Wall Effect on the Motion of a Rigid Body Immersed in a Free Molecular Flow

Motion of a rigid body immersed in a semi-infinite expanse of gas in a $d$-dimensional region bounded by an infinite plane wall is studied for free molecular flow on the basis of the free Vlasov equation under the specular boundary condition. We show that the velocity $V(t)$ of the body approaches its terminal velocity $V_{\infty}$ according to a power law $V_{\infty}-V(t)\approx Ct^{-(d-1)}$ by carefully analyzing the pre-collisions due to the presence of the wall. The exponent $d-1$ is smaller than $d+2$ for the case without the wall found in the classical work by Caprino, Marchioro and Pulvirenti~[Comm. Math. Phys., \textbf{264} (2006), pp. 167--189] and thus slower convergence rate results from the presence of the wall.


1.
Introduction. Consider a cylinder immersed in a semi-infinite expanse of gas in a quiescent equilibrium in a d-dimensional region bounded by an infinite plane wall ( Figure 1). We study the motion of this cylinder for free molecular flow on the basis of the free Vlasov equation (1) under the specular boundary condition (2). The cylinder is accelerated instantaneously to an initial velocity V 0 in the direction away from the plane wall and parallel to the axis of the cylinder and thereafter applied a constant force E in this direction. As the cylinder moves through the gas, a drag force D V (t) is exerted to the body from the surrounding gas and the velocity of the cylinder V (t) is determined dynamically by the balance of the applied force E and the drag force D V (t): dV (t)/dt = E − D V (t), where we assumed that the mass of the cylinder is unity.
In the long time limit, the velocity V (t) of the cylinder is expected to approach a terminal velocity V ∞ and our primary interest in this paper is the rate of approach to the terminal velocity. The pioneering work by Caprino, Marchioro and Pulvirenti [5] showed that in the absence of the wall, the asymptotic convergence rate obeys a power law t −(d+2) , that is, V ∞ − V (t) ≈ Ct −(d+2) . This is due to the dependence of the drag force on the past history of the motion. Note that if the drag force is solely determined from the instantaneous velocity of the cylinder, that is, D V (t) = D(V (t)) for some smooth and increasing function D(U ), the approach to the terminal velocity is exponential in time. Since molecules colliding with the 2 KAI KOIKE cylinder might had collisions previously, which we call pre-collisions, the drag force exerted to the cylinder depends on the whole history of the motion: {V (t) | t ≥ 0}.
We show in this paper that the presence of the wall modifies the asymptotic convergence rate to t −(d−1) , which is slower than the case without the wall. This is caused by the pre-collisions of the molecules with large horizontal velocities ξ 1 , satisfying in particular ξ 1 > sup{V (t) | t ≥ 0}. Note that there are no such precollisions in the absence of the wall. Our proof is based on the framework developed in [5] and the novelty of this work lies in establishing appropriate estimates for the pre-collisions due to the presence of the wall. Our results show that a distant obstacle, the plane wall in our case, may change the asymptotic behaviour substantially. This might be particularly important in the design and interpretation of results of laboratory experiments because vacuum chamber has a wall.
We consider in this paper both cases of 0 < V 0 < V ∞ and 0 < V ∞ < V 0 . In the absence of the wall, the case of 0 < V 0 < V ∞ is treated in [5] and the case of 0 < V ∞ < V 0 is treated in [4]. Note that the more difficult case of 0 = V ∞ < V 0 is also treated in [4] but we still are not able to treat this case in the presence of the wall. As in the case without the wall, there is a change in sign of V ∞ − V (t) in the case of 0 < V ∞ < V 0 . We note that in the absence of the wall, an almost necessary and sufficient condition for velocity reversal is given in [10]. Body shapes other than a cylinder can be considered as well. In the absence of the wall, general convex bodies are treated in [6] and the asymptotic convergence rate is shown to be t −(d+2) . Note that due to the presence of the wall, even if the body is convex the asymptotic convergence rate may be different from t −(d+2) . In the presence of the wall, general convex bodies or a U-shaped body treated in [12] may be included but we shall restrict ourselves to the case of a cylinder for simplicity. We note that for the model considered in this paper, the drag force D V (t) is time dependent even when the velocity V (t) is