the existence theorem of the stokes-neumann problem · 2010-04-28 · nasrin arab (casa tu/e) the...
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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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 2 / 21
Outline
1 Review
2 Stokes-Neumann problem
3 Solvability of the Stokes equations
4 conformal bijection
5 Main theorem
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 3 / 21
Review
Figure: General solution of Stokes equations.
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 4 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 5 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 5 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 5 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 6 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 6 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 6 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 6 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 6 / 21
Outline
1 Review
2 Stokes-Neumann problem
3 Solvability of the Stokes equations
4 conformal bijection
5 Main theorem
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 7 / 21
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”
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 8 / 21
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”
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 8 / 21
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”
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 8 / 21
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”
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 8 / 21
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”
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 8 / 21
Outline
1 Review
2 Stokes-Neumann problem
3 Solvability of the Stokes equations
4 conformal bijection
5 Main theorem
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 9 / 21
Solvability
Figure: Boundary condition.
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 10 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 11 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 11 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 11 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 11 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 11 / 21
Outline
1 Review
2 Stokes-Neumann problem
3 Solvability of the Stokes equations
4 conformal bijection
5 Main theorem
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 12 / 21
conformal bijection
Figure: conformal bijection.
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 13 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 14 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 14 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 14 / 21
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).
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 15 / 21
Operator J
The mapping J : L2(S1;R1; 1⊥)→ L2(S1;R1; 1⊥)
f1 7→ Jf1 = f2J2 = −I
Jf1(θ) = f2(θ) = 12π
> π−π
cot( 12 (θ − θ1))f1(θ1)dθ1
∂θJ = J∂θ,J(f1g1) = J ((Jf1)(Jg1)) + (Jf1)g1 + f1(Jg1).
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 15 / 21
Operator J
The mapping J : L2(S1;R1; 1⊥)→ L2(S1;R1; 1⊥)
f1 7→ Jf1 = f2J2 = −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).
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 15 / 21
Operator J
The mapping J : L2(S1;R1; 1⊥)→ L2(S1;R1; 1⊥)
f1 7→ Jf1 = f2J2 = −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).
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 15 / 21
Outline
1 Review
2 Stokes-Neumann problem
3 Solvability of the Stokes equations
4 conformal bijection
5 Main theorem
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 16 / 21
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)
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 17 / 21
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 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
and at the boundary Φ,X ∈ L2(S1;C)
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 17 / 21
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)
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 17 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 18 / 21
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
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 18 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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!
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 19 / 21
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].
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 20 / 21
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].
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 20 / 21
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].
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 20 / 21
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].
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 20 / 21
Thank you
Nasrin Arab (CASA Tu/e) The Existence theorem of the Stokes-Neumann Problem 28 April 2010 21 / 21