Global Solution to Enskog Equation with External Force in Infinite Vacuum
aa r X i v : . [ m a t h - ph ] M a y GLOBAL SOLUTION TO ENSKOG EQUATION WITH EXTERNAL FORCEIN INFINITE VACUUM
ZHENGLU JIANG
Abstract.
We first give hypotheses of the bicharacteristic equations corresponding to theEnskog equation with an external force. Since the collision operator of the Enskog equation ismore complicated than that of the Boltzmann equation, these hypotheses are more complicatedthan those given by Duan et al. for the Boltzmann equation. The hypotheses are very relatedto collision of particles of moderately or highly dense gases along the bicharacteristic curves andthey can be used to make the estimation of the so-called gain and loss integrals of the Enskogintegral equation. Then, by controlling these integrals, we show the existence and uniqueness ofthe global mild solution to the Enskog equation in an infinite vacuum for moderately or highlydense gases. Finally, we make some remarks on the locally Lipschitz assumption of the collisionfactors in the Enskog equation. Introduction
This paper is to consider the existence and uniqueness of the global mild solution to theEnskog equation with an external force in an infinite vacuum for moderately or highly densegases. Throughout this paper, R + represents the positive side of the real axis including itsorigin and R denotes a three-dimensional Euclidean space. In the presence of external forcesdepending on the time and space variables, the Enskog equation is as follows (see [7] or [10]): ∂f∂t + v · ∂f∂x + E ( t, x ) · ∂f∂v = Q ( f ) (1.1)where f = f ( t, x, v ) is a one-particle distribution function that depends on the time t ∈ R + , theposition x ∈ R and the velocity v ∈ R , E ( t, x ) is a vector-valued function which belongs to R and represents an external force with respect to the time and space variables, and Q is theEnskog collision operator whose form will be explained below.The collision operator Q is expressed by the difference between the gain and loss termsrespectively defined by Q + ( f )( t, x, v ) = Z R × S F + ( f ) f ( t, x, v ′ ) f ( t, x − aω, w ′ ) B ( v − w, ω ) dωdw, (1.2) Q − ( f )( t, x, v ) = Z R × S F − ( f ) f ( t, x, v ) f ( t, x + aω, w ) B ( v − w, ω ) dωdw. (1.3)Here and below everywhere, S = { ω ∈ S : ω ( v − w ) ≥ } is a subset of a unit sphere surface S in R , a is a positive constant that represents a diameter of hard sphere, ω is a unit vector alongthe line passing through the centers of the spheres at their interaction, B ( v − w, ω ) = ( v − w ) ω is the collision kernel and F ± are the collision factors. F ± are usually assumed to be twofunctionals of f, more precisely speaking, they depend on the density ρ ( t, x ) = R R f ( t, x, v ) dv at the time t and the point x. Date : November 23, 2018.2000
Mathematics Subject Classification.
Key words and phrases.
The Enskog equation; global solution.This work was supported by NSFC 10271121 and the Scientific Research Foundation for the ReturnedOverseas Chinese Scholars, the Education Ministry of China, and sponsored by joint grants of NSFC10511120278/10611120371 and RFBR 04-02-39026.
In equations (1.2) and (1.3), ( v, w ) and ( v ′ , w ′ ) are velocities before and after the collision,respectively. As for the Boltzmann equation, the conservation of both kinetic momentum andenergy of two colliding particles gives v + w = v ′ + w ′ , v + w = v ′ + w ′ , (1.4)which leads to their velocity relations v ′ = v − [( v − w ) ω ] ω, w ′ = w + [( v − w ) ω ] ω, (1.5)where ω ∈ S . We denote u = v − w, u k = ( uω ) ω and u ⊥ = u − u k (see [9] or [11]), thus gettinganother expression of (1.5) as follows: v ′ = v − u k , w ′ = v − u ⊥ . (1.6)Then the gain and loss terms (1.2) and (1.3) can be rechanged as Q + ( f )( t, x, v ) = Z R × S F + ( f ) f ( t, x, v − u k ) f ( t, x − aω, v − u ⊥ ) B ( u, ω ) dωdu, (1.7) Q − ( f )( t, x, v ) = Z R × S F − ( f ) f ( t, x, v ) f ( t, x + aω, v − u ) B ( u, ω ) dωdu. (1.8)If the factors F ± are set to be the same positive constant and the diameter a equal to zeroin the density variables, then the Enskog equation becomes the Boltzmann one that provides asuccessful description for dilute gases. The Boltzmann equation is no longer valid for moderatelyor highly dense gases. As a modification of the Boltzmann equation, the Enskog equationproposed by Enskog [10] in 1922 is usually used to explain the dynamical behavior of the densityprofile of moderately or highly dense gases.As