Existence of energy maximizing vortices in a three-dimensional quasigeostrophic shear flow with bounded height

Nonlinear Analysis: Real World Applications 11 (2010) 1589–1599

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Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa

Existence of energy maximizing vortices in a three-dimensional quasigeostrophic shear flow with bounded height Fariba Bahrami a , Jonas Nycander b,∗ , Robab Alikhani a a

Department of Mathematics, University of Tabriz, Tabriz, Iran

b

Department of Meteorology, Stockholm University, 106 91 Stockholm, Sweden

article

info

Article history: Received 4 December 2007 Accepted 13 March 2009 Keywords: Rearrangements Vortices Variational problems Quasigeostrophic three-dimensional equation

abstract The existence of an energy maximizer relative to a class of rearrangements of a given function is proved. The maximizers are stationary and stable solutions of the quasigeostrophic equation, which governs the time evolution of large-scale threedimensional geophysical flow in a vertically bounded domain. The background flow is unidirectional, with linear horizontal shear. The theorem proved implies the existence of a family of stationary and stable vortices that rotate in the same direction as the background shear. It extends an earlier theorem by Burton and Nycander, which is valid for a vertically unbounded domain. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction In this paper we will prove the existence of maximizers for a variational problem associated with large-scale geophysical flows. The basic equation governing such flows is the three-dimensional barotropic vorticity equation

∂Q + J (Q , Ψ ) = 0. ∂t Here Ψ (t , x, y, z ) is the stream function, J denotes the Jacobian, i.e. J (f , g ) =

(1)

∂f ∂g ∂g ∂f − ∂x ∂y ∂x ∂y

and Q is defined by Q = −∆Ψ , where ∆ is the three-dimensional Laplacian. Usually Q is called the potential vorticity (PV), with positive Q corresponding to cyclonic PV. The flow domain is infinite in the xy-plane, but bounded in the vertical, with the boundary conditions

∂Ψ = 0, ∂z

at z = 0 and z = H .

(2)

According to Eq. (1), Q is advected by a velocity field which is horizontally divergence-free. Therefore, Q (·, z , t ) remains in the set of rearrangements of the initial condition Q (·, z , 0). In the terminology of Burton and Nycander [1], Q (x, y, z , t ) is a stratified rearrangement of Q (x, y, z , 0). The dynamics defined by Eq. (1) also conserve the energy (to be defined below).

∗

Corresponding author. E-mail addresses: [email protected] (F. Bahrami), [email protected] (J. Nycander).

1468-1218/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2009.03.013

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F. Bahrami et al. / Nonlinear Analysis: Real World Applications 11 (2010) 1589–1599

A field Q that maximizes or minimizes the energy in a set of stratified rearrangements therefore defines a stationary and stable flow. This variational principle was used by Burton and Nycander [1] to prove the existence of a family of stationary and stable vortex solutions in a unidirectional background flow with linear horizontal and vertical shear. However, they assumed that the domain was vertically unbounded, while in virtually all applications the domain is bounded both from above and from below. Their result is therefore relevant only for vortices whose height is very small compared to the height of the atmosphere or the depth of the ocean. Since their proof relied on the convexity of the energy functional, it could not easily be extended to the bounded case, in which it has not been possible to decide whether the energy functional is convex. In the present paper we extend the result of Burton and Nycander [1] to the more realistic case of a vertically bounded domain. By extending Theorem 3 in Burton [2] to stratified rearrangements, we are able to do this without using convexity. Also, since the boundary condition at the vertical boundaries is not satisfied by a flow with non-vanishing vertical shear, we assume that the shear of the background flow is purely horizontal. Corresponding results for two-dimensional flow, proving the existence of stable vortices in a background shear flow, have earlier been obtained by Nycander [3] and Emamizadeh [4]. Similar existence proofs, based on a variational principle and rearrangement theory, have also been obtained for three-dimensional vortex rings [2], two-dimensional vortex couples [5], and vortices attached to a localized seamount [6,7]. A heuristic discussion of several of these cases has been given by Nycander [8]. 2. Notation and definitions Assuming that the background flow is unidirectional and has linear horizontal shear, we set

Ψ = ψ − c0 y2 . Substituting this expression into Eq. (1), we obtain

∂q + J (q, ψ − c0 y2 ) = 0, ∂t

(3)

where q is the PV anomaly associated with the vortex, and q = −∆ψ.

(4)

Eq. (3) conserves the energy, defined by E (q) =

1

Z

Z qKqdr −

2

Ω

Ω

c0 y2 q(r )dr ,

(5)

where r = (x, y, z ), Ω = R2 × (0, H ) and the operator K is defined by Kq(r ) =

1

Z

4π

Ω

G(r , r 0 )q(r 0 )dr 0

for r ∈ Ω .

Here G(r , r 0 ) denotes the Green’s function for −4π ∆ with homogeneous Neumann boundary condition on Ω , defined as follows: G(r , r 0 ) =

1 1 2 2

1

+

1

[(x − x0 )2 + (y − y0 )2 + (z − z 0 ) ] [(x − x0 )2 + (y − y0 )2 + (z + z 0 )2 ] 2 X 1 + 1 [(x − x0 )2 + (y − y0 )2 + (z − (2nH + z 0 ))2 ] 2 n∈Z \{0} ! 1

+ [(x −

x0 )2

+ (y −

y0 )2

+ (z + (2nH +

1

z 0 ))2 ] 2

−

1

|n|H

.

