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STEADY VORTEX FLOWS OBTAINED FROM A CONSTRAINED
VARIATIONAL PROBLEM
B. EMAMIZADEH and M. H. MEHRABI
Received 10 January 2001 and in revised form 24 August 2001
We prove the existence of steady two-dimensional ideal vortex flows occupying the first quadrant and containing a bounded vortex; this is done by solving a constrained variational problem. Kinetic energy is maximized subject to the vorticity, being a rearrangement of a prescribed function and subject to a linear constraint.
2000 Mathematics Subject Classification: 35J20, 76B47.
1. Introduction. In this paper, we prove the existence of steady two-dimensional ideal vortex flows occupying the first quadrant,Π+, containing a bounded vortex. This is done by solving a constrained variational problem. Such a flow will be described by a stream function ˆψ:Π+→R. At infinity we will have ˆψ→ −λx1x2which is the stream
function for an irrotational flow with velocity field−λ(x1,x2), whereλis not known
a priori. The vorticity is given by−∆ψˆ, where∆is the Laplacian, and−∆ψˆvanishes outside a bounded region. It will be shown that ˆψsatisfies the following semilinear partial differential equation:
−∆ψˆ=φ◦ψ,ˆ (1.1)
almost everywhere inΠ+ forφan increasing function, unknown a priori. In our re-sult the vorticity functionζ(= −∆ψ)ˆ is a rearrangement of a prescribed nonnegative, nontrivial functionζ0having bounded support, and theimpulse,, given by
(ζ):=
Π+x1x2ζ,
(1.2)
is a prescribed positive number. We prove that the variational problem, P (I) (see
Section 2), is solvable provided thatIis sufficiently large. Since the domain of interest
Π+ is unbounded, we first consider the problem over bounded sets,Π+(ξ,η), where Burton’s theory, related to constrained variational problems, can be applied. We then show that the maximizers are the same for all sufficiently largeΠ+(ξ,η).
Problems of this kind have been investigated by many authors; in particular we cite Badiani [1], Burton [2], Burton and Emamizadeh [3], Elcrat and Miller [7], Emamizadeh [8,9,10,11], Nycander [14] for theoretical results and Elcrat et al. [5,6] for numerical.
the reflections ofxabout thex1-axis,x2-axis, and the origin, respectively. For positive
ηandξwe set
Π+(η):=x∈Π+|x1x2< η
,
Π+(ξ,η):=x∈Π+|x1x2< η,max
x1,x2
< ξ. (2.1)
ForA⊂R2,|A|denotes the two-dimensional Lebesgue measure ofA.
For a measurable functionζ, the strong support ofζis defined by
supp(ζ)=x∈dom(ζ)|ζ(x) >0. (2.2)
To define the rearrangement class needed for our variational problem, we fix a non-negative, nontrivial functionζ0∈Lp(R2)which vanishes outside a bounded set. In
addition, we assume that
supp
ζ0=πa2, (2.3)
for somea >0. We say thatζis a rearrangement ofζ0if and only if
x|ζ(x)≥α=x|ζ0(x)≥α, (2.4)
for every positiveα. The set of rearrangements ofζ0which vanish outside bounded
subsets ofΠ+is denoted byᏲ. The set of functionsζ∈Ᏺthat satisfy(ζ)=I, for someI >0, is denoted byᏲ(I); and the set of functions inᏲ(I)that vanish outside
Π+(ξ,η)is denoted byᏲ(ξ,η,I); to ensure thatᏲ(ξ,η,I)≠∅, we present the following definition: letI1:= (ζ0∗), whereζ0∗is the Schwarz-symmetrisation ofζ0, and assume
that I > I1; we say thatΠ+(ξ,η) satisfies the hypothesisᏴ(I)if the following two conditions hold:
ξ≥η1/2, (2.5)
η≥4 maxa2,l(I), (2.6)
wherel(I):=(I−I1)/ζ01. Now it is immediate that ifΠ+(ξ,η)satisfiesᏴ(I), forI > I1, thenᏲ(ξ,η,I)≠∅. Indeed if we sett=l(I)1/2, then(ζ0∗)t(x):=ζ0∗(x1−t,x2−t)
belongs toᏲ(ξ,η,I).
The Green’s function for−∆onΠ+with homogeneous Dirichlet boundary condi-tions is denoted byG+, hence
G+(x,y)=21 πlog
x−yx−y
|x−y|x−y. (2.7)
Next we define the integral operatorK+
K+ζ(x)=
Π+G+(x,y)ζ(y)dy,
(2.8)
for measurable functionsζonR2, whenever the integral exists. TheKinetic energyis
defined by
Ψ(ζ)=
Π+ζK+ζ,
(2.9)
In this paper, we are concerned with constrained variational problems which are defined as follows. ForI > I1,
P (I): sup ζ∈Ᏺ(I)Ψ(ζ)
; (2.10)
and the corresponding solution set is denoted byΣ(I). IfI > I1andΠ+(ξ,η)satisfies Ᏼ(I), then we define the truncated variational problem
P (ξ,η,I): sup ζ∈Ᏺ(ξ,η,I)Ψ(ζ),
(2.11)
with the solution setΣ(ξ,η,I).
