This document discusses whether and when the entropy of a spatially homogeneous Boltzmann equation system with infinite initial entropy will become finite. It presents examples showing that for some initial distributions, the entropy remains infinite over time, while for others it becomes finite after a finite time. The main tool used to estimate the entropy is the Duhamel formula for the solution. For hard potential and hard sphere models, rules are given for determining whether the entropy will appear finite based on properties of the initial distribution and collision kernel.

1. Appearance of Entropy
for the Spatially Homogeneous Boltzmann Equation
Teng Li, Xuguang Lu
June 18, 2010
Abstract
In this paper, we study the spatially homogeneous Boltzmann equation with infinite initial entropy.
We discuss whether and when the entropy of the system will appear, i.e. get finite. Under the hard
potential or hard sphere model, substantial examples are given to show that appearance of entropy
does happen for some f0 and not for some others. The main tool we use to estimate H(f)(t) is the
Duhamel formula of the solution f(v, t).
1 Introduction
In the classical Boltzmann kinetic theory, there are four basic assumptions: the finiteness of
mass, momentum, energy and entropy. While the finite-mass, finite-momentum and finite-
energy assumptions are based on everyday experience, the finite-entropy assumption is of somewhat
less intuitive meaning. There is actually no reason to reject the existence of a gas system in the real
world to have infinite entropy.
On the other hand, systems with infinite entropy do exist and exhibit similar irreversible behavior,
too. See Abrahamsson’s work [1], Lu and Mouhot’s recent work [7] for detail. This is a theoretical
support for us to remove the finite-entropy assumption and consider the infinite entropy case.
In addition, the current finite-entropy assumption is actually finite-H-functional assumption. By
[7] and this paper, there are systems whose H-functionals keep infinite during all the way that they
converges to corresponding equilibriums in L1
sense. For these systems, other functionals need to
be defined in place of the H-functional to describe the irreversibility. From this point of view, the
finite-entropy, i.e. finite-H-functional assumption is even more doubtable.
According to the above discussion, we study the infinite entropy (i.e. H-functional) case in
this paper. Consider the spatially homogeneous Boltzmann equation
∂f
∂t
(v, t) = Q(f, f)(v, t), (v, t) ∈ R3
× (0, +∞) (1.1)
where Q is the collision integral operator given by
Q(f, f)(v, t) =
R3×S2
B(v − v∗, ω) f f∗ − ff∗ dωdv∗ (1.2)
Here
f = f(v , t), f∗ = f(v∗, t), f = f(v, t), f∗ = f(v∗, t)
v and v∗ denote the post-collisional velocities and are given by
v =
v + v∗
2
+
|v − v∗|
2
σ, v∗ =
v + v∗
2
−
|v − v∗|
2
σ, σ ∈ S2
(1.3)
where σ is a random parameter which together with the conservation laws uniquely determines v
and v∗ from v and v∗. The function B is the collision kernel giving the distribution of σ. In this
paper, we consider the interaction potentials of inverse power law, so B takes the form
B(z, σ) = b(θ)|z|β
, θ = arccos σ, z/|z| ∈ [0, π] (1.4)
1

2. The exponent β is related to the potentials of interacting particles, namely, the Maxwell model
(β = 0), the hard potentials (0 < β < 1) and the hard sphere model (β = 1). The nonnegative
function b(θ) is often assumed to satisfy the angular cutoff condition
0 < A0 := 2π
π
0
b(θ) sin θdθ < +∞ (1.5)
Entropy or the H-functional is defined as
H(f)(t) = H(f(·, t)) =
R3
f(v, t) log f(v, t)dv (1.6)
In this paper, we always assume the initial data 0 ≤ f0 ∈ L1
2(R3
) has infinite entropy, i.e.
H(f0) = +∞ (1.7)
then H(f)(t) will either keep infinite all the time or go back to finite in a finite time. We call the
second case appearance of entropy, and in this case, tools and results based on the finite-entropy
assumption might be used after a finite time.
The main result of this paper is that we proved both cases, i.e. appearance of entropy happens
and doesn’t happen, exist for the hard potential and hard sphere model. Substantial examples of
initial data f0 for both cases are given. Especially for radically symmetric and locally bounded initial
data f0 with infinite entropy, we can completely judge whether appearance of entropy will happen
from B and f0 by a set of rules.
The tool to estimate H(f)(t) is the Duhamel formula of the solution: suppose f(v, t) is a
conservative solution of equation (1.1), then for a.e. v ∈ R3
, we have
f(v, t) = f0(v) exp −
t
0
L(f)(v, τ)dτ +
t
0
Q+
(f, f)(v, τ) exp −
t
τ
L(f)(v, s)ds dτ(1.8)
holds for ∀t ≥ 0, where
L(f)(v, t) =
R3×S2
B(v − v∗, σ)f(v∗, t)dσdv∗ (1.9)
Q+
(f, f)(v, t) =
R3×S2
B(v − v∗, σ)f(v , t)f(v∗, t)dσdv∗ (1.10)
This paper is only a first study of the infinite entropy case. We are still unable to judge whether
appearance of entropy will happen for general initial data with infinite entropy. New functional should
be defined to quantitatively describe the irreversibility of gas systems whose entropy (H-functional)
keeps infinite all the time.
2 Maxwell Molecular Model
For the Maxwellian molecular model, it’s easy to show that entropy couldn’t appear for any initial
data with infinite entropy.
Theorem 2.1 For the Maxwell molecular model, i.e. the collision kernel B satisfies (1.4) and
(1.5) with β = 0. If the initial value 0 ≤ f0 ∈ L1
2(R3
) satisfies H(f0) = +∞, then for any conservative
solution f(v, t), we have
H(f)(t) = +∞ ∀t ≥ 0
Proof: First estimate f(v, t)’s lower bound. As β = 0
L(f)(v, t) = A0
R3
f(v∗, t)dv∗ = A0||f0||L1
then by the Duhamel formula (1.8), we have
f(v, t) ≥ f0(v) exp (−A0||f0||L1 t) := Cf0(v) (2.1)
Note that C doesn’t depend on v.
2

