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Appendix H

Solution of the Problem (Chapter 9)

Using the dimensionless expressions in Eqs. (9.28)–(9.38) and the following approximate expressions at interfaces z=0 and z=h₃,

${\frac{\partial p_{1}^{'}}{\partial z}|}_{z = h_{3}} = \frac{{p_{1} - p_{1}^{′}|}_{z = h_{3}}}{h_{1} / 2} and {\frac{\partial p_{2}^{'}}{\partial z}|}_{z = 0} = \frac{{p_{2}^{1}|}_{z = 0} - p_{2}}{h_{2} / 2}$ ${\frac{\partial p_{1}^{'}}{\partial z}|}_{z = h_{3}} = \frac{{p_{1} - p_{1}^{′}|}_{z = h_{3}}}{h_{1} / 2} and {\frac{\partial p_{2}^{'}}{\partial z}|}_{z = 0} = \frac{{p_{2}^{1}|}_{z = 0} - p_{2}}{h_{2} / 2}$

si1_e (H.1)

Eqs. (9.8)–(9.21) becomes

$ω_{1} \frac{\partial p_{D 1}}{\partial t_{D}} - w_{1} Δ_{D} p_{D 1} + 2 k_{v D 1} (p_{D 1} - {p_{D 1}^{'}|}_{z_{D} = 1}) = 0$ $ω_{1} \frac{\partial p_{D 1}}{\partial t_{D}} - w_{1} Δ_{D} p_{D 1} + 2 k_{v D 1} (p_{D 1} - {p_{D 1}^{'}|}_{z_{D} = 1}) = 0$

si2_e (H.2)

$ω_{2} \frac{\partial p_{D 2}}{\partial t_{D}} - w_{2} Δ_{D} p_{D 2} + 2 k_{v D 2} (p_{D 2} - {p_{D 2}^{'}|}_{z_{D} = 0}) = 0$ $ω_{2} \frac{\partial p_{D 2}}{\partial t_{D}} - w_{2} Δ_{D} p_{D 2} + 2 k_{v D 2} (p_{D 2} - {p_{D 2}^{'}|}_{z_{D} = 0}) = 0$

si3_e (H.3)

$p_{D j} \to 0 when t_{D} \to 0, j = 1, 2$ $p_{D j} \to 0 when t_{D} \to 0, j = 1, 2$

(H.4)

$p_{D j} \to 0 when r_{D} \to \infty, j = 1, 2$ $p_{D j} \to 0 when r_{D} \to \infty, j = 1, 2$

(H.5)

$w_{j} {(r_{D} \frac{\partial p_{D j}}{\partial r_{D}})}_{r_{D} = 1} = - q_{D j}, j = 1, 2$ $w_{j} {(r_{D} \frac{\partial p_{D j}}{\partial r_{D}})}_{r_{D} = 1} = - q_{D j}, j = 1, 2$

si6_e (H.6)

$p_{w D j} = {(p_{D j} - s_{j} r_{D} \frac{\partial p_{D j}}{\partial r_{D}})}_{r_{D} = 1}, j = 1, 2$ $p_{w D j} = {(p_{D j} - s_{j} r_{D} \frac{\partial p_{D j}}{\partial r_{D}})}_{r_{D} = 1}, j = 1, 2$

si7_e (H.7)

$q_{L D} = f (p_{w D 1}, p_{w D 2})$ $q_{L D} = f (p_{w D 1}, p_{w D 2})$

(H.8)

$q_{L D} = α_{D} (p_{w D 1} - p_{w D 2})$ $q_{L D} = α_{D} (p_{w D 1} - p_{w D 2})$

(H.9)

$q_{D 1} = 1 - q_{L D} - C_{D 1} \frac{d p_{w D 1}}{d t_{D}}$ $q_{D 1} = 1 - q_{L D} - C_{D 1} \frac{d p_{w D 1}}{d t_{D}}$

si10_e (H.10)

$q_{D 2} = q_{L D} - C_{D 2} \frac{d p_{w D 2}}{d t_{D}}$ $q_{D 2} = q_{L D} - C_{D 2} \frac{d p_{w D 2}}{d t_{D}}$

si11_e (H.11)

$k_{v D 1} h_{D 1} {\frac{\partial p_{D 1}^{'}}{\partial z_{D}}|}_{z_{D} = 1} = k_{v D 3} {\frac{\partial p_{D 3}^{'}}{\partial z_{D}}|}_{z_{D} = 1}$ $k_{v D 1} h_{D 1} {\frac{\partial p_{D 1}^{'}}{\partial z_{D}}|}_{z_{D} = 1} = k_{v D 3} {\frac{\partial p_{D 3}^{'}}{\partial z_{D}}|}_{z_{D} = 1}$

si12_e (H.12)

