limiting absorption principle for singular solutions to maxwell's … · 2016. 12. 14. ·...
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Limiting Absorption Principle for SingularSolutions to Maxwell’s Equations and Plasma
Heating
Ricardo Weder
Universidad Nacional Autónoma de México
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In collaboration with
Bruno Desprès.Laboratoire Jacques-Louis Lions, Université de Paris 6
Lise-Marie Imbert-GérardCourant Institute. New York University
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References
B. Desprès, L.-M. Imbert-Gerard, R. Weder,Hybrid resonanceof Maxwell’s equations in slab geometry, arXiv: 1210.0779[physics.plasm-ph], Journal de Mathématiques Pures etAppliquées 101 (2014) 623659.
B. Després, R. Weder, Hybrid resonance and long-timeasymptotic of the solution to Maxwell’s equations,arXiv:1507.04792[physics.plasm-ph], Physics Letters A 380(2016) 1284-1289.
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Related results
For the numerical implementation these results see,L.-M. Imbert-Gérard, Analyse mathématique et numérique deproblèmes d’ondes apparaissant dans les plasmasmagnétiques,PhD thesis, University Paris VI, 2013.
M. C. Pintos, B. Després, Constructive formulations of resonantMaxwell’s equations, 2016, hal-01278860.
For a local analysis around the resonance for analyticcoefficients see,B. Després, L-M Inbert-Gérard, O. Laffite, Singular solutions forthe plasma at resonance, 2015, hal-01097364.
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I will consider the heating of plasmas with electromagneticwaves for the cool plasma model, in the case of hybridresonances. This is a classical problem that has beenextensively studied in the plasma physics literature.However, in context of the ITER project, there is currentlyinterest in revisiting it with the purpose of obtaining precisemathematical methods that give a rigorous formula for theenergy absorbed by the plasma, as well as efficient numericalmethods.
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The ITER Project
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The mathematical difficulty lies on the fact that the solutionsthat heat the plasma are singular, and that it is necessary toconsider Maxwell’s equations in non-homogeneous andanisotropic media that do not belong to the classes that can beanalyzed with the standard methods.
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The Model
Let us consider Maxwell’s equations with a linear current andbulk magnetic field B0.
− 1c2∂tE +∇∧ B = µ0J, J = −eNeue,
∂tB +∇∧ E = 0,me∂tue = −e (E + ue ∧ B0)−meνue.
The velocity of the electrons is ue. The electronic density is Ne.There is a friction between the electrons and the ions withcollision frequency ν.
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The loss of energy in a domain Ω is given by,
ddt
∫Ω
(ε0 |E|2
2+|B|2
2µ0+
meNe |ue|2
2
)= −
∫ΩνmeNe |ue|2
plus boundary terms. Therefore
Q(ν) =
∫ΩνmeNe |ue|2
represents the total loss of energy of the electromagnetic fieldplus the electrons in function of the collision frequency ν. Sincethe energy loss is necessarily equal to what is gained by theions, it will be referred to as the heating.
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In typical applications in fusion plasmas the collision frequencyν is much smaller that the frequency of the electromagneticwaves, and it is of interest to compute the heating in the limitwhen ν → 0, Q(0+).
We will show that in certain conditions characteristic of thehybrid resonance in frequency domain, the heating does notvanish for vanishing collision friction, and moreover a largefraction of the energy of the incident wave, up to 95% fornormal incidence, can be transferred to the ions.
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Shifting to the frequency domain: ∂t = −iω, takingB0 = (|B0|,0,0) and eliminating the variables of the electron weobtain Maxwell’s equations ,
∇∧∇ ∧ E −(ω
c
)2εE = 0,
with the dielectric tensor
ε =
1− ωω2
p
ω(ω2−ω2c )
i ωcω2p
ω(ω2−ω2c )
0
−i ωcω2p
ω(ω2−ω2c )
1− ωω2p
ω(ω2−ω2c )
0
0 0 1− ω2p
ωω ,
where the cyclotron frequency is ωc = e|B0|
me, the plasma
frequency ωp =√
e2Neε0me
depends on the electronic density Ne,and ω = ω + iν is the complex pulsation.
