Research article Special Issues

Soil erosion control from trash residues at varying land slopes under simulated rainfall conditions


  • Trash mulches are remarkably effective in preventing soil erosion, reducing runoff-sediment transport-erosion, and increasing infiltration. The study was carried out to observe the sediment outflow from sugar cane leaf (trash) mulch treatments at selected land slopes under simulated rainfall conditions using a rainfall simulator of size 10 m × 1.2 m × 0.5 m with the locally available soil material collected from Pantnagar. In the present study, trash mulches with different quantities were selected to observe the effect of mulching on soil loss reduction. The number of mulches was taken as 6, 8 and 10 t/ha, three rainfall intensities viz. 11, 13 and 14.65 cm/h at 0, 2 and 4% land slopes were selected. The rainfall duration was fixed (10 minutes) for every mulch treatment. The total runoff volume varied with mulch rates for constant rainfall input and land slope. The average sediment concentration (SC) and sediment outflow rate (SOR) increased with the increasing land slope. However, SC and outflow decreased with the increasing mulch rate for a fixed land slope and rainfall intensity. The SOR for no mulch-treated land was higher than trash mulch-treated lands. Mathematical relationships were developed for relating SOR, SC, land slope, and rainfall intensity for a particular mulch treatment. It was observed that SOR and average SC values correlated with rainfall intensity and land slope for each mulch treatment. The developed models' correlation coefficients were more than 90%.

    Citation: Sachin Kumar Singh, Dinesh Kumar Vishwakarma, Salwan Ali Abed, Nadhir Al-Ansari, P. S. Kashyap, Akhilesh Kumar, Pankaj Kumar, Rohitashw Kumar, Rajkumar Jat, Anuj Saraswat, Alban Kuriqi, Ahmed Elbeltagi, Salim Heddam, Sungwon Kim. Soil erosion control from trash residues at varying land slopes under simulated rainfall conditions[J]. Mathematical Biosciences and Engineering, 2023, 20(6): 11403-11428. doi: 10.3934/mbe.2023506

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  • Trash mulches are remarkably effective in preventing soil erosion, reducing runoff-sediment transport-erosion, and increasing infiltration. The study was carried out to observe the sediment outflow from sugar cane leaf (trash) mulch treatments at selected land slopes under simulated rainfall conditions using a rainfall simulator of size 10 m × 1.2 m × 0.5 m with the locally available soil material collected from Pantnagar. In the present study, trash mulches with different quantities were selected to observe the effect of mulching on soil loss reduction. The number of mulches was taken as 6, 8 and 10 t/ha, three rainfall intensities viz. 11, 13 and 14.65 cm/h at 0, 2 and 4% land slopes were selected. The rainfall duration was fixed (10 minutes) for every mulch treatment. The total runoff volume varied with mulch rates for constant rainfall input and land slope. The average sediment concentration (SC) and sediment outflow rate (SOR) increased with the increasing land slope. However, SC and outflow decreased with the increasing mulch rate for a fixed land slope and rainfall intensity. The SOR for no mulch-treated land was higher than trash mulch-treated lands. Mathematical relationships were developed for relating SOR, SC, land slope, and rainfall intensity for a particular mulch treatment. It was observed that SOR and average SC values correlated with rainfall intensity and land slope for each mulch treatment. The developed models' correlation coefficients were more than 90%.



    Five basic quantities (voltage, charge, current, capacitance, and resistance) in electrostatics are involved in almost all applications. Electrostatics generally plays an important role in improving the performance of microelectro/mechanical systems (MEMS) and electron devices in the design stage. Many numerical methods (e.g., the finite difference method, the variational method, the moment method, the finite element method, and the boundary element method) were popularly used for engineering problems. Among diverse numerical techniques, the finite element method (FEM) and the boundary integral equation method (BIEM), as well as the boundary element method (BEM) become acceptable tools for engineers due to the increasing development of digital computing power. Here, we may focus on the mathematical study of the BIEM for electrostatics of two identical cylinders. Researchers have paid attention to the dual BEM paper of IEEE in 2003 [1], which has received nearly 5000 views in the Research Gate.

