Solutions of the Vlasov part

∫

𝛾𝑇•⁺

𝑓₊^𝛼𝜓𝑑𝛾𝛼−

∫

𝛾𝑇•⁻

𝒦_𝛼𝑓₊^𝛼+𝑔^𝛼

𝜓𝑑𝛾𝛼−

∫

Ω

∫

R³

𝑓˚𝛼𝜓(0)𝑑𝑣𝑑𝑥

for all𝜓∈Ψ𝑇•.

We explain in the following remark in what sense we can speak of traces 𝑓𝛼 + of 𝑓𝛼

in a weak solution concept.

Remark 1.2.1. If Definition 1.1.1.(ii) is satisfied,𝑓₊^𝛼is the trace of 𝑓𝛼in the following sense:

• As we have just seen, 𝑓₊^𝛼is the restriction of 𝑓^𝛼to𝛾𝑇⁺_•if 𝑓^𝛼 ∈𝐶¹ 𝐼_𝑇

•×Ω×R³

.

• There is no other𝑓˜𝛼 + ∈𝐿¹

loc

𝛾𝑇⁺_•

such that Definition 1.1.1.(ii) is satisfied as well, since for such 𝑓˜₊^𝛼we have

∫

𝛾𝑇•⁺

𝑓₊^𝛼− ˜𝑓₊^𝛼

𝜓𝑑𝛾𝛼 =0

for all𝜓∈𝐶^∞ 𝐼_𝑇

•×Ω×R³

with supp𝜓⊂ [0, 𝑇_•[ ×Ω×R³compact that vanish on𝛾𝑇⁻_•∪𝛾𝑇⁰_•. Consequently,𝑓˜₊^𝛼= 𝑓₊^𝛼.

1.2.2 Solutions of the Vlasov part

We give a brief introduction to the techniques and statements of Beals and Pro-topopescu [BP87], who used an approach via characteristics to tackle linear transport problems with initial-boundary conditions in a very general setting. Since we do not need the full statements of [BP87], we formulate those results in the way we will need them in our situation.

Throughout this subsection, let𝑇 >0,Ω ⊂R³ be an open, bounded set with𝐶^1,𝜅 -boundary for some𝜅 >0, andΣ𝑇 B^]0, 𝑇[ ×Ω×R³. Furthermore, let𝑌be a first order linear differential operator that is divergence free and whose coefficients are Lipschitz continuous onΣ𝑇. In accordance to our situation, we choose

𝑌B 𝜕^𝑡+

b^𝑣𝛼·𝜕^𝑥+𝐹·𝜕^𝑣.

Thus, the assumptions about𝑌here reduce to two conditions on𝐹, namely, that𝐹is Lipschitz continuous onΣ𝑇 and divergence free with respect to 𝑣. We additionally assume that𝐹is bounded onΣ𝑇. By Lipschitz continuity of𝐹, for each(𝑡, 𝑥, 𝑣) ∈Σ𝑇

there is a well-defined integral curve𝑠↦→ (𝑆, 𝑋 , 𝑉)(𝑠, 𝑡, 𝑥, 𝑣)satisfying 𝑑

𝑑𝑠𝑆=1, 𝑑

𝑑𝑠𝑋=^𝑉b_𝛼^, 𝑑

𝑑𝑠𝑉=𝐹(𝑠, 𝑋 , 𝑉), (𝑆, 𝑋 , 𝑉)(𝑡, 𝑡, 𝑥, 𝑣)=(𝑡, 𝑥, 𝑣).

This curve is defined as long as it remains inΣ𝑇and there is a corresponding maximal time interval𝐼 ⊂Rfor which it is defined. We define the length of this curve to be the

length of the maximal time interval for which the curve remains inΣ𝑇, that is to say, the length equals𝑠⁺−𝑠⁻where

𝑠⁺B^{sup{𝑠∈𝐼} | (𝑆, 𝑋 , 𝑉)(𝑠, 𝑡, 𝑥, 𝑣) ∈Σ𝑇}, 𝑠⁻Binf{𝑠∈𝐼 | (𝑆, 𝑋 , 𝑉)(𝑠, 𝑡, 𝑥, 𝑣) ∈Σ𝑇}.

The next assumption is that there is a finite upper bound to all lengths of such integral curves. This condition is trivially satisfied in our caseΣ𝑇 =]0, 𝑇[ ×Ω×R³since𝑇is an upper bound. The last assumption is that each integral curve has a left and right limit point, i.e.,

𝑠→𝑠lim⁻,𝑠>𝑠⁻(𝑆, 𝑋 , 𝑉)(𝑠, 𝑡, 𝑥, 𝑣),_𝑠→𝑠lim_+,𝑠

<𝑠⁺(𝑆, 𝑋 , 𝑉)(𝑠, 𝑡, 𝑥, 𝑣) ∈Σ𝑇.

These limits, if they exist, have to be elements of𝜕Σ𝑇. For their existence it is sufficient that 𝐹 is bounded by some constant 𝐶 > 0 since then both 𝑋¤ and 𝑉¤ are bounded because of

^𝑋¤

=

^𝑉b_𝛼

^≤1,

^𝑉¤

=|𝐹(𝑠, 𝑋 , 𝑉)| ≤𝐶.

Accordingly, we define 𝐷_𝑇⁻ (𝐷_𝑇⁺) to be the subset of 𝜕Σ𝑇 consisting of all such left (right) limits, often referred to as incoming (outgoing) sets. These sets are Borel sets since 𝐷_𝑇⁻ (𝐷_𝑇⁺) is the image of the open set Σ𝑇 under the continuous function that maps a point inΣ𝑇to the left (right) limit point of the integral curve passing through this point. Note that possibly 𝐷_𝑇^± are not disjoint and/or do not exhaust 𝜕Σ𝑇 but both𝐷_𝑇⁺∩𝐷_𝑇⁻ and𝜕Σ𝑇\ 𝐷_𝑇⁺∪𝐷_𝑇⁻

are negligible in the sense that the union of all associated integral curves inΣ𝑇has Lebesgue measure zero.

We proceed with the definition of the test function space corresponding to𝑌. Definition 1.2.2. LetΦ^𝑌_𝑇be the space of all measurable functions𝜙:Σ𝑇→Rwith the following three properties:

(i) 𝜙is continuously differentiable along each integral curve.

(ii) 𝜙and𝑌𝜙are bounded functions.

(iii) The support of𝜙is bounded and there is a positive lower bound to the lengths of the integral curves which meet the support of𝜙.