Using the operator K , we can invert Eq. (4):

ψ = Kq.

(6)

3. Statement of the main result Henceforth H denotes a fixed positive number and Ω = R2 × (0, H ). Points in R3 are denoted by r = (x, y, z ), r 0 = (x0 , y0 , z 0 ). The open box (−l, l) × (−l, l) × ( 1l , H − 1l ) is denoted by Ωl . We use the notation BR (r ) for the open ball at point r ∈ Rn with radius R > 0. For A ⊂ Rn ; n ∈ N, |A| denotes the Lebesgue measure of A. We say A ⊂ Rn , n = 2, 3 is dense

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in Rn at point r ∈ A if, for  > 0, B (r ) ∩ A has positive measure in Rn . (This definition is non-standard.) For a measurable function q on Ω the strong support (or simply support) of q is denoted by supp(q) = {r ∈ Ω |q(r ) > 0}. If f and g are measurable and non-negative functions that vanish outside sets of finite measure in Rn , n ∈ N, we say f is a rearrangement of g if

|{x ∈ Rn : f (x) ≥ α}| = |{x ∈ Rn : g (x) ≥ α}|, for α > 0. Some particular symmetric rearrangements have also been defined, such as decreasing rearrangement, Schwarz symmetrization and Steiner symmetrization. Here we make use of the doubly Steiner symmetric rearrangement (for the sake of convenience we say DSS in what follows; see [4]). The DSS of f is symmetric with respect to both axes x = 0 and y = 0, a decreasing function of x for x > 0 and fixed y, and a decreasing function of y for y > 0 and fixed x. Let q0 ∈ L∞ (Ω ) be a non-negative function which vanishes outside a set of finite measure such that for almost every z ∈ (0, H ) we have |supp(q(., z ))| < π R0 2 , for some positive R0 . The set of all rearrangements with bounded support of q0 on Ω is denoted FΩ (q0 ). The subset of FΩ (q0 ) comprising functions vanishing outside Ωl is denoted FΩl (q0 ), and the closed convex hull in L2 (Ω ) of FΩ (q0 ) is denoted FΩ (q0 )w . Now we can define

RΩ (q0 ) = {q ∈ L2 (Ω ) : q(., z ) ∈ FΩ (z ) (q0 (., z )) for a.e. real z },

=Ω (q0 ) = {q ∈ L2 (Ω ) : q(., z ) ∈ FΩ (z ) (q0 (., z ))w for a.e. real z }, where Ω (z ) = {(x, y) : (x, y, z ) ∈ Ω }. For l > R0 we can define RΩl (q0 ) and =Ωl (q0 ) as subsets of RΩ and =Ω comprising functions that vanish outside Ωl . Since q0 (., z ) ∈ L2 (Ωl (z )) these definitions are well defined, and we refer to elements of RΩ (q0 ) as stratified rearrangements of q0 . We denote the following maximization problem by P: P :

sup

q∈RΩ (q0 )

E (q).

(7)

The set of solutions of P is denoted by Σ . Similarly, for a box Ωl , we define P (l) as follows: P (l) :

sup

q∈RΩl (q0 )

E (q),

(8)

and the set of solutions of P (l) is denoted Σ (l). We are now in a position to state our main result. Theorem 1. Let q0 ∈ L∞ (Ω ) be non-negative and have bounded support. Let c0 be a positive number. Then there exists a maximizer q¯ ∈ Σ such that it is DSS and ψ := K q¯ satisfies

− ∆ψ = ϕ(ψ(x, y, z ) − c0 y2 , z )

(9)

almost everywhere in Ω , where ϕ(., z ) is increasing for almost every real z. 4. Preliminaries We begin by a result from Burton [2]. Lemma 1. For l > R0 and s ≥ 1 we have (i) kqks = kq0 ks , for q ∈ FΩl (q0 ); (ii) FΩl (q0 )w is weakly sequentially compact in Ls (Ωl );

(iii) FΩl (q0 )w = {q ∈ L1 (Ωl ) : rearrangement of q0 . Proof. See [2,9].

Rs 0

q4 (t )dt ≤

Rs 0

q0 4 (t )dt , 0 < s ≤ |Ωl | and

R

Ωl

q =

R

Ωl

q0 }, where q4 is the decreasing



Lemma 2. Let l > R0 and f , g : Ωl → R be non-negative and measurable functions. Suppose that, for almost all z ∈ (0, H ), for every α ∈ R, Gα (z ) = {(x, y)|g (x, y, z ) = α} has zero measure. Then ϕ : R × (0, H ) −→ R exists such that f¯ (x, y, z ) = ϕ(g (x, y, z ), z ) is a stratified rearrangement of f (x, y, z ) and ϕ is an increasing function for almost every z ∈ (0, H ). Proof. For almost every z ∈ (0, H ), let g 4 (t , z ), where t ∈ (0, ∞), be the decreasing rearrangement of g (., z ), defined by

|{t ∈ (0, 4l2 )|g 4 > α}| = |{(x, y) ∈ Ωl |g (x, y, z ) > α}| for α > 0. Since the level sets of g (., z ) have zero measure, g 4 (., z ) is strictly decreasing and therefore injective. Then we can follow the proof of Lemma (2.9) in [9]. 

