Existence of steady symmetric vortex patch in an open disk

EXISTENCE OF STEADY SYMMETRIC VORTEX PATCH IN AN OPEN DISK

arXiv:1807.02993v1 [math.AP] 9 Jul 2018

DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO Abstract. In this paper we construct a family of steady symmetric vortex patches for the incompressible Euler equations in an open disk. The result is obtained by studying a variational problem in which the kinetic energy of the fluid is maximized subject to some appropriate constraints for the vorticity. Moreover, we show that these vortex patches shrink to a given minimum point of the corresponding Kirchhoff-Routh function as the vorticity strength parameter goes to infinity.

1. Introduction In this paper, we prove an existence result of steady symmetric vortex patch for a planar ideal fluid in an open disk. More specifically, by maximizing the kinetic energy subject to some appropriate constraints for the vorticity, we construct such a flow in which the vorticity has the form ω λ = λI{ψλ >µλ } − λI{ψλ 0, then the Kirchhoff-Routh function can be written as H2 (x, y) = 2κ2 G(x, y) + κ2 h(x, x) + κ2 h(y, y), (2.5) (2) where (x, y) ∈ D and x 6= y. It is easy to see that lim H2 (x, y) = +∞,

|x−y|→0

lim

x→∂D or y→∂D

H2 (x, y) = +∞,

(2.6)

so H2 attains its minimum in D ×pD \ {(x, y) ∈ D × Dp| x 6= y}. Moreover, every √ √ 5 − 2(cosθ, sinθ), − 5 − 2(cosθ, sinθ)) for some minimum point of H2 has the form ( θ ∈ [0, 2π), see Proposition A.1 in the Appendix. Now we consider a steady ideal fluid with unit density moving in D with impermeability boundary condition, which is described by the following Euler equations:   in D, (v · ∇)v = −∇P (2.7) ∇·v =0 in D,  v · n = 0 on ∂D , where v = (v1 , v2 ) is the velocity field, P is the scalar pressure, and n(x) is the outward unit normal at x ∈ ∂D. To simplify The Euler equations (2.7), we define the vorticity ω := ∂1 v2 − ∂2 v1 . By using the identity 21 ∇|v|2 = (v · ∇)v + v⊥ ω, where J(v1 , v2 ) = (v2 , −v1 ) denotes clockwise rotation through π2 , the first equation of (1.1) can be written as 1 ∇( |v|2 + P ) − v⊥ ω = 0. (2.8) 2 Taking the curl in (2.8) we get the vorticity equation

∇ · (ωv) = 0. (2.9) To recover the velocity field from the vorticity, we define the stream function ψ by solving the following Poisson’s equation with zero Dirichlet boundary condition ( −∆ψ = ω in D, (2.10) ψ=0 on ∂D. Using the Green’s function we have

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

ψ(x) = Gω(x) :=

Z

G(x, y)ω(y)dy.

(2.11)

D

Since D is simply-connected and v · n on ∂D, it is easy to check that v can be uniquely determined by ω in the following way(see [18], Chapter 1, Theorem 2.2 for example) v = ∇⊥ ψ,

(2.12)

where ∇⊥ ψ := (∇ψ)⊥ = (∂2 ψ, −∂1 ψ). From the above formal discussion we get the following vorticity form of the Euler equations: ( ∇ · (ωv) = 0, v = ∇⊥ Gω.

(2.13)

In this paper, we interpret the vorticity equation (2.13) in the weak sense. Definition 2.1. We call ω ∈ L∞ (D) a weak solution of (2.13) if Z

D

for all ξ ∈ C0∞ (D).

