Erratum to: Wireless three-hop networks with stealing II: exact solutions through boundary value problems

Erratum to: Queueing Syst (2013) 74:235–272 DOI 10.1007/s11134-012-9332-8

Contrary to what we claimed in [5], the solution to the Riemann–Hilbert problem (4) considered in [5] for some domain $D_x$ is in general not the restriction to $D_x$ of the solution to the modified Riemann–Hilbert problem (6) in [5]. This occurs only when $D_x$ is a circle, which is not the case considered in that paper. To solve problem (4), we have to consider the conformal mapping $\gamma _x(x)$ from $D_x$ onto the unit disk $D$. We are thus led to consider a modified version of the Riemann–Hilbert problem (5) formulated in that paper.

Modified boundary value problem

To compute the function $P(x,0)$, by using arguments similar to those in [5], we consider the function $P_x(x)$ given by Eq. (20) in [5], namely

$$\begin{aligned} P_x(x) = P(x,0)-\frac{R_x}{x-x^*}+P(0,0). \end{aligned}$$

The function $P_x(x)$ is analytic in $D_x$ and satisfies for $x\in \partial D_x$ and $y=Y^*(x)$

$$\begin{aligned} \mathfrak {R}\left( i\frac{h_2(x,y)}{h_3(x,y)}P_x(x)\right) = \mathfrak {I}\left( \frac{R_xh_2(x,y)}{h_3(x,y)(x-x^*)} \right) . \end{aligned}$$

(1)

To solve this problem, we consider the conformal mapping $\gamma _x(x)$ from $D_x$ onto the unit disk. This conformal mapping can be chosen to preserve the symmetry with respect to the horizontal axis and to satisfy $\gamma _x(X^+(y_1))=-1$ and $\gamma _x(X^+(y_2))=1$. Moreover, by imposing the condition $\gamma _x(0)=0$, the conformal mapping $\gamma _x(x)$ is unique.

We are then led to consider the following problem on the unit circle: The function $P_x(c_x(u))$ is analytic in the unit disk $D$ and satisfies for $u$ on the unit circle $C$

$$\begin{aligned} \mathfrak {R}\left( i\frac{h_2(c_x(u),Y^*(c_x(u)))}{h_3(c_x(u),Y^*(c_x(u)))}P_x(c_x(u))\right) = \mathfrak {I}\left( \frac{R_xh_2(c_x(u),Y^*(c_x(u)))}{h_3(c_x(u),Y^*(c_x(u)))(c_x(u)-x^*)} \right) , \end{aligned}$$

where $c_x(u)$ is the inverse of $\gamma _x(x)$ and maps the unit disk onto the domain $D_x$.

The above Riemann–Hilbert problem is of the following form: Find a function which is analytic in $D$, continuous in the closure $\overline{D}$ of the unit disk, and satisfies

$$\begin{aligned} \mathfrak {R}(ia(u)f(u)) = c(u). \end{aligned}$$

Since, by construction, the function $f(u)$ is real on the real axis, it is possible to define by the reflection principle [3] the function $g(u) = \overline{f(1/\overline{u})}$, which is analytic in $\mathbb {C}\setminus \overline{D}$. The function $f(u)$ (resp. $g(u)$) is the restriction to $D$ (resp. $\mathbb {C}\setminus \overline{D}$) of the sectionally analytic function $F(u)$, which satisfies the following Riemann–Hilbert problem: Find a sectionally analytic function $F(u)$ with respect to the unit circle, bounded at infinity ($F(\infty ) = f(0)$) and such that for $u \in C$

$$\begin{aligned} a(u)F^i(u) -\overline{a(u)}F^e(u) = -2ic(u), \end{aligned}$$

where $F^i(u)$ (resp. $F^e(u)$) is the interior (resp. exterior) limit of the function $F(u)$ on the unit circle.