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We consider in this work ω 6= ωc that is the frequency is away ofthe cyclotron frequency.As already mentioned, physical values in fusion plasmas aresuch that we can assume that ν = 0 (However we will latercome back to this point) and take
ε(x) =
1− ω2
p
ω2−ω2c
i ωcω2p
ω(ω2−ω2c )
0
−i ωcω2p
ω(ω2−ω2c )
1− ω2p
ω2−ω2c
0
0 0 1− ω2pω2
.
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X-Mode in Slab Geometry
The hybrid resonance concerns the the upper-left block in thedielectric tensor ε, that corresponds to the modeE = (Ex ,Ey ,0), and Ex ,Ey , independent of z. In this case weget the system,
W = ∂xEy − ∂yEx ,
∂yW − αω2
c2 Ex − iδ ω2
c2 Ey = 0,−∂xW + iδ ω
2
c2 Ex − αω2
c2 Ey = 0,
where W = iωBz is the vorticity, and the coefficients are
α = 1−ω2
p
ω2 − ω2c
δ =ωcω
2p
ω(ω2 − ω2
c) ,
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xAntenna
x=−L
reflected wave
incident wave
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x
B0
x=−L
Ne
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To be more specific we consider the simplified 2D domain
Ω =
(x , y) ∈ R2, −L ≤ x , y ∈ R, L > 0.
We supplement the X-mode equations with a nonhomogeneous boundary condition
W + iλEy = g on the left boundary x = −L, λ > 0,
which models a given source, typically an radiating antenna. Inreal Tokamaks this antenna is used to heat or to probe theplasma. We will consider slab geometry also for the sake ofsimplicity, that is all coefficients α and δ are functions only ofthe variable x
∂yα = ∂yδ = 0.
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Other assumptions which correspond to the physical context ofidealized reflectometry or heating devices are the following. Weassume that
δ ∈ C1[−L,∞[, δ(0) 6= 0.
The coefficient α vanishes at a finite number of points andα(x) = 0 implies that α′(x) 6= 0. For the sake of simplicity wewill now suppose that α only vanishes at x = 0. More precisely
α ∈ C2[−L,∞[, α(0) = 0, α′(0) < 0, (H1)
and
α− ≤ α(x) ≤ α+, ∀x ∈ [−L,∞[,0 < r ≤∣∣∣∣α(x)
x
∣∣∣∣ , ∀x ∈ [−L,H], (H2),
where H > 0. We will also assume that the coefficients areconstant at large scale: there exists δ∞ and α∞ so that
δ(x) = δ∞ and α(x) = α∞ H ≤ x <∞, (H3)
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We assume that the problem is coercive at infinity ( thiscorresponds to the absorption of the waves at infinity)
α2∞ − δ2
∞ > 0, (H4)
An additional condition is defined by
4‖δ‖2∞H < r . (H5).
It expresses the fact that the length of the transition zonebetween x = 0 and x = H is small with respect to the otherparameters of the problem.
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x=−L
x
x=H
slope −r
δ
α
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PLANE WAVE SOLUTIONS
Let us consider first that α and δ are constant. Consider aplane wave
(Ex ,Ey ) = Rei(k1x+k2y), R ∈ C2.
We assume that c = 1 for simplicity. We set k = |k |d withd = (cos θ, sin θ) the direction of the wave. The phase velocityvϕ = ω
|k | is given by
v2ϕ =
α
α2 − δ2 .
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When the phase velocity is real we are in a propagating region,and when the phase velocity is pure imaginary we are in anon-propagating region. One distinguishes two cutoffs wherethe local phase velocity is infinite
Cutoff : α(x) = ±δ(x)
and one resonance where the phase velocity is null
Resonance : α(x) = 0.
This structure is characteristic of the hybrid resonance.
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Sign of the square of the phase velocity. In this exampleα = −x and δ = 1, so that v2
ϕ = x1−x2 .
