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Research article Special Issues

Periodic solution of a stage-structured predator-prey model incorporating prey refuge

  • Received: 29 January 2020 Accepted: 09 April 2020 Published: 22 April 2020
  • In this paper, we consider a delayed stage-structured predator-prey model incorporating prey refuge with Holling type Ⅱ functional response. It is assumed that prey can live in two different regions. One is the prey refuge and the other is the predatory region. Moreover, in real world application, we should consider the stage-structured model. It is assumed that the prey in the predatory region can divided by two stages: Mature predators and immature predators, and the immature predators have no ability to attack prey. Based on Mawhin's coincidence degree and novel estimation techniques for a priori bounds of unknown solutions to Lu = λNu, some sufficient conditions for the existence of periodic solution is obtained. Finally, an example demonstrate the validity of our main results.

    Citation: Weijie Lu, Yonghui Xia, Yuzhen Bai. Periodic solution of a stage-structured predator-prey model incorporating prey refuge[J]. Mathematical Biosciences and Engineering, 2020, 17(4): 3160-3174. doi: 10.3934/mbe.2020179

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  • In this paper, we consider a delayed stage-structured predator-prey model incorporating prey refuge with Holling type Ⅱ functional response. It is assumed that prey can live in two different regions. One is the prey refuge and the other is the predatory region. Moreover, in real world application, we should consider the stage-structured model. It is assumed that the prey in the predatory region can divided by two stages: Mature predators and immature predators, and the immature predators have no ability to attack prey. Based on Mawhin's coincidence degree and novel estimation techniques for a priori bounds of unknown solutions to Lu = λNu, some sufficient conditions for the existence of periodic solution is obtained. Finally, an example demonstrate the validity of our main results.



    Let p be a prime, f be a polynomial with k variable and Fp=Z/(p) be the finite field, where Z is the integer ring, and let

    N(f;p)=#{(x1,x2,,xk)Fkp|f(x1,x2,,xk)=0}.

    Many scholars studied the exact formula (including upper bound and lower bound) for N(f;p) for many years, it is one of the main topics in the finite field theory, the most elementary upper bounds was given as follows (see [14])

    N(f;p)pk1degf.

    Let ordp denote the p-adic additive valuation normalized such that ordpp=1. The famous Chevalley-Warning theorem shows that ordpN(f;p)>0 if n>degf. Let [x] denote the least integer x and e denote the extension degree of Fq/Fp. Ax (see [2]) showed that

    ordpN(f;q)e[ndegfdegf].

    In 1977, S. Chowla et al. (see [7]) investigated a problem about the number of solutions of a equation in finite field Fp as follow,

    x31+x32++x3k0,

    where p is a prime with p1mod 3 and xiFp, 1ik.

    Let Mk denotes the number of solutions of the above equation. They proved that

    M3=p2+d(p1),M4=p2+6(p2p),s=1Msxs=x1px+x2(p1)(2+dx)13px2pdx3,

    where d is uniquely determined by 4p=d2+27y2 and d1 mod 3.

    Myerson [12] extended the result in [2] to the field Fq and first studied the following equation over Fq,

    x31+x32++x3k0.

    Recently J. Zhao et al. (see [17]) investigated the following equations over field Fp,

       f1=x41+x42+x43,f2=x41+x42+x43+x44.

    And they give exact value of N(f1;p) and N(f2;p). For more general problem about this issue interested reader can see [6,9,10,11].

    In this paper, let A(k,p) denotes the number of solutions of the following equation in Fp,

    x61+x62++x6k0,

    where p is a prime with p1mod 3 and xiFp, 1ik, and for simplicity, in the rest of this paper, we assume there exists an integer z such that z32mod p, we use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid a lot of complicated calculations. We proved the following:

    Theorem 1. For any positive integer k1, we have the recurrence formula

    A(k+6,p)=5pA(k+4,p)+10dpA(k+3,p)+(46p2+5d2p+dp)A(k+2,p)+(2p2+120dp2+3d3p+d2p+dp)A(k+1,p)+(4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p)A(k,p)+pk+5pk+4(10dp+2d2)pk+364pk+5/2(429+121d+5d2)pk+22d2pk+3/2(3d3+130d2+12d+6)pk+1d4pk,

    with the initial condition

    A(1,p)=1,A(2,p)=4(p1)+p,A(3,p)=10d(p1)+p2,A(4,p)=56p(p1)+10d2(p1)+p3,A(5,p)=188dp(p1)+5d3(p1)+16dC(p)(p1)+p4,A(6,p)=p5+1400p2(p1)+(388d2+8d576)p(p1)+d2pd2,

    where d is uniquely determined by 4p=d2+27y2 and d1 mod 3, and C(p)=pa=1ep(a3).

