On a nonlinear parabolic problem arising in the quantum diffusive description of a degenerate fermion gas
aa r X i v : . [ m a t h . A P ] J un ON A NONLINEAR PARABOLIC PROBLEM ARISINGIN THE QUANTUM DIFFUSIVE DESCRIPTIONOF A DEGENERATE FERMION GAS ∗ LUIGI BARLETTI † AND
FRANCESCO SALVARANI ‡ Abstract.
This article studies, both theoretically and numerically, a nonlinear drift-diffusionequation describing a gas of fermions in the zero-temperature limit. The equation is considered on abounded domain whose boundary is divided into an “insulating” part, where homogeneous Neumannconditions are imposed, and a “contact” part, where non-homogeneous Dirichlet data are assigned.The existence of stationary solutions for a suitable class of Dirichlet data is proven by assuminga simple domain configuration. The long-time behavior of the time-dependent solution, for morecomplex domain configurations, is investigated by means of numerical experiments.
Key words. quantum drift-diffusion, fermions, nonlinear parabolic equations, mixed boundaryconditions
AMS subject classifications.
1. Introduction.
The derivation of quantum fluid equations from quantum ki-netic equations [5, 17, 23] is a natural problem in quantum statistical mechanics, in ex-actly the same way as the derivation of classical fluid equations from Boltzmann equa-tion is a standard topic in classical statistical mechanics [2, 3, 4, 8, 9, 11, 12, 14, 15].It allows, indeed, to clarify the relationships between two levels of descriptions and toobtain “quantum corrections” to the classical fluid equations, that are difficult (if notimpossible) to identify only on the grounds of physical intuition. This is particularlytrue in the case of a quantum system of identical particles, i.e. obeying either theBose-Einstein or the Fermi-Dirac statistics [6, 23].Quantum fluid equations, whose prototype are the Madelung equations [20], arenot only important from a theoretical point of view, but they are very interestingalso for applications, in particular in semiconductor devices modelling [5, 17, 16, 18].Indeed, the fluid description of a quantum system has many practical advantages: firstof all, it provides a description in terms of macroscopic variables with a direct physicalinterpretation (such as density, current, temperature); moreover, it allows to modelopen systems in a very natural way, by imposing suitable semiclassical boundaryconditions.In this article, we study a very specific model, namely the diffusive equation for adegenerate (i.e. at zero absolute temperature) gas of fermions in two space dimensions.The two-dimensional case is rather peculiar, since the third-order “quantum pressure”term, remarkably, vanishes [6, 23]. The resulting diffusive equation have hence theform of a purely semiclassical equation [19], although being (formally) exact up toorder ~ . Such equation (after a suitable rescaling of variables) reads as follows [6, 23]:(1) u t = ∇ · ( u ∇ ( u + V )) , ∗ This paper has been supported by INDAM (GNFM project “Quantum fluid-dynamics for systemsof identical particles: analytical and numerical study”), by the ANR project number ANR-14-ACHN-0030-01
Kimega and by the bilateral programme “Galileo”, project number G14-34. † Dipartimento di Matematica e Informatica “U. Dini”, Universit`a degli Studi di Firenze, Italy(luigi.barletti@unifi.it). ‡ CEREMADE, UMR CNRS 7534, Universit´e Paris-Dauphine, PSL Research University,France & Dipartimento di Matematica “F. Casorati”, Universit`a degli Studi di Pavia, Italy([email protected]). 1
L. BARLETTI AND F. SALVARANI where V is a given potential and ∇ = ( ∂/∂x , ∂/∂x ), and has therefore the form ofa two-dimensional “porous media” equation [24] endowed with a drift term. In thispaper we choose to work on Equation (1), not only because of its particularly neatform, but also because the use of two-dimensional models is natural in many instancesof semiconductor devices [17, 21].The mathematical study of Equation (1) has been partially developed in theliterature. In particular, existence and uniqueness of a weak solution in the evolu-tionary case has been proven in [1] and in [13], whereas an analysis of the long-timebehaviour has been provided in [10] and [7]. In [1], the time-dependent equation isendowed with mixed homogeneous Neumann and non-homogeneous Dirichlet bound-ary data; in [10], also the stationary equation is considered, but only in the case ofhomogeneous Neumann boundary data.However, a semiconductor device cannot be fully described by imposing onlyhomogeneous Dirichlet boundary conditions. Indeed, for describing real situations,a portion of the boundary (corresponding to metallic contacts) should be describedby using non-homogeneous Dirichlet data, whereas other regions of the boundary(corresponding to insulating boundaries) should be supplemented with homogeneousNeumann conditions [17, 21].The main aim of this article is hence to investigate, both theoretically and nu-merically, some aspects of Equation (1) by allowing both non-homogeneous Dirichletboundary data and homogeneous Neumann conditions.The structure of the article is the following. In the next section we will formulatethe mathematical problem and review some basic results on the evolutionary case.Then, in section 3, we will study the stationary case. Particular attention will begiven to the situation in which the domain is rectangular and both the data and thepotential depend only on one space variable, in which case solutions that depend onlyon that variable can be considered. A theorem of existence and uniqueness for suchsolutions, requiring some restrictions on the Dirichlet data, will be proven. Finally,in section 4, we will describe a numerical procedure for studying the initial-boundaryvalue problem and provide some numerical experiments.
