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GLOBAL A PRIORI ESTIMATES FOR THE INHOMOGENEOUS LANDAU EQUATION WITH MODERATELY SOFT POTENTIALS STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON 7 1 0 Abstract. We establish a priori upper bounds for solutions to the spatially inhomogeneous 2 Landauequationinthecaseofmoderatelysoftpotentials,witharbitraryinitialdata,underthe assumptionthatmass,energyandentropydensitiesstayundercontrol. Ourpointwiseestimates n decay polynomially in the velocity variable. We also show that if the initial data satisfies a a Gaussianupperbound,thisboundispropagated forallpositivetimes. J 7 2 ] 1. Introduction P A WeconsiderthespatiallyinhomogeneousLandauequation,akineticmodelfromplasmaphysics . that describes the evolution of a particle density f(t,x,v) 0 in phase space (see, for example, h ≥ [4, 13]). It is written in divergence form as t a m (1.1) ∂tf +v xf = v [a¯(t,x,v) vf]+¯b(t,x,v) vf +c¯(t,x,v)f, ·∇ ∇ · ∇ ·∇ [ where t [0,T ], x Rd, and v Rd. The coefficients a¯(t,x,v) Rd×d, ¯b(t,x,v) Rd, and 0 c¯(t,x,v)∈ R are give∈n by ∈ ∈ ∈ 1 ∈ v w w 5 (1.2) a¯(t,x,v):=ad,γ I wγ+2f(t,x,v w)dw, 1 ZRd(cid:18) − |w| ⊗ |w|(cid:19)| | − 2 (1.3) ¯b(t,x,v):=b wγwf(t,x,v w)dw, 8 d,γ 0 ZRd| | − 1. (1.4) c¯(t,x,v):=cd,γ wγf(t,x,v w)dw, 0 ZRd| | − 7 where γ is a parameter in [ d, ), and a , b , and c are constants. When γ = d, the d,γ d,γ d,γ 1 formula for c¯must be replac−ed b∞y c¯= c f. Equation (1.1) arises as the limit of the Bol−tzmann d,γ : v equation as grazing collisions predominate, i.e. as the angular singularity approaches 2 (see the i discussion in [2]). The case d = 3,γ = 3, corresponds to particles interacting by Coulomb X − potentials in small scales. The case γ [ d,0) is known as soft potentials, γ = 0 is known as r ∈ − a Maxwell molecules,andγ >0hardpotentials. Inthispaper,wefocusonmoderatelysoft potentials, which is the case γ ( 2,0). ∈ − We assume that the mass density, energy density, and entropy density are bounded above, and the mass density is bounded below, uniformly in t and x: (1.5) 0<m f(t,x,v)dv M , 0 0 ≤ZRd ≤ (1.6) v 2f(t,x,v)dv E , 0 ZRd| | ≤ (1.7) f(t,x,v)logf(t,x,v)dv H . 0 ZRd ≤ LSwaspartiallysupportedbyNSFgrantsDMS-1254332andDMS-1362525. SSwaspartiallysupportedbyNSF grantDMS-1246999. 1 2 STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON In the space homogeneous case, because of the conservation of mass and energy, and the mono- tonicity of the entropy, it is not necessary to make the assumptions (1.5), (1.6) and (1.7). It would suffice to require the initial data to have finite mass, energy and entropy. It is currently unclear whether these hydrodynamic quanitites will stay under control for large times and away fromequilibriuminthe spaceinhomogeneouscase. Thus,atthis point,it issimply anassumption we make. We now state our main results. Our first theorem makes no further assumption on the initial data f :R2d R beyond what is required for a weak solution to exist in [0,T ]. in 0 → Theorem 1.1. Let γ ( 2,0]. If f : [0,T ] R2d R is a bounded weak solution of (1.1) 0 ∈ − × → satisfying (1.5), (1.6), and (1.7), then there exists K >0 such that f satisfies 0 (1.8) f(t,x,v) K 1+t−d/2 (1+ v )−1, 0 ≤ | | for all (t,x,v) [0,T ] R2d. The constant(cid:16)K depend(cid:17)s on d, γ, m , M , E , and H . 