    For a pair of two conducting cylinders, there is a large amount of literature on charged cylinders [2,3,4]. Different solutions existed to the electrostatic problem of two identical parallel cylinders held at the same (symmetric) potential [2,3]. A note was given to show their equivalence, and the identities were confirmed [4]. Four distinct solutions for the potential distribution around two equal circular parallel conducting cylinders by [2,3,5,6] were demonstrated to be equivalent by Lekner [7] by ways of several identities. Here, we may try an alternative way of BIEM using degenerate kernels to revisit this problem. A degenerate kernel is based on the method of separation variables, but it separates the variables in the two-point kernel function. Although the BIE in conjunction with the available degenerate kernel can only solve simple geometries and the results may be obtained more directly by using the method of separation variables for the solution instead of the fundamental solution, the tool can explain the rank-deficiency mechanism in the BIE/BEM such as degenerate scale, degenerate boundary, spurious eigenvalues and fictitious frequency, which is meaningful to the BEM community. Besides, symmetric and anti-symmetric cases are both considered. Regarding the anti-symmetric electrostatic potential, Lebedev et al. [8] have provided a closed-form solution by using the bipolar coordinates. The solution is interestingly found to be the simplest method of fundamental solution (MFS) of two opposite strengths of sources at the two foci. It is not trivial to check the asymptotic behavior at infinity of the two cases, symmetric and anti-symmetric. Besides, whether the equilibrium of the boundary flux along the two cylinders is satisfied or not is also our main concern.

    Regarding the potential problem of a two-dimensional plane containing two circular boundaries, Chen and Shen [9] studied the multiply-connected Laplace problem. They found that a degenerate scale depends on the outer boundary. Chen et al. [10] solved the Laplace problem by using the BIEM in conjunction with the degenerate kernel to derive an analytical solution. It is found that a degenerate scale may occur due to the introduction of the logarithmic kernel for the two-dimensional case. Efficient techniques for the rank-deficiency of the BEM in electrostatic problems were proposed by Chyuan et al. [11]. Later, it was found that the special (degenerate) geometry happened to be the shape of unit logarithmic capacity. Kuo et al. [12] studied the degenerate scale for regular N-gon domains by using complex variables. Numerical implementation was also done by using the BEM. Kuo et al. [13] revisited the degenerate scale for an infinite plane problem containing two circular holes using the conformal mapping. Chen et al. [14] linked the logarithmic capacity in the potential theory and the degenerate scale in the BEM for two tangent discs. The logarithmic capacity of the line segment as well as the double degeneracy in the BIEM/BEM was studied by Chen et al. [15]. Due to the use of the two-dimensional fundamental solution in the BIEM, the solution space is expanded, and sometimes the corresponding matrix is rank deficient in the BEM. In other words, the integral operator of the logarithmic kernel is range deficient. A corresponding chart to show the rank deficiency and the null space of the integral operator of single, double layer potentials and their derivatives was given in [16,17,18], while the original one was provided in the face cover of the Strang book [19]. Fikioris et al. [20] solved rectangularly shielded lines by using the Carleman-Vekua method. In the mentioned paper [20], it is interesting to find that its formulation also needs a constraint [21] to ensure a unique solution. This outcome is similar to the paper of Chen et al. [22] using the Fichera's approach, where an additional constraint is also required.

    In this paper, we revisit two cylinders of electrostatics by using the BIE with the degenerate kernel of the bipolar coordinates. Both the symmetric and anti-symmetric specified potentials are considered. Besides, the logarithmic capacity is also discussed. The boundary potential and flux are expanded by using the Fourier series, while the fundamental solution is represented by using the degenerate kernel. The equilibrium of boundary flux and the asymptotic behavior at infinity are also examined. The solution space expanded using the BIEM is compared with the true solution space. After summarizing the single (circle and ellipse) and two cylinders, a conclusion for constructing the solution space can be made.

    First, we consider a conducting cylinder. The governing equation and the Dirichlet boundary condition are shown below:

    2u(x)=0, xD,u(x)=ˉu(x), xB,u(x)= ln |x|+O(1), x, (1)

    where 2, D and B are the Laplace operator, the domain of interest and the boundary, respectively. Furthermore, x is the position vector of a field point and ˉu(x) is the specified B.C. The integral formulation for the Laplace problem is derived from Green's third identity. The representation of the conventional integral equation for the domain point is written as

    2πu(x)=BT(s,x)ˉu(s)dB(s)BU(s,x)t(s)dB(s),xD, (2)

    where s is the position vector of a source point, U(s,x)= ln |xs|  is the fundamental solution, T(s,x)=U(s,x)ns, and t(x) is the unknown boundary flux. By moving the field point to the smooth boundary, Eq (2) becomes:

    πˉu(x)=C.P.V.BT(s,x)ˉu(s)dB(s)BU(s,x)t(s)dB(s),xB, (3)

    where the C.P.V. denotes the Cauchy principal value, and T(s,x)=U(s,x)/ns is the closed-form kernel. Once the field point x locates outside the domain, we obtain the null-field integral equation as shown below:

    0=BT(s,x)ˉu(s)dB(s)BU(s,x)t(s)dB(s),xDc. (4)

    where Dc is the complementary domain. By employing the proper degenerate kernel (U(s,x)) to represent the closed-form fundamental solution, the collocation point can be exactly located on the real boundary free of facing the singular integral. Equations (2) and (4) can be rewritten as:

    2πu(x)=BTdk(s,x)ˉu(s)dB(s)BUdk(s,x)t(s)dB(s),xDB (5)

    and

    0=BTdk(s,x)ˉu(s)dB(s)BUdk(s,x)t(s)dB(s),xDcB. (6)

    where Tdk(s,x) and Udk(s,x) are the corresponding degenerate kernels to represent T(s,x) and U(s,x), respectively. By setting the field point x=(ρ,ϕ) and the source point s=(R,θ) in the polar coordinates for a circular domain, the closed-form fundamental solution in Eqs (5) and (6) can be expressed by using the degenerate kernel form as shown below:

    Udk(s,x)={Ui(R,θ;ρ,ϕ)= ln Rm=11m(ρR)m cos m(θϕ),  Rρ,          (a)Ue(R,θ;ρ,ϕ)= ln ρm=11m(Rρ)m cos m(θϕ),  ρ>R,          (b). (7)

    and

    Tdk(s,x)={Ti(R,θ;ρ,ϕ)=(1R+m=1(ρmRm+1) cos m(θϕ)),  R>ρ,      (a)Te(R,θ;ρ,ϕ)=m=1(Rm1ρm) cos m(θϕ),  ρ>R.                 (b). (8)

    The unknown boundary flux t(s) is expanded in terms of Fourier series as shown below:

    t(s)=1Js(a0+n=1an cos (nθ)+n=1bn sin (nθ)), 0θ2π, (9)

    where Js=1 is the Jacobian term, a0, an and bn are unknown coefficients. The given boundary condition is

    ˉu(x)=v. (10)

    where v is a constant. By considering R=a in Eqs (6)–(8), the coefficient of the Fourier constant base is

    a ln a a0=v, (11)

    where a is the radius of the circular cylinder. Equation (11) indicates that the occurring mechanism of a degenerate scale is

     ln a=0. (12)

    When a=1, the coefficient of a0 cannot be determined. It results in a non-unique solution. This critical size is called a degenerate scale. In Rumely's book [23], the logarithmic capacity, cL, of a circle is equal to its radius. It is easily found that the special (degenerate) geometry happens to be the shape of unit logarithmic capacity. The discriminant Dp(a) of the degenerate scale in the BEM/BIEM for a circular boundary is written as

    Dp(a)= ln a. (13)

    If Dp(a)0, this size is an ordinary scale and there exists a unique solution. Otherwise, according to the Fredholm alternative theorem, there is no solution or infinite solutions. For an ordinary scale, the boundary flux, of the electrostatic field along the boundary, is obtained as

    t(x)=va ln a,xB. (14)

    The unique solution of electrostatic potential is obtained by

    u(x)=v ln ρ ln a, (15)

    as shown in Figure 1(a). Even though the electrostatic field along the boundary in Eq (14) is not in equilibrium, i.e. Bt(x)dB(x)0, the electrostatic field at infinity, Γ, would exist and satisfy the equilibrium condition together in total, B+Γt(x)dB(x)=0. If we normalize the potential on the cylinder to the unity, and let λ be the dimensionless ratio, the potential becomes

    u(x)=v+λud(x), (16)
    Figure 1.  Analytical solution of the electrostatic potential subject to conducting cylinders derived by using the null-field BIEM.

    where

    ud(x)= ln ρ ln a. (17)

    The solution by using the direct BIE of Eq (15) is the special case of Eq (16) by setting λ=vDp(a). When the size of the boundary is a degenerate scale, i.e., a=1 and Dp(a)= ln a=0, it has no solution if v0. If v=0, then the constant term in Eq (9), a0, is a free constant. The electrostatic potential yields

    u(x)=a0 ln ρ, (18)

    and Eq (16) would reduce to

    u(x)=λud(x), (19)

    and ud(x) in Eq (19) reduces to  ln ρ since  ln a=0. It is easy to find that a0 and λ are equivalent.