$k_{v D 2} h_{D 2} {\frac{\partial p_{D 2}^{'}}{\partial z_{D}}|}_{z_{D} = 0} = k_{v D 3} {\frac{\partial p_{D 3}^{'}}{\partial z_{D}}|}_{z_{D} = 0}$ $k_{v D 2} h_{D 2} {\frac{\partial p_{D 2}^{'}}{\partial z_{D}}|}_{z_{D} = 0} = k_{v D 3} {\frac{\partial p_{D 3}^{'}}{\partial z_{D}}|}_{z_{D} = 0}$

si13_e (H.13)

where $Δ_{D} = \frac{1}{r_{D}} (r_{D} \frac{\partial}{\partial r_{D}})$ $Δ_{D} = \frac{1}{r_{D}} (r_{D} \frac{\partial}{\partial r_{D}})$ is a dimensionless differential operator. Eqs. (9.23)–(9.26) for the middle layer becomes

$ω_{3} \frac{\partial p_{D 3}^{'}}{\partial t_{D}} - k_{v D 3} \frac{\partial^{2} p_{D 3}^{'}}{\partial z_{D}^{2}} = 0$ $ω_{3} \frac{\partial p_{D 3}^{'}}{\partial t_{D}} - k_{v D 3} \frac{\partial^{2} p_{D 3}^{'}}{\partial z_{D}^{2}} = 0$

si15_e (H.14)

$p_{D 3}^{'} \to 0 f o r t_{D} \to 0$ $p_{D 3}^{'} \to 0 f o r t_{D} \to 0$

(H.15)

${p_{D 1}^{'}|}_{z_{D} = 1} = {p_{D 3}^{'}|}_{z_{D} = 1}$ ${p_{D 1}^{'}|}_{z_{D} = 1} = {p_{D 3}^{'}|}_{z_{D} = 1}$

(H.16)

${p_{D 2}^{'}|}_{z_{D} = 0} = {p_{D 3}^{'}|}_{z_{D} = 0}$ ${p_{D 2}^{'}|}_{z_{D} = 0} = {p_{D 3}^{'}|}_{z_{D} = 0}$

(H.17)

After Laplace transformation, Eqs. (H.14), (H.15) become

$\frac{d^{2} {\bar{p}}_{D 3}^{'}}{d z_{D}^{2}} - υ^{2} {\bar{p}}_{D 3}^{'} = 0$ $\frac{d^{2} {\bar{p}}_{D 3}^{'}}{d z_{D}^{2}} - υ^{2} {\bar{p}}_{D 3}^{'} = 0$

si19_e (H.18)

where

$υ = \sqrt{ω_{3} σ / k_{v D 3}}$ $υ = \sqrt{ω_{3} σ / k_{v D 3}}$

(9.58)

The solution of Eq. (H.18) under boundary conditions Eqs. (H.16), (H.17) is

${\bar{p}}_{D 3}^{'} = [{{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} sinh υ z_{D} + {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} sinh υ (1 - z_{D})] / sinh υ$ ${\bar{p}}_{D 3}^{'} = [{{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} sinh υ z_{D} + {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} sinh υ (1 - z_{D})] / sinh υ$

(H.19)

Substituting Eq. (H.19) into the Laplace transforms of conditions Eqs. (H.12), (H.13) and noticing Eq. (H.1), we have

$[2 k_{v D 1} + υ k_{v D 3} \coth υ] {{\bar{p^{'}}}_{D 1}|}_{z_{D} = 1} - \frac{υ k_{v D 3}}{sinh υ} {{\bar{p^{'}}}_{D 2}|}_{z_{D} = 0} = 2 k_{v D 1} {\bar{p}}_{D 1}$ $[2 k_{v D 1} + υ k_{v D 3} \coth υ] {{\bar{p^{'}}}_{D 1}|}_{z_{D} = 1} - \frac{υ k_{v D 3}}{sinh υ} {{\bar{p^{'}}}_{D 2}|}_{z_{D} = 0} = 2 k_{v D 1} {\bar{p}}_{D 1}$

(H.20)

$- \frac{υ k_{v D 3}}{sinh υ} {{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} + (2 k_{v D 2} + υ k_{v D 3} \coth υ) {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} = 2 k_{v D 2} {\bar{p}}_{D 1}$ $- \frac{υ k_{v D 3}}{sinh υ} {{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} + (2 k_{v D 2} + υ k_{v D 3} \coth υ) {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} = 2 k_{v D 2} {\bar{p}}_{D 1}$

(H.21)

Solve Eqs. (H.20), (H.21) to obtain the following:

