A BOLTZMANN MODEL FOR SCHOOLING OF FISH Maria C. Carvalho - - PowerPoint PPT Presentation

A BOLTZMANN MODEL FOR SCHOOLING OF FISH

Maria C. Carvalho University of Lisbon Joint work with E. Carlen, P . Degond and B. Wennberg

A BOLTZMANN MODEL FOR SCHOOLING OF FISH – p. 1/49

The kinetic equation

The main concern of this paper is the following Boltzmann equation:

∂tf(t, x) =

π

−π

π

−π

• f(t, x′)f(t, x′ + y)g(x − x′ − y

2) − f(t, x)f(t, x + y)

• β(| sin(y/2)|)dx′

2π dy 2π .

(1)

The unknown f is a probability density on T1, giving e.g. the distribution of directions in a fish school, and g is a given probability density modeling the noise in the model. In our case β is just a constant or, perhaps more realistically, β(x) = |x|.

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The microscopic model

This kinetic equation has been rigorously derived as a limit as N → ∞ of an pair interaction driven N-particle system. These systems are defined as Markov jump processes in an N-fold product space TN, where jumps only involve two

• coordinates. The jumps are triggered by a Poisson clock

with a rate proportional to N, and the outcome of the jump is independent of the clock. A jump involves first a choice of a pair (j, k) from the set 1 ≤ j < k ≤ N, and then a transition

x → x′, independent of (j, k): x = (x1, ...., xj, ..., xk, ...., xN) → (x1, ...., x′

j, ..., x′ k, ...., xN) = x′ .

(2)

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This jump process we study here has the following state space and jump rules. The state space is the N-dimensional torus. We use coordinates zj = exp(ixj) ∈ C. The jumps take a pair (zj, zk) to

(z′

j, z′ k) = (

zj,keiXj, zj,keiXk) ,

(3)

where

• zj,k = zj + zk

|zj + zk|

is the mid point of the smallest interval on the circle limited by zj and zk, and Xj and Xk are independent and equally distributed angles.

A BOLTZMANN MODEL FOR SCHOOLING OF FISH – p. 4/49

• zjk

zj zk z′

j

z′

k

Figure 1: zjk = zj+zk

|zj+zk|

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It is clear from this picture the process produces

• correlations. It is also clear that to describe schooling of

fish, the process must produce correlations While this may seem inconsistent with propagation of chaos, it has been shown that propagation of chaos does hold for this model, and the kinetic equation we study has been rigorously derived from it (Carlen, Degond and Wennberg 2013).

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Our focus here is on the steady states of this equation. While the uniform distribution is always a steady state, no matter what the noise is, it is not always stable. Our main result is a proof of existence of stable, sharply peaked steady states.

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Fourier variables

Because all functions are periodic, it is natural to consider to rewrite the system in terms of the Fourier series. Multiplying our kinetic equation with a test function ϕ, integrating over [−π, π], and performing a change of variables gives

d dt

• S1 f(t, x)ϕ(x) dx

2π =

• −π,π3 f(t, x)f(t, x+y)g(z)˜

β(y) (ϕ(x + y/2 + z) − ϕ(x)) dx 2π dy 2π dz 2π ,

where ˜

β = 1 (the Maxwellian case) or ˜ β = | sin(y/2)| (the

“hard sphere“ case).

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Introduce

ak(t) =

π

−π

f(t, x)e−ikx dx 2π . γk = (2π)−1

π

−π

g(x)e−ikxdx, Γ(u) = (2π)−1

π

−π

˜ β(y)eiuydy,

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Consider the case in which g and f, are even, so that

γk = γ−k and ak = a−k. Then they can be written as

cosine-series,

g(x) = 1 + 2

∞

• k=1

γk cos(kx) , f(t, x) = 1 + 2

∞

• k=1

ak(t) cos(kx) .

Solving for the ak(t) gives us the evolution of f(t, x).

