一类由空间粗糙高斯噪声驱动分数阶动力学方程的性质研究.pdf

应用概率统计第 40 卷第 1 期2024 年 2 月Chinese Journal of Applied Probability and StatisticsFeb.,2024,Vol.40,No.1,pp.139-156doi:10.3969/j.issn.1001-4268.2024.01.009Some Properties of Fractional Kinetic Equation withGaussian Noise Rough in SpaceLU WeidongLIU Junfeng(School of Mathematics,Nanjing Audit University,Nanjing,211815,China)Abstract:In this article,we study a class of fractional kinetic equation driven by Gaussian noisewhich is white/colored in time and has the covariance of a fractional Brownian motion with Hurstindex H 0,x R;u(0,x)=1,x R,(1)Corresponding author,E-mail:.Received January 17,2022.Revised January 26,2023.140Chinese Journal of Applied Probability and StatisticsVol.40where the operators(I )/2and()/2are interpreted as the inverses of Besselpotential and Riesz potential,respectively.W=2W/(tx)(t,x)is a Gaussian noiseand is a positive constant.Apart from in the classical context of heat conduction,anequation of the form(1)with =0 and =2 named stochastic heat equation also arisingin neurophysiology14,15.If =0 and 0 6 2,it is the factional stochastic heatequation had been studied in 1012 and references therein.The It o-type probabilistic approach for stochastic partial differential equation(SPDE)was established in 15,where Walsh introduced martingale measures and defined stochas-tic integral with respect to the martingale measures,and then SPDEs driven by space-timewhite noise were investigated.In 16,Dalang extended Walshs stochastic integral andapplied it to solve SPDEs whose Greens function is not a function but a Schwartz dis-tribution.For SPDEs driven by a multiplicative Gaussian noise which is colored in time,the probabilistic approach based on martingale properties cannot be applied directly sincethe noise does not have martingale structure in time.An alternative approach is to applyMalliavin calculus to study the chaos expansion of the Skorohod solution,see,for instance,1721 for stochastic heat equations and 2224 for stochastic wave equations.This equation proposed in the FKE(1)is a generalization of the well-known stochasticheat equation =0,=2 which has been studied by many authors.We briefly describesome related recent developments on stochastic heat equations driven by(fractional)mul-tiplicative Gaussian noise,especially the fractional Gaussian noise with Hurst index lessthan 1/2.In 25,Balan,Jolis and Quer-Sardanyons proved the existence and uniquenessof a mild solution to stochastic heat equation in the linear case.That means the diffusionterm is changed into au+b with a,b R.The stochastic integral is understood in theIt os sense.The case of a general nonlinear coefficient satisfying some suitable conditionhas been studied by the authors in 26.They proved the existence and uniqueness of amild solution whose trajectories belong to a suitable space of trajectories.We would liketo mention that work19,20,in which the author studied the parabolic Anderson modelwith fractional Gaussian noise which is rough in