≤ T ∫ − T ∫ Ω$_{0}$ | ∂$_{t}$χ∂$_{t}$u | $^{2}$e 2 $^{sϕ}$dxdydt + C T ∫ − T ∫ Ω$_{0}$ ( | χ∂$_{t}$u | 2 + ∣ χ∂ 2 $_{t}$u ∣ 2 + s | χ∂$_{t}$u | 2 ) e 2 $^{sϕ}$dxdydt ≤ CM $^{2}$e2 sκ$_{1}$ + C ∫ Ω$_{1}$ ( | w$_{1}$ | 2 + | w$_{2}$ | 2 + s | w$_{1}$ | 2 e 2 $^{sϕ}$dxdydt. ∫ Ω$_{0}$ | χ ( y, 0) | 2 | i∂$_{t}$u ( x, y, 0) | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy ≤ CM $^{2}$e2 sκ$_{1}$ + C ∫ Ω$_{1}$ ( s | w$_{1}$ | 2 + | w$_{2}$ | 2 ) e 2 $^{sϕ}$dxdydt. ∫ Ω$_{0}$ | χ$_{0}$ ( | y − y$_{0}$ | ) | 2 | i∂$_{t}$u ( x, y, 0) | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy ≤ C s ∫ Ω$_{1}$ χ $^{2}$( y, t ) | f | $^{2}$e 2 sϕ ( x,y,t $^{)}$dxdydt + CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d 2 = C s Ω$_{1}$ χ 2 $_{0}$( t ) χ 2 $_{0}$( | y − y$_{0}$ | ) | f | $^{2}$e 2 sϕ ( x,y,t $^{)}$dxdydt + CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d 2 ≤ C s ∫ Ω$_{0}$ χ 2 $_{0}$( | y − y$_{0}$ | ) | f | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy + CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d $^{2}$, (4.21) f ( x, y ) = i∂$_{t}$u ( x, y, 0) R ( x, y, 0) , x ∈ D, | y | ≤ 2 L. (4.22) ∫ Ω$_{0}$ χ 2 $_{0}$( | y − y$_{0}$ | ) | f | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy ≤ C s ∫ Ω$_{0}$ χ 2 $_{0}$( | y − y$_{0}$ | ) | f | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy + CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d 2 29 Hence Then, applying (4.20), we obtain where we used | χ$_{0}$ ( t ) | ≤ 1 and e 2 sϕ ( x,y,t ) ≤ e 2 sϕ ( x,y, 0) for x ∈ D and y ∈ G$_{1}$. On the other hand, by substituting t = 0 in (2.1) and applying u ( x, y, 0) = 0 and R ( x, y, 0) = 0 , for x ∈ D and | y | ≤ 2 L, we get By applying (4.22) in (4.21), we have