The vorticity equations in a half plane with measures as initial data
aa r X i v : . [ m a t h . A P ] A p r THE VORTICITY EQUATIONS IN A HALF PLANE WITH MEASURES ASINITIAL DATA
K.ABEA bstract . We consider the two-dimensional Navier-Stokes equations subject to the Dirich-let boundary condition in a half plane for initial vorticity with finite measures. We study lo-cal well-posedness of the associated vorticity equations for measures with a small pure pointpart and global well-posedness for measures with a small total variation. Our constructionis based on an L -estimate of a solution operator for the vorticity equations associated withthe Stokes equations.
1. I ntroduction
We consider the Navier-Stokes equations in a half plane:(1.1) ∂ t u − ∆ u + u · ∇ u + ∇ p = , div u = R + × (0 , ∞ ) , u = ∂ R + × (0 , ∞ ) , u = u on R + × { t = } , for initial data u = t ( u , u ) ∈ L , ∞ σ ( R + ) with a finite measure ω = ∂ u − ∂ u ∈ M ( R + ),where L , ∞ σ ( R + ) = n f ∈ L , ∞ ( R + ) (cid:12)(cid:12)(cid:12) div f = R + , f ( x , = , x ∈ R o , and M ( R + ) denotes the space of finite real regular Borel measures on R + equipped with thetotal variation || · || M . Examples of such ω are vortex sheets and point sources of vorticities.A vortex sheet is a continuous measure supported on a smooth curve in the plane and a pointsource is a pure point measure. For the Cauchy problem, global-in-time solutions exist forsuch initial data [9], [21] (see also [4], [7], [23]), while for a half plane a few results isin known. As in R , initial velocity u ∈ L , ∞ σ satisfying ω ∈ M is represented by theBiot-Savart law u ( x ) = Z R + ∇ ⊥ x D ( x , y ) ω (d y ) , (1.2)where ∇ ⊥ = t ( ∂ , − ∂ ) and Date : April 9, 2019.2010
Mathematics Subject Classification.
Key words and phrases.
Vorticity equations, half plane, finite measures. D ( x , y ) = E ( x − y ) − E ( x − y ∗ ) , y ∗ = t ( y , − y ) , E ( x ) = − π log | x | . The right-hand side of (1.2) is an integral by the Borel measure ω . We write (1.2) by u = K ω . Since K ω = ∇ ⊥ E ∗ ω odd0 for a measure ω odd0 and the convolution ∗ in R , K actsas a bounded operator from M to L , ∞ . If the total variation of ω is small, u is small in L , ∞ . Hence for small ω ∈ M unique global-in-time solutions to (1.1) exist by a small dataresult in L , ∞ [24]. If the total variation of ω is large, even local well-poseness of (1.1) isunknown in general.We study the vorticity equations associated with (1.1):(1.3) ∂ t ω − ∆ ω + u · ∇ ω = R + × (0 , ∞ ) ,∂ ω − A ω = − ∂ p on ∂ R + × (0 , ∞ ) ,ω = ω on R + × { t = } , where A is the generator of the Poisson semigroup e sA g ( x ) = Z R P s ( x − y ) g ( y )d y , P s ( x ) = s π ( | x | + s ) , and p = p − p is a remainder from the harmonic pressure p = − R ∞ x e sA ∂ ω d s . By theFourier transform, we write A = − H ∂ with the Hilbert transform H (see Section 3 for thedefinition of H ). Since − ∆ u = ∇ ⊥ ω and H = − I , the boundary condition (1 . follows bytaking the tangential trace to (1 . .The vorticity equations (1.3) is studied in [26] by using a solution formula for that asso-ciated with the Stokes flow (i.e., u = , p = T ( t ) ω = Z R + W ( x , y , t ) ω ( y )d y , W ( x , y , t ) = Γ ( x − y , t ) − Γ ( x − y ∗ , t ) + H ∂ − ∂ ) ∂ E ∗ Γ ( x − y ∗ , t ) , Γ ( x , t ) = π t e −| x | / t . The formula (1.4) is written with A in [26]. We write it with H . Since the Hilbert transformis bounded on L q ( R ) for q ∈ (1 , ∞ ), T ( t ) is a bounded operator on L q ( R + ). For q =
1, thekernel W ( x , y , t ) is not integrable in R + for the x -variable (see Remarks A.2 (iii)). The for-mula (1.4) is available to represent vorticity of the Stokes flow, provided that the tangentialtrace of u vanishes, i.e., u ( x , = , x ∈ R . (1.5)By (1.2), this condition is equivalent to Z R + y | x − y | + y ω ( y )d y = , x ∈ R . (1.6)For example, for ω ∈ L q , q ∈ (1 , Z R + ∂ x E ( x − y ∗ ) ω ( y )d y vanishes by the Liouville theorem. Hence, T ( t ) ω → ω + H ∂ x − ∂ x ) Z R + ∂ x E ( x − y ∗ ) ω ( y )d y = ω in L q as t → . (We give a proof for (1.4) in Appendix A for the completeness.) If ω is integrable, (1.5)implies the zero total mass condition, Z R + ω ( y )d y = , (1.7)by integrating (1.6) by the x -variable [26].The condition (1.5) is not always satisfied for all ω ∈ M . For example, if ω is a pointmass, e.g., ω = κδ x for κ ∈ R and the Dirac measure δ x at x = t (0 , u = κ ∇ ⊥ D ( x , x ) does not vanish, i.e., u ( x , = κπ ( | x | + , x ∈ R . For ω ∈ M , the tangential trace u ( x ,
0) belongs to L ( R ) by (1.2). To study (1.3) formeasures ω ∈ M , we construct a di ff erent solution operator based on the Green matrix ofthe Stokes semigroup [34]. As is well known, the integral form of (1.1) is u ( t ) = S ( t ) u − Z t S ( t − s ) P ( u · ∇ u )d s , (1.8)where S ( t ) denotes the Stokes semigroup and P denotes the Helmholtz projection. Since P ( u · ∇ u ) = P ( ω u ⊥ ) for u ⊥ = t ( − u , u ), ( ω, u ) satisfies ω ( t ) = T ( t ) ω + Z t ∇ ⊥ · S ( t − s ) P ( ω u ⊥ )d s , u = K ω, (1.9)for T ( t ) = −∇ ⊥ · S ( t ) K . The equations (1.9) may be viewed as an integral form of the vorticityequations (1.3). Since S ( t ) u is defined for u ∈ L , ∞ σ , T ( t ) is defined for all ω ∈ M . Weshow that by the Green matrix of S ( t ), T ( t ) is represented by (1.10) T ( t ) ω = Z R + W ( x , y , t ) ω (d y ) , W ( x , y , t ) = Γ ( x − y , t ) + Γ ( x − y ∗ , t ) + Z y Z R Γ ( x − z ∗ , t ) ∂ z E ( z − y )d z − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y )d z , Γ ( r , t ) = π t ) / e − r / t . With the operators,(1.11) e t ∆ N ϕ = Z R + (cid:0) Γ ( x − y , t ) + Γ ( x − y ∗ , t ) (cid:1) ϕ ( y )d y , e t ∆ D ϕ = Z R + (cid:0) Γ ( x − y , t ) − Γ ( x − y ∗ , t ) (cid:1) ϕ ( y )d y , e t ∂ g = Z R Γ ( x − y , t ) g ( y )d y . ( − ∆ D ) − ω = Z R + D ( x , y ) ω (d y ) , T ( t ) is represented by(1.12) T ( t ) ω = e t ∆ N ω − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω − Γ ( x , t ) e t ∂ u ( · , . If (1.5) is satisfied, T ( t ) ω agrees with T ( t ) ω (see Theorem A.3). But the kernel W ( x , y , t )is di ff erent from W ( x , y , t ). The formulas (1.10) and (1.12) are available to represent vortic-ity of the Stokes flow even if (1.5) is not satisfied.An important property of the operator T ( t ) is the L -estimate || T ( t ) ω || ≤ C || ω || M , t > . (1.13)This follows from integrability of the kernel W ( x , y , t ) for the x -variable. We shall show thatthe kernel W ∗ ( x , y , t ) = Γ ( x − y ∗ , t ) + Z y Z R Γ ( x − z ∗ , t ) ∂ z E ( z − y )d z agrees with − G ∗ ( y , x , t ) for the Green matrix G i j ( x , y , t ) = Γ ( x − y , t ) δ i j + G ∗ i j ( x , y , t ) of S ( t )(see Section 2 for the definition of G i j ( x , y , t )). Note that in contrast to R , S ( t ) does notsatisfy the L -estimate [10], [30], i.e., S (1) v < L ( R + ) for some v ∈ L ∩ L ( R + ) , div v = , v ( x , = . On the other hand, since G i j ( x , y , t ) satisfies a Gaussian bound for the y -variable, the L ∞ -estimate || S ( t ) v || ∞ ≤ C || v || ∞ , t > , (1.14)holds [10], [34]. Since W ∗ ( x , y , t ) = − G ∗ ( y , x , t ), (1.13) is obtained similarly to (1.14) andis di ff erent from the L -estimate of S ( t ).The continuity at t = C ( R + ) = ( ϕ ∈ C ( R + ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) lim | x |→∞ ϕ ( x ) = ) . The space C is the pre-dual space of M [29]. We consider the vague (weak-star) topologyon M . Let δ , [0 , ∞ ) denote the Dirac measure on [0 , ∞ ) at x =
0, i.e., < δ , [0 , ∞ ) , ψ > = ψ (0) , ψ ∈ C [0 , ∞ ) , where < · , · > denotes the paring for M [0 , ∞ ) and C [0 , ∞ ). For ω ∈ M , we shall show that T ( t ) ω → ω − δ , [0 , ∞ ) u ( x ,
0) vaguely on M as t → . (1.15)Since ∇ ⊥ x D ( x , y ) = y = K δ , [0 , ∞ ) u ( · , ≡
0. Thus by normalizing ω ∈ M by˜ ω = ω − δ , [0 , ∞ ) u ( x , , (1.16)(1.15) is rephrased as T ( t ) ˜ ω = T ( t ) ω → ˜ ω vaguely on M as usual. Since T ( t ) ω becomesvaguely continuous by the normalization, we simply say that T ( t ) ω is vaguely continuouson M at t =
0. If ω is a continuous measure, t − / q T ( t ) ω tends to zero on L q for q ∈ (1 , ∞ ].If ω has a density (i.e., ω ∈ L ) and (1.5) is satisfied, the stronger convergence T ( t ) ω → ω on L holds. The condition (1.5) is necessary for the L -convergence since the trace of S ( t ) u vanishes for t > || S ( t ) u − u || L , ∞ ( R + ) + || S ( t ) u − u || L ( ∂ R + ) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R + ( T ( t ) ω − ω )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . || T ( t ) ω − ω || L ( R + ) , by the Biot-Savart law (1.2). The L -convergence of T ( t ) ω also implies the zero total massfor ω and continuity of S ( t ) u on L , ∞ .We construct solutions of the vorticity equations (1.3) for u ∈ L , ∞ σ satisfying ω ∈ M with a small pure point part. We say that a measure µ ∈ M is pure point (discrete) if there exists a countable set { x j } ⊂ R + and { κ j } ⊂ R such that µ = P j κ j δ x j . A measure µ ∈ M is called continuous if µ ( { x } ) = x ∈ R + . If the total variation of µ is finite, the set D = { x ∈ R + | µ ( { x } ) , } is countable. Hence, µ ∈ M is uniquely decomposed as µ = µ pp + µ cont with pure point µ pp and continuous µ cont by setting µ pp ( E ) = µ ( D ∩ E ) for Borel sets E ⊂ R + .Since t − / q T ( t ) ω , pp does not tend to zero as t →
0, we assume a smallness for ω , pp in orderto construct local-in-time solutions. If the total variation of ω ∈ M is small, we are able toconstruct small global-in-time solutions. Let BC ([0 , T ]; X ) (resp. BC w ([0 , T ]; X )) denote thespace of all bounded (resp. weakly-star) continuous functions from [0 , T ] to a Banach space X . We denote by BC ((0 , T ]; X ) the space of all bounded functions in [0 , T ], continuous in(0 , T ]. The main result of this paper is: Theorem 1.1. (i) There exists δ > such that for u ∈ L , ∞ σ satisfying ω ∈ M and || ω , pp || M ≤ δ , there exists T > and a unique ( ω, u ) satisfying (1.8), (1.9) and ω ∈ BC w ([0 , T ]; M ) , (1.17) u ∈ BC w ([0 , T ]; L , ∞ ) , (1.18) t − / q ω ∈ BC ((0 , T ]; L q ) , < q ≤ ∞ , (1.19) t / − / p u ∈ BC ((0 , T ]; L p ) , < p ≤ ∞ . (1.20) If ω is continuous, both values (1.19) and (1.20) vanish at t = . If in addition that ω ∈ L and u ( x , = , ( ω, u ) is strongly continuous at t = .(ii) There exists δ > such that for u ∈ L , ∞ σ satisfying ω ∈ M and || ω || M ≤ δ , thereexists a unique ( ω, u ) satisfying (1.8), (1.9), (1.17)-(1.20) for T = ∞ . Since (1.1) is globally well-posed for bounded initial data with finite Dirichlet integral [1],by replacing t ∈ (0 , T ] as an initial time, we have: Theorem 1.2.
