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Explicit computation of the electrostatic energy for an elliptical charged disc.
Sophie Laurens, Sébastien Tordeux
To cite this version:
Sophie Laurens, Sébastien Tordeux. Explicit computation of the electrostatic energy for an elliptical charged disc.. Applied Mathematics Letters, Elsevier, 2013, 26 (2), pp.301-305.
�10.1016/j.aml.2012.09.013�. �hal-00723234v2�
Explicit computation of the electrostatic energy for an elliptical charged disc
S. Laurensa,b, S. Tordeuxc
aMathematical Institute of Toulouse, 118 route de Narbonne, F-31400 Toulouse
bCERFACS, 42 Avenue Gaspard Coriolis F-31100 Toulouse
cINRIA Bordeaux Sud-Ouest-LMA, avenue de l'Université, F-64013 Pau
Abstract
This letter describes a method for obtaining an explicit expression for the elec- trostatic energy of a charged elliptical innitely thin disc. The charge distri- bution is assumed to be polynomial. Such explicit values for this energy are fundamental for assessing the accuracy of boundary element codes. The main tools used are an extension of Copson's method and a diagonalization, given by Leppington and Levine, of the single-layer potential operator associated with the electrostatic potential created by a distribution of charges on the elliptical disc.
1. Introduction
In recent years, integral equations have become an essential tool for solv- ing both industrial and scientic problems in electromagnetism and acoustics.
The assessment of the accuracy delivered by such codes, in particular in their handling of the singular integrals involved, is a major issue. Here, we present a method for deriving an analytical expression for the electrostatic energy of a charged elliptical innitely thin plate, providing a means for the validation of these codes.
Let us denote byA=
(x1, x2)∈R2 withx21/a2+x22/b2−1<0anda > b the ellipse with major and minor semi-axesaandb. Let f be the electrostatic potential generated by a density of chargesσdistributed over A:
f(x) = 1 4π
Z
A
σ(y)
|x−y|dsy for allxin A, (1) withx= (x1, x2). The units have been chosen such that the electric permittivity of air is1. The electrostatic energyIcan be expressed in either the two following forms:
Iσ= Z
A
f(x)σ(x)dsx= Z
A
1 4π
Z
A
σ(x)σ(y)
|x−y| dsxdsy. (2)
Email address: sophie.laurens@insa-toulouse.fr (S. Laurens)
We aim in this letter at proving and numerically illustrating the following the- orem, whereεis the eccentricity of the ellipseAgiven byε=p
1−b2/a2. Theorem 1.1. Let σ(x) =α0+α1x1/a+α2x2/b, withα∈R3, be the distri- bution of charges overA. The corresponding electrostatic energy is given by
Iσ= 8ab2 15π
5α20+α22
K(ε) + α21−α22K(ε)−E(ε) ε2
,
withK(ε)andE(ε) the complete elliptic integrals of the rst and second kind K(ε) =
Z π/2 0
dφ
p1−ε2sin2φ and E(ε) = Z π/2
0
q
1−ε2sin2φ dφ. (3)
2. Diagonalization of the electrostatic energy
Following [1], we consider the spheroidal coordinate system (θ, ϕ) giving a parametrization ofAin terms of the unit half-sphere
x1=asinθ cosϕ and x2=bsinθ sinϕ, with θ∈[0, π/2], ϕ∈[0,2π]. (4) The elemental area associated with the new variables is abcosθsinθ dθdϕ. In these spheroidal coordinates, the electrostatic potentialf dened in (1) and the electrostatic energy can also be written in terms ofθ andϕas
f(θ, ϕ) = ab 4π
Z π/2 0
Z 2π 0
g(θ′, ϕ′)
d(θ, ϕ, θ′, ϕ′)sinθ′dθ′dϕ′, Iσ=ab
Z π/2 0
Z 2π 0
f(θ, ϕ)g(θ, ϕ) sinθ dθdϕ,
withg(θ, ϕ) =σ(θ, ϕ) cosθand d(θ, ϕ, θ′, ϕ′)the distance separatingx fromy d=|x−y|expressed in the spheroidal coordinates (4).