Remark 1.2.3. • Here and in the following, the term𝑌 ℎ, where ℎ ∈ 𝐿¹

loc(Σ𝑇), is in general to be understood as a distribution, i.e.,

(𝑌 ℎ) 𝜑

=−

∫

Σ𝑇

𝜕^𝑡𝜑+

b^𝑣𝛼·𝜕^𝑥𝜑+𝐹·𝜕^𝑣𝜑ℎ 𝑑(𝑡, 𝑥, 𝑣), 𝜑∈𝐶^∞_𝑐 (Σ𝑇).

In Definition 1.2.2.(ii) or later in Definition 1.2.7.(i), this distribution is assumed to be given by a function onΣ𝑇.

• Because of Definition 1.2.2.(ii) and 1.2.2.(iii) we have𝜙, 𝑌𝜙∈𝐿^𝑞(Σ𝑇)for any𝜙∈Φ^𝑌_𝑇 and 1≤ 𝑞≤ ∞.

1.2 The Vlasov part 17

• Note that a function𝜙∈ Φ^𝑌_𝑇 only has to be continuously differentiable along each integral curve but may be discontinuous in other directions. Because of Defini-tion 1.2.2.(i) and 1.2.2.(ii) every 𝜙 ∈ Φ^𝑌_𝑇 can be extended to be continuous at the endpoints of each integral curve.

SinceΦ^𝑌_𝑇depends on𝐹, it cannot be suitable for the whole nonlinear system (VM), where𝐹is unknown. Thus, an important (technical) statement is that our test function spaceΨ𝑇_•, which is independent of𝐹, belongs toΦ^𝑌_𝑇after a cut-off in the time variable (if𝑇≤𝑇_•). This is verified in the following two lemmas, where we follow the proof of [Guo93, Lemma 2.1.].

Lemma 1.2.4. (i) For any 𝜄 > 0 there is a 𝛿 = 𝛿(𝜄) > 0 such that for all (𝑥, 𝑣) ∈ ˜𝛾⁻ satisfyingdist (𝑥, 𝑣),𝛾˜⁰

> 𝜄we haveb^𝑣𝛼·𝑛(𝑥) ≤ −𝛿.

(ii) For any𝜄 >0there is an𝜂=𝜂(𝜄)>0such that for any𝑥 ∈𝜕Ω,𝑦 ∈R³we have𝑦∈Ω if

^𝑦−𝑥

_{< 𝜂and} ^𝑦−𝑥

_{·𝑛(𝑥) ≤ −} 𝜄

^𝑦−𝑥 _<0.

Proof. As for part 1.2.4.(i), suppose the contrary. Then we can find a 𝜄 > 0 and a sequence(𝑥_𝑘, 𝑣_𝑘) ⊂ ˜𝛾⁻ with dist (𝑥_𝑘, 𝑣_𝑘),𝛾˜⁰ _{> 𝜄}

for𝑘 ∈ Nandb^𝑣𝑘,𝛼 ·𝑛(𝑥_𝑘) → 0 for 𝑘→ ∞. Without loss of generality we can assume that(𝑣_𝑘)is bounded: If|𝑣_𝑘| ≥1 let 𝑤_𝑘 B |𝑣^𝑣𝑘_𝑘|. Then,

0>^𝑤b^𝑘,𝛼·𝑛(𝑥_𝑘)=|

𝑤b_𝑘,𝛼|cos(](

𝑤b_𝑘,𝛼, 𝑛(𝑥_𝑘))) ≥ |

b^𝑣𝑘,𝛼|cos(](

b^𝑣𝑘,𝛼, 𝑛(𝑥_𝑘)))

=b^𝑣𝑘,𝛼·𝑛(𝑥_𝑘) →0 for𝑘→ ∞because of|

𝑤b_𝑘,𝛼| ≤ | b^𝑣𝑘,𝛼|.

Therefore,(𝑥_𝑘, 𝑣_𝑘) ⊂𝜕Ω×R³converges, after extracting a suitable subsequence, to some(𝑥, 𝑣) ∈𝜕Ω×R³. On the one hand, we have dist (𝑥, 𝑣),𝛾˜⁰_≥𝜄, and on the other handb^𝑣𝛼·𝑛(𝑥)=0 which is a contradiction.

The proof of part 1.2.4.(ii) exploits that 𝜕Ω is of class 𝐶^1,𝜅. Suppose that the assertion does not hold, i.e., we can find a𝜄 >0 and sequences(𝑥_𝑘) ⊂ 𝜕Ω, 𝑦_{𝑘 ⊂}

R³ with

^𝑦𝑘−𝑥_𝑘

_< ¹_{𝑘 and} ^𝑦𝑘−𝑥_{𝑘·𝑛(𝑥}

𝑘) ≤ −𝜄 ^𝑦𝑘−𝑥_𝑘

_{<0 but}^𝑦𝑘∉ Ω. We may assume that both sequences converge because of(𝑥_𝑘) ⊂𝜕Ωand 𝑦_𝑘⊂

𝜕Ω+𝐵₁. The limits of both sequences have to be the same; we call the limit𝑥 ∈𝜕Ω. Since𝑥_𝑘+𝑡 𝑦_𝑘−𝑥_𝑘∈

Ω for𝑡 >0 small enough and𝑦_𝑘 ∉ Ω, there has to be a𝑥˜_𝑘 ∈ _𝑥

𝑘, 𝑦_𝑘

∩𝜕Ω. Obviously we have| ˜𝑥_𝑘−𝑥_𝑘| < ¹𝑘 and

( ˜𝑥_𝑘−𝑥_𝑘) ·𝑛(𝑥_𝑘)= 𝑦_𝑘−𝑥_{𝑘·𝑛(𝑥}

𝑘)| ˜𝑥_𝑘−𝑥_𝑘| ^𝑦𝑘−𝑥_𝑘

≤ −𝜄| ˜𝑥_𝑘−𝑥_𝑘|<0. (1.2.5) Since𝜕Ωis compact and𝑥˜_𝑘 → 𝑥,𝑥_𝑘 → 𝑥for𝑘 → ∞,𝑥˜_𝑘,𝑥_𝑘, and𝑥 lie in the image of the same 𝐶^1,𝜅-chart 𝜑:R^{2 ⊃ 𝑊 →} 𝜕Ω if 𝑘is large enough. Let 𝑝˜_𝑘 B 𝜑⁻¹( ˜𝑥_𝑘), 𝑝_𝑘 B 𝜑⁻¹(𝑥_𝑘), and𝑝 B 𝜑⁻¹(𝑥). By continuity of𝜑⁻¹, both 𝑝˜_𝑘

and 𝑝_𝑘

converge to 𝑝. Thus, we may assume that𝑝˜_𝑘, 𝑝_𝑘∈𝐾_𝑝 B^{𝐵𝑟 𝑝} _⊂𝑊

for suitable𝑟 >0 and large𝑘. We expand the left-hand side of (1.2.5) to get the estimate

|( ˜𝑥_𝑘−𝑥_𝑘) ·𝑛(𝑥_𝑘)|

=

. Together with (1.2.5) and the fact that 𝜑⁻¹ is Lipschitz continuous on 𝜑 𝐾_𝑝

with some Lipschitz constant 𝐿_𝜑,𝐾

𝑝 >0—see proof below—, this yields for large𝑘 0< 𝜄

But this contradicts ^𝑝˜𝑘−𝑝_𝑘

^→_{0 for}^{𝑘→ ∞}_.