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Now, taking the choice ϕ(x) = e−ikx in the equation

d dt

• S1 f(t, x)ϕ(x) dx

2π =

• [−π,π]3 f(t, x)f(t, x+y)g(z)˜

β(y) (ϕ(x + y/2 + z) − ϕ(x)) dx 2π dy 2π dz 2π ,

and doing the integrals, one gets an evolution equation for the Fourier coefficients:

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Proposition 0.1. Suppose that g is even. Let ak(t) be the Fourier coefficients of a solution of our evolution equation which is an even probability density. Then, a0 = 1 and ak for k = 0 satisfy a−k = ak and solve the following system:

d dtak(t) = ( 2γkΓ(k/2) − Γ(0) − Γ(k) ) ak(t) +

k−1

• n=1

(γkΓ(n − k/2) − Γ(n)) an(t)ak−n(t) +

∞

• n=k+1

(2γkΓ(n − k/2) − Γ(n) − Γ(n − k)) an(t)an−k(t)

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The function Γ(u), which is to be evaluated only on half-integer points, is

Γ(u) = sin(πu) πu

which is       

1

when

u = 0

when

u ∈ Z \ {0} 2(−1)ℓ π(2ℓ + 1)

when

u = ℓ + 1/2

in the Maxwellian case, which is the case ˜

β(1) ≡ 1.

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We will mostly be concerned with the Maxwellian case, but the equations for the Fourier coefficients can be worked out also in the hard-sphere case, when ˜

β(y) = | sin(y/2)|. In this

case,

Γ(u) = 2 − 4u sin(πu) π − 4πu2

which is       

2/(π(1 − 4u2))

when

u ∈ Z 1/π

when

u = ±1/2 2(−1)ℓℓ + (−1)ℓ − 1 2πℓ2 + 2πℓ

when

u = ℓ + 1/2, ℓ = 0, −1 ,

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Finitely many modes in the noise

Notice that since Γ(j) = 0 for all j ∈ N, if it is the case that

γk := g(k) =

π

−π

g(x)e−ikxdx 2π = 0,

the equation for ak(t) reduces to

d dtak(t) = −Γ(0)ak(t) = −ak(t)

The important conclusion is that if the noise distribution g has only finitely many Fourier modes, so that for some integer N, γk = 0 for |k| > N, then for all k such that |k| > N,

ak(t) tends to zero exponentially fast. Only the modes with |k| ≤ N can be present in a steady state.

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Now we rewrite our evolution equation, which is

d dtak(t) = ( 2γkΓ(k/2) − Γ(0) − Γ(k) ) ak(t) +

k−1

• n=1

(γkΓ(n − k/2) − Γ(n)) an(t)ak−n(t) +

∞

• n=k+1

(2γkΓ(n − k/2) − Γ(n) − Γ(n − k)) an(t)an−k(t)

in a manner to make this evident, specializing to the Maxwellian case

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The Maxwellian case equation

Finally our equation for the Maxwellian case is

d dtak(t) = ( 2γkΓ(k/2) − 1 ) ak(t) +

k−1

• n=1

γkΓ(n − k/2)an(t)ak−n(t) +

∞

• n=k+1

2γkΓ(n − k/2)an(t)an−k(t)

(4)

We now have our equation in a form that is adapted to the study of stability and existence of steady states. We begin with the stability.

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The stability analysis

To investigate the stability of the uniform density, let

f(x, t) = 1 + εF(x, t), and let bk(t), k ∈ Z be the Fourier

coefficients of F(x, t). Then b0 = 0, and for k = 0,

d dtbk(t) = bk(t) (2γkΓ(k/2) − Γ(0) − Γ(k)) + O(ε)

Hence the linearized stability may be determined by analysing separately λk = (2γkΓ(k/2) − Γ(0) − Γ(k)). Thus each Fourier mode is an eigenfunction of the linearized operator and for k = 0 the eigenvalue is

λk = 2γkΓ(k/2) − Γ(0) − Γ(k) .

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In the Maxwellian case.

Γ(0) + Γ(k) = 1

for all positive integers k, and

Γ(k/2) = 2 sin

kπ

2

• kπ

.

Therefore,

λk = γk 4 sin

kπ

2

• kπ

− 1 .

Whatever the choice of g, |γk| ≤ 1, so that

k > 1 ⇒ λk < 0 .

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This leads to:

Theorem 0.2. In the Maxwellian case, there is at most one linearly unstable mode at the uniform density, namely the k = 1 mode. The uniform density is linearly stable if and only if

γ1 := g(1) ≤ π 4. (∗)

When there is strict inequality in (∗), the uniform density is (non-linearly) stable.

This theorem points to the special role that a1 :=

f(1) will

play in the proof of existence of nonuniform steady states. We would expect that whenever the noise function g is such that (∗) is violated, there will exist a stable, non-uniform steady state.

