A Solution of the Spherical Poisson-Boltzmann Equation
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1 International Journal of Mathematical Analysis Vol. 1, 018, no. 1, 1-7 HIKARI Ltd, A Solution of the Spherical Poisson-Boltzmann quation. onseca Universidad Nacional de Colombia Grupo de Ciencia de Materiales y Superficies Departamento de ísica Bogotá, Colombia Copyright c 018. onseca. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, we used a field and coordinate transformations and the Jacobi lliptic functions, in order to solve the Spherical Poisson-Boltzman (SPB) equation. We find several families of solutions. Keywords: Spherical Poisson-Boltzman, Jacobi lliptic functions 1 Introduction The spherical Poisson-Boltzmann equation (SPBq) gives the electrical potential distribution for a charged spherical electrolyte [1]-[]. SPB equation is a mean field theory that gives into account of the electrostatic interactions between the charges in an ionic solution [3]. Also, a such enormous theoretical effort has been done over the years, in order to provide analytical solutions to nonlinear partial differential equations. Among the most popular developed methods we can find the so called solitary wave methods. We employ the Jacobi lliptic functions [4] and [5] solitary wave methods. Spherical Poisson-Boltzmann equation We start from the spherical Poisson-Boltzmann equation [1]-[]:
2 . onseca ρ = d ρ dr + dρ r dr = κ e ρ (1) Where κ = λ 1 D, and λ D is known as the Debye-Hückel screening length [1]-[]. Using the transformation [7] And [6] r = e ζ, And replacing in eq. (1) Now defining the variables Then ρ = ψ ln (r) + ln () () d dr = e ζ d dζ, d dr = e ζ ( d dζ d dζ ) (3) ψ dζ + ψ dζ = κ e ψ (4) v = e ψ (5) dψ dζ = 1 dv v dζ (6) And replacing in eq. (4) d ψ dζ = 1 v (dv dζ ) + 1 d v (7) v dζ 3 Solution 1 v d v dζ (dv dζ ) + v dv dζ v κ v 3 = 0 (8) We suppose a solution given by the Jacobi elliptic functions, [4] and [5]. They hold the next relations sn (ζ, k) + cn (ζ, k) = 1, k sn (ζ, k) + dn (ζ, k) = 1 (9) dn (ζ, k) k cn (ζ, k) = k, k sn (ζ, k) + cn (ζ, k) = dn (ζ, k) (10)
3 Spherical Poisson-Boltzmann equation 3 Then, we suppose the next solution k = 1 k (11) dv v = Acn(aζ, k); = aasn(aζ, k)dn(aζ, k); (1) dζ d v dζ = Aa ( k cn 3 (aζ, k) (1 k )cn(aζ, k)) Then, replacing eqs. (1) in eq. (8) A a k cn 4 (aζ, k) A a (1 k )cn (aζ, k) (13) a A sn (aζ, k)dn (aζ, k) aa cn(aζ, k)sn(aζ, k)dn(aζ, k) A cn (aζ, k) κ A 3 cn 3 (aζ, k) = 0 quating the left hand side of eq. (13) to zero, we get: Then ( a sn(aζ, k)dn(aζ, k) acn(aζ, k))a sn(aζ, k)dn(aζ, k) = 0 (14) asn(aζ, k)dn(aζ, k) = cn(aζ, k) ( a (1 k ) )A cn (aζ, k) = 0 a 1, = ± (k 1) (15) And So, using eqs. (9) and (14), we get ( a k cn(aζ, k) Aκ )A cn 3 (aζ, k) = 0 (16) cn(aζ, k) = Aκ a k cn (aζ, k) = A κ 4 a 4 k 4 Solving for sn (aζ, k) k sn 4 (aζ, k) sn (aζ, k) + A κ 4 a 6 k 4 = 0 (17)
