SOME ELEMENTARY INEQUALITIES IN GAS DYNAMICS EQUATION
V. A. KLYACHIN, A. V. KOCHETOV, AND V. M. MIKLYUKOV
Received 12 January 2005; Accepted 25 August 2005
We describe the sets on which difference of solutions of the gas dynamics equation satisfy some special conditions. By virtue of nonlinearity of the equation the sets depend on the solution gradient quantity. We show double-ended estimates of the given sets and some properties of these estimates.
Copyright © 2006 V. A. Klyachin et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Main results
n(cid:2)
(cid:4)
(cid:4)
= 0,
Consider the gas dynamics equation
(cid:3) σ
(cid:3) |∇ f |
c=1
(1.1) fxi ∂ ∂xi
(cid:5)
(cid:6)1/(γ−1)
where
(1.2) . t2 σ(t) = 1 − γ − 1 2
(cid:7)
(cid:8)
Here γ is a constant, −∞ < γ < +∞. This equation describes the velocity potential of a steady-state flow of ideal gas in the adiabatic process. In the case n = 2 the parameter γ characterizes the flow of substance. For different values γ it can be a flow of gas, fluid, plastic, electric or chemical field in different mediums, and so forth (see, e.g., [1, Section 2], [2, Section 15, Chapter IV]). For γ = 1 ± 0 we assume
− 1 2
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 21693, Pages 1–29 DOI 10.1155/JIA/2006/21693
(1.3) σ(t) = exp . t2
2 Some elementary inequalities in gas dynamics equation
⎛
⎝
(cid:11)
The case of γ = −1 is known as the minimal surface equation (Chaplygin’s gas):
⎞ ⎠ = 0.
∇ f 1 + |∇ f |2
div (1.4)
(cid:14)
(cid:16)
(cid:15)
For γ = −∞, (1.1) becomes the Laplace equation. In general, a solution of (1.1) with a function σ of variables (x1,...,xn) is called σ- harmonic function. Such functions were studied in many works (see., e.g., [3, 4] and liter- ature quoted therein). We set Ωγ = Rn for γ ≤ 1,
(1.5) ξ ∈ Rn : |ξ| < for γ > 1. Ωγ = 2 γ − 1
n(cid:2)
n(cid:2)
(cid:3)
(cid:4)(cid:3)
(cid:4)2 ≤
(cid:4) ,
The following inequalities were crucial in previous analysis of solutions to (1.1) for γ = −1 (see [5–9]):
(cid:3) σ
|ξ|
(cid:3) |η|
(cid:3) ξi − ηi
(cid:4) ξi − σ
(cid:4) ηi
i=1
i=1
n(cid:2)
n(cid:2)
(cid:3)
(cid:4)(cid:3)
(cid:4) ,
(1.6) c1 ξi − ηi ξ,η ∈ Ωγ,
(cid:3) σ
(cid:3) |ξ|
|η|
(cid:3) σ
(cid:3) |ξ|
(cid:4)2 ≤ c2
(cid:4) ξi − σ
(cid:4) ηi
(cid:4) ξi − σ
(cid:4) (cid:3) |η| ηi
i=1
i=1
(1.7) ξi − ηi ξ,η ∈ Ωγ.
Here ξ = (ξ1,ξ2,...,ξn), η = (η1,η2,...,ηn) and c1 > 0, c2 > 0 are constants not depend- ing on ξ and η.
(cid:4)
(cid:17)
=
In general, the latter inequalities are valid only on subsets of Ωγ × Ωγ with c1 and c2 depending on these subsets. The purpose of the present paper is to describe that depen- dence. Introduce the sets
(cid:4)
(cid:18) (ξ,η) ∈ Ωγ × Ωγ : ξ,η satisfy (1.6) (cid:17)
=
(cid:3) c1 (cid:3) c2
(1.8) (cid:2)γ , (cid:18) (1.9) . (cid:3)γ (ξ,η) ∈ Ωγ × Ωγ : ξ,η satisfy (1.7)
(cid:14)
(cid:16)
(cid:15)
Generally, the sets (cid:2)γ(c1) and (cid:3)γ(c2) have a complicated structure. We will describe them by comparing with canonical sets of the “simplest form.” We set Σγ = {x ∈ R : x ≥ 0} for γ ≤ 1 and
(1.10) x ∈ R : 0 ≤ x < for γ > 1. Σγ = 2 γ − 1
V. A. Klyachin et al. 3
γ on Σγ × Σγ by
γ and I +
⎧ ⎪⎨
For every γ ∈ R, define the functions I −
⎪⎩
⎧ ⎪⎨
if x (cid:7)= y, I − γ (x, y) = if x = y, xσ(x) − yσ(y) x − y σ(x) + σ (cid:8)(x)x (1.11)
⎪⎩ 1
if x2 + y2 > 0, xσ(x) + yσ(y) x + y I + γ (x, y) = if x = y = 0.
