6-19

Chapter 6

Boundary-Value Problems

6.4 The Alternating Direction Implicit (ADI) Method

When the partial differential equation 2u = 0 is solved by the finite difference method, the resulting coefficient matrix is spare. The sparseness increases as the number of nodes

increases. If there are 21 nodes, 81% of the coefficients are zeros; if there are 105 nodes,

96% are zeros. The system of equations for the one-dimensional case always has a

tridiagonal coefficient matrix for which the efficient Thomas algorithm can be used. The ADI

method can be applied for the two or three-dimensional system to get a tridiagonal

coefficient matrix. We will use a two dimensional example of the Laplace equation

2u = 0

Using finite difference, the value at the node (i, j) for iteration (m+1) is given as

)1(,m

jiu = 4

1[ )( 1,

m

jiu + )(

1,

m

jiu + )(

,1

m

jiu + )(

,1

m

jiu ]

We now add and subtract )(,m

jiu from this equation to yield

)1(,m

jiu = )(

,

m

jiu + 4

1[ )( 1,

m

jiu + )(

1,

m

jiu + )(

,1

m

jiu + )(

,1

m

jiu 4)(

,

m

jiu ]

or equivalently

)1(,m

jiu )(

,

m

jiu = 4

1{[ )( 1,

m

jiu 2)(

,

m

jiu + )(

1,

m

jiu ] + [)(

,1

m

jiu 2)(

,

m

jiu + )(

,1

m

jiu ]}

Each iteration is considered to be a two-step procedure wherein the first step advances to the

(m+2

1) level and the second step to the (m+1) level.

First step:

)2/1(,m

jiu )(

,

m

jiu = 4

1{[ )2/1( 1,

m

jiu 2)2/1(

,

m

jiu + )2/1(

1,

m

jiu ] + [)(

,1

m

jiu 2)(

,

m

jiu + )(

,1

m

jiu ]}

Second step:

)1(,m

jiu )2/1(

,

m

jiu = 4

1{[ )2/1( 1,

m

jiu 2)2/1(

,

m

jiu + )2/1(

1,

m

jiu ] + [)1(

,1

m

jiu 2)1(

,

m

jiu + )1(

,1

m

jiu ]}

6-20

The ADI method produces a tridiagonal set of equations at the (m+1/2) level. The equations

can be solved along all rows of the grid, one row at a time. Once, all nodes have been

elevated to the (m+1/2) level, a similar procedure for the column of nodes is applied. A two-

step iteration is completed when the new values )1(,m

jiu are calculated.

Example 6.4-1 _____________________________________________________

Assuming two dimensional, steady-state conduction, determine the temperature of nodes 1,

2, 3, and 4 in the square shape subjected to the uniform temperature shown.

50 200

100

300

50 200

100

300

1 2

3 4

1,0 1,1 1,2 1,3

2,0 2,1 2,2 2,3

0,1 0,2

3,1 3,2

Figure 6.4-1 The nodes in a two dimensional, steady-state conduction.

Solution

The two-dimensional heat conduction equation for steady state, no heat generation, and k

independent of temperature is given as

2

2

x

u

+

2

2

y

u

= 0

The ADI method will be applied to solve this problem

)1(,m

jiu )(

,

m

jiu = 4

1{[ )( 1,

m

jiu 2)(

,

m

jiu + )(

1,

m

jiu ] + [)(

,1

m

jiu 2)(

,

m

jiu + )(

,1

m

jiu ]}

Let )( 1,1mu = 100, )( 2,1

mu = 150, )( 1,2mu = 150, and )( 2,2

mu = 250

First step calculation with row 1 and row 2:

)2/1(,m

jiu )(

,

m

jiu = 4

1{[ )2/1( 1,

m

jiu 2)2/1(

,

m

jiu + )2/1(

1,

m

jiu ] + [)(

,1

m

jiu 2)(

,

m

jiu + )(

,1

m

jiu ]}

Node (1,1): )2/1( 1,1mu 100 =

4

1{[ )2/1( 2,1

mu 2 )2/1( 1,1mu + 50] + [100 2(100) + 150]}

6-21

1.5 )2/1( 1,1mu 0.25 )2/1( 2,1

mu = 125

Node (1,2): )2/1( 2,1mu 150 =

4

1{[200 2 )2/1( 2,1

mu + )2/1( 1,1mu ] + [100 2(150) + 250]}

0.25 )2/1( 1,1mu + 1.5 )2/1( 2,1

mu = 212.5

The two nodes in row 1 are solved with the following results

)2/1( 1,1mu = 110, and )2/1( 2,1

mu = 160

Node (2,1) )2/1( 1,2mu 150 =

4

1{[ )2/1( 2,2

mu 2 )2/1( 1,2mu + 50] + [100 2(150) + 300]}

1.5 )2/1( 1,2mu 0.25 )2/1( 2,2

mu = 187.5

Node (2,2) )2/1( 2,2mu 250 =

4

1{[200 )2/1( 2,2

mu + )2/1( 1,2mu ] + [150 2(250) + 300]}

0.25 )2/1( 1,2mu + 1.5 )2/1( 2,2

mu = 287.5

The two nodes in row 2 are solved with the following results

)2/1( 1,2mu = 161.43, and )2/1( 2,2

mu = 218.57

Second step calculation with column 1 and column 2:

)1(,m

jiu )2/1(

,

m

jiu = 4

1{[ )2/1( 1,

m

jiu 2)2/1(

,

m

jiu + )2/1(

1,

m

jiu ] + [)1(

,1

m

jiu 2)1(

,

m

jiu + )1(

,1

m

jiu ]}

Node (1,1) )1( 1,1mu 110 =

4

1{[160 2(110)+ 50] + [100 2 )1( 1,1

mu + )1( 1,2mu ]}

1.5 )1( 1,1mu 0.25 )1( 1,2

mu = 132.5

Node (2,1) )1( 1,2mu 161.43 =

4

1{[218.57 2(161.43)+ 50] + [ )1( 1,1

mu 2 )1( 1,2mu + 300]}

0.25 )1( 1,1mu + 1.5 )1( 1,2

mu = 222.86

The two nodes in column 1 are solved with the following results

6-22

)1( 1,1mu = 116.33, and )1( 1,2

mu = 167.96

Node (1,2) )1( 2,1mu 160 =

4

1{[200 2(160)+ 110] + [100 2 )1( 2,1

mu + )1( 2,2mu ]}

1.5 )1( 2,1mu 0.25 )1( 2,2

mu = 182.5

Node (2,2) )1( 2,2mu 218.57 =

4

1{[200 2(218.57)+ 161.43] + [ )1( 2,1

mu )1( 2,2mu + 300]}

0.25 )1( 2,1mu + 1.5 )1( 2,2

mu = 274.64

The two nodes in column 2 are solved with the following results

)1( 2,1mu = 156.53, and )1( 2,2

mu = 209.18

Table 6.4-1 lists the temperatures before and after one ADI iteration.

Table 6.4-1 )(

,

m

jiu (left side) and )1(

,

m

jiu (right-side)

i\j 0 1 2 3 i\j 0 1 2 3

0 100 100 0 100 100 1 50 100 150 200 1 50 116.33 156.53 200 2 50 150 250 200 2 50 167.96 209.18 200 3 300 300 3 300 300