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The next lemma is crucial. It extends Theorem 3 in [2] to stratified rearrangements. Lemma 3. Let l > R0 and f0 , g ∈ L2 (Ωl ) be non-negative functions. Suppose there is a function ϕ(., z ) such that f ∗ (x, y, z ) = ϕ(g (x, y, z ), z ) is a stratified rearrangement of f0 , where ϕ(., z ) is increasing for almost every z ∈ (0, H ). Then f ∗ is the unique R maximizer of the functional h., g i relative to =Ωl (f0 ), where hf , g i = Ω fg. l

Proof. With only a slight modification in the proof of Lemma 3 in [2], we can prove that f ∗ maximizes the functional h., g i relative to =Ωl (f0 ). It remains to prove that this maximizer is unique. Let O ⊂ (0, H ) be such that |(0, H ) \ O| = 0 and all of the lemma’s conditions are satisfied on O. For z ∈ O we have f ∗ (x, y, z ) = ϕ(g (x, y, z ), z ),

a.e. on Ωl (z ),

(10)

where ϕ(., z ) : R → R is increasing and f ∗ (., z ) ∈ FΩl (z ) (f0 (., z )). From Theorem 3 in [2], f ∗ (., z ) is an unique maximizer of the functional h., g (., z )i on FΩl (f0 (., z ))w . Therefore,

Z Ωl (z )

(f ∗ (x, y, z ) − f (x, y))g (x, y, z )dxdy ≥ 0 for f ∈ FΩl (f0 (., z ))w .

(11)

Suppose the claim in the lemma is false. Then there exists another maximizer, say, f¯ ∈ =Ωl (f0 ). This, along with the fact that f ∗ is a maximizer too, yields

Z Ωl

(f ∗ − f¯ )g =

H

Z

Z Ωl (z )

0

(f ∗ (x, y.z ) − f¯ (x, y, z ))g (x, y, z )dxdydz = 0.

(12)

Since f¯ ∈ =Ωl (f0 ), then f¯ (., z ) ∈ FΩl (f0 (., z ))w . By (11), we deduce

Z Ωl (z )

(f ∗ (x, y, z ) − f¯ (x, y, z ))g (x, y, z )dxdy ≥ 0.

(13)

By comparing (12) and (13), we have

Z Ωl (z )

(f ∗ (x, y, z ) − f¯ (x, y, z ))g (x, y, z )dxdy = 0.

Then f¯ (., z ) is a maximizer for the functional h., g (., z )i on FΩl (f0 (., z ))w . But this contradicts the uniqueness of f ∗ (., z ). Therefore, we have f ∗ (x, y, z ) = f¯ (x, y, z ), for almost every z ∈ (0, H ). This completes the proof of Lemma 3.



The following lemma estimates the asymptotic behavior of the Green’s function G. Lemma 4. Let r = (x, y, z ), r 0 = (x0 , y0 , z 0 ) ∈ Ω , and let ρ = (x, y) and ρ 0 = (x0 , y0 ). Then

(i) G(r , r 0 ) ≤

2

|r − r 0 |

4

+

1 2 2

[|ρ − ρ 0 |2 + (z − (2H − z 0 )) ] (ii) G(r , r ) ≤ C1 − C2 ln |ρ − ρ 0 |, for |ρ − ρ 0 | > 2H ,

+C

∞ X 1 n =2

n2

0

where C and C1 , C2 are constants independent of r , r 0 , and C2 is positive. Proof. For (i), from the definition of the Green’s function we obtain G(r , r ) ≤

2

0

|r − r 0 |

+

4 1

[|ρ − ρ 0 |2 + (z − (2H − z 0 ))2 ] 2

−

2 H

+

∞ X

4

n =2

[|ρ − ρ 0 |2 + (2nH − z − z 0 )2 ] 2

1

−

2 nH

! . (14)

Since 0 < z , z 0 < H, we get (i) by a simple calculation (notice that the two terms in the series of Eq. (14) cancel to order 1/n). For (ii), we use the following inequalities: ln(N + 1) ≤

N X 1 n =1

n

≤ ln N + 1,

for N ≥ 1.

(15)

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Let |ρ − ρ 0 | = a, where a ≥ 0. Since z + z 0 < 2H, we deduce from (14) 2

G(r , r ) ≤ 0

+

|r − r 0 | 2

=

|r −

+

r 0|

4 a 4 a

2

−

+2

H

+2

∞ X

2 1

n =2

∞ X

[a2 + (2(n − 1)H )2 ] 2 ! 2

1 2 2

−

[a2 + 4n2 H ]

n=1

1

nH

−

1

!

nH

.

(16)

Since every term in the series is negative we can omit some of the terms. Let N be the largest integer satisfying N ≤

a 2H

.

(17)

By (16) we then have G(r , r 0 ) ≤

2

|r − r 0 |

+

4 a

+2

N  X 2

a

n =1

−

1



nH

.

(18)

From (15) and (18) we obtain G(r , r 0 ) ≤

2

|r − r 0 |

+

4 a

+

2 H

−

2 H

ln

 a  2H

.