We are now in a position to state our main result.
Theorem2.1. There existsI0>0such that ifI > I0thenP (I)has a solution, that is, Σ(I)≠∅; ifζis a solution andψ:=K+ζthen the following semilinear elliptic partial differential equation holds
−∆ψ=φ◦ψ−λx1x2
, a.e. inΠ+, (2.12)
whereφis an increasing function andλ >0, both unknown a priori. Furthermore,I0 can be chosen to ensure that the vortex core, the strong support ofζ, avoids∂Π+.
3. Preliminary results. We present some lemmas that are used in the proof of
Theorem 2.1. We begin by stating a lemma from Burton’s theory, see for example, Burton and McLeod [4].
Lemma3.1. LetΩbe a nonempty open set inRn. Let1≤p <∞andp∗denote the conjugate exponent ofp. Forζ∈Lp(µ)letᏲ(Ω)denote the set of rearrangements ofζ
onΩ. Let
ᏸ:=
1≤|α|≤m
Ꮽα(x)Ᏸα (3.1)
be anmth-order linear partial differential operator, whose coefficientsᏭα are
finite-valued measurable functions onΩ, having no0th-order term, and suppose that there exists a compact, symmetric, positive linear operatorK:Lp(Ω)→Lp∗(Ω)such that if ζ∈Lp(Ω), thenKζ∈Lp∗(Ω)∩Wm,1
loc (Ω)andᏸKζ=ζalmost everywhere inΩ. Define
ˆ
Ψ(ζ):=
ΩζKζ, ζ∈L
p(Ω). (3.2)
Letw∈Lp∗(Ω)∩Wm,1
loc (Ω)be such thatᏸwis essentially constant, and define
᐀(ζ):=
Ωwζ, ζ∈L
p(Ω). (3.3)
Letb∈R. Then
(i) Ifb∈᐀(Ᏺ(Ω))then
sup ˆΨ᐀−1(b)∩Ᏺ(Ω)=sup ˆΨ᐀−1(b)∩Ᏺ(Ω)w, (3.4)
(ii) Ifbis, relatively, interior to᐀(Ᏺ(Ω)), and ifζis a maximizer forΨ relative to
᐀−1(b)∩Ᏺ(Ω), then there exist scalarλand an increasing functionφsuch that
ζ=φ◦Kζ+λw, a.e. inΩ. (3.5)
Remark3.2. It is clear that ifI > I1 andΠ+(ξ,η)satisfiesᏴ(I)then, byLemma 3.1(i),Σ(ξ,η,I)≠∅.
Before stating the next result we give the following definition: forI > I1,
σ (I):=infΨ(ζ)|ζ∈Σ(ξ,η,I),for someΠ+(ξ,η)satisfyingᏴ(I). (3.6)
We point out thatσ (I)=Ψ(ζ)ˆ for some ˆζ∈Σ(ξ0,η0,I), whereΠ+(ξ0,η0)is the
mini-mal region that satisfiesᏴ(I).
Lemma3.3. Letσ be as defined in (3.6), then
lim
I→∞σ (I)= ∞. (3.7)
Proof. LetI > I1and sett=l(I)1/2. IfΠ+(ξ,η)satisfiesᏴ(I), then(ζ0∗)t∈Ᏺ(ξ,η,I) and therefore, according to the last remark, we have
σ (I)≥Ψζ0∗
t
. (3.8)
Now applying same method as in Burton [2, Lemma 12], we obtainΨ((ζ0∗)t)≥klogt, for all sufficiently larget, hence largeI. Thus our claim is done.
LetI > I1andΠ+(ξ,η)satisfiesᏴ(I). We set
M(ξ,η,I):=(ζ,φ,λ)|ζ∈Σ(ξ,η,I)for someφ,λ∈R
such thatζ=φ◦K+ζ−λx1x2
a.e. inΠ+(ξ,η). (3.9)
Note that under the conditions imposed onξ,η,Iand in view ofLemma 3.1(ii) the set M(ξ,η,I)is nonempty. The following two inequalities are standard, see Burton [2]
K+ζ(x)≤Nmin
x1,x2
, (3.10)
∇K+ζ(x)≤N, (3.11)
for everyx∈Π+and everyζ∈Ᏺ, whereNis a universal constant.
