Optimized Parameters of Optimized Schwarz Waveform Relaxation Methods for The Heat Equation

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Optimized Parameters of Optimized Schwarz Waveform Relaxation Methods for The Heat Equation

Minh-Binh Tran

To cite this version:

Minh-Binh Tran. Optimized Parameters of Optimized Schwarz Waveform Relaxation Methods for

The Heat Equation. 2011. �hal-00589252�

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Methods for The Heat Equation

INTERNAL REPORT

Minh-Binh TRAN

Laboratoire Analyse G´ eom´ etrie et Applications Institut Galil´ ee, Universit´ e Paris 13, France

Email: binh@math.univ-paris13.fr

April 28, 2011

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Chapter 1 Introduction

The Schwarz domain decomposition methods is a procedure to parallelize and solve partial differential equations numerically, where each iteration involves the solutions of the original equations on smaller subdomains. It was original proposed by H. A. Schwarz [7] in 1870 as a technique to prove the existence of a solution to the Laplace equation on a domain which is a combination of a rectangle and a circle. The idea was then used by P. L. Lions [4], [5], [6] as parallel algorithms in solving partial differential equations. Since then, many kind of domain decomposition methods have been developped, to im- prove the performance of the classical domain decomposition method. One of the main streams in this direction is to replace the Dirichlet transmission condition by Robin and Ventcell transmission conditions and then calculate the convergence rates. Using different transmissions condition gives different convergence rates and we need to optimize to get the best transmission conditions, the methods are then called the optimized Schwarz methods. In [1]

and [2], D. Bennequin, M. Gander and L. Halpern show that the problem of optimizing the convergence rates is in fact a new class of best approximation problems and suggest a new method to solve this class of problems. The au- thors consider the model problem of optimizing the convergence factors for advection-diffusion equations. In this report, we use their methods to check the results announced in [3] and then extend the results to optimized Robin and Ventcell transmission conditions for 2 dimensional heat equations.

3

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Optimized Schwarz Waveform Relaxation Methods For The One Dimensional Heat

Equation

2.1 Optimized Schwarz Waveform Relaxation Methods For The One Dimensional Heat Equation With Robin Transmission Con- dition

In this section, we consider the optimized Schwarz waveform relaxation method. The algorithm is



 



 



(∂

t

− ν∂

xx

)u

^k₁

= f in Ω

1

× (0, T ), u

^k₁

(x, 0) = u

0

(x) in Ω

1

,

(∂

x

+

_2ν^p

)u

^k₁

(L, .) = (∂

x

+

_2ν^p

)u

^k₂⁻¹

(L, .) in (0, T ),

(2.1.1)



 



 



(∂

t

− ν∂

xx

)u

^k₂

= f in Ω

2

× (0, T ), u

^k₂

(x, 0) = u

0

(x) in Ω

2

,

(∂

x

−

_2ν^p

)u

^k₂

(0, .) = (∂

x

−

_2ν^p

)u

^k₁⁻¹

(0, .) in (0, T ).

4

(7)

Let h

L

and h

0

be given in oH

³⁴

(0, T ). Let (e

1

, e

2

) be the solution in H

^3,³²

(Ω

1

× (0, T )) × H

^3,³²

(Ω

2

× (0, T )) of the problem



 



 



(∂

t

− ν∂

xx

)e

1

= 0 in Ω

1

× (0, T ), e

1

(x, 0) = 0 in Ω

1

,

(∂

x

+

_2ν^p

)e

1

(L, .) = h

L

in (0, T ),

(2.1.2)



 



 



(∂

t

− ν∂

xx

)e

2

= 0 in Ω

2

× (0, T ), e

2

(x, 0) = 0 in Ω

2

,

(∂

x

−

_2ν^p

)e

2

(0, .) = h

0

in (0, T ).

We have

Fe

1

(x, ω) = 2ν

√ 4iων + p Fh

L

exp(

r iω

ν (x − L)), (2.1.3) and

Fe

2

(x, ω) = − 2ν

√ 4iων + p Fh

0

exp( − r iω

ν x). (2.1.4) We have also

F(g

D

(h

L

, h

0

)) = F((∂

x

e

2

+ p

2ν e

2

)(L, .), (∂

x

e

1

− p

2ν e

1

)(0, .)). (2.1.5) We have

∂

x

Fe

2

(L, ω) + p

2ν Fe

2

(L, ω) = − 2ν 2 √

iων + p ( − r iω

ν ) exp( − r iω

ν L)Fh

0

− 2ν 2 √

iων + p p

2ν exp( − r iω

ν L)Fh

0

= 2 √

iων − p 2 √

iων + p exp( − r iω

ν L)Fh

0

, (2.1.6) and

∂

x

Fe

1

(0, ω) − p

2ν Fe

1

(0, ω) = 2ν 2 √

iων + p r iω

ν exp( − r iω

ν L)Fh

L

− 2ν 2 √

iων + p p

2ν exp( − r iω

ν L)Fh

0

= 2 √

iων − p 2 √

iων + p exp( − r iω

ν L)Fh

L

. (2.1.7)

5

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Therefore

F(g

²_D

(h

L

, h

0

)) = ( 2 √

iων − p 2 √

iων + p )

²

exp( − 2L r iω

ν )F(h

L

, h

0

). (2.1.9) Consequently

| F(g

²_D

(h

L

, h

0

)) | = | ( 2 √

iων − p 2 √

iων + p )

²

exp( − 2L r iω

ν ) || F(h

L

, h

0

) | (2.1.10)

= exp( − L q

2|ω|

ν

)

⁽

√

2|ω|ν−p)²+2|ω|ν (

√

2|ω|ν+p)²+2|ω|ν

| F(h

L

, h

0

) | . Thus for k ∈ N ,

| Fg

^2k_D

(h

L

, h

0

) | (ω) = exp( − kL r 2 | ω |

ν )( ( p

2 | ω | ν − p)

²

+ 2 | ω | ν ( p

2 | ω | ν + p)

²

+ 2 | ω | ν )

^k

| F(h

L

, h

0

) | (ω).

Therefore

|| g

^2k_D

h

L

||

_H³4(R)

=

Z

+∞

−∞

(1 + | ω |

²

)

³⁴

| Fg

^2k_D

h

L

(ω) |

²

dω

=

Z

+∞

−∞

(1 + | ω |

²

)

³⁴

exp( − 2kL r 2 | ω |

ν ) ×

× ( ( p

2 | ω | ν − p)

²

+ 2 | ω | ν ( p

2 | ω | ν + p)

²

+ 2 | ω | ν )

^2k

| Fh

L

(ω) |

²

dω.