The solution to this problem, when it exists, is given by [2]

$$\begin{aligned} F(u) = \frac{\phi ^i(u)}{2i\pi } \int _{C}\frac{C(z)}{\phi ^{(i)}(z)(z-u)}dz + Q(u) \phi (u), \end{aligned}$$

where $Q(u)$ is a polynomial, which can be determined by using the conditions at infinity, the function $C(u)$ is given by

$$\begin{aligned} C(u) = -2i \frac{c(u)}{a(u)}, \end{aligned}$$

and the function $\phi (u)$ is defined by

$$\begin{aligned} \phi (u) = \left\{ \begin{array}{ll} \displaystyle \exp \left( \frac{1}{2i\pi }\int _C \log \left( z^{-\kappa } \frac{\overline{a(z)}}{a(z)} \right) \frac{dz}{z-u} \right) , &{} u \in D,\\ \displaystyle \frac{1}{u^\kappa }\exp \left( \frac{1}{2i\pi }\int _C \log \left( z^{-\kappa } \frac{\overline{a(z)}}{a(z)} \right) \frac{dz}{z-u} \right) , &{} u \in \mathbb {C}\setminus \overline{D}, \end{array}\right. \end{aligned}$$

with $\kappa$ denoting the index of the Riemann–Hilbert problem and $\phi ^{(i)}(u)$ being the interior limit of the function $\phi (u)$ on the unit circle. When $\kappa <0$, the solution to the Riemann–Hilbert problem exists and is unique if and only if for $k=0, \ldots ,|\kappa |-1$

$$\begin{aligned} \int _{C}\frac{z^k C(z)}{\phi ^{(i)}(z)} dz = 0; \end{aligned}$$

in that case, the polynomial $P(u)\equiv 0$. When $\kappa =0$, the solution is unique and $P(u)$ is a constant.

Conformal map

We determine in this section the conformal mapping $\gamma _x(x)$ from the domain $D_x$ onto the unit disk.

Let $w_x(x)$ be the conformal gluing function for the shift transformation $t_x(x)$ and the contour $\partial D_x$, where

$$\begin{aligned} t_x(x) \mathop {=}\limits ^{def} \overline{x} = \frac{Y^*(x)}{(1-p)x}. \end{aligned}$$

The function $w_x(x)$ is solution to the Carleman problem

$$\begin{aligned} w_x(t_x(x)) = w_x(x) \end{aligned}$$

for $x \in \partial D_x$. As proved in [8], the function $w_x(x)$ conformally transforms the domain $D_x$ onto the complex plane deprived of the positively oriented arc $L_x$ joining the points $w_x(X^+(y_2))$ and $w_x(X^+(y_1))$.

For the case under consideration, the conformal gluing function $w_x(t)$ is given in [1] and [7, Remark 16] by

$$\begin{aligned} w_x(t) = \frac{t}{(t-x_2)\left( t-\sqrt{\frac{1+p}{(1-p)^2x_2}}\right) ^2}. \end{aligned}$$

It is worth noting that $w_x(x)$ is real for $x\in \partial D_x$ and that

$$\begin{aligned} w_x(X^+(y_1))<w_x(X^+(y_2)). \end{aligned}$$

Using the fact that for reals $a<b$ the function $v(a,b;x)$ defined by

$$\begin{aligned} v(a,b;x) = \frac{\sqrt{\frac{x-a}{x-b}}-1}{\sqrt{\frac{x-a}{x-b}}+1} \end{aligned}$$

conformally maps the complex plane deprived of the segment $[a,b]$ onto the unit disk, the conformal mapping $\gamma _x(x)$ is given by

$$\begin{aligned} \gamma _x(x) = v(w_x(X^+(y_1)),w_x(X^+(y_2));w_x(x)). \end{aligned}$$

The function $w_x(x)$ takes value in $[w_x(X^+(y_1)),w_x(X^+(y_2))]$ for $x\in [x_3,x_4]$. It can then be shown that the equation with degree three $w_x(t)=w_x(x)$ for $x\in [x_3,x_4]$ has three solutions, namely $x$, $Y^*(x)/((1-p)x)$ and $Y_*(x)/((1-p)x)$. Since $Y^\pm ([x_3,x_4])= \partial D_x$, we deduce that the conformal mapping $\gamma _x(x)$ can be analytically continued in the whole of $\mathbb {C}\setminus [x_3,x_4]$ with an algebraic singularity at point $x_3$.

