EXISTENCE OF A STEADY FLOW WITH A BOUNDED VORTEX IN AN UNBOUNDED DOMAIN

EXISTENCE OF A STEADY FLOW WITH

A BOUNDED VORTEX IN AN

UNBOUNDED DOMAIN

B. Emamizadeh

*

Department of Mathematics, Iran University of Science and Technology, Tehran, Islamic Republic of Iran

Abstract

We prove the existence of steady 2-dimensional flows, containing a bounded

vortex, and approaching a uniform flow at infinity. The data prescribed is the

rearrangement class of the vorticity field. The corresponding stream function

satisfies a semilinear elliptic partial differential equation. The result is proved by

maximizing the kinetic energy over all flows whose vorticity fields are

rearrangements of a prescribed function.

* _{E-mail: [email protected]}

Introduction

In this paper we prove the existence of steady 2-dimensioal ideal fluid flows occupying (the first quadrant) and containing a bounded vortex. Such a flow will be described by a stream function

+

∏

→ ∏₊ :

ψ . At

infinity we will have which is the stream

function for an irrotational flow with velocity field . The vorticity is given by

2 1x

x λ

ψ →−

(x1,−x2

−λ ) −Δψ, where Δ is

the Laplacian, and −Δψ vanishes outside a bounded region avoiding the boundary of ∏₊. We will show that

ψ satisfies the following semilinear elliptic partial differential equation:

ψ φ ψ= o

Δ

− ,

almost everywhere in ∏₊, where φ is an increasing function, unknown a prior. In our result the vorticity

Keywords: Rearrangements, Vorticity, Irrotational flows, Elliptic partial differential equations, Variational problem

field is a rearrangement of a prescribed

non-negative function having bounded support. The existence theorem is proved by maximizing a functional over the set of rearrangements of vanishing outside bounded sets in . This variational principle was adapted by Burton [1] from one for vortex rings in 3 dimensions, proposed by Benjamin [2].

( ψ ζ =−Δ )

0

ζ

0

ζ

+

∏

Lack of compactness caused by the unboundedness of the domain of interest is the motivation to use the strategy proposed by Benjamin [2].

Notation and Definitions

Henceforth p denotes a real number in (2,∞) and

(

1

)

/

:= −

∗ _{p p}

p . The upper and the right half planes are

designated by and , respectively, and the first quadrant by . Generic points of

u

∏ ∏_r

+

∏ 2 _{are denoted by}

(x₁,x₂),y (y₁,y₂),z (z₁,z₂

x= = = ), etc. For x∈ 2 _we

let x,x,x denote the reflections of x about the x1-axis, x2-axis and the origin, respectively. For ξ >0 we define

( ) {= ∈

∏₊ξ : x 2 _{< <ξ < <ξ}} 2 1 ,0

The ball centered at x with radius r is denoted B_r( )x , when the origin is the center we write BBr.

Here we deal with three different Green’s functions,

namely, the Green’s functions for −Δ with

homogeneous Dirichlet boundary conditions on , and ; it is a standard result that these functions are given as follows

u ∏

r

∏ ∏₊

(

)

(

)

(

)

log , , , ,

2 1 : ,

, , , ,

log 2

1 : ,

, , , ,

log 2

1 : ,

y x y

x y x y x

y x y x y

x G

y x y

x y

x y x y

x G

y x y

x y

x y x y

x G

r r

u u

≠ ∏ ∈ −

− − − =

≠ ∏ ∈ −

− =

≠ ∏ ∈ −

− =

+

+ _π

π π

respectively. For measurable functions ζ on 2 _we

define the following integral operators

( )

(

) ( )

( )

(

) ( )

( )

∫

(

) ( )

∫

∫

+

∏ +

+

∏ ∏

= = =

, ,

:

, ,

:

, ,

:

dy y y x G x

K

dy y y x G x

K

dy y y x G x

K

r u

r r

u u

ζ ζ

ζ ζ

ζ ζ

whenever the integrals exit.

For a measurable set A in 2_{, we use A to denote}

the 2-dimensional Lebesgue measure of A, and A for the topological closure. The strong support of a measurable function ζ , denoted supp( , is defined as follows

)

ζ

( ): { dom( ) ( ) 0}

suppζ = x∈ ζ ζ x > .

