Explicit computation of the electrostatic energy for an elliptical charged disc

0.13 MB·

by  S. Laurens

Explicit computation of the electrostatic energy for an
elliptical charged disc
S. Laurensa,b, S. Tordeuxc
4
aMathematical Institute of Toulouse, 118 route de Narbonne, F-31400 Toulouse
1 bCERFACS, 42 Avenue Gaspard Coriolis F-31100 Toulouse
0 cINRIA Bordeaux Sud-Ouest-LMA, avenue de l’Université,F-64013 Pau
2

n
a
J

6 Abstract

] This letter describes a method for obtaining an explicit expression for the elec-
A
trostatic energy of a charged elliptical infinitely thin disc. The charge distri-
N bution is assumed to be polynomial. Such explicit values for this energy are
. fundamental for assessing the accuracy of boundary element codes. The main
h
t toolsusedareanextensionofCopson’smethodanda diagonalization,givenby
a
Leppington and Levine, of the single-layer potential operator associated with
m
the electrostatic potential created by a distribution of charges on the elliptical
[
disc.

1
v
1. Introduction
3
1
In recent years, integral equations have become an essential tool for solv-
1
1 ing both industrial and scientific problems in electromagnetism and acoustics.
. The assessment of the accuracy delivered by such codes, in particular in their
1
handling of the singular integrals involved, is a major issue. Here, we present
0
4 a method for deriving an analytical expression for the electrostatic energy of a
1 charged elliptical infinitely thin plate, providing a means for the validation of
v: these codes.
i
X LetusdenotebyA= (x ,x ) R2 with x2/a2+x2/b2 1<0 and a>b
r 1 2 ∈ 1 2 −
the ellipse with major and minor semi-axes a and b. Let f be the electrostatic
a
potential generated by a(cid:8)density of charges σ distributed over A: (cid:9)
1 σ(y)
f(x)= ds for all x in A, (1)
y
4π A x y
Z | − |
withx=(x ,x ). Theunitshavebeenchosensuchthattheelectricpermittivity
1 2
ofairis1. TheelectrostaticenergyI canbeexpressedineitherthetwofollowing
forms:
1 σ(x)σ(y)
I = f(x)σ(x)ds = ds ds . (2)
σ A x A 4π A x y x y
Z Z Z | − |
Email address: [email protected] (S.Laurens)
Preprint submitted toApplied Mathematics Letters January 7, 2014
We aim in this letter at proving and numerically illustrating the following the-
orem, where ε is the eccentricity of the ellipse A given by ε= 1 b2/a2.
−
Theorem 1.1. Let σ(x) = α +α x /a+α x /b, with α Rp3, be the distri-
0 1 1 2 2
bution of charges over A. The corresponding electrostatic en∈ergy is given by
8ab2 K(ε) E(ε)
I = 5α2+α2 K(ε)+ α2 α2 − ,
σ 15π 0 2 1− 2 ε2
(cid:20) (cid:21)
(cid:0) (cid:1) (cid:0) (cid:1)
with K(ε) and E(ε) the complete elliptic integrals of the first and second kind
π/2 dφ π/2
K(ε)= and E(ε)= 1 ε2sin2φdφ. (3)
Z0 1−ε2sin2φ Z0 q −
p
2. Diagonalization of the electrostatic energy
Following [1], we consider the spheroidal coordinate system (θ,ϕ) giving a
parametrizationof A in terms of the unit half-sphere
x =a sinθ cosϕ and x =b sinθ sinϕ, with θ [0,π/2], ϕ [0,2π]. (4)
1 2
∈ ∈
The elemental area associated with the new variables is abcosθsinθdθdϕ. In