(19)

Thus, for |ρ − ρ 0 | > 2H there exist C1 , C2 such that (ii) in the lemma is satisfied.



The following lemma clarifies the compactness of operator K . Lemma 5. Let l > R0 ; then K : L2 (Ωl ) → L2 (Ωl ) is compact. Proof. Since the domain is bounded, we obtain from Lemma 4, part (i), and the similar estimate for the lower bound of G:

|Kq(r )| ≤ C kqk2 ,

for q ∈ L2 (Ωl ),

(20)

where C is a constant depending on the measure of Ωl . Since Kq is a distribution solution of −∆u = q on Ω , then, by Agmon’s 2,2 results [10], Kq ∈ Wloc (Ω ) and −∆Kq = q is satisfied almost everywhere on Ω . From Theorem 9.11 in [11] we have

kKqk2,2,Ωl ≤ C 0 (kKqk2,Ωl+1 + kqk2,Ωl+1 ),

(21)

where C 0 depends on Ωl . Since kKqk1,2,Ωl ≤ kKqk2,2,Ωl , and (20) implies that kKqk1,2,Ωl ≤ C 00 kqk2 for a constant C 00 depending on l, this proves the boundedness of K : L2 (Ωl ) → W 1,2 (Ωl ). The compactness of K is obtained by compact embedding from W 1,2 (Ωl ) to L2 (Ωl ).  Lemma 6. Let q ∈ Lp (Ω ), p > 3, vanishing outside Ωl for some large enough l > 0, be DSS relative to x, y for almost every z ∈ (0, H ), and let Ψ = Kq − c0 y2 . Then Ψ ∈ C 1 (Ωl ) and ∂ Ψ /∂ y 6= 0 for all y 6= 0. 2,p

Proof. Let q ∈ Lp (Ω ) for some p > 3; then Ψ ∈ Wloc (Ω ), and from Sobolev embedding we have Ψ ∈ C 1 (Ω ). We show that ∂ Ψ /∂ y < 0 for y > 0; it then follows by symmetry that ∂ Ψ /∂ y < 0 for y > 0. We have

∂ ∂ Ψ (x, y, z ) = Kq(x, y, z ) − 2c0 y. ∂y ∂y From the definition of the Green’s function, we obtain 1 ∂ Kq(x, y, z ) = − ∂y 4π

+

Z " Ωl

y − y0 3 2 2

[(x − x0 )2 + (y − y0 )2 + (z − z 0 ) ]

+

y − y0 3

[(x − x0 )2 + (y − y0 )2 + (z + z 0 )2 ] 2

y − y0

X

3

[(x − x0 )2 + (y − y0 )2 + (z − (2nH + z 0 ))2 ] 2 !# y − y0 + q(r 0 )dr 0 . 3 [(x − x0 )2 + (y − y0 )2 + (z + (2nH + z 0 ))2 ] 2 n∈Z \{0}

We consider the first term of the above expression: I (x, y, z ) =

y − y0

Z Ωl

3

[(x − x0 )2 + (y − y0 )2 + (z − z 0 )2 ] 2

q(r 0 )dr 0 .

(22)

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F. Bahrami et al. / Nonlinear Analysis: Real World Applications 11 (2010) 1589–1599

We show that I (x, y, z ) ≥ 0 for all y > 0, and hence that I (x, y, z ) ≤ 0 for all y < 0. The similar inequalities for the other terms of Eq. (22) can be proved analogously. If y ≥ l, then y − y0 ≥ 0 for y0 ∈ supp(q(r 0 )); hence I ≥ 0. Fix α so that 0 < α < l. We then have I (x, α, z ) =

=

α − y0

Z

3

[(x − x0 )2 + (α − y0 )2 + (z − z 0 )2 ] 2 Z H Z l Z 2α−l α − y0 Ωl

q(r 0 )dr 0

3

−l

−l

0

H

Z

[(x − x0 )2 + (α − y0 )2 + (z − z 0 )2 ] 2 α − y0

Z lZ

α

−l

2α−l

q(r 0 )dr 0

q(r 0 )dr 0 3 [(x − x0 )2 + (α − y0 )2 + (z − z 0 )2 ] 2 Z HZ lZ l α − y0 + q(r 0 )dr 0 3 −l α [(x − x0 )2 + (α − y0 )2 + (z − z 0 )2 ] 2 0 = I1 + I2 + I3 .

+

0

If −l < y0 < 2α − l, then α − y0 > −α + l > 0; hence I1 ≥ 0. Denote the reflection of r = (x, y, z ) ∈ Ω about the line y = α by rα = (xα , yα , zα ); that is, rα = (x, 2α − y, z ). By the change of variables r 0 = rα0 we then obtain H

Z

Z lZ

I2 = − −l

0

α − y0

l

α

[(x −

x0 )2

+ (α −

y0 )2

+ (z −

3

z 0 )2 ] 2

q(rα0 )dr 0 .

Now if α < y < l, then −y < < y , and since q(x, y, z ) is DSS relative to x, y for almost every z ∈ (0, H ), then q(rα0 ) ≥ q(r 0 ) for almost every z 0 ∈ (0, H ). Therefore 0

0

H

Z

Z lZ

I2 + I3 = 0

−l

y0α

0

α − y0

l

α

[(x −

x0 ) 2

+ (α −

y0 )2

+ (z −

3

z 0 )2 ] 2

(q(r 0 ) − q(rα0 ))dr 0 ≥ 0.