we know, there are global solutions for the Boltzmann equation in the absence of externalforces not only in an infinite vacuum but also with large initial data. Existence of such vacuumsolutions was first considered by Illner & Shinbrot [12] and later by Bellomo & Toscani [4].The global existence of solution was shown by DiPerna & Lions [8] for the Boltzmann equationwith the large data. There are also some similar results about the Enskog equation withoutany effect of external forces. For example, Polewczak [15] and Arkeryd [1] gave their differentexistence proofs of global-in-time solutions to the Enskog equation without external forces forboth near-vacuum and large data, respectively. It is worth mentioning that some early workson the existence of global solutions to the Enskog equation were given by Cercignani and/orArkeryd (e.g. [2], [3], [5], [6]). Recently, Duan et al. [9] proved the existence and uniqueness ofa global mild solutions for the Boltzmann equation with external forces in an infinite vacuumand many relevant works can be found in the reference. Now there is not yet such similar resultfor the Enskog equation in the presence of external forces. The aim of this paper is to extendthe result to the case of the Enskog equation, that is, to show the existence and uniqueness ofsuch vacuum solution to the Enskog equation (1.1) with (1.2) and (1.3). In Section 2 hypothesesof the external forces are given and a Banach space and its operators are constructed. Thesehypotheses are very related to collision of particles of moderately or highly dense gases alongthe bicharacteristic curves and they are much more complicated than those given by Duanet al. mentioned above for the Boltzmann equation since the collision operator of the Enskogequation is fairly more complicated than that of the Boltzmann equation. In spite of this, thetwo examples shown by Duan et al. in [9] satisfy our hypotheses. Then the estimation of theso-called gain and loss integrals is made in Section 3. An existence and uniqueness theorem ofglobal mild solution to the Enskog equation in an infinite vacuum is given in Section 4 and someremarks on the assumption of the factors F ± are finally made in Section 5. LOBAL SOLUTION TO ENSKOG EQUATION 3 Hypotheses and Operators
In this section we first give some constructive hypotheses of the external forces with the help ofthe bicharacteristic equations of the Enskog equation with these forces and then build a Banachspace and its operators relative to the Enskog integral equation.Let us begin with considering the bicharacteristic equations of the Enskog equation (1.1) dXds = V, dVds = E ( s, X ) , ( X, V ) | s = t = ( x, v ) . (2.1)Suppose that such a vector-valued force function E ( t, x ) allows the above system (2.1) to havea global-in-time smooth solution denoted by[ X ( s ; t, x, v ) , V ( s ; t, x, v )] (2.2)for any fixed ( t, x, v ) ∈ R + × R × R , and that there exist three functions α i ( s ; t, x, v ) ( i = 1 , , α ( s ; t, x, v ) > s > , α (0; t, x, v ) ≥ , α ( s ; t, x, v ) > , α ( s ; t, x, v ) ≥ , (2.3) X (0; s, X ( s ; t, x, v ) + ξ, V ( s ; t, x, v ) − η ) = X (0; t, x, v ) + α ( s ; t, x, v ) η + α ( s ; t, x, v ) ξ, (2.4) V (0; s, X ( s ; t, x, v ) + ξ, V ( s ; t, x, v ) − η ) = V (0; t, x, v ) − α ( s ; t, x, v ) η − α ( s ; t, x, v ) ξ, (2.5)either α ( s ; t, x, v ) ≡ { α ( s ; t, x, v ) , α ( s ; t, x, v ) } /α ( s ; t, x, v ) ≤ τ , (2.6)min { ( α ( s ; t, x, v )) α ( s ; t, x, v ) , α ( s ; t, x, v ) α ( s ; t, x, v ) , α ( s ; t, x, v ) } ≥ α > , (2.7)for any s ∈ R + and ( ξ, η ) ∈ R × R when any point ( t, x, v ) is fixed in R + × R × R , where α , e and τ are three fixed positive constants independent of s and ( t, x, v ) , and α ( s ; t, x, v ) isdenoted by α ( s ; t, x, v ) ≡ α ′ ( s ; t, x, v ) α ( s ; t, x, v ) − α ( s ; t, x, v ) α ′ ( s ; t, x, v ) , (2.8)here α ′ i ( s ; t, x, v ) ( i = 1 ,
