Research article

pth moment exponential stability and convergence analysis of semilinear stochastic evolution equations driven by Riemann-Liouville fractional Brownian motion

  • Many works have been done on Brownian motion or fractional Brownian motion, but few of them have considered the simpler type, Riemann-Liouville fractional Brownian motion. In this paper, we investigate the semilinear stochastic evolution equations driven by Riemann-Liouville fractional Brownian motion with Hurst parameter H<1/2. First, we prove the pth moment exponential stability of mild solution. Then, based on the maximal inequality from Lemma 10 in [1], the uniform boundedness of pth moment of both exact and numerical solutions are studied, and the strong convergence of the exponential Euler method is established as well as the convergence rate. Finally, two multi-dimensional examples are carried out to demonstrate the consistency with theoretical results.

    Citation: Xueqi Wen, Zhi Li. pth moment exponential stability and convergence analysis of semilinear stochastic evolution equations driven by Riemann-Liouville fractional Brownian motion[J]. AIMS Mathematics, 2022, 7(8): 14652-14671. doi: 10.3934/math.2022806

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  • Many works have been done on Brownian motion or fractional Brownian motion, but few of them have considered the simpler type, Riemann-Liouville fractional Brownian motion. In this paper, we investigate the semilinear stochastic evolution equations driven by Riemann-Liouville fractional Brownian motion with Hurst parameter H<1/2. First, we prove the pth moment exponential stability of mild solution. Then, based on the maximal inequality from Lemma 10 in [1], the uniform boundedness of pth moment of both exact and numerical solutions are studied, and the strong convergence of the exponential Euler method is established as well as the convergence rate. Finally, two multi-dimensional examples are carried out to demonstrate the consistency with theoretical results.



    The key to solving the general quadratic congruence equation is to solve the equation of the form x2amodp, where a and p are integers, p>0 and p is not divisible by a. For relatively large p, it is impractical to use the Euler criterion to distinguish whether the integer a with (a,p)=1 is quadratic residue of modulo p. In order to study this issue, Legendre has proposed a new tool-Legendre's symbol.

    Let p be an odd prime, the quadratic character modulo p is called the Legendre's symbol, which is defined as follows:

    (ap)={1, if a is a quadratic residue modulo p;1, if a is a quadratic non-residue modulo p;0, if pa.

    The Legendre's symbol makes it easy for us to calculate the level of quadratic residues. The basic properties of Legendre's symbol can be found in any book on elementary number theory, such as [1,2,3].

    The properties of Legendre's symbol and quadratic residues play an important role in number theory. Many scholars have studied them and achieved some important results. For examples, see the [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21].

    One of the most representative properties of the Legendre's symbol is the quadratic reciprocal law:

    Let p and q be two distinct odd primes. Then, (see Theorem 9.8 in [1] or Theorems 4–6 in [3])

    (pq)(qp)=(1)(p1)(q1)4.

    For any odd prime p with p1mod4 there exist two non-zero integers α(p) and β(p) such that

    p=α2(p)+β2(p). (1)

    In fact, the integers α(p) and β(p) in the (1) can be expressed in terms of Legendre's symbol modulo p (see Theorems 4–11 in [3])

    α(p)=12p1a=1(a3+ap)andβ(p)=12p1a=1(a3+rap),

    where r is any integer, and (r,p)=1, (rp)=1, (p)=χ2 denote the Legendre's symbol modulo p.

    Noting that Legendre's symbol is a special kind of character. For research on character, Han [7] studied the sum of a special character χ(ma+ˉa), for any integer m with (m,p)=1, then

    |p1a=1χ(ma+ˉa)|2=2p+(mp)p1a=1χ(a)p1b=1(b(b1)(a2b1)p),

    which is a special case of a general polynomial character sums N+Ma=N+1χ(f(a)), where M and N are any positive integers, and f(x) is a polynomial.

