Long-time behavior of second order linearized Vlasov-Poisson equations near a homogeneous equilibrium
MManuscript submitted to doi:10.3934/xx.xx.xx.xxAIMS’ JournalsVolume X , Number , XX pp. X–XX
LONG-TIME BEHAVIOR OF SECOND ORDER LINEARIZEDVLASOV-POISSON EQUATIONS NEAR A HOMOGENEOUSEQUILIBRIUM.
Joackim Bernier ∗ Univ Rennes, INRIA, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France
Michel Mehrenberger
Aix Marseille Univ, CNRS, Centrale Marseille, I2M, Marseille, France (Communicated by the associate editor name)
Abstract.
The asymptotic behavior of the solutions of the second order lin-earized Vlasov-Poisson system around homogeneous equilibria is derived. Itprovides a fine description of some nonlinear and multidimensional phenomenasuch as the existence of
Best frequencies . Numerical results for the 1 D × D and2 D × D Vlasov-Poisson system illustrate the effectiveness of this approach. Introduction.
We consider potentials φ = φ ( t, x ) : R × T d → R and distributionfunctions f = f ( t, x, v ) : R × T d × R d → R satisfying the Vlasov-Poisson system ∂ t f + v · ∇ x f − ∇ x φ · ∇ v f = 0∆ x φ = n ( f ) − (cid:82) R d f d vf ( t = 0) = f . (VP)Here periodic boundary conditions being used, T d is a d dimensional torus: thereexist L , . . . , L d > T d = ( R /L Z ) × · · · × ( R /L d Z ) . Furthermore, doingan assumption of neutrality, we only consider solutions of (VP) such that n ( f ) = 1 L . . . L d (cid:90) (cid:90) T d × R d f d x d v. In this paper, we aim at exhibiting nonlinear and multidimensional phenomena ofsolutions of (VP), pursuing a first preliminary work [4] on this subject. Beyond theirphysical interest, these phenomena can be relevant to evaluate the performances andthe qualitative properties of numerical methods.Since the very first developments of numerical methods for solving VP (we referto [15], for a review; the literature is particularly huge in 1 D × D and we can Mathematics Subject Classification.
Primary: 35Q83, 65Z05; Secondary: 44A10.
Key words and phrases.
Dispersion relations, Best frequency, Landau damping, Vlasov-Poisson,Laplace transform.This work was granted access to the HPC resources of Aix-Marseille Universit´e financed bythe project Equip@Meso (ANR-10-EQPX-29-01) of the program ”Investissements d’Avenir” su-pervised by the Agence Nationale de la Recherche. This work has been carried out within theframework of the EUROfusion Consortium and has received funding from the Euratom researchand training programme 2014-2018 and 2019-2020 under grant agreement No 633053. The viewsand opinions expressed herein do not necessarily reflect those of the European Commission. ∗ Corresponding author: Joackim Bernier. a r X i v : . [ m a t h . NA ] S e p JOACKIM BERNIER ∗ AND MICHEL MEHRENBERGER mention [14], as one of the earliest works in 2 D × D ), the numerical solutionsare compared to the solutions of the Vlasov-Poisson system linearized around ahomogeneous equilibria f eq ≡ f eq ( v ) . It consists in looking for solutions of (VP)of the type (cid:26) f = f eq + εgφ = 0 + εψ. Neglecting second order terms, g is formally a solution of ∂ t g + v · ∇ x g − ∇ x ψ · ∇ v f eq = 0 , ∆ x ψ + (cid:90) R d g d v = 0 ,g ( t = 0) = g . (VPL)This equation being linear and homogeneous, it is natural to try to solve it realizinga Fourier transform we respect to the variable x . Thus, we get ∂ t (cid:98) g + i ( v · k ) (cid:98) g − i (cid:98) ψ ( k · ∇ v ) f eq = 0 , −| k | (cid:98) ψ + (cid:90) R d (cid:98) g d v = 0 , (cid:98) g ( t = 0) = (cid:98) g , (VPLF)where k ∈ (cid:98) T d = (2 π/L ) Z × · · · × (2 π/L d ) Z and the Fourier transform with respectto the space variable is defined for u ∈ L ( T d ) and k ∈ (cid:98) T d by (cid:98) u ( k ) = d (cid:89) j =1 L j − (cid:90) T d u ( x ) e − ik · x d x. It is relevant to notice on (VPLF) that there is no energy exchange between spacemodes at the linear level. In other words, if g is a solution of (VPL) such that g ≡ (cid:98) g ( v ) e ik · x then it is of the form g ( t, x, v ) = (cid:98) g ( t, v ) e ik · x . As a consequence,the linear analysis is not well suited to describe multidimensional phenomena thatcould be confronted with numerical simulations.Since (VPLF) is linear and autonomous, it is natural to solve it with the Laplacetransform . This transform is defined for functions u : R ∗ + → C such that there exists λ ∈ R satisfying ue − λt ∈ L ∞ ( R ∗ + ) and z ∈ C such that (cid:61) z > λ by L [ u ]( z ) = (cid:90) ∞ u ( t ) e izt d t. Thus, it can be proven that solutions of (VPLF) are given by g ( t, x, v ) = (cid:88) k ∈ (cid:99) T d e ik · ( x − vt ) (cid:98) g ( k, v ) + i (cid:90) t e ik · ( x − v ( t − s )) (cid:98) ψ ( s, k ) k · ∇ v f eq ( v )d s, (1)and for (cid:61) z large enough L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) = N k ( z ) D k ( z ) =: M k ( z ) . (2)where N k and D k are holomorphic functions defined when (cid:61) z is large enough by N k ( z ) = − i | k | (cid:90) R d (cid:98) g ( k, v ) v · k − z d v and D k ( z ) = 1 − | k | (cid:90) k · ∇ v f eq ( v ) v · k − z d v. (3) it can be noticed that every function depending only on v is an equilibrium of (VP). ECOND ORDER DISPERSION RELATIONS 3
Thus to get a solution g of (VPL) by (1) we just have to solve the equation (2)(called dispersion relation ) determining an inverse Laplace transform.Up to some strong assumptions on f eq and (cid:98) g ( k ) (precised later), it can be proventhat N k and D k are entire functions and that, for all λ ∈ R , the number of zeros of D k with an imaginary part larger than λ is finite (see Remark 2). Thus, using theformula L [ t m e − iωt ]( z ) = i m +1 m !( z − ω ) m +1 , ω ∈ C , m ∈ N , (4)and realizing precise estimates of remainder terms, we can prove that (2) has ananalytic solution (cid:98) ψ whose analytic expansion is given, for all λ ∈ R , by (cid:98) ψ ( t, k ) = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ P ω,k ( t ) e − iωt + O ( e λt ) , (5)where P ω,k is the polynomial such that M k ( z ) = z → ω L [ P ω,k ( t ) e − iωt ]( z ) + O (1) is theexpansion of M k ( z ) in ω .Such an analysis was first realized by Landau [10], in 1946. It has been donerigorously and generalized in 1986 by Degond [7]. It gave a partial explanation tothe phenomenon of Landau damping . This latter corresponds to the dynamic of(VP) when for all k ∈ (cid:99) T d , D k ( z ) does not vanish if (cid:61) z ≥
0. In this case, the electricpotential goes exponentially fast to zero as t goes to + ∞ . In 2011, Mouhot andVillani proved the existence of this phenomenon for the nonlinear Vlasov-Poissonequation (VP) in [11].As we have just seen, due to the absence of energy exchange between the spacesmodes at the linear level, the linearization is not relevant to explain really mul-tidimensional phenomena. Furthermore, of course, it can not explain nonlinearphenomena. This motivates thus the study of the dynamic of the second order term in the expansion of f as powers of ε . More precisely, we look for a solution of (VP)under the form (cid:26) f = f eq + εg + ε h + o ( ε ) ,φ = 0 + εψ + ε µ + o ( ε ) , where h ( t = 0) ≡
0. Neglecting the third order terms, it can be proven formallythat ( h, v ) is a solution of ∂ t h + v · ∇ x h − ∇ x µ · ∇ v f eq = ∇ x ψ · ∇ v g, ∆ x µ + (cid:90) R d h d v = 0 ,h ( t = 0) = 0 . (VPL2)We recognize the linearized Vlasov-Poisson equations, with an initial condition equalto zero but with a source term. In that case, we refer to Denavit [8], for one of thefirst works on the subject, in 1965. Different second order oscillations appear andhave been studied by physicists (see for example [13] and references therein; thereare many references especially from the 1960s and 1970s). Our aim is to make herea rigorous mathematical study of the asymptotical behavior of the solutions of theseequations, which has, to the best of our knowledge, not already been performed.This expansion in power of ε is quite natural because if we assume that f , thesolution of (VP), is a smooth function of ε then g and h are just its first and secondTaylor coefficients around ε = 0. We admit that there is a priori no reason thatthis second order approximation provides a more accurate approximation of f for JOACKIM BERNIER ∗ AND MICHEL MEHRENBERGER long times than the usual first order approximation. Nevertheless, numerical resultssuggest that this second order approximation is relevant for long times both if theequilibrium is stable or not (see [4], [13] and Section 5). Furthermore, to the bestof our knowledge, the asymptotic expansion of the solutions of (VPL2) we realizein this paper (see Theorem 1.4) is much more precise than what is known for theasymptotic behavior of the solution of (VP) (see [5],[11]).Here, as for the linear case, we solve (VPL2) using a Laplace transform for thetime variable and a Fourier transform for the space variable. More precisely, somecalculations prove that (VPL2) is equivalent to h ( t, x, v ) = (cid:88) k ∈ (cid:99) T d i (cid:90) t e ik · ( x − v ( t − s )) (cid:98) µ ( s, k ) k · ∇ v f eq ( v )d s + (cid:90) t (cid:92) ∇ x ψ · ∇ v g ( s, k, v ) e ik · ( x − v ( t − s )) d s, and when (cid:61) z is large enough L [ (cid:98) µ ( t, k )] ( z ) = N k ( z ) D k ( z ) =: M k ( z ) , (6)where D k is given by (3) and N k is a meromorphic function on C explicitly known.As previously, there is just to invert a Laplace transform to solve (VPL2). Asfor the linear case, a precise study of M k and its poles gives a solution µ of (6) andan asymptotic expansion of the form ∀ λ ∈ R , (cid:98) µ ( t, k ) = (cid:88) M k ( ω )= ∞(cid:61) ω ≥ λ Q ω,k ( t ) e − iωt + O ( e λt ) . (7)where Q ω,k is the polynomial such that M k ( z ) = z → ω L [ Q ω,k ( t ) e − iωt ]( z ) + O (1).The poles of M k are of two kinds: they can be zeros of D k (generating the samefrequencies as at the first order) or poles of N k . The study of the poles is technicalbecause N k is defined from the solution of (VPL). However, the asymptotic expan-sion of ψ (see (5)) enables a decomposition of N k in more elementary terms whosepoles can be determined.In order to give an intuition of these poles, we consider a term that is veryrepresentative of this decomposition: N ( rep ) k ( z ) = L (cid:104) F ( rep ) k ( t ) (cid:105) ( z )where F ( rep ) k ( t ) = e − i ( ω t + ω t ) (cid:90) (cid:90) ≤ τ ≤ s ≤ t e i ( ω τ + ω s ) F [ f eq ]( τ k + sk ) d s d τ (8)with k , k ∈ (cid:99) T d \ k + k = k , ω , ω ∈ C are such that D k ( ω ) = D k ( ω ) = 0 and F [ f eq ] is the Fourier transform of f eq . The later being definedfor u ∈ L ( R d ) and ξ ∈ R d by F [ u ]( ξ ) = (cid:90) R d u ( v ) e − iv · ξ d v. but slightly simplified. ECOND ORDER DISPERSION RELATIONS 5
Since N ( rep ) k is the Laplace transform of F ( rep ) k ( t ), it can be proven that its polesare given by the asymptotic expansion of F ( rep ) k ( t ) with the formula (4). As it issuggested by the formula (8), the behavior of this later is quite different if the setof the points ( τ, s ) such that τ k + sk = 0 is a line segment ( resonant case ) or apoint ( non-resonant case ).In the non-resonant case, there exists a constant c > ≤ τ ≤ s ≤ t, | τ k + sk | ≥ cs. So, assuming that f eq is regular enough so that F [ f eq ]( ξ ) decreases faster than anyexponential as | ξ | goes to + ∞ (for example like a Gaussian), we can prove thatthe integral in (8) converges faster than any exponential as t goes to + ∞ . As aconsequence, we get a constant a ∈ C such that ∀ λ ∈ R , F ( rep ) k ( t ) = ae − i ( ω t + ω t ) + O ( e λt ) . In the resonant case, there exists γ ∈ (0 ,
1) such that k = − γk . Realizing a natural change of coordinates in (8), we get F ( rep ) k ( t ) = (cid:90) t (cid:90) (1 − γ ) s − γs e − i ( ω ( t − τ − γs )+ ω ( t − s )) F [ f eq ]( τ k ) d τ d s. Thus, assuming that f eq is regular enough so that F [ f eq ]( ξ ) decrease faster thanany exponential as | ξ | goes to + ∞ , we have F ( rep ) k ( t ) = (cid:18)(cid:90) t e − i ( ω ( t − γs )+ ω ( t − s )) d s (cid:19) (cid:18)(cid:90) R e iω τ F [ f eq ]( τ k )d τ (cid:19) − e − it ( ω + ω ) (cid:90) ∞ (cid:90) τ ≥ (1 − γ ) s or τ< − γs e i ( ω ( τ + γs )+ ω s ) F [ f eq ]( τ k ) d τ d s + e − it ( ω + ω ) (cid:90) ∞ t (cid:90) τ ≥ (1 − γ ) s or τ< − γs e i ( ω ( τ + γs )+ ω s ) F [ f eq ]( τ k ) d τ d s, and we can prove that the third term decreases faster than any exponential. Thus,this decomposition provides the following asymptotic expansion ∀ λ ∈ R , F ( rep ) k ( t ) = ae − it ( ω + ω ) + be − itω b + O ( e λt ) , where a, b ∈ C and ω b = (1 − γ ) ω = ( | k | / | k | ) ω is the Best frequency (accordingto [13]).As suggested by this sketch of proof, we can prove that M k have three kinds ofpoles. More precisely, if ω is a pole of M k it satisfies one of the following conditions(I) ω is a zero of D k ,(II) ω = ω + ω where D k ( ω ) = D k ( ω ) = 0 and k + k = k ,(III) ω = ( | k | / | k | ) ω where D k ( ω ) = 0 and there exists γ ∈ (0 ,
1) such that k = γk .We recall that these poles drive the asymptotic behavior of (cid:98) µ ( k ) through formula(7). The frequencies (I) and (II) have already been identified in our preliminarywork on this subject [4], but not the frequency (III). We emphasize that all thethree type of frequencies are listed in [13], which makes our analysis coherent withthe physics litterature. JOACKIM BERNIER ∗ AND MICHEL MEHRENBERGER
To conclude this presentation, we are going to state a precise theorem giving theasymptotic behavior of the solutions of (VPL2). To this end, we need to introducesome notations.
