Linear Boltzmann Equation and Fractional Diffusion

Consider the linear Boltzmann equation of radiative transfer in a half-space, with constant scattering coefficient $\sigma$. Assume that, on the boundary of the half-space, the radiation intensity satisfies the Lambert (i.e. diffuse) reflection law with albedo coefficient $\alpha$. Moreover, assume that there is a temperature gradient on the boundary of the half-space, which radiates energy in the half-space according to the Stefan-Boltzmann law. In the asymptotic regime where $\sigma\to+\infty$ and $1-\alpha\sim C/\sigma$, we prove that the radiation pressure exerted on the boundary of the half-space is governed by a fractional diffusion equation. This result provides an example of fractional diffusion asymptotic limit of a kinetic model which is based on the harmonic extension definition of $\sqrt{-\Delta}$. This fractional diffusion limit therefore differs from most of other such limits for kinetic models reported in the literature, which are based on specific properties of the equilibrium distributions (heavy tails) or of the scattering coefficient as in [U. Frisch-H. Frisch: Mon. Not. R. Astr. Not. 181 (1977), 273-280].


Introduction
The diffusion approximation for the linear Boltzmann equation has been known for a long time and in very different contexts, such as nuclear engineering (see chapter IX in [23]), or radiative transfer (see chapter III.2 in [21]).There are several proofs of the validity of the diffusion approximation, involving rather different mathematical methods: stochastic processes [15,20,5], Hilbert expansion [17,7], or chapter XXI, §5 in [10], moment method [2,3] . . .However, there are several situations where the same type of scaling limits of the linear Boltzmann equation lead to fractional (or more generally nonlocal) diffusion equation.A first class of linear Boltzmann equations leading to a fractional diffusion equation includes situations where the scattering rate is not uniformly large over the whole energy range of the particle system considered.See [12] for an example in radiative transfer.
A second class of linear Boltzmann equations leading to a fractional diffusion equation includes the case of collision integrals whose equilibrium solutions have infinite second order moments: see for instance [4,18,19,1].
In the present work, we give an example of a completely different type of asymptotic limit of a linear Boltzmann equation leading to a fractional diffusion equation.
Before describing more precisely the physical problem considered in the present paper, we recall the following well-known fact from classical analysis, which the reader might find convenient to keep in mind.
We assume that true absorption and emission are negligible effects in this medium.We assume that the only process of physical importance in the medium is scattering (of Thomson type), which means that the elementary scattering process at the level of particles is independent of, and does not modify the frequency of photons.The scattering coefficient σ > 0 and scattering transition probability (also called "phase function" in §3 of [8]) p ≡ p(ω, ω ′ ) are assumed to be independent of the position variable z.Under these assumptions, the radiative intensity (integrated in the frequency variable) at the position z and in the direction ω, henceforth denoted f ≡ f (z, ω), satisfies the radiative transfer equation (see chapter I in [8]): On the plane of equation y = 0 (the boundary of the half-space), we assume that the radiation field obeys the Lambert reflection law, with albedo α (see formula (81) in §47 on p. 146 in [8]).On the other hand, we assume that the temperature at the point x of the plane of equation y = 0 is T (x), so that the surface emits a radiative intensity acT (x) 4 in all directions at the point x, according to the Stefan-Boltzmann law.The constant a is , where k B is the Boltzmann constant, c is the speed of light in the vacuum, and is the reduced Planck constant.
In other words, the radiative intensity f satisfies the boundary condition 1.3.Scaling and Other Assumptions.We are concerned with the following limit of the radiative transfer equation ( 1) with the boundary condition ( 2): (a) σ ≫ 1 (high scattering regime); (b) 0 < 1 − α ≪ 1 (high albedo surface).We shall adopt the following mathematical setting.For any integer d ≥ 1, set Z := T d × (0, +∞); points in Z are denoted by z = (x, y) with x ∈ T d and y > 0. Accordingly, unit tangent vectors to Z are denoted by 2 ω = (ω x , ω y ) ∈ S d , with ω x ∈ R d and ω y ∈ R. One has We shall henceforth use the following notation: and We assume that condition (b) is realized by taking the albedo coefficient α of ∂Z of the form α := κσ 1 + κσ where κ > 0 is a constant, and we set S(x) := (1 + κσ)aT (x) 4 .
Since β 2 = π, the boundary condition ( 2) is recast as Observing that the right hand side of (3) is independent of ω, this boundary condition implies that f (x, 0, ω) is independent of ω for ω y > 0, so that Substituting this expression on the left hand side of (3), we arrive at the following equivalent formulation of the boundary condition (3): We further assume that p ≡ p(ω, ω ′ ) is a measurable function defined a.e. on S d × S d satisfying the following condition: (5) We shall denote by L the bounded linear operator defined on L ∞ (S d ) by the formula 2 We denote by S d the unit sphere in R d+1 , and by B d the unit ball in R d .We also denote by With these elements of notation, the boundary value problem for the radiative transfer equation (1) with boundary condition (2) takes the form (6) where The purpose of the present work is to study the boundary value problem (6) in the asymptotic regime σ → +∞.