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The Budden problem
Consider the case where the solution is independent of y , whatfor the plane waves corresponds to normal incidence. In thissituation the Maxwell equations can be solved analytically insome cases which helps a lot to understand the singularity ofthe general problem. Let us consider that α = −x andδ(x) =
√x2 − x
4 + 1 > 0.Denote
ω(x) = −ex/2 + x e−x/2 Ei(x),
where Ei(x) is the Exponential-integral function,
Ei(x) = ln |x |+∞∑
j=1
x j
j · j!.
It follows that,
ω(x) = −1 + x ln |x |+ O(|x |), |x | → 0.
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The singular solution is given by,
Ey = ω ⇒ Ey ∈ L∞]− ε, ε[.
Ex = i
√x2 − x
4 + 1
xω(x) ≈ −i
1x, x → 0.
W (x) = ω′(x).
The problem now is that 1x is not a distribution, and then the
singular solution is not really a solution in mathematical senseat x = 0. It has to be regularized.This can be done in many different ways, in general 1
x can bereplaced by
P.V.1x
+N∑
j=1
aj δ(j)D (x),
for some arbitrary coefficients aj . So, which possible realizationwe choose? The answer to this question has to come from thephysics
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We will solve this problem, for general profiles that are notnecessarily solvable, by considering first the case where thereis a small friction between the electrons and the ions.
For this purpose we replace α(x) by α(x) + iν, ν 6= 0. In thiscase the solution is regular and uniquely defined, and we takethe limit as the friction goes to zero, ±ν ↓ 0. This also gives usa explicit formula to compute the heating of the plasma.The answer for the Budden profile is quite simple.Denote,
Ei,±(x) = ln±(x)+
∞∑j=1
x j
j j!,
where ln± are the branches of the logarithm, so that,
ln±(x) = ln |x | ± iπ, x < 0.
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We designate,
ω±(x) := −e−x/2 + xex/2Ei,∓(x).
We define, (UB,± = (E±x ,E±y ,W±))
UB,± =
(−P.V.
1x∓ iπδD(x) + uB,±(x), vB,±(x),wB,±(x)
),
where,
uB,±(x) :=1x
√x2 − x
4+ 1 ω±(x) +
1x,
vB,±(x) := −iω±(x),
wB,±(x) :=ddx
vB,±(x).
The UB,+ solution heats the plasma and UB,− cools the plasma.
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General Profiles.The Differential System
We replace α(x) by α(x) + iν, ν 6= 0.We denote by g the Fourier transform of g,
g(θ) :=
∫R
g(y) e−iθy dy .
We apply the Fourier transform and we obtain the system:W θ,ν + iθUθ,ν − d
dx V θ,ν = 0,iθW θ,ν − (α(x) + iν)Uθ,ν − iδ(x)V θ,ν = 0,− d
dx W θ,ν + iδ(x)Uθ,ν − (α(x) + iν)V θ,ν = 0.
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By elimination Uθ,ν one gets a system of two coupled ordinarydifferential equations
ddx
(V θ,ν
W θ,ν
)= Aθ,ν(x)
(V θ,ν
W θ,ν
)with
Aθ,ν(x) =
(θδ(x)α(x)+iν 1− θ2
α(x)+iνδ(x)2
α(x)+iν − α(x)− iν − θδ(x)α(x)+iν
)
In the case ν 6= 0 the matrix is non singular for all x , whichgives a meaning to the regularized problem. One notices thematrix is singular for ν = 0.
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General Profiles. The Integral Equation
Let us denote by (Aν ,Bν) the two fundamental solutions of themodified equation
−u′′ − (α(x) + iν)u = 0,
with the usual normalization
Aν(0) = 1, A′ν(0) = 0 and Bν(0) = 0, B′ν(0) = 1.
Let us denote Dθz the operator
Dθzh(z) = iθ∂zh(z)− iδ(z)h(z).
Let us define the kernel
kν(x , z) = Bν(z)Aν(x)− Bν(x)Aν(z).