    Remark. Our method is suitable to calculus the number of solutions of the following equation in Fp,

    xt1+xt2++xtk0,

    where p satisfied a certain congruence conditions, and t is any nature number.

    Our Theorem 2 can be deduced from Theorem 1 and the theory of the Difference equations.

    Theorem 2. Let ti (1ik) be the real root of the below equation with multiplicity si (1ik) respectively, and ρje±iwj (1jh) be the complex root of the below equation with multiplicity rj (1jh) respectively,

    x6=5px4+10dpx3+(46p2+5d2p+dp)x2+(2p2+120dp2+3d3p+d2p+dp)x+(4p5+2d2p4+64p7/2+381p3+2d2p5/2+(129d2+11d+6)p2+d4p).

    We have

    A(n,p)=ki=1sia=1Ciansiatni+hj=1rjb=1Djbnrjbρnjcosnwj+hj=1rjb=1Ejbnrjbρnjsinnwj,

    where Cia,Djb,Ejb, are determined by

    A(6,p)=ki=1sia=1Cia6siat6i+hj=1rjb=1Djb6rjbρ6jcos6wj+hj=1rjb=1Ejb6rjbρ6jsin6wj,A(5,p)=ki=1sia=1Cia5siat5i+hj=1rjb=1Djb5rjbρ5jcos5wj+hj=1rjb=1Ejb5rjbρ5jsin5wj,A(4,p)=ki=1sia=1Cia4siat4i+hj=1rjb=1Djb4rjbρ4jcos4wj+hj=1rjb=1Ejb4rjbρ4jsin4wj,A(3,p)=ki=1sia=1Cia3siat3i+hj=1rjb=1Djb3rjbρ3jcos3wj+hj=1rjb=1Ejb3rjbρ3jsin3wj,A(2,p)=ki=1sia=1Cia2siat2i+hj=1rjb=1Djb2rjbρ2jcos2wj+hj=1rjb=1Ejb2rjbρ2jsin2wj,A(1,p)=ki=1sia=1Ciati+hj=1rjb=1Djbρjcoswj+hj=1rjb=1Ejbρjsinwj. (1.1)

    Before we prove these lemmas, we give some notations, χ2 denotes the second-order character of Fp, χ denotes the third-order character of Fp, ψ denotes the sixth order character of Fp.

    ep(x)=e2πixp,τ(χ)=pm=1χ(m)ep(m),G(χ,m)=pa=1χ(a)ep(am).

    We call G(χ,m) the Gauss sum, and we have the following:

    G(χ,m)=τ(χ)¯χ(m),(m,p)=1. (2.1)

    And also we have

    |τ(χ)|=p, (2.2)

    where χ is a primitive character of Fp. And let G(m,6;p)=p1a=0ep(ma6). For the property of the exponential sum and the general Gauss sum, interested readers can see [1,4,5,8,13,15].

    Lemma 1. Let p be a prime with p1mod 3. Then for any third-order character χ of Fp, we have the identity

    τ3(χ)+τ3(¯χ)=dp,

    where d is uniquely determined by 4p=d2+27y2 and d1mod 3.

    Proof. For the proof of this lemma see [3].

    Lemma 2. Let χ be a third-order character of Fp with p1mod 3, and C(p)=τ(χ)+τ(¯χ), then C(p)=pa=1ep(a3).

    Proof.

    A=τ(χ)+τ(¯χ)=pa=1(1+χ(a)+¯χ(a))e(ap)=pa=1e(a3p).

    Lemma 3. Let p1mod 6, 2z3mod p for some z, and let χ be a third-order character of Fp, ψ be a sixth-order character of Fp, then we have the identity

    τ(ψ)=τ2(χ)p.

    Proof. This is Lemma 3 in [16].

    Lemma 4. As the definition above, we have the identity

    G(m,6;p)=pχ2(m)+¯x2pψ(m)+x2p¯ψ(m)+¯xχ(m)+x¯χ(m),

    where (m,p)=1 and x=τ(χ).

    Proof. Firstly we have the identity

    1+χ2(m)+χ(m)+¯χ(m)+ψ(m)+¯ψ(m)={6, if  ma6 mod p;0,otherwise.