Using the Lebesgue dominated convergence theorem with the notice that (

⁽

√

2|ω|ν−p)²+2|ω|ν (

√

2|ω|ν+p)²+2|ω|ν

)

^2k

< 1, we can see that {|| g

^2k_D

h

L

||

_H³4(R)

} converges to 0 when k tends to ∞ . Similarly, {|| g

^2k_D

h

0

||

_H³4(R)

} converges to 0 when k tends to ∞ .

For k ∈ N ,

| Fg

^2k_D

(h

L

, h

0

) | (ω) = exp( − kL r 2 | ω |

ν )( ( p

2 | ω | ν − p)

²

+ 2 | ω | ν ( p

2 | ω | ν + p)

²

+ 2 | ω | ν )

^k

| F(h

L

, h

0

) | (ω).

6

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We define the convergence factor by ρ(ω; p, L) = exp( − L

r 2 ω

ν ) (

_ν^p

− p

2

^ω_ν

)

²

+ 2

^ω_ν

(

_ν^p

+ p

2

^ω_ν

)

²

+ 2

^ω_ν

. (2.1.11) Put

^ω_ν

= ¯ ω,

_ν^p

= ¯ p and ¯ ρ(¯ ω; ¯ p, L) = ρ(ω; p, L), we need to consider the following min-max problem

min

p¯∈R

max

¯

ω∈[¯ωmin,¯ωmax]

ρ(¯ ¯ ω; ¯ p, L), (2.1.12) where ¯ ω

min

=

_{2T ν}^π

, ¯ ω

max

=

_2∆tν^π

. Since

exp( − L √

2¯ ω) ( | p ¯ | − √

2¯ ω)

²

+ 2¯ ω ( | p ¯ | + √

2¯ ω)

²

+ 2¯ ω ≤ exp( − L √

2¯ ω) ( −| p ¯ | − √

2¯ ω)

²

+ 2¯ ω ( −| p ¯ | + √

2¯ ω)

²

+ 2¯ ω , so the minimum is attained for ¯ p ≥ 0, we only need to consider problem (2.1.12) in the case ¯ p ≥ 0, or the following problem

min

p¯≥0

max

¯

ρ(¯ ¯ ω; ¯ p, L). (2.1.13) We have the following theorems for the overlapping case (L > 0)

Theorem 2.1.1. We suppose that L is small and ω ¯

max

is large.

a) For L(¯ ω

max

)

³⁴

small (which means L ∽ C (¯ ω

max

)

⁻³⁴⁻^γ

, γ > 0), problem (2.1.13) has a unique solution

¯

p

_∗

∽ 2 √

2(¯ ω

min

ω ¯

max

)

¹⁴

then

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

∽ 1 − 2 √

2( ω ¯

min

¯ ω

max

)

¹⁴

,

where the asymptotic expansion is based on the scale of (ω

max

)

⁻¹

.

b) For L(¯ ω

max

)

³⁴

large (which means L ∽ C(¯ ω

max

)

⁻³⁴^+γ

, γ > 0) and L <

ν q

10√ 2−12

¯

ωmin

, the problem (2.1.13) has a unique solution

¯

p

_∗

∽ (4¯ ω

min

)

¹³

(L)

⁻¹³

, then

|| ρ(ω, ¯ p ¯

_∗

, L) ||

∞

∽ 1 − 4( ω ¯

min

2 )

¹⁶

(L)

¹³

, where the asymptotic expansion is based on the scale of L.

7

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¯

p

_∗

∽ 2 √ π

(2T C

2

ν

²

)

¹⁴

∆x

⁻¹⁴

, then

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

∽ 1 − 2 √ 2C

1 4

2

2T

¹⁴

∆x

⁴¹

+ O(∆x

¹²

).

For L = C

1

∆x, ∆t = C

2

∆x

²

, ∆x ≤ min

_π²_2T(C1ν⁻¹)⁴ 4C2³

¹₂

,

⁽¹⁰_π(C^√²₁⁻_ν₋^14)2T1)²

¹₂

, then problem (2.1.13) has a unique solution

¯

p

_∗

∽ 4π 2T C

1

ν

¹₃

∆x

⁻¹³

, then

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

∽ 1 − 4 πC

₁²

ν

⁻¹

4T

¹₆

∆x

¹³

+ O(∆x

²³

).

Remark 2.1.1.

8

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Figure 2.1.1

Figure 2.1.2

9

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the two boundaries to get p ¯

_∗

(figure 2.2).

For the non-overlapping case, we have the following result

Theorem 2.1.3. Problem (2.1.13) has one and only one solution which is given by ω ¯ = ¯ ω

min

and ω ¯ = ¯ ω

max

; p ¯ = 2

^√^π

(2T ν²)¹⁴

∆t

⁻¹⁴

. min

p¯

max

¯

ω

ρ ¯ ∽ 1 − ( 32C

1

T )

¹⁴

∆t

¹⁴

.

10

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2.1.1 Proof of the Theorems in the Overlapping Case

Putting

h

L

(¯ p) = max

¯

ρ(¯ ¯ ω, p, L) = ¯ || ρ(¯ ω; ¯ p, L) ||

∞

,

we recall that (¯ p

^∗

, h

L

(¯ p

^∗

)) is a strict local minimum of h

L

(¯ p) if and only if there exists ǫ positive such that for all ¯ p in (¯ p

^∗

− ǫ, p ¯

^∗

+ ǫ), we have h

L

(¯ p) >

h

L

(¯ p

^∗

).

In order to prove the theorems, we need the following lemma as in [1]

Lemma 2.1.1. If (¯ p

^∗

, h

L

(¯ p

^∗

)) is a strictly local minimum of h

L

(¯ p), then it is the global minimum of h

L

(¯ p) and p ¯

^∗

is the unique solution of (2.1.13).

Proof of Lemma 2.1.1

We denote D (z

0

, δ) = { z ∈ C , |

_z+z^z⁻^z⁰0

| < δ } , and D

_δ^L

= { p | h

L

(p) ≤ δ } . We first prove that D

^L_δ

is a convex set. Let ¯ p

1

and ¯ p

2

be to elements of D

^L_δ

, we have that

|| exp( − L √ i ω) ¯

¯ p1

ν

− 2 √ i¯ ω

¯

p

1

+ 2 √

i ω ¯ ||

∞

≤ δ.

Thus ∀ ω ¯ ∈ [¯ ω

min

, ω ¯

max

],

| exp( − L √

i¯ ω) p ¯

1

− 2 √ i¯ ω

¯ p

1

+ 2 √

i ω ¯ | ≤ δ.

Hence

exp( − L r ω ¯

2 ) | p ¯

1

− 2 √ i ω ¯

¯

p

1

+ 2 √

i ω ¯ | ≤ δ.

