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Solvability of the matrix equation AX2=B with semi-tensor product

  • Received: 01 May 2020 Revised: 01 August 2020 Published: 19 October 2020
  • Primary: 15A06; Secondary: 15A03, 15A04

  • We investigate the solvability of the matrix equation AX2=B in which the multiplication is the semi-tensor product. Then compatible conditions on the matrices A and B are established in each case and necessary and sufficient condition for the solvability is discussed. In addition, concrete methods of solving the equation are provided.

    Citation: Jin Wang, Jun-E Feng, Hua-Lin Huang. Solvability of the matrix equation AX2=B with semi-tensor product[J]. Electronic Research Archive, 2021, 29(3): 2249-2267. doi: 10.3934/era.2020114

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  • We investigate the solvability of the matrix equation AX2=B in which the multiplication is the semi-tensor product. Then compatible conditions on the matrices A and B are established in each case and necessary and sufficient condition for the solvability is discussed. In addition, concrete methods of solving the equation are provided.



    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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