${{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} = \frac{1}{G_{0}} [(1 + \frac{k_{v D 3} υ \coth υ}{2 k_{v D 2}}) {\bar{p}}_{D 1} + \frac{k_{v D 3} υ csch υ}{2 k_{v D 1}} {\bar{p}}_{D 2}]$ ${{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1} = \frac{1}{G_{0}} [(1 + \frac{k_{v D 3} υ \coth υ}{2 k_{v D 2}}) {\bar{p}}_{D 1} + \frac{k_{v D 3} υ csch υ}{2 k_{v D 1}} {\bar{p}}_{D 2}]$

si24_e (H.22)

${{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} = \frac{1}{G_{0}} [\frac{k_{v D 3} υ csch υ}{2 k_{v D 2}} {\bar{p}}_{D 1} + (1 + \frac{k_{v D 3} υ \coth υ}{2 k_{v D 1}}) {\bar{p}}_{D 2}]$ ${{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0} = \frac{1}{G_{0}} [\frac{k_{v D 3} υ csch υ}{2 k_{v D 2}} {\bar{p}}_{D 1} + (1 + \frac{k_{v D 3} υ \coth υ}{2 k_{v D 1}}) {\bar{p}}_{D 2}]$

si25_e (H.23)

where

$G_{0} = 1 + \frac{k_{v D 3} υ \coth υ}{2} (\frac{1}{k_{v D 1}} + \frac{1}{k_{v D 2}}) + \frac{ω_{3} σ k_{v D 3}}{4 k_{v D 1} k_{v D 2}}$ $G_{0} = 1 + \frac{k_{v D 3} υ \coth υ}{2} (\frac{1}{k_{v D 1}} + \frac{1}{k_{v D 2}}) + \frac{ω_{3} σ k_{v D 3}}{4 k_{v D 1} k_{v D 2}}$

si26_e (9.57)

The Laplace transform of Eqs. (H.2), (H.3) is

$ω_{1} σ {\bar{p}}_{D 1} - w_{1} Δ_{D} {\bar{p}}_{D 1} + 2 k_{v D 1} ({\bar{p}}_{D 1} - {{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1}) = 0$ $ω_{1} σ {\bar{p}}_{D 1} - w_{1} Δ_{D} {\bar{p}}_{D 1} + 2 k_{v D 1} ({\bar{p}}_{D 1} - {{\bar{p}}_{D 1}^{'}|}_{z_{D} = 1}) = 0$

(H.24)

$ω_{2} σ {\bar{p}}_{D 2} - w_{2} Δ_{D} {\bar{p}}_{D 2} + 2 k_{v D 2} ({\bar{p}}_{D 2} - {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0}) = 0$ $ω_{2} σ {\bar{p}}_{D 2} - w_{2} Δ_{D} {\bar{p}}_{D 2} + 2 k_{v D 2} ({\bar{p}}_{D 2} - {{\bar{p}}_{D 2}^{'}|}_{z_{D} = 0}) = 0$

(H.25)

Substituting Eqs. (H.22), (H.23) into Eqs. (H.24), (H.25) results in the following:

$- w_{1} Δ_{D} {\bar{p}}_{D 1} + G_{1} {\bar{p}}_{D 1} - G {\bar{p}}_{D 2} = 0$ $- w_{1} Δ_{D} {\bar{p}}_{D 1} + G_{1} {\bar{p}}_{D 1} - G {\bar{p}}_{D 2} = 0$

(H.26)

$- w_{2} Δ_{D} {\bar{p}}_{D 2} - G {\bar{p}}_{D 1} + G_{2} {\bar{p}}_{D 2} = 0$ $- w_{2} Δ_{D} {\bar{p}}_{D 2} - G {\bar{p}}_{D 1} + G_{2} {\bar{p}}_{D 2} = 0$

(H.27)

where

$G_{1} = ω_{1} σ + (υ \coth υ + \frac{ω_{3} σ}{2 k_{v D 2}}) \frac{k_{v D 3}}{G_{0}}$ $G_{1} = ω_{1} σ + (υ \coth υ + \frac{ω_{3} σ}{2 k_{v D 2}}) \frac{k_{v D 3}}{G_{0}}$

si31_e (9.54)

$G_{2} = ω_{2} σ + (υ \coth υ + \frac{ω_{3} σ}{2 k_{v D 1}}) \frac{k_{v D 3}}{G_{0}}$ $G_{2} = ω_{2} σ + (υ \coth υ + \frac{ω_{3} σ}{2 k_{v D 1}}) \frac{k_{v D 3}}{G_{0}}$

si32_e (9.55)