Therefore

| p ¯

1

− 2 √ i¯ ω

¯ p

1

+ 2 √

i¯ ω | ≤ δ exp(L r ω ¯

2 ).

This means ¯ p

1

∈ D (2 √

i¯ ω, δ exp(L q

¯ ω 2

)).

Similarly, we have also ¯ p

2

∈ D (2 √

i ω, δ ¯ exp(L q

¯ ω

2

)).

According to Lemma 2.1 in [1], D (z

0

, δ) is the interior of the circle with center at

^1+δ₁₋_δ²²

z

0

and radius

_|₁₋^2δ_δ²_|

| z

0

| and the exterior otherwise.

11

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If δ exp(L

^ω^¯₂

)) ≥ 1, using Lemma 2.1 in [1], we can see that for ¯ p

1

, ¯ p

2

≥ 0, θ ∈ [0, 1], we have θ p ¯

1

+ (1 − θ) ¯ p

2

∈ D (2 √

iω, δ exp(L q

¯ ω

2

)). Thus for θ ∈ [0, 1], we have θ p ¯

1

+ (1 − θ) ¯ p

2

∈ D

_δ^L

.

Therefore D

_δ^L

is convex.

Suppose that (¯ p

^∗

, h

L

(¯ p

^∗

)) is a strictly local minimum of h

L

(¯ p), we prove that it is a global minimum of h

L

(¯ p). Suppose the contrary that there exists (¯ p

^∗∗

, h

L

(¯ p

^∗∗

)) such that h

L

(¯ p

^∗

) ≥ h

L

(¯ p

^∗∗

). Then there exists a convex neighborhood U of ¯ p

^∗

, such that ∀ s ∈ U , s 6 = ¯ p

^∗

and h

L

(s) > h

L

(¯ p

^∗

). Since

¯

p

^∗∗

∈ D

_h^L_L_(¯_p∗∗)

⊂ D

^L_h_L_(¯_p∗)

,we have that ∀ θ ∈ [0, 1], θ p ¯

^∗

+ (1 − θ)¯ p

^∗∗

∈ D

^L_h_L_(¯_p∗)

. For θ small enough, we have that θ p ¯

^∗∗

+ (1 − θ)¯ p

^∗

∈ U . This is a contradic- tion.

Thus ¯ p

^∗

is the unique solution of (2.1.13).

12

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Proof of theorem 2.1.1

Case 1: For L(¯ ω

max

)

³⁴

large and L < q

10√ 2−12

¯ ωmin

.

Firstly, we will prove that || ρ(¯ ¯ ω, p, L) ¯ ||

∞

= max { ρ(¯ ¯ ω

min

, p, L), ρ(¯ ¯ ω

2

, p, L) ¯ } when ¯ p is closed enough to ¯ p

_∗

= (4¯ ω

min

)

¹³

L

⁻¹³

.

We have that

¯

p

ω¯

(¯ ω, p, L) = ¯ −

√ 2

2 exp( − L √

2¯ ω) 16L ω ¯

²

− 16¯ p ω ¯ + L¯ p

⁴

+ 4¯ p

³

(4¯ ω + 2 √

2¯ ω p ¯ + ¯ p

²

)

²

. We consider the function:

f(¯ ω) = 16L ω ¯

²

− 16¯ p¯ ω + L¯ p

⁴

+ 4¯ p

³

.

This is a quadratic equation in ¯ ω. We will prove that ∆

^′

= 64¯ p

²

− 16L(L p ¯

⁴

+ 4¯ p

³

) = 16¯ p

²

(4 − 4L p ¯ − L

²

p ¯

²

) > 0. Since ¯ p is closed to ¯ p

_∗

, we only need to prove that 4 − 4L¯ p

_∗

− L

²

p ¯

²_∗

> 0, or L¯ p

_∗

< 2 √

2 − 2.

Since L <

q

10√ 2−12

¯

ωmin

, we have that

L¯ p

_∗

= L(4¯ ω

min

)

¹³

(L)

⁻¹³

= (4¯ ω

min

)

¹³

L

²³

< (4¯ ω

min

)

¹³

( 10 √ 2 − 14

¯ ω

min

)

¹³

= 2 √ 2 − 2.

Thus ∆

^′

> 0.

Therefore the equation f = 0 has the following solutions:

¯

ω

1

= 2¯ p − p ¯ p

4 − 4L¯ p − L

²

p ¯

²

4L = 2¯ p − p ¯ p

4 − 4L p ¯ − L

²

p ¯

²

4L ,

¯

ω

2

= 2¯ p + ¯ p p

4 − 4L¯ p − L

²

p ¯

²

4L = 2¯ p + ¯ p p

4 − 4L¯ p − L

²

p ¯

²

4L .

We will prove that in this case ¯ ω

max

> ω ¯

2

(¯ p). In order to do that, we only need to prove that ¯ ω

max

> ω ¯

2

(¯ p

_∗

).

Since L(¯ ω

max

)

³⁴

is large, then L > (

_2¯^ω^¯_ω^min3

max

)

¹⁴

, we have L

⁴

> ω ¯

min

2¯ ω

_max³

. Thus

¯

ω

_max³

> ω ¯

min

2 L

⁻⁴

.

13

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This means that in order to prove ¯ ω

max

> ω ¯

2

(¯ p

_∗

), we only have to prove that

16L¯ ω

²_max

− 16¯ p

_∗

ω ¯

max

+ L¯ p

⁴_∗

+ 4¯ p

³_∗

> 0.

This inequality is equivalent to

16L¯ ω

²_max

− 16(4¯ ω

min

)

¹³

L

⁻¹³

ω ¯

max

+ L(4¯ ω

min

)

⁴³

(L)

⁻⁴³

+ 16¯ ω

min

L

⁻¹

> 0, or

¯

ω

_max²

L

²³

)

³

− 2( ω ¯

min

2 )

¹³

ω ¯

max

L

²³

+ ( ω ¯

min

2 )

⁴³

L

²³

+ ¯ ω

min

=

= ¯ ω

max

L

²³

[¯ ω

max

L

⁴³

− 2( ω ¯

min

2 )

¹³

] + ( ω ¯

min

2 )

⁴³

+ ¯ ω

min

> 0.

This is true.

Hence ¯ ω

max

> ω ¯

2

(¯ p

_∗

) and ¯ ω

max

> ω ¯

2

(¯ p) for ¯ p closed enough to ¯ p

_∗

. Therefore || ρ(¯ ¯ ω, p, L) ¯ ||

∞

= max { ρ(¯ ¯ ω

min

, p, L), ¯ ρ(¯ ¯ ω

2

, p, L) ¯ } .