Definition 1.1.
Let E ( R d ) be the subspace of the Schwartz space S ( R d ), of func-tions f , whose Fourier transform, F f , extends to an entire function on C d andsuch that ∃ α ∈ (0 , π , ∀ β ∈ (0 , α ) , ∀ λ ∈ R , sup x ∈ R d sup θ ∈ ( − β,β ) e λ | x | | F f ( e iθ x ) | < ∞ , (9)where | · | denotes the canonical Hermitian norm of C d . Remark 1.
Most of our results require that f eq ∈ E ( R d ) and v (cid:55)→ (cid:98) g ( k, v ) ∈ E ( R d ). This assumption is probably not optimal but it is crucial in our proof inorder to invert easily some Laplace transforms (see Theorem 3.1 and Lemma 3.2).Furthermore the space E ( R d ) contains most of the usual functions used in Vlasov-Poisson simulations. For example, the Maxwellian functions belong to this space.Appendix 6.1 provides many examples and details about this space. Remark 2.
Assuming that f eq ∈ E ( R d ) and v (cid:55)→ (cid:98) g ( k, v ) ∈ E ( R d ), D k and N k areentire functions and for all λ ∈ R , the number of zeros of D k with an imaginarypart larger than λ is finite (proof will be given in Corollary 3 and Proposition 7).Appendix 6.3 provides an algorithm to computate the zeros of D k . Definition 1.2. If k ∈ (cid:99) T d , n k,ω denotes the multiplicity of ω as zero of D k , i.e. n k,ω = max { m ∈ N | ∀ (cid:96) < m, D ( (cid:96) ) k ( ω ) = 0 } . Most of the result of this paper will require that g is supported on a finitenumber of spatial modes whose set is denoted K ⊂ (cid:99) T d \ { } . More precisely, theyrequire the following assumption Assumption 1.3.
There exists K , a finite part of (cid:99) T d \ { } such that ∀ x ∈ T d , g ( x, v ) = (cid:88) k ∈ K e ik · x (cid:98) g ( k, v ) , with v (cid:55)→ (cid:98) g ( k, v ) ∈ E ( R d ) . This assumption seems clearly not optimal but it is general enough to exhibit therelevant phenomena and it corresponds to the usual initial data used for numericalsimulations. Furthermore, it simplifies most of the proof avoiding several problemsof convergences.We can now state the main result of this paper: the following theorem proves theexistence of smooth solutions of (VPL) and (VPL2) and describes their asymptoticbehavior.
Theorem 1.4.
Let f eq ∈ E ( R d ) and g be a function satisfying Assumption 1.3.Then there exist two C ∞ functions ψ, µ : R ∗ + × T d → R and two continuous functions g, h : R + × T d × R d → R , C ∞ on R ∗ + × T d × R d , such that ( g, ψ, h, µ ) is solution of (VPL) and (VPL2) .Furthermore, if λ ∈ R , ψ is a linear combination of functions of the two types J ( t, x ) = t m e i ( k · x − ωt ) and R ( t, x ) = r ( t ) e i ( k · x − iλt ) where k ∈ K , D k ( ω ) = 0 , (cid:61) ω ≥ λ and ≤ m < n k,ω and r is a bounded analyticfunction on R ∗ + . ECOND ORDER DISPERSION RELATIONS 7
Similarly, µ is a linear combination of functions of the four types J ( t, x ) = t m e i ( k · x − ωt ) I ( t, x ) = t (cid:96) e i ( k · x − ( ω + ω ) t ) B ( t, x ) = t p e i ( k · x − | k || k | ω t ) d k k R ( t, x ) = r ( t ) e i ( k · x − iλt ) where k = k + k , r is a bounded analytic function on R ∗ + and k , k ∈ K satisfy D k ( ω ) = D k ( ω ) = D k ( ω ) = 0 k · k (cid:54) = 0 and (cid:16) d k k (cid:54) = 0 ⇐⇒ ∃ γ ∈ (0 , , k = γk (cid:17) (cid:61) ω ≥ λ and (cid:16) (cid:61) ω + (cid:61) ω ≥ λ or | k || k | (cid:61) ω ≥ λ (cid:17) m < n k,ω , (cid:96) < n k ,ω + n k ,ω − σ k ,k ω ,ω , p < n k ,ω + 1 + ν k ,k ω ,ω , with σ k ,k ω ,ω , ν k ,k ω ,ω some non negative integers equal to zero in the non degeneratecases (see Remark 5 for details). Remark 3.
We have e − iωt = e (cid:61) ( ω ) t − i (cid:60) ( ω ) t , so that all the terms of the sum exceptmaybe the remainder R are of the form e ik · x P ( t ) e ˜ λt + iβt , with P ( t ) ∈ C [ t ]. If one ofthis term satisfies ˜ λ < λ , it can be put in the remainder term R . Remark 4.
Taking λ decreasing to −∞ makes the sum larger, but it always remainfinite, for a fixed λ , since K, K + K are finite together with the zeros (see Remark2). We warn the reader that, a priori, the expansion does not converge as λ goes to −∞ . Remark 5.
It may exist some degenerate cases for which the four types of functionsintroduced in the second part of Theorem 1.4 are non distinct. In such a case, thenumbers σ k ,k ω ,ω and ν k ,k ω ,ω do not vanish and we have σ k ,k ω ,ω = ( n k,ω + ω − D k ( ω + ω )=0 + 2 · d k k (cid:54) =0 and ω + ω = | k || k | ω and ν k ,k ω ,ω = ( n k, | k || k | ω − D k ( | k || k | ω )=0 where P denotes the characteristic function of the property P . Remark 6.
In the case where d k k (cid:54) = 0, which we will call resonant case , where the Best frequency , that is the term B appears, p can a priori be ≥
1. For the J and I terms, the multiplicity can be equal to one, corresponding to m = (cid:96) = 0. Remark 7.
It is quite direct to extend the classical linear analysis of the VlasovPoisson with 1 specie to the case of multi-species charged particles (see § § Remark 8.
It may be interesting to try to extend this second order analysis tonon-homogeneous equilibria. However such an analysis seems much more involved.Indeed, in this case, to carry out an analysis of the linearized equation similar to theanalysis of the homogeneous case, it is usual to use action-angle variables (see forexample [3],[9]). However, this change of coordinates generates some singularitiesand boundary effects leading to an algebraic decay of the electric field.
JOACKIM BERNIER ∗ AND MICHEL MEHRENBERGER
The fifth section of this paper is devoted to some numerical experiments. Theyprincipally aim at highlighting the Best’s waves because most of the other phe-nomena associated with second order terms have been studied numerically in theproceedings [4]. Unlike the linear case, it seems that there is no elementary wayto determine a priori the coefficients associated with the asymptotic expansion of µ . Indeed, they depend non trivially on the solution of (VPL) (and not only on itsasymptotic expansion). Consequently, we use here least squares procedures, whichpermit to have a simple and quick way to find these coefficients.There are some difficulty arising of these computations because, as we comparethe solution of the second order expansion to the solution of (VP), this gives aconstraint on ε and the final time that should be small enough. As we have seen,the final time should also not be too small in order to be in the asymptotic regime,and this is also true for ε (which is here put to the square, as we consider secondorder expansion) due to the limits imposed by machine precision.We admit that for the numerical checking of codes, second order terms have notgained much popularity, maybe as the linear terms generally already give the mainphenomena. We emphasize that we are here able to identify the contributions ofthe different frequencies, and thus do an effective comparison with, as already told,multidimensional and nonlinear features. Some remarks about the notations . In order to keep proofs as readable as possible,we do some classical abuses of notation for integral transforms. For example, theFourier transform on R d is always associated with the variable v , it means that if u ∈ L ( R d ) then F [ u ] and F [ u ( v )] denotes the same functions. Similarly, if u is afunction of t, x, v then F [ u ]( t, x, ξ ) denotes F [ v (cid:55)→ u ( t, x, v )]( t, x, ξ ). Similarly, t isassociated with L , z with L − , ξ with F − , x with u (cid:55)→ (cid:98) u and k with ( u (cid:55)→ (cid:98) u ) − . Outline of the work . In Section 2, we derive some integral equations (called disper-sion relations) satisfied by solutions ψ, µ of (VPL) and (VPL2). Then we provethat it is enough to solve these dispersion relations to get solutions for (VPL) and(VPL2). The next two sections are devoted to the resolution of these dispersionrelations and to the asymptotic expansions of their solutions: Section 3 is for thefirst order expansion and Section 4 is for the second order expansion. Finally inSection 5, we give some numerical results.2.
Derivation of the dispersion relations.
Dispersion relations for first and second order.
In the following propo-sitions, we give the dispersion relations , that are obtained through Fourier andLaplace transforms. Note that we have an expression for both the electric poten-tials ψ resp. µ and the distribution function g resp. h of the first resp. second orderdispersion relations. Proposition 1.
Assume f eq ∈ E ( R d ) and g satisfies Assumption 1.3. Assumethere exists a C ∞ function on R ∗ + × T d , denoted ψ , and there exists λ > such that e − λ t ψ ( t, x ) is bounded on R ∗ + × T d . Furthermore, assume that, for all k ∈ (cid:99) T d \ { } , L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) is a solution of L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) D k ( z ) = − i | k | (cid:90) R d (cid:98) g ( k, v ) v · k − z d v. (10) ECOND ORDER DISPERSION RELATIONS 9 for (cid:61) z > λ .If we define g by g ( t, x, v ) = (cid:88) k ∈ K e ik · ( x − vt ) (cid:98) g ( k, v ) + i (cid:90) t e ik · ( x − v ( t − s )) (cid:98) ψ ( s, k ) k · ∇ v f eq ( v ) d s, (11) then g is a C ∞ function on R ∗ + × T d × R d , continuous on R + × T d × R d and ( g, ψ ) is a solution of (VPL) . Proposition 2.
Assume f eq ∈ E ( R d ) and g satisfies Assumption 1.3. Assumethere exists a solution of (VPL) as in Proposition (1) . Assume there exists a C ∞ function on R ∗ + × T d , denoted µ , and there exists λ > λ such that e − λ t ψ ( t ) isbounded on R ∗ + × T d . Furthermore, assume that, for all k ∈ (cid:99) T d \ { } , L [ (cid:98) µ ( t, k )] ( z ) is a solution of L [ (cid:98) µ ( t, k )] ( z ) D k ( z ) = − i | k | (cid:90) R d L (cid:104) (cid:92) ∇ x ψ · ∇ v g ( t, k, v ) (cid:105) ( z ) v · k − z d v. (12) for (cid:61) z > λ .If we define h by h ( t, x, v ) = (cid:88) k ∈ K + K i (cid:90) t e ik · ( x − v ( t − s )) (cid:98) µ ( s, k ) k · ∇ v f eq ( v ) d s + (cid:90) t (cid:92) ∇ x ψ · ∇ v g ( s, k, v ) e ik · ( x − v ( t − s )) d s, then h is a C ∞ function on R ∗ + × T d × R d , continuous on R + × T d × R d and ( h, µ ) is a solution of (VPL2) . A general linearized Vlasov-Poisson equation.
In order to prove Propo-sitions 1 and 2, as (VPL) and (VPL2) share the same structure, we focus on ageneral linearized Vlasov Poisson equation ∂ t g ( t, x, v ) + v · ∇ x g ( t, x, v ) − ∇ x u ( t, x ) · ∇ v f eq ( v ) = S ( t, x, v ) , ∆ x u ( t, x ) + (cid:82) g ( t, x, v )d v = 0 , g (0 , x, v ) = g ( x, v ) . (VPLG)In the following proposition, we derive a general dispersion relation satisfied by u .We first do not consider the coupling with the Poisson equation. Proposition 3.
Assume g ∈ C ( T d × R d ) , f eq ∈ C ( R d ) and S ( t, x, v ) ∈ C ( R ∗ + × T d × R d ) and there exist λ > , d ∈ C ( R d ) ∩ L ( R d ) satisfying ∀ ( t, k, v ) ∈ R ∗ + × (cid:99) T d × R d , e − λ t | (cid:98) S ( t, k, v ) | + | (cid:98) g ( k, v ) | + |∇ v f eq ( v ) | ≤ d ( v ) . Assume there exists u ∈ C ( R ∗ + × T d ) such that e − λ t u ( t ) is bounded on R ∗ + × T d .Assume there exists a continuous function g ∈ C ( R ∗ + × T d × R d ) , continuous on R + × T d × R d such that g is solution of the Vlasov equation, i.e. for all ( t, x, v ) ∈ R ∗ + × T d × R d (cid:26) ∂ t g ( t, x, v ) + v · ∇ x g ( t, x, v ) − ∇ x u ( t, x ) · ∇ v f eq ( v ) = S ( t, x, v ) , g (0 , x, v ) = g ( x, v ) . (13) If λ > λ then for all k ∈ (cid:99) T d , there exists C > , ∀ v ∈ R d , sup t ∈ R + | e − λt (cid:98) g ( t, k, v ) | ≤ C d ( v ) . (14) ∗ AND MICHEL MEHRENBERGER
Furthermore, for all z ∈ C with (cid:61) ( z ) > λ we have L (cid:20)(cid:90) R d (cid:98) g ( t, k, v ) d v (cid:21) ( z ) = − i (cid:90) R d (cid:98) g ( k, v ) v · k − z d v + L [ (cid:98) u ( t, k )]( z ) (cid:90) R d k · ∇ v f eq ( v ) v · k − z d v − i (cid:90) R d L (cid:104) (cid:98) S ( t, k, v ) (cid:105) ( z ) v · k − z d v. (15) Proof.
