= H ± 1 (U ± * ; ÿ * ; (∇ÿ) * ; (∇ ± ) * ) in ± × [0; T ]; U ± * ( x; 0) = U ± 0 * ( x) in ± ; (u + * ; v + * ; w + * ) • | @ ×[0;T ] = 0; G(U + * ; U - * ; ÿ * ; @ t ÿ * ; (∇ÿ) * )| M0×[0;T ]
where
Then we use them to solve the initial-boundary value problem (4.2) -(4.5) to get (U ±; 1 * ; ÿ 1 * ). Then again by (4.1) we obtain (U ±; 1 ; ÿ 1 ). On the other hand, we start from (U ±; 0 ; ÿ 0 ) to solve the boundary value problem (4.6) -(4.9) to obtain ±; 1 .
Once we get (U ±;k ; ÿ k ; ±;k ); we will compute (U ±;k * ; ÿ k * ; ±;k * ) by (4.1). And then using (U ±;k * ; ÿ k * ; ±;k * ); we solve (4.2) -(4.5) to get (U ±;k+1 * ; ÿ k+1 * ). Then again by (4.1), we obtain (U ±;k+1 ; ÿ k+1 ). By using (U ±;k ; ÿ k ) to solve (4.6) -(4.9), we will get ±;k+1 . Recall that we use Newton's iteration (4.5), i.e. * -U ±; 0 * , then U * satisÿes
where in (4.22) we have used estimate (4.11).
with the boundary conditions similar to (4.7) -(4.9).
Where in (4.22), C(U +; 0 * ; U -; 0 * ; ÿ 0 * ; @ t ÿ 0 * ;
) is a smooth matrix function of its arguments, the expression (V +;k * ; V -;k * ; k * ; @ t k * ) 2 is a simpliÿed notation for a quadratic function of all its variables.
* -ÿ k * and ±;k+1 = ±;k+1 -±;k satisfy the boundary value problem
4.27) G(U +;k * ; U -;k * ; ÿ k * )(V +;k+1 * ; V -;k+1 * ; k+1 * ) = r k * -r k-1 * + (G(U +;k-1 * ; U -;k-1 * ; ÿ k-1 * ) -G(U +;k * ; U -;k * ; ÿ k * ))(U +;k * ; U -;k * ; ÿ k * ) ≡ g k on M 0 × [0; T ]: (4.28) From (4.27) we readily get the energy estimate (4.25). If we can show that |F k | 2 0;Á;T 6 C 1 (|||U +;k