So there remains to show the Lipschitz continuity of𝜑⁻¹on𝜑 𝐾_𝑝

. This relies on the fact that, since𝜑is a chart, the function

𝐺:𝐾_𝑝×𝜕𝐵₁ →R^{, 𝐺 𝑝,˜} 𝛿𝑝

is continuous and positive so that it is bounded from below by some positive constant 𝑐>0. For𝑥, 𝑥˜ ∈𝜑 𝐾_𝑝

also converge to the same limit due to the continuity of𝜑⁻¹. But this contradicts (1.2.6). Hence,𝜑⁻¹is Lipschitz continuous on 𝜑 𝐾_𝑝

and the proof is finished.

Lemma 1.2.5. For each𝜓 ∈Ψ𝑇we have𝜓

1.2 The Vlasov part 19 Since𝜓

_Σ

𝑇 obviously satisfies Definition 1.2.2.(i) and 1.2.2.(ii), we only have to take care about Definition 1.2.2.(iii). First note that, since the support of𝜓 is compact in [0, 𝑇[ ×Ω×R³, there is a 0 ≤ 𝑠₀ < 𝑇such that 𝜓(𝑡, 𝑥, 𝑣)= 0 for𝑠₀ ≤ 𝑡 <𝑇,𝑥 ∈ Ω, 𝑣∈R³.

We consider an integral curve which meets supp𝜓 = supp𝜓

Σ𝑇. This curve can be written as𝑠↦→ (𝑆, 𝑋 , 𝑉)(𝑠, 𝑠⁻, 𝑥, 𝑣)and remains inΣ𝑇for a maximal time interval ]𝑠⁻, 𝑠⁺[ ⊂ ]0, 𝑇[so that(𝑠⁻, 𝑥, 𝑣) ∈𝐷_𝑇⁻. Obviously it holds that𝑠⁻ ≤ 𝑠₀. We have to find a positive lower bound for 𝑠⁺−𝑠⁻that does not depend on𝑠⁻,𝑠⁺,𝑥, and𝑣. In the following, let𝑠 ∈ ]𝑠⁻, 𝑠⁺[.

Case 1. If

dist (𝑠⁻, 𝑥, 𝑣),supp𝜓_≥ ^𝑑0

2

we can find an 𝑠 such that(𝑆, 𝑋 , 𝑉)(𝑠, 𝑠⁻, 𝑥, 𝑣) ∈ supp𝜓 since the curve meets the support of𝜓. By

^𝑋¤

^≤_{1 and} ^𝑉¤

^≤_supΣ

𝑇|𝐹|we have 𝑑₀

2 ≤dist (𝑠⁻, 𝑥, 𝑣),supp𝜓 ≤ |(𝑆, 𝑋 , 𝑉)(𝑠, 𝑠⁻, 𝑥, 𝑣) − (𝑠⁻, 𝑥, 𝑣)|

≤ q

2+ k𝐹k²_∞(𝑠−𝑠⁻) so that ^𝑑0

2

√

2+k𝐹k²_∞ is such a desired lower bound in this case.

Case 2. The more complicated case is

dist (𝑠⁻, 𝑥, 𝑣),supp𝜓

< 𝑑₀ 2. Since{𝑇} ×Ω×R³and𝛾𝑇⁺do not intersect𝐷_𝑇⁻, we have

𝐷_𝑇⁻⊂𝛾𝑇⁻∪𝛾𝑇⁰ ∪

{0} ×Ω×R³ .

Clearly, it holds that(𝑠⁻, 𝑥, 𝑣)∉𝛾𝑇⁰ because of dist (𝑠⁻, 𝑥, 𝑣),supp𝜓

< 𝑑₀

2 <𝑑₀ ≤dist supp𝜓,𝛾𝑇⁰

_. If(𝑠⁻, 𝑥, 𝑣) ∈ {0} ×Ω×R³we have

dist(𝑥,𝜕Ω)=dist (𝑠⁻, 𝑥, 𝑣),{0} ×𝜕Ω×R³

≥dist supp𝜓,{0} ×𝜕Ω×R³₋

dist (𝑠⁻, 𝑥, 𝑣),supp𝜓

>𝑑₀−𝑑₀ 2 = 𝑑₀

2 . Thus,𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣) ∈Ωfor 0≤𝑠<minn_𝑑

0 2 , 𝑇o

again because of ^𝑋¤

^≤1. Therefore, a positive lower bound to the length of the integral curve in this case is minn_𝑑

0 2 , 𝑇o

.

Finally, suppose(𝑠⁻, 𝑥, 𝑣) ∈𝛾𝑇⁻. First note that dist (𝑥, 𝑣),𝛾˜⁰

=dist (𝑠⁻, 𝑥, 𝑣),𝛾𝑇⁰

_≥

dist supp𝜓,𝛾𝑇⁰

₋

dist (𝑠⁻, 𝑥, 𝑣),supp𝜓

>𝑑₀− 𝑑₀ 2 = 𝑑₀

2 . (1.2.7)

Let𝛿=𝛿_𝑑

0 2

and𝜂=𝜂 ^𝛿₂

according to Lemma 1.2.4. We claim that

𝑚B^min (

𝑇−𝑠₀,𝜂

2, 𝛿

9

2k𝐹k_𝐿∞(Σ𝑇;R³⁾+1 )

is such a positive lower bound (to the length of the integral curve) we search for.

Indeed, we firstly have[𝑠⁻, 𝑠⁻+𝑚] ⊂ [0, 𝑇]due to𝑠⁻ ≤𝑠₀. Secondly, let 𝑠B^{sup{𝑠∈ ]𝑠−, 𝑠−+𝑚] |𝑋(˜𝑠, 𝑠−, 𝑥, 𝑣) ∈Ωfor all𝑠˜∈ ]𝑠−, 𝑠[}.}

Because of

|𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣) −𝑥| ≤𝑠−𝑠⁻ < 𝜂 and

(𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣) −𝑥) ·𝑛(𝑥)=

∫ ^𝑠

𝑠⁻

𝑉b_𝛼(𝜏, 𝑠⁻, 𝑥, 𝑣)𝑑𝜏

!

