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Research article

Improved sliding mode control for induction motor based on twisting algorithm


  • Received: 18 October 2024 Revised: 06 December 2024 Accepted: 06 January 2025 Published: 14 January 2025
  • In this paper, the output feedback tracking issue of induction motors is resolved by applying the sliding mode approach. We designed and implemented two robust sliding mode (SM) techniques to achieve high-performance control of induction motor drive; the second-order sliding mode (SOSM) approach using the twisting algorithm was compared with the classical sliding mode control. The method of decoupling electromagnetic torque and rotor flux for the induction motor was derived from the rotor field orientation control in the synchronous reference frame. The objective of the proposed methods is to control the rotor speed and the square of the rotor flux separately, in order to obtain robust control against disturbances and parametric uncertainties, and at the same time minimize the chattering phenomenon—the most significant drawback in the actual implementation of this technique. The stability of the proposed first-order sliding mode control was confirmed using Lyapunov stability theory. The availability and effectiveness of the proposed techniques were demonstrated through experimental results. The comparison between the results of the two proposed methods shows that the second-order sliding mode control using the twisting algorithm not only guarantees the same robustness and dynamic performances of traditional first-order sliding mode control but also achieves the reduction of the chattering phenomenon.

    Citation: Hadda Benderradji, Samira Benaicha, Larbi Chrifi Alaoui. Improved sliding mode control for induction motor based on twisting algorithm[J]. AIMS Electronics and Electrical Engineering, 2025, 9(1): 81-98. doi: 10.3934/electreng.2025005

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  • In this paper, the output feedback tracking issue of induction motors is resolved by applying the sliding mode approach. We designed and implemented two robust sliding mode (SM) techniques to achieve high-performance control of induction motor drive; the second-order sliding mode (SOSM) approach using the twisting algorithm was compared with the classical sliding mode control. The method of decoupling electromagnetic torque and rotor flux for the induction motor was derived from the rotor field orientation control in the synchronous reference frame. The objective of the proposed methods is to control the rotor speed and the square of the rotor flux separately, in order to obtain robust control against disturbances and parametric uncertainties, and at the same time minimize the chattering phenomenon—the most significant drawback in the actual implementation of this technique. The stability of the proposed first-order sliding mode control was confirmed using Lyapunov stability theory. The availability and effectiveness of the proposed techniques were demonstrated through experimental results. The comparison between the results of the two proposed methods shows that the second-order sliding mode control using the twisting algorithm not only guarantees the same robustness and dynamic performances of traditional first-order sliding mode control but also achieves the reduction of the chattering phenomenon.



    Diophantine equation is a classical problem in number theory. Let [α] denote the integer part of the real number α and N be a sufficiently large integer. In 1933, Segal [27,28] firstly studied additive problems with non-integer degrees, and proved that there exists a k0(c)>0 such that the Diophantine equation

    [xc1]+[xc2]++[xck]=N (1.1)

    is solvable for k>k0(c), where c>1 is not an integer. Later, Deshouillers [5] improved Segal's bound of k0(c) to 6c3(logc+14) with c>12. Further Arkhilov and Zhitkov [1] refined Deshouillers's result to 22c2(logc+4) with c>12. Afterwards, many results of various Diophantine equations were established (e.g., see [7,10,14,16,17,18,19,21,25,41,42]). In particular, Laporta [17] in 1999 showed that the equation

    [pc1]+[pc2]=N (1.2)

    is solvable in primes p1, p2 provided that 1<c<1716 and N is sufficiently large. Recently, the range of c in (1.2) was enlarged to 1<c<1411 by Zhu [40]. Kumchev [15] showed that the equation

    [mc]+[pc]=N (1.3)

    is solved for almost all N provided that 1<c<1615, where m is an integer and p is a prime. Afterwards, the range of c in (1.3) was enlarged to 1<c<1711 by Balanzario, Garaev and Zuazua [3].

