Numerical Methods for PDEs
Alternating direction implicit (ADI) schemes
(Lecture 11, Week 4)
Markus Schmuck
Department of Mathematics and Maxwell Institute for Mathematical Sciences
Heriot-Watt University, Edinburgh
Edinburgh, February 4, 2013
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
1/ 7
Outline
1
The ADI method
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
2/ 7
A more practical 2D scheme: The ADI method
For u smooth enough, the Taylor expansion of u(xj + hx , yl , tn ) admits
an Exponential operator notation
[
]
∂u hx2 ∂ 2 u hx3 ∂ 3 u
u(xj +hx , yl , tn ) = u + hx
+
+
+ ...
∂x
2! ∂x 2
3! ∂x 3
(xj ,yl ,tn )
[
]
2
2
3
3
h ∂
h ∂
∂
= 1 + hx
+ x
+ x
+ . . . u |(xj ,yl ,tn )
∂x
2! ∂x 2
3! ∂x 3
(
)
∂
= exp hx
u |(xj ,yl ,tn ) .
∂x
Similarly (missing out the steps in between) we can write
(
)
∂
u(xj , yl + hy , tn ) = exp hy
u |(xj ,yl ,tn )
∂y
and
(
)
∂
u(xj , yl , tn + k ) = exp k
u |(xj ,yl ,tn )
∂t
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
3/ 7
With the help of this notation, we derive a different implicit scheme for
ut = uxx + uyy .
Suppose that u(x, y , t) is a smooth solution of the PDE. Taylor
expanding u(x, y , tn + k ) about (x, y , tn ) gives
(
)
∂
u(x, y , tn +k ) = exp k
u |(x,y ,tn ) .
∂t
Use ea = ea/2 · ea/2 to rewrite this as
(
)
(
)
k ∂
k ∂
u|t=tn+k = exp
exp
u|t=tn
2 ∂t
2 ∂t
and hence
(
)
(
)
k ∂
k ∂
exp −
u|t=tn+k = exp
u|t=tn .
2 ∂t
2 ∂t
|
{z
} |
{z
}
new time-level
old time-level
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
[∗]
4/ 7
With the help of this notation, we derive a different implicit scheme for
ut = uxx + uyy .
Suppose that u(x, y , t) is a smooth solution of the PDE. Taylor
expanding u(x, y , tn + k ) about (x, y , tn ) gives
(
)
∂
u(x, y , tn +k ) = exp k
u |(x,y ,tn ) .
∂t
Use ea = ea/2 · ea/2 to rewrite this as
(
)
(
)
k ∂
k ∂
u|t=tn+k = exp
exp
u|t=tn
2 ∂t
2 ∂t
and hence
(
)
(
)
k ∂
k ∂
exp −
u|t=tn+k = exp
u|t=tn .
2 ∂t
2 ∂t
|
{z
} |
{z
}
new time-level
old time-level
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
[∗]
4/ 7
We now use the fact that u solves the PDE to write
(
)
k ∂
exp ±
u
2 ∂t
=
=
(
[
])
k ∂2
∂2
exp ±
u
+
2 ∂x 2
∂y 2
)
(
)
(
k ∂2
k ∂2
exp ±
exp ±
u
2 ∂x 2
2 ∂y 2
(using ea+b = ea · eb ). Plugging this into [∗] gives
(
)
(
)
k ∂2
k ∂2
exp −
exp −
u|t=tn+k
2 ∂x 2
2 ∂y 2
)
(
)
(
k ∂2
k ∂2
= exp
exp
u|t=tn .
2 ∂x 2
2 ∂y 2
We now chop the exponentials at first order (e±q ≈ 1 ± q) to get
(
)(
)
(
)(
)
k ∂2
k ∂2
k ∂2
k ∂2
1−
1−
u|t=tn+k ≈ 1 +
1+
u|t=tn ,
2 ∂x 2
2 ∂y 2
2 ∂x 2
2 ∂y 2
and with second central differences in space leads to
(
ry ) n+1 (
ry ) n
rx ) (
rx ) (
= 1 + δx2 1 + δy2 wj,l
,
1 − δx2 1 − δy2 wj,l
2
2
2
2
where rx , ry defined as before.
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
5/ 7
The scheme
(
ry ) n+1 (
ry ) n
rx )(
rx )(
= 1 + δx2 1 + δy2 wj,l
,
1 − δx2 1 − δy2 wj,l
2
2
2
2
it is much easier and quicker to use than(it looks )as it splits into two
r
n+1
:
stages with an intermediate quantity vj,l := 1 − 2y δy2 wj,l

)
)
(
(
r
n 
Stage 1: 1 − r2x δx2 vj,l
= 1 + 2y δy2 wj,l

ADI scheme
)
)
(
(

r

Stage 2: 1 − y δ 2 w n+1 = 1 + rx δ 2 v
2 y
j,l
2 x
j,l
y
y
The name ADI comes from this idea of alternately solving along the
x−direction and y −direction.
x
x
PDEs, Lecture
11 right)
Stage Numerical
1 (on Methods
left), for
Stage
2 (on
M. Schmuck (Heriot-Watt University)
6/ 7
)
(
The two step splitting solves the full scheme: Applying 1 − r2x δx2
to Stage 2 gives
(
ry ) n+1 (
rx ) (
rx ) (
rx )
1 − δx2 1 − δy2 wj,l
= 1 − δx2 1 + δx2 vj,l
2
2
2
2
(
rx 2 ) (
rx 2 )
=
1 + δx 1 − δx vj,l (difference operators commute (Why?))
2
2
(
ry 2 ) n
rx 2 ) (
=
1 + δx 1 + δy wj,l by Stage 1.
2
2
Advantages:
Faster than 2D θ-method: Matrices are tridiagonal and involve
only J or L unknowns with operations of O(JL) compared to
O(J 3 L3 ) (for the θ-scheme).
Exercise for the remaining part of today’s lecture:
Show that the 2D ADI scheme is unconditionally stable.
The generalisation to 3D (with 2 intermediate variables) of the 2D
ADI is not unconditionally stable!
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
7/ 7
)
(
The two step splitting solves the full scheme: Applying 1 − r2x δx2
to Stage 2 gives
(
ry ) n+1 (
rx ) (
rx ) (
rx )
1 − δx2 1 − δy2 wj,l
= 1 − δx2 1 + δx2 vj,l
2
2
2
2
(
rx 2 ) (
rx 2 )
=
1 + δx 1 − δx vj,l (difference operators commute (Why?))
2
2
(
ry 2 ) n
rx 2 ) (
=
1 + δx 1 + δy wj,l by Stage 1.
2
2
Advantages:
Faster than 2D θ-method: Matrices are tridiagonal and involve
only J or L unknowns with operations of O(JL) compared to
O(J 3 L3 ) (for the θ-scheme).
Exercise for the remaining part of today’s lecture:
Show that the 2D ADI scheme is unconditionally stable.
The generalisation to 3D (with 2 intermediate variables) of the 2D
ADI is not unconditionally stable!
M. Schmuck (Heriot-Watt University)
Numerical Methods for PDEs, Lecture 11
7/ 7