To define the rearrangement classes needed for our variational problem we fix a non-negative, non-trivial function ζ0∈Lp( 2)) which vanishes outside a bounded

set; in addition we assume

( )

2

0

suppζ =πa (1)

for some . The set F comprises the functions (which we call rearrangements of ζ

0 > a

0) that vanish outside

bounded subsets of and that are equimeasurable to

ζ

+

∏

0. A function ζ is said to be equimeasurable to ζ0

whenever

( )

{x∈∏₊ζ x ≥α}={x∈∏₊ζ₀( )x ≥α},

for every positive α. It is well known that if ζ∈F then

s s ζ0

ζ = ,

for s∈

[ ]

1,∞ . The subset of F comprising functions vanishing outside ∏₊( )ξ is designated byF( )ξ , where it is assumed that _ξ≥aπ1/2 to ensure _{F( )ξ ≠0/}.

Next we define the kinetic energy. For _v∈_Lp

(

∏₊

)

having bounded support and λ∈ , we define

( )

( )

∫

∫

+ +

∏

∏ +

= ℑ

= Ψ

ν ν

ν ν ν

2 1 :

, 2

1 :

x x

K

and the kinetic energy

( )ν ( )ν λ ( )ν

λ = − ℑ

Ψ : Ψ ,

whenever the integrals exist. Now we are in a position to define the variational problem

( )

_λ

( )

ζ

ζ

λ Ψ

F

∈

sup :

P .

The set of solutions of ( )Pλ is denoted by . For

the truncated variational problem

λ

∑

2 / 1

π

ξ >a (Pλ( )ξ ) is

defined by

( )

(

)

( )

( )

ζ

ξ λ

ζ

λ Ψ

ξ

F

∈

sup :

P ,

and ∑λ( )ξ denotes the set of solutions. Proofs of Some Lemmas

Lemma 1. Let ζ ∈Lp

(

∏+

)

vanish outside a bounded

set. Then (i) _{K ∈C}1(

+ζ 2).

(ii) ∇K₊ζ( )x ≤Cζ _p, for every x∈ 2_{, where C} depends on supp( )ζ and p.

(iii) K+ζ( )x ≤Cmin{x1,x2}ζ _p, for every , where C is the constant in (ii).

+

∏ ∈ x

Proof. (i) is an immediate consequence of a result about Newtonian potentials of densities with compact support. Specifically, let

( )

π ζ

2 1 = x

N _e

∫

² _x−yζe

( )

ydy

1 log

denote the Newtonian potential of the zero-extension of

ζ, to all of 2_{. Since and ζ has compact support}

we can apply Lemmas A.7 an A.9 in [3] to deduce that (

2 > p

1

C

( )

π ζ

2 1 =

∇N _e x

∫

²∇x _x−y e

( )

y dy, ∀x∈

1

log ζ 2_{. (2)}

Clearly we have

( )x N ( )x N ( )x N ( )x N ( )x

K₊ζ_e = ζ_e + ζ_e − ζ_e − ζ_e .

Hence, _{K ∈C}1(

+ζ 2). For (ii) it obviously suffices to

show

( )≤ ∀ ∈

∇Nζ_e x Cζ _p, x 2_{. (3)}

where C is a constant depending on supp( )ζ and p. To do this, we use (2) to deduce

( )≤

∇Nζe x

∫

_{² −}

( )

y dy ∀x∈ y

x ,

1 _ζ 2_.

Now let us fix x∈ 2 _{and denote the}

Schwartz-rearrangement of ζ , about x, by . Therefore, by a standard inequality, see for example [4], we obtain

∗

ζ

( )

( )

( )

∫

∗

− ≤

∇

x B e

l

dy y y x x

N ζ

π

ζ 1

2

1 _,

where

( )

(

_{supp /}

)

1/2

:= ζ π

l .