First, applying a space Fourier transform to (13), we get for all ( t, k, v ) ∈ R ∗ + × (cid:99) T d × R d ∂ t (cid:98) g ( t, k, v ) + iv · k (cid:98) g ( t, k, v ) − i (cid:98) u ( t, k ) k · ∇ v f eq ( v ) = (cid:98) S ( t, k, v ) . (16)Consequently, applying Duhamel formula, we get for all ( t, k, v ) ∈ R + × (cid:99) T d × R d (cid:98) g ( t, k, v ) = e − ik · vt (cid:98) g ( k, v ) + i (cid:90) t e − ik · v ( t − s ) (cid:98) u ( s, k ) k · ∇ f eq ( v )d s + (cid:90) t e − ik · v ( t − s ) (cid:98) S ( s, k, v )d s. (17)So, we deduce, there exist M, C > λ > λ then | (cid:98) g ( t, k, v ) | ≤ | (cid:98) g ( k, v ) | + (cid:90) t e λ s M | k ||∇ v f eq ( v ) | d s + (cid:90) t e λ s e − λ s | (cid:98) S ( s, k, v ) | d s, ≤ d ( v ) + te λ t (1 + | k | M ) d ( v ) , ≤ Ce λt d ( v ) . We deduce of this last estimation, that for any fixed v ∈ R d and for any λ > λ , e − λt (cid:98) g ( t, k, v ) is continuous and bounded on R + . Consequently, we can apply aLaplace transform on (16) and get for all z ∈ C such that (cid:61) z > λ and v ∈ R d , − iz L [ (cid:98) g ( t, k, v )]( z ) − (cid:98) g ( k, v ) + iv · k L [ (cid:98) g ( t, k, v )]( z ) − i L [ (cid:98) u ( t, k )]( z ) k · ∇ v f eq ( v )= L (cid:104) (cid:98) S ( t, k, v ) (cid:105) ( z ) . Since (cid:61) z >
0, this relation can be divided by i ( v · k − z ) to get for all v ∈ R d L [ (cid:98) g ( t, k, v )]( z ) = − i (cid:98) g ( k, v ) v · k − z + L [ (cid:98) u ( t, k )]( z ) k · ∇ v f eq ( v ) v · k − z − i L (cid:104) (cid:98) S ( t, k, v ) (cid:105) ( z ) v · k − z . Finally we conclude this proof integrating with respect to v and applying FubiniTheorem (with the control (14)) to get for all z ∈ C with (cid:61) z > λ L (cid:20)(cid:90) R d (cid:98) g ( t, k, v )d v (cid:21) ( z ) = (cid:90) R d L [ (cid:98) g ( t, k, v )] ( z )d v. If we want to get a closed equation on u , we have to use Poisson equation∆ x u ( t, x ) = − (cid:90) R d g ( t, x, v )d v. (18)Formally, applying a space Fourier transform and a Laplace transform we would get | k | L [ (cid:98) u ( t, k )] = L (cid:20)(cid:90) R d (cid:98) g ( t, k, v )d v (cid:21) . ECOND ORDER DISPERSION RELATIONS 11
Consequently, applying (15), we should get the following dispersion relation L [ (cid:98) u ( t, k )] ( z ) D k ( z ) = − i | k | (cid:90) R d (cid:98) g ( k, v ) v · k − z d v − i | k | (cid:90) R d L (cid:104) (cid:98) S ( t, k, v ) (cid:105) ( z ) v · k − z d v, (19)where D k is defined by (3). Proposition 4.
Assume g ∈ C ( T d × R d ) , f eq ∈ C ( R d ) and S ( t, x, v ) ∈ C ( R ∗ + × T d × R d ) and there exist λ > , d ∈ C ( R d ) ∩ L ( R d ) satisfying ∀ ( t, k, v ) ∈ R ∗ + × (cid:99) T d × R d , e − λ t | (cid:98) S ( t, k, v ) | + | (cid:98) g ( k, v ) | + |∇ v f eq ( v ) | ≤ d ( v ) . Assume there exists u ∈ C ( R ∗ + × T d ) such that e − λ t u ( t ) is bounded on R ∗ + × T d .Furthermore, assume that, for all k ∈ (cid:99) T d \ { } , L [ (cid:98) u ( t, k )] ( z ) is a solution of (19) for (cid:61) z > λ . Assume there exists a finite set K ⊂ (cid:99) T d such that ∀ t ∈ R , ∀ v ∈ R d , k ∈ (cid:99) T d \ K ⇒ (cid:98) g ( k, v ) = (cid:98) S ( t, k, v ) = (cid:98) u ( t, k ) = 0 . If we define g by g ( t, x, v ) = (cid:88) k ∈ K e ik · ( x − vt ) (cid:98) g ( k, v ) + i (cid:90) t e ik · ( x − v ( t − s )) (cid:98) u ( s, k ) k · ∇ f eq ( v ) d s + (cid:90) t e ik · ( x − v ( t − s )) (cid:98) S ( s, k, v ) d s, (20) then g ∈ C ( R ∗ + × T d × R d ) is continuous on R + × T d × R d and ( g , u ) is a solutionof (VPLG) .Proof. By construction of g through Duhamel formula (20), g is obviously a con-tinuous function on R + × T d × R d and C on R ∗ + × T d × R d . Furthermore, we mayverify by a straightforward calculation that g is solution of the Vlasov equation(13). Consequently, we just have to prove that g , u is solution of Poisson equation(18).However u and g satisfy assumptions of Proposition 3, so we can apply it. Conse-quently, we know that if λ > λ then e − λt (cid:82) (cid:98) g ( t, k, v )d v is continuous and boundedand that its Laplace transform satisfies (15). But since L [ (cid:98) u ( t, k )] is a solution of thedispersion relation (10), we deduce that for all z ∈ C such that (cid:61) z > λ we have | k | L [ (cid:98) u ( t, k )] ( z ) = L (cid:20)(cid:90) R d (cid:98) g ( t, k, v )d v (cid:21) ( z ) . But it is well known that Laplace transform is injective on continuous functionswith an exponential order (i.e. bounded by an exponential function), see Theorem1 . . t > | k | (cid:98) u ( t, k ) = (cid:90) R d (cid:98) g ( t, k, v )d v. Since space Fourier transform is also injective on regular functions, we have proventhat u , g is a solution of Poisson equation (18). ∗ AND MICHEL MEHRENBERGER
Proof of Propositions 1 and 2.
We now apply Proposition 4 for the proofof Propositions 1 and 2.
Proof of Proposition 1.
First, observe that if k ∈ (cid:99) T d \ ( { K } ∪ { } ) then for any t >
0, we have (cid:98) ψ ( t, k ) = 0. Indeed, since L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) is a solution of (10), wehave D k ( z ) L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) = 0 . But, we have proven in Lemma 6 that D k ( z ) (cid:54) = 0 if (cid:61) z is large enough. Con-sequently, L (cid:104) (cid:98) ψ ( t, k ) (cid:105) ( z ) = 0 if (cid:61) z is large enough. So we deduce by a classicalcriterion about Laplace transform (see Theorem 1 . . (cid:98) ψ ( t, k ) = 0.We observe on (11) that g is clearly a C ∞ function on R ∗ + × T d × R d . Finally, wejust need to apply Proposition 4 to prove that ( g, ψ ) is a solution of (VPL). Proof of Proposition 2.
Let S be defined by S ( t, x, v ) = ∇ x ψ ( t, x ) · ∇ v g ( t, x, v ) . By construction, it is a C ∞ function on R ∗ + × T d × R d . Since, space Fourier transformof ψ is supported by K (see proof of Proposition 1), its space Fourier transformis supported by K + K . Furthermore, since λ > λ , we can construct, by astraightforward estimation, a continuous function d ∈ C ( R d ) ∩ L ( R d ) such that ∀ v ∈ R d , e − λ t | (cid:98) S ( t, k, v ) | ≤ d ( v ) . In particular, this estimation proves that the right member of (12) is well definedif (cid:61) z ≥ λ .Now, as in Proposition 1, we can first prove that space Fourier transform of µ issupported by ( K + K ) ∪ { } , then observe that h is a C ∞ function and concludethat ( h, µ ) is a solution of (VPL) by Proposition 4.3. Resolution and expansion of the linearized equation.
Introduction and statement of the result.
In Proposition 1, we haveproved that it is enough to solve dispersion relation (10) to get a solution ( g, ψ ) tolinearized Vlasov-Poisson equation (VPL). So the aim of this section is to solve thisdispersion relation introducing most of the theoretical tools useful in the resolutionof the second order relation (12). In particular, many of them deal with analyticfunction defined on open sectors , denoted Σ α , with α ∈ (0 , π ), and defined byΣ α = { re iβ | − α < β < α and r > } . The result we are going to establish in this section is the following.
Proposition 5.
Assume f eq ∈ E ( R d ) and g satisfies Assumption 1.3. For all λ ∈ R , for all k ∈ K , for all zero point ω of D k there exists a polynomial, denoted P k,ω , whose degree is strictly smaller than the multiplicity of ω , α ∈ (0 , π ) andthere exists r k,λ an analytic and bounded function on Σ α such that the followingexpansion defines a solution of the dispersion relation (10) ∀ t ∈ R ∗ + , ∀ x ∈ T d , ψ ( t, x ) = (cid:88) k ∈ K (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ P k,ω ( t ) e i ( k · x − ωt ) + e ik · x e λt r k,λ ( t ) . ECOND ORDER DISPERSION RELATIONS 13
This proposition will be proven at the end of this section. First, we introducesome notations and many useful theoretical tools.3.2.
Definition of N k and theoretical tools. The right member of the dispersionrelation (10) is very important in our study. We denote it N k ( z ). More precisely,it is an analytic function defined, when (cid:61) z > N k ( z ) = − i | k | (cid:90) (cid:98) g ( k, v ) v · k − z d v. In the first part of this proof we study the regularity and the behavior of D k and N k . However, we need to introduce some classical results on Laplace transform.First, consider the following Theorem that is very useful to invert Laplace trans-forms and to control it. Theorem 3.1. (Analytic representation)Let α ∈ (0 , π ) , λ ∈ R and q : i ( λ , ∞ ) → C . The following assertions areequivalent:(i) There exists a holomorphic function f : Σ α → C such that ∀ < β < α, sup z ∈ Σ β | e − λ z f ( z ) | < ∞ and ∀ λ > λ , q ( iλ ) = L [ f ]( iλ ) . (ii) The function q has a holomorphic extension (cid:101) q : iλ + i Σ α + π → C such that ∀ < γ < α, sup z ∈ iλ + i Σ γ + π | ( z − iλ ) (cid:101) q ( z ) | < ∞ . Proof.
See Theorem 2 . . Remark 9.
Note that if e − λ t f ( t ) is bounded on R ∗ + then for λ > λ , e − λt f ( t ) ∈ L ( R + ) and so L [ f ] is well defined for (cid:61) ( z ) > λ , which is the set iλ + i Σ π .There is a direct corollary of the proof of Theorem 3.1 that is useful in our study. Corollary 1.
Assume that conclusion of Theorem 3.1 holds. Then for all < γ <β < α , we have sup z ∈ iλ + i Σ γ + π | ( z − iλ ) (cid:101) q ( z ) | ≤ β − γ ) sup z ∈ Σ β | e − λ z f ( z ) | . Then, we observe that D k and N k are defined through a integral operator whosekernel is v · k − z . The following lemma links this operator to more classical ones. Lemma 3.2.
Let f ∈ L ( R d ) and k ∈ R d \ { } then for all z ∈ C with (cid:61) z > (cid:90) R d f ( v ) k · v − z d v = i L [ F [ f ]( kt )] ( z ) . Proof.
First, remark that since f ∈ L ( R d ), t → F [ f ]( kt ) = (cid:82) R d f ( v ) e − itv · k d v is acontinuous and bounded function, so its Laplace transform is well defined if (cid:61) z > F ( t ) = (cid:90) R d f ( v ) k · v − z e − i ( k · v − z ) t d v. Since f ∈ L ( R d ), it is a regular function and we have F (cid:48) ( t ) = − i (cid:90) R d f ( v ) e − i ( k · v − z ) t d v = − i F [ f ]( kt ) e izt . ∗ AND MICHEL MEHRENBERGER
But, since (cid:61) z > F ( t ) goes to 0 when t goes to + ∞ . Consequently,we get F (0) = − (cid:90) ∞ F (cid:48) ( t )d t = i (cid:90) ∞ F [ f ]( kt ) e izt d t = i L [ F [ f ]( kt )] ( z ) . Estimations for D k and N k . With Lemma 3.2, we can write D k and N k asLaplace transforms. So, in the following proposition, we can prove their analyticityusing the analytic representation theorem (Theorem 3.1). In particular, we proveand extend Remark 2. Proposition 6. If f eq ∈ E ( R d ) then • for all k ∈ R d \ { } , D k is an entire function, • there exists α ∈ (0 , π ) such that for all < γ < α and for all λ ∈ R thereexists C > satisfying ∀ k ∈ R d \ { } , ∀ z ∈ i | k | λ + i Σ γ + π , | D k ( z ) − | ≤ C | k || z − i | k | λ | . Proof.
