MATH 676
–
Finite element methods in
scientific computing
Wolfgang Bangerth, Texas A&M University
http://www.dealii.org/
Wolfgang Bangerth
Lecture 31.65:
Nonlinear problems
Part 4: Fixed point/Picard iteration
for the minimal surface equation
http://www.dealii.org/
Wolfgang Bangerth
The minimal surface equation
Consider the minimal surface equation:
(
−∇⋅
A
∇u = f
2
√ 1+|∇ u|
u = g
)
in Ω
on ∂Ω
where we choose
2
Ω=B1 (0)⊂ℝ ,
f =0,
g=sin (2 π( x + y ))
Goal: Solve this numerically with a Picard iteration method.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
General approach: To solve
L(u) = f
by Picard iteration, we need to rewrite L(u) as
L(u) = G(u)u
and then repeatedly solve
G(u k )uk + 1=f
Note 1: Here, G(u) is in general a differential operator.
Note 2: This is a linear problem in uk+1.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Here: To solve
(
−∇⋅
A
∇u = f
2
√ 1+|∇ u|
u = g
)
in Ω
on ∂Ω
by Picard iteration, we need to repeatedly solve
(
−∇⋅
A
∇ uk + 1 = f
2
√ 1+|∇ uk|
u k +1 = g
)
in Ω
on ∂Ω
Note: This is a linear problem in uk+1.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Question: With
(
L(u) = −∇⋅
A
∇u
2
√ 1+|∇ u|
)
why choose
(
G(u)w = −∇⋅
A
∇w
2
√ 1+|∇ u|
)
? We could also choose
G(u)w =
[ (
)]
L(u)
A
w
w = −∇⋅
∇
u
u
u
√ 1+|∇ u|2
Answer: Yes, but this is no differential equation in w. It may
or may not converge.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Question: With
(
L(u) = −∇⋅
A
∇u
2
√ 1+|∇ u|
)
why choose
(
G(u)w = −∇⋅
A
∇w
2
√ 1+|∇ u|
)
? We could also choose
(
G(u)w = −∇⋅
A
∇u
√ 1+∇ u⋅∇ w
)
Answer: Yes, but this is not linear in w. We don't know how
to solve G(u)w = f.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Question: How to choose G(u) ?
Answers:
●
●
●
We often have different choices for G(u)
We choose so that
– G(u) is linear
– The equation G(u) w = f is a well-posed PDE
– The operator G(u) has a “convenient” structure
There is often an element of “experience” involved
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Remember: To solve
A
−∇⋅
∇u = f
2
√ 1+|∇ u|
u = g
(
)
in Ω
on ∂Ω
by Picard iteration, we need to repeatedly solve
(
−∇⋅
A
∇ uk + 1 = f
2
√ 1+|∇ uk|
u k +1 = g
)
in Ω
on ∂Ω
Notes: (i) This is linear uk+1. (ii) It is well-posed if the
coefficient is bounded from below.
(iii) It is “like” the
Poisson equation.
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Picard iteration: Repeatedly solve
(
−∇⋅
A
∇ uk + 1 = f
2
√ 1+|∇ uk|
u k +1 = g
)
in Ω
on ∂Ω
Question: What should our initial guess u0 be?
Answers:
●
As close as possible to the exact solution!
●
Here, u0=0 will suffice (even though this doesn't satisfy
boundary conditions).
http://www.dealii.org/
Wolfgang Bangerth
Picard iteration
Picard iteration: Repeatedly solve
(
−∇⋅
A
∇ uk + 1 = f
2
√ 1+|∇ uk|
u k +1 = g
)
in Ω
on ∂Ω
Question: Wouldn't the solution of
be better?
−∇⋅( A ∇ u0 ) = f
u0 = g
in Ω
on ∂ Ω
Answer: Yes! But if u0=0 then u1 satisfies
(√
−∇⋅
A
1+0
2
)
∇ u1 = f
u1 = g
http://www.dealii.org/
in Ω
on ∂ Ω
Wolfgang Bangerth
Picard iteration
Remember: To solve
A
−∇⋅
∇u = f
2
√ 1+|∇ u|
u = g
(
)
in Ω
on ∂Ω
by Picard iteration, we need to repeatedly solve
(
−∇⋅
A
∇ uk + 1 = f
2
√ 1+|∇ uk|
u k +1 = g
)
in Ω
on ∂Ω
Let's see if we can morph step-6 to do this for us.
http://www.dealii.org/
Wolfgang Bangerth
MATH 676
–
Finite element methods in
scientific computing
Wolfgang Bangerth, Texas A&M University
http://www.dealii.org/
Wolfgang Bangerth