The Dirichlet problem for − ∆ φ = e − φ in an infinite sector . Application to plasma equilibria

We consider here a nonlinear elliptic equation in an unbounded sectorial domain of the plane. We prove the existence of a minimal solution to this equation and study its properties.We infer from this analysis some asymptotics for the stationary solution of an equation arising in plasma physics. © 2014 Elsevier Ltd. All rights reserved. 1. Setting of the problem Solving elliptic PDE in unbounded domains of Rn such as half-spaces occurs naturally when using some blow-up argument to analyze the properties of a particular solution of a PDE in a bounded domain. We refer for instance to [1] where the analysis of the properties of the solution uε to ε1uε + f (uε) = 0, uε = 0 on the boundary (1.1) in a neighborhood of a point of the boundary, leads naturally to the study of an elliptic PDE in the half-space. The nonlinear elliptic PDE 1u + f (u) = 0, (1.2) has been widely studied in domains as half-spaces or cylindrical unbounded domains. We refer here to the articles [2–5] which have been instrumental for any later results concerning the symmetry and monotonicity properties of solutions. In the literature, there are various results concerning the properties of bounded solutions to these equations, mainly using consequences of themaximumprinciple as themoving planemethodor the slidingmethod (see [6,7], . . . ).We also refer to [8] where the properties of solutions in a quarter-space have been studied using tools from infinite-dimensional dynamical systems. ∗ Corresponding author. Tel.: +33 3 22 82 75 06. E-mail addresses: olivier.goubet@u-picardie.fr (O. Goubet), simon.labrunie@univ-lorraine.fr (S. Labrunie). http://dx.doi.org/10.1016/j.na.2014.08.015 0362-546X/© 2014 Elsevier Ltd. All rights reserved. 116 O. Goubet, S. Labrunie / Nonlinear Analysis 119 (2015) 115–126 Let us nowdescribe the equationwe are interested in. ConsiderΩ a sectorial domain ofR2 defined in polar coordinates as: Ω = {x(r, θ) ∈ R2; |θ | < θ0 ≤ π}. We shall sometimes denote this set as Ω[θ0] when we need to specify the opening. We are interested in the non-negative solutions to the problem: −1φ = e in Ω, φ = 0 on ∂Ω. (1.3) Our motivation here comes from the article [9], where the authors study stationary solutions to the Vlasov–Poisson system in a polygon and link them to those of a non-linear elliptic equation. The singular limit of the latter while some scaling parameter converges towards 0 leads to (1.3). Our aim in analyzing this equation is to provide more insight on the solutions to the original Vlasov–Poisson equation. More specifically, we shall look for two types of solutions to (1.3). Local variational solutions satisfy φ ∈ H1(O) for any bounded open set O ⊂ Ω . By the Trudinger inequality, this implies  O exp(φ2) < +∞, and thus e ∈ L2(Ω). In this case the Dirichlet condition holds in the sense of the usual trace theory: 


a b s t r a c t
We consider here a nonlinear elliptic equation in an unbounded sectorial domain of the plane. We prove the existence of a minimal solution to this equation and study its properties. We infer from this analysis some asymptotics for the stationary solution of an equation arising in plasma physics.