Hence I1 + I2 + I3 ≥ 0. Similar inequalities can be proved in the same way for the other terms in Eq. (22); hence, for all y > 0, (∂/∂ y)Kq(x, y, z ) ≤ 0. Since −2c0 y < 0, then (∂/∂ y)Ψ (x, y, z ) < 0 for all y > 0. The inequality (∂/∂ y)Ψ (x, y, z ) > 0 for y < 0 follows by symmetry, since q is DSS.  5. Existence in a bounded domain In this section we consider the existence of maximizers for the energy functional in a bounded domain. Lemma 7. Let l > R0 , q0 ∈ L∞ (Ωl ) be non-negative and have compact support. Let c0 be a positive number. Then there exists a maximizer q¯ ∈ Σ (l) such that ψ := K q¯ satisfies

− ∆ψ = ϕ(ψ(x, y, z ) − c0 y2 , z )

(23)

almost everywhere in Ωl , where ϕ(., z ) is increasing for almost every real z ∈ ( , H − ). Also, the solution is DSS. 1 l

1 l

Proof. Because of the compactness of K , the energy functional is weakly sequentially continuous on L2 (Ωl ). Then, by Lemma 1(ii), there exists q¯ ∈ =Ωl (q0 ) such that it maximizes the functional E on =Ωl (q0 ). Fix q ∈ =Ωl (q0 ). Then, for t ∈ [0, 1], we have q¯ + t (q − q¯ ) ∈ =Ωl (q0 ), since =Ωl (q0 ) is convex. By calculating the first variation of E at q¯ , we obtain E (¯q + t (q − q¯ )) = E (¯q) + t hDE (¯q), q − q¯ i + o(t )

(24)

as t → 0+ . Here DE (¯q) denotes the Gâteaux-derivative of E at q¯ . This, in turn, along with the fact that q¯ is a maximizer, yields hDE (¯q), q − q¯ i ≤ 0. Thus, q¯ maximizes hDE (¯q), .i with respect to =Ωl (q0 ). Since DE (¯q) = K q¯ − c0 y2 , then q¯ solves the following variational problem: max

q∈=Ωl (q0 )



K q¯ − c0 y2 , q .



(25)

Let qˆ be the DSS of q¯ . From Lemma 1(iii), qˆ ∈ =Ω (q0 ), and by Riesz inequality [12], qˆ solves (25). Since q0 ∈ L∞ (Ω ), then, using Lemma 1(iii) again we have qˆ ∈ L∞ . This means that we can suppose the solution q¯ of the variational problem (25) to be DSS in L∞ (Ω ). For z ∈ (0, H ), let Gα (z ) := {(x, y) ∈ (−l, l) × (−l, l) : K q¯ (x, y, z ) − c0 y2 = α}, where α is an arbitrary real number. Let z ∈ (0, H ) be a fixed number. If |Gα (z )| > 0 for some α > 0, then by Lemma 7.7 in [11], ∂ Ψ /∂ y = 0. However, this contracts Lemma 6. Therefore |Gα (z )| = 0 for α > 0. Setting g = K q¯ − c0 y2 , Lemma 2 then

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shows that there exists ϕ(., z ) such that qˆ = ϕ(g (x, y, z ), z ) is a stratified rearrangement of q0 . We can also use Lemma 3 to deduce that qˆ is the unique maximizer of the variational problem (25); hence q¯ = qˆ and q¯ is a maximizer of E on RΩl (q0 ). This completes the existence part of the proof. Since −∆K q¯ = q¯ almost everywhere on Ωl , we have

− ∆K q¯ = ϕ(K q¯ − c0 y2 , z ) a.e. on Ωl . 

(26)

Remark. Let q ∈ Σ (l) for l > R0 and put Ψ (r ) = Kq(r ) − c0 y2 . From the proof of Lemma 6 we know that the level sets of Ψ (., z ) have zero measure for z ∈ (0, H ). Let z0 be such that (26) is satisfied for z = z0 . Suppose supp(q(., z0 )) is dense in R2 at the point (x0 , y0 , z0 ), then, by continuity of Ψ and Lemma 3.8 in [4], we have

Ψ (x, y, z0 ) < Ψ (x0 , y0 , z0 ),

(27)

for almost every (x, y) ∈ Ωl (z0 ) \ supp(q(., z0 )). 6. Extension to an unbounded domain Lemma 7 was proved on a bounded domain Ωl , and the solution could depend on the choice of l. In the following lemmas we investigate the dependence of the solutions on l to show that, if l is chosen large enough, it ceases to have any influence whatever on the variational problem, P (l). Lemma 8. There exists l∗ > R0 such that for l > l∗ and q¯ ∈ Σ (l) we have

Z Ωl∗

q¯ >

1 2

kq0 k1 .

Proof. Suppose that the R R assertion1 in the lemma is false. Then for any l > R0 there exists ¯l > l and q¯ ∈ Σ (¯l) such that 1 q¯ ≤ 2 kq0 k1 . Hence Ω \Ω q¯ ≥ 2 kq0 k1 . By Lemma 7, q¯ is DSS. Therefore, by Lemma 5 in [13], ψ = K q¯ is DSS, and we have Ω l

l

ψ(r ) ≤ ψ(0),

(28)

where 4π ψ(0) =

Z Ωl

G(0, r 0 )¯q(r 0 )dr 0 +

Z Ω \Ωl

G(0, r 0 )¯q(r 0 )dr 0 .