    For an elliptical case, we naturally utilize the elliptic coordinates to solve the problem in the BIE. The relation between the Cartesian coordinates and the elliptic coordinates is given below:

    x=c cosh ξ cos η,y=c sinh ξ sin η. (20)

    where c is the focal length. By separating the source point and the field point in the elliptic coordinates [24] to represent the closed-form fundamental solution, we have

    Udk(s,x)= ln |xs|={Ui(ξs,ηs;ξx,ηx)=ξs+ ln c2m=12memξs cosh m ξx cos m ηx cos m ηs                       m=12memξs sinh m ξx sin m ηx sin m ηs,  ξsξx,  (a)Ue(ξs,ηs;ξx,ηx)=ξx+ ln c2m=12memξx cosh m ξs cos m ηx cos m ηs m=12memξx sinh m ξs sin m ηx sin m ηs,  ξs<ξx,   (b) (21)
    Tdk(s,x)=U(s,x)ns={Ti(ξs,ηs;ξx,ηx)=1Js(1+2m=1emξs cosh m ξx cos m ηx cos m ηs +2m=1emξs sinh m ξx sin m ηx sin m ηs),    ξs>ξx,  (a)Te(ξs,ηs;ξx,ηx)=1Js(2m=1emξx sinh m ξs cos m ηx cos m ξs+2m=1emξx cosh m ξs sin m ηx sin m ηs),    ξs<ξx. (b) (22)

    where Js=c cosh 2ξs sin 2ηs+ sinh 2ξs cos 2ηs. The unknown boundary flux t(s) is expanded in terms of generalized Fourier series. We have

    t(s)=1Js(a0+n=1an cos (nηs)+n=1bn sin (nηs)), 0ηs2π, (23)

    where a0, an and bn are unknown coefficients. The given boundary condition is

    ˉu(x)=v. (24)

    By substituting Eqs (21a), (22a), (23) and (24) into Eq (6), the coefficient of the Fourier constant base is

    (ξ0+ ln c2)a0=v, (25)

    Equation (25) indicates that the occurring mechanism of a degenerate scale is

    ξ0+ ln c2=0. (26)

    Equation (26) yields the degenerate scale of a+b2=1, where a and b are the semi-major and semi-minor axes of an ellipse, respectively. According to Eq (25), the discriminant of a degenerate scale in the BEM/BIEM is obtained

    De(c,ξ0)=ξ0+ ln c2= ln (a+b2). (27)

    In Rumely's book [23], the logarithmic capacity of an ellipse is equal to a+b2. According to Eqs (13) and (27), the logarithmic capacity, cL, and the discriminant, De(), satisfy the relation,

    cL=eDe(). (28)

    The relationship of the discriminant, logarithmic capacity, and degenerate scale are summarized in Table 1. If De(c,ξ0) is not equal to zero, this size is an ordinary scale with a unique solution. Otherwise, according to the Fredholm alternative theorem, it has no solution or infinite solution. For an ordinary scale, the boundary flux is obtained by

    t(x)=vDe(c,ξ0),xB. (29)
    Table 1.  Relationship of the discriminant, logarithmic capacity and degenerate scale.

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    The unique solution of electrostatic potential is

    u(x)=(ξx+ ln c2)(vDe(c,ξ0)), (30)

    as shown in Figure 1(b). Even though the boundary flux in Eq (29) is not in equilibrium, i.e., Bt(x)dB(x)0, the electrostatic field at infinity, Γ, would exist and satisfy the equilibrium condition together in total, i.e., B+Γt(x)dB(x)=0. If we normalize the potential on the cylinder to the unity, and let λ be the dimensionless ratio, the potential becomes

    u(x)=v+λud(x), (31)

    where

    ud(x)=(ξx+ ln c2)(ξ0+ ln c2)=ξxξ0. (32)

    A neat formula of ud(x) could be defined as ud(x)=De(ξx)De(ξ0). The solution by using the direct BIE of Eq (30) is the special case of Eq (31), if λ=vDe(c,ξ0).

    When the size of the boundary is a degenerate scale, De(c,ξ0)=ξ0+lnc2=0, it is no solution if v0. If v=0, the constant term in Eq (23), a0, is a free constant. The electrostatic potential yields

    u(x)=(ξx+ ln c2)a0, (33)

    and Eq (16) reduces to

    u(x)=λud(x), (34)

    where ud(x) in Eq (32) reduces to ξx+ ln c2 since De(ξ0) is equal to zero. It is easy to find that a0 and λ are equivalent. In addition, the degenerate scale in the BEM/BIEM is due to the logarithmic kernel.

