On the Boltzmann equation: global solutions in one spatial dimension

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On the Boltzmann equation: global solutions in one spatial dimension Department of Mathematics & Statistics Colloque de mathématiques de Montréal Centre de Recherches Mathématiques November 11, 2005

Collaborators Andrei Biryuk Vlad Panferov

Summary The Boltzmann equation for f (x, v, t) 0 with x R 1 /Z 1, v R 3 The macroscopic density ρ(x, t) = f (x, v, t) dv v R 3 The entropy relative to a Maxwellian M(v) = m ( ) a 3/2 π e a v u 2 is H(f M) = (f log( fm ) ) f + M dvdx x T 1 v R 3 Main result: There exists a constant C 0 such that if ρ 0 (x) L x and for some M 0, H(f 0 M 0 ) C 0 then global strong solutions exist.

Outline The Boltzmann equation Global existence results Uniqueness Properties of propagation Main ideas of the proof

Outline The Boltzmann equation Global existence results Uniqueness Properties of propagation Main ideas of the proof

Outline The Boltzmann equation Global existence results Uniqueness Properties of propagation Main ideas of the proof

Outline The Boltzmann equation Global existence results Uniqueness Properties of propagation Main ideas of the proof

Outline The Boltzmann equation Global existence results Uniqueness Properties of propagation Main ideas of the proof

Boltzmann equation The classical Boltzmann equation t f + v x f = Q(f, f ), f (x, v, 0) = f 0 (x, v) (1) Phase space coordinates (x, v) R 3 x R 3 v. Phase space distribution function f (x, v, t). In case Q = 0 the solution is f (x, v, t) = f 0 (x tv, v) = Φ t (f 0 )(x, v) Free streaming flow of the equation (1) linearized about f = 0 Q(f, f ) is the collision operator

Boltzmann equation The classical Boltzmann equation t f + v x f = Q(f, f ), f (x, v, 0) = f 0 (x, v) (1) Phase space coordinates (x, v) R 3 x R 3 v. Phase space distribution function f (x, v, t). In case Q = 0 the solution is f (x, v, t) = f 0 (x tv, v) = Φ t (f 0 )(x, v) Free streaming flow of the equation (1) linearized about f = 0 Q(f, f ) is the collision operator

Boltzmann equation The classical Boltzmann equation t f + v x f = Q(f, f ), f (x, v, 0) = f 0 (x, v) (1) Phase space coordinates (x, v) R 3 x R 3 v. Phase space distribution function f (x, v, t). In case Q = 0 the solution is f (x, v, t) = f 0 (x tv, v) = Φ t (f 0 )(x, v) Free streaming flow of the equation (1) linearized about f = 0 Q(f, f ) is the collision operator

portrait of Ludwig Boltzmann (1844-1906) 29/09/05 http://www.sil.si.edu/digitalcollectio...-identity/thumbnails/tnsil14-b5-06.jpg #1

portrait of J. C. Maxwell (1831-1879) 03/10/05 http://www.egr.msu.edu/~bohnsac3/nano/figures/maxwe #1

Collision operator The collision operator of Maxwell and Boltzmann Q(f, f )(x, v) = (f (x, v )f (x, v ) f (x, v)f (x, v )) R 3 v S 2 σ K ds σ dv = Q + (f, f )(x, v) Q (f, f )(x, v) (2) Velocities before v, v and after v, v a binary collision satisfy v + v = v + v and (v v )/ v v = σ S 2 The collision kernel K = K( v v, (v v ) v v σ)

Collision operator The collision operator of Maxwell and Boltzmann Q(f, f )(x, v) = (f (x, v )f (x, v ) f (x, v)f (x, v )) R 3 v S 2 σ K ds σ dv = Q + (f, f )(x, v) Q (f, f )(x, v) (2) Velocities before v, v and after v, v a binary collision satisfy v + v = v + v and (v v )/ v v = σ S 2 The collision kernel K = K( v v, (v v ) v v σ)

Collision operator The collision operator of Maxwell and Boltzmann Q(f, f )(x, v) = (f (x, v )f (x, v ) f (x, v)f (x, v )) R 3 v S 2 σ K ds σ dv = Q + (f, f )(x, v) Q (f, f )(x, v) (2) Velocities before v, v and after v, v a binary collision satisfy v + v = v + v and (v v )/ v v = σ S 2 The collision kernel K = K( v v, (v v ) v v σ)

Macroscopic quantities macroscopic mass density ρ(x, t) = f (x, v, t) dv v R 3 macroscopic momentum density ρu(x, t) = v f (x, v, t) dv v R 3 macroscopic energy density ρe(x, t) = v 2 f (x, v, t) dv v R 3