3. Now let’s estimate H(f)(t).
f(v,t)>1
f(v, t) log f(v, t)dv
(2.1)
≥ C
f(v,t)>1
f0(v) log f(v, t)dv
(2.1)
≥ C
f(v,t)>1
f0(v) log f0(v) + log C dv
= C
f(v,t)>1
f0(v) log f0(v)dv + C log C
f(v,t)>1
f0(v)dv
As it’s well-known, H(f0) = +∞ implies the first term on the right hand side is +∞. The second
term is obviously finite, so the value of the right hand side is +∞. Consequently, the left hand side
is +∞, so is H(f)(t).
3 Hard Potential and Hard Sphere model
In this section, we study the hard potential and hard sphere model (0 < β ≤ 1). Examples are
given to show that appearance of entropy happens for some initial data (Example 3.1 and Example
3.2) and doesn’t happen for some others (Example 3.3). Especially for those f0 that are radically
symmetric and locally bounded, we can judge whether appearance of entropy will happen by a set of
rules (Theorem 3.1).
Define two related functions of f0
E1(f0)(t) :=
R3
f0(v) log f0(v) exp −L1|v|β
t dv (3.1)
E2(f0)(t) :=
R3
f0(v) log f0(v) exp −L2|v|β
t dv (3.2)
where L1 = 4πA0||f0||L1 , L2 = A0||f0||L1
β
are two constants. Note that
E1(f0)(0) = E2(f0)(0) = H(f0) = +∞
and both E1(f0)(t) and E2(f0)(t) are decreasing.
Theorem 3.1 Consider the hard potential and hard sphere model, i.e the collision kernel B
satisfies (1.4) and (1.5) with 0 < β ≤ 1. Additionally assume
0 ≤ b(θ) ≤ b∞ (3.3)
where b∞ > 0 is a constant. If the initial value 0 ≤ f0 ∈ L1
2(R3
) satisfies H(f0) = +∞, then for any
conservative solution f(v, t), we have
(1) if ∃ T2 > 0 s.t. E2(f0)(T2) = +∞, then for any conservative solution f(v, t), we have
H(f)(t) = +∞ ∀t ≤ T2
(2) Further, if f0 is radically symmetric and locally bounded and ∃ T1 > 0 s.t. E1(f0)(T1) <
+∞, then for any conservative solution f(v, t), we have
H(f)(t) < +∞ ∀t ≥ T1
Remark 3.1 It’s easy to see that if f0 is radically symmetric and locally bounded, at least one of
case (1) and case (2) will happen. Therefore, Theorem 3.1 can completely determine whether entropy
will appear.
3