Hence, by an application of Hölder’s inequality we derive

( )

_{( ) p}

p

x

B p

e

l

dy y x x

N ζ

π ζ

∗

∗ ⎟⎟

⎠ ⎞ ⎜

⎜ ⎝ ⎛

− ≤

∇

∫

/ 1

1 2

1

. (4)

Elementary calculations yield

( )x _{x y} dy C

B_{l −} p ≤

∫

∗

1 _,

where C depends only on l and p. So we derive (3). Now to derive (iii) we fix x∈∏₊. Since _{K ∈C}1(

+ζ 2)

and vanishes on the boundary of , we can apply the Mean Value Theorem to obtain

+

∏

( )x K ( )x K (x ) x K

( )

x K₊ζ = ₊ζ − ₊ζ ₁,0 ≤ ₂∇ ₊ζ ˆ , where is a point on the segment joining xˆ x to (x₁,0). Whence from (ii) we deduce that K+ζ( )x ≤Cx2ζ _p.

Similarly, one can show K+ζ( )x ≤Cx1ζ _p, so we

derive (iii) as desired.◊

Lemma 2. Let U be an open and bounded subset of . Then for every

+

∏ q_≥ K Lp

( )

U _→Lq

( )

U

+: ,

1 is a

compact linear operator, in the sense that if { }ζn is a

sequence of functions, bounded in _Lp

(

∏₊

)

_and vanishing outside U, then the restrictions to U of the Kζn’s have a subsequence converging in the q-norm.

Proof. The well-defindness of K Lp

( )

U _→Lq

( )

U

+:

follows from Lemma 1(iii). The linearity of K+ follows

from the definition. To show compactness of K+ it

suffices to show that K _:Lp

( )

U _→W1,2

( )

U

+ is

bounded, since then by an application of the Sobolev Embedding Theorem we derive the desired result. Let us now fix _ζ∈_Lp

(

∏₊

)

_{that vanishes outside U. Then} by applying Lemma 1(ii), (iii) we infer K₊ζ ₂≤Cζ _p and ∇K₊ζ ₂≤Cζ _p, hence

( )U p

W C

K₊ζ 1,2 ≤ ζ ,

here C stands for different constants. So we are done.◊

The next lemma is an immediate consequence of Lemma 7 in [1].

Lemma 3.Let ζ∈Lp

(

∏+

)

vanishes outside a bounded

set. Then

( )=

( )

( )=

( )

→∞

∇ −

+ −

+ x O x K x O x x

K _ζ 2 , _ζ 1 , _.

Lemma 4. Let q and U be as in Lemma 2. Then

( )

U L

( )

U L

K p _→ q

+: is strictly positive, that is, for

every non-trivial function vanishing

outside U,

(

∏+

∈_Lp

ζ

)

)

0 >

∫

∏+ +

ζ

ζK .

Proof. Let us fix vanishing outside U.

Then, from Lemma 3(i) in [1], we have

(

∏₊ ∈_Lp

ζ

( )_u

D′∏

= Δ

− Kuζ ζ, in ,

that is, in the sense of distributions. Hence, we also have

(∏₊)

′ =

Δ

− K_uζ ζ, inD ,

since K+ζ( )x =Kuζ( )x −Kuζ( )x for all x∈ 2. Now by Agmon’s regularity theory [5] we deduce that

(

p loc

W

K _∈ 2,

+ζ 2). In particular, K+ζ∈Wloc2,p

( )

∏+ .

ζ ζ = Δ

− K_u ,

almost everywhere in . Next we let

; since the boundary of

+

∏

( )= ∩∏+

Ω R : B_R Ω( )R is

Lipschitz we can apply the weak Divergence Theorem, see for example [6], to obtain

( ) ( ) ( )

(

)

(

)

,

2

∫

∫

∫

Ω + + Ω ∇ + = ∂Ω + ∂ +

−

R n

R

RζKζ Kζ γ Kζ γ Kζ dσ

where γstands for the trace operator on ∂Ω( )R and n denotes the unit outward normal vector to ∂Ω( )R . Since K+ζ∈C1

(

∏+

)

we have

(

)

(

)

( )

∫

( )

(

)

(

)

∫

∂ΩRγ K+ζ γ ∂nK+ζ dσ= ∂ΩR K+ζ ∂nK+ζ dσ .

Therefore

( )

∫

( )

∫

( )

(

)

(

)

∫

Ω + + Ω ∇ + = ∂Ω + ∂ +

−

R n

R

R ζK ζ K ζ K ζ K ζ dσ

2

.

