where Equalities (6.3) and (6.6) show that − 2 πa$_{ii}$ (0) = q$_{i}$ (0 $_{, ξ}$′ $_{i}$) , (2 ≤ i ≤ n ) . Then by Lemma 5, we have q$_{i}$ (0 $_{, ξ}$′ $_{i}$) → 0 as $_{ξ}$′ i → 0 and thus a$_{ii}$ (0) = 0 , (2 ≤ i ≤ n ) . In order to complete the proof, let $_{ξ}$′ $_{ij}$(1) = (0 , · · · , 0 ︸ ︷︷ ︸ i − 2 , 1 , 0 , · · · , 0 ︸ ︷︷ ︸ , j − i − 1 1 , 0 , · · · , 0 ︸ ︷︷ ︸ n − j ) ∈ R n − $^{1}$, i = j, and $_{ξ}$′ ij = $_{εξ}$′ $_{ij}$(1) ∈ G ′ . In this case, by a$_{ii}$ (0) = 0, a$_{ij}$ (0) = a$_{ji}$ (0) and (6.1), we have for $_{ξ}$′ = $_{ξ}$′ $_{ij}$, where and the indices i, j ∈ { 2 , ..., n } are fixed. Then, in the same way as above, equation (6.7) and Lemma 6 yield where $^{U}$± $_{q}$(0 $_{, ξ}$′ $_{ij}$) = ∂$_{x}$ 1 q ( 0 , ± 0 $_{, ξ}$′ ij ) . From (6.7), by virtue of Lemmata 2, 3 we have Hence, q$_{i}$ (0 $_{, ξ}$′ $_{i}$) = −$^{∞}$ ∫ −∞ ∂$_{ξ}$ i ∂$_{x}$ i p (0 $_{, η, ξ}$′ $_{i}$θ$_{1}$ ) dη. ∂$_{η}$ ( ∂$_{x}$ 1 q ( 0 $_{, η, ξ}$′ ij )) = − 4 πδ ( η ) a$_{ij}$ (0) ε 2 + F$_{2}$$_{ij}$ ( 0 $_{, η, ξ}$′ ij ) (6.7) F$_{2}$$_{ij}$ (0 $_{, η, ξ}$′ $_{ij}$) = ε ( ∂$_{x}$ i p (0 $_{, η, ξ}$′ $_{ij}$) + ∂$_{x}$ j p (0 $_{, η, ξ}$′ $_{ij}$)) U$_{+}$$_{q}$ (0 $_{, ξ}$′ $_{ij}$) − $^{U}$− $_{q}$(0 $_{, ξ}$′ $_{ij}$) = − 4 πa$_{ij}$ (0) ε $^{2}$, (6.8) U$_{+}$$_{q}$ (0 $_{, ξ}$′ $_{ij}$) = −$^{∞}$ ∫ 0 F$_{2}$$_{ij}$ ( 0 $_{, η, ξ}$′ ij ) dη, $^{U}$− $_{q}$(0 $_{, ξ}$′ $_{ij}$) = 0 ∫ −∞$^{F$_{2}$$_{ij}$}$ ( 0 $_{, η, ξ}$′ ij ) dη . U$_{+}$$_{q}$ (0 $_{, ξ}$′ $_{ij}$) − $^{U}$− $_{q}$(0 $_{, ξ}$′ $_{ij}$) = − ε ∞ ∫ −∞$^{(}$$^{∂$_{x}$}$$^{i}$$^{p}$$^{(0}$ $_{, η, ξ}$′ $_{ij}$) + ∂$_{x}$ j p (0 $_{, η, ξ}$′ $_{ij}$)) dη. (6.9) 16