Macroscopic conservation laws Conservation of mass: t ρ(x, t) + x F ρ (x, t) = 0 with density flux F ρ = v R v f (x, v, t) dv 3 Conservation of momentum: t ρu(x, t) + x F ρu (x, t) = 0 with momentum flux Conservation of energy: F ρu = v R 3 v v f (x, v, t) dv t ρe(x, t) + x F ρe (x, t) = 0 with energy flux F ρe = v R 3 v 2 v f (x, v, t) dv NB The closure problem

One dimensional geometry We take x R 1 and v R 3, which is known as the slab geometry. It is the setting of a narrow shock tube, or a situation with spanwise constant macroscopic quantities. Furthermore, we ask for periodic spatial boundary conditions; that is x R 1 and v R 3. NB: Periodic boundary conditions, or perturbations of a state of thermodynamic equilibrium (a constant Maxwellian background) are more difficult problems than setting f L 1 (R 3 x R 3 v) in the background of a vacuum.

Conserved quantities Total Mass M(f ) = ρ(x, t) dx = T 1 Total Momentum I(f ) = ρu(x, t) dx = T 1 Total Energy E(f ) = ρe(x, t) dx = T 1 T 1 T 1 T 1 f (x, v, t) dvdx v R 3 v f (x, v, t) dvdx v R 3 v 2 f (x, v, t) dvdx v R 3 These quantities are constants of motion for the Boltzmann equation.

Relative entropy Define the relative entropy of a phase space distribution function f (x, v) with respect to a constant Maxwellian distribution by H(f M) = (f log( f ) M ) f + M dvdx T 1 v R 3 Proposition (1) Positivity properties of the relative entropy 1. H(f M) 0 2. Let m := v R M dv. The Csiszár Kullback Pinsker inequality 3 is equivalent to f (x, v) M(v) dvdx 4mH(f M) + H(f M) T 1 v R 3

Relative entropy Define the relative entropy of a phase space distribution function f (x, v) with respect to a constant Maxwellian distribution by H(f M) = (f log( f ) M ) f + M dvdx T 1 v R 3 Proposition (1) Positivity properties of the relative entropy 1. H(f M) 0 2. Let m := v R M dv. The Csiszár Kullback Pinsker inequality 3 is equivalent to f (x, v) M(v) dvdx 4mH(f M) + H(f M) T 1 v R 3

The H-theorem Proposition (2) The relative entropy H(f M) decreases along the flow of the Boltzmann equation t H(f M) 0 Remarks: The following are equivalent: 1. f L log(l) and x v (1 + v 2 ) f dvdx < + ) 2. f L log for some Maxwellian M(v) ( f M

The f norm Definition (3) For functions f (x, v) on phase space define the norm f := sup f (x pv, v) dv (3) x T 1 R 3 p R + v Denote by X the space of functions for which this is finite. The function space X L x (L 1 v). NB: The streaming flow Φ t (f )(x, v) = f (x tv, v) is not continuous on L x (L 1 v) but it is so on X

The f norm Definition (3) For functions f (x, v) on phase space define the norm f := sup f (x pv, v) dv (3) x T 1 R 3 p R + v Denote by X the space of functions for which this is finite. The function space X L x (L 1 v). NB: The streaming flow Φ t (f )(x, v) = f (x tv, v) is not continuous on L x (L 1 v) but it is so on X

First global result Theorem (4) Assume that the collision kernel K satisfies hypothesis (H1). Then there is a constant K 0 such that if initial data f 0 for the Boltzmann equation (1) satisfies f 0 < + and H(f 0 M 0 ) 1 4πK 0 (4) for some Maxwellian M 0, then the solution f (, t) X for all t R +. More specifically the theorem asserts that there is an estimate on the L x norm of the density ρ(x, t). When strict inequality holds in (4) β such that ρ(x, t) L x f (x, v, t) C 0 exp( t/β)

First global result Theorem (4) Assume that the collision kernel K satisfies hypothesis (H1). Then there is a constant K 0 such that if initial data f 0 for the Boltzmann equation (1) satisfies f 0 < + and H(f 0 M 0 ) 1 4πK 0 (4) for some Maxwellian M 0, then the solution f (, t) X for all t R +. More specifically the theorem asserts that there is an estimate on the L x norm of the density ρ(x, t). When strict inequality holds in (4) β such that ρ(x, t) L x f (x, v, t) C 0 exp( t/β)

Hypotheses on the collision kernel The result depends upon a newly described smoothing property of the collision kernel K. ( K(r, ξ) = K v v, (v v ) σ ) v v Hypothesis (H1) Large relative velocity interactions are soft, while small relative velocity interactions are hard: (H1) K 0 w 0 K(r, ξ) 1 + w log 1+ε (w + 1) NB This hypothesis is related to one appearing in Cercignani s work.