Main Result
2.1.Heuristic approach.In order to gain some intuition on this problem, we apply the Hilbert expansion method.This method is named after Hilbert on the basis of [14], where it has been used for the first time on the Boltzmann equation in the context of the kinetic theory of gases.Its application to the linear Boltzmann equation is discussed in detail in [17,7].
Thus, we seek f σ as the formal power series (7) ) for all integers j, m ≥ 0 .
Substituting this ansatz in the radiative transfer equation leads to the sequence of integral equations: Order O(σ): Order O(1): Proof.Because of the last equality in (5), one has R ⊂ Ker L.
For each φ ∈ L 2 (S d ), one has (8) The second equality above follows from the symmetry of p in (5), while the third equality is obtained by adding the right hand sides of the first and the second equalities.
If φ ∈ Ker L, then Averaging both sides of this equality in ω ′ implies that Thus φ is a.e. a constant, and this proves that Ker L = { constants }.Finally where the inequality follows from the Cauchy-Schwarz inequality applied to the inner integral, and the last equality from (8) with φ = ψ.Applying the Cauchy-Schwarz inequality to the last right hand side, one obtains which implies that L ≤ 2.
We conclude from this lemma the solution to the equation at order O(σ): Next we study the equation at order O(1) for f 1 .We shall need the following additional assumption on p: Denoting by K the linear operator defined by (10) (Kφ)(ω) := Lemma 2.2.The operator L is bounded and of Fredholm type on L 2 (S d ; C d ), and satisfies Proof.The operator K is a self-adjoint Hilbert-Schmidt operator on L 2 (S d ; C d ) since its integral kernel p is a symmetric real-valued square-integrable function by ( 5)- (9).In particular K is a compact operator on L 2 (S d ; C d ) (see section 6.4.2 and Theorem 6.12 in [6]).This implies that L is a self-adjoint, bounded operator on L 2 (S d ; C d ) of Fredholm type (see section 6.4.1 in [6]).
Fredholm's alternative (Theorem 6.6 in [6]) states that and we conclude the proof with the characterization of Ker L in Lemma 2.1.
Since ω = 0, Fredholm's alternative implies that there exists a unique function Combining Lemmas 2.1 and 2.2 leads to the following result for the solution of the equation at order O(1): Order O(1): Higher order terms in the expansion are found in the same way, by solving successively for f j+1 the integral equations We shall not pursue this line of investigation, since only f 0 and f 1 will be used in the present section.
At this point, we introduce a last assumption on the scattering transition probability p, specifically, we require that p is rotationally invariant: for each Q in the orthogonal group O d+1 (R), (12) p(Qω, Under this assumption, it has been proved in [7] that the diffusion matrix Besides Notice that this inequality is strict: otherwise, one would have 3 See formulas ( 40)-(44) on p. 624 in [7].Actually, one can deduce from equation (42) in [7] that Ω is of the form Ω(ω) = λω: see Lemma 3 in [11], or Lemma 8 in Appendix 1 of [13].With Ω of this form, one immediately concludes that ω ⊗ Ω = λ d+1 I.
This would imply that 0 = which is obviously a contradiction.
From the equation at order O(σ), i.e.
Otherwise, one would have ω With the expression for f 1 obtained above, this equality takes the form Moreover, if the Hilbert expansion ( 7) holds at the boundary, then we deduce from the equivalent formulation (4) of the boundary condition in ( 6) that ( 14) Returning to the original variables, and to the case of physical interest where d = 2, we recall that the radiation pressure P(x) at the point x of ∂Z is defined by the identity 1 c S 2 ω ⊗ ωf (x, 0, ω) dω = P(x)I .
(See for instance formula (1.19) in [21]).Therefore, to leading order in 1/σ, one has Since 4π 3c ρ 0 is an harmonic extension of P, applying the formula for the fractional diffusion operator recalled in section 1.1 leads to the following equation for the radiation pressure field on the boundary: Assumption ( 12) is verified by most of the scattering transition probabilities p used in practice, such as (see formulas (30)-( 31)-(33) in §3 or formula (192) in §16 of [8]).More generally, all scattering transition probabilities of the form ( 16) under the normalizing condition 12).In that case (since the integral the right hand side of this last equality must be invariant under the substitution ω ′ → −ω ′ ), so that Lω = ω , and therefore Ω(ω) = ω .
In that case ω • Ω = 1 , and the diffusion coefficient in the usual diffusion approximation of the linear Boltzmann equation is In particular, the equation ( 15) satisfied by the radiation pressure field on the boundary ∂Z is, in this case, 2.2.The Limit Theorem.The analysis based on Hilbert's expansion presented in the previous section is only formal.A rigorous analysis of the problem based on the moment method for kinetic models leads to the following result.
(a) For each σ > 0, the boundary value problem (6) has a unique solution this solution satisfies and ) for all j = 1, . . ., d.(b) In the limit as σ → +∞, one has and is the weak solution of the boundary value problem In the limit as σ → +∞, one has is the solution to the fractional diffusion equation 3. Proof of Theorem 2.3

