On Liouville type theorems for the steady Navier-Stokes equations in R 3

Transcription

1 On Liouville type theorems for the steady Navier-Stokes equations in R 3 arxiv: v [math.ap] 6 Apr 06 Dongho Chae and Jörg Wolf Department of Mathematics Chung-Ang University Seoul , Republic of Korea dchae@cau.ac.kr and Department of Mathematics Humboldt University Berlin Unter den Linden 6, 00 Berlin, Germany jwolf@math.hu-berlin.de Abstract In this paper we prove three different Liouville type theorems for the steady Navier-Stokes equations in R 3. In the first theorem we improve logarithmically the well-known L R 3 ) result. In the second theorem we present a sufficient condition for the trivially of the solutionv = 0) in terms of the head pressure, Q = v +p. The imposed integrability condition here has the same scaling property as the Dirichlet integral. In the last theorem we present Fubini type condition, which guarantee v = 0. AMS Subject Classification Number: 35Q30, 76D05 keywords: steady Navier-Stokes equations, Liouville type theorems Introduction We consider the following stationary Navier-Stokes equations equations NS) on R 3. { v )v = p+ v, NS) divv = 0, wherevx) = v x),v x),v 3 x)) andp = px) forallx R 3. Thesystem isequipped with the boundary condition: vx) 0 as x +..)

2 In addition to.) one usually also assume following finiteness of the Dirichlet integral. R 3 v < +..) A long standing open question is if any weak solution ofns) satisfying the conditions.) and.) is trivial namely, v = 0 on R 3 ). We refer the book by Galdi[]) for the details on the motivations and historical backgrounds on the problem and the related results. As a partial progress to the problem we mention that the condition v L R 3 ) implies that v = 0 see Theorem X..5, pp.7 []). As shown in [], a different condition v L 6 5R 3 ) also imply v = 0. Another interesting progress, which shows that a solution v BMO R 3 ) to NS), satisfying.) is trivial is obtained very recently by Seregin in [6]. For the case of plane flows the problem is solved by Gilbarg and Weinberger in[3], while the special case of the axially symmetric 3D flows without swirl is studied recently by Korobkov, M. Pileckas and R. Russo in [5]see also [4]). In this paper we present three theorems, which present sufficient conditions to guarantee the triviality of the solution to NS). In the first theorem below we improve the above mentioned L -result logarithmically. Theorem.. Let v L loc R3 ) be a distributional solution to NS) such that { v log + ) } < +..3) v R 3 Then v 0. For discussion of the next theorem we introduce the head pressure, Q = v +p, which has an important role in the study of the stationary Euler equations via the Bernoulli theorem. It is knownsee e.g. Theorem X.5., pp. 688 []) that under the condition.)-.) we have px) p 0 as x +, where p 0 is a constant, which implies that Qx) 0 as x +.4) after re-defining Q p 0 as the new head pressure. Our second theorem below assumes integrability of Q to conclude the triviality of v. Theorem.. Let v,p) be a smooth solution to NS) satisfying.4). Let us set M := sup x R 3 Qx). Then, we have the following inequality. R3 Q log em ) α ω α > 0..5) Q Q α R 3 Moreover, suppose there holds the boundary conditions.),.4) and R3 Q log em ) α ) = o as α 0,.6) Q Q α then v = 0 on R 3.

3 Remark.. Since Q = Q, and Q has the same scaling as the velocity the integral Q has the same scaling property as the Dirichlet integral 4 Q R 3 Q in.). Our third result concerns on the Fubini type condition for suitable function Φx, y) for x,y) R 3 R 3 to guarantee the triviality of the solution to NS). Theorem.3. Let v be a smooth solution to NS) on R 3 satisfying.) and set ω = curlv. Suppose there exists q [ 3,3) such that x Lq R 3 ). We set Φx,y) := ωx) x y) vy) ωy)).7) 4π x y 3 for all x,y) R 3 R 3 with x y. Then, it holds Φx,y) dy+ Φx,y) < x,y) R 3 R 3..8) R 3 R 3 Furthermore, if there holds Φx,y)dy = R 3 R 3 then, v = 0 on R 3. R 3 R 3 Φx,y)dy,.) Remark.. One can show that if ω L 5R 3 ) is satisfied together with.), then.) holds, and therefore v is trivial. Although this result follows immediately by applying the L -result together with Sobolev inequality and the Calderon-Zygmund inequality, v L C v L 5 C ω L 5. The above theorem provides us with different proof of this. In order to check this result we first recall the estimate of the Riesz potential on R 3 [8]), I α f) L q C f L p, q = p α, p < q < +,.0) 3 where I α f) := C R 3 fy) x y 3 αdy, 0 < α < 3 for a positive constant C = Cα). Applying.0) with α =, we obtain by the 3