time.They mainly concerned on the cor-responding solvability,Feynman-Kacs moment formula and intermittency of the model.While the authors in 23 studied the intermittency for the hyperbolic Anderson modelswith(fractional)Gaussian noise rough noise in space.The stochastic heat and wave e-quations driven by Gaussian noise which is white in time and rough in space were alsostudied in 21,24,2628.For the FKE(1)driven by Gaussian noise,there exists several works,for example47,13,29 and etc.In this paper,we will mainly focus on intermittency properties andH older continuity of the solutions to the FKE(1)driven by Gaussian noiseW.One canNo.1LU W.D.,LIU J.F.:Fractional Kinetic Equation with Gaussian Noise Rough in Space141see the precise definition ofW in Section 2.For the topics on the intermittency for thesolution to SPDE,one can consult the references,17,23,3033 and etc.There are somedifferences between this study and the papers pointed above.Firstly,the FKE(1)isdriven by a fractional Gaussian noise which is rough in space.Secondly,thanks to theused Gaussian noise and the techniques of Malliavin calculus,the upper and lower boundsfor the moments of the solution are checked.Therefore one can deduce the intermittencyproperty for the solution.A brief outline of this article is stated as follow.In Section 2 we recall some prelim-inaries we needed in this paper,such as the Gaussian noiseW and some basic elementsof Malliavin calculus.Section 3 is devoted to prove the existence and uniqueness of thesolution to the FKE(1).In Section 4,we establish the moment bounds for the solution tothe FKE(1)and prove it is weakly intermittent.We study the H older continuity of thesolution to the FKE(1)with respect to the time and space variables in Section 5.Furtherdescriptions of the Green function and proofs of some technical lemmas are given in theAppendix.2The PreliminariesThis section is devoted to recall the Gaussian noiseW and some basic elements ofMalliavin calculus used in this paper are recalled in the appendix.Denote by S(R+R)the Schwartz space.For,S(R+R),let H be the completion of S(R+R)under the inner product,H=cHR2+R0(r s)F(s,)()F(r,)()(d)drds,with cH=(2H+1)sin(H)/(2),0()is a non-negative definite function and(d)=|12Hd with H (0,1/2).Here F(s,)()is the Fourier transform of with respectto the space variable.That is,for any S(R+R),F(s,)()=Reix(s,x)dx.In particular,if is a measurable function such that F(s,)()is also a measurablefunction andR2+R0(r s)|F(s,)()|F(r,)()|(d)drds 0,P)be a complete probability space.Define W=W(),H a isonormal Gaussian process with the covariance function EW()W()=,H,H,and we also denote W()=R+R(t,x)W(dt,dx).We call W()the Wienerintegral of with respect to W.Moreover,the Gaussian family W=W(),H coincides with the linear span of the Gaussian processes W(t,x),(t,x)R+R with thecovariance function EW(t,x)W(s,y)=21(|x|2H+|y|2H|xy|2H)t0s00(uv)dudv,and in particular,W(t,x)=W(I0,t0,x)with the convention I0,tx,0=I0,t0,xfor142Chinese Journal of Applied