γ are continuous on the closing of Σγ × Σγ and they
γ and I +
γ (ε) = {(ξ,η) ∈ Ωγ × Ωγ : I − γ (ε) = {(ξ,η) ∈ Ωγ ×Ωγ : I −
γ (|ξ|, |η|) ≥ ε}, W + γ (|ξ|, |η|) ≤ ε}, V +
γ (ε) = γ (ε) =
{(ξ,η) ∈ Ωγ ×Ωγ : I + {(ξ,η) ∈ Ωγ × Ωγ : I +
γ (|ξ|, |η|) ≥ ε}, V − γ (|ξ|, |η|) ≤ ε}.
Note that the functions I − are infinitely differentiable at each inner point of Σγ × Σγ. For arbitrary ε ≥ 0 we put W −
Also we will need the sets Dγ = {(ξ,ξ) ∈ Ωγ × Ωγ}, Qγ = {(ξ,η) ∈ Ωγ × Ωγ : ξσ(|ξ|) = ησ(|η|)}. The main result of our paper are the following theorems.
(cid:4)
(cid:4)
(cid:3) W −
(cid:3) W +
∀ε ∈ (0,1),
⊂ (cid:2)γ(ε) ⊂
γ (ε) ∪ Dγ
γ (ε) ∪ Dγ
Theorem 1.1. For every γ ∈ R,
(1.12) (cid:2)γ(ε) = Dγ ∀ε ∈ [1,+∞).
(cid:4)
(cid:4)
∀(cid:2) ∈ (0,1),
Theorem 1.2. (a) If γ ∈ (−∞, −1], then
(cid:3) V +
(cid:3) V −
⊂ (cid:3)γ(ε) ⊂
γ (ε) ∪ Dγ
γ (ε) ∪ Dγ
(1.13)
(1.14) (cid:3)γ(ε) = R2n ∀ε ∈ [1,+∞).
(cid:4)
(cid:4)
(cid:3) V −
∀ε ∈ (0,1),
⊂ (cid:3)γ(ε) ⊂
(cid:3) γ (ε) ∩ W − V +
γ (ε) ∪ Qγ
(b) If γ ∈ (−1,+∞), then
γ (0) W −
γ (0) ⊂ (cid:3)γ(ε) ∀ε ∈ [1,+∞).
(1.15)
Relation (1.14) was first proved for γ = −1 and ε = 1 in [5] and later repeatedly in [6–9].
2. Properties of σ
Consider the equation
(2.1) θ(cid:8)(t) = ε,
4 Some elementary inequalities in gas dynamics equation
where θ(t) = tσ(t) and ε is an arbitrary parameter. It is easy to verify that for γ (cid:7)= 1, (2.1) can be rewritten in the following form:
(2.2) σ(t) + ε = 0. 2 γ − 1 σ 2−γ(t) − γ + 1 γ − 1
(cid:4)
(cid:23) (cid:24) (cid:24) (cid:25) 2
For arbitrary ε ∈ (0,1) we set
(cid:3) 1 − εγ−1 γ − 1 (cid:26)
−2lnε.
if γ (cid:7)= 1, rγ(ε) = (2.3)
r1(ε) =
(cid:11)
Observe that rγ(ε) ∈ Σγ for every γ ∈ R and every ε ∈ (0,1). The following assertions hold. (1) Let γ ∈ R. Then the domain of σ is the set Σγ. Moreover, σ(0) = 1, σ(+∞) = 0 2/(γ − 1)) = 0 for γ > 1. for γ ≤ 1 and σ( (2) For each γ ∈ R we have
(2.4) 0 < σ(t) ≤ 1 ∀t ∈ Σγ.
(3) Let γ ∈ R. Then σ (cid:8)(0) = 0 and
(2.5) σ (cid:8)(t) < 0 ∀t > 0, t ∈ Σγ.
(4) If γ ∈ (−∞, −1], then
θ(cid:8)(+∞) = 0, θ(cid:8)(0) = 1, (2.6) θ(cid:8)(t) > 0 ∀t ∈ [0,+∞).
(cid:28)
(cid:27)(cid:15)
= 0,
(5) If γ ∈ (−1,+∞), then
(cid:27)
(cid:28)
θ(cid:8)(0) = 1, θ(cid:8)
2 γ + 1 (cid:15)
(cid:15)
(2.7) , 0, θ(cid:8)(t) > 0 ∀t ∈ 2 γ + 1
θ(cid:8)(t) < 0 ∀t > , t ∈ Σγ. 2 γ + 1
V. A. Klyachin et al. 5
Moreover,
(cid:27)(cid:15)
= 0
if γ ∈ (−1,1], θ(cid:8)(+∞) = 0 (cid:28)
(cid:28)
(cid:27)(cid:15)
θ(cid:8) if γ ∈ (1,2), 2 γ − 1
= −2
(cid:28)
(cid:27)(cid:15)
− 0
(2.8) θ(cid:8) if γ = 2, 2 γ − 1
= −∞ if γ ∈ (2,+∞).
θ(cid:8) 2 γ − 1
(6) If γ ∈ (−∞, −1] ∪ [2,+∞), then θ(cid:8)(cid:8)(0) = 0 and
(2.9) θ(cid:8)(cid:8)(t) < 0 ∀t > 0, t ∈ Σγ.
