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Research article

Efficient estimation combining exponential and ln functions under two phase sampling

  • In this study, we propose the combination of exponential and ln ratio type estimator to estimate the mean of Y (Study Variable) by incorporating two auxiliary variables in two phase sampling scheme. Under simple random sampling without replacement, the illustration for mean square error and mathematical comparisons are presented. Several approaches are available in literature to estimate the study variable by using information on the variable of interest. The performance of our proposed estimator is compared with other ratio type estimators theoretically and empirically. It is observed that ratio and exponential ratio estimators considered by various researchers and usual unbiased estimator is less efficient than our proposed estimator. An efficiency comparison is also given using five data sets and simulation studies for checking the merits of our proposed estimator and outcomes are sound and moderately illuminating in comparison to different estimators.

    Citation: Yasir Hassan, Muhammad Ismai, Will Murray, Muhammad Qaiser Shahbaz. Efficient estimation combining exponential and ln functions under two phase sampling[J]. AIMS Mathematics, 2020, 5(6): 7605-7623. doi: 10.3934/math.2020486

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  • In this study, we propose the combination of exponential and ln ratio type estimator to estimate the mean of Y (Study Variable) by incorporating two auxiliary variables in two phase sampling scheme. Under simple random sampling without replacement, the illustration for mean square error and mathematical comparisons are presented. Several approaches are available in literature to estimate the study variable by using information on the variable of interest. The performance of our proposed estimator is compared with other ratio type estimators theoretically and empirically. It is observed that ratio and exponential ratio estimators considered by various researchers and usual unbiased estimator is less efficient than our proposed estimator. An efficiency comparison is also given using five data sets and simulation studies for checking the merits of our proposed estimator and outcomes are sound and moderately illuminating in comparison to different estimators.


    The classical Thomas-Fermi problem for the neutral atom is a second-order non-linear ordinary differential equation, named after Llewellyn Thomas and Enrico Fermi [1,2,3,4,5] which can be derived by applying the Thomas-Fermi model to atoms. The Thomas-Fermi model assumes that all electrons are subject to the same conditions and energy conservation law, and has potential energy eΦ[6] so when assuming that the potential is spherically symmetric, then the charge density ρ and the potential energy are related through the Poisson's equation

    1rd2dr2(rΦ(r))+4πρ(r)=0, (1.1)

    where is the Planck's constant, r is the distance from the nucleus, and ρ is given by

    ρ=13π23(2m)3/2[eΦ(r)]3/2, (1.2)

    where e is the electronic charge and m is the mass. Substituting in the above equation yields

    1rd2dr2(rΦ(r))4e3π(2m)3/2[eΦ(r)]3/2=0, (1.3)

    with the corresponding boundary conditions:

    limr0rΦ(r)=eZ, (1.4)

    where Z is the atomic number, and

    limrΦ(r)=0. (1.5)

    By introducing the following transformation r=μx for some appropriate parameter μ and y=rΦ(r)eZ, we arrive at the so-called differential equation of the Thomas-Fermi equation

    y=x12y32, 0<x<, (1.6)
    y(0)=1, y()=0. (1.7)

    This equation models the charge distribution of a neutral atom as a function of the radius x. It should be noted here that the basic Thomas-Fermi (TF) model for ions is subject to the boundary conditions y(0)=1 and y(x0)=0, where x0>0 is the dimensionless ion size which measures the boundary radius and satisfies the relation x0y(x0)=q for some ionization factor q. When the nuclear charge equals the number of (bound) electrons, then q=0 which occurs when x0, and the the problem, in this case, describes the neutral atom model case [7]. The TF equation (1.6) has connections to other important partial differential equations, for example, it is considered a special case of the well-known Poisson equation, and can also be viewed as an Euler-Lagrange equation associated with the Fermi energy [8]. The Thomas-Fermi model has deep connections to the quantum gravity theory where it is reformulated at the Planck scale [9,10,11].

    The equation has a particular solution yp(x), which satisfies the boundary condition y0 as x, but not the initial condition y(0)=1. This particular solution is yp(x)=144x3.