(29)

From (28) and (29) and Lemma 4 we deduce 4π ψ(r ) ≤ C + (C1 − C2 ln l)

Z

q¯ dr 0 Ω \Ωl

1

≤ C + (C1 − C2 ln l) kq0 k1 , 2

(30)

where we assumed l > 2H and l > exp(C1 /C2 ), so that C1 − C2 ln l is negative. Here C depends only on q0 , and C1 and C2 are as in Lemma 4. From (30) and the definition of the energy functional we deduce the following upper bound for the maximum energy on RΩ¯l : E (¯q) ≤

  kq0 k1 1 C + (C1 − C2 ln l) kq0 k1 . 8π 2

Let q∗ ∈ RΩ¯l (q0 ) be such that q∗ (., z ) is a Schwartz symmetrization of q0 (., z ) for almost every z ∈ (0, H ). (This means that

q∗ (., z ) is spherical and decreasing with respect to ρ = (x2 + y2 )1/2 .) We can then choose l so large that E (¯q) < E (q∗ ), which is in contradiction to q¯ being an energy maximizer.  Lemma 9. There exists l∗ such that, if q is a DSS solution of P (l) for l > l∗ , then, for l∗ < l0 < l, supp(q(., z )) is not dense in R2 at either of the points Tl0 (z ) = (l0 , 0) or T¯l0 (z ) = (−l0 , 0) for almost every z ∈ (0, H ). ∗ Proof. Since we are considering √ ∗ DSS solutions, it suffices to prove the lemma only for Tl0 . To do this, let l be as in Lemma 8 and l > R where R = 2l . Let q be a DSS solution of P (l), and set ψ = Kq. We proceed to derive a lower bound for ψ(r 0 ) − ψ(l, 0, z 0 ) for r 0 ∈ BR (0, 0, z 0 ). This is done in two steps. We first find a lower bound for ψ(0, 0, z 0 ) − ψ(l, 0, z 0 ), and then a lower bound for ψ(r 0 ) − ψ(0, 0, z 0 ). By the notation

G0 (x, y, z , z 0 ) := G((0, 0, z 0 ), (x, y, z )), Gl (x, y, z , z 0 ) := G((l, 0, z 0 ), (x, y, z )),

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we have

ψ(0, 0, z 0 ) − ψ(l, 0, z 0 ) = Kq(0, 0, z 0 ) − Kq(l, 0, z 0 ) Z 1 [G0 − Gl ]q(r )dr . = 4π Ωl

(31)

Note that G0 (x, y, z , z 0 ) = Gl (l − x, y, z , z 0 ).

(32)

We define S1 = {r ∈ Ωl |x > 0}, S2 = {r ∈ Ωl |x < 0,

p

(x2 + y2 ) < R},

S3 = Ωl − (S1 ∪ S2 ). Let us show that the integral over S1 in (31) is positive. To do this, we introduce the following sets:



+

S1 = S1− =



r ∈ S1 |x > r ∈ S1 |x <

l



2 l



2

, .

We obtain

Z

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr Z Z = [G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr + [G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr

S1

S

−

S

−

+ S1

Z1

  [G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r ) + [Gl (x, y, z , z 0 ) − G0 (x, y, z , z 0 )]q(l − x, y, z ) dr

= Z1 = − S1

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )][q(r ) − q(l − x, y, z )]dr

≥ 0,

(33)

where both factors in the last integral are positive on S1− , since G0 and q are both decreasing functions of x. (For q, this follows from being DSS.) In the integral over S1+ we performed a variable substitution from x to l − x, and used Eq. (32). We then estimate the integral over S2 :

Z

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr Z ≥ [G0 (R, z , z 0 ) − Gl (l, z , z 0 )]q(r )dr ,

S2

S2

where G0 (R, z , z 0 ) is obtained by substituting R2 for x2 + y2 in G0 (x, y, z , z 0 ), and Gl (l, z , z 0 ) by substituting l2 for (l − x)2 + y2 in Gl (x, y, z , z 0 ). Using G0 (R, z , z 0 ) > A, where A is a constant depending on R, and Gl (l, z , z 0 ) ≤ C1 − C2 ln l from Lemma 4, we obtain

Z

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr ≥ (C + C2 ln l)β(l),

(34)

S2

where C is a constant depending on R, C2 is the same as in Lemma 4, and

β(l) =

Z

q(r )dr = S2

1 2

Z

q(r )dr . BR,H

Here BR,H = {(x, y, z )|x2 + y2 < R2 , 0 < z < H }. From Lemma 8, we obtain

β(l) ≥

1 4

kq0 k1 .

(35)

F. Bahrami et al. / Nonlinear Analysis: Real World Applications 11 (2010) 1589–1599

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Finally, we estimate the integral over S3 . Clearly,

Z

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(x, y, z )dxdydz ≥ 0.

(36)

S3

From Eqs. (33), (34) and (36), we obtain

ψ(0, 0, z 0 ) − ψ(l, 0, z 0 ) ≥ ≥

Z

1 4π

[G0 (x, y, z , z 0 ) − Gl (x, y, z , z 0 )]q(r )dr S2

1

(C + C2 ln l)β(l).

