Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 9 (2016), 75–82
Research Article
A Laplace type problem for three lattices with
non-convex cell
Giuseppe Caristia,∗, Maria Pettineob , Marius Stokac
a
University of Messina, Department S.E.A.M., via dei Verdi, 75 98122, Messina, Italy.
b
University of Palermo, Department of Mathematics and Informatics, Via Archirafi, 34 90123 Palermo, Italy.
c
Sciences Accademy of Turin, Via Maria Vittoria, 3, 10123 Torino, Italy.
Communicated by C. Vetro
Abstract
In this paper we consider three lattices with cells represented in Fig. 1, 3 and 5 and we determine the
c
probability that a random segment of constant length intersects a side of lattice. ⃝2016
All rights reserved.
Keywords: Geometric probability, stochastic geometry, random sets, random convex sets and integral
geometry.
2010 MSC: 60D05, 52A22.
1. Cell without obstacles.
(1)
Let ℜ1 (a) be the lattice with the fundamental cell C0 rappresented in Fig. 1.
By Fig. 1 we obtain that
(1)
areaC0 = 5a2 .
(1.1)
We want to compute the probability that a random segment s of constant length l intersects a side of
(1)
lattice, i.e. the probability Pint that the segment s intersects a side of fundamental cell C0 .
The position of segment s is determined by the center and by the angle φ that it formed with the line
BC.
∗
Giuseppe Caristi
Email addresses: gcaristi@unime.it (Giuseppe Caristi), maria.pettineo@unipa.it (Maria Pettineo),
marius.stoka@gmail.com (Marius Stoka)
Received 2015-07-03
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
A
H
a
E
D
a
a
a
2a
76
G
2a
F
a
C0(1)
B
C
3a
Fig. 1
To compute the probability Pint we consider the limiting positions, for a fixed value of φ, of segment s. We
obtain the Fig.2
A
A2
a1 ϕ
A1 ϕ
a11
A3
h2
ϕ
H
E
H1 h10
E1
E2
ϕ
h6
a7
E3
D
ϕ
D1
h8
a8
a10
G
a2
a6
a9
h9
F
G1
F2
a5
F1
! (1)
C
0 (ϕ )
B1
C2
C3
a4
a3
ϕ h3
C1
B
C
Fig. 2
and the formula
b (1) (φ) = C (1) −
C
0
0
11
∑
areaai (φ) .
i=1
b (1) (φ) we have that
To compute C
0
areaa1 (φ) = areaa4 (φ) = areaa7 (φ) =
areaa2 (φ) = areaa5 (φ) = al cos φ −
l2
sin 2φ,
4
l2
sin 2φ,
4
3al
l2
sin φ − sin 2φ,
2
4
al
l2
areaa6 (φ) = areaa11 (φ) =
sin φ − sin 2φ,
2
4
areaa3 (φ) =
(1.2)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
areaa9 (φ) =
areaa8 (φ) =
areaa10 (φ) =
77
al
sin φ,
2
al
l2
cos φ − sin φ,
2
4
al
cos φ.
2
We can write that
A1 (φ) =
11
∑
areaai (φ) = 3al (sin φ + cos φ) −
i=1
3l2
sin 2φ.
4
(1.3)
Replacing this formula in relation (1.2) we obtain
b (φ) = areaC − A1 (φ) .
areaC
0
0
(1)
(1)
(1.4)
(1)
Denoting with M1 , the set of all segments s that they have center in the cell C0 , and with N1 the set of
all segments s entirely conteined in the cell Co , we have [2]:
µ (N1 )
,
µ (M1 )
Pint = 1 −
(1.5)
where µ is the Lebesgue measure in the euclidean plane.
To compute the measure µ (M1 ) and µ (N1 ) we use the kinematic measure of Poincarè [1]:
dk = dx ∧ dy ∧ dφ,
where x, y are
[ the
] coordinate of center of s and φ the fixed angle.
π
For φ ∈ 0, 2 we have
∫ π(
∫ ∫
∫ π
)
2
2
π
(1)
(1)
dφ
areaC0 dφ = areaC0 .
µ (M1 ) =
} dxdy =
{
(1)
2
0
(x,y)ϵC0
0
Then, considering formula (1.4) we can write
∫
µ (N1 ) =
π
2
∫ ∫
dφ
{
0
∫
=
=
0
=
}
b (1) (φ)
(x,y)ϵC
0
π
2
[
]
b (1) (φ) dφ
areaC
0
π
2
[
]
(1)
areaC0 − A1 (φ) dφ
0
∫
(1.6)
π
(1)
areaC0 −
2
∫
π
2
[A1 (φ)] dφ.