4. Remark 3.2 The constants L1, L2 in Theorem 3.1 are from the upper and lower bound estimate
of L(f)(v, t) in the following Lemma 3.1.
The idea of proving Theorem 3.1 is as below: using the Duhamel formula (1.8) to give the upper
and lower bound of the solution f(v, t) for each t. Then this bounds will help infer that whether
H(f)(t) is finite. To estimate the first term on the right hand side of (1.8), we need to estimate the
functional L(f)(v, t) in the decaying factor exp −
t
0
L(f)(v, τ)dτ (Lemma 3.1). The specialty of
the estimate of L(f)(v, t) here is that it holds for both finite and infinite entropy cases.
Lemma 3.1 Let the collision kernel B satisfies (1.4) and (1.5) with 0 < β ≤ 1. Assume the initial
data 0 ≤ f0 ∈ L1
2(R3
), then for any conservative solution f(v, t), we have
L(f)(v, t) ≤ L2(1 + |v|2
)β/2
∀v ∈ R3
, ∀t ≥ 0 (3.4)
where L2 = A0||f0||L1
β
. If we further assume that f0 is radically symmetric, i.e. f0(v) = ˜f0(|v|), then
for any conservative solution f(v, t), we have
L(f)(v, t) ≥ L1|v|β
∀v ∈ R3
, ∀t ≥ 0 (3.5)
where L1 = 4πA0||f0||L1 .
Proof: First, let’s prove (3.4). Using the following inequality
|v − v∗| ≤ (1 + |v|2
)1/2
(1 + |v∗|2
)1/2
∀v, v∗ ∈ R3
we have
L(f)(v, t) =
R3×S2
B(v − v∗, σ)f(v∗)dσdv∗
= A0
R3
|v − v∗|β
f(v∗)dv∗
≤ A0(1 + |v|2
)β/2
R3
(1 + |v∗|2
)β/2
f(v∗)dv∗
= A0||f0||L1
β
(1 + |v|2
)β/2
So (3.4) is proved.
Next assume f0 is radically symmetric, then f(v, t) is also radically symmetric, i.e. f(v, t) =
˜f(|v|, t). Let’s prove (3.5). We have
L(f)(v, t) = A0
R3
|v − v∗|β
f(v∗)dv∗
v∗=ru
= A0
+∞
0
r2 ˜f(r)dr
S2
|v − ru|β
du
u=cos θ· v
|v|
+sin θ·ω
= A0
+∞
0
r2 ˜f(r)dr
π
0
(|v|2
+ r2
− 2|v|r cos θ)
β
2 sin θdθ
S1(v)
dω
= 2πA0
+∞
0
r2 ˜f(r)dr
π
0
(|v|2
+ r2
− 2|v|r cos θ)
β
2 d(− cos θ)
− cos θ=t
= 2πA0
+∞
0
r2 ˜f(r)dr
1
−1
(|v|2
+ r2
+ 2|v|rt)
β
2 dt
=
πA0
(1 + β
2
)|v|
+∞
0
r((r + |v|)β+2
− |r − |v||β+2
) ˜f(r)dr
=
πA0
(1 + β
2
)|v|
+∞
0
rφ(r) ˜f(r)dr (3.6)
where
φ(r) = (r + |v|)β+2
− |r − |v||β+2
, r ≥ 0
For r ∈ [0, |v|],
φ (r) = (β + 2) (r + |v|)β+1
+ (|v| − r)β+1
4

5. φ (r) = (β + 2)(β + 1) (r + |v|)β
− (|v| − r)β
For r ∈ [|v|, +∞),
φ (r) = (β + 2) (r + |v|)β+1
− (r − |v|)β+1
φ (r) = (β + 2)(β + 1) (r + |v|)β
− (r − |v|)β
So φ (r) ≥ 0 ∀r ≥ 0, and φ−(|v|) = φ+(|v|). Consequently, φ is convex on the interval [0, +∞).
Therefore,
φ(r) ≥ φ(0) + φ (0)r = 2(β + 2)|v|β+1
r ∀r ≥ 0
Put it back into (3.6), we have
L(f)(v, t) ≥ 4πA0|v|β
+∞
0
r2 ˜f(r)dr = 4πA0||f0||L1 |v|β
This is exactly (3.5), so the proof is complete.
Proof of Theorem 3.1 (1): Assume ∃ T2 > 0 s.t. E2(f0)(T2) = +∞. For any arbitrarily
given t ≤ T2, let’s prove H(f)(t) = +∞. First, applying Lemma 3.1 on the right hand side of the
Duhamel formula (1.8), yielding
f(v, t) ≥ f0(v) exp −L2(1 + |v|2
)β/2
t ∀v ∈ R3
(3.7)
As we know, H(f)(t) is finite when and only when the following integral is finite:
f(v,t)>1
f(v, t) log f(v, t)dv
We have
f(v,t)>1
f(v, t) log f(v, t)dv
(3.7)
≥
f(v,t)>1
f0(v) exp −L2(1 + |v|2
)β/2
t log f(v, t)dv
(3.7)
≥
f(v,t)>1
f0(v) exp −L2(1 + |v|2
)β/2
t log f0(v) − L2(1 + |v|2
)β/2
t dv
=
f(v,t)>1
f0(v) log f0(v) exp −L2(1 + |v|2
)β/2
t dv
−L2t
f(v,t)>1
f0(v) exp −L2(1 + |v|2
)β/2
t (1 + |v|2
)β/2
dv (3.8)
The second term on right hand side of (3.8) is obviously finite, so to prove H(f)(t) = +∞, we only
need to prove the first term on the right hand side of (3.8) equals to +∞.
The first term on the right hand side of (3.8) can be split into the following three terms:
f(v,t)>1
f0(v) log f0(v) exp −L2(1 + |v|2
)β/2
t dv =
R3
−
f(v,t)≤1,f0≤1
−
f(v,t)≤1,f0>1
(3.9)
The absolute value of the second term on the right hand side of (3.9) is bounded from above by
f0≤1
f0(v)| log f0(v)|dv < +∞
so is finite. According to (3.7), we have
f(v, t) ≤ 1 ⇒ f0(v) ≤ exp L2(1 + |v|2
)β/2
t
5