Lemma3.4. ForI > I1we define
Λ(I):=supλ|(ζ,φ,λ)∈M(ξ,η,I)for someζ,φ
and someΠ+(ξ,η)satisfyingᏴ(I). (3.12)
Then,lim supI→∞Λ(I)≤0.
Proof. Assume that the assertion of the lemma is not true and seek a contradiction. Hence, to this end we suppose that there existsβ∈(0,∞]such that lim supI→∞Λ(I)= β. Hence there existsΛ>0 such that the set
S:=I|Λ(I) >Λ (3.13)
M(ξ,η,I) such that Π+(ξ,η)satisfies Ᏼ(I) and Λ(I)≥λ >Λ>0. Observe that by taking I sufficiently large we can ensure the existence of ξ1 such that Π+(ξ,η)⊇ Π+(ξ1,a)and|Π+(ξ1,a)| ≥πa2= |supp(ζ)|. Now define the set
U:=x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥ −λa. (3.14)
Then,Π+(ξ1,a)⊆Uand|U| ≥ |supp(ζ)|. Sinceζis essentially an increasing function
ofK+ζ−λx1x2onΠ+(ξ,η)we deduce that supp(ζ)⊆U.
Next we show that there exists a constantC >0, independent ofI∈S, such that forx∈supp(ζ)we havex1x2≤C. From (3.10) we observe that for a sufficiently large
k >0
K+ζ(x)≤Λ2x1x2, (3.15)
for allζ∈Ᏺand allxfor which min{x1,x2} ≥k. We next define
S1:=
x∈Π+|minx1,x2
≥k,
S2:=x∈Π+|minx1,x2< k, x1< α, x2< α,
S3:=x∈Π+|minx1,x2< k, maxx1,x2≥α,
(3.16)
whereα:=max{2N/λ,k}. First considerx∈supp(ζ)∩S1; then
−λa≤K+ζ(x)−λx1x2≤Λ
2x1x2−λx1x2<− λ
2x1x2, (3.17) where the first inequality follows from supp(ζ)⊆Uand the second one from (3.15); whencex1x2<2a. Next, considerx∈supp(ζ)∩S2; then we have
x1x2< α2≤
max 2N
Λ ,k 2
, (3.18)
sinceλ >Λ. Finally, considerx∈supp(ζ)∩S3; then an application of (3.10) yields
that
−λa≤K+ζ(x)−λx1x2
≤Nminx1,x2
−λx1x2
=N ααmin
x1,x2
−λx1x2
≤N
αx1x2−λx1x2 ≤N2λ
Nx1x2−λx1x2 = −λ2x1x2,
(3.19)
hencex1x2≤2a. Therefore, from above argument, it is clear that a constantC >0,
as required, exists. This, in turn, implies that
I= (ζ):=
Π+x
1x2ζ≤Cζ01. (3.20)
Lemma3.5. ForI > I1we define
A(I):=inf ess inf x∈supp(ζ)
K+ζ(x)−λx1x2
|(ζ,φ,λ)∈M(ξ,η,I)
for someΠ+(ξ,η)and someφ
,
(3.21)
whereΠ+(ξ,η)is to satisfyᏴ(I). Then,lim infI→∞A(I)≥0.
Proof. Fix >0. By definition ofA(I)there existsΠ+(ξ,η), satisfyingᏴ(I), and (ζ,φ,λ)∈M(ξ,η,I)such that
A(I)+≥ ess inf x∈supp(ζ)
K+ζ(x)−λx1x2
. (3.22)
Note that by increasingI, the size ofΠ+(ξ,η)increases as well, hence there is no loss of generality if we assume thatΠ+(ξ,η)contains the square D:=[0,2a]×[0,2a], sinceIwill eventually tend to infinity. Forx∈Dwe have
K+ζ(x)−λx1x2≥ −4a2Λ(I)+, (3.23)
whereΛ(I)+ denotes the positive part ofΛ(I), sinceK+ζis nonnegative. From this, we infer that
D⊆x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥ −4a2Λ(I)+
. (3.24)
Hence
x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥ −4a2Λ
(I)+>supp(ζ), (3.25)
since 4a2>|supp(ζ)|. Sinceζ is essentially an increasing function ofK
+ζ−λx1x2
onΠ+(ξ,η), we then deduce that
supp(ζ)⊆x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥ −4a2Λ(I)+
, (3.26)
hence, by applying (3.22), we obtainA(I)+≥ −4a2Λ(I)+. Therefore, fromLemma 3.4 we have
lim inf
I→∞ A(I)+≥0. (3.27)
Sincewas arbitrary, we derive the desired conclusion.