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PropositionAny triplet (U,V ,W ) = (Ex ,Ey ,W ) solution of the regularizedsystem of differential equations admits the following integralrepresentation. One first chooses an arbitrary reference pointG ∈ [−L,∞[.
The x component of the electric field is solution of theintegral equation
U(x)−∫ x
G
DθxDθzkν(x , z)
α(x) + iνU(z)dz =
F θ,ν(x)
α(x) + iν,
where the right hand side is
F θ,ν(x) = aGDθxAν(x) + bGDθxBν(x).
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The y component of the electric field is recovered as
V (x) = aGAν(x) + bGBν(x) +
∫ x
GDθzkν(x , z)U(z)dz,
and the vorticity is recovered as
W (x) = aGA′ν(x) + bGB′ν(x) +
∫ x
G∂xDθzkν(x , z)U(z)dz.
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The two complex numbers (aG,bG) solve the linear systemaGAν(G) + bGBν(G) = V (G),aGA′ν(G) + bGB′ν(G) = W (G).
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So, we have to solve the integral equation,
U(x)−∫ x
G
DθxDθzkν(x , z)
α(x) + iνU(z)dz =
F θ,ν(x)
α(x) + iν,
When ν 6= 0 this is a standard integral equation of thesecond-class.However, when ν = 0 this is a singular integral equation of thethird class that has been seldom studied. Some importantcontributions are E. Picard, 1911, D. Hilbert 1912, G. R. Bartand R. L. Warnock 1973
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The integral kernelDθxDθzk0(x , z)
α(x)
and the right-hand sideF θ,0(x)
α(x)
are singular because α(0) = 0. If G 6= 0, in the integrationdomain D
θxDθz kµ(x ,z)α(x) is not even bounded, and if G = 0 it is
bounded but not Hölder continuous.
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A singular Limiting Absorption Principle
The solution to this problem is rather involved. The main ideasare as follows.Let us denote by Uθ,ν
2 ; =(
Eθ,νx ,Eθ,ν
y ,W θ,ν), the, properly
normalized, solution that goes to zero as x →∞.
Uθ,ν2 (0) =
1iν.
Then,
Theorem
For some Cθp and Cθ which are continuous with respect to θ∥∥∥∥Uθ,ν2 −
1α(·) + iν
∥∥∥∥Lp(−L,H)
≤ Cθp , 1 ≤ p <∞,
∥∥∥Uθ,ν2
∥∥∥H1
loc[−L,0)∪(0,H]≤ Cθ.
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This allows us to pass to the limit ν → 0
Theorem
The solution Uθ,ν2 admits a limit in the sense of distributions for
ν = 0+ as follows:
Uθ,ν2 → Uθ,+
2 =
(P.V .
1α(x)
+iπ
α′(0)δD + uθ,+2 , vθ,+2 , wθ,+
2
)where uθ,+2 , vθ,+2 ,wθ,+
2 ∈ L2(−L,∞) and δD is the Dirac mass atthe origin.
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Theorem
The singular solution Uθ,+2 is the unique solution of the following
system.Find a triplet(uθ,+2 , vθ,+2 ,wθ,+
2 ) ∈ L2(−L,∞)⊕ L2(−L,∞)⊕ L2(−L,∞) whichsatisfies the system,
wθ,+2 − d
dxvθ,+2 + iθuθ,+2 = −iθP.V .
1α(x)
+θπ
α′(0)δD,
iθwθ,+2 − αuθ,+2 − iδ(x)vθ,+2 = 1,
− ddx
wθ,+2 + iδ(x) uθ,+2 − α(x)vθ,+2 = −iP.V .
δ(x)
α(x)+δ(0)π
α′(0)δD.
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RemarkWe observe the similarity with the standard limiting absorptionprinciple in scattering theory. In scattering theory the solutionsobtained by the limiting absorption principle are characterizedas the unique solutions that satisfy the radiation condition, i.e.,they are uniquely determined by the behavior at infinity. Here,the singular solutions are uniquely determined by their behaviorat +∞ and by their singular part P.V . 1
α(x) ±iπ
α′(0)δD Note that itis natural that we have to specify the singularity at x = 0because our equations are degenerate at x = 0.