·𝑛(𝑥)

=(𝑠−𝑠⁻)

b^𝑣𝛼·𝑛(𝑥) +

∫ ^𝑠

𝑠⁻

∫ 𝜏 𝑠⁻

_𝑑 𝑑𝑠^𝑉b_𝛼

(𝑙, 𝑠⁻, 𝑥, 𝑣)𝑑𝑙𝑑𝜏·𝑛(𝑥)

≤ −𝛿(𝑠−𝑠⁻) +9

2k𝐹k_𝐿∞(Σ𝑇;R³⁾·1

2(𝑠−𝑠⁻)² ≤ −𝛿

2(𝑠−𝑠⁻) ≤ −𝛿

2|𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣) −𝑥| (which also implies𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣)≠ 𝑥 since−^𝛿

2(𝑠−𝑠⁻) <0) by (1.2.7) and

𝑑b^𝑣𝛼,𝑖 𝑑𝑣_𝑗

^≤

3 2, 𝑖, 𝑗 = 1,2,3, we have 𝑋(𝑠, 𝑠⁻, 𝑥, 𝑣) ∈ Ω and thus 𝑠 = 𝑠⁻+𝑚. This completes the proof.

We should remark that the three conditions on𝜓∈Ψ𝑇in (1.1.1) are really necessary:

Let𝜄 >0 be small and, for simplicity, take𝐹 =0. Firstly, if we allow a test function𝜓 that does not vanish before time𝑇and has support on𝛾𝑇⁻, we can find an integral curve enteringΣ𝑇on𝛾𝑇⁻∩supp𝜓at time𝑠⁻ =𝑇−𝜄. Secondly, if we allow a test function 𝜓 with support on𝛾𝑇⁰, then for some(𝑡, 𝑥, 𝑣) ∈ 𝛾𝑇⁰—such that in a neighborhood of 𝑥 there are no common points of Ω and the tangent space of 𝜕Ω at 𝑥—the curves (𝑆, 𝑋 , 𝑉)(𝑠) = (𝑠, 𝑥−𝜄𝑛(𝑥) + (𝑠−𝑡)

b^𝑣𝛼, 𝑣), defined for all 𝑠 ∈ [0, 𝑇], will meet the support of𝜓. Thirdly, if we allow a test function𝜓 with support on{0} ×𝜕Ω×R³ we can find an integral curve meeting the support of𝜓, (its𝑋-coordinate) starting at time 0 near𝜕Ωand leavingΩat time𝜄. In all three cases, there will be no positive lower bound to the length of these curves.

1.2 The Vlasov part 21 Conversely, these restrictions cause no problems for later considerations. Firstly, we do not want to test a solution of (1.2.1) at time𝑇since we are interested in an initial, and not final, value problem. Secondly, we only want to impose a boundary condition on 𝛾⁻ and not on 𝛾⁰. Thirdly, proper initial data of the distribution function have to satisfy the boundary condition at time 0 a priori so that this property need not be tested, and{0} ×𝜕Ω×R³is even a null set with respect to𝑑𝛾𝛼.

We now proceed with some important results of [BP87]. There, the main idea is to use the “identifications”

Σ𝑇 ≈_{(𝑠, 𝑧) |𝑧∈𝐷}−

𝑇,0<𝑠 <𝑙(𝑧) , 𝑌≈ 𝑑 𝑑𝑠,

where𝑙(𝑧)is the length of the integral curve corresponding to𝑧. The first important result is the following property which is closely related to Green’s identity; see [BP87, Proposition 7].

Proposition 1.2.6. There are two unique Borel measures𝜈^±on𝐷^±

𝑇such that

∫

Σ𝑇

𝑌𝜙𝑑(𝑡, 𝑥, 𝑣)=

∫

𝐷⁺_𝑇𝜙𝑑𝜈⁺−

∫

𝐷_𝑇⁻𝜙𝑑𝜈⁻ for all𝜙∈Φ^𝑌_𝑇.

We have to define the space of functions in which we search for solutions of some initial-boundary problem.

Definition 1.2.7. For 1 ≤ 𝑝 < ∞let𝐸^𝑝(Σ𝑇;𝑌)be the space of functions 𝑓 ∈ 𝐿^𝑝(Σ𝑇) with the following two properties:

(i) 𝑌 𝑓 ∈𝐿^𝑝(Σ𝑇).

(ii) There is a trace of 𝑓 on𝐷^±

𝑇, i.e., a pair of functions 𝑓^± ∈𝐿^𝑝 𝐷^±

𝑇, 𝑑𝜈^±

satisfying the extended Green’s identity

∫

Σ𝑇

𝜙𝑌 𝑓 +𝑓 𝑌𝜙_{𝑑(𝑡, 𝑥, 𝑣)}

=

∫

𝐷⁺ 𝑇

𝑓⁺𝜙𝑑𝜈⁺−

∫

𝐷⁻ 𝑇

𝑓⁻𝜙𝑑𝜈⁻

for all𝜙∈Φ^𝑌_𝑇.

Note that a trace in the sense as stated above is unique and that all terms are well-defined according to Remark 1.2.3.

Lemma 1.2.8. Let 1 ≤ 𝑝 < ∞, 𝑓 ∈ 𝐸^𝑝(Σ𝑇;𝑌) and 𝑤 ∈ 𝐶^∞(Σ𝑇) ∩𝐶¹_𝑏 Σ𝑇

. Then, 𝑤 𝑓 ∈𝐸^𝑝(Σ𝑇;𝑌)and 𝑤 𝑓^±

=𝑤 𝑓^±. Proof. Because of

𝑌 𝑤 𝑓 𝜑

=−

∫

Σ𝑇

𝜕^𝑡𝜑+

b^𝑣𝛼·𝜕^𝑥𝜑+𝐹·𝜕^𝑣𝜑𝑤 𝑓 𝑑(𝑡, 𝑥, 𝑣)

=−

∫

Σ𝑇

𝜕^𝑡 𝑤𝜑₊

b^𝑣𝛼·𝜕^𝑥 𝑤𝜑_+𝐹·

𝜕^𝑣 𝑤𝜑 𝑓 𝑑(𝑡, 𝑥, 𝑣)

+

∫

Σ𝑇

(𝜕^𝑡𝑤+

b^𝑣𝛼·𝜕^𝑥𝑤+𝐹·𝜕^𝑣𝑤)𝑓𝜑𝑑(𝑡, 𝑥, 𝑣)

=

∫

Σ𝑇

𝑌 𝑓_𝑤

𝜑𝑑(𝑡, 𝑥, 𝑣) +

∫

Σ𝑇

(𝑌𝑤)𝑓𝜑𝑑(𝑡, 𝑥, 𝑣)

for any 𝜑 ∈ 𝐶^∞_𝑐 (Σ𝑇), it holds that𝑌 𝑤 𝑓

= 𝑤𝑌 𝑓 + 𝑓 𝑌𝑤 ∈ 𝐿^𝑝(Σ𝑇). Now let𝜙 ∈ Φ^𝑌_𝑇. We have𝑤𝜙 ∈ Φ^𝑌_𝑇 since Definition 1.2.2.(i) and 1.2.2.(ii) are satisfied because of the regularity of𝑤and Definition 1.2.2.(iii) is satisfied because of supp 𝑤𝜙 _⊂

supp𝜙.