6. so
f(v,t)≤1,f0>1
f0(v) log f0(v) exp −L2(1 + |v|2
)β/2
t dv
≤
1<f0≤exp(L2(1+|v|2)β/2t)
f0(v) log f0(v) exp −L2(1 + |v|2
)β/2
t dv
≤
1<f0≤exp(L2(1+|v|2)β/2t)
f0(v)L2(1 + |v|2
)β/2
t exp −L2(1 + |v|2
)β/2
t dv < +∞
that is to say, the third term on the right hand side of (3.9) is also finite. Thus (3.9)’s value is finite
if and only if the first term on its right hand side is finite. Therefore, to prove H(f) = +∞, we only
need to prove the first term of (3.9)’s right hand side equals to +∞.
Denote the first term on the right hand side of the (3.9) as
I =
R3
f0(v) log f0(v) exp −L2(1 + |v|2
)β/2
t dv
Using the following inequality
|v|β
≤ (1 + |v|2
)β/2
≤ 1 + |v|β
, 0 < β ≤ 1
we have
exp(−L2t)E2(f0)(t) ≤ I ≤ E2(f0)(t)
which shows that I is finite if and only if E2(f0)(t) is. Thus to prove H(f)(t) = +∞, we only need
to prove E2(f0)(t) = +∞.
As E2(f0)(T2) = +∞, and t ≤ T2, we know E2(f0)(t) = +∞. Therefore we have H(f)(t) =
+∞ ∀t ≤ T2. (1) is proved.
To prove Theorem 3.1 (2), we need the following two lemmas to estimate the second term on the
right hand side of (1.8).
Lemma 3.2 Let the collision kernel B satisfies (1.4) and (1.5) with 0 < β ≤ 1 and (3.3). Assume
the initial data 0 ≤ f0 ∈ L1
2(R3
) satisfies the following condition
R3
f0(v)
|v|
dv < +∞ (3.10)
then for any conservative solution f(v, t) with f(·, 0) = f0, we have
I(t) :=
R3
f(v, t)
|v|
dv < +∞ ∀t ≥ 0
and the upper bound estimate
I(t) ≤ (I(0) + C1t)eC2t
∀t ≥ 0 (3.11)
where
C1 = 16
√
2π(
√
3
2
)β
b∞||f0||2
L1 , C2 = 8
√
2π3
β
2 b∞||f0||L1 (3.12)
Proof: Let
φε(v) =
1
ε + |v|2
∀ε > 0
Obviously, φε is bounded, so integrals in the following equations are all well-defined.
R3
f(v, t)φε(v)dv ≤
R3
f0(v)φε(v)dv +
t
0
ds
R3×R3×S2
b(θ)|v − v∗|β
f f∗
1
ε + |v|2
dσdvdv∗
≤
R3
f0(v)φε(v)dv + b∞
t
0
ds
R3×R3
ff∗|v − v∗|β
dvdv∗
S2
dσ
ε + |v |2
6

7. Calculate the inner integral
S2
dσ
ε + |v |2
h= v+v∗
|v+v∗|
=
S2
dσ
ε + |v|2+|v∗|2
2
+ |v+v∗||v−v∗|
2
< σ, h >
=
π
0
sin θdθ
ε + |v|2+|v∗|2
2
+ |v+v∗||v−v∗|
2
cos θ S1(h)
dω
= 2π
1
−1
dt
ε + |v|2+|v∗|2
2
− |v+v∗||v−v∗|
2
t
= 2π · −
4
|v + v∗||v − v∗|
ε +
|v|2 + |v∗|2
2
−
|v + v∗||v − v∗|
2
t
t=1
t=−1
=
8π
ε + |v|2+|v∗|2
2
+ |v+v∗||v−v∗|
2
+ ε + |v|2+|v∗|2
2
− |v+v∗||v−v∗|
2
≤
8π
ε + |v|2+|v∗|2
2
+ |v+v∗||v−v∗|
2
≤
8
√
2π
ε + |v|2 + |v∗|2
thus
R3
f(v, t)φε(v)dv ≤
R3
f0(v)φε(v)dv + 8
√
2πb∞
t
0
ds
R3×R3
ff∗|v − v∗|β
ε + |v|2 + |v∗|2
dvdv∗
As
|v − v∗|β
ε + |v|2 + |v∗|2
≤
(
√
ε + |v| + |v∗|)β
ε + |v|2 + |v∗|2
Cauchy Ineq.
≤
[3(ε + |v|2
+ |v∗|2
)]
β
2
ε + |v|2 + |v∗|2
=
3
β
2
(ε + |v|2 + |v∗|2)
1−β
2
≤
3
β
2
(ε + |v|2)
1−β
2
thus
R3
f(v, t)φε(v)dv ≤
R3
f0(v)φε(v)dv + 8
√
2π · 3
β
2 b∞||f0||L1
t
0
ds
R3
f(v, s)
(ε + |v|2)
1−β
2
dv
Without losing generality, assume ε ∈ (0, 1
2
], then
R3
f(v, s)
(ε + |v|2)
1−β
2
dv =
|v|≤ 1
2
f(v, s)
(ε + |v|2)
1−β
2
dv +
|v|> 1
2
f(v, s)
(ε + |v|2)
1−β
2
dv
≤
|v|≤ 1
2
f(v, s)
(ε + |v|2)
1
2
dv + 4
1−β
2
|v|> 1
2
f(v, s)dv
≤
R3
f(v, s)φε(v)dv + 21−β
||f0||L1
7