The solution constructed in Theorem 1.1 (i) is global, i.e., ( ω, u ) satisfies(1.8), (1.9), (1.17)-(1.20) for all T > . Theorem 1.2 implies global well-posedness of (1.1) for ω ∈ M with a small pure pointpart (e.g., ω , pp ≡ ff use by the Navier-Stokesflow with boundary. On the other hand, smallness conditions are assumed in Theorems1.1 (ii) and 1.2 for the pure point part ω , pp in order to construct global-in-time solutions.Existence for ω ∈ M with large ω , pp is unknown even if ω is a point mass, i.e., ω = κδ x for x ∈ R + and large κ ∈ R . For the Stokes flow, κ W ( x , x , t ) defined by (1 . is an exactsolution for ω = κδ x .For the Cauchy problem, global-in-time solutions of (1.1) exist for all ω ∈ M ( R ) bya priori estimates of vorticity [9], [21]. The uniqueness for ω ∈ M ( R ) with small ω , pp is proved in [21] based on an integral form of the vorticity equations. See also [23]. Theuniqueness for ω ∈ M ( R ) with large ω , pp is more di ffi cult. For u = (2 π ) − x ⊥ | x | − ∈ L , ∞ ( R ) with ω = δ ∈ M ( R ), there exists a forward self-similar solution of (1.1) in R ,called the Lamb-Oseen vortex: Ω ( x , t ) = t Ω x √ t ! , U ( x , t ) = √ t U x √ t ! , where Ω ( x ) = π e −| x | / , U ( x ) = x ⊥ π | x | (cid:16) − e −| x | / (cid:17) . The uniqueness for ω = κδ and large κ ∈ R is proved in [16] by using a relative entropyfor the self-similar transform of ω . See also [14] for an alternative proof. The uniquenessrelates to the asymptotic formulalim t →∞ t − / q || ω − κ Ω || q = , ω ∈ M ( R ) , κ = Z R ω (d y ) , ≤ q ≤ ∞ . (1.21)The formula (1.21) is studied in [18] for ω with a small total variation and extended in [8]for small κ (see also [15], [17]). For large κ , (1.21) is proved in [16]. The uniqueness forgeneral ω ∈ M ( R ) with large ω , pp is proved in [13].For the half plane, initial data u of homogeneous of degree − u = u ( x , = u ( x ) = κ ( θ ) x | x | , for some κ ( θ ), due to the boundary condition. Here, ( r , θ ) is the polar coordinate. Obviously, ω ( x ) = κ ( θ ) | x | − < M . Hence, any forward self-similar solutions of (1.1) in R + do notsatisfy the initial condition ω ∈ M , in contrast to R . As noted in [16], there are forwardself-similar solutions in R such that ω < M . For the half plane, existence of small forwardself-similar solutions follows from a result in L , ∞ [24].It is an interesting question whether solutions for u ∈ L , ∞ σ satisfying ω ∈ M tend tozero as time goes to infinity. For the Stokes flow, we havelim t →∞ t − / q || T ( t ) ω || q = , ω ∈ M , ≤ q ≤ ∞ . (1.22)See Theorem 4.7. If the total variation of ω is small, t − / q ω is globally bounded in L q byTheorem 1.1 (ii). It is unknown whether t − / q ω tend to zero as t → ∞ . If u ∈ L , wehave lim t →∞ || u || = L ∞ ( −∞ , L ).This paper is organized as follows. In Section 2, we prove the Biot-Savart law (1.2).In Section 3, we prove the formulas (1.10), (1.12) and a kernel estimate for W ( x , y , t ). In Section 4, we study continuity of T ( t ) at time zero. We also prove the asymptotic formula(1.22). In Section 5, we prove Theorems 1.1 and 1.2. In Appendix A, we give a proof forthe formula (1.4). 2. T he S tokes flow on L , ∞ In this section, we prove the Biot-Savart law (1.2) for solenoidal vector fields u ∈ L , ∞ ( R + ) with a finite Borel measure ω = −∇ ⊥ · u on R + (Lemma 2.3). We define allfunction spaces used in the subsequent sections.2.1. Solenoidals in L , ∞ . We recall the Lorentz space [36], [5], [3]. For a measurablefunction f in R + we set a distribution function m ( t , f ) and a decreasing rearrangement f ∗ ( t )by m ( t , f ) = (cid:12)(cid:12)(cid:12)(cid:12)n x ∈ R + | | f ( x ) | > t o(cid:12)(cid:12)(cid:12)(cid:12) , f ∗ ( t ) = inf { s ∈ (0 , ∞ ) | m ( s ) < t } , t > , where | E | denotes the Lebesgue measure for a measurable set E ⊂ R + . For p ∈ (1 , ∞ ), wedefine L p , q ( R + ) by the space of all measurable functions f such that || f || p , q = Z ∞ ( t / p f ∗ ( t )) q d tt ! / q < ∞ , ≤ q < ∞ , || f || p , ∞ = sup t > t / p f ∗ ( t ) < ∞ , q = ∞ . The space L p , q agrees with L p if q = p and L p , q ⊂ L p , q for 1 ≤ q ≤ q ≤ ∞ [5, p.16]. Inparticular, L p ⊂ L p , ∞ . A function f belongs to L p , ∞ if and only ifsup E | E | − + / p Z E | f ( x ) | d x < ∞ . This becomes an equivalent norm to || · || p , ∞ [20]. The space L p , q is a quasi-normed Banachspace and agrees with the real interpolation space ( L p , L p ) θ, q , i.e.,( L p , L p ) θ, q = L p , q , < p < p < p < ∞ , / p = (1 − θ ) / p + θ/ p . By a duality theorem [5, 3.7.1 Theorem],( L p , q ) ∗ = ( L p , L p ) ∗ θ, q = ( L p ′ , L p ′ ) θ, q ′ = L p ′ , q ′ , ≤ q < ∞ , where p ′ denotes the conjugate exponent to p . We denote by C ∞ c ( R + ) the space of all smooth functions with compact support in R + . Since C ∞ c is dense in L p ∩ L p , C ∞ c is also dense in L p , q for 1 ≤ q < ∞ [5, 3.4.2 Theorem (b)]. In the sequel, we do not distinguish the space ofscaler and vector-valued functions.We set the subspaces of L p by L p σ ( R + ) = C ∞ c ,σ ( R + ) ||·|| Lp , C ∞ c ,σ ( R + ) = n f ∈ C ∞ c ( R + ) (cid:12)(cid:12)(cid:12) div f = o , G p ( R + ) = n ∇ Φ ∈ L p ( R + ) (cid:12)(cid:12)(cid:12) Φ ∈ L ( R + ) o . The space L p is decomposed into the direct sum L p ( R + ) = L p σ ( R + ) ⊕ G p ( R + ) . We call P : L p −→ L p σ the Helmholtz projection operator (e.g., [6]). The space L p σ agreeswith the space of all L p -solenoidal vector fields in R + , i.e., L p σ ( R + ) = n f ∈ L p ( R + ) (cid:12)(cid:12)(cid:12) div f = R + , f ( x , = , x ∈ R o . (2.1)The normal trace f ( x ,
0) is understood in the Sobolev space of a negative order W − / p ′ , p ′ ( R ) = W − / p , p ( R ) ∗ [32, II. 1.2.3 Lemma], [12, Theorem II 10.2]. Indeed, for f ∈ L p satisfyingdiv f = f ( x , =
0, set f = P f + ( I − P ) f = f + ∇ Φ . Since ∆Φ = ∂ Φ ( x , = ∇ Φ ≡ f = P f ∈ L p σ by the Liouville theorem. Thisimplies (2.1).Following [28], [24], [37], we define the L p , q -solenoidal space. For the two interpolationpairs { L p , L p } and { L p σ , L p σ } , P : L p i −→ L p i σ , i = ,
1, is bounded and surjective. Since( L p , L p ) θ, q and ( L p σ , L p σ ) θ, q are exact interpolation spaces of type θ [3, 7.23 THEOREM], P : L p , q = ( L p , L p ) θ, q −→ ( L p σ , L p σ ) θ, q ⊂ L p , q is bounded and surjective. We set L p , q σ : = P L p , q = ( L p σ , L p σ ) θ, q . Since C ∞ c ,σ is dense in L p σ ∩ L p σ , C ∞ c ,σ is dense in L p , q σ for 1 ≤ q < ∞ . Moreover, we have L p , q σ ( R + ) = n f ∈ L p , q ( R + ) (cid:12)(cid:12)(cid:12) div f = R + , f ( x , = , x ∈ R o . (2.2)Since L p , q σ = ( L p σ , L p σ ) θ, q ⊂ L p σ + L p σ by the definition of the real interpolation, the right-hand side of (2.2) is larger than the left-hand side. The converse inclusion follows in thesame way as (2.1).For p = ∞ , we define L ∞ σ by the space of all f ∈ L ∞ satisfying div f = f ( x , = The Biot-Savart law.
We recall the space of finite real regular Borel measures on R + [29], [11]. Let B be a Borel σ -algebra on R + (i.e., the σ -algebra generated by open setsin R + ). We say that µ : B −→ [ −∞ , ∞ ] is a singed Borel measure if µ is countably additive.For positive µ ≥ E ∈ B is called outer (resp. inner) regular if µ ( E ) = inf (cid:8) µ ( U ) | E ⊂ U , U : open (cid:9) (resp. µ ( E ) = sup { µ ( K ) | K ⊂ E , K : compact } ). If all E ∈ B are outer and inner regular, µ is called regular. By the Jordan decomposition µ = µ + − µ − for µ + , µ − ≥
0, we set the totalvariation measure | µ | = µ + + µ − . A signed measure µ is called regular if | µ | ≥ M ( R + ) the space ofall signed regular Borel measures on R + equipped with the norm || µ || M = | µ | ( R + ). We set C ( R + ) = ( ϕ ∈ C ( R + ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) lim | x |→∞ ϕ ( x ) = ) . The space C ∞ c ( R + ) is dense in C ( R + ). By the Riesz representation theorem, M ( R + ) = C ( R + ) ∗ . The weak-star topology of M ( R + ) is called vague topology [11]. In the sequel, we write M = M ( R + ) by omitting the symbol R + . Proposition 2.1.
