Uniqueness of the 2D Euler equation on a corner domain with non-constant vorticity around the corner
aa r X i v : . [ m a t h . A P ] S e p UNIQUENESS OF THE 2D EULER EQUATION ON A CORNERDOMAIN WITH NON-CONSTANT VORTICITY AROUND THECORNER
SIDDHANT AGRAWAL AND ANDREA R. NAHMOD Abstract.
We consider the 2D incompressible Euler equation on a corner domain Ω withangle νπ with < ν <
1. We prove that if the initial vorticity ω ∈ L (Ω) ∩ L ∞ (Ω) andif ω is non-negative and supported on one side of the angle bisector of the domain, thenthe weak solutions are unique. This is the first result which proves uniqueness when thevelocity is far from Lipschitz and the initial vorticity is nontrivial around the boundary. Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Notation and Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . .
53 Basic Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction
We are interested in the well-posedness problem for the 2D incompressible Euler equationon a domain Λ u t + ( u · ∇ ) u = −∇ P in Λ , ∇ · u = 0 in Λ ,u · n = 0 on ∂ Λ . (1)Here u is the velocity, P is the pressure and n is the outward unit normal. The vorticity is ω = ∇ × u = ∂ x u − ∂ x u and satisfies the transport equation ω t + u · ∇ ω = 0 in Λ . (2) A.N. is funded in part by NSF DMS-1800852 and the Simons Foundation Collaborations Grant onWave Turbulence (Nahmod’s Award ID 651469).
One can recover the velocity from the vorticity by the Bio-Savart law u = ∇ ⊥ ∆ − ω where∆ is the Dirichlet Laplacian and ∇ ⊥ = ( − ∂ x , ∂ x ). The 2D Euler equation has severalconserved quantities, chief among them being k ω ( · , t ) k L p (Λ) for any 1 ≤ p ≤ ∞ . This isused in an essential way to prove any kind of global well-posedness result.The study of the well-posedness problem for the 2 D Euler equation has a long history.There are two important considerations to keep in mind while talking about the well-posedness problem: one is the regularity of the initial vorticity and the other is the regularityof the boundary. Let us first consider the case of both the vorticity and boundary beingregular enough. Global well-posedness for strong solutions in smooth domains was provedby Wolibner [28] and H¨older [13] (see also [23, 16]). One of the most important works in thewell-posedness theory is the work of Yudovich [15] who established global well-posednessfor weak solutions on smooth domains for initial data ω ∈ L (Λ) ∩ L ∞ (Λ) (see also [2, 25]).The uniqueness result of Yudovich used the Eulerian formulation and relied on the Calder´onZygmund inequalities k∇ u ( · , t ) k L p (Λ) ≤ Cp k ω ( · , t ) k L p (Λ) for all p ∈ [2 , ∞ ) . (3)Later on Marchioro and Pulvirenti [22] gave a different proof of uniqueness by using theLagrangian formulation which relied on the log-Lipschitz nature of the velocitysup x,y ∈ Λ | u ( x, t ) − u ( y, t ) || x − y | max {− ln | x − y | , } ≤ C k ω ( · , t ) k L (Λ) ∩ L ∞ (Λ) . (4)These estimates hold for C , domains but may not hold for less regular domains (see [14]).For the case of initial vorticity being less regular, global existence of weak solutions wasproved by DiPerna and Majda [9] for ω ∈ L ( R ) ∩ L p ( R ) for p > ω ∈ H − ( R ) ∩ M + ( R ) (here M + is the space of positive Radon measures). Uniquenessis not expected in general in this case and this is a major open problem (see the works[26, 27, 5, 4]).For the case of boundary being less regular, global existence of weak solutions for boundedconvex domains was proved by Taylor [24] and for arbitrary simply connected boundeddomains (and exterior domains) was proved by Gerard-Varet and Lacave [10, 11]. Bothresults prove existence for initial vorticity ω ∈ L (Λ) ∩ L p (Λ) or ω ∈ H − (Λ) ∩ M + (Λ).However even for ω ∈ L (Λ) ∩ L ∞ (Λ) the question of uniqueness is a major open problem.It is important to note that if the domain is less regular, then the uniqueness question doesnot become simpler even if the initial vorticity is assumed to be smooth, as the regularityof the vorticity can be destroyed at a later time (see [17, 1]).There have been some recent works that establish uniqueness for rough domains withinitial vorticity ω ∈ L (Λ) ∩ L ∞ (Λ). One strategy used was to identify domains rougherthan C , which satisfy either (3) or (4) and use this to prove uniqueness. This was firstachieved by Bardos, Di Plinio and Temam [3] for rectangle domains and for C domainswhich allow corners of angle π/m for m ∈ N , m ≥
2. Later Lacave, Miot and Wang [19]proved uniqueness for C ,α domains with a finite number of acute angled corners, and thenDi Plinio and Temam [8] proved uniqueness for C , domains with finitely many acuteangled corners. Note that for angles bigger than π/
2, the estimates (3) and (4) fail to hold
D EULER UNIQUENESS 3 and uniqueness is open in general. Another strategy used to prove uniqueness is to prove itfor initial vorticity which is constant around the boundary. The idea behind this strategyis that if the vorticity is constant around the boundary, then the uniqueness proof of [22]works, as in this case one only needs the estimate (4) for x, y ∈ K where K ⊂ Λ is a compactset outside of which the vorticity is constant. The strategy thus reduces to showing that ifthe vorticity is initially constant around the boundary, then it remains constant for latertimes. Lacave [18] proved that if the domain is C , with finitely many corners with anglesgreater than π/ ω is constant around the boundary and has a definite sign, then ω remains constant around the boundary and the weak solutions are unique. Lacave andZlatoˇs [20] proved the same result but removed the restriction of definite sign on ω and thecorners are allowed to be of any angle in (0 , π ). Recently Han and Zlatoˇs [12] generalizedthe results of [18, 20] to more general domains which include all convex domains.In this paper we are interested in the uniqueness question for a domain which does notsatisfy (3) or (4) and has non-constant initial vorticity around the boundary. In this case,the methods used previously to prove uniqueness cannot work and new ideas are needed.Let us first state our result. Fix < ν < + be the domainsΩ = n re iθ ∈ C (cid:12)(cid:12)(cid:12) r > < θ < νπ o Ω + = n re iθ ∈ C (cid:12)(cid:12)(cid:12) r > < θ < νπ o . Let the initial vorticity be ω := ω ( · , ω ∈ L (Ω) ∩ L ∞ (Ω) along with supp( ω ) ⊂ Ω + and ω ≥ . (5)We can now state our main result. Theorem 1.1.
Consider the Euler equation in Ω with initial vorticity ω satisfying (5) .Then there exists a unique Yudovich weak solution in the time interval [0 , ∞ ) with thisinitial data. See § ω ) ⊂ (cid:8) re iθ ∈ C (cid:12)(cid:12) r ≥ ≤ θ ≤ β ( ν ) νπ (cid:9) for some β ( ν ) > . Similarly the assumption of ω ≥ x ( t ) ∈ Ω be the position of the particlewhich starts at the corner i.e. x (0) = 0. From (26), (27) and (35) we see that heuristically dxdt = x ν − x (0) = 0 . (6)Observe that as < ν <
1, the function x ν − is not Lipschitz and hence one cannot usethe Picard-Lindel¨of theorem to prove uniqueness of this ODE. Indeed one sees that thisODE has several solutions. However as the vorticity is non-negative, it turns out that theflow automatically chooses the solution with the property that the particle moves to the SIDDHANT AGRAWAL AND ANDREA R. NAHMOD right i.e. x ( t ) > t >
0. With this constraint the ODE has a unique solution, namely x ( t ) = (cid:2)(cid:0) ν − ν (cid:1) t (cid:3) ν ν − .So essentially any method employed to prove Theorem 1.1 has to be strong enough that itcan prove the uniqueness of the above ODE (with the constraint x ( t ) > t > Step 1 : Consider two solutions x ( t ) and x ( t ) of (6) with x i ( t ) > t > i = 1 ,
2. Prove that given 0 < ǫ <
T > i = 1 , x i ( t ) ≥ (cid:2)(cid:0) ν − ν (cid:1) (1 − ǫ ) t (cid:3) ν ν − for t ∈ [0 , T ].This can be proven by observing that dx i dt ≥ (1 − ǫ ) x ν − . Integrating this inequality weget the required estimate. Step 2 : Consider the energy E ( t ) = | x ( t ) − x ( t ) | . By a simple inequality we prove inLemma 3.2 part (1), we obtain dE dt ≤ (cid:12)(cid:12)(cid:12) x ν − − x ν − (cid:12)(cid:12)(cid:12) . ν | x − x | ν − = E ν − and E (0) = 0 . Hence by integration we get E ( t ) . ν t ν ν − in the time interval [0 , T ]. Step 3 : Consider the energy E ( t ) = t − α E ( t ) = t − α | x ( t ) − x ( t ) | . Observe that if0 < α < ν ν − then by step 2, E ( t ) → t → + . Now dEdt ≤ (cid:18) − αt (cid:19) t − α | x ( t ) − x ( t ) | + t − α (cid:12)(cid:12)(cid:12) x ν − − x ν − (cid:12)(cid:12)(cid:12) = (cid:26)(cid:18) − αt (cid:19) + (cid:12)(cid:12)(cid:12)(cid:12) x ν − − x ν − x − x (cid:12)(cid:12)(cid:12)(cid:12)(cid:27) t − α | x ( t ) − x ( t ) |≤ (cid:26)(cid:18) − αt (cid:19) + (cid:18) ν − (cid:19) max n x ν − , x ν − o(cid:27) t − α | x ( t ) − x ( t ) | . Now using step 1, we get dEdt ≤ (cid:26)(cid:18) − αt (cid:19) + (cid:18) ν − (cid:19)(cid:20)(cid:18) ν − ν (cid:19) (1 − ǫ ) t (cid:21) − (cid:27) t − α | x ( t ) − x ( t ) | = (cid:26) − α + 1 − ν (2 ν − − ǫ ) (cid:27) t − α +1 | x ( t ) − x ( t ) | . As ν > − ν , we see that we can suitably choose α and ǫ so that dEdt ≤
0. Hence E ( t ) = 0in [0 , T ] and this proves x ( t ) = x ( t ) for t ∈ [0 , T ]. Step 4 : Uniqueness for t ≥ T follows from the Picard-Lindel¨of theorem by observingthat there exists a c > x ( t ) , x ( t ) ≥ c for all t ≥ T . D EULER UNIQUENESS 5