Thus, it holds that

∫

Σ𝑇

𝜙𝑌 𝑤 𝑓_{+𝑤 𝑓 𝑌}

𝜙_{𝑑(𝑡, 𝑥, 𝑣)}

=

∫

Σ𝑇

𝑤𝜙𝑌 𝑓 +𝑓 𝑌 𝑤𝜙 𝑑(𝑡, 𝑥, 𝑣)

=

∫

𝐷⁺ 𝑇

𝑓⁺𝑤𝜙𝑑𝜈⁺−

∫

𝐷⁻ 𝑇

𝑓⁻𝑤𝜙𝑑𝜈⁻,

which proves the assertion.

In the following it is convenient to split𝐷_𝑇^±as follows:

𝐷_±^𝑇B (𝑡, 𝑥, 𝑣) ∈𝐷_𝑇^±|0<𝑡<𝑇 ,

𝐷₀ B (𝑡, 𝑥, 𝑣) ∈𝐷⁻_𝑇 |𝑡=0 , 𝐷^𝑇_𝑇 B (𝑡, 𝑥, 𝑣) ∈𝐷_𝑇⁺ |𝑡=𝑇 , so that𝐷⁻

𝑇 =𝐷₋^𝑇∪𝐷₀and𝐷⁺

𝑇 =𝐷₊^𝑇∪𝐷^𝑇

𝑇. Note that𝐷₀does not depend on𝑇(in the sense that any 0<𝑇˜ <𝑇yields the same set𝐷₀). According to this decomposition we write𝑑𝜈⁻=𝑑𝜈⁻

_𝐷𝑇

−,𝑑𝜈0=𝑑𝜈⁻ _𝐷

0,𝑑𝜈⁺=𝑑𝜈⁺ _𝐷𝑇

+,𝑑𝜈^𝑇=𝑑𝜈⁺ _𝐷𝑇

𝑇, 𝑓₋= 𝑓⁻ _𝐷𝑇

−, 𝑓₀= 𝑓⁻ _𝐷

0, 𝑓₊= 𝑓⁺

_𝐷𝑇

+, and 𝑓_𝑇= 𝑓⁺ _𝐷𝑇

𝑇. We have

{0} ×Ω×R^{3 ⊂𝐷}0 ⊂ {0} ×Ω×R^{3, {𝑇} ×Ω×}R^3⊂𝐷𝑇^𝑇⊂ {𝑇} ×Ω×R^3, 𝛾𝑇⁻⊂𝐷^𝑇₋ ⊂𝛾𝑇⁻∪𝛾𝑇⁰, 𝛾𝑇⁺ ⊂𝐷₊^𝑇⊂𝛾𝑇⁺∪𝛾𝑇⁰.

Therefore, we can identify 𝐿^𝑝-functions on 𝐷₀ (or𝐷^𝑇_𝑇) with𝐿^𝑝-functions onΩ×R³ since

Ω×R³

\ Ω×R³

has(𝑥, 𝑣)-Lebesgue measure zero. Additionally, we may write 𝑓(0)and 𝑓(𝑇)instead of 𝑓₀and 𝑓_𝑇pointing out that we may evaluate 𝑓 at time 0 and𝑇in some sense.

For each𝜓∈Ψ𝑇we have

∫

Σ𝑇

𝑌𝜓𝑑(𝑡, 𝑥, 𝑣)=−

∫

Ω

∫

R³

𝜓(0)𝑑𝑣𝑑𝑥+

∫ ^𝑇

0

∫

𝜕Ω

∫

R³

𝜓b^𝑣𝛼·𝑛 𝑑𝑣𝑑𝑆_𝑥𝑑𝑡

=−

∫

Ω

∫

R³

𝜓(0)𝑑𝑣𝑑𝑥+

∫

𝛾𝑇⁺

𝜓𝑑𝛾𝛼−

∫

𝛾⁻𝑇

𝜓𝑑𝛾𝛼.

1.2 The Vlasov part 23 This shows that𝑑𝜈0 =𝑑(𝑥, 𝑣)on𝐷₀and𝑑𝜈^± =𝑑𝛾𝛼on𝛾𝑇^±. With an analog reasoning (consider test functions 𝜓˜(𝑡, 𝑥, 𝑣) = 𝜓(𝑇−𝑡, 𝑥, 𝑣),𝜓 ∈ Ψ𝑇) we conclude that 𝑑𝜈^𝑇 = 𝑑(𝑥, 𝑣)on𝐷_𝑇^𝑇as well.

We proceed with a definition of some properties of operators.

Definition 1.2.9. Let 𝑂 be an operator between two function spaces on subsets of someR^𝑛, whose first component we call time. 𝑂is called

(i) local in time if𝑂(𝑢𝑣)=𝑢𝑂(𝑣)for all continuous functions𝑢that only depend on time and all possible𝑣;

(ii) nonnegative if𝑂(𝑣) ≥0 for all𝑣 ≥0.

Now we are ready to state the following result regarding the unique solvability of linear transport problems with initial-boundary conditions; see [BP87, Proposition 1, Theorems 1 and 2].

Proposition 1.2.10. Let1 ≤ 𝑝 < ∞, ℎ ∈ 𝐿^∞(Σ𝑇),𝐹:Σ𝑇 → R be Lipschitz continuous, differentiable with respect to𝑣, and divergence free with respect to𝑣, and𝑌=𝜕^𝑡+

b^𝑣𝛼·𝜕^𝑥+𝐹·𝜕^𝑣. (i) For all 𝑓 ∈𝐸^𝑝(Σ𝑇;𝑌)we have

∫

𝐷^𝑇 𝑇

^𝑓𝑇

𝑝𝑑𝜈^𝑇+

∫

𝐷^𝑇 +

^𝑓+

𝑝𝑑𝜈⁺+𝑝

∫

Σ𝑇

ℎ ^𝑓

𝑝𝑑(𝑡, 𝑥, 𝑣)

=

∫

𝐷₀

^𝑓₀

𝑝𝑑𝜈0+

∫

𝐷^𝑇₋

^𝑓−

𝑝𝑑𝜈⁻+𝑝

∫

Σ𝑇

sign 𝑓 ^𝑓

𝑝−1(𝑌+ℎ)𝑓 𝑑(𝑡, 𝑥, 𝑣). (1.2.8)