Set K µ ( x ) = Z R + ∇ ⊥ x D ( x , y ) µ ( dy ) . (2.3) Then, K : M −→ L , ∞ , L q −→ L p , q ∈ (1 , , / p = / q − / , L , −→ C , (2.4) is bounded.Proof. For a measure µ ∈ M , we set a measure ¯ µ on R by¯ µ ( E ) = µ ( E ∩ R + ) for Borel sets E ⊂ R . By the reflection ¯ µ ∗ ( E ) = ¯ µ ( E ∗ ) and E ∗ = { x ∗ | x ∈ E } , we set µ odd = ¯ µ − ¯ µ ∗ . By changing the variable, K µ ( x ) = Z R + ( ∇ ⊥ x E ( x − y ) − ∇ ⊥ x E ( x − y ∗ )) µ (d y ) = Z R ∇ ⊥ x E ( x − y ) µ odd (d y ) = ∇ ⊥ E ∗ µ odd . Since µ odd ⊥ E ∗ µ odd is bounded from M ( R ) to L , ∞ ( R ) [21, Lemma 2.2 (i)], (2 . follows.For µ ∈ L q , q ∈ (1 , µ odd ( x ) = µ ( x , x ) , x ≥ , − µ ( x , − x ) , x < . (2.5)Since µ odd ⊥ E ∗ µ odd is bounded from L q ( R ) to L p ( R ) [21], (2 . follows. Since C ∗ = M and ( L , ) ∗ = L , ∞ , µ odd ⊥ E ∗ µ odd is bounded from L , ( R ) to C ( R ). Thus(2 . follows. (cid:3) Proposition 2.2.
Set T µ ( x ) = π Z R y | x − y | + y µ ( dy ) . (2.6) Then, T : M ( R + ) −→ L ( R ) , L q ( R + ) −→ L s , q ( R ) , < q < , / (2 s ) = / q − / , L , ( R + ) −→ C ( R ) , (2.7) is bounded.Proof. By integrating (2.6) directly, (2 . follows. Since ( K µ ) ( x , = T µ ( x ), (2 . follows from (2 . . By applying the general Marcinkiewicz interpolation theorem [5, 5.3.2Theorem] to (2 . and (2 . , (2 . follows. (cid:3) Lemma 2.3. (i) For u ∈ L , ∞ σ (resp. u ∈ L p σ ) satisfying ω ∈ M (resp. ω ∈ L q , q ∈ (1 , , / p = / q − / ), u = K ω holds.(ii) For ω ∈ M and u = K ω , set ˜ ω = ω − δ , [0 , ∞ ) u ( x , by the Dirac measure δ , [0 , ∞ ) on [0 , ∞ ) at x = . Then, K ˜ ω = K ω . Proof.
Since u ∈ L , ∞ σ satisfies ∇ · u = u ( x , =
0, there exists a stream function ψ such that u = ∇ ⊥ ψ . Since 0 = u ( x , = − ∂ ψ ( x , ψ ( x , = u = K ω = ∇ ˜ ψ ∈ L , ∞ (resp. ˜ u ∈ L p ). Since − ∆ ψ = ω and ψ ( x , = ϕ = ψ − ˜ ψ satisfies − ∆ ϕ = ϕ ( x , =
0. Since ∂ ϕ ∈ L , ∞ (resp. ∂ ϕ ∈ L p ), applying theLiouville theorem implies ∂ ϕ ≡
0. Hence, ∂ ϕ ≡
0. Since ∇ ϕ → | x | → ∞ and ϕ ( x , = ϕ ≡ u = K ω . This proves (i).We prove (ii). By (2 . , u ( y , ∈ L ( R ). For fixed x ∈ R + and y ∈ R ( y , x ),observe that ∇ ⊥ x D ( x , y ) = π ( x − y ) ⊥ | x − y | + | x − y | − ( x − y ∗ ) ⊥ | x − y | + | x + y | ! ∈ C [0 , ∞ )as a function of y ∈ [0 , ∞ ). Let < · , · > denote the pairing for M [0 , ∞ ) and C [0 , ∞ ). Since ∇ ⊥ x D ( x , y ) = y =
0, it follows that K δ , [0 , ∞ ) u ( · , = Z R < δ , [0 , ∞ ) , ∇ ⊥ D ( x , y ) > u ( y , y = . We proved K ˜ ω = K ω . (cid:3) The Stokes semigroup.
We define the Stokes semigroup S ( t ) u = Z R + G ( x , z , t ) u ( z )d z , (2.9)by the Green matrix G = ( G i j ) ≤ i , j ≤ ,(2.10) G i j ( x , z , t ) = Γ ( x − z , t ) δ i j + G ∗ i j ( x , z , t ) , G ∗ i j ( x , z , t ) = − Γ ( x − z ∗ , t ) δ i j − − δ j ) ∂ x j Z x Z R ∂ x i E ( x − w ) Γ ( w − z ∗ , t )d w , and the Kronecker delta δ i j [33, p.336]. The function G ∗ satisfies the pointwise estimate | ∂ st ∂ kx ∂ mz G ∗ ( x , z , t ) | ≤ Ce − cz / t t s + m / ( x + t ) k / ( | x − z ∗ | + t ) ( k + m + / , (2.11)for k = ( k , k ), m = ( m , m ) and s ≥ C [33, Proposition 2.5]. By(2.11), S ( t ) satisfies the L p − L r -estimate || ∂ st ∂ kx S ( t ) u || r ≤ Ct | k | / + s + / p − / r || u || p , t > , u ∈ L p σ , < p ≤ r < ∞ . (2.12)See [6, Proposition 4.1] for p ∈ (1 , ∞ ) and [10], [34] for p = ∞ . The estimate (2.12)implies that S ( t ) is a bounded analytic semigroup on L p σ . Since L p , q σ = ( L p σ , L p σ ) θ, q for1 < p < p < p < ∞ , by applying an interpolation theorem [3, 7.23 Theorem], S ( t ) is also a bounded analytic semigroup on L p , q σ for 1 ≤ q ≤ ∞ . Since C ∞ c ,σ is dense in L p , q σ for 1 ≤ q < ∞ , S ( t ) is a C -semigroup on L p , q σ . By the duality ( L p ′ , ) ∗ = L p , ∞ , S ( t ) u isweakly-star continuous on L p , ∞ at t =
0. Moreover, we have || ∂ st ∂ kx S ( t ) u || r ≤ Ct | k | / + s + / p − / r || u || p , ∞ , t > , u ∈ L p , ∞ σ , p < r ≤ ∞ . (2.13)The estimate (2.13) follows from (2.12) by taking 1 < p < p < p < r and applying aninterpolation theorem [3, 7.23 Theorem] for ∂ st ∂ kx S ( t ) : L p i σ −→ L r , i = , || ∂ S ( t ) P f || p ≤ Ct / || f || p , t > , f ∈ L p , < p < ∞ . (2.14)The adjoint operator satisfies || S ( t ) P ∂ F || p ≤ Ct / || F || p , t > , F ∈ L p , < p < ∞ , (2.15)where ∂ = ∂ kx indiscriminately denotes the spatial derivatives | k | =
1. The operators ∂ S ( t ) P and S ( t ) P ∂ are understood as one operators acting on L p . It still acts as a bounded operatorfor p = p = ∞ even if P is unbounded. Lemma 2.4.
The operators ∂ S ( t ) P and S ( t ) P ∂ are uniquely extendable to bounded opera-tors on L and C together with || ∂ S ( t ) P f || ≤ Ct / || f || , t > , f ∈ L , (2.16) || S ( t ) P ∂ F || ∞ ≤ Ct / || F || ∞ , t > , F ∈ C . (2.17) Proof.
Let ( · , · ) denote the pairing for L and C . By integration by parts, observe that( ∂ S ( t ) P f , F ) = − ( f , S ( t ) P ∂ F ) , f , F ∈ C ∞ c ( R + ) . (2.18)Since (2.17) holds for F ∈ C ∞ c ( R + ) [34], we estimate | ( ∂ S ( t ) P f , F ) | = | ( f , S ( t ) P ∂ F ) | ≤ Ct / || f || L ( R + ) || F || L ∞ ( R + ) . By taking a supremum for F ∈ C ∞ c ( R + ), we obtain (2.16) for f ∈ C ∞ c ( R + ). By takingthe closure in L , ∂ S ( t ) P is uniquely extendable to a bounded operator on L together with(2.16) and (2.18). By (2.18), (2.17) holds for F ∈ C ∞ c ( R + ). By taking the closure in C , S ( t ) P ∂ is uniquely extendable to C together with (2.17). (cid:3) Remark . The estimate || ∂ S ( t ) f || H ( R + ) ≤ Ct / || f || L ( R + ) , t > , f ∈ L , div f = , f ( x , = , (2.19)is known to hold [19] for the Hardy space H ( R + ). The estimate (2.16) holds even if f isnot solenoidal. On the other hand, (2.16) is weaker than (2.19) since H ( R + ) ⊂ L ( R + ).3. V orticity associated with the S tokes flow We first derive (1.12) by calculating a kernel of T ( t ) = −∇ ⊥ · S ( t ) K . The explicit form ofthe kernel (1.10) follows from a computation of the kernel of ( H ∂ − ∂ ) ∂ ( − ∆ D ) − . A keyfact is that the Hilbert transform of the Poisson kernel P s is the conjugate Poisson kernel Q s ,i.e., HP s = Q s , s >
0. By using this fact, we calculate the discontinuous kernel H ∂ E .3.1. The Hilbert transform.
To prove (1.12), we use the Hilbert transform [36, ChapterIII-VI]. For a rapidly decreasing function ϕ , we set the Fourier transformˆ ϕ ( ξ ) = F ϕ ( ξ ) = Z R e − ix ξ ϕ ( x )d x , i = √− . For a tempered distribution ϕ , the Fourier transform is defined by( ˆ ϕ, ψ ) = ( ϕ, ˆ ψ )by the pairing ( · , · , ) and rapidly decreasing functions ψ . We define the Hilbert transform H by c H ϕ ( ξ ) = − i ξ | ξ | ˆ ϕ ( ξ ) . (3.1)The operator H satisfies H = − I . It acts as a bounded operator on L q ( R ) for q ∈ (1 , ∞ ) [36].We set the Poisson kernel P s and the conjugate Poisson kernel Q s by P s ( x ) = s π ( x + s ) , Q s ( x ) = x π ( x + s ) . Their Fourier transforms areˆ P s ( ξ ) = e − s | ξ | , ˆ Q s ( ξ ) = − i ξ | ξ | e − s | ξ | . We set the Poisson semigroup by d e sA ϕ = e − s | ξ | ˆ ϕ , i.e., e sA ϕ ( x ) = ( P s ∗ ϕ )( x ) = π Z R s ( x − y ) + s ϕ ( y )d y . (3.2)By di ff erentiating d e sA ϕ , we have c A ϕ = −| ξ | ˆ ϕ = i ξ | ξ | ( i ξ ) ˆ ϕ = − [ H ∂ ϕ. (3.3)Since [ e sA H ϕ = [ Q s ∗ ϕ, (3.4)the Hilbert transform is represented by H ϕ ( x ) = lim s → ( Q s ∗ ϕ )( x ) = lim s → π Z R x − y ( x − y ) + s ϕ ( y )d y . (3.5)We use the kernels P s and Q s to calculate the Hilbert transform of ∇ E . Since(3.6) ∂ x E ( x ) = − x π ( x + x ) = − Q x ( x ) ,∂ x E ( x ) = − x π ( x + x ) = − P x ( x ) , and d HP s = c Q s , s >
0, we have(3.7) H ∂ x E ( x ) = ∂ x E ( x ) , x > , − ∂ x E ( x ) , x < . To prove (1.22), we use the Hardy space H ( R ) [35]. See also [27]. A tempered distribution ϕ belongs to H ( R ) if ϕ + ( x ) = sup t > | ( e t ∂ ϕ )( x ) | ∈ L ( R ) . The quasi-norm for ϕ ∈ H ( R ) is defined by || ϕ || H ( R ) = || ϕ + || L ( R ) . The space H ( R ) is smaller than L ( R ). Indeed, ϕ ∈ L ( R ) belongs to H ( R ) if and only if H ϕ ∈ L ( R ) and the quasi-norm || · || H ( R ) is equivalent to || ϕ || H ( R ) (cid:27) || ϕ || L ( R ) + || H ϕ || L ( R ) . If H ϕ ∈ L ( R ), c H ϕ = − i ξ | ξ | − ˆ ϕ ( ξ ) is continuous for ξ ∈ R . Hence, ϕ ∈ H ( R ) implies0 = ˆ ϕ (0) = Z R ϕ ( x )d x . Note that Γ ( x , = (4 π ) − / e −| x | / < H ( R ) since R R Γ ( x , x =
1. On the other hand, ∂ Γ ( x , ∈ H ( R ) since ∂ Γ + ( x , = sup t > | e t ∂ ∂ Γ ( x , | = sup t > | ∂ Γ ( x , t ) | = sup ((cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | ∂ Γ ( z , | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | z | ≤ | x | ) ≤ C + | x | , x ∈ R . This implies A Γ ( x , = − H ∂ Γ ( x , ∈ L ( R ) . (3.8)We use (3.8) to prove (1.22) in Section 4.3.2. Solution formulas.