Setting of the problem
Solving elliptic PDE in unbounded domains of R n such as half-spaces occurs naturally when using some blow-up argument to analyze the properties of a particular solution of a PDE in a bounded domain. We refer for instance to [1] where the analysis of the properties of the solution u ε to ε u ε + f (u ε ) = 0, u ε = 0 on the boundary (1.1) in a neighborhood of a point of the boundary, leads naturally to the study of an elliptic PDE in the half-space.
The nonlinear elliptic PDE has been widely studied in domains as half-spaces or cylindrical unbounded domains. We refer here to the articles [2][3][4][5] which have been instrumental for any later results concerning the symmetry and monotonicity properties of solutions. In the literature, there are various results concerning the properties of bounded solutions to these equations, mainly using consequences of the maximum principle as the moving plane method or the sliding method (see [6,7], . . . ). We also refer to [8] where the properties of solutions in a quarter-space have been studied using tools from infinite-dimensional dynamical systems.
Let us now describe the equation we are interested in. Consider Ω a sectorial domain of R 2 defined in polar coordinates as: Ω = {x(r, θ ) ∈ R 2 ; |θ| < θ 0 ≤ π }. We shall sometimes denote this set as Ω[θ 0 ] when we need to specify the opening. We are interested in the non-negative solutions to the problem: − ϕ = e −ϕ in Ω, ϕ = 0 on ∂Ω. (1.3) Our motivation here comes from the article [9], where the authors study stationary solutions to the Vlasov-Poisson system in a polygon and link them to those of a non-linear elliptic equation. The singular limit of the latter while some scaling parameter converges towards 0 leads to (1.3). Our aim in analyzing this equation is to provide more insight on the solutions to the original Vlasov-Poisson equation.
More specifically, we shall look for two types of solutions to (1.3). Local variational solutions satisfy ϕ ∈ H 1 (O) for any bounded open set O ⊂ Ω. By the Trudinger inequality, this implies  O exp(ϕ 2 ) < +∞, and thus e −ϕ ∈ L 2 (Ω). In this case the Dirichlet condition holds in the sense of the usual trace theory:  Ω ∇ϕ · ∇v =  Ω e −ϕ v, ∀v ∈ H 1 0 (Ω) with bounded support.  Ω e −ϕ v, ∀v ∈ H 2 ∩ H 1 0 (Ω) with suitable bounded support. (1.5) The trace is defined in a very weak sense on any bounded subset of each side of ∂Ω, by an immediate generalization of [10].
Anyway, as we are interested in non-negative solutions, there automatically holds e −ϕ For both types of solutions, there obviously holds: Since 0 is a subsolution to the problem (1.3), the existence of a solution is equivalent to that of a non-negative supersolution to the problem. We will develop this in the sequel. Furthermore, any non-negative solution to (1.3) in Ω[θ 0 ] is a supersolution in a smaller sector Ω[θ 1 ], θ 1 < θ 0 . Therefore, the existence of a solution in the split plane Ω[π ] implies the solvability of (1.3) in any sector. By symmetry, it is enough to solve the mixed Dirichlet-Neumann problem in the upper half-plane (1.6) Glueing this ϕ * to its even reflection with respect to the axis [θ = 0] yields a solution to (1.3) on Ω[π ].

Remark 1.2.
It is worth pointing out that there exists no (non-negative, very weak) solution to (1.3) in the whole plane.
The article is written as follows. In Section 2 we construct a supersolution to the equation in the split plane. For this purpose we use a constructive method which relies on complex analysis. In Section 3 we discuss some properties, such as monotonicity, symmetry or regularity, of the minimal positive solution; this minimal solution is relevant for the Physics of the original problem. In Section 4 we prove the non-uniqueness of solutions and list some of their properties. Eventually, we discuss the application to the stationary Vlasov-Poisson system in last section. In this last section, we also show the link between this asymptotic and the boundary blow-up solutions for u = e u , see [11][12][13][14][15]. Boundary blow-up (or large) solutions were introduced in the seminal articles [16,17].