4 Hölder inequality, ωx) ωy) vy) Φx,y) dy dy R 3 R 3 R 3 R x y 3 { )5 ωx) ωy) vy) 5 dy R 3 R 3 R x y 3 C ω L 5 C ω L 5 )7 ω 7 v 7 R 3 )5 ω 5 R 3 R3 C ω v L5 L5 C ω 3 L5 Thus, by the Fubini-Tonelli theorem,.) holds. v ) < +, Proof of the main theorems ) 4 }4 Below we use the notation A B if there exists an absolute constant κ such that A κb.. Proof of Theorem. Definition.. Let φ C R) be an N-function, i.e. φ is an even function such that lim 0 φ ) = 0, and lim φ ) = +. We say φ belongs to the class Np 0,p ) < p 0 p < + ) if for all 0 p 0 )φ ) φ ) p )φ )..) Remark.. It is well known that φ Np 0,p ) implies for all 0 We now define for q > φ) φ ) p φ)..) We easily calculate, φ q ) = 0 ξ q log +ξ ξ dξ, 0. φ q q ) = log +, φ q ) = q ) q = φ q ) log + q )+ + q log + + +)log + ). 4

5 Observing that +)log +, we get for all 0 log q )φ q ) φ q ) q +log) )φ q )..3) This shows that φ Nq,q +log) ), and according to.) it holds Thus,.) is equivalent to φ q ) φ q R 3 φ Lemma.. For any constant a > q ) = log +..4) v ) < +..5) we have loga + log +..6) Proof In case a we immediately get loga + log +. For the reverse we get for all 0 <, loga + loga ) log3 + log +, and for all > loga + loga+log3 loga+log3 log log +, which proves the claim. In case a < we see that loga + log +. On the other hand, we may choose 0 > 0, such that log + 0 = + log ) loga. 0 loga Then for 0 we obtain loga + = loga +log + [ + log ) loga = log loga + log log+loga For > 0 we easily see that loga + Whence, the claim. = ] log +. + log loga ) log + 0 log loga = log loga log + log log loga log + log = log loga + log log+loga 5. +log +. 0

6 Lemma.. For all k N log + log +k,.7) k where the hidden constants depend on q and k only. Proof In fact having + k +) k k + k k ) k + k ) along with Lemma., we obtain log +k klog + k This proves the claim. log k +k k log +k k. Lemma.3. Let f L R 3 ). Then for every ε > 0, there exists R > ε, such that ε f logr..8) B R \B R/ Proof Assume the assertion of the lemma is not true. Then there exists ε > 0 such that for all R ε.8) does not hold. This implies for all k N with k ε f ε klog. B k\b k However the sum of right-hand side from k = N to is infinite which clearly contradicts to f L R 3 ). Thus, the assumption is not true and therefore the assertion of the lemma holds. In view of.3) we easily see that v L loc R 3 ). By using a standard mollifying argument we verify that v W, loc R3 ), and therefore v C R 3 ) and p C R 3 ). In particular, we have for all ζ Cc R 3 ) v ζ = v ζ+ v v ζ+ pv ζ..) R 3 R 3 R 3 R 3 On the basis of.) we have the following Caccioppoli-type inequality. { } v R + v 3..0) B R B R Proof of.0): Let R r < ρ R. Into.) we insert a off function ζ Cc B ρ) such that ζ on B r, 0 ζ in R 3 and ζ + ζ ρ r). This together with Hölder s inequality and Young s inequality immediately gives v ρ r) v +ρ r) v 3 +ρ r) p p Bρ v B r B ρ B ρ B ρ ρ r) {+ v 3 }+ρ r) p p Bρ v. B R B ρ.) 6