Compare with Boltzmann s collision kernel Power law potential interactions between molecules for 1 < p are in the form γ V( q 1 q 2 ) = q 1 q 2 p In this case, the classical Boltzmann collision kernel has the form K(r, ξ) = b(ξ) v v β where < β = β(p) 1. For hard spheres β = 1, For the Maxwell molecule case p = 5 it turns out that β = 0. NB: For all but the hard spheres case, b(ξ) diverges nonintegrably as ξ ±1. Grad cutoffs appear here, where they truncate grazing collisions.

A stronger hypothesis on the collision kernel gives a better result Hypothesis (H2) Small relative velocity interactions are absent. (H2) (i) K(r, ξ) = 0 for r < R 1 (ii) K(r, ξ) β(ξ) sup ξ ( 1,1) ( K(r, ξ) ) for R1 < r, where β(ξ) is positive on a set of positive measure. Theorem (5) Assume that hypotheses (H1) and (H2) hold, and that f 0 < + and M(f 0 ) + E(f 0 ) < + (5) then for all t R + solutions to (1) exist and have Lx density ρ(x, t). NB These solutions could have infinite entropy macroscopic

A stronger hypothesis on the collision kernel gives a better result Hypothesis (H2) Small relative velocity interactions are absent. (H2) (i) K(r, ξ) = 0 for r < R 1 (ii) K(r, ξ) β(ξ) sup ξ ( 1,1) ( K(r, ξ) ) for R1 < r, where β(ξ) is positive on a set of positive measure. Theorem (5) Assume that hypotheses (H1) and (H2) hold, and that f 0 < + and M(f 0 ) + E(f 0 ) < + (5) then for all t R + solutions to (1) exist and have Lx density ρ(x, t). NB These solutions could have infinite entropy macroscopic

Uniqueness in X Solutions f (x, v, t) of the Boltzmann equation (1) such that ρ(x, t) = f (x, v, t) dv C([0, T] : Lx ) R 3 v are known as strong solutions. Solutions in the class X are unique. However the more general property of weak/strong uniqueness holds.

Uniqueness in X Solutions f (x, v, t) of the Boltzmann equation (1) such that ρ(x, t) = f (x, v, t) dv C([0, T] : Lx ) R 3 v are known as strong solutions. Solutions in the class X are unique. However the more general property of weak/strong uniqueness holds.

Uniqueness in X Solutions f (x, v, t) of the Boltzmann equation (1) such that ρ(x, t) = f (x, v, t) dv C([0, T] : Lx ) R 3 v are known as strong solutions. Solutions in the class X are unique. However the more general property of weak/strong uniqueness holds.

Dissipative solutions Definition (P.-L. Lions (1994)) A nonnegative f (x, v, t) C([0, T]; L 1 (T 1 R 3 )) is a dissipative solution of the Boltzmann equation (1) if for all satisfying g(x, v, t) L ([0, T]; L x (T 1 ; L 1 v(r 3 ))) t g + v x g Q(g, g) = E(g), in the sense of distributions, then f obeys the inequality t f g dv + x v f g dv Q(g, f g) sgn(f g) dv E(g) sgn(f g) dv

Weak/strong uniqueness Theorem (6) Given initial data satisfying (4), that is f 0 < + and H(f 0 M 0 ) 1 4πK 0 (6) any other dissipative solution starting with the same initial data f 0 must coincide with the strong solution for all t R + In the case where hypothesis (H2) also holds, for uniqueness we only require f 0 and M(f 0 ) + E(f 0 ) < +

Weak/strong uniqueness Theorem (6) Given initial data satisfying (4), that is f 0 < + and H(f 0 M 0 ) 1 4πK 0 (6) any other dissipative solution starting with the same initial data f 0 must coincide with the strong solution for all t R + In the case where hypothesis (H2) also holds, for uniqueness we only require f 0 and M(f 0 ) + E(f 0 ) < +

Moments and derivatives This question has to do with the evolution of smoothness and moment properties of the phase space distribution function f (x, v, t), once the basic property of being a strong solution is established. Moments M κ (f ) = sup x T 1 R 3 vκ f (x, v, t) dv v Derivatives in v Q λ (f ) = sup x T 1 R 3 λ v f (x, v, t) dv v Derivatives in x P µ (f ) = sup x T 1 R 3 µ x f (x, v, t) dv v

Propagation of moments and derivatives Theorem (7) Given initial data f 0 (x, v) satisfying (4) (or in the case where hypothesis (H2) also holds, (5)). If in addition for integers k, l, m the data satisfies sup v κ v λ x µ f 0 (x, v) dvdx < + v R 3 κ k, λ l, µ m x T 1 then for all t > 0 sup v κ v λ x µ f (x, v, t) dvdx ϕ klm (t) < + v R 3 κ k, λ l, µ m x T 1 The growth rate is bounded by c klm exp(exp( t/β)).