    Arnold Sommerfeld used this particular solution and provided an approximate solution that can satisfy the other boundary condition [4]:

    ys(x)=yp(x)(1+yp(x)λ1/3)λ2/2, (1.8)

    where λ1=0.772 and λ2=7.772. This solution predicts the correct solution accurately for large x but still fails near the origin. A considerable amount of literature is devoted to the numerical solutions for the classical version of this problem [12,13,14,15,16]. In [12], numerical solution was obtained using the variational principle. J. Boyd [13] obtained a numerical solution using rational Chebyshev functions. The authors in [14,15] obtained numerical solutions using a spectral method based on the fractional order of rational Bessel functions. Pikulin [16] used a semi-analytical numerical method to compute the solution. Furthermore, different methods such as homotopy analysis and iterative methods are used to investigate the approximate solutions to this problem, see, e.g., [17,18,19,20,21].

    However, a free boundary value issue is also implemented to approach the initial Thomas-Fermi equation. As a result, the free boundary value issue is changed into a nonlinear boundary value problem that is defined on a closed interval. An adaptive approach is used to tackle the issue utilizing the moving mesh finite element method [22].

    The present paper investigates the generalized boundary value problem of the Thomas-Fermi equation

    {y+f(x,y)=0, 0<x<,y(0)=1, y()=0, (1.9)

    where

    f(x,y)=y(yx)pp+1, p>0, 0<x<, (1.10)

    and we assume that 0y(x)1.

    The Thomas-Fermi equation is a special case of this equation when p=1. As pointed out in [23], this generalized TF equation is related to non-integrable Abel equations, and therefore no closed solutions are possible for any case of this type of equation.

    In this paper, we aim to provide an analytic approximate solution in explicit form for problem (1.9) with (1.10). In Section 2, we establish a theorem that provides the lower and upper bounds of the solution y and guarantees the existence of the solution to this problem as well as a theorem on the uniqueness of the solution. In Section 3, we present analytic approximate solutions in different explicit forms to this problem. Also, an interesting variation of the Adomian decomposition method (ADM) [24,25,26,27,28,29,30,31,32,33] is presented, which allows the determination of the solution in an easily-computed series. In Section 4, we carry out an analysis of the solution and compare it with other numerical solutions.

    We first prove a result on the double inequalities for the lower and upper bounds of the solution y, which is an important tool in the proof of the existence of the solution to problem (1.9) with (1.10).

    Theorem 2.1. The generalized boundary value problem of TF equation (1.9) with (1.10) has at least one solution yC2[0,) such that

    y1yy2 on [0,), (2.1)

    where y1(x)=1(x+1)1+52, y2(x)=x((K(0,x)K(1,x))), and K is the modified Bessel function of the second kind.

    Proof. Using the following inequality:

    (1x)r<(1+1x)r, r=pp+1<1, x>0, (2.2)

    we obtain

    (1x)r<1+1x, r=pp+1<1, x>0, (2.3)

    which will be helpful later.

    In view of 0y(x)1, it follows that

    (yx)pp+1(1+1x)ypp+1, for 0y1, (2.4)

    and with ypp+11, we have

    (yx)pp+11+1x, for 0y1. (2.5)

    Consequently,

    y(yx)pp+1(1+1x)y, for 0y1. (2.6)

    Hence,

    f(x,y)=y(yx)pp+1(1+1x)y, for 0y1. (2.7)

    On the other hand, in view of the solution y remains in the interval [0,1] and since pp+1<1, we have

    ypp+1y. (2.8)

    Using now x<x+1 to obtain 1x>1x+1. Hence, (1x)pp+1>1(x+1)pp+1. Thus,

    (yx)pp+11(x+1)pp+1 (2.9)

    or

    (yx)pp+1yx+1. (2.10)

    It can be checked easily that

    y(yx)pp+1y(x+1)2, for 0y1. (2.11)

    Hence,

    f(x,y)=y(yx)pp+1y(x+1)2,  for 0y1. (2.12)

