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CCM, Part III (4)

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CCM, Part III (4)

Daniele Boffi

Dipartimento di Matematica “F. Casorati”

Universit` a di Pavia [email protected]

http://www-dimat.unipv.it/boffi/

March 2004

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Parabolic problems

Heat equation

∂u(t)

∂t − ∆u(t) = f (t)

Variational formulation: for each t, find u(t) ∈ V = H 0 1 (Ω) s.t.

 ∂u(t)

∂t , v



+ a(u(t), v) = (f (t), v) ∀v ∈ V

Space semidiscretization. Take V h ⊂ V and, for each t, look for u h (t) ∈ V h such that

 ∂u h (t)

∂t , v h



+ a(u h (t), v h ) = (f (t), v h ) ∀v h ∈ V h

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Parabolic problems (cont’ed)

Fully discretized problem

I Explicit Euler

u n+1 h − u n h

∆t , v h

!

+ a(u n h , v h ) = (f n , v h ) ∀v h ∈ V h

I Implicit Euler

u n+1 h − u n h

∆t , v h

!

+ a(u n+1 h , v h ) = (f n+1 , v h ) ∀v h ∈ V h

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Parabolic problems (cont’ed)

θ-method (somewhat inbetween explicit and implicit) 0 ≤ θ ≤ 1

u n+1 h − u n h

∆t , v h

!

+(1 − θ)a(u n h , v h ) + θ a(u n+1 h , v h ) =

(1 − θ)(f n , v h ) + θ(f n+1 , v h ) ∀v h ∈ V h In one space dimension (finite differences, and f = 0)

u n+1 i − u n i

∆t = 1

h 2 (1 − θ)(u n i+1 − 2u n i + u n i−1 )+

θ (u n+1 i+1 − 2u n+1 i + u n+1 i−1 )

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Parabolic problems (cont’ed)

If u(0) = sin(πx) then the solution in [0, π] with homogeneous Dirichlet boundary conditions is

u(t) = sin(πx) exp(−π 2 t)

In particular, it goes to zero as t → +∞

Study of discrete (absolute) stability Discrete solution has the form

u n i = α n sin(πih)

Stability condition |α| ≤ 1

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Parabolic problems (cont’ed)

u n+1 i − u n i = k

h 2 (1 − θ)(u n i+1 − 2u n i + u n i−1 )+

θ(u n+1 i+1 − 2u n+1 i + u n+1 i−1 ) u n i = α n sin(πih)

Some trigonometry

sin π(i + 1)h − 2 sin(πih) + sin π(i − 1)h = 2 sin(πih) cos(πh) − 2 sin(πih) =

sin(πih)(−4 sin 2 (πh/2)) Hence

α − 1 = k

h 2 ((1 − θ) + θα)(−4 sin 2 (πh/2))

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Parabolic problems (cont’ed)

Finally

α = 1 − (1 − θ)w

1 + θw = 1 − w 1 + θw with w = 4 h k

2

sin 2 (πh/2) ≥ 0

Condition |α| ≤ 1 equivalent to

w(1 − 2θ) ≤ 2

I 1/2 ≤ θ ≤ 1 inconditionally stable

I 0 ≤ θ < 1/2 stability condition k

h 2 ≤ 1

2(1 − 2θ)

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End of Part IV

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