0 0 0 0 0 0 ∈ × Note that even though we work with a bounded weak solution f, none of the constants in our estimates depend on f . Note also that our estimate does not depend on T . We use a L∞ 0 k k definition of weak solution for which the estimates in [8] apply, since that is the main tool in our proofs. We will show in Theorem 4.3 that an estimate of the form (1.8) cannot hold with a power of (1+ v )lessthan (d+2),whichalsoimplies thereisnoa priori exponentialdecay. Onthe other | | − hand, if f satisfies a Gaussian upper bound in the velocity variable, this bound is propagated: in Theorem 1.2. Let f :[0,T ] R2d R be a bounded weak solution of the Landau equation (1.1) 0 such that f (x,v) C e−α|v|2×, for s→ome C >0 and a sufficiently small α>0. Then in 0 0 ≤ f(t,x,v) C e−α|v|2, 1 ≤ where C depends on C , α, d, γ, m , M , E and H . The value of α must be smaller than some 1 0 0 0 0 0 α >0 that depends on γ, d, m , M , E and H . 0 0 0 0 0 This estimate is also independent of T . As a consequence of Theorem 1.2, we will show 0 in Theorem 5.4 that in this regime, f is uniformly Ho¨lder continuous on [t ,T ] R2d for any 0 0 × t (0,T ). 0 0 ∈ Note that under some formal asymptotic regime, the hydrodynamic quantities of the inhomo- geneous Landau equations converge to solutions of the compressible Euler equation [3], which is known to develop singularities in finite time. Should we expect singularities to develop in finite time for the inhomogeneous Landau equation as well? That question seems to be out of reach with current techniques. A more realistic project is to provethat the solutions stay smooth for as long as the hydrodynamic quantities stay under control (as in (1.5), (1.6) and (1.7)). The results in this paper are an important step forward in that program. 1.1. Related work. It was established in [14] that solutions to (1.1) become C∞ smooth in all threevariablesconditionallytothesolutionbeingawayfromvacuum,boundedinH8 (inthed=3 case)andhavinginfinitelymanyfinitemoments. Itwouldbeconvenienttoextendthisconditional regularityresulttohavelessstringentassumptions. Inparticular,theassumptions(1.5),(1.6)and (1.7) are a much weaker assumption, which is also in terms of physically relevant hydrodynamic quantities. In [8], the authors show how their local Ho¨lder continuity result for linear kinetic equationswithroughcoefficientscanbeappliedtosolutionsoftheLandauequationprovidedthat (1.5), (1.6) and (1.7) hold and in addition the solution f is assumed to be bounded. While we also assume boundedness of f, our results do not quantitatively rely on this and in addition tell us some information about the decay