7 Using Hölder s inequality,, Young s inequality and consulting Theorem III.3., Theorem III.5. of [], we estimate the last integral involving the pressure as follows ρ r) B ρ p p Bρ v ) ρ r) v 3 + v 3 v 3 3 B ρ B ρ B ρ ) ρ / ρ r) v v 3 B ρ B R ) 3 δ +ρ r) v +ρρ r) B ρ δ B ρ v +ρ r) ) 3 v 3 B R { + B R v 3 ) 3 +ρ r) }. B R v 3 B R v 3 Inserting this inequality into the right-hand side of.), we arrive at { } v ρ r) + v 3 +δ v..) B r B R B ρ In.) taking δ > 0 sufficiently small, and applying a well known iteration argument, we obtain.0). This completes the proof of.0). Proof of Theorem. Let ε > 0 be arbitrarily chosen, but fixed. Thanks to Lemma.3, in view of.8) we may choose R ε such that B R \B R/ φ v ) ε logr..3) Let ζ Cc B R) be a cut off function such that 0 ζ in B R, ζ on B R/, and ζ R, ζ R. Then from.) we deduce v R v +R v 3 +R p p BR\B v R/ B R/ B R \B R/ B R \B R/ B R \B R/.4) Using Hölder s inequality, Young s inequality and consulting [], we estimate the last 7

8 integral involving the pressure as follows R p p BR\B v R/ B R \B R/ ) R 3 v + v 3 3 v 3 B R \B R/ B R \B R/ B R \B R/ ) ) R v v 3 3 +R v 3 B R B R \B R/ B R \B R/ )3 R 3 v 4 +R v 3 B R B R \B R/ R 8 +R 6 v +R v 3. B R B R \B R/ Once more using Hölder s inequality along with Young s inequality we easily find R v R +R v 3. B R \B R/ B R \B R/ Inserting the last two inequalities into the right-hand side of.4), we arrive at v R v 3 +R 8 +R 6 v..5) B R/ B R \B R/ B R We now estimate the last integral on the right-hand side of.5) by means of.0). This implies v R v 3 +R 8 +R 7 6 v 3..6) B R/ B R \B R/ B R By our assumption.3) we know that v L q R 3 ) for all q >. This follows from standard regularity theory of the steady Navier-Stokes equations e.g. see [7]). For < q < 54 we find with the help of Jensen s inequality R 8 +R 7 3q 6 8 +R 7 q 6 v q 0 as R +..7) B R v 3 R ) 3 8

9 Noting that φ 3/ is convex, applying Jensen s inequality, we get ) ) 8 φ3 v 3 = φ3 R v 3 7RmeasB ) B R \B R/ B R \B R/ R v 3 ) φ3 B R \B R/ B R \B R/ v log +R v 3 R v 3 We split the integral on the right-hand side into two parts by setting A = {x B R \B R/ v 3 εr }, A = {x B R \B R/ v 3 > εr }.. Firstly, we easily see that A v ε 3 log +R v 3 R v 3. Secondly, with help of Lemma. and recalling that R ε we have in A 4logR logr +log ε log + v 3 v 3 +log = logr ε log + v. v log +εr εr With this estimate along with.3) we get v A log +R v 3 v log R v 3 A logr logr log B R \B R/ φ B R \B R/ v ) ε. v log + v v Accordingly, φ3 ) 8 v 3 ε. 7RmeasB ) B R \B R/ Thus, in view.6) together with the estimates we have just obtained we are able to chose a sequence R k + as k +, such that v 0 as k +, B Rk / which yields v = 0 and therefore v const = 0.