(cid:28)
(cid:27)(cid:15)
= 0,
(7) If γ ∈ (−1,2), then
(cid:27)
(cid:28)
θ(cid:8)(cid:8) θ(cid:8)(cid:8)(0) = 0,
6 γ + 1 (cid:15)
(cid:15)
(2.10) , 0, θ(cid:8)(cid:8)(t) < 0 ∀t ∈ 6 γ + 1
θ(cid:8)(cid:8)(t) > 0 ∀t > , t ∈ Σγ. 6 γ + 1
(cid:4) ,
(8) For every γ ∈ R and every (cid:2) ∈ (0,1), (2.1) has a unique positive solution sγ(ε) ∈ (0,rγ(ε)) and
(cid:29) 0,sγ(ε)
(2.11) θ(cid:8)(t) > ε ∀t ∈ θ(cid:8)(t) < ε ∀t > sγ(ε), t ∈ Σγ.
(cid:15)
Moreover, for every γ > −1 and (cid:2) ∈ (0,1),
(2.12) . sγ(ε) < 2 γ + 1
(9) Let γ ∈ R. Then for all x, y ∈ Σγ, x2 + y2 > 0,
γ (x, y) < 1.
(2.13) I − γ (x, y) ≤ I +
6 Some elementary inequalities in gas dynamics equation
⎧
(cid:5)
(cid:6)(2−γ)/(γ−1)
Proof. The proof of assertions (1)–(7) follows from the equalities
−t
(cid:7)
⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩
−t exp
⎧
(cid:6)(cid:5)
(cid:6)(2−γ)/(γ−1)
if γ (cid:7)= 1, t2 (cid:8) σ (cid:8)(t) = if γ = 1, t2 1 − γ − 1 2 − 1 2
(cid:8)
(cid:5) 1 − γ + 1 2 (cid:4)
⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩
(cid:3) 1 − t2
⎧
− 1 2 (cid:6)(cid:5)
(cid:6)(3−2γ)/(γ−1)
−t
if γ (cid:7)= 1, t2 1 − γ − 1 2 t2 (cid:7) (2.14) θ(cid:8)(t) = exp if γ = 1, t2
(cid:8)
(cid:5) 3 − γ + 1 t2 2 (cid:7) (cid:4)
⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎩
(cid:3) t2 − 3
− 1 2
(cid:4)
(cid:4)
(cid:4)
(cid:4)
if γ (cid:7)= 1, t2 1 − γ − 1 2 θ(cid:8)(cid:8)(t) = exp t if γ = 1. t2
= ε = θ(cid:8)
= σ
(cid:3) rγ(ε)
(cid:3) sγ(ε)
(cid:3) sγ(ε)
(cid:3) sγ(ε)
(cid:3) sγ(ε)
(2.15) Let γ ∈ R and (cid:2) ∈ (0,1). Suppose that sγ(ε) ∈ Σγ satisfies (2.1). We have (cid:4) . < σ σ + sγ(ε)σ (cid:8)
From this sγ(ε) < rγ(ε). Next, using assertions (4)–(7), we obtain assertion (8). We prove assertion (9). Let x, y ∈ Σγ, x2 + y2 > 0. If x = y, then
γ (x, y) < 1.
γ (x, y) = σ(x) + xσ (cid:8)(x) < σ(x) = I + I −
(2.16)
Suppose that x > y. Since
(2.17) σ(x) < σ(y),
= σ(x)
γ (x, y) = xσ(x) − yσ(y) I −
we obtain
≤ xσ(x) − yσ(x) x − y
= I +
x − y
γ (x, y)
= xσ(x) + yσ(x) x + y
≤ xσ(x) + yσ(y) x + y
(2.18)
= σ(y) ≤ 1.
< xσ(y) + yσ(y) x + y
(cid:2)
The case x < y is analogous.
γ (ε), V −
γ (ε)
γ (ε), W +
3. Properties of W −
γ (ε), and V + γ (ε), V −
γ (ε).
γ (ε), W +
γ (ε) and V + We say that a set G ⊂ Rn is linearly connected if any pair of points x, y ∈ G can be joined
Here we study the sets W −
on D by an arc.
V. A. Klyachin et al. 7
γ (ε) = ∅ for every γ ∈ R and ε > 1. γ (1) = {0} for every γ ∈ R.
γ (ε) for every γ ∈ R and ε ∈ (0,1). γ (ε) = Ωγ × Ωγ for every γ ∈ R and ε ≥ 1.
γ (ε) for every γ ∈ R and ε ∈ (0,1). γ (ε) is linearly connected for every γ ∈ R and ε ∈ (0,1). γ (0) is linearly connected for every γ > −1. γ (ε) is linearly connected for every γ ∈ R and ε ∈ (0,1).
The following assertions hold.
(cid:17)
(cid:18)
(1) W − γ (ε) = W + (2) W − γ (1) = W + (3) W − γ (0) = R4 for every γ ≤ −1. γ (0) = Ωγ × Ωγ for every γ ∈ R. (4) W + (5) W − γ (ε) ⊂ W + (6) V − γ (ε) = V + (7) V − γ (0) = ∅ for every γ ≤ −1. γ (0) = ∅ for every γ ∈ R. (8) V + γ (ε) ⊂ V − (9) V + (10) The set W − (11) The set W − (12) The set W + (13) For every γ ∈ R and ε ∈ (0,1), we have
⊂ W −
γ (ε).