The next two results can be proved similarly to Burton [2, Lemmas 8 and 9]; they bear some resemblance to Pohazaev-type identities proved in Friedman and Turking-ton [12] for 3-dimensional vortex rings. We add that, contrary to Burton [2], we can give a direct proof, using the weak divergence theorem (see, e.g., Grisvard [13]) for
Lemma 3.6below without referring to any density theorems.
Lemma3.6. Let2< p <∞, letζ∈Lp(Π
+)have bounded support, and letψ:=K+ζ. Then
Π+(x·∇ψ)ζ=
Lemma3.7. Let2< p <∞, letζ∈Lp(Π
+)be nonnegative, nontrivial and vanish outside the square D(ξ) :=[0,ξ]×[0,ξ], for some ξ >0. Let λ ∈R, and let ψ := K+ζ−λx1x2. Suppose thatζ=φ◦ψalmost everywhere inD(ξ)for some increasing functionφ, and thatφhas a nonnegative indefinite integralF. Then
2
D(ξ)F◦ψ−2λ
D(ξ)x1x2ζ=
∂D(ξ)(F◦ψ)
x·n, (3.29)
wherenis the outward unit normal, and consequently
D(ξ)F◦ψ≥λ
D(ξ)x1x2ζ. (3.30)
If additionallyF(s)=0for somes≤β, then
D(ξ)ζK+ζ≥ 2λ
D(ξ)x1x2ζ+βζ1.
(3.31)
Lemma3.8. ForI > I1we define
µ(I):=inf sup x∈Π+(ξ,η)
K+ζ(x)−λx1x2
|(ζ,φ,λ)∈M(ξ,η,I)
for someΠ+(ξ,η)satisfyingᏴ(I),and someφ
.
(3.32)
ThenlimI→∞µ(I)= ∞.
Proof. It clearly suffices to show that
lim inf
I→∞ µ(I)= ∞. (3.33)
LetI > I1and consider(ζ,φ,λ)∈M(ξ,η,I)for someΠ+(ξ,η)satisfyingᏴ(I). Since K+ζ(x)−λx1x2≥Λ(I)for almost everyx∈supp(ζ), we may assume thatφ(s)=0
for−∞< s < A(I). Now write
F(s)=
s
−∞φ, (3.34)
for allsin the domain ofφ. Now, byLemma 3.7, we have
Π+ζ
K+ζ−λx1x2=2Ψ(ζ)−λI
=1 2
2Ψ(ζ)−2λI−A(I)ζ1+Ψ(ζ)+
1
2A(I)ζ1 ≥Ψ(ζ)+12A(I)ζ1
≥σ (I)+12A(I)ζ1.
(3.35)
Hence
sup Π+(ξ,η)
K+ζ(x)−λx1x2
≥ζσ (I)
1+
1
Therefore
µ(I)≥σ (I)ζ
1+
1
2A(I). (3.37)
Thus by applying Lemmas3.4and3.5we obtain (3.33).
Lemma3.9. There existsI2> I1such that
A(I)≥aN, I≥I2. (3.38)
Proof. ByLemma 3.7there existsI2> I1such that
µ(I)≥7aN, I≥I2; (3.39)
moreover by takingI2sufficiently large we can ensure that ifI≥I2, then anyΠ+(ξ,η) satisfyingᏴ(I), also satisfies
Π+(ξ,η)\Π+
ξ,η2≥πa2. (3.40)
To see it, observe that in general we have
Π+(ξ,η)=η1+logξ
2
η
, (3.41)
for anyΠ+(ξ,η)satisfying (2.5); therefore
Π+(ξ,η)\Π+
ξ,η2≥12(1−log 2)η. (3.42)
Hence, in view of (2.6), for sufficiently largeI we derive (3.40). Now, fixI ≥I2 and
consider(ζ,φ,λ)∈M(ξ,η,I)for someΠ+(ξ,η)satisfyingᏴ(I). SinceK+ζ−λx1x2∈
C(Π+(ξ,η)), it attains its maximum at, say,z∈Π+(ξ,η). Now from the definition of µ(I)and (3.10) we infer that
µ(I)≤K+ζ(z)−λz1z2≤Nmin
z1,z2
−λz1z2; (3.43)
and applying (3.39), we obtain
7aN≤Nminz1,z2
−λz1z2. (3.44)
Clearly, ifλ≥0 we obtain min{z1,z2} ≥7a. Ifλ <0, then
7aN≤Nminz1,z2
−λη, (3.45)
or
Nminz1,z2
≥7aN+λη. (3.46)
Case 1. When λη≥ −2aN, then Nmin{z1,z2} ≥5aN, hence min{z1,z2} ≥5a.
Therefore, whenλ≥0, or whenλ <0, andλη≥ −2aNwe find that min{z1,z2} ≥5a.