Construction of a solution in the split plane
We begin with a construction inherited from complex analysis. It is worth pointing out that this construction method works for any sectorial domain. Let z 2 = Φ(z 1 ) or z 1 = Ψ (z 2 ) be a conformal mapping between the complex z 1 and z 2 -planes, and let D 1 , D 2 be two domains conformally mapped to one another. Suppose we are given two functions w 1 , w 2 on D 1 and D 2 respectively, which are transformed into each other by the formulas: w 2 (z 2 ) = log |Ψ ′ (z 2 )| + w 1 (Ψ (z 2 )); (2.1) w 1 (z 1 ) = log |Φ ′ (z 1 )| + w 2 (Φ(z 1 )). (2.2) Using ∆ = 4 ∂ z ∂z and the fact that the logarithm of the modulus of an analytic function is harmonic, one easily checks the following lemma. Lemma 2.2. Let ∆ i , i = 1, 2 be the Laplacian in the z i -plane, and let w 1 , w 2 be related by (2.1) or (2.2). Then, w 1 satisfies ∆ 1 w 1 = 4e 2w 1 in D 1 if, and only if, w 2 satisfies ∆ 2 w 2 = 4e 2w 2 in D 2 .
In this section, we denote (x i , y i ) and (r i , θ i ) the Cartesian and polar coordinates in the z i -plane. We choose D 1 as the upper half-plane [0 < θ 1 < π] = [y 1 > 0], and we introduce the mixed Dirichlet-Neumann problem . Consider now the conformal mapping The half-plane is mapped by Φ onto the disk D 2 centered at 1 2i and of radius 1 2 ; the negative and positive real half-axes Γ D it is worth observing that this function is non-negative and singular at z 2 = 0 (corresponding to z 1 = ∞) only. On the Neumann half-circle Γ N 2 , we compute as follows. In polar coordinates, we have: Parametrizing the boundary as , we observe that: (cos ϑ, sin ϑ) = (− sin 2θ 2 , cos 2θ 2 ).
Proof. Computing as above, we see that u 2 solves the problem and u 2 is bounded above on supp v. On the other hand, the variational formulation of (2.7) reads is bounded above on the support of v. One finds as usual (w k . By the monotone convergence theorem, w k 2 → w 2 and e 2w k 2 → e 2w 2 in L p (K), for any compact K ⊂ D 2 \ {0} and p < ∞. This implies ∆ 2 w 2 = 4e 2w 2 in the sense of distributions in D 2 . Furthermore, one can consider boundary conditions in a very weak sense [10]. Summing up, w 2 solves the problem: Using the conformal mapping Φ, the function w 1 defined

The minimal solution
From the previous section, we have constructed a solution ϕ * to (1.6), which satisfies the bounds (cf. (2.9)): Thus, ϕ * , extended by reflection to Ω[π ], is bounded on any bounded subset of Ω[π ]. We will provide a better version of this upper bound in the sequel.

A limiting process
Consider the truncated domain The proof of the following result is standard.

Lemma 3.1. There exists a unique variational solution u R to the problem
This solution is positive and is symmetric with respect to θ  → −θ .
Let us observe that, for R ′ ≥ R, both u R ′ and ϕ * are supersolutions to (3.2). Therefore, for any Passing to the limit while R goes to the infinity, setting u(x) = sup R u R (x) we construct this way a solution u to the original problem. This solution is symmetric with respect to θ = 0. Moreover, u is the minimal solution, in the sense that any (non-negative) local variational solution ϕ to (1.3) satisfies u ≤ ϕ; ( 1 ) therefore, u is unique. Similarly, any non-negative supersolution ϕ of local variational regularity satisfies u ≤ ϕ. From (3.1), we deduce that there exists C ≥ 0 such that We can compute the minimal solution in the half-plane Ω[π /2], which reduces to a 1D problem on the half-line, that is 2 log bounded by above in that case.

Regularity
We introduce the following function spaces on a domain O: In the first two lines, O is a bounded Lipschitz domain. The second characterization of N(O), proved in [10] for polygons, can be extended to curvilinear polygons 2 such as Ω R , whose boundary is composed of smooth sides that meet at corners. To express the regularity of solutions and give asymptotic expansions, we shall generally make use of the parameter: We introduce the well-known primal and dual harmonic singularities in Ω: From [10], we know the following facts. • If the angle at the tip of the sector is salient or flat, then: , for all s < 1 + α; any w ∈ Φ 2 admits the regular-singular decomposition: As a first step, we prove that the minimal solution is a local variational solution. 1. If the sector is salient of flat (θ 0 ≤ π /2), u ∈ C 1 (Ω R ).
To prove Λ > 0, we return to the solution u R on Ω R . Let Λ R be its singularity coefficient; it controls the dominant behavior of u R near the corner, u R ∼ Λ R S(r, θ ) as r → 0, uniformly in θ [9]. It is given by the formula [10]: By the monotone convergence theorem, we know that , and (as in Proposition 3.5), χ B/2 u R → χ B/2 u ∈ Φ 2 (Ω B ). As the singularity coefficient is a continuous linear form on Φ 2 , one infers Λ R → Λ.