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Representation of the Solution
Recall the boundary condition
W + iλEy = g on the left boundary x = −L, λ > 0,
We have the following representation of the singular solution E+x
E+y
W +
(x , y) =1
2π
∫R
g(θ)
τ θ,+
P.V . 1α(x) + iπ
α′(0)δD + uθ,+2
vθ,+2wθ,+
2
eiθydθ.
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The Heating
Where we denote,
τ θ,+ := limν↓0
(W θ,ν
2 (−L) + iλV θ,ν2 (−L)
).
Moreover, the heating coefficient is given by,
Q = ωε0 limν→0+
ν
∫|Eν
x (x , y)|2dxdy =ωε0
2
∫R
∣∣g(θ)∣∣2
|α′(0)| |τ θ,+|2dθ.
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Large Time Asymptotics
We consider the evolution of wave packets. We denote,
Pϕ(t) :=
∫R
e−iωt A(ω)(
Uθ,+2,ω , ϕ
)dω,
where A(ω) is an integrable function with compact support. Thefunction ϕ is continuously differentiable on R, with values in C3,and has compact support. We made explicit the dependence inω of the singular solution Uθ,+
2,ω .We have that,
limt→∞
Pϕ(t) = 0.
This shows that our singular solution properly describes thephysics of our problem: for large times, in weak sense, theincoming solution tends to zero in any compact region of space,hence part of the solution is reflected to minus infinite and therest is absorbed by the plasma in the region (−L,∞).
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Idea of the Proofs
We construct a basis of solutions to the Maxwell equations withν 6= 0. We denote by
U := (U,V ,W ) = (Ex ,Ey ,W )
a general solution to Maxwell’s equations.The regular solution
Uθ,ν1 =
(Uθ,ν
1 ,V θ,ν1 ,W θ,ν
1
), Uθ,ν
1 (0) = 0,
V θ,ν1 (0) = iθ, and W θ,ν
1 (0) = iδ(0) (6= 0).
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PROPOSITION
The regular solution Uθ,ν1 is uniformly bounded with respect to
ν: for any interval θ ∈ [θ−, θ+] and any H ∈]L,∞[, there exists aconstant independent of ν such that∥∥∥Uθ,ν
1
∥∥∥L∞(−L,H)
+∥∥∥V θ,ν
1
∥∥∥L∞(−L,H)
+∥∥∥W θ,ν
1
∥∥∥L∞(−L,H)
≤ C.
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This is true because as in this case F θ,ν(0) = 0, the integralequation
U(x)−∫ x
0
DθxDθzkν(x , z)
α(x) + iνU(z)dz =
F θ,ν(x)
α(x) + iν,
is a standard integral equation of the second kind. Furthermore,Uθ,ν
1 extends by continuity to a regular solution for ν = 0
However, for ν 6= 0 the regular solution increases exponentiallyand this is also true for ν = 0, at least for |θ| small enough.
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The Singular Solution
Uθ,ν2 = (Uθ,ν
2 ,V θ,ν2 ,W θ,ν
2 ),
is built with two requirements: It is exponentially decreasing atinfinity, and
Uθ,ν2 (0) =
1i ν.
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To ensure that these conditions can be satisfied, we observethat since for x ≥ H, α = α∞, δ = δ∞ are constant, any solutionsatisfies,
U = c+ R+ eλθ,νx + c−R− e−λ
θ,νx , x ≥ H,
where c± ∈ C,R± ∈ C3, and λθ,ν has a positive real partbecause of the coercivity assumption.Consider a third solution
Uθ,ν3 = (Uθ,ν
3 ,V θ,ν3 ,W θ,ν
3 )(x) = R−e−λθ,νx , x ≥ H,
and it is smoothly extended to −L ≤ x ≤ H so that Uθ,ν3 is a
solution
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It turns out that since δ(0) 6= 0, we have that Uθ,ν3 (0) 6= 0 and
we can take,
Uθ,ν2 = c−Uθ,ν
3 , c− =1
iνUθ,ν3 (0)
.
