
Let M2(F2) be the ring of matrices of order 2×2 over finite field F2 and ω∈M2(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 xt−ωi,i=0,1,2 and (xt−ωj)(xt−ωk), 0≤j<k≤2 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 ω∈M2(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 xt−ωi,i=0,1,2 and (xt−ωj)(xt−ωk), 0≤j<k≤2 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)=−α∂F(c)∂u. |
Other higher order derivatives can be obtained with ease, and its Taylor series solution is
u(x)=u(a)+(x−a)u′(a)+12!(x−a)2u″(a)+13!(x−a)3u‴(a)+...+1N!(x−a)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(β−β2)−c2(β−β1)(β−β2)−(β−β1), |
and its terminal value can be calculated as
u(b,c3)=β3. |
For a given small threshold, ε, |β−β3|≤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]:
d2udx2−u1+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)3−1. |
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=R2c1−R1c2R2−R1=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.
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).
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×(1−1.2550)−1×(1−0.6726)(1−1.2550)−(1−0.6726)=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.
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×(1−1.0058)−0.7810×(1−0.6726)(1−1.0058)−(1−0.6726)=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+(1−c)x2. | (0.12) |
Equation (0.12) meets all boundary conditions.
The residual equation is
R(x)=d2udx2−u1+u. |
It is easy to find that
R(0)=2(1−c)−c1+c. |
We choose two guesses:
c1=0.5,c2=1. |
We obtain the following residuals
R1(0)=2(1−0.5)−0.51+0.5=23, |
R2(0)=2(1−1)−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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