Monotonicity
We can now prove a monotonicity property. Standard results in the literature assert that bounded solutions to this type of PDE have some monotonicity properties. Our minimal solution is not bounded, see Remark 3.2.
Proof. Consider the half-domain Ω + R := [0 < θ < θ 0 and r < R]. The function v := ∂ θ u R is solution to and v ≤ 0 on the boundary. Actually, v = 0 on the lower ray [θ = 0] and on the arc of circle [r = R], while v ≤ 0 on the upper ray [θ = θ 0 ]. By the maximum principle, 4 v < 0 in Ω + R . The results follow promptly.
Remark 3.13. We reckon that the map θ  → u(r, θ ) is concave, but we do not have a proof of this fact.

Non-uniqueness
For any µ = (µ − , µ + ) ∈ (R + ) 2 , the function H µ := µ − S * + µ + S is non-negative and harmonic in Ω, and vanishes on the boundary: it is a subsolution to (1.3). We shall construct a solution in the form ϕ µ = H µ + v µ . If such a solution exists, The corresponding problem in the truncated domain and passing to the limit we obtain a solution to (4.1). Thus we have constructed a solution ϕ µ = H µ + v µ to (1.3), which is bounded as Because of (3.3), there holds if µ − > 0, ϕ µ (r, θ ) ∼ 1 π µ − r −α cos(αθ ) as r → 0, θ ̸ = ±θ 0 fixed; if µ + > 0, ϕ µ (r, θ ) ∼ µ + r α cos(αθ ) as r → +∞, θ ̸ = ±θ 0 fixed. Thus, solutions corresponding to different µ are distinct. They are local variational solutions if µ − = 0, and very weak solutions if µ − > 0 and the sector is reentrant (α < 1). In a flat or salient sector (α ≥ 1) the solutions corresponding to µ − > 0 are not L 2 in a neighborhood of the origin, thus they do not qualify as very weak solutions. Remark 4.1. One may wonder if there exists a solution to (1.3) that is not non-negative. Using the maximum principle in unbounded domains of R 2 (see [4]) we can prove that any solution of (1.3) that is bounded from below is non-negative.

Local regularity near the tip of the sector
Using the tools of Proposition 3.5 (localization, bootstrapping, and regularity theory for the linear Poisson-Dirichlet problem), one obtains the following results. 1. Any solution to (1.3) belongs to C ∞ (K), for any compact subset K ⊂ Ω such that the origin 0 ̸ ∈ K.
The second term belongs to H s (Ω R ) for s < 1 + α, and the first is treated as above.
Proof. We only prove the last claim; the others are similar. By definition, a variational solution belongs to H 1 (Ω R ) for all finite R; computing as in (3.6) we see that χ B ϕ ∈ Φ 2 (Ω 2B ), hence the decomposition. As any solution ϕ ≥ 0 is larger than the minimal solution u, its singularity coefficient λ is larger than that of u, namely Λ. If ϕ only is a very weak solution, one finds ∆( As B is arbitrary, we deduce that for any R ϕ(r, θ ) = λ * S * (r, θ ) +φ(r, θ ),φ ∈ H 1 (Ω R ).

Unboundedness
We know from Remark 3.2 that the minimal solution in a flat or reentrant sector is unbounded, hence all non-negative solutions are unbounded. Actually, the same holds in a salient sector.
 .  This solution can be computed explicitly: On the other hand, ϕ is a supersolution to (4.5), and ϕ ≥ ϕ on D. In particular which is (4.4).

