Journal of Computational and Applied Mathematics

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1 Journal of Computational and Applied Mathematics 35 (0) Contents lists available at ScienceDirect Journal of Computational and Applied Mathematics journal homepage: Analysis and algorithms for the computation of the excited states of helium Zhonghai Ding a,, Goong Chen b a Department of Mathematical Sciences, University of Nevada, Las Vegas, Las Vegas, NV , USA b Department of Mathematics and Institute for Quantum Studies, Texas A&M University, College Station, TX 77843, USA a r t i c l e i n f o a b s t r a c t Article history: Received 9 October 008 MSC: primary 65N5 secondary 70G75 8Q05 Keywords: Helium atom Schrödinger equation Excited states Dimensional scaling Numerical algorithm In this paper, we study a dimensionally scaled helium atom model for excited states of helium. The mathematical analysis of the corresponding effective energy potential is presented. Two simple numerical algorithms are developed for the computation of the excited states of helium. Comparison between our numerical results and those in the existing literature is given to indicate the accuracy and efficiency of the proposed algorithms. 00 Elsevier B.V. All rights reserved.. Introduction The study of helium atom and helium-like ions, which are composed of one nucleus and two electrons, plays an important role in the development of the wave-mechanical quantum theory of Heisenberg and Schrödinger during the mid 90 s []. Since the first experiment in [] in 963, the study of doubly excited states of helium has attracted the interest of both physicists and mathematicians. Let e denote the electron s charge, m denote the electron s mass, r k = (x k, y k, z k ) denote the position vector of the kth electron (k =, ), and r = (r, r ). The electronic wave function Ψ (r) is governed by the (steady-state) Schrödinger equation [3] ] [ h m ( ) V(r) Ψ (r) = E e Ψ (r), (.) where k = x k, k =,. y k z k h is the Planck constant; E e is the energy level of the electronic structure; and V(r) is the Coulomb potential energy containing the attracting and repelling potentials, given by V(r) = Ze r Ze r e r Corresponding author. Tel.: addresses: dingz@unlv.nevada.edu, Zhonghai.Ding@unlv.edu (Z. Ding), gchen@math.tamu.edu (G. Chen) /$ see front matter 00 Elsevier B.V. All rights reserved. doi:0.06/j.cam

2 04 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) where r = r r and Ze is the nuclear charge. Ψ (r) can be interpreted as the probability density that gives the probability per unit volume of finding the electrons at r. By using the atomic unit a 0 = h /me for length, where a 0 is the radius of the first Bohr orbit, and the Hartree E h = e /a 0 for the unit of energy, one obtains [ ] ( ) V(r) Ψ (r) = EΨ (r), (.) where the Coulomb potential V(r) is now given by V(r) = Z r Z r r. Eq. (.) constitutes an eigenvalue problem. We point out that Eq. (.) is not separable, and does not have close-form solutions. Define the Hamiltonian operator H by HΨ (r) = [ ] ( ) V(r) Ψ (r). Let H (R 6 ) denote the usual Sobolev space W, (R 6 ). For any Ψ, Ψ H (R 6 ), use the Dirac bra-ket notation and the physicist s convention to write Ψ Ψ = Ψ (r)ψ R 6 (r)dr, where Ψ denotes the complex conjugate of Ψ, and [ ] Ψ H Ψ = ( Ψ Ψ Ψ Ψ ) Ψ V(r)Ψ dr. R 6 Let σ (H), σ p (H) and σ e (H) denote the spectrum, point spectrum and essential spectrum of H respectively. It is well-known [4] that σ (H) = σ p (H) σ e (H), where σ e (H) = [0, ) and σ p (H) (, 0). Each λ σ p (H) is an eigenvalue of H, and H has infinitely many negative eigenvalues which accumulate at 0. The lowest eigenvalue and the corresponding eigenfunction can be obtained by solving the minimization problem inf Ψ H Ψ. Ψ Ψ = Ψ H (R 6 ) The solution Φ 0 attaining the minimum of Ψ H Ψ is called the ground state, and the associated value E 0 = Ψ 0 H Ψ 0 is called the ground state energy. Excited states Ψ k with successively higher energy levels may be obtained sequentially for k =,,..., through E k = Ψ k H Ψ k = min Ψ Ψ = Ψ W k Ψ H Ψ, where W k = {Ψ H (R 6 ) Ψ Ψ j = 0, j = 0,,,..., k }. The computation of excited states of helium is challenging due to the high space dimension, and the singularities and electronic interaction in the Coulomb potential. There are two types of methods in the computational physics literature for calculating the energies of ground and excited states: the Monte Carlo methods coupled with cluster expansion, configuration interactions, and density functions; and the perturbation expansion methods using the Hylleraas-type wave functions coupled with finite discrete spectrum and complex rotations. All of these numerical methods have the following unavoidable issues: (a) numerical algorithms are complex; (b) computations are extremely intensive; and (c) a physically appealing picture of chemical bonds is largely lost. There remains a need for understanding electronic structures in some relatively simple way that are capable of describing ground and excited states of atoms and molecules with good accuracy. Physical parameters, such as masses and charges of particles, often disappear from dynamical equations when variables such as distances and energies are expressed in terms of dimensionless ratios. In that case, there is no free parameter that can be used to set up a perturbation or interpolation calculation. Then a quantity like a dimension, that otherwise would be considered as given or fixed, may be treated as variable, in order to provide a perturbation parameter. The idea of a dimensional scaling (D-scaling) technique was suggested first by the Field medalist Witten [5] and developed essentially by Herschbach [6]. It provides an innovative approach to study many-particle systems, and is now a well-established technique in chemical physics (see [7,8], and the references therein). Using the D-scaling with interpolation and /D expansion, Herschbach [6] studied the ground state of helium. The accuracy of the Herschbach s work was improved by in [9] by using the weighted truncation and Padé approximation. Using the dimensional perturbation, Carzoli et al. [0,] extended Herschbach s method to study the excited states of helium. However, the /D expansion converges slowly and the entire procedure, which uses the moments of coupled harmonic oscillators and the Padé approximation, seems to be quite intricate. In a recent paper in [], a quantum number D-scaling (.3) (.4)