constant. This feature is seen also in the case of a V-shaped body analyzed in [11]. For they only treated the time dependency of the drag force when the velocity V (t) is constant, our result is the first case to obtain a precise asymptotic convergence rate for a model with such feature. Other boundary conditions, for example the Maxwell boundary condition, are treated in [1,9] for the case without the wall. The cases with the wall and boundary conditions other than the specular one are left for future study. For other related mathematical and numerical studies, we refer the book by Buttà, Cavallaro and Marchioro [3] and the references therein. We briefly mention here that the approach to the terminal velocity is algebraic also for a (non-stationary) Stokes fluid [2,7,8]. This is also caused by the dependence of the drag force on the past history of the motion.
2. Mathematical Formulation of the Problem. Consider a cylinder of radius R and height h in a d-dimensional Euclidean space R d . We assume that d ≥ 2. Writing the coordinate of the space as x = (x 1 , x ⊥ ) ∈ R × R d−1 , this cylinder is described by the set is the x 1 -coordinate of the left end of the cylinder and is allowed to vary in time ( Figure 1). This cylinder is placed in a d-dimensional region bounded by an infinite plane wall which is the half plane R d Figure 1. A two dimensional picture of a cylinder immersed in a semi-infinite expanse of gas in a region bounded by an infinite plane wall is shown. A constant force E is applied in the direction of the axis of the cylinder and a drag force D V (t) is exerted to the cylinder from the surrounding gas. The balance between the applied force E and the drag force D V (t) determines the dynamics of the cylinder: The right and the left boundary of C V (t) are written C ± V (t) respectively: Note that when we use V as a subscript, this means that the subscripted quantity may depend on the whole history of the motion, that is, The state of the gas is described by the distribution function f (x, ξ, t), where x is the position variable and moves through R d + \C V (t) and ξ = (ξ 1 , ξ ⊥ ) ∈ R × R d−1 is the velocity variable of the constituent molecules. We assume that the region is initially occupied by a semi-infinite expanse of ideal monatomic gas in a quiescent equilibrium of pressure p 0 and temperature T 0 , that is, the distribution function f is initially the Maxwell-Boltzmann distribution with pressure p 0 and temperature T 0 .
where β 0 = (2R g T 0 ) −1 and R g is the gas constant. We consider the gas to be in the free molecular regime. This means that the time evolution of the distribution function f obeys the free Vlasov equation For the boundary condition, we impose the specular boundary condition.
Here V = (V (t), 0) ∈ R × R d−1 and n is the outward unit normal vector to ∂C V (t). A constant force E > 0 is applied to the cylinder in the direction away from the plane wall and parallel to the axis of the cylinder. A drag force D V (t) is exerted to the body from the surrounding gas and is given by See [5] for the derivation of the formula. Note that there is no contribution to the drag force from the lateral boundary C S V (t) of the cylinder. Therefore the dynamics of the cylinder is described by the equations Here we assumed that the mass of the cylinder is unity. The initial conditions are where L and V 0 are positive constants. For a fixed velocity ξ, the Vlasov equation (1) is a constant coefficient transport equation and is therefore solvable by the method of characteristics. We write the set of (x, ξ) representing incoming molecules into C + V (t) and C − V (t) by , we denote the backward characteristics starting from (x, ξ) by (x(s; x, ξ, t), ξ(s; x, ξ, t)), where 0 ≤ s ≤ t. We may hereafter write x(s) = x(s; x, ξ, t) and ξ(s) = ξ(s; x, ξ, t) for notational convenience. More precisely, the definition of the characteristics (x(s), ξ(s)) are given as follows. Define the first pre-collision time τ 1 = τ 1 (x, ξ, t) by where x ∨ y = max(x, y). Here we use the convention that the supremum of the empty set equals −∞. For τ 1 ≤ s ≤ t, the characteristics (x(s), ξ(s)) are defined by If τ 1 = 0, we have thus defined the characteristics (x(s), ξ(s)) for 0 ≤ s ≤ t. If τ 1 > 0, we define the reflected velocity ξ (τ 1 ) as follows.
In particular, D 0 (U ) grows at least linearly in U .
By Lemma 2.1, the equation is uniquely solvable for V ∞ given E > 0. Therefore we can and will consider V ∞ to be the parameter of the problem instead of E. We prove in the sequel that V ∞ is the terminal velocity, that is, lim t→∞ V (t) = V ∞ .