Now from Lemma 3 we infer

(

)

(

)

( ) 0

lim

∫

∂ =

Ω

∂ + +

∞

→ R n

R Kζ Kζ dσ .

Moreover, since

∫

is finite and

+

∏ ζK+ζ ∇K+ζ is

bounded in 2 _{we may apply the Lebesgue Dominated}

Convergence Theorem to conclude

( )

( ) .

lim

, lim

2 2

∫

∫

∫

∫

+ +

∏ +

Ω +

∞ →

∏ +

Ω +

∞ →

∇ = ∇

=

ζ ζ

ζ ζ ζ ζ

K K

K K

R R

R R

Therefore, we derive

∫

∫

+

+ ∏ +

∏ + = ∇

2

ζ ζ

ζK K and we

are done.◊

The following lemma is a result from Burton’s theory [7].

Lemma 5. Suppose Φ_:Lp

(

∏₊

( )

ξ

)

→ _{is a weakly} sequentially continuous, strictly convex functional. Then

attains a maximum relative to

( )ζ

Φ ζ∈F( )ξ , for

. Moreover, if is a maximizer and

2 / 1

π

ξ≥a ζˆ _ψ∈

, the subdifferential of Φ at , then

) ˆ (ζ Φ

∂ ζˆ

ψ φ ζˆ= _o ,

almost everywhere in , for some increasing

function

( )ξ

+

∏

φ.

Lemma 6.For every λ>0 and _ξ≥aπ1/2, the problem

( )ξ

λ

P is solvable. Moreover, if ζ∈∑_λ( )ξ , then

(K ζ λx₁x₂)

φ

ζ = _{o +} − , (5)

almost everywhere in , for some increasing

function

+

∏

φ.

Proof. Let us being by noting that _K+:Lp

(

∏₊

( )

ξ

)

→ is a symmetric operator, that is,

( ) (∏₊ξ

∗ p

L )

(

₊

)

∏ +

∏ + =

∫

∀ ∈ ∏

∫

+ +

p

L w K

w w

K ν, ν,

ν ,

which readily follows from the symmetry of G+. Since K+ is compact, strictly positive and symmetric it follows

that Ψλ, defined on the set of functions in Lp

(

∏+

)

vanishing outside ∏₊( )ξ , is strictly convex and weakly sequentially continuous. Now by applying Lemma 5 we deduce that P_λ( )ξ is solvable. Next we show that if

( )ξ

ζ ∈∑_λ , then . For this

purpose we consider

( )ζ λ

ζ − ∈∂Ψ_λ

+ x1x2

K

(

∏+

∈_Lp

ζ

)

which vanishes

outside ∏₊( )ξ , then we need to show that

( )

( )

∫

(

)

(

)

+

∏ − + −

+ Ψ ≥

Ψ_λζ _λ ζ ζ ζ K ζ λx₁x₂ ,

or equivalently

(

−

) (

−

)

≥0

∫

∏+ +

ζ ζ ζ

ζ K ,

but this is true since K+ is strictly positive. Therefore,

again by Lemma 5, existence of an increasing function

φ is ensured so that (5) holds.◊

Results and Discussion

In this section we present our main result (see the theorem below). We begin with some technical lemmas.

Lemma 7.Let λ>0. Then there exists such that

( )λ >0

R

( )− ≤ ≥ ( ) ∈F +ζ x λx1x2 0, x R λ , ζ

K .

Proof. Let us fix x∈∏₊ and ζ∈_F; we assume

{₁, ₂}≥α

min x x for some α>0 to be determined later.

According to Lemma 1(ii) there exists ,

independent of

0 > M

Hence

( ) λ { }( λα)

ζ − ≤ −

+ x xx min x x M

K _{1 2 1}, ₂ .

Thus if we assume α≥M/λ, then .

Hence we can take .◊

( )− _{1 2}≤0

+ x xx

K ζ λ

( )λ M/λ

R =

Lemma 8. Suppose _ζ∈Lp( 2_{) is a non-negative} function which is spherically decreasing and vanishes outside Ba. Then there exists a positive constant k such

that

t k

K t

t ≥ log

∫

∏+ +

ζ

ζ ,

for all sufficiently large t, where ζ_t( )x :=ζ(x₁−t,x₂−t).