Next, we will prove that ¯ p

_∗

is an asymptotic solution to the equation

¯

ρ(¯ ω

min

, p, L) = ¯ ¯ ρ(¯ ω

2

, p, L). Let ¯ p is a number closed to ¯ p

_∗

, and suppose that

¯

p has the form ¯ p ∽ C(

^L_ν

)

⁻^γ

, we have that

¯

ρ(¯ ω

min

, p, L) = exp( ¯ − L √

2¯ ω

min

) ( √

2¯ ω

min

− CL

⁻^γ

)

²

+ 2¯ ω

min

( √

2¯ ω

min

+ CL

⁻^γ

)

²

+ 2¯ ω

min

= exp( − L √

2¯ ω

min

) 4¯ ω

min

− 2 √

2¯ ω

min

CL

⁻^γ

+ C

²

L

⁻^2γ

4¯ ω

min

+ 2 √

2¯ ω

min

CL

⁻^γ

+ C

²

L

⁻^2γ

= exp( − L ν

√ 2ω

min

ν)

4ωminν

C²

(

^L_ν

)

^2γ

−

²^√^2¯_C^ω^min

L

^γ

+ 1

4¯ωmin

C²

L

^2γ

+

²^√^2¯_C^ω^min

L

^γ

+ 1

∽ 1 − L √

2¯ ω

min

− 4 √ 2¯ ω

min

C L

^γ

+ 16¯ ω

min

C

²

L

^2γ

. We also have that

¯

ρ(¯ ω

2

, p, L) = exp( ¯ − L √

2¯ ω

2

) ( √

2¯ ω

2

− CL

⁻^γ

)

²

+ 2¯ ω

2

( √

2¯ ω

2

+ CL

⁻^γ

)

²

+ 2¯ ω

2

.

14

(17)

We have

exp( − L √

2¯ ω

2

) = exp( − s

2L

²

2¯ p + ¯ p p

4 − 4L¯ p − L

²

p ¯

²

4L )

= exp( − s

2¯ pL + ¯ pL p

4 − 4L¯ p − L

²

p ¯

²

2 )

∽ 1 − p

2¯ pL ∽ 1 − √

2CL

¹⁻²^γ

, and

( √

2¯ ω

2

− p) ¯

²

+ 2¯ ω

2

ν ( √

2¯ ω

2

+ ¯ p)

²

+ 2¯ ω

2

= 1 −

^√_2¯^p^¯_ω2

+

_4¯^p^¯_ω²₂

1 +

^√_2¯^p^¯_ω₂

+

_4¯^p^¯_ω²₂

∽ 1 − 2¯ p

√ 2¯ ω

2

. Moreover, we have that

¯

√ p 2¯ ω

2

= p ¯

q

2¯p+¯p

√

4−4L¯p−L²p¯² 2L

=

s 2¯ pL 2 + p

4 − 4¯ pL − p ¯

²

L

²

∽ r C

2 L

¹⁻²^γ

. Therefore

( √

2¯ ω

2

− p) ¯

²

+ 2¯ ω

2

( √

2¯ ω

2

+ ¯ p)

²

+ 2¯ ω

2

∽ 1 − √

2CL

¹^−γ²

+ CL

¹⁻^γ

. Thus

¯

ρ(¯ ω

2

, p, L) ¯ ∽ (1 − √

2CL

¹^−γ²

+ 2CL

¹⁻^γ

)(1 − √

2CL

¹^−γ²

+ CL

¹⁻^γ

∽ 1 − 2 √

2CL

¹⁻²^γ

+ 5CL

¹⁻^γ

.

Equilibrate ¯ ρ(¯ ω

2

, p, L) and ¯ ¯ ρ(¯ ω

min

, p, L), we get ¯ ¯ p

_∗

.

Finally, we prove that this ¯ p

_∗

is a stricly local minimum of || ρ(¯ ω, p, L) ¯ ||

∞

. We have that

∂

p¯

¯

ρ(¯ ω

2

, p ¯

_∗

, L) = 4 √ 2 √

¯

ω

2

exp ( − L √ 2 √

¯ ω

2

) (¯ p

²_∗

+ 2 √

2 √

¯

ω

2

ν p ¯

_∗

+ 4¯ ω

2

ν)

²

(¯ p

²_∗

− 4¯ ω

2

) ω ¯

2

¯ p < 0, and

∂

p¯

¯

ρ(¯ ω

min

, p ¯

_∗

, L) = 4 √ 2 √

¯

ω

min

exp ( − L √ 2 √

¯ ω

min

) (¯ p

²_∗

+ 2 √

2 √

¯

ω

min

p ¯

_∗

+ 4¯ ω

min

)

²

(¯ p

²_∗

− 4¯ ω

min

) > 0.

15

(18)

cording to Lemma 2.1.1 it is also the global minimum. And

¯

p

_∗

∽ (4¯ ω

min

)

¹³

L

⁻¹³

,

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

∽ 1 − 4( ω ¯

min

2 )

¹⁶

L

¹³

. Case 2: L¯ ω

3

max4

is small

In this case, we can see that ¯ ω

2

> ω ¯

max

. Thus

|| ρ(¯ ¯ ω, p, L) ¯ ||

∞

= max { ρ(¯ ¯ ω

min

, p, L), ¯ ρ(¯ ¯ ω

max

, p, L) ¯ } .

As in the previous case, we will prove that ¯ p

_∗

is a solution of the equation

¯

ρ(¯ ω

min

, p, L) = ¯ ¯ ρ(¯ ω

max

, p, L). ¯

Let ¯ p be a number closed enough to ¯ p

_∗

. We have that

¯

ρ(¯ ω

min

, p, L) = exp( ¯ − L √

2¯ ω

min

) ( √

2¯ ω

min

− p) ¯

²

+ 2¯ ω

min

( √

2¯ ω

min

+ ¯ p)

²

+ 2¯ ω

min

ν

∽ (1 − L √

2¯ ω

min

+ L

²

2¯ ω

min

)(1 − 4

√ 2¯ ω

min

¯

p + 16¯ ω

min

¯ p

²

).

Since L¯ ω

3

max4

is small, then L < (

_2¯^ω^¯_ω^min³

max

)

¹⁴

, we have that L √

2¯ ω

min

< ( ω ¯

min

2¯ ω

_max³

)

¹⁴

√

2¯ ω

min

= 2

¹⁴

( ω ¯

min

¯ ω

max

)

³⁴

,

and r

¯ ω

min

¯ p

²

∽

r ω ¯

min

4 √

¯

ω

min

ω ¯

max

= 1 2 ( ω ¯

min

¯ ω

max

)

¹⁴

. Therefore

¯

ρ(¯ ω

min

, p, L) ¯ ∽ 1 − 4 √ 2¯ ω

min

¯

p .