Since f ∈ S ( R d ), we have k · ∇ v f eq ∈ L ( R d ). Consequently, D k is welldefined as an analytic function on { z ∈ C | (cid:61) z > } . Furthermore, we can applyLemma 3.2 to get for (cid:61) z > D k ( z ) = 1 − i | k | L [ F [ k · ∇ v f eq ] ( kt )] ( z ) . Then we define e k = k | k | and we get by the change of variable t (cid:48) = | k | t L [ F [ k · ∇ v f eq ] ( kt )] ( z )= (cid:90) ∞ F [ | k | e k · ∇ v f eq ] ( | k | e k t ) e i z | k | | k | t d t = (cid:90) ∞ F [ e k · ∇ v f eq ] ( e k t (cid:48) ) e i z | k | t (cid:48) d t (cid:48) , so that D k ( z ) = 1 − i | k | L [ F [ e k · ∇ v f eq ] ( e k t )] (cid:18) z | k | (cid:19) . (21)Now, using Theorem 3.1, we are going to prove this Laplace transform defines anentire function and we are going to control it.Since f eq ∈ E ( R d ), it extends to an analytic function and there exists α ∈ (0 , π )such that for all β ∈ (0 , α ) and for all λ ∈ R , there exist C > ∀ z ∈ Σ β R d , | F [ f eq ]( z ) | ≤ C | e − λz | . Consequently, we get ∀ z ∈ Σ β , | F [ e k · ∇ v f eq ] ( e k z ) | = | iz F [ f eq ]( z ) | ≤ C | z | e − λ (cid:60) z . Finally, we have proven that for all λ ∈ R , there exists a constant M > e k ) such that ∀ z ∈ Σ β , | F [ e k · ∇ v f eq ] ( e k z ) | ≤ M | e − λ z | . Applying Theorem 3.1 and its corollary, we have proven that L [ F [ e k · ∇ v f eq ] ( e k t )] ECOND ORDER DISPERSION RELATIONS 15 is an entire function and that for all γ ∈ (0 , α ) and all λ ∈ R , there exists M > β = α + γ ) such that ∀ z ∈ iλ + i Σ γ + π , |L [ F [ e k · ∇ v f eq ] ( e k t )] ( z ) | ≤ M sin( α − γ ) 1 | z − iλ | . Finally, we deduce directly the result from formula (21): | D k ( z ) − | = (cid:12)(cid:12)(cid:12)(cid:12) i | k | L [ F [ e k · ∇ v f eq ] ( e k t )] (cid:18) z | k | (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M | k | sin( α − γ ) 1 | z | k | − iλ | = M | k | sin( α − γ ) 1 | z − i | k | λ | . Corollary 2. If f eq ∈ E ( R d ) then there exists α ∈ (0 , π ) such that for all λ ∈ R and γ ∈ (0 , α ) , there exists c > such that for all k ∈ R d \ { } we have { z ∈ C | D k ( z ) = 0 } ⊂ i | k | λ − (cid:18) D (0 , c | k | ) ∪ i Σ π − γ (cid:19) . Proof.
Indeed, we have either z ∈ i | k | λ + i Σ γ + π , so that 1 = | D k ( z ) − | ≤ C | k || z − i | k | λ | and thus z ∈ i | k | λ − D (0 , C | k | ). Otherwise, z ∈ C \ (cid:8) i | k | λ + i Σ γ + π (cid:9) ,that is z = i | k | λ + ire iδ , δ ∈ [ π + γ, π − π − γ ], and thus π − δ ∈ [ − π + γ, − γ + π ],which leads to z = i | k | λ − ire − i ˜ δ , with ˜ δ = π − δ ∈ [ − π + γ, − γ + π ]. Corollary 3. If f eq ∈ E ( R d ) then for all k ∈ (cid:99) T d \ { } , D k is an entire functionand for all λ ∈ R , { ω ∈ C | D k ( ω ) = 0 and (cid:61) ω ≥ λ } is a finite set.Proof. In Proposition 6, we have proved that D k is an entire function and it canbe directly deduced from its Corollary 2 that { z ∈ C | D k ( z ) = 0 and (cid:61) ω ≥ λ } isbounded. Consequently, since zero points of D k are isolated, it is a finite set.It is very natural to adapt this result to N k . More precisely, we deduce thefollowing proposition. Proposition 7.
For all k ∈ (cid:99) T d \ { } , N k is an entire function and there exists α ∈ (0 , π ) such that for all λ ∈ R and for all β ∈ (0 , α ) , we have sup z ∈ iλ + i Σ β + π | N k ( z ) || z − iλ | < ∞ . Proof.
Since v (cid:55)→ (cid:98) g ( k, v ) ∈ S ( R d ), N k defines naturally an analytic function for (cid:61) z >
0. Furthermore, applying Lemma 3.2 we know that for (cid:61) z > N k ( z ) = 1 | k | L [ F [ (cid:98) g ( k, v )] ( kt )] ( z ) . Since v (cid:55)→ (cid:98) g ( k, v ) ∈ E ( R d ), t (cid:55)→ F [ (cid:98) g ( k, v )] ( kt ) is an entire function and there ex-ists α ∈ (0 , π ) such that for all β ∈ (0 , α ) and all λ ∈ R , t (cid:55)→ | e λ t F [ (cid:98) g ( k, v )] ( kt ) | is bounded on Σ β . Consequently, we deduce of Theorem 3.1, that for all λ ∈ R , N k has a holomorphic extension on iλ + i Σ α + π such that ∀ β ∈ (0 , α ) , sup z ∈ iλ + i Σ γ + π | ( z − iλ ) N k ( z ) | < ∞ . Finally, since N k is analytic on iλ + i Σ α + π for all λ ∈ R , it is an entire function. ∗ AND MICHEL MEHRENBERGER
A theoretical tool for the control of L − [ N k /D k ] . Now, we introduce ageneral criterion to invert Laplace transform and get an asymptotic expansion.
Lemma 3.3.
Let R ∈ H ( C ) be an entire function and N be a meromorphic functiondefined on C . If there exists α ∈ (0 , π ) such that ∃ C > , sup z ∈ i Σ α + π | zR ( z ) | + | zN ( z ) | < C, then for any λ ∈ R , there exist β ∈ (0 , α ) and a function r ∈ H (Σ β ) analytic andbounded on Σ β such that if (cid:61) z is large enough then N ( z )1 − R ( z ) = L (cid:88) ω ∈ Z (cid:61) ω ≥ λ P ω ( t ) e − iωt + e λt r ( t ) ( z ) , where Z is the set of poles of N ( z )1 − R ( z ) , P ω = n ω − (cid:88) k =0 a k +1 ,ω ( − i ) k +1 k ! X k is the polynomial whose coefficients are defined by the expansion of N − R in z = ωN ( z )1 − R ( z ) = z → ω n ω (cid:88) j =1 a j,ω ( z − ω ) j + O (1) . Remark 10.
In the application, for the proof of Proposition 5, N will be entire(thus meromorphic), but for the second order case, in the next section, we will reallyneed that N is meromorphic. Proof of Lemma 3.3.
Many geometrical objects are going to be introduced in thisproof. The reader can refer to Figure 1 to an illustration of these constructions.First observe that to prove the lemma, we can assume that λ is negative enough.In particular we assume that λ < − C .By construction, if | z | ≥ C and z ∈ i Σ α + π then | − R ( z ) | ≥ −| R ( z ) | > − C | z | ≥ .Consequently, all zero points of (1 − R ) belong to D (0 , C ) ∪ − i Σ π − α (note that − i Σ π − α \{ } = (cid:0) i Σ α + π (cid:1) c ). Since the poles of N lie on − i Σ π − α , the poles of N ( z )1 − R ( z ) lie on D (0 , C ) ∪ − i Σ π − α . In particular, the set of its poles with an imaginary partlarger than or equal to λ is finite (see Figure 1).Now consider the following rational fraction Q ( z ) = (cid:88) ω ∈ Z (cid:61) ω ≥ λ n ω (cid:88) j =1 a j,ω ( z − ω ) j . We introduce r > D (0 , C ) ∪ (cid:0) { z ∈ C | (cid:61) z ≥ λ } ∩ − i Σ π − α (cid:1) ⊂ D ( iλ, r ) . Now, we observe that there exists β ∈ (0 , α ) such that N − R − Q is a continuousfunction on iλ + i Σ β + π . Indeed, it is a meromorphic function whose poles lie on { z ∈ C | (cid:61) z < λ } ∩ − i Σ π − α and are isolated, and thus we can choose such β (smallenough). Consequently, there exists M > ∀ z ∈ D ( iλ, r ) ∩ (cid:0) iλ + i Σ β + π (cid:1) , (cid:12)(cid:12)(cid:12)(cid:12) N ( z )1 − R ( z ) − Q ( z ) (cid:12)(cid:12)(cid:12)(cid:12) < M ≤ M r | z − iλ | . ECOND ORDER DISPERSION RELATIONS 17
Furthermore since zN ( z ) is bounded on i Σ α + π , there exists M > ∀ z ∈ iλ + i Σ β + π , | N ( z ) | ≤ M | z − iλ | . Indeed, we distinguish the case z ∈ i Σ α + π ∩ D c (0 , | λ | ), for which there exists C > | N ( z ) | ≤ C | z | ≤ C | z − iλ | | z − iλ || z | ≤ C | z − iλ | | λ | | λ | ≤ M | z − iλ | , and the case z ∈ (cid:0) iλ + i Σ β + π (cid:1) ∩ (cid:0) ( i Σ α + π ) c ∪ D (0 , | λ | ) (cid:1) which is a bounded setensuring | z − iλ || N ( z ) | ≤ M . Consequently, by construction of r , we get | − R ( z ) | ≥ , when | z | ≥ C and z ∈ i Σ α + π [so, in particular when z ∈ i Σ α + π ∩ D c ( iλ, r ) ∩ (cid:0) iλ + i Σ β + π (cid:1) ] and | − R ( z ) | ≥ c , with c >
0, when z ∈ (cid:0) i Σ α + π (cid:1) c ∩ (cid:0) iλ + i Σ β + π (cid:1) (which is a boundedset) [so, in particular when z ∈ (cid:0) i Σ α + π (cid:1) c ∩ D c ( iλ, r ) ∩ (cid:0) iλ + i Σ β + π (cid:1) ] and thus ∀ z ∈ (cid:0) iλ + i Σ β + π (cid:1) ∩ D c ( iλ, r ) , (cid:12)(cid:12)(cid:12)(cid:12) N ( z )1 − R ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ max(2 , /c ) M | z − iλ | . Finally, since Q is a rational fraction whose poles lie on D ( iλ, r ) and vanishing as z goes to ∞ , the function z → ( z − iλ ) Q ( z ) is bounded on D c ( iλ, r ) and thus thereexists M > ∀ z ∈ (cid:0) iλ + i Σ β + π (cid:1) ∩ D c ( iλ, r ) , | Q ( z ) | ≤ M | z − iλ | . Then we get a constant M > ∀ z ∈ iλ + i Σ β + π , (cid:12)(cid:12)(cid:12)(cid:12) N ( z )1 − R ( z ) − Q ( z ) (cid:12)(cid:12)(cid:12)(cid:12) < M | z − iλ | . Applying Theorem 3.1 to N − R − Q , we get an analytic and bounded function t → e − λt e λt r ( t ) = r ( t ) on Σ γ (with γ = β ), such that L (cid:2) e λt r ( t ) (cid:3) ( z ) = N ( z )1 − R ( z ) − Q ( z ) . To conclude the proof of Lemma 3.3, we just need to determine the invert Laplacetransform of Q . But we get, by straightforward calculation, Q ( z ) = L (cid:88) ω ∈ Z (cid:61) ω ≥ λ P ω ( t ) e − iωt ( z ) . Proof of Proposition 5.
Finally we can prove the result stated at the be-ginning of this section.
Proof of Proposition 5.
In Proposition 6 and 7 we have proven that we can applyLemma 3.3 with D k = 1 − R and N = N k , taking λ = 0. But the result of thislemma is exactly the expansion of Proposition 5. ∗ AND MICHEL MEHRENBERGER
Figure 1.
An illustration of the geometrical constructions intro-duced in the proof of Lemma 3.3.
Remark 11.
As we use only λ = 0 in Proposition 6 and 7 for the proof ofProposition 5, we may wonder of we could use a weaker assumption on f eq for gettingthe estimate on D k for example. Indeed, that estimate derives from Theorem 3.1for λ = 0 and so the weaker assumption could be the hypothesis of (i) in Theorem3.1 for λ = 0. However, we also need to have that D k is entire, and there we haveused Theorem 3.1 for all λ ∈ R .4. Resolution and expansion of the second order equation.
Introduction and statement of the result.
In Proposition 2, we haveproven that it is enough to solve dispersion relation (12) to get a solution ( h, µ ) tosecond order linearized Vlasov-Poisson equation (VPL2). So this section is devotedto the resolution of this second order dispersion relation, following the strategyestablished for the first order dispersion relation in the previous section, by provingthe following proposition, which permits to complete the proof of our main result,Theorem 1.4.
Proposition 8.