3 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) method was proposed to capture the excited states of helium, where the principal and orbital quantum numbers were included in dimension parameters for the first time in the literature. This method does not require calculation of high order /D corrections in order to obtain high-angular momentum states. However, the effective potential for the excited states of helium obtained in [] is not consistent with the known results given in [9,6,3]. Furthermore, it has been noted [4,] that five quantum numbers (two principal quantum numbers, two angular momentum quantum numbers for electrons, and a combined angular momentum quantum number) are required to characterize the doubly excited states of helium. In a recent paper [5] on three-body problems in atomic and molecular quantum mechanics, we have proposed a quantum number D- scaling method involving five quantum numbers to study the excited states of helium and obtained a dimensionally scaled helium atom model. The aim of this paper is to study the dimensionally scaled helium atom model obtained in [5] and to propose numerical algorithms for the computation of excited states of helium. The organization of this paper is as follows. In Section, for the benefit of the readers, we introduce briefly the idea of D-scaling methods through the example of the hydrogen atom [7], and the dimensionally scaled helium atom model. In Section 3, we present the mathematical analysis of the effective energy potential corresponding to excited states of helium. As an important byproduct of our analysis, it is proved rigorously that the dimensionally scaled helium atom model obtained in [] can capture only the so-called S-states of the helium atom. Based on the mathematical analysis given in Section 3, two simple numerical algorithms are proposed in Section 4 to compute excited states of helium. The comparison between our numerical results and those in the existing literature is also given to indicate the accuracy and efficiency of the dimensionally scaled helium atom model.. A dimensionally scaled helium atom model D-scaling offers a useful computational approach to study multi-particle systems. There are four typical steps in various D-scaling methods [7, Chapter ]. () generalize the problem to D-dimension by writing the usual kinetic energy term in the Schrödinger equation as = 3 D. x i= i x i= i () transform the problem to a suitably scaled space to remove the major D-dependence of the quantity to be determined; (3) evaluate the scaled quantity at one or more special D-values, such as D, where the computation is relatively easy ; and (4) obtain an approximation for D = 3 by relating it to the special D-values, usually by some interpolation or extrapolation procedures. We now illustrate the idea of D-scaling through the example of the hydrogen atom. Example.. The electronic wave function Ψ of the hydrogen atom is governed by the Schrödinger equation 3 Z Ψ (r) = EΨ (r), r x i= i (.) where r = (x, x, x 3 ) R 3, r = r, and Z is the nuclear charge. The hydrogen atom is the simplest atom whose ground state energy and excited state energies can be obtained by solving (.) directly [6, Chapter 8]. Generalize Eq. (.) to D-dimension: D Z Ψ D (r) = EΨ D (r), (.) r x i= i where r = (x, x,..., x D ) R D. Introduce the D-dimensional hyperspherical coordinates (r, θ, θ,..., θ D ) by x = r cos θ sin θ sin θ 3 sin θ D, x = r sin θ sin θ sin θ 3 sin θ D, x 3 = r cos θ sin θ 3 sin θ 4 sin θ D, x 4 = r cos θ 3 sin θ 4 sin θ 5 sin θ D,. x D = r cos θ D sin θ D, x D = r cos θ D, where r 0, 0 θ π, 0 θ j π, j =, 3,..., D. The Jacobian volume factor is given by J = r D sin θ sin θ 3 sin 3 θ 4 sin D θ D. (.3)

4 044 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) The D-dimensional Laplacian now becomes where D = D x i= i K D (r) r D = K D (r) L D r, r r D r (.4) (.5) and L D is the generalized orbital angular momentum operator given by L = θ Thus Eq. (.) becomes Let, L k = sin k θ k K D (r) L D r Ψ D (r, Ω D ) = R(r)Y(Ω D ), sin k θ k θ k θ k L k, k D. (.6) sin θ k Ψ D Z r Ψ D = E D Ψ D. (.7) where Ω D denotes the collective angular variables. One can verify that Y(Ω D ) is an eigenfunction of L D [7]: L D Y(Ω D) = l(l D )Y(Ω D ), where l is the orbital angular momentum quantum number, l = 0,,,.... Thus Eq. (.7) reduces to [ K l(l D ) D(r) Z ] Ψ r D = E D Ψ D, (.8) r where we have used Ψ D = Ψ D (r) to replace R(r). Note that the radial part of the Jacobian volume factor (.3) is J D = r D. Introduce the following transformation Ψ D = r (D)/ Φ D, (.9) which is motivated by the need to normalize the Jacobian factor to unity for the radial probability density, Φ D = J D Ψ D. Thus (.8) becomes [ where r Λ(Λ ) r Z r ] Φ D = E D Φ D, (.0) Λ = l (D 3). (.) Eq. (.0) is the radial Schrödinger equation in D-dimension. In (.), there is an inter-dimensional degeneracy [8]: increasing the space dimension by implies increasing the quantum number l by. By scaling variables r and E as r s = 4 (D ) r, (.) (D ) E s = E D, 4 Eq. (.0) becomes [ (D ) r s Λ(Λ ) (D ) r s By letting D, we obtain Z Φ = E s Φ. r s r s (.3) Z r s ] Φ D = E s Φ D. (.4)

5 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) The problem of calculating the ground state energy of hydrogen is reduced to a simple algebraic problem of minimizing the effective potential E s = r s Z r s. (.5) A simple calculation gives Ẽ = min E(r s ) = Z / at r s = /Z, which coincides with the exact ground state energy of hydrogen. The success of the D-scaling analysis on the hydrogen atom hinges very much on (.) and (.3), due to [6]. We call (.) and (.3) the Herschbach hydrogenic D-scaling, which is a key ingredient in various D-scaling methods. The interpolation or extrapolation procedure is not necessary for the hydrogen atom, but essential for other atomic and molecular structures. More examples of D-scaling can be found in [7] and the references therein. Next, we introduce the dimensionally scaled helium atom model obtained in [5]. Let (r j, θ j, φ j ) denote the spherical coordinates of r j for j =,. The Coulomb potential in (.) is then given by V(r, r, θ, θ, φ) = Z r Z r r, where r = [r r r r (cos θ cos θ sin θ sin θ cos φ)] /, and φ = φ φ is usually called the dihedral angle. If E in (.) is an eigenvalue, then (.) admits at least one eigenstate Ψ which depends only on five variables: r, r, θ, θ, and φ. Thus, to compute the energies of excited states of helium, we consider only those excited states depending on r, r, θ, θ, and φ. On the space of eigenstates that depend on r, r, θ, θ, and φ, the kinetic term in (.) reduces to = r r r r r sin θ θ r r sin θ θ r r r r sin θ sin θ sin θ θ θ r sin θ φ. (.6) Introduce the following generalized kinetic energy term, = K γ (r ) K γ (r ) L r α (θ ) L r α (θ ) L r sin θ r sin β (φ), (.7) θ where K γj (r j ) = r γ j j r j L λ (ψ) = sin λ ψ r γ j j ψ r j sin λ ψ ψ for j =,, and for λ = α, α or β, where γ, γ, α, α, β are different parameters tending toward. The five parameters are expressed as a product of a quantum number and a term of space dimension D according the following quantization postulate: γ = n (D ), γ = n (D ), α = l (D ), α = l (D ), (.8) β = (L )(D 3), where n, n, l, l, L are quantum numbers: and n j = the principal quantum number of electron j for j =,, l j = the angular momentum quantum number of electron j for j =,, L = the combined angular momentum quantum number of both electrons, nj =,, 3,... l j = 0,,,..., n j, L = l l,..., l l. To recover in (.6), one can make a first order /D-expansion by adding a second term in (.8), (.9) γ = n (D ) 6( n )/D, α = l (D ) 3( l )/D, β = (L )(D 3). γ = n (D ) 6( n )/D, α = l (D ) 3( l )/D, (.0)

6 046 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) As D, (.8) and (.0) are equivalent asymptotically. The parameterizations (.8) augment the usual D-dimensional space for to a much larger space which will be called the quantum-number dimensional space. The corresponding Jacobian factor is given by J = r γ rγ (sin θ ) α (sin θ ) α (sin φ) β. By substituting (.7) into (.), we have the Schrödinger equation for the helium atom in the quantum-number dimensional space, [ K γ (r ) K γ (r ) L r α (θ ) L r α (θ ) r sin θ = E D Ψ D. Introduce the following transformation Ψ D = r γ / r γ / (sin θ ) α / (sin θ ) α / (sin φ) β/ Φ D r sin θ ] L β (φ) V(r, r, θ, θ, φ) Ψ D which is based on the probability density in the quantum-number dimensional space given by Φ D = J Ψ D. Then Eq. (.) becomes [ ] Φ r r r θ r θ r sin θ r sin θ φ D [ γ γ 8r r sin θ γ γ 8r r sin θ α 4α cot θ α 4α cot θ 8r β 4β By using the Herschbach hydrogenic D-scaling, 8 8r (.) ] cot φ Φ D V(r, r, θ, θ, φ)φ D = E D Φ D. (.) (D ) (D ) r r, r r, E D (D ) E D, (.3) and letting D, Eq. (.) becomes [ n n l cot θ r r r l cot θ r r sin θ r sin θ (L ) ] cot φ Φ V(r, r, θ, θ, φ)φ = E Φ. (.4) Thus, we obtain the effective potential for excited state n, l, n, l, L of helium, E = j= n j r j l j r j cot (L ) θ j cot φ sin θ j r j V(r, r, θ, θ, φ), (.5) where r > 0, r > 0, 0 θ, θ π, and 0 φ π. More discussion on the effective potential given by (.5) and its comparison with the known results in the literature [6,9,] can be found in [5]. A single electron has spin angular momentum /. For a two-electron system, the total spin angular momentum can be either 0 or. The state of total spin 0 is a singlet while the state of total spin is triplet. The singlet and triplet of helium constitute, respectively, what is called parahelium and orthohelium. The energies of excited states of helium can be computed by minimizing or max-minimizing E(r, r, θ, θ, φ), min E(r, r, θ, θ, φ) r,r,θ,θ,φ (.6) which gives the triplet (or orthohelium) states, and max φ min E(r, r, θ, θ, φ), r,r,θ,θ (.7) which gives the singlet (or parahelium) states.