16

(19)

We can suppose that 4¯ ω

min

< ω ¯

max

, then 2(¯ ω

min

ω ¯

max

)

¹⁴

< √

2¯ ω

max

. Thus

¯ p < √

2¯ ω

max

. Hence

¯

ρ(¯ ω

max

, p, L) = exp( ¯ − L √

2¯ ω

max

) ( √

2¯ ω

max

− p) ¯

²

+ 2¯ ω

max

( √

2¯ ω

max

+ ¯ p)

²

+ 2¯ ω

max

= exp( − L √

2¯ ω

max

) (1 −

^√_2¯_ω^p^¯_max

)

²

+ 1 (1 +

^√_2¯_ω^p^¯

max

)

²

+ 1

= exp( − L √

2¯ ω

max

) 1 −

^√_2¯_ω^p^¯_max

+

_4¯_ω^p^¯²

max

1 +

^√_2¯_ω^p^¯

max

+

_4¯_ω^p^¯²

max

∽ (1 − L √

2¯ ω

max

+ L

²

2¯ ω

max

)(1 − 2¯ p

√ 2¯ ω

max

+ p ¯

²

¯ ω

max

).

Since L ∽ C ω ¯

⁻

3 4+γ max

, then L √

2¯ ω

max

< C ( ω ¯

min

2¯ ω

_max³

)

⁴¹

√

2¯ ω

max

ω ¯

_max^γ

= C (¯ ω

min

)

¹⁴

2

¹⁴

ω ¯

⁻

1 4+γ max

. We have also

¯ p

²

¯

ω

max

∽ 4 √

¯

ω

min

ω ¯

max

¯ ω

max

= 4

r ω ¯

min

¯ ω

max

. Therefore

¯

ρ(¯ ω

max

, p, L) ¯ ∽ 1 − 2¯ p

√ 2¯ ω

max

.

Equilibrate the two asymptotic expansion ¯ ρ(¯ ω

max

, p, L) and ¯ ¯ ρ(¯ ω

2

, p, L) we ¯ have the equation

2¯ p

√ 2¯ ω

max

∽ 4 √ 2¯ ω

min

¯

p .

Thus ¯ p ∽ 2(¯ ω

min

ω ¯

max

)

¹⁴

or ¯ p

_∗

is an asymptotic solution of the equation

¯

ρ(¯ ω

max

, p, L) = ¯ ¯ ρ(¯ ω

2

, p, L). Using the same argument as in the previous ¯ section we have that this ¯ p

_∗

is a global minimum of || ρ(¯ ¯ ω, p, L) ¯ ||

∞

. And

|| ρ(¯ ¯ ω

min

, p ¯

_∗

, L) ||

∞

∽ 1 − 2¯ p

_∗

√ 2¯ ω

max

∽ 1 − 4(¯ ω

min

ω ¯

max

)

¹⁴

√ 2¯ ω

max

∽ 1 − 2 √

2( ω ¯

min

¯ ω

max

)

¹⁴

.

17

(20)

¯

p is closed to ¯ p

_∗

. We have that

∂

ω¯

p(¯ ¯ ω; ¯ p, L) = −

√ 2

2 exp( − L √

2¯ ω) 16L¯ ω

²

− 16¯ p¯ ω + L p ¯

⁴

+ 4¯ p

³

(4¯ ω + 2 √

2¯ ω p ¯ + ¯ p

²

)

²

.

The function 16L¯ ω

²

− 16¯ p ω ¯ + L¯ p

⁴

+ 4¯ p

³

, is a quadratic function of ¯ ω and it has ∆

^′

(¯ p) = 16¯ p

²

(4 − 4L¯ p − L

²

p ¯

²

). We will prove that ∆

^′

(¯ p) > 0. Since ¯ p is closed to ¯ p

_∗

, we only need to prove that ∆

^′

(¯ p

_∗

) > 0. We have that

∆x < (17 − 12 √

2)2T C

2

π

²

(C

1

ν

⁻¹²

)

⁴

¹₃

. This implies

∆x < ( √

2 − 1)

⁴³

(2T C

2

)

¹³

( √

πC

1

ν

⁻¹²

)

⁴³

. Therefore

2 √

πνC

1

ν

⁻¹

(2T C

2

)

¹⁴

∆x

³⁴

< 2 √ 2 − 2, or

L¯ p

_∗

< 2 √ 2 − 2.

Hence ∆

^′

(¯ p

_∗

) > 0.

Therefore the equation f = 0 has the following solutions:

¯

ω

1

= 2¯ p − p ¯ p

4 − 4L¯ p − L

²

p ¯

²

4L = 2¯ p − p ¯ p

4 − 4L p ¯ − L

²

p ¯

²

4L ,

¯

ω

2

= 2¯ p + ¯ p p

4 − 4L¯ p − L

²

p ¯

²

4L = 2¯ p + ¯ p p

4 − 4L¯ p − L

²

p ¯

²

4L .

We prove that for ¯ p closed to ¯ p

_∗

, we also have

_2L^p^¯

>

_∆tν^π

, which implies

¯

ω

2

>

_∆tν^π

. In fact, we only need to prove that

_2L^p^¯^∗

>

_∆tν^π

. Since

∆x < C

₂³

π

²

(C

1

ν

⁻¹²

)

⁴

2T ,

18

(21)

we have

2√ πν

(2T C2)¹⁴

∆x

⁻¹⁴

2C

1

∆x > π C

2

∆x , or

_2L^p^¯

>

_∆tν^π

.

Therefore

|| ρ(¯ ¯ ω, p, L) ¯ ||

∞

= max { ρ( ¯ π

2T ν ; ¯ p, L), ρ( ¯ π

∆tν ; ¯ p, L) } .

Using the same argument as in Theorem 2.1.1, we have that problem (2.1.13) has a unique solution

¯

p

_∗

∽ 2 √ π

(2T C

2

)

¹⁴

∆x

⁻¹⁴

, then

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

= 1 − 2 √ 2C

1 4

2

(2T )

¹⁴

∆x

⁴¹

+ O(∆x

¹²

).

Case 2: L = C

1

∆x, ∆t = C

2

∆x

²

, ∆x ≤ min

_π²_2T(C1ν⁻¹)⁴ 4C₂³

¹₂

,

⁽¹⁰_π(C^√²₁⁻_ν₋^14)2T1)²

¹₂

. Firstly, we prove that || ρ( ¯

_{2T ν}^π

, p, L) ¯ ||

∞

= max { ρ( ¯

_{2T ν}^π

, p, L), ¯ ρ(¯ ¯ ω

2

, p, L) ¯ } when ¯ p is closed to ¯ p

_∗

.