2) represent the derivative with respect to s. For their understandingof their hypotheses of the external force in the Boltzmann equation, Yuan et al. [9] took thefollowing two examples: E ( t, x ) = E ( t ) and E ( t, x ) = e x + E ( t ) with e being a positiveconstant. Since the Enskog equation is much more complicated than the Boltzmann one, ourhypotheses of the external forces for the Enskog equation are much more complicated than thosegiven by Yuan et al. mentioned above for the Boltzmann equation. In spite of this, the abovetwo examples satisfy the above hypotheses of the external force. These examples are obviouslysuitable for our explanation of the corresponding constructive conditions whether the Boltzmannequation or the Enskog one is considered.We give the five conditions (2.3)-(2.7) in order to get the following inequalities | X (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k ) | + | X (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ ) | ≥ | X (0; t, x, v ) | + | X (0; t, x, v ) + α ( s ; t, x, v ) u − aα ( s ; t, x, v ) ω | and | V (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k ) | + | V (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ ) | ≥ | V (0; t, x, v ) | + | V (0; t, x, v ) − α ( s ; t, x, v ) u + aα ( s ; t, x, v ) ω | along their bicharacteristic curves after collision of particles of moderately or highly dense gases.The above two inequalities can be used to control the gain and loss integral terms of the Enskogintegral equation along the bicharacteristic curves. Therefore this form of the external forces ispertinent to collision of particles of moderately or highly dense gases along the bicharacteristiccurves so that the gain and loss integral terms can be estimated.We below give a representation of mild solution to the Enskog equation. Let us first introducea notation f defined as f ( s ; t, x, v ) = f ( s, X ( s ; t, x, v ) , V ( s ; t, x, v )) ZHENGLU JIANG for any measurable function f on R + × R × R . Obviously, it can be found that f ( t ; t, x, v ) = f ( t, x, v ) and that f (0; t, x, v ) = f (0 , X (0; t, x, v ) , V (0; t, x, v )) . Along the bicharacteristic curves, the Enskog equation (1.1) can be also written as dds f ( s ; t, x, v ) = Q ( f ) ( s ; t, x, v ) , which leads to the following integral equation f ( t, x, v ) = f ( X (0; t, x, v ) , V (0; t, x, v )) + Z t Q ( f ) ( s ; t, x, v ) ds, (2.9)where f ( x, v ) ≡ f (0 , x, v ) . A function f ( t, x, v ) is called global mild solution to the Enskogequation (1.1) if f ( t, x, v ) satisfies the above integral equation (2.9) for almost every ( t, x, v ) ∈ R + × R × R . Then we construct a subset M of a Banach space C ( R + × R × R ) , which has the propertythat every element f = f ( s, x, v ) ∈ M if and only if there exists a positive constant c such that f satisfies | f ( t, x, v ) | ≤ ch ( X (0; t, x, v )) m ( V (0; t, x, v ))where h ( x ) = e − p | x | , m ( v ) = e − qv , (2.10)for any fixed p and q in (0 , + ∞ ) . It follows that M is a Banach space when it has a norm of thefollowing form || f || = sup t,x,v {| f ( t, x, v ) | h − ( X (0; t, x, v )) m − ( V (0; t, x, v )) } . In particular, || f (0 , x, v ) || = sup x,v {| f (0 , x, v ) | h − ( x ) m − ( v ) } . The initial data f ≡ f (0 , x, v ) isbounded in L ( R × R ) . This implies that the total mass of the system is finite. Hence themean free path is sufficiently large if the finite total mass is sufficiently small. This is exactlythe requirement on the Enskog equation with external forces in an infinite vacuum, which issimilar to one considered by Illner & Shinbrot [12] for the Boltzmann equation. It is worthmentioning that in other cases there are many different classes of functions which can be takenas the choice of h ( x ) and m ( v ) (see [9], [11]). For example, in the case of the external forcesdepending only on the time t, one can also choose h ( x ) = (1 + | x | ) − p and m ( v ) = e − qv for anyfixed p ∈ (1 / , + ∞ ) and q ∈ (0 , + ∞ ); when p > / , the initial data f ≡ f (0 , x, v ) is boundedin L ( R × R ); when 1 / < p ≤ / , the initial total mass might be infinite; this choice of both h ( x ) and m ( v ) is not suitable for using our method considered in this paper to deal with thisexistence problem of vacuum solutions to the Enskog equation with external forces depending onthe time and space variables, however, this choice can be applied to the case of the Boltzmannequation in the presence of external forces [9].To give global existence, it is necessary to study the properties of the collision operator ina Banach space. To do this, by (1.7) and (1.8), Q ( f ) ( s ; t, x, v ) can be first rewritten as thedifference between the gain and loss terms of other two forms Q + ( f ) ( s ; t, x, v ) = Z R × S F + ( f ) f ( s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k ) × f ( s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ ) B ( u, ω ) dωdu, (2.11) Q − ( f ) ( s ; t, x, v ) = Z R × S F − ( f ) f ( s, X ( s ; t, x, v ) , V ( s ; t, x, v )) × f ( s, X ( s ; t, x, v ) + aω, V ( s ; t, x, v ) − u ) B ( u, ω ) dωdu. (2.12) LOBAL SOLUTION TO ENSKOG EQUATION 5