Remark 2.
The asymptotic convergence rate is t −(d−1) in our case -the case with the plane wall; it is t −(d+2) without the plane wall [5]. So the presence of the wall delays the convergence to the terminal velocity V ∞ . This is because the presence of the plane wall strengthens the drag force D V (t).
An explanation for this is as follows. Consider a molecule with velocity ξ impinging on the left side of the cylinder at time t: (x, ξ) ∈ I − V (t). Suppose for simplicity that ξ 1 > V ∞ . If there is no plane wall, then no pre-collisions occur because V (t) < V ∞ from inequality (13). So we have ξ 0 = ξ in this case. On the other hand, if the plane wall is present, then the molecule might have several pre-collisions. Let us assume for simplicity that there are only two pre-collisions: τ 2 > 0 and τ 3 = 0. Since V (t) < V ∞ , the first pre-collision is with the plane wall (at s = τ 1 ); and the second pre-collision is with the cylinder (at s = τ 2 ). See Figure 2.
is smaller if the plane wall is present: The momentum transfer from the surrounding gas to the left side of the cylinder is smaller. This means that the drag force D V (t) is strengthened by the presence of the plane wall.
Remark 2. The asymptotic convergence rate is t (d 1) in our case -the case with the plane wall; it is t (d+2) without the plane wall [5]. So the presence of the wall delays the convergence to the terminal velocity V 1 . This is because the presence of the plane wall strenghthens the drag force D V (t). An explanation for this is as follows. Consider a molecule with velocity ⇠ impinging on the left side of the cylinder at time t: (x, ⇠) 2 I V (t). Suppose for simplicity that ⇠ 1 > V 1 . If there is no plane wall, then no pre-collisions occur because V (t) < V 1 from inequality (13). So we have ⇠ 0 = ⇠ in this case. On the other hand, if the plane wall is present, then the molecule might have several pre-collisions. Let us assume for simplicity that there are only two pre-collisions: is smaller if the plane wall is present: The momentum transfer from the surrounding gas to the left side of the cylinder is smaller. This means that the drag force D V (t) is strengthened due to the presence of the plane wall. Remark 3. Let (X(t), V (t)) be any solution to the problem. If we further assume that is a positive constant. Note that t 0 grows infinitely as ! +0 since L (d 1) ⌧ . These are proved in the appendix (Sections 5.1 and 5.2).

Remark 4.
Under an additional assumption that L (d 1) ⌧ , we can refine inequality (14) to where A = A ( 0 , p 0 , R, V 1 , d),t =t( 0 , p 0 , R, V 1 , d) are positive constants and The precise meaning of the notation L (d 1) ⌧ is explained in the appendix (Section 5).
γ. These are proved in the appendix (Sections 5.1 and 5.2).
wherec =c(β 0 , p 0 , R, V ∞ , d) is a positive constant. This is also proved in the appendix (Section 5.3). If we fix γ and let L → ∞, inequalities (12) and (15) become These are exactly the same estimates obtained in the case without the plane wall [5].
is a solution to the problem. See Remark 8 after Proposition 6.
We can also prove the following theorems.
to the problem. Moreover, any solution (X(t), V (t)) to the problem satisfies the following inequalities: and L ∈ [L 0 , ∞), any solution (X(t), V (t)) to the problem satisfies the inequality We only prove Theorems 3.1 and 3.2. Theorems 3.3 and 3.4 can be proved similarly by the method developed in this paper. 4. Proof of the Theorems.

4.1.
Strategy of the Proof. We first briefly describe the strategy of our proof. First, we define a function space Ω(γ, L, A * , B * , V ∞ ) as follows. In the following, A * and B * are positive constants.
is Lipschitz continuous in t, W (0) = V 0 and satisfies for t ≥ 0 the following inequalities. where with the initial conditions Here K W (t) is defined by By Lemma 2.1, we have We note here that r ± W (t) are computed via the characteristics (x(s), ξ(s)) determined from the dynamics of the cylinder described by (X W (t), W (t)).

4.2.
Analysis of the Characteristics for (x, ξ) ∈ I + W (t). Let us take W from Ω(γ, L, A * , B * , V ∞ ). 2 To obtain an estimate for r + W (t), we analyze the characteristics starting from (x, ξ) ∈ I + W (t). We define the modified first pre-collision timeτ 1 byτ Note the difference between τ 1 (definition (6)) andτ 1 : τ 1 takes into account the plane wall at x 1 = 0 andτ 1 do not. Ifτ 1 > 0 (and not just τ 1 > 0), we have See Figure 3. Introducing a function we see that this is equivalent to Furthermore, the following inequality holds.
See Figure 3 again.

