Finally, we have where As a result, (4.17) yields Now, by using the fact that χ ( y, − T ) = 0 for y ∈ G$_{1}$ by (4.12), we can write S$_{3}$ = C ∫ ∂D$_{+}$ × G$_{1}$ × ( − T,T ) 2 ∑ k =1 s | ∂$_{ν}$w$_{k}$ | 2 e 2 $^{sϕ}$dS$_{x}$ dydt ≤ Ce cs ∫ ∂D$_{+}$ × G$_{2}$ × ( − T,T ) 2 k =1 ∣ ∂$_{ν}$∂ k t u ∣ 2 dS$_{x}$dydt := Ce $^{cs}$d $^{2}$, (4.19) d 2 = ∫ ∂D$_{+}$ × G$_{2}$ × ( − T,T ) 2 ∑ k =1 ∣ ∂$_{ν}$∂ k t u ∣ 2 dS$_{x}$dydt. ∫ Ω$_{1}$ 2 ∑ k =1 ( s |∇ $_{x}$w$_{k}$ | 2 + s |∇ $_{y}$w$_{k}$ | 2 + s 3 | w$_{k}$ | 2 ) e 2 $^{sϕ}$dxdydt ≤ C ∫ Ω$_{1}$ χ $^{2}$f $^{2}$e2 $^{sϕ}$dxdydt + CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d $^{2}$. (4.20) ∫ Ω$_{0}$ | χ ( y, 0) | 2 | i∂$_{t}$u ( x, y, 0) | 2 e 2 sϕ ( x,y, $^{0)}$dxdy = 0 − T ∂$_{t}$   ∫ Ω$_{0}$ χ 2 | ∂$_{t}$u ( x, y, t ) | 2 e 2 sϕ ( x,y,t $^{)}$dxdy   dt = 0 ∫ − T ∫ Ω$_{0}$ ( 2 χ∂$_{t}$χ | ∂$_{t}$u | 2 + χ $^{2}$∂$_{t}$ ( | ∂$_{t}$u | 2 ) + χ 2 | ∂$_{t}$u | $^{2}$2 s∂$_{t}$ϕ ) e 2 sϕ ( x,y,t $^{)}$dxdydt = 0 ∫ − T ∫ Ω$_{0}$ ( 2 χ∂$_{t}$χ | ∂$_{t}$u | 2 + 2 χ $^{2}$Re ∂ 2 t u∂$_{t}$ ¯ u + χ 2 | ∂$_{t}$u | $^{2}$2 s∂$_{t}$ϕ ) e 2 sϕ ( x,y,t $^{)}$dxdydt ≤ T ∫ − T ∫ Ω$_{0}$ | χ∂$_{t}$u | 2 e 2 $^{sϕ}$dxdydt + T ∫ − T ∫ D × G$_{1}$ | ∂$_{t}$χ∂$_{t}$u | 2 e 2 $^{sϕ}$dxdydt + T ∫ − T ∫ D × G$_{1}$ | χ∂$_{t}$u | 2 e 2 $^{sϕ}$dxdydt + T ∫ − T ∫ Ω$_{0}$ ∣ χ∂ 2 t u ∣ 2 e 2 $^{sϕ}$dxdydt + s T ∫ − T ∫ Ω$_{0}$ | χ∂$_{t}$u | 2 e 2 $^{sϕ}$dxdydt 28