    Thus, from (2.7) and (2.12), we obtain

    G2(x,y)f(x,y)=y(yx)pp+1G1(x,y), for 0y1, (2.13)

    where

    G1(x,y)=1(x+1)2y and G2(x,y)=(1+1x)y, for 0y1. (2.14)

    For comparison purposes, we have the following linear boundary value problems:

    {y1+G1(x,y1)0, 0<x<,y1(0)=1, y1()=0, (2.15)

    and

    {y2+G2(x,y2)0, 0<x<,y2(0)=1, y2()=0. (2.16)

    Then, suitable comparison problems are

    {y11(1+x)2y1=0, 0<x<,y1(0)=1, y1()=0, (2.17)

    and

    {y2(1+1x)y2=0, 0<x<,y2(0)=1, y2()=0. (2.18)

    To find the solution y1 of problem (2.17), we write

    (1+x)2y1y1=0. (2.19)

    Let ξ=x+1. Thus this equation becomes

    ξ2y1(ξ)y1(ξ)=0. (2.20)

    The substitution ξ=et leads to a constant coefficient linear equation

    y1(t)y1(t)y1(t)=0. (2.21)

    Thus,

    y1(x)=C1(x+1)1+52+C2(x+1)152, (2.22)

    where C1 and C2 are two constants. Using the boundary conditions y1(0)=1, y1()=0 to find C1=1 and C2=0. This gives

    y1(x)=1(x+1)1+52. (2.23)

    To find the solution y2 of problem (2.18), we bring back the form of the confluent hypergeometric equation with parameters a and b [34,35,36]:

    xy+(bx)yay=0, (2.24)

    which has a regular singularity at 0 and an irregular one at infinity; and whose solution is 1F1(a;c;x).

    The equation of problem (2.18) can be simply written as

    y2(1+22x)y2=0. (2.25)

    We introduce the change of variables ξ=2x and y2(x)=v(ξ). Then

    4ξv(ξ)(ξ+2)v(ξ)=0. (2.26)

    The transformation v(ξ)=ξeξ2w(ξ) leads to

    ξw(ξ)+(2ξ)w(ξ)32w(ξ)=0, (2.27)

    which is the confluent hypergeometric equation with parameters a=3/2 and b=2.

    Thus the general solution of problem (2.18) is given in terms of the modified Bessel functions as

    y2(x)=x(c1(I(0,x)+I(1,x))+c2(K(0,x)K(1,x))), (2.28)

    where I and K are the modified Bessel functions of the first and second kind, respectively. c1 and c2 are arbitrary constants, which can be determined from the boundary conditions. Indeed, to satisfy these conditions y2(0)=1 and y2()=0, we get c1=0 and c2=1, and so the required solution is given by

    y2(x)=x((K(0,x)K(1,x))). (2.29)

    For small x, we have

    y2(x)1+x(ln(x2)+γ)+O(x2) as x0+. (2.30)

    Hence, the condition y2(0)=1 is satisfied.

    We are now able to apply the method of upper and lower solutions. For more details about this technique, we refer the reader to (Chapter 7, [37]), which is applicable when f(x,y) has a singularity at x=0 and the Lipschitz constants L1(x)=1(x+1)2 and L2(x)=(1+1x) are functions of the independent variable x and continuous everywhere except for L2(x) at x=0.

    It should be noted here that y1 and y2 are both twice continuously differentiable and satisfy the above differential inequalities functions (2.15) and (2.16) on (0,) with y1<y2. Furthermore, the function f(x,y) is continuous and bounded in

    S={(x,y): 0x<, y1yy2}. (2.31)

    This completes the proof.

    To show the variation of these two extremum functions, we present in Figure 1 the variation of the upper and lower functions in terms of the independent variable x. In addition, the extremum functions are independent of the parameters p, which makes them the optimum functions for all kinds of Thomas-Fermi equations.

    Figure 1.  The variation of the extremum functions y1(x) and y2(x) versus the independent variable x. Lower function: red dashed line; Upper function: blue solid line.