4π

(37)

The second step is to find a lower bound for ψ(x0 , y0 , z 0 ) − ψ(0, 0, z 0 ). To do this we use the following general inequality, valid for r , s, t ∈ R3 : 1

1

−

|r − s|

|r − t |

1

≥

|r − t | + |t − s| |t − s| . ≥− |r − t |2

1

−

|r − t | (38)

We assume henceforth that ρ = (x, y) and ρ 0 = (x0 , y0 ) for r , r 0 ∈ R3 . We have

ψ(x0 , y0 , z 0 ) − ψ(0, 0, z 0 ) =

Z

1 4π

Ωl

[G − G0 ]q(r )dr .

We estimate G − G0 term by term using the definition of the Green’s function and the inequality (38). (For example, when estimating the difference involving the first term in the series of the Green’s function we use (38) with s = (x0 , y0 , 2nH + z 0 ) and t = (0, 0, 2nH + z 0 ).) This gives

ψ(x0 , y0 , z 0 ) − ψ(0, 0, z 0 ) ≥ −

Z

1 4π

Ωl

F (r 0 , r )q(r )dr ,

where

p F (r , r ) = 0

x0 2 + y0 2

p

x2 + y2 + (z 0 − z )2

+

"

p

+

X n∈Z \{0}

x0 2 + y0 2

x2 + y2 + (z 0 + z )2

p

x0 2 + y0 2

x2 + y2 + (z − (2nH + z 0 ))2

+

x0 2 + y0 2

x2 + y2 + (z + (2nH + z 0 ))2

# .

(39)

By Riesz’ inequality [12], we obtain

ψ(x , y , z ) − ψ(0, 0, z ) ≥ − 0

0

0

0

1 4π

Z

F (r 0 , r )q∗ (r )dr , BR,H

where the domain of the integral could be restricted since supp(q∗ (r )) ⊂ BR,H . By applying Hölder’s inequality, we deduce

ψ(x0 , y0 , z 0 ) − ψ(0, 0, z 0 ) ≥ −

1 4π

Z

F (r 0 , r )dr .

kq0 k∞

(40)

BR,H

Therefore, from (37), (35) and (40) we deduce 2

ψ(x0 , y0 , z 0 ) − ψ(l, 0, z 0 ) − c0 y0 " # Z 1 kq0 k1 0 ≥ (C + C2 ln l) − kq0 k∞ F (r , r )dr − c0 R2 , 4π 4 BR,H

(41)

0 0 0 02 02 2 F (r ) ∈ L1 (BR,H ) such that Rfor r 0 ∈ BR (0, R0, z ). Since 00 < z < H 0and (x + y ) ≤ R , then from (39) there exists F (r , r )dr ≤ F (r )dr for r ∈ BR (0, 0, z ). Hence, we can choose l large enough that Ψ (x0 , y0 , z 0 ) − Ψ (l, 0, z 0 ) >  on a subset with positive measure of BR (0, 0) \ supp(q(., z 0 )) for some  > 0. (Here BR (0, 0) is an open disc in R2 .) Therefore, by (27), supp(q(., z 0 )) cannot be dense in R2 at points Tl (z 0 ) = (l, 0, z 0 ) for almost every z 0 ∈ (0, H ). Note that the choice of l is independent of r 0 . Then, there exists l∗ such that for DSS solutions of P (l), say q, supp(q(., z 0 )) cannot be dense in R2 at Tl (z 0 ) for almost every z 0 for l > l∗ . Let l > l∗ and q ∈ Σ (l) be DSS. Fixing l0 so that l∗ < l0 < l, we can prove that supp(q(., z 0 )) cannot be dense in R2 at Tl0 (z 0 ) for almost every z 0 by replacing l by l0 in the proof above. 

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F. Bahrami et al. / Nonlinear Analysis: Real World Applications 11 (2010) 1589–1599

Lemma 10. There exists ˆl such that if q is a DSS solution of P (l) for l > ˆl, then supp(q(., z )) is not dense in R2 at the points Ql0 (z ) = (0, l0 ) and Q¯ l0 (z ) = (0, −l0 ) for almost every z ∈ (0, H ) and ˆl < l0 ≤ l. Proof. Since q is DSS, we only need to consider Ql0 (z ). Fix l > 2R0 and let A = {r ∈ Ω |R0 < r 0 ∈ A we have 4π

Z Ωl (z )

0

H

Z

H

Z

1

Ψ (r 0 ) =

G(r , r 0 )q(r )dr − c0 y0

x2 + y2 < 2R0 }; then for

2

√

G(2 2l, z , z 0 )kq0 (., z )k1 dz − 4R0 2 c0 ,

≥

p

(42)

0

where G(s, z , z 0 ) is obtained by substituting s2 for (x − x0 )2 + (y − y0 )2 in G(r , r 0 ). Also, using the inequality (20) obtained from Lemma 4, we get

Ψ (Ql (z 0 )) = Kq(Ql (z 0 )) − c0 l2

≤ C (l)kq0 k2 − c0 l2 ,

(43)

where, from Lemma 4, C (l)/l → 0, l → ∞. Since 2

√

lim

G(2 2l, z , z 0 )

l→∞

l2

= 0,

there exists ˆl sufficiently large that for all l ≥ ˆl we have

Ψ (Ql (z )) < Ψ (r 0 ) ∀r 0 ∈ A, ∀l ≥ ˆl.