(ii) Let moreover𝔎: 𝐿^𝑝 𝐷₊^𝑇, 𝑑𝜈⁺ _{→ 𝐿}𝑝 𝐷^𝑇₋, 𝑑𝜈⁻

be a bounded linear operator, that is local in time and has operator norm less than1, and𝑔₀ ∈𝐿^𝑝(𝐷₀),𝑔₋∈ 𝐿^𝑝 𝐷^𝑇₋, 𝑑𝜈⁻

. Then the problem

𝑌 𝑓 =0 onΣ𝑇, (1.2.9a)

𝑓₀ =𝑔₀ on𝐷₀, (1.2.9b)

𝑓₋ = 𝔎𝑓₊+𝑔₋ on𝐷^𝑇₋ (1.2.9c)

has a unique solution𝑓 ∈𝐸^𝑝(Σ𝑇;𝑌). Here,(1.2.9a)holds pointwise almost everywhere (cf. Definition 1.2.7.(i) and Remark 1.2.3), and(1.2.9b)and (1.2.9c)hold pointwise almost everywhere (with respect to the corresponding measures) in the sense of trace (cf. Definition 1.2.7.(ii)). Moreover, the solution is nonnegative if𝔎, 𝑔₀, and 𝑔₋ are nonnegative.

Here and in the following, for functions the property “nonnegative” usually means

“nonnegative almost everywhere”. We want to express, in some way, the theorem above in words that fit to our problem (1.2.1), that is to say, we should somehow replace 𝐷₀,𝐷^𝑇₋ (and so on) byΩ×R³,𝛾𝑇⁻(and so on). Moreover, we search for solutions of (1.2.1) on𝐼_𝑇

• instead of solutions on some time interval [0, 𝑇]. To this end, we first have to define what we call a strong solution of (1.2.1). From now on, the force term 𝐹shall satisfy the following condition.

Condition 1.2.11. 𝐹:𝐼_𝑇

• ×Ω×R^{3 →}R³ is Lipschitz continuous and bounded onΣ𝑇

for any𝑇 ∈ 𝐼_𝑇•, and moreover differentiable and divergence free (both) with respect to𝑣.

Definition 1.2.12. Assume that 1 ≤ 𝑝 < ∞,𝒦: 𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

→ 𝐿^𝑝

lt

𝛾𝑇⁻_•, 𝑑𝛾𝛼

, 𝑓˚ ∈ 𝐿^𝑝 Ω×R³

, and 𝑔 ∈ 𝐿^𝑝

lt

𝛾𝑇⁻•

, 𝑑𝛾𝛼

. We call a function 𝑓:𝐼_𝑇

• ×Ω×R^{3 →}Ra strong solution of (1.2.1) if:

(i) 𝜒^𝑇𝑓 ∈𝐸^𝑝(Σ𝑇;𝑌)for all 0<𝑇∈𝐼_𝑇

•. (ii) For all𝜓 ∈Ψ𝑇_•it holds that

∫ ^𝑇• 0

∫

Ω

∫

R³

𝜕^𝑡𝜓+

b^𝑣𝛼·𝜕^𝑥𝜓+𝐹·𝜕^𝑣𝜓_{𝑓 𝑑𝑣𝑑𝑥𝑑𝑡}

=

∫

𝛾⁺𝑇•

𝑓₊𝜓𝑑𝛾𝛼−

∫

𝛾⁻𝑇•

𝒦𝑓₊+𝑔

𝜓𝑑𝛾𝛼−

∫

Ω

∫

R³

𝑓˚𝜓(0)𝑑𝑣𝑑𝑥.

Note that, for each 0<𝑇 ∈𝐼_𝑇

•, at first only a trace of𝜒^𝑇𝑓 is defined. By uniqueness, for another 𝐼_𝑇

• 3 𝑇⁰ ≥ 𝑇, the traces of 𝜒^𝑇0𝑓 and 𝜒^𝑇𝑓 coincide on the common time interval[0, 𝑇]. Thus, we may write 𝑓_±, which is defined on all of𝐼_𝑇

•, and may drop the dependence on some𝑇.

Proposition 1.2.13. Let1 ≤ 𝑝 <∞,𝐹 satisfy Condition 1.2.11, and𝒦:𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

→ 𝐿^𝑝

lt

𝛾𝑇⁻_•, 𝑑𝛾𝛼

be a linear operator, that is local in time and such that there is a0 ≤ 𝑘₀ < 1 satisfying

k𝒦𝑐k_𝐿𝑝(_𝛾𝑇^−,𝑑_𝛾𝛼) ≤𝑘₀k𝑐k_𝐿𝑝(_𝛾𝑇^+,𝑑_𝛾𝛼) for all 𝑐 ∈ 𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

, 0 < 𝑇 ∈ 𝐼_𝑇

•. Furthermore, let 𝑓˚ ∈ 𝐿^𝑝 Ω×R³

and 𝑔 ∈ 𝐿^𝑝

lt

𝛾𝑇⁻_•, 𝑑𝛾𝛼

. Then:

(i) There is exactly one strong solution of (1.2.1)in the sense of Definition 1.2.12.

(ii) This solution is nonnegative if𝒦, 𝑓˚, and𝑔are nonnegative.

Proof. Let 0<𝑇∈𝐼_𝑇•and define

𝑔^𝑇₋:𝐷^𝑇₋ →R^{, 𝑔𝑇}−(𝑡, 𝑥, 𝑣)=

(𝑔(𝑡, 𝑥, 𝑣), (𝑡, 𝑥, 𝑣) ∈𝛾𝑇⁻, 0, otherwise;

𝑔₀:𝐷₀→R^{, 𝑔}0(0, 𝑥, 𝑣)= 𝑓˚(𝑥, 𝑣).

Note that the latter definition makes sense since, as mentioned above,𝐷₀ coincides with{0} ×Ω×R³up to a negligible set. We have