    In 1995, Laporta and Tolev [18] considered the equation

    [pc1]+[pc2]+[pc3]=N (1.4)

    with prime variables p1,p2,p3. Denote the weighted number of solutions of Eq (1.4) by

    R(N)=[pc1]+[pc2]+[pc3]=N(logp1)(logp2)(logp3). (1.5)

    They established the following asymptotic formula

    R(N)=Γ3(1+1c)Γ(3c)N3c1+O(N3c1exp(log13δN))

    for any 0<δ<13 and 1<c<1716. Afterwards, the range of c was enlarged to 1<c<1211 by Kumchev and Nedeva [16], to 1<c<258235 by Zhai and Cao [39], and to 1<c<137119 by Cai [4].

    In this paper, we first show a more general result related to (1.5) by proving the following theorem.

    Theorem 1.1. Let N be a sufficiently large integer. Then for 1<c<3+3κλ3κ+2, we have

    R(N)=Γ3(1+1c)Γ(3c)N3c1+O(N3c1exp(log13δN)) (1.6)

    for any 0<δ<13, where (κ,λ) is an exponent pair, and the implied constant in the Osymbol depends only on c.

    Choosing (κ,λ)=BA2BABABABAB(0,1)=(81242,132242) in Theorem 1.1, we can immediately get the following corollary, which further improves the result of Cai [4].

    Corollary 1.2. Under the notations of Theorem 1.1, for 1<c<837727 the asymptotic formula (1.6) follows.

    It is easy to verify that the range of c in Corollary 1.2 is larger than one of Cai's result. Our improvement mainly derives from more accurate estimates of exponential sums by combining Van der Corput's method, exponent pairs and some elementary methods. Also the estimates of exponential sums has lots of applications in problems including automorphic forms (e.g., see [8,11,12,20,22,24,29,30,31,32,33,34,35,36,37,38]).

    Notation. Throughout the paper, N always denotes a sufficiently large integer. The letter p, with or without subscripts, is always reserved for primes. Let ε(0,1010(3+3κλ3κ+2c)). We denote by {x} and x the fraction part of x and the distance from x to the nearest integer, respectively. Let 1<c<3+3κλ3κ+2 and

    P=N1c,  τ=P1cε,  e(x)=e2πix,  S(α)=pP(logp)e(α[pc]).

    To prove Theorem 1.1, we need the following lemmas.

    Lemma 2.1 ([9,Lemma 5]). Suppose that zn is a sequence of complex numbers, then we have

    |Nn2Nzn|2(1+NQ)Qq=0(1qQ)Re(Nn2Nq¯znzn+q),

    where Re(t) and ¯t denote the real part and the conjugate of the complex number t, respectively.

    Lemma 2.2. Suppose that |x|>0 and c>1. Then for any exponent pair (κ,λ) and Ma<b2M, we have

    anbe(xnc)(|x|Mc)κMλκ+M1c|x|.

    Proof. We can get this lemma from [6,(3.3.4)].

    Lemma 2.3 ([2,Lemma 12]). Suppose that t is not an integer and H3. Then for any α(0,1), we have

    e(α{t})=|h|Hch(α)e(ht)+O(min(1,1Ht)),

    where

    ch(α)=1e(α)2πi(h+α).

    Lemma 2.4 ([9,Lemma 3]). Suppose that 3<U<V<Z<X, and {Z}=12, X64Z2U, Z4U2, V332X. Further suppose that F(n) is a complex valued function such that |F(n)|1. Then the sum

    Xn2XΛ(n)F(n)

    can be decomposed into O(log10X) sums, each of which either of type {I}:

    Mm2Ma(m)Nn2NF(mn)

    with N>Z, where a(m)mε and XMNX, or of type {II}:

    Mm2Ma(m)Nn2Nb(n)F(mn)

    with UMV, where a(m)mε,b(n)nε and XMNX.

    Lemma 2.5. Let f(t) be a real value function and continuous differentiable at least three times on [a,b](1a<b2a), |f(x)|Δ>0, then we have

    a<nbe(f(n))aΔ16+Δ13.

    Moreover, if 0<c1λ1c2λ1, |f(x)|λ1a1, then we have

    a<nbe(f(n))a12λ121+λ11;

    if c2λ112, then we have

    a<nbe(f(n))λ11.

    Proof. The first result was proved by Sargos [26]. And the remaining two results were due to Jia [13].