2. The mathematical problem.
Let Ω ⊂ R be a bounded domain, with apiecewise smooth boundary ∂ Ω, and let Γ D and Γ N be two non-empty subsets of ∂ Ω,such that Γ D ∩ Γ N = ∅ and ∂ Ω = Γ D ∪ Γ N . The gas of fermions in the zero-temperature limit is described by a density function u : R + × Ω → R + , whose time evolution is governed by the following nonlineardrift-diffusion equation [6]:(2) ( u t + ∇ · J = 0 ,J = − u ∇ ( u + V ) , ( t, x ) ∈ R + × Ω . Here V : Ω → R is a given potential which, for the sake of simplicity, will be assumedto be continuously differentiable. The problem is supplemented with the initial datum(3) u (0 , x ) = u in ( x ) , x ∈ Ωand mixed boundary conditions(4) u ( t, x ) = u D ( x ) , ( t, x ) ∈ R + × Γ D J ( t, x ) · n x = 0 , ( t, x ) ∈ R + × Γ N , NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS u in ( x ) > x ∈ Ω, u D ( x ) > x ∈ Γ D and n x is the outwardnormal to Γ N , at x ∈ Γ N .Equation (2), with non-negative mixed Neumann/Dirichlet boundary conditions(3) and (4) has been studied in several works, and its main properties have beenanalyzed. In particular, existence, uniqueness and preservation of the cone of non-negative functions for problem (2)–(3)–(4) are the specialization to the uncoupledcase (without reaction terms) of the results obtained in Ref. [13]. More precisely, byadapting to our case the theorems proven in sections 2 and 3 of Ref. [13], we can statethe following: Theorem 1.
Let u in ∈ L p (Ω) , u D ∈ L p (0 , T ; W ,p (Ω)) ∩ L ∞ ((0 , T ) × Ω) , p ≥ ,and ∂ t u D ∈ L (0 , T ; L ∞ (Ω)) , with non-negative u in and u D . Then, there exists aunique non-negative weak solution u ∈ L p (0 , T ; W ,p (Ω)) + L p (0 , T ; Ξ) of the initial-boundary value problem (2) – (3) – (4) , where Ξ = { ξ ∈ W ,p (Ω) : ξ = 0 on Γ D } . Another approach for proving this result consists in transforming Equation (2) intothe equation(5) ∂ t b ( u ) = 12 ∆ u + ∇ · ( b ( u ) ∇ V )by means of the transformatiion(6) b ( z ) := sign( z ) p | z | , i.e. z = u sign( u ) , and then by using the existence and uniqueness theory of Alt and Luckhaus, whichstudied the class of nonlinear parabolic equations written above, with boundary con-ditions of type (4), in [1].In the next section we shall investigate the stationary case and prove (at least for aparticular class of initial/boundary conditions) a theorem of existence and uniquenessof the stationary solution. Despite of the general result on the evolutionary equation,we are able to prove a sufficient condition which guarantees the well-posedness of thestationary problem only for a restricted class of “supercritical” Dirichlet data (seeTheorem 11).It is worth to remark that the well-posedness of the initial value problem forEquation (2) with only Neumann conditions (Γ D = ∅ ) has been proven by Carrilloet Al. [10]. Then, the origin of troubles is clearly in the non-homogeneous Dirichletconditions.
3. The stationary problem.
The study of non-negative stationary solutionsof Equation (2), satisfying the prescribed mixed boundary conditions (4), leads toconsider the problem(7) ∇ · [ u ∇ ( u + V )] = 0 , x ∈ Ω , with boundary conditions(8) u ( x ) = u D ( x ) , x ∈ Γ D ,u ∇ ( u + V ) · n x = 0 , x ∈ Γ N . L. BARLETTI AND F. SALVARANI
A peculiar feature ofthe stationary problem is the lack of uniqueness of the stationary solution, as shownin the following proposition.
Proposition 2.