Propagation of moments and derivatives Theorem (7) Given initial data f 0 (x, v) satisfying (4) (or in the case where hypothesis (H2) also holds, (5)). If in addition for integers k, l, m the data satisfies sup v κ v λ x µ f 0 (x, v) dvdx < + v R 3 κ k, λ l, µ m x T 1 then for all t > 0 sup v κ v λ x µ f (x, v, t) dvdx ϕ klm (t) < + v R 3 κ k, λ l, µ m x T 1 The growth rate is bounded by c klm exp(exp( t/β)).

Main ideas of the proof Using Duhamel s principle and the streaming flow, rewrite the Boltzmann equation as an integral equation f (x, v, t) = f 0 (x tv, v) + t 0 Q ( x (t s)v, v, s ) ds Solutions satisfy the maximum principle: f 0 (x, v) 0 implies f (x, v, t) 0 for t > 0 Drop the Q term for the inequality 0 f (x, v, t) f 0 (x tv, v) + Integrate in v R 3 0 ρ(x, t) f 0 (x tv, v) dv+ v t 0 t 0 Q +( x (t s)v, v, s ) ds v Q +( x (t s)v, v, s ) dvds (7)

The integrand in (7) involves a spherical integral through the part of the collision operator Q + Q +( x (t s)v, v, s ) dv v = f (x (t s)v, v )f (x (t s)v, v ) = v v S 2 σ K( v v, (v v ) v v σ) ds σdv dv f (x (t s)v, v)f (x (t s)v, v ) v v S 2 σ K( v v, (v v ) v v σ) ds σdv dv where v = 1 2 ((v + v ) + v v σ)

A smoothing property In one dimensional geometries, the integral over σ S 2 converts to a spatial integral over the interval [v min, v max ] where v min = 1 2 ((v + v ) e 1 v v ) v max = 1 2 ((v + v ) e 1 + v v ) Changing variables in the integral (assume for simplicity that K(r, ξ) = 0 for r > R) v Q +( x (t s)v, v, s ) dv 1 R(t s) 4πK 0 ρ 2 (x + z, s) dz t s R(t s)

Two estimates of the integrand The first estimate of the integrand is simply through ρ L 1 R(t s) 4πK 0 ρ 2 (x + z, s) dz 8πK 0 R ρ 2 L t s R(t s) The second estimate uses the relative entropy H(f M) 1 R(t s) t s R(t s) 4πK 0 ρ 2 (x + z, s) dz 4πK 0H(f M) t s ρ L log( ρ L ) Each one individually gives rise to an estimate which doesn t forbid blowup in finite time.

The integral inequality Estimate the integrand by optimizing the two estimates Denote f (x, v, t) = ϕ(t) and α = C 0 K 0 H(f 0 M) ϕ(t) ϕ(0) + t 0 { ( α ) ϕ(s) } min c 1 ϕ 2 (s), t s + c 2 ds (8) log(ϕ(s)) Theorem (8) Global solutions of (8) depend only upon α 1. The constant c 1 depends upon ε, K 0 and H(f 0 M), while c 2 depends upon the initial data.

The Bony functional The proof of the second theorem under hypothesis (H2) is based on the Bony functional b(t) = f (x, v, t)f (x, v, t) K( v v, σ (v v )/ v v ) v v 2 ds σ dvdv dx which has the property that 0 b(t) dt < + Using similar ideas in the case where (H2) holds, the integral inequality is t { ( ) } 1 ϕ(t) ϕ(0) + min c 1 ϕ 2 (s), t s + 1 c 3 b(s) ds (9) 0

global existence Proposition (9) Given b(t), the maximal solution ϕ(t) of the integral inequality (9) is a locally bounded function of t. Furthermore the quantity f (, t) ϕ(t). This implies global existence of a strong solution f (x, v, t) However no rate of growth is available from this method, and indeed there may not be any quantitative rate.

Future directions Theorem (5) is restricted to one dimensional geometries. It is an open question whether Theorem (4) is so constrained. Can we relax the conditions (H1) on the collision kernel, possibly by using energy conservation. Does a similar theorem hold for nonelastic collisions (P. Degond). One of the difficult tendencies is for particles to coagulate (this is avoided precisely through hypothesis (H2) in our second category of results). Use Boltzmann-like kinetic equations to study homogeneous forms of dispersive nonlinear Hamiltonian partial differential equations