$G = \frac{υ k_{v D 3}}{G_{0} sinh υ}$ $G = \frac{υ k_{v D 3}}{G_{0} sinh υ}$

si33_e (9.56)

Considering Eq. (H.5), the solution of Eqs. (H.26), (H.27) has the form

${\bar{p}}_{D 1} = A K_{0} (λ r_{D})$ ${\bar{p}}_{D 1} = A K_{0} (λ r_{D})$

(H.28)

${\bar{p}}_{D 2} = D K_{0} (λ r_{D})$ ${\bar{p}}_{D 2} = D K_{0} (λ r_{D})$

(H.29)

where K₀ is the modified Bessel function of zero order, and A and D are arbitrary functions of Laplace variable σ.

Substituting Eqs. (H.28), (H.29) into Eqs. (H.26), (H.27), we have

$(G_{1} - λ^{2} w_{1}) A - G D = 0$ $(G_{1} - λ^{2} w_{1}) A - G D = 0$

(H.30)

$- G A + (G_{2} - λ^{2} w_{2}) D = 0$ $- G A + (G_{2} - λ^{2} w_{2}) D = 0$

(H.31)

The eigenequation is

$λ^{4} - (\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}) λ^{2} + \frac{G_{1} G_{2} - G^{2}}{w_{1} w_{2}} = 0$ $λ^{4} - (\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}) λ^{2} + \frac{G_{1} G_{2} - G^{2}}{w_{1} w_{2}} = 0$

si38_e (H.32)

The two eigenvalues are

$λ_{1, 2} = {[\frac{1}{2} (\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}) \pm \sqrt{\frac{1}{4} {(\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}})}^{2} + \frac{G^{2} - G_{1} G_{2}}{w_{1} w_{2}}}]}^{\frac{1}{2}}$ $λ_{1, 2} = {[\frac{1}{2} (\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}) \pm \sqrt{\frac{1}{4} {(\frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}})}^{2} + \frac{G^{2} - G_{1} G_{2}}{w_{1} w_{2}}}]}^{\frac{1}{2}}$

si39_e (9.51)

From Eqs. (H.31), (H.30), we find

$D = β_{2} A f o r eigenvalue λ_{1}$ $D = β_{2} A f o r eigenvalue λ_{1}$

and

$A = β_{1} D f o r eigenvalue λ_{2}$ $A = β_{1} D f o r eigenvalue λ_{2}$

where

$β_{1} = G / (G_{1} - λ_{2}^{2} w_{1})$ $β_{1} = G / (G_{1} - λ_{2}^{2} w_{1})$

(9.52)

$β_{2} = G / (G_{2} - λ_{1}^{2} w_{2})$ $β_{2} = G / (G_{2} - λ_{1}^{2} w_{2})$

(9.53)

Using

$λ_{1}^{2} + λ_{2}^{2} = \frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}$ $λ_{1}^{2} + λ_{2}^{2} = \frac{G_{1}}{w_{1}} + \frac{G_{2}}{w_{2}}$

si44_e

we have

$β_{2} = - β_{1} w_{1} / w_{2}$ $β_{2} = - β_{1} w_{1} / w_{2}$

(H.33)

So the solution is

${\bar{p}}_{D 1} = A \frac{K_{0} (λ_{1} r_{D})}{λ_{1} K_{1} (λ_{1})} + D β_{1} \frac{K_{0} (λ_{2} r_{D})}{λ_{2} K_{1} (λ_{2})}$ ${\bar{p}}_{D 1} = A \frac{K_{0} (λ_{1} r_{D})}{λ_{1} K_{1} (λ_{1})} + D β_{1} \frac{K_{0} (λ_{2} r_{D})}{λ_{2} K_{1} (λ_{2})}$

si46_e (H.34)

and

${\bar{p}}_{D 2} = A β_{2} \frac{K_{0} (λ_{1} r_{D})}{λ_{1} K_{1} (λ_{1})} + D \frac{K_{0} (λ_{2} r_{D})}{λ_{2} K_{1} (λ_{2})}$ ${\bar{p}}_{D 2} = A β_{2} \frac{K_{0} (λ_{1} r_{D})}{λ_{1} K_{1} (λ_{1})} + D \frac{K_{0} (λ_{2} r_{D})}{λ_{2} K_{1} (λ_{2})}$

si47_e (H.35)

where K₁ is the modified Bessel function of the first order. The Laplace transforms of Eqs. (H.6), (H.7) are

$w_{j} {(r_{D} \frac{d {\bar{p}}_{D j}}{d r_{D}})}_{r_{D} = 1} = - {\bar{q}}_{D j}, j = 1, 2$ $w_{j} {(r_{D} \frac{d {\bar{p}}_{D j}}{d r_{D}})}_{r_{D} = 1} = - {\bar{q}}_{D j}, j = 1, 2$

si48_e (H.36)