(3.1) (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ sγ(ε), |η| ≤ sγ(ε)
(cid:17)
Here sγ(ε) is a unique positive solution of (2.1). (14) For every γ ∈ R and ε ∈ (0,1), we have
(cid:18) .
γ (ε) ⊂
(3.2) W − (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ rγ(ε), |η| ≤ rγ(ε)
(cid:14)
(cid:16)
(cid:15)
(cid:15)
(15) If γ > −1, then
⊂ W −
γ (0).
(3.3) , |η| ≤ (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ 2 γ + 1 2 γ + 1
(cid:18)
(cid:17)
(16) For every γ ∈ R and ε ∈ (0,1), we have
⊂ W +
γ (ε).
(3.4) (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ rγ(ε), |η| ≤ rγ(ε)
(cid:17)
(17) For every γ ∈ R and ε ∈ (0,1), we have
(cid:18) .
γ (ε) ⊂
(3.5) V − (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≥ sγ(ε) or |η| ≥ sγ(ε)
(cid:18)
(cid:17)
(18) For every γ ∈ R and ε ∈ (0,1), we have
⊂ V −
γ (ε).
(3.6) (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≥ rγ(ε) or |η| ≥ rγ(ε)
(cid:14)
(cid:16)
(cid:15)
(cid:15)
(19) If γ > −1, then
γ (0) ⊂
(3.7) V − or |η| ≥ . (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≥ 2 γ + 1 2 γ + 1
(cid:18) .
γ (ε) ⊂
(20) For every γ ∈ R and ε ∈ (0,1), we have (cid:17) (3.8) V + (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≥ rγ(ε) or |η| ≥ rγ(ε)
8 Some elementary inequalities in gas dynamics equation
γ (ε). To prove the statement, it is sufficient to show that W −
Proof of assertions (1)–(9). The proof follows from assertions (4) and (9) of Section 2. (cid:2)
γ (ε) be arbitrary. Let (cid:4)(cid:8), (cid:4)(cid:8)(cid:8) be the segments with the endpoints 0, ζ (cid:8) and 0, ζ (cid:8)(cid:8), respectively. Denote by (cid:4)(cid:8) ∪ (cid:4)(cid:8)(cid:8) the double curve which consists of two segments (cid:4)(cid:8) and (cid:4)(cid:8)(cid:8). Then this double curve will join the points ζ (cid:8), ζ (cid:8)(cid:8) and it will lie on W −
γ (ε). Assume that Iγ(x, y) ≥ (cid:2). As above, for the case x > y we obtain
Proof of assertions (10)–(12). We prove assertion (10). Fix γ ∈ R, ε ∈ (0,1), and a nonzero point ζ = (ξ,η) ∈ W − γ (ε) contains the segment (cid:4) = {(ξt,ηt) : 0 ≤ t ≤ 1} with the endpoints 0 and ζ. Indeed, let ζ (cid:8),ζ (cid:8)(cid:8) ∈ W −
γ (x, y) ≤ σ(x) < σ(y).
(3.9) ε ≤ I −
From this x, y ∈ [0,rγ(ε)]. The case x < y is analogous. Suppose that x = y. Then
γ (x, y) = θ(cid:8)(x) = σ(x) + xσ (cid:8)(x) ≤ σ(x) = σ(y),
(3.10) ε ≤ I −
and consequently x, y ∈ [0,rγ(ε)]. Thus if Iγ(x, y) ≥ (cid:2), then x, y ∈ [0,rγ(ε)]. Further we will need the function
(cid:3) σ(x) − ε
(cid:4) .
(3.11) μ(x) = x
It is easy to see that for all x, y ∈ [0,rγ(ε)], x (cid:7)= y,
(3.12) I − γ (x, y) = ε ⇐⇒ μ(x) = μ(y).
Define the monotonicity intervals of μ. Since
(3.13) μ(cid:8)(x) = θ(cid:8)(x) − ε,
(cid:4)
from assertion (8) of Section 2 it follows that the function μ is strictly increasing on [0,sγ(ε)] and strictly decreasing on [sγ(ε),rγ(ε)]. Moreover,
= 0.
(cid:3) rγ(ε)
(3.14) μ(0) = μ
γ (x, y) = ε and x = y, then x = y = sγ(ε). Consequently for each x ∈ [0,rγ(ε)] there is a unique number y ∈ [0,rγ(ε)], satisfying (3.12). Therefore there exists the function g : [0,rγ(ε)] → [0,rγ(ε)] such that for all x, y ∈ [0,rγ(ε)],
Note that if I −
(3.15) I − γ (x, y) = ε ⇐⇒ y = g(x).
In addition
(cid:29) 0,sγ(ε)
(cid:4) , (cid:30) ,
sγ(ε) < g(x) ≤ rγ(ε) (3.16) if x ∈ 0 ≤ g(x) < sγ(ε) if x ∈ (cid:3) sγ(ε),rγ(ε)
V. A. Klyachin et al. 9
(cid:4)
(cid:4)
as well as
= 0.