10 . Proof of Theorem. Proof of Theorem. Let us denote the vorticity ω = curlv. Then, it is wellknown that from NS) that the following equation holds true. Q v Q = ω..8) Under the condition.4) we have Qx) 0 for all x R 3 by the maximum principle applied to.8). Moreover, by the maximum principle again, either Qx) 0 on R 3, or Qx) < 0 for all x R 3. Indeed, any point x 0 R 3 such that Qx 0 ) = 0 is a point of local maximum, which is not allowed unless Q 0 by the maximum principle. Let Qx) 0 on R 3, then without the loss of generality we may assume Qx) > 0 for all x R 3. We set sup x R 3 Q = M > 0. Let f CR). For λ [0,M) we set D λ = {x R 3 Qx) > λ}. Then, we compute ) Qx) fqx))v Q = v fq)dq D λ D λ 0 ) Qx) ) Qx) = div v fq)dq = fq)dq v νds D λ 0 D λ 0 λ λ = fq)dq v νds = fq)dq divv = 0..) 0 D λ 0 D λ where ν = Q/ Q is the outward unit normal vector on D λ. For λ 0,M) we Integrate.8) over D λ. Then, using the fact.), we have ω Q = Q = D λ D λ D λ ν ds = Q ds..0) D λ Using the co-area formula, we obtain Q Dλ log em ) α = Q Q = M λ M λ { M λ log em q q q log em q log em λ = α ω, α R 3 Q D q Q ) α log em ) α dsdq Q Q dsdq D q ) α dq Q ds D λ ) } α ω D λ where we used.0) in the fourth line. Passing λ 0, and applying the monotone convergence theorem, we obtain.5). Next, we assume.6) holds. We consider a 0

11 standard cut-off function σ C0 [0, )) such that σs) = if s <, and σs) = 0 if s >, and 0 σs) for < s <. For each α 0,) we define σ α x) := σ α Qx)) C0 R 3 ) by { σ α x) = σ 3 log em ) } α. Qx) We note that σ α x) =, if Qx) Me 3 ) α, 0 < σ α x) <, if Me 3 α < Qx) < Me 3 α ), σ α x) = 0, if Qx) Me 3 α. We multiply.8) by σ α, and integrate it over R 3. Then, the convection term vanishes by.). Let α > 0 be fixed. For all α > α we have ω σ α x) ω σ α x) = Qσ α x) R 3 R 3 R 3 {Me 3 Q = 3α α < Qx) <Me 3) α log em ) { α σ 3 log em ) } α } Q Q Qx) {Me 3 3α sup σ Q s) s α < Qx) <Me 3) α log em ) α } Q Q 0 as α 0. Hence, we have shown R 3 ω σ α x) = 0 for all α > 0, which implies that ω = 0 on R 3. This, combined with the fact div v = 0 implies that v is a harmonic function on R 3. The boundary condition, together with the Liouville theorem for harmonic function, leads us to conclude v = 0 on R 3..3 Proof of Theorem.3 We first establish integrability conditions on the vector fields for the Biot-Savart s formula in R 3. Proposition.. Letξ = ξx) = ξ x),ξ x),ξ 3 x)) andη = ηx) = η x),η x),η 3 x)) be smooth vector fields on R 3. Suppose there exists q [,3) such that η L q R 3 ). Let ξ solve ξ = η,.) under the boundary condition; either ξx) 0 as x +,.)