    Theorem 2.2. The generalized boundary value problem of TF equation (1.9) with (1.10) has at most one solution yC2[0,).

    Proof. To obtain an important result on the uniqueness, we assume that ˉy1 and ˉy2 are two different solutions to problem (1.9) with (1.10). Then,

    {ˉy1=g(x,ˉy1), 0<x<,ˉy1(0)=1, ˉy1()=0, (2.32)

    and

    {ˉy2=g(x,ˉy2), 0<x<,ˉy2(0)=1, ˉy2()=0, (2.33)

    where g(x,y)=y(yx)pp+1.

    Consider the positive function h(x)=12(ˉy1ˉy2)2. Thus h vanishes at zero and infinity. Therefore, if is not identically zero it must have a positive maximum at a point ˉx, where ˉx>0. Thus, its graph is concave down at ˉx>0, and we have

    h(ˉx)=[12(ˉy1ˉy2)2]x=ˉx0. (2.34)

    Since

    [12(ˉy1ˉy2)2]x=ˉx=(ˉy1(ˉx)ˉy2(ˉx))(ˉy1(ˉx)ˉy2(ˉx))+(ˉy1(ˉx)ˉy2(ˉx))2, (2.35)

    or

    [12(ˉy1ˉy2)2]x=ˉx=(ˉy1(ˉx)ˉy2(ˉx))(ˉy1(ˉx)ˉy2(ˉx)). (2.36)

    Hence,

    (ˉy1(ˉx)ˉy2(ˉx))(ˉy1(ˉx)ˉy2(ˉx))0. (2.37)

    From (2.32) and (2.33), we have

    ˉy1(x)ˉy2(x)=g(x,ˉy1)g(x,ˉy2). (2.38)

    Applying the mean value theorem to the function g with respect to ˉy, we obtain

    ˉy1(x)ˉy2(x)=gy(x,ˉy)(ˉy1(x)ˉy2(x)), (2.39)

    where 0ˉy1<ˉy<ˉy21.

    On the other hand, differentiating the function g(x,y) with respect to y, we obtain

    gy(x,y)=(2p+1p+1)(yx)pp+1. (2.40)

    Hence,

    gy(ˉx,ˉy)=(2p+1p+1)(ˉy(ˉx)ˉx)pp+1, ˉx>0. (2.41)

    Consequently,

    ˉy1(ˉx)ˉy2(ˉx)=(2p+1p+1)(ˉy(ˉx)ˉx)pp+1(ˉy1(ˉx)ˉy2(ˉx)). (2.42)

    Substituting this into (2.37), we obtain

    (2p+1p+1)(ˉy(ˉx)ˉx)pp+1(ˉy1(ˉx)ˉy2(ˉx))20, (2.43)

    which contradicts the assumption that 2p+1p+1>0, ˉy(ˉx)ˉx>0, ˉx>0 and (ˉy1(ˉx)ˉy2(ˉx))2>0. So h(x)=12(ˉy1ˉy2)20. This shows the uniqueness of the solution and completes the proof of the theorem.

    We conclude here based on Theorem 2.1, which may offer advantages in finding out lower and upper solutions of our problem (1.9) with (1.10) in explicit forms such that y1yy2 on [0,), and consequently we should expect y to take similar explicit forms in the whole region with the corresponding boundary conditions.

    To obtain an approximate solution y to problem (1.9) with (1.10), we first make the following approximation.

    A possible linear approximation of a function f(x) at x=x0 may be obtained using the equation of the tangent line

    f(x)f(x0)+f(x0)(xx0). (3.1)

    If we choose f(x)=βx, β>0 and x0=1β, then

    βx1+βx2, (3.2)

    when x is close enough to x0=1β. Hence,

    βx(1+βx)24. (3.3)

    Substituting (3.3) into the nonlinear term of the ODE of problem (1.9), we obtain

    y22pp+1βpp+1(1+βx)2pp+1ypp+1+1=0. (3.4)

