Caffarelli-Kohn-Nirenberg Theory

2022-08-08   Varnothing

Introduction

The C-K-N theory, originated from the paper in 1982, study the singular behavior of weak solutions of the incompressible Navie-Stokes equations in 3D with unit viscosity:

(NS){tv+(v)vνΔv+p=fv=0In this lecture, we adapt the compactness method by Lin's paper to prove the whole theory. 

Iteration Estimate

Now we give a significant iteration to verify the ϵregularity criterion. Consider the following quantities:
(1)ϕ(z,r,v)=r|vQz,r|,ψβ,γ(z,r,f)=r1+βfL2,2γ+1(Qz,r);ψ(z,r,v)=r23vL3(Q(z,r)),ψ(z,r,p)=r43pL32(Q(z,r));φ(z,r,v)=(1r2Q(z,r)|v(v)Q(z,r)|3)13=r23v(v)z,rL3(Q(z,r)),φ(z,r,p)=(1r2Q(z,r)|p(p)B(x,r)|32)23=r43p(p)z,rL32(Q(z,r)).For convenience, we'd like to omit z if z=0, and denote φ(r,v,p):=φ(r,v)+φ(r,p), etc. It is easy to check those dimensions:
ϕ(r,v)=ϕ(1,vr),ψβ,γ(r,f)=r(γβ)ψβ,γ(1,fr),ψ/φ(r,v,p)=ψ/φ(1,vr,pr);

(u)BrLq(Br)uLq(Br).

The estimate below is necessary for the following iteration.

C>0,0<β,γ2,θ(0,12],ϵ2,r2(0,1] such that if rr2, (v,p,f) is a suitable weak solution of (NS) in Qr and
ϕ(r,v)1,φ(r,v,p)+ψ(r,f)ϵ2,then ϕ(θr,v)1 and φ(θr,v,p)Cθ1+23(φ(r,v,p)+ψ(r,f)).
Notice that the result can be generalized to z0 case.
The first claim is clear:
ϕ(θr,v)=θr|Qθr|1Qθr(vvr)+θrvrCθr|Qθr|1131vvrL3(Qθr)+θrvrCθ23φ(r,v)+θϕ(r,v)Cϵ2θ23+121if ϵ2θ23/(2C). As for the latter, we prove it by contradiction: if not, then C>0,β<γ,θ(0,12], ϵ2,r2>0(where we set both as 1n), if rn1n, (vn,pn) are SWS of (NS) in Qrn such that
ψ(rn,vn)1,ϵn:=φ(rn,vn,pn)+ψ(rn,fn)1n,then φ(θrn,vn,pn)>Cθ1+23(φ(rn,vn,pn)+ψ(rn,fn))=Cϵnθ1+23.
Before the contradiction, first we take a scaling by following formulation:
(bn,un,qn,gn):=((vnrn)1,ϵn1(vnrn(vnrn)1),ϵn1(pnrn(pnrn)1),ϵn1fnrn).Then the homogeneity of ψ and φ transform the above relations as1
(2)|bn|1,φ(1,un,qn)+rn(γβ)ψ(1,gn)=un3+qn32+rn(γβ)gn2,2λ+1=1,and the claim: C>0,φ(θ,un,qn)Cθ1+23(φ(1,un,qn)+rn(γβ)ψ(1,gn))=Cθ1+23. The remained work is to show φ(θ,un),φ(θ,qn)θ1+23, which counter the claim immediately.