ω∇⊥ Gω · ξdx = 0

(2.14)

It should be noted that if ω ∈ L∞ (D), then by the regularity theory for elliptic equations ψ ∈ C 1,α (D) for some α ∈ (0, 1), so (2.14) makes sense for all ω ∈ L∞ (D). The following lemma from [10] gives a criterion for weak solutions of (2.13). Lemma A. Let ω ∈ L∞ (D) and ψ = Gω. Suppose that ω = f (ψ) a.e. in D for some monotonic function f : R → R, then ω is a weak solution of (2.13). Now we can state our main result. Theorem 2.2. Let κ be a positive number. Then there exists a λ0 > 0 such that for any λ > λ0 , there exists ω λ ∈ L∞ (D) satisfying

(1) ω λ is a weak solution of (2.13); (2) ω λ = ω1λ + ω2λ , where ω1λ = λI{ψλ >µλ } and ω2λ = −λI{ψλ 0 sufficiently small such that √Bδ (Pi ) ⊂⊂ D for i = 1, 2 and √

5−2

Bδ (P1 ) ∩ Bδ (P2 ) = ∅. For example, we can choose δ = . In the rest of this paper, 2 we use Bi to denote Bδ (Pi ) for i = 1, 2 for simplicity. For λ > 0 sufficiently large, we define the vorticity class K λ as follows: Z  λ ∞ K := ω ∈ L (D) | ω = ω1 + ω2 , supp(ωi ) ⊂ Bi for i = 1, 2, ω1 (x)dx = κ, (3.1) D 0 ≤ ω1 ≤ λ, ω1 (x) = ω1 (¯ x) and ω2 (x) = ω1 (˜ x) for x ∈ D .

It is easy to check that for any ω ∈ K λ , ω is even in x1 and odd in x2 . It is also clear that Kλ is not empty if λ > 0 is large enough. The kinetic energy of the fluid with vorticity ω is Z Z 1 E(ω) = G(x, y)ω(x)ω(y)dxdy, ω ∈ K λ . (3.2) 2 D D Integrating by parts we also have Z Z Z 1 1 1 2 ψ(x)ω(x)dx = |∇ψ(x)| dx = |v(x)|2 dx. E(ω) = 2 D 2 D 2 D

(3.3)

In the rest of this section we consider the maximization of E on Kλ and study the properties of the maximizer. 3.1. Existence of a maximizer. First we show the existence of a maximizer of E on K λ . Lemma 3.1. There exists ω λ ∈ Kλ such that E(ω λ ) = supω∈Kλ E(ω). Proof. The proof can be divided into three steps. Step 1: E is bounded from above on K λ . In fact, since |ω|L∞ (D) ≤ λ for any ω ∈ K λ , we have Z Z Z Z 1 1 2 E(ω) = G(x, y)ω(x)ω(y)dxdy ≤ λ |G(x, y)|dxdy ≤ Cλ2 2 D D 2 D D

for some absolute constant C. Here we use the fact that G ∈ L1 (D × D). This gives sup E(ω) < +∞.

ω∈Kλ

Step 2: K λ is closed in the weak∗ topology of L∞ (D), or equivalently, for any sequence {ωn } ⊂ Kλ and ω ∈ L∞ (D) satisfying Z Z lim ωn (x)φ(x)dx = ω(x)φ(x)dx, ∀ φ ∈ L1 (D), (3.4) n→+∞

λ

we have ω ∈ K .

D

D

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

To prove this, we first show that supp(ω) ⊂ ∪2i=1 Bi .

(3.5)

In fact, for any φ ∈ C0∞ (D \ ∪2i=1 Bi ), by (3.4) we have Z Z ω(x)φ(x)dx = lim ωn (x)φ(x)dx = 0, n→+∞

D

D

∪2i=1 Bi .

which means that ω = 0 a.e. in D \ Now we define ωi = ωIBi for i = 1, 2. It is obvious that ω = ω1 + ω2 . By choosing φ = IB1 in (3.4), we have Z Z Z Z ωn (x)dx = κ. (3.6) ωn (x)φ(x)dx = lim ω1 (x)dx = ω(x)φ(x)dx = lim D

n→+∞

D

n→+∞

D

B1

It is also easy to show that 0 ≤ ω1 ≤ λ in D. In fact, suppose that |{ω1 > λ}| > 0, then we can choose ε0 , ε1 > 0 such that |{ω1 > λ + ε0 }| > ε1 . Denote S = {ω1 > λ + ε0 } ⊂ B1 , then by choosing φ = IS in (3.4) we have Z (ω1 − ωn )(x)dx ≥ ε0 |S| ≥ ε0 ε1 . S