    For the solution y of the approximate equation (3.4), by Theorem 2.1, we expect that the solution y can be obtained in the form

    y=(1+βx)m, (3.5)

    where β>0 and m<0 are two parameters to be determined. Inserting the ansatz given by (3.5) into Eq (3.4), we obtain

    m(m1)β2(1+βx)m2=22pp+1βpp+1(1+βx)mpp+1+m2pp+1. (3.6)

    If we assume that m2=mpp+1+m2pp+1, that is m=2p, then, we derive the following relation between the parameters

    2pp+2p=22pp+1βp+2p+1, (3.7)

    that is

    β=22pp+2(2pp+2p)p+1p+2. (3.8)

    Thus the first analytic approximate solution to the generalized TF equation is given by

    y1(x;p)=1(1+βx)2p, where β=22pp+2(2pp+2p)p+1p+2. (3.9)

    The term (βx)pp+1 can be approximated by (1+βx)pp+1 for sufficiently large values of βx; that is,

    (βx)pp+1(1+βx)pp+1. (3.10)

    Substituting (3.10) into the nonlinear term of the ODE of problem (1.9), we obtain

    (m2+m)β2(1+βx)2=βpp+1(1+βx)(m+1)pp+1. (3.11)

    It follows that m=1+2p and β=[(p+2p)2+p+2p]p+1p+2.

    Thus, the second analytic approximate solution to the generalized TF equation for x large is given as

    y2(x;p)=1(1+βx)1+2p, where β=[(p+2p)2+p+2p]p+1p+2. (3.12)

    For x near 1, we can substitute x1 in the denominator of the nonlinear term of the ODE of problem (1.9) to find

    (m2+m)β2(1+βx)2=(1+βx)(m+1)pp+1. (3.13)

    It follows that m=2+2p and β=[(2p+2p)2+2p+2p]12.

    Thus, the third analytic approximate solution to the generalized TF equation for x near 1 is given as

    y3(x;p)=1(1+βx)2+2p, where β=[(2p+2p)2+2p+2p]12. (3.14)

    Thus, our approximate solutions can be obtained by direct approaches.

    In this section, we consider an interesting variation of the modified Adomian decomposition method (ADM) [24,25,26,27,28,29,30,31,32,33,34], which permits the determination of the solution of nonlinear initial-boundary value problem (1.9) with (1.10).

    Rewrite the ODE of problem (1.9) with (1.10) in Adomian's operator-theoretic form

    Ly=xpp+1N(y), 0<x<, (3.15)

    where L=d2dx2 and N(y)=y2p+1p+1.

    Applying L1 to both sides of Eq (3.15) and using the initial condition y(0)=1, we obtain

    y=1+Bx+x0x0[xpp+1N(y)]dxdx, (3.16)

    where B=y(0) is an unknown constant to be determined by using the boundary condition y()=0.

    According to the Adomian decomposition method [24,25,26,27,28,29,30,31,32,33], assuming the decomposition

    y=n=0yn and N(y)=y2p+1p+1=n=0An, (3.17)

    where An are the Adomian polynomials [24,25,33]. Thus, Eq (3.16) becomes

    n=0yn=1+Bx+x0x0[xpp+1n=0An]dxdx. (3.18)

    We identify

    y0=1, y1=Bx and n=2yn=x0x0[xpp+1n=0An]dxdx. (3.19)

    Hence, a new recurrence relation for yn, n0, is established as

    {y0=1,y1=Bx,yn+2=x0x0[xpp+1An]dxdx, (3.20)

    where the Adomian polynomials An [24,25,33] for the N(y)=y2p+1p+1 term are

    {A0(y0)=y2p+1p+10,A1(y0,y1)=2p+1p+1y1ypp+10,A2(y0,y1,y2)=2p+1p+1y2ypp+10+12!2p+1p+1pp+1y21yp+2p+10,... (3.21)

    The first few components of the solution yn, n0 are given by

    {y0=1,y1=Bx,y2=(p+1)2p+2xp+2p+1,y3=B(p+1)(2p+1)(p+2)(2p+3)x2p+3p+1,y4=(p+1)3(2p+1)2(p+2)2(p+3)x2p+4p+1+B2p(2p+1)2(3p+4)(2p+3)x3p+4p+1,... (3.22)