(cid:3) sγ(ε)
= sγ(ε),
(cid:3) rγ(ε)
(3.17) g g g(0) = rγ(ε),
γ (x, y) is infinitely differentiable at each point of [0, rγ(ε)] ×
Note that the function I −
(cid:4)(cid:3)
(cid:4)(cid:4)
(cid:4)
(cid:3)
(cid:4)
(cid:4)
−
− θ
− ε
(cid:3) x0
= θ(cid:8)
= θ(cid:8)
(cid:7)= 0,
(cid:3) x0, y0
(cid:3)
(cid:4)(cid:3)
(cid:4)
(cid:4)(cid:4)
(cid:3) x0 x0 − y0 (cid:4) (cid:3)
(cid:4)
−
− θ
− ε
(cid:3) x0
= θ(cid:8)
= θ(cid:8)
(cid:7)= 0.
(cid:3) x0, y0
(cid:3) (cid:3) x0 − y0 θ x0 (cid:4)2 (cid:3) x0 − y0 (cid:3) (cid:4) (cid:3) θ (cid:4)2 y0 − x0
[0,rγ(ε)]. Fix arbitrary x0, y0 ∈ [0,rγ(ε)], x0 (cid:7)= y0, satisfying (3.15). We have (cid:4) y0 I − γ ∂ ∂x (3.18) y0 y0 y0 − x0 (cid:3) I − γ ∂ ∂y y0 y0 − x0
(cid:31)
−1(cid:31)
(cid:4)
(cid:4)(cid:4)
(cid:3)
(cid:4)(cid:4)
(cid:4) − ε (cid:4)(cid:4)
= −
Using the implicit function theorem, we obtain
(cid:3) x0
(cid:3) x0,g
(cid:3) x0
(cid:3) x0
= θ(cid:8) (cid:3) θ(cid:8) g
− ε
(cid:3) x0 (cid:3) x0
(3.19) . g (cid:8) x0,g I − γ I − γ ∂ ∂y ∂ ∂x
(cid:4)
By assertion (8) of Section 2, (3.16), it follows that
(cid:3) x0
(3.20) g (cid:8) < 0.
γ (ε).
Thus the function y = g(x) is strictly decreasing on [0,rγ(ε)].
We prove that the segment (cid:4) lies in W − Indeed, assume that |ξ| ≤ |η| and for some t ∈ (0,1),
(cid:4) (cid:3) |ξt|, |ηt|
(cid:4)
! !
(3.21) < ε. I − γ
= ε
! !ηt0
(3.22) Then there is a number t0 ∈ (0,1) such that ! (cid:3)! !, !ξt0 I − γ
γ (0,0) = 1 > ε, we have |η| ≤ g(|ξ|). We deduce
(cid:4)
(cid:4)
! !
and hence |ηt0| = g(|ξt0|). Since 0 < rγ(ε) = g(0) and I −
≤ g
= |η| ≤ g
(cid:3) |ξ|
(cid:4) .
(cid:3) |ξ| t0
(cid:3)! !ξt0 t0
g (3.23)
γ (ε) contains (cid:4). The proof of assertion (11) is analogous. Now we prove assertion (12). We fix γ ∈ R, ε ∈ (0,1), and a nonzero point ζ = (ξ,η) ∈ γ (ε). As above, to prove this statement, it is sufficient to show that W + γ (ε) contains the
From this t0 ≥ 1 and we arrive at a contradiction. The case |ξ| > |η| is analogous. Thus W −
(cid:4)
(cid:4)
(cid:4) (cid:3) |ηt|
|ξ|σ
(cid:3) |η|
=
W + segment (cid:4). We have
(cid:4) (cid:3) |ξt|, |ηt|
≥ ε
|ξ|σ(|ξt|) + |η|σ |ξ| + |η|
(cid:3) |ξ| + |η|σ |ξ| + |η|
(3.24) > I + γ
(cid:2)
γ (ε) contains (cid:4).
for all t ∈ (0,1). Thus W +
10 Some elementary inequalities in gas dynamics equation
(cid:17)
(cid:18)
Proof of assertions (13), (15), (17), and (19). Let
(3.25) . (ξ,η) ∈ (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ sγ(ε), |η| ≤ sγ(ε)
(cid:4)
(cid:4)
By assertion (8) of Section 2 it follows that
(cid:3) |ξ|
≥ ε,
(cid:3) |η|
≥ ε.
(3.26) θ(cid:8) θ(cid:8)
(cid:4)
(cid:4)
(cid:4)
Suppose that |ξ| = |η|. We have
(cid:3) |ξ|, |η|
(cid:3) |ξ|
= θ(cid:8)
(cid:3) |η|
= θ(cid:8)
≥ ε.
(3.27) I − γ
γ (ε).
From this (ξ,η) ∈ W − Assume that |ξ| < |η|. Using the well-known Lagrange mean value theorem, we obtain
(cid:4) (cid:3) |ξ|, |η|
= θ(cid:8)(c),
|ξ| < c < |η|.
(3.28) I − γ
By assertion (8) of Section 2,
(3.29) θ(cid:8)(c) > ε.
γ (ε). The case |ξ| > |η| is analogous.
Therefore (ξ,η) ∈ W −
The proof of assertion (15) is analogous. Assertion (17) follows from assertion (13), (cid:2) and assertion (19) follows from assertion (15).