Probability and StatisticsVol.40 x 2)andu(t,x)satisfiesu(t,x)=1+t0RGts(x y)u(s,y)W(ds,dy),(2)whereGt(x):(t,x)R+Ris the Green function to the FKE(1)and the differentialW(ds,dy)is used for the Skorohod integral.From 23,24,31 together with Equation(2),intuitively,the solution to the FKE(1)should be given by a series of iterated integralu(t,x)=1+t0RdGts1(x z1)W(ds1,dz1)+2t0RdGts2(x z2)s20RdGs2,s1(z1 z2)W(ds1,dz1)W(ds2,dz2)+.More precisely,we want to prove that the solution to the FKE(1)has the following formu(t,x)=n=0In(efn)(,t,x),(t,x)R+R,(3)where the kernel fn=fn(,t,x),(t,x)R+R,n 0 is given byf0(t,x)=1,fn(t1,x1,tn,xn;t,x)=nGttn(x xn)Gt2t1(x2 x1)10t1tnt,(4)and we denote byefn(t1,x1,tn,xn;t,x)=Sym(fn(t1,x1,tn,xn;t,x),where“Sym”denotes the symmetrization with respect to n variables(t1,x1),(tn,xn).Note thatefn(t1,x1,tn,xn;t,x)=nn!SnGtt(n)(x(n+1)x(n)Gt(2)t(1)(x(2)x(1),(5)where Sndenotes the permutation of the set 1,2,n such that 0 t(1)t(2)t(n)1,fn(,t,x)Hnwithfn(,t,x)defined by(4),then the FKE(1)has a solution if and only if the seriesn0In(fn(,t,x)converges inL2().In this setting,the solution to the FKE(1)isgiven by,for allx R,u(t,x)=n0In(fn(,t,x),(t,x)R+R.The main result of this section is stated as follows.Theorem 3Assume that+(1,2,1/2 H0 min3/2(+),1andH (3 2H0)/4,1/2),then the FKE(1)has a unique square integrable mild solutionu=u(t,x),(t,x)R+Rin the sense of Definition 1.ProofTo show the existence and uniqueness of the solution to the FKE(1),itsuffices to prove that for any(t,x)R+R,we haveE|u(t,x)|2=n=0n!efn(,t,x)2Hn 0,denote byn(t,x)=E|In(efn(,t,x)|2=n!efn(,t,x)2Hn(6)withefngiven by(5).Since we assume that 0(s)|s|2H02for H0(1/2,1),throughoutthe rest of this paper,we will simply assume that 0(s)=|s|2H02in this case.Now let usanalyze the quantities n(t,x)defined by(6).Fix t R+and x R.Setefn(s,z;t,x)=efn(s1,z1,sn,zn;t,x).We have the following expressionn!efn(,t,x)2Hn=n!Rn0,t2nFefn(s,t,x)()Fefn(r,t,x)()nj=1|sj rj|2H02dsdr(d)withFefn(s,t,x)()=nn!eix(1+2+n)nj=1FGs(j+1)s(j)(1)+(2)+(j),where we use the convention sn+1=t and ds=ds1ds2dsn,the differentials dr and(d)are defined similarly.Then,by using Cauchy inequality,one obtains thatn!efn(,t,x)2Hn6 CnH0n!0,tnRn|Fefn(s,t,x)()|2(d)1/(2H0)ds2H0144Chinese Journal of Applied Probability and StatisticsVol.40=CnH02n(n!)12H0Tn(t)Rnnj=1e2|s(j+1)s(j)|jk=1(k)|(1+|jk=1(k)|2)/2(d)1/(2H0)ds2H0(7)with Tn(t)=s=(s1,s2,sn);0 s1 s2 sn t.Recall that(d)=nj=1|j|12Hdj,using a change of variables j=1+2+jwith j=1,2,n,one getsn!efn(,t,x)2Hn6CnH02n(n!)12H0Tn(t)(Rnnj=1e2|sj+1sj|j|(1+|j|2)/2nj=1|j j1|12Hd1dn)1/(2H0)ds2H0.