The proof of Theorem 1.1 closely follows the above strategy. The analogs of step 1-4 areProposition 4.6, Proposition 5.1, proof of main Theorem 1.1 and Lemma 4.7 respectively.The analog of the Picard-Lindel¨of theorem is the uniqueness proof given in Sec 2.3 of [22].The assumptions on the initial vorticity (5) are imposed so that the uniqueness problem forthe Euler equation behaves in a similar manner to uniqueness problem of the ODE (6). Theassumption ω ≥ ω ) ⊂ Ω + ensures that essentially the particles in the support of the vorticity move away from thecorner and this is shown in Proposition 4.6. Both of these properties are the analogs of theconstraint x ( t ) > t > § § § § Notation and Preliminaries
Let H = { ( x , x ) ∈ C | x > } denote the upper half plane and we will identify R ≃ C .Let H + = { ( x , x ) ∈ C | x > x > } and denote a ball of radius r by B r ( z ) = B ( z , r ) = { z ∈ C | | z − z | < r } . For z , z ∈ H , let [ z , z ] denote the line segment con-necting z and z . We define the function φ : [0 , ∞ ) → R as φ (0) = 0 and for x > φ ( x ) = x max {− ln( x ) , } . (7)Observe that φ is a continuous increasing function on R with x ≤ φ ( x ) for all x ≥ φ is a concave function on the interval [0 , / c ≥ φ ( cx ) ≤ cφ ( x ) for all x ≥ f ∈ L ∞ ( C ) and let z , · · · , z n ∈ C be n distinct complex numbers. If α , · · · , α n ≥ ≤ r, R ≤ ∞ we define I (( z , α ) , · · · , ( z n , α n ) : ( f, r, R )) = Z A | s − z | α · · · | s − z n | α n | f ( s ) | ds, (8)where A = B ( z , r ) c ∩ · · · ∩ B ( z n , r ) c ∩ B ( z , R ) ∩ · · · ∩ B ( z n , R ). Observe that the set A isthe set of all s ∈ C with distance to the set { z , · · · , z n } between r and R .We write a . b if there exists a universal constant C > a ≤ Cb . We write a . η b if there exists a constant C = C ( η ) > η so that a ≤ Cb . Similardefinitions for . η ,η , . η ,η ,η etc. We write a ≈ b if a . b and b . a . Similarly we write a ≈ η b if a . η b and b . η a etc. In this paper we fix the angle of the domain Ω as νπ (with SIDDHANT AGRAWAL AND ANDREA R. NAHMOD / < ν <
1) and we will suppress the dependence of constants on ν as it shows up quitefrequently.Let us now derive the equation of the flow. As we are only interested in the flow in Ωand domains which smoothly approximate Ω, we derive the equation only for such domains.Let Λ be a domain homeomorphic to H with Λ being homeomorphic to H . If the Green’sfunction of the domain Λ is G Λ ( x, y ), then the kernel of the Biot-Savart law is K Λ ( x, y ) := ∇ ⊥ x G Λ ( x, y ) with ∇ ⊥ x = ( − ∂ x , ∂ x ). Let Ψ : Λ → H be a Riemann map and observe that Ψextends continuously to Λ by Caratheodary’s theorem. Fix z ∈ Λ and let f : Λ \{ z } → C be defined as f ( z ) = 12 π ln Ψ( z ) − Ψ( z )Ψ( z ) − Ψ( z ) ! . (9)Clearly f is holomorphic and we have for z , z ∈ Λ, z = z G Λ ( z , z ) = 12 π ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Ψ( z ) − Ψ( z )Ψ( z ) − Ψ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Re { f ( z ) } . Hence K Λ ( z , z ) = Re (cid:18) − ∂ x f ( z ) ∂ x f ( z ) (cid:19) = Re( − if x ( z )) + i Re( f x ( z ))= if z ( z ) . Then from (9) we have K Λ ( z , z ) = (cid:18) i π (cid:19) Ψ z ( z ) (cid:20) z ) − Ψ( z ) − z ) − Ψ( z ) (cid:21) . (10)If ω ( · , t ) is the vorticity at time t , then from the Biot-Savart law we see that u ( z , t ) = Z Λ K Λ ( z , z ) ω ( z , t ) dz . Now the equation for the flow X : Λ × [0 , ∞ ) → Λ is given by dX ( x, t ) dt = u ( X ( x, t ) , t ) = Z Λ K Λ ( X ( x, t ) , z ) ω ( z, t ) dz. Hence we have dX ( x, t ) dt = (cid:16) i π (cid:17) Ψ z ( X ( x, t )) Z Λ (cid:20) X ( x, t )) − Ψ( z ) − X ( x, t )) − Ψ( z ) (cid:21) ω ( z, t ) dz. Define the function b : Λ × [0 , ∞ ) → C as b ( x, t ) = (cid:16) i π (cid:17) Z Λ (cid:20) x ) − Ψ( z ) − x ) − Ψ( z ) (cid:21) ω ( z, t ) dz. (11) D EULER UNIQUENESS 7
Hence the equation for X can be written as dX ( x, t ) dt = b ( X ( x, t ) , t )Ψ z ( X ( x, t )) . (12)We now convert the flow equation above in Λ to a flow equation on H . For x ∈ Λ, let y ∈ H be given by y = Ψ( x ). Consider the flow Y : H × [0 , ∞ ) → H given by Y ( y, t ) = Ψ( X ( x, t )) (13)and define e b : H × [0 , ∞ ) → C as e b ( y, t ) = b ( x, t ). Then e b ( Y ( y, t ) , t ) = b ( X ( x, t ) , t ) and wehave dY ( y, t ) dt = e b ( Y ( y, t ) , t ) (cid:12)(cid:12) Ψ z ◦ Ψ − ( Y ( y, t )) (cid:12)(cid:12) . (14)We can write a simple formula for e b . As y = Ψ( x ) we see that e b ( y, t ) = b ( x, t ) = (cid:16) i π (cid:17) Z Λ (cid:20) x ) − Ψ( z ) − x ) − Ψ( z ) (cid:21) ω ( z, t ) dz = (cid:16) i π (cid:17) Z Λ (cid:20) y − Ψ( z ) − y − Ψ( z ) (cid:21) ω ( z, t ) dz. Next, we change variables by setting s = Ψ( z ) with s ∈ H and observe that ds = | Ψ z ( z ) | dz and hence dz = (cid:12)(cid:12) Ψ z ◦ Ψ − ( s ) (cid:12)(cid:12) − ds . Defining e ω : H × [0 , ∞ ) → R as e ω ( s, t ) = ω ( z, t ) we get e b ( y, t ) = (cid:16) i π (cid:17) Z H (cid:20) y − s − y − s (cid:21)e ω ( s, t ) (cid:12)(cid:12) Ψ z ◦ Ψ − ( s ) (cid:12)(cid:12) − ds. (15)3. Basic Estimates
In this section we collect some basic estimates we use throughout the paper.
Lemma 3.1.
Let
T, R, c > and let y : [0 , T ] → R be such that | y ( t ) | ≤ R for all t ∈ [0 , T ] and satisfy (cid:12)(cid:12)(cid:12)(cid:12) dydt (cid:12)(cid:12)(cid:12)(cid:12) ≤ cφ ( y ( t )) y (0) = y > where φ is given by (7) . Then { y (0) } e ct . R y ( t ) . R { y (0) } e − ct for all t ∈ [0 , T ] . Proof.
We only prove y ( t ) . R { y (0) } e − ct since the other estimate is proved similarly. Wehave dydt ≤ cy max {− ln( y ) , } ≤ cy {− ln( y ) + 1 + ln( R + 1) } . Therefore d ln( y ) dt ≤ c {− ln( y ) + 1 + ln( R + 1) } . SIDDHANT AGRAWAL AND ANDREA R. NAHMOD
Now multiplying by e ct we obtain d ( e ct ln( y )) dt ≤ e ct c (1 + ln( R + 1)) . Integrating the above inequality we get e ct ln( y ( t )) − ln( y (0)) ≤ ( e ct − R + 1)) ≤ e ct (1 + ln( R + 1)) . Hence ln( y ( t )) ≤ e − ct ln( y (0)) + 1 + ln( R + 1)and so y ( t ) . R { y (0) } e − ct . (cid:3) Lemma 3.2.
Let ν > and let a, b ∈ C be non-zero complex numbers satisfying thecondition ≤ arg( a ) , arg( b ) < min (cid:8) π, πν (cid:9) . Then we have(1) If < ν < then | a ν − b ν | ≈ ν | a − b | min n | a | ν − , | b | ν − o ≈ ν | a − b | min n | a | ν − , | b | ν − , | a − b | ν − o . (2) If < ν < ∞ then | a ν − b ν | ≈ ν | a − b | max n | a | ν − , | b | ν − o ≈ ν | a − b | max n | a | ν − , | b | ν − , | a − b | ν − o . Proof.
We only prove it for 0 < ν < < ν < ∞ is similar. Without lossof generality | a | ≤ | b | .(a) Case 1: | a | ≤ | b | . We have | a ν − b ν | ≈ ν | b | ν ≈ ν | a − b || b | ν − . In this case | b | ν − = min n | a | ν − , | b | ν − o ≈ ν min n | a | ν − , | b | ν − , | a − b | ν − o .(b) Case 2: | b | ≤ | a | ≤ | b | and (cid:12)(cid:12) a − bb (cid:12)(cid:12) ≤ . Hence we have | a ν − b ν | = | b | ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) a − bb + 1 (cid:19) ν − (cid:12)(cid:12)(cid:12)(cid:12) . Using the binomial theorem we see that | a ν − b ν | ≈ ν | b | ν (cid:12)(cid:12)(cid:12)(cid:12) a − bb (cid:12)(cid:12)(cid:12)(cid:12) = | a − b || b | ν − . In this case | b | ν − = min n | a | ν − , | b | ν − o = min n | a | ν − , | b | ν − , | a − b | ν − o .(c) Case 3: | b | ≤ | a | ≤ | b | and < (cid:12)(cid:12) a − bb (cid:12)(cid:12) <
2. Observe that12 < (cid:12)(cid:12)(cid:12) ab − (cid:12)(cid:12)(cid:12) < ⇒ (cid:12)(cid:12)(cid:12)(cid:16) ab (cid:17) ν − (cid:12)(cid:12)(cid:12) ≈ ν . D EULER UNIQUENESS 9
Hence we have | a ν − b ν | = | b | ν (cid:12)(cid:12)(cid:12)(cid:16) ab (cid:17) ν − (cid:12)(cid:12)(cid:12) ≈ ν | b | ν ≈ ν | a − b || b | ν − . In this case | b | ν − = min n | a | ν − , | b | ν − o ≈ ν min n | a | ν − , | b | ν − , | a − b | ν − o . (cid:3) Lemma 3.3.
Let n ≥ and let < R ≤ ∞ . Let z , · · · , z n ∈ C be n distinct complexnumbers and let f ∈ L ∞ ( C ) . Let d min = min {| z i − z j | | ≤ i, j ≤ j, i = j } > and let ≤ r ≤ d min / . If α , · · · , α n > , d min = | z − z | > , then for I defined as in (8) wehave the following estimate I (( z , α ) , · · · , ( z n , α n ) : ( f, r, R )) . α , ··· ,α n k f k ∞ n X i =1 Z d min / r | x | ( α i − dx ! Y j = i, ≤ j ≤ n | z j − z i | − α i + I (( z , α + α ) , ( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . Proof.
Clearly we can assume that k f k ∞ >
0. If r = 0 and max { α , · · · , α n } ≥ ∞ and there is nothing to prove. Hence we assume thateither r > { α , · · · , α n } <
2. Now as 0 ≤ r ≤ d min / ≤ i ≤ n Z B ( z i ,r ) c ∩ B ( z i ,d min / | s − z | α · · · | s − z n | α n | f ( s ) | ds . α , ··· ,α n k f k ∞ Y j = i, ≤ j ≤ n | z j − z i | − α i Z B ( z i ,r ) c ∩ B ( z i ,d min / | s − z i | α i ds . α , ··· ,α n k f k ∞ Y j = i, ≤ j ≤ n | z j − z i | − α i Z d min / r | x | ( α i − dx. Summing these up we get I (( z , α ) , · · · , ( z n , α n ) : ( f, r, R )) . α , ··· ,α n k f k ∞ n X i =1 Z d min / r | x | ( α i − dx ! Y j = i, ≤ j ≤ n | z j − z i | − α i + I (( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . Now by the weighted AM-GM inequality we have1 | s − z | α + α + 1 | s − z | α + α & α ,α | s − z | α | s − z | α . Hence I (( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . α , ··· ,α n I (( z , α + α ) , ( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R ))+ I (( z , , ( z , α + α ) , ( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . Now we observe that 1 | s − z | ≤ | s − z | for all s ∈ B ( z , | z − z | / c and as d min = | z − z | we obtain I (( z , , ( z , α + α ) , ( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . α , ··· ,α n I (( z , α + α ) , ( z , α ) , · · · , ( z n , α n ) : ( f, d min / , R )) . Hence proved. (cid:3)
Lemma 3.4.