8. where we applied ε + |v|2
< 1 at the first inequality. Thus
R3
f(v, t)φε(v)dv
≤
R3
f0(v)φε(v)dv + 16
√
2π(
√
3
2
)β
b∞||f0||2
L1 t + 8
√
2π · 3
β
2 b∞||f0||L1
t
0
ds
R3
f(v, s)φε(v)dv
:=
R3
f0(v)φε(v)dv + C1t + C2
t
0
ds
R3
f(v, s)φε(v)dv
By the Gronwall’s inequality, we have
R3
f(v, t)φε(v)dv ≤
R3
f0(v)φε(v)dv + C1t eC2t
Note that when ε → 0+, φε ↑ 1/|v|. So take the limit ε → 0+, then we have
I(t)
Fatou
≤ lim
ε→0+ R3
f(v, t)φε(v)dv
≤ lim
ε→0+
(
R3
f0(v)φε(v)dv + C1t)eC2t
Mono. Cvg.
= (I(0) + C1t)eC2t
By (3.10), we know the proof is complete.
Lemma 3.3 Let the collision kernel B satisfies (1.4) and (1.5) with 0 < β ≤ 1 and (3.3). Assume
the initial data 0 ≤ f0 ∈ L1
2(R3
) satisfies (3.10) and is radically symmetric, i.e. f0(v) = ˜f0(|v|), then
for any conservative solution f(v, t), we have
Q+
(f, f)(v, t) ≤
q(t)
|v|2−β
∀t ≥ 0, ∀v ∈ R3
(3.13)
where
q(t) =
16π2
(2 + 22−β
β)b∞||f0||L1
β
(I(0) + C1t)eC2t
(3.14)
where C1, C2 is defined as C1, C2 in Lemma 3.2.
Proof: Since f0 is radically symmetric, f(v, t) is also radically symmetric, i.e. f(v, t) = ˜f(|v|, t).
In the following proof, we omit the parameter t in f(v, t) if it doesn’t make any confusion. Using the
Carleman formula and the radical symmetry, we have
Q+
(f, f)(v, t) ≤ b∞
R3×S2
|v − v∗|β
f f∗dσdv∗
Carleman
= 4b∞
R3
˜f(|v − x|)
|x|
dx
R2(x)
˜f(|v − y|)
|x − y|1−β
dy
|x−y|≥|x|
≤ 4b∞
R3
˜f(|v − x|)
|x|2−β
dx
R2(x)
˜f(|v − y|)dy
Let vx is the projection of v in the direction of x, i.e
vx = v,
x
|x|
x
|x|
8