Let us suppose that u D ≡ . If u ∈ H (Ω) is a non-negativeweak solution of the boundary value problem (7) - (8) , then, for every x ∈ Ω , either u = 0 or ∇ ( u + V ) = 0 in x .Proof . We multiply Equation (7) by u + V and then integrate with respect to x in Ω. After integrating by parts, we obtain that Z Ω u |∇ ( u + V ) | d x = 0 , because of the boundary conditions (8) with u D = 0. Since u ≥
0, by assumption,the statement follows.This simple result shows that, depending on V , the homogeneous stationary prob-lem may have non-unique solution. For example, if V ≤ K ⊂ Ω, then both u ≡ u ( x ) = ( , x ∈ Ω \ K, − V ( x ) , x ∈ K are non-negative solutions of (7). Moreover, if Γ D = ∅ , then u = γ − V is a strictlypositive solution for all constants γ such that γ > V ( x ) for all x ∈ Ω (in particular, if V is allowed to go to + ∞ , no positive solution may exist at all). We now examine the sta-tionary problem in a particular case, which is basically one-dimensional. Let Ω =(0 , × ( a, b ) ⊂ R , letΓ D = { , } × [ a, b ] , Γ N = [0 , × { a, b } , and let u = u ( x , x ), so that the Dirichlet boundary conditions read as follows: u (0 , x ) = u ( x ) , u (1 , x ) = u ( x ) , x ∈ [ a, b ] , and the Neumann conditions are imposed on the two other sides of the rectangle.Moreover, let us assume that ∂V∂x = ∂u ∂x = ∂u ∂x = 0 , i.e. both the potential V and the Dirichlet data u D only depend on the variable x .Then, clearly, u ( x , x ) = u ( x ) is a solution of the stationary problem provided that u ( x ) solves the one-dimensional problem(9) ddx (cid:20) u ( x ) ddx ( u ( x ) + V ( x )) (cid:21) = 0 , x ∈ [0 , ,u (0) = u , u (1) = u , where we recall that V : [0 , → R is assumed to be continuously differentiable. Weare going to prove that problem (9) has a unique positive solution for positive Dirichletdata, u > u >
0, subject to suitable restrictions (see Theorem 11).
NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS x the one-dimensional variable x ; then, recall that x ≡ x . From (9) we immediately obtain that a constant c ∈ R exists such that(10) u ( x ) [ u ′ ( x ) + V ′ ( x )] = c, x ∈ [0 , c = 0 corresponds to allsituations in which either u ( x ) = 0 or u ′ ( x ) = − V ′ ( x ), for some x ∈ [0 , c = 0 and, consequently, we are forced to consider strictly positive Dirichlet data. If c = 0, then any regular solution of (10) cannot vanish in [0 ,
1] and then we can assume u to be strictly positive in [0 ,
1] and write the differential equation in the normal form(11) u ′ ( x ) = cu ( x ) − V ′ ( x ) . Let us first consider the case of constant V ′ , for which we have an explicit represen-tation of the solution of the Cauchy problem for the ODE (11). We denote by W : [ − /e, + ∞ ) → [ − , + ∞ ) and ˜ W : [ e, + ∞ ) → [1 , + ∞ ) , respectively, the inverse functions of f ( x ) = xe x , and g ( x ) = e x /x ( W is known as the Lambert W-function [22]). It is readily proven that the functions W and ˜ W satisfythe differential equations(12) W ′ ( y ) = W ( y ) y (cid:0) W ( y ) + 1 (cid:1) and ˜ W ′ ( y ) = ˜ W ( y ) y (cid:0) ˜ W ( y ) − (cid:1) . Lemma 3.
Let u > , c > and α ∈ R . Then, the solution of the Cauchyproblem (13) u ′ ( x ) = cu ( x ) − α, u ( x ) = u , is given by u ( x ) = φ ( x − x , u , c, α ) , where (14) φ (∆ x, u , c, α ) = cα n W h(cid:16) αu c − (cid:17) e − α c ∆ x + αu c − io , if α > , q u + 2 c ∆ x, if α = 0 , cα (cid:26) − ˜ W (cid:20)(cid:16) − αu c (cid:17) − e α c ∆ x +1 − αu c (cid:21)(cid:27) , if α < .Moreover, for fixed x > x , we have that φ ( x, u , c, α ) is strictly increasing with respectto u and c , and strictly decreasing with respect to α .Proof . The fact that φ satisfies (13) comes straightforwardly from (12). Then wejust prove the growth properties with respect to α and c . Although they could bechecked directly on expression (14), it is easier to use the differential equation satisfiedby φ , i.e.(15) ∂φ∂x = cφ − α. L. BARLETTI AND F. SALVARANI
Defining ψ = ∂φ/∂c we obtain(16) ∂ψ∂x = − c ψφ + 1 φ and, since φ | x =0 = u for all α ∈ R , we also have(17) ψ | x =0 = 0 . If x ∈ (0 ,
1] existed such that ψ ≤ , x ), then from (16) one would obtain ∂ψ∂x ≥ φ > . in (0 , x ) (where the fact that φ > ψ > , x ). Then ψ > , φ is strictly increasing with respect to c > φ is strictly decreasing with respect to α ∈ R can be carried outanalogously.In addition to the properties listed in Lemma 3 let us also note that φ (∆ x, u , c, α ),as ∆ x increases from 0 to + ∞ :(i) increases monotonically from u to + ∞ , if α ≤ u to the asymptotic value c/α , if α > u < c/α ;(iii) is identically equal to c/α , if α > u = c/α ;(iv) decreases monotonically from u to the asymptotic value c/α , if α > u > c/α .Lemma 3 will allow us to prove properties of the more general differential equation(11) by approximating V ′ ( x ) with piecewise constant functions. Lemma 4.
