So, no matter which kind of anti-periodic boundary value conditions we shall propose, first of all, we must ensure that the limits exist taken on the left-hand side of the formula on the anti-periodic boundary value conditions when variable tends to zero.
However, instead of the usual Dirichlet boundary value condition (1.4), we may conjecture that only a partial boundary value condition begin{aligned} u x,t)=0,quad (x,t inSigma_{2}times 0,T), end{aligned} (1.9) is required.
The main point is that, to assure the posedness of the solutions to (1.1), instead of the whole boundary value condition (1.11), we can require only a partial boundary value condition, u x,t) = 0, quad (x,t) inSigma _{p} times (0,T), (1.12) where (Sigma _{p}) is a subset of ∂Ω.
Multiplying equation (2.7) by u 2, integrating the result over S R and using the boundary value condition (2.9), we have γ ∬ S R ( D 3 u 2 ( x, t ) ) 2 d x d t = ∬ S R Φ ( x, t ) u 2 ( x, t ) d x d t. (2.13).
Proof of Proposition 3 Multiplying both sides of (12) by x ( t ) and integrating from a to b by parts and using the boundary value condition (13), we can obtain ∫ a b x ( 2 n ) ( t ) x ( t ) d t = ( − 1 ) n ∫ a b ( x ( n ) ( t ) ) 2 d t = − ∑ k = 0 n ∫ a b p k ( t ) x ( k ) ( t ) x ( t ) d t.
where C is a positive constant, i = 0, 1, 2. Proof Noticing the condition (2.8) and the boundary value condition (2.9), we use the Poincaré inequality and interpolation method (see Chapter 5 in [12]) and get ∬ S R u 2 2 d x d t ≤ C R 2 ∬ S R ( D u 2 ) 2 d x d t, ∬ S R ( D u 2 ) 2 d x d t ≤ ε R 2 ∬ S R u 2 2 d x d t + C R 4 ∬ S R ( D 3 u 2 ) 2 d x d t, (2.11).
Integrating the above inequalities over ( 0, T ) and noticing (3.13), we see that the terms of left hand side in these inequalities can all be estimated by ε ∬ Q T ( D 6 u σ ) 2 d x d t and a constant number C. Then by the boundary value condition and (3.10), we have ∬ Q T ( D 4 u σ ) 2 d x d t ≤ ε ∬ Q T ( D 6 u σ ) 2 d x d t + C, (3.14).