    For the single elliptical cylinder, the degenerate kernel is expanded in terms of the generalized form as

    Udk(s,x)= ln |xs|={Ui(ξs,ηs;ξx,ηx)=De(ξs)m=1αm(ξs,ηs;ξx,ηx),  ξsξx,  (a)Ue(ξs,ηs;ξx,ηx)=De(ξx)m=1αm(ξx,ηx;ξs,ηs),  ξs<ξx,   (b) (35)
    Tdk(s,x)=U(s,x)ns={Ti(ξs,ηs;ξx,ηx)=1Js(D'eξs)m=1βm(ξs,ηs;ξx,ηx)),  ξs>ξx,  (a)Te(ξs,ηs;ξx,ηx)=1Jsm=1βm(ξs,ηs;ξx,ηx),    ξs<ξx, (b) (36)

    where ξ and η are the radial and angular directions, respectively, De() is the constant function for ηs and other term is α(). The unknown boundary flux t(s) is expanded in terms of the generalized Fourier series as shown below:

    t(s)=1Js(a0+n=1an cos (nηs)+n=1bn sin (nηs)), 0ηs2π, (37)

    where a0, an and bn are unknown coefficients. By substituting Eqs (35a), (36a), (37) and the boundary condition (Eq (10)) into Eq (6), the coefficient of the Fourier constant base is

    De(ξ0)a0=vD'e(ξ0), (38)

    If D(ξ0)0, then the unique solution of electrostatic potential is

    u(x)=vDe(ξ0)De(ξx), (39)

    and

    ud(x)=De(ξx)De(ξ0). (40)

    When the size of the boundary is a degenerate scale, De(ξ0)=0, there is no solution if v0. If v=0, the constant term in Eq (37), a0, is a free constant. The electrostatic potential yields

    u(x)=(De(ξx)De(ξ0))a0, (41)

    and Eq (40) reduces to

    u(x)=λud(x), (42)

    where ud(x) in Eq (40) is reduced to De(ξx), since De(ξ0) is equal to zero. By using the generalized form of Eqs (35) and (36), the analytical and neat form of ud(x) for the single elliptical cylinder is derived. It is easy to find that a0 and λ are equivalent. The generalized potential and the solution by using the BIEM are compared in Table 2.

    Table 2.  Comparison of the general solution and the BIE solution for single or double cylinders.

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    In this section, we consider two circular cylinders of electrostatics. The Dirichlet boundary conditions of two circular cylinders are given by

    ul(x)=v1 and ur(x)=v2,xB, (43)

    where ul(x) and ur(x) are potentials of the left and right circular boundaries, respectively, x is the position vector of the field point, B is the boundary, and v1 and v2 are specified constant potentials. The original problem can be decomposed into a symmetric problem and an anti-symmetric problem as shown below:

    ul(x)=ur(x)=v,xB, symmetry BC, (44)

    and

    ul(x)=ur(x)=v,xB. anti-symmetry BC.  (45)

    Since the problem contains two circular boundaries, we naturally employ the bipolar coordinates to express the closed-form fundamental solution. The relation between the Cartesian coordinates and the bipolar coordinates is shown below:

    x=c sinh η cosh η cos ξ,y=c sin ξ cosh η cos ξ. (46)

    where η and ξ are the radial and angular coordinates, respectively, c is the half distance between the two foci of the bipolar coordinates. By separating the source point and the field point in the bipolar coordinates [9] for the closed-form fundamental solution, we have

    Udk(s,x)= ln |xs|=
    { ln (2c)+ηsm=11m{em(ηsηx) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},0>ηsηx ln (2c)+ηxm=11m{em(ηxηs) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},0>ηx>ηs ln (2c)m=11m{em(ηxηs) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},ηx>0>ηs ln (2c)ηsm=11m{em(ηxηs) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},ηxηs>0 ln (2c)ηxm=11m{em(ηsηx) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},ηs>ηx>0 ln (2c)m=11m{em(ηsηx) cos [m(ξxξs)]emηx cos (mξx)emηs cos (mξs)},ηs>0>ηx (47)
    Tdk(s,x)=U(s,x)ns=
    {1Js{1+m=1[em(ηsηx) cos [m(ξxξs)]emηs cos (mξs)]},0>ηsηx1Jsm=1{em(ηxηs) cos [m(ξxξs)]emηs cos (mξs)},0>ηx>ηs1Jsm=1{em(ηxηs) cos [m(ξxξs)]emηs cos (mξs)},ηx>0>ηs1Js{1+m=1[em(ηxηs) cos [m(ξxξs)]emηs cos (mξs)]},ηxηs>01Jsm=1{em(ηsηx) cos [m(ξxξs)]emηs cos (mξs)},ηs>ηx>0 1Jsm=1{em(ηsηx) cos [m(ξxξs)]emηs cos (mξs)},ηs>0>ηx (48)

    where x=(ηx,ξx), s=(ηs,ξs) and Js=c/[cosh(ηs)cos(ξs)].