for large velocities. The local estimates for parabolic kinetic equations with rough coefficients play an important roleinthiswork. LocalL∞ estimateswereobtainedin[16]usingMoseriteration,andlocalHo¨lder GLOBAL ESTIMATES FOR THE INHOMOGENEOUS LANDAU EQUATION 3 estimates were proven in [21, 22] using a weak Poincar´e inequality. A new proof was given in [8] using a version of De Giorgi’s method. Classical solutions for (1.1) have so far only been constructed in a close-to-equilibrium setting: seetheworkofGuo[10]andMouhot-Neumann[15]. Asuitablenotionofweaksolution,forgeneral initial data, was constructed by Alexandre-Villani [2, 19]. The global L∞ estimate we prove in Theorem 1.1 is similar to an estimate in [18] for the Boltzmann equation. The techniques in the proof are completely different. The propagation of Gaussian bounds that we give in Theorem 1.2 is reminiscent of the result in [7]. That result is for the space-homogeneous Boltzmann equation with cut-off, which is in some sense the opposite of the Landau equation in terms of the angular singularity in the cross section. In order to keep track of the constants for parabolic regularization estimates (as in [8]) for large velocities, we describe a change of variables in Lemma 4.1. This change of variables may be useful in other contexts. It is related to one mentioned in the appendix of [12] for the Boltzmann equation. For the homogeneousLandau equation,which arises when f is assumed to be independent of x in (1.1), the theory is more developed. The C∞ smoothing is establishedfor hard potentials in [6] and for Maxwell molecules in [20], under the assumption that the initial data has finite mass and energy. Propagation of Lp estimates in the case of moderately soft potentials was shown in [23] and[1]. Globalupper bounds ina weightedL1(L3)space wereestablishedin [5], evenfor γ = 3, t v − as a consequence of entropy dissipation. Global L∞ bounds that do not depend on f and that in do not degenerate as t were derived in [17] for moderately soft potentials, and this result → ∞ also implies C2 smoothing by standard parabolic regularity theory. Note that in the space homogeneous case our assumptions (1.5), (1.6) and (1.7) hold for all t > 0 provided that the initial data has finite mass, energy and entropy. Both Theorems 1.1 and 1.2 are new results even in the space homogeneous case. The previous results for soft potentials do not address the decay of the solution for large velocities. 1.2. Organization of the paper. In Section 2, we establish precise bounds on the coefficients a¯, ¯b, and c¯ in (1.1). In Section 3, we derive the local estimates we will use to prove Theorem 1.1, starting from the Harnack estimate of [8]. Section 4 contains the proof of Theorem 1.1 and a propagatinglower bound thatimplies the exponent of(1+ v ) in (1.8) cannotbe arbitrarilyhigh. | | In Section 5, we prove Theorem 1.2 and the Ho¨lder estimate, Theorem 5.4. In Appendix A, we derive a convenient maximum principle for kinetic Fokker-Planck equations. 