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An example

Before turning to the existence of stable non-uniform steady states, let us look at a concrete example. Consider a family

• f distributions g(y) defined as the periodization of 1

τ ρ(y τ ),

where ρ is a given, even, probability density on R:

gτ(y) = 2π

∞

• j=−∞

1 τ ρ

y − 2πj

τ

• Then:

A BOLTZMANN MODEL FOR SCHOOLING OF FISH – p. 21/49

γk =

π

−π

e−iky2π

∞

• j=−∞

1 τ ρ(y − 2πj τ ) dy 2π =

∞

−∞

e−iτkyρ(y) dy = ρ(τk) .

An example is ρ(x) =

1 √ 2πe−x2/2 which gives

• ρ(τk) = e−(τk)2/2.

In any case

lim

τ→∞

ρ(τk) = 0 and lim

τ→0

ρ(τk) = 1

Therefore

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When τ is small, the angles added to the average of the two directions is small. When τ is large, the angles are also large, and gτ converges to the uniform distribution when τ → ∞. In the latter case, γk → 0 when τ → ∞, and

λk → −Γ(0) − Γ(k)

when

τ → ∞

(5)

λk → 2Γ(k/2) − Γ(0) − Γ(k)

when

τ → 0

(6)

Consequently, for large enough τ, all the λk will be negative, and hence, in that case the uniform distribution f(x) ≡ 1 is stable.

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A more explicit example

To proceed further, we must make a more explicit choice for the noise.

Hypothesis 0.3. We assume that g = gγ1 is a family of noise distributions with a finite number of non-zero Fourier coefficients: for some N < ∞,

gγ1(x) = 1 + 2γ1 cos x + 2

N

• k=2

γk(γ1) cos kx, ∀x ∈] − π, π].

with C2 functions γ1 ∈ [0, 1] → γk(γ1) ∈ [−1, 1] and with γ2 such that

γ2(γ1) > 0.

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An explicit example with which we have carried out computations is a convex combination of a uniform distribution and a Fejer kernel

gλ(x) =(1 − λ) + λ 1 N

sin(Nx/2)

x/2

2

.

For such a noise distribution, we have γk = λ(N − k)/N for

1 ≤ k < N. Therefore, this family can be put in the

framework of Hypothesis 0.3 if we link λ to γ1 by λ =

N N−1γ1.

In the numerical computations that follow, we use N = 9. In this seeing we prove:

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Theorem 0.4. Consider a one-parameter family of noise functions gγ1 satisfying Hypothesis 0.3. Then: (i) The uniform distribution, with Fourier coefficients

a0 = 1, ak = 0 (k ≥ 1) is stationary. It is stable for γ1 < π/4 and

unstable for γ1 > π/4. (ii) In an interval π

4 < γ1 < γmax there is another invariant solution to

the dynamic problem, with Fourier coefficients a0 = 1,

a1 =

• 12(γ1 − π/4)

πγ2(π/4) + O((γ1 − π/4)3/2), ...., ak = 0 (k > N) .

(iii) This solution is stable with a leading eigenvalue

λ(γ1) = 1 − 8

π(γ1 − π/4) + O((γ1 − π/4)3/2).

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The following figure shows the Fourier coefficient a1 as a function of the parameter γ1.

7 Π 32 Π 4 9 Π 32

0.6 0.3 0.3 0.6

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This figure exhibits a typical pitchfork bifurcation pattern. The order parameter a1 is identically zero as long as γ1 is less than the critical value γ1c = π/4 and the associated uniform equilibrium is stable. When γ1 becomes larger than the critical value γ1c a second branch of non-uniform equilibria starts. This branch is stable while the branch of uniform equilibria becomes

• unstable. In fact the non-uniform equilibria forms a

continuum, because the system is rotationally invariant, and therefore, if f is a non-isotropic equilibrium, then any

f(eiθ0x) with θ0 ∈]0, 2π[ is another equilibrium. This feature

is represented by the lower branch in the diagram.

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In physical terms, the system exhibits a symmetry-breaking second-order phase transition as γ1 crosses γ1c. From the point (ii) of the theorem, it appears that the critical exponent is 1/2, i.e. the order parameter behaves like

a1 ∼ (γ1 − γ1c)1/2 when γ1

≥

→ γ1c.

