The Existence theorem of the Stokes-Neumann Problem · 2010-04-28 · Nasrin Arab (CASA Tu/e) The...

The Existence theorem of the Stokes-Neumann Problem

Nasrin Arab

CASA Tu/e

28 April 2010

Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 1 / 21

Overview

1 Review

2 Stokes-Neumann problem

3 Solvability of the Stokes equations

4 conformal bijection

5 Main theorem


Outline

1 Review




5 Main theorem


Review

Figure: General solution of Stokes equations.


The Stokes equation in R2

Stokes equations ∆v − ∇p = 0∇ · v = 0

x ∈ G.

we can rewrite it as ∇ · T = 0∇ · v = 0

x ∈ G.

whereT := −pI + ∇v + (∇v)T




x ∈ G.


x ∈ G.





x ∈ G.


x ∈ G.



General solutions of Stokes equations without regardingboundary conditions

Holomorphic representation

TheoremIf p(x), v(x) solves the Stokes equation on G, then there exists a pair ofanalytic functions z 7−→ ϕ(z), χ(z) on G, such that

v1 + iv2 = −ϕ + zϕ′ + χ′

−4p = T11 + T22 = −8Reϕ′ ⇒ p = 2Reϕ′

T · n = 2i dds (zϕ′ + ϕ + χ′)

Vice versa

The holomorphic representation of a solution by ϕ, χ is unique ifϕ(0) = χ(0) = 0





v1 + iv2 = −ϕ + zϕ′ + χ′

−4p = T11 + T22 = −8Reϕ′ ⇒ p = 2Reϕ′

T · n = 2i dds (zϕ′ + ϕ + χ′)

Vice versa






v1 + iv2 = −ϕ + zϕ′ + χ′

−4p = T11 + T22 = −8Reϕ′ ⇒ p = 2Reϕ′

T · n = 2i dds (zϕ′ + ϕ + χ′)

Vice versa






v1 + iv2 = −ϕ + zϕ′ + χ′

−4p = T11 + T22 = −8Reϕ′ ⇒ p = 2Reϕ′

T · n = 2i dds (zϕ′ + ϕ + χ′)

Vice versa






v1 + iv2 = −ϕ + zϕ′ + χ′

−4p = T11 + T22 = −8Reϕ′ ⇒ p = 2Reϕ′

T · n = 2i dds (zϕ′ + ϕ + χ′)

Vice versa



Outline

1 Review




5 Main theorem


Stokes-Neumann problem

Stokes equation with Neumann boundary condition∇ · T (x) = 0 , x ∈ G∇ · v(x) = 0 , x ∈ GT (x) · n(x) = f (x) , x ∈ ∂G

on the prescribed boundary stress field x 7−→ f(x) ∈ R2 we put condition on f

f (x(s)) =ddsK1(s)n(s) + K2(s)t(s)

∫∂G

K1(s)ds = 0∫∂GK1(s)n(s) + K2(s)t(s)ds = 0

”What about (non)-uniqueness of the Stokes-Neumann problem”






∫∂G








∫∂G








∫∂G








∫∂G




Outline

1 Review




5 Main theorem


Solvability

Figure: Boundary condition.


necessary condition in terms of ϕ, χ

we want to find analytic ϕ, χ : G −→ C, such that at the boundary ∂G

T · n(s) = 2idds

(z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s))) = −idds

K(s)z(s). (1)

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

dds

(z(s)ϕ(z(s)) + χ(z(s))

)+ ϕ(z(s))˙z(s) − ϕ(z(s))z(s) = −

12

K(s)

∫∂G

K1(s)ds = 0




T · n(s) = 2idds


K(s)z(s). (1)

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

dds

(z(s)ϕ(z(s)) + χ(z(s))

)+ ϕ(z(s))˙z(s) − ϕ(z(s))z(s) = −

12

K(s)

∫∂G

K1(s)ds = 0




T · n(s) = 2idds


K(s)z(s). (1)

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

dds

(z(s)ϕ(z(s)) + χ(z(s))

)+ ϕ(z(s))˙z(s) − ϕ(z(s))z(s) = −

12

K(s)

∫∂G

K1(s)ds = 0




T · n(s) = 2idds


K(s)z(s). (1)

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

dds

(z(s)ϕ(z(s)) + χ(z(s))

)+ ϕ(z(s))˙z(s) − ϕ(z(s))z(s) = −

12

K(s)

∫∂G

K1(s)ds = 0




T · n(s) = 2idds


K(s)z(s). (1)

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

dds

(z(s)ϕ(z(s)) + χ(z(s))

)+ ϕ(z(s))˙z(s) − ϕ(z(s))z(s) = −

12

K(s)

∫∂G

K1(s)ds = 0


Outline

1 Review




5 Main theorem


conformal bijection

Figure: conformal bijection.