We have that

∂

ω¯

p(¯ ¯ ω, p, L) = ¯ −

√ 2

2 exp( − L √

2¯ ω) 16L¯ ω

²

− 16¯ p¯ ω + L p ¯

⁴

+ 4¯ p

³

(4¯ ω + 2 √

2¯ ω p ¯ + ¯ p

²

)

²

.

Similar as in the previous case, we prove that ∆

^′

(¯ p

_∗

) = 16¯ p

²_∗

(4 − 4L¯ p

_∗

− L

²

p ¯

²_∗

) > 0.

We have that

∆x ≤ (10 √

2 − 14)2T π(C

1

ν

⁻¹

)

²

¹₂

. This leads to

4π(C

1

ν

⁻¹

)

²

2T ∆x

²

< 8(5 √

2 − 7).

Thus 4π(C

1

ν

⁻¹

)

²

2T ∆x

²

< 8(5 √

2 − 7) = 8( √

2 − 1)

³

.

19

(22)

We prove that in this case, ¯ ω

2

<

_∆tν^π

. We have

∆x ≤ π

²

2T (C

1

ν

⁻¹

)

⁴

4C

₂³

¹₂

. Hence

2π

T (C

1

ν

⁻¹

)

⁴

∆x

²

≤ π

³

C

₂³

. Thus

(

_{2T C}^4πν²₁

)

¹³

∆x

⁻¹³

C

1

∆x ≤ π

C

2

∆x

⁻²

. Therefore

_L^p^¯

≤

∆tν^π

, which means ¯ ω

2

<

_∆tν^π

. Therefore

|| ρ(¯ ¯ ω, p, L) ¯ ||

∞

= max { ρ( ¯ π

2T ν ; ¯ p, L), ρ(¯ ¯ ω

2

; ¯ p, L) } .

Equilibrate ¯ ρ(

_{2T ν}^π

, p, L) and ¯ ¯ ρ(¯ ω

2

, p, L), we get ¯ ¯ p

_∗

and using the same argument of the previous case, we can conclude that problem (2.1.13) has a unique solution

¯

p

_∗

∽ 2π T C

1

¹₃

∆x

⁻¹³

, then

|| ρ(¯ ¯ ω, p ¯

_∗

, L) ||

∞

= 1 − 4 πC

₁²

4T

¹6

∆x

³¹

+ O(∆x

²³

).

2.1.2 Proof of the Theorems in the Non-Overlapping Case

Proof of Theorem 2.1.3

Since ω ∈ [

_2T^π

,

_∆t^π

], then x =

^√^2ων_p

belongs to [ q

2πν p²2T

, q

2πν p²∆t

].

Thus

ρ(ω) = f(x) = 2x

²

− 2x + 1

2x

²

+ 2x + 1 . (2.1.14)

20

(23)

We have

f

^′

(x) = 4(2x

²

− 1)

(2x

²

+ 2x + 1)

²

. (2.1.15) We have the following cases

Case 1: q

2νπ

p²2T

≥

^√¹₂

. We have that

max

ω

ρ = f (

r 2πν

p

²

∆t ), (2.1.16)

Since q

2νπ

p²2T

≥

^√¹₂

, we have that q

2νπ

p²∆t

≥

^√¹₂

q

2T

∆t

>

√¹ 2

. Thus

min

p

max

ω

ρ = f ( 1

√ 2 r 2T

∆t ), (2.1.17)

when ω =

_∆t^π

and p = 2 p

_νπ

2T

. Case 2: q

2νπ

p²∆t

≤

^√¹₂

. We have that

max

ω

ρ = f ( r πν

p

²

T ), (2.1.18)

Since q

2νπ

p²∆t

≤

^√¹₂

, we have that q

_νπ

p²T

≤

^√¹₂

q

∆t

2T

<

^√¹₂

. Thus

min

p

max

ω

ρ = f ( 1

√ 2 r ∆t

2T ) = f ( 1

√ 2 r 2T

∆t ), (2.1.19) when ω =

_2T^π

and p = 2 p

_νπ

∆t

. Case 3: q

2νπ p²∆t

>

√¹

2

> q

2νπ p²2T

.

We can see that if x > y and 2xy > 1, f(x) > f (y); and if x > y and 2xy < 1, f (x) < f (y); and if x > y and 2xy = 1, f(x) = f (y).

max

ω

ρ = max { f(

r 2πν p

²

2T ), f (

r 2πν

p

²

∆t ) } . (2.1.20) Case 3.1: q

2νπ p²∆t

. q

2νπ

p²2T

≥

¹₂

, or q

2νπ p²∆t

. q

2νπ

p²∆t

≥

¹₂

q

2T

∆t

, or q

2νπ p²∆t

≥

√1

2

(

^2T_∆t

)

¹⁴

>

√¹ 2

.

max

ω

ρ = f (

r 2πν

p

²

∆t ). (2.1.21)

21

(24)

when ω =

_∆t

and p = 2

(∆t2T)⁴¹

. Case 3.2: q

2νπ p²∆t

. q

νπ

p²T

≤

¹₂

, or q

νπ p²T

. q

νπ

p²T

≤

¹₂

q

∆t

2T

, or q

νπ p²T

≤

√1

2

(

^∆t_2T

)

¹⁴

<

^√¹₂

.

max

ω

ρ = f ( r πν

p

²

T ). (2.1.23)

Thus min

p

max

ω

ρ = f ( 1

√ 2 ( ∆t

2T )

¹⁴

) = f ( 1

√ 2 ( 2T

∆t )

¹⁴

) < f ( 1

√ 2 r 2T

∆t ), (2.1.24) when ω =

_2T^π

and p = 2

^√^νπ

(∆t2T)¹⁴

.

22

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2.2 Optimized Schwarz Waveform Relaxation Methods For One Dimensional Heat Equa- tion With First Order Transmission Con- dition

In this chapter, we consider the following algorithm



 



 



(∂

t

− ν∂

xx

)u

^k₁

= f in Ω

1

× (0, T ),

u

^k₁

(x, 0) = u

0

(x) in Ω

1

,

(∂

x

+

_2ν^p

+ 2q∂

t

)u

^k₁

(L, .) = (∂

x

+

_2ν^p

+ 2q∂

t

)u

^k₂⁻¹

(L, .) in (0, T ),

(2.2.1)



 



 



(∂

t

− ν∂

xx

)u

^k₂

= f in Ω

2

× (0, T ),

u

^k₂

(x, 0) = u

0

(x) in Ω

2

,

(∂

x

−

_2ν^p

− 2q∂

t

)u

^k₂

(0, .) = (∂

x

−

_2ν^p

− 2q∂

t

)u

^k₁⁻¹

(0, .) in (0, T ).