    Lemma 2.6 ([23,Lemma 2]). Let M>0, N>0, um>0, υn>0, Am>0, Bn>0 (1mM,1nN). Let also Q1 and Q2 be given non-negative numbers, Q1Q2. Then there is one q such that Q1qQ2 and

    Mm=1Amqum+Nn=1BnqυnMm=1Nn=1(AυnmBumn)1um+υn+Mm=1AmQum1+Nn=1BnQυn2.

    Lemma 2.7 ([36,Lemma 5]). Let f(x), g(x) be algebraic functions in [a,b], |f(x)|1R, f(x)1RU, U1, |g(x)|G, |g(x)|GU1. [α,β] is the image of [a,b] under the mapping y=f(x). nu is the solution of f(n)=u.

    bu={1,α<u<β,12,u=αNoru=βN.

    Then we have

    a<nbg(n)e(f(n))=α<uβbug(nu)|f(nu)|e(f(nu)unu+18)+O(Glog(βα+2)+G(ba+R)u1)+O(Gmin(R,1α)+Gmin(R,1β)).

    Lemma 2.8 ([13,Lemma 3]). Suppose that xN, f(x)P, and f(x)Δ. Then we have

    nNmin(D,1f(n))(P+1)(D+Δ1)log(2+Δ1).

    Lemma 2.9. For 0<α<1 and any exponent pair (κ,λ), we have

    T(α,X)=X<n2Xe(α[nc])Xκc+λ1+κlogX+XαXc.

    Proof. Throughout the proof of this lemma, we write H=Xκc+1λ+κ1+κ for convenience. Using Lemma 2.3 we can get

    T(α,X)=|h|Hch(α)X<n2Xe((h+α)nc)+O((logX)X<n2Xmin(1,1H||nc||)).

    Then by the expansion

    min(1,1H||θ||)=h=ahe(hθ),

    where

    |ah|=min(log2HH, 1|h|, Hh2),

    we have

    X<n2Xmin(1,1H||nc||)h=|ah||X<n2Xe(hnc)|Xlog2HH+1hH1h((hXc)κXλκ+XhXc)+hHHh2((hXc)κXλκ+XhXc)Xκc+λ1+κlogX,

    where we estimated the sum over n by Lemma 2.2.

    In a similar way, we have

    |h|Hch(α)X<n2Xe((h+α)nc)=c0(α)X<n2Xe(αnc)+1hHch(α)X<n2Xe((h+α)nc)Xκc+λ1+κlogX+XαXc.

    Then this lemma follows.

    Lemma 2.10 ([42,Lemma 2.1]). Suppose that f(n) is a real-valued function in the interval [N,N1], where 2N<N12N. If 0<c1λ1|f(n)|c2λ112, then we have

    N<nN1e(f(n))λ11.

    If |f(j)(n)|λ1Nj+1(j=1,2), then we have

    N<nN1e(f(n))λ11+N12λ121.

    If |f(j)(n)|λ1Nj+1(j=1,2,3,4,5,6), then we have

    N<nN1e(f(n))λ11+Nλλκ1,

    where (κ,λ) is any exponent pair.

    Lemma 2.11 ([9,Lemma 6]). Suppose that 0<a<b2a and R is an open convex set in C containing the real segment [a,b]. Suppose further that f(z) is analytic on R. f(x) is real for real xR. f(z)M for zR. There is a constant k>0 such that f(x)kM for all real xR. Let f(b)=α and  f(a)=β, and define xυ for each integer υ in the range α<υ<β by f(xυ)=υ. Then we have

    a<nbe(f(n))=e(18)α<υβ|f(xυ)|12e(f(xυ)υxυ)+O(M12+log(2+M(ba))).

