{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

14a ie t n tu u t u xx n yy jj k x x jk y y x n x y x

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 (u x y n x y t jk = txxyy ) +1 2 + : : : ]: n jk = n y jk 5.2. Operator Splitting 9 Thus, the local discretization error of the ADI method is n jk =( ) n jk CN + O( t2) = O( x2) + O( y2) + O( t2) which is the same order as that of the Crank-Nicolson method. The stability of (5.1.5) can be analyzed by the von Neumann method. The twodimensional form of the discrete Fourier series is U= n ;1 ;1 XX J jk p K =0 =0 A e2 n pq ( i pj=J + qk=K ): (5.1.8a) q Substituting into (5.1.5 and proceeding as in one dimension, we nd A = (M ) A0 n (5.1.8b) n pq pq pq where A0 is a Fourier component of the initial data and M is the ampli cation factor. Again, following the one-dimensional analysis, we verify that jM j 1 for all positve r and r hence, the Peaceman-Rachford version of the ADI method (5.1.5) is unconditionally stable. pq pq pq x y 5.2 Operator Splitting Methods The ADI approach is often di cult to extend to problems on non-rectangular domains, to nonlinear problems, and to problems having mixed derivatives such as u . The dimensional reduction developed for the ADI method can be viewed as an approximate factorization of the di erential or discrete operator. Let us motivate the factorization by rst examining the ordinary di erential equation xy dy = (a + b)y dt which, of course, has the solution y(t) = e ( + ) y(0) = e e y(0): ta b ta tb The latter form suggests that the solution of the initial value problem may be obtained by rst solving dy=dt = by to time t with y(0) prescribed as initial data, and then solving 10 Multi-Dimensional Parabolic Problemss dy=dt = ay subject to the initial condition e y(0). This interpretation, however, does tb not extend to vector systems of the form dy = (A + B)y dt unless A and B commute. Thus, we may write the solution of the vector problem as y(t) = e (A+B) y(0) t where t2 e C = I + tC + 2! C2 + : : : : t However, y(t) = et(A+B)y(0) = etAetBy(0) 6 unless AB = BA. Nevertheless, let's push on and consider a linear partial di erential equation u = u = ( 1 + 2 )u L t L (5.2.1) L where L is a spatial di erential operator that has been split into the sum of L1 and L2. We'll think of L1 as being associated with x derivatives and L2 as being associated with y derivatives, but this is not necessary. Any splitting will do. The solution of the linear partial di erential equation can also be written as the exponential u(x y t) = e L u(x y 0) t when L is independent of t. The interpretation of the exponential of the operator follows from a Taylor's series expansion of u in powers of t, i.e., L 2 @ u(x y t) = u(x y 0) + tu (x y 0) + t u (x...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online