^𝑔

𝑇

−

_𝐿𝑝(𝐷^𝑇₋,𝑑𝜈⁻)= _𝜒𝑇𝑔

_𝐿𝑝(𝛾⁻,𝑑𝛾𝛼), ^𝑔₀

_𝐿𝑝(𝐷₀)=

𝑓˚ _𝐿𝑝(Ω×R³⁾

1.2 The Vlasov part 25 so that 𝑔₋^𝑇∈𝐿^𝑝 𝐷^𝑇₋, 𝑑𝜈^±

and𝑔₀∈𝐿^𝑝(𝐷₀). Furthermore, forℎ ∈𝐿^𝑝 𝐷₊^𝑇, 𝑑𝜈⁺ let ℎ:𝛾𝑇⁺• →R^{, ℎ(𝑡, 𝑥, 𝑣)=}

(ℎ(𝑡, 𝑥, 𝑣), (𝑡, 𝑥, 𝑣) ∈𝛾𝑇⁺, 0, otherwise and

𝔎𝑇ℎ:𝐷₋^𝑇→R^{, (𝔎𝑇}ℎ)(𝑡, 𝑥, 𝑣)=

( 𝒦ℎ

(𝑡, 𝑥, 𝑣), (𝑡, 𝑥, 𝑣) ∈𝛾𝑇⁻, 0, otherwise. Because of

k𝔎𝑇ℎk_𝐿𝑝(𝐷^𝑇₋,𝑑𝜈⁻)= ^{𝜒𝑇𝒦ℎ}

_𝐿𝑝(_𝛾𝑇⁻,𝑑𝛾𝛼) ≤𝑘₀ ^𝜒𝑇ℎ

_𝐿𝑝(_𝛾𝑇⁺,𝑑𝛾𝛼)=𝑘₀kℎk_𝐿𝑝(𝐷^𝑇₊,𝑑𝜈⁺) we conclude that 𝔎𝑇 maps 𝐿^𝑝 𝐷₊^𝑇, 𝑑𝜈⁺

to 𝐿^𝑝 𝐷₋^𝑇, 𝑑𝜈⁻

and has operator norm less than 1. Moreover,𝔎𝑇is local in time. Thus, by Proposition 1.2.10.(ii) there is a solution of (1.2.9) (with𝔎𝑇, 𝑔₀, 𝑔^𝑇₋ given). By uniqueness and𝐷₋^𝑇 ⊂ 𝐷₋^𝑇0 for𝑇 ≤ 𝑇⁰, such a solution (and its trace) does not depend on𝑇, whence there is a function 𝑓 such that 𝜒^𝑇𝑓 ∈ 𝐸^𝑝(Σ𝑇;𝑌)is the unique solution of (1.2.9) for any given𝑇. Now take𝜓 ∈ Ψ𝑇_•

and 0<𝑇 <𝑇_•such that𝜓(𝑡, 𝑥, 𝑣)=0 if𝑡 ≥𝑇. By Lemma 1.2.5 we have𝜓

Σ𝑇 ∈Φ^𝑌_𝑇. Applying the definition of trace and using the properties of𝜓, this leads to

∫

Σ𝑇•

𝑓 𝑌𝜓𝑑(𝑡, 𝑥, 𝑣)=

∫

Σ𝑇

𝑓 𝑌𝜓𝑑(𝑡, 𝑥, 𝑣)

=−

∫

𝐷₀

𝑓𝜓𝑑𝜈0+

∫

𝐷₊^𝑇

𝑓⁺𝜓𝑑𝜈⁺−

∫

𝐷₋^𝑇

𝔎𝑇𝑓⁺+𝑔^𝑇_{−𝜓𝑑𝜈}

−

=−

∫

Ω

∫

R³

𝑓˚𝜓(0)𝑑𝑣𝑑𝑥+

∫

𝛾⁺𝑇

𝑓⁺𝜓𝑑𝛾𝛼−

∫

𝛾𝑇⁻

𝔎𝑇𝑓⁺+𝑔₋^𝑇 𝜓𝑑𝛾𝛼

=−

∫

Ω

∫

R³

𝑓˚𝜓(0)𝑑𝑣𝑑𝑥+

∫

𝛾⁺𝑇•

𝑓⁺𝜓𝑑𝛾𝛼−

∫

𝛾𝑇•⁻

𝒦𝑓⁺+𝑔_𝜓𝑑𝛾

𝛼

and the proof of part 1.2.13.(i) is complete.

Part 1.2.13.(ii) follows from the fact that𝔎𝑇,𝑔₀, and𝑔^𝑇₋are nonnegative if𝒦,𝑓˚, and 𝑔are nonnegative, and Proposition 1.2.10.(ii).

We now turn to the special situation that𝒦=𝑎𝐾, where𝐾is the reflection operator.

According to (0.2), 𝐾 𝑓 is defined for any function 𝑓 (that is defined on a subset of R^×𝜕Ω×R³) and is self inverse, i.e., 𝐾² = id. Its restriction to 𝐿^𝑝

lt

𝛾𝑇⁺•

, 𝑑𝛾𝛼

yields an operator 𝐾: 𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

→ 𝐿^𝑝

lt

𝛾𝑇⁻_•, 𝑑𝛾𝛼

for any 1 ≤ 𝑝 ≤ ∞. Its inverse is 𝐾:𝐿^𝑝

lt

𝛾𝑇⁻_•, 𝑑𝛾𝛼

→𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

.

Lemma 1.2.14. For any𝑇 >0,1≤𝑝 <∞, and any𝑎, 𝑏 ∈ 𝐿^∞ 𝛾𝑇⁻

,𝑐∈ 𝐿^𝑝 𝛾𝑇⁺, 𝑑𝛾𝛼 , and ℎ ∈𝐿^𝑝 𝛾⁻𝑇, 𝑑𝛾𝛼

we have

∫

𝛾𝑇⁻

𝑏|𝑎𝐾𝑐+ℎ|^𝑝𝑑𝛾_𝛼=

∫

𝛾𝑇⁺

𝐾𝑏|𝑐𝐾𝑎+𝐾 ℎ|^𝑝𝑑𝛾_𝛼.

If additionally𝑎₀B ^{k𝑎k𝐿∞}(_𝛾𝑇⁻)<1and𝑎, 𝑏, 𝑐, ℎare nonnegative, the estimate

∫

𝛾𝑇⁻

𝑏|𝑎𝐾𝑐+ℎ|^𝑝𝑑𝛾𝛼 ≤ 𝑎₀

∫

𝛾⁺𝑇

(𝐾𝑏)𝑐^𝑝𝑑𝛾𝛼+ (1−𝑎₀)^1−𝑝

∫

𝛾𝑇⁻

𝑏 ℎ^𝑝𝑑𝛾𝛼

holds.