Estimation of the collision integrals can be then made by use of a similar argument to thatdeveloped in the previous work (see [9], [15], [18]). According to (2.11) and (2.12), we in facthave to estimate the following two integrals: I g ≡ Z t Z R × S h ( X (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k )) × m ( V (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k )) h ( X (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ )) × m ( V (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ )) B ( u, ω ) dωduds, (2.13) I l ≡ Z t Z R × S h ( X (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ))) × m ( V (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ))) h ( X (0; s, X ( s ; t, x, v ) + aω, V ( s ; t, x, v ) − u )) × m ( V (0; s, X ( s ; t, x, v ) + aω, V ( s ; t, x, v ) − u )) B ( u, ω ) dωduds. (2.14)Here I g and I l are called the gain and loss integrals respectively. Once the estimation of theintegrals is finished, the global existence result may be obtained by constructing a contractivemap from a Banach space to itself. Therefore it is one of the best important to estimate theabove two integrals, especially the gain one. It will be discussed in the next section.We finally denote an operator J on M by J ( f ) = f ( X (0; t, x, v ) , V (0; t, x, v )) + Z t Q ( f ) ( s ; t, x, v ) ds. (2.15)It will be proved in Section 4 that J is indeed a contractive map on a Banach space. This iswhat we need. 3. Estimation of the Gain and Loss Integrals
In this section the so-called gain and loss integrals are estimated by use of a similar device toone given in [15] for the Enskog equation in the absence of external forces. To do this, we firstintroduce the preliminary lemmas which will be used below.
Lemma 3.1.
Let any z ∈ R , s ∈ R + , ( u k , u ⊥ ) ∈ R × R with u k u ⊥ = 0 and ω ∈ S with u k ω ≥ . Then | z ± su k | + | z ± su ⊥ ∓ aω | ≥ | z | + | z ± s ( u k + u ⊥ ) ∓ aω | (3.1) for any fixed real number a ∈ R + . Lemma 3.2.
Let p > and ( z, u ) ∈ R × R with u = 0 . Then Z + ∞ e − p | z + su | ds ≤ √ π √ p | u | . (3.2) Lemma 3.3.
Let q > , − < γ ≤ and z ∈ R . Then Z R | u | γ − e − q | z − u | du ≤ πγ + 2 + πq / . (3.3) Lemma 3.4.
Let any ( s, z, u ) ∈ R × R × R and ω ⊥ be a unit vector perpendicular to ω ∈ S . Then | z + su + hω | ≥ | zω ⊥ + suω ⊥ | (3.4) for any h ∈ R . The proof of Lemmas 3.1 and 3.4 is easily given. Lemmas 3.2 and 3.3 can be obtained fromthe transformation of integral variables.
ZHENGLU JIANG
Lemma 3.5.
Assume that three functions α i ( s ; t, x, v ) ( i = 1 , , ) satisfy the external forceconditions (2.3) and (2.7), and that ˜ I l ( z , z , t, x, v ) ≡ Z t Z R × S | uw | e − p | z + α ( s ; t,x,v ) u + aα ( s ; t,x,v ) ω | × e − q | z − α ( s ; t,x,v ) u − aα ( s ; t,x,v ) ω | dudωds for any fixed real number a ∈ R + . Then ˜ I l ( z , z , t, x, v ) ≤ ˜ I lpq (3.5) for any ( z , z ) ∈ R × R , ( t, x, v ) ∈ R + × R × R , where ˜ I lpq is a positive constant dependingonly on p and q, p > , q > . Proof.
First let us fix ( z , z ) ∈ R × R and ( t, x, v ) ∈ R + × R × R . Put ¯ u = α ( s ; t, x, v ) u + aα ( s ; t, x, v ) ω. Then ¯ uω = α ( s ; t, x, v )( uω ) ω + aα ( s ; t, x, v ) for ω ∈ S . By (2.3), it thusfollows that˜ I l ( z , z , t, x, v ) ≤ Z t Z R × S | ¯ uω | ( α ( s ; t, x, v )) − × e − p | z + α s ; t,x,v ) α s ; t,x,v ) ¯ u + a e α ( s ; t,x,v ) α s ; t,x,v ) ω | e − q | z − ¯ u | d ¯ udωds, (3.6)where e α ( s ; t, x, v ) = α ( s ; t, x, v ) − α ( s ; t, x, v ) α ( s ; t, x, v ) . Take τ = α ( s ; t,x,v ) α ( s ; t,x,v ) . Then, by (2.7), dτds = α ( s ; t,x,v )( α ( s ; t,x,v )) > . By replacing the integral variable s with the new variable τ and usingLemma 3.4, the estimation of the integral on the right side of (3.6) thus gives˜ I l ( z , z , t, x, v ) ≤ Z + ∞ Z R × S | ¯ uω | ( α ( s ; t, x, v )) − ( α ( s ; t, x, v )) − × e − p | z ω ⊥ + τ ¯ uω ⊥ | e − q | z − ¯ u | d ¯ udωdτ ≤ α Z R × S | ¯ uω | (cid:26)Z + ∞ e − p | z ω ⊥ + τ ¯ uω ⊥ | dτ (cid:27) e − q | z − ¯ u | d ¯ udω, (3.7)where ω ⊥ is a unit vector perpendicular to ω. The last inequality in (3.7) comes from (2.7) andFubini’s theorem. By estimation of the integral on the right side of the last inequality, it thenfollows that˜ I l ( z , z , t, x, v ) ≤ α √ π √ p Z R × S | ¯ uω || ¯ uω ⊥ | e − q | z − ¯ u | d ¯ udω ≤ α π / √ p (cid:18)
43 + 1 q / (cid:19) , (3.8)where Lemmas 3.2 and 3.3 are used. The proof of Lemma 3.5 is hence completed. (cid:3) Lemma 3.6.