    So we have

    G(m,6;p)=p1a=0(1+χ2(a)+χ(a)+¯χ(a)+ψ(a)+¯ψ(a))ep(ma)=G(χ2,m)+G(ψ,m)+G(¯ψ,m)+G(χ,m)+G(¯χ,m)

    By (2.1) and Lemma 3, we have

    G(m,6;p)=τ(χ2)χ2(m)+τ(¯ψ)ψ(m)+τ(ψ)¯ψ(m)+τ(¯χ)χ(m)+τ(χ)¯χ(m)=pχ2(m)+¯x2pψ(m)+x2p¯ψ(m)+¯xχ(m)+x¯χ(m). (2.3)

    By (2.3), we complete the proof of our lemma.

    Next we let,

    Gn(m,6;p)=an+bnχ2(m)+cnψ(m)+dn¯ψ(m)+enχ(m)+fn¯χ(m). (2.4)

    Then we have following Lemma 5.

    Lemma 5. Let an,bn,cn,dn,en,fn are defined as above, then we have that an,bn,cn,dn,en,fn are uniquely determined by n, where n1.

    Proof. By the orthogonality of characters of Fp, we have

    \begin{eqnarray} \sum\limits_{a = 1}^{p-1}\chi\left(a\right) = \left\{ \begin{array}{ll} p-1, & \text{ if}\ χ = χ_0; \\ 0, & \text{otherwise.}\end{array}\right. \end{eqnarray} (2.5)

    By (2.4) and (2.5) we have

    \begin{eqnarray} \sum\limits_{m = 1}^{p-1}G^n(m, 6;p)& = &(p-1)a_n+b_n\sum\limits_{m = 1}^{p-1}\chi_2\left(m\right)+c_n\sum\limits_{m = 1}^{p-1}\psi\left(m\right)+d_n\sum\limits_{m = 1}^{p-1}\overline{\psi}\left(m\right)\\ &&+e_n\sum\limits_{m = 1}^{p-1}\chi\left(m\right)+f_n\sum\limits_{m = 1}^{p-1}\overline{\chi}\left(m\right)\\ & = &(p-1)a_n. \end{eqnarray}

    So we have

    \begin{eqnarray} &&a_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}G^n(m, 6;p). \end{eqnarray} (2.6)

    By the same method, we have

    \begin{eqnarray} &&b_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\chi_2(m)G^n(m, 6;p), \\ &&c_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\overline{\psi}(m)G^n(m, 6;p), \\ &&d_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\psi(m)G^n(m, 6;p), \\ &&e_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\overline{\chi}(m)G^n(m, 6;p), \\ &&f_n = \frac{1}{p-1}\sum\limits_{m = 1}^{p-1}\chi(m)G^n(m, 6;p). \end{eqnarray}

    So now it is easy to see the conclusion of the lemma.

    Lemma 6. The sequences \{a_n\} , \{b_n\} , \{c_n\} , \{d_n\} , \{e_n\} , \{f_n\} are defined above, then they satisfied the following recurrence formulae ( n\geq0 ):

    \begin{eqnarray} &&a_{n+1} = \sqrt{p}b_n+\frac{\overline{x^2}}{\sqrt{p}}d_n+\frac{x^2}{\sqrt{p}}c_n+xe_n+\overline{x}f_n, \end{eqnarray} (2.7)
    \begin{eqnarray} &&b_{n+1} = \sqrt{p}a_n+\frac{\overline{x^2}}{\sqrt{p}}e_n+\frac{x^2}{\sqrt{p}}f_n+xd_n+\overline{x}c_n, \end{eqnarray} (2.8)
    \begin{eqnarray} &&c_{n+1} = \sqrt{p}f_n+\frac{\overline{x^2}}{\sqrt{p}}a_n+\frac{x^2}{\sqrt{p}}e_n+xb_n+\overline{x}d_n, \end{eqnarray} (2.9)
    \begin{eqnarray} &&d_{n+1} = \sqrt{p}e_n+\frac{\overline{x^2}}{\sqrt{p}}f_n+\frac{x^2}{\sqrt{p}}a_n+xc_n+\overline{x}b_n, \end{eqnarray} (2.10)
    \begin{eqnarray} &&e_{n+1} = \sqrt{p}d_n+\frac{\overline{x^2}}{\sqrt{p}}c_n+\frac{x^2}{\sqrt{p}}b_n+xf_n+\overline{x}a_n, \end{eqnarray} (2.11)
    \begin{eqnarray} &&f_{n+1} = \sqrt{p}c_n+\frac{\overline{x^2}}{\sqrt{p}}b_n+\frac{x^2}{\sqrt{p}}d_n+xa_n+\overline{x}e_n, \end{eqnarray} (2.12)

    with the initial condition

    a_0 = 1, b_0 = c_0 = d_0 = e_0 = f_0 = 0.