Application to plasma equilibria
In [9] the authors study the properties of the solution φ to the problem in a bounded polygonal (or curvilinear polygonal) domain Υ 1 of R 2 . The mass parameter M is a data of the problem; the normalization factor κ is an unknown which ensures that the constraint  Υ 1 ρ = M is satisfied. The external potential φ e is fixed; it belongs to L ∞ (Υ 1 ).
We shall need the following assumptions. 2. If the domain Υ 1 has reentrant corners, it is contained in its local tangent cone at any of them.
The second condition is automatically satisfied when Υ 1 is a straight polygon with only one reentrant corner: then, it can be described as Υ 1 = Υ C 1 \ (R 2 \ Ω) ⊂ Ω, where Υ C 1 is the convex envelope of Υ 1 , and Ω is the tangent cone at the reentrant corner.
We say that the domain Υ 1 is regular if it has no reentrant corner, i.e., it is smooth where it is not convex and convex where it is not smooth. Otherwise, it is singular. To simplify the exposition, we shall only consider singular domains with one reentrant corner of opening 2θ 0 = π /α, 1/2 < α < 1, and such that the two sides that meet there are locally straight.
In that case, the decomposition of φ with respect to regularity writes as in Section 3.2: φ =φ + λ χ (r) r α cos(αθ ),φ ∈ H 2 ∩ H 1 0 (Υ 1 ), (5.2) assuming that the reentrant corner is located at 0 and the axes are suitably chosen. Notice, however, that these conditions are not essential as long as assumption (2) above is satisfied. If the sides are not locally straight, the expression of the singular part is modified; the case of multiple reentrant corners is treated by localization. We shall generally discuss the cases of regular and singular domains together; statements about singularity coefficients are void for a regular domain. The goal of this section is to study the behavior of the coefficients κ and λ as M → +∞. In [9] it was proved that: M κ → 0 and λ κ → 0 and λ → +∞.
We want to refine the estimates (5.3) by obtaining explicit growth rates.

Getting rid of the external potential
For a fixed external potential φ e , the parameters M and κ are strictly increasing functions of each other [9]. Thus, the problem (5.1) can be parametrized by κ, even though M is the significant variable. For fixed κ and φ e , we provisionally Similarly, we know [9] that for two given external potentials (φ 1 e , φ 2 e ) and normalization factors (κ 1 , κ 2 ) the corresponding solutions (φ 1 , φ 2 ) and their singularity coefficients satisfy Therefore, we shall mostly concentrate on the case φ e = 0. Using the physical scaling κ = ϵ −2 , with ϵ → 0, we shall denote by φ ϵ the solution to (5.1) with φ e = 0, i.e.: the corresponding mass will be written M ϵ ; and λ ϵ is the singularity coefficient of φ ϵ .

Limit of φ ϵ and λ ϵ
Now we assume that the domain Υ 1 is singular, and we set the origin at the reentrant corner. First, we prove by some blow-up argument: Proposition 5.1. For any ξ in the unbounded sectorial domain Ω, the rescaled function v ϵ (ξ) = φ ϵ (ϵξ), defined for ϵ small enough, converges towards u(ξ), where u is the minimal solution to (1.3) in Ω.
Proof. Let Γ 1 be the union of the two sides that meet at the reentrant corner 0, and R the radius of the circle centered at 0 that is tangent to ∂Υ 1 \ Γ 1 ; in other words we consider R to be the maximum of r such that Ω r = Ω[θ 0 ] ∩ {|x| < r} is included in Υ 1 .
We now perform the blow-up argument by a dilation of factor 1 ϵ . We obtain that v ϵ is the solution to the following problem in the dilated domain Υ 1  (5.8) and the result follows promptly from the results of Section 3.1.