Assume that three functions α i ( s ; t, x, v ) ( i = 1 , , ) satisfy the external forceconditions (2.3), (2.6) and (2.7), and put ˜ I g ( z , z , t, x, v ) ≡ Z t Z R × S | uω | e − p | z + α ( s ; t,x,v ) u − aα ( s ; t,x,v ) ω | × e − q | z − α ( s ; t,x,v ) u + aα ( s ; t,x,v ) ω | dudωds, where a ≥ . Then ˜ I g ( z , z , t, x, v ) ≤ ˜ I gpq (3.9) for any ( z , z ) ∈ R × R , ( t, x, v ) ∈ R + × R × R , where ˜ I gpq is a positive constant dependingonly on p and q, p > , q > . LOBAL SOLUTION TO ENSKOG EQUATION 7
Proof.
Let us fix ( z , z ) ∈ R × R and ( t, x, v ) ∈ R + × R × R . Note that α ( s ; t, x, v ) > u = ¯ u k + ¯ u ⊥ , where ¯ u k = α ( s ; t, x, v ) u k − aα ( s ; t, x, v ) ω and ¯ u ⊥ = α ( s ; t, x, v ) u ⊥ . Then ¯ uω = α ( s ; t, x, v )( uω ) − aα ( s ; t, x, v ) for ω ∈ S . Thus˜ I g ( z , z , t, x, v ) ≤ Z t Z R × S ( | ¯ uω | + aα ( s ; t, x, v ))( α ( s ; t, x, v )) − × e − p | z + α s ; t,x,v ) α s ; t,x,v ) ¯ u − a e α ( s ; t,x,v ) α s ; t,x,v ) ω | e − q | z − ¯ u | d ¯ udωds, where e α ( s ; t, x, v ) = α ( s ; t, x, v )) − α ( s ; t, x, v ) α ( s ; t, x, v ) . To prove this lemma, it suffices toconsider the case of max { α ( s ; t, x, v ) , α ( s ; t, x, v ) } /α ( s ; t, x, v ) ≤ τ in (2.3). By repeating asimilar integral estimation to one given in Lemma 3.5, the estimate of the integrals on the rightside of the above inequality gives˜ I g ( z , z , t, x, v ) (a) ≤ Z τ Z R × S ( | ¯ uω | + aα ( s ; t, x, v ))( α ( s ; t, x, v )) − ( α ( s ; t, x, v )) − × e − p | z ω ⊥ + τ ¯ uω ⊥ | e − q | z − ¯ u | d ¯ udωdτ (b) ≤ α Z R × S (cid:26) | ¯ uω | Z + ∞ e − p | z ω ⊥ + τ ¯ uω ⊥ | dτ + aτ (cid:27) e − q | z − ¯ u | d ¯ udω (c) ≤ α Z R × S (cid:26) √ π √ p | ¯ uω || ¯ uω ⊥ | + aτ (cid:27) e − q | z − ¯ u | d ¯ udω (d) ≤ π α (cid:18) √ π √ p + aτ (cid:19) ( 43 + 1 q / ) , (3.10)for α ( s ; t, x, v ) = 0 , where ω ⊥ is a unit vector perpendicular to ω, (a) is obtained by first makingthe transformation τ = α ( s ; t,x,v ) α ( s ; t,x,v ) and then using (2.6) and Lemma 3.4, (b) is given by (2.7), (c)is obtained by Lemma 3.2 and (d) results from Lemma 3.3. This hence completes the proof ofLemma 3.6. (cid:3) By Lemmas 3.5 and 3.6, we can hence give the estimates of the gain and loss integrals asfollows.
Lemma 3.7.