7 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Analysis of the effective potential as a min max problem From (.5), the effective potential for excited state n, l, n, l, L of helium is given by E = n j l j cot (L ) θ r j r j cot Z φ (3.) j sin θ j r j r where j= r = [r r r r (cos θ cos θ sin θ sin θ cos φ)] /, r j j= r > 0, r > 0, 0 θ, θ π, and 0 φ π. For the helium atom, Z =. From (.6) and (.7), we need to determine the critical points of E which correspond to the singlet and triplet states of helium. Note that the ground state of helium which corresponds to the global minimum of E is a singlet state. Since E(r, r, θ, θ, φ) is symmetric with respect to φ = π on R [0, π] [0, π], we will consider only the critical points of E(r, r, θ, θ, φ) in D = R [0, π] [0, π]. Note that for any (r, r, θ, θ, φ) D, E n j Z = nj. (3.) r j r j n j j= r j j= Thus, E(r, r, θ, θ, φ) is bounded below on D. Define M = M is obviously nonempty. j= n j= j (r, r, θ, θ, φ) D E(r, r, θ, θ, φ) < and n j= j j= Z > 0. r j r Theorem 3.. For any (r, r, θ, θ, φ) M, there exists a unique t R such that E(t r, t r, θ, θ, φ) = min E(tr, tr, θ, θ, φ) < 0. t R For any (r, r, θ, θ, φ) M, either E(tr, tr, θ, θ, φ) = for any t R, or E(tr, tr, θ, θ, φ) is a monotone decreasing function of t on R. Proof. Let (r, r, θ, θ, φ) M be given. Define g : R R by Note that and g(t) = E(tr, tr, θ, θ, φ) = n j l j cot (L ) θ t r j r j cot φ Z. j sin θ j t r j r g (t) = t 3 g (t) = 3 t 4 j= j= j= n j n j r j r j l j r j l j r j r j cot (L ) θ j cot φ sin θ j r j cot (L ) θ j cot φ sin θ j One can verify easily that g (t) = 0 admits a unique solution n j l j cot θ t r r j (L) cot φ r sin j= j j j θ j = > 0 Z r j r satisfying g (t ) > 0, and g(t ) = j= n j r j j= l j r j j= Z r j r r j < 0. cot θ j (L) cot φ r sin j θ j j= t j= t 3 j= Z, r j r Z. r j r

8 048 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Thus, t R is unique, and E(t r, t r, θ, θ, φ) = min E(tr, tr, θ, θ, φ) < 0. t R Let (r, r, θ, θ, φ) M be given. Then either or E(r, r, θ, θ, φ) =, j= j= Z 0. r j r For case (3.3), one has either n j l j cot (L ) θ r j r j cot φ =, j sin θ j or j= Z =. r j r r j (3.3) (3.4) Thus, g(t) = E(tr, tr, θ, θ, φ) = for any t R. For case (3.4), the expression of g (t) indicates that g (t) < 0 for any t R. Thus, g(t) = E(tr, tr, θ, θ, φ) is a monotone decreasing function on R. From Theorem 3., we have the following. Corollary 3.. If (r, r, θ, θ, φ) is a critical point of E in D and E(r, r, θ, θ, φ) <, then (r, r, θ, θ, φ) M, and E(r, r, θ, θ, φ) < 0, and n j l j cot (L ) θ r j r j cot Z φ =. j sin θ j r j r j= r j j= Since E(r, r, θ, θ, φ) is bounded below on D, it follows from Theorem 3. that inf E(r, r, θ, θ, φ) < 0. (r,r,θ,θ,φ) D The next theorem indicates that the infimum value is attainable. Theorem 3.3. There exists at least one global minimum (r, r, θ, θ, φ ) D such that Proof. Let E(r, r, θ, θ, φ ) = γ = inf E(r, r, θ, θ, φ). (r,r,θ,θ,φ) D inf E(r, r, θ, θ, φ) < 0. (r,r,θ,θ,φ) D Assume {(r n, r n, θ n, θ n, φ n )} n= D is a sequence such that γ E(r n, r n, θ n, θ n, φ n ) γ n. (3.5) Note that {θ n } [0, π], {θ n } [0, π], and {φ n } [0, π]. By the Balzano Weierstrass Theorem [0], there exist subsequences of {θ n }, {θ n }, and {φ n }, denoted again by themselves, such that θ n θ, θ n θ, and φ n φ, as n, and θ, θ, φ [0, π]. By (3.), there exists a δ 0 > 0 such that r n δ 0, r n δ 0, n. There are only three possible cases for {r n } and {r n }: (A) both {r n } and {r n } are unbounded; (B) only one of {r n } and {r n } is unbounded; (C) both {r n } and {r n } are bounded above.

9 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) If case (A) is true, then there exist subsequences of {r n } and {r n }, denoted again by themselves, such that r n and r n as n. Thus, lim E(r n, r n, θ n, θ n, φ n ) = 0. n Hence by (3.5), γ = 0, contradicting the fact that γ < 0. Thus, case (A) is impossible. If case (B) is true, then assume, for instance, {r n } is unbounded and {r n } is bounded. Thus there exist subsequences of {r n } and {r n }, denoted again by them themselves, such that r n and r n r as n. Then it follows from (3.5) that γ = lim E(r n, r n, θ n, θ n, φ) n = n l cot θ (L ) r r r sin θ cot φ Z r. Then, for any r > 0, one has Note that γ E(r, r, θ, θ, φ ) n = γ l cot θ (L ) r r r sin θ cot φ Z r r r r r (cos θ cos θ sin θ sin θ cos φ ) / = γ Z n l cot θ (L ) r r r r sin θ cot φ r r r r r (cos θ cos θ sin θ sin θ cos φ ) / = γ r (r ). lim (r ) = Z > 0. r By choosing r large enough, one has γ E(r, r, θ, θ, φ ) < γ, again a contradiction. Thus, case (B) is also impossible. Therefore, case (C) is true, that is, both {r n } and {r n } are bounded. By the Balzano Weierstrass Theorem, there exist subsequences of {r n } and {r n }, denoted again by themselves, such that r n r and r n r, as n. Thus, there exists at least one global minimum (r, r, θ, θ, φ ) D such that E(r, r, θ, θ, φ ) = inf E(r, r, θ, θ, φ). (r,r,θ,θ,φ) D If (r, r, θ, θ, φ) D is a critical point of E, then it must satisfy namely, E r = 0, E r = 0, E θ = 0, [n (L ) r 3 l cot θ sin θ [n (L ) r 3 l cot θ sin θ ] cot φ Z r ] cot φ Z r E θ = 0, E φ = 0, r r (cos θ cos θ sin θ sin θ cos φ) r 3 r r (cos θ cos θ sin θ sin θ cos φ) r 3 = 0, (3.6) = 0, (3.7) l cot θ r csc (L ) θ cot θ r csc θ cot φ r r ( sin θ cos θ cos θ sin θ cos φ) = 0, (3.8) r 3