12 or ξ L s R 3 ) for some s [, )..3) Then, the solution of.) is given by ξx) = x y) ηy) dy x R 3..4) 4π R x y 3 3 Proof Let σ C0 R 3 ) be the cut-off function ) defined in the proof of Theorem.. For each R > 0 we define σ R x) := σ. Given ǫ > 0 we denote B ǫ y) = {x x R R 3 x y < ǫ}. Let us fix y R 3 and ǫ 0, R). We multiply.) by σ R x y), x y and integrate it with respect to the variable x over R 3 \B ǫ y). Then, { x y >ǫ} ξσ R x y = Since x y = 0 on R3 \B ǫ y), one has ξσ R 3 ) x y = xi ξσ R xi x y i=0 3 i=0 { x y >ǫ} 3 ) ξ xi σ R xi x y )) + ξ σ 3 R x y + i=0 xi ξσ R xi x y { x y =ǫ} σ R ηy)..5) x y i=0 ) ξ xi xi σ R. x y Therefore, applying the divergence theorem, and observing ν σ R = 0 on B ǫ y), we have ξσ R { x y >ǫ} x y = ν ξ { x y =ǫ} x y ds ξ x y ds + ξ σ R x y ds { x y >ǫ} { x y >ǫ} x y) σ R ξ x y 3 ds.6) where ν ) denotes the outward normal derivative on B ǫ y). Passing ǫ 0, one can easily compute that ξ σ R RHS of.6) 4πξy) + R x y x y) σ R ξ 3 R x y 3 3 := I +I +I 3.7) Next, using the formula σ R η x y = ) σr η σ R η + x y) ησ R, x y x y x y 3

13 and using the divergence theorem, we obtain the following representation for the right hand side of.5). { x y >ǫ} σ R η x y = { x y >ǫ} σ R η x y + { x y =ǫ} ν { x y >ǫ} ) η ds x y x y) ησ R x y 3,.8) where we denoted ν = y x, the outward unit normal vector on B y x ǫy). Passing ǫ 0, we easily deduce σ R η RHS of.8) R x y x y) ησ R 3 R x y 3 3 := J +J as ǫ 0..) We now pass R for each term of.7) and.) respectively below. Under the boundary condition.) we estimate: ξx) σ R x y) I {R x y R} x y σ ) L 3 sup ξx) R R x y R x y 3 σ L R R dr r ) 3 {R x y R} sup ξx) 0 R x y R ) 3 {R x y R} as R by the assumption.), while under the condition.3) we have ξx) σ R x y) I {R x y R} x y σ )s L s ξ R L s x y s s R 3 s σ L ξ L s 0 as R. Similarly, under.) ξx) σ R x y) I 3 {R x y R} x y σ L sup ξx) R R x y R σ L R R dr r ) 3 {R x y R} {R x y R} sup ξx) 0 R x y R ) 3 {R x y R} ) 3 x y 3 3

14 as R, while under the condition.3) we estimate ξx) σ R x y) I 3 {R x y R} x y σ L ξ R L s R 3 s σ L ξ L s 0 {R x y R} x y s s )s s as R. Therefore, the right hand side of.6) converges to 4πξy) as as R. For J,J we estimate σ R η J x y {R x y R} σ L η L R q R x y R) σ L η L q R x y R)R q 0 {R x y R} )q q x y q q as R. In passing R in J of.), in order to use the dominated convergence theorem, we estimate x y) ηy) R x y 3 3 η η { x y <} x y + { x y } x y := J +J..30) J is easy to handle as follows. J η L B y)) For J we estimate J η L q { x y <} R 3 η q ) q x y = 4π η L B y)) < +.3) { x y >} )q r q q dr x y q q )q q < +,.3) if < q < 3. In the case of q = we estimate simply J η η L..33) { x y >} Estimates of.30)-.33) imply x y) ηy) x y 3 < +. R 3 4

15 Summarising the above computations, one can pass first ǫ 0, and then R + in.5), applying the dominated convergence theorem, to obtain finally.4). Corollary.. Let v be a smooth solution to NS) satisfying.) such that ω L q R 3 ) for some q [ 3,3). Then, we have vx) = x y) ωy) dy,.34) 4π R x y 3 3 and ωx) = x y) vy) ωy)) dy..35) 4π R x y 3 3 Proof Taking curl of the defining equation of the vorticity, v = ω, using divv = 0, we have v = ω, which provides us with.34) immediately by application of Proposition.. In order to show.35) we recall that, using the vector identity v = v )v+v v), one can rewrite NS) as v ω = p+ ) v + v. Taking curl on this, we obtain ω = v ω). The formula.35) is deduced immediately from this equations by applying the proposition.. For the allowed rage of q we recall the Sobolev and the Calderon-Zygmund inequalities[8]), v 3q v L q ω Lq, < q < 3,.36) L 3 q which imply v ω L 3q 6 q R 3 ) if ω L q R 3 ). We also note that 3 only if 3q < 3. 6 q q < 3 if and Proof of Theorem.3 Under the hypothesis.) and ω L q R 3 ) with q [ 3,3) both of the relations.34) and.35) are valid. We first prove the following. Claim: For each x,y R 3 and 0 ωx) = Φx,y)dy Φx,y) dy < +,.37) R 3 R 3 0 = Φx,y) Φx,y) < +..38) R 3 R 3 5