On the other hand, by (3.4) Z Z lim (ω1 − ωn )(x)dx = lim (ω − ωn )(x)φ(x)dx = 0, n→+∞

n→+∞

S

D

which is a contradiction. So we have ω1 ≤ λ. Similarly we can prove ω1 ≥ 0. To finish Step 2, it suffices to show that ω is even in x1 and odd in x2 . For fixed x ∈ D ∩ {x1 > 0}, define φ = πs12 IBs (x) − πs12 IBs (¯x) , where s > 0 is sufficiently small. Since ωn is even in x1 for each n and φ is odd in x1 , by (3.4) we have Z Z ω(y)φ(y)dy = lim ωn (y)φ(y)dy = 0, (3.7) D

n→+∞

D

which means that

Z Z 1 1 ω(y)dy = ω(y)dy. |Bs (x)| Bs (x) |Bs (¯ x)| Bs (¯x) By Lebesgue differential theorem, for a.e. x ∈ D ∩ {x1 > 0}, we have Z 1 ω(y)dy = ω(x), lim s→0+ |Bs (x)| Bs (x)

(3.8)

(3.9)

and

Z 1 lim ω(y)dy = ω(¯ x). s→0+ |Bs (¯ x)| Bs (¯x) Combining (3.8), (3.9) and (3.10) we get ω(x) = ω(¯ x) a.e. x ∈ D ∩ {x1 > 0},

which means that ω is even in x1 . Similarly we can prove that ω is odd in x2 .

(3.10)

(3.11)

EXISTENCE OF STEADY SYMMETRIC VORTEX PATCH IN AN OPEN DISK

7

From all the above arguments we know that ω ∈ K λ . Step 3: E is weak∗ continuous on K λ , that is, for any sequence {ωn } ⊂ K λ such that ωn → ω weakly∗ in L∞ (D) as n → +∞, we have lim E(ωn ) = E(ω).

n→+∞

In fact, by (3.4) as n → +∞ we know that ωn → ω weakly in L2 (D), then ψn → ψ weakly in W 2,2 (D) thus strongly in L2 (D), where ψn = Gωn and ψ = Gω. So we get as n → +∞ Z Z 1 1 E(ωn ) = ψn (x)ωn (x)dx → ψ(x)ω(x)dx = E(ω). 2 D 2 D Now we finish the proof of Lemma 3.1 by using the standard maximization technique. By Step 1, we can choose a maximizing sequence {ωn } ⊂ K λ such that lim E(ωn ) = sup E(ω)

n→+∞ λ

ω∈K λ ∞

Since K is bounded thus weakly∗ compact in L (D), we can choose a subsequence {ωnk } such that as k → +∞ ωnk → ω λ weakly∗ in L∞ (D) for some ω λ ∈ L∞ (D). By Step 2, we have ω λ ∈ K λ . Finally by Step 3 we have E(ω λ ) = lim E(ωnk ) = sup E(ω), k→+∞

ω∈K λ

which is the desired result.  3.2. Profile of ω λ . Since ω λ ∈ K λ , we know that ω λ has the form ω λ = ω1λ + ω2λ with ω1λ and ω2λ satisfying (1) Rsupp(ωiλ) ⊂ Bi for R i =λ 1, 2, λ (2) D ω1 (x)dx = − D ω2 (x)dx = κ, (3) ω1λ(x) = ω1λ (¯ x), ω1λ(x) = −ω2λ (˜ x) for any x ∈ D. λ In fact, we can prove that ω has a special form. Lemma 3.2. There exists µλ ∈ R depending on λ such that ω1λ = λI{ψλ >µλ }∩B1

and where ψ λ := Gω λ.