9. thus
R2(x)
˜f(|v − y|)dy =
R2(x)
˜f(|vx + v − vx − y|)dy
˜y=v−vx−y
=
R2(x)
˜f(|vx + ˜y|)d˜y
=
R2(x)
˜f( |vx|2 + |˜y|2)d˜y
=
+∞
0
ρ ˜f( |vx|2 + ρ2)dρ
S1(x)
dω
r=
√
|vx|2+ρ2
= 2π
+∞
|vx|
r ˜f(r)dr
Lemma 3.2(3.11)
≤ 2π(I(0) + C1t)eC2t
so
Q+
(f)(v, t) ≤ 8πb∞(I(0) + C1t)eC2t
R3
˜f(|v − x|)
|x|2−β
dx
Calculate the integral on the right hand side
R3
˜f(|v − x|)
|x|2−β
dx =
R3
˜f(|x|)
|v − x|2−β
dx
=
+∞
0
r2 ˜f(r)dr
π
0
sin θdθ
(|v|2 + r2 − 2|v|r cos θ)
2−β
2 S1(v)
dω
= 2π
+∞
0
r2 ˜f(r)dr
1
−1
dt
(|v|2 + r2 + 2|v|rt)
2−β
2
=
2π
β|v|
+∞
0
r ˜f(r)(||v| + r|β
− ||v| − r|β
)dr
thus
Q+
(f)(v, t) ≤
16π2
b∞
β
·
(I(0) + C1t)eC2t
|v|
+∞
0
r ˜f(r)((|v| + r)β
− ||v| − r|β
)dr
Now let’s estimate the integrand on the right hand side. Define
φ(x) = (x + r)β
− |x − r|β
, x ∈ [0, +∞)
For 0 ≤ x ≤ r,
φ(x) = (x + r)β
− (r − x)β
0<β≤1
≤ (2x)β
=
2β
x
x1−β
≤
2β
r
x1−β
For r < x < 2r, φ(x) monotonously decreasing, so
φ(x) ≤ φ(r) = (2r)β
=
(2r)β
x1−β
x1−β
x<2r
≤
(2r)β
(2r)1−β
x1−β
=
2r
x1−β
For x ≥ 2r,
φ(x) = (x + r)β
− (x − r)β
=
β
ξ1−β
· 2r
where x − r < ξ < x + r, so
φ(x) ≤
β
(x − r)1−β
· 2r
x−r≥ x
2
≤
22−β
βr
x1−β
Combining the above three cases, we have
φ(x) ≤
(2 + 22−β
β)r
x1−β
∀x ∈ [0, +∞)
9

10. thus
Q+
(f)(v, t) ≤
16π2
(2 + 22−β
β)b∞
β
·
(I(0) + C1t)eC2t
|v|2−β
+∞
0
r2 ˜f(r)dr
=
16π2
(2 + 22−β
β)b∞||f0||L1
β
·
(I(0) + C1t)eC2t
|v|2−β
:=
q(t)
|v|2−β
The proof is complete.
The proof of Theorem 3.1 (2): Since f0 is radically symmetric, each conservative solution
f(v, t) will also be radically symmetric, i.e. f(v, t) = ˜f(|v|, t). By the assumption in (2), ∃ T1 >
0 s.t. E1(f0)(T1) < +∞. Let’s use the Duhamel formula (1.8) of the solution f(v, t). As f0 is
locally bounded, so it satisfies (3.10), thus satisfies the condition of Lemma 3.3, so
f(v, t)
Lemma 3.1, Lemma 3.3
≤ f0(v)e−L1|v|β
t
+
C(t)
|v|2−β
∀v ∈ R3
, ∀t ≥ 0 (3.15)
where
L1 = 4πA0||f0||L1 , C(t) =
t
0
q(τ)dτ (3.16)
where q(·) is defined as in Lemma 3.3. Notice that q(t) is locally bounded on [0, +∞), so C(t) <
+∞ ∀t ≥ 0.
For any t ≥ T1, for any R > 0, we have
H(f)(t) =
+∞
0
r2 ˜f(r, t) log ˜f(r, t)dr ≤
+∞
0, ˜f(r,t)>1
r2 ˜f(r, t) log ˜f(r, t)dr
≤
R
0, ˜f(r,t)>1
r2 ˜f(r, t) log ˜f(r, t)dr +
+∞
R, ˜f(r,t)>1
r2 ˜f(r, t) log ˜f(r, t)dr
= IR + JR
Let’s prove both IR and JR are finite.
Consider IR. As f0 is locally bounded, we know ∃ MR < +∞ s.t. ˜f0(r) ≤ MR ∀r ∈ [0, R]. So
by (3.15), for ∀r ∈ [0, R], we have
˜f(r, t) ≤ ˜f0(r) +
C(t)
r2−β
≤ MR +
C(t)
r2−β
≤
MRR2−β
+ C(t)
r2−β
:=
C2(R, t)
r2−β
where 0 < C2(R, t) < +∞ is a constant that is dependent on R, t but not on r. Therefore,
IR =
R
0, ˜f(r,t)>1
r2 ˜f(r, t) log ˜f(r, t)dr
≤
R
0, ˜f(r,t)>1
r2 C2(R, t)
r2−β
log ˜f(r, t)dr
≤ C2(R, t)
R
0, ˜f(r,t)>1
rβ
log
C2(R, t)
r2−β
dr
= C2(R, t) log C2(R, t)
R
0, ˜f(r,t)>1
rβ
dr − (2 − β)
R
0, ˜f(r,t)>1
rβ
log rdr
In the square brackets on the right hand side of the above equation, the first term is obviously
finite. The second term is finite because for ∀R > 0, ∀β > 0, rβ
| log r| is Lebesgue integrable on
[0, R] and consequently Lebesgue integrable on its subset {r ∈ [0, R] | ˜f(r, t) > 1}. So we have
IR < +∞ ∀R > 0.
Consider JR. Notice that t ≥ T1 > 0, so there is R large enough such that
C(t)/R2−β
≤ 1/2 (3.17)
2 exp(−L1Rβ
t) < 1 (3.18)
10

