5.5 Alternating Direction Implicit (ADI)
As motivation consider the two variable Poisson equation
− u xx − u yy = f .
Or
− u xx = f + u yy .
Fix y, and this is a sequence of odes with respect to x.
− u yy = f + u xx
Fix x, and this is a sequence of odes with respect to y.
Solve the following sequences of odes:
1
− uxx m + 2 = f + u yy m
1
− u yy m +1 = f + uxx m + 2
Or, applying the FDM and solve a sequence of tridiagonal systems:
1
1
1
−ui -1, j m + 2 +2ui, j m + 2 − ui +1, j m + 2
∆x 2
−ui , j -1
m +1
m +1
+ 2ui , j
∆y 2
+
−ui , j -1 m +2ui, j m −ui , j +1m
m +1
− ui , j +1
= fi, j −
∆y 2
−ui -1, j
m+
1
2
m+
+ 2ui, j
∆x 2
1
2
= fi , j
− ui +1, j
m++
1
2
.
→ 
→  There is a sequence of tridiagonal solves that are independent of j.
 
→ 
↑ ↑ ↑ There is a sequence of tridiagonal solves that are independent of i.
2
Matrix Form.
A = H +V
 Aˆ

H=
O





ˆA
 2 -1

ˆA =  -1 O O


O O

2I -I


V =  - I O O

O O
ADI Splittings.
A = (α I + H ) - (α I -V )
A = (α I + V ) - (α I - H )
where α ∈¡ ,or α is a diagonal matrix.
ADI Algorithm.
A = H + V and Ax = d
for m = 0, maxm
solve (α I + H ) x
m+
1
2
= d + (α I -V )x m
solve (α I + V ) x m +1 = d + (α I - H ) x
test for convergence.
m+
1
2
3
Proposition 8.
Let A = H + V and α >0.
If A,H,V are SPD, then for all x0, xm+1→ x, where Ax = d.
Proof.
1
m+ 

u m +1 = (α I + V ) -1  d + (α I - H ) u 2 


-1
= (α I + V )  d + (α I - H )(α I + H )-1{d + (α I -V )u m }
ˆ m + g where
= Hu
Hˆ =  (α I + V ) -1 (α I - H )(α I + H )-1 (α I -V )  and
g = (α I + V )-1 [ d + (α I - H )(α I + H )-1 d ].
ˆ
ˆ < 1 for some norm.
Show ρ (H)<1
or H
Note the following:
ˆ α I+V) -1 = (α I-H)(α I+H) -1 (α I-V)(α I+V) -1
(α I+V)H(
= FG where
F ≡ (α I-H)(α I+H) -1
G ≡ (α I-V)( α I+V)-1.
ˆ σ (FG)
σ (H)=
Find
so that F < 1 and G < 1.
Consequently, FG ≤ F G < 1.
4
We will use the following facts:
1. F
2
Fx
= sup
x
x
2
2
= ρ (F) where
2
≡ (x T x)1/2 and F is symmetric.
In order to show this, represent x by an
orthonormal basis of eigenvectors of F.
2. H, V symmetric implies that F and G are symmetric.
3. Eigenvalues of F and G have the form
α −λ
where
α +λ
Hx = λx or
Vx = λx
λ >0 because H and V are SPD.
Therefore,
F 2 = (α I-H)( α I+H)-1
α−λ
<1
α+λ
2
< 1 and G
2
; λ >0 and α ≥ 0
α − λ

α−λ
= α + λ
α+λ
− α − λ
 α + λ
α ≥λ
α < λ.
< 1 because
5
Application to 3D Problem.
−u xx − u yy − u zz = f
Apply the FDM and form
A = H + V + W where W is associated with the z direction.
A = (α I + H ) - (α I - V -W )
= (α I + V ) - (α I - H - W )
= (α I + W ) - (α I - H -V ).
From these three splittings one can form the three stage ADI
algorithm, and prove an analogue to Proposition 8.