3.1.
Step 1.Consider the boundary value problem Its unique solution is given by the method of characteristics Setting θ := e −(λ+σ)y/ωy we find that, for a.e.(z, ω) . This inequality implies, on the one hand, that, for each H ∈ L ∞ (T d ), the map λ+σ .Indeed, denoting by h 1 and h 2 corresponding to Q 1 and Q 2 respectively, one has , as a consequence of the previous inequality.Indeed, by linearity h 1 −h 2 is a solution of the boundary value problem above with source term Q 1 − Q 2 and boundary data H = 0.
Using assumption (5), we see that the operator K defined in ( 10) is a bounded operator on L ∞ (S d ), satisfying In other words, there exists a unique solution g ∈ L ∞ (Z × S d ) to the boundary value problem (19) λg and this solution satisfies the bound

Indeed, since
Kg L ∞ ≤ g L ∞ , the inequality above would imply that Step 2. With step 1, for each λ > 0 and σ > 0, we have constructed a linear map ) defined by the formula T λ,σ H = g , where g is the solution to the boundary value problem (19).We have also proved that In terms of the operator K defined in (10), one has obviously In other words, A λ,σ,β is a contraction in the Banach space L ∞ (T d ) with Lipschitz constant ≤ β 1+β σ λ+σ < 1.By the fixed point theorem, there exists a unique The solution to the boundary value problem (20) is given by the formula The fixed point F satisfies in particular the bound Therefore which is uniform in λ > 0. Now we consider the boundary value problem ( 22) of which we seek a solution by passing to the limit in f λ as λ → 0. On account of the uniform bound ( 21), the Banach-Alaoglu theorem implies the existence of a sequence λ n → 0 such that One has obviously For each φ ∈ W 1,1 (T d ), one has By passing to the limit in the boundary condition In particular for a.e.ω ∈ S d such that ω y > 0.
The uniform bound (21) obviously implies that the solution to the boundary value problem (22) satisfies the bound

3.4.
Step 4. In this step, we check that, for each S ∈ L ∞ (T d ), there exists at most one weak solution f of the boundary value problem By linearity, it is enough to show that, if g ∈ L ∞ (T d ×(0, +∞)×S d ) is a solution to the boundary value problem then g = 0, a.e.. Denote the sequence of Fourier coefficients of g in the y variable by For each k and for a.e.ω ∈ S d , the function y → g(k, y, ω) belongs therefore to W 1,∞ (0, +∞).Multiplying both sides of the differential equation above by ĝ(k, y, ω) leads to the identity Taking the real part of the right hand side of the equality above, we arrive at the identty

