k=0,\dots,n-1,$		(2.23b)
	$\displaystyle\int x^{j}q_{2k+1}(x)d\mu(x)=0,\ \ j=1,\dots,k-1,\ \ k=0,\dots,n-1.$		(2.23c)

Proof.

By the first approximation formula of Theorem 10 we have

W(z)Q_{2k+1}(z)-P_{2k+1}(z)=\mathcal{O}(\frac{1}{z^{k}}),\ \ \ \ \ k=0,\dots,n-1.

(2.24)

Therefore,

z^{j}(W(z)Q_{2k+1}(z)-P_{2k+1}(z))=\mathcal{O}(\frac{1}{z^{k-j}}),\ \ \ j=0,\dots,k-2.

Since $W(z)$ is analytic at $z=\infty$ so is the remainder and its order there is $\mathcal{O}(\frac{1}{z^{2}})$ . We can therefore use the residue calculus. To this end we choose a circle $\Gamma$ whose radius is large enough to contain the support of $\mu$ defined in Theorem 3. An elementary contour integration implies then

\frac{1}{2\pi i}\int_{\Gamma}\int\frac{z^{j}Q_{2k+1}(z)}{z-x}d\mu(x)dz=0,\ \ \ \ j=0,\dots,k-2.

Consequently, applying Fubini’s theorem and Cauchy’s residue theorem we have

\int x^{j}Q_{2k+1}(x)d\mu(x)=0,\ \ j=0,\dots,k-2,\ \ k=0,\dots,n-1.

Similarly, from the second approximation formula of Theorem 10 we have

W(z)Q_{2k}(z)-P_{2k}(z)=\mathcal{O}(\frac{1}{z^{k+1}}),\ \ \ \ \ k=0,\dots,n-1.

(2.25)

Thus,

\int x^{j}Q_{2k}(x)d\mu(x)=0,\ \ j=0,\dots,k-1,\ \ k=0,\dots,n-1.

Also we have

\frac{W(z)Q_{2k+1}(z)}{z}-\frac{P_{2k+1}(z)}{z}=\mathcal{O}(\frac{1}{z^{k+1}}),\ \ \ \ \ k=0,\dots,n-1.

(2.26)

Since $Q_{2k+1}[0]=0$ and $P_{2k+1}[0]=1$ , we have

\frac{1}{2\pi i}\int_{\Gamma}\int\frac{z^{-1}Q_{2k+1}(z)}{z-x}d\mu(x)dz-1=0.

Hence,

\int x^{-1}Q_{2k+1}(x)d\mu(x)=1,\ \ \ k=0,\dots,n-1.

Finally, using $Q_{2k+1}(x)=x\,q_{2k+1}(x)$ we obtain the claim. ∎

Proposition 12.


	$\displaystyle P_{2k}(z)=\int\frac{zQ_{2k}(z)-xQ_{2k}(x)}{x(z-x)}\,\,d\mu(x),$		(2.27a)
	$\displaystyle P_{2k+1}(z)=\int\frac{zQ_{2k+1}(z)-xQ_{2k+1}(x)}{x(z-x)}\,\,d\mu(x).$		(2.27b)

Proof.

From the second approximation formula of theorem 10 we have

W(z)Q_{2k}(z)=P_{2k}(z)+\mathcal{O}(\frac{1}{z^{k+1}}),\ \ \ z\rightarrow\infty.

By Theorem 3 $W(z)=\gamma+\int\frac{1}{z-x}d\mu(x)$ which implies

P_{2k}(z)=\gamma Q_{2k}(z)+\int\frac{Q_{2k}(z)-Q_{2k}(x)}{z-x}d\mu(x).

Recalling that $\gamma=\int\frac{1}{x}d\mu(x)$ and performing elementary manipulations we obtain the claim. The proof for $P_{2k+1}$ is analogous. ∎

Let $c_{j}=\int x^{j}d\mu(x),\,j\in\mathbb{Z}$ be the $j$ th moment of the measure $d\mu(x)$ and let $I=[I_{ij}]=[c_{i+j}]$ be the Hankel matrix of the moments of the measure $d\mu(x)$ appearing in Theorem 3. We note that the shift $\gamma=c_{-1}$ . In addition, let us denote by $\Delta^{i}_{k}$ the minor of the submatrix of $I$ of size $k\times k$ which starts at $I_{0i}$ . By convention, $\Delta_{0}^{j}=1,\,j\in\mathbb{Z}$ .