11. so for ∀r ≥ R s.t. ˜f(r, t) > 1, we have
˜f(r, t)
(3.15)(3.17)
≤ ˜f0(r) exp(−L1rβ
t) +
1
2
< ˜f0(r) exp(−L1rβ
t) +
˜f(r, t)
2
Consequently,
˜f(r, t) ≤ 2 ˜f0(r) exp(−L1rβ
t) ∀r ∈ {r ∈ [R, +∞) : ˜f(r, t) > 1} (3.19)
Therefore,
JR =
+∞
R, ˜f(r,t)>1
r2 ˜f(r, t) log ˜f(r, t)dr
(3.19)
≤ 2
+∞
R, ˜f(r,t)>1
r2 ˜f0(r) exp(−L1rβ
t) log ˜f(r, t)dr
(3.19)
≤ 2
+∞
R, ˜f(r,t)>1
r2 ˜f0(r) exp(−L1rβ
t) log 2 ˜f0(r) exp(−L1rβ
t) )dr
(3.18)
≤ 2
+∞
R, ˜f(r,t)>1
r2 ˜f0(r) exp(−L1rβ
t) log ˜f0(r)dr
t≥T1
≤ 2
+∞
0
r2 ˜f0(r) exp(−L1rβ
T1)| log ˜f0(r)|dr
E1(f0)(T1)<+∞
< +∞
So we get ∃ R > 0 s.t. JR < +∞. The proof is complete.
Theorem 3.1 gives for all f0 that is radically symmetric and locally bounded a way to judge whether
entropy will appear. Now we construct a concrete example of f0, to illustrate that appearance
of entropy does happen sometimes. According to Theorem 3.1, to construct such an example is
equivalent to find some 0 ≤ f0 ∈ L1
2(R3
) such that
H(f0) =
R3
f0(v) log f0(v)dv = +∞
but ∃ T1 > 0,
E1(f0)(T1) =
R3
f0(v) log f0(v) exp(−L1|v|β
T1)dv < +∞
Example 3.1 Let the collision kernel B satisfies (1.4) and (1.5) with 0 < β ≤ 1 and (3.3). Define
f0(v) = ˜f0(|v|) =



0 |v| ∈ [0, 2)
nn2
n5(log n)2 |v| ∈ [n, n + n−n2
], n = 2, 3, ...
0 |v| ∈ (n + n−n2
, n + 1), n = 2, 3, ...
then 0 ≤ f0 ∈ L1
2(R3
) and H(f0) = +∞ and for any conservative solution f(v, t), we have
H(f)(t) < +∞ ∀t > 0
A special property of this example is that entropy appears instantaneously after the time t = 0.
Proof: f0 is obviously radically symmetric and locally bounded, so Theorem 3.1 can be applied.
11

12. First, we prove that 0 ∈ L1
2(R3
).
||f0||L1
2
= 4π
+∞
0
r2
(1 + r2
) ˜f0(r)dr
= 4π
+∞
2
r2
(1 + r2
) ˜f0(r)dr
≤ 8π
+∞
2
r4 ˜f0(r)dr
= 8π
n≥2
n+ 1
nn2
n
r4 ˜f0(r)dr
≤ 128π
n≥2
n4 ˜f0(n)
1
nn2
= 128π
n≥2
1
n(log n)2
< +∞
Second, we prove H(f0) = +∞. By the definition of f0, we know ∃ N s.t. ˜f0(n) ≥ 1 ∀n ≥ N,
so
H(f0) = 4π
+∞
0
r2 ˜f0(r) log ˜f0(r)dr
= 4π
+∞
2
r2 ˜f0(r) log ˜f0(r)dr
= 4π
n≥2
n+ 1
nn2
n
r2 ˜f0(r) log ˜f0(r)dr
≥ 4π
n≥2
n2 ˜f0(n) log ˜f0(n)
1
nn2
= 4π
n≥2
1
n3(log n)2
(log(nn2
) − log(n5
(log n)2
))
= 4π[
N−1
n=2
1
n3(log n)2
(log(nn2
) − log(n5
(log n)2
))
+
n≥N
1
n log n
−
n≥N
1
n3(log n)2
(5 log n + 2 log(log n))]
In the square brackets on the right hand side of the above equation, the first summation only has
finite terms, so is finite; the second summation equals to +∞ as is well-known; the third summation
is bounded from above by a series in the form of
n≥N
C/n3
, so it’s also finite. Therefore H(f0) = +∞.
Now let’s prove
E1(f0)(t) < +∞ ∀t > 0 (3.20)
12