Let x = 0 < x < · · · < x n = 1 be a partition of [0 , and let V ′ n ( x ) be a piecewise constant function, taking the value α i in the interval ( x i , x i +1 ) , suchthat V ′ n → V ′ uniformly in [0 , , as n → + ∞ . Let u ( x ) be the solution of the Cauchyproblem (18) u ′ ( x ) = cu ( x ) − V ′ ( x ) , x ∈ [0 , , u (0) = u > , with c > , and let v n ( x ) be defined by v n ( x ) = φ ( x, u , c, α ) , x ∈ [0 , x ] ,v n ( x ) = φ ( x − x i , v n ( x i ) , c, α i ) , x ∈ ( x i , x i +1 ] , i = 1 , , . . . , n − (in other words, v n solves v ′ n ( x ) = cv n ( x ) − α i in [ x i , x i +1 ] , taking as initial valuein each interval the final value of the preceding interval, starting with u in the firstinterval). Then v n → u uniformly in [0 , , as n → + ∞ .Proof . Note that u and v n are continuous solutions of the integral equations, u ( x ) = u + Z x cu ( y ) dy − V ( x ) + V (0)and v n ( x ) = u + Z x cv n ( y ) dy − V n ( x ) + V (0) , NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS V n ( x ) := V (0) + R x V ′ n ( y ) dy is a piecewise linear ap-proximation of V ( x )). Then, | u ( x ) − v n ( x ) | ≤ c Z x | u ( y ) − v n ( y ) || u ( y ) v n ( y ) | dy + | V ( x ) − V n ( x ) | . We note that u ( x ) can be decreasing only in the set { x ∈ [0 , | V ′ ( x ) > u ( x ) > c/V ′ ( x ) } . Since u (0) = u >
0, we have that u ( x ) > min { u , c/V ′ + } , where V ′ + denotes the maximum positive value of V ′ ( x ) in [0 , v n (with a lower bound that can be supposed to be independent on n ). Then, ǫ > u ( x ) v n ( x ) ≥ ǫ for all x ∈ [0 ,
1] and, therefore, | u ( x ) − v n ( x ) | ≤ cǫ Z x | u ( y ) − v n ( y ) | dy + | V ( x ) − V n ( x ) | . Since V n ( x ) → V ( x ) uniformly in [0 , Lemma 5.
Let u as in the previous lemma and let α M = max ≤ x ≤ V ′ ( x ) , α m = min ≤ x ≤ V ′ ( x ) . Then, (19) v ( x ) ≤ u ( x ) ≤ v ( x ) , x ∈ [0 , , where v ( x ) = φ ( x, u , c, α M ) , v ( x ) = φ ( x, u , c, α m ) . Proof . Let V ′ n and v n be as in the previous lemma, and assume, without loss ofgenerality, that α m ≤ V ′ n ( x ) ≤ α M for all x ∈ [0 , , x ], wehave v ( x ) ≤ v n ( x ) ≤ v ( x ), because φ ( x, u , c, α M ) ≤ φ ( x, u , c, α ) ≤ φ ( x, u , c, α m ) , as φ is increasing for decreasing α (Lemma 3). Then, in each of the successive intervalswe still have v ( x ) ≤ v n ( x ) ≤ v ( x ), because, a fortiori , φ ( x, v ( x i ) , c, α M ) ≤ φ ( x, v n ( x i ) , c, α ) ≤ φ ( x, v ( x i ) , c, α m ) , being φ also increasing with respect to the initial value. Thus v ( x ) ≤ v n ( x ) ≤ v ( x )in the whole interval [0 , v n → u uniformly in [0 ,
1] theinequalities are also true for u , which proves our claim. Note that the result ofLemma 5 implies, in particular, that u ( x ) > x ∈ [0 , Lemma 6.