Theorem 13.

Suppose we are given an arbitrary discrete measure $d\mu(x)=\sum_{j}a_{j}\delta_{\lambda_{j}},\quad\lambda_{j}>0,$ and define its shifted Stieltjes transform

W(z)=\gamma+\int\frac{1}{z-x}d\mu(x),\qquad\gamma=\int\frac{1}{x}d\mu(x).

Let $k$ be a natural number such that $0\leq k\leq n-1$ . Then there exists a unique solution to the approximation problem stated in Theorem 10 whose polynomials satisfy degree requirements dictated by Corollary 7:


	$\displaystyle Q_{2k}(z)=\frac{1}{\Delta_{k}^{1}}\begin{vmatrix}1&z&z^{2}&\dots&z^{k}\\ c_{0}&c_{1}&c_{2}&\dots&c_{k}\\ c_{1}&c_{2}&c_{3}&\dots&c_{k+1}\\ \vdots&\vdots&\vdots&&\vdots\\ c_{k-1}&c_{k}&c_{k+1}&\dots&c_{2k-1}\end{vmatrix},\ \$		(2.28a)
	$\displaystyle Q_{2k+1}=\frac{1}{\Delta_{k+1}^{0}}\begin{vmatrix}z&z^{2}&z^{3}&\dots&z^{k+1}\\ c_{1}&c_{2}&c_{3}&\dots&c_{k+1}\\ \vdots&\vdots&\vdots&&\vdots\\ c_{k}&c_{k+1}&c_{k+2}&\dots&c_{2k}\end{vmatrix},\ \$		(2.28b)
	$\displaystyle P_{2k}(z)=\int\frac{zQ_{2k}(z)-xQ_{2k}(x)}{x(z-x)}\,\,d\mu(x),$		(2.28c)
	$\displaystyle P_{2k+1}(z)=\int\frac{zQ_{2k+1}(z)-xQ_{2k+1}(x)}{x(z-x)}\,\,d\mu(x).$		(2.28d)

Proof.

The proof is based on the work of Stieltjes [8]. We give only a complete argument for $Q_{2k}$ , the proof for $Q_{2k+1}(z)$ is analogous. Both proofs rely on Proposition 11.

We write $Q_{2k}(z)=\sum_{i=0}^{k}q_{i}z^{i}$ . Now, using orthogonality conditions and the fact that $Q_{2k}[0]=1$ we have

\int x^{j}(1+\sum_{i=1}^{k}q_{i}x^{i})d\mu(x)=0,\ \ j=0,1,\dotsc,k-1,\ \ \ k=0,\dots,n-1.

Recall that $c_{j}=\int x^{j}d\mu(x)$ . Then we have

\sum_{i=1}^{k}c_{i+j}q_{i}=-c_{j},\ \ j=0,1,\dotsc,k-1.

Hence we obtain the system

Bq=-c,

where

B=\begin{pmatrix}c_{1}&c_{2}&\dots&c_{k}\\ c_{2}&c_{3}&\dots&c_{k+1}\\ \vdots&\vdots&&\vdots\\ c_{k}&c_{k+1}&\dots&c_{2k-1}\end{pmatrix},\ \ q=(q_{1},q_{2}\dotsc,q_{k})^{T},\ \ c=(c_{0},c_{1},\dotsc,c_{k-1})^{T}.

Thus, Cramer’s rule implies that

Q_{2k}(z)=\frac{1}{\Delta_{k}^{1}}\begin{vmatrix}1&z&z^{2}&\dots&z^{k}\\ c_{0}&c_{1}&c_{2}&\dots&c_{k}\\ c_{1}&c_{2}&c_{3}&\dots&c_{k+1}\\ \vdots&\vdots&\vdots&&\vdots\\ c_{k-1}&c_{k}&c_{k+1}&\dots&c_{2k-1}\end{vmatrix}.

The formulas for $P_{2k}$ and $P_{2k+1}$ were already obtained in Proposition 12. ∎

Using Theorem 8 we arrive at the inversion formulas

Corollary 14.