Let z , z ∈ C be such that z = z and let f ∈ L ( C ) ∩ L ∞ ( C ) . Then | z − z | I (( z , , ( z ,
1) : ( f, , ∞ )) . k f k L ∩ L ∞ φ ( | z − z | ) . Proof.
We see from Lemma 3.3 that I (( z , , ( z ,
1) : ( f, , ∞ )) . k f k ∞ + I (( z ,
2) : ( f, | z − z | / , ∞ )) . k f k ∞ + I (( z ,
2) : ( f, | z − z | / , I (( z ,
2) : ( f, , ∞ )) . k f k ∞ max {− ln( | z − z | ) , } + k f k . k f k L ∩ L ∞ max {− ln( | z − z | ) , } . (cid:3) Lemma 3.5.
Let z , z ∈ C be nonzero complex numbers with z = z and let f ∈ L ( C ) ∩ L ∞ ( C ) . If < ν < then | z − z | I ((0 , − ν ) , ( z , , ( z ,
1) : ( f, , ∞ )) . ν k f k L ∩ L ∞ min (cid:8) | z | ν − , | z | ν − (cid:9) φ ( | z − z | ) . We also have the estimate | z − z | I ((0 , − ν ) , ( z , , ( z ,
1) : ( f, , ∞ )) . ν k f k L ∞ (cid:16) (cid:8) | z | ν − , | z | ν − (cid:9)(cid:17) φ ( | z − z | ) . Proof.
Let d min = min {| z | , | z | , | z − z |} >
0. We prove this in cases.
Case 1: d min = min {| z | , | z |} D EULER UNIQUENESS 11
Without loss of generality we can assume that d min = | z | . Hence | z | ≤ | z − z | ≤ | z | and so by the weighted AM-GM inequality and Lemma 3.3 we have I ((0 , − ν ) , ( z , , ( z ,
1) : ( f, , ∞ )) . ν I ((0 , − ν ) , ( z ,
1) : ( f, , ∞ )) + I (( z , − ν ) , ( z ,
1) : ( f, , ∞ )) . ν k f k ∞ | z | ν − + I ((0 , − ν ) : ( f, | z | / , ∞ ))+ k f k ∞ | z − z | ν − + I (( z , − ν ) : ( f, | z − z | / , ∞ )) . ν k f k ∞ | z | ν − + k f k ∞ | z − z | ν − . ν k f k L ∞ min n | z | ν − , | z | ν − o . Case 2: d min = | z − z | In this case we see that | z | ≤ | z | ≤ | z | . Hence by Lemma 3.3 we have I ((0 , − ν ) , ( z , , ( z ,
1) : ( f, , ∞ )) . ν k f k ∞ (cid:16) | z − z | ν +1 | z | − + | z | ν − (cid:17) + I ((0 , − ν ) , ( z ,
2) : ( f, | z − z | / , ∞ )) . ν k f k ∞ | z | ν − + I ((0 , − ν ) , ( z ,
2) : ( f, | z − z | / , | z | / I ((0 , − ν ) , ( z ,
2) : ( f, | z | / , ∞ )) . Now observe that from the weighted AM-GM inequality we have I ((0 , − ν ) , ( z ,
2) : ( f, | z | / , ∞ )) . ν I ((0 , − ν ) : ( f, | z | / , ∞ )) + I (( z , − ν ) : ( f, | z | / , ∞ )) . ν k f k ∞ | z | ν − . Hence we have I ((0 , − ν ) , ( z , , ( z ,
1) : ( f, , ∞ )) . ν k f k ∞ | z | ν − + I ((0 , − ν ) , ( z ,
2) : ( f, | z − z | / , | z | / . ν k f k ∞ | z | ν − + | z | − I ((0 , − ν ) : ( f, | z − z | / , | z | / | z | ν − I (( z ,
2) : ( f, | z − z | / , | z | / . ν k f k ∞ | z | ν − + | z | ν − I (( z ,
2) : ( f, | z − z | / , | z | / . ν k f k ∞ | z | ν − + | z | ν − I (( z ,
2) : ( f, | z − z | / , | z | ν − I (( z ,
2) : ( f, , | z | / . ν k f k ∞ min n | z | ν − , | z | ν − o max {− ln | z − z | , } + | z | ν − I (( z ,
2) : ( f, , | z | / . We now easily see that | z | ν − I (( z ,
2) : ( f, , | z | / . ν | z | ν − k f k . Now I (( z ,
2) : ( f, , | z | / | z | ≥ | z | ν − ln( | z | / . ν | z | ≥
2. Therefore | z | ν − I (( z ,
2) : ( f, , | z | / . ν k f k ∞ . Hence proved. (cid:3) Weak Solutions
We now give the definition of Yudovich weak solutions and prove their existence for thedomain Ω. We prove the existence of weak solutions in Ω as the previous existence resultsdo not apply directly. The existence proof of Taylor [24] and Gerard-Varet and Lacave[10, 11] are either for bounded domains or for exterior domains. We modify the existenceproof for R as given in the book by Majda and Bertozzi [21] to prove existence of weaksolutions in Ω. Even though the method for proving existence of weak solutions is quitestandard, we include it for the sake of completeness.For the definition of weak solution we closely follow the definition as given in [10, 11].Consider a domain Λ homeomorphic to H with Λ being homeomorphic to H . Note that inthis paper the domain Λ will be either Ω or smooth approximations of Ω and the definitionbelow is tailored to these domains. For more general domains a slightly different definitionas compared to the one below may be needed. We say that ( u, ω ) is in the Yudovich classin the time [0 , T ) if u ∈ L ∞ loc ([0 , T ); L loc (Λ)) , ω = ∇ × u ∈ L ∞ ([0 , T ); L (Λ) ∩ L ∞ (Λ)) , and u ( · , t ) ∈ C (Λ) with lim R →∞ sup | x |≥ R | u ( x, t ) | = 0 for a.e. t ∈ [0 , T ) . (16)Now let G c (Λ) = (cid:8) h ∈ L c (Λ) (cid:12)(cid:12) h = ∇ p for some p ∈ H loc (Λ) (cid:9) . Consider initial data ( u , ω ) satisfying u ∈ C (Λ) , ω = ∇ × u ∈ L (Λ) ∩ L ∞ (Λ) , lim R →∞ sup | x |≥ R | u ( x ) | = 0 and Z Λ u · h = 0 ∀ h ∈ G c (Λ) . (17) Definition 4.1.
We say that ( u, ω ) is a Yudovich weak solution to the Euler equation (1)with initial condition ( u , ω ) in the time interval [0 , T ), if ( u, ω ) is in the Yudovich class(16) and satisfies Z T Z Λ ω ( ∂ t ϕ + u · ∇ ϕ ) dx dt = − Z Λ ω ϕ ( · , dx ∀ ϕ ∈ C ∞ c (Λ × [0 , T )) , (18)and for all a.e. t ∈ [0 , T ) we have Z Λ u ( · , t ) · h = 0 ∀ h ∈ G c (Λ) . (19) D EULER UNIQUENESS 13
Note that we have given the definition of Yudovich weak solutions as weak solutions to thetransport equation. It can be shown that this is equivalent to the definition of weak solutionto the Euler equation, see Remark 1.2 of [11]. Now for < ν < (cid:8) re iθ ∈ C (cid:12)(cid:12) r > < θ < νπ (cid:9) and the Riemann map Ψ : Ω → H given byΨ( z ) = z ν . We want to prove the existence of Yudovich weak solutions for this domainand understand the properties of the flow map.4.1. Existence of weak solutions
In this section we prove the existence of Yudovich weak solutions in Ω. For 0 ≤ ǫ ≤ ǫ = { z ν | Im( z ) > ǫ } and so Ω = Ω and Ω ǫ are smooth domains for ǫ > ǫ ⊂ Ω. We see that the Riemann maps Ψ ǫ : Ω ǫ → H and Ψ − ǫ : H → Ω ǫ aregiven by Ψ ǫ ( z ) = ( z ν − iǫ ) Ψ − ǫ ( z ) = ( z + iǫ ) ν (20)From (10) we see that K Ω ǫ ( z , z ) = (cid:18) i πν (cid:19) z ( ν − ) z ν − z ν − z ν − z ν + 2 iǫ . (21)Observe that for Ω ǫ and all ǫ ≥
0, the condition (19) is equivalent to the condition that ∇ · u = 0 in Ω ǫ and that u · n = 0 on ∂ Ω ǫ .We now prove some basic properties of the velocity on such domains and also show thatthe velocity has to be given by the Biot-Savart law. Lemma 4.2.