ω2λ = −λI{ψλ µλ }∩B1 . Choose α, β ∈ L∞ (D) satisfying  R R α, β ≥ 0, D α(x)dx = D β(x)dx,    supp(α), supp(β) ⊂ B ∩ {x > 0}, 1 1 (3.12) λ  α = 0 in D \ {ω ≤ λ − a},  1   β = 0 in D \ {ω1λ ≥ a},

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

where a is a small positive number. Now we define a family of test functions ωs = ω λ + s(g1 − g2 ) for s > 0 sufficiently small, where ˜¯ ) g1 (x) = α(x) + α(¯ x) − α(˜ x) − α(x

and ˜¯ ). g2 (x) = β(x) + β(¯ x) − β(˜ x) − β(x

Note that g1 , g2 are both even in x1 and odd in x2 , and supp(g1), supp(g2) ⊂ B1 ∪ B2 . It is easy check that ωs ∈ K λ if s is sufficiently small(depending on α, β and a). So we have Z Z dE(ωs ) λ 0≥ = g1 (x)ψ (x)dx − g2 (x)ψ λ (x)dx. (3.13) ds s=0+ D D On the other hand , Z Z Z λ λ ψ λ (x)(g1 − g2 )(x)dx g1 (x)ψ (x)dx − g2 (x)ψ (x)dx = D D D1 Z Z Z λ λ ψ λ (x)(g1 − g2 )(x)dx, ψ (x)(g1 − g2 )(x)dx + ψ (x)(g1 − g2 )(x)dx + + D4

D3

D2

(3.14)

where D1 = B1 ∩ {x1 > 0}, D2 = B1 ∩ {x1 < 0}, D3 = B2 ∩ {x1 < 0}, D4 = B2 ∩ {x1 > 0}.

Since ψ λ is also even in x1 and odd in x2 (this can be proved directly using the symmetry of the Green’s function), we have Z Z λ ψ (x)(g1 − g2 )(x)dx = ψ λ (x)(g1 − g2 )(x)dx ZD2 ZD1 (3.15) λ λ ψ (x)(g1 − g2 )(x)dx. ψ (x)(g1 − g2 )(x)dx = = D4

D3

Combining (3.13), (3.14) and (3.15) we conclude that Z Z λ ψ λ (x)g2 (x)dx, ψ (x)g1 (x)dx ≤ or equivalently

Z

D1

(3.16)

D1

D1

λ

ψ (x)α(x)dx ≤

Z

ψ λ (x)β(x)dx.

(3.17)

D1

By the choice of α and β, inequality (3.16) holds if and only if sup {ω λ 0}∩D

1

(3.18)

EXISTENCE OF STEADY SYMMETRIC VORTEX PATCH IN AN OPEN DISK

9

Combining the continuity of ψ λ in {ω λ < λ} ∩ D1 , we obtain ψλ =

sup {ω λ 0}∩D1

ψλ .

(3.19)

Now we define µλ as follows µλ :=

sup {ω λ 0}∩D

(3.20)

1

a.e. in {ψ λ < µλ } ∩ D1 , a.e. in {ψ λ > µλ } ∩ D1 .

(3.21)

On {ψ1λ = µλ } ∩ D1 , ψ λ is a constant, so we have ∇ψ λ = 0, then ω λ = −∆ψ λ = 0. Hence we conclude that ( ω λ = 0 a.e. in {ψ λ ≤ µλ } ∩ D1 , (3.22) ω λ = λ a.e. in {ψ λ > µλ } ∩ D1 . Finally by the symmetry of ω λ and ψ λ we get the desired result.  3.3. Asymptotic behavior of ω λ as λ → +∞. Now we give some asymptotic estimates on ω λ as λ → +∞. We will use C to denote various various numbers not depending on λ. pκ . Suppose that ω λ is obtained as in Lemma 3.1. Then Lemma 3.3. Let ε = λπ

1 2 (1) E(ω λ) ≥ − 2π κ ln ε − C; 1 κ ln ε − C; (2) µλ ≥ − 2π (3) there exists a positive number R > 1, not depending on λ, such that diam(supp(ωiλ)) < RεR for i = 1, 2; R (4) κ1 D xω1λ (x)dx → x1 , − κ1 D xω2λ(x)dx → x2 , as λ → +∞, where x1 , x2 satisfies x1 ∈ B1 , x2 ∈ B2 and H2 (x1 , x2 ) = min(x,y)∈D×D H2 (x, y).