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Let us consider a general solution U = (U,V ,W ) of the integralequation with prescribed data in H under the form
V (H) = aH and W (H) = bH .
Let us introduce the compact notation
‖H‖ = |aH |+ |bH | .
PROPOSITIONThere exists a constant Cθ with continuous dependence withrespect to θ such that
|U(x)| ≤ Cθ√r2x2 + ν2
‖H‖, 0 < x ≤ H.
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Let us define
Q(U) = V θ,ν1 (H)W (H)−W θ,ν
1 (H)V (H).
This quantity is the Wronskian of the current solution U againstthe first basis function. It is therefore independent of theposition H which is used to evaluate Q(U). Note that
|Q(U)| ≤ Cθ||H||.
To pursue the analysis, we begin by rewriting the integralequation for U in a way that shows that the various singularitiesof the equation can be recombined under a more convenientform.
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(α(x) + iν)U(x) = amθ,ν(x)x + bnθ,ν(x)x + Q(U)
−∫ H
xDθxDθzkν(x ,0)U(z)dz
+
∫ x
0
(DθxDθzkν(x , z)−DθxDθzkν(x ,0)
)U(z)dz, ∀x ∈ [−L,∞[.
Where, |a| ≤ Cθ5‖H‖, |b| ≤ Cθ
6‖H‖ are bounded uniformly withrespect to ν and mθ,ν(x),nθ,ν(x) are bounded functions.
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Setting u(x) = U(x)− Q(U)α(x)+iν , this equation can be written as:
u(x) +DθxDθzkν(x ,0)
α(x) + iν
∫ H
xu(z)dz
=x
α(x) + iνpθ,ν(x)−Q(U)
DθxDθzkν(x ,0)
α(x) + iν
∫ H
x
1α(z) + iν
dz
where
pθ,ν(x) = amθ,ν(x) + bnθ,ν(x)
+1x
∫ x
0
(DθxDθzkν(x , z)−DθxDθzkν(x ,0)
)U(z)dz
‖pθ,ν‖L∞(0,H) ≤ Cθ‖H‖, ∀ν ∈ [0,1].
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As a consequence one has
PROPOSITION
For all 1 ≤ p <∞, there exists a constant Cθp independent of ν
and which depends continuously on θ such that∥∥∥∥U − Q(U)
α(·) + iν
∥∥∥∥Lp(0,H)
≤ Cθp‖H‖.
Now we have to cross the singularity at x = 0.
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Dθ,ν(x) := −DθxDθzkν(x ,0)
α(x) + iν
∫ H
0
1α(z) + iν
dz
+
∫ x
0
DθxDθzkν(x , z)
α(x) + iν1
α(z) + iνdz
which is nothing than the integral part of our reorganizedintegral equation where U is replaced by the function 1
α(·)+iν .
PROPOSITIONLet 1 ≤ p <∞. One has∥∥∥Dθ,ν
∥∥∥Lp(−L,H)
≤ Cθp
where the constant depends continuously on θ and does notdepend on ν.
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Setting u(x) = U(x)− Q(U)α(x)+iν we write our integral equation as
u(x)−∫ x
0
DθxDθzkν(x , z)
α(x) + iνu(z)dz
=x
α(x) + iνpθ,ν(x)−Q(U)Dθ,ν(x)−
∫ H
0
DθxDθzkν(x ,0)
α(x) + iνu(z)dz.
Here pθ,ν(x) = amθ,ν(x) + bnθ,ν(x), so that‖pθ,ν‖L∞(−L,H) ≤ Cθ‖H‖ over the whole interval (−L,H).The left-hand side is a non singular integral operator of thesecond kind with a bounded kernel. The right-hand side isbounded in Lp with a continuous dependence with respect to‖H‖.This gives us the following proposition
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PROPOSITION
For all 1 ≤ p <∞, there exists a constant Cθp independent of ν
such that ∥∥∥∥U − Q(U)
α(·) + iν
∥∥∥∥Lp(−L,H)
≤ Cθp‖H‖.