ThusΠ+(ξ,η)must contain at least a quadrant ofB4a(z), denoted byQ. Forx∈Q, by the mean value inequality, we have
K+ζ(x)−λx1x2≥K+ζ(x) ≥K+ζ(z)−4aN
=K+ζ(z)−λz1z2−4aN+λz1z2
≥µ(I)−4aN+λz1z2
≥µ(I)−4aN+λη ≥7aN−4aN−2aN =aN.
(3.47)
This means that
Q⊆x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥aN. (3.48)
Case2. Whenλη <−2aN, then forx∈Π+(ξ,η)\Π+(ξ,η/2)we have
K+ζ(x)−λx1x2≥ −λx1x2>−λη
2 ;
Π+(ξ,η)\Π+
ξ,η 2
⊂x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥aN.
(3.49)
From (3.40) and the fact that|Q| =4πa2, we infer that
x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥aN≥supp(ζ). (3.50)
Sinceζis an increasing function ofK+ζ−λx1x2onΠ+(ξ,η), we derive
supp(ζ)⊆x∈Π+(ξ,η)|K+ζ(x)−λx1x2≥aN, (3.51)
modulo a set of zero measure, from which we obtain (3.38).
Lemma 3.10. Let b >0, let 2< p <∞ and0< γ <1. Then there exist positive constantsM1,M2, andM3such that
K+ζ(x)≤M1x1x2−1(ζ)+M2x1x2−1(ζ)log
25x1x2
4|x| +M3
x1x2 −γ
(ζ)γζ1−γ p ,
(3.52)
for everyx∈Π+such thatmin{x1x2} ≥b/2and every nonnegativeζ∈Lp(Π+)that vanishes outside a set of measureπb2.
Proof. Fixx∈Π+such thatν:=min{x1x2} ≥b/2. Fory∈Π+we define
Thus
K+ζ(x)=21 π
Π+
logαβ
ρδζ(y)dy
=21 π
Bν/2(x)
logαβ
ρδζ(y)dy
+ 1 2π
Π+\Bν/2(x)
logαβ
ρδζ(y)dy,
(3.54)
whereBν/2(x)denotes the ball centered atxwith radiusν. From the identity
α2β2=ρ2δ2+16x1x2y1y2, (3.55)
we obtain
Π+\Bν/2(x)
logαβ
ρδζ(y)dy= 1 2
Π+\Bν/2(x)
log
1+16x1x2y1y2
ρ2δ2
ζ(y)dy
≤8x1x2
Π+\Bν/2(x)
y1y2
ρ2δ2ζ(y)dy
≤32x1x2
ν2|x|2
Π+\Bν/2(x)
y1y2ζ(y)dy
≤32x1x2 −1
(ζ),
(3.56)
where the first inequality follows from the fact that log(1+x)≤x, forx≥0. To estimateBν/2(x)log(αβρ−1δ−1)ζ(y)dy, we note that fory∈B
ν/2(x)we have
α≤x−x+x−y=2x2+ρ <
5
2x2. (3.57)
Similarly,β <5/2x1. Therefore
Bν/2(x)
logαβ
ρδζ(y)dy≤
Bν/2(x)
log25x1x2
4ρ|x| ζ(y)dy
=log25x1x2 4|x|
Bν/2(x)
ζ(y)dy
+
Bν/2(x)
log1
ρζ(y)dy.
(3.58)
Observe that fory∈Bν/2(x)we havey1y2≥x1x2/4, hence
Bν/2(x)
ζ(y)dy≤4x1x2 −1
Bν/2(x)
y1y2ζ(y)dy≤4
x1x2 −1
On the other hand, if we let ˆζdenote the Schwarz-symmetrisation ofζ:=ζχBν/2(x),
whereχBν/2(x)is the characteristic function ofBν/2(x)inΠ+, aboutx; then by a
stan-dard inequality (see, e.g., [3]) and Hölder’s inequality we obtain
Bν/2(x)
log1
ρζ(y)dy≤
Bν/2(x)
log1
ρζ(y)dyˆ
≤
Bˆb(x) log1 ρ τdy 1/τ ζˆ , (3.60)
where ˆb:= |supp(ζχBν/2(x))|(≤b),:=p/(1+pγ−γ)andτ is the conjugate
expo-nent of. It is elementary to show that
Bˆb(x) log1
ρ
τdy≤C, (3.61)
whereC is a constant independent ofx. Next observe that=γ+(1−γ)pand γ <1, hence applying the standard interpolation inequality yields
ζˆ ≤ζˆ
γ
1 ζˆ
(1−γ)p
p , (3.62)
or
ζˆ ≤ζˆ
γ
1ζˆ
(1−γ)p/
p =ζˆ
γ
1ζˆ 1−γ
p . (3.63)
Therefore, we obtain
ζˆ ≤4γ
x1x2 −γ
(ζ)γζ1−γ
p . (3.64)
Finally from (3.56), (3.58), (3.60), (3.61), and (3.64) we derive (3.52).