Proof. We compute

∫

𝛾𝑇⁻

𝑏|𝑎𝐾𝑐+ℎ|^𝑝𝑑𝛾𝛼

=

∭

𝛾𝑇⁻

𝑏(𝑡, 𝑥, 𝑣)|𝑎(𝑡, 𝑥, 𝑣)𝑐(𝑡, 𝑥, 𝑣−2(𝑣·𝑛(𝑥))𝑛(𝑥)) +ℎ(𝑡, 𝑥, 𝑣)|^𝑝|

b^𝑣𝛼·𝑛(𝑥)|𝑑𝑣𝑑𝑆_𝑥𝑑𝑡

=

∭

𝛾𝑇⁺

𝑏(𝑡, 𝑥, 𝑣−2(𝑣·𝑛(𝑥))𝑛(𝑥))|𝑎(𝑡, 𝑥, 𝑣−2(𝑣·𝑛(𝑥))𝑛(𝑥))𝑐(𝑡, 𝑥, 𝑣) +ℎ(𝑡, 𝑥, 𝑣−2(𝑣·𝑛(𝑥))𝑛(𝑥))|^𝑝|−

b^𝑣𝛼·𝑛(𝑥)|𝑑𝑣𝑑𝑆_𝑥𝑑𝑡

=

∫

𝛾𝑇⁺

𝐾𝑏|𝑐𝐾𝑎+𝐾 ℎ|^𝑝𝑑𝛾𝛼.

using the change of variables𝑣 ↦→𝑣−2(𝑣·𝑛(𝑥))𝑛(𝑥). Note that the determinant of the corresponding Jacobian equals−1 since the map is a reflection. As for the second statement, we estimate

∫

𝛾𝑇⁻

𝑏|𝑎𝐾𝑐+ℎ|^𝑝𝑑𝛾𝛼=

∫

𝛾⁺𝑇

𝐾𝑏|𝑐𝐾𝑎+𝐾 ℎ|^𝑝𝑑𝛾𝛼

≤

∫

𝛾⁺𝑇

𝐾𝑏

^{𝑎0𝑐+ (1−𝑎0)(1−𝑎0)}

−1𝐾 ℎ

𝑝

𝑑𝛾𝛼

≤𝑎₀

∫

𝛾𝑇⁺

(𝐾𝑏)𝑐^𝑝𝑑𝛾𝛼+ (1−𝑎₀)

∫

𝛾𝑇⁺

𝐾𝑏

^(1−𝑎0)⁻¹𝐾 ℎ

𝑝

𝑑𝛾𝛼

=𝑎₀

∫

𝛾𝑇⁺

(𝐾𝑏)𝑐^𝑝𝑑𝛾𝛼+ (1−𝑎₀)^1−𝑝

∫

𝛾𝑇⁻

𝑏 ℎ^𝑝𝑑𝛾𝛼

using the convexity of the𝑝-th power and the first statement.

1.2 The Vlasov part 27 Then there is a unique, nonnegative strong solution 𝑓 ∈ 𝐿^∞

lt 𝐼_𝑇•; 𝐿¹∩𝐿^∞

in the sense that the conditions of Definition 1.2.12 are satisfied for all1≤𝑝 <∞. Moreover, we have the estimates

(1−𝑎₀)^1𝑝 nonnegative𝐶¹-function𝜃=𝜃(𝑣)onR³that only depends on|𝑣|, is monotonically increasing in |𝑣|(i.e.,𝜃(𝑣) =𝜃˜(|𝑣|)for some monotonically increasing𝜃˜ ∈ 𝐶¹(R^≥0)with𝜃˜⁰(0)=0), linear and local in time. We have

k𝒦𝑐k_𝐿𝑝(_𝛾𝑇⁻,𝑑𝛾𝛼)=k𝑐𝐾𝑎k_𝐿𝑝(_𝛾𝑇⁺,𝑑𝛾𝛼) ≤𝑎₀k𝑐k_𝐿𝑝(_𝛾𝑇⁺,𝑑𝛾𝛼)

for all𝑐∈ 𝐿^𝑝

lt

𝛾𝑇⁺_•, 𝑑𝛾𝛼

and 0<𝑇 ∈𝐼_𝑇

•by Lemma 1.2.14. Thus, by Proposition 1.2.13 there is a unique solution for this 𝑝 in the sense of Definition 1.2.12. Since 𝑝 was arbitrary, it follows that, for all 1≤𝑝 <∞, 𝑓 ∈𝐿^𝑝

lt(Σ𝑇•)and 𝑓_± ∈𝐿^𝑝

lt

𝛾𝑇^±_•, 𝑑𝛾_𝛼

, and all conditions of Definition 1.2.12 are satisfied.

Let 0 < 𝑇 ∈ 𝐼_𝑇

• and recall that in the proof of Proposition 1.2.13 the solution on [0, 𝑇]was given by a solution of (1.2.9) with𝔎𝑇,𝑔₀,𝑔₋^𝑇. Thus, 𝑓₋ = 𝔎𝑇𝑓₊+𝑔^𝑇₋ =0 on 𝐷₋^𝑇\𝛾𝑇⁻. Applying Proposition 1.2.10.(i) (withℎ =0), dropping negligible terms in (1.2.8), and using Lemma 1.2.14 we arrive at

∫

altogether and therefore 𝑓₋ ∈ 𝐿¹ that it is nonnegative and only depends on|𝑣|since𝛽is nonnegative and only depends on|𝑣|. Proceeding similarly as before, we define 𝑓˜(𝑡, 𝑥, 𝑣)B𝛽𝜄(𝑣)𝑓(𝑡, 𝑥, 𝑣) ≥0,

1.2 The Vlasov part 29 Taking the limit𝜄→0 does not cause any problem because we have

_𝛽𝜄⁻_𝛽 finite. Hence, (1.2.14) holds with𝜄removed. Next we insert the definition of𝛽and drop the terms where|𝑣| ≥𝑅on the left-hand side to get

since𝜃is monotonically increasing in|𝑣|. Note that it is important that∇𝛽vanishes for |𝑣| > 𝑅. This proves (1.2.11) for 0 < 𝑅 < ∞. Because of ∇𝜃 ∈ 𝐿^𝑞 R³;R³

is a bounded function.

As for (1.2.13), let 0<𝑅≤ ∞and first derive the following key estimate:

∫

where we set𝑟B ∫

𝐵_𝑅𝑣⁰_𝛼𝑓(𝑇, 𝑥, 𝑣)𝑑𝑣¹₄

∈ [0,∞[and used (1.2.10); if𝑟=0, the second step makes no sense but clearly both the left-hand side and the right-hand side of (1.2.15) are zero in this case. Note that the integral on the left-hand side exists for almost all(𝑇, 𝑥) ∈𝐼_𝑇

• ×Ωby Fubini’s theorem and that the first estimate above holds trivially if𝑟 >𝑅by 𝑓 ≥0 and is an equality if𝑟≤ 𝑅. Taking both sides of (1.2.15) to the power ⁴₃ and then integrating overΩyields (1.2.13).

Remark 1.2.16. The𝐿^∞-spaces on𝛾𝑇^± with respect to𝑑𝛾𝛼 and the standard surface measure are the same and the respective norms coincide since null sets with respect to 𝑑𝛾𝛼 are null sets with respect to the standard surface measure and vice versa by b^𝑣𝛼·𝑛(𝑥)>0 (<0) on𝛾𝑇⁺(𝛾𝑇⁻). Consequently, from now on we will (mostly) not point out the measure in the denotation of such𝐿^∞-spaces and simply write𝐿^∞ 𝛾𝑇^±

.

Im Dokument The Relativistic Vlasov-Maxwell System with External Electromagnetic Fields (Seite 25-40)