    The boundary condition of the symmetry problem is shown in Eq (44). The unknown boundary densities on the two circular cylinders can be expanded by using the generalized Fourier series as shown below:

    tM(s)={1Js(al0+n=1aln cos n ξs+n=1bln sin n ξs), ηs < 0,sBl,1Js(ar0+n=1arn cos n ξs+n=1brn sin n ξs), ηs0,sBr, (49)

    where al0,aln,bln,ar0,arn and brn are unknown coefficients of the generalized Fourier series. By substituting Eqs (47a), (47f), (48a), (48f), (44) and (49) into Eq (6), and collocating the null-field point on the left boundary, Bl, we have

    2πvπ{2( ln (2c)η0)al0+n=11nenη0alnn=11n(2enη0al0+aln) cos n ξxn=11nbln sin n ξx}π{2 ln (2c)ar0+n=11nenη0arn+n=11n[(2enη0ar0e2nη0arn) cos n ξxe2nη0brn sin n ξx]}=0. (50)

    Similarly, substituting Eqs (47c), (47d), (48c), (48d), (44) and (49) into Eq (6), and collocating the null-field point on the right boundary, Br, we have

    π{2 ln (2c)al0+n=11nenη0aln+n=11n[(e2nη0aln+2enη0al0) cos n ξxe2nη0bln sin n ξx]}2πvπ{2( ln (2c)η0)ar0+n=11nenη0arn+n=11n[(2enη0ar0arn) cos n ξxbrn sin n ξx]}=0. (51)

    By adding Eqs (50) and (51) together, we obtain

    (4 ln (2c)2η0)(al0+ar0)+2n=11nenη0(aln+arn)+n=11n[(1+e2nη0)(aln+arn)+4enη0(al0+ar0)] cos n ξxn=11n[(1+e2nη0)(bln+brn)] sin n ξx=4v (52)

    After comparing the coefficient of generalized Fourier bases, we have

    {(4ln(2c)2η0)(al0+ar0)+2n=11nenη0(aln+arn)=4v,n=1, 2, 3...1n(1+e2nη0)(aln+arn)+4enη0(al0+ar0)=0,n=1, 2, 3...1n(1+e2nη0)(bln+brn)=0,n=1, 2, 3... (53)

    By similarly subtracting Eq (50) from Eq (51), we have

    π{2η0al0+n=11n[(1+e2nη0)aln cos n ξx+(1+e2nη0)bln sin n ξx]}+π{2η0ar0+n=11n[(1e2nη0)arn cos n ξx+(1e2nη0)brn sin n ξx]}=0 (54)

    After comparing the coefficient of generalized Fourier bases, we have

    {2η0(al0ar0)=0,n=1, 2, 3...1n(1+e2nη0)(alnarn)=0, n=1, 2, 3...1n(1+e2nη0)(blnbrn)=0, n=1, 2, 3... (55)

    In order to solve the coefficients al0 and ar0, we need to define a discriminant as shown below:

    Db(c,η0)=2 ln (2c)η0+n=11n4e2nη0(1+e2nη0). (56)

    For the case of two cylinders, Rumely [23] employed the complex variable to derive the logarithmic capacity, as shown in Table 1. Since the logarithmic capacity is not a closed-form or an exact formula, the postulate in Eq (28) for the case of two cylinders could not be analytically verified at present. If Db0, the geometry of the problem is an ordinary scale, Eqs (53) and (55) yield the coefficients as shown below:

    al0=ar0=vDb(c,η0)aln=arn=2enη0(1+e2nη0)al0,n=1,2,3...bln=brn=0,n=1,2,3... (57)

    Substituting Eqs (47b), (47f), (48b), (48f), (44) and the obtained unknown boundary densities into Eq (5) for the field solution of ηx<0, we have the unique solution

    u(x)=vDb(c,η0)((2 ln (2c)+ηx+n=14ne2nη01+e2nη0)n=12n(enη0enηx+enηxenη0+enη0enηx) cos (nξx)),η0ηx<0. (58)

    Similar substitution of Eqs (47c), (47e), (48c), (48e), (44) and the obtained unknown densities into Eq (5), the field solution for ηx0 yields

    u(x)=vDb(c,η0)((2 ln (2c)ηx+n=14ne2nη01+e2nη0)n=12n(enη0enηx+enηxenη0+enη0enηx) cos (nξx)),η0ηx0. (59)

    It is found that Eqs (58) and (59) show the symmetry solution. All potentials are shown in Figure 1(c). This solution will be compared and discussed with that of Darevski [2] later.

