Finally, we rewrite our inequality with z instead of u . By the relation we see that there exist positive constants C$_{9}$ = C$_{9}$ ( γ ) and s$_{0}$ > s$_{2}$ ( γ ) such that for all s > s$_{0}$, We complete the proof of Theorem 1. Since u itself does not satisfy (3.6), in order to apply Proposition 1, we have to introduce a cut-off function. Moreover, we need to introduce several notations. We set By x$_{0}$ / ∈ D , we see that r > 0. We choose ρ > 1 sufficiently large so that By (4.3) and the assumption on L, we have | z | 2 = e 2 sϕ | u | $^{2}$, | ∂$_{ν}$z | 2 = | ∂$_{ν}$u | $^{2}$e 2 sϕ on ∂D$_{+}$ × G × ( − T, T ) , |∇ $_{x}$ue $^{sϕ}$| 2 = |∇ $_{x}$z − sλϕe $^{sϕ}$u ∇ $_{x}$ψ | 2 ≤ 2 |∇ $_{x}$z | 2 + 2 s $^{2}$λ$^{2}$ϕ2 |∇ $_{x}$ψ | 2 | z | $^{2}$, |∇ $_{y}$ue $^{sϕ}$| 2 = |∇ $_{y}$z − sλϕe $^{sϕ}$u ∇ $_{y}$ψ | 2 ≤ 2 |∇ $_{y}$z | 2 + 2 s $^{2}$λ$^{2}$ϕ2 |∇ $_{y}$ψ | 2 | z | 2 , | L$_{0}$u | 2 ≤ 2 | Lu | 2 + 2 ∣ n ∑ i =1 a$_{i}$ ( x, y, t ) u$_{x}$$_{i}$ + m ∑ j =1 b$_{j}$ ( x, y, t ) u$_{y}$$_{j}$ + a$_{0}$ ( x, y, t ) u ( x, y, t ) ∣ , ∫ Ω ( s |∇ $_{y}$u | 2 + s |∇ $_{x}$u | 2 + s 3 | u | $^{2}$) e 2 $^{sϕ}$dxdydt ≤ C$_{9}$ ∫ Ω | Lu | $^{2}$e 2 $^{sϕ}$dxdydt + C$_{9}$ ∫ ∂D$_{+}$ × G × ( − T,T ) s | ∂$_{ν}$u | $^{2}$e 2 $^{sϕ}$dS$_{x}$ dydt. ˜ r = max x ∈ D | x − x$_{0}$ | , (4.1) r = min x ∈ D | x − x$_{0}$ | . (4.2) r = max x ∈ D | x − x$_{0}$ | , (4.1) ˜ r r < ρ. (4.3) αL 2 ρ 2 < r 2 < ˜ r 2 < αL $^{2}$. (4.4) 23 4. PROOF OF THEOREM 1