    In [8], Du and Li introduced a special character sums C(χ,m,n,c;p) in the following form:

    C(χ,m,n,c;p)=p1a=0p1b=0χ(a2+nab2nb+c)e(mb2ma2p),

    and studied the asymptotic properties of it. They obtained

    p1c=1|C(χ,m,n,c;p)|2k={p2k+1+k23k22p2k+O(p2k1),ifχ is the Legendre symbol modulo p;p2k+1+k23k22p2k+O(p2k1/2),ifχ is a complex character modulo p.

    Recently, Yuan and Zhang [12] researched the question about the estimation of the mean value of high-powers for a special character sum modulo a prime, let p be an odd prime with p1mod6, then for any integer k0, they have the identity

    Sk(p)=13[dk+(d+9b2)k+(d9b2)k],

    where

    Sk(p)=1p1p1r=1Ak(r),
    A(r)=1+p1a=1(a2+rˉap),

    and for any integer r with (r,p)=1.

    More relevant research on special character sums will not be repeated. Inspired by these papers, we have the question: If we replace the special character sums with Legendre's symbol, can we get good results on p1mod4?

    We will convert β(p) to another form based on the properties of complete residues

    β(p)=12p1a=1(a+nˉap),

    where ˉa is the inverse of a modulo p. That is, ˉa satisfy the equation xa1modp for any integer a with (a,p)=1.

    For any integer k0, G(n) and Kk(p) are defined as follows:

    G(n)=1+p1a=1(a2+nˉa2p)andKk(p)=1p1p1n=1Gk(n).

    In this paper, we will use the analytic methods and properties of the classical Gauss sums and Dirichlet character sums to study the computational problem of Kk(p) for any positive integer k, and give a linear recurrence formulas for Kk(p). That is, we will prove the following result.

    Theorem 1. Let p be an odd prime with p1mod4, then we have

    Kk(p)=(4p+2)Kk2(p)8(2α2p)Kk3(p)+(16α416pα2+4p1)Kk4(p),

    for all integer k4 with

    K0(p)=1,K1(p)=0,K2(p)=2p+1,K3(p)=3(4α22p),

    where

    α=α(p)=p12a=1(a+ˉap).

    Applying the properties of the linear recurrence sequence, we may immediately deduce the following corollaries.

    Corollary 1. Let p be an odd prime with p1mod4. Then we have

    1p1p1n=111+p1a=1(a2+nˉa2p)=16α2p28α28p2+14p16α416α2p+4p1.

    Corollary 2. Let p be an odd prime with p1mod4. Then we have

    1p1p1n=1p1m=0(1+p1a=1(a2+nˉa2p))e(nm2p)=p.

    Corollary 3. Let p be an odd prime with p1mod4. Then we have

    1p1p1n=1p1m=0[1+p1a=1(a2+nˉa2p)]2e(nm2p)=(4α22p)p.

    Corollary 4. Let p be an odd prime with p1mod8. Then we have

    p1n=1(1+p1a=1(a2+nˉa2p))p1m=0e(nm4p)=p(1+B(1))p,

    where

    B(1)=p1m=0e(m4p).

    If we consider such a sequence Fk(p) as follows: Let p be a prime with p1mod8, χ4 be any fourth-order character modulo p. For any integer k0, we define the Fk(p) as

    Fk(p)=p1n=11Gk(n),

    we have

    Fk(p)=116α416α2p+4p1Fk4(p)(4p+2)16α416α2p+4p1Fk2(p)+4(4α22p)16α416α2p+4p1Fk1(p).

    Lemma 1. Let p be an odd prime with p1mod4. Then for any fourth-order character χ4modp, we have the identity

    τ2(χ4)+τ2(¯χ4)=2pα,

    where

    τ(χ4)=p1a=1χ4(a)e(ap)

    denotes the classical Gauss sums, e(y)=e2πiy,i2=1, and α is the same as in the Theorem 1.

    Proof. See Lemma 2.2 in [9].

    Lemma 2. Let p be an odd prime. Then for any non-principal character ψ modulo p, we have the identity

    τ(ψ2)=ψ2(2)τ(χ2)τ(ψ)τ(ψχ2),

    where χ2=(p) denotes the Legendre's symbol modulo p.

    Proof. See Lemma 2 in [12].