Assume f eq ∈ E ( R d ) and g satisfies Assumption 1.3. Considerthe solution ( g, ψ ) of (VPL) given by Proposition 5 and Proposition 1. Then there ECOND ORDER DISPERSION RELATIONS 19 exists a solution µ of the dispersion relation (12) whose expansion is detailed inTheorem 1.4. Definition of N k and N k (the right hand side). We will first look forthe right hand side of the second order dispersion relation, that was N k in the firstorder case.Let k ∈ ( K + K ) \ { } . By looking at the second order dispersion relation (12),we can assume, without loss of generality, that there exist k , k ∈ K such that k + k = k and (cid:92) ∇ x ψ · ∇ v g ( t, k, v ) = i (cid:98) ψ ( t, k ) k · ∇ v (cid:98) g ( t, k , v ) . Consequently, we can determine more precisely the right member of (12). However,to be rigorous we need to prove that our integrals are convergent. Indeed, as inProposition 2, there exists λ ∈ R such that for any λ > λ , we can construct acontinuous and integrable function d ∈ C ( R d ) ∩ L ( R d ) such that ∀ ( t, v ) ∈ R ∗ + × R d , | (cid:98) ψ ( t, k ) | e − λt + e − λt | k · ∇ v (cid:98) g ( t, k , v ) | ≤ d ( v ) . Consequently, if (cid:61) z > λ , we can apply Lemma 3.2 to prove that (cid:90) R d L (cid:104) i (cid:98) ψ ( t, k ) k · ∇ v (cid:98) g ( t, k , v ) (cid:105) ( z ) v · k − z d v = i L (cid:104) F (cid:104) L (cid:104) i (cid:98) ψ ( t, k ) k · ∇ v (cid:98) g ( t, k , v ) (cid:105) ( z ) (cid:105) ( kt ) (cid:105) ( z )= i (cid:90) ∞ (cid:90) R d (cid:90) ∞ i (cid:98) ψ ( t, k ) k · ∇ v (cid:98) g ( t, k , v ) e izt d t e − i ( kτ ) · v d v e izτ d τ = − i (cid:90) ∞ (cid:90) ∞ (cid:98) ψ ( t, k )( k · k ) τ F (cid:98) g ( t, k , kτ ) e iz ( τ + t ) d t d τ = − i ( k · k ) (cid:90) ∞ (cid:90) s (cid:98) ψ ( t, k )( s − t ) F (cid:98) g ( t, k , k ( s − t ))d t e izs d s = − i ( k · k ) L (cid:20)(cid:90) t (cid:98) ψ ( τ, k )( t − τ ) F (cid:98) g ( τ, k , k ( t − τ ))d τ (cid:21) ( z ) , where we have used the change of variable τ + t ← s and the notation F (cid:98) g ( t, k, ξ ) = F [ (cid:98) g ( t, k, v )]( ξ ) . Furthermore, using definition of g (see (11)), we can precise F (cid:98) g ( τ, k , k ( t − τ )).Indeed, we start fromˆ g ( t, k, v ) = e − ik · vt (cid:98) g ( k, v ) + i (cid:90) t e − ik · v ( t − s ) (cid:98) ψ ( s, k ) k · ∇ v f eq ( v )d s, so that we have F (cid:98) g ( t, k , ξ ) = F [ (cid:98) g ( k , v )]( ξ + k t ) + i (cid:90) t (cid:98) ψ ( s, k ) F [ k . ∇ v f eq ] ( ξ + k ( t − s ))d s. Consequently, we get F (cid:98) g ( τ, k , k ( t − τ )) = F [ (cid:98) g ( k , v )]( kt + ( k − k ) τ ) − i (cid:90) τ (cid:98) ψ ( s, k ) F [ k · ∇ v f eq ] ( k ( t − τ ) + k ( τ − s ))d s. ∗ AND MICHEL MEHRENBERGER
So we have two numerators to study for this dispersion relation. On the onehand, we have N k ( z ) := − k · k | k | L (cid:2) F k ( t ) (cid:3) ( z ) , (22)with F k ( t ) := (cid:90) t (cid:98) ψ ( τ, k )( t − τ ) F [ (cid:98) g ( k , v )]( kt + ( k − k ) τ )d τ . On the other hand, we have N k ( z ) := i k · k | k | L (cid:2) F k ( t ) (cid:3) ( z ) , (23)with F k ( t ) = (cid:90) t (cid:98) ψ ( τ, k )( t − τ ) (cid:90) τ (cid:98) ψ ( s, k ) F [ k · ∇ v f eq ] ( k ( t − τ ) + k ( τ − s ))d s d τ . So with these notations,the dispersion relation (8) may be written as, for all z suchthat (cid:61) z > λ , L [ (cid:98) µ ( t, k )] ( z ) D k ( z ) = N k ( z ) + N k ( z ) . (24)4.3. Estimates for N k and N k . We are going to apply the same strategy as forthe resolution of the first order dispersion relation. It will be solve using Lemma3.3. The denominator has been studied in Proposition 6. The following lemmadescribes the regularity and the behavior of the numerators N k and N k . Lemma 4.1.
The function N k , N k have a meromorphic continuation and thereexist (cid:101) λ ∈ R and β ∈ (0 , π ) such that sup z ∈ i (cid:101) λ + i Σ β + π | z − i (cid:101) λ | (cid:2) |N k ( z ) | + |N k ( z ) | (cid:3) < ∞ . • If there exists γ ∈ (0 , such k = − γk then the poles of N k are the points ω ∈ C such that ω = ω | k || k | where D k ( ω ) = 0 and its multiplicity is smaller thanor equal to n k ,ω + 1 . N k has two kinds of poles. On the one hand, there arethe points ω ∈ C such that ω = ω + ω where D k ( ω ) = D k ( ω ) = 0 . On theother hand, there are the points ω ∈ C such that ω = ω | k || k | where D k ( ω ) = 0 .The multiplicity of a pole belonging to the two families is smaller than or equal to n k ,ω + n k ,ω + 1 . Else the multiplicity of a pole of the first kind is smaller thanor equal to n k ,ω + n k ,ω − and the multiplicity of a pole of the second kind issmaller than or equal to n k ,ω + 1 . • Else N k is an entire function and the poles of N k are the points ω ∈ C such that ω = ω + ω where D k ( ω ) = D k ( ω ) = 0 and its multiplicity is smaller than orequal to n k ,ω + n k ,ω − . Now, we focus on proving Lemma 4.1. However, using analytic representationTheorem 3.1, it is directly deduced of the two following lemmas (Lemma 4.2 andLemma 4.3) involving properties of F k and F k . Lemma 4.2.
For all λ ∈ R there exist β ∈ (0 , π ) and an analytic and boundedfunction on Σ β denoted r such that ECOND ORDER DISPERSION RELATIONS 21 • if k = − γk , γ ∈ (0 , , then for all t > F k ( t ) = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ R ω ( t ) e − iω | k || k | t + e λt r ( t ) , (25) with R ω a polynomial of degree smaller than or equal to n k ,ω , • else, for all t > F k ( t ) = e λt r ( t ) . (26) Lemma 4.3.
For all λ ∈ R there exist β ∈ (0 , π ) and an analytic and boundedfunction on Σ β denoted r such that • if k = − γk , γ ∈ (0 , , then for all t > F k ( t ) = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ R ω k ,k ( t ) e − iω | k || k | t + (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ Q k ,k ω ,ω ( t ) e − i ( ω + ω ) t + e λt r ( t ) , (27) with Q k ,k ω ,ω a polynomial of degree smaller than or equal to n k ,ω + n k ,ω − and R ω k ,k a polynomial of degree smaller than or equal to n k ,ω (if there exist ω , ω such that ω + ω = ω | k || k | the maximal possible degree of Q k ,k ω ,ω and R ω k ,k is n k ,ω + n k ,ω ) • else, for all t > F k ( t ) = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ Q k ,k ω ,ω ( t ) e − i ( ω + ω ) t + e λt r ( t ) , (28) with Q k ,k ω ,ω a polynomial of degree smaller than or equal to n k ,ω + n k ,ω − . Remark 12.
In Lemma 4.1, we need that the inequality is true for a given ˜ λ ,in order to apply Lemma 3.3. However, applying Theorem 3.1, we deduce fromLemmae 4.2 and 4.3 that the inequality is true for all ˜ λ ∈ R . On the other hand,we have needed that Lemma 4.2 and 4.3 are true for all λ ∈ R , in order to provethat N k and N k are meromorphic.We are going to prove these lemmas distinguishing the non resonant case fromthe resonant case (when there exists γ ∈ (0 ,
1) such that k = − γk ). In orderto get proofs as clear as possible we do not prove that the remainder term can beextended on complex cones and we only control them on R ∗ + . Indeed, there are noreal issues to extend them and the arguments to control them on Σ α or R ∗ + arethe same. Furthermore, the notations induced for the complex extensions are quiteheavy and so do not help to understand the ideas. However, in the first proof, togive an example, we prove the analytic extension and we really estimate it.4.4. Proof of Lemma 4.2 in the non-resonant case.
Since we are studyingthe non-resonant case, there exists δ > ∀ θ ∈ (0 , , δ ≤ | (1 − θ ) k + θk | . Indeed, in the resonant case there exists γ ∈ (0 ,
1) such that k = − γk = γ ( k − k ),so that (1 − γ ) k + γk = 0. We have proven in Proposition 5 that there exists ∗ AND MICHEL MEHRENBERGER α ∈ (0 , π ) such that (cid:98) ψ ( t, k ) extends to an analytic function on Σ α and that thereexists λ ∈ R and M > ∀ z ∈ Σ α , | (cid:98) ψ ( z, k ) | ≤ M e λ (cid:60) z . Furthermore, since v (cid:55)→ (cid:98) g ( k , v ) ∈ E ( R d ), its Fourier transform extends to an entirefunction on C d and we can assume (choosing α small enough) that for all λ ∈ R there exists C λ > ∀ z ∈ Σ α R d , | F [ (cid:98) g ( k , v )] ( z ) | ≤ C λ e λ |(cid:60) z | . (29)Now observe that by a change of variable, F k ( t ) can be written as F k ( t ) = t (cid:90) (1 − θ ) (cid:98) ψ ( θt, k ) F [ (cid:98) g ( k , v )]( t ((1 − θ ) k + θk ))d θ. Consequently, F k ( t ) naturally extends to an analytic function on Σ α . Now, we haveto control F k ( z ) e − λz on Σ α for any λ ∈ R . Indeed, we have, for z ∈ Σ α , as we canassume λ ≤ | F k ( z ) e − λz | ≤ | z | e − λ (cid:60) z (cid:90) | (cid:98) ψ ( θz, k ) | (1 − θ ) | F [ (cid:98) g ( k , v )]( z ((1 − θ ) k + θk )) | d θ ≤ C λ M | z | (cid:90) e ( λ θ − λ ) (cid:60) z (1 − θ ) e λ (cid:60) z | (1 − θ ) k + θk | d θ ≤ C λ M | z | e ( | λ |− λ + δλ ) (cid:60) z ≤ C λ M (cid:18) (cid:60) z cos α (cid:19) e ( | λ |− λ + δλ ) (cid:60) z . So this quantity is bounded uniformly with respect to z ∈ Σ α if λ < λ −| λ | δ . Proof of Lemma 4.2 in the resonant case.
As explained before, fromnow, we do not pay attention to the analytic extension anymore. First, we use theresonance to give a more adapted expression of F k F k ( t ) = (cid:90) t (cid:98) ψ ( τ, k )( t − τ ) F [ (cid:98) g ( k , v )]( k ((1 − γ ) t − τ ))d τ = (cid:90) (1 − γ ) t − γt (cid:98) ψ ((1 − γ ) t − s, k )( γt + s ) F (cid:98) g ( k , k s )d s, making the change of variable s ← (1 − γ ) t − τ . We want to expand ψ , so we intro-duce the dependency of F k with respect to t (cid:55)→ (cid:98) ψ ( t, k ) by denoting F k [ (cid:98) ψ ( t, k )]( t ).Consequently, using the expansion of ψ of Proposition 5, for any λ ∈ R , we get F k [ (cid:98) ψ ( t, k )]( t ) = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ F k [ P k ,ω ( t ) e − iω t ]( t ) + F k [ e λ t r k ,λ ( t )]( t ) , where r k ,λ is a bounded function on R ∗ + .First, we are going to control the remainder term F k [ e λ t r k ,λ ( t )]( t ). Using thesame control of the Fourier transform as previously (see (29)), we have, as we can ECOND ORDER DISPERSION RELATIONS 23 assume λ , λ ≤ e − λt | F k [ e λ t r k ,λ ( t )]( t ) |≤ (cid:107) r k ,λ (cid:107) L ∞ e − λt (cid:90) (1 − γ ) t − γt e λ [(1 − γ ) t − s ] ( γt + s ) | F [ (cid:98) g ( k , v )]( k s ) | d s ≤ (cid:107) r k ,λ (cid:107) L ∞ C λ e − λt (cid:90) (1 − γ ) t − γt e λ [(1 − γ ) t − s ] ( γt + s ) e λ | k || s | d s ≤ (cid:107) r k ,λ (cid:107) L ∞ C λ e [(1 − γ ) λ − λ ] t (cid:90) R ( γt + s ) e ( λ | k |− λ ) | s | d s. So this quantity is bounded uniformly with respect to t ∈ R ∗ + if (1 − γ ) λ < λ and λ | k | < λ .Now, we are going to study one leading term of the type F k [ t n e − iω t ]( t ). So, weare doing a new expansion. F k [ t n e − iω t ]( t ) = e − iω (1 − γ ) t (cid:90) (1 − γ ) t − γt ((1 − γ ) t − s ) n e iω s ( γt + s ) F [ (cid:98) g ( k , v )]( k s )d s = n +1 (cid:88) j =0 b j t j e − iω (1 − γ ) t (cid:90) (1 − γ ) t − γt s n +1 − j e iω s F [ (cid:98) g ( k , v )]( k s )d s, where b , . . . , b n +1 are real numbers. Here we recognise the leading terms of (25)since, by construction, 1 − γ = | k || k | .Then, observe that since the right integral is convergent (see (29)), there exists A ∈ C such that for any t ∈ R ∗ + , we have (cid:90) (1 − γ ) t − γt s n +1 − j e iω s F [ (cid:98) g ( k , v )]( k s )d s = A − (cid:90) − γt −∞ s n +1 − j e iω s F [ (cid:98) g ( k , v )]( k s )d s − (cid:90) + ∞ (1 − γ ) t s n +1 − j e iω s F [ (cid:98) g ( k , v )]( k s )d s. The complex number A is the leading term of this integral whereas the other onesare remainder terms. So we just have to control them. Indeed, we have (cid:12)(cid:12)(cid:12)(cid:12) e − λt t j e − iω (1 − γ ) t (cid:90) − γt −∞ s n +1 − j e iω s F [ (cid:98) g ( k , v )]( k s )d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ e − λt t j e (cid:61) ω (1 − γ ) t (cid:90) ∞ γt s n +1 − j e (cid:61) ω s e λ | k | s d s ≤ C λ (cid:90) ∞ e − λ sγ (cid:18) sγ (cid:19) j e (cid:61) ω (1 − γ ) sγ s n +1 − j e (cid:61) ω s e λ | k | s d s, as we can assume λ ≤ t ≤ sγ . Consequently, it is bounded uniformlywith respect to t if | k | γλ < λ − (cid:61) ω . The estimation of the third integral can berealized with the same ideas.As we have a term in t n +1 , we see that R ω is of degree ≤ n k ,ω , since P k ,ω isof degree ≤ n k ,ω − ∗ AND MICHEL MEHRENBERGER
Proof of Lemma 4.3 in the non resonant case.