(8)Next we shall use the arguments in,for example,23,24 to bound the termnj=1|jj1|12Hin(8).Let Anbe a set of cardinality 2n1consisting of multi-indices =(1,2,n)with the following properties:|=nj=1j=n(1 2H),1 1 2H,2(1 2H),j 0,1 2H,2(1 2H)for j=2,3,n 1 and n 0,1 2H.Using the inequality(a+b)6 a+bfor some 0 0,we have|j j1|12H6(|j|+|j1|)2H6|j|2H+|j1|2H.We also need the followingfact:for any finite set S and positive numbers(ai)iSand(bi)iS,ni=1(ai+bi)=IS(iIai)(iS/Ibi).Hence,we havenj=1|j j1|12H6|1|12Hnj=2|j j1|12H6Annj=1|j|j.Then using the notation sn+1=t,we obtain thatn!efn(,t,x)2Hn6CnH02n(n!)12H0AnTn(t)(nj=1Re2|sj+1sj|j|(1+|j|2)/2|j|jdj)1/(2H0)ds2H0.Then it follows thatn!efn(,t,x)2Hn6 CnH02n(n!)2H01AnTn(t)nj=1(sj+1sj)(1+j)/2(+)H0ds2H0.Let j=(1+j)/2H0(+),j=1,2,n.Note that if H0(+)(34H)/2,that the quantities j 1 with j=1,2,n hold.Moreover,|=n(1 H)/H0(+).Then we obtain thatn!efn(,t,x)2Hn6 CnH02n(n!)2H01nj=1(1+j)(n1 (1 H)/H0(+)+1)tn(1H)/H0(+)+n2H0No.1LU W.D.,LIU J.F.:Fractional Kinetic Equation with Gaussian Noise Rough in Space1456Cn2n(n!)2H01t2nH0(1H)/(+)(an+1)2H0Cn2nt2nH0(1H)/(+)(n!)12(1H)/(+)a2H0an+H0nH0(1a),(9)as n where we denote by a=1 (1 H)/H0(+)and b=1.Notice that thereexists some 1 such that n6 a2H0an+H0nH0(1a)6 nfor all n 1.Hence thereexists some positive constants C,C1,C2such thatE|u(t,x)|26n=0Cn2ntn2H02(1H)/(+)(n!)12(1H)/(+)6 C1expC2|2(+)/+2(1H)t2H0(+)2(1H)/+2(1H).(10)If H0=1/2,that is 0(t)=(t),and we haven!efn(,t,x)2Hn=n!Rn0,tn|Fefn(s,t,x)()|2ds(d)=1n!0,tnRne2|s(j+1)s(j)|jk=1(k)|(1+|jk=1(k)|2)/2nj=1|j|12Hdds=Tn(t)Rne2(sj+1sj)|j|(1+|j|2)/2nj=1|j j1|12Hdds.The last term in the above equation equals the right-hand of Equation(7).Hence oncan follow the similar lines in(8)to(9),we can get the following estimate for the secondmoment for the solution to the FKE(1)with H0=1/2 as followsE|u(t,x)|2=n=0n!efn(,t,x)2Hn6 C1expC2|2(+)/+2(1H)t.Then the proof of this theorem is complete.?4Moment Bounds and IntermittencyIn this section,we firstly prove the lower and upper moment bounds for the p-thmoment of the solution to the FKE(1)and then deduce the weakly intermittency.Themain result in this section is stated as follows.Theorem 4Under the conditions in Theorem 3,forp 2,there exist two positiveconstantsC1,C2such thatC1expC1|2(+)/+2(1H)t2H0(+)2(1H)/+2(1H)6 u(t,x)p6 C2expC2p(+)/+2(1H)|2(+)/+2(1H)t2H0(+)2(1H)/+2(1H).(11)146Chinese Journal of Applied Probability and StatisticsVol.40ProofWe shall prove(11)with H0(1/2,1).The case H0=1/2,i.e.0(t)=(t)is similar.So we omit the details.Now let us assume that H0(1/2,1).From the proofof Theorem 3,we haven!efn(,t,x)2Hn=n!Rn0,t2nFefn(s,t,x)()Fefn(r,t,x)()nj=1|si rj|2H02dsdr(d)=2nn!Rn0,t2nnj=1e(sj+1sj)|1+2+j|(1+|1+2+j|2)/2nj=1e(rj+1rj)|1+2+j|(1+|1+2+j|2)/2nj=1|sj rj|2H02dsdr(d)=2nn!RnTn(t)Tn(t)nj=1e(sj+1sj)|j|(1+|j|2)/2nj=1e(rj+1rj)|j|(1+|j|2)/2nj=1|sj rj|2H02dsdrnj=1|j j1|12Hd.By using the inequality|j j1|12H|j|12Hon Bn=(1,2,n)Rn:10,26 0,3 0,46 0,.n!efn(,t,x)2Hn 2nn!BnTn(t)Tn(t)nj=1e(sj+1sj)|j|(1+|j|2)/2nj=1e(rj+1rj)|j|(1+|j|2)/2nj=1|sj rj|2H02dsdrnj=1|j|12Hd.Then one obtainsn!efn(,t,x)2Hn 2nn!tn(2H02)Rn+?Tn(t)nj=1e(sj+1sj)|j|(1+|j|2)/2ds?2nj=1|j|12Hd,(12)where the last inequality holds due to|sj rj|6 t for j=1,2,n and the fact thatthe integral with respect to is nonnegative.Now let us denote byAn(t)=Rn+?Tn(t)nj=1e(sj+1sj)|j|(1+|j|2)/2ds?2nj=1|j|12Hd.