γ (ε). Assume that |ξ| = |η|. We have
(cid:4)
(cid:4)
(cid:4)
(cid:3)
Proof of assertions (14) and (18). Let (ξ,η) ∈ W −
(cid:4) (cid:3) |ξ|, |η|
(cid:3) |ξ|
= θ(cid:8)
(cid:3) |ξ|
= σ
(cid:4) (cid:3) |ξ|
≤ σ
|ξ|
= σ
(cid:3) |η|
(cid:4) .
(3.30) + |ξ|σ (cid:8) ε ≤ I − γ
(cid:4)
(cid:4)
(cid:3)
Then the inequalities
|ξ|
= σ
(cid:3) |η|
≥ ε
(3.31) σ
imply
|ξ| = |η| ≤ rγ(ε).
(3.32)
(cid:17)
(cid:18)
Hence
(3.33) (ξ,η) ∈ . (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ rγ(ε), |η| ≤ rγ(ε)
(cid:4)
(cid:4)
(cid:4)
(cid:4)
(cid:3)
|ξ|σ
(cid:3) |η|
|ξ|σ
(cid:3) |ξ|
≤
=
(cid:4) (cid:3) |ξ|, |η|
(cid:3) |ξ|
= σ
Now we assume that |ξ| > |η|. We have (cid:4)
|η|
(cid:4) .
(cid:3) |ξ| − |η|σ |ξ| − |η|
(cid:3) |ξ| − |η|σ |ξ| − |η|
< σ ε ≤ I − γ
(3.34)
(cid:17)
(cid:18)
From this
(3.35) (ξ,η) ∈ . (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ rγ(ε), |η| ≤ rγ(ε)
V. A. Klyachin et al. 11
(cid:2)
The case |ξ| < |η| is analogous. Assertion (18) follows from assertion (14).
(cid:17)
(cid:18)
Proof of assertions (16) and (20). Let
(3.36) . (ξ,η) ∈ (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ rγ(ε), |η| ≤ rγ(ε)
(cid:4)
(cid:4)
(cid:3)
Then
|ξ|
≥ ε,
(cid:3) |η|
≥ ε.
(3.37) σ σ
(cid:4)
(cid:3)
(cid:3)
Suppose |ξ| = |η|. Then
(cid:4) |ξ|, |η|
= σ
|ξ|
≥ ε.
(3.38) I + γ
γ (ε). Assume that |ξ| > |η|. We have
(cid:4)
(cid:4)
(cid:4)
(cid:3)
(cid:4)
|ξ|σ
(cid:3) |η|
|ξ|σ
(cid:3) |ξ|
≥
=
Hence (ξ,η) ∈ W +
(cid:4) (cid:3) |ξ|, |η|
(cid:3) |ξ|
= σ
≥ ε.
|ξ| + |η|σ |ξ| + |η|
(cid:4) (cid:3) |ξ| + |η|σ |ξ| + |η|
(3.39) I + γ
γ (ε). The case |ξ| < |η| is analogous.
From this (ξ,η) ∈ W +
(cid:2)
= {(ξ,η) ∈ Ωγ × Ωγ : I −
γ (|ξ|, |η|) < 0}, U + γ
γ (ε) ∩ Pγ), F −
γ ) ∪ Qγ ∪ (V +
γ ) ∪ Qγ ∪ (V +
γ (ε) = (V −
γ (ε) ∩ U +
Assertion (20) follows from assertion (16).
γ (ε) ∩ U + γ (ε) ∩ U − γ ).
"
(cid:3)
# ,
4. Proofs of main theorems Introduce the sets Hγ = {(ξ,η) ∈ Ωγ × Ωγ : |ξ| = |η|,ξ (cid:7)= η}, Gγ = {(ξ,η) ∈ Ωγ × Ωγ : = {(ξ,η) ∈ Ωγ × Ωγ : I − |ξ| (cid:7)= |η|}, U − γ γ (|ξ|, |η|) > 0}, Pγ = {(ξ,η) ∈ Ωγ × Ωγ : |ξ|σ(|ξ|) = |η|σ(|η|), ξσ(|ξ|) (cid:7)= ησ(|η|)}, γ (ε) = (V + γ (ε) ∩ Pγ) F+ ∪ (V + For any ξ,η ∈ Rn, their inner product is denoted by (cid:17)ξ,η(cid:18). Obviously inequalities (1.6) and (1.7) with some constant ε > 0 can be written as
# ,
(4.1)
(cid:3) |ξ| σ ! !2 ≤ ε
(cid:4) ξ − σ (cid:3) " |ξ|
(cid:4) η,ξ − η |η| (cid:4) (cid:4) (cid:3) η,ξ − η |η| ξ − σ
! !σ
(cid:3) |ξ|
(4.2) ε|ξ − η|2 ≤ (cid:4) (cid:4) (cid:3) |η| ξ − σ η σ
|ξ − η|2 = |ξ|2 + |η|2 − 2|ξ||η|cosϕ,
(cid:4)(cid:4)
(cid:4)
"
(cid:3)
#
respectively. Let ϕ be the angle between the vectors ξ and η. Then
|η|
= σ
(cid:3) σ
|ξ||η|cosϕ,
(cid:3) |ξ| ! !σ
(cid:4) ξ − σ (cid:4) (cid:3) ξ − σ
|ξ|
(cid:4) η,ξ − η ! (cid:4) (cid:3) !2 = σ 2 |η| η
(cid:3) |ξ| (cid:3) |ξ|
(cid:4) |ξ|2 + σ (cid:4) |ξ|2 + σ 2
(cid:3) |η| (cid:3) |η|
(cid:4) |η|2 − (cid:4) |η|2 − 2σ
(cid:3) |ξ| (cid:3) |ξ|
(cid:3) |η| (cid:4) |ξ||η|cosϕ.