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The next figure shows the noise function g and the corresponding stationary solution f when γ1 = π/4 + 0.1.

Π 4 Π 2 3 Π 4

Π

1 2 3 4

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To prove this theorem, we must solve

ak = 2γkΓ(k/2)ak + γk

k−1

• n=1

Γ(n − k/2)anak−n + 2γk

∞

• n=k+1

Γ(n − k/2)anan−k

(7)

for k ≥ 1. Note that γk is a factor for all terms in the right hand side, implying that if g only has finitely many terms in the Fourier series, only the corresponding terms are nonzero in f. Thus under our hypothesis, the system of equations is a fixed point system in finitely many real variables, and its the bifurcation can be analyzed using the Implicit Function Theorem.

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The hard sphere case

In the ”hard spheres” case, we find that

2Γ(1/2) − Γ(0) − Γ(1) = 2/(3π) ,

and that for k > 1,

2Γ(k/2) − Γ(0) − Γ(k) = − 4

• 2k4 − 4 sin

kπ

2

• k3 + k2 + sin

kπ

2

• k
• (k2 − 1) (4k2 − 1) π

A similar, but more complicated analysis follows.

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Because Γ is an even function, it is enough to consider

k ≥ 2, and in that case the numerator is larger than 4

• 2k4 − 4 sin

kπ

2

• k3 + k2 + sin

kπ

2

• k
• ≥

4

• 2k4 − 4k3 + k2 − k
• ≥

4(k2 − k) > 0

and hence we may deduce that λk < 0 for k > 1 also in this

• case. If γk changes sign the calculation is more

complicated, but the result is the same: it is only the first Fourier mode of the solution f that may cause instability of the uniform stationary states.

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Stationary solutions in more generality

Recall our kinetic equation for the Maxwellian case written in terms of the Fourier coefficients

d dtak(t) = ( 2γkΓ(k/2) − 1 ) ak(t) +

k−1

• n=1

γkΓ(n − k/2)an(t)ak−n(t) +

∞

• n=k+1

2γkΓ(n − k/2)an(t)an−k(t)

For a stationary solution the coefficients are constant and the left hand side is 0. This permits us to write the steady state equation as a fixed point equation.

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ak = 2γkΓ(k/2)ak +

k−1

• n=1

γkΓ(n − k/2)anak−n +

∞

• n=k+1

2γkΓ(n − k/2)anan−k

If g is a positive trigonometric polynomial, so that for some finite N,

k > N ⇒ γk = 0 , then k > N ⇒ ak = 0 ,

and we have a finite dimensional problem.

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In any case, we can write the steady state equation as a fixed point equation

{ak} = Φ({ak}) ,

that must be satisfied by the sequence {ak} of Fourier coefficients of any steady state. We now introduce some notation to enable us to write Φ in a more useful form. With γk =

g(k) and Γ(u) = sin(πu)/(πu), we define Gi,j = γi+j 1 − 2γi+jΓ

• i+j

2

Γ i − j

2

• ,

which is defined for i, j ∈ Z.

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Then our fixed point equation becomes:

ak =

k−1

• j=1

Gk−j,jak−jaj + 2

∞

• j=1

Gk+j,−jak+jaj .

One easily sees

Gi,j = Gj,i, Gi,j = G−i,−j,

and

Gj,j = γ2j .

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Also, in the Maxwellian case,

Gi,j = 0

when

(i − j) = 0

is even , whereas for j − i odd, Gi,j satisfies

|Gi,j| ≤ |γi+j|

• 1 −

4|γi+j| (π|i + j|)

• 2

π|i − j|

By the top line, when k = 2m, our equation reduces to

a2m = γ2m(a2m−1)2 .

Iterating , we find a2m =

m−1

• j=0

(γ2m−j)2ja2m

1 .

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In fact, we can write every coefficient ak in terms of a1, though only when k is a power of 2 will this be a polynomial in a1 (and then it is a monomial). We take the value of a1 as an order parameter R, and seek an expression

ak(R) =

∞

• n=0

pk,nRk+2n .

As we have just seen, when k = 2m, we have such an expression with

p2m,n =

m−1

• j=0

(γ2m−j)2jδ0,n .

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Construction of invariant densities

Our program is to solve for a double sequence pk,n so that with

ak(R) =

∞

• n=0

pk,nRk+2n .

