# Exact Solutions of Fragmentation Equations with General Fragmentation Rates and Separable Particles Distribution Kernels

Authors: S. C. Oukouomi Noutchie; E. F. Doungmo Goufo
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

We make use of Laplace transform techniques and the method of characteristics to solve fragmentation equations explicitly. Our result is a breakthrough in the analysis of pure fragmentation equations as this is the first instance where an exact solution is provided for the fragmentation evolution equation with general fragmentation rates. This paper is the key for resolving most of the open problems in fragmentation theory including “shattering” and the sudden appearance of infinitely many particles in some systems with initial finite particles number.


---

## Body

## 1. Preliminaries

Fragmentation models occur in a large variety of situations including the study of stellar fragments in astrophysics, rock fracture, polymer degradation, DNA fragmentation, aggregates breakage, breakup of solid drugs in organisms, and liquid droplet decay. The theoretical framework of fragmentation dynamics can be traced back to papers by Melzak [1] (analytically) and Filippov [2] (probabilistically). In the 1980s, a systematic investigation of fragmentation models was undertaken, mainly by Ziff and his students, for example [3, 4]. They provided analytical solutions to a large class of equations of the form
(1)
∂
∂
t
u
(
t
,
x
)
=
-
a
(
x
)
u
(
t
,
x
)
+
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
(
t
,
y
)
d
y
,
x
≥
0
,
t
>
0
,
with power rates a
(
x
)
=
x
α, α
∈
(
-
∞
,
∞
) and where b
(
x
∣
y
) represented the distribution of particle mass x spawned by the breakage of a particle of mass y
>
x. In their setting, b
(
x
∣
y
) was given by a power law
(2)
b
(
x
∣
y
)
=
(
ν
+
2
)
x
ν
y
ν
+
1
,
with ν
∈
(
-
2,0
] (see also [5] for a more insight regarding this case). Note that the density of particles having mass x at time t is denoted by u
(
t
,
x
). Additionally in absence of any other mechanism, the mass of all daughter particles is equal to the mass of the parent. This “local” conservation mass principle is given by
(3)
∫
0
y
‍
x
b
(
x
∣
y
)
d
x
=
y
.In a similar way, the amount of particles created by a particle of sizey is given by
(4)
n
(
y
)
=
∫
0
y
‍
b
(
x
∣
y
)
d
x
and n
(
y
) may be infinite.Formal conservation principles(5)
d
d
t
M
(
t
)
=
∫
0
∞
‍
∂
∂
t
u
(
t
,
x
)
x
d
x
=
0
,
d
d
t
N
(
t
)
=
∫
0
∞
‍
∂
∂
t
u
(
t
,
x
)
d
x
=
∫
0
∞
‍
a
(
x
)
(
n
(
x
)
-
1
)
u
(
t
,
x
)
d
x
can be obtained through the means of local conservation principles (3) and (4) and the integration of (1).This paper extends the class of power law rates to any positive and continuous function on(
0
,
∞
) and it is assumed that b is separable and can be written as
(6)
b
(
x
∣
y
)
=
β
(
x
)
γ
(
y
)
,
where
(7)
γ
(
y
)
=
y
∫
0
y
‍
s
β
(
s
)
d
s
in order to satisfy the local mass conservation rule. Moreover, it is assumed that β is any continuous function on (
0
,
∞
). A generalization of the power law b described in (2) can be found in (6) which has the advantage of allowing the number of daughter particles,
(8)
n
(
y
)
=
y
∫
0
y
‍
β
(
s
)
d
s
∫
0
y
‍
s
β
(
s
)
d
s
,
to depend on the parent size y [6].Due to the inability of getting exact solutions in fragmentation models, various authors have used several functional analytic approaches to investigate the dynamics of the system. These methods include semigroup theory [7–10], perturbation theory [11–13], approximation techniques [14], and probabilistic methods [2, 15]. The efficiency of these methods is limited as these problems are reformulated in abstract spaces that are norm dependent and the overall behavior of the dynamics changes radically as different metrics are included in the system. The Laplace transforms as extensively discussed in [16, 17] will play a key role in our investigations. Next we introduce a theorem and a definition that are instrumental in the analysis required in the obtention of exact solutions.Theorem 1 (see [6]).
Assume thatlim
⁡
x
→
0
+
a
(
x
) exists (finite or infinite). Then the fragmentation equation 1 is conservative if and only if there exists δ
>
0 such that b
(
x
∣
x
)
/
a
(
x
)
∉
L
1
(
[
0
,
δ
]
).