    Lemma 3. Let p be a prime with p1mod4, then for any integer n with (n,p)=1 and fourth-order character χ4modp, we have the identity

    p1a=1(a2+nˉa2p)=1χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)).

    Proof. For any integer a with (a,p)=1, we have the identity

    1+χ4(a)+χ2(a)+¯χ4(a)=4,

    if a satisfies ab4modp for some integer b with (b,p)=1 and

    1+χ4(a)+χ2(a)+¯χ4(a)=0,

    otherwise. So from these and the properties of Gauss sums we have

    p1a=1(a2+nˉa2p)=p1a=1(a2p)(a4+np)=p1a=1χ2(a4)χ2(a4+n)=p1a=1(1+χ4(a)+χ2(a)+¯χ4(a))χ2(a)χ2(a+n)=p1a=1(1+χ4(na)+χ2(na)+¯χ4(na))χ2(na)χ2(na+n)=p1a=1χ2(a)χ2(a+1)+p1a=1χ4(na)χ2(a)χ2(a+1) (2)
    +p1a=1χ2(na)χ2(a)χ2(a+1)+p1a=1¯χ4(na)χ2(a)χ2(a+1)=p1a=1χ2(1+ˉa)+p1a=1χ4(na)χ2(a)χ2(a+1)+p1a=1χ2(n)χ2(a+1)+p1a=1¯χ4(na)χ2(a)χ2(a+1).

    Noting that for any non-principal character χ,

    p1a=1χ(a)=0

    and

    p1a=1χ(a)χ(a+1)=1τ(ˉχ)p1b=1p1a=1ˉχ(b)χ(a)e(b(a+1)p).

    Then we have

    p1a=1χ2(1+ˉa)=1,p1a=1χ2(a+1)=1,
    p1a=1χ4(a)χ2(a)χ2(a+1)=1τ(χ2)p1b=1p1a=1χ2(b)χ4(a)χ2(a)e(b(a+1)p)=1τ(χ2)p1b=1¯χ4(b)e(bp)p1a=1χ4(ab)χ2(ab)e(abp) (3)
    =1τ(χ2)τ(¯χ4)τ(χ4χ2).

    For any non-principal character ψ, from Lemma 2 we have

    τ(ψ2)=ψ2(2)τ(χ2)τ(ψ)τ(ψχ2). (4)

    Taking ψ=χ4, note that

    τ(χ2)=p,  τ(χ4)τ(¯χ4)=χ4(1)p,

    from (3) and (4), we have

    p1a=1χ4(a)χ2(a)χ2(a+1)=¯χ42(2)τ(χ24)τ(χ2)τ(¯χ4)τ(χ2)τ(χ4)=χ2(2)τ(χ2)τ2(¯χ4)τ(χ4)τ(¯χ4)=χ2(2)pτ2(¯χ4)χ4(1)p (5)
    =χ2(2)τ2(¯χ4)χ4(1)p.

    Similarly, we also have

    p1a=1¯χ4(a)χ2(a)χ2(a+1)=χ2(2)τ2(χ4)χ4(1)p. (6)

    Consider the quadratic character modulo p, we have

    (2p)=χ2(2)={1,if p±1mod8;1,if p±3mod8. (7)

    And when p1mod8, we have χ4(1)=1; when p5mod8, we have χ4(1)=1. Combining (2) and (5)–(7) we can deduce that

    p1a=1(a2+nˉa2p)=1χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)).

    This prove Lemma 3.

    Lemma 4. Let p be an odd prime with p1mod4. Then for any integer k4 and n with (n,p)=1, we have the fourth-order linear recurrence formula

    Gk(n)=(4p+2)Gk2(n)+8(p2α2)Gk3(n)+[(4α22p)2(2p1)2]Gk4(n),

    where

    α=α(p)=12p1a=1(a3+ap)=p12a=1(a+ˉap),

    (p)=χ2 denotes the Legendre's symbol.

    Proof. For p1mod4, any integer n with (n,p)=1, and fourth-order character χ4 modulo p, we have the identity

    χ44(n)=¯χ44(n)=χ0(n),  χ24(n)=χ2(n),

    where χ0 denotes the principal character modulo p.