The next step consists in introducing a well chosen spectral basis for the half- sphere involving the even Legendre functionsQmn normalized by
Z π 0
Qmn (cosθ)Qmn′(cosθ) sinθdθ=δn,n′. (5) This basis yields a block diagonalization of the convolution operator (see [1])
1 d = 1
√ab
∞
X
n=0 n
X
m=−n n−meven
n
X
m′=−n n−m′even
dnmm′Qmn (cosθ)Qmn′(cosθ′)ei(mϕ−m′ϕ′),
withdnmm′ =Qmn (0)Qmn′(0) 2n+ 1
Z 2π 0
ei(m−m′)ϕ qb
acos2ϕ+absin2ϕ
dϕ. (6)
2
The functionsf andgcan be expanded in this basis as u(θ, ϕ) =
∞
X
n=0 n
X
m=−n n−meven
umnQmn (cosθ)eimϕ withu=f or g (7)
with
umn = 1 π
Z π/2 0
Z 2π 0
u(θ, ϕ)Qmn(cosθ)e−imϕsin(θ)dθdϕ. (8) Due to (6), coecientsfnmare related togmn by
fnm =
√ab 4
n
X
m′=−n n−m′even
dnmm′gnm′′. (9)
Moreover, the orthogonal properties of the spectral basis yield Iσ = π
4(ab)3/2 X
n,m,m′
dnmm′gmngmn′,
where we have lightened the notation by making the range of the summation index implicit. Indicesn, n′ are varying from0 to∞, andm, m′ are such that
|m| ≤n, |m′| ≤n′, with n−m andn−m′ even. Substituting expression (6) fordnmm′ and introducing the eccentricity of the ellipse ε, we get
Iσ=π ab2 X
n,m,m′
gnmgmn′ Qmn (0)Qmn′(0) 2n+ 1
Z π/2 0
cos(m−m′)ϕ
p1−ε2cos2ϕdϕ. (10) 3. Electrostatic energy for an ane distribution of charges
This section is dedicated to the calculation of the electrostatic energy gene- rated by an ane density of charges.
Proof of Theorem 1.1. Letσ0(x) = 1, σ1(x) = x1/a andσ2(x) = x2/b. In view of the symmetry ofAwith respect tox1andx2, we have
Z
A
1 4π
Z
A
σi(x)σj(y)
|x−y| dsxdsy = 0 fori6=j.
Consequently, the electrostatic energyIσ can be expanded as Iσ=α20Iσ0+α21Iσ1+α22Iσ2
The result will follow from the computation ofIσ0,Iσ1 andIσ2. 3.1. Computation ofIσ0
For σ(x) =σ0(x) = 1, the function g(θ, ϕ) = cosθ does not depend on ϕ.
The gm coecients are independent from a and b, and since gm = 0 for all
m 6= 0, the function g can be expanded as g(θ) =
+∞
X
n=0
gn0Q0n(cosθ) . Due to (10), the electrostatic energyIσ0 depends only onaandband is given by
Iσ0(a, b) =π ab2
+∞
X
n=0
|g0n|2 Q0n(0)2
2n+ 1
Z π/2 0
dϕ
p1−ε2cos2ϕ =κ ab2K(ε), (11) withε =p
1−b2/a2 and κa constant depending neither ona nor onb. The constant κis deduced from the classical case of an unit circle which has been detailed for example in [2]
Iσ0(1,1) = Z
C
1 4π
Z
C
1
|x−y|dsxdsy= 4/3. (12) Comparing (11) and (12), this yields toκ= 8/3πsinceK(0) =π/2. Therefore,
Iσ0 = 8
3πab2K(ε). (13)
3.2. Computation ofIσ1
Forσ(x) =σ1(x) =x1/a, the functiongis given byg(θ, ϕ) = sinθcosθcosϕ.
In that case, thegmn are zero except for|m|= 1. By denition of the Legendre functions, we have Q−n1 = −Q1n. As the function g is even, it emerges that g−n1=−gn1, and thus
Iσ1=πab2X
n
g1nQ1n(0)2
2n+ 1
"
2 Z π/2
0
dϕ
p1−ε2cos2ϕ+ 2 Z π/2
0
cos 2ϕ dϕ p1−ε2cos2ϕ
# . Due to (3), it emerges that
Iσ1(a, b) =κ ab2 K(ε)−E(ε)
ε2 . (14)
To determine the constant κ, we consider again the case of an unit circle. In this case,Iσ1 can be explicitly computed (see the Appendix A), and is given by Iσ1(1,1) = 2/15. Evaluating (14) ata=b= 1we get
Iσ1(1,1) =κπ
4 since lim
ε→0
K(ε)−E(ε) ε2 = π
4. (15)
It follows thatκ= 8/15πand therefore we have Iσ1 = 8
15πab2K(ε)−E(ε)
ε2 . (16)
4
3.3. Computation ofIσ2
ForIσ2, we considerσ(x) =σ2(x) =x2/b, meaning thatg(θ, ϕ) = sinθcosθsinϕ.