First, operating thechange of variable τ (cid:48) = t − τ , s (cid:48) = t − s , we can write F k as F k ( t ) = (cid:90) ≤ τ (cid:48) ≤ s (cid:48) ≤ t (cid:98) ψ ( t − τ (cid:48) , k ) (cid:98) ψ ( t − s (cid:48) , k ) τ (cid:48) F [ k · ∇ v f eq ] ( kτ (cid:48) + k ( s (cid:48) − τ (cid:48) ))d s (cid:48) d τ (cid:48) , since, if 0 ≤ s ≤ τ ≤ t we get 0 ≤ t − τ ≤ t − s and t − τ ≤ t − s ≤ t , that is0 ≤ τ (cid:48) ≤ s (cid:48) ≤ t . In order to get notations general enough but compact, we denote u = F [ k . ∇ v f eq ]. Since u ∈ E ( R d ), for all λ ∈ R there exists a constant C λ > ∀ ξ ∈ R d , ∀ t > , | u ( tξ ) | ≤ C λ e λ t | ξ | . (30)We define, for continuous functions φ , φ with an exponential order, a bilinearoperator q by q [ φ , φ ]( t ) = q [ φ ( t ) , φ ( t )]( t ) = (cid:90) ≤ τ ≤ s ≤ t φ ( t − τ ) φ ( t − s ) τ u ( kτ + k ( s − τ ))d s d τ . With these notations, we have F k ( t ) = q [ (cid:98) ψ ( t, k ) , (cid:98) ψ ( t, k )]( t ) . Consequently, using the expansions of (cid:98) ψ ( t, k ) and (cid:98) ψ ( t, k ) established in Propo-sition 5, we get for λ , λ ∈ R , F k = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ q [ P k ,ω e − iω t , P k ,ω e − iω t ] + q [ e λ t r k ,λ , (cid:98) ψ ( t, k )]+ q [ (cid:98) ψ ( tk ) , e λ t r k ,λ ] − q [ e λ t r k ,λ , e λ t r k ,λ ] (31)where r k ,λ and r k ,λ are respectively bounded by constants C λ and C λ .Furthermore, we can also assume that there exists λ ∈ R and M > ∀ t > , | (cid:98) ψ ( t, k ) | + | (cid:98) ψ ( t, k ) | ≤ M e λ t . Finally, since we are treating the non-resonant case, we may assume that thereexists δ > ∀ ≤ τ ≤ s, δs ≤ | τ k + ( s − τ ) k | . (32)So first, we are going to control the remainder terms of (31). For example, weconsider q [ e λ t r k ,λ , (cid:98) ψ ( t, k )]. So, if t > λ < λ <
0, we have e − λt | q [ e λ t r k ,λ , (cid:98) ψ ( t, k )]( t ) |≤ C λ M C λ e ( λ + λ − λ ) t (cid:90) ≤ τ ≤ s ≤ t e − λ τ − λ s τ e λ | kτ + k ( s − τ ) | d s d τ ≤ C λ M C λ e ( λ + λ − λ ) t (cid:90) ≤ τ ≤ s ≤ t e − λ τ − λ s + λ δs τ d s d τ ≤ C λ M C λ t e ( λ + λ − λ ) t (cid:90) s> e − λ s − λ s + λ δs d s. Realizing a decomposition of the form q [ a + b , a + b ] = q [ a , a ] + q [ b , a + b ] + q [ a + b , b ] − q [ b , b ] . ECOND ORDER DISPERSION RELATIONS 25
So, this quantity is bounded uniformly with respect to t > λ < λ − λ and λ < λ + λ δ < λδ . Similarly, we could prove that if λ is chosen negative enoughthen we could control q [ (cid:98) ψ ( t, k ) , e λ t r k ,λ ]( t ) e − λt uniformly with respect to t , andalso q [ e λ t r k ,λ , e λ t r k ,λ ].Now, we consider a generic leading terms of (31) of the type q [ t n e − iω t , t n e − iω t ].So first, we can expand it q [ t n e − iω t , t n e − iω t ]( t )= (cid:90) ≤ τ ≤ s ≤ t ( t − τ ) n e − iω ( t − τ ) ( t − s ) n e − iω ( t − s ) τ u ( kτ + k ( s − τ ))d s d τ = n (cid:88) j =0 n (cid:88) j =0 (cid:0) b j ,j e − i ( ω + ω ) t t n − j + n − j × (cid:90) ≤ τ ≤ s ≤ t τ j +1 s j e iω τ + iω s u ( kτ + k ( s − τ ))d s d τ (cid:1) , where b ∈ R (cid:74) ,n (cid:75) × (cid:74) ,n (cid:75) are some real coefficients.We observe that this last integral converge when t goes to + ∞ . Indeed, we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) s τ j +1 s j e iω τ + iω s u ( kτ + k ( s − τ ))d τ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ s j + j +2 e ( | ω | + | ω | ) s e λ δs ∈ L ( R + ) , if δλ < −| ω | − | ω | . Consequently, there exists a complex constant A ∈ C such that (cid:90) ≤ τ ≤ s ≤ t τ j +1 s j e iω τ + iω s u ( kτ + k ( s − τ ))d s d τ = A − (cid:90) ≤ τ ≤ st ≤ s τ j +1 s j e iω τ + iω s u ( kτ + k ( s − τ ))d s d τ . This complex number A generates the term of frequency ω + ω in (28). So wejust need to prove that the other term is a remainder term controlling it. Indeed,we have e − λt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − i ( ω + ω ) t t n − j + n − j (cid:90) ≤ τ ≤ st ≤ s τ j +1 s j e iω τ + iω s u ( kτ + k ( s − τ ))d s d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ e − λt e (cid:61) ( ω + ω ) t t n − j + n − j (cid:90) ≤ τ ≤ st ≤ s s j + j +1 e −(cid:61) ω τ −(cid:61) ω s e λ δs d s d τ ≤ C λ e − λt e (cid:61) ( ω + ω ) t t n − j + n − j (cid:90) t ≤ s s j + j +2 e |(cid:61) ω | s −(cid:61) ω s e λ δs d s ≤ C λ (cid:90) s> e |(cid:61) ( ω + ω ) | s − λs s n +2+ n e |(cid:61) ω | s −(cid:61) ω s e λ δs d s, as λ can be supposed ≤
0, and this last quantity is finite if λ is negative enough( λ < λ −|(cid:61) ( ω + ω ) |−|(cid:61) ω | + (cid:61) ω δ ).Concerning the degree, we see that it is ≤ n k ,ω − n k ,ω −
1, since n ≤ n k ,ω − n ≤ n k ,ω −
1, which corresponds to what is expected. ∗ AND MICHEL MEHRENBERGER
Proof of Lemma 4.3 in the resonant case.
We consider now the last case,which is the most complex. We keep the notations of the previous subsection butwe need a new expression of q adapted to the resonance: q [ φ , φ ]( t ) = (cid:90) ≤ τ ≤ s ≤ t φ ( t − τ ) φ ( t − s ) τ u ( k [(1 − γ ) τ − γ ( s − τ )])d s d τ , = (cid:90) t (cid:90) s φ ( t − τ ) φ ( t − s ) τ u ( k [ τ − γs ])d τ d s, = (cid:90) t (cid:90) (1 − γ ) s − γs φ ( t − τ − γs ) φ ( t − s )( τ + γs ) u ( k τ )d τ d s. The term τ + γs is quite heavy for our estimations, so we introduce a last notation q ml [ φ , φ ]( t ) = (cid:90) t (cid:90) (1 − γ ) s − γs φ ( t − τ − γs ) φ ( t − s ) τ l s m u ( k τ )d τ d s. Consequently, we can expand q [ φ , φ ] as follow q [ φ , φ ] = q [ φ , φ ] + γq [ φ , φ ] . We also introduce a new expansion of F k more adapted to the resonance F k = (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ q [ P k ,ω e − iω t , P k ,ω e − iω t ]+ (cid:88) D k ( ω )=0 (cid:61) ω ≥ λ q [ P k ,ω e − iω t , e λ t r k ,λ ] + q [ e λ t r k ,λ , (cid:98) ψ ( t, k )] . Now, we are going to study each one of the terms of this expansion.
Last term .First, we control the last remainder term, q [ e λ t r k ,λ , (cid:98) ψ ( t, k )]. Indeed, if t > e − λt | q ml [ e λ t r k ,λ , (cid:98) ψ ( t, k )]( z ) |≤ C λ M C λ e − λt t l + m (cid:90) t (cid:90) (1 − γ ) s − γs e λ [ t − τ − γs ] e λ ( t − s ) e λ | k || τ | d τ d s ≤ C λ M C λ (cid:18)(cid:90) R e − λ τ + λ | k || τ | d τ (cid:19) t l + m e ( − λ + λ + λ ) t (cid:18)(cid:90) t e − γsλ − λ s d s (cid:19) ≤ C λ M C λ (cid:18)(cid:90) R e − λ τ + λ | k || τ | d τ (cid:19) t l + m +1 e ( − λ + λ + λ ) t e − γtλ − λ t ≤ C λ M C λ (cid:18)(cid:90) R e ( − λ + λ | k | ) | τ | d τ (cid:19) t l + m +1 e ( − λ + λ (1 − γ )) t So this last quantity is bounded uniformly with respect to t if λ and λ are chosennegative enough. More precisely, we need (1 − γ ) λ < λ and λ | k | < λ . Realizing a decomposition of the form: q [ a + b , a + b ] = q [ a , a ] + q [ a , b ] + q [ b , a + b ] . ECOND ORDER DISPERSION RELATIONS 27
Second term .Now, we study the behavior of the second kind of term in the expansion of F k .Expanding P k ,ω ( t − τ − γs ), we can write q [ P k ,ω e − iω t , e λ t r k ,λ ]( t ) as a linearcombination of term of the type t j q ml +1 [ e − iω t , e λ t r k ,λ ]( t ) and t j q m +1 l [ e − iω t , e λ t r k ,λ ]( t ) , with j + l + m ≤ deg P k ,ω .Let t >
0, then we have q ml [ e − iω t , e λ t r k ,λ ]( t )= e − iω t (cid:90) t (cid:90) (1 − γ ) s − γs e iω τ e iω γs e λ ( t − s ) r k ,λ ( t − s ) τ l s m u ( k τ )d τ d s. So, using (30),we introduce R − ( s ) = (cid:90) − γs −∞ e iω τ τ l u ( k τ )d τ and R + ( s ) = (cid:90) ∞ (1 − γ ) s e iω τ τ l u ( k τ )d τ and A = (cid:90) R e iω τ τ l u ( k τ )d τ and B λ ,p = (cid:90) ∞ e − iω γs s p e λ s r k ,λ ( s )d s, (33)where B λ ,p is well defined if λ is negative enough (i.e. λ < − γ | λ | ). Consequently,we get (since 1 − γ = | k || k | ) q ml [ e − iω t , e λ t r k ,λ ]( t )= Ae − iω t (cid:90) t e iω γs s m e λ ( t − s ) r k ,λ ( t − s )d s + (cid:90) t e iω γs e λ ( t − s ) s m r k ,λ ( t − s ) ( R − ( s ) + R + ( s )) d s = Ae − iω | k || k | t (cid:90) t e − iω γs ( t − s ) m e λ s r k ,λ ( s )d s + (cid:90) t e iω γs e λ ( t − s ) s m r k ,λ ( t − s ) ( R − ( s ) + R + ( s )) d s = m (cid:88) p =0 C pm t m − p AB λ ,p e − iω | k || k | t + m (cid:88) p =0 C pm t m − p Ae − iω | k || k | t (cid:90) ∞ t e − iω γs s p e λ s r k ,λ ( s )d s + (cid:90) t e iω γs e λ ( t − s ) s m r k ,λ ( t − s ) ( R − ( s ) + R + ( s )) d s, where C pm = (cid:0) mp (cid:1) is a binomial coefficient. Here there are three kinds of terms. Thefirst one is one of expected leading term. The two others are remainder terms. Sowe have to control them. ∗ AND MICHEL MEHRENBERGER
First, we control the second kind of term. If t > (cid:12)(cid:12)(cid:12)(cid:12) e − λt e − iω | k || k | t (cid:90) ∞ t e − iω γs s p e λ s r k ,λ ( s )d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ e − λt e (cid:61) ω | k || k | t (cid:90) ∞ t e (cid:61) ω γs s p e λ s d s ≤ C λ (cid:90) s> s p e [ |(cid:61) ω | + | λ | + λ ] s s p e λ s d s. So this last quantity is finite if λ is negative enough.Then we control the last kind of term. If t > (cid:12)(cid:12)(cid:12)(cid:12) e − λt (cid:90) t e iω γs e λ ( t − s ) s m r k ,λ ( t − s ) R − ( s )d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ C λ e − λt (cid:90) t e −(cid:61) ω γs e λ ( t − s ) s m (cid:90) ∞ γs e −(cid:61) ω τ τ l e λ | k | τ d τ d s ≤ C λ C λ t m e ( λ − λ + |(cid:61) ω | γ ) t (cid:90) τ> τ l e (cid:16) λ | k |−(cid:61) ω + | λ | γ (cid:17) τ d τ . So this last quantity is bounded uniformly with respect to t if λ < λ − |(cid:61) ω | γ and λ | k | < (cid:61) ω ) − | λ | γ . Of course, we could control the other remainder term (with R + ) in a similar way.Concerning the degree, it is smaller or equal than the degree of P k ,ω , that is ≤ n k ,ω −
1, as j + m ≤ deg P k ,ω . This is for the moment one degree less thanwhat is expected in the Lemma 4.3. Remark 13.
Note the term B λ ,p in (33) is not explicit, as it relies on a remainderterm of the first order dispersion relation. It is worth mentioning that this termcontributes to the second order expansion, and not as a remainder term. First term .Finally we study the first kind of terms in the expansion of F k . These termsare of the type q [ P k ,ω e − iω t , P k ,ω e − iω t ]. By a straightforward calculation, as inthe previous case, it can be extended as a linear combination of terms of the type t j q ml +1 [ e − iω t , t n e − iω t ] and t j q m +1 l [ e − iω t , t n e − iω t ] with j + l + m = deg P k ,ω and n ≤ deg P k ,ω .In order to pursue the proof for this first kind of terms, in the following elementarylemma, we introduce a useful algebraic decomposition. It is proven in Appendix6.2. Lemma 4.4.
For all n, m ∈ N , for all ω ∈ C , there exists Q m,n,ω , R m,n,ω ∈ C [ X ] such that ∀ t > , (cid:90) t e iωs s m ( t − s ) n d s = Q m,n,ω ( t ) e iωt + R m,n,ω ( t ) . If ω (cid:54) = 0 then deg Q m,n,ω = m and deg R m,n,ω = n . If ω = 0 then Q m,n,ω = 0 and deg R m,n,ω = m + n + 1 . Remark 14.