We prove the formulas (1.12) and (1.10). Let u ∈ L , ∞ σ satisfy ω ∈ M . Since u = K ω by Lemma 2.3 (i), it follows from (2.9) that S ( t ) u = Z R + G ( x , z , t ) Z R + ∇ ⊥ z D ( z , y ) ω (d y ) ! d z = Z R + Z R + G ( x , z , t ) ∇ ⊥ z D ( z , y )d z ! ω (d y ) . By taking the rotation, we have(3.9) T ( t ) ω = −∇ ⊥ · S ( t ) K ω = Z R + W ( x , y , t ) ω (d y ) , W ( x , y , t ) = −∇ ⊥ x · Z R + G ( x , z , t ) ∇ ⊥ z D ( z , y )d z ! . Proposition 3.1. (3.10) W ( x , y , t ) = − ∂ x Z R + (cid:0) Γ ( x − z , t ) − Γ ( x − z ∗ , t ) (cid:1) ∂ z D ( z , y ) dz − ∂ x Z R + (cid:0) Γ ( x − z , t ) − Γ ( x − z ∗ , t ) (cid:1) ∂ z D ( z , y ) dz + Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y ) dz . Proof.
By (2.10), G ( x , z , t ) = Γ ( x − z , t ) − Γ ( x − z ∗ , t ) − Z x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w , G ( x , z , t ) = , G ( x , z , t ) = − Z x Z R ∂ x ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w , G ( x , z , t ) = Γ ( x − z , t ) − Γ ( x − z ∗ , t ) . It follows from (3.9) that(3.11) W ( x , y , t ) = ∂ x Z R + G ( x , z , t ) ∂ z D ( z , y )d z − Z R + G ( x , z , t ) ∂ z D ( z , y )d z ! − ∂ x Z R + G ( x , z , t ) ∂ z D ( z , y )d z ! = Z R + (cid:0) ∂ x G ( x , z , t ) − ∂ x G ( x , z , t ) (cid:1) ∂ z D ( z , y )d z − Z R + ∂ x G ( x , z , t ) ∂ z D ( z , y )d z . Since ∂ x G ( x , z , t ) = ∂ x ( Γ ( x − z , t ) − Γ ( x − z ∗ , t )) − Z x Z R ∂ x ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w − w → x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w = − ∂ x ( Γ ( x − z , t ) − Γ ( x − z ∗ , t )) + ∂ x G ( x , z , t ) − w → x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w , we have ∂ x G ( x , z , t ) − ∂ x G ( x , z , t ) = − ∂ x ( Γ ( x − z , t ) − Γ ( x − z ∗ , t )) + w → x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w . It follows from (3.11) that W ( x , y , t ) = − Z R + ∂ x (cid:0) Γ ( x − z , t ) − Γ ( x − z ∗ , t ) (cid:1) ∂ z D ( z , y )d z − Z R + ∂ x ( Γ ( x − z , t ) − Γ ( x − z ∗ , t )) ∂ z D ( z , y )d z + Z R + lim w → x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w ! ∂ z D ( z , y )d z To prove (3.10), it su ffi ces to show thatlim w → x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w = Γ ( x + z , t )( A Γ )( x − z , t ) . (3.12)Since ∂ x E ( x ) = − − Q x ( x ) by (3.6) and Γ ( x , t ) = Γ ( x , t ) Γ ( x , t ), integration by partsyields Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w = − Z R ∂ w ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w = Z R ∂ x E ( x − w ) ∂ w Γ ( w − z ∗ , t )d w = − Γ ( w + z , t ) Z R Q x − w ( x − w ) ∂ w Γ ( w − z , t )d w = − Γ ( w + z , t )( Q x − w ∗ ∂ Γ )( x − z , t ) . By (3.5) and (3.3), we have( Q x − w ∗ ∂ Γ )( x − z , t ) → ( H ∂ Γ )( x − z , t ) = − ( A Γ )( x − z , t ) as w → x . We proved (3.12). The proof is complete. (cid:3)
Proposition 3.2.
Set e t ∆ N , e t ∆ D and e t ∂ by (1.11). Then,e t ∆ D ∂ ϕ = ∂ e t ∆ N ϕ, (3.13) e t ∆ N ∂ ϕ = ∂ e t ∆ D ϕ − Γ ( x , t ) e t ∂ ϕ ( · , . (3.14) Proof.
By integration by parts, e t ∆ N ∂ ϕ = Z R + ( Γ ( x − y , t ) + Γ ( x − y ∗ , t )) ∂ y ϕ ( y )d y = − Z R + ∂ y ( Γ ( x − y , t ) + Γ ( x − y ∗ , t )) ϕ ( y )d y + "Z R ( Γ ( x − y , t ) + Γ ( x − y ∗ , t )) ϕ ( y )d y y = ∞ y = = ∂ x Z R + ( Γ ( x − y , t ) − Γ ( x − y ∗ , t )) ϕ ( y )d y − Γ ( x , t ) Z R Γ ( x − y , t ) ϕ ( y , y . Thus (3.14) holds. In a similar way, (3.13) follows. (cid:3)
Lemma 3.3.
The formulas (1.10) and (1.12) hold for u ∈ L , ∞ σ satisfying ω ∈ M.Proof.
We substitute ϕ ( z , z ) = ∂ z D ( z , y ) into e t ∆ N ∂ ϕ = Z R + (cid:0) Γ ( x − z , t ) + Γ ( x − z ∗ , t ) (cid:1) ∂ z ϕ ( z )d z . Since ∂ z D ( z , y ) = ∂ z (cid:0) E ( z − y ) − E ( z − y ∗ ) (cid:1) = − π z − y | z − y | − z + y | z − y ∗ | ! , we observe that ϕ ( z , = y π ( | z − y | + y ) = P y ( z − y ) . It follows from (3.14) that e t ∆ N ∂ z D = ∂ x e t ∆ D ∂ z D − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y )d z . (3.15)By (3.10) and (3.15), we have W ( x , y , t ) = − ∂ x e t ∆ D ∂ z D − ∂ x e t ∆ D ∂ z D + Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y )d z = − e t ∆ N ∂ z D − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y )d z − e t ∆ D ∂ z D + Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y )d z . Since − ∆ z D ( z , y ) = δ y with the Dirac measure δ y in R + at z = y , we have (3.16) W ( x , y , t ) = e t ∆ N δ y + ( e t ∆ N − e t ∆ D ) ∂ z D − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y )d z + Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y )d z . Since F ( A Γ ∗ ∂ z D ) = −| ξ | b Γ d ∂ z D = F ( Γ ∗ A ∂ z D ) , we have Z R ( A Γ )( x − z , t ) ∂ z D ( z , y )d z = Z R Γ ( x − z , t )( A ∂ z D )( z , y )d z . Hence,2 Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y )d z = Z R + Γ ( x + z , t ) Γ ( x − z , t )( A ∂ z D )( z , y )d z = Z R + Γ ( x − z ∗ , t )( A ∂ z D )( z , y )d z = ( e t ∆ N − e t ∆ D ) A ∂ z D . Since A = − H ∂ by (3.3), this implies( e t ∆ N − e t ∆ D ) ∂ z D + Z R + Γ ( x + z , t )( A Γ )( x − z , t ) ∂ z D ( z , y )d z = ( e t ∆ N − e t ∆ D )( ∂ z D + A ∂ z D ) = − ( e t ∆ N − e t ∆ D )( H ∂ z − ∂ z ) ∂ z D . It follows from (3.16) that(3.17) W ( x , y , t ) = e t ∆ N δ y − ( e t ∆ N − e t ∆ D )( H ∂ z − ∂ z ) ∂ z D − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y )d z . Since u ( z , = π Z R + y | z − y | + y ω (d y ) = Z R + P y ( z − y ) ω (d y ) , by (1.2), we have Z R + Z R Γ ( x − z , t ) P y ( z − y )d z ! ω (d y ) = Z R Γ ( x − z , t ) Z R + P y ( z − y ) ω (d y ) ! d z = Z R Γ ( x − z , t ) u ( z , z = e t ∂ u ( · , . Thus, (1.12) follows by integrating (3.17) by the measure ω ∈ M .To prove (1.10), we set R ( z ) = ( H ∂ z − ∂ z ) ∂ z E ( z ) , Π ( z , y ) = ( H ∂ z − ∂ z ) ∂ z D ( z , y ) = R ( z − y ) − R ( z − y ∗ ) . By (3.7), R ( z ) = ∂ z (cid:0) H ∂ z E ( z ) − ∂ z E ( z ) (cid:1) = , z > , − ∂ z E ( z ) , z < . This implies R ( z − y ∗ ) = z , y ∈ R + and Π ( z , y ) = R ( z − y ) = , y < z , − ∂ z E ( z − y ) , < z < y . Hence ( e t ∆ N − e t ∆ D )( H ∂ z − ∂ z ) ∂ z D = Z R + Γ ( x − z ∗ , t ) Π ( z , y )d z = − Z y Z R Γ ( x − z ∗ , t ) ∂ z E ( z − y )d z . By (3.17), (1.10) follows. (cid:3)
A kernel estimate.
We give a pointwise estimate for W ( x , y , t ). Lemma 3.4. (i) Set (3.18) W ( x , y , t ) = Γ ( x − y , t ) + W ∗ ( x , y , t ) + W tr ( x , y , t ) , W ∗ ( x , y , t ) = Γ ( x − y ∗ , t ) + Z y Z R Γ ( x − z ∗ , t ) ∂ z E ( z − y ) dz , W tr ( x , y , t ) = − Γ ( x , t ) Z R Γ ( x − z , t ) P y ( z − y ) dz . Then, W ( x , y , t ) = λ W ( λ x , λ y , λ t ) , λ > , (3.19) W ∗ ( x , y , t ) = − G ∗ ( y , x , t ) , (3.20) | ∂ st ∂ kx ∂ my W ∗ ( x , y , t ) | ≤ Ce − cx / t t s + k / ( y + t ) m / ( | x − y ∗ | + t ) ( k + m + / , (3.21) for k = ( k , k ) , m = ( m , m ) , and s ≥ .(ii) Set ˜ W ( x , y , t ) = Z y Z R Γ ( x − z ∗ , t ) ∂ z E ( z − y ) dz . (3.22) Then, − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω = Z R + ˜ W ( x , y , t ) ω ( dy ) , ω ∈ M . (3.23) Proof.