16 Proof of.8): Decomposing the integral and using the Hölder inequality, we estimate ) vy) ωy) vy) ωy) Φx,y) dy ωx) dy + dy R 3 { x y } x y { x y >} x y dy ωx) v L B x)) ω L B x)) x y + ωx) v L 3q 3 q ω L q ωx) v L B x)) ω L B x)) + ωx) ω L q { x y } { x y } )4q 6 r q 6 3q q 3 dr dy x y 6q 4q 6 )4q 6 3q < +,.3) where we used.36) and the fact that q 6 3q 3 < if 3 < q < 3. In the case q = 3 we estimate, instead, Φx,y) dy ωx) R 3 { x y } We also have Φx,y) vy) ωy) R 3 ) vy) ωy) vy) ωy) dy + dy x y { x y >} x y ωx) v L B x)) ω L B x)) + ωx) v L 3 ω L 3 < +..40) { x y } ωx) x y + vy) ωy) ω L B y)) + vy) ωy) ω L q vy) ωy) ω L B y)) + vy) ωy) ω L q { x y >} { x y >} r )q q q dr ) ωx) x y x y q q )q q < +.4) where we used the fact that < if 3 q < 3. From.35) we immediately q obtain ) x y) vy) ωy)) Φx,y)dy = ωx) dy R 4π 3 R x y 3 3 = ωx) 0, x R 3.4) and combining this with.3), we deduce.37). On the other hand, using.34), we find Φx,y) = ωx) x y) vy) ωy)) R 4π 3 R x y 3 3 ) ωx) x y) = vy) ωy) 4π R x y 3 3 = vy) vy) ωy) = 0.43) 6

17 for all y R 3, and combining this with.4), we have proved.38). This completes the proof of the claim. If.) holds, then from.4) and.43) provide us with ωx) = Φx,y)dy = Φx,y)dy = 0. R 3 R 3 R 3 R 3 R 3 Hence, ω = 0 on R 3..44) We remark parenthetically that in deriving.44) it is not necessary to assume that R 3 ωx) < +, andthereforewedo notneed torestrict ourselves toω L R 3 ). Hence, from.34) and.44), we conclude v = 0 on R 3. Acknowledgements Chae was partially supported by NRF grants and , while Wolf has been supported supported by the German Research Foundation DFG) through the project WO88/-; 644. References [] D. Chae, Liouville-type theorems for the forced Euler equations and the Navier- Stokes equations, Comm. Math. Phy., 36, 04), pp [] G. P. Galdi, An Introduction to the Mathematical Theory of the Navier-Stokes Equations: Steady-State Problems, nd ed., Springer, 0). [3] D. Gilbarg and H. F. Weinberger, Asymptotic Properties of Steady Plane Solutions of the Navier-Stokes Equations with Bounded Dirichlet Integral, Ann. Scuola Norm. Sup. Pisa, 4), 5, pp [4] G. Koch, N. Nadirashvili, G. Seregin and V. Šverák, Liouville theorems for the Navier-Stokes equations and applications, Acta Math., 03, no., 00), pp [5] M. Korobkov, M. Pileckas and R. Russo, The Liouville Theorem for the Steady State Navier-Stokes Problem for Axially Symmetric 3D Solutions in Absence of Swirl, J. Math. Fluid Mech., 7, 05), no., pp [6] G. Seregin, Liouville type theorem for stationary Navier-Stokes equations, preprint 06). [7] G. Seregin, Lecture Notes on Regularity Theory for the Navier-Stokes Equations, World Scientific, Oxford University 06). [8] E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Univ. Press 70). 7