    Hence,

    y=1+Bx+(p+1)2p+2xp+2p+1+B(p+1)(2p+1)(p+2)(2p+3)x2p+3p+1+.... (3.23)

    It remains now to apply the second boundary condition y0 as x to the function y(x). This boundary condition cannot be applied directly to the series (3.23). Recall that it is customary to combine the series solutions obtained by the decomposition method with the Padé approximants to provide an effective tool to treat boundary value problems on an infinite or semi-infinite interval [33]. To illustrate this, we choose p=1. For convenience, we list below, by using (3.21), few terms of the Adomian polynomials An

    {A0(y0)=1,A1(y0,y1)=32y1,A2(y0,y1,y2)=32y2+38y21,A3(y0,y1,y2)=32y3+34y1y2116y21,... (3.24)

    The first few components of the solution yn, n0, are given by

    {y0=1,y1=Bx,y2=43x32,y3=25Bx52,y4=13x3+370B2x72,... (3.25)

    Hence,

    y=1+Bx+43x32+25Bx52+13x3+370B2x72+215Bx4+.... (3.26)

    Setting x12=ξ into (3.26), we obtain

    y=1+Bξ2+43ξ3+25Bξ5+13ξ6+370B2ξ7+215Bξ8+..., (3.27)

    which is indeed the same approximation of y that obtained by Baker in 1930 [5] and Wazwaz [33]. In applying the boundary condition y()=0 to the diagonal Padé approximants P10,10=[10/10], we obtain the approximation for the initial slope B=y(0)=1.588077, which is a very good approximation to accuracy 105 comparing to the value obtained by Parand et al. as 1.588071 [14]. These values are also in good agreement with the obtained numerical value yn(0)=1.564036 for p=1.

    We are now in the position to explore some mathematical results and investigate the numerical treatment of the boundary value problem (1.9) with (1.10). In Figure 2, we present the different solutions of problem (1.9) with (1.10) with the particular case p=1. The first approximation (solid blue line) is in good agreement with the numerical solution and Sommerfeld's approximation. On the other hand, the third approximation (black dash-dotted line) is in good agreement with the numerical solution for small values of the independent variable x. While the second approximation diverges slightly from the other solutions for small and intermediate values of the independent variable x. All solutions coincide together for large values of x. Due to the potential limits of the numerical volume [38,39,40], we chose the maximum value of the independent variable as x=14. The numerical solution is obtained, using the Maple software, and the available mid-rich sub-method, which is a midpoint method with the same enhancement schemes. So, the midpoint sub-methods are capable of handling harmless end-point singularities that the trapezoid sub-methods cannot. For the enhancement schemes, Richardson extrapolation is generally faster, but deferred corrections use less memory on difficult problems [41,42].

    Figure 2.  The variation of the different solutions y(x) of (1.9) with (1.10) versus the independent variable x. Numerical solution: black long dashed line; Sommerfeld's solution: green dotted line; first approximation: blue solid line; second approximation: red dashed line; third approximation: black dash-dotted line. All solutions are obtained for p=1.

    In addition, we present in Tables 1 and 2 a comparison between the numerical solution and different proposed approximations for the case p=1, for small and large values of the independent variables x. These numerical values show clearly that the first and third approximations agree very well with the numerical solution in all ranges of the independent value x.

    Table 1.  Comparison between different approximations and numerical solutions of problem (1.9) with (1.10) for the case p=1, and small values of the independent variable x. yn= Numerical solution, ys= Sommerfeld's approximation, y1;1= First solution, y2;1= Second approximation and y3;1= Third approximation.
    x yn ys y1;1 y2;1 y3;1
    .100000 .890589 .836423 .910357 .944876 .915349
    .200000 .800549 .740601 .832265 .893735 .839461
    .300000 .725548 .666917 .763802 .846210 .771278
    .400000 .662283 .606766 .703443 .802028 .709884
    .500000 .608242 .556122 .649967 .760838 .654476
    .600000 .561517 .512617 .602373 .722429 .604368
    .700000 .520665 .474709 .559820 .686559 .558968
    .800000 .484586 .441319 .521616 .653027 .517735
    .900000 .452445 .411651 .487194 .621639 .480250
    1.00000 .423598 .385104 .456075 .592235 .446096