The functions Γ ( x , t ), ∇ E ( x ) and P s ( x ) satisfy the scaling properties Γ ( x , t ) = λ Γ ( λ x , λ t ) , ∇ E ( x ) = λ ( ∇ E )( λ x ) , P s ( x ) = λ P λ s ( λ x ) , λ > . By the changing variable, we observe that W ∗ ( λ x , λ y , λ t ) = Γ ( λ ( x − y ∗ ) , λ t ) + Z λ y Z R Γ ( λ x − z ∗ , λ t ) ∂ z E ( z − λ y )d z = Γ ( λ ( x − y ∗ ) , λ t ) + Z y Z R Γ ( λ ( x − w ∗ ) , λ t )( ∂ E )( λ ( w − y )) λ d w = λ − Γ ( x − y ∗ , t ) + Z y Z R Γ ( x − w ∗ , t ) ∂ w E ( w − y )d w ! = λ − W ( x , y , t ) , W tr ( λ x , λ y , λ t ) = − Γ ( λ x , λ t ) Z R Γ ( λ x − z , λ t ) P λ y ( z − λ y )d z = − Γ ( λ x , λ t ) Z R Γ ( λ ( x − w ) , λ t ) P λ y ( λ ( w − y )) λ d w = λ − W tr ( x , y , t ) . Thus (3.19) holds. To prove (3.20), we observe from (2.10) that G ∗ ( x , z , t ) = − Γ ( x − z ∗ , t ) − Z x Z R ∂ x E ( x − w ) Γ ( w − z ∗ , t )d w . By replacing x and z = y , we have G ∗ ( y , x , t ) = − Γ ( y − x ∗ , t ) − Z y Z R ∂ y E ( y − w ) Γ ( w − x ∗ , t )d w . We change the variable w to z . Since Γ ( y − x ∗ , t ) = Γ ( x − y ∗ , t ) and ∂ y E ( y − z ) = − | y − z | − | y − z | π | y − z | = ∂ z E ( z − y ) , it follows from (3.18) that G ∗ ( y , x , t ) = − Γ ( y − x ∗ , t ) − Z y Z R ∂ y E ( y − z ) Γ ( z − x ∗ , t )d z = − Γ ( x − y ∗ , t ) − Z y Z R ∂ z E ( z − y ) Γ ( x − z ∗ , t )d z = − W ∗ ( x , y , t ) . We proved (3.20). The estimate (3.21) follows from (3.20) and (2.11). Since W ( x , y , t ) =Γ ( x − y , t ) + Γ ( x − y ∗ , t ) + ˜ W ( x , y , t ) + W tr ( x , y , t ), (3.23) follows from (1.12). (cid:3) Remarks . (i) The operator T ( t ) : ω ω ( · , t ) is a solution operator of the heat equa-tion,(3.24) ∂ t ω − ∆ ω = Ω × (0 , ∞ ) ,∂ω∂ n + A ω = ∂ Ω × (0 , ∞ ) ,ω = ω on Ω × { t = } , for the half plane Ω = R + , where ∂/∂ n = n ·∇ denotes the normal derivative and n = t (0 , − ∂ t v − ∆ v + ∇ q = , div v = Ω × (0 , ∞ ) , v = ∂ Ω × (0 , ∞ ) , v = v on Ω × { t = } . Since − ∆ v = ∇ ⊥ ω and the pressure q satisfies the Neumann problem − ∆ q = R + ,∂ q ∂ x = ∆ v = ∂ ω on ∂ R + , q is represented as q = − Z ∞ x e sA ∂ ω d s by the Poisson semigroup e sA defined by (3.2). Hence, ∂ q = − Z ∞ x e sA ∂ ω d s = Z ∞ x e sA A ω d s = − e x A A ω. Thus taking the tangential trace to (3 . implies0 = lim x → ( ∂ t v + ∂ ω + ∂ q ) = ( ∂ − A ) ω. (ii) The formula (1.12) gives a solution to (3.24). Since A = − H ∂ by (3.3), (1.12) isrepresented by T ( t ) ω = e t ∆ N ω + ( e t ∆ N − e t ∆ D ) A ( ∂ − A )( − ∆ D ) − ω − Γ ( x , t ) e t ∂ u ( · , . (3.26)By (3.13) and (3.14), we have( ∂ − A )( e t ∆ N − e t ∆ D ) ϕ = − ( e t ∆ N − e t ∆ D )( ∂ + A ) ϕ − Γ ( x , t ) e t ∂ ϕ ( · , . We substitute ϕ = A ( ∂ − A )( − ∆ D ) − ω . Since ( ∂ + A ) ϕ = − A ω and ϕ ( x , = A ∂ ( − ∆ D ) − ω + ∂ ( − ∆ D ) − ω = Au ( x , + ∂ u ( x , = Au ( x , ∂ − A )( e t ∆ N − e t ∆ D ) A ( ∂ − A )( − ∆ D ) − ω = ( e t ∆ N − e t ∆ D ) A ω − Γ ( x , t ) e t ∂ Au ( · , . Multiplying ( ∂ − A ) by (3.26) implies ( ∂ − A ) T ( t ) ω = ( ∂ − A ) e t ∆ N ω + ( e t ∆ N − e t ∆ D ) A ω − Γ ( x , t ) e t ∂ Au ( · , − ∂ x − A )( Γ ( x , t ) e t ∂ u ( · , = ( ∂ − A ) e t ∆ N ω + A ( e t ∆ N − e t ∆ D ) ω − ∂ x Γ ( x , t ) e t ∂ u ( · , . Since ∂ x Γ (0 , t ) =
0, sending x → ∂ − A ) T ( t ) ω = x =
0. We prove theconvergence to initial data (1.15) in Lemma 4.1.(iii) We are able to write the vorticity equations by (3.24) even for domains Ω by using theoperators A = − H ∂ tan and H : g ∂ tan q , associated with the Neumann problem − ∆ q = Ω ,∂ q ∂ n = g on ∂ Ω . Here, ∂ tan = n · ∇ ⊥ = n ⊥ · ∇ for the unit outward normal vector field n = t ( n , n ) on ∂ Ω and n ⊥ = t ( − n , n ). Since div v = − ∆ v = ∇ ⊥ ω , the pressure q solves the Neumannproblem for g = ∆ v · n = − ∂ tan ω [22], [25]. Hence, − ∂ tan q = Hg = − H ∂ tan ω = A ω. Thus multiplying − n ⊥ by (3 . and taking the trace implies0 = − n ⊥ · ∇ ⊥ ω − n ⊥ · ∇ q = ∂ω∂ n − ∂ tan q = ∂ω∂ n + A ω. The operator T ( t ) = −∇ ⊥ · S ( t ) K can be defined also for domains. For example, if Ω isbounded and simply-connected, the Biot-Savart law u = K ω is available and we are ableto define T ( t ) in the same way as the half plane. It is an interesting question whether the L -estimate (1.13) holds for domains. The L ∞ -estimate (1.14) is still valid for boundeddomains [2], while the L -boundedness of S ( t ) has been an open question [10, Remark 5.2].4. T he semigroup associated with vorticity We study continuity of T ( t ) ω as t → ω ∈ M , (ii) continuous measures ω ∈ M and (iii) ω ∈ L satisfying u ( x , = t → ∞ . Continuity in the vague topology.
We shall show that T ( t ) forms a (not stronglycontinuous) bounded analytic semigroup on M . Lemma 4.1. || ∂ st ∂ kx T ( t ) ω || q ≤ Ct − / q + | k | / + s || ω || M , t > , ω ∈ M , ≤ q ≤ ∞ . (4.1) T ( t + s ) = T ( t ) T ( s ) t , s ≥ . (4.2) T ( t ) ω → ω − δ , [0 , ∞ ) u ( x , varguely on M as t → . (4.3) Set u = K ω and ˜ ω = ω − δ , [0 , ∞ ) u ( · , for ω ∈ M. Then,T ( t ) ˜ ω = T ( t ) ω → ˜ ω varguely on M as t → . (4.4) Proof.
We show (4.1) for q = s = k =
0. The case q ∈ (1 , ∞ ], s ≥ | k | ≥ T ( t ) ω = e t ∆ N ω − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω − Γ ( x , t ) e t ∂ u ( · , . By (2 . , we have || u || L ( R ) ≤ || ω || M . By estimating the kernels for e t ∆ N and e t ∂ definedby (1.11), we see that the first and third terms satisfy the desired estimate. We estimate thesecond term. By (3.23) and (3.21), we observe that | ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω | . Z R + e − cx / t | x − y ∗ | + t | ω | (d y ) . Since Z R + e − cx / t | x − y ∗ | + t d x = Z ∞ e − cx / t Z R d z z + | x + y | + t d x = C Z ∞ e − cx / t ( | x + y | + t ) / d x ≤ C Z ∞ e − cx / t ( | x | + t ) / d x = C ′ , we obtain || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || ≤ C || ω || M , t > . We proved (4.1). Since u = K ω = K ( −∇ ⊥ ) · u for u ∈ L , ∞ σ with ω ∈ M by Lemma 2.3 (i), T ( t + s ) = −∇ ⊥ · S ( t + s ) K = −∇ ⊥ · S ( t ) K ( −∇ ⊥ ) · S ( s ) K = T ( t ) T ( s ) . Thus (4.2) holds. To prove (4.3), we take ϕ ∈ C and set( T ( t ) ω , ϕ ) = ( e t ∆ N ω , ϕ ) − (cid:16) ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω , ϕ (cid:17) − (cid:16) Γ ( · , t ) e t ∂ u ( · , , ϕ (cid:17) = I ( t ) + II ( t ) + III ( t ) , with the pairing ( · , · ) for M and C . Let ϕ even be the even extension of ϕ , i.e.,(4.5) ϕ even ( x ) = ϕ ( x , x ) , x ≥ ,ϕ ( x , − x ) , x < . Observe that e t ∆ N ϕ = Z R + (cid:0) Γ ( x − y , t ) + Γ ( x − y ∗ , t ) (cid:1) ϕ ( y )d y = Z R Γ ( x − y , t ) ϕ even ( y )d y = : e t ∆ ϕ even . Since ϕ even ∈ C ( R ), it follows that || e t ∆ N ϕ − ϕ || L ∞ ( R + ) ≤ || e t ∆ ϕ even − ϕ even || L ∞ ( R ) → t → . By the Fubini’s theorem, I ( t ) = ( e t ∆ N ω , ϕ ) = Z R + Z R + (cid:0) Γ ( x − y , t ) + Γ ( x − y ∗ , t ) (cid:1) ω (d y ) ! ϕ ( x )d x = Z R + Z R + (cid:0) Γ ( x − y , t ) + Γ ( x − y ∗ , t ) (cid:1) ϕ ( x )d x ! ω (d y ) = Z R + Z R + (cid:0) Γ ( y − x , t ) + Γ ( y − x ∗ , t ) (cid:1) ϕ ( x )d x ! ω (d y ) = ( ω , e t ∆ N ϕ ) → ( ω , ϕ ) as t → . By (3.23), II ( t ) = Z R + Z R + ˜ W ( x , y , t ) ω (d y ) ! ϕ ( x )d x = Z R + Z R + ˜ W ( x , y , t ) ϕ ( x )d x ! ω (d y ) . Since ˜ W ( x , y , t ) satisfies (3.21), we estimate (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R + ˜ W ( x , y , t ) ϕ ( x )d x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Z R + e − cx / t ( | x − y ∗ | + t ) | ϕ ( x ) | d x . Z ∞ e − cx / t (( x + y ) + t ) / || ϕ || L ∞ ( R ) ( x )d x = : ρ ( y , t ) . Observe that sup t > || ρ || ∞ ( t ) ≤ C for some C > t → ρ ( y , t ) = y > | II ( t ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R + Z R + ˜ W ( x , y , t ) ϕ ( x )d x ! ω (d y ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Z R + ρ ( y , t ) | ω | (d y ) → t → . It remains to show that lim t → III ( t ) = − Z R u ( x , ϕ ( x , x . (4.6)We set η ( x , t ) = Z ∞ Γ ( x , t ) ϕ ( x , x )d x , x ∈ R . (4.7)Since ϕ ∈ C ( R + ), we have lim | x |→∞ ϕ ( x , x ) = x ≥
0. Hence, lim | x |→∞ η ( x , t ) = η ( · , t ) ∈ C ( R ). It follows that η ( x , t ) = Z ∞ π t ) / e −| x | / t ϕ ( x , x )d x = π / Z ∞ e −| z | ϕ ( x , t / z )d z → ϕ ( x ,
0) uniformly for x ∈ R as t → . Thus, η ( · , t ) → ϕ ( · ,
0) on C ( R ) as t →
0. By (4.7),
III ( t ) = − Z R + Γ ( x , t ) (cid:16) e t ∂ u ( · , (cid:17) ( x , t ) ϕ ( x , x )d x = − Z R (cid:16) e t ∂ u ( · , (cid:17) ( x , t ) η ( x , t )d x . Since e t ∂ u ( · , → u ( · ,
0) on L ( R ), sending t → K ˜ ω = K ω by Lemma 2.3 (ii), T ( t ) ˜ ω = −∇ ⊥ · S ( t ) K ˜ ω = −∇ ⊥ · S ( t ) K ω = T ( t ) ω . Thus(4.4) holds. (cid:3) Remark . The kernel estimate (3.21) also implies || ∂ st ∂ kx T ( t ) ω || q ≤ Ct / r − / q + | k | / + s || ω || r , t > , ω ∈ L r , < r < , r ≤ q ≤ ∞ , (4.8)satisfying u ( x , =
0. The trace is understood in L s , r ( R ), 1 / (2 s ) = / r − /
2, by (2 . . A convergence of the L q -norm. The function t − / q T ( t ) ω is bounded in L q by (4.1).Furthermore, we have: Lemma 4.3.
