http://dx.doi.org/10.12988/ams.2015.5157
A Laplace Type Problem for Irregular Lattice
with Fundamental Cell Composed by Three
Triangles and a Trapetium
D. BarillaDepartment of Economic Science and Quantitative Method University of Messina - Italy
F. Grasso
Department Cognitive Sciences Educational and Cultural Studies University of Messina - Italy
Copyright c 2015 D. Barilla and F. Grasso. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribu-tion, and reproduction in any medium, provided the original work is properly cited.
Abstract
In the previous papers, [1], [2], [3], [4],[5], [6], [7], [8], [9], [10], [11], [12], [13] and [14] the authors studies some Laplace problem for differ-ent lattices. In this paper paper we determine the probability that a random segment of constant length intersects a side of the lattice with fundamental cell represented in fig. 1.
A B C D E F a a a a a a G fig.1
With the rotations of this figure we can write that C0 = C01∪ C02∪ C03∪ C04
and [
ABE = [AEB = \DEF = π
5, \CBE = \DEB = \EDF = \EF D = 2π 5 , \ BCD = \CDE = 3π 5 , BAG = [[ EAG = 3π 4 , (1) |BG| = |EG| = a cosπ 5, |BE| = 2a cos π 5, |AG| = a sin π 5, (2) |DF | = a 2 cosπ5, areaC0 = a2 2 sin π 5 2 cosπ 5 + 1 2 . (3)
We want to compute the probability that a segment s of random position and of constant length l, with l < 4 cosa π
5
, intersects a side of the lattice < (a), i.e. the probability Pint that s intersects a side of the fundamental cell C0.
The position of the segment s is determined by center and by the angle ϕ that it formed with the line CF .
In order to compute the probability Pint we consider the limit positions of
the segment s, for a fixed value of ϕ, in the cells C0i, (i = 1, 2, 3, 4). We have
A B C D E F G A1 A 2 A3 A' A'' A4 B' B'' B1 B2 G ' G 1 G2 G 3 a1 a2 a4 a3 b1 b2 b3 b4 b5 E '' E ' E1 E2 E3 E4 E5 F1 F2 F ' D ' D1 D2 D '' C ' ϕ c1 c2 c3 c 4 c5 c6 d1 d2 d3 d 4 d5 Cˆ03(ϕ ) Cˆ01(ϕ ) Cˆ02(ϕ ) Cˆ04(ϕ ) fig.2 and the relations
area bC01(ϕ) = areaC01− 5 X j=1 areaaj(ϕ) , (4) area bC02(ϕ) = areaC02− 5 X j=1 areabj(ϕ) , (5) area bC03(ϕ) = areaC03− 5 X j=1 areacj(ϕ) , (6) area bC04(ϕ) = areaC04− 5 X j=1 areadj(ϕ) . (7)
By fig.1 and fig.2 we have that: areaa4(ϕ) = l2 4 sin 2ϕ, areaa3(ϕ) = al 2 cos π 5sin ϕ − l2 4 cos ϕ, areaa1(ϕ) = l2cos ϕ sin π5 − ϕ 2 cos π5 ,
areaa2(ϕ) = al 2 sin π 5 − ϕ −l 2cos ϕ sin π 5 − ϕ 2 cosπ5 , areaa5(ϕ) = al 2 sin π 5 cos ϕ − l2 4 sin 2ϕ − l2cos ϕ sin π5 − ϕ 2 cosπ5 . We obtain that: area bC01 = areaC01− A1(ϕ) , where A1(ϕ) = al sin π 5 cos ϕ − l2 4tg π 5 (1 + cos 2ϕ) . In order to compute area bC2(ϕ) we have that:
areab4(ϕ) = l2sin ϕ sin π 5 + ϕ 2 sinπ5 , areab3(ϕ) = al 2 cos π 5sin ϕ − l2sin ϕ sin π 5 + ϕ 2 sinπ5 , areab1(ϕ) = l2cos ϕ sin π 5 + ϕ 2 cosπ5 , areab2(ϕ) = al 2 sin π 5cos ϕ − l2cos ϕ sin π 5 + ϕ cosπ5 , areab5(ϕ) = al 2 sin π 5 + ϕ − l2 2 sin2π5 1 − cos2π 5 cos 2ϕ + sin2π5 2 sin 2ϕ . We obtain that: area bC02 = areaC02− A2(ϕ) , where A2(ϕ) = al sin π 5 + ϕ − l 2 2 sin2π5 1 − cos2π 5 cos 2ϕ + 1 2sin 2π 5 sin 2ϕ .