1.3. Notation. We say a constant is universal if it depends only on d, γ, m , M , E , and H . 0 0 0 0 ThenotationA.B meansthatA CB forauniversalconstantC,andA B meansthatA.B and B . A. We will let z = (t,x,≤v) denote a point in R Rd Rd. Fo≈r any z = (t ,x ,v ), + 0 0 0 0 × × define the Galilean transformation (t,x,v):=(t +t,x +x+tv ,v +v). Sz0 0 0 0 0 We also have −1(t,x,v):=(t t ,x x (t t )v ,v v ). Sz0 − 0 − 0− − 0 0 − 0 For any r >0 and z =(t ,x ,v ), let 0 0 0 0 Q (z ):=(t r2,t ] x: x x (t t )v <r3 B (v ), r 0 0 0 0 0 0 r 0 − ×{ | − − − | }× andQ =Q (0,0,0). Theshift andthe scalingofQ correspondtothesymmetriesofthe left- r r Sz0 r hand side of (1.1). We will sometimes write ∂ or ∂ , and these will alwaysrefer to differentiation i ij in v. 4 STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON 2. The coefficients of the Landau equation In this section we review various estimates of the coefficients a¯,¯b and c¯in (1.1). In calculating these upper and lower bounds, the dependence of f on t and x is irrelevent, so in this section we will write f(v) and a¯(v), etc. Lemma 2.1. Let γ [ 2,0), and assume f satisfies (1.5), (1.6), and (1.7). Then there exist constants c and C dep∈en−ding on d, γ, m , M , E , and H , such that for unit vectors e Rd, 0 0 0 0 ∈ (1+ v )γ, e Sd−1, (2.1) a¯ (v)e e c | | ∈ ij i j ≥ ((1+ v )γ+2, e v =0, | | · and (1+ v )γ+2, e Sd−1, (2.2) a¯ (v)e e C | | ∈ ij i j ≤ ((1+ v )γ, e v = v , | | · | | where a¯ (v) is defined by (1.2). ij Proof. The lower bounds (2.1) are proven in [17, Lemma 3.1]. For the upper bounds, the formula (1.2) implies w e 2 a¯ (v)e e =a 1 · wγ+2f(v w)dw ij i j d,γ ZRd −(cid:18) |w| (cid:19) !| | − . wγ+2f(v w)dw ZRd| | − = v z γ+2f(z)dz ZRd| − | . (v γ+2+ z γ+2)f(z)dz ZRd | | | | .M (1+ v γ+2)+E , 0 0 | | since 0 γ+2 2. The ≤above bo≤und is valid for all e Sd−1. If e is parallel to v, then ∈ w e 2 (v z) e 2 1 · wγ+2f(v w)dw = 1 − · v z γ+2f(z)dz ZRd −(cid:18) |w| (cid:19) !| | − ZRd −(cid:18) |v−z| (cid:19) !| − | = v z 2 (v z e)2 v z γf(z)dz ZRd(cid:16)| − | − | |− · (cid:17)| − | = z 2 (z e)2 v z γf(z)dz ZRd | | − · | − | (cid:0) (cid:1) = z 2sin2θ v z γf(z)dz, ZRd| | | − | where θ is the angle between v and z. Let R= v /2. If z B (v), then sinθ v z /v , and R | | ∈ | |≤| − | | | z 2sin2θ v z γf(z)dz z 2 v −2 v z γ+2f(z)dz | | | − | ≤ | | | | | − | ZBR(v) ZBR(v) v γ | | z 2f(z)dz .E v γ. ≤ 2γ+2 | | 0| | ZBR(v) If v z R= v /2, then v z γ . v γ, and we have | − |≥ | | | − | | | z 2sin2θ v z γf(z)dz . v γ z 2f(z)dz .E v γ. 0 ZRd\BR(v)| | | − | | | ZRd\BR(v)| | | | GLOBAL ESTIMATES FOR THE INHOMOGENEOUS LANDAU EQUATION 5 (cid:3) In the proof of Theorem 1.1, we will need to keep track of how the bounds on ¯b and c¯in the next two lemmas depend on the local L∞ norm of f. In Lemma 2.2 and Lemma 2.3, f L∞(A) k k means f(t,x, ) for any set A Rd. L∞(A) k · k ⊆ Lemma 2.2. Let f satisfy (1.5), (1.6), and (1.7). Then c¯(v) defined by (1.4) satisfies 2d (1+ v )γ(1+ f )−γ/d, − γ <0, c¯(v). | | k kL∞(Bρ(v)) d+2 ≤ (1+|v|)−2−2γ/d 1+kfkL∞(Bρ(v)) −γ/d, −d<γ < d−+2d2, where the constants depend on d,γ,M0,(cid:0)and E0, and (cid:1) 1, v <2, ρ= | | (v −2/d, v 2. | | | |≥ Proof. Assume first v 2. Let r := v −2/d(1+ f )−1/d <ρ. Consider | |≥ | | k kL∞(Bρ(v)) I = wγf(v w)dw, I = wγf(v w)dw, 1 2 | | − | | − ZBr ZB|v|/2\Br I = wγf(v w)dw. 3 ZRd\B|v|/2| | − We have I . f rd+γ . v −2−2γ/d f −γ/d . 1 k kL∞(Bρ(v)) | | k kL∞(Bρ(v)) I .rγ v −2 v w2f(v w)dw .E v −2−2γ/d(1+ f )−γ/d. 2 | | | − | − 0| | k kL∞(Bρ(v)) Z B|v|/2 Finally, for w v /2, we have wγ . v γ, and | |≥| | | | | | I . v γ f(v w)dw M v γ. 3 0 | | ZRd\B|v|/2 − ≤ | | Thus c¯(v).(1+ f )−γ/d v −2−2γ/d+ v γ for v >2. k kL∞(Bρ(v)) | | | | | | 2d When γ d, − , 2 2γ/d>γ and we get ∈ − d+2 − − (cid:18) (cid:19) c¯(v).(1+ f )−γ/d v −2−2γ/d. k kL∞(Bρ(v)) | | 2d When γ − ,0 , γ > 2 2γ/d and we get ∈ d+2 − − (cid:20) (cid:19) c¯(v).(1+ f )−γ/d v γ. k kL∞(Bρ(v)) | | This completes the proof in the case v >2. | | For v 2, γ ( d,0], and any R (0,1] we have that | |≤ ∈ − ∈ wγf(v w)dw = wγf(v w)dw+ wγf(v w)dw, RZd | | − ZBR| | − ZRd\BR| | − .Rd+γ f +RγM . k kL∞(B1(v)) 0 Choosing R=(1+ f )−1/d, we then have k kL∞(B1(v)) c¯(v).(Rd+γ f +RγM ).(1+ f )−γ/d, k kL∞(B1(v)) 0 k kL∞(B1(v)) for v 2, completing the proof. (cid:3) | |≤ 6 STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON Lemma 2.3. Let f satisfy (1.5), (1.6), and (1.7). Then¯b(v) definedby (1.3)satisfies the estimate (1+ v )γ+1(1+ f )−(γ+1)/d, γ [ 2, 1), (2.3) ¯b(v) . | | k kL∞(Bρ(v)) ∈ − − | |  (1+ v )γ+1, γ [ 1,0]  | | ∈ − where the constants depend on d,γ,M , and E , and  0 0 1, v <2, ρ= | | (v −2/d, v 2. | | | |≥ Proof. Taking norms, we have ¯b(v) . w1+γf(v w)dw. | | ZRd| | − 2d If γ [ 2, 1), then 0 > 1+γ 1 − , and the conclusion follows from Lemma 2.2. If ∈ − − ≥ − ≥ d+2 γ [ 1,0], we have ∈ − ¯b(v) . (v γ+1+ v wγ+1)f(v w)dw | | ZRd | | | − | − . v γ+1M +E(1+γ)/2M(1−γ)/2 .(1+ v )γ+1. | | 0 0 0 | | (cid:3) 3. Local estimates Inthissectionwerefinethelocalestimatesin[16]and[8]forlinearkineticequationswithrough coefficients. Essentially, we start from their results and apply scaling techniques to improve the local L∞ estimates. We will need the following technical lemma. See [11, Lemma 4.3] for the proof. Lemma 3.1. Let η(r) 0 be bounded in [r ,r ] with r 0. Suppose for r r < R r , we 0 1 0 0 1 ≥ ≥ ≤ ≤ have A η(r) θη(R)+ +B ≤ (R r)α − for some θ [0,1) and A,B,α 0. Then there exists c(α,θ)>0 such that for any r r <R 0 ∈ ≥ ≤ ≤ r , there holds 1 A η(r) c(α,θ) +B . ≤ (R r)α (cid:18) − (cid:19) Proposition 3.2. If g(t,x,v) 0 is a weak solution of ≥ (3.1) ∂ g+v g = (A g)+B g+s t x v v v ·∇ ∇ · ∇ ·∇ in Q , with 1 0<λI A(t,x,v) ΛI, (t,x,v) Q , 1 ≤ ≤ ∈ B(t,x,v) Λ, (t,x,v) Q , 1 | |≤ ∈ s L∞(Q ), 1 ∈ then (3.2) supg ≤C kgkL∞t,xL1v(Q1)+kskL∞(Q1) , Q1/2 (cid:16) (cid:17) with C depending only on d,λ, and Λ. GLOBAL ESTIMATES FOR THE INHOMOGENEOUS LANDAU EQUATION 7 Proof. It is provenin [8] that