By a simple modification of Burton [2, Lemma 1] we get the following lemma.
Lemma3.11. Letζbe a nonnegative measurable function onΠ+, lett >0. Letζt
be the function, defined onΠ+, obtained by translatingζ along the diagonal ofΠ+, diag(Π+),√2tunits, that is,
ζt
x1,x2 := ζ
x1−t,x2−t
, x1≥t, x2≥t
0, 0< x1< t,0< x2< t.
(3.65)
Then
Π+ζtK+ζt≥
Π+ζK+ζ.
(3.66)
Lemma3.12. Let2< p <∞andζ∈Lp(Π
+)be a nonnegative, nontrivial function which vanishes outsideΠ+(h)for someh >0. Then
K+ζ(x)≤ 4hx1x2 πx12−x22
ζ1+Nminx1,x2, (3.67)
Proof. Fixx∈Π+\diag(Π+)and define
U(x):=y∈Π+ y12−y22
−x2
1−x22<x12−x22 1/2
. (3.68)
Next we decomposeζas follows:ζ:=ζ1+ζ2, where
ζ1(y):=
ζ(y), y
∈Π+(h)∩U(x),
0, otherwise. (3.69)
Again by settingα:= |x−y|,β:= |x−y|,ρ:= |x−y|,δ:= |x−y|, we obtain
K+ζ2(x)=
1 4π
Π+
logα
2β2
ρ2δ2ζ2(y)dy
= 1 4π
Π+
log
1+16x1x2y1y2
ρ2δ2
ζ2(y)dy
≤4hx1x2
π
Π+\U(x) 1
ρ2δ2ζ2(y)dy.
(3.70)
In view of the following identity:
ρ2δ2=y2 1−y22
−x21−x22 2
+4x1x2−y1y2 2
, (3.71)
we infer that ify∈Π+\U(x), thenρ2δ2>|x12−x22|. This, in conjunction with (3.70),
yields
K+ζ2(x)≤
4hx1x2
πx21−x22
ζ1. (3.72)
Finally, recalling (2.5) we obtain
K+ζ1(x)≤Nmin
x1,x2
. (3.73)
SinceK+ζ(x)=K+ζ1(x)+K+ζ2(x), (3.67) follows from (3.72) and (3.73).
Remark3.13. Under the hypotheses ofLemma 3.12withbreplaced byaand an additional assumption, namely, (ζ) ≥1 we can show the existence of a positive constantPsuch that
K+ζ(x)≤Px1x2−γ(ζ), (3.74)
provided that min{x1,x2} ≥a/2 andζ∈Ᏺ. Clearly, the truth of (3.74) emerges from
4. Proof of Theorem2.1. We first show that, forIsufficiently large, there exists a positive constantR(I)such that ifΠ+(ξ,η)is sufficiently large (satisfyingᏴ(I)) and ζ∈Σ(ξ,η,I), then
supp(ζ)⊂Π+R(I), (4.1)
modulo a set of zero measure. FromLemma 3.3, there existsI3> I1such that ifI > I3,
then
σ (I) >5
2aNζ01. (4.2)
FixI > I3and considerζ∈Σ(ξ,η,I)for someΠ+(ξ,η)satisfyingᏴ(I). From (4.2) and definition ofσ, we infer that
5
2aNζ1≤Ψ(ζ)≤ 1
2ζ1x∈supsupp(ζ)K+ζ(x),
(4.3)
thus
sup
x∈supp(ζ)K+ζ(x)≥
5aN. (4.4)
SinceK+ζ∈C(R2), it attains its maximum relative to supp(ζ)atz, say. Therefore, by applying (4.4), we obtain
5aN≤K+ζ(z)≤Nminz1,z2
, (4.5)
whence min{z1,z2} ≥5a. Without loss of generality, we may assume that(ζ)≥1,
hence, by (3.74) we obtain
5aN≤K+ζ(z)≤P Iz1z2−γ, (4.6)
so
z1z2≤
P I 5aN
γ
. (4.7)
Now we define
R(I):=max P I 5aN
γ
,25a2. (4.8)
ThenV:= {x∈Π+|x1x2≤R(I),min{x1,x2} ≥5a}is not empty andz∈V. Note that
at least a quadrant ofB4a(x), for everyx∈V, is contained inΠ+(R(I))and, in fact, contained inΠ+(ξ1,R(I))for someξ12> R(I). ByΠt+(ξ1,R(I))we denote the