### 1.1. Laplace Transforms

Definition 2.
The Laplace transform of a piecewise continuous functionf
(
t
), 0
≤
t
<
+
∞ is the function F
(
s
)
=
L
{
f
(
t
)
} defined by
(9)
F
(
s
)
=
∫
0
∞
‍
e
-
s
t
f
(
t
)
d
t
.
The inverse Laplace transform of F
(
s
) is f
(
t
), f
(
t
)
=
L
-
1
(
F
(
s
)
).


## 1.1. Laplace Transforms

Definition 2.
The Laplace transform of a piecewise continuous functionf
(
t
), 0
≤
t
<
+
∞ is the function F
(
s
)
=
L
{
f
(
t
)
} defined by
(9)
F
(
s
)
=
∫
0
∞
‍
e
-
s
t
f
(
t
)
d
t
.
The inverse Laplace transform of F
(
s
) is f
(
t
), f
(
t
)
=
L
-
1
(
F
(
s
)
).


## 2. Solvability of the Fragmentation Equation

In this section, we use Laplace transform to solve the fragmentation equation(10)
∂
∂
t
u
(
t
,
x
)
=
-
a
(
x
)
u
(
t
,
x
)
+
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
(
t
,
y
)
d
y
,
x
≥
0
,
t
>
0
,
u
(
0
,
x
)
=
u
0
(
x
)
.Letu
~
(
s
,
x
)
=
L
[
u
(
t
,
x
)
]. Clearly, we have that
(11)
L
{
∂
∂
t
u
(
t
,
x
)
}
=
s
u
~
(
s
,
x
)
-
u
0
(
x
)
,
L
{
a
(
x
)
u
(
t
,
x
)
}
=
a
(
x
)
u
~
(
s
,
x
)
,
L
{
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
(
t
,
y
)
d
y
}
=
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
~
(
s
,
y
)
d
y
.
Making use of (10) and (11), we obtain the following equation:
(12)
s
u
~
(
s
,
x
)
-
u
0
(
x
)
=
-
a
(
x
)
u
~
(
s
,
x
)
+
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
~
(
s
,
y
)
d
y
;
that is,
(13)
u
0
(
x
)
=
(
s
+
a
(
x
)
)
u
~
(
s
,
x
)
-
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
~
(
s
,
y
)
d
y
.
Viewing s as a parameter, this is similar to the resolvent equation solved in 2010 (Banasiak and Noutchie). The solution reads as
(14)
u
~
(
s
,
x
)
=
u
0
(
x
)
s
+
a
(
x
)
+
β
(
x
)
s
+
a
(
x
)
e
-
ξ
s
(
x
)
×
∫
x
∞
‍
a
(
y
)
γ
(
y
)
s
+
a
(
y
)
e
ξ
s
(
y
)
u
0
(
y
)
d
y
,
where
(15)
ξ
s
(
x
)
=
∫
1
x
‍
a
(
η
)
b
(
η
∣
η
)
s
+
a
(
η
)
d
η
.
The solution u
(
t
,
x
) of (10) is the inverse Laplace transform of u
~
(
s
,
x
). Clearly,
(16)
L
-
1
{
u
0
(
x
)
s
+
a
(
x
)
}
=
u
0
(
x
)
L
-
1
{
1
s
+
a
(
x
)
}
=
u
0
(
x
)
e
-
t
a
(
x
)
,
L
-
1
{
β
(
x
)
s
+
a
(
x
)
e
-
ξ
s
(
x
)
∫
x
∞
‍
a
(
y
)
γ
(
y
)
s
+
a
(
y
)
e
ξ
s
(
y
)
u
0
(
y
)
d
y
}
=
∫
x
∞
a
(
y
)
b
(
x
∣
y
)
u
0
(
y
)
L
-
1
×
{
1
s
+
a
(
x
)
1
s
+
a
(
y
)
e
{
ξ
s
(
y
)
-
ξ
s
(
x
)
}
}
d
y
=
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
0
(
y
)
L
-
1
{
Θ
(
s
,
x
,
y
)
}
d
y
,
where
(17)
Θ
(
s
,
x
,
y
)
=
1
s
+
a
(
x
)
1
s
+
a
(
y
)
exp
⁡
{
∫
x
y
‍
a
(
η
)
b
(
η
∣
η
)
s
+
a
(
η
)
d
η
}
.Therefore, the solution of the fragmentation equation(18)
∂
∂
t
u
(
t
,
x
)
=
-
a
(
x
)
u
(
t
,
x
)
+
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
(
t
,
y
)
d
y
,
x
≥
0
,
t
>
0
,
u
(
0
,
x
)
=
u
0
(
x
)
is given by
(19)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
a
(
x
)
+
∫
x
∞
‍
a
(
y
)
b
(
x
∣
y
)
u
0
(
y
)
L
-
1
{
Θ
(
s
,
x
,
y
)
}
d
y
.