Let ≤ ǫ ≤ and let g ∈ L (Ω ǫ ) ∩ L ∞ (Ω ǫ ) . If v ( x ) = R Ω ǫ K Ω ǫ ( x, y ) g ( y ) dy then(1) k v k ∞ . k g k / ∞ k g k / + k g k . k g k L ∩ L ∞ .(2) For z , z ∈ Ω ǫ we have | v ( z ) − v ( z ) | . k g k L ∩ L ∞ | z − z | min n | z | ν − , | z | ν − o + k g k L ∩ L ∞ φ ( | z − z | ) . Hence v is continuous on Ω ǫ and for any compact set K ⊂ Ω ǫ , there exists C K > depending only on K and ν such that sup z ,z ∈ K | v ( z ) − v ( z ) | φ ( | z − z | ) ≤ C K k g k L ∩ L ∞ . Therefore the velocity in the interior is log-Lipschitz.(3) lim R →∞ sup | x |≥ R | v ( x ) | = 0 .(4) Suppose f ∈ C (Ω ǫ ) is such that ∇ × f = g in Ω ǫ , ∇ · f = 0 in Ω ǫ , f · n = 0 on ∂ Ω ǫ and we have lim R →∞ sup | x |≥ R | f ( x ) | = 0 . Then f = v .Proof. We prove each statement individually. (1) From (21) we see that for z ∈ Ω ǫ we have v ( z ) = (cid:18) i πν (cid:19) z ( ν − ) Z Ω ǫ " z ν − z ν − z ν − z ν + 2 iǫ g ( z ) dz. (22)Observe that for z , z ∈ Ω ǫ we have | ( z ν − iǫ ) − ( z ν + iǫ ) | ≥ | ( z ν − iǫ ) − ( z ν − iǫ ) | = | z ν − z ν | . (23)Hence from Lemma 3.2 we have | v ( z ) | . | z | ν − Z Ω ǫ | z − z | max n | z | ν − , | z | ν − o | g ( z ) | dz . Z Ω ǫ | z − z | | g ( z ) | dz . Z B ( z ) ∩ Ω ǫ | z − z | | g ( z ) | dz + Z B ( z ) c ∩ Ω ǫ | z − z | | g ( z ) | dz . k g k L ( B ( z ) ∩ Ω ǫ ) + k g k . k g k / ∞ k g k / + k g k . (24)Note that all the above estimates are independent of ǫ .(2) We fix z , z ∈ Ω ǫ . Now using (22),(23) and Lemma 3.2 we see that | v ( z ) − v ( z ) | . | z ν − − z ν − | Z Ω ǫ | z − z | max n | z | ν − , | z | ν − o | g ( z ) | dz + | z | ν − Z Ω ǫ | z ν − z ν || z ν − z ν || z ν − z ν | | g ( z ) | dz . | z − z | min n | z | ν − , | z | ν − o k g k L ∩ L ∞ + | z | ν − Z Ω ǫ | z − z | max n | z | ν − , | z | ν − o | g ( z ) || z − z | max n | z | ν − , | z | ν − o | z − z | max n | z | ν − , | z | ν − o dz . | z − z | min n | z | ν − , | z | ν − o k g k L ∩ L ∞ + | z | ν − Z Ω ǫ | z − z || g ( z ) || z − z || z − z || z | ν − dz. Now observe that Z Ω ǫ | z − z || g ( z ) || z − z || z − z || z | ν − dz = | z − z | I ((0 , ν − , ( z , , ( z ,
1) : ( g Ω ǫ , , ∞ )) . D EULER UNIQUENESS 15
Hence using the first estimate of Lemma 3.5 we see that | v ( z ) − v ( z ) | . | z − z | min n | z | ν − , | z | ν − o k g k L ∩ L ∞ + | z | ν − k g k L ∩ L ∞ min n | z | − ν , | z | − ν o φ ( | z − z | ) . | z − z | min n | z | ν − , | z | ν − o k g k L ∩ L ∞ + φ ( | z − z | ) k g k L ∩ L ∞ . (3) Let r > g = g {| x |≤ r } and g = g − g . Let v ( x ) = R Ω ǫ K Ω ǫ ( x, y ) g ( y ) dy and v ( x ) = R Ω ǫ K Ω ǫ ( x, y ) g ( y ) dy . For z ∈ Ω ǫ with | z | ≥ r we see from (24) that | v ( z ) | . Z Ω ǫ | z − z | | g ( z ) | dz . r k g k . r k g k . Also from part (1) of this lemma we have | v ( z ) | . k g k / ∞ k g k / + k g k . k g k / ∞ k g k / + k g k . Hence sup | x |≥ r | v ( x ) | . r k g k + k g k / ∞ k g k / + k g k . As k g k → r → ∞ , we are done.(4) As v ( x ) = R Ω ǫ K Ω ǫ ( x, y ) g ( y ) dy and K Ω ǫ ( x, y ) = ∇ ⊥ x G Ω ǫ ( x, y ), where G Ω ǫ is theGreen’s function of Ω ǫ , we see that ∇ · v = 0, ∇ × v = g and v · n = 0 on ∂ Ω ǫ .From part (2) of this lemma we have v ∈ C (Ω ǫ ) and from part (3) we also see thatlim R →∞ sup | x |≥ R | v ( x ) | = 0. Hence v satisfies all the properties satisfied by f .Now let p = f − v . As ∇ · p = 0 and ∇ × p = 0 we see that p is a holomorphicfunction on Ω ǫ . Let P : H → C be defined as P ( z ) = p (Ψ − ǫ ( z ))(Ψ − ǫ ) z ( z ) . Observe that P is a holomorphic function on H . From (20) we see that for z ∈ H we have (Ψ − ǫ ) z ( z ) = ν ( z + iǫ ) ν − . Now as p · n = 0 on ∂ Ω ǫ we see that P is realvalued on R \{ } (and on R if ǫ > P to a homomorphic function on C \{ } . As p ∈ C (Ω ǫ ) we see thatlim z → zP ( z ) = 0 and hence P can be extended to a holomorphic function on C .As lim R →∞ sup | x |≥ R | p ( x ) | = 0, we see that P is a bounded entire function on C which goes to 0 at infinity. Hence P = 0 and so p = 0. (cid:3) Now let ( u ǫ , ω ǫ ) be a Yudovich weak solution in Ω ǫ in the time interval [0 , T ). For thissolution, the flow X ǫ : Ω ǫ × [0 , T ) → Ω ǫ is defined by dX ǫ ( x, t ) dt = u ǫ ( X ǫ ( x, t ) , t ) X ǫ ( x,
0) = x. (25) We see from Lemma 4.3 that this map is well defined.
From Lemma 4.2 part (2) we see that the velocity is log-Lipschitz in the interior of Ω ǫ and hence this ODE can be solved uniquely as long as X ǫ ( x, t ) ∈ Ω ǫ . We first recall thequantities related to the flow for the domain Ω ǫ . We see from (11),(12) and (20) that dX ǫ ( x, t ) dt = 1 ν b ǫ ( X ( x, t ) , t ) X ǫ ( x, t ) ν − , (26)where b ǫ : Ω × [0 , ∞ ) → C is defined as b ǫ ( z , t ) := (cid:18) i π (cid:19) Z Ω ǫ " z ν − z ν − z ν − z ν + 2 iǫ ω ǫ ( z, t ) dz. (27)Similarly we define the flow Y ǫ : H × [0 , T ) → H as Y ǫ ( y, t ) := Ψ ǫ ( X ǫ ( x, t )), where y = Ψ ǫ ( x ).Similarly define e b ǫ : H × [0 , ∞ ) → C as e b ǫ ( y, t ) := b ǫ ( x, t ). Hence from (14) we have dY ǫ ( y, t ) dt = 1 ν e b ǫ ( Y ǫ ( y, t ) , t ) | Y ǫ ( y, t ) + iǫ | − ν . (28)Defining e ω ǫ : H × [0 , ∞ ) → R as e ω ǫ ( s, t ) := ω ǫ ( z, t ), where s = Ψ ǫ ( z ), we get from (15) e b ǫ ( y, t ) = (cid:16) iν π (cid:17) Z H (cid:20) y − s − y − s (cid:21)e ω ǫ ( s, t ) | s + iǫ | ν − ds. (29)We now show that the flow X ǫ always remains in the domain Ω ǫ and hence the maps X ǫ : Ω ǫ × [0 , T ) → Ω ǫ and Y ǫ : H × [0 , T ) → H are well defined. The following lemma isanalogous to similar statements proven in [18, 20, 12]. Lemma 4.3.
Let ≤ ǫ ≤ and let ( u ǫ , ω ǫ ) be a Yudovich weak solution in the domain Ω ǫ in the time interval [0 , T ) with initial vorticity ω ∈ L (Ω ǫ ) ∩ L ∞ (Ω ǫ ) . Let R > and let x ∈ Ω ǫ with | x | ≤ R . Then there exists constants c, C , C , C , C > depending only on R, T and ess sup t ∈ [0 ,T ) k ω ǫ ( · , t ) k L ∩ L ∞ so that C { Im ( Y ǫ ( y , } e ct ≤ Im ( Y ǫ ( y , t )) ≤ C { Im ( Y ǫ ( y , } e − ct and also C d ( X ǫ ( x , , ∂ Ω ǫ ) ν e ct ≤ d ( X ǫ ( x , t ) , ∂ Ω ǫ ) ≤ C d ( X ǫ ( x , , ∂ Ω ǫ ) νe − ct . Proof.
In this proof we will let
C >
R, T and on ess sup t ∈ [0 ,T ) k ω ǫ ( · , t ) k L ∩ L ∞ and we write a . C b instead of a ≤ Cb .We will first prove the estimate for Y ǫ ( y, t ) and then translate that information into anestimate for X ǫ ( x, t ). Let y = Ψ ǫ ( x ). Now as u ǫ is bounded from Lemma 4.2 we see from(25) that for all t ∈ [0 , T ) we have | X ǫ ( x , t ) | . C | Y ǫ ( y , t ) | . C . (30)For y ∈ H we see from (29) that e b ǫ ( y, t ) = (cid:16) iν π (cid:17) Z H (cid:20) y − s − y − s (cid:21)e ω ǫ ( s, t ) | s + iǫ | ν − ds. D EULER UNIQUENESS 17
ThereforeIm( e b ǫ ( y, t ))= (cid:16) ν π (cid:17) Re (cid:26)Z H (cid:20) y − s − y ) − s + 1Re( y ) − s − y − s (cid:21)e ω ǫ ( s, t ) | s + iǫ | ν − ds (cid:27) . Hence we see that | Im( e b ǫ ( y, t )) | . | Im( y ) | Z H e ω ǫ ( s, t ) | s + iǫ | ν − | y − s || Re( y ) − s | ds . k ω ǫ ( · , t ) k ∞ Z R | Im( y ) || ( y + iǫ ) − s || (Re( y ) + iǫ ) − s || s | − ν ds. Observe that if we let z = y + iǫ and z = Re( y ) + iǫ , then | Im( y ) | = | z − z | . Henceusing the definition of I from (8) we see that | Im( e b ǫ ( y, t )) | . k ω ǫ ( · , t ) k ∞ | z − z | I ((0 , − ν ) , ( z , , ( z ,
1) : (1 , , ∞ )) . Thus using the second estimate of Lemma 3.5 and observing that | y + iǫ | ≥ | Re( y ) + iǫ | weobtain | Im( e b ǫ ( y, t )) | . k ω ǫ ( · , t ) k ∞ (cid:16) | y + iǫ | ν − (cid:17) φ ( | Im( y ) | ) . Now from (28) we get d Im { Y ǫ ( y , t ) } dt = 1 ν Im (cid:8)e b ǫ ( Y ǫ ( y , t ) , t ) (cid:9) | Y ǫ ( y , t ) + iǫ | − ν and so (cid:12)(cid:12)(cid:12)(cid:12) d Im { Y ǫ ( y , t ) } dt (cid:12)(cid:12)(cid:12)(cid:12) . k ω ǫ ( · , t ) k ∞ φ ( | Im( Y ǫ ( y , t )) | ) (cid:16) | Y ǫ ( y , t ) + iǫ | ν − (cid:17) | Y ǫ ( y , t ) + iǫ | − ν . k ω ǫ ( · , t ) k ∞ φ ( | Im( Y ǫ ( y , t )) | ) (cid:16) | Y ǫ ( y , t ) + iǫ | − ν (cid:17) . (31)Now using (30) we get (cid:12)(cid:12)(cid:12)(cid:12) d Im { Y ǫ ( y , t ) } dt (cid:12)(cid:12)(cid:12)(cid:12) . C φ ( | Im( Y ǫ ( y , t )) | ) . Consequently from Lemma 3.1 there exists c = c ( R, T, ess sup t ∈ [0 ,T ) k ω ǫ ( · , t ) k L ∩ L ∞ ) > t ∈ [0 , T ) we have { Im( Y ǫ ( y , } e ct . C Im( Y ǫ ( y , t )) . C { Im( Y ǫ ( y , } e − ct . (32)This proves the first part of the lemma. Now from the definition of Y ǫ and using (20) we see that X ǫ ( x, t ) = ( Y ǫ ( y, t ) + iǫ ) ν .Therefore from Lemma 3.2 we see that d ( X ǫ ( x , t ) , ∂ Ω ǫ ) = min s ∈ R | ( Y ǫ ( y , t ) + iǫ ) ν − { s + iǫ } ν |≈ min s ∈ R | ( Y ǫ ( y , t ) − s || ( Y ǫ ( y , t ) + iǫ ) | ν − ≈ | ( Y ǫ ( y , t ) + iǫ ) ν − { Re( Y ǫ ( y , t )) + iǫ } ν | . Furthermore we see from Lemma 3.2 that { Im( Y ǫ ( y , t )) }| Y ǫ ( y , t ) + iǫ | ν − . C d ( X ǫ ( x , t ) , ∂ Ω ǫ ) . C { Im( Y ǫ ( y , t )) } ν . Hence { Im( Y ǫ ( y , t )) } . C d ( X ǫ ( x , t ) , ∂ Ω ǫ ) . C { Im( Y ǫ ( y , t )) } ν . In particular for t = 0 we have { Im( Y ǫ ( y , } . C d ( X ǫ ( x , , ∂ Ω ǫ ) . C { Im( Y ǫ ( y , } ν . Combing these two estimates with (32) we obtain for all t ∈ [0 , T ) d ( X ǫ ( x , , ∂ Ω ǫ ) ν e ct . C d ( X ǫ ( x , t ) , ∂ Ω ǫ ) . C d ( X ǫ ( x , , ∂ Ω ǫ ) νe − ct . (cid:3) We are now ready to prove the existence of Yudovich weak solutions in Ω.