• ur fixed point equation is satisfied for all k ≥ 1.

Then the requirement a1 = R leads to

R = a1(R) =

∞

• n=0

p1,nR1+2n .

Solving this equation will then determine the value of R, and hence determine f, as we now explain. Finally, we shall need to show that f is non-negative, so that it is a probability density.

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The next lemma translates our fixed point equation for the

ak ak =

k−1

• j=1

Gk−j,jak−jaj + 2

∞

• j=1

Gk+j,−jak+jaj . (∗∗)

into a fixed point equation for the pk,n.

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Lemma 0.5. For each k ∈ N, let {pk,n} be a sequence such that

∞

n=0 pk,nzk+2n convergences for |z| ≤ 1. For −1 < R < 1, define

a−k(R) = ak(R) =

∞

• n=0

pk,nRk+2n.

Then the {ak(R)} satisfy (∗∗) for each k ≥ 2 if and only if the numbers

{pk,n}, k ≥ 1 and n ≥ 0 satisfy pk,n =

k−1

• j=1

n

• ℓ=0

Gk−j,jpk−j,ℓpj,n−ℓ + 2

n

• j=1

n−j

• ℓ=0

Gk+j,−jpk+j,ℓpj,n−(j+ℓ).

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Note that when n = 0, the second sum is zero; the summation is over an empty index set.

• Proof. Substitute

ak(R) =

∞

• n=0

pk,nRk+2n

into (∗∗); i.e.,

ak(R) =

k−1

• j=1

Gk−j,jak−j(R)aj(R) + 2

∞

• j=1

Gk+j,−jak+j(R)aj(R) .

and equate coefficient of like powers of R. .

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One might hope based on

R = a1(R) =

∞

• n=0

p1,nRk+2n

that

p1,n = δn,0 . (∗)

However this is inconsistent with the equation for pk,n provided by the lemma. Indeed, this formula show that

p1,1 = 0 since both sums are empty in this case.

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However, if we use the ansatz (∗) to define pk,n, we can then use the equation provided by the lemma to recursively define pk,n, first for k = 2, then k = 3, etc. The next lemma shows that this procedure gives us coefficients ak(R) that satisfy our fixed point equation for all

k > 1:

Lemma 0.6. If we define p1,n = δn,0, and then use the fixed point equation for {pk,n} to define pk,n for all k ≥ 2 and all n ≥ 0, we obtain under mild conditions on {γk}, a sequence of analytic functions

{ak(R)}, for k ≥ 1 (with a1(R) = R), such that for all |R| < 1, and for

all k ≥ 2,

ak(R) =

k−1

• j=1

Gk−j,jak−j(R)aj(R) + 2

∞

• j=1

Gk+j,−jak+j(R)aj(R) .

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It remains to patch things up for k = 1. For k = 1,

pk,n =

k−1

• j=1

n

• ℓ=0

Gk−j,jpk−j,lpj,n−l + 2

n

• j=1

n−j

• ℓ=0

Gk+j,−jpk+j,ℓpj,n−(j+l).

reduces to

p1,n = 2

n

• j=1

n−j

• ℓ=0

G1+j,−jp1+j,ℓpj,n−(j+ℓ)

We now use the {pk,n} that we have on the right to define new coefficients

p1,n. This can now be computed

recursively.

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We compute

• p1,2 = 2G3,−2p3,0p2,0
• p1,3 = 2[G2,−1p2,0p1,2 + G3,−2p3,1p2,0 + G4,−3p4,0p3,0].

This gives us a new function

a1(R) such that

• a1(R) = 2

∞

• j=1

G1+j,−ja1+j(R)aj(R) .

This becomes our k = 1 equation, and all is consistent, if and only if

• a1(R) = a1(R) = R .

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Positivity and radius of convergence

Suppose for example that one has a constant C such that

|pk,n| ≤ C for all k, n. This is implied by mild conditions on g

and hence the γk. Then by the geometric series formula,

|ak| ≤ CRk 1 1 − R2

and then

1 2|f(z) − 1| ≤

∞

• k=1

|ak| ≤ C R 1 − R 1 1 − R2

This is less than 1/2 for all R sufficiently small, and this ensures positivity of f.

A BOLTZMANN MODEL FOR SCHOOLING OF FISH – p. 48/49

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