For continuous measures ω ∈ M, lim t → t − / q || T ( t ) ω || q = , < q ≤ ∞ . (4.9) Proof.
We set u = K ω and ω = T ( t ) ω . Since the trace of v = S ( t ) u vanishes for t > v = K ω implies Z R + y | x − y | + y ω ( y , t )d y = , x ∈ R , t > . Thus applying (4.8) yields t − / q || T ( t ) ω || q = t − / q || T ( t / T ( t / ω || q . t − / r || T ( t / ω || r , < r < , r ≤ q ≤ ∞ . It su ffi ces to show (4.9) for q ∈ (1 , T ( t ) ω = e t ∆ N ω − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω − Γ ( x , t ) e t ∂ u ( · , . We shall show that lim t → t − / r || e t ∆ N ω || q = . (4.10)We set a measure ω on R by ω ( E ) = ω ( E ∩ R + )for Borel sets E ⊂ R . By the reflection ω ∗ ( E ) = ω ( E ∗ ) and E ∗ = { x ∗ | x ∈ E } , we define ω even0 = ω + ω ∗ . By changing the variable, we see that e t ∆ N ω = Z R + (cid:0) Γ ( x − y , t ) + Γ ( x − y ∗ , t ) (cid:1) ω (d y ) = Z R Γ ( x − y , t ) ω even0 (d y ) = e t ∆ ω even0 . Since ω is continuous, so is ω even0 . Since lim t → t − / q || e t ∆ ω even0 || q = ω even0 [21, Lemma 4.4], (4.10) follows. Since u ( · , ∈ L ( R ) by (2 . , we have t − / q || Γ ( · , t ) e t ∂ u ( · , || L q ( R + ) . t / − / q ) || e t ∂ u ( · , || L q ( R ) → t → . It remains to show that lim t → t − / q || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || q = . (4.11)We may assume that ω ≥ ω ( R + ) = ω ( ∪ n ≥ { y ≥ / n } ) = lim n →∞ ω ( { y ≥ / n } ) , by continuity of the measure from below, for ε > δ > ω ( { y < δ } ) = ω ( R + ) − ω ( { y ≥ δ } ) ≤ ε. We use the kernel representation (3.23). Since ˜ W ( x , y , t ) satisfies (3.21), we estimate Z R + | ˜ W ( x , y , t ) | q d x . Z R + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − cx / t | x − y ∗ | + t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q d x . Z ∞ e − cqx / t (cid:0) ( x + y ) + t (cid:1) q − / d x . t / ( y + t ) q − / . It follows from (3.23) that t − / q || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || q . t − / q Z R + Z R + | ˜ W ( x , y , t ) | q d x ! / q | ω | (d y ) . Z R + t − / q ( y + t ) − / q | ω | (d y ) = Z { y ≥ δ } + Z { y <δ } . t − / q δ − / q || ω || M + ε. Hence, lim t → t − / q || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || q ≤ C ε for some constant C > ε > (cid:3) Continuity in L . We prove the continuity of T ( t ) ω in L for ω ∈ L satisfying u ( x , = Proposition 4.4.
For ω ∈ L satisfying u ( x , = for u = K ω , there exists a sequence { u , m } such that (4.12) ω , m = −∇ ⊥ · u , m ∈ L , spt ω , m ⊂ R + ,ω , m → ω on L as m → ∞ . Proof.
Let u and ω be the zero extensions of u and ω to R . Since u ( x , = ω = −∇ ⊥ · u in the sense of distribution. We set u , m ( x , x ) = u ( x , x − / m ). Then, ω , m = −∇ ⊥ · u , m satisfies the desired property. (cid:3) Lemma 4.5.
For ω ∈ L satisfying u ( x , = for u = K ω , T ( t ) ω → ω on L ast → .Proof. By (1.12), T ( t ) ω = e t ∆ N ω − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω . Since e t ∆ N ω → ω on L , it su ffi ces to showlim t → || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || = . (4.13)We first show (4.13) under the additional assumption spt ω ⊂ R + . We take δ > ω ⊂ { x > δ } . It follows from (3.23) and (3.21) that || ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω || . Z R + Z R + e − cx / t ( | x − y ∗ | + t ) | ω ( y ) | d y d x . Z ∞ Z R + e − cx / t (( x + y ) + t ) / | ω ( y ) | d y d x . t / δ || ω || → t → . Thus, (4.13) holds. If spt ω ⊂ R + , we take a sequence { u , m } satisfying (4.12) and estimate || T ( t ) ω − ω || ≤ || T ( t )( ω − ω , m ) || + || T ( t ) ω , m − ω , m || + || ω , m − ω || ≤ C || ω , m − ω || + || T ( t ) ω , m − ω , m || . Since spt ω , m ⊂ R + , lim m →∞ || T ( t ) ω − ω || ≤ C || ω − ω , m || . Since the right-hand side tends to zero as m → ∞ by (4.12), the desired result follows. (cid:3) The asymptotic formula.
We prove the asymptotic formula (1.22).
Proposition 4.6. lim t →∞ || e t ∆ µ || = , µ ∈ L ( R ) , Z R µ ( x ) dx = . (4.14) Proof.
The assertion is well known (e.g., [27, Lemma 3.3 (i)]). Since the total mass of µ iszero,( e t ∆ µ )( x ) = Z R ( Γ ( x − y , t ) − Γ ( x , t )) µ ( y )d y = t − Z R (cid:16) Γ (cid:16) ( x − y ) / t / , (cid:17) − Γ (cid:16) x / t / , (cid:17)(cid:17) µ ( y )d y . Integrating e t ∆ µ by x = t / z and applying the dominated convergence theorem yield || e t ∆ µ || ≤ Z R t − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R Γ (cid:16) ( x − y ) / t / , (cid:17) − Γ (cid:16) x / t / , (cid:17) µ ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d x = Z R (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R Γ (cid:16) z − y / t / , (cid:17) − Γ ( z , µ ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d z ≤ Z R Z R (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:16) z − y / t / , (cid:17) − Γ ( z , (cid:12)(cid:12)(cid:12)(cid:12) d z ! | µ ( y ) | d y → t → ∞ . (cid:3) Theorem 4.7. lim t →∞ t − / q || T ( t ) ω || q = , ω ∈ M , ≤ q ≤ ∞ . (4.15) Proof.
By (4.1) and (4.2), we estimate t − / q || T ( t ) ω || q = t − / q || T ( t / T ( t / ω || q . || T ( t / ω || . It su ffi ces to show (4.15) for q =
1. Since T ( t ) ω ∈ L and S ( t ) u = { x = } for t > ω ∈ L and u ( x , = u = K ω , i.e.,0 = u ( x , = π Z R + y | x − y | + y ω ( y )d y , x ∈ R . By integrating u ( x ,
0) by the x -variable, we have Z R + ω ( y )d y = . (4.16)By (1.12), T ( t ) ω = e t ∆ N ω − ( e t ∆ N − e t ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω . Let ω even0 be the even extension of ω defined by (4.5). By (4.16), ω even0 ∈ L ( R ) and R R ω even0 d x =
0. Since e t ∆ N ω = e t ∆ ω even0 , it follows from (4.14) thatlim t →∞ || e t ∆ N ω || = . (4.17)We set ω = T ( t ) ω and consider the scaling ω λ ( x , t ) = λ ω ( λ x .λ t ) , λ > . Since || ω λ || ( t ) = || ω || ( λ t ), observe thatlim t →∞ || T ( t ) ω || = lim λ →∞ || ω || ( λ ) = lim λ →∞ || ω λ || (1) . By the scaling property of the kernel (3.19), ω λ = T ( t ) ω ,λ . Hence, ω λ ( · , = e ∆ N ω ,λ − ( e ∆ N − e ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω ,λ . By (4.17), lim λ →∞ || e ∆ N ω ,λ || = lim λ →∞ || e λ ∆ N ω || = . It su ffi ces to show thatlim λ →∞ || ( e ∆ N − e ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω ,λ || = . (4.18)We use the kernel representation (3.23). We shall show that Z R + | ˜ W ( x , y , | d x ≤ C η ( y ) , y ∈ R + , (4.19)for some constant C > η ( y ) = Z y ρ Γ ( ρ, ρ + y Z ∞ y Γ ( ρ, ρ, y ≥ . The function η is bounded continuous in [0 , ∞ ) and satisfies η (0) =
0. The convergence(4.18) follows from (4.19) since || ( e ∆ N − e ∆ D )( H ∂ − ∂ ) ∂ ( − ∆ D ) − ω ,λ || = Z R + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R + ˜ W ( x , y , ω ,λ ( y )d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d x . Z R + η ( y ) λ | ω ( λ y ) | d y = Z R + η ( z /λ ) | ω ( z ) | d z → λ → ∞ . To prove (4.19), we use the shorthand notation ˜ W ( x , y ) = ˜ W ( x , y ,
1) and Γ ( x ) = Γ ( x , ∂ x E ( x ) = − − Q x ( x ) by (3.6),˜ W ( x , y ) = Z y Z R Γ ( x − z ∗ ) ∂ z E ( z − y )d z = ∂ x Z y Z R Γ ( x − z ) Γ ( x + z ) ∂ z E ( z − y )d z = − ∂ x Z y Z R Γ ( x − z ) Γ ( x + z ) Q y − z ( z − y )d z = − ∂ x Z y Γ ( x + z )( Q y − z ∗ Γ )( x − y )d z . Since Q s ∗ Γ = e sA H Γ and A = − H ∂ by (3.4) and (3.3),˜ W ( x , y ) = − ∂ x Z y Γ ( x + z ) e ( y − z ) A H Γ d z = Z y Γ ( x + z ) e ( y − z ) A A Γ d z . Since A Γ ∈ L ( R ) by (3.8), we estimate Z R + | ˜ W ( x , y ) | d x . Z ∞ Z y Γ ( x + z ) || e ( y − z ) A A Γ || L ( R ) d z ! d x . Z y Z ∞ Γ ( x + z )d x ! d z = Z y Z ∞ z Γ ( ρ )d ρ ! d z . By changing the order of the integrals, Z y Z ∞ z Γ ( ρ )d ρ ! d z = Z y Z ρ Γ ( ρ )d z ! d ρ + Z ∞ y Z y Γ ( ρ )d z ! d ρ = Z y ρ Γ ( ρ )d ρ + y Z ∞ y Γ ( ρ )d ρ = η ( y ) . Thus (4.19) holds. The proof is now complete. (cid:3)
5. A pplications to the N avier -S tokes flow We prove Theorems 1.1 and 1.2.5.1.
Local well-posedness.