(1.7)
0
The formulas (1.5), (1.6) and (1.7) give us
(1)
Pint
By (1.3), we have that
=
∫
2
(1)
πareaC0
∫
0
π
2
π
2
[A1 (φ)] dφ.
(1.8)
0
[A1 (φ)] dφ = 6al −
3l2
.
4
Replacing in (1.8) the relations (1.1) and (1.9) we obtain that
( )2
12 l
3
l
(1)
pe =
−
.
5π a 10π a
(1.9)
(1.10)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
78
2. Cell with obstacles triangular
(2)
rappresented in Fig. 3
E
m E2
Let ℜ2 (a, m) be the lattice with the fundamental cell C0
A
H
m A2 a − 2m H 2 m
m
A1
D2 m
D
m
m
m
H1
E1
D1
a−m
a−m
a
2a − 2m
2a − 2m
F
G
C0(2)
B1
C1
m
m
B
m B2
C2 m C
3a − 2m
Fig. 3
where m < a2 . By Fig. 3 we have that
(2)
areaC0 (φ) = 5a2 − 3m2 .
(2.1)
(2)
We want to compute the probability Pint that a random segment s of constant length l intersects a side of
(2)
cell C0 .
As in the paragraph 1, we have Fig .4
A
A2
a14
ϕ
A1a1 A4
A3
H3
h2
H2
a13
ϕ
H
E
H1
E1 a9 E
4
E3h10
H 4 h12
E2
a8
a11
h11 ϕ
D4
a6
F
F1
G1
! (2)
C
0 (ϕ )
B1 h3a
C3
C4 a5 C1
B3
3
ϕ h4
B
D1
F2
G
B4
D
D2
a7
D3
a10
a12
a2
h8ϕ
a4
B2
C2
C
Fig. 4
and
b (2) (φ) = areaC (2) −
areaC
0
0
14
∑
areaai (φ) .
i=1
We have that
areaa1 (φ) = areaa5 (φ) = areaa9 (φ) =
lm
(sin φ − cos φ) ,
2
(2.2)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
79
lm
l2
cos φ − sin 2φ,
2
4
lm
areaa7 (φ) = areaa13 (φ) =
(sin φ + cos φ) ,
2
3al
sin φ − lm sin φ,
2
al
sin φ,
2
al
l2
cos φ − sin 2φ,
2
4
al
lm
cos φ −
cos φ,
2
2
al
areaa14 (φ) =
sin φ − lm sin φ.
2
areaa2 (φ) = areaa6 (φ) = al cos φ −
areaa3 (φ) =
areaa4 (φ) =
areaa11 (φ) =
areaa10 (φ) =
areaa12 (φ) =
areaa8 (φ) =
We can write,
A2 (φ) =
14
∑
areaai (φ) = 3al (sin φ + cos φ) −
i=1
3l2
3lm
sin 2φ −
cos φ.
4
2
(2.3)
As in the paragraph 1, we can write,
(2)
Pint = 1 −
µ (N2 )
.
µ (M2 )
(2.4)
We have that:
∫
π
2
µ (M2 ) =
∫ ∫
dy
∫
} dxdy
{
(2)
(x,y)ϵC0
0
(
π
2
=
(2)
)
areaC0
0
dφ =
π
(2)
areaC0
2
(2.5)
and, considering (2.2) and (2.3),
∫
µ (N2 ) =
π
2
} dxdy
{
b (2)
(x,y)ϵC
dφ
0
∫
=
0
∫
∫ ∫
=
0
0
π
2
[
(2)
areaC0
By (2.3) we have that
∫
π
2
π
2
(
(2)
]
π
(2)
− A2 (φ) dφ = areaC0 −
2
[A2 (φ)] dφ = 6al −
0
)
areaC0
dφ
∫
π
2
[A2 (φ)] dφ.
(2.6)
0
3l2
− 2lm.
4
(2.7)
The relations (2.1), (2.4), (2.5), (2.6) and (2.7) give us
(2)
Pint
2l
=
2
π (5a − m2 )
(
)
3l
6a − − 2m .