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The Transversality Condition
We apply the results above to the second basis function forwhich
Q(Uθ,ν2 ) = V θ,ν
1 (x)W θ,ν2 (x)−W θ,ν
1 (x)V θ,ν2 (x) = 1 ∀x .
In this case we have that,∥∥∥∥Uθ,ν2 − 1
α(·) + iν
∥∥∥∥Lp(−L,H)
≤ Cθp
(∣∣∣V θ,ν2 (H)
∣∣∣+∣∣∣W θ,ν
2 (H)∣∣∣) ,
for 1 ≤ p <∞.
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PROPOSITION
There exists a maximal value θthresh > 0 such that: If4 ‖δ‖2∞H < r and |θ| < θthresh, then Uθ
1 increases exponentiallyat infinity.
We define:
σ(θ, ν) = V θ,ν1 (H)W θ,ν
3 (H)−W θ,ν1 (H)V θ,ν
3 (H).
The transversality condition is defined:
σ(θ,0) 6= 0.
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PROPOSITION
Assume the transversality condition. There exists a constant Cθ
independent of ν and continuous with respect to θ such that∣∣∣V θ,ν2 (H)
∣∣∣+∣∣∣W θ,ν
2 (H)∣∣∣ ≤ Cθ.
Proof: The pair (v ,w) = (V θ,ν2 (H),W θ,ν
2 (H)) is solution of thelinear system
−vW θ,ν1 (H) + wV θ,ν
1 (H) = 1,vW θ,ν
3 (H)− wV θ,ν3 (H) = 0.
Weder singular LAP
The determinant of this linear system is equal to the value ofthe function −σ(θ, ν). So the transversality conditionestablishes that
det
(−W θ,ν
1 (H) V θ,ν1 (H)
W θ,ν3 (H) −V θ,ν
3 (H)
)= −σ(θ, ν) 6= 0.
Therefore the solution of the linear system is given by,
v = −V θ,ν
3 (H)
σ(θ, ν), w = −
W θ,ν3 (H)
σ(θ, ν)
and then, it is bounded uniformly with respect to ν.
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Theorem
Assume the same transversality condition. The second basisfunction satisfies the following estimates for some Cθ
p and Cθ
which are continuous with respect to θ∥∥∥∥Uθ,ν2 − 1
α(·) + iν
∥∥∥∥Lp(−L,H)
≤ Cθp , 1 ≤ p <∞,
∥∥∥Uθ,ν2
∥∥∥H1
loc([−L,0)∪(0,H])≤ Cθ.
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The space Xθ,+
The basis of solutions on the limit ν → 0+, is given by: Xθ,+where
Xθ,+ = Span
Uθ1,U
θ,+2
⊂ H1
loc ((−L,∞) \ 0) .
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The space Xθ,−
It is of course possible do all the analysis with negative ν < 0and to study the limit ν → 0−. The first basis function is exactlythe same. The second basis function is chosen exponentiallydecreasing at infinity and such that
iνUθ,ν2 = 1 ν < 0.
We also have that,∥∥∥∥Uθ,ν2 − 1
α(·) + iν
∥∥∥∥Lp(−L,H)
≤ Cθp , −1 ≤ ν < 0.
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We have that,
Uθ,ν2 → Uθ,−
2 =
(P.V .
1α(x)
− iπα′(0)
δD + uθ,−2 , vθ,−2 , wθ,−2
)where uθ,−2 , vθ,−2 ,wθ,−
2 ∈ L2(−L,∞) and δD is the Dirac mass atthe origin.
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As in the case of ν ↓ 0, the singular solution Uθ,−2 is the unique
solution of the following system.Find a triplet (uθ,−2 , vθ,−2 ,wθ,−
2 ) ∈ L2(−L,∞)3 which satisfies thesystem,
wθ,−2 − d
dxvθ,−2 + iθuθ,−2 = −iθP.V .