transla-tion ofΠ+(ξ1,R(I))along diag(Π+),√2tunits. Observe that the family of translations {Πt
+(ξ1,R(I))}0≤t≤t0, wheret0:=(I/ζ01)1/2, is uniformly contained inΠ+(ξ2,η2),
for someξ2andη2(in fact we can takeξ2=ξ1+t0). From now on we assume thatξ > ξ2
andη > η2. Since a quadrant ofB4a(z), designated byQ, is contained inΠ+(R(I))we can apply the mean value inequality and (2.5) to deduce that
where the last inequality is obtained from (4.4). To seek a contradiction we assume thatE:=supp(ζ)\Π+(R(I))has a positive measure and writeζ=ζ0+ζ1, where
ζ1:=ζχE. (4.10)
Since|Q| =4πa2>|supp(ζ)| =πa2, there exists a measure preserving bijection,
denoted by T, fromE onto a subset of Q\supp(ζ), say G, see Royden [15]. Now define
ζ2:=ζ1◦T−1, (4.11)
on the range ofT and zero elsewhere, that is,
ζ2=
ζ1◦T−1
χim(T ), (4.12)
where im(T )is the range ofT, and letζ:=ζ0+ζ2. Clearlyζ∈Ᏺ(ξ,η). We show that
(ζ) <(ζ):
ζ=
Π+x
1x2ζ0+
Π+x
1x2ζ2
=
Π+x
1x2ζ0+
Π+x
1x2ζ1◦T−1
=
Π+x
1x2ζ0+
E
x1x2◦Tζ1
<
Π+x
1x2ζ0+
Π+x
1x2ζ1
= (ζ).
(4.13)
On the other hand, we have
Ψζ−Ψ(ζ)=
Π+
ζ2−ζ1K+ζ+Ψζ2−ζ1>
Π+
ζ2−ζ1K+ζ, (4.14)
sinceK+is strictly positive, see Emamizadeh [10]. Hence
Ψζ−Ψ(ζ) >
Π+ζ
2K+ζ−
{x∈Π+|x1x2>R(I)}
ζ1K+ζ
≥aN
Π+ζ
2−
{x∈Π+|x1x2>R(I)}
ζ1K+ζ,
(4.15)
by (4.9). Now we proceed to estimate{x∈Π+|x1x2>R(I)}ζ1K+ζ. For this purpose we set
supp(ζ)=J1∪J2, (4.16)
where
J1:= x∈supp(ζ)|x1x2> R(I), min
x1,x2
≥a2
,
J2:= x∈supp(ζ)|x1x2> R(I), min
Ifx∈J1, then by (3.74)
K+ζ(x)≤P Ix1x2 −γ
≤P IR(I)−γ. (4.18)
On the other hand, ifx∈J2then by (2.5)
K+ζ(x)≤Nminx1,x2≤a
2. (4.19)
Therefore, ifx∈supp(ζ1)
K+ζ(x)≤max P IR(I)−γ,aN2
. (4.20)
Assume thatR(I)is large enough to ensure
aN−max P IR(I)−γ,aN 2
>0. (4.21)
Therefore, we obtain
Ψζ−Ψ(ζ)≥aN−max P IR(I)−γ,aN 2
ζ11>0. (4.22)
This implies thatΨ(ζ) >Ψ(ζ). Finally, we defineζto be the function obtained by translatingζalong diag(Π+)so that(ζ)=I. If we denote the amount of translation byt, then it is clear thattis the bigger root of the following algebraic equation:
ζ 1t2+2
Π+
x1+x2ζ
t+
Π+x
1x2ζ=I. (4.23)
Note thattdepends onζ; but we are able to find a uniform bound, independent ofζ, as follows. Solving (4.23) fortyields
t= −
Π+
x1+x2ζ+
Π+
x1+x2ζ 2
−ζ1
ζ−I
1/2 ζ
1
<ζ1
I−ζ1/2<
I
ζ 1
1/2
,
(4.24)
as desired. Note that the choices ofξ2 and η2 ensure thatζ∈Ᏺ(ξ,η,I). Now, by Lemma 3.11we have
Ψζ≥Ψζ>Ψ(ζ). (4.25)
This is a contradiction to the maximality ofζ. Therefore we have been able to show that ifI > I3, then there existsR(I)given by (4.8) such that ifΠ+(ξ,η)is sufficiently large (ξ≥ξ2andη≥η2) andζ∈Σ(ξ,η,I), then, for almost everyx∈supp(ζ), (4.1)
holds.