${\bar{p}}_{w D j} = {({\bar{p}}_{D j} - s_{j} r_{D} \frac{d {\bar{p}}_{D j}}{d r_{D}})}_{r_{D} = 1}, j = 1, 2$ ${\bar{p}}_{w D j} = {({\bar{p}}_{D j} - s_{j} r_{D} \frac{d {\bar{p}}_{D j}}{d r_{D}})}_{r_{D} = 1}, j = 1, 2$

si49_e (H.37)

Substituting Eqs. (H.34), (H.35) into Eqs. (H.36), (H.37) results in the following:

$w_{1} [A (σ) + D (σ) β_{1}] = {\bar{q}}_{D 1}$ $w_{1} [A (σ) + D (σ) β_{1}] = {\bar{q}}_{D 1}$

(9.44)

$w_{2} [A (σ) β_{2} + D (σ)] = {\bar{q}}_{D 2}$ $w_{2} [A (σ) β_{2} + D (σ)] = {\bar{q}}_{D 2}$

(9.45)

$A (σ) [K (λ_{1}) + s_{1}] + D (σ) β_{1} [K (λ_{2}) + s_{1}] = {\bar{p}}_{w D 1}$ $A (σ) [K (λ_{1}) + s_{1}] + D (σ) β_{1} [K (λ_{2}) + s_{1}] = {\bar{p}}_{w D 1}$

(9.46)

$A (σ) β_{2} [K (λ_{1}) + s_{2}] + D (σ) [K (λ_{2}) + s_{2}] = {\bar{p}}_{w D 2}$ $A (σ) β_{2} [K (λ_{1}) + s_{2}] + D (σ) [K (λ_{2}) + s_{2}] = {\bar{p}}_{w D 2}$

(9.47)

where function K(x) is defined by

$K (x) = \frac{K_{0} (x)}{x K_{1} (x)} \approx ln \frac{2}{γ x}$ $K (x) = \frac{K_{0} (x)}{x K_{1} (x)} \approx ln \frac{2}{γ x}$

si54_e (9.59)

The Laplace transform of Eqs. (H.8)–(H.11) gives

${\bar{q}}_{D 1} = \frac{1}{σ} - {\bar{q}}_{L D} - σ C_{D 1} {\bar{p}}_{w D 1}$ ${\bar{q}}_{D 1} = \frac{1}{σ} - {\bar{q}}_{L D} - σ C_{D 1} {\bar{p}}_{w D 1}$

(9.48)

${\bar{q}}_{D 2} = {\bar{q}}_{L D} - σ C_{D 2} {\bar{p}}_{w D 2}$ ${\bar{q}}_{D 2} = {\bar{q}}_{L D} - σ C_{D 2} {\bar{p}}_{w D 2}$

(9.49)

${\bar{q}}_{L D} = \bar{f} (p_{w D 1}, p_{w D 2})$ ${\bar{q}}_{L D} = \bar{f} (p_{w D 1}, p_{w D 2})$

(9.50a)

${\bar{q}}_{L D} = α_{D} ({\bar{p}}_{w D 1} - {\bar{p}}_{w D 2})$ ${\bar{q}}_{L D} = α_{D} ({\bar{p}}_{w D 1} - {\bar{p}}_{w D 2})$

(9.50b)

Eqs. (9.44)–(9.50) provides the solution of the problem.

H.1 When Leakage Function f Is Known

Solving A(σ) and D(σ) from Eqs. (9.44), (9.45) and using Eqs. (9.48)–(9.50a), (H.33), we have

$A (σ) = [\frac{1}{σ} - \bar{f} - σ C_{D 1} {\bar{p}}_{w D 1} + β_{2} (\bar{f} - σ C_{D 2} {\bar{p}}_{w D 2})] / (1 - β_{1} β_{2}) w_{1}$ $A (σ) = [\frac{1}{σ} - \bar{f} - σ C_{D 1} {\bar{p}}_{w D 1} + β_{2} (\bar{f} - σ C_{D 2} {\bar{p}}_{w D 2})] / (1 - β_{1} β_{2}) w_{1}$

si59_e (H.38)

$D (σ) = [\bar{f} - σ C_{D 2} {\bar{p}}_{w D 2} + β_{1} (\frac{1}{σ} - \bar{f} - σ C_{D 1} {\bar{p}}_{w D 1})] / (1 - β_{1} β_{2}) w_{2}$ $D (σ) = [\bar{f} - σ C_{D 2} {\bar{p}}_{w D 2} + β_{1} (\frac{1}{σ} - \bar{f} - σ C_{D 1} {\bar{p}}_{w D 1})] / (1 - β_{1} β_{2}) w_{2}$

si60_e (H.39)