Let us denote by u ( x, c ) , with x ∈ [0 , and c > , the solution of theCauchy problem (18) . Then, u ( x, c ) is strictly increasing with respect to c , for everyfixed x ∈ (0 , . L. BARLETTI AND F. SALVARANI
Proof . We resort again to the uniformly approximating sequence defined inLemma 4, that we now denote by v n ( x, c ) in order to stress the dependence on c .We can assume, as in Lemma 5, that α m ≤ V ′ n ( x ) ≤ α M (independently on n ).From definition (14) it is apparent that φ (∆ x, u , c, α ) is a C -function of c > α ∈ R . However, if we prove that ∂φ∂c is actuallycontinuous with respect to α also at α = 0, then ∂φ∂c has a (positive) lower boundwhen α varies in [ α m , α M ] and, therefore, for any given ∆ x ∈ (0 , u > c > µ > ∂∂c φ (∆ x, u , c, α i ) ≥ µ, for all i = 0 , , . . . , n − ∂φ∂c (and of φ as well) as α → + . For α > φ (which is now seen as a function of c and α ) is the first one in (14),i.e. φ ( c, α ) = cα { W [ g ( c, α )] } , where g ( c, α ) = (cid:16) αu c − (cid:17) e − α c ∆ x + αu c − . By using (12) we easily get ∂∂α { W [ g ( c, α )] } = 2 αW [ g ( c, α )]( u + 2 c ∆ x − αu ∆ x ) αcu − c . Since W [ g ( c, α )] → − α → + , by applying de l’Hˆopital’s theorem we obtainlim α → + c α { W [ g ( c, α )] } = u + 2 c ∆ x so that(21) lim α → + φ ( c, α ) = lim α → + cα { W [ g ( c, α )] } = q u + 2 c ∆ x, which, according to (14), proves the continuity of φ for α → + .Coming to the c -derivative, after straightforward calculations, we have (for α > ∂φ∂c ( c, α ) = 1 α { W [ g ( c, α )] } + αW [ g ( c, α )]( αu ∆ x − c ∆ x − u ) c { W [ g ( c, α )] } ( αu − c ) . By using (21) we obtain thereforelim α → + ∂φ∂α ( c, α ) = ∆ x p u + 2 c ∆ x = ∂∂c q u + 2 c ∆ x, which proves the continuity of ∂φ∂c as α → + .The continuity of ∂φ∂c (and φ ) for α → − is proven in the same way by using theexpression of φ for α < µ only depends on α m and α M , and does not depend onthe sequence of the α i ’s. Then, from the definition of v n (see Lemma 4) we have that v n ( x, c + ∆ c ) − v n ( x, c ) ≥ µ ∆ c, NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS x ∈ (0 ,
1] and ∆ c ≥ µ independent on n . Then, passingto the limit for n → ∞ we obtain that u ( x, c ) is strictly increasing with respect to c for every 0 < x ≤ Lemma 7.
Referring to Lemma 6 for the notations, we have that u ( x, c ) convergesuniformly to a continuous limit u ( x, as c → + .Proof . It is not difficult to show that(22) lim c → + φ (∆ x, u , c, α ) = max { u − αx, } , uniformly with respect to ∆ x , u and α . Then, still assuming α m ≤ V ′ n ( x ) ≤ α M ,we obtain that lim c → + v n ( x, c ) = v n ( x, x and n , where v n ( x,
0) is a continuous, piecewise linear, limit function. Since c v n ( x, c ) (seen asa sequence in c ) is uniformly Cauchy with respect to x and n , it is straightforward toprove that c u ( x, c ) is in turn uniformly Cauchy (with respect to x ) and, therefore,converges uniformly to a continuous limit u ( x, Definition 8.
To any given data ( u , V ) of the Cauchy problem (18) we associatethe set of points < x ≤ y < x ≤ y < · · · < x n = y n = 1 by means of the following recursive rule (where V = V (0) ):1. we start by putting x = sup (cid:8) x ∈ [0 , (cid:12)(cid:12) u + V − V ( ξ ) > ∀ ξ ∈ [0 , x ) (cid:9) ; if x = 1 , then we put y = 1 and the procedure ends with n = 1 , otherwisewe proceed to step 2.2. If x < we put y = sup (cid:8) x ∈ [ x , (cid:12)(cid:12) − V ′ ( ξ ) ≤ ∀ ξ ∈ [ x , x ) (cid:9) ; if y = 1 , then the procedure ends with n = 1 , otherwise we proceed to step 3.3. If y i − < we put x i = sup (cid:8) x ∈ ( y i − , (cid:12)(cid:12) V ( y i − ) − V ( ξ ) > ∀ ξ ∈ ( y i − , x ) (cid:9) ; if x i = 1 , then we put y i = 1 and the procedure ends with n = i , otherwise weproceed to step 4.4. If x i < we put y i = sup (cid:8) x ∈ [ x i , (cid:12)(cid:12) − V ′ ( ξ ) ≤ ∀ ξ ∈ [ x i , x ) (cid:9) ; if y i = 1 , then the procedure ends with n = i , otherwise we increment theindex i and repeat steps 3 and 4 until we find x i = 1 or y i = 1 for some i .(For the sake of simplicity we can assume that V changes sign a finite numberof times, so that the procedure stops at a finite n .)Finally, we define the continuous function (23) U ( x ) = u + V − V ( x ) , x ∈ [0 , x ]0 , x ∈ [ x i , y i ] ,V ( y i ) − V ( x ) , x ∈ [ y i , x i +1 ] . L. BARLETTI AND F. SALVARANI
Lemma 9 (
Asymptotic behaviour for c → + ). Let u be the solution of theCauchy problem (18) . Then, lim c → + u ( x ) = U ( x ) , where U is the function defined in Definition 8.Proof . We divide the proof into three recursive steps. Step 1.
Let ǫ > be arbitrarily small and let x ∈ [0 , x − ǫ ]. From Definition 8 wehave that η > u + V − V ( ξ ) ≥ η for all ξ ∈ [0 , x ]. Then, from theobvious inequality u ( x ) = u + V − V ( x ) + Z x cu ( ξ ) dξ > u + V − V ( x ) , we obtain u ( x ) − u − V + V ( x ) ≤ cxη , x ∈ [0 , x − ǫ ] , which proves that lim c → + u ( x ) = U ( x ), for all x ∈ [0 , x ). Since the limit has to bea continuous function (Lemma 7), from the arbitrariness of ǫ we also havelim c → + u ( x ) = U ( x ) = 0 . Step 2.