10 050 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) l r cot θ csc (L ) θ (L ) r sin θ r (L ) r sin θ cot θ csc θ cot φ r r ( cos θ sin θ sin θ cos θ cos φ) r 3 cot φ csc φ r r sin θ sin θ sin φ r 3 = 0, (3.9) = 0. (3.0) To find excited states of the helium atom, we need to solve the system of nonlinear algebraic equations (3.6) (3.0). Due to the complexity of this system, we will develop a numerical algorithm to compute the excited states of helium. Before developing such a numerical algorithm, we need to gain more information on the critical points of E(r, r, θ, θ, φ). Theorem 3.4. If (r, r, θ, θ, φ) D is a critical point of E, 0 < θ < π, 0 < θ < π, and 0 < φ < π, then (a) π < φ < π. (b) If θ π and θ π, then either {0 < θ < π, π < θ < π} or { π < θ < π, 0 < θ < π }. Proof. Assume (r, r, θ, θ, φ) D is a critical point of E, 0 < θ < π, 0 < θ < π, and 0 < φ < π. (r, r, θ, θ, φ) must satisfy system (3.6) (3.0). Rewrite Eq. (3.0) as (L ) (L ) cos φ r r sin θ sin θ sin 4 φ = 0. r sin θ r sin θ r 3 Note that 0 < θ < π and 0 < θ < π. Thus, we have cos φ < 0, which implies claim (a) because of 0 < φ < π. By adding Eqs. (3.8) and (3.9), we have l (L ) cos cot θ l (L ) cos φ cot θ φ r r sin(θ θ )( cos φ) = 0. r r sin 3 θ r r sin 3 θ r 3 Note that 0 < θ < π and 0 < θ < π. If 0 < θ, θ < π, then cos θ sin 3 θ > 0, cos θ sin 3 θ > 0, sin(θ θ ) > 0. Thus, the above equation will not hold. Similarly, if π < θ, θ < π, then cos θ sin 3 θ < 0, cos θ sin 3 θ < 0, sin(θ θ ) < 0. Thus, the above equation will also not hold. Therefore, one has the assertion (b). Note that θ = π and θ = π are the trivial solutions of Eqs. (3.8) and (3.9). In this case, Eqs. (3.6), (3.7) and (3.0) become n r 3 (L ) cot φ Z r r cos φ r [r r = 0, r (3.) r cos φ] 3/ r 3 n (L ) cot φ Z r r cos φ r 3 r 3 r [r r = 0, r (3.) r cos φ] 3/ (L ) (L ) cot φ csc r r sin φ φ [r r = 0. r (3.3) r cos φ] 3/ r r Solutions of system (3.) (3.3) correspond, in fact, to the critical points of E 0 (r, r, φ) = n n (L ) Z cot φ Z r r r r r r [r r r r cos φ] / in D 0 = R [0, π]. If L is replaced by L, E 0(r, r, φ) is exactly the effective energy potential obtained in [], which, as pointed out in [5], is not consistent with the known results given in [9,6,3]. Note that E 0 (r, r, φ) = E(r, r, π/, π/, φ). Thus, E 0 (r, r, φ) is bounded below. Define M 0 = (r, r, φ) D 0 E(r, r, φ) <, and j= Z r j [r r r r cos φ] > 0 / By repeating the same argument as in Theorems 3. and 3.3, we have the following theorems..

11 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Theorem 3.5. For any (r, r, φ) M 0, there exists a unique t R such that E 0 (t r, t r, φ) = min E(tr, tr, φ) < 0. t R For any (r, r, φ) M 0, either E(tr, tr, φ) = for any t R, or E(tr, tr, φ) is a monotone decreasing function of t on R. Theorem 3.6. There exists at least one global minimum (r, r, φ ) D 0 such that E 0 (r, r, φ ) = inf (r,r,φ) D 0 E 0 (r, r, φ). If (r, r, φ) is a critical point of E 0 in D 0, then (r, r, π/, π/, φ) is a critical point of E in D. Note that all critical points of E 0 in D 0 correspond to the solutions of the system (3.) (3.3). From Eq. (3.3), it follows that π < φ < π if (r, r, φ) is a critical point of E 0 in D 0. By applying Theorem 3.6, the system (3.) (3.3) admits at least one solution. Numerical examples have indicated that E 0 has only one critical point in D 0. Thus, we study the uniqueness of critical points of E 0 in D 0. To address the uniqueness of critical points of E 0 in D 0, we replace (r, r ) by polar coordinates (r, θ), and let Ẽ 0 (r, θ, φ) = E 0 (r cos θ, r sin θ, φ) = n r cos θ n (L ) cot φ sin θ r sin θ cos θ Z r cos θ sin θ r[ sin θ cos φ] /. Restrict the domain of Ẽ 0 to D0 = R (0, π ) ( π, π). If (r, θ, φ) D0 is a critical point of Ẽ 0, then it must satisfy that is, Ẽ 0 = 0, r r 3 n r cos θ n sin θ cos 3 θ n Z r Ẽ 0 θ = 0, Ẽ 0 φ = 0, n (L ) sin θ r 3 cos θ (L ) sin 3 θ sin θ cos θ cos θ sin θ r cot φ sin θ cos θ Z r cot φ cos θ sin θ sin 3 θ cos θ sin θ cos 3 θ = 0, (3.4) r [ sin θ cos φ] / cos θ cos φ = 0, (3.5) r[ sin θ cos φ] 3/ (L ) r sin θ cos θ cot φ sin θ sin φ csc φ = 0. (3.6) r[ sin θ cos φ] 3/ From (3.6), we have r = (L ) cos φ sin 3 θ cos 3 θ sin 4 φ [ sin θ cos φ]3/. (3.7) By substituting (3.7) into (3.4) and (3.5) and simplifying them, one obtains and sin 4 φ(n sin6 θ cos θ n cos6 θ sin sin θ) (L ) cos φ sin φ(cos 4 θ sin θ cos sin 4 θ) Z(L ) cos φ(sin 3 θ cos 3 θ)( sin θ cos φ) 3/ (L ) cos φ(cos 4 θ sin θ cos θ sin 4 θ) = 0, (3.8) sin 4 φ(n sin4 θ cos θ n sin θ cos 4 θ) (L ) sin φ cos φ sin θ cos θ (L ) cos φ sin θ cos θ Z(L ) cos φ sin θ cos θ( sin θ cos φ) 3/ (L ) cos φ sin θ cos θ = 0. (3.9) By multiplying sin θ to both sides of Eq. (3.9), subtracting the resulting equation from Eq. (3.8), and dividing both sides of the resulting equation by cos θ, we obtain n sin4 φ cos θ sin 3 θ (L ) cos φ sin φ cos θ sin 3 θ (L ) cos φ sin θ cos θ Z(L ) cos φ( sin θ cos φ) 3/ (L ) cos φ cos 3 θ = 0. (3.0)