Let I g and I l be the same integrals as defined by (2.13) and (2.14), respectively.In the integrals, a is a positive constant. Suppose that all the five external force conditions(2.3)–(2.7) hold for these functions X ( s ; t, x, v ) and V ( s ; t, x, v ) defined by the solution (2.2) tothe system (2.1), and that h ( x ) and m ( v ) are the same as in (2.10). Then it follows that I g ≤ Kh ( X (0; t, x, v )) m ( V (0; t, x, v )) , (3.11) I l ≤ Kh ( X (0; t, x, v )) m ( V (0; t, x, v )) , (3.12) for any ( t, x, v ) ∈ R + × R × R and some positive constant K. Proof.
Let us first estimate the loss integral. By using (2.4) and (2.5), the loss integral (2.14)can be rewritten as I l = Z t Z R × S | uω | e − p | X (0; t,x,v ) | e − p | X (0; t,x,v )+ α ( s ; t,x,v ) u + aα ( s ; t,x,v ) ω | × e − q | V (0; t,x,v ) | e − q | V (0; t,x,v ) − α ( s ; t,x,v ) u − aα ( s ; t,x,v ) ω | dudωds = h ( X (0; t, x, v )) m ( V (0; t, x, v )) e I l ( X (0; t, x, v ) , V (0; t, x, v ) , t, x, v )) . (3.13)It follows from Lemma 3.5 that I l ≤ h ( X (0; t, x, v )) m ( V (0; t, x, v )) e I lpq . (3.14) ZHENGLU JIANG
Then the estimation of the gain integral will be made below. Similarly, by using (2.4) and(2.5), the gain integral (2.13) is as follows: I g = Z t Z R × S | uω | e − p | X (0; t,x,v )+ α ( s ; t,x,v ) u k | e − p | X (0; t,x,v )+ α ( s ; t,x,v ) u ⊥ − aα ( s ; t,x,v ) ω | × e − q | V (0; t,x,v ) − α ( s ; t,x,v ) u k | e − q | V (0; t,x,v ) − α ( s ; t,x,v ) u ⊥ + aα ( s ; t,x,v ) ω | dudωds. (3.15)Using Lemma 3.1, we have I g ≤ e − p | X (0; t,x,v ) | e − q | V (0; t,x,v ) | Z t Z R × S | uω | e − p | X (0; t,x,v )+ α ( s ; t,x,v ) u − aα ( s ; t,x,v ) ω | × e − q | V (0; t,x,v ) − α ( s ; t,x,v ) u + aα ( s ; t,x,v ) ω | dudωds = h ( X (0; t, x, v )) m ( V (0; t, x, v )) e I g ( X (0; t, x, v ) , V (0; t, x, v ) , t, x, v ) . (3.16)It follows from Lemma 3.6 that I g ≤ h ( X (0; t, x, v )) m ( V (0; t, x, v )) e I gpq . (3.17)The proof of Lemma 3.7 is hence completed. (cid:3) Existence and Uniqueness
In this section we show a result about the existence and uniqueness of such vacuum solutionto the Enskog equation in presence of external forces. To do this, we first define a set M R by M R = { f : || f || ≤ R, f ∈ C ( R + × R × R } (4.1)and then assume that F ± are two functionals on M R such that the so-called locally Lipschitzcondition | F ± ( f ) − F ± ( g ) | ≤ L ( R ) || f − g || (4.2)holds for any f, g ∈ M R where M R is defined by (4.1) and L ( R ) is a positive nondecreasingfunction on R + . Thus we can get the following lemma.
Lemma 4.1.
Suppose that all the five conditions (2.3)–(2.7) hold for any X ( s ; t, x, v ) and V ( s ; t, x, v ) defined by the solution (2.2) to the system (2.1), and that the factors F ± in thecollision integrals Q ± ( f ) ( s ; t, x, v ) defined by (2.11) and (2.12) are two functionals satisfyingthe inequality (4.2). Let h ( x ) and m ( v ) be the same as in (2.10). In the collision integrals, a isa positive constant. Then the following inequalities hold: Z t | Q + ( f ) ( s ; t, x, v ) | ds ≤ C ( R ) h ( X (0; t, x, v )) m ( V (0; t, x, v )) || f || , Z t | Q − ( f ) ( s ; t, x, v ) | ds ≤ C ( R ) h ( X (0; t, x, v )) m ( V (0; t, x, v )) || f || for any f ∈ M R , where C ( R ) is a positive nondecreasing function on R + . Proof.