    According to Lemma 3,

    p1a=1(a2+nˉa2p)=1χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)),
    G(n)=1+p1a=1(a2+nˉa2p).  

    We have

    G(n)=χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)), (8)
    G2(n)=[χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4))]2=12χ2(n)1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4))+1p(χ2(n)τ4(¯χ4)+χ2(n)τ4(χ4)+2p2)=12χ2(n)1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4))+1p(χ2(n)(τ4(¯χ4)+τ4(χ4))+2p2).

    According to Lemma 1, we have

    (τ2(χ4)+τ2(¯χ4))2=τ4(¯χ4)+τ4(χ4)+2p2=4pα2.

    Therefore, we may immediately deduce

    G2(n)=12(χ2(n)(G(n)+χ2(n))+1p(χ2(n)(τ4(¯χ4)+τ4(χ4))+2p2)=12χ2(n)(G(n)+χ2(n)) (9)
    +1p[χ2(n)((τ2(¯χ4)+τ2(χ4))22p2)+2p2]=2p12χ2(n)G(n)+(4α22p)χ2(n),
    G3(n)=[χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4))]3=(2p12χ2(n)G(n)+(4α22p)χ2(n))G(n) (10)
    =(4α22p)χ2(n)G(n)+(2p+3)G(n)(4p2)χ2(n)2(4α22p)

    and

    [G2(n)(2p1)]2=[χ2(n)(4α22p)2χ2(n)G(n)]2,

    which implies that

    G4(n)=(4p+2)G2(n)+8(p2α2)G(n)+[(4α22p)2(2p1)2]. (11)

    So for any integer k4, from (8)–(11), we have the fourth-order linear recurrence formula

    Gk(n)=Gk4(n)G4(n)=(4p+2)Gk2(n)+8(p2α2)Gk3(n)+[(4α22p)2(2p1)2]Gk4(n).

    This proves Lemma 4.

    In this section, we will complete the proof of our theorem.

    Let p be any prime with p1mod4, then we have

    K0(p)=1p1p1n=1G0(n)=p1p1=1. (12)
    K1(p)=1p1p1n=1G1(n)=1p1p1n=1(χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))=0, (13)
    K2(p)=1p1p1n=1G2(n)=1p1p1n=1(χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))2=2p+1, (14)
    K3(p)=1p1p1n=1G3(n)=1p1p1n=1(χ2(n)+1p(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))3=3(4α22p). (15)

    It is clear that from Lemma 4, if k4, we have

    Kk(p)=1p1p1n=1Gk(n)=(4p+2)Kk2(p)8(2α2p)Kk3(p)+(16α416pα2+4p1)Kk4(p). (16)

    Now Theorem 1 follows (12)–(16). Obviously, using Theorem 1 to all negative integers, and that lead to Corollary 1.

    This completes the proofs of our all results.

    Some notes:

    Note 1: In our theorem, know n is an integer, and (n,p)=1. According to the properties of quadratic residual, χ2(n)=±1, χ4(n)=±1.

    Note 2: In our theorem, we only discussed the case p1mod8. If p3mod4, then the result is trivial. In fact, in this case, for any integer n with (n,p)=1, we have the identity

    G(n)=1+p1a=1(a2+nˉa2p)=1+p1a=1(a4p)(a4+np)=1+p1a=1(ap)(a+np)=1+p1a=1(a2+nap)=1+p1a=1(1+nˉap)=p1a=0(1+nap)=0.

    Thus, for all prime p with p3mod4 and k1, we have Kk(p)=0.

    The main result of this paper is Theorem 1. It gives an interesting computational formula for Kk(p) with p1mod4. That is, for any integer k, we have the identity

    Kk(p)=(4p+2)Kk2(p)8(2α2p)Kk3(p)+(16α416pα2+4p1)Kk4(p).

    Thus, the problems of calculating a linear recurrence formula of one kind special character sums modulo a prime are given.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are grateful to the anonymous referee for very helpful and detailed comments.

    This work is supported by the N.S.F. (11971381, 12371007) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (22JSY007).

    The authors declare no conflicts of interest.



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