    Lemma 3.1. Let P56XP, H=X1(1+2κ)c+λ2+2κ and ch(α) denote complex numbers such that ch(α)(1+|h|)1. Then uniformly for α(τ,1τ), we have

    SI=|h|Hch(α)Mm2Ma(m)Nn2Ne((h+α)(mn)c)X(1+2κ)c+λ2+2κ+2ε (3.1)

    for any a(m)mε, where (κ,λ) is any exponent pair, XMNX and MY with Y=min{X1,X2,X3,X4,X5,X6,X7,X8},

    X1=X152(1+2κ)c+λ2+2κc2112, X2=X5211(1+2κ)c+λ2+2κ4c11811,X3=X318(1+2κ)c+λ2+2κc8238, X4=X2(1+2κ)c+2λ1+κ3,X5=X4(1+2κ)c+4λ1+κ467,X6=X167(1+2κ)c+λ1+κ257,X7=X203(1+2κ)c+λ1+κ343,X8=X73(1+2κ)c+λ1+κ113.

    Proof. It is easy to deduce that

    SIMεhHKh,

    where Kh=mM|nNe((α+h)(mn)c)|. According to Hölder's inequality, we have

    K4hM3mM|nNe((α+h)(mn)c)|4. (3.2)

    Let zn=zn(m,α)=(α+h)(mn)c. Suppose that Q, J are two positive integers such that 1QNlog1X, 1JNlog1X. For the inner sum in (3.2), applying Lemma 2.1 twice, we can get

    K4hX4Q2+X4J+X3JQJj=1Qq=1|Eq,j|, (3.3)

    where

    Eq,j=mMN<n2Nqje(znzn+q+zn+q+jzn+j). (3.4)

    Let Δ(nc;q,j)=(n+q+j)c(n+q)c(n+j)c+nc, G(m,n)=(α+h)mcΔ(nc;q,j). Then znzn+qzn+j+zn+q+j=G(m,n). Thus we have

    Eq,j=mne(G(m,n)). (3.5)

    For any t1,0, we have

    Δ(nt;q,j)=t(t1)qjnt2+O(Nt3qj(q+j)), (3.6)

    then

    Gn=c(c1)(c2)(α+h)qjmcnc3(1+O(q+jN))

    and

    2Gn2=c(c1)(c2)(c3)(α+h)qjmcnc4(1+O(q+jN)). (3.7)

    If c(c1)(c2)(α+h)qjMcNc31100, by Lemma 2.5 we have

    mne(G(m,n))MN3((α+h)qjMcNc)1.

    From now we always suppose that c(c1)(c2)(α+h)qjMcNc31100. By Lemma 2.7 we have

    N<n2Njqe(G(m,n))=e(18)α<υ<β|2Gn2(m,nυ)|12e(G(m,nυ)υnυ)+R(m,q,j),

    where

    Gn(m,nυ)=υ,β=Gn(m,N),α=Gn(m,2Nqj),R=N4[(h+α)qjXc]1,υ(h+α)qjMcNc3,R(m,q,j)=O(logX+RN1+min(R,max(1α,1β))). (3.8)

    By Lemma 2.8, the contribution of R(m,q,j) to E(q,j) is

    MlogX+MRN1+mMmin(R,1α)+mMmin(R,1β)MlogX+X3c[(h+α)qjM2]1+[(h+α)qj]12MXc21logX. (3.9)

    Then we only need to deal with the following exponential sum

    mMα<υ<β|2Gn2(m,nυ)|12e(G(m,nυ)υnυ)=υmIυ|2Gn2(m,nυ)|12e(G(m,nυ)υnυ),

    where Iυ is a subinterval of [M,2M]. For a fixed υ, we define Δλ=Δ(nλυ;q,j), where λ is an arbitrary real number. We take the derivative of m in (3.8) and get

    nυ=cΔc1(c1)mΔc2. (3.10)

    It follows from (3.7) that

    ddm(2Gn2(m,nυ))=c2(h+α)mc1Δc2((c1)Δ2c2(c2)Δc1Δc3).

    Recalling (3.6), we can get

    ddm(2Gn2(m,nυ))=c2(c1)(c2)(c3)(h+α)qjmc1nc4υ(1+O(q+jN)).

    Thus for m, |2Gn2(m,nυ)|12 is monotonic. Let g(m)=G(m,nυ(m))υnυ(m). Then we have

    g(m)=c(α+h)mc1Δc,g(m)=c(α+h)c1(c1)2ΔcΔc2c2Δ2c1m2cΔc2=c(α+h)(c1)g1(m)g2(m)g0(m),g(m)=c(α+h)(c1)(g1g2)g0g0(g1g2)g20,

    where

    g1=((c1)2cΔc1Δc2+(c1)2(c2)ΔcΔc3)nυ(m),g2=2c2(c1)Δc1Δc2nυ(m),g0=(2c)m1c(c1)Δc2((c1)Δ2c2+cΔc1Δc3).