Limit of M ϵ
We now state and prove a result that complements the numerical evidence in [9].
Proposition 5.2. Let |∂Υ 1 | be the perimeter of Υ 1 . When ϵ → 0, there holds: Proof. Consider φ ϵ solution to (5.6). The main idea is to write M ϵ =  Υ 1 − φ ϵ = −  ∂Υ 1 ∂ n φ ϵ , and to derive an equivalent to ∂ n φ ϵ as ϵ → 0. Unfortunately, in a non-smooth domain this equivalent is not uniform along ∂Υ 1 , and cannot be straightforwardly integrated on this boundary; thus some technicalities are needed to overcome the difficulty.
For almost every point x 0 on the boundary of Υ 1 (except the corners), we have the interior and exterior sphere condition: there is a (small) ball B that is included in Υ 1 (resp. R 2 \ Υ 1 ) and tangent at x 0 . We begin with a lower bound for M ϵ . Fix η > 0 small enough. We consider ∂Υ 1,η the set of x 0 ∈ ∂Υ 1 such that there exists a ball B of radius η, included in Υ 1 and tangent at x 0 .
If Υ 1 is a straight polygon and 2θ c ∈ (0, π ) is the angle at the corner c, the maximum radius of an interior sphere tangent at a nearby point x 0 is |x 0 − c| tan θ c ; see Fig. 1 left. Elsewhere, this radius is bounded below by a constant. We deduce that there exists a constant K such that |∂Υ 1,η | ≥ |∂Υ 1 | − K η. This extends to a curvilinear polygon by diffeomorphism; anyway, |∂Υ 1 | − |∂Υ 1,η | → 0 as η → 0. Fix x 0 ∈ ∂Υ 1,η , and let B be the ball defined above. Solving the equation v = e v in B, v = −2 log ϵ on ∂B, (5.13) we easily find that w ϵ is a subsolution to (5.13), so w ϵ (x) ≤ v(x) for any x in B, and then ∂ n w ϵ (x 0 ) ≥ ∂ n v(x 0 ).
We now compute the normal derivative of the solution to v = e v in a given ball. Up to a translation assume that the ball is centered at the origin. The solution is radially symmetric, i.e. we solve (rv ′ (r)) ′ = re v(r) , (5.14) for r < η; and ∂ n v = v ′ (η) ≥ 0. Multiplying this equation by rv ′ (r) and integrating between 0 and η, we find: re v(r) dr = η 2 ϵ 2 − 2ηv ′ (η), (5.15) using once more re v(r) = (rv ′ (r)) ′ and the fact that v = −2 log ϵ on the boundary. We infer from this equality: v ′ (η) ≥ 2ηϵ −2  2η 2 ϵ −2 + 2 . (5.16) We then have  We then let η go to zero to obtain the lower bound.
We now proceed to the upper bound. If x 0 is not a corner, we have another small ball B 1 included in R 2 \Υ 1 that is tangent at x 0 . Introduce another ball B 2 , with the same center as B 1 , and large enough to have Υ 1 ⊂ B 2 . Then consider the annulus N = B 2 \ B 1 that contains Υ 1 , and solve the boundary-value problem: v = e v in N, v = −2 log ϵ on ∂N. The solution v ≤ −2 log ϵ, thus it appears as a subsolution to (5.12), and v ≤ w ϵ in Υ 1 ; at x 0 , we have ∂ n v ≥ ∂ n w ϵ .
The rest of the proof amounts to compute the normal derivative of the solution to v = e v in an annulus of radii a < b. Once again, the solution is radially symmetric, i.e. we solve (5.14) for a < r < b. In the case of (5.19), we know that there exists c ∈ (a, b) such that v ′ (c) = 0, and that v ′ (a) ≤ 0. Multiplying (5.14) by rv ′ (r) and integrating between r = a and c, ϵ . This estimate provides us with a sharp bound of ∂ n w, but it blows up near the reentrant corner in the singular case: in this case a → 0. From Fig. 1 middle, we see that the maximum radius of an exterior sphere tangent at x 0 is |x 0 |/| tan θ 0 |, if the reentrant corner is located at 0.
We overcome this difficulty as follows. For η > 0, consider a mollified/rounded domainΥ 1 = Υ 1 \ B(0, η), see Fig. 1 right. The maximum radius of the exterior sphere is equal to η on the rounded part of the boundary. As shown above, it is at least η/| tan θ 0 | on the remaining part of the sides that meet at the reentrant corner; elsewhere, it is bounded below by a constant.