Theorem 4.4.
Consider an initial data ( u , ω ) satisfying (17) in the domain Ω . Thenthere exists a Yudovich weak solution ( u, ω ) in domain Ω in the time interval [0 , ∞ ) in thesense of (18) and (19) .Proof. We closely follow the existence proof of weak solutions in R as given in Chapter8 of [21]. Observe that it is enough to prove the existence in the time interval [0 , T ) forarbitrary T >
0. By restricting ω to compact sets and by convolution, we see that thereexists initial vorticities ( ω ) ǫ ∈ C ∞ c (Ω ǫ ) ⊂ C ∞ c (Ω) such that for all 0 < ǫ ≤ k ( ω ) ǫ k L ∞ (Ω ǫ ) ≤ k ω k L ∞ (Ω) , k ( ω ) ǫ k L (Ω ǫ ) ≤ k ω k L (Ω) and k ( ω ) ǫ − ω k L (Ω) → ǫ → . Now by modifying the existence proof as given in Chapter 4 of [6], we have a uniquesmooth solution ( u ǫ , ω ǫ ) in Ω ǫ in the time interval [0 , T ) with initial vorticity ( ω ) ǫ . Letthe corresponding flows be X ǫ : Ω ǫ × [0 , T ) → Ω ǫ , then from the transport equation we seethat ω ǫ ( x, t ) = ( ω ) ǫ ( X − ǫ ( x, t )). As X ǫ ( · , t ) and X − ǫ ( · , t ) are measure preserving, we seethat for all 1 ≤ p ≤ ∞ we have k ω ǫ ( · , t ) k L p (Ω ǫ ) = k ( ω ) ǫ k L p (Ω ǫ ) ≤ k ω k L p (Ω) . Step 1:
Let K ⊂ Ω be a compact set and let 0 < ǫ ≤ K ⊂ Ω ǫ for all 0 < ǫ ≤ ǫ . From using the fact that the velocity is uniformly bounded by Lemma 4.2 followed byLemma 4.3, we see that there exists a compact set K ⊂ Ω such that for all 0 < ǫ ≤ ǫ and x ∈ K and t , t ∈ [0 , T ) we have that X ǫ ( X − ǫ ( x, t ) , t ) , X − ǫ ( X ǫ ( x, t ) , t ) ∈ K . Similarly D EULER UNIQUENESS 19 there also exists compact sets K , K ⊂ Ω such that for all 0 < ǫ ≤ ǫ and t , t ∈ [0 , T ) wehave for x ∈ K X ǫ ( X − ǫ ( x, t ) , t ) , X − ǫ ( X ǫ ( x, t ) , t ) ∈ K and similarly for x ∈ K we have X ǫ ( X − ǫ ( x, t ) , t ) , X − ǫ ( X ǫ ( x, t ) , t ) ∈ K . These sets will be useful to prove estimates for the maps X and X − below. Now fromLemma 4.2 observe that for all z , z ∈ K and 0 < ǫ ≤ ǫ we have d | X ǫ ( z , t ) − X ǫ ( z , t ) | dt ≤ | u ǫ ( X ǫ ( z , t ) , t ) − u ǫ ( X ǫ ( z , t ) , t ) |≤ C K , k ω k L ∩ L ∞ φ ( | X ǫ ( z , t ) − X ǫ ( z , t ) | ) . Hence from Lemma 3.1 there exists γ , γ > K , T and k ω k L ∩ L ∞ sothat for all z , z ∈ K and t ∈ [0 , T ) | z − z | γ ≤ | X ǫ ( z , t ) − X ǫ ( z , t ) | ≤ | z − z | γ . (33)Therefore for all z , z ∈ K and t ∈ [0 , T ) | z − z | γ ≤ (cid:12)(cid:12) X − ǫ ( z , t ) − X − ǫ ( z , t ) (cid:12)(cid:12) ≤ | z − z | γ . (34)As the velocity is bounded by Lemma 4.2, we have for all x ∈ K and all t , t ∈ [0 , T ) | X ǫ ( x, t ) − X ǫ ( x, t ) | . K, k ω k L ∩ L ∞ | t − t | . Now let X ∗ ǫ ( x, t ; τ ) = X ǫ ( X − ǫ ( x, t ) , t − τ ) denote the backward particle trajectories with X ∗ ǫ ( x, t ; t ) = X − ǫ ( x, t ) and which satisfies the ODE dX ∗ ǫ ( x, t ; τ ) dτ = − u ǫ ( X ∗ ǫ ( x, t ; τ ) , t − τ ) X ∗ ǫ ( x, t ; 0) = x. Observe that for x ∈ K we have X ∗ ǫ ( x, t ; τ ) ∈ K . Hence from the above equation, (34)and Lemma 4.2 we see that for all x ∈ K and all 0 ≤ t ≤ t < T we have (cid:12)(cid:12) X − ǫ ( x, t ) − X − ǫ ( x, t ) (cid:12)(cid:12) = (cid:12)(cid:12) X − ǫ ( x, t ) − X − ǫ ( X ∗ ǫ ( x, t ; t − t ) , t ) (cid:12)(cid:12) ≤ | x − X ∗ ǫ ( x, t ; t − t ) | γ . k ω k L ∩ L ∞ | t − t | γ . Step 2:
Using these estimates we see that for 0 < ǫ ≤ ǫ the restricted functions X ǫ , X − ǫ : K × [0 , T ) → Ω form equicontinuous families. Hence by Arzela Ascoli and a diagonalizationargument and passing to a subsequence we get continuous functions
X, X − : Ω × [0 , T ) → Ωsuch that X ǫ → X and X − ǫ → X −
10 SIDDHANT AGRAWAL AND ANDREA R. NAHMOD uniformly on compact subsets of Ω × [0 , T ). Hence for all t ∈ [0 , T ) the function X ( · , t ) :Ω → Ω is a homeomorphism. As X ǫ ( · , t ) and X ǫ ( · , t ) are measure preserving, we see thatfor any f ∈ C c (Ω ǫ ) ⊂ C c (Ω) Z Ω f ( X ǫ ( x, t ))1 x ∈ Ω ǫ dx = Z Ω ǫ f ( X ǫ ( x, t )) dx = Z Ω ǫ f ( x ) dx = Z Ω f ( x ) dx. Hence by letting ǫ → X ( · , t ) and X − ( · , t )are also measure preserving.We can finally define ω : Ω × [0 , T ) → R as ω ( x, t ) := ω ( X − ( x, t )) and u ( x, t ) := R Ω K Ω ( x, y ) ω ( y, t ) dy . It is easy to see that ( u, ω ) is in the Yudovich class (16) by usingthat fact that X − ( · , t ) is measure preserving and from Lemma 4.2.Let us extend the function ω ǫ ( · , t ) : Ω ǫ → R to ω ǫ ( · , t ) : Ω → R by zero. We then claimthat for any t ∈ [0 , T ) we have k ω ǫ ( · , t ) − ω ( · , t ) k L (Ω) → ǫ → . To see this observe that for any fixed 0 < ǫ <
1, for all 0 < ǫ ≤ ǫ and x ∈ Ω ǫ we have | ω ǫ ( x, t ) − ω ( x, t ) | = (cid:12)(cid:12) ( ω ) ǫ ( X − ǫ ( x, t )) − ω ( X − ( x, t )) (cid:12)(cid:12) ≤ (cid:12)(cid:12) ( ω ) ǫ ( X − ǫ ( x, t )) − ( ω ) ǫ ( X − ǫ ( x, t )) (cid:12)(cid:12) + (cid:12)(cid:12) ( ω ) ǫ ( X − ǫ ( x, t )) − ( ω ) ǫ ( X − ( x, t )) (cid:12)(cid:12) + (cid:12)(cid:12) ( ω ) ǫ ( X − ( x, t )) − ω ( X − ( x, t )) (cid:12)(cid:12) . Hence using the fact that k ( w ) ǫ − ω k L (Ω) → ǫ → X − ǫ ( · , t ) and X − ( · , t ) aremeasure preserving and the fact that X − ǫ ( · , t ) → X − ( · , t ) uniformly on compact subsetsof K , we see that k ( ω ǫ ( · , t ) − ω ( · , t )) Ω ǫ k L (Ω) → ǫ → . As k ω ( · , t ) Ω ǫ − ω ( · , t ) k L (Ω) → ǫ →
0, the claim is proved.We extend u ǫ ( · , t ) : Ω ǫ → C to u ǫ ( · , t ) : Ω → C by zero. We now claim that for any fixed t ∈ [0 , T ) we have u ǫ ( x, t ) → u ( x, t ) a.e. x ∈ Ω. To see this observe that from (21) we havefor z ∈ Ω ǫ u ǫ ( z , t ) = (cid:18) i πν (cid:19) z ( ν − ) Z Ω ǫ " z ν − z ν − z ν − z ν + 2 iǫ ω ǫ ( z, t ) dz. As ω ǫ ( · , t ) = 0 on Ω \ Ω ǫ we have u ǫ ( z , t ) = (cid:18) i πν (cid:19) z ( ν − ) Z Ω " z ν − z ν − z ν − z ν + 2 iǫ ( ω ǫ ( z, t ) − ω ( z, t )) dz + (cid:18) i πν (cid:19) z ( ν − ) Z Ω " z ν − z ν − z ν − z ν + 2 iǫ ω ( z, t ) dz. D EULER UNIQUENESS 21
We see that the second term converges to u ( z , t ) by dominated convergence. The firstterm can be easily controlled by a similar computation as done in Lemma 4.2; that is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) i πν (cid:19) z ( ν − ) Z Ω " z ν − z ν − z ν − z ν + 2 iǫ ( ω ǫ ( z, t ) − ω ( z, t )) dz (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Z Ω | z − z | | ω ǫ ( z, t ) − ω ( z, t ) | dz . k ω ǫ ( · , t ) − ω ( · , t ) k L (Ω ∩ B ( z )) + k ω ǫ ( · , t ) − ω ( · , t ) k L (Ω) which goes to 0 as ǫ → Step 3:
Let us now show that ( u, ω ) is a weak solution to the Euler equation (18). Let ϕ ∈ C ∞ c (Ω × [0 , T )). Then there exists ǫ > ϕ ) ⊂ Ω ǫ × [0 , T ). Hence forall 0 < ǫ ≤ ǫ we see that Z T Z Ω ω ǫ ( ∂ t ϕ + u ǫ · ∇ ϕ ) dx dt = − Z Ω ( ω ) ǫ ϕ ( · , dx. Now observe that Z T Z Ω ω ǫ ( ∂ t ϕ + u ǫ · ∇ ϕ ) dx dt = Z T Z Ω ( ω ǫ − ω )( ∂ t ϕ + u ǫ · ∇ ϕ ) dx dt + Z T Z Ω ω ( ∂ t ϕ + u ǫ · ∇ ϕ ) dx dt. The second term converges to Z T Z Ω ω ( ∂ t ϕ + u · ∇ ϕ ) dx dt by dominated convergence. The first term can be controlled by using the fact that u ǫ arebounded by Lemma 4.2 (cid:12)(cid:12)(cid:12)(cid:12)Z T Z Ω ( ω ǫ − ω )( ∂ t ϕ + u ǫ · ∇ ϕ ) dx dt (cid:12)(cid:12)(cid:12)(cid:12) . ϕ, k ω k L ∩ L ∞ Z T k ω ǫ ( · , t ) − ω ( · , t ) k L (Ω) dt which goes to zero by dominated convergence. We also see that as ( ω ) ǫ → ω in L (Ω) wehave − Z Ω ( ω ) ǫ ϕ ( · , dx → − Z Ω ( ω ) ϕ ( · , dx. Thus ( u, ω ) satisfies (18). Now for any h ∈ G c (Ω) we see that Z Ω ǫ u ǫ ( · , t ) · h = Z Ω u ǫ ( · , t ) · h = 0 . Consequently by using the fact that u ǫ are bounded by Lemma 4.2, we get from dominatedconvergence that Z Ω u ( · , t ) · h = 0 . Hence proved. (cid:3)
Lemma 4.5.