Similar as in the previous chapter, we consider the following problem



 



 



∂

t

e

1

− ν∂

xx

e

1

= 0 in Ω

1

× (0, T ),

e

1

(x, 0) = u

0

(x) in Ω

1

,

(∂

x

+

_2ν^p

+ 2q∂

t

)e

^k₁

(L, .) = (∂

x

+

_2ν^p

+ 2q∂

t

)e

2

(L, .) in (0, T ),

(2.2.2)



 



 



∂

t

e

2

− ν∂

xx

e

2

= 0 in Ω

2

× (0, T ),

e

2

(x, 0) = u

0

(x) in Ω

2

,

(∂

x

−

_2ν^p

− 2q∂

t

)e

2

(0, .) = (∂

x

−

_2ν^p

− 2q∂

t

)e

1

(0, .) in (0, T ).

From (2.2.2), we have that

iωFe

1

− ν∂

xx

Fe

1

= 0.

Therefore

Fe

1

= C

1

exp(

r iω

ν x) + C

2

exp( − r iω

ν x),

23

(26)

Fe

1

= C

1

exp( iω ν x).

From (2.2.2), we have that

∂

x

Fe

1

(L, ω) + p

2ν e

₁

(L, ω) + 2qiωFe

1

(L, ω) = Fh

L

(ω).

Thus

(C

1

r iω ν + C

1

p

2ν + C

1

2qωi) exp(

r iω

ν L) = Fh

L

(ω).

Hence

C

1

√ 4νωi + p + 4qωνi

2ν = Fh

L

exp( − r iω

ν L).

Thus

C

1

= 2ν

√ 4νωi + p + 4qωνi Fh

L

exp( − r iω

ν L).

Therefore

Fe

1

= 2ν

√ 4νωi + p + 4qωνi Fh

L

exp exp(

r iω

ν (x − L)).

From (2.2.2), we have that

iωFe

2

− ν∂

xx

Fe

2

= 0.

Therefore

Fe

2

= D

1

exp(

r iω

ν x) + D

2

exp( − r iω

ν x), where Re( q

iω ν

) ≥ 0.

Since x ∈ (0, ∞ ) and Fe

2

(x, .) ∈ L

²

( R ), we have D

1

= 0. Thus Fe

2

= D

2

exp( −

r iω ν x).

24

(27)

From (2.2.2), we have that

∂

x

Fe

2

(0, ω) − p

2ν Fe

2

(0, ω) − 2q

ν iωFe

2

(0, ω) = h

0

(ω).

Thus

(D

2

r iω ν − D

2

p

2ν − D

2

2qωi) = Fh

0

(ω).

Hence

D

2

√ 4νωi − p − 4qωi

2 = Fh

0

.

Thus

D

2

= 2ν

√ 4νωi − p − 4qωνi Fh

0

. Therefore

Fe

2

= 2ν

√ 4νωi − p − 4qωνi Fh

0

exp(

r

− iω ν x).

Similar as in the previous chapter, we can define the convergence factor as

ρ(ω, p, q, L) = | 2 √

iων − p − 4qωνi 2 √

iων + p + 4qωνi exp( − √ iων L

ν ) |

²

.

Put ¯ ω =

^ω_ν

, ¯ p =

^p_ν

and ¯ ρ(¯ ω, p, q, L) = ¯ ρ(ω, p, q, L), we need to solve the problem

min

¯

p,q∈R

max

¯

ρ(¯ ¯ ω, p, q, L). ¯ (2.2.3) Similar as in the previous chapter, we only need to solve the following problem

min

¯

p,q≥0

max

¯

ρ(¯ ¯ ω, p, q, L). ¯ (2.2.4) We have the following theorems for the Overlapping Case

Theorem 2.2.1. If we fix ω ¯

min

and ω ¯

max

, then for L small satisfying that L

⁸⁵

ω ¯

max

is small and L¯ ω

max

is not small, the problem (2.2.4) has a unique solution

¯ p

_∗

∽ √

2¯ ω

3 8

min

ω ¯

1

max8

,

25

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Theorem 2.2.2.

Case 1: For ∆x small enough L = C

1

∆x, ∆t = C

2

∆x, ω ¯ ∈ [

_{2T ν}^π

,

_C^π₂_ν

∆x

⁻¹

].

There exists a unique pair (¯ p

_∗

, q

_∗

) such that min

p,q¯ ≥0

max

_ω_¯_∈_[ ^π

2T ν,_∆tν^π ]

ρ(¯ ¯ ω, p, q, L) = ¯ max

ω¯∈[_{2T ν}^π ,_∆tν^π ]

ρ(¯ ¯ ω, p ¯

_∗

, q

_∗

, L). Then

¯ p

_∗

∽ √

2( π

⁴

T

³

C

2

ν

⁴

)

¹⁸

∆x

⁻¹⁸

, q

_∗

∽ √

2( π

⁴

ν

⁴

T C

₂³

)

⁻¹⁸

∆x

⁸³

, min

¯

p,q≥0

max

¯

ω∈[_{2T ν}^π ,_∆tν^π ]

ρ(¯ ¯ ω, p, q, L) ¯ ∽ 1 − 4( C

2

2T )

¹⁸

∆x

¹⁸

+ O(∆x

¹⁴

).

Case 2: For ∆x small enough L = C

1

∆x, ∆t = C

2

∆x

²

, ω ¯ ∈ [

_{2T ν}^π

,

_C₂^π_ν4

∆x

⁻²

].

There exists a unique pair (¯ p

_∗

, q

_∗

) such that min

p,q¯ ≥0

max

ω∈[_{2T ν}^π ,_∆tν^π ]

ρ(¯ ¯ ω, p, q, L) = ¯ max

_ω_¯_∈_[ ^π

2T ν,_∆tν^π ]

ρ(¯ ¯ ω, p ¯

_∗

, q

_∗

, L). Then

¯

p

_∗

∽ (4 π

²

ν

²

4T

²

(C

1

ν

⁻¹

)

⁻¹

)

¹⁵

∆x

⁻¹⁵

, q

_∗

∽ (16(C

1

ν

⁻¹

)

³

2T

πν )

¹⁵

∆x

⁵³

, min

¯

p,q≥0

max

¯

ω∈[_{2T ν}^π ,_∆tν^π ]

ρ(¯ ¯ ω, p, q, L) ¯ ∽ 1 − 2

²¹¹⁰

( πν

2T )

¹⁰¹

(C

1

ν

⁻¹

)

¹⁵

∆x

¹⁵

+ O(∆x

²⁵

).

