Leibman Method

8/3/2019 Leibman Method

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Liebmann’s method

• Rearrange the approximation to give ui,j in terms of neighbouring values

1

1 −4 1

1

ui,j ⇒ uk+1i,j =

1

4 uki−1,j + uk

i+1,j + uki,j−1 + uk

i,j+1

1 4 1

4 −20 4

1 4 1

ui,j ⇒ ui,j =1

20

4uk

i−1,j + 4uki+1,j + 4uk

i,j−1 + 4uki,j+1+

uki−1,j−1 + uk

i−1,j+1 + uki+1,j−1 + uk

i+1,j+1

• Incorporate boundary conditions as before

• Creates iteration scheme

EMAT33040 ANA 2001-02 3.7 Department of Engineering Mathematics



Refining Liebmann’s method

•Compute uk+1

i,j in sequence

• Some terms on right hand side are known at k + 1 time step — previously computed

• Improve speed of convergence by using them

1

1 −4 1

1

ui,j ⇒ uk+1i,j = 1

4

uk+1i−1,j + uk

i+1,j + uk+1i,j−1 + uk

i,j+1

1 4 1

4 −20 4

1 4 1

ui,j ⇒ ui,j = 120

4u

k+1

i−1,j + 4u

k

i+1,j + 4u

k+1

i,j−1 + 4u

k

i,j+1+uk+1i−1,j−1 + uk

i−1,j+1 + uki+1,j−1 + uk

i+1,j+1

• Disadvantage: slow convergence




SOR method

• SOR = Successive Over Relaxation

• Consider standard central difference based iteration

uk+1i,j =

1

4 uk+1i−1,j + uk

i+1,j + uk+1i,j−1 + uk

i,j+1= uk

i,j + 14

uk+1i−1,j + uk

i+1,j − 4uki,j + uk+1

i,j−1 + uki,j+1

• Second term is ‘residual’; relaxes to zero at convergence

•Multiply ‘residual’ term by factor ω to ‘overrelax’

uk+1i,j = uk

i,j +ω

4

uk+1i−1,j + uk

i+1,j − 4uki,j + uk+1

i,j−1 + uki,j+1




Using SOR

•Solve Laplace’s equation using SOR

uxx + uyy = 0, 0 < x < 20, 0 < y < 10

u(x, 0) = 0

u(x, 10) = 0

u(0, y) = 0

u(20, y) = 100

• Note that ω = 1 is Liebmann’s method

• Stop when maximum change in u is < 10−3

21 interior points 105 interior points

(mx = 3,my = 7) (mx = 15,my = 7)

ω No. of iterations ω No. of iterations

1.00 20 1.00 70

1.10 15 1.10 58

1.20 13 1.20 46

1.30 12 1.30 35

1.40 15 1.40 29

1.50 18 1.50 26

1.60 23 1.60 28

1.70 36

• Find an optimal value of ω — increases with grid spacing




Choosing ω for SOR

• Consider a rectangular region with Dirichlet boundary conditions

•Optimal value of ω is the smaller root of

cos

π

mx + 1

+ cos

π

my + 1

2ω2 − 16ω + 16 = 0

• Can solve to give

ωopt = 4

2 +√

4− c2, c = cos

π

mx + 1

+ cos

π

my + 1

• Previous example

mx my ωopt

7 3 1.2668

15 7 1.5325

• Non Dirichlet boundary conditions use results from first few iterations — see textbooks




Neumann boundary conditions

• Modification for Neumann boundary conditions follows parabolic case

•Consider only one edge

∂u

∂x(0, y) = ln(x)

• Boundary grid points included in numerical scheme

•On the boundary

∇2u ∼ 1

∆2

1

1 −4 1

1

u0,j

• u−1,j is included via approximation for derivative

∂u

∂x(0, yj) =

u1,j − u−1,j

2∆= ln(yj)

•Change of +1-diagonal entry, and extra components of right hand side vector


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Leibman Method