1α(x)
− θπ
α′(0)δD,
iθwθ,−2 − αuθ,−2 − iδ(x)vθ,−2 = 1,
− ddx
wθ,−2 + iδ(x) uθ,−2 − α(x)vθ,−2 = −iP.V .
δ(x)
α(x)− δ(0)π
α′(0)δD.
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Passing to the limit ν → 0−, it defines the limit space Xθ,−
Xθ,− = Span
Uθ1,U
θ,−2
⊂ H1
loc ((−L,∞) \ 0) .
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Comparison of the limits
The first basis function Uθ1 is independent of the sign and
belongs to Xθ,+ ∩ Xθ,−. Since the limit equation and thenormalization at x = H are the same, we readily observe thatthe limits of the second basis functions are identical for 0 < x
Uθ,+2 (x) = Uθ,−
2 (x) 0 < x .
PROPOSITIONOne has
Uθ,+2 (x)− Uθ,−
2 (x) =−2iπα′(0)
Uθ1(x) x < 0. (1)
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Numerical Results
Lise-Marie Imbert-Gérard.Analyse mathématique et numérique de problèmes d’ondesapparaissant dans les plasmas magnétiques,PhD thesis, University Paris VI, 2013.
The numerical tests show a fast pointwise convergence of thenumerical solution to the exact one, except at the origin ofcourse. Moreover our numerical tests show that a large part ofthe incoming energy of the wave may be absorbed by theheating, around 90 % in some cases.
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This is for example the case for the Fourier mode θ = 0, withL = 2, c = ω = 2 and ωc = 1, so that α = 1− 2δ. We considerthe profiles
α(x) =
1, −L ≤ x ≤ −1,−x , −1 ≤ x ≤ 3,−3, 3 ≤ x <∞,
δ(x) =
0, −L ≤ x ≤ −1,x+1
2 , −1 ≤ x ≤ 3,2, 3 ≤ x <∞,
which satisfy additionally |α∞| > |δ∞| and the fact that theelectronic density is increasing from the left to the right.
We compute numerically the singular solution U0,ν2 taking
ν = 10−3 as a small regularization parameter.
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The efficiency of the heating is defined as the ratio of theheating Q over the incoming energy. In this case ourcalculations show an efficiency of around 95%. Anothercalculation in oblique incidence θ = cos π
4 shows an efficiencystill around 76,7%. These values indicate a high efficiency.
The numerical results below were obtained with H = 2,α(x) = −x , x ≤ 2, α(x) = −2, x ≥ 2, δ(x) = 0.25, θ = 1.5 Thereal part of the singular solution (Uθ,ν
2 ,V θ,ν2 ,W θ,ν
2 ) is plotted.The real part of (−x + iν)−1 is also represented, and evidencesthe blow up of the solution at the resonance as ν goes to zero.
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−10 −5 0 5−60
−40
−20
0
20
40
60
µ=10−2
Re v2
Re w2
Re u2
Re 1/(−x+imu)
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−10 −5 0 5−600
−400
−200
0
200
400
600
µ=10−3
Re v2
Re w2
Re u2
Re 1/(−x+imu)
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−10 −5 0 5−6000
−4000
−2000
0
2000
4000
6000
µ=10−4
Re v2
Re w2
Re u2
Re 1/(−x+imu)
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Below the real parts of the three components of the singularsolution (Uθ,ν
2 ,V θ,ν2 ,W θ,ν
2 ) are represented. The scale is nowfixed to show the strong convergence observed on]− 10,5[\0 : the convergence observed is strong everywherebut at the resonance point x = 0.
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−10 −5 0 5−60
−40
−20
0
20
40
60
µ=10−2
Re v2
Re w2
Re u2
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−10 −5 0 5−60
−40
−20
0
20
40
60
µ=10−3
Re v2
Re w2
Re u2
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−10 −5 0 5−60
−40
−20
0
20
40
60
µ=10−4
Re v2
Re w2
Re u2
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THANKS FOR YOUR ATTENTION
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