if g(t,x,v) solves (3.1) weakly with A, B, and s as in the statement of the proposition, then kgkL∞(Q1/2) ≤C kgkL2(Q1)+kskL∞(Q1) , with C depending on d,λ, and Λ. Since kgk(cid:0)L2(Q1) ≤ √ωdkgkL∞t,xL2v(cid:1)(Q1), where ωd = Ld(B1), we also have (3.3) kgkL∞(Q1/2) ≤C kgkL∞t,xL2v(Q1)+kskL∞(Q1) . (cid:16) (cid:17) To replace kgkL∞t,xL2v(Q1) with kgkL∞t,xL1v(Q1), we use an interpolation argument. For 0 < r ≤ 1, define g (t,x,v):=g(r2t,r3x,rv), s (t,x,v):=s(r2t,r3x,rv), (3.4) r r A (t,x,v):=A(r2t,r3x,rv), B (t,x,v):=B(r2t,r3x,rv), r r and note that g satisfies r (3.5) ∂ g +v g = (A g )+rB g +r2s t r x r v r v r r v r r ·∇ ∇ · ∇ ·∇ in Q . Since r 1, we may apply (3.3) to g , which gives 1 r ≤ 1 (3.6) kgkL∞(Qr/2) ≤C rd/2kgkL∞t,xL2v(Qr)+r2kskL∞(Qr) , (cid:18) (cid:19) for any r (0,1]. Now, for θ,R (0,1), apply (3.6) in Q (z) for each z Q to obtain (1−θ)R θR ∈ ∈ ∈ 1 kgkL∞(QθR) ≤C [(1 θ)R]d/2kgkL∞t,xL2v(QR)+R2kskL∞(QR) (cid:18) − (cid:19) 1 ≤C [(1 θ)R]d/2kgkL∞t,xL2v(QR)+kskL∞(Q1) . (cid:18) − (cid:19) By the Ho¨lder and Young inequalities, we have 1 1/2 1/2 g C g g + s k kL∞(QθR) ≤ [(1 θ)R]d/2k kL∞(QR)k kL∞t,xL1v(QR) k kL∞(Q1) (cid:18) − (cid:19) 1 1 ≤ 2kgkL∞(QR)+C [(1 θ)R]dkgkL∞t,xL1v(QR)+kskL∞(Q1) . (cid:18) − (cid:19) Define η(ρ)= g for ρ (0,1]. Then for any 0<r <R 1, we have k kL∞(Qρ) ∈ ≤ 1 C η(r)≤ 2η(R)+ (R r)dkgkL∞t,xL1v(Q1)+CkskL∞(Q1). − Applying Lemma 3.1, we obtain C η(r)≤ (R r)dkgkL∞t,xL1v(Q1)+CkskL∞(Q1). − Let R 1 and set r = 1 to conclude (3.2). (cid:3) → − 2 Lemma 3.3. Let g(t,x,v) solve (3.1) weakly in Q (z ) for some z R2d+1 and R>0, with R 0 0 ∈ 0<λI A(t,x,v) ΛI, (t,x,v) Q , R ≤ ≤ ∈ B(t,x,v) Λ/R, (t,x,v) Q , R | |≤ ∈ s L∞(Q ). R ∈ Then the improved estimate (3.7) g(t0,x0,v0)≤C kgkL2/∞t(,xdL+1v2()QR)kskLd/∞(d(+Q2R))+R−dkgkL∞t,xL1v(QR) holds, with C depending only on d(cid:16),λ, and Λ. (cid:17) 8 STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON Proof. By applying the change of variables t t x x (t t )v v v 0 0 0 0 0 (t,x,v) − , − − − , − 7→ R2 R3 R (cid:18) (cid:19) to g and s, we may suppose (t ,x ,v )=(0,0,0) and R=1. 0 0 0 Forr (0,1]to be determined,wemakethe transformation(3.4)asinthe proofofProposition ∈ 3.2 and get a function g satisfying (3.5) in Q . Then Proposition 3.2 implies r 1 g(0,0,0)≤C kgrkL∞t,xL1v(Q1)+kr2srkL∞(Q1) (cid:16) (cid:17) =C r−dkgkL∞t,xL1v(Qr)+r2kskL∞(Qr) (cid:16) (cid:17) ≤C r−dkgkL∞t,xL1v(Q1)+r2kskL∞(Q1) . (cid:16) (cid:17) If kgkL∞t,xL1v(Q1) ≤kskL∞(Q1), then the choice r=(kgkL∞t,xL1v(Q1)/kskL∞(Q1))1/(d+2) implies g(0,0,0) C g 2/(d+2) s d/(d+2). ≤ k kL∞t,xL1v(Q1)k kL∞(Q1) Ontheotherhand,ifkskL∞(Q1) ≤kgkL∞t,xL1v(Q1),thechoicer =1impliesg(0,0,0)≤CkgkL∞t,xL1v(Q1), so we have g(0,0,0)≤C kgkL2/∞t(,xdL+1v2()Q1)kskLd/∞(d(+Q21))+kgkL∞t,xL1v(Q1) (cid:16) (cid:17) in both cases. (cid:3) 4. Global estimates In this section, we prove global upper bounds for solutions f of (1.1). Our bounds depend only on the estimates on the hydrodynamic quantities (1.5), (1.6) and (1.7). Our bound does not