    From the above formulas we can obtain

    g(m)(h+α)qjM1Xc2.

    Using Lemma 2.5 and partial summation we can get

    mMυ|2Gn2(m,nυ)|12e(G(m,nυ)υnυ)(M((h+α)qjM1Xc2)16+((h+α)qjM1Xc2)13)×(h+α)qjMcNc3((h+α)qjMcNc4)12((h+α)qj)23M116X2c343+((h+α)qj)16M43Xc613. (3.11)

    By (3.5), (3.9) and (3.11), we have

    Eq,jlog1XM+((h+α)qjM2)1X3c+((h+α)qj)12MXc21+((h+α)qj)23M116X2c343+((h+α)qj)16M43Xc613. (3.12)

    Inserting (3.12) into (3.3), we obtain

    K4hlog1XX4Q2+X4J+MX3+((h+α)QJM2)1X6c+((h+α)QJ)12MXc2+2+((h+α)QJ)23M116X2c3+53+((h+α)QJ)16M43Xc6+83.

    Then choosing optimal J[0,Nlog1X] and Q[0,Nlog1X] and using Lemma 6 twice we can get

    Khlog3XB(h),

    where

    B(h)=X56+(α+h)114M17Xc14+57+(α+h)112M1148Xc12+1724+(α+h)130M415Xc30+1115+X34M14+(α+h)14X1c4+X2328M18+X2532M732+X1720M340+X1114M314.

    Recalling the definitions of H and Y, we have

    SIlog3XMεHB(H)X(1+2κ)c+λ2+2κ+2ε,

    and Lemma 3.1 is proved.

    Lemma 3.2. Let P56XP, H=X1(1+2κ)c+λ2+2κ, F=(h+α)Xcand ch(α) denote complex numbers such that ch(α)(1+|h|)1. Then uniformly for α(τ,1τ), we have

    SII=|h|Hch(α)Mm2Ma(m)Nn2Nb(n)e((h+α)(mn)c)X(1+2κ)c+λ2+2κ+2ε, (3.13)

    for any a(m)mε, b(n)nε, where (κ,λ) is any exponent pair, XMNX and

    max{X3(1+2κ)c+λ1+κF1,X42(1+2κ)c+2λ1+κ,X26533(1+2κ)c+λ1+κF539}MX(1+2κ)c+λ1+κ1.

    Proof. Taking Q=[X2(1+2κ)c+λ1+κlog1X], then we have Q=o(N). By Cauchy's inequality and Lemma 2.1, we have

    |SII|2mM|a(m)|2mM|nNb(n)e(f(mn))|2M2N2log2A+2BXQ+MNlog2AXQQq=1Eq, (3.14)

    where Eq=nN|b(n+q)b(n)||mMe(G(mn))| and G(m,n)=G(m,n,q)=(h+α)mcΔ(n,q;c), Δ(n,q;c)=(n+q)cnc.

    If |Gm|103Mq2, by Lemma 2.10 we have

    EqnN|b(n+q)b(n)|(MNFq+(FqMN)12M12)nN(|b(n+q)|2+|b(n)|2)(MNFq+Mq)MNqlog2BX (3.15)

    noting that MXF.

    Now we suppose |G/m|>103Mq2. By Lemma 11 we get

    mMe(G(m,n))MN1/2(Fq)1/2|r1(n)rr2(n)φ(n,r)e(s(r,n))|+logX+MN1/2(Fq)1/2,

    where s(r,n)=G(M(r,n),n)rm(r,n), φ(r,n)=(Fq)12MN12|2G(m(r,n),n)m2|12 and

    r1(n)=Gm(M,n),  r2(n)=Gm(2M,n).