Substituting Eqs. (H.38), (H.39) into Eqs. (9.46), (9.47) and solving ${\bar{p}}_{w D 1}$ ${\bar{p}}_{w D 1}$ for ${\bar{p}}_{w D 2}$ ${\bar{p}}_{w D 2}$ and, we have:

${\bar{p}}_{w D 1} = \frac{M_{4} M_{5} - M_{2} M_{6}}{M_{1} M_{4} - M_{2} M_{3}}$ ${\bar{p}}_{w D 1} = \frac{M_{4} M_{5} - M_{2} M_{6}}{M_{1} M_{4} - M_{2} M_{3}}$

si63_e (9.60)

and

${\bar{p}}_{w D 2} = \frac{M_{1} M_{6} - M_{3} M_{5}}{M_{1} M_{4} - M_{2} M_{3}}$ ${\bar{p}}_{w D 2} = \frac{M_{1} M_{6} - M_{3} M_{5}}{M_{1} M_{4} - M_{2} M_{3}}$

si64_e (9.61)

where

$M_{1} = w_{1} w_{2} (1 - β_{1} β_{2}) + σ C_{D 1} w_{2} \{K (λ_{1}) + s_{1} - β_{1} β_{2} [K (λ_{2}) + s_{1}]\}$ $M_{1} = w_{1} w_{2} (1 - β_{1} β_{2}) + σ C_{D 1} w_{2} \{K (λ_{1}) + s_{1} - β_{1} β_{2} [K (λ_{2}) + s_{1}]\}$

(9.62)

$M_{2} = w_{1} β_{1} σ C_{D 2} [K (λ_{2}) - K (λ_{1})]$ $M_{2} = w_{1} β_{1} σ C_{D 2} [K (λ_{2}) - K (λ_{1})]$

(9.63)

$M_{3} = w_{1} β_{1} σ C_{D 1} [K (λ_{2}) - K (λ_{1})]$ $M_{3} = w_{1} β_{1} σ C_{D 1} [K (λ_{2}) - K (λ_{1})]$

(9.64)

$M_{4} = w_{1} w_{2} (1 - β_{1} β_{2}) + σ C_{D 2} w_{1} \{K (λ_{2}) + s_{2} - β_{1} β_{2} [K (λ_{1}) + s_{2}]\}$ $M_{4} = w_{1} w_{2} (1 - β_{1} β_{2}) + σ C_{D 2} w_{1} \{K (λ_{2}) + s_{2} - β_{1} β_{2} [K (λ_{1}) + s_{2}]\}$

(9.65)

$M_{5} = w_{2} [K (λ_{1}) + s_{1}] [\frac{1}{σ} + (β_{2} - 1) \bar{f}] + w_{1} β_{1} [K (λ_{2}) + s_{1}] [\bar{f} + β_{1} (\frac{1}{σ} - \bar{f})]$ $M_{5} = w_{2} [K (λ_{1}) + s_{1}] [\frac{1}{σ} + (β_{2} - 1) \bar{f}] + w_{1} β_{1} [K (λ_{2}) + s_{1}] [\bar{f} + β_{1} (\frac{1}{σ} - \bar{f})]$

si69_e (9.66)

$M_{6} = w_{2} β_{2} [K (λ_{1}) + s_{2}] [\frac{1}{σ} + (β_{2} - 1) \bar{f}] + w_{1} [K (λ_{2}) + s_{2}] [\bar{f} + β_{1} (\frac{1}{σ} - \bar{f})]$ $M_{6} = w_{2} β_{2} [K (λ_{1}) + s_{2}] [\frac{1}{σ} + (β_{2} - 1) \bar{f}] + w_{1} [K (λ_{2}) + s_{2}] [\bar{f} + β_{1} (\frac{1}{σ} - \bar{f})]$

si70_e (9.67)

H.2 When Leakage Function $f = α_{D} (p_{w D 1}, p_{w D 2})$ $f = α_{D} (p_{w D 1}, p_{w D 2})$

Substituting Eqs. (9.50b), (H.38), (H.39) into Eqs. (9.46), (9.47) and solving for ${\bar{p}}_{w D 1}$ ${\bar{p}}_{w D 1}$ and ${\bar{p}}_{w D 2}$ ${\bar{p}}_{w D 2}$ , we have