Solution of the corrected boundary value problem for $P(x,0)$

By using the above results, Eq. (34) in [5] should then read for $x\in D_x$ (when $p\le p^*$, the index $\kappa _x=0$)

$$\begin{aligned} \varphi ^{(1)}_x(x) = \exp \left( \frac{1}{2i\pi } \int _{\partial D_x} \frac{\log (A_x(z))}{z-x}k_x(x,z)dz \right) , \end{aligned}$$

(2)

where the kernel $k_x(x,z)$ is given by

$$\begin{aligned} k_x(x,z) = \frac{(z-x)\gamma _x'(z)}{\gamma _x(z)-\gamma _x(x)}. \end{aligned}$$

The kernel $k_x(x,z)$ is defined for $x \in D_x$ and when $x=z\in D_x$, $k_x(x,z)=1$.

The function $\varphi ^{(1)}_x(x)$ can be continued by setting

$$\begin{aligned}&\varphi ^{(1)}_x(x)= \nonumber \\&\quad \left\{ \begin{array}{l} \displaystyle \exp \left( \frac{1}{2i\pi } \int _{\partial D_x} \frac{\log (A_x(z))}{z-x}k_x(x,z)dz \right) , \text{ for } x \in {D_x},\\ \displaystyle A_x(x) \exp \left( \frac{1}{2i\pi } \int _{\partial D_x} \frac{\log (A_x(z))}{z-x}k_x(x,z)dz \right) , \text{ for } x \in \overline{D_x}^c\cap V_x\cap D(0,x_3), \end{array}\right. \nonumber \\ \end{aligned}$$

(3)

where $V_x$ is the domain of analyticity of the function $\log (A_x(x))$.

Similarly, Eq. (37) in [5] should read

$$\begin{aligned}&\varphi ^{(2)}_x(x) = \nonumber \\&\quad \left\{ \begin{array}{l} \displaystyle \exp \left( \frac{1}{2i\pi } \int _{\mathcal {C}_x} \frac{\log (\mathcal {A}_x(z))}{z-x}k_x(x,z)dz \right) , \text{ for } x \in \overline{D_x},\\ \displaystyle \mathcal {A}_x(x)\exp \left( \frac{1}{2i\pi } \int _{\partial D_x} \frac{\log (\mathcal {A}_x(z))}{z-x}k_x(x,z)dz\right) , \text{ for } x \in \overline{D_x}^c\cap \mathcal {V}_x\cap D(0,x_3). \end{array}\right. \end{aligned}$$

(4)

Eq. (35) is erroneous.

The function $P(x,0)$ is eventually given by

$$\begin{aligned} P(x,0)&= \frac{R_x}{x-x^*} + B_x(x)+ \frac{\varphi _x(x)}{2i\pi }\int _{\partial D_x} \frac{B_x(z)-B_x(x)}{\phi ^i_x(z)(z-x)}k_x(x,y)dz\nonumber \\&-P(0,0) +K_x\varphi _x(x), \end{aligned}$$

(5)

where

$$\begin{aligned} \varphi _x(x)=\left\{ \begin{array}{ll} \varphi ^{(1)}_x(x), &{}\quad \hbox {when}\,p \le p^*, \\ \varphi ^{(2)}_x(x), &{}\quad \hbox {when}\,p >p^*, \end{array}\right. \end{aligned}$$

(6)

with $\varphi _x^{(1)}(x)$ as in (3) and $\varphi _x^{(2)}(x)$ as in (4), $K_x$ is some constant, and the function $B_x(x)$ defined by Eq. (31) in [5].