Let ( u, ω ) be a Yudovich weak solution with initial vorticity ω in the domain Ω in the time interval [0 , T ) . Then(1) The map X ( · , t ) : Ω → Ω is a homeomorphism for each t ∈ [0 , T ) and the functions X, X − : Ω × [0 , T ) → Ω are continuous.(2) ω ( x, t ) = ω ( X − ( x, t )) for a.e. ( x, t ) ∈ Ω × [0 , T ) (3) If ( t n ) ∞ n =1 is a sequence in [0 , T ) with t n → t ∈ [0 , T ) , then k ω ( · , t n ) − ω ( · , t ) k → as n → ∞ .(4) The functions b : Ω × [0 , T ) → C , e b : H × [0 , T ) → C and u : Ω × [0 , T ) → C arebounded continuous functions and the ODE (25) for ǫ = 0 is true pointwise for all ( x, t ) ∈ Ω × [0 , T ) .Proof. We prove the statements sequentially.(1) As X is defined as the solution to the ODE (25) for ǫ = 0 and as the velocityis locally log-Lipschitz from Lemma 4.2, we see that X is continuous as long as X ( x, t ) ∈ Ω. Now from Lemma 4.3 we see that X ( x, t ) ∈ Ω for all ( x, t ) ∈ Ω × [0 , T )and hence X : Ω × [0 , T ) → Ω is continuous. From the same argument as theone used to derive (33) we see that X ( · , t ) : Ω → Ω is one to one. Now usingthis together with Lemma 4.2 and Lemma 4.3, we see that X ( · , t ) is onto Ω andhence X ( · , t ) : Ω → Ω is a homeomorphism. Hence X − : Ω × [0 , T ) → Ω is alsocontinuous.(2) By using Lemma 3.1 in [12] by Han and Zlatoˇs, we directly get that ω ( x, t ) = ω ( X − ( x, t )) for a.e. ( x, t ) ∈ Ω × [0 , T ).(3) If K ⊂ Ω is a compact set then by a similar argument as the one used in Theorem 4.4we see that the restricted functions X − ( · , t n ) : K → Ω form an equicontinuousfamily. As X − ( · , t n ) → X − ( · , t ) pointwise, this implies that X − ( · , t n ) → X − ( · , t )uniformly on compact sets of Ω. We now get that k ω ( · , t n ) − ω ( · , t ) k → ω by a function g ǫ ∈ C c (Ω) in L (Ω) and passing to the limit.(4) Recall that b is given by the formula (27) with ǫ = 0. From a similar computationas in Lemma 4.2 we see that k b k L ∞ (Ω ǫ ) . k ω k L (Ω ǫ ) ∩ L ∞ (Ω ǫ ) . Now if ( z , t ) , ( z , t ) ∈ Ω × [0 , T ) then from Lemma 3.2 and the calculations ofLemma 4.2 we see that | b ( z , t ) − b ( z , t ) | ≤ | b ( z , t ) − b ( z , t ) | + | b ( z , t ) − b ( z , t ) | . φ ( | z − z | ) min n | z | − ν , | z | − ν o k ω k L ∩ L ∞ + Z Ω | z − z || z | ν − | ω ( z, t ) − ω ( z, t ) | dz. D EULER UNIQUENESS 23
Now by using the weighted AM-GM inequality and using the definition of φ from(7) we get | b ( z , t ) − b ( z , t ) | . max {− ln( | z − z | ) , }| z − z | − ν k ω k L ∩ L ∞ + Z Ω | z − z | ν | ω ( z, t ) − ω ( z, t ) | dz + Z Ω | z | ν | ω ( z, t ) − ω ( z, t ) | dz . max {− ln( | z − z | ) , }| z − z | − ν k ω k L ∩ L ∞ + k ω ( · , t ) − ω ( · , t ) k L ν +12 ν − + k ω ( · , t ) − ω ( · , t ) k L . Hence b : Ω × [0 , T ) → C is bounded and continuous and hence e b : H × [0 , T ) → C isalso bounded and continuous. From Lemma 4.2 we already know that u is boundedand from (21) and (27) we see that u ( z, t ) = ν z ν − b ( z, t ) and hence u : Ω × [0 , T ) → C is also continuous. As the velocity is continuous on Ω × [0 , T ), the ODE (25) for ǫ = 0 is true pointwise for all ( x, t ) ∈ Ω × [0 , T ). (cid:3) Properties of the flow
From now on we will only consider flows on the domain Ω and so we will only be concernedwith equations (25), (26), (27) , (28) and (29) for ǫ = 0. Let us now assume that the initialvorticity ω ∈ L (Ω) ∩ L ∞ (Ω) satisfisfies ω ≥
0. Note that if ω ≡ < R Ω ω ( s ) ds < ∞ . Observe that b (0 ,
0) = e b (0 ,
0) = (cid:16) iν π (cid:17) Z H (cid:20) − s + 1 s (cid:21)e ω ( s ) | s | ν − ds = ν π Z H Im( s ) e ω ( s ) | s | ν − ds. Hence 0 < b (0 , < ∞ and it is completely determined by ω . Define b := b (0 , > . (35)The next proposition quantifies the property that the support of the vorticity moves awayfrom the corner for a short period of time. This is proved in part (3) of the propositionbelow. This is the analog of step (1) of the proof of the uniqueness of the ODE (6) in theintroduction. Proposition 4.6.
Let ( u, ω ) be a Yudovich weak solution in the domain Ω in the time inter-val [0 , ∞ ) with initial vorticity ω ∈ L (Ω) ∩ L ∞ (Ω) satisfying ω ≥ and < R Ω ω ( s ) ds < ∞ . Let X : Ω × [0 , ∞ ) → Ω be the flow map of the solution. Let ǫ > be such that < ǫ < min (cid:8) b , (cid:9) . Then there exists T > and < R < / such that for all t ∈ [0 , T ] we have(1) For all x ∈ Ω ∩ B R (0) we have | b ( x, t ) − b | < ǫ .(2) For all x ∈ Ω ∩ B R (0) we have | X ( x, t ) | ≤ R and for all x ∈ Ω ∩ B R (0) c we have | X ( x, t ) | ≥ R . (3) For all x ∈ Ω + ∩ B R (0) we have X ( x, t ) ∈ Ω + and | X ( x, t ) | ≥ (cid:20) (2 ν − b − ǫ ) tν (cid:21) ν ν − . (4) For all x ∈ Ω + ∩ B R (0) we have | X ( x, t ) | & | x | ǫ Proof.
We will define
T > Y ( y, t ) = X ( x, t ) ν in the upper half plane.(1) From Lemma 4.5 we know that b : Ω × [0 , ∞ ) → C is a continuous function. As0 < b < ∞ there exists T > < R < /
10 such that | b ( x, t ) − b | < ǫ for all x ∈ Ω ∩ B R (0) and t ∈ [0 , T ].Now let R ∗ := ( R ) ν and hence 0 < R ∗ < /
10 and we have | e b ( y, t ) − b | < ǫ forall y ∈ H ∩ B (0 , ν R ∗ ) and t ∈ [0 , T ], where e b was defined in the paragraph after(27).(2) As the velocity is bounded from Lemma 4.2, there exists a constant C > (cid:12)(cid:12)(cid:12) dX ( x,t ) dt (cid:12)(cid:12)(cid:12) ≤ C for all x ∈ Ω and t ∈ [0 , ∞ ). Letting T = R C > t ∈ [0 , T ] we have | X ( x, t ) − X ( x, | ≤ C t ≤ R . Now as X ( x,
0) = x we see that for all x ∈ Ω ∩ B R (0) c we have | X ( x, t ) | ≥ R .Similarly for all x ∈ Ω ∩ B R (0) we have | X ( x, t ) | ≤ R ≤ / y ∈ H ∩ B R ∗ (0) c and t ∈ [0 , T ] we have | Y ( y, t ) | ≥ ( R ) ν = R ∗ / ν . Similarly we have | Y ( y, t ) | ≤ ( R ) ν ≤ / y ∈ H ∩ B R ∗ (0)and t ∈ [0 , T ].(3) Let T = min { T , T } >
0. From (28) and part (1) and (2) of this proposition wesee that for all y ∈ H + ∩ B R ∗ (0) and t ∈ [0 , T ], we have d Re { Y ( y, t ) } dt = 1 ν Re( e b ( Y ( y, t ) , t )) | Y ( y, t ) | − ν ≥ ( b − ǫ ) ν | Re Y ( y, t ) | − ν ≥ . This says that the particle is moving to the right and hence Y ( y, t ) ∈ H + . We canquantify exactly how much it moves to the right by integrating and soRe { Y ( y, t ) } ν − ν − − Re { Y ( y, } ν − ν − ≥ ( b − ǫ ) ν t. As Re { Y ( y, } = Re( y ) ≥ y ∈ H + ∩ B R ∗ (0) we obtain | Y ( y, t ) | ≥ Re { Y ( y, t ) } ≥ (cid:20) (2 ν − b − ǫ ) tν (cid:21) ν − . (36)Hence | X ( x, t ) | ≥ h (2 ν − b − ǫ ) tν i ν ν − for all x ∈ Ω + ∩ B R (0). D EULER UNIQUENESS 25 (4) Let x ∈ Ω ∩ B R (0) and y = x ν . As | Y ( y, t ) | . y ∈ H ∩ B R ∗ (0) and t ∈ [0 , T ], we see from (31) in Lemma 4.3 that (cid:12)(cid:12)(cid:12)(cid:12) d Im { Y ( y, t ) } dt (cid:12)(cid:12)(cid:12)(cid:12) . φ ( | Im( Y ( y, t )) | ) (cid:16) | Y ( y, t ) | − ν (cid:17) k ω k ∞ . φ ( | Im( Y ( y, t )) | ) k ω k ∞ . Hence from Lemma 3.1 we see that there exists C = C ( k ω k L ∩ L ∞ ) > y ∈ H ∩ B R ∗ (0) and t ∈ [0 , T ] we haveIm( Y ( y, t )) & { Im( Y ( y, } e C t . Let T > < e C T < ǫ and let T = min { T , T } >
0. Then forall y ∈ H + ∩ B R ∗ (0) in the time t ∈ [0 , T ] we haveIm { Y ( y, t ) } & (Im { Y ( y, } ) ǫ . We can now prove the required estimate. For y ∈ H + ∩ B R ∗ (0) satisfying Re( y ) ≤ Im( y ) we see that for all t ∈ [0 , T ] we have | Y ( y, t ) | ≥ Im { Y ( y, t ) } & (Im { Y ( y, } ) ǫ & (Im( y )) ǫ & | y | ǫ . For y ∈ H + ∩ B R (0) satisfying Re( y ) ≥ Im( y ) we see that for all t ∈ [0 , T ] we have d Re { Y ( y, t ) } dt ≥ . Hence | Y ( y, t ) | ≥ Re( Y ( y, t )) ≥ Re( y ) ≥ | y |√ & | y | ǫ and thus | X ( x, t ) | & | x | ǫ . We define T ∗ = T and the proof is complete. (cid:3) We now prove that around the corner the flow moves to the right for all time and particlesin Ω + cannot come very close to the origin. We need this to prove uniqueness for all timein Theorem 1.1 and not just for a short time. The following lemma is the analog of provingthat x ( t ) , x ( t ) ≥ c for t ≥ T in step (4) of the proof of uniqueness of the ODE (6) inthe introduction. Proving this lemma would be immediate by a continuity argument if weknew that X : Ω × [0 , ∞ ) → Ω extends continuously to X : Ω × [0 , ∞ ) → Ω. We suspectthat this is true but do not have an argument for it. As we do not know this property, theproof of the following lemma becomes a little more involved.