4.3.
Non-Negativity of r + W (t) and its Upper Bound. We prove here the nonnegativity of r + W (t) and obtain its upper bound. First, we derive estimates for W (t) − ξ 1 . In the following, C represents a positive constant depending only on β 0 , p 0 , R, V ∞ and d, which might change from place to place.
We next obtain an upper bound for r + W (t). Proposition 2. Let W ∈ Ω(γ, L, A * , B * , V ∞ ). Then Proof. We define two sets A ≤t/2 and A >t/2 as follows.

WALL EFFECT ON THE MOTION OF A RIGID BODY 13
By Lemma 4.2, we have Next we derive an estimate for II. By Lemma 4.3, we have These estimates prove inequality (26).
The lemma above shows a necessary condition for τ 2 > 0. We give below a sufficient condition. This will be needed when we derive a lower bound for r − W (t) in Section 4.9.
If the following conditions are satisfied, then τ 2 > 0.
Proof. Condition (iii) implies that By the intermediate value theorem, there exists s ∈ (0, t) such that Let σ 2 be the largest s ∈ (0, t) satisfying equation (30). If we can show that we can easily see that τ 2 = σ 2 ∈ (0, t) and the lemma is proved. To prove inequality (31), note that by conditions (i) and (ii), we have and by the definition of σ 2 , we have These imply inequality (31).

4.5.
Non-Negativity of r − W (t) and its Upper Bound. First we prove the nonnegativity of r − W (t).
We next prove an upper bound for r − W (t).
Proof. We split r − W (t) into two parts.
By inequality (17), it follows that For II, we use Lemma 4.4. Note that since ξ 1 > V ∞ , we have ξ 01 = ±ξ 1 if τ 1 = 0 or τ 2 = 0 and there is no contribution to the integral defining II. Therefore we can assume that τ 2 > 0. Thus equation (28) and Lemma 4.4 imply that These estimates for I and II prove the proposition. 4.6. Proof of V W ∈ Ω. Let W ∈ Ω = Ω(γ, L, A * , B * , V ∞ ). We prove here that V W ∈ Ω for sufficiently small γ and sufficiently large L. Two positive constants A + = A + (β 0 , p 0 , R, V ∞ , d) and B + = B + (β 0 , p 0 , R, V ∞ , d) are chosen appropriately and set equal to A * and B * respectively. Proposition 5. There exist positive constants γ 0 , L 0 , A + and B + depending only on β 0 , p 0 , R, V ∞ and d such that for all γ ∈ (0, γ 0 ] and L ∈ [L 0 , ∞), W ∈ Ω = Ω(γ, L, defined by equations (19), (20) and (21). Moreover, inequality (17) is strict, that is, Proof. Let W ∈ Ω(γ, L, A * , B * , V ∞ ). First, note that by the non-negativity of r ± W (t) (Propositions 1 and 3) and Lemma 2.1, we have Therefore, inequality (18) is proved for V W (t). We next prove inequality (17). By Propositions 2 and 4, we have for both cases d ≥ 3 and d = 2 Splitting the integral at t/2, we obtain for someC =C(β 0 , p 0 , R, V ∞ , d) > 0. Now we take A + = B + = A * = B * = 2C. Note that A + and B + depends only on β 0 , p 0 , R, V ∞ and d. Finally, we take γ 0 small enough and L 0 large enough so that Note that this smallness and largeness depend only on the above mentioned parameters. With this γ 0 and L 0 , we have for all γ ∈ (0, γ 0 ] and L ∈ [L 0 , ∞) This proves inequality (17) for V W (t) and the inequality holds strictly. Note that V W (t) is Lipschitz continuous because r ± W (t) are bounded. It remains to show that V W (t) ≥ V ∞ /2. Take γ 0 small enough and L 0 large enough so that