  1. φ(θ,un)θ1+23: Notice (un,qn) is suitable weak solution of following system on Q1:
    {tunΔun+(bn+ϵnun)un+qn=gn,un=0.Moreover, the uniform bound (2) implies
    bnRb,unL3u,qnL32q,gnL2,2γ+10. Now we attempt to derive a strong Lq-convergence for un by Aubin-Lion lemma: The local energy inequality implies2
    unL10/3(Q7/8)unLL2L2H˙1(Q7/8)1,and following calculation implies tunL43(Hσ1)1:
    |tunζ||unζ|+|unζ|+|ununζ|unL2H˙1ζH˙1+unL2H˙1ζL2+unLL212unL2L632ζL4L2ζL4(Hσ1).Since H1KL2(Hσ1), together with unL2H11 and tunL43(Hσ1)1, we see {un} is precompact in L2L2 and thus unL2u. Moreover, since unL10/31, so the Holder interpolation implies3 unLq(Q7/8)u,q[2,10/3).
    Consequently, (u,q) solves4 following system in Q78:
    {tuΔu+bu+q=0,u=0.Following we show that u is Holder continuous:
    (tΔ+b)(×u)=0×uL(Q5/6)u=0uLL100(Q4/5),qL32,Δq(t)=0qL3/2CuLL100,Δu?tuL3/2L(Q4/5).It comes uCα,α2(Q3/4) for α2=113/2α=23. Then the Campanato characterization in parabolic version gives out following bound since θ3/4:
    θ323αQθ|u(u)θ|31unL3uθ323αQθ|un(un)θ|31,which implies φ(θ,u)θ1+α.
  2. φ(θ,qn)1: notice Δqn=ϵn(unun)+gn=ϵn(ij(vivj))+gn on Q7/8, so we split qn=q~n+hn where
    Δq~n=ϵnζ(ij(vivj))+ζgn,ζD(Q78) cutoff on Q34;Δhn=Δ(qnq~n)=0 on Q34.We estimate q~n by Riesz potential, and hn by properties of harmonic function:5
    q~nL3(Q7/8)=I2((ϵnζ(ij(uniunj))+ζgn))L3/2(Q7/8)ϵn2Γ(uniunj)L3/2(Q7/8)+gnL32L2(Q7/8)ϵnunL3(Q(7/8))2+gnL2,2γ+1(Q7/8)ϵn+rnγβ;
    Qθ|hn(hn)θ|32=θ20Bθ|hn(hn)θ|32=θ20Bθ(θhnlip(Qθ))32θ3+329/160hnL(B3/4)32θ3+329/160hnL1(B3/4)3/2θ3+329/160hnL3/2(B3/4)3/2θ3+32Q3/4|hn|3/2θ3+32Q3/4(|qn|3/2+|q~n|3/2)θ3+32.And then the estimate qn follows:
    Qθ|qn(qn)θ|32Qθ|hn(hn)θ|32+Qθ|q~n(q~n)θ|32θ3+32+(ϵn+rnγβ)3/2θ3+32if ϵn+rnγβθ3. Consequently, we have φ(θ,qn)θ1+23.

Combine together, we get φ(θ,un,pn)θ1+23.

ϵcriterion

For any γ(0,2],αmin{23,γ}, there are ϵ1>0,θ(0,12),r1>0 such that if Rr1,fL2,2γ+1(QR),(v,p,f) is a suitable weak solution of (NS) in QR, and
ψ(R,v,p)ϵ1.Then vCα,α2(QθR).
First we set following relations to apply the above lemma:
β=12(α+γ);θ(0,12],Cθ23+θβθα;(ϵ2,r2)(θ,β,γ)(γ,α);r1=min{r2,(ϵ22fL2,2λ+1(QR))11+β}.Then for any zQθR,r[θ(1θ)R,(1θ)R], denote Ψ(z,r):=φ(z,r,v,p)+ψ(z,r,f), we attempt to show
(3)Ψ(z,θkr)θ(1+α)kϵ2,k0,which means for any ρ(0,(1θ)R](ρ=θkr),
φ(z,ρ,v)Ψ(z,ρ)(ρr)1+αϵ2(ρθr0)1+αϵ2ρ1+α.That is say, vCα,α2(QθR) by Campanato characterization. Then the remained work is to verify (3). Indeed, the condition implies k=0 case:
ϕ(z,r)1;?Ψ(z,r)=φ(z,r,v,p)+ψ(z,r,f)cψ(z,r,v,p)+ψ(z,r,f)c(R/r)43ψ(R,v,p)+(r/r0)1+βψ(z,r0,f)cθϵ1+ϵ22ϵ2for ϵ1 small enough. Moreover, the iteration follows by the lemma:
ϕ(z,θkr)=1;Ψ(z,θkr)=φ(z,θkr,v,p)+ψ(z,θkr,f)Cθ1+23(φ(z,θk1r,v,p)+ψ(z,θk1r,f))+θ1+βψ(z,θk1r,f)θ1+αΨ(z,θk1r)θ(1+α)kϵ2.Thus the claim finished.
For any s[1,], there exists ϵ>0, such that any (v,p,f) a suitable weak solution of (NS) in ΩT, v is regular at zΩT if one of following conditions holds:

  1. 3q+2s[1,2],lim infr0r(3q+2s1)v(v)BrLsLq(Q(z,r))ϵ;
  2. 3q+2s[2,3],lim infr0r(3q+2s2)vLsLq(Q(z,r))ϵ;
It is convenient to denote homogenous quantities:
A(r)=1rvLL2(Qr)2,B(r)=1rvL2L2(Qr)2;C(r)=1r2vL3(Qr)3,C~(r)=1r2v(v)BrL3(Qr)3,D(r)=1r2pL32(Qr)32;G1(r)=1rv(v)BrLsLq(Qr),G2(r)=1rvLsLq(Qr).For above quantities, there are rough dominations for k1,
A/B/C/C~/D/G(r)ckA/B/C/C~/D/G(kr).To process further, we'd like to show the following estimates for iteration: rρ2,
(4)C(r)(rρ)C(ρ)+(ρr)2C~(ρ),D(r)(rρ)D(ρ)+(ρr)2C~(ρ);C~(r)A1s(r)E11s(r)G(r),(A+E)(r)1+(C+D)(2r).And all above estimates give out: rρ4,
C~(ρ2)A1s(ρ2)E11s(ρ2)G(ρ2)(1sA(ρ2)+(11s)E(ρ2))G(ρ2)(A+E)(ρ2)G(ρ2)(1+(C+D)(ρ))G(ρ);(C+D)(r)(rρ)(C+D)(ρ2)+(ρr)2C~(ρ2);(rρ)(C+D)(ρ)+(ρr)2(1+C(ρ)+D(ρ))G(ρ).Set the constant as c>0, then we choose θ(0,1/4) so that cθ<1/4. By assumption, there is r0>0,rr0,G(r)<θ2ϵ11+8c. Then the estimate indicates:
(C+D)(θr)12(C+D)(r)+θ2(1+(C+D)(r))G(r)12(C+D)(r)+ϵ14?,(C+D)(θkr)12k(C+D)(r)+ϵ12,r<r0.Then for k big enough(depends on r), we have (C+D)(θkr)ϵ1. Let R=θkr, then the above criterion shows v is regular near 0. Following are the verification of (4):

  1. C(r)(rρ)C(ρ)+(ρr)2C~(ρ):
    C(r)1r2Qr|v(v)Bρ|3+1r2Qr|(v)Bρ|3(ρr)2(1ρ2Qρ|v(v)Bρ|3)+rρ(1ρ2Qρ|v|3)(ρr)2C~(ρ)+rρC(ρ);6
  2. D(r)(rρ)D(ρ)+(ρr)2C~(ρ): Since Δp=ij(vivj), then we decompose p=p~+h, where Δp~=ζ(ij(vivj)),ζ cutoff on Bρ/2. Then
    p~L32(Bρ)Γ(ij(vi(v)Bρi)(vj(v)Bρj))v(v)BρL3(Bρ),and since |h|32 is sub-harmonic\footnote{Sub-harmonicity is preserved under convex function.}, it comes
    D(r)1r2Qr|p~|32+1r2Qr|h|321r2Qρ|p~|32+rρ3Bρ|h|32(rρ)(1ρ2Bρ|p|32)+(rρ3+1r2)(Bρ|p~|32)(rρ)D(ρ)+(ρr)2C~(ρ).
  3. C~(r)A1s(r)E11s(r)G(r): Apply Holder interpolation:
    C~(r)=1r2v(v)BrL3(Qr)31r2v(v)BrLL223v(v)BrL2L622sv(v)rLsLq(1rvLL22)1s(1rvL2L22)11sv(v)BrLsLqA1s(r)E11s(r)G(r);The second condition is similar: For (q,s)=(3,1),
    v(v)BrL3(Br)3v(v)BrL3(Br)2vL3(Br)+r32v(v)BrL2(Br)3vL2(Br)2vL3(Br).Integrating in time and apply Holder interpolation :
    C~(r)=r2v(v)BrL3(Br)3L1(r2,0)r2vL2(Br)2vL3(Br)L1(r2,0)r2vLL2(Qr)2vL1L3(Qr)=A(r)G2(r).
  4. (A+E)(r)1+(C+D)(2r): We take ϕD(Q2r) cutoff on Qr, then the local energy inequality gives:
    vLL2(Qr)2+vL2(Qr)2v0L2(Br)+Q2r|v|2(tϕ+Δϕ)+(|v|2+2p)|v|ϕ1+r2vL2(Q2r)2+r2vL3(Q2r)3+r1vpL1(Q2r).Take r1 at both hand, we get that
    A(r)+B(r)1+(r2vL3(Q2r)3)32+r2vL3(Q2r)3+(r2vL3(Q2r)3)13(r2pL32(Q2r)32)231+C(2r)+D(2r).7

Singularity

Appendix

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Reference

  1. A new proof of CKN theory by Lin, 1998.

  1. un=n(vnrn(vnrn)1)=n(vnrn(vnrn)1)n((vnrn(vnrn)1)1)=un(un)1. Similar for qn.
     

  2. \qst is the initial data for each un determined? 

  3. umunqumun2θumun10/31θumun2θ,q=θ2+1θ10/3. 

  4. Particularly, it is a suitable weak solution which is preseved by weak limit. Details see Lin's Paper, Theorem 2.2. 

  5. this is reached by following estimate:
    I2(ζgn)L32(B7/8)I2(ζgn)+I2gn3/2I1gn6+I2gn152gn2+g5/4gn2.Here we will encounter critical case q=1 for gq if no amplification by cutoff function. 

  6. Here Qr|(v)Bρ|3=r20|Br||(v)Bρ|(rρ)3ρ20|Bρ||(v)Bρ|(rρ)3Bρ|v|3. 

  7. Notice vL2(Q2r)|Q|1213vL3(Q2r)r56vL3(Q3r)

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