Proof. The proof is exactly the same as the one in [13], we omit it here therefore.



4. Proof for Theorem 2.2 In this section we prove Theorem 2.2. The key point is to show that the maximizer ω λ obtained in Lemma 3.1 satisfies the condition in Lemma A. Proof of Theorem 2.2. First, we show that the support of ωiλ shrinks to Pi for i = 1, 2. By (3) and (4) of Lemma 3.3, supp(ωiλ) shrinks to xi , where (x1 , x2 ) is a minimum point of H2 in D × D. Then by Proposition A.1 in the Appendix, there exists some θ ∈ [0, 2π) such that q q √ √ x1 = 5 − 2(cosθ, sinθ), x2 = − 5 − 2(cosθ, sinθ).

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

But by the symmetry of ω λ (see the definition of K λ ), we have cosθ = 0, sinθ > 0, so θ = π2 . That is, q q √ √ 5 − 2), x2 = (0, − 5 − 2). (4.1) x1 = (0, Second, we show that ω λ is a weak solution of (2.13). To begin with, we show that |ψ λ | ≤ C on ∂B1 ∪ ∂B2 ,

(4.2)

where C is a positive number not depending on λ. In fact, by (3)(4) of Lemma 3.3 and (4.1), we have dist(suppω1λ, ∂Bi ) > δ0 , i = 1, 2, (4.3) for some δ0 ∈ (0, 1) not depending on λ if λ is sufficiently large. Then for x ∈ ∂B1 , Z λ G(x, y)ω λ(y)dy| |ψ (x)| = | ZD Z λ G(x, y)ω1 (y)dy + G(x, y)ω2λ(y)dy| =| D Z Z ZD 1 λ λ h(x, y)ω1 (y)dy + G(x, y)ω2λ(y)dy| − ln |x − y|ω1 (y)dy − =| 2π D D Z Z Z D 1 λ λ ln δ0 ω1 (y)dy| + |h(x, y)|ω1 (y)dy + |G(x, y)ω2λ(y)|dy ≤ |− λ λ 2π D supp(ω1 ) supp(ω2 ) κ ≤ − ln δ0 + C. 2π (4.4) Here we use the fact that |h| ≤ C on ∂B1 × supp(ω1λ) and |G| ≤ C on ∂B1 × supp(ω2λ). Similarly |ψ λ | ≤ C on ∂B2 . So (4.2) is proved. By Lemma 3.2, ω λ has the form ω λ = λI{ψλ >µλ }∩B1 − λI{ψλ µλ } ∩ B1 = {ψ λ > µλ }, {ψ λ < −µλ } ∩ B2 = {ψ λ < −µλ } (4.7)

provided λ is sufficiently large. So in fact ω λ has the form

or equivalently,

ω λ = λI{ψλ >µλ } − λI{ψλ µλ ,   0, t ∈ [−µλ , µλ ], f (t) =   −λ, t < −µλ .

11

(4.10)

Then by Lemma A, ω λ is a weak solution of (2.13). Last, combining the other properties of ω λ obtained in Section 2 we finish the proof of Theorem 2.2.  5. Non-symmetric Case In this section, we briefly discuss the existence of steady non-symmetric vortex patch. Theorem 5.1. Let κ1 , κ2 be two real numbers such that κ1 > 0, κ2 < 0. Then there exists a λ0 > 0 such that for any λ > λ0 , there exists ω λ ∈ L∞ (D) satisfying (1) ω λ = ω1λ + ω2λ, where ω1λ = λI{ψλ >µλ1 } and ω2λ = −λI{ψλ 0, q < 0 depending only on κκ12 . Proof. The construction of ω λ here is similar to the one of the symmetric case in Section 2. For simplicity we only sketch the proof. First we choose (P, Q) as a minimum point of the corresponding Kirchhoff-Routh function H2 (x, y) := −2κ1 κ2 G(x, y) + κ1 2 h(x, x) + κ2 2 h(y, y), x, y ∈ D, x 6= y.