Proof. Clearly we can assume t≥

(

1+ 2

)

a. Now we observe that there exists β>0 and such that for all x with

a b< < 0

b

x≤ we have ζ( )x ≥β . Let Bb( )t denote the ball centered at ( )t,t with radius b and consider x∈B_a( )t and y∈B_b( )t . Hence if we set

, : , : , : ,

: 2 3 4

1 = x−y γ = x−y γ = x−y γ = x−y

γ

then it is clear that

a t a

a t a

t 2 , 2 2 , 2 , 2 2 2

2 _{2 3 4}

1≥ − γ ≥ − γ ≤ γ ≤ +

γ .

Therefore,

( )

(

(

)

)

(

)

(

t a

)

a a t b

dy a t a

a t x

K

t B t

b

+ −

= +

−

≥

∫

+

2 log 2

2 2 2 2

2 2 log 2

2 2

) (

2

β

π β ζ

Hence

(

)

(

t a

)

a a t b

K t

t ≥ − ₊

∫

∏₊ + ₂ log ₂ 2 4

2

πβ ζ

ζ ,

and we are done.◊

An immediate consequence of Lemma 8 is the following

Corollary. We have

( )Ψ

( )

=+∞

∈ ∞

→ ζ

ξ ζ ξ

F sup

lim .

Lemma 9. There exists and such

that, if , and

0 0>

λ 1/2

0 π

ξ >a

0

0<λ≤λ ξ≥ξ₀ ζ_λ,_ξ is a maximizer of

( )ζ

λ

Ψ relative to ζ∈F( )ξ then

( )

( )

{

}

2

2 1

, x xx 0 a

K

x∈∏+ξ +ζ_λξ −λ > ≥π .

Proof. Let us fix α>0, ε>0. Then according to the

Corollary there exists such that if

then

2 / 1

0 π

ξ >a ξ ≥ξ₀

( )ξ ( )ζ α ε

ζ∈F Ψ ≥ +

sup . In particular,

( )ξ ( )ζ α ε

ζ∈ 0 Ψ ≥ +

sup _F . Since Ψ is a real-valued

functional on _Lp

(

∏₊

( )

ξ₀

)

_{which is weakly sequentially} continuous and strictly convex we can apply Lemma 5 to ensure existence of ζˆ∈F

(

ζ0

)

such that

( )

ζ = ζ ( )ζ Ψ

( )

ζ

Ψ _∈

0 sup ˆ

F , whence

( )

ζ ≥α+ε

Ψ ˆ . (6)

Now choose λ₀>0 such that λ₀ℑ

( )

ζˆ <ε. Since

( ) ( ) ( )

ζ ζ λ ζ

λ ˆ :=Ψ ˆ − ℑ ˆ

Ψ we can use (6) to obtain

( )

ζˆ α, 0 λ λ0

λ ≥ < ≤

Ψ .

This shows that

( ) ( ) , 0 0, 0

sup_ζ∈Fξ Ψ_λ ζ ≥α <λ≤λ ξ≥ξ . (7)

Next we set α=3aCζ0 p ζ0₁, where C is the

constant in Lemma 1(iii). Hence from (7) we have

( ) ( ) 3 0 01

supζ∈Fξ Ψλ ζ ≥ aCζ p ζ , (8)

for all and . Now we fix ,

and let

0

0<λ≤λ ξ≥ξ₀ 0<λ≤λ₀

0

ξ

ξ ≥ ζ_λ,_ξ denote a maximizer of Ψλ( )ζ

relative to ζ∈F

( )

ξ . Then we have

( )

( )

( )

⎟⎠

⎞ ⎜

⎝

⎛ ₋

≤

Ψ ₊

∏+

2 1 ,

1 0

, ₂

1

sup K ζ x λxx ζ

ζ _λξ

ξ ξ

λ

λ .