    Proof. We only prove (2.7), the rest can be proved in the same way. By Lemma 5, we know a_n is unique determined by n . We can compare the coefficient of the equation

    G^{n+1}(m, 6;p) = G^{n}(m, 6;p)G(m, 6;p).

    We have

    a_{n+1} = \sqrt{p}b_n+\frac{\overline{x^2}}{\sqrt{p}}d_n+\frac{x^2}{\sqrt{p}}c_n+xe_n+\overline{x}f_n.

    So we complete the proof of the lemma.

    Lemma 7. Let a_n is defined as above, then we have

    \begin{eqnarray} &&a_0 = 1, a_1 = 0, a_2 = 5p, a_3 = 10dp, a_4 = 56p^2+10d^2p, \\&&a_5 = 188dp^2+5d^3p+16dpC(p). \end{eqnarray}

    Proof. By Lemma 4 and after some elementary calculations we have

    \begin{eqnarray} G^2(m, 6;p)& = &5p+2dp^{1/2}\chi_2(m)+4p^{1/2}x\psi(m)+4p^{1/2}\overline{x}\overline{\psi}(m)\\ &&+(p^{-1}\overline{x^4}+3x^2)\chi(m)+(p^{-1}x^4+3\overline{x^2})\overline{\chi}(m), \\ G^3(m, 6;p)& = &10dp+(16p^{3/2}+dp^{1/2})\chi_2(m)+(15p\overline{x}+2dx^2+p^{-1}x^5)\chi(m)\\ &&+(15px+2d\overline{x^2}+p^{-1}\overline{x^5})\overline{\chi}(m)\\ &&+(4p^{-1/2}x^4+12p^{1/2}\overline{x^2}+2dp^{1/2}x)\psi(m)\\ &&+(4p^{-1/2}\overline{x^4}+12p^{1/2}x^2+2dp^{1/2}\overline{x})\overline{\psi}(m), \\ G^4(m, 6;p)& = &60p^2+9d^2p+dp+48dp^{3/2}\chi_2(m)\\ &&+(p^{-2}x^8+17\overline{x^4}+46px^2+16dp)\chi(m)\\ &&+(p^{-2}\overline{x^8}+17x^4+46p\overline{x^2}+16dp)\overline{\chi}(m)\\ &&+(56p^{3/2}x+4dp^{-1/2}x^4+12dp^{1/2}\overline{x^2}+8p^{-1/2}\overline{x^5})\psi(m)\\ &&+(56p^{3/2}\overline{x}+4dp^{-1/2}\overline{x^4}+12dp^{1/2}x^2+8p^{-1/2}x^5)\overline{\psi}(m), \\ G^5(m, 6;p)& = &188dp^2+5d^3p+16dpC(p)\\ &&+(52d^2p^{3/2}+208p^{5/2}+16dp^{1/2}(x^2+\overline{x^2}))\chi_2(m)\\ &&+(p^{-2/5}x^{10}+p^{-3/2}\overline{x^8}+4dp^{-1/2}\overline{x^5}+71p^{1/2}x^4\\&&+(46p^{3/2}+16p^{1/2})x^2\\ &&+(129p^{3/2}+10d^2p^{1/2})\overline{x^2}+60dp^{3/2}x+16dp^{3/2})\psi(m)\\ &&+(p^{-2/5}\overline{x^{10}}+p^{-3/2}x^8+4dp^{-1/2}x^5+71p^{1/2}\overline{x^4}\\&&+(46p^{3/2}+16p^{1/2})\overline{x^2}\\ &&+(129p^{3/2}+10d^2p^{1/2})x^2+60dp^{3/2}\overline{x}+16dp^{3/2})\overline{\psi}(m)\\ &&+(8p^{-1}\overline{x^7}+p^{-1}x^7+25x^5+52dpx^2\\&&+(28dp+46p^2)x+16d\overline{x^4}+112p^2\overline{x})\chi(m)\\ &&+(8p^{-1}x^7+p^{-1}\overline{x^7}+25\overline{x^5}+52dp\overline{x^2}\\&&+(28dp+46p^2)\overline{x}+16dx^4+112p^2x)\overline{\chi}(m), \end{eqnarray}

    and comparing the above formulae with (2.6), we have

    a_0 = 1, a_1 = 0, a_2 = 5p, a_3 = 10dp, a_4 = 60p^2+9d^2p+dp, a_5 = 188dp^2+5d^3p+16dpC(p).\\