Lemma 4.7.
Let ( u, ω ) be a Yudovich weak solution in the domain Ω in the time interval [0 , ∞ ) with initial vorticity ω ∈ L (Ω) ∩ L ∞ (Ω) satisfying ω ≥ and < R Ω ω ( s ) ds < ∞ .Let X : Ω × [0 , ∞ ) → Ω be the flow map of the solution and let < T < T . Then thereexists c > such that | X ( x, t ) | ≥ c > for all x ∈ Ω + and t ∈ [ T , T ] . Proof.
Put ǫ = min { b , } > R, T > R ∗ = R ν and let T = min { T, T } >
0. Define δ > δ = (cid:20) (2 ν − b − ǫ ) T ν (cid:21) ν − > . Hence from (36) we see that for all y ∈ H + ∩ B R ∗ (0) and we haveRe { Y ( y, T ) } ≥ δ > . Let R > y ∈ H ∩ B (0) and t ∈ [0 , T ] we have Y ( y, t ) , Y − ( y, t ) ∈ B R (0). Similarly let R > R be large enough so that for all y ∈ H ∩ B R (0) and t ∈ [0 , T ] we have Y ( y, t ) , Y − ( y, t ) ∈ B R (0).Now observe that e b : H × [0 , T ] → C is continuous from Lemma 4.5. Also observe thatfor y ∈ R we have from (29) e b ( y, t ) = (cid:16) iν π (cid:17) Z H (cid:20) y ) − s − y ) − s (cid:21)e ω ( s, t ) | s | ν − ds = ν π Z H Im( s ) | Re( y ) − s | e ω ( s, t ) | s | ν − ds> . Hence there exists 0 < δ < e b ( y, t )) ≥ y ∈ [ − R , R ] × [0 , δ ] and t ∈ [0 , T ]. From Lemma 4.3 we see that there exists 0 < δ < min { δ , R ∗ / } such that forall y ∈ [ − R , R ] × (0 , δ ] and t ∈ [0 , T ] we have Y ( y, t ) , Y − ( y, t ) ∈ [ − R , R ] × (0 , δ ].Now again from Lemma 4.3 we see that there exists 0 < r < min { δ, δ , R ∗ / , } such thatfor all y ∈ H ∩ B r (0) and t ∈ [0 , T ] we have Y ( y, t ) , Y − ( y, t ) ∈ [ − R , R ] × (0 , δ ]. R r R ∗ δ r < δr < δ < δ < r < δ < R ∗ < < R < R Figure 1.
Particle flow around the boundary
We now claim that for all y ∈ H + and t ∈ [ T , T ] we have | Y ( y, t ) | ≥ r . We showthis via contradiction. Suppose y ∈ H + and t ∈ [ T , T ] be such that Y ( y , t ) ∈ B r (0).Then by definition of r we have y ∈ [ − R , R ] × (0 , δ ]. Hence we have that Y ( y , t ) ∈ [ − R , R ] × (0 , δ ] for all t ∈ [0 , T ]. Hence from (28) we have d Re { Y ( y , t ) } dt = 1 ν Re( e b ( Y ( y , t ) , t )) | Y ( y , t ) | − ν ≥ . D EULER UNIQUENESS 27
Now if y ∈ H + ∩ B R ∗ (0) then we see that Re { Y ( y , T ) } ≥ δ > r . Hence by the aboveestimate we have Re { Y ( y , t ) } ≥ Re { Y ( y , T ) } > r and hence | Y ( y , t ) | > r . On theother hand if y / ∈ H + ∩ B R ∗ (0) but satisfies y ∈ [ − R , R ] × (0 , δ ] and y ∈ H + , thenfrom the fact that δ < R ∗ / y ) ≥ R ∗ / > r . Hence we similarly obtainRe { Y ( y , t ) } > r and hence | Y ( y , t ) | > r , which is a contradiction. The lemma nowfollows by setting c = r ν . (cid:3) Energy Estimate
In this section we will ignore the dependence of constants on ν and k ω k L ∩ L ∞ . So wewrite a . b instead of a . ν, k ω k L ∩ L ∞ b .We now consider two Yudovich weak solutions ( u , ω ) and ( u , ω ) in Ω in the timeinterval [0 , ∞ ) with the same initial vorticity ω satisfying (5). Let X , X : Ω × [0 , ∞ ) → Ωbe the corresponding flows of the solutions. Let b , b : Ω × [0 , ∞ ) → C be the correspondingfunctions from (26),(27) so that dX ( x, t ) dt = 1 ν b ( X ( x, t ) , t ) X ( x, t )( ν − ) dX ( x, t ) dt = 1 ν b ( X ( x, t ) , t ) X ( x, t )( ν − ) . Consider the energy E ( t ) = Z Ω | X ( x, t ) − X ( x, t ) || ω ( x ) | dx. We first prove that this energy cannot grow too fast near t = 0. Note that if the domain is C , or if the corner has an angle of νπ with 0 < ν ≤ , then the energy E ( t ) is sufficient toprove uniqueness and one can directly show that E ( t ) = 0 for all t ≥ E ( t ) is not sufficient to prove uniqueness in our case, wecan still gain some useful information out of it. Proposition 5.1.
Let ( u , ω ) and ( u , ω ) be two Yudovich weak solutions in Ω in the timeinterval [0 , ∞ ) with the same initial vorticity ω satisfying (5) . Let X , X : Ω × [0 , ∞ ) → Ω be the corresponding flows of the solutions. Then for any α > satisfying < α < ν ν − there exists constants C, T > such that for all t ∈ [0 , T ] we have E ( t ) ≤ Ct α . Proof.
Observe that if ω ≡
0, then the result is obviously true. So we can assume that0 < R Ω ω ( s ) ds < ∞ . We first define constants depending on α, ν and on b (see (35)).Define p = αα − > ǫ = 12 min (cid:26) ν − α (2 ν − α (2 ν − , b (2 ν − − ν ) , (cid:27) > . (37)Hence p > ν > < p < ǫ (1 − ν ) ν . Also 0 < b + ǫb − ǫ < − ν ) and 0 < ǫ < min { b , } .We will use these inequalities in the upcoming computation. Observe that as the velocity is bounded by Lemma 4.2, we see that for all x ∈ Ω and t ∈ [0 , ∞ ) we have | X ( x, t ) − X ( x, t ) | . t . Hence there exists T ∗ > x ∈ Ωand t ∈ [0 , T ∗ ] we have | X ( x, t ) − X ( x, t ) | ≤ / . Again using the fact that the velocity is bounded we get dE ( t ) dt ≤ Z Ω (cid:12)(cid:12)(cid:12)(cid:12) dX ( x, t ) dt − dX ( x, t ) dt (cid:12)(cid:12)(cid:12)(cid:12) | ω ( x ) | dx . E ( t ) . t . Thus there exists T ∗ > t ∈ [0 , T ∗ ] we have E ( t ) ≤
110 min { , k ω k } . Now using the ǫ from (37) in Proposition 4.6 for the flows X ( · , t ) , X ( · , t ) we get constants R , R and T , T so that Proposition 4.6 is satisfied. Let R = min { R , R } > , T = min { T ∗ , T ∗ , T , T } > . (38)Now for t ∈ [0 , ∞ ) we have dE ( t ) dt ≤ Z Ω (cid:12)(cid:12)(cid:12)(cid:12) dX ( x, t ) dt − dX ( x, t ) dt (cid:12)(cid:12)(cid:12)(cid:12) | ω ( x ) | dx ≤ ν Z Ω (cid:12)(cid:12)(cid:12) b ( X ( x, t ) , t ) X ( x, t ) ν − − b ( X ( x, t ) , t ) X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx ≤ ν Z Ω | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx + 1 ν Z Ω | b ( X ( x, t ) , t ) − b ( X ( x, t ) , t ) || X ( x, t ) | ν − | ω ( x ) | dx = I + II. (39)
Controlling I : We first control the first term I in (39). Using Lemma 4.5,Lemma 3.2 andProposition 4.6 we have for all t ∈ [0 , T ] I = 1 ν Z Ω + | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx . Z Ω + ∩ B R (0) (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx + Z Ω + ∩ B cR (0) (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx . Z Ω + ∩ B R (0) | x | ( ν − ǫ ) | X ( x, t ) − X ( x, t ) || ω ( x ) | dx + Z Ω + ∩ B R (0) c | R | ( ν − | X ( x, t ) − X ( x, t ) || ω ( x ) | dx . R (cid:13)(cid:13) | X ( x, t ) − X ( x, t ) || ω ( x ) | (cid:13)(cid:13) L p (Ω + ∩ B R (0)) (cid:13)(cid:13)(cid:13) | x | ( ν − ǫ ) (cid:13)(cid:13)(cid:13) L q ( B R (0)) + E ( t ) , D EULER UNIQUENESS 29 where p + q = 1. For | x | ( ν − ǫ ) ∈ L q ( B R (0)) we need (cid:16) ν − (cid:17) (1 + ǫ ) q > − ⇐⇒ ν − ǫ (1 − ν ) ν > − (cid:18) − p (cid:19) ⇐⇒ ǫ (1 − ν ) ν > p , which is satisfied by the choice of p and ǫ given in (37). Now as | X ( x, t ) − X ( x, t ) | ≤ / E ( t ) ≤ /
10 and k ω k ∞ < ∞ , we see that for all t ∈ [0 , T ] we have I . α,b ,R E ( t ) p + E ( t ) . α,b ,R E ( t ) p . (40) Controlling II : We now control the second term II in (39). From the definition of b in(27) we see that | b ( X ( x, t ) , t ) − b ( X ( x, t ) , t ) | . Z Ω (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − s ν − X ( x, t ) ν − s ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | ω ( s, t ) | ds + Z Ω (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − s ν − X ( x, t ) ν − s ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | ω ( s, t ) | ds + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Ω X ( x, t ) ν − s ν ( ω ( s, t ) − ω ( s, t )) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Ω X ( x, t ) ν − s ν ( ω ( s, t ) − ω ( s, t )) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = II + II + II + II . Now by Lemma 3.2 we see that II + II . Z Ω (cid:12)(cid:12)(cid:12) X ( x, t ) ν − X ( x, t ) ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − s ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − s ν (cid:12)(cid:12)(cid:12) | ω ( s, t ) | ds . Z Ω | X ( x, t ) − X ( x, t ) | max n | X ( x, t ) | ν − , | X ( x, t ) | ν − o | ω ( s, t ) | ds | X ( x, t ) − s || X ( x, t ) − s | max n | X ( x, t ) | ν − , | s | ν − o max n | X ( x, t ) | ν − , | s | ν − o . Z Ω | X ( x, t ) − X ( x, t ) || X ( x, t ) − s || X ( x, t ) − s || s | ν − | ω ( s, t ) | ds. If z = X ( x, t ), z = X ( x, t ) and f = | ω ( · , t ) | Ω then we see that the above integral equals | z − z | I ((0 , ν − , ( z , , ( z ,