thesespheroidalcoordinates,the electrostaticpotentialf definedin(1)andthe
electrostatic energy can also be written in terms of θ and ϕ as
ab π/2 2π g(θ′,ϕ′)
′ ′ ′
f(θ,ϕ)= sinθ dθ dϕ,
4π d(θ,ϕ,θ′,ϕ′)
 Z0 Z0
π/2 2π
 Iσ =ab f(θ,ϕ)g(θ,ϕ)sinθdθdϕ,
Z0 Z0
with g(θ,ϕ)=σ(θ,ϕ) cosθ and d(θ,ϕ,θ′,ϕ′) the distance separating x from y
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 functions Qm normalized by
n
π
Qmn (cosθ)Qmn′(cosθ)sinθdθ =δn,n′. (5)
Z0
This basis yields a block diagonalizationof the convolution operator (see [1])
∞ n n
d1 = √1ab dnmm′Qmn (cosθ)Qmn′(cosθ′)ei(mϕ−m′ϕ′),
nX=0n−mXm=−evnenn−mmX′=′−evnen
Qm(0)Qm′(0) 2π ei(m−m′)ϕ
with dnmm′ = n 2n+n1 Z0 b cos2ϕ+ asin2ϕdϕ. (6)
a b
q
2
The functions f and g can be expanded in this basis as
∞ n
u(θ,ϕ)= umQm(cosθ)eimϕ with u=f or g (7)
n n
n=0 m=−n
X n−Xmeven
with
1 π/2 2π
um = u(θ,ϕ)Qm(cosθ)e−imϕsin(θ)dθdϕ. (8)
n π n
Z0 Z0
Due to (6), coefficients fm are related to gm by
n n
fnm = √4ab n dnmm′gnm′′. (9)
n−mmX′=′−evnen
Moreover,the orthogonal properties of the spectral basis yield
π
Iσ = 4(ab)3/2 dnmm′gnmgnm′,
n,m,m′
X
where we have lightened the notation by making the range of the summation
′ ′
index implicit. Indices n,n are varying from 0 to , and m,m are such that
′ ′ ′ ∞
m n, m n, with n m and n m even. Substituting expression (6)
| | ≤ | | ≤ − −
for dn and introducing the eccentricity of the ellipse ε, we get
mm′
Qm(0)Qm′(0) π/2 cos(m m′)ϕ
I =πab2 gmgm′ n n − dϕ. (10)
σ n,Xm,m′ n n 2n+1 Z0 1−ε2cos2ϕ
p
3. Electrostatic energy for an affine distribution of charges
This section is dedicated to the calculation of the electrostatic energy gene-
rated by an affine density of charges.
Proof of Theorem 1.1. Let σ (x) = 1, σ (x) = x /a and σ (x) = x /b. In
0 1 1 2 2
view of the symmetry of A with respect to x and x , we have
1 2
1 σ (x)σ (y)
i j
ds ds = 0 for i=j.
x y
A 4π A x y 6
Z Z | − |
Consequently, the electrostatic energy I can be expanded as
σ
I =α2I +α2I +α2I
σ 0 σ0 1 σ1 2 σ2
The result will follow from the computation of I , I and I .
σ0 σ1 σ2
3.1. Computation of I
σ0
For σ(x) = σ (x) = 1, the function g(θ,ϕ) = cosθ does not depend on ϕ.
0
The gm coefficients are independent from a and b, and since gm = 0 for all
n n
3
+∞
m = 0, the function g can be expanded as g(θ) = g0Q0 (cosθ) . Due to
6 n n
n=0
X
(10), the electrostatic energy I depends only on a and b and is given by
σ0
+∞ g0 2 Q0 (0) 2 π/2 dϕ
I (a,b)=πab2 | n| n =κab2K(ε), (11)
σ0 nX=0 2n(cid:0) +1 (cid:1) Z0 1−ε2cos2ϕ
p
with ε = 1 b2/a2 and κ a constant depending neither on a nor on b. The
−
constant κ is deduced from the classical case of an unit circle which has been
p
detailed for example in [2]
1 1
I (1,1) = ds ds =4/3. (12)
σ0 C 4π C x y x y
Z Z | − |
Comparing (11) and(12), this yields to κ=8/3π since K(0)=π/2. Therefore,
8
I = ab2K(ε). (13)
σ0 3π
3.2. Computation of I
σ1
Forσ(x)=σ (x)=x /a,thefunctiongisgivenbyg(θ,ϕ)=sinθcosθcosϕ.
1 1
In that case, the gm are zero except for m =1. By definition of the Legendre
functions, we havenQ−1 = Q1. As th|e |function g is even, it emerges that
g−1 = g1, and thus n − n
n − n