For u ∈ L , ∞ σ satisfying ω ∈ M , we set ω = ω , pp + ω , cont . (5.1)Since t − / q T ( t ) ω , cont → L q for q ∈ (1 ,
2) as t → . and (2.12) implythat u , cont = K ω , cont satisfieslim t → t / − / p ( || S ( t ) u , cont || p + t / ||∇ S ( t ) u , cont || p ) = , < p < ∞ . Thus by (2.13) and (2 . , there exists a constant C > t → t / − / p ( || S ( t ) u || p + t / ||∇ S ( t ) u || p ) ≤ C || ω , pp || M , < p < ∞ . (5.2)We set a sequence { u j } by(5.3) u j + = u − Z t S ( t − s ) P ( u j · ∇ u j )d s , u = S ( t ) u . By taking the rotation to (5.3),(5.4) ω j + = ω + Z t ∇ ⊥ · S ( t − s ) P ( ω j u ⊥ j )d s , u j = K ω j ,ω = T ( t ) ω . For T >
0, set(5.5) N j = sup < t ≤ T t − / q || ω j || q , ≤ q < , L j = sup < t ≤ T t / − / p (cid:16) || u j || p + t / ||∇ u j || p (cid:17) , < p < ∞ . By the Sobolev inequality || u j || ∞ . || u j || − / pp ||∇ u j || / pp ,sup < t ≤ T t / || u j || ∞ ≤ CL j . (5.6) Proposition 5.1. L j + ≤ L + C L j , (5.7) N j + ≤ N + C L j N j , (5.8) with some constant C > .Proof. We set r = p /
2. Applying (2.12) and (2.15) implies || u j + || p ≤ || u || p + Z t || S (( t − s ) / S (( t − s ) / P div u j u j || p d s ≤ || u || p + C Z t d s ( t − s ) / r − / p + / || u j || r d s ≤ || u || p + CL j Z t d s ( t − s ) / p + / s − / p = || u || p + C ′ t / − / p L j . We estimate ||∇ u j + || p ≤ ||∇ u || p + Z t / ||∇ S ( t − s ) P div ( u j u j ) || p d s + Z tt / ||∇ S ( t − s ) P u j · ∇ u j || p d s . Applying (2.12) and (2.15) yields Z t / ||∇ S (( t − s ) / S (( t − s ) / P div ( u j u j ) || p d s ≤ CL j Z t / d s ( t − s ) + / p s − / p = C ′ t − / p L j , Z tt / ||∇ S ( t − s ) P u j · ∇ u j || p d s ≤ C Z tt / t − s ) / + / r − / p || u j · ∇ u j || r d s ≤ CL j Z tt / t − s ) / + / p s / − / p d s = C ′ t − / p L j . We obtained ||∇ u j + || p ≤ ||∇ u || p + Ct − / p L j . Thus (5.7) holds. By (2.14), (2.16), (5.5) and (5.6), we estimate || ω j + || q ≤ || ω || q + Z t ||∇ ⊥ · S ( t − s ) P ( ω j u ⊥ j ) || q d s ≤ || ω || q + C Z t t − s ) / || ( ω j u ⊥ j ) || q d s ≤ || ω || q + C ′ L j N j Z t d s ( t − s ) / s / − / q = || ω || q + C ′′ t − / q L j N j . Thus (5.8) holds. (cid:3)
Proposition 5.2.
There exits a constant δ > such that for ω ∈ M satisfying || ω , pp || M ≤ δ , there exists T > and a unique ( ω, u ) satisfying (1.8), (1.9) and ω ∈ BC w ([0 , T ]; M ) , (5.9) u ∈ BC w ([0 , T ]; L , ∞ ) , (5.10) t − / q ω ∈ BC ((0 , T ]; L q ) , < q < , (5.11) t / − / p u , t − / p ∇ u ∈ BC ((0 , T ]; L p ) , < p < ∞ . (5.12) Proof.
It follows from (5.2) that lim T → L ≤ C || ω , pp || M . (5.13)We take δ = (8 C C ) − . By (5.13), lim T → L ≤ (8 C ) − . We take T > L ≤ (4 C ) − . Then, (5.7) yields L j + ≤ L , j ≥ . (5.14)Since (5.8) and (5.14) imply N j + ≤ N + C L j N j ≤ N + C L N j ≤ N + N j , we have N j + ≤ N , j ≥ . (5.15)Thus ( ω j , u j ) satisfies t − / q ω j ∈ BC ((0 , T ]; L q ) , < q < , (5.16) t / − / p u j ∈ BC ((0 , T ]; L p ) , < p < ∞ . (5.17)We show that ω j ∈ BC w ([0 , T ]; M ) , (5.18) u j ∈ BC w ([0 , T ]; L , ∞ ) . (5.19) Since ω j is bounded on M by (5.14), u j = K ω j is bounded on L , ∞ by (2 . . We shall showthe weak-star continuity at t =
0. The function T ( t ) ω is vaguely continuous on M at t = ϕ ∈ C ∞ c ( R + ). Let ( · , · ) denote the paring for M and C .It follows from (5.5), (5.6) and (2.12) that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t ( ∇ ⊥ · S ( t − s ) P ( ω j u ⊥ j ) , ϕ )d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t (cid:16) ω j u ⊥ j , S ( t − s ) P ∇ ⊥ · ϕ (cid:17) d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z t || ω j || ( s ) || u j || ∞ ( s ) || S ( t − s ) P ∇ ⊥ · ϕ || ∞ d s ≤ N j L j Z t s / || S ( t − s ) P ∇ ⊥ · ϕ || ∞ d s . N j L j sup <ρ ≤ t ρ / || S ( ρ ) P ∇ ⊥ · ϕ || ∞ → t → . Since C ∞ c ( R + ) is dense in C ( R + ), ω j is vaguely continuous on M at t =
0. This proves(5.18).We prove (5.19). The function S ( t ) u is weakly-star continuous on L , ∞ . We take anarbitrary ϕ ∈ C ∞ c ( R + ). Applying (5.5), (5.6) and (2.12) implies (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t ( S ( t − s ) P ( ω j u ⊥ j ) , ϕ )d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t ( ω j u ⊥ j , S ( t − s ) P ϕ )d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N j L j Z t s / || S ( t − s ) P ϕ || ∞ d s . N j L j sup <ρ ≤ t ρ / || S ( ρ ) P ϕ || ∞ → t → . Since C ∞ c ( R + ) is dense in L , ( R + ), u j is weakly-star continuous on L , ∞ . We proved (5.19).We estimate ω j + − ω j , u j + − u j and obtain(5.20) lim j →∞ sup < t ≤ T (cid:16) t − / q || ω j + − ω j || q + t / − / p || u j + − u j || p + t − / p ||∇ ( u j + − u j ) || p (cid:17) = , < q < , < p < ∞ . Since t − / q ω j converges in BC ((0 , T ]; L q ) for q ∈ (1 ,
2) and t / − / p u j , t − / p ∇ u j convergein BC ((0 , T ]; L p ) for p ∈ (2 , ∞ ), respectively, the limit ( ω, u ) satisfies (5.11) and (5.12).Sending j → ∞ to (5.3) and (5.4) implies (1.8) and (1.9). The weak-star continuity (5.9)and (5.10) follows in the same way as (5.18) and (5.19).The uniqueness follows by estimating the di ff erence of two solutions w = u − ˜ u . By(5.14), (5.13) and δ = (8 C C ) − , the constructed solution satisfiessup < s ≤ T s / − / p || u || p ≤ C δ . Since w satisfies w = − Z t S ( t − s ) P div ( wu + ˜ uw )d s , in the same way as the proof of Proposition 5.1, we estimate || w || p ≤ Z t || S ( t − s ) P div ( wu + ˜ uw ) || p d s ≤ C t / − / p sup < s ≤ t s / − / p || w || p ! ( sup < s ≤ t s / − / p || u || p ! + sup < s ≤ t s / − / p || ˜ u || p !) ≤ C t / − / p sup < s ≤ t s / − / p || w || p ! C δ ≤ t / − / p sup < s ≤ t s / − / p || w || p ! . Thus, w ≡ (cid:3) Proposition 5.3. t − / q ω ∈ BC ((0 , T ]; L q ) , ≤ q ≤ ∞ , (5.21) t / − / p u , t ∇ u ∈ BC ((0 , T ]; L ∞ ) . (5.22) Proof.
The property (5.21) follows from (5.22) and (5.11). By (5.12) and the Sobolev em-bedding, t / u ∈ BC ((0 , T ]; L ∞ ). We estimate ||∇ ( u − S ( t ) u ) || ∞ ≤ Z t / ||∇ S ( t − s ) P div ( uu ) || ∞ d s + Z tt / ||∇ S ( t − s ) P ( u · ∇ u ) || ∞ d s . Since t / − / r u , t − / r ∇ u ∈ BC ((0 , T ]; L r ) for r ∈ (1 , ∞ ) by (5.12), it follows from (2.12)and (2.15) that Z t / ||∇ S ( t − s ) P div ( uu ) || ∞ d s = Z t / ||∇ S (( t − s ) / S (( t − s ) / P div ( uu ) || ∞ d s . Z t / t − s ) / + / r || S (( t − s ) / P div ( uu ) || r d s . Z t / t − s ) + / r || u || r d s . sup < s ≤ t s − / r || u || r ! Z t / d s ( t − s ) + / r s − / r . sup < s ≤ t s − / r || u || r ! t . Applying (2.12) yields Z tt / ||∇ S ( t − s ) P ( u · ∇ u ) || ∞ d s = Z tt / ||∇ S (( t − s ) / S (( t − s ) / P u · ∇ u || ∞ d s . Z tt / t − s ) / + / r || u || ∞ ||∇ u || r d s . sup < s ≤ t s / || u || ∞ ! sup < s ≤ t s − / r ||∇ u || r ! Z tt / d s ( t − s ) / + / r s / − / r . sup < s ≤ t s / || u || ∞ ! sup < s ≤ t s − / r ||∇ u || r ! t . Since t ∇ S ( t ) u ∈ BC ((0 , ∞ ); L ∞ ) by (2.13), t ∇ u ∈ BC ((0 , T ]; L ∞ ). We proved (5.22). (cid:3) Proof of Theorem 1.1 (i). If ω is continuous (i.e., ω , pp ≡ t → t − / q || T ( t ) ω || q = , < q < , lim t → t / − / p ( || S ( t ) u || p + t / ||∇ S ( t ) u || p ) = , < p < ∞ . Thus L → N → T →
0. By (5.14) and (5.15), the sequence ( ω j , u j ) in the proofof Proposition 5.2 satisfylim t → t − / q || ω j || q = , < q < , lim t → t / − / p ( || u j || p + t / ||∇ u j || p ) = , < p < ∞ . Since t − / q ω j and t / − / p u j , t − / p ∇ u j converge in BC ([0 , T ]; L q ) and BC ([0 , T ]; L p ), thelimit ( ω, u ) satisfies(5.23) lim t → t − / q || ω || q = , < q < , lim t → t / − / p ( || u || p + t / ||∇ u || p ) = , < p < ∞ . From the proof of Proposition 5.3, (5.23) holds also for q ∈ [2 , ∞ ] and p ∈ [2 , ∞ ].If ω ∈ L and u ( x , = T ( t ) ω ∈ BC ([0 , T ]; L ) by Lemma 4.5. Applying (2.16)yields || ω − T ( t ) ω || ≤ Z t ||∇ ⊥ · S ( t − s ) P ( ω u ⊥ ) || d s . Z t t − s ) / || ω || || u || ∞ d s . sup < s ≤ t || ω || ! sup < s ≤ t s / || u || ∞ ! → t → . Thus, ω ∈ BC ([0 , T ]; L ) follows. By (2 . , u ∈ BC ([0 , T ]; L , ∞ ). The proof is complete. (cid:3) Global well-posedness.
It remains to show Theorems 1.1 (ii) and 1.2.
Proof of Theorem 1.1 (ii).
For u ∈ L , ∞ σ satisfying ω ∈ M , (2.13), (2 . and (4.1) yieldsup < t < ∞ t / − / p (cid:16) || S ( t ) u || p + t / ||∇ S ( t ) u || p (cid:17) ≤ C || ω || M , < p < ∞ , sup < t < ∞ t − / q || T ( t ) ω || q ≤ C || ω || M , ≤ q < . We set a sequence ( ω j , u j ) by (5.3), (5.4) and take T = ∞ in (5.5). Then, (5.6)-(5.8) holds.We assume || ω || M ≤ δ for δ = (8 C C ) − . Then, L ≤ (4 C ) − and (5.14)-(5.20) holdsfor T = ∞ . Thus the limit ( ω, u ) satisfies (1.8), (1.9), (5.9)-(5.12), (5.21) and (5.22). Theuniqueness follows in the same way as the proof of Proposition 5.2. (cid:3) Proof of Theorem 1.2.
