The Conjugate Gradient Method - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 224

5.4 Remark. In the Uzawa algorithm (5.4), u k + 1 and λ k + 1 are independent of u k .
In the iteration (5.7), u k + 1 and λ k + 1 are independent of λ k .
The assertion about the Uzawa algorithm follows directly from the defini-
tion (5.4) of the algorithm. The other assertion is a consequence of the following
formula which is equivalent to (5.7):
u k + 1
λ k + 1

CB t
B

− 1 f − (A − C)u k
g

.

= ( 5 . 8 )

Bramble and Pasciak [1988] took a completely different approach. By employing a
different metric for the indefinite problem, they were able to get a preconditioning
in almost the same way as in the positive definite case.

Problems

5.5 Consider the special case A = I , and compare the condition number of
BA − 1 B t with that of the squared matrix. In particular, show that the Uzawa algo-
rithm is better than the gradient method for the squared matrix.

5.6 For the case m n , the restriction can be eliminated indirectly. Let F be an

m × m matrix with FF t = BB t , e.g., say F stems from the Cholesky decompo-

sition of BB t . In the special case A = I , we have the triangular decomposition:
I
BF

IB t

− F t

IB t
B

.

=

How can we construct a corresponding triangular decomposition for the matrix in
(5.2) if a decomposition A = L t L is known?

5.7 Show that κ(BA − 1 B t ) ≤ κ(A)κ(BB t ) .

5.8 For the saddle point problem (5.2), the norm · A is obviously the natural
norm for the u components. Show that the norm · BA − 1 B t is then the natural one

for the λ components in the following sense: the inf-sup condition holds for the
mapping B t

m n

: R → R with the constant β = 1.

5.9 Verify the block Cholesky decomposition for the matrix
AB t
B

A 0
BI

A − 1

− BA − 1 B t AB t

0

=

0 0 0 I

appearing in the saddle point problem. What is the connection between this fac-
torization and the computation of the reduced equation (5.3)? In addition, prove
that the inverse has the following decomposition:
AB t
B

− 1 A − 1 ,

− A − 1 B t S − 1 BA − 1 A − 1 B t S − 1

=

S − 1 BA − 1 − S − 1 0

where S = BA − 1 B t is the Schur complement.