Solution of the corrected boundary value problem for $P(0,y)$

For determining the function $P(0,y)$, the conformal mapping $\gamma _y(y)$ from the domain $D_y$ onto the unit disk is required, which is given by (see [1, 10])

$$\begin{aligned} \gamma _y(y) = \frac{yk_y\left( \frac{\eta }{1+p}\right) -(1+p)\eta k_y\left( \frac{y}{1+p}\right) }{yk_y\left( \frac{\eta }{1+p}\right) +(1+p)\eta k_y\left( \frac{y}{1+p}\right) }, \end{aligned}$$

where

$$\begin{aligned} k_y(y)=(\alpha -y)\sqrt{\frac{1-p}{1+p} -(1-p)^2 \alpha ^2y} \end{aligned}$$

with $\alpha = Y^*(x_2)/(1+p)$ and $\eta$ is any constant in $\left( 0,\alpha \right)$. In the following, we take $\eta = \alpha /2$. By using similar arguments as those for the conformal mapping $\gamma _x(x)$, we can prove that the conformal mapping $\gamma _y(y)$ is analytic in $D(0,y_3)$ with continuous limits on the boundary.

Equation (51) in [5] should read

$$\begin{aligned} \phi _y(y) = \exp \left( \frac{1}{2i\pi }\int _{\partial D_y} \frac{\log \left( A_y(z) \right) }{z-y} k_y(y,z)dz \right) , \end{aligned}$$

(7)

where the kernel $k_y(y,z)$ is defined by

$$\begin{aligned} k_y(y,z) = \frac{(z-y)\gamma '_y(z)}{\gamma _y(z)-\gamma _y(y)}. \end{aligned}$$

The interior limit of $\phi _y(y)$ on the contour $\partial D_y$ is defined by

$$\begin{aligned} \phi _y^{(i)}(y) = \exp \left( \frac{1}{2}\log (A_y(y)) + \frac{1}{2i\pi }\oint _{\partial D_y} \frac{\log \left( A_y(z) \right) }{z-y} k_y(y,z)dz \right) . \end{aligned}$$

(8)

The restriction to $D_y$ of the function $\phi _y(y)$ coincides with the restriction to $D_y$ of the function $\varphi _y(y)$ defined by

$$\begin{aligned} \varphi _y(y) = \left\{ \begin{array}{ll} \phi _y(y), &{}\quad \mathrm{if}\,y\in D_y, \\ \displaystyle A_y(y)\phi _y(y), &{}\quad \mathrm{if}\, y\notin D_y, \end{array}\right. \end{aligned}$$

(9)

which is a meromorphic function in $\mathbb {C}\setminus [y_3,\infty )$ with a pole at the point $y^{**}$.

The function $P(0,y)$ is given in $\mathbb {C}\setminus [y_3,y_4]$ by

$$\begin{aligned} P(0,y)&= \frac{R_y}{y-y^{**}} -P(0,0) +B_y(y) + \frac{\varphi _y(y)}{2i\pi }\int _{\partial D_y} \frac{B_y(z)-B_y(y)}{\phi ^i_y(z)(z-y)}k_y(z,y)dz \nonumber \\&+\,K_y \varphi _y(y), \end{aligned}$$

(10)

where $B_y(y)$ is given by Eq. (50) in [5] and $K_y$ is a constant as in Eq. (56) in [5].

Asymptotic analysis

In spite of these modifications, the asymptotic analysis given in Sect. 6 of [5] is correct; the factors $\kappa _1$ and $\kappa _2$ should be computed by using the new formulas giving $P(0,y^*)$ and $P(0,0)$ depending on the conformal mapping $\gamma _y(y)$.

The same flaw occurs in [6] but the asymptotic analysis is also correct.