Now we average both sides of this equality in ω:
At this point, we recall the following property of L.
Lemma 3.1.Under assumptions (5)-( 9) on the function p, the operator L satisfies the following properties.For each φ ∈ L 2 (S d ; C), one has Moreover, there exists µ > 0 such that The proof of this lemma is deferred until the end of the present step.
Thus, for each k ∈ Z d , the function is bounded and nonincreasing on (0, +∞), and applying Lemma 3.1 shows that 2σµ Next observe that ( 23) In view of the previous estimate, the first term on the right hand side of this equality belongs to L 1 (0, +∞) since |ω y | ≤ 1, while the second term is bounded by the Cauchy-Schwarz inequality: Hence the second term on the right hand side of ( 23) belongs to L 2 (0, +∞), and since the first term is also bounded in L ∞ (0, +∞) because g is assumed to belong to L ∞ (Z × S d ), we conclude that Thus, for each Y > 0, one has Moreover, the right hand side of this inequality satisfies by the Cauchy-Schwarz inequality.Summarizing, we have proved that, for all k ∈ Z d and all In particular, since L1 = 0, one has The second equality above implies that, for a.e.(k, y) ∈ Z d × (0, +∞), the function ω → ĝ(k, y, ω) is a.e.constant on S d .Hence for all ω ∈ S d , and a.e.(k, y) ∈ Z × (0, +∞) .
The discussion above shows the existence and uniqueness of the solution f σ ∈ L ∞ (Z × S d ) to the boundary value problem (6) in the case where β = κσ; besides we have seen at the end of Step 3 that If moreover S ∈ W 1,∞ (T d ), we may apply the results of Steps 1-4 to ∂ xj S; this shows that and satisfies the bound Notice that this bound is uniform in σ > 0. In other words, statement (a) in Theorem 2.3 is implied by steps 1-4 with β = κσ.
Proof of Lemma 3.1.The formula for φLφ is (8) where φ is replaced with φ and ψ with φ.
On the other hand, we already know from Lemma 2.2 that L is self-adjoint and of Fredholm type on L 2 (S d ; C d ).The Fredholm alternative implies that the continuous, one-to-one linear map is onto.By Banach's open mapping theorem (see Theorem 2.6 and Corollary 2.7 in [6]), this map is bicontinuous, which implies the existence of the positive constant µ.

3.5.
Step 5. Set arguing as at the end of step 2, we find that and then By the same token Let χ(y) := (1 − y) 2 + , so that χ ∈ C 1 ([0, ∞)) has its support equal to [0, 1], and set g σ (x, y) := F σ (x)χ(y) ; define further the function It is easily seen that h σ is a solution to the boundary value problem Because of the regularity of h σ in x, we deduce from the equation that the function z → h σ (z, ω) belongs to W 1,∞ (T d × (0, +∞)).Thus, we can multiply both sides of the first equation in (26) by 2σh σ and integrate in ω, to obtain After integrating in x ∈ T d both sides of this equality, we see that By construction y) and g σ (x, y) = 0 , for all y > 1 .On the other hand, Parseval's theorem implies that Proceeding as in step 4 and applying Lemma 3.1, we see that For all k ∈ Z d , the function is bounded and Lipschitz continuous on (0, +∞), and nonincreasing on (1, +∞).Hence With the decomposition we conclude that for all k ∈ Z d .(Notice that we do not seek a uniform in σ estimate in L 2 for fσ − fσ at this stage in the argument, although this is our ultimate goal.)Since this function is nonincreasing on (1, +∞) for all k ∈ Z d , we find that 0 = lim Thus, for each Y > 1, one has , 0, ω) = 0 for ω y > 0 by construction, and on the other hand, as explained above, Notice that the first integral on the right hand side of (27) involves only y ∈ [0, 1] since g(x, y) = 0 for all y > 1 by construction.By the Cauchy-Schwarz inequality, for all Y > 1, one has , so that Observe that and since this inequality holds for all Y > 0, we conclude that according to Step 3, and Therefore, there exists a sequence σ n → +∞ such that In particular, the fact that J n is bounded in L 2 (Z × S d ) implies that (28) f (z, ω) = ρ(z) for a.e.(z, ω) ∈ Z × S d .
In order to compute J, observe that, since L1 = 0, Since the linear operator L is continuous on L 2 (S d ), one has and since we conclude that Beside, since J n (z, •) ∈ (Ker L) ⊥ for all n and a.e.z ∈ Z, one has also Applying the Fredholm alternative to L, we conclude that Notice that the formula above for J and the L 2 bound on J n imply that ∇ρ ∈ L 2 (Z) .
Let now φ ∈ C ∞ c (T d × (0, +∞)).Multiplying both sides of the equation for f σn in (6) by σ n φ(z) + Ω(ω) • ∇φ(z) and integrating both sides in (z, ω), we obtain (30) 2), one has Hence the third term on the left hand side of (30) is 0, while the first and last terms on the left hand side of (30) combine to give The first integral on the left hand side is simplified by using Green's formula: σ n ωf σn φ(x, 0) dx .
Next we use the boundary condition verified by f σn , i.e.
or equivalently , ω y > 0 .Since the right hand side of this identity is independent of ω, one has also (31) In other words On the other hand, arguing as in Step 3 shows that By Ascoli-Arzelà theorem, we conclude that Summarizing, we have proved that One easily checks that this is the variational formulation of the boundary value problem (17).
This proves statement (b) of Theorem 2.3.