Then we have by inequality (17) that
. By Propositions 2, 4 and 5, the map W → V W maps K into itself. The Arzelà-Ascoli theorem implies that K is a compact subset of C b ([0, ∞)). 3 We prove next that the map K W → V W ∈ K is continuous in the topology of C b ([0, ∞)).
Proof. We only treat r + W (t) here because r − W (t) can be handled similarly. Let (x, ξ) ∈ I + W (t). Note that for sufficiently large j, we have ξ 1 < W j (t). Take m to be the integer satisfying τ m > 0 and τ m+1 = 0 for the dynamics given by W (t). Let τ n,j denote the n-th pre-collision time for the dynamics given by W j (t). We now prove that for sufficiently large j, we have τ m,j > 0 and τ m+1,j = 0. Moreover, we show that τ n,j → τ n for 1 ≤ n ≤ m.
We first treat the case of m = 0, the case without pre-collisions. We first note that m = 0 implies ξ 1 < X(t) + h t because otherwise there would be a pre-collision at the plane wall. Note that we also have ξ 1 < X j (t) + h t for sufficiently large j, where X j (t) = L + t 0 W j (s) ds. In order to have m = 0, we have two possibilities, that is, the characteristic curve x(s) never catches up the cylinder in the x 1 -direction (i); or it does catch up the cylinder in the x 1 -direction but escapes in the x ⊥ -direction (ii). Expressed in equations, these are (i): (33) Here σ 1 is the largest s ∈ (0, t) satisfying ξ 1 = W s,t , which exists by inequality (32). For the case (i), we have for sufficiently large j. Hence there also are no pre-collisions for the dynamics given by W j (t). For the case (ii), we have inf 0≤s<t W j s,t < ξ 1 for sufficiently large j. Let σ 1,j be the largest s ∈ (0, t) satisfying ξ 1 = W j s,t . By the definition of σ 1,j , we have ξ 1 < W j s,t for s ∈ (σ 1,j , t), Hence for any convergent subsequence {σ 1,j } of {σ 1,j }, we have ξ 1 < W s,t for s ∈ (σ 1 * , t), where σ 1 * is the limit of the subsequence. We excluded the equality in the first inequality because it only happens on a measure theoretically negligible set. 4 This shows that σ 1 * = σ 1 and hence σ 1,j → σ 1 . By inequality (33), we have for sufficiently large j. Thus there also are no pre-collisions for the dynamics given by W j (t).
We consider next the case of m = 1. We first treat the case of x 1 (τ 1 ) = 0. In this case, we have Let η 1 be the largest s ∈ (0, t) satisfying ξ 1 = W s,t . Note that we have Define σ 1,j by the equation Assuming that τ 1 > 0, we have σ 1,j > 0 for sufficiently large j. Note that τ 1 = 0 implies ξ 1 = (X(t) + h)/t and therefore this case is measure theoretically negligible. Note also that for sufficiently large j. Let η 1,j be the largest s ∈ (0, t) satisfying ξ 1 = W j s,t . A similar argument as in the case of m = 0 shows that η 1,j → η 1 . Hence inequality (34) implies |x ⊥ − (t − η 1,j )ξ ⊥ | > R for sufficiently large j. These show that τ 1,j = σ 1,j > 0 and that x 1 (τ 1,j ) = 0. For a simple geometrical reason, there are no pre-collisions after a pre-collision at the plane wall. Hence τ 2,j = 0. We next treat the case of x(τ 1 ) ∈ C + W (τ 1 ). In this case, we have inf 0≤s<t W s,t < ξ 1 Therefore we have inf 0≤s<t W j s,t < ξ 1 for sufficiently large j. Let σ 1,j be the largest s ∈ (0, t) satisfying ξ 1 = W j s,t . A similar argument as in the case of m = 0 shows that σ 1,j → τ 1 . By inequality (35), we have |x ⊥ − (t − σ 1,j )ξ ⊥ | < R for sufficiently large j. This shows that τ 1,j = σ 1,j > 0. Lastly, an argument similar to the case of m = 0 shows that τ 2,j = 0 for sufficiently large j. Note also that since τ 1,j → τ 1 , we have Therefore we can extend the argument employed here to treat general m.
Let ξ 0,j = ξ j (τ m,j ). Here the characteristics (x j (s), ξ j (s)) are defined using W j . From what we showed above, we have ξ 0,j → ξ 0 a.e. (x, ξ) ∈ I + W (t). Hence by the Lebesgue convergence theorem, we conclude that Remark 8. A similar argument as in the proof above shows that r ± W (t) are continuous in t. Therefore V (t) is necessarily continuously differentiable if (X(t), V (t)) is a solution to the problem.
Proof. Duhamel's formula implies that We show here that as j → ∞ uniformly in t ≥ 0. All other terms are similarly treated or are easier to treat. Now for any ε > 0, take T > 0 such that for all t ≥ T t T e −C+(t−s) r + Wj (s) + r + W (s) ds < ε/2. This is possible because by Proposition 2, r + Wj (t) and r + W (t) decay as t → ∞ uniformly in j. Now by the Lebesgue convergence theorem and Proposition 6, there exists N = N (T ) ∈ N such that for all j ≥ N T 0 r + Wj (s) − r + W (s) ds < ε/2.
for t ≤ T . This proves (36) and Now applying Schauder's fixed point theorem shows that there exists a fixed point V ∈ K for the map W → V W . This proves the first part of Theorem 3.1.