σ
+ σ (cid:4) (cid:3) |η| σ (4.3)
12 Some elementary inequalities in gas dynamics equation
We set
(cid:4)
(cid:4)(cid:4)
Υ(ϕ) = |ξ|2 + |η|2 − 2|ξ||η|cosϕ,
(cid:3) σ
|ξ||η|cosϕ,
Φ(ϕ) = σ (4.4)
(cid:3) |ξ| (cid:3) |ξ|
(cid:4) |ξ|2 + σ (cid:4) |ξ|2 + σ 2
(cid:3) |η| (cid:3) |η|
(cid:4) |η|2 − (cid:4) |η|2 − 2σ
(cid:3) |ξ| (cid:3) |ξ|
(cid:3) |η| (cid:3) |η|
(cid:4) |ξ||η|cosϕ.
Ψ(ϕ) = σ 2 + σ (cid:4) σ
(cid:4)
Proof of Theorem 1.1. Fix γ ∈ R and ε > 0. It is clear that inequality (4.1) holds for all (ξ,η) ∈ Dγ.
(cid:3) |η|
= σ
(cid:4) .
(4.5) Let (ξ,η) ∈ (cid:2)γ(ε) ∩ Hγ. In this case inequality (4.1) is rewritten in the form (cid:3) |ξ| ε ≤ σ
Obviously
γ ((cid:2)) ∩ Hγ.
(cid:4)
(cid:4)
⊂
⊂
(4.6) (cid:2)γ((cid:2)) ∩ Hγ = W +
(cid:3) W +
(cid:4) .
(cid:3) (cid:2)γ(ε) ∩ Hγ
γ (ε) ∩ Hγ
γ (ε) ∩ Hγ
(cid:28)
(cid:27)
(4.7) Using assertion (5) of Section 3, we see that (cid:3) W −
(cid:4)(cid:4)(cid:3)
(cid:3)
(cid:3)
(cid:3) |η|
− σ
|ξ|
(cid:4) |ξ||η|sinϕ
=
|ξ|2 − |η|2 Υ2(ϕ)
Let (ξ,η) ∈ Gγ. Then Υ(ϕ) > 0 and after simple calculations we find (cid:4) σ (4.8) . ∂ ∂ϕ Φ(ϕ) Υ(ϕ)
(cid:4)(cid:4)(cid:3)
(cid:4)
It is clear that
(cid:3) σ
(cid:4) (cid:3) |η|
(cid:3) |ξ|
− σ
|ξ|2 − |η|2
(4.9) > 0.
(cid:28)
=
Therefore (cid:27)
(cid:4)
(cid:4)(cid:4)
(cid:3)
min ϕ∈[0,π] Φ(ϕ) Υ(ϕ)
(cid:4) |ξ|2 + σ
|η|
(cid:3) |η|
|ξ||η|
= σ
(cid:4) (cid:3) |ξ|, |η| ,
= I − γ
(cid:3) |ξ| (cid:4)2
(cid:3) (cid:4) |η|2 − σ (cid:3) |ξ| − |η|
(cid:27)
(cid:28)
=
Φ(0) Υ(0) (cid:3) |ξ| + σ
(cid:4)(cid:4)
(cid:3)
max ϕ∈[0,π] Φ(ϕ) Υ(ϕ)
(cid:4) |ξ|2 + σ
|η|
(cid:3) |η|
|ξ||η|
= σ
(cid:3) |ξ|, |η|
(cid:4) .
= I + γ
(cid:4) (cid:3) |ξ| (cid:4)2
(cid:4) (cid:3) |η|2 + σ (cid:3) |ξ| + |η|
Φ(π) Υ(π) (cid:3) |ξ| + σ
(4.10)
(cid:3)
#
" σ
|ξ|
(cid:4) ξ − σ
(cid:4) η,ξ − η
≤
Thus for all (ξ,η) ∈ Gγ,
(cid:4) (cid:3) |ξ|, |η|
(cid:3) |ξ|, |η|
(cid:4) .
≤ I + γ
(cid:3) |η| |ξ − η|2
(4.11) I − γ
V. A. Klyachin et al. 13
(cid:4)
(cid:4)
⊂
⊂
This implies
(cid:3) W −
(cid:3) W +
(cid:4) .
(cid:3) (cid:2)γ(ε) ∩ Gγ
γ (ε) ∩ Gγ
γ (ε) ∩ Gγ
(4.12)
(cid:2)
From this, by (4.7), and assertions (1), (2) of Section 3 we obtain (1.12).