4 4. onseca sn (aζ, k) = 1 ± 1 4k A κ 4 a 6 k 4 (18) k Again, using eqs. (16), (18) and (9), we have 1 ± 1 4k A κ 4 a 6 k 4 k + A κ 4 a 4 k 4 = 1 (19) a 8 k 4 4a k A κ 4 = (k a 4 k a 4 k A κ 4 ) (0) If we define in eq. (0) = 4κ 8, = (4a k κ 4 + 8a 4 k κ 4 8a 4 k 4 κ 4 ) and G = +3a 8 k 4 8a 8 k 6 + 4a 8 k 8, we get solving for A A 4 + A + G = 0 (1) A 1 = 4G () A = 4G (3) A 3 = + 4G (4) A 4 = + 4G (5) So, we find eight families of solutions v = A i cn(a j ζ, k) (6)
5 Spherical Poisson-Boltzmann equation 5 4 Solution Also, we suppose a solution dv v = Asn(aζ, k); = aacn(aζ, k)dn(aζ, k); (7) dζ d v dζ = Aa (k sn 3 (aζ, k) (1 + k )sn(aζ, k)) Then, replacing eq. (7) in eq. (8) A a k sn 4 (aζ, k) a (1 + k )A sn (aζ, k) (8) a A cn (aζ, k)dn (aζ, k) + aa sn(aζ, k)cn(aζ, k)dn(aζ, k) A sn (aζ, k) κ A 3 sn 3 (aζ, k) = 0 quating the left hand side of eq. (8) to zero, we get: Also ( a cn(aζ, k)dn(aζ, k) + asn(aζ, k))a cn(aζ, k)dn(aζ, k) = 0 (9) acn(aζ, k)dn(aζ, k) = sn(aζ, k) ( a (1 + k ) )A sn (aζ, k) = 0 a 1, = ±i (k + 1) (30) And (a k sn(aζ, k) Aκ )A sn 3 (aζ, k) = 0 (31) sn(aζ, k) = Aκ a k sn (aζ, k) = A κ 4 a 4 k 4 So, using eqs. (9), (31) and eq. (9), we have cn (aζ, k) = Using eqs. (31), (3) and (9), we get A κ 4 a 6 k 4 (1 A κ 4 a 4 k ) (3) A 4 A + G = 0 (33)
6 6. onseca where, we define = κ 8, = (a k κ 4 (1 + k + a k ) and G = a 8 k 6 in eq. (33), and solving A 5 = 4G (34) A 6 = 4G (35) A 7 = + 4G (36) A 8 = + 4G (37) So, we get eight families of solutions v = A i sn(a j ζ, k) (38) 5 Conclusions We solved the spherical Poisson Boltzmann equation using the Jacobi lliptic functions. We obtain sixteen families of solutions. As a future work, we can extend the method to investigate other charge configurations and other geometric symmetries. ρ(r) = ln (A i cn(a j ln (r), k)) ln (r) + ln () (39) ρ(r) = ln (A i sn(a j ln (r), k)) ln (r) + ln () (40) Acknowledgements. This research was supported by Universidad Nacional de Colombia in Hermes project (3501).
7 Spherical Poisson-Boltzmann equation 7 References [1] D. Andelman, Introduction to electrostatics in soft and biological matter, Chapter in Soft Condensed Matter Physics in Molecular and Cell Biology, d. by W. Poon and D. Andelman, Taylor & rancis, New York, 006, [] D. Andelman, in Handbook of Physics of Biological Systems, Vol. I, Chap. 1, edited by R. Lipowsky and. Sackmann, lsevier, Amsterdam, 1994, 603. [3] D. McQuarrie, Statistical Mechanics, University Science Books, 000. [4] Zuntao u, Shikuo Liu, Shida Liu, Qiang Zhaoa, New Jacobi elliptic function expansion and new periodic solutions of nonlinear wave equations, Physics Letters A, 90 (001), [5] Zhenya Yan, Jacobi elliptic function solutions of nonlinear wave equations via the new sinh-gordon equation expansion method. J. Phys. A: Math. Gen., 36 (003), [6] S. Chandrasekhar, An Introduction to the Study of Stellar Structure, Dover Publications, New York, [7] M.A. Soliman and Y. Al-Zeghayer, Approximate Analytical Solution for the Isothermal Lane mden quation in a Spherical Geometry, Revista Mexicana de Astronomía y Astrofísica, 51 (015), Received: December 1, 017; Published: January 5, 018
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