The fact that the degree of R m,n,ω can change contains the discussionon the multiplicity. Indeed, it will be applied for ω = γω + ω which is equal tozero when ω + ω = | k || k | ω , since | k || k | = | k + k || k | = (1 − γ ). ECOND ORDER DISPERSION RELATIONS 29
Furthermore, using the previous constructions, we introduce B ( t ) = (cid:90) s> e iω γs e iω s ( t − s ) n s m ( R − ( s ) + R + ( s )) d s ∈ C n [ t ] . Now, if t >
0, we have q ml [ e − iω t , t n e − iω t ]( t )= (cid:90) t (cid:90) (1 − γ ) s − γs e − iω ( t − τ − γs ) e − iω ( t − s ) ( t − s ) n τ l s m u ( k τ )d τ d s = A (cid:90) t e − iω ( t − γs ) e − iω ( t − s ) ( t − s ) n s m d s + (cid:90) t e − iω ( t − γs ) e − iω ( t − s ) ( t − s ) n s m ( R − ( s ) + R + ( s )) d s = Ae − i ( ω + ω ) t (cid:104) Q m,n,γω + ω ( t ) e i ( γω + ω ) t + R m,n,γω + ω ( t ) (cid:105) + B ( t ) e − i ( ω + ω ) t − (cid:90) ∞ t e − iω ( t − γs ) e − iω ( t − s ) s m ( R − ( s ) + R + ( s )) d s =( AR m,n,γω + ω + B ( t )) e − i ( ω + ω ) t + AQ m,n,γω + ω ( t ) e − iω | k || k | t − (cid:90) ∞ t e − iω ( t − γs ) e − iω ( t − s ) ( t − s ) n s m ( R − ( s ) + R + ( s )) d s. Finally we just have to prove that this last integral is a remainder term. Indeed,we have (cid:12)(cid:12)(cid:12)(cid:12) e − λt (cid:90) ∞ t e − iω ( t − γs ) e − iω ( t − s ) ( t − s ) n s m R − ( s )d s (cid:12)(cid:12)(cid:12)(cid:12) ≤ C λ t n e ( − λ + (cid:61) ω + (cid:61) ω ) t (cid:90) ∞ t e − ( γ (cid:61) ω + (cid:61) ω ) s s m (cid:90) ∞ γs e (cid:61) ω τ τ l e λ | k | τ d τ d s ≤ C λ (cid:90) ∞ t e − γs (cid:90) ∞ γs τ n + l + m γ n + m e (1+ (cid:61) ω + |− λ + (cid:61) ω (cid:61) ω | + | γ (cid:61) ω (cid:61) ω | γ + λ | k | ) τ d τ d s ≤ C λ (cid:90) s> e − γs d s (cid:90) τ> τ n + l + m γ n + m e (1+ (cid:61) ω + |− λ + (cid:61) ω (cid:61) ω | + | γ (cid:61) ω (cid:61) ω | γ + λ | k | ) τ d τ . So this last quantity is finite if λ is negative enough.Concerning the degree, we consider first the case γω + ω (cid:54) = 0. As B is ofdegree ≤ n and R m,n,γω + ω is of degree ≤ n . So we get, as j ≤ deg P k ,ω and n ≤ deg P k ,ω , that Q k ,k ω ,ω is of degree ≤ n k ,ω − n k ,ω −
1, which is the expectedvalue. Now, as we can have a q m +1 l term, leading to Q m +1 ,n,γω + ω which is of degree ≤ m + 1 and as m can be chosen ≤ deg P k ,ω , R ω k ,k is of degree ≤ n k ,ω , whichis now the expected value. We consider finally the case γω + ω = 0, so that theterms e − i ( ω + ω ) t and e − iω | k || k | t are the same. The terms of highest degree is then R m +1 ,n,γω + ω which is here of degree ≤ m + n +2, that is ≤ n k ,ω − n k ,ω − Proof of Proposition 8.
Proof of Proposition 8.
In Proposition 6 and 4.1 we have proven that we can applyLemma 3.3 with 1 − R ( z ) = D k ( z + i (cid:101) λ ) and N ( z ) = N k ( z + i (cid:101) λ ) + N k ( z + i (cid:101) λ ), taking ∗ AND MICHEL MEHRENBERGER λ = (cid:101) λ/ | k | in Proposition 6: we get from Proposition 6 and Lemma 4.1 ∀ z ∈ i Σ γ + π , | zR ( z ) | ≤ C | k | , sup z ∈ i Σ β + π | zN ( z ) | < ∞ . But the result of this lemma is that for all λ ∈ R , we have N ( z )1 − R ( z ) = L (cid:20) (cid:88) ω pole of N − R (cid:61) ω ≥ λ P ω ( t ) e − iωt + e λt r ( t ) (cid:21) ( z ) , with a function r ∈ H (Σ ˜ β ) analytic and bounded on Σ ˜ β , for (cid:61) z large enough, withsome ˜ β satisfying 0 < ˜ β < γ < β and P ω is the polynomial such that N ( z )1 − R ( z ) = z → ω L [ P ω ( t ) e − iωt ] + O (1) . Thus, we have N k ( z ) + N k ( z ) D k ( z ) = L (cid:20) (cid:88) ω pole of N − R (cid:61) ω ≥ λ P ω ( t ) e − iωt + e λt r ( t ) (cid:21) ( z − i (cid:101) λ )= L (cid:20) (cid:88) ω pole of N − R (cid:61) ω ≥ λ P ω ( t ) e − i ( ω + i (cid:101) λ ) t + e ( λ + (cid:101) λ ) t r ( t ) (cid:21) ( z )= L (cid:20) (cid:88) ω pole of N k + N kDk (cid:61) ( ω ) ≥ λ + (cid:101) λ P ω − i (cid:101) λ ( t ) e − iωt + e ( λ + (cid:101) λ ) t r ( t ) (cid:21) ( z )So, defining µ by (cid:98) µ ( t, k ) = (cid:88) D ( ω )=1 (cid:61) ( ω ) ≥ λ + (cid:101) λ P ω − i (cid:101) λ ( t ) e − iωt + e ( λ + (cid:101) λ ) t r ( t ) , we get (24), which is (12). We finally have the expansion of Theorem 1.4. Concern-ing the multiplicity, if one pole is common to N k + N k and D − k we have to sum upthe multiplicity, leading to add n k,ω + ω − (cid:96) and n k, | k || k | ω − p . The other concerns about the multiplicity follow from Lemmae 4.2and 4.3, and the condition k · k (cid:54) = 0 directly follows from the factor k · k in frontof (22) and (23). Note also that R ∗ + ⊂ Σ (cid:101) β , so that r is bounded on R ∗ + as stated inTheorem 1.4.5. Numerical results.
Simulations have already been performed for multi-speciesand multi-dimensional simulations in [4], highlighting the relevance of second orderexpansion. We focus here more specifically on exhibiting a case where the Bestfrequency, that corresponds to the terms B in Theorem 1.4, appears. ECOND ORDER DISPERSION RELATIONS 31
First example.
We consider the one dimensional case ( d = 1 and L = 2 π )and solve numerically (VP) with a Semi-Lagrangian scheme and an adapted 6-thorder splitting [6]. 1 D periodic centered Lagrange interpolation of degree 17 is usedin both x and v directions and the periodic Poisson solver is solved with fast Fouriertransform.Initial condition is f ( x, v ) = f eq ( v ) + εg ( x, v ) , with f eq ( v ) = e − v / , g ( x, v ) = cos(2 x ) e − v / (2 σ ) + cos(3 x ) e − v / (2 σ ) and σ = 2 / , σ = √ π/ ε = 0 . v ∈ [ − v max , v max ], with v max = 10. Numerical parameters are: thenumber of uniform cells in x (resp. v ) that are N x (resp. N v ) and the time step∆ t ∈ R ∗ + , leading to a grid which will be referred as N x × N v × ∆ t grid.The first Fourier mode (cid:98) E ,num ( t ) of the electric field E := −∇ Φ is computed fromthe simulation at each time step t = t n = n ∆ t , using a discrete Fourier transform.We first compute the zeros of D k = D − k (see Remark 15), for | k | = 1 , , ω , ± (cid:39) ± . − . i,ω , ± (cid:39) ± . − . i,ω , ± (cid:39) ± . − . i. The second frequency of the mode 1 is ω (2)1 , ± (cid:39) ± . − . i . Suchzeros can be computing with a symbolic calculus software. An example using Mapleis provided in the Appendix. Here the modes that are initialized are k , k ∈{± , ± } . The main term is for k = k + k = ±
1, with k = ∓ k = ± ω ± has the greatest imaginary part among the ω k + k , with k , k ∈ {± , ± } .For having k = − γk , with γ ∈ (0 , k = ± k = ∓
3, sothat the Best frequencies ω b, ± of greatest imaginary part are defined by ω b, ± = | k + k || k | ω , ± = ω , ± . In order to see such term, we have to remove the main part coming from ω ± . Theprocedure is detailed as follows. From Theorem 1.4, we look here for (cid:60) ( (cid:98) E ,num )( t ) (cid:39) (cid:60) (cid:0) ze − iω t + ( z + tz ) e − iω b t (cid:1) , with z, z , z ∈ C , with ω = ω , + or ω = ω , − , as it leads to the same value, and similarly for ω b .We estimate z by using a least square procedure: we first define χ ( y ) = (cid:88) t min ≤ t j ≤ t max (cid:16) (cid:60) (cid:0) ye − iω t j (cid:1) − (cid:60) ( (cid:98) E ,num )( t j ) (cid:17) and then define z by minimizing this quantity, that is, χ ( z ) = min y ∈ C χ ( y ), whichis explicitely given by as solution of A T A (cid:20) (cid:60) ( z ) (cid:61) ( z ) (cid:21) = A T b, A = [ (cid:60) ( e − iω t j ) j ; −(cid:61) ( e − iω t j ) j ] , b = (cid:60) ( (cid:98) E ,num )( t j ) j , with A a matrix given by its 2 columns and b a vector, all the three vectors beingindexed by j that goes through all the values such that t min ≤ t j ≤ t max . ∗ AND MICHEL MEHRENBERGER -20 -18 -16 -14 -12 -10 -8 -6
0 5 10 15 20mode 1mode 1 - leading mode 1Bestmode 1 (ref)mode 1 - leading mode 1 (ref)Best (ref)
Figure 2.
Time evolution of |(cid:60) ( (cid:98) E ,num )( t ) | (mode 1), |(cid:60) (cid:16) (cid:98) E ,num ( t ) − ze − iω t (cid:17) | (mode 1 - leading mode 1) and |(cid:60) (cid:0) ( z + t j z ) e − iω b t j (cid:1) | (Best), for coarse 128 × × . × × . t min , t max ] = [17 . ,
35] and[˜ t min , ˜ t max ] = [1 . , .
5] are used for the least square procedures.Once z is found, we estimate z and z using again a least square procedure onthe remainder: defining this time˜ χ ( y , y ) = (cid:88) ˜ t min ≤ t j ≤ ˜ t max (cid:16) (cid:60) (cid:0) ( y + t j y ) e − iω b t j (cid:1) − (cid:60) (cid:16) (cid:98) E ,num ( t j ) − ze − iω t (cid:17)(cid:17) ,z and z are obtained by minimizing this quantity, that is,˜ χ ( z , z ) = min y ,y ∈ C ˜ χ ( y , y ) . Again the solution is explicitely given, the matrix A being here A = [ (cid:60) ( e − iω b t j ) j ; −(cid:61) ( e − iω b t j ) j ; (cid:60) ( t j e − iω b t j ) j ; −(cid:61) ( t j e − iω b t j ) j ] . On Figure 2, we represent the time evolution of the real part of the first Fouriermode |(cid:60) ( (cid:98) E ,num )( t ) | in absolute value, together with |(cid:60) ( (cid:98) E ,num )( t ) − (cid:60) (cid:0) ze − iω t (cid:1) | ,that is the quantity where we have removed the main part (it is a term J in Theorem1.4); the latter is compared to |(cid:60) (cid:0) ( z + t j z ) e − iω b t j (cid:1) | that corresponds to the Bestterm. The parameters t min , t max , ˜ t min and ˜ t max are chosen properly so that, in thecorresponding interval, the approximation is valid. Note that a too low value is notgood, as the expansion is only asymptotic and we consider only one term which is ECOND ORDER DISPERSION RELATIONS 33 the main term asymptotically. A too high value is also not good, as we have to facewith the round off or numerical error and the nonlinear behavior (note that we donot solve here the second linearized equation but the full nonlinear equation). Weobserve a well agreement, which is even better, by refining the grid, so that we canclaim that we have exhibited the Best frequency in the numerical results, which isfully coherent with the theoretical results.5.2.
Another case where the Best frequency is almost dominant on aspatial mode.
Now we consider again d = 1 (dimension 1), but we change thespatial length of the domain L = 20 π , and take f eq ( v ) = e − v / , g ( x, v ) = cos( x ) e − v / (2 σ ) + cos(0 . x ) e − v / (2 σ ) and σ = 2 / , σ = √ π/ ε = 0 . k , k ∈ {± , ± . } . We now need to know (we already have the value of ω , ± fromthe previous subsection) ω . , ± (cid:39) ± . . · − i,ω . , ± (cid:39) ± . − . · − i,ω . , ± (cid:39) ± . − . i,ω . , ± (cid:39) ± . − . i. The second frequency of the mode 0 . ω (2)0 . , ± (cid:39) ± . − . i . Thepossible values of k = k + k are in the set {± . , ± . , ± . , ± } . The first orderexpansion already gives a term that is not damped (the imaginary part is almostequal to zero). We also have terms on the second order expansion that are notdamped (for k = ± . k = ± .
9, one canlook at |(cid:60) ( (cid:98) E . ,num )( t ) | . From Theorem 1.4, we look thus here for an approximationof ε − (cid:60) ( (cid:98) E . ,num )( t ) in the form E ( t, z ) = (cid:60) (cid:16) z e − iω . t + ( z t + z ) e − i . ω t + z e − i ( ω + ω . , − ) t + z e − i ( ω + ω . , + ) t (cid:17) , with z = ( z , z , z , z , z ) ∈ C , using again ω (cid:96) = ω (cid:96), + or ω (cid:96) = ω (cid:96), − , for (cid:96) ∈ R , as itleads to the same result. In order to estimate z , we computemin y ∈ C (cid:88) t min ≤ t j ≤ t max (cid:16) e λt j (cid:60) (cid:16) E ( t j , y ) − ε − (cid:98) E . ,num ( t j ) (cid:17)(cid:17) , that is attained for y = z , by using the least square method as previously. Note thatwe add here the weight e λt , with λ = 0 .
