Then we choose s$_{2}$ = s$_{2}$ ( γ ) > 0 such that ∀ s > s$_{2}$ all the terms of X$_{2}$ and X$_{6}$ can be absorbed into ‖ P + s z ‖ 2 L $^{2}$(Ω) , ‖ P − s z ‖ 2 L 2 $_{(Ω)}$, C$_{5}$ ‖∇ $_{x}$z ‖ 2 L 2 $_{(Ω)}$, C$_{5}$ ‖∇ $_{y}$z ‖ 2 L $^{2}$(Ω) and C$_{5}$s 3 ‖ z ‖ 2 L 2 $_{(Ω)}$. Therefore ∫ Ω | P$_{s}$z + is∂$_{t}$ϕz | 2 dxdydt ≤ 2 ∫ Ω | P$_{s}$z | $^{2}$dxdydt + C$_{6}$s 2 ∫ Ω | z | $^{2}$dxdydt, taking s > 0 sufficiently large, we have Since z = 0 on Γ$_{x}$ ∪ Γ$_{y}$ , ∇ $_{y}$z = 0 and ∇ $_{x}$z = ( ∂$_{ν}$z ) · ν on Γ$_{x}$ , all the integrations on Γ$_{y}$ vanish in (3.28), we have We obtain C$_{7}$ ∫ Ω | P$_{s}$z | $^{2}$dxdydt ≥ ∫ Ω ∣ P + s z ∣ $^{2}$dxdydt + ∫ Ω ∣ P − s z ∣ $^{2}$dxdydt + s ∫ Ω |∇ $_{x}$z | $^{2}$dxdydt + s ∫ Ω |∇ $_{y}$z | $^{2}$dxdydt + s 3 Ω | z | $^{2}$dxdydt + B$_{0}$. B$_{0}$ = − 4 Re ∫ Γ$_{x}$ sγϕ ( ∂$_{ν}$z ) ∇ $_{x}$ψ · ∇ $_{x}$zdS$_{x}$ dydt + ∫ Γ$_{x}$ 2 sγϕ ( ∂$_{ν}$ψ ) |∇ $_{x}$z | $^{2}$dS$_{x}$ dydt = − 8 ∫ Γ$_{x}$ sγϕ | ∂$_{ν}$z | $^{2}$( x − x$_{0}$ ) · νdS$_{x}$dydt + 4 Γ$_{x}$ sγϕ | ∂$_{ν}$z | $^{2}$( x − x$_{0}$ ) · νdS$_{x}$dydt = − 4 Γ$_{x}$ sγϕ | ∂$_{ν}$z | $^{2}$( x − x$_{0}$ ) · νdS$_{x}$dydt ≥ − 4 ∫ Γ$_{x}$ ∩{ ( x − x$_{0}$ ) · ν ≥ 0 } sγϕ | ∂$_{ν}$z | $^{2}$( x − x$_{0}$ ) · νdS$_{x}$dydt. ∫ Ω ∣ P + s z ∣ $^{2}$dxdydt + ∫ Ω ∣ P − s z ∣ $^{2}$dxdydt + s ∫ Ω |∇ $_{x}$z | $^{2}$dxdydt + s ∫ Ω |∇ $_{y}$z | $^{2}$dxdydt + s 3 ∫ Ω | z | $^{2}$dxdydt. ≤ C$_{8}$ Ω | P$_{s}$z | $^{2}$dxdydt + C$_{8}$s ∂D$_{+}$ × G × ( − T,T ) | ∂$_{ν}$z | $^{2}$dS$_{x}$ dydt. 22