depend on an upper bound of the initial data. We also get that the solution will have certain polynomial decay in v for t>0. From Lemma 2.1, we see that the bounds on a¯ (t,x,v) degenerate as v . In the first ij | | → ∞ lemma, we show how to change variables to obtain an equation with uniform ellipticity constants independent of v . | | Lemma 4.1. Let z = (t ,x ,v ) R R2d be such that v 2, and let T be the linear 0 0 0 0 + 0 ∈ × | | ≥ transformation such that v 1+γ/2e, e v =0 0 0 Te= | | · (v0 γ/2e, e v0 = v0 . | | · | | Let T˜(t,x,v)=(t,Tx,Tv), and define (t,x,v):= T˜(t,x,v) Tz0 Sz0 ◦ =(t +t,x +Tx+tv ,v +Tv). 0 0 0 0 Then, (a) There exists a constant C >0 independent of v Rd B such that for all v B , 0 2 1 ∈ \ ∈ C−1 v v +Tv C v . 0 0 0 | |≤| |≤ | | (b) If f (t,x,v):=f( (t,x,v)), then f satisfies T Tz0 T (4.1) ∂ f +v f = [A(z) f ]+B(z) f +C(z)f t T x T v v T v T T ·∇ ∇ ∇ ·∇ GLOBAL ESTIMATES FOR THE INHOMOGENEOUS LANDAU EQUATION 9 in Q for any 0<R<min √t ,c v −1−γ/2 , where c is a universal constant, and R 0 1 0 1 { | | } λI A(z) ΛI, ≤ ≤ v 1+γ/2 1+ f(t,x, ) −(γ+1)/d, γ [ 2, 1), | 0| k · kL∞(Bρ(v)) ∈ − − B(z) . | | v 1+γ/2,(cid:0) (cid:1) γ [ 1,0],  0 | | ∈ − v γ 1+ f(t,x, ) −γ/d, −2d γ <0, C(v) . | 0| k · kL∞(Bρ(v)) d+2 ≤ | | |v0|−2(cid:0)−2γ/d 1+kf(t,x,·)kL∞(cid:1)(Bρ(v)) −γ/d, −2<γ < d−+2d2, with λ and Λuniversal, and(cid:0) ρ.1+ v0 −2/d. (cid:1) | | Proof. Since v 1 and v >2, 0 | |≤ | | v v 1+γ/2 v Tv v +Tv v + Tv v + v 1+γ/2. 0 0 0 0 0 0 0 | |−| | ≤| |−| |≤| |≤| | | |≤| | | | Thus, (a) follows since γ ( 2,0). ∈ − For (b), by direct computation, f satisfies (4.1) with T A(z)=T−1a¯( (z))T−1, B(z)=T−1¯b( (z)), C(z)=c¯( (z)). Tz0 Tz0 Tz0 In order to keep the proof clean, let us write a¯ and A instead of a¯ ( (z)) and A (z) for the ij ij ij Tz0 ij rest of the proof. Fix z = (t,x,v) Q , and let v˜=v +Tv. From part (a), we know that v˜ v . Applying R 0 0 ∈ | | ≈| | Lemma 2.1, we have that for any unit vector e, (1+ v )γ, e=v˜/v˜, (4.2) a¯ e e . | 0| | | ij i j ((1+ v0 )γ+2, e Sd−1. | | ∈ and, (1+ v )γ, e Sd−1, (4.3) a¯ e e & | 0| ∈ ij i j ((1+ v0 )γ+2, e v˜=0. | | · Our first step is to verify that we can switch v˜ for v in (4.2) and (4.3). 0 Let us start with (4.2). This is where the assumption v <R C v −1−γ/2 plays a role. We 1 0 | | ≤ | | can choose c so as to ensure that Tv 1. Since v = v˜ Tv and using the fact that a¯ is 1 0 ij | | ≤ − positive definite, a¯ (v ) (v ) 2a¯ v˜v˜ +2a¯ (Tv) (Tv) C v 2+γ. ij 0 i 0 j ij i j ij i j 0 ≤ ≤ | | Let e =v /v . The computation above tells us that a¯ (e ) (e ) . v γ. 0 0 0 ij 0 i 0 j 0 | | | | Let us now turn to (4.3). We will show that (4.4) a¯ w w &(1+ v )γ+2 w2 if w v =0. ij i j 0 0 | | | | · Note that(1+ v )2+γ and(1+ v )γ arecomparablewhen v issmall,sowe onlyneedto verify 0 0 0 | | | | | | (4.4) for w v = 0 and v arbitrarily large. For such vector w, we write w = ηv˜+w′ with 0 0 · | | w′ v˜ = 0. Since v˜ v = Tv 1, we have η = w v˜/v˜2 = w (v˜ v )/v˜2 w v˜−2. 