12 05 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) By multiplying cos θ to both sides of Eq. (3.9), subtracting the resulting equation from Eq. (3.8), and dividing both sides of the resulting equation by sin θ, we obtain n sin4 φ cos 3 θ sin θ (L ) cos φ sin φ cos 3 θ sin θ (L ) cos φ sin θ cos θ Z(L ) cos φ( sin θ cos φ) 3/ (L ) cos φ sin 3 θ = 0. (3.) Note that the system of Eqs. (3.0) (3.) admits two trivial solutions (θ, φ) = (0, π/) and (π/, π/). However, for r R, (r, 0, π/) and (r, π/, π/) cannot be critical points of Ẽ 0 because they are not in the domain of Ẽ 0. To avoid this awkward situation and to focus on the uniqueness of nontrivial solutions of system (3.0) (3.) in Ω = [0, π/] [π/, π], we rewrite system (3.0) (3.) as [ ] [ ] a a sin 4 φ sin θ cos θ = 0, a a (L ) (3.) cos φ where a = n sin θ, a = cos φ sin φ cos θ sin 3 θ cos φ sin θ cos θ Z[ sin θ cos φ] 3/ cos 3 θ, a = n cos θ, a = cos φ sin φ cos 3 θ sin θ cos φ sin θ cos θ Z[ sin θ cos φ] 3/ sin 3 θ. Eq. (3.) admits a nontrivial solution (θ, φ) if and only if [ ] a a det = 0, a a that is, cos φ sin φ sin 3 θ cos 3 θ(n n) Z[ sin θ cos φ]3/ (n cos θ n sin θ) cos φ sin θ cos θ(n cos θ n sin θ) (n cos4 θ n sin4 θ) = 0. (3.3) Theorem 3.7. Let n = n > 0 be given. Then system (3.0) (3.) admits a unique nontrivial solution (π/4, φ ) in Ω. Consequently, Ẽ 0 (r, θ, φ) has a unique critical point in D0, and E 0 (r, r, φ) has a unique critical point in D 0. Proof. Let n = n = n > 0 be given. If (θ, φ) Ω is a nontrivial solution of the system (3.0) (3.), then (θ, φ) must satisfy (3.3). One then has Z[ sin θ cos φ] 3/ (cos θ sin θ) cos φ sin θ cos θ(cos θ sin θ) (cos 4 θ sin 4 θ) = 0. By simplifying this equation, one has Then either or (cos θ sin θ){z[ sin θ cos φ] 3/ cos φ sin θ cos θ(cos θ sin θ) (cos θ sin θ)} = 0. cos θ sin θ = 0, (3.4) Z[ sin θ cos φ] 3/ cos φ sin θ cos θ(cos θ sin θ) (cos θ sin θ) = 0. (3.5) Since 0 < θ < π/, one has < sin θ cos θ. Note that Z = and π/ < φ < π. Thus, one has Z[ sin θ cos φ] 3/ cos φ sin θ cos θ(cos θ sin θ) (cos θ sin θ) = Z[ sin θ cos φ] 3/ (cos θ sin θ)( cos φ sin θ cos θ) Z[ sin θ cos φ] 3/ (cos θ sin θ)( cos φ sin θ cos θ) = {Z[ sin θ cos φ] / (cos θ sin θ)}( cos φ sin θ) Z > 0. Thus, Eq. (3.5) admits no solution. From Eq. (3.4), one obtains θ = π/4. When θ = π/4, Eqs. (3.0) and (3.) are identical. By substituting θ = π/4 into Eq. (3.0), one has n sin 4 φ (L ) [cos φ sin φ 4 Z cos φ( cos φ) 3/ cos φ cos φ] = 0.

13 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Denote the left-hand side of this equation by F(φ). Note that F(φ) is continuously differentiable on [π/, π], and F(π/) = n > 0, F(π) = (L ) (6Z 4) < 0. Then there exists at least one φ (π/, π) such that F(φ) = 0. Note that Z = and F (φ) = 4n sin 3 φ cos φ (L ) [sin φ cos φ 4 Z sin φ( cos φ) 3/ 6 Z cos φ sin φ( cos φ) / 4 cos φ sin φ L sin φ] sin φ[0 (L ) ( 4 Z 0 4 )] = (L ) sin φ(4 Z 8) < 0. Thus, F(φ) is monotonically decreasing on [π/, π]. Hence there is a unique φ (π/, π) such that F(φ ) = 0. Thus, the system (3.0) (3.) admits a unique nontrivial solution (π/4, φ ) in Ω. Consequently, Ẽ 0 (r, θ, φ) has a unique critical point in D0. Hence, E 0 (r, r, φ) has a unique critical point in D 0. To study the uniqueness of nontrivial solutions of system (3.0) (3.) when n n, denote, for simplicity, the lefthand sides of Eqs. (3.0) and (3.) by F (θ, φ) and F (θ, φ) respectively. One can verify easily that both F (θ, φ) and F (θ, φ) are continuously differentiable on Ω, and F θ (θ, φ) = n sin4 φ cos θ sin 4 θ 3n sin4 φ cos 3 θ sin θ (L ) [ sin φ cos φ cos θ sin 4 θ 3 cos φ sin φ cos 3 θ sin θ 3Z cos φ( sin θ cos φ) / cos θ cos φ cos 3 θ cos φ sin θ cos θ 3 cos φ cos θ sin θ], F φ (θ, φ) = 4n sin3 φ cos φ cos θ sin 3 θ (L ) [ cos 3 φ sin φ cos θ sin 3 θ cos φ sin 3 φ cos θ sin 3 θ Z sin φ( sin θ cos φ) 3/ 3Z sin φ cos φ sin θ cos θ( sin θ cos φ) / sin φ cos φ sin θ cos θ L sin φ cos 3 θ], F θ (θ, φ) = n sin4 φ cos 4 θ sin θ 3n sin4 φ cos θ sin 3 θ (L ) [ sin φ cos φ cos 4 θ sin θ 3 cos φ sin φ cos θ sin 3 θ 3Z cos φ( sin θ cos φ) / cos θ cos φ sin 3 θ cos φ sin θ cos θ 3 cos φ cos θ sin θ], F φ (θ, φ) = 4n sin3 φ cos φ cos 3 θ sin θ (L ) [ cos 3 φ sin φ cos 3 θ sin θ cos φ sin 3 φ cos 3 θ sin θ Z sin φ( sin θ cos φ) 3/ 3Z sin φ cos φ sin θ cos θ( sin θ cos φ) / sin φ cos φ sin θ cos θ sin φ sin 3 θ]. Note that for any φ (π/, π), cos φ < 0, sin φ > 0 and sin φ cos φ. Thus, for any θ [0, π/] and φ (π/, π), F φ (θ, φ) (L ) sin φ[ cos φ sin φ cos θ sin 3 θ Z( sin θ cos φ) 3/ cos φ sin θ cos θ cos 3 θ] (L ) sin φ(z cos θ sin 3 θ sin θ cos θ cos 3 θ). By using cos θ sin θ and 0 sin θ cos θ on [0, π/], we have F φ (θ, φ) (L ) sin φ Z < 0, θ [0, π/], φ (π/, π). (3.6) Similarly, we can obtain F φ (θ, φ) (L ) sin φ Z < 0, θ [0, π/], φ (π/, π). (3.7) Lemma 3.8. Eqs. (3.0) and (3.) define respectively the unique continuously differentiable functions φ = φ (θ) and φ = φ (θ) on [0, π/], satisfying φ (0) = φ (π/) = π/ and φ (0) = φ (π/) = π/. Proof. We consider Eq. (3.0) first. When θ = 0 or π/, it is easy to check that Eq. (3.0) admits a unique solution φ = π/. For θ (0, π/), define F θ : [π/, π] R by F θ (φ) = F (θ, φ). Note that F θ (φ) is continuously differentiable on [π/, π], F θ (π/) = n cos θ sin 3 θ > 0 and F θ (π) (L ) (Z ) < 0.