It can be first found from the assumption (4.2) of the two functionals F ± that thereexists a positive constant ˜ L ( R ) = L ( R ) R + | F + (0) | + | F − (0) | such that | F ± ( f ) | ≤ ˜ L ( R ) for any f ∈ M R . It follows from (2.11) and (2.12) that Z t Q + ( f ) ( s ; t, x, v ) ds ≤ ˜ L ( R ) Z t Z R × S || f || h ( X (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k )) × m ( V (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ) − u k )) h ( X (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ )) × m ( V (0; s, X ( s ; t, x, v ) − aω, V ( s ; t, x, v ) − u ⊥ )) uωdωduds, (4.3) Z t Q − ( f ) ( s ; t, x, v ) ds ≤ ˜ L ( R ) Z t Z R × S || f || h ( X (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ))) × m ( V (0; s, X ( s ; t, x, v ) , V ( s ; t, x, v ))) h ( X (0; s, X ( s ; t, x, v ) + aω, V ( s ; t, x, v ) − u )) LOBAL SOLUTION TO ENSKOG EQUATION 9 × m ( V (0; s, X ( s ; t, x, v ) + aω, V ( s ; t, x, v ) − u )) uωdωduds. (4.4)By (3.11) and (3.12), (4.3) and (4.4) give Z t Q + ( f ) ( τ, x, v ) dτ ≤ ˜ L ( R ) Kh ( X (0; t, x, v )) m ( V (0; t, x, v )) || f || , Z t Q − ( f ) ( τ, x, v ) dτ ≤ ˜ L ( R ) Kh ( X (0; t, x, v )) m ( V (0; t, x, v )) || f || . Take C ( R ) = ˜ L ( R ) K. It obviously follows that Lemma 4.1 holds. (cid:3)
Then we can get the following theorem.
Theorem 4.2.
Suppose that all the five external force conditions (2.3)–(2.7) hold for thesefunctions X ( s ; t, x, v ) and V ( s ; t, x, v ) defined by the solution (2.2) to the system (2.1), andthat the factors F ± in the collision integrals Q ± ( f )( t, x, v ) defined by (1.2) and (1.3) are twofunctionals satisfying the inequality (4.2). In the collision integrals, a is a positive constant.Then there exists a positive constant R such that the Enskog equation (1.1) with (1.2) and(1.3) has a unique non-negative global mild solution f = f ( t, x, v ) ∈ M R through a non-negativeinitial data f = f ( x, v ) when sup t,x,v { f ( X (0; t, x, v ) , V (0; t, x, v )) h − ( X (0; t, x, v )) m − ( V (0; t, x, v )) } is sufficiently small, where h ( x ) and m ( v ) are the same as in (2.10). Theorem 4.2 shows that there exists a unique global mild solution to the Enskog equation(1.1) given by (1.2) and (1.3) with the initial data near vacuum if a suitable assumption of theexternal force is given. As in [13], we below give our proof of Theorem 4.2.
Proof.
By (2.15) and Lemma 4.1, we have | J ( f ) | h − ( X (0; t, x, v )) m − ( V (0; t, x, v )) ≤ | f ( X (0; t, x, v ) , V (0; t, x, v )) | h − ( X (0; t, x, v )) m − ( V (0; t, x, v )) + 2 C ( R ) || f || ≤ R/ C ( R ) R for any f ∈ M R and f with || f || ≤ R/ . Since C ( R ) is a positive nondecreasing function on R + , it follows that || J ( f ) || ≤ R for sufficiently small R. Therefore J is an operator from M R toitself for sufficiently small R. Similarly, it can be also found that J is a contractive operator on M R for some suitably small R. Thus there exists a unique element f ∈ M R such that f = J ( f ) , i.e., (2.9) holds. It then follows from the same argument as the one in [12] (or see [14], [19]) thatif f ( x, v ) ≥ f ( t, x, v ) ≥ . Hence the proof of Theorem 4.2 is finished. (cid:3) Remarks on the Assumption of the Factors F ± In this section we make some remarks on the locally Lipschitz assumption (4.2) of the factorsof F ± appearing in Theorem 4.2 given in the previous section.We begin with a different kind of locally Lipschitz condition of F ± . It was originally given byPolewczak [15] as follows: | F ± ( f ) − F ± ( g ) | ≤ L ( R ) (cid:12)(cid:12)(cid:12)(cid:12)Z R f ( t, x, v ) dv − Z R g ( t, x, v ) dv (cid:12)(cid:12)(cid:12)(cid:12) (5.1)holds for any f = f ( t, x, v ) , g = g ( t, x, v ) ∈ M R where M R is defined by (4.1) and L ( R ) is apositive nondecreasing function on R + . Note that assumptions (2.4) and (2.5) have the followingproperties: X (0; t, x + ξ, v ) − X (0; t, x, v ) = α ( s ; t, x, v ) ξ (5.2)and V (0; t, x, v + η ) − V (0; t, x, v ) = α ( s ; t, x, v ) η (5.3) for any ( ξ, η ) ∈ R × R when any point ( t, x, v ) is fixed in R + × R × R . By (5.2) and (5.3),we have ∂X (0; t, x, v ) ∂x = ∂V (0; t, x, v ) ∂v = α ( s ; t, x, v ) , (5.4)thus giving ∂X (0; t, x, v ) ∂x = ∂V (0; t, x, v ) ∂v ≥ α > L ( R ) = L ( R ) R R m ( v ) dv/α . Then, by use of (5.5), (5.1)gives (4.2). This means that (5.1) is a stronger assumption than (4.2) when the external forcessatisfy the assumptions (2.3)-(2.7). Therefore the locally Lipschitz assumption (4.2) is at leastmathematically very useful in a more general case.Now let us recall the Enskog equation for our further understanding the locally Lipschitzassumption (4.2) of the factor F ± under the assumptions (2.3)-(2.7) of the external forces. TheEnskog equation can be roughly divided into two classes: the standard and the revised one. In thestandard Enskog equation (see [7], [10], [15]), F ± are defined by a geometrical factor Y whichis a contact-value pair correlation function of the hard-sphere system at uniform equilibriumand depends on the density ρ ( t, x ) , i.e., they are given by F + = Y ( ρ ( t, x − aω/ F − = Y ( ρ ( t, x + aω/ . For a fairly rare uniform gas of one particle with mass m, it can be found in [7]that the value of Y is approximatively expressed by Y ( ρ ( t, x )) = [1 − bρ ( t, x ) / / [1 − bρ ( t, x )]where b = 2 πa / (3 m ) . It can be easily shown that in this case the factor F ± = Y satisfies (5.1)and is locally Lipschitz as defined in (4.2) with the external forces of the assumptions (2.3)-(2.7)for R ∈ (0 , R ] , where R is some suitably small positive constant. Generally, the dependenceof the function Y on the local density ρ ( t, x ) is of the form Y ( ρ ( t, x )) = 1 + + ∞ X i =1 b i [2 πa ρ ( t, x ) / i , (5.6)where b i are given in terms of the virial coefficients B i appearing in the equation of state for thehard sphere system. We cannot know whether the series (5.6) converges when one of the twodifferent locally Lipschitz conditions (4.2) and (5.1) is satisfied. Even if this series converges,we cannot yet know whether one of the assumptions (4.2) and (5.1) of F ± holds for any factor Y of the above form. Of course, if F ± = Y is a factor defined by the form (5.6) with finiteterms, then (5.1) holds and so F ± satisfy (4.2) when the external forces are of the assumptions(2.3)-(2.7).In the case of the revised Enskog equation (see [1], [16], [20]), F ± are expressed by a contact-value pair correlation function G of the hard-sphere system at non-uniform equilibrium. Theform of G is given by the Mayer cluster expansion in terms of local density ρ ( t, x ) . The function G depends on the position x, the vector x ± aω and the density ρ ( t, x ) , i.e., F + = G ( x, x − aω, ρ ( t, x ))and F − = G ( x, x + aω, ρ ( t, x )) . In fact, G ( x, y, ρ ( t, x )) = exp( − β Φ( | x − y | )) e G ( ρ ( t, x )) , where β is a positive constant, Φ( | x − y | ) is a potential of two interaction spheres at the positions x and y, and e G is a functional of the following form [16] e G ( ρ ( t, x )) = 1 + Z V (12 | ρ (3) dx + 12 Z Z V (12 | ρ (3) ρ (4) dx dx + · · · + 1( k − Z Z · · · Z V (12 | · · · k ) ρ (3) ρ (4) · · · ρ ( k ) dx dx · · · dx k + · · · , (5.7)here ρ ( k ) = ρ ( t, x k ) and V (12 | · · · k ) is the sum of all the graphs of k labeled points whichare biconnected when the Mayer factor exp( − β Φ( | x i − x j | )) − LOBAL SOLUTION TO ENSKOG EQUATION 11 yet know whether one of the assumptions (4.2) and (5.1) of F ± holds for any functional e G ofthe above form.The assumption (4.2) or (5.1) of F ± is satisfied by some geometric factors present in thetruncated Enskog equations in both standard and revised cases. In the standard case thegeometric factor Y considered above can be assumed to be of the truncated form (5.6) with finiteterms while in the revised case an obvious example is that the functional e G can be truncatedto be a positive constant. It can be found that these factors satisfy the two assumptions of F ± when one assumes that both L ( R ) and L ( R ) are two positive constant functions on R + . Therefore the above two assumptions are completely suitable for our understanding evolutionsof moderately or highly dense gases by use of our investigation of the properties of the Enskogequation.
Acknowledgement.
The author would like to thank the referees of this paper for their valuable com-ments and suggestions on this work.
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