## 3. Applications

In this section, we assume that(20)
a
(
x
)
=
x
α
+
1
,
α
∈
(
-
∞
,
∞
)
,
b
(
x
∣
y
)
=
(
2
+
ν
)
x
ν
y
ν
+
1
,
with ν
∈
(
-
2,0
]. We have
(21)
∫
x
y
‍
a
(
η
)
b
(
η
∣
η
)
s
+
a
(
η
)
d
η
=
(
2
+
ν
)
∫
x
y
‍
η
α
s
+
η
α
+
1
d
η
=
2
+
ν
α
+
1
ln
⁡
{
s
+
y
α
+
1
s
+
x
α
+
1
}
;
it follows that
(22)
exp
⁡
{
∫
x
y
‍
a
(
η
)
b
(
η
∣
η
)
s
+
a
(
η
)
d
η
}
=
{
s
+
y
α
+
1
s
+
x
α
+
1
}
γ
,
where
(23)
γ
=
2
+
ν
α
+
1
.
Thus
(24)
Θ
α
,
ν
(
s
,
x
,
y
)
=
(
s
+
y
α
+
1
)
γ
-
1
(
s
+
x
α
+
1
)
γ
+
1
=
{
1
s
+
x
α
+
1
}
γ
+
1
{
s
+
y
α
+
1
}
γ
-
1
.
Therefore, the solution u
(
t
,
x
) is given by
(25)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
α
+
1
+
(
2
+
ν
)
∫
x
∞
‍
{
x
y
}
ν
y
α
u
0
(
y
)
L
-
1
×
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
d
y
.


### 3.1. Caseα
=
-
3 and ν
=
0

We want to solve the following equation:(26)
∂
∂
t
u
(
t
,
x
)
=
-
x
-
2
u
(
t
,
x
)
+
2
∫
x
∞
‍
y
-
3
u
(
t
,
y
)
d
y
,
u
(
0
,
x
)
=
u
0
(
x
)
.We haveγ
=
-
1; it follows that
(27)
Θ
-
3,0
(
s
,
x
,
y
)
=
{
1
s
+
x
-
2
}
0
{
s
+
y
-
2
}
-
2
=
{
s
+
y
-
2
}
-
2
.
Thus
(28)
L
-
1
{
Θ
-
3,0
(
s
,
x
,
y
)
}
=
L
-
1
{
(
s
+
y
-
2
)
-
2
}
=
t
e
-
t
y
-
2
.
Therefore,
(29)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
-
2
+
2
t
∫
x
∞
‍
y
-
3
e
-
t
y
-
2
u
0
(
y
)
d
y
.