Conformal bijection

Φ(ζ) = ϕ(Ω(ζ)), X(ζ) = χ(Ω(ζ)), ζ = eiθ

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

⇓

Ω(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0


Conformal bijection

Φ(ζ) = ϕ(Ω(ζ)), X(ζ) = χ(Ω(ζ)), ζ = eiθ

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

⇓

Ω(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0


Conformal bijection

Φ(ζ) = ϕ(Ω(ζ)), X(ζ) = χ(Ω(ζ)), ζ = eiθ

z(s)ϕ′(z(s)) + ϕ(z(s)) + χ′(z(s)) = −1

2 K(s)zϕ(0) = χ(0) = 0Imϕ′(0) = 0

⇓

Ω(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0


Operator J

The mapping J : L2(S1;R1; 1⊥)→ L2(S1;R1; 1⊥)

f1 7→ Jf1 = f2

J2 = −IJf1(θ) = f2(θ) = 1

2π

> π−π

cot( 12 (θ − θ1))f1(θ1)dθ1

∂θJ = J∂θ,J(f1g1) = J ((Jf1)(Jg1)) + (Jf1)g1 + f1(Jg1).


Operator J


f1 7→ Jf1 = f2J2 = −I

Jf1(θ) = f2(θ) = 12π

> π−π

cot( 12 (θ − θ1))f1(θ1)dθ1



Operator J


f1 7→ Jf1 = f2J2 = −IJf1(θ) = f2(θ) = 1

2π

> π−π

cot( 12 (θ − θ1))f1(θ1)dθ1



Operator J


f1 7→ Jf1 = f2J2 = −IJf1(θ) = f2(θ) = 1

2π

> π−π

cot( 12 (θ − θ1))f1(θ1)dθ1



Outline

1 Review




5 Main theorem


The Existence Theorem

TheoremLet K1,K2 : ∂G −→ R be given.a. θ 7−→ |∂θΩ(eiθ)|K1(s(θ)) ∈ L2(S1;R1; 1⊥).b. θ 7−→ |∂θΩ(eiθ)|K2(s(θ)) ∈ L2(S1;R1).

c. θ 7−→ |∂θΩ(eiθ)| is bounded on S1.

d. θ 7−→ |∂θ∂θΩ(eiθ)| is bounded on S1.

Then there exist unique Φ,X : D −→ C , such thatΩ(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0

and at the boundary Φ,X ∈ L2(S1;C)



TheoremLet K1,K2 : ∂G −→ R be given.a. θ 7−→ |∂θΩ(eiθ)|K1(s(θ)) ∈ L2(S1;R1; 1⊥).b. θ 7−→ |∂θΩ(eiθ)|K2(s(θ)) ∈ L2(S1;R1).c. θ 7−→ |∂θΩ(eiθ)| is bounded on S1.


Then there exist unique Φ,X : D −→ C , such that

Ω(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0




TheoremLet K1,K2 : ∂G −→ R be given.a. θ 7−→ |∂θΩ(eiθ)|K1(s(θ)) ∈ L2(S1;R1; 1⊥).b. θ 7−→ |∂θΩ(eiθ)|K2(s(θ)) ∈ L2(S1;R1).c. θ 7−→ |∂θΩ(eiθ)| is bounded on S1.


Then there exist unique Φ,X : D −→ C , such thatΩ(ζ)∂θΦ(ζ) + ∂θΩ(ζ)Φ(ζ) + ∂θX(ζ) = − 1

2 |∂θΩ(ζ)|K(s(θ)).Φ(0) = X(0) = 0ImΦ′(0) = 0



Proof

Proof.