${\bar{p}}_{w D 1} = \frac{L_{5} L_{4} - L_{2} L_{6}}{σ (L_{1} L_{4} - L_{2} L_{3})}$ ${\bar{p}}_{w D 1} = \frac{L_{5} L_{4} - L_{2} L_{6}}{σ (L_{1} L_{4} - L_{2} L_{3})}$

si74_e (9.68)

and

${\bar{p}}_{w D 2} = \frac{L_{1} L_{6} - L_{3} L_{5}}{σ (L_{1} L_{4} - L_{2} L_{3})}$ ${\bar{p}}_{w D 2} = \frac{L_{1} L_{6} - L_{3} L_{5}}{σ (L_{1} L_{4} - L_{2} L_{3})}$

si75_e (9.69)

where

$\begin{array}{l} L_{1} & = w_{1} w_{2} (1 - β_{1} β_{2}) + w_{2} [K (λ_{1}) + s_{1}] [α_{D} (1 - β_{2}) + σ C_{D 1}] \\ - β_{1} w_{1} [K (λ_{2}) + s_{1}] [α_{D} (1 - β_{1}) - σ β_{1} C_{D 1}] \end{array}$ $\begin{array}{l} L_{1} & = w_{1} w_{2} (1 - β_{1} β_{2}) + w_{2} [K (λ_{1}) + s_{1}] [α_{D} (1 - β_{2}) + σ C_{D 1}] \\ - β_{1} w_{1} [K (λ_{2}) + s_{1}] [α_{D} (1 - β_{1}) - σ β_{1} C_{D 1}] \end{array}$

si76_e (9.70)

$\begin{array}{c} L_{2} = - w_{2} [K (λ_{1}) + s_{1}] [α_{D} (1 - β_{2}) - σ β_{2} C_{D 2}] \\ + β_{1} w_{1} [K (λ_{2}) + s_{1}] [α_{D} (1 - β_{1}) + σ C_{D 2}] \end{array}$ $\begin{array}{c} L_{2} = - w_{2} [K (λ_{1}) + s_{1}] [α_{D} (1 - β_{2}) - σ β_{2} C_{D 2}] \\ + β_{1} w_{1} [K (λ_{2}) + s_{1}] [α_{D} (1 - β_{1}) + σ C_{D 2}] \end{array}$

si77_e (9.71)

$\begin{array}{c} L_{3} = w_{2} β_{2} [K (λ_{1}) + s_{2}] [α_{D} (1 - β_{2}) + σ C_{D 1}] \\ - w_{1} [K (λ_{2}) + s_{2}] [α_{D} (1 - β_{1}) - σ β_{1} C_{D 1}] \end{array}$ $\begin{array}{c} L_{3} = w_{2} β_{2} [K (λ_{1}) + s_{2}] [α_{D} (1 - β_{2}) + σ C_{D 1}] \\ - w_{1} [K (λ_{2}) + s_{2}] [α_{D} (1 - β_{1}) - σ β_{1} C_{D 1}] \end{array}$

si78_e (9.72)

$\begin{array}{l} L_{4} & = w_{1} w_{2} (1 - β_{1} β_{2}) - w_{2} β_{2} [K (λ_{1}) + s_{2}] [α_{D} (1 - β_{2}) - σ β_{2} C_{D 2}] \\ + w_{1} [K (λ_{2}) + s_{2}] [α_{D} (1 - β_{1}) + σ C_{D 2}] \end{array}$ $\begin{array}{l} L_{4} & = w_{1} w_{2} (1 - β_{1} β_{2}) - w_{2} β_{2} [K (λ_{1}) + s_{2}] [α_{D} (1 - β_{2}) - σ β_{2} C_{D 2}] \\ + w_{1} [K (λ_{2}) + s_{2}] [α_{D} (1 - β_{1}) + σ C_{D 2}] \end{array}$

si79_e (9.73)

$L_{5} = w_{2} \{K (λ_{1}) + s_{1} - β_{1} β_{2} [K (λ_{2}) + s_{1}]\}$ $L_{5} = w_{2} \{K (λ_{1}) + s_{1} - β_{1} β_{2} [K (λ_{2}) + s_{1}]\}$

(9.74)

$L_{6} = w_{1} β_{1} [K (λ_{2}) - K (λ_{1})]$ $L_{6} = w_{1} β_{1} [K (λ_{2}) - K (λ_{1})]$

(9.75)

H.3 When Leakage Function f Is Unknown

Adding Eqs. (9.44), (9.45) and using Eqs. (9.48), (9.49), (H.33), we have

$w_{1} (1 - β_{1}) A (σ) + w_{2} (1 - β_{2}) D (σ) = \frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})$ $w_{1} (1 - β_{1}) A (σ) + w_{2} (1 - β_{2}) D (σ) = \frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})$

(9.81)