(44)

Also,

|A(z 0 )| = 4π R0 2 − π R0 2 = 3π R0 2 > |supp(q0 (., z 0 ))|, where A(z 0 ) = {r ∈ A|z = z 0 }. Thus, there exists B(z 0 ) ⊆ A(z 0 ) such that supp(q(., z 0 )) ∩ B(z 0 ) = ∅, |B(z 0 )| > 0. From (44) we have Ψ (r 0 ) > Ψ (Ql (z 0 )) for all r 0 ∈ B(z 0 ). Then by (27) supp(q(., z 0 )) cannot be dense in R2 at Ql (z 0 ) for almost every z 0 ∈ (0, H ). The argument for Ql0 (z 0 ) = (0, l0 , z 0 ), ˆl < l0 < l, is similar. Since the choice of ˆl is independent of z 0 ∈ (0, H ), this completes the proof.  7. Proof of theorem Let l∗ be as in Lemma 9 and ˆl as in Lemma 10, and assume q to be a DSS solution of P (l) for l > max{l∗ , ˆl}. We claim that supp(q) cannot be dense at the point (0, l, z ) in R3 for z ∈ (0, H ). To prove this, suppose supp(q) is dense at (0, l, z ). Then, for all δ > 0, |Bδ (0, l, z ) ∩ supp(q)| > 0. Therefore, we can find ˆl < l0 < l such that supp(q(., z 0 )) is dense in R2 at (0, l0 , z 0 ) for z 0 ∈ I with |I | > 0, which contradicts Lemma 10. A similar argument using Lemma 9 implies that supp(q) is not dense at the point (l, 0, z ). Let ¯l = max{l∗ , ˆl}; then these results, together with the fact that q is DSS, imply that supp(q) ⊂ Ω¯l for l > ¯l. This completes the existence part of the theorem. Moreover, by Lemma 7, we have

−∆ψ = ϕ(ψ(x, y, z ) − c0 y2 , z ) almost everywhere on Ω¯l , where ϕ(., z ) is increasing for almost every z ∈ (0, H ). To extend this to hold on Ω , let q¯ be an energy maximizer on Ωl for l > ¯l. Then there exists κ > 0 so that Ψ > κ almost everywhere on supp(¯q). By the definition of Ψ , we can find l > ¯l so that Ψ ≤ κ outside Ωl . Define

ϕ( ¯ t, z) =

 ϕ(t , z ), 0,

t >κ otherwise.

Clearly ϕ¯ is an increasing function and q¯ = ϕ¯ ◦ (Ψ , z ) almost everywhere on Ω .



Acknowledgements The first author thanks Dr. Behrouz Emamizadeh for useful conversations and wishes to acknowledge grants from the University of Tabriz, as well as the International Meteorology Institute at Stockholm university which made possible extended visits to the Department of Meteorology at Stockholm University. We are grateful to the anonymous reviewer for comments that led to substantial improvement of the paper.

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References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

G.R. Burton, J. Nycander, Stationary vortices in three-dimensional quasi-geostrophic shear flow, J. Fluid Mech. 389 (1999) 255–274. G.R. Burton, Rearrangements of functions, maximization of convex functionals, and vortex rings, Math. Ann. 276 (1987) 225–253. J. Nycander, Existence and stability of stationary vortices in a uniform shear flow, J. Fluid Mech. 287 (1995) 119–132. B. Emamizadeh, Steady vortex in a uniform shear flow of an ideal fluid, Proc. Roy. Soc. Edinburgh 130A (2000) 801–812. G.R. Burton, Steady symmetric vortex pairs and rearrangements, Proc. Roy. Soc. Edinburgh 108A (1988) 269–290. J. Nycander, B. Emamizadeh, Variational problem for vortices attached to seamounts, Nonlinear Anal. TMA 55 (2003) 15–24. F. Bahrami, J. Nycander, Existence of energy minimizing vortices attached to a flat-top seamount, Nonlinear Anal. RWA 8 (2007) 288–294. J. Nycander, Stable vortices as maximum or minimum energy flows, in: O.U. Velasco Fuentes, Julio Sheinbaum, Jose Luis, Ochoa Torre (Eds.), Nonlinear Processes in Geophysical Fluid Dynamics, Kluwer, Dordrecht, 2003. G.R. Burton, Variational problems on classes of rearrangements and multiple configurations for steady vortices, Ann. Inst. H. Poincaré – Anal. Non Linéaire 6 (1989) 295–319. S. Agmon, The Lp approach to the Dirichlet problem, Ann. Scuola Norm. Pisa Cl. Sci. 13 (1956) 405–448. D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, 2nd ed., Springer-Verlag, Berlin, New York, 1983. E.H. Lieb, M. Loss, Analysis, in: Graduate Studies in Mathematics, vol. 14, American Mathematical Society, Providence, RI, 1997. G.R. Burton, B. Emamizadeh, A constrained variational problem for steady vortices in a shear flow, Comm. Partial Differential Equations 24 (1999) 1341–1365.