and Since and for 0 < α, β < 1 , we have Then, we see that Since the signs of the terms of |∇ $_{x}$z | 2 and |∇ $_{y}$z | 2 are different, we need to perform 17 d$_{3}$ : = d$_{3}$ ( ψ ) = ( d$_{1}$ ( ψ )) 2 + Δ$_{y}$ ( d$_{2}$ ( ψ )) − Δ$_{x}$ ( d$_{2}$ ( ψ )) , d$_{4}$ : = d$_{4}$ ( ψ ) = d$_{1}$ ( ψ ) d$_{2}$ ( ψ ) + ∇ $_{y}$ψ · ∇ y ( d$_{2}$ ( ψ )) − ∇ $_{x}$ψ · ∇ $_{x}$( d$_{2}$ ( ψ )) , d$_{5}$ : = d$_{5}$ ( ψ ) = ∇ $_{y}$ψ · ∇ $_{y}$( d$_{2}$ ( ψ )) − ∇ $_{x}$ψ · ∇ $_{x}$( d$_{2}$ ( ψ )) . ∫ Ω 4 sγ $^{2}$ϕ |∇ $_{y}$ψ · ∇ $_{y}$z − ∇ $_{x}$ψ · ∇ $_{x}$z | 2 dxdydt ≥ 0 | x − x$_{0}$ | 2 − α $^{2}$| y | 2 − β 2 | t | 2 ≥ δ 2 $_{0}$, d 2 = 16( | x − x$_{0}$ | 2 − α $^{2}$| y − y$_{0}$ | $^{2}$)2 ≥ 16( | x − x$_{0}$ | 2 − α $^{2}$| y − y$_{0}$ | 2 − β 2 | t | $^{2}$) 2 ≥ 16 δ 2 $_{0}$. 2 Re ( P + s z, P − s z ) L $^{2}$(Ω) ≥ − ∫ Ω 8 sαγϕ |∇ $_{y}$z | $^{2}$dxdydt + ∫ Ω 8 sγϕ |∇ $_{x}$z | $^{2}$dxdydt +64 δ 2 0 ∫ Ω s $^{3}$γ$^{4}$ϕ$^{3}$| z | $^{2}$dxdydt + B$_{0}$ + X$_{1}$ + X$_{2}$. (3.29)