ηare large enough thenλcan not be too negative. For this purpose let ξ≥ξ2and
η≥max{h,η2},ξ2andη2are as above, where
h:=Nλ∗−1+1R(I), λ∗:= −3aN
R(I), (4.26)
such thatΠ+(ξ,η)satisfiesᏴ(I). We show that
λ > λ∗. (4.27)
To seek a contradiction suppose thatλ≤λ∗. Without loss of generality we may assume thatR(I)≥1. Letx∈W:= {y∈Π+(ξ,η)|y1y2> h}. Then
K+ζ(x)−λx1x2>−λx1x2= |λ|x1x2>|λ|h
= |λ|Nλ∗−1+1R(I) >N+|λ|R(I). (4.28)
Now consider x ∈ supp(ζ). If max{x1,x2} ≥ 1, then min{x1,x2} ≤ x1x2, hence
min{x1,x2} ≤R(I). If, however, max{x1,x2}<1 then min{x1,x2}<1≤R(I).
There-fore in either case we have min{x1,x2} ≤R(I). This, in turn, implies that
K+ζ(x)−λx1x2≤Nmin
x1,x2
−λx1x2<
N+|λ|R(I), (4.29)
whence
sup x∈supp(ζ)
K+ζ(x)−λx1x2≤N+|λ|R(I). (4.30)
ThereforeK+ζ(x)−λx1x2 takes greater values on a nonempty subset ofΠ+(ξ,η), namelyW, than its supremum on supp(ζ). This is impossible, sinceζis essentially an increasing function ofK+ζ(x)−λx1x2onΠ+(ξ,η). Hence we derive (4.27). For the rest of the proof we fixI > I0:=max{I1,I2,I3}. Letξ > ξ2,η > h(as above) be such that Π+(ξ,η)satisfiesᏴ(I). Consider(ζ,φ,λ)∈M(ξ,η,I). Now fixx∈supp(ζ)\diag(Π+) such that min{x1,x2}< a/6. Then by Lemmas3.9and3.12, in conjunction with (4.27),
aN≤K+ζ(x)−λx1x2
≤ 4R(I)x1x2
πx2 1−x22
ζ1+Nmin
x1,x2
−λ∗x1x2
≤ 4R(I)x1x2
πx2 1−x22
ζ1+Nminx1,x2−λ∗R(I)
≤ 4R(I)x1x2
πx21−x22
ζ1+aN
6 + aN
3 .
(4.31)
Hence
x2 1−x22<
8R(I)ζ01
aπN . (4.32)
To summarise, we have shown that ifx∈supp(ζ) is such that min{x1,x2}> a/6,
thenx∈Π+(R(I))∩ {y∈Π+|min{y1,y2}> a/6}; otherwisexsatisfies (4.32). This
Now considerζ∈Σ(I). Then there exists ˆξ >0 such that supp(ζ) is a compact subset ofD(ξ)ˆ :=(0,ξ)ˆ ×(0,ξ)ˆ and, according toLemma 3.1,
ζ=φ◦K+ζ−λx1x2
, a.e. inDξˆ, (4.33)
for some increasing functionφandλ∈R. Note that fromLemma 3.9
κ:=ess supK+ζ(x)−λx1x2|x∈supp(ζ)≥aN >0. (4.34)
Since the level sets ofK+ζ−λx1x2, on supp(ζ), have zero measure, in particular we
have
x∈supp(ζ)|K+ζ−λx1x2=κ=0. (4.35)
Therefore
K+ζ−λx1x2> κ, a.e. in supp(ζ). (4.36)
Thus we may suppose thatφ(s)=0 fors≤κ. Now if we defineF(s):=0sφ(t)dt, then Lemma 3.7yields
2
D(ξ)ˆF◦ψ− 2λI=
∂D(ξ)ˆ(F◦ψ)
x·n, (4.37)
whereψ:=K+ζ−λx1x2. We claim that forx∈∂D(ξ)ˆ we haveψ≤κ. Otherwise, by
the continuity ofψwe can findB(x)such thatB(x)∩supp(ζ)has positive measure, since supp(ζ)is a compact subset ofD(ξ)ˆ, andψ(s) > κ fors∈B(x); but this is a contradiction to (4.33). Therefore, ifx∈∂D(ξ)ˆ we haveF◦ψ(x)=0. Hence from (4.37) we deduce thatλ >0, as required.
Now fixx∈supp(ζ). Sinceλ >0, we can employLemma 3.9to obtain
aN≤K+ζ(x)−λx1x2< K+ζ(x)≤Nminx1,x2. (4.38)
Thus min{x1,x2} ≥a. This proves the vortex core avoids∂Π+. The validity of (2.12) is established as in Emamizadeh [11].
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B. Emamizadeh: Department of Mathematics, Iran University of Science and
Technology, Narmak16844, Tehran, Iran
Current address: Institute for Studies in Theoretical Physics and Mathematics, Niavaran Building, Niavaran Square, Tehran, Iran
M. H. Mehrabi: Department of Mathematics, Iran University of Science and