3.7.
Step 7. Set Y = T d × (0, 1).The results obtained in Step 6 imply that By applying Cessenat's trace theorem (Theorem 1 in [9]), we find that (32) This last point requires additional explanations.At this point, we use Cessenat's notation in [9].Observe that, for all x ∈ T d , the exit time from Y , starting from y in the direction ω, is Choosing the arbitrary parameter K = 1 in the definition of the measure dξ on p. 832 in [9], viz. dξ This observation justifies that the convergence holds as stated in (32).Since ρ ∈ L ∞ (0, +∞; L 2 (T d )) and ∇ρ ∈ L 2 (Z) , one has in particular ρ Y ∈ H 1 (Y ), so that Since ∇ρ ∈ L 2 (Z) and div(∇ρ) = 0 , we deduce from the trace theorem of J.-L.Lions for the space H(Y, div) of vector fields in L 2 (Y ; R d+1 ) with divergence in L 2 (Y ) (see Lemma 20.2 in [22]) that On the other hand and div z (σ n ωf σn ) = 0 .Applying again the Lions trace theorem in H(Y, div), we conclude that Since ∇ρ ∈ L 2 (Z), we find that Since ρ is a harmonic extension of R = ρ y=0 , we deduce from Kwasnicki's Theorem 1.1 (j) that This concludes the proof of statement (c) of Theorem 2.3.

Open Problems
The result obtained in the present paper suggests various questions, still open at the time of this writing.
(a) Can one extend Theorem 2.3 to other kinetic models -for instance, to the linearized Boltzmann equation, or even to the linearized BGK model?For instance, one could consider the Boltzmann equation for a vapor in a half-space over its liquid, condensed phase, with a linear combination of diffuse reflection and condensation or evaporation at the boundary, assuming that the Knudsen number in the vapor is small, and that the accomodation coefficient is close to one.At present, the proof of Theorem 2.3 is based on the maximum principle, which the linearized Boltzmann equation does not satisfy.Before treating the case of the linearized Boltzmann equation, it would be necessary to have a proof of Theorem 2. We recall that the linear map u → U is continuous from H This defines ∂U/∂n ∂Ω as an element of H −1/2 (∂Ω), and shows that the linear map Λ is continuous from H 1/2 (∂Ω) to H −1/2 (∂Ω).Observe that Λ is self-adjoint since Then ∆ z f = 0 on Ω if and only if or equivalently r 2 f ′′ n (r) + drf ′ n (r) − λ n f n (r) = 0 , n ∈ N , r > 0 .This differential equation has a two-dimensional space of solutions over the half-line (0, +∞), viz.

3. 3 .
Step 3.Call f λ the unique solution of the boundary value problem(20) obtained in step 2, for which we have obtained the bound(21) Since this function is moreover nonincreasing, we conclude that 0 = lim Y →+∞ ω y |ĝ| 2 (k, Y ) ≤ ω y |ĝ| 2 (k, y) for all y > 0 and all k ∈ Z d .

2 γ
3 based on an L 2 energy estimate.(b) Can one obtain in this way powers of the Laplacian other than (−∆) 1/2 ?The harmonic extension result recalled in section 1.1 suggests that one should seek a linearized Boltzmann equation leading to the diffusion operatory −2 ∆ x + γ 2 c γ/2 γ ∂ 2 y .The scattering operator of such a linearized Boltzmann equation must be strongly anisotropic; besides, the scattering coefficient should vanish either as y → +∞ or as y → 0 depending on whether γ ∈ (0, 1) or γ ∈ (1, 2).(c) Can one extend the result in Theorem 2.3 to other domains than half-spaces?Assuming that one can derive from the linearized Boltzmann equation in a bounded (spatial) domain Ω with smooth boundary the diffusion problem the trace ρ ∂Ω must satisfy the equationρ ∂Ω + κΛρ ∂Ω = S ,where Λ is the Dirichlet-to-Neumann operator, defined as follows.For u ∈ H 1/2 (∂Ω), we set Λu := ∂U ∂n ∂Ω where U ∈ H 1 (Ω) is the unique solution of −∆U = 0 in Ω , U ∂Ω = u .