In the interval [ x , y ] we have that lim c → + u ( x ) = 0 (from the previousstep) and, by definition, α m = min x ∈ [ x ,y ] V ′ ( x ) ≥
0. Then, from Lemma 5lim c → + u ( x ) ≤ lim c → + p u ( x ) + 2 c ( x − x ) = 0 , for all x ∈ [ x , y ]. Step 3.
In the subsequent intervals of the form [ y i , x i +1 ] or [ x i , y i ], we repeat the sameproofs of the steps 1 and 2 (respectively), with the only difference that U ( y i ) = 0instead of U (0) = u > η > V ( y i ) − V ( ξ ) ≥ η for all ξ ∈ [ y i + ǫ, x i +1 − ǫ ]).Note that the limiting process c → + selects one particular solution among theinfinitely many solutions of the case c = 0, for which we have seen that there is notuniqueness (see section 3.1). Definition 10.
We define the critical values for the Dirichlet data u > and u > as follows: (24) u crit0 = V M − V , u crit1 = V M − V , where we put V = V (0) , V = V (1) and V M = max x ∈ [0 , V ( x ) . If a boundary term is greater than the corresponding critical value, then it is said tobe supercritical, otherwise it is said to be subcritical.
We can finally prove that the solution of the the Dirichlet problem (9) exists andis unique, provided that the Dirichlet data u > u > NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS Theorem 11. If u > u crit0 , then the Dirichlet problem (9) has a unique, strictlypositive, solution u for all u > .If u > u crit1 , then the Dirichlet problem (9) has a unique, strictly positive, solution u for all u > .Proof . Consider first the case u > u crit0 and assume, temporarily, that c > c u (1), obtained by solving the Cauchy problem(18) is strictly monotone. Moreover, from Lemma 5, we have thatlim c → + ∞ u (1) ≥ lim c → + ∞ v (1) = + ∞ , since it is easy to see that φ ( x, u , c, α ) → + ∞ as c → + ∞ , for all x > α ∈ R .Moreover, since u > u crit0 , from Definition 8 and Lemma 9 we have thatlim c → + u ( x ) = U ( x ) = u + V − V ( x ) > , i.e., in particular,lim c → + u (1) = u − ∆ V > , where ∆ V = V − V . What we have proven so far is that the left-Dirichlet datum u > u crit0 can be uniquelylinked to any right-Dirichlet datum u = u (1) ∈ [ u − ∆ V, + ∞ ), by solving the Cauchyproblem (18) with a suitable c ≥
0. To go below the threshold u − ∆ V we have toconsider negative values of c . However, the case c < c > u satisfies u ′ ( x ) = c/u ( x ) − V ′ ( x ), then ˜ u ( x ) = u (1 − x )satisfies ˜ u ′ ( x ) = − c ˜ u ( x ) − ˜ V ′ ( x ) , where ˜ V ( x ) = V (1 − x ) . In other words, the forward Cauchy problem with c < c >
0. Hence, if we fix the right-Dirichlet datum 0 < u
0, we can consider the Cauchy problem(25) ˜ u ′ ( x ) = | c | ˜ u ( x ) − ˜ V ′ ( x ) , x ∈ [0 , , ˜ u (0) = u , that is the backward Cauchy problem for u with right-Cauchy datum u . What wewant to prove is that u can be linked to u with a suitable choice of c <
0. i.e.,that c < u (1) = u (0) = u . Since the datum u is not necessarilysupercritical for problem (25), then in the limit c → − we have in general that ˜ u tends to the asymptotic solution ˜ U ( x ) (with the obvious definition) and, according tothe theory we already know, for ˜ u (1) there are three possibilities:lim c → − ˜ u (1) = u − ∆ ˜ V , (that is u + ∆ V ),0 ,V ( y ) − V (0) , for some point y ∈ (0 , u supercritical for problem (25) we have lim c → − ˜ u (1) = u + ∆ V < u (since we have assumed u < u − ∆ V ). In the second case, obviously,lim c → − ˜ u (1) = 0 < u and in the third case u > − V + V ( y ) = lim c → − ˜ u (1) , L. BARLETTI AND F. SALVARANI (because u is supercritical). In each of the three cases, therefore,lim c → − u (0) = lim c → − ˜ u (1) < u and then (owing to the monotonic growth with respect to | c | ), u = u (0) for a suitable c <
0. In conclusion, we have shown that the left-Dirichlet datum u > u crit0 can beuniquely linked to any right-Dirichlet datum u = u (1) >
0, by solving the Cauchyproblem (18) with a suitable c ∈ R .The proof of the second part of the claim, i.e. the case u > u crit1 , is completelyequivalent to the proof of the first part since, as we have just shown, it suffices tochange c into − c .