(5.1)

Let P = (0, p), Q = (0, q) with p > 0, q < 0. In this case, P, Q are uniquely determined by κκ21 (see Proposition A.1 in the Appendix). Now we choose δ > 0 sufficiently small such that Bδ (P ), Bδ (Q) ⊂⊂ D and Bδ (P ) ∩ Bδ (Q) = ∅. Consider the maximization of E on the following class  M λ := ω ∈ L∞ (D) | ω = ω1 + ω2 , supp(ω1) ⊂ Bδ (P ), supp(ω2) ⊂ Bδ (Q), Z ωi (x)dx = κi , 0 ≤ sgn(κi )ωi ≤ λ, for i = 1, 2, ω(x) = ω(¯ x) for x ∈ D .

(5.2)

D

Then by repeating the procedures in Section 2, we can prove that there exists a maximizer ω λ and this maximizer satisfies (1)-(5) in Theorem 5.1 if λ is sufficiently large. 

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

Appendix A. Minimum Points of H2 In this appendix we give location of the minimum points of the function H2 . Proposition A.1. Let D = {x ∈ R2 | |x| < 1}, κ1 > 0, κ2 < 0 be two real numbers, and γ = − κκ21 . Denote M the set of all minimum points of H2 , where H2 (x, y) := −2κ1 κ2 G(x, y) + κ1 2 h(x, x) + κ2 2 h(y, y), x, y ∈ D, x 6= y.

(A.1)

M = {(P, Q) ∈ D × D | P = p(cosθ, sin θ), Q = q(cosθ, sinθ), θ ∈ [0, 2π)}.

(A.2)

Then there exists p ∈ (0, 1), q ∈ (−1, 0) depending only on γ, such that If γ = 1, then

p = −q = Proof. First, it is easy to see that lim H2 (x, y) = +∞,

|x−y|→0

q √

5 − 2.

lim

x→∂D or y→∂D

H2 (x, y) = +∞,

(A.3)

so H2 attains its minimum in D × D \ {(x, y) ∈ D × D | x 6= y}, or equivalently, M is not empty. For x, y ∈ D, x 6= y, H2 (x, y) = −2κ1 κ2 G(x, y) + κ1 2 h(x, x) + κ2 2 h(y, y)

|x − y| 1 2 1 ∗ 2 1 ln ) − κ ( ln |x||x − x |) − κ ln |y||y − y∗ |) 1 2( ∗ 2π |y||x − y | 2π 2π 2 2γ ∗ 2γ κ |y| |x − y | = 1 ln , π |x − y|2γ |x||y|γ 2 |y − y∗ |γ 2 (A.4) = −2κ1 κ2 (−

where x = |x|x 2 and y = |y|y 2 . So it suffices to consider the minimum points of T (x, y), where T (x, y) := By using the polar coordinates,

|y|2γ |x − y∗ |2γ . |x − y|2γ |x||y|γ 2 |y − y∗ |γ 2

(1 + ρ21 ρ22 − 2ρ1 ρ2 cos(θ1 − θ2 ))γ , (ρ21 + ρ22 − 2ρ1 ρ2 cos(θ1 − θ2 ))γ (1 − ρ21 )(1 − ρ22 )γ 2 where x = ρ1 (cosθ1 , sinθ1 ) and y = ρ2 (cosθ2 , sinθ2 ). Now we show that if (x, y) is a minimum point of H2 , then T (x, y) :=

θ1 − θ2 = π.

In fact, it is not hard to check that for fixed ρ1 , ρ2 , T is strictly increasing in cos(θ1 − θ2 ), so at any minimum point we must have cos(θ1 − θ2 ) = −1.