Ω1∂θΦ1 + Ω2∂θΦ2 + (∂θΩ1)Φ1 + (∂θΩ2)Φ2 + ∂θX1 = − 1

2 |∂θΩ|K1

Ω2∂θΦ1 −Ω1∂θΦ2 − (∂θΩ2)Φ1 + (∂θΩ1)Φ2 − ∂θX2 = − 12 |∂θΩ|K2

use the operator JΩ1∂θΦ1 + Ω2J∂θΦ1 + Ω1Φ1 + Ω2JΦ1 + ∂θX1 = − 1

2 |Ω′|K1

Ω2∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + Ω1JΦ1 − J∂θX1 = −12 |Ω

′|K2


Proof

Proof.

Ω1∂θΦ1 + Ω2∂θΦ2 + (∂θΩ1)Φ1 + (∂θΩ2)Φ2 + ∂θX1 = − 1

2 |∂θΩ|K1

Ω2∂θΦ1 −Ω1∂θΦ2 − (∂θΩ2)Φ1 + (∂θΩ1)Φ2 − ∂θX2 = − 12 |∂θΩ|K2

use the operator JΩ1∂θΦ1 + Ω2J∂θΦ1 + Ω1Φ1 + Ω2JΦ1 + ∂θX1 = − 1

2 |Ω′|K1

Ω2∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + Ω1JΦ1 − J∂θX1 = −12 |Ω

′|K2


Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

]

[JΩ1 −Ω1J] ∂θΦ1 +[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1

Then condition d. implies that k is Hilbert-Schmidt. so it is compact.The operator L is bijection!


Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]

∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1



Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]

k(Φ1) + L(Φ1) = −14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1



Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]

k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1



Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1



Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1

Then condition d. implies that k is Hilbert-Schmidt. so it is compact.

The operator L is bijection!


Proof.

JΩ1∂θΦ1 −Ω1J∂θΦ1 − Ω2Φ1 + JΩ1Φ1 = −14

[J(|Ω′|K1) + |Ω′|K2

][JΩ1 −Ω1J] ∂θΦ1 +

[JΩ1 − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]∂θ

([JΩ1 −Ω1J] Φ1

)+

[Ω1J − Ω2

]Φ1 = −

14

[J(|Ω′|K1) + |Ω′|K2

]k(Φ1) + L(Φ1) = −

14

[J(|Ω′|K1) + |Ω′|K2

]k can be written

k(Φ1) = −1

2π∂θ

∫ π

−πcos(

θ − θ1

2)Ω1(θ) −Ω1(θ1)

sin( θ−θ12 )

Φ1(θ1)dθ1



Fredholm alternative

Part of the result states that a non-zero complex number in the spectrum of acompact operator is an eigenvalue.

orLet K(x, y) be an integral kernel, and considerthe homogeneous equation,λφ(x) −

∫ ba K(x, y)φ(y) dy = 0

and the inhomogeneous equationλφ(x) −

∫ ba K(x, y)φ(y) dy = f (x).

The Fredholm alternative states that,∀ 0 , λ ∈ C, either the first equation has a non-trivial solution, or the secondequation has a solution for all f (x).A sufficient condition for this theorem to hold is for K(x, y) to be squareintegrable on the rectangle [a, b] × [a, b].



Part of the result states that a non-zero complex number in the spectrum of acompact operator is an eigenvalue.orLet K(x, y) be an integral kernel, and considerthe homogeneous equation,λφ(x) −

∫ ba K(x, y)φ(y) dy = 0


∫ ba K(x, y)φ(y) dy = f (x).





∫ ba K(x, y)φ(y) dy = 0


∫ ba K(x, y)φ(y) dy = f (x).

The Fredholm alternative states that,∀ 0 , λ ∈ C, either the first equation has a non-trivial solution, or the secondequation has a solution for all f (x).

A sufficient condition for this theorem to hold is for K(x, y) to be squareintegrable on the rectangle [a, b] × [a, b].




∫ ba K(x, y)φ(y) dy = 0


∫ ba K(x, y)φ(y) dy = f (x).



Thank you


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The Existence theorem of the Stokes-Neumann Problem · 2010-04-28 · Nasrin Arab (CASA Tu/e) The...

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