    Thus we have

    EqMN1/2(Fq)1/2nN|b(n+q)b(n)||r1(n)<rr2(n)φ(n,r)e(s(r,n))|+Nlog2B+1X+MN3/2(Fq)1/2log2BX. (3.16)

    So it suffices to bound the sum

    Σ1=nN|b(n+q)b(n)||r1(n)<rr2(n)φ(n,r)e(s(r,n))|.

    Let T=[Fq3/M2N] and R=Fq/MN. By Cauchy's inequality and Lemma 2.1 again we get

    Σ12nN|b(n+q)b(n)|2nN|r1(n)<rr2(n)φ(n,r)e(s(r,n))|2N2R2log4BXT+NRlog4BXTΣ2, (3.17)

    where

    Σ2=Tt=1|nNr1(n)<rr2(n)tφ(n,r+t)φ(n,r)e(s(r+t)s(r,n))|

    and where we used the estimate

    nN|b(n+q)b(n)|2nN(|b(n+q)|4+|b(n)|4)Nlog4BX.

    It is easy to check that 10<T=o(R).

    Recall that s(r,n)=G(m(r,n),n)rm(r,n), where m(r,n) denotes the solution of

    Gm(m,n)=r.

    It is easy to deduce that

    sr(r,n)=Gmmrm(r,n)rmr=m(r,n).

    So we can obtain

    H(n):=Hr,t,q(n)=s(r+t,n)s(r,n)=r+trsu(u,n)du=r+trm(u,n)du,

    which implies that |H(j)|tMNj, (j=0,1,2,3,4,5,6). Denote by I(r,t) the interval N<n2N, r1(n)<nr2(n)t. Then we have

    Σ2Tt=1rR|nI(r,t)φ(n,r+t)φ(n,r)e(s(r+t,n)s(r,n))|.

    Thus using partial summation, we get

    Σ2Tt=1rR(tMN)κNλRMκNλκT1+κNR (3.18)

    with the exponent pair (κ,λ), if we note that MX26533(1+2κ)c+λ1+κF539. From (3.15)–(3.18) we get that for any 1qQ,

    EqMNlog2B+1Xq+Nlog2B+1X+MN32(Fq)12log2BX. (3.19)

    Now this lemma follows from inserting (3.19) into (3.14).

    Lemma 3.3. For τα1τ, we have

    S(α)P(1+2κ)c+λ2+2κ+4ε,

    where (κ,λ) is any exponent pair.

    Proof. Throughout the proof of this lemma, we write H=X1(1+2κ)c+λ2+2κ for convenience. By a dissection argument we only need to prove that

    X<n2XΛ(n)e(α[nc])X(1+2κ)c+λ2+2κ+3ε (3.20)

    holds for P56XP and τα1τ. According to Lemma 2.3, we have

    X<n2XΛ(n)e(α[nc])=|h|Hch(α)X<n2XΛ(n)e((h+α)nc)+O((logX)X<n2Xmin(1,1Hnc)). (3.21)

    By the expansion

    min(1,1Hnc)=h=ahe(hnc),

    where

    |ah|min(log2HH,1|h|,Hh2),

    we get

    X<n2Xmin(1,1Hnc)h=ah|X<n2Xe(hnc)|Xlog2HH+1hH1h((hXc)12+XhXc)+hHHh2((hXc)12+XhXc)X(1+2κ)c+λ2+2κlogX, (3.22)

    where we estimated the sum over n by Lemma 2.2 with the exponent pair (12,12).

    Let R=max{X3(1+2κ)c+λ1+κF1,X42(1+2κ)c+2λ1+κ,X26533(1+2κ)c+λ1+κF539}. Recall the definition of Y in Lemma 3.1. Let U=R, V=X(1+2κ)c+λ1+κ1, Z=[XY1]+12. By Lemma 4 with F(n)=e((h+α)nc), then we reduce the estimate of

    |h|Hch(α)X<n2XΛ(n)e((h+α)nc)

    to the estimates of sums of type I

    SI=|h|Hch(α)M<m2Ma(m)N<n2Ne((h+α)(mn)c), N>Z,

    and sums of type II

    SII=|h|Hch(α)M<m2Ma(m)N<n2Nb(n)e((h+α)(mn)c), U<M<V,

    where a(m)mε, b(n)nε, XMNX. By Lemma 3.1, we get

    SIX(1+2κ)c+λ2+2κ+2ε. (3.23)

    By Lemma 3.2, we get

    SIIX(1+2κ)c+λ2+2κ+3ε. (3.24)

    From (3.23) and (3.24) we can obtain

    |h|Hch(α)X<n2XΛ(n)e((h+α)nc)X(1+2κ)c+λ2+2κ+3ε. (3.25)

    Now (3.20) follows from (3.21), (3.22) and (3.25). Thus we complete the proof of this Lemma.

