Research article

Weight distributions of a class of skew cyclic codes over M2(F2)

  • Received: 25 February 2025 Revised: 08 May 2025 Accepted: 14 May 2025 Published: 20 May 2025
  • MSC : 11T71, 94B05, 94B15

  • Let M2(F2) be the ring of matrices of order 2×2 over finite field F2 and ωM2(F2) be a cubic primitive root of unity. For any even positive integer t, the weight distributions of the skew cyclic codes of length 3t with parity check polynomials xtωi,i=0,1,2 and (xtωj)(xtωk), 0j<k2 were determined.

    Citation: Zhen Du, Chuanze Niu. Weight distributions of a class of skew cyclic codes over M2(F2)[J]. AIMS Mathematics, 2025, 10(5): 11435-11443. doi: 10.3934/math.2025520

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  • Let M2(F2) be the ring of matrices of order 2×2 over finite field F2 and ωM2(F2) be a cubic primitive root of unity. For any even positive integer t, the weight distributions of the skew cyclic codes of length 3t with parity check polynomials xtωi,i=0,1,2 and (xtωj)(xtωk), 0j<k2 were determined.



    Nonlinear equations arise always in electroanalytical chemistry with complex and esoteric nonlinear terms[1,2], though there are some advanced analytical methods to deal with nonlinear problems, for examples, the Gamma function method[3], Fourier spectral method[4], the reproducing kernel method[5], the perturbation method[6], the homotopy perturbation method[7,8], He's frequency formulation[9,10,11] and the dimensional method[12], chemists are always eager to have a simple one step method for nonlinear equations. This paper introduces an ancient Chinese algorithm called as the Ying Buzu algorithm[13] to solve nonlinear differential equations.

    We first introduce the Taylor series method[14]. Considering the nonlinear differential equation:

    d2udx2+F(u)=0. (0.1)

    The boundary conditions are

    dudx(a)=α, (0.2)
    u(b)=β. (0.3)

    If u(a) is known, we can use an infinite Taylor series to express the exact solution[14]. We assume that

    u(a)=c. (0.4)

    From (0.1), we have

    u(a)=F(u(a))=F(c),
    u(a)=F(c)uu(a)=αF(c)u.

    Other higher order derivatives can be obtained with ease, and its Taylor series solution is

    u(x)=u(a)+(xa)u(a)+12!(xa)2u(a)+13!(xa)3u(a)+...+1N!(xa)Nu(N)(a),

    the constant c can be determined by the boundary condition of (0.3).

    The Ying Buzu algorithm[15,16] was used to solve differential equations in 2006[13], it was further developed to He's frequency formulation for nonlinear oscillators[13,17,18,19,20,21,22,23] and Chun-Hui He's algorithm for numerical simulation[24].

    As c in (0.4) is unknown, according to the Ying Buzu algorithm[13,15,16], we can assume two initial guesses:

    u1(a)=c1,u2(a)=c2. (0.5)

    where c1 and c2 are given approximate values.

    Using the initial conditions given in (0.2) and (0.5), we can obtain the terminal values:

    u(b,c1)=β1,u(b,c2)=β2.

    According to the Ying Buzu algorithm[6,7,8,9,10,11,12], the initial guess can be updated as

    u(a)est=c3=c1(ββ2)c2(ββ1)(ββ2)(ββ1),

    and its terminal value can be calculated as

    u(b,c3)=β3.

    For a given small threshold, ε, |ββ3|ε, we obtain u(a)=c3 as an approximate solution.

    Here, we take Michaelis Menten dynamics as an example to solve the equation. Michaelis Menten reaction diffusion equation is considered as follows[25,26]:

    d2udx2u1+u=0. (0.6)

    The boundary conditions of it are as follows:

    dudx(0)=0,u(1)=1. (0.7)

    We assume

    u(0)=c.

    From (0.6), we have

    u(0)=c1+c,
    u(0)=0, (0.8)
    u(4)=c(1+c)3.

    The 2nd order Taylor series solution is

    u(x)=u(0)+u(0)1!x+u(0)2!x2=c+c2(1+c)x2.

    In view of the boundary condition of (0.7), we have

    u(1)=c+c2(1+c)=1, (0.9)

    solving c from (0.9) results in

    c=0.7808.

    So we obtain the following approximate solution

    u(x)=0.7808+0.2192x2.

    Similarly the fourth order Taylor series solution is

    u(x)=c+c2!(1+c)x2+c4!(1+c)3x4.

    Incorporating the boundary condition, u(1)=1, we have

    c+c2!(1+c)+c4!(1+c)3=1. (0.10)

    We use the Ying Buzu algorithm to solve c, and write (0.10) in the form

    R(c)=c+c2(1+c)+c24(1+c)31.

    Assume the two initial solutions are

    c1=0.8,c2=0.5.

    We obtain the following residuals

    R1(0.8)=0.0279,R2(0.5)=0.3271.

    By the Ying Buzu algorithm, c can be calculated as

    c=R2c1R1c2R2R1=0.0279×0.5+0.3271×0.80.0279+0.3271=0.7764.

    The exact solution of (0.10) is

    c=0.7758.

    The 4th order Taylor series solution is

    u(x)=0.7758+0.2192x2+0.0057x4.

    Figure 1 shows the Taylor series solutions, which approximately meet the requirement of the boundary condition at x=1.

    Figure 1.  Taylor series solution.

    Now we use the Ying Buzu algorithm by choosing two initial guesses

    u1(0)=0.5,u2(0)=1,

    which lead to u1=0.6726 and u2=1.2550, respectively, see Figure 2 (a) and (b).

    Figure 2.  The shooting processes with different initial guesses.

    It is obvious that the terminal value at x=1 deviates from u(1)=1 for each guess, according to the Ying Buzu algorithm, the initial guess can be updated as

    u3(0)=0.5×(11.2550)1×(10.6726)(11.2550)(10.6726)=0.7810. (0.11)

    The shooting process using (0.11) results in

    u3(1)=1.0058,

    which deviates the exact value of u(1)=1 with a relative error of 0.5%, see Figure 3.

    Figure 3.  The shooting processes with an updated initial guess of u(0)=0.7810..

    We can continue the iteration process to obtain a higher accuracy by using two following two guesses u1(0)=0.5, u3(0)=0.7810:

    u4(0)=0.5×(11.0058)0.7810×(10.6726)(11.0058)(10.6726)=0.7761.

    Using this updated initial value, the shooting process leads to the result

    u(1)=1.0001,

    so the approximate u(0)=0.7761 has only a relative error of 0.01%.

    The above solution process couples the numerical method, and the ancient method can also be solved independently.

    We assume that solution is

    u(x)=c+(1c)x2. (0.12)

    Equation (0.12) meets all boundary conditions.

    The residual equation is

    R(x)=d2udx2u1+u.

    It is easy to find that

    R(0)=2(1c)c1+c.

    We choose two guesses:

    c1=0.5,c2=1.

    We obtain the following residuals

    R1(0)=2(10.5)0.51+0.5=23,
    R2(0)=2(11)11+1=12.

    The Ying Buzu algorithm leads to the updated result:

    c=c2R1(0)c1R2(0)R1(0)R2(0)=23×1+12×0.523+12=0.7857.

    The relative error is 1.2%, and the process can continue if a higher accuracy is still needed.

    The ancient Chinese algorithm provides a simple and straightforward tool to two-point boundary value problems arising in chemistry, and it can be used for fast insight into the solution property of a complex problem.

    The authors declare that they have no conflicts of interest to this work.



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