4. Numerical simulations.
Let x = ( x , x ) and let Ω the square in R definedas Ω = { x ∈ R : 0 ≤ x ≤ ≤ x ≤ } . We denote moreover ∂ Ω := Γ = X i =1 Γ i , where Γ := { x ∈ R : 0 < x ≤ x = 0 } ;Γ := { x ∈ R : x = 1 and 0 < x ≤ } ;Γ := { x ∈ R : 0 ≤ x < x = 1 } ;Γ := { x ∈ R : x = 0 and 1 / ≤ x < } ;Γ := { x ∈ R : x = 0 and 0 ≤ x < / } and consider the unknowns u : R + × Ω → R and J : R + × Ω → R .We now introduce, for n = 0 , . . . , N , a semi-discrete weak formulation of Equation(2) for the semi-discrete density unknowns u n ( x ) = u ( n ∆ t, x ) and the semi-discreteflux J n ( x ) = J ( n ∆ t, x ), where ∆ t > φ ∈ H (Ω) and ψ ∈ H (Ω) × H (Ω). We approximate the weak formulationof Equation (2) by means of the following coupled system(26) Z Ω t ( u n − max { u n − , } ) φ dx + Z ∂ Ω φJ n · n x dS − Z Ω J n · ∇ φ dx = 0 , Z Ω J n · ψ dx + 12 π Z Ω max { u n − , } [ ∇ u n · ψ ] dx + Z Ω u n [ ∇ V · ψ ] dx = 0 , where n x is the outward normal with respect to Ω starting from a point x ∈ ∂ Ω.The boundary term (i.e. the integral on ∂ Ω in the formulation written above)will be treated in agreement with the different boundary conditions specified for eachnumerical simulation.The “positive part” term in the weak formulation (26) is pleonastic at the contin-uous level, since the solution of the problem is known to be non negative by Theorem1. At the discrete level, this strategy helps in controlling the non-negativity of thenumerical solution.
NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS V and of the boundary conditions, andhave been obtained by using the Finite Element Method.From the semi-discrete formulation, by using quadratic P Lagrangian elementson a triangular mesh, we obtain a linear system whose size is given by twice thenumber of vertices and the number of mid-edges in the triangulation. The system issolved by a multi-frontal Gauss LU factorization.The simulations are written in
FreeFem++ . The mesh discretization used in oursimulations is composed by 19514 triangles, with 9940 vertices.
We consider here the boundary conditions(27) J | Γ = J | Γ = 0 , u | Γ = u , u | Γ ∪ Γ = u . and assume that both the initial datum and the potential only depend on x . Then,as already discussed in section 3.2 in the stationary case, the two-dimensional problemreduces to the one-dimensional problem(28) ∂u∂t = ∂∂x (cid:20) u ∂∂x ( u + V ) (cid:21) = 0 , x ∈ [0 , , t > ,u (0 , t ) = u , u (1 , t ) = u , u ( x ,
0) = u in ( x ) , where the initial datum, u in ( x ), and the (constant) Dirichlet data, u and u , arepositive. In this subsection we report a set of numerical simulations showing that thesolution to Equation (28) tends asymptotically to the stationary solution discussedin section 3.2 (see, in particular, Theorem 11). In Figure 1 the spatial profile ofthe solution along the direction x (recall that the solution is homogeneous in thedirection x ) is shown at different instants of time. This first set of simulations hasbeen performed with the potential(29) V ( x ) = sin(2 πx ) . Note that, according to Definition 10, for such potential we have u crit0 = u crit1 = 1 . The initial datum u in ( x ) is chosen as a linear function interpolating the values u and u and the evolution of such datum towards the asymptotic, stationary solution(dotted black curve) is illustrated for different choices of u and u .In panels (a), (b), (c) and (d) we have chosen, respectively,( u , u ) = (1 . , . , ( u , u ) = (1 . , . , ( u , u ) = (1 . , . , ( u , u ) = (0 . , . . Then, in cases (a) and (b) both data are supercritical while, in case (c), u is subcriticaland, in case (d), u is subcritical. We recall that Theorem 11 guarantees the well-posedness of the stationary problem if at least one of the Dirichlet data is supercritical.In case (b) since the potential (29) is compatible with equal Dirichlet data, theasymptotic solution is exactly u ( x , ∞ ) = U ( x ) = u − V ( x ) (corresponding to c = 0, see definition (23) and Lemma 9). In cases (a) and (d), since u > u , we4 L. BARLETTI AND F. SALVARANI (a) (b)(c) (d)
Fig. 1 . Evolution towards the stationary state in the 1D case with sinusoidal potential (29).In each panel the initial datum is the straight line and the asymptotic solution is the dotted blackcurve. The other lines correspond to t = 0 . , t = 0 . , t = 0 . , t = 0 . . In panels (a) and (b)the two Dirichlet data are both supercritical (in particular, in panel (b) the asymptotic solution is u − V ( x ) ). In panels (c) and (d), one of the data is subcritical (respectively, u and u ). have that c > u < u , we have that c < V = 0). In the three cases (a), (c) and (d), therefore, theasymptotic solution is not U ( x ) (which corresponds to u = u , in the present casewhere ∆ V = 0). Nevertheless, it is interesting to see that the asymptotic curves inFigure 1 tend to have the same character as U ( x ), i.e. regular in the supercriticalcases and piecewise regular in the subcritical cases (however, as long as t is finite, thesolution is everywhere regular). We