We also have Ψλ

( )

ζλ,ξ ≥3aCζ0 p ζ0 ₁, from (8),

hence

( ) 2K ,

( )

x x1x2 3aC 0 p

1

sup ζ_λξ λ ζ

ξ ⎟⎠>

⎞ ⎜

⎝

⎛ ₋

+ ∏+

. (9)

Since 1_2K+ζ_λ,ξ

( )

_x −λ_x1x2∈

(

∏₊

( )

ξ

)

_{, it attains its} maximum at

(

0

)

∈∏₊( )ξ

2 0 1,x

x , say. Whence, by Lemma

( )

( )

(

)

{

,

}

. min 2 , 2 1 2 1 sup 0 0 2 0 1 0 2 0 1 , 2 1 , p x x C x x K x x x K ζ ζ λ

ζ_{λξ λξ}

ξ ≤ ≤ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₋ + + ∏+

Therefore, from (9) we infer .

Now we define the set

{

x ,x

}

6a 2a

min 0

2 0

1 ≥ >

{

}

0

2 0 1 2 0 2 2 0 1

1 , ,

: x x x x x B x x

S = ∈∏₊ < < ∩ _a

(

)

)

,

where

(

0 denotes the ball centered at

2 0 1 2 x ,x

B a

(

x10,x02

)

with radius 2a; clearly S⊂∏₊( )ξ . Consider x∈S, then

( )

( )

0

2 0 1 , 2 1

, x xx 1/2K x x ,x

K₊ζ_λξ −λ ≥ ₊ζ_λξ −λ . (10)

On the other hand, by an application of the Mean Value Theorem and Lemma 1(ii),

( )

(

)

( )

(

)

, 2 , ˆ , 0 0 2 0 1 , 0 2 0 1 , , p aC x x x x K x x K x K ζ ζ ζ

ζλξ λξ λξ

≤ − ∇ ≤ − _{+ +} +

where is a point on the segment joining xˆ x to

(

0

)

2 0 1,x

x ,

whence

( )

x K

(

x x

)

aC _p

K 0 ₀

2 0 1 ,

, ζ , 2 ζ

ζ_λξ ≥ _{+ λξ} −

+ .

This, in turn, implies

( )

x K

(

x x

)

aC _p

K 0 ₀

2 0 1 ,

, ζ , 2 ζ

ζ_λξ ≥ _{+ λξ} −

+ . (11)

Thus from (8), (10) and (11) we infer

( )

(

)

. 2 3 , 2 1 0 0 0 0 2 0 1 0 0 2 0 1 , 2 1 , p p p p aC aC aC x x aC x x K x x x K ζ ζ ζ λ ζ ζ λ

ζ_{λξ λξ}

= − ≥ − − ≥ − ₊ +

Therefore, S⊆supp

(

K+ζλ,ξ( )x −λx1x2

)

. Hence

( )

(

)

2

2 1 ,

suppK+ζλξ x −λxx ≥ S =πa , as desired.◊

Theorem. There exists such that , for

. Moreover, if and , then ψ

satisfies the following elliptic partial differential equation

0 0 >

λ ∑λ ≠0/

) , 0 ( λ₀

λ∈ ζ ∈∑λ ψ:=K+ζ

( − ) ∏₊ =

Δ

− ψ φ_oψ λx₁x₂ , a.e.in , (12)

for some increasing function φ, unknown a priori.

Proof. Let ξ0 and λ0 be as in Lemma 9. If we fix λ∈

(0, λ0), then by Lemma 7 there exists R(λ) > 0 such that

( )

( )

(

)

∈F.

∏ ∏ ∈ ≤ − + + + ζ λ λ ζ_λξ

, \ , 0 2 1 , R x x x x K (13)

Next we define ξ∗:=max

{

ξ0,R

( )

λ

}

; then according to

Lemma 6, has a maximizer relative to

, say ) (ζ λ Ψ ) ( ∗ ∈ ξ

ζ F ζ_λ,ξ∗. For simplicity we write

. We claim that . To prove this,

suppose and consider

∗ =ζ_λξ

ζˆ: _, ζˆ∈∑_λ

∗

≥ξ

l ζ ∈∑_λ(l). We will first

show that

( )

( )

∗ + Π ⊆ ξ ζ

supp , (14)

modulo a set of measure zero. By Lemma 6 there sxists an increasing function φ such that

(

K ζ λx1x2

)

φ

ζ = _{o +} − , (15)

almost everywhere in . Next we observe that

since

) (l

+

∏

φ is increasing,

( )

φ −1

(

0,∞

]

, the pre-image of

(0,∞

]

under φ , is an interval, say I, of the form (c,∞)

or

[

c,∞), by assuming that φ takes on the value +∞

on the interval

(

− _{( )} ∞

)

+

∏ ∞ + x1x2 _{, l},

K ζ λ . This, along

with (15), implies

( ) (

K x₁x₂

)

1

( )

I

suppζ = ₊ζ −λ − ,

modulo a set of measure zero in ∏₊(l). Hence

(

)

1

( )

2

2

1x I a

x

K₊ζ −λ − =π . On the other hand, from

Lemma 9, we have

(

)

1

(

)

2

2

1x 0, a

x

K₊ζ −λ − ∞ ≥π .