g1Q1(0) 2 π/2 dϕ π/2 cos2ϕ dϕ
I =πab2 n n 2 +2 .
σ1 Xn (cid:0) 2n+1(cid:1) " Z0 1−ε2cos2ϕ Z0 1−ε2cos2ϕ#
p p
Due to (3), it emerges that
K(ε) E(ε)
I (a,b)=κab2 − . (14)
σ1 ε2
To determine the constant κ, we consider again the case of an unit circle. In
this case, I canbe explicitly computed (see the Appendix A), andis givenby
σ1
I (1,1)=2/15. Evaluating (14) at a=b=1 we get
σ1
π K(ε) E(ε) π
I (1,1)=κ since lim − = . (15)
σ1 4 ε→0 ε2 4
It follows that κ=8/15π and therefore we have
8 K(ε) E(ε)
I = ab2 − . (16)
σ1 15π ε2
4
3.3. Computation of I
σ2
ForI ,weconsiderσ(x)=σ (x)=x /b,meaningthatg(θ,ϕ)=sinθcosθsinϕ.
σ2 2 2
We still have gm =0 except for m =1, but in that case, g−1 =g1. Thus
n | | n n
+∞ g1Q1(0) 2 π/2 dϕ π/2 cos2ϕ dϕ
I =πab2 n n 2 2
σ2 nX=0(cid:0) 2n+1(cid:1) " Z0 1−ε2cos2ϕ − Z0 1−ε2cos2ϕ#
Moreover,both integrals I and I pare equal on C by symmpetry. We obtain
σ1 σ2
8 K(ε) E(ε)
I = ab2 K(ε) − . (17)
σ2 15π − ε2
(cid:18) (cid:19)
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 with P1 continuous elements. It can be ob-
servedthatthetwovaluescoincideatleastuptothefourthdecimaldigit. Table
3showsthemaximumrelativeerrorforeachofthecasesofTables1and2cases,
which is less than 0.35 per mil.
a 0.5 0.7 0.9 1.1 1.3 1.5
Icomp 0.1666 0.2741 0.3939 0.5234 0.6608 0.8048
σ0
Iexact 0.1666 0.2741 0.3939 0.5234 0.6608 0.8048
σ0
Icomp 10−1 0.0417 0.1455 0.3651 0.7535 1.3715 2.2781
Iσex1act ×10−1 0.0417 0.1456 0.3656 0.7543 1.3717 2.2800
σ1 ×
Icomp 10−2 0.4167 0.6280 0.8426 1.0585 1.2748 1.4910
Iσex2act ×10−2 0.4167 0.6280 0.8427 1.0586 1.2748 1.4911
σ2 ×
Table 1: Exactand computedvaluesof theelectrostatic energy,Iσexiact andIσcoimp,for
an elliptical disc with minor axis b=0.5, given by (13), (16), (17) .
a 0.75 0.9 1.05 1.2 1.35 1.5
Icomp 2.7734 3.6157 4.5162 5.4706 6.4761 7.5313
σ
Iexact 2.7736 3.6159 4.5165 5.4708 6.4763 7.5316
σ
Table 2: Affinedensityofchargesσ=x1+2x2+3foranelliptical discwithb=0.5.
I I I I
σ0 σ1 σ2 σ
εrel 0.0550/00 0.1950/00 9.1 10−40/00 0.3340/00
×
Table 3: Maximum valueof therelative error εrel =max|Iσexiact−Iσcoimp| in permil
for an elliptical disc for b=0.5 and a=0.5:0.05:1.5.
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
Let C be the circle with radius 1. We aim here at computing the integral
1 x y
1 1
IC = dsxdsy
4π C x y
Z | − |
This integral is rewritten in polar coordinates (r, φ for x and ρ, φ′ for y) as
1 1 2π 1 2π rcosφρcosφ′
′
IC = rdrdφρdρdφ
4π r2+ρ2 2rρcos(φ φ′)
Z0 Z0 Z0 Z0 − −
p
and evaluated using to the formula (3.4.5) of [3, p70]
2π cosφdφ 4cosφ′ min(ρ,r) t2dt
= .
Z0 ρ2+r2−2rρcos(φ−φ′) ρr Z0 ρ2−t2√r2−t2
This leadps to p
1 1 1 min(ρ,r) rρt2 2π
IC = drdρdt cos2φ′dφ′.
π Z0 Z0 Z0 ρ2−t2√r2−t2 Z0
p 2π
This integral is symmetric in ρ and r, and since cos2φ′dφ′ =π, we have
Z0
1 1 ρ 1 r 2
IC =2 t2 drdρdt= .
Zt=0 Zρ=t ρ2−t2 Zr=ρ √r2−t2 15
p
References
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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 electrified disc, Proceedings of the Edin-
burgh Mathematical Society (Series 2) 8 (01) (1947) 14–19.
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