EXISTENCE OF STEADY SYMMETRIC VORTEX PATCH IN AN OPEN DISK

13

To finish the proof it suffices to consider the minimization of the following function: (1 + ρ1 ρ2 )2γ R(ρ1 , ρ2 ) := (ρ1 + ρ2 )2γ (1 − ρ21 )(1 − ρ22 )γ 2

for ρ1 , ρ2 ∈ (0, 1). Case 1: γ = 1. In this simple case, R(ρ1 , ρ2 ) becomes R(ρ1 , ρ2 ) :=

(1 + ρ1 ρ2 )2 , (ρ1 + ρ2 )2 (1 − ρ21 )(1 − ρ22 )

ρ1 , ρ2 ∈ (0, 1).

By direct calculation, we obtain

∂ρ1 R = 0 ⇔ ρ1 ρ2 + ρ21 + ρ31 ρ2 + 2ρ22 = 1,

(A.5)

∂ρ2 R = 0 ⇔ ρ2 ρ1 + ρ22 + ρ32 ρ1 + 2ρ21 = 1.

(A.6)

Subtracting the two expressions in (A.5) and (A.6) we get (1 + ρ1 ρ2 )(ρ21 − ρ22 ) = 0, which gives ρ1 = ρ2 . Now we can see that ρ1 satisfies ρ41 + 4ρ21 − 1 = 0,

p√ so ρ1 = ρ2 = 5 − 2. Case 2: γ > 0 is arbitrary. In this case, we show that R(ρ1 , ρ2 ) has a unique minimum point for ρ1 , ρ2 ∈ (0, 1). Existence is obvious since we have proved that M is not empty. Now we show the uniqueness. In fact, it suffices to prove that the critical point of R in (0, 1) × (0, 1) is unique. Direct calculation gives ∂ρ1 R = 0 ⇔ ρ1 ρ2 + (1 + γ)ρ21 + γρ22 + ρ31 ρ2 + (1 − γ)ρ21 ρ22 − γ = 0, ∂ρ2 R = 0 ⇔ γρ1 ρ2 + ρ21 + ρ22 + (γ − 1)ρ21 ρ22 + γ 2 ρ22 + γρ1 ρ32 − 1 = 0.

For simplicity we write

F1 (ρ1 , ρ2 ) := ρ1 ρ2 + (1 + γ)ρ21 + γρ22 + ρ31 ρ2 + (1 − γ)ρ21 ρ22 − γ and F2 (ρ1 , ρ2 ) := γρ1 ρ2 + ρ21 + ρ22 + (γ − 1)ρ21 ρ22 + γ 2 ρ22 + γρ1 ρ32 − 1.

It is not hard to check that F1 , F2 are both strictly monotonic in ρ1 for fixed ρ2 and in ρ2 for fixed ρ1 if ρ1 , ρ2 ∈ (0, 1), which means that the system F1 (ρ1 , ρ2 ) = 0, F2 (ρ1 , ρ2 ) = 0 has most one solution. In other words, under the condition θ1 − θ2 = π, H2 has a unique minimum point, which completes the proof.  Acknowledgements: D. Cao was supported by NNSF of China (grant No. 11331010) and Chinese Academy of Sciences by grant QYZDJ-SSW-SYS021. G. Wang was supported by NNSF of China (grant No.11771469).

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DAOMIN CAO, GUODONG WANG, AND BIJUN ZUO

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EXISTENCE OF STEADY SYMMETRIC VORTEX PATCH IN AN OPEN DISK

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School of Mathematics and Information Science, Guangzhou Univeristy, Guangzhou 510405, Guangdong and Institute of Applied Mathematics, AMSS, Chinese Academy of Science, Beijing 100190, People’s Republic of China E-mail address: [email protected] Institute of Applied Mathematics, Chinese Academy of Science, Beijing 100190 and University of Chinese Academy of Sciences, Beijing 100049, People’s Republic of China E-mail address: [email protected] Institute of Applied Mathematics, Chinese Academy of Science, Beijing 100190 and University of Chinese Academy of Sciences, Beijing 100049, People’s Republic of China E-mail address: [email protected]