### 3.2. Caseα
=
-
2 and ν
=
0

We want to solve the following equation:(30)
∂
∂
t
u
(
t
,
x
)
=
-
x
-
1
u
(
t
,
x
)
+
2
∫
x
∞
‍
y
-
2
u
(
t
,
y
)
d
y
,
u
(
0
,
x
)
=
u
0
(
x
)
.We haveγ
=
-
2; it follows that
(31)
Θ
-
2,0
(
s
,
x
,
y
)
=
{
1
s
+
x
-
1
}
-
1
{
s
+
y
-
1
}
-
3
=
s
+
y
-
1
-
y
-
1
+
x
-
1
(
s
+
y
-
1
)
3
=
1
(
s
+
y
-
1
)
2
+
(
x
-
1
-
y
-
1
)
(
s
+
y
-
1
)
3
.Thus(32)
L
-
1
{
Θ
-
2,0
(
s
,
x
,
y
)
}
=
L
-
1
{
1
(
s
+
y
-
1
)
2
}
+
(
x
-
1
-
y
-
1
)
L
-
1
{
1
(
s
+
y
-
1
)
3
}
=
t
e
-
t
y
-
1
+
(
x
-
1
-
y
-
1
)
e
-
t
y
-
1
t
2
2
.Therefore,(33)
u
(
x
,
t
)
=
e
-
t
/
x
u
0
(
x
)
+
2
t
∫
x
∞
‍
e
-
t
/
y
y
2
e
-
t
/
y
u
0
(
y
)
d
y
+
t
2
∫
x
∞
‍
e
-
t
/
y
y
2
(
1
x
-
1
y
)
u
0
(
y
)
d
y
.


### 3.3. General Casea
(
x
)
=
x
α
+
1 and b
(
x
∣
y
)
=
(
2
+
ν
)
(
x
ν
/
y
ν
+
1
)

We want to solve(34)
∂
∂
t
u
(
t
,
x
)
=
-
x
α
+
1
u
(
t
,
x
)
+
(
2
+
ν
)
x
ν
∫
x
∞
‍
y
α
-
ν
u
(
t
,
y
)
d
y
,
x
≥
0
,
t
>
0
,
u
(
0
,
x
)
=
u
0
(
x
)
.
From the previous section, the solution of this equation is
(35)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
α
+
1
+
(
2
+
ν
)
×
∫
x
∞
‍
{
x
y
}
ν
y
α
u
0
(
y
)
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
d
y
.
Note that
(36)
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
=
L
-
1
{
{
1
s
+
x
α
+
1
}
γ
+
1
{
s
+
y
α
+
1
}
γ
-
1
}
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
=
L
-
1
{
{
1
s
+
x
α
+
1
}
γ
+
1
{
s
+
y
α
+
1
}
γ
-
1
}
=
t
exp
⁡
⁡
(
-
t
x
α
+
1
)
F
1
1
×
(
1
-
γ
;
2
;
t
(
x
α
+
1
-
y
α
+
1
)
)
.
It follows that
(37)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
α
+
1
+
(
2
+
ν
)
t
exp
⁡
(
-
t
x
α
+
1
)
×
∫
x
∞
‍
{
x
y
}
ν
F
1
1
(
1
-
γ
;
2
;
t
(
x
α
+
1
-
y
α
+
1
)
)
×
y
α
u
0
(
y
)
d
y
.We recover the results of Ziff and his students [3, 4, 18].


## 3.1. Caseα
=
-
3 and ν
=
0

We want to solve the following equation:(26)
∂
∂
t
u
(
t
,
x
)
=
-
x
-
2
u
(
t
,
x
)
+
2
∫
x
∞
‍
y
-
3
u
(
t
,
y
)
d
y
,
u
(
0
,
x
)
=
u
0
(
x
)
.We haveγ
=
-
1; it follows that
(27)
Θ
-
3,0
(
s
,
x
,
y
)
=
{
1
s
+
x
-
2
}
0
{
s
+
y
-
2
}
-
2
=
{
s
+
y
-
2
}
-
2
.
Thus
(28)
L
-
1
{
Θ
-
3,0
(
s
,
x
,
y
)
}
=
L
-
1
{
(
s
+
y
-
2
)
-
2
}
=
t
e
-
t
y
-
2
.
Therefore,
(29)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
-
2
+
2
t
∫
x
∞
‍
y
-
3
e
-
t
y
-
2
u
0
(
y
)
d
y
.