Let ( ω, u ) be a local-in-time solution in [0 , t ] constructed in Theo-rem 1.1 (i). We take an arbitrary T >
0. Since u ( · , t ) ∈ L ∞ , ∇ u ( · , t ) ∈ L and u ( · , t ) = { x = } , by [1, Remark 6.4], u is extendable to a global-in-time solution in [ t , T ] satisfying u , ( t − t ) / ∇ u ∈ BC w ([ t , T ]; L ∞ ) , ∇ u ∈ BC w ([ t , T ]; L ) , and u ( t ) = S ( t − t ) u ( t ) − Z tt S ( t − s ) P ( u · ∇ u )d s , t ≤ t ≤ T . (5.25)Taking the rotation yields ω ( t ) = T ( t − t ) ω ( t ) + Z tt ∇ ⊥ · S ( t − s ) P ( ω u ⊥ )d s , t ≤ t ≤ T . (5.26)Since (1.18) and (1.20) hold in (0 , t ], we have t / u , t ∇ u ∈ BC ((0 , T ]; L ∞ ) , t / ∇ u ∈ BC ((0 , T ]; L ) . (5.27)By the H ¨oler’s inequality, t − / p ∇ u ∈ BC ((0 , T ]; L p ) , < p ≤ ∞ . (5.28)By (5.25) and (5.28), (1.20) follows. We take q ∈ (1 , || ω || q ( t ) . || ω || q ( t ) + Z tt t − s ) / || ω || q ( s ) || u || q ( s )d s . || ω || q ( t ) + Z tt t − s ) / s / − / q d s . Thus (1.19) holds. By (1.20) and (1.19), || ω u ⊥ || ≤ || ω || p ′ || u || p . t / , < t ≤ T . By (5.26) and (2.16), (1.17) follows. Since (1.17) implies (1.18) by (2 . , we proved (1.17)-(1.20) for T > (cid:3) A ppendix A. A solution formula under the trace zero condition
We show that the formula (1.4) gives a solution to (3.24).
Theorem A.1.
Let q ∈ (1 , . Let ω ∈ L q satisfy u ( x , = for u = K ω . Then, ω = T ( t ) ω satisfies (3.24) and T ( t ) ω → ω in L q as t → .Proof. By (1.4),(A.1) W ( x , y , t ) = Γ ( x − y , t ) − Γ ( x − y ∗ , t ) + H ∂ − ∂ ) ∂ E ∗ Γ ( x − y ∗ , t ) . T ( t ) ω = e t ∆ D ω + H ∂ − ∂ ) Z R + ( ∂ E ∗ Γ )( x − y ∗ , t ) ω ( y )d y . Since ( ∂ t − ∆ x ) W ( x , y , t ) = ω = T ( t ) ω satisfies the heat equation. Since A = − H ∂ , itfollows that( ∂ − A )( H ∂ − ∂ ) ∂ E ∗ Γ ( z , t ) = − ( ∂ − A )( ∂ + A ) ∂ E ∗ Γ ( z , t ) = ∂ Γ ( z , t ) . Multiplying ∂ − A by W ( x , y , t ) yields ( ∂ − A ) W ( x , y , t ) = ( ∂ − A )( Γ ( x − y , t ) − Γ ( x − y ∗ , t )) + ∂ − A )( H ∂ − ∂ ) ∂ E ∗ Γ ( x − y ∗ , t ) = ( ∂ − A )( Γ ( x − y , t ) − Γ ( x − y ∗ , t )) + ∂ Γ ( x − y ∗ , t ) = ∂ ( Γ ( x − y , t ) + Γ ( x − y ∗ , t )) − A ( Γ ( x − y , t ) − Γ ( x − y ∗ , t )) → x → . Thus, ( ∂ − A ) ω = { x = } . We shall show the convergence to initial data. We observethat Z R + ∂ x E ( x − y ∗ ) ω ( y )d y = , x ∈ R + . (A.2)The left-hand side belongs to L p , 1 / p = / q − /
2, by (2 . and is harmonic in R + , vanishingon { x = } by u ( x , =
0. Thus, (A.2) follows from the Liouville theorem. With theoperators ( − ∆ D ) − and( − ∆ N ) − ω = Z R + ( E ( x − y ) + E ( x − y ∗ )) ω ( y )d y , (A.2) is represented as ∂ (cid:16) ( − ∆ N ) − − ( − ∆ D ) − (cid:17) ω = . (A.3)Since ( ∂ i E ∗ Γ )( z , t ) = Z ∞ t ∂ i Γ ( z , s )d s , i = , , (A.4)we have2 Z R + ( ∂ E ∗ Γ )( x − y ∗ , t ) ω ( y )d y = Z R + Z ∞ t ∂ Γ ( x − y ∗ , s )d s ! ω ( y )d y = Z ∞ t ∂ ( e s ∆ N − e s ∆ D ) ω d s = ∂ (cid:16) ( − ∆ N ) − e t ∆ N − ( − ∆ D ) − e t ∆ D (cid:17) ω = ∂ (cid:16) ( − ∆ N ) − ( e t ∆ N − I ) − ( − ∆ D ) − ( e t ∆ D − I ) (cid:17) ω . Since H and ∇ ( − ∆ N ) − , ∇ ( − ∆ D ) − are bounded on L q ( R ) and L q ( R + ), it follows from(A.1) that || T ( t ) ω − ω || q . || e t ∆ N ω − ω || q + || e t ∆ D ω − ω || q → t → . The proof is complete. (cid:3)
Remarks
A.2 . (i) The formula(A.5) W ( x , y , t ) = Γ ( x − y , t ) + Γ ( x − y ∗ , t ) − A ( ∂ + A ) E ∗ Γ ( x − y ∗ , t ) , T ( t ) ω = e t ∆ N ω − A ( ∂ + A ) Z R + E ∗ Γ ( x − y ∗ , t ) ω ( y )d y , is obtained in [26, Theorem 3.1]. Since A = − H ∂ and( H ∂ − ∂ ) ∂ E = H ∂ ∂ E + ∂ E − ∆ E = − A ∂ E − A E + δ = − A ( ∂ + A ) E + δ , we have W ( x , y , t ) = Γ ( x − y , t ) − Γ ( x − y ∗ , t ) + H ∂ − ∂ ) ∂ E ∗ Γ ( x − y ∗ , t ) = Γ ( x − y , t ) + Γ ( x − y ∗ , t ) − A ( ∂ + A ) E ∗ Γ ( x − y ∗ , t ) . (ii) We are able to derive (A.5) from (3.24). Indeed, for a solution ω of (3.24), ( ∂ − A ) ω satisfies the heat equation subject to the Dirichlet boundary condition, i.e.,( ∂ − A ) ω = e t ∆ D ( ∂ − A ) ω . Since (3.13) implies e t ∆ D ( ∂ − A ) ω = ∂ e t ∆ N ω − Ae t ∆ D ω = ( ∂ − A ) e t ∆ N ω + A ( e t ∆ N − e t ∆ D ) ω , we have ∆ ω = ( ∂ + A )( ∂ − A ) ω = ( ∂ + A ) e t ∆ D ( ∂ − A ) ω = ( ∂ + A )( ∂ − A ) e t ∆ N ω + ( ∂ + A ) A ( e t ∆ N − e t ∆ D ) ω = ∆ e t ∆ N ω + ( ∂ + A ) A ( e t ∆ N − e t ∆ D ) ω . Integrating ∂ t ω = ∆ ω on (0 , t ) yields ω = e t ∆ N ω + A ( ∂ + A ) Z t ( e s ∆ N − e s ∆ D ) ω d s . Since (A.3) holds for ω satisfying u ( x , = u = K ω , A ( ∂ + A ) Z ∞ ( e s ∆ N − e s ∆ D ) ω d s = − H ∂ ( ∂ − H ∂ )(( − ∆ N ) − − ( − ∆ D ) − ) ω = − ( H ∂ ∂ + ∂ )(( − ∆ N ) − − ( − ∆ D ) − ) ω = − ( H ∂ ∂ − ∂ + ∆ )(( − ∆ N ) − − ( − ∆ D ) − ) ω = − ( H ∂ − ∂ ) ∂ (( − ∆ N ) − − ( − ∆ D ) − ) ω = . Hence by (A.4), we have A ( ∂ + A ) Z t ( e s ∆ N − e s ∆ D ) ω d s = − A ( ∂ + A ) Z ∞ t ( e s ∆ N − e s ∆ D ) ω d s = H ∂ ( ∂ + A ) Z ∞ t ( e s ∆ N − e s ∆ D ) ω d s = H ( ∂ + A ) Z ∞ t Z R + ∂ Γ ( x − y ∗ , t ) ω ( y )d y ! d s = H ( ∂ + A ) Z R + ∂ E ∗ Γ ( x − y ∗ , t ) ω ( y )d y = − A ( ∂ + A ) Z R + E ∗ Γ ( x − y ∗ , t ) ω ( y )d y . Thus (A.5) follows.(iii) The kernel W ( x , y , t ) has an explicit form W ( x , y , t ) = Γ ( x − y , t ) + Γ ( x − y ∗ , t ) + Z R + ∂ z E ( z ) Γ ( x − y ∗ − z , t )d z . (A.6)Since A ( ∂ + A ) E ( z ) = − ∂ ( H ∂ + ∂ ) E ( z ) = − ∂ E ( z ) z > , z < , by (3.7) and − A ( ∂ + A ) E ∗ Γ ( z , t ) = − Z R A ( ∂ + A ) E ( w ) Γ ( z − w )d w = Z R + ∂ E ( w ) Γ ( z − w , t )d w , (A.6) follows from (A.5). Since the kernel (A.6) is not integrable for the x -variable in R + , theformula (A.5) does not give a L -bound for T ( t ) ω and ω ∈ M . However, if u ( x , = u = K ω , T ( t ) ω agrees with T ( t ) ω by the following Theorem A.3. Hence T ( t ) ω belongs to L for such ω by (4.1). Theorem A.3.
For ω ∈ M (resp. ω ∈ L q , q ∈ (1 , ) satisfying u ( x , = for u = K ω ,T ( t ) ω = T ( t ) ω . (A.7) Proof.
Since (A.2) holds by u ( x , = H ∂ x E ( x ) = ∂ x E ( x ) for x ∈ R + by (3.7),multiplying H by (A.2) implies Z R + ∂ x E ( x − y ∗ ) ω (d y ) = , x ∈ R + . (A.8)We use the indicator function χ ( z ) = , z ≥ , , z < , to observe that Z R + ∂ z E ( z ) Γ ( x − y ∗ − z , t )d z = ∂ E χ ∗ Γ ( x − y ∗ , t ) = Z R ∂ z E ( z − y ∗ ) χ ( z + y ) Γ ( x − z , t )d z = Z R ∂ w E ( w ∗ − y ∗ ) χ ( − w + y ) Γ ( x − w ∗ , t )d w . Since χ ( − w + y ) = w > y and ∂ x E ( x ) = − x − x π | x | = ∂ x E ( x ∗ ) , changing the variable yeilds Z R ∂ w E ( w ∗ − y ∗ ) χ ( − w + y ) Γ ( x − w ∗ , t )d w = Z y Z R ∂ w E ( w ∗ − y ∗ ) Γ ( x − w ∗ , t )d w + Z −∞ Z R ∂ w E ( w ∗ − y ∗ ) Γ ( x − w ∗ , t )d w = Z y Z R ∂ z E ( z − y ) Γ ( x − z ∗ , t )d z + Z ∞ Z R ∂ z E ( z − y ∗ ) Γ ( x − z , t )d z . We integrate the both sides by the measure ω . It follows from (A.8) that Z R + Z R + ∂ z E ( z ) Γ ( x − y ∗ − z , t )d z ! ω (d y ) = Z R Z y Z R ∂ z E ( z − y ) Γ ( x − z ∗ , t )d z ! ω (d y ) . It follows from (A.6) that T ( t ) ω = Z R + W ( x , y , t ) ω (d y ) = e t ∆ N ω + Z R + Z y Z R ∂ z E ( z − y ) Γ ( x − z ∗ , t )d z ! ω (d y ) = T ( t ) ω . The proof is complete. (cid:3) A cknowledgements This work is partially supported by JSPS through the Grant-in-aid for Young Scientist (B)17K14217, Scientific Research (B) 17H02853 and Osaka City University Strategic ResearchGrant 2018 for young researchers. R eferences [1] K. Abe. Global well-posedness of the two-dimensional exterior Navier-Stokes equations for non-decayingdata.
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