0 0 · | − | | | ≤ | | | · | | | | · − | | | ≤ | || | Moreover, w′ w. | |≈| | Since a¯ is positive definite, ij a¯ (√2ηv˜ w′/√2) (√2ηv˜ w′/√2) 0, ij i j − − ≥ then we have 1 a¯ w w a¯ w′w′ η2a¯ v˜v˜ ij i j ≥ 2 ij i j − ij i j c(1+ v )γ+2 (1+ v )γ w2 &(1+ v )γ+2 w2, 0 0 0 ≥ | | − | | | | | | | | (cid:0) (cid:1) 10 STEPHENCAMERON,LUISSILVESTRE,ANDSTANLEYSNELSON as desired. Let w Rd be arbitrary. We will estimate A w w from above. Writing w = µe +w˜, with ij i j 0 ∈ w˜ e=0. · A w w = v −γ µ2a¯ (e ) (e ) +2µv −1a¯ (e ) w˜ + v −2a¯ w˜ w˜ , ij i j 0 ij 0 i 0 j 0 ij 0 i j 0 ij i j | | | | | | and using that a¯ is positive d(cid:0)efinite, (cid:1) ij A w w 2v −γ µ2a¯ (e ) (e ) + v −2a¯ w˜ w˜ , ij i j 0 ij 0 i 0 j 0 ij i j ≤ | | | | C µ2+(cid:0)w˜ 2 =:Λw2. (cid:1) ≤ | | | | This establishes upper bou(cid:0)nd Aij (cid:1) ΛI for some Λ>0. { }≤ Now we will prove the lower bound for A . Again, we write w =µe +w˜ with e w˜ =0. We ij 0 0 · need to analyze the quadratic form associated with the coefficients a¯ more closely. From (4.3), ij we have that for some universal constant c>0, cv γ(µ2+ w˜ 2) a¯ w w =µ2a¯ (e ) (e ) +2µa¯ (e ) w˜ +a¯ w˜ w˜ . 0 ij i j ij 0 i 0 j ij 0 i j ij i j | | | | ≤ Moreover,(4.2) implies that there is a universal constant δ >0 so that cv γ(µ2+ w˜ 2) δµ2a¯ (e ) (e ) +δ v −2a¯ w˜ w˜ . 0 ij 0 i 0 j 0 ij i j | | | | ≥ | | Subtracting the two inequalities above, (1 δ)µ2a¯ (e ) (e ) +2µa¯ (e ) w˜ +(1 δ v −2)a¯ w˜ w˜ 0. ij 0 i 0 j ij 0 i j 0 ij i j − − | | ≥ Thesameinequalityholdsifwereplacew =µe +w˜withw=(1 δ/2)−1/2µe +(1 δ/2)1/2 v −1w˜, 0 0 0 − − | | therefore 1 δ − µ2a¯ (e ) (e ) +2µv −1a¯ (e ) w˜ +(1 δ/2)(1 δ v −2)v −2a¯ w˜ w˜ 0. ij 0 i 0 j 0 ij 0 i j 0 0 ij i j 1 δ/2 | | − − | | | | ≥ − Recalling the formula above for A w w , and replacing it in the left hand side, we get ij i j 1 δ A w w 1 − v −γµ2a¯ (e ) (e ) 1 (1 δ/2)(1 δ v −2) v −2−γa¯ w˜ w˜ 0. ij i j 0 ij 0 i 0 j 0 0 ij i j − − 1 δ/2 | | − − − − | | | | ≥ (cid:18) − (cid:19) (cid:0) (cid:1) Therefore, using (4.3) and (4.4), 1 δ A w w 1 − v −γµ2a¯ (e ) (e ) + 1 (1 δ/2)(1 δ v −2) v −2−γa¯ w˜ w˜ , ij i j 0 ij 0 i 0 j 0 0 ij i j ≥ − 1 δ/2 | | − − − | | | | (cid:18) − (cid:19) λ(µ2+ w˜ 2), (cid:0) (cid:1) ≥ | | for some universal constant λ>0. This establishes the lower bound A λI. ij { }≥ To derive the bound on B(z), Lemma 2.3 and conclusion (a) imply B(z) . T−1 ¯b( (z)) | | k k| Tz0 | (1+ v )γ/2+1(1+ f )−(γ+1)/d, γ ( 2, 1), . | 0| k kL∞(Bρ′(v˜)) ∈ − − ((1+ v0 )γ/2+1, γ [ 1,0], | | ∈ − where ρ′ = v˜−2/d. From the triangle inequality, we have that B (v˜) B (v ), with ρ . ρ′ ρ 0 | | ⊂ (1+ v )−2/d +R(1+ v )(γ+2)/2 1+(1+ v )−2/d. The bound on C(z) follows in a similar 0 0 0 | | | | ≤ | | manner, using Lemma 2.2. (cid:3) The key lemma in the proof of Theorem 1.1 is the following pointwise estimate on f: Lemma 4.2. Let γ ( 2,0], T > 0, and let f : [0,T ] R2d R solve the Landau equation 0 0 + ∈ − × → (1.1) weakly. If f(t,x,v) K(1+t−d/2)(1+ v )−α ≤ | |

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