    Lemma 8. Let a_n , b_n , c_n , d_n , e_n , f_n are defined as above, then we have

    \begin{eqnarray} a_6& = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2+(2p^2+120dp^2+3d^3p+d^2p+dp)a_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)a_0\\ b_6& = &5pb_4+10dpb_3+(46p^2+5d^2p+dp)b_2+(2p^2+120dp^2+3d^3p+d^2p+dp)b_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)b_0\\ c_6& = &5pc_4+10dpc_3+(46p^2+5d^2p+dp)c_2+(2p^2+120dp^2+3d^3p+d^2p+dp)c_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)c_0\\ d_6& = &5pd_4+10dpd_3+(46p^2+5d^2p+dp)d_2+(2p^2+120dp^2+3d^3p+d^2p+dp)d_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)d_0\\ e_6& = &5pe_4+10dpe_3+(46p^2+5d^2p+dp)e_2+(2p^2+120dp^2+3d^3p+d^2p+dp)e_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)e_0\\ f_6& = &5pf_4+10dpf_3+(46p^2+5d^2p+dp)f_2+(2p^2+120dp^2+3d^3p+d^2p+dp)f_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}+(129d^2+11d+6)p^2+d^4p)f_0 \end{eqnarray}

    Proof. We only proof the first formula, the rest can be proof in the same way. By Lemma 6, we have

    \begin{eqnarray} a_6& = &\sqrt{p}b_5+\frac{\overline{x^2}}{\sqrt{p}}d_5+\frac{x^2}{\sqrt{p}}c_5+xe_5+\overline{x}f_5\\ & = &5pa_4+2dp^{1/2}b_4+4p^{1/2}\overline{x}c_4+4p^{1/2}xd_4\\&&+(3\overline{x^2}+p^{-1}x^4)e_4+(3x^2+p^{-1}\overline{x^4})f_4\\ & = &5pa_4+10dpa_3+(d^2p^{1/2}+12p^{3/2})b_3+(2dp^{1/2}\overline{x}\\&&+8p^{1/2}x^2+p^{-1/2}\overline{x^4})c_3\\ &&+(2dp^{1/2}x+8p^{1/2}\overline{x^2}+p^{-1/2}x^4)d_3+(11px+\overline{x^2}+p^{-1}\overline{x^5})e_3\\ &&+(11p\overline{x}+x^2+p^{-1}x^5)f_3\\ & = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2+(25dp^{3/2}+2p^{3/2})b_2\\ &&+(p^{-3/2}\overline{x^7}+2p^{-1/2}x^5+p^{-1/2}\overline{x^4}+42p^{3/2}\overline{x}+2dp^{1/2}x^2\\&&+(d^2+1)p^{1/2}\overline{x})c_2\\ &&+(p^{-3/2}x^7+2p^{-1/2}\overline{x^5}+p^{-1/2}x^4+42p^{3/2}x+2dp^{1/2}\overline{x^2}\\&&+(d^2+1)p^{1/2}x)d_2\\ &&+(10x^4+(32p+d^2)\overline{x^2}+(4dp+p)x)e_2\\&&+(10\overline{x^4}+(32p+d^2)x^2+(4dp+p)\overline{x})f_2\\ & = &5pa_4+10dpa_3+(46p^2+5d^2p+dp)a_2\\&&+(2p^2+120dp^2+3d^3p+d^2p+dp)a_1\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)a_0. \end{eqnarray}

    So we complete the proof of this lemma.

    Lemma 9. Let a_n is defined as above, then for any integer n\geq0 , we have

    \begin{eqnarray} a_{n+6}& = &5pa_{n+4}+10dpa_{n+3}+(46p^2+5d^2p+dp)a_{n+2}\\&&+(2p^2+120dp^2+3d^3p+d^2p+dp)a_{n+1}\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)a_n. \end{eqnarray}

    Proof. By (2.4) and Lemma 8, we have

    \begin{eqnarray} G^{6}(m, 6;p)& = &5pG^{4}(m, 6;p)+10dpG^{3}(m, 6;p)+(46p^2+5d^2p+dp)G^{2}(m, 6;p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)G(m, 6;p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p). \end{eqnarray}