    If Db(c,η0)=0, a degenerate scale occurs. When the constant potential v0, it yields no solution. When the constant potential v=0, it yields infinite solutions. Equations (53) and (55) yield the coefficients as shown below:

    ar0=al0=k,aln=arn=4enη0(1+e2nη0)k,n=1,2,3...bln=brn=0,n=1,2,3... (60)

    where k is an arbitrary constant. In case of a degenerate scale, η0 becomes

    η0=2 ln (2c)+n=11n4e2nη0(1+e2nη0) (61)

    Substituting Eqs (47b), (47f), (48b), (48f), (44) and the obtained boundary unknown densities into Eq (5) for the field solution of ηx<0, we have the infinite solution,

    u(x)=((η0+ ηx)n=12n(enη0enηx+enηxenη0+enη0enηx) cos (nξx))k, η0ηx<0. (62)

    Similar substitution of Eqs (47c), (47e), (48c), (48e), (44) and the obtained boundary unknown densities into Eq (5), the field solution for ηx0 yields the infinite solution,

    u(x)=((η0ηx)n=12n(enη0enηx+enηxenη0+enη0enηx) cos (nξx))k, η0ηx0. (63)

    Equations (62) and (63) also indicate symmetry.

    Similarly, we consider the anti-symmetry problem. The coefficient of generalized Fourier bases in Eqs (45) and (49) satisfy

    {(2 ln (2c)η0+4n=1e2nη0n(1+e2nη0))(al0+ar0)=0,1n(1+e2nη0)(aln+arn)+4enη0(al0+ar0)=0, n=1, 2, 3...1n(1+e2nη0)(bln+brn)=0, n=1, 2, 3... (64)

    and

    {2η0(al0ar0)=4v,1n(1+e2nη0)(alnarn)=0, n=1, 2, 3...,1n(1+e2nη0)(blnbrn)=0, n=1, 2, 3... (65)

    We also find the discriminant, Db(c,η0) in Eq (64). If Db(c,η0)0, the geometry of the problem is an ordinary scale. Equations (64) and (65) yield the coefficients as shown below:

    al0=ar0=vη0aln=arn=0,n=1,2,3...bln=brn=0,n=1,2,3... (66)

    Substituting Eqs (47b), (47f), (48b), (48f), (45) and the obtained boundary unknown densities into Eq (5) for the field solution of ηx<0, we have

    u(x)=vηxη0, ηx < 0. (67)

    Substituting Eqs (47c), (47e), (48c), (48e), (45) and the obtained unknown densities into Eq (5) for the field solution of ηx0, we also have

    u(x)=vηxη0, ηx0. (68)

    All potentials are shown in Figure 1(d). The solution in Eq (68) matches well with that of Lebedev et al. [8]. From the viewpoint of the MFS, this solution is the simplest one since only two sources with opposite strengths are required to locate the two foci.

    If Db(c,η0)=0, a degenerate scale occurs. Fortunately, it doesn't result in no solution whether v is equal to zero or not as shown in the boundary condition of Eq (45). Equations (64) and (65) yield the coefficients as shown below:

    al0+ar0=2k,aln=arn=4enη0(1+e2nη0)k,n=1,2,3...bln=brn=0,n=1,2,3... (69)

    where k is an arbitrary constant. For a degenerate scale case, η0 satisfies Db(c,η0)=0, i.e.