Solving Eqs. (9.47), (9.81), we have

$A = \{{\bar{p}}_{w D 2} w_{2} (1 - β_{2}) - [\frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})] [K (λ_{2}) + s_{2}]\} / L_{7}$ $A = \{{\bar{p}}_{w D 2} w_{2} (1 - β_{2}) - [\frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})] [K (λ_{2}) + s_{2}]\} / L_{7}$

si83_e (H.40)

$D = \{β_{2} [\frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})] [K (λ_{1}) + s_{2}] - {\bar{p}}_{w D 2} (1 - β_{1}) w_{1}\} / L_{7}$ $D = \{β_{2} [\frac{1}{σ} - σ (C_{D 1} {\bar{p}}_{w D 1} + C_{D 2} {\bar{p}}_{w D 2})] [K (λ_{1}) + s_{2}] - {\bar{p}}_{w D 2} (1 - β_{1}) w_{1}\} / L_{7}$

si84_e (H.41)

where

$L_{7} = (1 - β_{2}) w_{2} β_{2} [K (λ_{1}) + s_{2}] - w_{1} (1 - β_{1}) [K (λ_{2}) + s_{2}]$ $L_{7} = (1 - β_{2}) w_{2} β_{2} [K (λ_{1}) + s_{2}] - w_{1} (1 - β_{1}) [K (λ_{2}) + s_{2}]$

(H.42)

Substituting Eqs. (H.40), (H.41) into Eq. (9.46) and solving for ${\bar{p}}_{w D 1}$ ${\bar{p}}_{w D 1}$ , we have

$\begin{array}{l} {\bar{p}}_{w D 1} & = [K (λ_{1}) + s_{1}] \{(1 - β_{2}) w_{2} {\bar{p}}_{w D 2} - (\frac{1}{σ} - σ C_{D 2} {\bar{p}}_{w D 2}) [K (λ_{2}) + s_{2}]\} / L_{8} \\ + β_{1} [K (λ_{2}) + s_{1}] \{β_{2} [K (λ_{1}) + s_{2}] (\frac{1}{σ} - σ C_{D 2} {\bar{p}}_{w D 2}) - w_{1} (1 - β_{1}) {\bar{p}}_{w D 2}\} / L_{8} \end{array}$ $\begin{array}{l} {\bar{p}}_{w D 1} & = [K (λ_{1}) + s_{1}] \{(1 - β_{2}) w_{2} {\bar{p}}_{w D 2} - (\frac{1}{σ} - σ C_{D 2} {\bar{p}}_{w D 2}) [K (λ_{2}) + s_{2}]\} / L_{8} \\ + β_{1} [K (λ_{2}) + s_{1}] \{β_{2} [K (λ_{1}) + s_{2}] (\frac{1}{σ} - σ C_{D 2} {\bar{p}}_{w D 2}) - w_{1} (1 - β_{1}) {\bar{p}}_{w D 2}\} / L_{8} \end{array}$

si87_e (9.82)

where

$\begin{array}{l} L_{8} & = (1 - β_{2}) w_{2} β_{2} [K (λ_{1}) + s_{2}] - w_{1} (1 - β_{1}) [K (λ_{2}) + s_{2}] \\ - σ C_{D 1} \{[K (λ_{1}) + s_{1}] [K (λ_{2}) + s_{2}] - β_{1} β_{2} [K (λ_{2}) + s_{1}] [K (λ_{1}) + s_{2}]\} \end{array}$ $\begin{array}{l} L_{8} & = (1 - β_{2}) w_{2} β_{2} [K (λ_{1}) + s_{2}] - w_{1} (1 - β_{1}) [K (λ_{2}) + s_{2}] \\ - σ C_{D 1} \{[K (λ_{1}) + s_{1}] [K (λ_{2}) + s_{2}] - β_{1} β_{2} [K (λ_{2}) + s_{1}] [K (λ_{1}) + s_{2}]\} \end{array}$

si88_e (9.83)

From Eqs. (9.45), (9.49), (9.50a), we have

$\bar{f} = w_{2} [A (σ) β_{2} + D (σ)] + σ C_{D 2} {\bar{p}}_{w D 2}$ $\bar{f} = w_{2} [A (σ) β_{2} + D (σ)] + σ C_{D 2} {\bar{p}}_{w D 2}$

(9.84)

where A(σ) and D(σ) are determined by Eqs. (H.40), (H.41).

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Table of Contents for Appendix H: Solution of the Problem (Chapter 9)

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H.1 When Leakage Function f Is Known

H.3 When Leakage Function f Is Unknown

Table of Contents for
Appendix H: Solution of the Problem (Chapter 9)