To compute area bC03(ϕ) we have that
areac4(ϕ) =
l2sin ϕ sin 2π 5 − ϕ
2 sin 2π5 ,
areac3(ϕ) = al 2 sin ϕ − l2sin ϕ sin 2π 5 − ϕ 2 sin2π5 , areac1(ϕ) = l2sin ϕ sin 2π 5 + ϕ 2 sin 2π5 , areac2(ϕ) = al 2 sin 2π 5 + ϕ −l 2sin ϕ sin 2π 5 + ϕ 2 sin2π5 , areac5(ϕ) = al 2 sin 2π 5 − ϕ −l 2sin ϕ sin 2π 5 − ϕ 2 sin2π5 , areac6(ϕ) = al cos π 5sin ϕ − l2sin ϕ sin 2π 5 + ϕ 2 sin2π5 . We obtain that area bC03(ϕ) − A3(ϕ) , where A3(ϕ) = al 2 2 sin2π 5 cos ϕ + 1 + 2 cosπ 5 sin ϕ − l 2 2 sin 2ϕ. To compute now the area bC04(ϕ).
aread4(ϕ) = l2sin ϕ sin 2π 5 + ϕ 2 sin2π5 , aread3(ϕ) = al sin(ϕ) 4 cosπ5 − l sin 2π5 + ϕ sin2π5 , aread1(ϕ) = l2sin 2π 5 − ϕ sin 2π 5 + ϕ 2 sinπ 5 , aread2(ϕ) = al 2 sin 2π 5 − ϕ −l 2sin 2π 5 − ϕ sin 2π 5 + ϕ 2 sinπ5 , aread5(ϕ) = al 2 sin 2π 5 + ϕ − l 2sin 2π 5 − ϕ sin 2π 5 + ϕ 2 sinπ5 − l2sin ϕ sin 2π 5 + ϕ 2 sin2π5 .
We obtain that: area bC04(ϕ) = areaC04− A4(ϕ) , where A4(ϕ) = al sin2π 5 cos ϕ + sin ϕ 4 cos2π5 − l2 4 sin2π5 2 cosπ 5 − cos 2π 5 cos 2ϕ + sin2π 5 sin 2ϕ + 4 cos 2 π 5 + 1 .
Denoting with Mi, (i = 1, 2, 3, 4), the set of segments s that have the center
in the cell C0iand with Ni the set of the segments s completely in C0i, we have
that [16]: Pint = 1 − P4 i=1µ (Ni) P4 i=1µ (Mi) , (8)
where µ is the Lebesgue measure in the euclidean plane.
In order to compute µ (Mi) and µ (Ni) we use the kinematic measure of
Poincar´e [15]:
dK = dx ∧ dy ∧ dϕ,
where x, y are the coordinates of middle point of s and ϕ the fixed angle. We can write: µ (Mi) = π 5 Z 0 dϕ Z Z {(x,y)∈C0i} dxdy = π 5 Z 0 (areaC0i) dϕ = π 5areaC0i, (i = 1, 2, 3, 4) and µ (Ni) = π 5 Z 0 dϕ Z Z {(x,y)∈ bC0i} dxdy = π 5 Z 0 h area bC0i(ϕ) i dϕ = π 5 Z 0 [areaC0i− Ai(ϕ)] dϕ = π 5areaC0i− π 5 Z 0 [Ai(ϕ)] dϕ, (i = 1, 2, 3, 4) .
By these two relations give us:
4 X i=1 µ (Mi) = π 5areaC0, (9)
and 4 X i=1 µ (Ni) = π 5areaC0− π 5 Z 0 " 4 X i=1 Ai(ϕ) # dϕ. (10) follow that: 4 X i=1 Ai(ϕ) = al sinπ 5 2 + 3 cosπ 5 cos ϕ +8 cos 2 π 5 + 2 cos π 5 + 1 4 cosπ5 sin ϕ − l2 4 tgπ 5 − 3ctg 2π 5 + 1 sinπ5
cos 2ϕ + 2 sin 2ϕ + 4 cos2 π 5 + tg π 5 + 3 sin2π5 , then π 5 Z 0 " 4 X i=1 Ai(ϕ) # dϕ = al 9 4− 3 cos 3 π 5 − 4 cos 2 π 5 − 9 2cos π 5 + 1 4 cos π5 − l2 4 9 2+ cos π 5 − 6 cos 2 π 5 + π 5 4 cos2π 5 + tg π 5 + 3 sin2π5 . (11) We obtain that: Pint = 10 πa2sinπ 5 2 cos π 5 + 1 2 al 9 4− 3 cos 3 π 5 − 4 cos 2 π 5 − 9 2cos π 5+ 1 4 cosπ5 −l 2 4 9 2 + cos π 5 − 6 cos 2 π 5 + π 5 4 cos2 π 5 + tg π 5 + 3 sin2π5 .
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