Remark 2.2.1.

26

(29)

Figure 2.2.1.

Figure 2.2.2.

27

(30)

maximum point ω ¯

2

, ω ¯

4

of ρ) on the graph. In the first case ¯ ω ¯

4

> ω ¯

max

, we equilibrate the two boundaries and the maximal point ω ¯

2

to get (¯ p

_∗

, q

_∗

).

We have the following theorem for the nonoverlapping case Theorem 2.2.3. The equation (2.2.4) has a unique solution

¯ p

_∗

= √

2¯ ω

3 8

min

ω ¯

1

max8

= √

2( πν

⁻¹

2T )

³⁸

(πν

⁻¹

)

¹⁸

∆t

⁻¹⁸

, and

q

_∗

= √

2(ω

min

ν)

⁻¹⁸

(ω

max

ν)

⁻³⁸

= √ 2( πν

2T )

⁻¹⁸

(πν)

⁻³⁸

∆t

³⁸

. Then, we have that

¯

min

p,q≥0

max

¯

ρ(¯ ω, p, q) = ¯ max

¯

ρ(¯ ω, p ¯

_∗

, q

_∗

) ∽ 1 − 4¯ ω

1 8

min

ω ¯

⁻

1

max8

∽ 1 − 4( 1

2T )

¹⁸

∆t

¹⁸

.

2.2.1 Proof of the Theorems in the Overlapping Case

In this section, we will consider the problem of optimizing (¯ p, q) in the overlapping case.

Putting

h

L

(¯ p, q) = max

¯

ρ(¯ ¯ ω, p, q, L) = ¯ || ρ(¯ ¯ ω, p, q, L) ¯ ||

∞

,

we call that (¯ p

^∗

, q

^∗

, h

L

(¯ p

^∗

, q

^∗

)) is a strictly local minimum of h

L

(¯ p, q) if and only if there exists ǫ

1

, ǫ

2

positive such that for all (¯ p, q) in (¯ p

^∗

− ǫ

1

, p ¯

^∗

+ ǫ

1

) × (q

^∗

− ǫ

2

, q

^∗

+ ǫ

2

), we have h

L

(¯ p, q) < h

L

(¯ p

^∗

, q

^∗

).

In order to prove the theorems, we need the following lemma:

Lemma 2.2.1. If (¯ p

^∗

, q

^∗

, h

L

(¯ p

^∗

, q

^∗

)) is a strictly local minimum of h

L

(¯ p, q), then it is the global minimum of h

L

(¯ p, q) and (¯ p

^∗

, q

^∗

) is the unique solution of (2.2.4).

28

(31)

Proof of Lemma 2.2.1

We denote D (z

0

, δ) = { z ∈ C , |

^zz+z⁻^z⁰0

| < δ } , and D

_δ^L

= { (¯ p, q) | h

L

(¯ p, q) ≤ δ } . We first prove that D

^L_δ

is a convex set. Let (¯ p

1

, q

1

) and (¯ p

2

, q

2

) be to elements of D

^L_δ

, we have that

|| 2 √

i¯ ω − p ¯ − q ωi ¯ 2 √

i ω ¯ + ¯ p + q ωi ¯ exp( − √ i ω ¯ L

ν ) ||

∞

≤ √ δ.

Thus ∀ ω ¯ ∈ [¯ ω

min

, ω ¯

max

],

| 2 √

i ω ¯ − p ¯ − q ωi ¯ 2 √

i¯ ω + ¯ p + q ωi ¯ exp( − r ω ¯

2 L) | ≤ √ δ.

Hence

| 2 √

i ω ¯ − p ¯ − q ωi ¯ 2 √

i¯ ω + ¯ p + q ωi ¯ | exp( − r ω ¯

2 L) ≤ √ δ.

Therefore

| 2 √

i ω ¯ − p ¯ − q ωi ¯ 2 √

i ω ¯ + ¯ p + q ωi ¯ | ≤ √

δ exp(L r ω ¯

2 ).

This means ¯ p

1

+ q

1

ωi ¯ ∈ D (2 √ i¯ ω, √

δ exp(L q

¯ ω 2

)).

Similarly, we have also ¯ p

2

+ q

2

ωi ¯ ∈ D (2 √ i ω, ¯ √

δ exp(L q

¯ ω 2

)).

If √

δ exp(L q

¯ ω

2

) < 1, using Lemma 2.1 in [1], we can see that D (2 √ i ω, ¯

√ δ exp(L q

¯ ω

2

)) is convex. Thus for θ ∈ [0, 1], we have θ(¯ p

1

, q

2

)+(1 − θ)(¯ p

2

, q

2

)

∈ D

_δ^L

. If √

δ exp(L q

¯ w

2

) ≥ 1, using Lemma 2.1 in [1], we can see that for

¯

p

1

, ¯ p

2

, q

1

, q

2

≥ 0, θ ∈ [0, 1], we have θ(¯ p

1

+ q

1

ωi) + (1 − θ)(¯ p

2

+ q

2

ωi) ¯

∈ D (2 √ i ω, ¯ √

δ exp(L q

¯ ω

2

)). Thus for θ ∈ [0, 1], we have θ(¯ p

1

, q

1

) + (1 − θ)(¯ p

2

, q

2

) ∈ D

_δ^L

.

Therefore D

_δ^L

is convex.

Suppose that (¯ p

^∗

, q

^∗

, h

L

(¯ p

^∗

, q

^∗

)) is a strictly local minimum of h

L

(¯ p, q ), we prove that it is a global minimum of h

L

(¯ p, q). Suppose the contrary that there exists (¯ p

^∗∗

, q

^∗∗

, h

L

(¯ p

^∗∗

)) such that h

L

(¯ p

^∗

, q

^∗

) ≥ h

L

(¯ p

^∗∗

, q

^∗∗

). Then there exists a convex neighborhood U of (¯ p

^∗

, q

^∗

), such that ∀ s ∈ U, s 6 = (¯ p

^∗

, q

^∗

) and h

L

(s) > h

L

(¯ p

^∗

, q

^∗

). Since (¯ p

^∗∗

, q

^∗∗

) ∈ D

_h^L

L(¯p^∗∗,q^∗∗)

⊂ D

^L_h

L(¯p^∗,q^∗)

,we have

29

(32)

30

Optimized Parameters of Optimized Schwarz Waveform Relaxation Methods for The Heat Equation

HAL Id: hal-00589252

https://hal.archives-ouvertes.fr/hal-00589252

Submitted on 28 Apr 2011

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