48 and then we look for all the coefficients inone step. The choice of the value of λ is coherent with the fact that from Theorem1.4, the function e λt (cid:16) ε − (cid:60) ( (cid:98) E . ,num )( t ) − E ( t, z ) (cid:17) should be bounded. Numericalresults are shown on Figure 3. We use t min = 0 and t max = 30 for the coarse grid andhave increased t max to 35 for the fine grid (for the fine grid, we could even increasethis value, which was not possible for the coarse grid: the results were worse, asthe solution is not precise enough for the coarse grid on late times, as shown onFigure 3). For the fine grid, we could also not really increase further than around t max = 50, as we are limited, with nonlinear effects, convergence and/or machineprecision; we have also preferred not to go until t max = 50, as it leads to a worsermatching, since the least square procedure tends to match for values around 50,where the matching is less good. We could also change the initial time, but it hasnot so much impact, as it was the case for the previous subsection, since we have ∗ AND MICHEL MEHRENBERGER added here a weight function in the least square procedure. We emphasize that wecan again exhibit the Best frequency and also the two other types of frequencies,which are all in the same range, for this example. In order to get this results, wenote that we had to adapt he strategy concerning the least square method thatwas presented for the first example; this is due to the fact the several modes arein a similar range, and it was not easy to use the first procedure (used for the firstexample) to catch the different frequencies.The values of z are given here for coarse and fine mesh: z , coarse (cid:39) . − . i, z , fine (cid:39) . − . i,z , coarse (cid:39) . . i, z , fine (cid:39) . . i,z , coarse (cid:39) − . . i, z , fine (cid:39) − . . i,z , coarse (cid:39) . . i, z , fine (cid:39) . . i,z , coarse (cid:39) . . i, z , fine (cid:39) . . i. A D case. Looking for Best frequencies in 2D.
Finally, we focus on a 2 D case. Here wecan write k = k + k with k j = (cid:18) m j πL , n j πL (cid:19) , j = 1 , , m j , n j ∈ Z . Now if k = γk , with γ ∈ (0 , m + m = γm , n + n = γn , which leads to 1 − γ = − m m = − n n , if m (cid:54) = 0 and n (cid:54) = 0. If m or m = 0, we get m = m = 0, and similarly for n and n . In order to have a ”real” 2 D case, we can suppose that m (cid:54) = 0 and n (cid:54) = 0.We have − m m = − n n = 1 − γ = pq , p, q ∈ N , p < q, p ∧ q = 1 . So we obtain − m q = pm , and thus m = (cid:96)q , (cid:96) ∈ Z ∗ and − m = (cid:96)p together with n = ˜ (cid:96)q , ˜ (cid:96) ∈ Z ∗ and − n = ˜ (cid:96)p . Note that we then have k · k = γ | k | (cid:54) = 0.5.3.2. A D test case with Best frequency. We choose here L = L = L , m = n = 3 , m = n = −
2, so that k = (3 ,
3) 2 πL , k = ( − , −
2) 2 πL , k + k = (1 ,
1) 2 πL , and | k | = 3 √ πL , | k | = 2 √ πL , | k + k | = √ πL . We will write ω (cid:96) , instead of ω (cid:96), + or ω (cid:96), − , when we can either use ω (cid:96), + or ω (cid:96), − .We will need for this subsection and the next one, the following values (note thatthe values are here not the same as in the one dimensional case, since the dispersionrelation is not the same, as we have considered here a normalized Maxwellian): ECOND ORDER DISPERSION RELATIONS 35
Figure 3.
Time evolution of • |(cid:60) ( ε − (cid:98) E ,num )( t ) | (simu) vs | z e − iω . t | (approx1), • |(cid:60) (cid:16) ε − (cid:98) E ,num ( t ) − z e − iω . t (cid:17) | (simu1)vs |(cid:60) (cid:0) ( z t + z ) e − i . ω t (cid:1) | (approx2), • |(cid:60) (cid:16) ε − (cid:98) E ,num ( t ) − z e − iω . t − ( z t + z ) e − i . ω t (cid:17) | (simu2) vs |(cid:60) (cid:0) z e − i ( ω + ω . , − ) t + z e − i ( ω + ω . , + ) t (cid:1) | (ap-prox3),for coarse (top) 128 × × . × × . t min , t max ] = [0 ,
30] for the coarse grid and [ t min , t max ] = [0 ,
35] forthe refined grid. ∗ AND MICHEL MEHRENBERGER ω √ / , ± (cid:39) ± . − . · − i,ω √ / , ± (cid:39) ± . − . i,ω √ / , ± (cid:39) ± . − . i,ω √ / , ± (cid:39) ± . − . i,ω √ / , ± (cid:39) ± . − . i. Also, the second frequency of the mode √ /
10 is ω (2) √ / , ± (cid:39) ± . − . i .The main frequencies that intervene on the spatial mode (1 , πL are ω √ πL , ± ,and ω √ πL , ± (the last one is the Best frequency). In that case, we expect a similarbehavior as the test case of the first subsection.We use here f eq ( v ) = π e − v / − v / , with x = ( x , x ) , v = ( v , v ), togetherwith g ( x, v ) = (cos(0 . x + 0 . x ) + cos(0 . x + 0 . x )) f eq ( v ) , taking L = 20 π . We solve again numerically (VP) with a Semi-Lagrangian schemeand an adapted 6-th order splitting [6] (here d = 2). The parameter ε is alwaysfixed to ε = 10 − .We take v , v ∈ [ − , x and x directions, 64 cells in v and v di-rections; time step is fixed to ∆ t = 0 .
1, leading to a 32 × × × × . ρ ( t, x , x ) = (cid:82) R f d v (com-puted from trapezoidal rule): we define (cid:98) ρ (cid:96) ,(cid:96) ,num ( t ) the Discrete Fourier Transformof the charge density at time t = t n = n ∆ t . Results are given on Figure 4. Theleast square procedure is here applied to minimize:min y ∈ C (cid:88) t min ≤ t j ≤ t max (cid:0) e λt j (cid:60) (cid:0) E ( t j , y ) − ε − (cid:98) ρ , ,num ( t j ) (cid:1)(cid:1) , with E ( t, y ) = (cid:60) (cid:16) y e − iω √ / t + ( y t + y ) e − i ω √ / t (cid:17) , and is attained for y j = z j , j = 1 , ,
3, where the z j are given in Figure 4. We clearlysee on Figure 4, that the Best frequency ω √ πL is needed: with the combination ofthe main frequency ω √ πL the simulated mode (cid:98) ρ , ,num is accurately asymptoticallydescribed. In that case, we see both frequencies are useful; the main frequency isnot enough as we can see it on Figure 4. Indeed both modes (main and Best) areshown (they are shifted towards bottom of the Figure in order to see them better),and we see that the combination of the modes is needed to describe the simulatedmode.5.3.3. A D test case without Best frequency. Now, if we change and take n =2 , n = −
1, we have no more Best frequency, and the main frequencies that inter-vene on the same spatial mode (1 , πL are ω √ πL , ± and ω √ πL , ± + ω √ πL , ± (thesefrequencies were defined in the previous subsection), as we have this time k = (3 ,
2) 2 πL , k = ( − , −
1) 2 πL , k + k = (1 ,
1) 2 πL , and | k | = √
13 2 πL , | k | = √ πL , | k + k | = √ πL , ECOND ORDER DISPERSION RELATIONS 37
Figure 4.
A 2 D -case with Best frequency: time evolution of • |(cid:60) ( ε − (cid:98) ρ , ,num )( t ) | (simu) • |(cid:60) (cid:16) z e − iω √ / t + ( z t + z ) e − i ω √ / t (cid:17) | (approx) • − |(cid:60) (cid:16) z e − iω √ / t (cid:17) | (main mode /1e3) • − |(cid:60) (cid:16) ( z t + z ) e − i ω √ / t (cid:17) | (Best mode /1e3)The parameters λ = 0 .
09 and [ t min , t max ] = [0 ,
60] are used for the least squareprocedure to fit (simu) by (approx) and leads to z (cid:39) . . i , z (cid:39) − . − . i and z (cid:39) . − . i .the initial data being changed to g ( x, v ) = (cos(0 . x + 0 . x ) + cos(0 . x + 0 . x )) f eq ( v ) , and we have still L = 20 π . For the least square procedure, we consider the mini-mization problemmin y ∈ C (cid:88) t min ≤ t j ≤ t max (cid:0) e λt j (cid:60) (cid:0) E ( t j , y ) − ε − (cid:98) ρ , ,num ( t j ) (cid:1)(cid:1) , with E ( t, y ) = (cid:60) (cid:16) y e − iω √ / t + y e − i ( ω √ / , + + ω √ / , − ) t + y e − i ( ω √ / , + + ω √ / , + ) t (cid:17) , attained for y j = z j , j = 1 , ,
3, where the z j are given in Figure 5. We remarkhere that we have some unexpected frequency at the beggining which might be in-terpreted as a Best frequency (the simu-first approx curve), but such one is dampedand we get the right asymptotic behavior, which shows that we cannot get a Bestfrequency in the asymptotic limit, which is fully consistant with Theorem 1.4. ∗ AND MICHEL MEHRENBERGER
Figure 5.
A 2 D -case without Best frequency: time evolution of • |(cid:60) ( ε − (cid:98) ρ , ,num )( t ) | (simu) • |(cid:60) (cid:16) z e − iω √ / t (cid:17) | (first approx) • |(cid:60) (cid:16) ε − (cid:98) ρ , ,num ( t ) − z e − iω √ / t (cid:17) | (simu - first approx) • |(cid:60) (cid:16) z e − i ( ω √ / , + + ω √ / , − ) t + z e − i ( ω √ / , + + ω √ / , + ) t (cid:17) | (approx2)The parameters λ = 0 .
05 and [ t min , t max ] = [0 , z (cid:39) . . i , z (cid:39) − . − . i and z (cid:39) − . − . i .6. Appendix.
Some remarks about space E ( R d ) . The aim of this subsection is to presentsome tools to construct explicit examples of functions of E ( R d ) (characterized by(9)).The Gelfand-Shilov spaces S βα ( R d ) provide many useful examples of functions of E ( R d ). They are defined, for α, β >
0, by S βα ( R d ) := { f ∈ S ( R d ) | ∃ ε, C > , ∀ v, ξ ∈ R d , | f ( v ) | ≤ Ce − (cid:15) | v | α and | F f ( ξ ) | ≤ Ce − (cid:15) | ξ | β } . Many details about these spaces can be found in [12], in particular these spaces arestable by multiplication by a polynomial or a trigonometric polynomial, derivationand the natural action of the affine group of R d . Furthermore, we obviously have F S βα ( R d ) = S αβ ( R d ). ECOND ORDER DISPERSION RELATIONS 39
Proposition 9. If ν ∈ (0 , , then S − νν ( R d ) ⊂ E ( R d ) .Proof. It is a direct corollary of Proposition 6 . . Example 6.1. • | v | cos( v − v ) e − v − ( v + v ) ∈ S ( R ) ⊂ E ( R ), • If k ∈ N ∗ then e − v k ∈ S − k k ( R ) ⊂ E ( R ) (see [12]).To get some other example, we remark that E ( R d ) is clearly stable by multiplica-tion by a trigonometric polynomial, derivation and the natural action of the affinegroup of R d . Furthermore, it enjoys the following tensor product property. Proposition 10. If d , d ∈ N ∗ then E ( R d ) ⊗ E ( R d ) ⊂ E ( R d + d ) . Example 6.2. ∂ v e − v − ( v − v ) ∈ E ( R ).6.2. An algebraic decomposition.
The aim of this subsection is to prove Lemma4.4.
Proof of Lemma 4.4. If ω = 0, we get the result by expanding the polynomial( t − s ) n . So we suppose now that ω (cid:54) = 0. Since we recognize a convolution product,we apply a Laplace transform. So we get L (cid:20)(cid:90) t e iωs s m ( t − s ) n d s (cid:21) ( z ) = L [ t m ] ( z + ω ) L [ t n ] ( z ) = n ! m !( − i ) n + m +2 ( z + ω ) m +1 z n +1 . We can apply a partial fraction decomposition to get some complex coefficients( a j ) j =0 ,...,n and ( b j ) j =0 ,...,m such that n ! m !( − i ) n + m +2 ( z + ω ) m +1 z n +1 = n (cid:88) j =0 a j z j +1 + m (cid:88) j =0 b j ( z + ω ) j +1 . Consequently, we have L (cid:20)(cid:90) t e iωs s m ( t − s ) n d s (cid:21) ( z ) = L n (cid:88) j =0 a j i j +1 j ! t j + m (cid:88) j =0 b j i j +1 j ! t j e iωt ( z ) , Since the Laplace transform characterizes the continuous functions with an expo-nential order (see Theorem 1 . . Computation of the zeros.
We have used the following Maple code to com-pute the zeros of D k . We recall from Lemma 3.2 that for (cid:61) ( z ) > D k ( z ) = 1 − | k | (cid:90) k · ∇ v f eq ( v ) v · k − z d v = 1 − i | k | L [ F [ k · ∇ v f eq ( v )]( kt )] ( z )= 1 − i | k | L [ ik · ( kt ) F [ f eq ( v )]( kt )] ( z ) = 1 + L [ t F [ f eq ( v )]( kt )] ( z )= 1 + (cid:90) ∞ t F [ f eq ( v )]( kt ) e izt d t = 1 + (cid:90) ∞ t (cid:90) R d f eq ( v ) e itk · v d ve izt d t. We can write v = v (cid:107) + v ⊥ , with v (cid:107) the component of v along k and v ⊥ perpendicularto k ,(when d ≥ D k ( z ) = 1 + (cid:90) ∞ t (cid:90) ( R d ) (cid:107) (cid:32)(cid:90) ( R d ) ⊥ f eq ( v (cid:107) + v ⊥ )d v ⊥ (cid:33) e itk · v (cid:107) d v (cid:107) e izt d t. ∗ AND MICHEL MEHRENBERGER
Remark 15.
Note that D − k ( z ) = D k ( z ), if F [ f eq ( v )] is an even function. with(inttrans):with(RootFinding):Digits:=20:feq:=exp(-((v)^2)/2); REFERENCES [1] W. Arendt, C. J.K. Batty and M. Hieber,
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