The solution to the inverse problem is given by: $e^{2 x_{k^{'}}} = \frac{Δ_{k + 1}^{- 1}}{Δ_{k}^{1}}, \sum_{j = 0}^{k} m_{j^{'}} = \frac{2}{Δ_{k + 1}^{0}} | \begin{matrix} c_{- 1} & c_{0} & c_{1} & \dots & c_{k - 1} \\ c_{1} & c_{2} & c_{3} & \dots & c_{k + 1} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ c_{k} & c_{k + 1} & c_{k + 2} & \dots & c_{2 k} \end{matrix} |, k = 0, 1, \dots, n - 1,$

Proof.

By Theorem 8 $e^{x_{k^{\prime}}}=P_{2k}[0]$ . From Theorem 13 we obtain

P_{2k}[0]=\int\frac{Q_{2k}(x)}{x}d\mu(x),

which, in view of the determinantal formula presented in Theorem 13, implies the first formula. Likewise, to get the second formula we note

P_{2k+1}[1]=P_{2k+1}^{\prime}(0)=\int\frac{q_{2k+1}(x)}{x}d\mu(x),

and use Theorems 13 and 8 to justify the claim.

∎

Remark 15.

Although we are not going to give a complete argument here, it is easy to show using for example the technique of Lemma 5.4 in [1] that all determinants appearing in the inversion are strictly positive. Also, it is worth noticing that all these determinants are in fact minors of the moment matrix $I$ .

Example 16.

Here we solve the inverse problem for $n=1$ and $n=2$ . If $n=1$ , by corollary 14 we get

e^{2x_{1}}=\frac{\Delta_{1}^{-1}}{\Delta_{0}^{1}}=c_{-1}.

x_{1}=\frac{1}{2}\log{c_{-1}}=\frac{1}{2}\log(\frac{a_{1}}{\lambda_{1}})

(2.29)

and

m_{1}=\frac{2c_{-1}}{c_{0}}=\frac{2}{\lambda_{1}}.

(2.30)

If n=2, we have

\begin{split}&e^{2x_{1}}=\frac{\Delta_{2}^{-1}}{\Delta_{1}^{1}}=\frac{c_{-1}c_{1}-c_{0}^{2}}{c_{1}},\\ &e^{2x_{2}}=\frac{\Delta_{1}^{-1}}{\Delta_{0}^{1}}=c_{-1}.\end{split}

Therefore,

\begin{split}&x_{1}=\frac{1}{2}\log\frac{c_{-1}c_{1}-c_{0}^{2}}{c_{1}}=\frac{1}{2}\log\frac{a_{1}a_{2}(\lambda_{1}-\lambda_{2})^{2}}{\lambda_{1}\lambda_{2}(a_{1}\lambda_{1}+a_{2}\lambda_{2})},\\ &x_{2}=\frac{1}{2}\log{c_{-1}}=\frac{1}{2}\log\frac{a_{1}\lambda_{2}+a_{2}\lambda_{1}}{\lambda_{1}\lambda_{2}}.\end{split}

(2.31)

Also, we have

\begin{split}&m_{1}+m_{2}=\frac{2}{\Delta_{2}^{0}}(c_{-1}c_{2}-c_{0}c_{1})=\frac{2(c_{-1}c_{2}-c_{0}c_{1})}{c_{0}c_{2}-c_{1}^{2}},\\ &m_{2}=\frac{2c_{-1}}{c_{0}}=\frac{2(a_{1}\lambda_{2}+a_{2}\lambda_{1})}{\lambda_{1}\lambda_{2}(a_{1}+a_{2})}.\end{split}

(2.32)

Therefore,

m_{1}=\frac{2c_{1}(c_{-1}c_{1}-c_{0}^{2})}{c_{0}(c_{0}c_{2}-c_{1}^{2})}=\frac{2(a_{1}\lambda_{1}+a_{2}\lambda_{2})}{\lambda_{1}\lambda_{2}(a_{1}+a_{2})}.

(2.33)

These formulas (after shifting $\lambda_{j}\rightarrow-\lambda_{j},\,a_{j}\rightarrow 2a_{j}\lambda_{j}$ ) are identical to what is given on page 246 in [1].

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