## 3.2. Caseα
=
-
2 and ν
=
0

We want to solve the following equation:(30)
∂
∂
t
u
(
t
,
x
)
=
-
x
-
1
u
(
t
,
x
)
+
2
∫
x
∞
‍
y
-
2
u
(
t
,
y
)
d
y
,
u
(
0
,
x
)
=
u
0
(
x
)
.We haveγ
=
-
2; it follows that
(31)
Θ
-
2,0
(
s
,
x
,
y
)
=
{
1
s
+
x
-
1
}
-
1
{
s
+
y
-
1
}
-
3
=
s
+
y
-
1
-
y
-
1
+
x
-
1
(
s
+
y
-
1
)
3
=
1
(
s
+
y
-
1
)
2
+
(
x
-
1
-
y
-
1
)
(
s
+
y
-
1
)
3
.Thus(32)
L
-
1
{
Θ
-
2,0
(
s
,
x
,
y
)
}
=
L
-
1
{
1
(
s
+
y
-
1
)
2
}
+
(
x
-
1
-
y
-
1
)
L
-
1
{
1
(
s
+
y
-
1
)
3
}
=
t
e
-
t
y
-
1
+
(
x
-
1
-
y
-
1
)
e
-
t
y
-
1
t
2
2
.Therefore,(33)
u
(
x
,
t
)
=
e
-
t
/
x
u
0
(
x
)
+
2
t
∫
x
∞
‍
e
-
t
/
y
y
2
e
-
t
/
y
u
0
(
y
)
d
y
+
t
2
∫
x
∞
‍
e
-
t
/
y
y
2
(
1
x
-
1
y
)
u
0
(
y
)
d
y
.


## 3.3. General Casea
(
x
)
=
x
α
+
1 and b
(
x
∣
y
)
=
(
2
+
ν
)
(
x
ν
/
y
ν
+
1
)

We want to solve(34)
∂
∂
t
u
(
t
,
x
)
=
-
x
α
+
1
u
(
t
,
x
)
+
(
2
+
ν
)
x
ν
∫
x
∞
‍
y
α
-
ν
u
(
t
,
y
)
d
y
,
x
≥
0
,
t
>
0
,
u
(
0
,
x
)
=
u
0
(
x
)
.
From the previous section, the solution of this equation is
(35)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
α
+
1
+
(
2
+
ν
)
×
∫
x
∞
‍
{
x
y
}
ν
y
α
u
0
(
y
)
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
d
y
.
Note that
(36)
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
=
L
-
1
{
{
1
s
+
x
α
+
1
}
γ
+
1
{
s
+
y
α
+
1
}
γ
-
1
}
L
-
1
{
Θ
α
,
ν
(
s
,
x
,
y
)
}
=
L
-
1
{
{
1
s
+
x
α
+
1
}
γ
+
1
{
s
+
y
α
+
1
}
γ
-
1
}
=
t
exp
⁡
⁡
(
-
t
x
α
+
1
)
F
1
1
×
(
1
-
γ
;
2
;
t
(
x
α
+
1
-
y
α
+
1
)
)
.
It follows that
(37)
u
(
t
,
x
)
=
u
0
(
x
)
e
-
t
x
α
+
1
+
(
2
+
ν
)
t
exp
⁡
(
-
t
x
α
+
1
)
×
∫
x
∞
‍
{
x
y
}
ν
F
1
1
(
1
-
γ
;
2
;
t
(
x
α
+
1
-
y
α
+
1
)
)
×
y
α
u
0
(
y
)
d
y
.We recover the results of Ziff and his students [3, 4, 18].


## 4. Concluding Remarks

In this paper, we successfully used Laplace transforms and the methods of characteristics to solve an open problem in applied mathematics derived over 60 years ago. We extended the work of Ziff and his students and provided the full solution of fragmentation equations with general explosion rates. This work enables the computation of removal rates and shattering in fragmentation models and provides a general framework for understanding particles distributions in fragmentation processes as time evolves. In particular, it enables a complete classification of shattering regimes.



---

*Source: 789769-2014-07-08.xml*