1) : ( f, , ∞ )). Hence by the first estimate of Lemma 3.5 weget II + II . φ ( | X ( x, t ) − X ( x, t ) | ) min n | X ( x, t ) | − ν , | X ( x, t ) | − ν o . Now let us concentrate on II and II . As ω ( X ( s, t ) , t ) = ω ( s ) from Lemma 4.5 and asthe mapping X ( · , t ) is measure preserving (and its inverse as well), we see from the changeof variable s X ( s, t ) on the first term of II Z Ω X ( x, t ) ν − s ν ω ( s, t ) ds = Z Ω X ( x, t ) ν − X ( s, t ) ν ω ( s ) ds. By doing a similar change of variable for the second term of II as well, we see that II = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z Ω X ( x, t ) ν − X ( s, t ) ν ω ( s ) ds − Z Ω X ( x, t ) ν − X ( s, t ) ν ω ( s ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Now by a similar computation for II and by using Lemma 3.2 we now see that II + II . Z Ω (cid:12)(cid:12)(cid:12) X ( s, t ) ν − X ( s, t ) ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − X ( s, t ) ν (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X ( x, t ) ν − X ( s, t ) ν (cid:12)(cid:12)(cid:12) | ω ( s ) | ds . Z Ω | X ( s, t ) − X ( s, t ) || X ( x, t ) − X ( s, t ) || X ( x, t ) − X ( s, t ) || X ( x, t ) | ν − | ω ( s ) | ds. Combining all these estimates we get from (39) that, II = 1 ν Z Ω | b ( X ( x, t ) , t ) − b ( X ( x, t ) , t ) || X ( x, t ) | ν − | ω ( x ) | dx . Z Ω | II + II + II + II || X ( x, t ) | ν − | ω ( x ) | dx . Z Ω φ ( | X ( x, t ) − X ( x, t ) | ) | ω ( x ) | dx + Z Ω (cid:26)Z Ω | X ( s, t ) − X ( s, t ) || X ( x, t ) − X ( s, t ) || X ( x, t ) − X ( s, t ) | | ω ( s ) | ds (cid:27) | ω ( x ) | dx. Now by using Fubini and using the change of variable X ( x, t ) x while observing thatthis is measure preserving, we obtain II . Z Ω φ ( | X ( x, t ) − X ( x, t ) | ) | ω ( x ) | dx + Z Ω (cid:26)Z Ω | X ( s, t ) − X ( s, t ) || x − X ( s, t ) || x − X ( s, t ) | (cid:12)(cid:12) ω ( X − ( x, t )) (cid:12)(cid:12) dx (cid:27) | ω ( s ) | ds. Now if z = X ( s, t ), z = X ( s, t ) and f = (cid:12)(cid:12) ω ( X − ( · , t )) (cid:12)(cid:12) Ω then we see that the in-ner integral in the second term equals | z − z | I (( z , , ( z ,
1) : ( f, , ∞ )). Hence using D EULER UNIQUENESS 31
Lemma 3.4 and the fact that (cid:13)(cid:13) ω ( X − ( · , t )) (cid:13)(cid:13) L ∩ L ∞ = k ω k L ∩ L ∞ we get II . Z Ω φ ( | X ( x, t ) − X ( x, t ) | ) | ω ( x ) | dx. As | X ( x, t ) − X ( x, t ) | ≤ /
10 for all x ∈ Ω and t ∈ [0 , T ] and as φ ( x ) is a concave functionin [0 , / II . k ω k φ (cid:18) E ( t ) k ω k (cid:19) . Now as E ( t ) ≤ (1 /
10) min { , k ω k } in the interval [0 , T ], we use the formula of φ from(7) to see that II . E ( t )( − ln( E ( t )) + ln( k ω k )) . φ ( E ( t )) . (41)We can now use the estimates (40) and (41) in the equation (39) to see that dE ( t ) dt . α,b ,R E ( t ) p . Hence by integrating we see that there exists a C > E ( t ) − p ≤ C t. We get the result by observing that α = pp − . (cid:3) We are now ready to prove our main result.
Proof of Theorem 1.1. If ω ≡ < R Ω ω ( s ) ds < ∞ . Let ( u , ω ) and ( u , ω ) be two Yudovich weak solutionsin Ω in the time interval [0 , ∞ ) with the same initial vorticity ω satisfying (5). Let X , X : Ω × [0 , ∞ ) → Ω be the corresponding flows of the solutions.Let α = ν − so that α satisfies the conditions of P roposition .
1. Let E ( t ) = t − α E ( t ) = t − α Z Ω | X ( x, t ) − X ( x, t ) || ω ( x ) | dx. From Proposition 5.1 we see that lim t → + E ( t ) = 0. We now use this energy to proveuniqueness in a time interval [0 , T ∗ ] for some T ∗ > ǫ, R, T > α = ν − given in (37) and (38). For all t ∈ [0 , T ] we see that dE ( t ) dt = t − α (cid:26)(cid:18) − αt (cid:19) E ( t ) + dE ( t ) dt (cid:27) . From the estimates (39), (41) and the computation for (40) obtained in the proof of Propo-sition 5.1 we get (cid:18) − αt (cid:19) E ( t ) + dE ( t ) dt ≤ (cid:18) − αt (cid:19) E ( t ) + I + II ≤ ((cid:18) − αt (cid:19) E ( t ) + 1 ν Z Ω + ∩ B R (0) | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx ) + 1 ν Z Ω + ∩ B cR (0) | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx + II . b ,R ((cid:18) − αt (cid:19) E ( t ) + 1 ν Z Ω + ∩ B R (0) | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx ) + E ( t ) + φ ( E ( t )) . Now for any z , z ∈ H + we see that min z ∈ [ z ,z ] | z | ≥ √ min {| z | , | z |} . From Proposi-tion 4.6 part (3) we see that for x ∈ Ω + ∩ B R (0), both X ( x, t ) , X ( x, t ) ∈ Ω + ⊂ H + . Hencefrom Proposition 4.6 part (3) we have (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) ≤ (cid:18) ν − (cid:19) | X ( x, t ) − X ( x, t ) | max z ∈ [ X ( x,t ) ,X ( x,t )] | z | ν − ≤ ( √ − ν (cid:18) ν − (cid:19) | X ( x, t ) − X ( x, t ) | max n | X ( x, t ) | ν − , | X ( x, t ) | ν − o ≤ (cid:18) ν − (cid:19) | X ( x, t ) − X ( x, t ) | (cid:20) (2 ν − b − ǫ ) tν (cid:21) − ≤ − ν ) ν (2 ν − b − ǫ ) t | X ( x, t ) − X ( x, t ) | . Hence from Proposition 4.6 part (1) and (2) we have (cid:18) − αt (cid:19) E ( t ) + 1 ν Z Ω + ∩ B R (0) | b ( X ( x, t ) , t ) | (cid:12)(cid:12)(cid:12) X ( x, t ) ν − − X ( x, t ) ν − (cid:12)(cid:12)(cid:12) | ω ( x ) | dx ≤ (cid:18) − αt (cid:19) Z Ω + ∩ B R (0) | X ( x, t ) − X ( x, t ) || ω ( x ) | dx + 2(1 − ν )( b + ǫ )(2 ν − b − ǫ ) t Z Ω + ∩ B R (0) | X ( x, t ) − X ( x, t ) || ω ( x ) | dx ≤ (cid:18) − α + 2(1 − ν )( b + ǫ )(2 ν − b − ǫ ) (cid:19) t Z Ω + ∩ B R (0) | X ( x, t ) − X ( x, t ) || ω ( x ) | dx. This term is non-positive as α = ν − and by the choice of ǫ from (37) we have 0 < b + ǫb − ǫ < − ν ) . Hence for all t ∈ [0 , T ] we have dE ( t ) dt . b ,R t − α φ ( E ( t )) . D EULER UNIQUENESS 33
Now as 0 ≤ E ( t ) ≤ /
10 in t ∈ [0 , T ] we have dE ( t ) dt . b ,R − t − α E ( t ) ln( E ( t )) . Now let β > < α < β < ν ν − . From Proposition 5.1 we see that thereexists a constant C > E ( t ) ≤ C t β for all t ∈ [0 , T ]. Hence E ( t ) ≤ C t β − α .Let T ∗ = min n T, (10 C ) − β − α ) o > t ∈ [0 , T ∗ ] we have 0 ≤ E ( t ) ≤ C t β − α ≤ /
10, and thus for all t ∈ [0 , T ∗ ] we have dE ( t ) dt . b ,R − t − α E ( t ) (cid:8) ln( t − α E ( t )) + ln( t α ) (cid:9) . β,C ,b ,R φ ( E ( t )) − t − α E ( t ) ln( C t β − α ) . β,C ,b ,R φ ( E ( t )) . Hence by Lemma 3.1 we have E ( t ) = 0 for t ∈ [0 , T ∗ ]. This implies that for a.e. x ∈ supp( ω ) and t ∈ [0 , T ∗ ] we have X ( x, t ) = X ( x, t ). Hence from Lemma 4.5 we seethat for a.e. t ∈ [0 , T ∗ ] we have supp( ω ( · , t )) = supp( ω ( · , t )) a.e. and that for a.e. x ∈ supp( ω ( · , t )) we have ω ( x, t ) = ω ( x, t ). All in all, we have that ω ( x, t ) = ω ( x, t )for a.e. ( x, t ) ∈ Ω × [0 , T ∗ ] and hence X ( x, t ) = X ( x, t ) a.e. ( x, t ) ∈ Ω × [0 , T ∗ ].To complete the proof we will show the uniqueness for any arbitrary large time interval[0 , T ′ ] where T ′ > T ∗ . From Lemma 4.7 we see that there exists c > x ∈ Ω + and t ∈ [ T ∗ , T ′ ] we have | X i ( x, t ) | ≥ c > i = 1 ,
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