    We multiple G^{n}(m, 6;p) to the both side of the above formula, we have

    \begin{eqnarray} G^{n+6}(m, 6;p)& = &5pG^{n+4}(m, 6;p)+10dpG^{n+3}(m, 6;p)\\&&+(46p^2+5d^2p+dp)G^{n+2}(m, 6;p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)G^{n+1}(m, 6;p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)G^n(m, 6;p). \end{eqnarray}

    By Lemma 5, we can compare the coefficient of the above equation, we have

    \begin{eqnarray} a_{n+6}& = &5pa_{n+4}+10dpa_{n+3}+(46p^2+5d^2p+dp)a_{n+2}\\&&+(2p^2+120dp^2+3d^3p+d^2p+dp)a_{n+1}\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\&&+(129d^2+11d+6)p^2+d^4p)a_n. \end{eqnarray}

    In the formula below, we always let k\geq1 . By the following formula,

    \begin{eqnarray} \sum\limits_{a = 0}^{p-1}e_p\left(ma\right) = \left\{ \begin{array}{ll} p, & \text{ if }\ p\mid m; \\ 0, & \text{ otherwise, }\end{array}\right. \end{eqnarray}

    we have

    \begin{eqnarray} A(k, p)& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}\sum\limits_{x_1 = 0, x_2 = 0, \cdot\cdot\cdot, x_k = 0}^{p-1}e_p(m(x_1^6+x_2^6+\cdot\cdot\cdot+x_k^6))\\& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}G^k(m, 6;p). \end{eqnarray} (3.1)

    By (8), we have

    \begin{eqnarray} A(k, p)& = &\frac{1}{p}\sum\limits_{m = 0}^{p-1}G^k(m, 6;p)\\ & = &\frac{1}{p}(\sum\limits_{m = 1}^{p-1}G^k(m, 6;p)+p^k)\\ & = &\frac{1}{p}((p-1)a_k+p^k) = \frac{p-1}{p}a_k+p^{k-1}. \end{eqnarray} (3.2)

    So by Lemma 9, we have

    \begin{eqnarray} A(k+6, p)-p^{k+5}& = &5p(A(k+4, p)-p^{k+3})+10dp(A(k+3, p)-p^{k+2})\\ &&+(46p^2+5d^2p+dp)(A(k+2, p)-p^{k+1})\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)(A(k+1, p)-p^k)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\ &&+(129d^2+11d+6)p^2+d^4p)(A(k, p)-p^{k-1}). \end{eqnarray}

    So we have

    \begin{eqnarray} A(k+6, p)& = &5pA(k+4, p)+10dpA(k+3, p)\\&&+(46p^2+5d^2p+dp)A(k+2, p)\\ &&+(2p^2+120dp^2+3d^3p+d^2p+dp)A(k+1, p)\\ &&+(-4p^5+2d^2p^4+64p^{7/2}+381p^3+2d^2p^{5/2}\\ &&+(129d^2+11d+6)p^2+d^4p)A(k, p)\\ &&+p^{k+5}-p^{k+4}-(10dp+2d^2)p^{k+3}-64p^{k+5/2}\\&&-(429+121d+5d^2)p^{k+2}\\ &&-2d^2p^{k+3/2}-(3d^3+130d^2+12d+6)p^{k+1}-d^4p^k. \end{eqnarray}

    And by Lemma 7 and (3.2), we have the initial conditions

    \begin{eqnarray} &&A(1, p) = 1, A(2, p) = 4(p-1)+p, A(3, p) = 10d(p-1)+p^2, \\ &&A(4, p) = 56p(p-1)+10d^2(p-1)+p^3, \\ &&A(5, p) = 188dp(p-1)+5d^3(p-1)+16dC(p)(p-1)+p^4.\\ &&A(6, p) = p^5+1400p^2(p-1)+(388d^2+8d-576)p(p-1)+d^2p-d^2. \end{eqnarray}

    So we complete the proof of the theorem.

    The main purpose of this paper is using analytic methods to give a recurrence formula of the number of solutions of an equation over finite field. And we give an expression of the number of solutions of the above equation by the root of sixth degree polynomial. We use analytic methods to give a recurrence formula for the number of solutions of the above equation. And our method is based on the properties of the Gauss sum. It is worth noting that we used a novel method to simplify the steps and avoid complicated calculations.

    The author thanks to referees for very important recommendations and warnings which improved the paper.

    The author declares that there is no competing interest.



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