We note that where we choose µ > 0 later and By adding (3.29) and (3.31) we have since z = 0 on Γ$_{x}$ ∪ Γ$_{y}$. We multiply (3.30) by − sγ (4 α + µ ), then we have B$_{1}$ = − ∫ Γ$_{y}$ γϕ ( ∂$_{ν}$ψ ) | z | 2 dS$_{y}$dxdt + 2 Re ∫ Γ$_{y}$ ( ∂$_{ν}$z ) ( ϕz ) dS$_{y}$dxdt − 2 Γ$_{y}$ sγϕ $^{2}$( ∂$_{ν}$ψ ) | z | 2 dS$_{y}$dxdt + ∫ Γ$_{x}$ γϕ ( ∂$_{ν}$ψ ) | z | 2 dS$_{x}$dydt − 2 Re ∫ Γ$_{x}$ ( ∂$_{ν}$z ) ( ϕz ) dS$_{x}$dydt + 2 Γ$_{x}$ sγϕ $^{2}$( ∂$_{ν}$ψ ) | z | 2 dS$_{x}$dxdt = 0 , − 2 Re ∫ Ω (4 α + µ ) ( P + s z + P − s z ) sγϕzdxdydt = 2 ∫ Ω (4 α + µ ) sγϕ |∇ $_{y}$z | 2 dxdydt − 2 ∫ Ω (4 α + µ ) sγϕ |∇ $_{x}$z | 2 dxdydt + X$_{5}$ + X$_{6}$, (3.31) X$_{5}$ = − 2 ∫ Ω (4 α + µ ) s $^{3}$γ$^{3}$ϕ$^{3}$d$_{2}$ ( ψ ) | z | 2 dxdydt, X$_{6}$ = 2 Im ∫ Ω sγ (4 α + µ ) ∂$_{t}$zzϕdxdydt − ∫ Ω (4 α + µ ) sγ $^{2}$ϕ | z | 2 d$_{1}$ ( ψ ) dxdydt − Ω (4 α + µ ) s $^{2}$γ$^{3}$ϕ | z | 2 d$_{2}$ ( ψ ) dxdydt − 2 ∫ Ω (4 α + µ ) s $^{2}$γ$^{3}$ϕ2 | z | 2 d$_{2}$ ( ψ ) dxdydt. 2 Re ( P + s z, P − s z ) L $^{2}$(Ω) − 2 Re ∫ Ω (4 α + µ ) sγ ( P + s z + P − s z ) ϕzdxdydt ≥ 2 µ ∫ Ω sγϕ |∇ $_{y}$z | 2 dxdydt + (8 − 8 α − 2 µ ) ∫ Ω sγϕ |∇ $_{x}$z | 2 dxdydt +64 δ 2 0 ∫ Ω s $^{3}$γ$^{4}$ϕ$^{3}$| z | $^{2}$dxdydt + B$_{0}$ + X$_{1}$ + X$_{2}$ + X$_{5}$ + X$_{6}$. (3.32) 20