14 054 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Thus, there exists at least one φ (π/, π) such that F θ (φ) = 0. By (3.6), F θ (φ) = F (θ, φ) < 0, φ on (π/, π). Therefore, F θ (φ) = 0 admits a unique solution in (π/, π). For each θ [0, π/], denoted the solution of F θ (φ) = 0 by φ = φ (θ). Thus, φ (0) = φ (π/) = π/. Note that F (θ, φ) is continuously differentiable on Ω and F φ (θ, φ) < 0 on [0, π/] (π/, π). By applying the Implicit Function Theorem, φ (θ) is continuously differentiable on [0, π/]. By a very similar argument, one can show that Eq. (3.) defines a unique continuously differentiable function φ = φ (θ) on [0, π/] satisfying φ (0) = π/ and φ (π/) = π/. Lemma 3.9. Let n > 0 and n > 0 be given. Then the system (3.0) (3.) admits at least one nontrivial solution (θ, φ) in Ω. Proof. By Lemma 3.8, Eqs. (3.0) and (3.) define respectively the unique continuously differentiable functions φ = φ (θ) and φ = φ (θ) on [0, π/], satisfying φ (0) = φ (π/) = π/ and φ (0) = φ (π/) = π/. By applying the implicit differentiation to F (θ, φ ) = 0 and F (θ, φ ) = 0, we have F dφ dθ = θ (θ, φ ) F φ (θ, φ ), dφ dθ = F θ (θ, φ ) F φ (θ, φ ). If θ 0, then φ (θ) π and φ (θ) π. Thus, F (θ, φ ) = n sin3 θ (Z )(L ) cos φ o(cos φ ), F (θ, φ ) = n sin θ Z(L ) cos φ o(cos φ ). From F (θ, φ ) = 0 and F (θ, φ ) = 0, we have n cos φ = (Z )(L ) sin3 θ o(sin 3 θ), (3.8) n cos φ = Z(L ) sin θ o(sin θ). (3.9) By using (3.8), we obtain Thus, F θ (θ, φ ) = 3n sin θ O(sin 4 θ), F φ (θ, φ ) = (Z )(L ) O(sin 4 θ). dφ dθ = 3n (Z )(L ) sin θ O(sin 4 θ). (3.30) By using (3.9), we obtain Thus, F θ (θ, φ ) = n sin θ O(sin3 θ), F φ (θ, φ ) = Z(L ) O(sin 3 θ). dφ dθ = n Z(L ) sin θ O(sin3 θ). From (3.30) and (3.3), there exists a δ > 0 such that dφ dθ (θ) > dφ dθ (θ) > 0, 0 < θ < δ. Note that φ (0) = φ (0) = π/. Therefore φ (θ) > φ (θ), 0 < θ < δ. (3.3) (3.3)

15 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) If θ π, then φ (θ) π and φ (θ) π. Thus, F (θ, φ ) = n cos θ Z(L ) cos φ o(cos φ ), F (θ, φ ) = n cos3 θ (Z )(L ) cos φ o(cos φ ). From F (θ, φ ) = 0 and F (θ, φ ) = 0, we have n cos φ = Z(L ) cos θ o(cos θ), (3.33) n cos φ = (Z )(L ) cos3 θ o(cos 3 θ). (3.34) By using (3.33), we obtain Thus, F θ (θ, φ ) = n cos θ O(cos3 θ), F φ (θ, φ ) = Z(L ) O(cos 3 θ). dφ dθ = n Z(L ) cos θ O(cos3 θ). By using (3.34), we obtain (3.35) Thus, F θ (θ, φ ) = 3n cos θ O(cos 4 θ), F φ (θ, φ ) = (Z )(L ) O(cos 4 θ). dφ dθ = 3n (Z )(L ) cos θ O(cos 4 θ). (3.36) From (3.35) and (3.36), there exists a δ > 0 such that dφ dθ (θ) < dφ π (θ) < 0, dθ δ < θ < π. Note that φ (π/) = φ (π/) = π/. Therefore φ (θ) < φ (θ), π δ < θ < π. Note that φ (θ) and φ (θ) are continuously differentiable on [0, π/]. It follows from (3.3) and (3.37) that there exists at least one θ (0, π/) such that φ (θ) = φ (θ). Thus, the system (3.0) (3.) admits at least one nontrivial solution (θ, φ) in Ω. Lemma 3.0. Let n > 0 and n > 0 be given. If (θ, φ) is a nontrivial solution of the system (3.0) (3.) in Ω, then 0 < θ < π 4 if n > n, and π 4 < θ < π if n < n. Proof. Assume (θ, φ) Ω to be a nontrivial solution of the system (3.0) (3.), then 0 < θ < π and π < φ < π. It follows from Eqs. (3.0) and (3.) that Then where n n = cos φ sin φ cos 3 θ sin 3 θ cos φ sin θ cos 3 θ Z( sin θ cos φ) 3/ cos θ cos 4 θ cos φ sin φ cos 3 θ sin 3 θ cos φ sin 3 θ cos θ Z( sin θ cos φ) 3/ sin θ sin 4 θ. n n = n (cos θ sin θ) h, sin θ h h = cos φ sin θ cos θ(sin θ cos θ) Z( sin θ cos φ) 3/ (cos θ sin θ), h = cos φ sin φ cos 3 θ sin θ cos φ sin θ cos θ Z( sin θ cos φ) 3/ sin 3 θ. (3.37) (3.38)

16 056 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Note that for any (θ, φ) Ω, h = Z[ sin θ cos φ] 3/ (cos θ sin θ)( sin θ cos θ cos φ) Z( sin θ cos φ) ( sin θ cos θ cos φ) = (Z )( sin θ cos φ) Z, h Z cos 3 θ sin θ sin cos θ sin 3 θ Z cos θ sin θ sin θ cos θ sin 3 θ Z sin θ cos θ Z where we have used the inequality cos θ sin θ on [0, π/]. Thus, it follows from Eq. (3.38) that 0 < θ < π 4 if n > n, and π 4 < θ < π if n < n. Lemma 3.. Let n > 0 be given. Then there exists a ε > 0 such that for any n (n ε, n ε), the system of Eqs. (3.0) (3.) admits a unique nontrivial solution in Ω. Proof. By Theorem 3.7, system (3.0) (3.) admits a unique nontrivial solution (π/4, φ ) in Ω if n = n. Let n > 0 be given, and n > 0 be a parameter. Denote F (θ, φ) by F (θ, φ, n ) and F (θ, φ) by F (θ, φ, n ) respectively. When n = n, we have F θ F φ F θ F φ π 4, φ, n π 4, φ, n π 4, φ, n π 4, φ, n = n 4 sin4 φ (L ) cos φ sin φ (L ) cos φ 3(L ) cos φ, 4 = n (L ) sin 3 φ cos φ (L ) cos 3 φ sin φ Z(L ) sin φ ( 5 cos φ )( cos φ ) (L ) sin φ cos φ = n 4 sin4 φ (L ) cos φ sin φ (L ) cos φ 3(L ) cos φ, 4 = n (L ) sin 3 φ cos φ (L ) cos 3 φ sin φ Z(L ) sin φ ( 5 cos φ )( cos φ ) (L ) sin φ cos φ (L ) sin φ, (L ) sin φ. Note that F θ π 4, φ, n = F θ π 4, φ, n, F π φ 4, φ, n = F π φ 4, φ, n. Thus, F π det θ 4, φ, n F π θ 4, φ, n F π φ 4, φ, n F π φ 4, = F π θ 4, F π φ, n φ 4, φ, n. φ, n From the proof of Theorem 3.7, φ is the unique solution of F (π/4, φ) = 0 in (π/, π), that is, Thus, n 4 sin4 φ (L ) cos φ sin φ Z(L ) cos φ ( cos φ ) 3 4 (L ) cos φ (L ) cos φ = 0. n 4 sin4 φ (L ) cos φ sin φ = Z(L ) cos φ ( cos φ ) 3 (L ) cos φ (L ) cos φ. 4