We still havegnm= 0except for|m|= 1, but in that case,gn−1=g1n. Thus
Iσ2=πab2
+∞
X
n=0
g1nQ1n(0)2
2n+ 1
"
2 Z π/2
0
dϕ
p1−ε2cos2ϕ−2 Z π/2
0
cos 2ϕ dϕ p1−ε2cos2ϕ
#
Moreover, both integralsIσ1 andIσ2 are equal onCby symmetry. We obtain Iσ2 = 8
15π ab2
K(ε)−K(ε)−E(ε) ε2
. (17)
4. Numerical tests and conclusion
Tables 1 and 2 give a comparison of the exact values given by an analyt- ical expression with numerical approximate values obtained by the boundary element code CESC of CERFACS withP1 continuous elements. It can be ob- served that the two values coincide at least up to the fourth decimal digit. Table 3 shows the maximum relative error for each of the cases of Tables 1 and 2 cases, which is less than0.35per mil.
a 0.5 0.7 0.9 1.1 1.3 1.5
Iσcomp0 0.1666 0.2741 0.3939 0.5234 0.6608 0.8048 Iσexact0 0.1666 0.2741 0.3939 0.5234 0.6608 0.8048 Iσcomp1 ×10−1 0.0417 0.1455 0.3651 0.7535 1.3715 2.2781 Iσexact1 ×10−1 0.0417 0.1456 0.3656 0.7543 1.3717 2.2800 Iσcomp2 ×10−2 0.4167 0.6280 0.8426 1.0585 1.2748 1.4910 Iσexact2 ×10−2 0.4167 0.6280 0.8427 1.0586 1.2748 1.4911 Table 1: Exact and computed values of the electrostatic energy,Iσexacti andIσcompi , for
an elliptical disc with minor axisb= 0.5, given by (13), (16), (17) .
a 0.75 0.9 1.05 1.2 1.35 1.5
Iσcomp 2.7734 3.6157 4.5162 5.4706 6.4761 7.5313 Iσexact 2.7736 3.6159 4.5165 5.4708 6.4763 7.5316
Table 2: Ane density of chargesσ=x1+ 2x2+ 3for an elliptical disc withb= 0.5.
Iσ0 Iσ1 Iσ2 Iσ
εrel 0.0550/00 0.1950/00 9.1×10−40/00 0.3340/00
Table 3: Maximum value of the relative errorεrel= max|Iσexacti −Iσcompi |in per mil for an elliptical disc forb= 0.5anda= 0.5 : 0.05 : 1.5.
Acknowledgements
The authors would like to express their thanks to A. Bendali (INSA) for fruitful discussions and M. Fares (CERFACS) for the numerical computations achieved with the CERFACS code CESC. Part of this work was supported by the French National Research Agency under grant no. ANR−08−SYSC−001.
Appendix A. The case of an unit circle disc
LetCbe the circle with radius1. We aim here at computing the integral IC = 1
4π Z
C
x1y1
|x−y| dsxdsy
This integral is rewritten in polar coordinates (r, φforxandρ,φ′ fory) as IC = 1
4π Z 1
0
Z 2π 0
Z 1 0
Z 2π 0
rcosφ ρcosφ′
pr2+ρ2−2rρcos(φ−φ′)rdrdφ ρdρdφ′ and evaluated using to the formula(3.4.5) of [3, p70]
Z 2π 0
cosφ dφ
pρ2+r2−2rρcos(φ−φ′) = 4 cosφ′ ρr
Z min(ρ,r) 0
t2dt pρ2−t2√
r2−t2. This leads to
IC = 1 π
Z 1 0
Z 1 0
Z min(ρ,r) 0
rρ t2 pρ2−t2√
r2−t2 drdρdt Z 2π
0
cos2φ′dφ′.
This integral is symmetric inρandr, and sinceZ 2π 0
cos2φ′dφ′=π, we have
IC= 2 Z 1
t=0
t2 Z 1
ρ=t
ρ pρ2−t2
Z 1 r=ρ
√ r
r2−t2 dr dρ dt= 2 15. References
[1] F. Leppington, H. Levine, Reexion and transmission at a plane screen with periodically arranged circular or elliptical apertures, J. Fluid Mech 61 (1973) 109127.
[2] S. Laurens, S. Tordeux, A. Bendali, M. Fares, R. Kotiuga, Lower and up- per bounds for the Rayleigh conductivity of a perforated plate, submitted, http://hal.archives-ouvertes.fr/hal-00686438.
[3] E. Copson, On the problem of the electried disc, Proceedings of the Edin- burgh Mathematical Society (Series 2) 8 (01) (1947) 1419.
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