4
For m = 0 this probability become
pe
(2)
12 l
3
=
−
5π a 10π
( )2
l
.
a
3. Cell with obstacles circular sectors
(3)
Let ℜ3 (a, m) be the lattice with the fundamental cell C0
rappresented in Fig. 5
(2.8)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
A
m A2 a − 2m H 2 m
E
H
m
A1
m E2 a − 2m D2 m
D
m
m
m
H1
E1
D1
a−m
2a − 2m
80
a−m
2a − 2m
F
a
G
C0(3)
B1
C1
m
m
B
m B2
C2 m C
3a − 2m
Fig. 5
where m < a2 . By Fig. 5 we have that
(3)
areaC0 = 5a2 −
3πm2
.
2
(3.1)
(3)
We want to compute the probability Pint that a random segment s of constant length l intersects a side of
(3)
cell C0. .
As in the paragraph 1, we have the Fig. 6,
A
a2
a11
a1
E
H2
H
A2 A5
ϕ
a17
A1ϕ A4
H1
H 3 a16
A3
ϕ
h3
H 4 h15
a15
a12
h14
D
D3 a9
D1
D4
F2
h4
a4
a8
F
a14
F1
! (3)
C
0 (ϕ )
B4
B
E1 ϕ E
4
E5
h13
G1
M
D2
ϕ
ϕh10 a10
G
a3
B1
E2 E3
B3
ϕ h5
C4
C3 a
C1 7
a5
C5 C2
a6
B2
C
Fig. 6
and the formula
b (3) (φ) = areaC (3) −
areaC
0
0
17
∑
i=1
We have that
areaa1 (φ) + areaa2 (φ) = areaa6 (φ) + areaa7 (φ)
= areaa11 (φ) + areaa12 (φ)
l2
πm2
=
sin 2φ −
,
4
4
areaai (φ) .
(3.2)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
81
lm
l2
cos φ − sin 2φ,
2
4
lm
m2
areaa9 (φ) = areaa16 (φ) =
(sin φ + cos φ) −
(π − 2) ,
2
4
lm
l2
3al
sin φ −
sin φ − sin 2φ,
2
2
4
al
l2
cos φ − sin 2φ,
2
4
al
lm
cos φ −
cos φ,
2
2
al
sin φ,
2
lm
l2
al
sin φ −
sin φ − sin 2φ.
2
2
4
areaa3 (φ) = areaa8 (φ) = al cos φ −
areaa4 (φ) =
areaa5 (φ) =
areaa13 (φ) =
areaa15 (φ) =
areaa14 (φ) =
areaa10 (φ) + areaa17 (φ) =
We can write that
17
∑
A3 (φ) =
areaai (φ) = 3al (sin φ + cos φ) −
i=1
3 (π − 1) m2
3l2
sin 2φ −
.
4
2
(3.3)
Replacing this relation in (3.2) we have that
b (φ) = areaC − A3 (φ) .
areaC
0
0
(3)
(3)
(3.4)
As in the paragraph 1, we can write that
Pint = 1 −
µ (N3 )
.
µ (M3 )
(3.5)
[
]
For φ ∈ 0, π2 we have that
∫
π
2
µ (M3 ) =
∫
∫ ∫
dφ
{
0
(3)
π
2
} dxdy =
(x,y)ϵC0
0
(
)
π
(3)
(3)
areaC0 dφ = areaC0
2
and considering (3.4),
∫
µ (N3 ) =
dφ
{
} dxdy
b (3) (φ)
(x,y)ϵC
0
∫
=
∫
∫ ∫
π
2
0
[
π
2
0
(3)
areaC0
]
=
0
π
2
(
)
(3)
areaC0 dφ
π
(3)
− A3 (φ) dφ = areaC0 −
2
∫
π
2
[A3 (φ)] dφ.
0
(3.6)
From relation (3.3) follows that
∫
π
2
[A3 (φ)] dφ = 6al −
0
3l2 3π (π − 1) m2
−
.
4
4
The (3.1), (3.5) and (3.7) give us
(3)
Pint
[
]
3l2 3π (π − 1) m2
)
12al −
−
.
=
2
2
2
π 5a2 − 3πm
2
(
1
(3.7)
G. Caristi, M. Pettineo, M. Stoka, J. Nonlinear Sci. Appl. 9 (2016), 75–82
For m = 0 this probability become
pe
(3)
12 l
3
=
−
5π a 10π
( )2
l
.
a
82
(3.8)
The relation (1.10), (2.8) and (3.8) give us the evident equality
pe(1) = pe(2) = pe(3) .
References
[1] H. Poincaré, Calcul des probabilités, ed.2, Gauthier Villars, Paris, (1912). 1
[2] M. Stoka, Probabilités géométriques de type Buffon dans le plan euclidien, Atti Acc. Sci. torino, 110 (1975-1976),
53–59. 1