4.8.
Proof of the Second Part of Theorem 3.1. Let (X(t), V (t)) be any solution to the problem. Define T by Here we use the convention that the infimum of the empty set equals +∞. It is obvious that T > 0. Suppose that T < +∞. By the definition of T , we have for t ≤ T and By inequality (37), taking γ 0 small enough and L 0 large enough, we have V (t) ≥ V ∞ /2 for all γ ∈ (0, γ 0 ], L ∈ [L 0 , ∞) and t ≤ T . Using this, we can prove as in the proof of Propositions 1 and 3 that r ± V (t) ≥ 0 for t ≤ T . Hence from the equation Hence inequality (13) holds for t ≤ T . Now Proposition 5 5 shows that which is a contradiction. Therefore T = +∞. Hence inequality (37) holds for t ≥ 0. Taking γ 0 small enough and L 0 large enough if necessary, it follows that V (t) ≥ V ∞ /2 for t ≥ 0. This shows that r ± V (t) ≥ 0 and V (t) < V ∞ for t ≥ 0. As a consequence, inequality (13) holds for t ≥ 0. This concludes the proof of Theorem 3.1.

4.9.
Lower Bound for r − W (t). We prove here a lower bound for r − W (t). Proposition 8. Let W ∈ Ω(γ, L, A * , B * , V ∞ ). Then we have Proof. Let (x, ξ) ∈ I − W (t) and suppose that conditions (i), (ii) and (iii) in Lemma 4.5 are satisfied. Then τ 2 > 0 by Lemma 4.5. Since Hence for t > L/2, we have 4.10. Proof of Theorem 3.2. Let (X(t), V (t)) be any solution to the problem. By Theorem 3.1, we see that V ∈ Ω(γ, L, A + , B + , V ∞ ). Hence by Proposition 8, we have for t > L is a positive constant. This concludes the proof of Theorem 3.2.

5.
Appendix. We prove here several assertions made in Remarks 3 and 4. In this appendix, we consider L = L γ to be a function of γ and write L −(d−1) γ to mean L −(d−1) γ = o(γ) as γ → +0. In the following, (X(t), V (t)) is a solution to the problem, and K V (t) is defined by equation (21) using V in place of W .
γ, then V (t) > V 0 for t > 0. First, by equations (4), (11) and Now V ∈ Ω(γ, L, A + , B + , V ∞ ) by Theorem 3.1, and we can use Propositions 2 and 4 to obtain upper bounds for r + V (t) and r − V (t). These and K V (t) ≥ C + show that for γ sufficiently small. Note that we used and V (t) > V 0 for t > 0 follows from this.

5.3.
Proof of a Refined Lower Bound of V ∞ − V (t). Finally, we give a proof of the refined lower bound (15) of V ∞ − V (t), under an additional assumption that L −(d−1) γ. As in [5], define s 0 by We prove below a lower bound Heret =t(β 0 , p 0 , R, V ∞ , d) is a positive constant and t 1 is given by wherec =c(β 0 , p 0 , R, V ∞ , d) is a positive constant. To obtain this, we first prove the following: If we taket sufficiently large andc > 0 sufficiently small, (i)-(v) below hold.
Finally, by Proposition 3 and inequality (41), we have Therefore, by choosing a positive constant A − = A − (β 0 , p 0 , R, V ∞ , d) appropriately, we have V ∞ − V (t) ≥ γe C−t + γ 4 A − t d+2 for t 1 > t >t. If t > L in addition, a similar argument as in Section 4.10 shows This proves inequality (15).