     | Show Table
    DownLoad: CSV
    Table 2.  Comparison between different approximations and numerical solutions of (1.9) with (1.10) for the case p=1, and large values of the independent variable x. yn= Numerical solution, ys= Sommerfeld's approximation, y1;1= First solution, y2;1= Second approximation and y3;1= Third approximation.
    x yn ys y1;1 y2;1 y3;1
    1. .423598 .385104 .456075 .592235 .446096
    2. .242734 .220660 .259910 .379212 .227968
    3. .156335 .142841 .167656 .257243 .128316
    4. .107979 .0993388 .117042 .182448 .0776398
    5. .781469e-1 .725516e-1 .863148e-1 .134052 .496894e-1
    6. .584026e-1 .549358e-1 .662721e-1 .101365 .332598e-1
    7. .445657e-1 .427789e-1 .524786e-1 .784985e-1 .230930e-1
    8. .343581e-1 .340689e-1 .425827e-1 .620228e-1 .165308e-1
    9. .264419e-1 .276399e-1 .352432e-1 .498537e-1 .121427e-1
    10. .199817e-1 .227745e-1 .296497e-1 .406706e-1 .911860e-2
    11. .144286e-1 .190161e-1 .252895e-1 .336111e-1 .698002e-2
    12. .940720e-2 .160612e-1 .218248e-1 .280956e-1 .543328e-2
    13. .465612e-2 .137020e-1 .190263e-1 .237236e-1 .429216e-2
    14. 0. .117932e-1 .167334e-1 .202137e-1 .343549e-2

     | Show Table
    DownLoad: CSV

    Now, we can explore other interesting cases with p1, to show the efficiency of the suggested approximations and their validity ranges. In Figure 3, we present the different solutions of (1.9) with (1.10) with the particular cases p=2,3. The first and third approximations (solid blue line, black dash-dotted line) are in good agreement with the numerical solution for small values of the independent variable x. On the other hand, the third approximation remains in good agreement with the numerical solutions, while the first approximation diverges from the numerical solution by increasing the parameter p. The second approximation is still larger than all approximations over the small and intermediate domains of x.

    Figure 3.  The variation of the different solutions y(x) of (1.9) with (1.10) versus the independent variable x. Numerical solution: black long dashed line; first approximation: blue solid line; second approximation: red dashed line; third approximation: black dash-dotted line. (a) for p=2 and (b) for p=3.

    Our overall findings demonstrate that it is possible to acquire a good approximation to the generalized TF equation. The charge distribution of a neutral atom as a function of radius x is also well-known to be described by this equation if and only if y(x) approaches zero as x grows in size. Solutions with y(x)=0 at a finite x are used to mimic positive ions. For solutions where y(x) becomes significant and positive as x increases significantly, it can be viewed as a model of a compressed atom, where the charge is squeezed into a smaller region. These broad comments are adequately supported by our plots. The proposed investigation might be useful in dense media where quantum gravity's effects could be felt strongly.

    The goal of this study is to solve the generalized TF equation which governs several physical issues, such as quantum systems, that naturally differ significantly from Fermi or Bose statistics, as well as some astrophysical or cosmological contexts where quantum electrostatics may exhibit more intertwined screening effects. The TF equation is modeled in this investigation as a singular boundary value problem with an upper and lower solution theory. The existence-construction of the aforementioned upper-lower solutions is also explored. Excellent approximations are proposed and the obtained results are in good agreement with those obtained numerically. We anticipate that the approximation solutions we have presented will be useful in assisting with the investigation of the TF model-governed physics issues.

    The authors extend their appreciation to the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University for funding this work through Research Group no. RG-21-09-14.

    The authors declare that they have no competing interests.



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