    η0=2 ln (2c)+n=11n4e2nη0(1+e2nη0) (70)

    Substituting Eqs (47b), (47f), (48b), (48f), (45) and the obtained boundary unknown densities into Eq (5) for the field solution of ηx<0, we have

    u(x)=vη0ηx+k( ln (2 cosh ηx2 cos ξx)η0+2n=11nenη0 cosh (nηx) cosh (nη0) cos (nξx)), η0ηx0. (71)

    Similarly substituting Eqs (47c), (47e), (48c), (48e), (45) and the obtained unknown densities into Eq (5), we obtain the field solution

    u(x)=vη0ηx+k(ln(2 cosh ηx2 cos ξx)η0+2n=11nenη0 cosh (nη0) cosh (nηx) cos (nξx)), η0ηx > 0. (72)

    Equations (71) and (72) destroy the anti-symmetry due to the second part of k. To obey the anti-symmetry solution, k should be zero. In other words, this k part in the solution of Eqs (71) and (72) also disobey the bounded potential at infinity. This solution for a free constant, k, will be compared with that of Lekner [4] later.

    According to the solution of Lekner [4], the general solution space of the symmetry problem in Eq (44) is expressed as follows:

    u(x)=v+λud(x) (73)

    where

    ud(x)= ln (2 cosh ηx2 cos ξx)η0+n=12nenη0 cosh (nηx) cosh (nη0) cos (nξx). (74)

    By using the identity equation,

     ln ( cosh ηx cos ξx)=ηxm=12memηx cos m ξx ln 2, (75)

    the solution by using the direct BIE of Eq (59) is rewritten as

    u(x)=vDb(c,η0)( ln (2 cosh ηx2 cos ξx)+n=12nenη0 cosh (nηx) cosh (nη0) cos (nξx)(2 ln (2c)+n=11n4e2nη01+e2nη0)),η0ηx > 0. (76)

    Equation (59) is the special case of Eq (73), if λ=vDb(c,η0). When the size of the boundary is a degenerate scale, i.e., Db(c,η0)=0, the BIE solution does not exist if v0. If v=0, the constant term in Eq (57), ar0, is a free constant. The electrostatic potential is obtained by

    u(x) =( ln (2 cosh ηx2 cos ξx)η0+n=12nenη0 cosh (nηx) cosh (nη0) cos (nξx))ar0, η0ηx > 0, (77)

    and Eq (73) can be reduced to

    u(x)=λud(x), (78)

    since v is zero. It is easy to find that ar0 and λ are equivalent.

    Similarly, the general solution space of the anti-symmetry problem in Eq (45) is expressed as follows:

    u(x)=vηxη0+λud(x) (79)

    where

    ud(x)= ln (2 cosh ηx2 cos ξx)η0+n=12nenη0 cosh (nηx) cosh (nη0) cos (nξx). (80)

    Lebedev et al. [8] considered the condition at the infinity, u(x)=0,x, the solution of Eq (79) would reduce to only

    u(x)=vηxη0. (81)

    It is the reason why the solution of Eq (68) by using the BIEM is a special case of Eq (79) for λ=0. When the size of the boundary is a degenerate scale, i.e., Db(c,η0)=0, it yields infinite solutions. Since the sum of constant terms, ar0 and al0, in Eq (69) is a free constant, k, the electrostatic potential yields

    u(x)=vηxη0( ln (2 cosh ηx2 cos ξx)η0+2n=11nenη0 cosh (nη0) cosh (nηx) cos (nξx))k, η0ηx > 0. (82)

    It is easy to find that k and λ are equivalent. To sum up, the free constant, λ and ud(x) in the general solution by Lekner [4] are similar to the constant term in the boundary potential and the obtained BIE solution for the degenerate scale by Chen et al. [9], respectively. The obtained BIE solution for the degenerate case yields nontrivial boundary flux even though the boundary potential is trivial. The generalized potential and the available solutions by using the BIEM for the problem containing two cylinders are compared with each other in Table 3.

    Table 3.  Comparison of the general solution and the BIE solution for the double cylinder subject to the symmetrical or anti-symmetrical condition.

     | Show Table
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    This paper investigates the solution space for the electrostatics of two cylinders using the BIEM. Both the symmetric and anti-symmetric cases are considered. Flux equilibrium on the cylindrical boundaries and the asymptotic behavior at infinity is also examined. Moreover, on the base of the Fredholm alternative theorem, the relation of unique solution and the degenerate scale in the BIEM is linked. The logarithmic capacity and the discriminant are also linked by using an exponential relation. Besides, the degenerate scale is also related. Not only two cylinders but also a single one (circle or ellipse) are considered. Finally, the results are compared with those derived by other researchers. Linkage and agreement are made.

    The authors wish to thank the financial supports from the National Science and Technology Council, Taiwan under Grant No. MOST 111-2221-E-019-009-MY3 for National Taiwan Ocean University.

    The authors declare there is no conflict of interest.



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