(13)For the function inside the above integral,one can obtain with some constant c,0e(sj+1sj)|j|(1+|j|2)/2 e(sj+1sj)(1+|j|2)(+)/2 ec,(sj+1sj)ec,(sj+1sj)|j|+.Thus one can rewrite the An(t)as follows:An(t)Rn+?Tn(t)nj=1ec,(sj+1sj)ec,(sj+1sj)|j|+ds?2nj=1|j|12HdNo.1LU W.D.,LIU J.F.:Fractional Kinetic Equation with Gaussian Noise Rough in Space147:=ec,tBn(t),where we denote by Bn(t)=Rn+?Tn(t)nj=1ec,(sj+1sj)|j|+ds?2nj=1|j|12Hd.Making the change of variables sj=sj/t and j=jt1/(+),thus we have the scalingAn(t)ec,tBn(t)=ec,tt2n1(1H)/(+)Bn(1).(14)Now we shall estimate EBn()where is an exponential random time with parameter1.By Fubinis theorem and Jensens inequality,we obtainEBn()=0etBn(t)dt=Rn+0et?Tn(t)nj=1ec,(sj+1sj)|j|+ds?2dtnj=1|j|12HdRn+?0etTn(t)nj=1ec,(sj+1sj)|j|+dsdt?2nj=1|j|12Hd.(15)Applying the changes of variables rj=sj+1sjwith j=0,1,n and s0=0,sn+1=t,we have0etTn(t)nj=1ec,(sj+1sj)|j|+dsdt=Rn+1+e(r0+r1+rn)nj=1ec,rj|j|+dr0drn=nj=1R+erj(1+c,|j|+)drj.Denote by E=R+er(1+c,|+)dr.Thus we can compute E as follows:E=R+er(1+c,|+)dr=er(1+c,|+)1+c,|+?0=11+c,|+.This yields that0etTn(t)nj=1ec,(sj+1sj)|j|+dsdt Cnj=111+c,|j|+.Substituting this estimate into(15),one getsEBn()CnRn+nj=1(11+c,|j|+)2|j|12Hdj:=(C)n,where we denote by C=CR+(1+c,|j|+)2|j|12Hd 2nn!tn(2H02)An(t)(C)n2nn!ec,tt2nH0(1H)/(+)1E2n1(1H)/(+)(C)n2nn!ec,tt2nH0(1H)/(+)1(2n1 (1 H)/(+)+1)(C)n2nn!ec,tt2nH0(1H)/(+)(n!)21(1H)/(+)1aan+1/2n1/2a/2,where we denote by a=21 (1 H)/(+)and b=1.Also noting that there exists 1 such that n6 aan+1/2n1/2a/26 n,we obtain that n!efn(,t,x)2Hn(C)n2nec,ttn2H02(1H)/(+)/(n!)12(1H)/(+).Then we haveu(t,x)p u(t,x)2n=0(C)n2ntn2H02(1H)/(+)(n!)12(1H)/(+)1/2 C1ec,texpC1|2(+)/+2(1H)t2(+)H02(1H)/+2(1H),for some C1 0,c,0.This concludes the proof of the lower bound in(11).For the upper bound in(11),using the Minskowskis inequality,the equivalence ofthe p-norms on a fixed Wiener chaos space Hn,estimates(9)and(10),we haveu(t,x)p6n=0In(efn(,t,x)p6n=0(p 1)n/2In(efn(,t,x)26 C2expC2p(+)/+2(1H)|2(+)/+2(1H)t2(+)H02(1H)/+2(1H),for some C2 0.This concludes the proof of the upper bound in(11).Thus we canconclude the proof of this theorem.?Let us now state the theorem related to the weakly intermittency of the solutionu(t,x)to the FKE(1),which can be immediately derived from Theorem 4.There existseveral ways to express this phenomenon of intermittency.The physical interpretation ofthis property is that,as time t becomes large,the paths of the random field u exhibit veryhigh peaks concentrated on small spatial islands.Let us firstly recall the definition of theweak intermittency by recalling the upper and lower p-th moment Lyapunov exponents(see,30,33,35 and etc).Definition 5For eachp 2,let us denote byL(p)and respectivelyL(p)the upperand lowerp-th moment Lyapunov exponents of the solutionu(t,x)to the FKE(1)which isdefined byL(p)=limsupt1r(t)supxRlnE|u(t,x)|p,L(p)=liminft1r(t)infxRlnE|u(t,x)|p,No.1LU W.D.,LIU J.F.:Fractional Kinetic Equation with Gaussian Noise Rough in Space149where we assume that the limit above is independent ofxandr(t)0.Then,if the smallestintegere pofL(p)0exists,that ise p=infp N:L(p)0