Whence c≥0, and this implies supp

( )

ζ ⊆supp(K₊ζ − modulo a set of measure zero in . Finally, according to (13), we also have

) 2 1x

x

λ ∏₊(l)

− +ζ K supp(

( )

∗ + ∏ ⊆ ξ

λx₁x₂) , hence we derive (14). Now from (14) we infer Ψ_λ(ζˆ)≥Ψ_λ(ζ) and this, in turn, implies that

) (l

λ

ζ ∈∑ . Since is arbitrary we deduce that

.

∗

≥ξ l

λ

ζˆ∈∑

To derive (12) we use Lemma 6 once again to ensure existence of an increasing function such that φˆ

(

ˆ 1 2

)

ˆ

ˆ _{φ K ζ λxx}

almost everywhere in . We obtain (12) by a modification process, that is, define

( )

∗ +

∏ ξ

( )

( )

( )

⎩ ⎨ ⎧

≤

> ∈

=

. 0 , 0

, 0 , ˆ ,

ˆ :

s

s dom s s

s φ φ

φ

therefore, clearly, we have

(

ˆ 1 2

)

ˆ _{φ K ζ λxx}

ζ = _{o +} − ,

almost everywhere in ∏+. As required.◊

Let us conclude with the following

Remark. A close inspection of the proofs of the Theorem and Lemma 9 confirms that if ζ∈∑λ, then

( )

{

x K

( )

x x₁x₂ 2aC _{0 p}

suppζ ⊆ ∈∏+ +ζ −λ ≥ ζ

}

,

modulo a set of measure zero. Hence for almost every we have

( )ζ

supp ∈ x

( ) ( )

{ , } , min

2

0 2 1 2

1 0

p p

x x C

x K x x x K aC

ζ ζ

λ ζ ζ

≤ ≤ −

≤ _{+ +}

where in the last inequality we have used Lemma 1(iii). Therefore, for almost every x∈supp( )ζ

{x,x } 2a

min _{1 2} ≥ .

This shows that the vortex core essentially avoids the boundary of ∏+.

Similar problems have been considered in Emamizadeh [8,9].

Acknowledgements

The author would like to thank his graduate students M. H. Mehrabi, V. Roomi, A. Assari, H. R. Keshavarz, M. Hosseini, H. Rashtizadeh for fruitful discussions concerning this problem.

References

1. Burton, G. R. Steady symmetric vortex pairs and rearrangements. Proceedings of the Royal Society of Edinburgh, 108A, 369-290, (1988).

2. Benjamin, T. B. The alliance of practical and analytical insights into the nonlinear problems of fluid mechanics, Applications of methods of functional analysis to problems in mechanics. Lecture Notes in Mathematics 503, Springer-Verlag, 8-29, (1979).

3. Fraenkel, L. E. An Introduction to Maximum Principles and Symmetry in Elliptic Problems: Cambridge Uni-versity Press. (In the press).

4. Hardy, G. H. Littlewood, Polya, G. Inequalities. Cam-bridge University Press, (1952).

5. Agmon, S. The Lp _{approaching to the Dirichlet problem.}

Ann. Scuola Norm. Pisa Cl. Sci., 13(3), 405-448, (1959). 6. Grisvard, P. Singularities in Boundary Value Problems.

Masson, Springer-Verlag, (1992).

7. Burton, G. R. Rearrangements of functions, maximization of convex functionals and vortex rings. Math. Ann., 276, 225-253, (1987).

8. Burton, G. R. Emamizadeh, B. A constrained variational problem for steady vortices in a shear flow. Commun. Partial Diff. Eq., 24(7/8), 1341-65, (1999).