    It is easy to see that

    R(N)=1ττS3(α)e(αN)dα=ττS3(α)e(αN)dα+1ττS3(α)e(αN)dα=R1(N)+R2(N). (4.1)

    Following the argument of Laporta and Tolev [18,pages 928–929], we can get that

    R1(N)=Γ3(1+1c)Γ(3c)N3c1+O(N3c1exp(log13δN)) (4.2)

    for 1<c<32 and any 0<δ<13, where the implied constant in the Osymbol depends only on c.

    Let

    S(α,X)=X<p2Xe(α[pc])logp, T(α,X)=X<n2Xe(α[nc]).

    We can get

    R2(N)=1ττS3(α)e(αN)dα(logX)maxP56X0.5P|1ττS2(α)S(α,X)e(αN)dα|+P116log2P, (4.3)

    where we used

    1ττ|S2(α)|dα10|S2(α)|dαPlog2P. (4.4)

    Now, we start to estimate the absolute value on the right hand in (4.3). By Cauchy's inequality we have

    |1ττS2(α)S(α,X)e(αN)dα|=|X<p2X(logp)1ττS2(α)e(α[pc]αN)dα|X<p2X(logp)|1ττS2(α)e(α[pc]αN)dα|(logX)X<n2X|1ττS2(α)e(α[nc]αN)dα|X12(logX)(X<n2X|1ττS2(α)e(α[nc]αN)dα|2)12=X12(logX)(1ττ¯S2(β)e(βN)dβ1ττS2(α)T(αβ,X)e(αN)dα)12X12(logX)(1ττ|S(β)|2dβ1ττ|S(α)|2|T(αβ,X)|dα)12. (4.5)

    Then we have

    1ττ|S(α)|2|T(αβ,X)|dατ<α<1τ|αβ|Xc|S(α)|2|T(αβ,X)|dα+τ<α<1τ|αβ|>Xc|S(α)|2|T(αβ,X)|dα. (4.6)

    By Lemma 3.3, we have

    τ<α<1τ|αβ|Xc|S(α)|2|T(αβ,X)|dαXmaxα(τ,1τ)|S(α)|2|αβ|Xc1dαX1cP(1+2κ)c+λ1+κ+8ε, (4.7)

    where we used the trivial bound T(α,X)X. By Lemma 2.9, Lemma 3.3 and (4.4), we get

    τ<α<1τ|αβ|>Xc|S(α)|2|T(αβ,X)|dατ<α<1τ|αβ|>Xc|S(α)|2(Xκc+λ1+κlogX+X|αβ|Xc)dαXκc+λ1+κ(logX)1ττ|S(α)|2dα+maxα(τ,1τ)|S(α)|2|αβ|>XcX|αβ|XcdαXκc+λ1+κPlog3P+X1cP(1+2κ)c+λ1+κ+9ε. (4.8)

    Thus, combining (4.6)–(4.8) we obtain

    1ττ|S(α)|2|T(αβ,X)|dαXκc+λ1+κPlog3P+X1cP(1+2κ)c+λ1+κ+9ε. (4.9)

    By (4.3), (4.5) and (4.9), we can obtain

    R2(N)P3cε. (4.10)

    Now putting (4.1), (4.2) and (4.10) into together, we have

    R(N)=Γ3(1+1c)Γ(3c)N3c1+O(N3c1exp(log13δN))

    follows for any 0<δ<13, where the implied constant in the Osymbol depends only on c. Thus we complete the proof of Theorem 1.1.

    The authors would like to thank the referees for their many useful comments. This work is supported by National Natural Science Foundation of China (Grant Nos. 11801328 and 11771256).

    The authors declare no conflict of interest.



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