13. where E1(f0)(t) is defined in Theorem 3.1. For any fixed t > 0, we have
R3
f0(v)| log f0(v)| exp(−L1|v|β
t)dv
= 4π
+∞
0
r2 ˜f0(r)| log ˜f0(r)| exp(−L1rβ
t)dr
= 4π
+∞
n=0
n+1
n
r2 ˜f0(r)| log ˜f0(r)| exp(−L1rβ
t)dr
= 4π
+∞
n=2
nn2
n5(log n)2
log
nn2
n5(log n)2
n+n−n2
n
r2
exp(−L1rβ
t)dr
≤ 4π
+∞
n=2
nn2
n5(log n)2
log
nn2
n5(log n)2
(2n)2
exp(−L1nβ
t)n−n2
≤ 4π
+∞
n=2
4 exp(−L1nβ
t)
n3(log n)2
n2
log n + |5 log n + 2 log(log n)|
= 16π
+∞
n=2
exp(−L1nβ
t)
n log n
+
+∞
n=2
exp(−L1nβ
t)|5 log n + 2 log(log n)|
n3(log n)2
(3.21)
Since we are considering the case β > 0, t > 0, the first term on the right hand side of (3.21) is
obviously finite; the second term would be finite even if there wasn’t the factor exp(−L1nβ
t), so
(3.21) is finite. Note that the left hand side of (3.21) is the absolute integral of E1(f0)(t), so (3.20)
holds. Then by Theorem 3.1 (2) we know H(f)(t) < +∞ ∀t > 0. The proof is complete.
Remark 3.3 From the proof of Example 3.1, one can see that the key of the appearance of entropy
is the series +∞
n=2
exp(−L1nβ
t)
n log n
. From t = 0 to t > 0, this series becomes convergent instantaneously.
Also, for the Maxwell Molecular Model, as β = 0, the decaying effect of the factor exp(−L1nβ
t) will
fail, thus the appearance of entropy never happen, which is consistent with the conclusion of Theorem
3.1.
The initial data defined by Example 3.1 vanishes on most of the places in the velocity space,
which makes this example look unnatural. Thus we give the following Example 3.2, which is positive
everywhere on R3
but does lead to appearance of entropy. In fact, Example 3.2 is obtained by adding
a Maxwellian distribution onto Example 3.1.
Example 3.2 Let the collision kernel B satisfies (1.4) and (1.5) with 0 < β ≤ 1 and (3.3). Define
f0(v) = g0(v) + exp(−|v|2
) ∀v ∈ R3
where g0 is defined as f0 in Example 3.1. Then H(f0) = +∞ and for any conservative solution f(v, t),
we have
H(f)(t) < +∞ ∀t > 0
Proof: It’s obvious that f0 ∈ L1
2(R3
).
For the entropy, we have
H(f0) =
R3
(g0(v) + e−|v|2
) log(g0(v) + e−|v|2
)dv
=
R3
g0(v) log(g0(v) + e−|v|2
)dv +
R3
e−|v|2
log(g0(v) + e−|v|2
)dv
≥
R3
g0(v) log g0(v)dv +
R3
e−|v|2
log(e−|v|2
)dv
According to Example 3.1, the first term on the right hand side of the above equation equals to +∞;
the second term is obviously finite, so H(f0) = +∞.
Consider the function E1 defined in Theorem 3.1. Using the convexity of function x log x, we can
prove E1(·) has the following convex-cone property: for any t ≥ 0
E1(F)(t) < +∞, E1(G)(t) < +∞ =⇒ E1(λF + µG)(t) < +∞ ∀λ, µ ≥ 0 (3.22)
13

14. According to Example 3.1, we have E1(g0)(t) < +∞ ∀t > 0, and it’s obvious that E1(e−|v|2
)(t) <
+∞ ∀t > 0, so by (3.22) we have E1(f0)(t) < +∞ ∀t > 0. Consequently, by Theorem 3.1 (2) we
know H(f)(t) < +∞ ∀t > 0
Now using Theorem 3.1 (1), we give an example that entropy doesn’t appear.
Example 3.3 Define
f0(v) = ˜f0(|v|) =
e2n
|v| ∈ ( 1
2n−1 − 1
2ne2n , 1
2n−1 ], n = 1, 2, 3, ...
0 extra
Proof: It can be proved that ||f0||L1 = 1 and H(f0) = +∞. It’s easy to see that f0 is supported
on the compact set {v ∈ R3
| |v| ≤ 1}, so it must hold that f0 ∈ L1
2(R3
) and E1(f0)(t) = +∞ ∀t > 0.
So the proof is complete by applying Theorem 3.1 (1).
4 Conclusion and Future Direction
Finite entropy could appear for some systems with infinite initial entropy. For some others, the
entropy will keep infinite all the time. In the latter case, new functionals need to be defined in place
of the H-functional to describe the system’s irreversible behavior.
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