consider again Equation (2) with the conditions (27) and the followingdata: and boundary conditions: u in = 2 − x , u = 2 , u = 1 . For this experiment, we have chosen the potential V ( x ) = − x . Since u crit0 = 0 , u crit1 = 1 , the boundary condition u in this simulation is supercritical. In Figure 2, we plotthe current versus time that passes through the boundaries Γ (blue dotted line) and NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS Fig. 2 . Currents J L (blue dotted line) and J R (red continuous line) versus time under theaction of the potential V ( x ) = − x . Γ ∪ Γ (red continuous line), i.e. we visualize the time evolution of J R := Z Γ J · n x dx and J L := Z Γ ∪ Γ J · n x dx respectively.We note that, after a transient period, both currents at the extremities of thedevice tend monotonically to the common value 3.11.In the second numerical simulation, we have left unchanged the initial and bound-ary conditions, but we have chosen a different potential, namely V ( x ) = − x + e − x . This potential is non-linear, but is still monotone. Here u crit0 = 0 , u crit1 = 2 − /e ≈ . , hence the boundary condition u is supercritical.In Figure 3, we plot the current versus time that passes through the boundariesΓ (blue dotted line) and Γ ∪ Γ (red continuous line) respectively.We note a different behaviour of J L and J R with respect to the previous simu-lation in the transient period before reaching the equilibrium: J L is still a monotoneincreasing function, but J R is no longer monotone. At time t = 1 . We now consider Equation (2) under the action of the potential V ( x ) = e − ( x − . . Here the critical values of the problem are u crit0 = u crit1 = 1 − e / ≈ . . L. BARLETTI AND F. SALVARANI
Fig. 3 . Currents J L (blue dotted line) and J R (red continuous line) versus time under theaction of the potential V ( x ) = − x + e − x . We compare the time evolution of the current at the extremities of the device intwo situations. The first one, whose results are plotted in Figure 4, has been obtainedwith the initial data and the boundary conditions u in = 2 − x , u = 2 , u = 1 . Fig. 4 . Currents J L (blue dotted line) and J R (red continuous line) versus time under theaction of the potential V ( x ) = e − ( x − . and small gap between the boundary data in Γ and Γ . The second situation, whose result are plotted in Figure 5, has been obtained byimposing the following initial data and the boundary conditions: u in = 6 − x , u = 6 , u = 1 . The numerical results of Figures 4 and 5 show that the stationary current flowingthrough the device in the case of a small density gap differ considerably from the one
NONLINEAR DIFFUSION PROBLEM FOR A FERMION GAS Fig. 5 . Currents J L (blue dotted line) and J R (red continuous line) versus time under theaction of the potential V ( x ) = e − ( x − . and wide gap between the boundary data in Γ and Γ . that is obtained in the case of a wide density gap (numerically we get a value whichis close to 1.36 in the first case and the value 16.84 in the second). Moreover, in thefirst case, the convergence speed to the asymptotic state is slower than in the secondcase.We have finally computed, for a ∈ [0 , a , which is a parameter that controls the gap of the boundary databetween the ends Γ and Γ , through the choice of the following initial and boundaryconditions: u in = . a (1 − x ) , u = . a, u = . . Both boundary data are supercritical.
Fig. 6 . Asymptotic current at the ends of the device under the action of the potential V ( x ) = e − ( x − . versus the gap between the boundary data in Γ and Γ . The results, plotted in Figure 6, show that the current is zero when a = 0 and8 L. BARLETTI AND F. SALVARANI that it is strictly monotone. The profile is parabolic with a good approximation. Thissimulation has been very time-consuming, since we had to repeat, for each element ofthe discretization of a ∈ [0 , a is ∆ a = 0 .
05, and the final time of each simulation has been t = 2 .
5. Thediscrepancies observed between the numerical values of J R ( t = 2 .
5) and J L ( t = 2 . a , beyond the resolution of Figure 6. We conclude the description of our simulations with a test-case that is widely used in the literature [21]. Consider the system defined by (2)under the action of the linear potential V ( x ) = 1 − x . In this case, the initial data and boundary conditions are u in ( x ) = cos( πx ) + 2 , and J | Γ = J | Γ = J | Γ = 0 , u | Γ = 1 , u | Γ = 3 . Note that, even though u in and V depend only on the first variable x , the problemis genuinely two-dimensional because of the boundary conditions.In Figure 7, we show the time evolution of the unknown density u . At time t = 1the system has already reached an almost stationary configuration.We also show in Figure 8 the time evolution of the L -norm of the density u . Notethat the total mass of the problem is not conserved in time, but has a non-monotonebehaviour around the value k u in k L (Ω) = 2. REFERENCES[1]
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