17 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Then F θ π 4, φ, n = Z(L ) cos φ ( cos φ ) 3 (L ) cos φ (L ) cos φ [ = (L ) cos φ Z( cos φ ) 3 cos φ ] [ (L ) cos φ Z( cos φ ) cos φ ] By (3.7), we have F π φ 4, φ, n < 0. Thus, F π det θ 4, φ, n F π θ 4, φ, n (L ) cos φ (Z ) > 0. F π φ 4, φ, n F π φ 4, 0. φ, n By the implicit function theorem [0], there exists an ε > 0 such that for any n (n ε, n ε), the system of Eqs. (3.0) (3.) admits a unique nontrivial solution (θ(n ), φ(n )) in Ω satisfying (θ(n ), φ(n )) = ( π, 4 φ ). Theorem 3.. Let n > 0 and n > 0 be given. Then the system of Eqs. (3.0) (3.) admits a unique nontrivial solution in Ω. Consequently, Ẽ 0 (r, θ, φ) has a unique critical point in D0, that is, E 0 (r, r, φ) has a unique critical point in D 0. Proof. Let n > 0 and n > 0 be given. The case of n = n was discussed in Theorem 3.7. Assume n n. We discuss the case n < n only. The discussion of case n > n is very similar. By Lemma 3., the system of Eqs. (3.0) (3.) admits at least one nontrivial solution in Ω. By Lemma 3.9, the system of Eqs. (3.0) (3.) is equivalent to the equation φ (θ) = φ (θ). Hence the equation φ (θ) = φ (θ) admits at least one solution in (0, π/). We need to show that the equation φ (θ) = φ (θ) admits one unique solution in (0, π/). Assume the equation φ (θ) = φ (θ) admits two solutions θ a and θ b in (0, π/) and θ a θ b. Since n < n, by Lemma 3.0, we have π 4 < θ a < π and π 4 < θ b < π. Introduce a parameter n > 0, and let n in Eq. (3.0) be replaced by n. The system of Eqs. (3.0) and (3.) can be written as F (θ, φ, n) = 0, F (θ, φ, n ) = 0. (3.39) By denoting φ (θ) by φ (θ, n) and φ (θ) by φ (θ, n ), the equation φ (θ) = φ (θ) is then replaced by the following equation φ (θ, n) = φ (θ, n ). By Lemmas 3.8 and 3.9 and the above assumption, Eq. (3.40) admits two solutions θ (n) and θ (n) in (0, π/) satisfying θ (n ) = θ a and θ (n ) = θ b, which are continuous functions of n. By Lemma 3., there exists a ε > 0 such that for any n (n ε, n ε), Eq. (3.40) admits a unique nontrivial solution in (0, π/). Thus, one must have θ (n) = θ (n) for any n (n ε, n ε). Since θ a θ b, there must exist a n satisfying n < n n ε, and a small δ > 0, such that π 4 < θ (n) θ (n) < π, n (n δ, n ), π (3.4) 4 θ (n) = θ (n) < π, n [n, n ]. Let θ = θ (n ) = θ (n ) and φ = φ (θ, n ) = φ (θ, n ). Note that for any θ (0, π/) and n > 0, F (θ, φ (θ, n), n) = 0. Differentiate this equation with respect to n and let n = n in the resulting equation, where F n (θ, φ, n ) F φ (θ, φ, n ) φ n (θ, n ) = 0, F n (θ, φ, n ) = n sin 4 φ cos θ sin 3 θ 0. (3.40)

18 058 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) By using (3.6), we have φ n (θ, n ) = F (θ, φ, n ) n F φ (θ, φ, n ) 0. Note that θ (n) = θ (n) when n n n. By applying the implicit differentiation to Eq. (3.40) with respect to n on [n, n ] and letting n = n in the resulting equation, we obtain φ n (θ, n ) φ θ (θ, n ) dθ dn (n ) = dφ dθ (θ, n ) dθ dn (n ). Hence dφ dθ (θ, n ) φ θ (θ, n dθ ) dn (n ) = φ n (θ, n ) 0. Thus, we have Note that and dφ dθ (θ, n ) φ θ (θ, n ) 0. φ θ (θ, n ) = dφ dθ (θ, n ) = F θ (θ, φ, n ) F φ (θ, φ, n ), F θ (θ, φ, n ) F φ (θ, φ, n ). By using (3.6) and (3.7), we then have F θ (θ, φ, n F ) φ (θ, φ, n ) det F θ (θ, φ F 0., n ) φ (θ, φ, n ) By applying the implicit function theorem [0], there exists an η > 0 such that for any n (n η, n η), the system (3.39) admits a unique nontrivial solution (θ(n), φ(n)) in Ω satisfying (θ(n ), φ(n )) = (θ, φ ), which contradicts (3.4). Therefore Eq. (3.40) admits a unique solution θ = θ(n) in (0, π/) when n < n. Hence the system (3.0) (3.) admits a unique nontrivial solution in Ω. Consequently, Ẽ 0 (r, θ, φ) has a unique critical point in D0, hence, E 0 (r, r, φ) has a unique critical point in D 0. We wish to point out that Theorem 3. indicates that the effective energy potential E 0 (r, r, φ) with three quantum numbers [5,] can capture only the excited S-states of helium. 4. Computation of ground and excited states of helium As pointed in Section, developing a reliable and efficient numerical algorithm for the computation of ground and excited states of helium has been an active research topic in the existing literature. Salomonson and Öster [] proposed a numerical method using the finite discrete spectrum. Ho and Bhatia [] proposed later a complex rotation method based on the Hylleraas-type wave functions. Lindroth [3] applied the complex rotation method with the method proposed in []. Bürgers et al. [4] used the complex rotation method with the perimetric coordinates. All of these methods use the perturbation expansion and require a delicate and significant amount of computation work. Based on the mathematical analysis given in Section 3, we propose in this section two simple numerical algorithms to compute the ground and excited states of helium. The ground and excited states of helium correspond to those critical points of E(r, r, θ, θ, φ) specified by (.6) and (.7). By Theorem 3.3, E(r, r, θ, θ, φ) has at least one global minimum in D. By Theorems 3.6 and 3., E 0 (r, r, φ) = E(r, r, π/, π/, φ) admits a unique critical point (r, r, φ ) in D 0, which is in fact the global minimum of E 0 (r, r, φ). At a critical point of a functional, the number of negative eigenvalues of the Hessian matrix is called the Morse index of the functional at that critical point. If a critical point is not degenerate, then it is a local minimum if its Morse index is equal to 0, and it is a saddle point if its Morse index is greater than or equal to. Note that E is bounded below on D. If the Morse index of E at (r, r, π/, π/, φ ) is greater than or equal to one, then E will have at least two other critical points, which correspond to the local minima of E. If the Morse index of E at (r, r, π/, π/, φ ) is 0, then there are two cases:

19 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) (a) (r, r, π/, π/, φ ) is the global minimum of E; and (b) (r, r, π/, π/, φ ) is a local minimum of E. Our numerical examples have indicated that case (a) is always true. In the numerical algorithm to be given next, we compute first the global minimum (r, r, φ ) of E 0 (r, r, φ) in D 0 by the steepest descent method, which is based on Theorems 3.6 and 3.. Then we check the Morse index of E at (r, r, π/, π/, φ ). If it is greater than or equal to, then we compute the local minima of E in D by the steepest descent method. Steepest Descent Algorithm Step. Choose an initial guess (r 0, r 0, φ 0 ) R (π/, π) such that E 0(r 0, r 0, φ 0 ) < 0; Step. Find t > 0 such that E 0 (t r 0, t r 0, φ 0 ) = min t (0, ) E 0(tr 0, tr 0, φ 0 ) and replace (r 0, r 0, φ 0 ) by (t r 0, t r 0, φ 0 ); Step 3. Compute E 0 (r 0, r 0, φ 0 ); Step 4. If E 0 (r 0, r 0, φ 0 ) ϵ, then set (r 0, r 0, φ 0 ) = (r 0, r 0, φ 0 ) E 0 (r 0, r 0, φ 0 ), and goto Step ; Step 5. Compute the Morse index I of E at (r 0, r 0, π/, π/, φ 0 ); Step 6. If I = 0, then output (r 0, r 0, π/, π/, φ 0 ), and stop; else, choose an initial guesses (r 0, r 0, θ 0, θ 0, φ 0 ) D satisfying either or 0 < θ 0 < π, π < θ 0 < π, E(r 0, r 0, θ 0, θ 0, φ 0 ) < 0, π Step 7. Find t > 0 such that < θ 0 < π, 0 < θ 0 < π, E(r 0, r 0, θ 0, θ 0, φ 0 ) < 0; E(t r 0, t r 0, θ 0, θ 0, φ 0 ) = min t (0, ) E(tr 0, tr 0, θ 0, θ 0, φ 0 ) and replace (r 0, r 0, θ 0, θ 0, φ 0 ) by (t r 0, t r 0, θ 0, θ 0, φ 0 ); Step 8. Compute E(r 0, r 0, θ 0, θ 0, φ 0 ); Step 9. If E(r 0, r 0, θ 0, θ 0, φ 0 ) ϵ, then set (r 0, r 0, θ 0, θ 0, φ 0 ) = (r 0, r 0, θ 0, θ 0, φ 0 ) E(r 0, r 0, θ 0, θ 0, φ 0 ), and goto Step 7; else, output (r 0, r 0, θ 0, θ 0, φ 0 ), and stop. Remark 4.. Steps and 7 in the algorithm are based on Theorems 3. and 3.5 respectively, and expedite the convergence of the algorithm. The choice of θ and θ in Step 6 is based on Theorem 3.4, and generates two different local minima of E in D. When either l or l equals 0, E(r, r, θ, θ, φ) will have at least two local minima on the boundary of D. In most cases, the Steepest Descent Algorithm finds usually two local minima and one saddle point of E with I. To find those critical points of E of Morse index, we propose an algorithm based on the Mountain Pass Theorem due to [5]. The Mountain Pass Theorem has been used to prove the existence of critical points of functionals satisfying a condition called the Palais Smale (PS) condition, which occurs repeatedly in the critical point theory. We say that a functional J satisfies the PS condition if any sequence {v n } H for which J(v n ) is bounded and J (v n ) 0 possesses a convergent subsequence. We include the Mountain Pass Theorem for the benefit of readers. Theorem 4. ([5]). Let X be a real Banach space, B ρ = {v X v < ρ} and B ρ = {v E v = ρ}. Assume that J C (X, R) satisfies the PS condition and J(0) = 0, and assume that (a) there are constants ρ, α > 0 such that J B ρ α, (b) there is an e X \ B ρ such that J(e) 0. Then J possesses a critical value c α. Moreover, c can be characterized as c = inf max J(u), g Γ u g([0,]) where Γ = {g C([0, ], X) g(0) = 0, g() = e}.

20 060 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Table Singly excited states of helium. State n, l, n, l, L r r θ ( ) E E a,b (%) s5s s5s s5p s5p s5d s5d s5f s5f s6s s6s s6p s6p s6d s6d s6f s6f s7s s7s s7p s7p s7d s7d s7f s7f s8s s8s s8p s8p s8d s8d s8f s8f a Bürgerset al. [4]. b NIST [6]. S, 0, 5, 0, S, 0, 5, 0, P, 0, 5,, P, 0, 5,, D, 0, 5,, D, 0, 5,, F, 0, 5, 3, F, 0, 5, 3, S, 0, 6, 0, S, 0, 6, 0, P, 0, 6,, P, 0, 6,, D, 0, 6,, D, 0, 6,, F, 0, 6, 3, F, 0, 6, 3, S, 0, 7, 0, S, 0, 7, 0, P, 0, 7,, P, 0, 7,, D, 0, 7,, D, 0, 7,, F, 0, 7, 3, F, 0, 7, 3, S, 0, 8, 0, S, 0, 8, 0, P, 0, 8,, P, 0, 8,, D, 0, 8,, D, 0, 8,, F, 0, 8, 3, F, 0, 8, 3, From the proof of Theorem 3.3, it is straightforward to deduce that E(r, r, θ, θ, φ) satisfies the PS condition. Assume P D and P D are two local minima of E. We propose the following algorithm. Mountain Pass Algorithm Step. Choose an initial guess Q 0 = (r 0, r 0, θ 0, θ 0, φ 0 ) D satisfying 0 < θ 0 < π, 0 < θ 0 < π, π < φ 0 < π, E(Q 0 ) < 0. Step. Find Q (P Q 0 Q0 P ) such that E(Q ) = max E(Q ); Q (P Q 0 Q0 P ) Step 3. Compute E(Q ); Step 4. If E(Q ) ϵ, then set Q 0 = Q E(Q ), and goto Step ; else, output Q, and stop. Remark 4.. Since E is not degenerate, the critical point obtained by the Mountain Pass Algorithm has the Morse index. Remark 4.3. By choosing Q 0 satisfying or 0 < θ 0 < π/, π/ < θ 0 < π, π < φ 0 < π, E(Q 0 ) < 0, π/ < θ 0 < π, 0 < θ 0 < π/, π < φ 0 < π, E(Q 0 ) < 0, which is based on Theorem 3.4, the Mountain Pass Algorithm may find two different saddle points of E with the Morse index.

21 Z. Ding, G. Chen / Journal of Computational and Applied Mathematics 35 (0) Table Doubly excited S states (l = l = 0) of helium. State n, n, L r r θ ( ) E E a (%) ss s3s s3s s4s s4s s5s s5s s6s s6s 3s3s 3s4s 3s4s 3s5s 3s5s 3s6s 3s6s a Bürgerset al. [4]. S,, S, 3, S, 3, S, 4, S, 4, S, 5, S, 5, S, 6, S, 6, S 3, 3, S 3, 4, S 3, 4, S 3, 5, S 3, 5, S 3, 6, S 3, 6, Table 3 Doubly excited states of helium. State n, l, n, l, L r r θ ( ) E E a (%) sp sp pp pp s3p s3p p3s p3p p3p s3d s3d p3d p3d 3p3p 3d3s 3d3s 3d3d 3d3p a Lindroth [3]. 3 P o, 0,,, P o, 0,,, P e,,,, D e,,,, P o, 0, 3,, P o, 0, 3,, P o,, 3, 0, P e,, 3,, D e,, 3,, D e, 0, 3,, D e, 0, 3,, D o,, 3,, F o,, 3,, P e 3,, 3,, D e 3,, 3, 0, D e 3,, 3,, F e 3,, 3,, D o 3,, 3,, By using the algorithms proposed in this paper, we computed three sets of data for singly and doubly excited states of helium, where π = () l l is the parity (o: odd; e: even), and θ (in degrees) denotes the angle between r and r. The singly excited states of helium are given in Table. By comparing with the reference values from [4,6], we have an excellent approximation for the singly excited states with the relative error being less than 0.%. The doubly excited states of helium are given in Tables and 3. The data are compared with the reference values from [4,3]. Variances between the existing data and ours do exist while there are no universally accepted exact values. Nevertheless, the overall agreement is quite favorable. Acknowledgements This research is supported in part by the ONR Grants N and N , and the TITF initiative from Texas A&M University. Part of this work was completed while both authors were visiting the Taida Institute for Mathematical Sciences, National Taiwan University, and the Institute of Mathematics, Academia Sinica, Taipei, Taiwan, in the summer of 007, and the Institute for Mathematical Sciences, National University of Singapore, in the summer of 008. Their hospitality is gratefully acknowledged. References [] G. Tanner, K. Richter, J. Rost, The theory of two electron atoms: between ground state and complete fragmentation, Rev. Modern Phys. 7 (000) [] R.P. Madden, K. Codling, New autoionizing atomic energy levels in He, Ne, and Ar, Phys. Rev. Lett. 0 (963)