(cid:4)
(cid:4)
(cid:3)
(cid:3)
Proof of Theorem 1.2. (a) We fix γ ≤ −1 and ε > 0. It is clear that inequality (4.2) holds for all (ξ,η) ∈ Dγ. Let (ξ,η) ∈ (cid:3)γ(ε) ∩ Hγ. In this case inequality (4.2) becomes
|ξ|
= σ
|η|
≤ ε.
(4.13) σ
Then
γ (ε) ∩ Hγ.
(cid:4)
(cid:4)
(cid:3)
⊂
⊂
(4.14) (cid:3)γ(ε) ∩ Hγ = V +
(cid:4) .
(cid:3) (cid:3)γ(ε) ∩ Hγ
γ (ε) ∩ Hγ
γ (ε) ∩ Hγ
(4.15) Using assertion (9) of Section 3, we see that (cid:3) V + V −
(cid:3)
(cid:4)2
Let (ξ,η) ∈ Gγ. Then by the inequality
(cid:3) σ
(cid:4) |ξ| − σ
(cid:4) (cid:3) |η| |η|
|ξ|
(cid:4)(cid:4)(cid:3)
(cid:4)(cid:4)
(cid:4)
(cid:4)
(cid:3)
(cid:5)
(cid:6)
(4.16) Ψ(ϕ) ≥
|ξ||η|sinϕ
− |η|2σ 2
(cid:3) |η|
(cid:3) |η|
|ξ|2σ 2
− σ
=
(cid:3) |ξ| Ψ2(ϕ)
and by assertion (4) of Section 2, we conclude that Ψ(ϕ) > 0 for all ϕ ∈ [0,π]. After simple calculations, we obtain (cid:3) |ξ| σ (4.17) . ∂ ∂ϕ Φ(ϕ) Ψ(ϕ)
(cid:4)(cid:4)(cid:3)
(cid:4)
(cid:4)(cid:4)
(cid:3)
(cid:3)
(cid:3) σ
|ξ|
− σ
|η|
|ξ|2σ 2
(cid:3) |ξ|
− |η|2σ 2
(cid:3) |η|
By assertions (3) and (4) of Section 2, it follows that (cid:4) (4.18) < 0.
(cid:28)
=
Therefore (cid:27)
(cid:3)
(cid:3)
(cid:4)
=
(cid:4) ,
min ϕ∈[0,π] Φ(ϕ) Ψ(ϕ)
= σ σ 2
(cid:4) |ξ|2 + σ (cid:4) |ξ|2 + σ 2
(cid:3) (cid:4) |ξ| |η|2 + σ (cid:4) (cid:3) |ξ| |η|2 + 2σ
|η| (cid:3) |η|
(cid:3) |η| (cid:3) |η|
(cid:4)(cid:4) |ξ||η| (cid:4) |ξ||η|
(cid:27)
(cid:28)
=
Φ(π) Ψ(π) (cid:3) |ξ| (cid:3) |ξ| + σ (cid:4) σ 1 (cid:3) |ξ|, |η| I + γ
(cid:4)
(cid:3)
(cid:3)
=
(cid:4) .
max ϕ∈[0,π] Φ(ϕ) Ψ(ϕ)
= σ σ 2
(cid:4) |ξ|2 + σ (cid:4) |ξ|2 + σ 2
(cid:4) (cid:3) |ξ| |η|2 − σ (cid:4) (cid:3) |ξ| |η|2 − 2σ
|η| (cid:3) |η|
(cid:4)(cid:4) (cid:3) |ξ||η| |η| (cid:4) (cid:3) |ξ||η| |η|
Φ(0) Ψ(0) (cid:3) |ξ| (cid:3) |ξ| + σ (cid:4) σ 1 (cid:3) |ξ|, |η| I − γ (4.19)
14 Some elementary inequalities in gas dynamics equation
(cid:3)
#
(cid:4) ≤
≤
Thus for all (ξ,η) ∈ Gγ,
(cid:4) .
" |ξ| σ ! (cid:3) !σ
(cid:4) ξ − σ (cid:4) ξ − σ
(cid:3) |η| (cid:3) |η|
(cid:4) η,ξ − η ! (cid:4) !2 η
|ξ|
(4.20) 1 (cid:3) |ξ|, |η| 1 (cid:3) |ξ|, |η| I − γ I + γ
(cid:4)
(cid:4)
⊂
⊂
This implies that
(cid:3) V +
(cid:3) V −
(cid:4) .
(cid:3) (cid:3)γ(ε) ∩ Gγ
γ (ε) ∩ Gγ
γ (ε) ∩ Gγ
(4.21)
(cid:3)
(cid:4) σ
(cid:3) |ξ|
(cid:4) |ξ|2(1 − cosϕ),
(cid:3) |ξ| (cid:4)
(cid:4) |ξ||η|cosϕ = 2σ 2 (cid:4)(cid:4)
From this, by (4.15) and assertion (6) of Section 3, we obtain (1.13) and (1.14). (b) We fix γ > −1 and ε > 0. It is clear that the inequality (4.2) holds for all (ξ,η) ∈ Qγ. By assertion (5) of Section 2, Qγ (cid:7)= Dγ. Let (ξ,η) ∈ (cid:3)γ(ε) ∩ Pγ. Similarly we establish that Pγ (cid:7)= Hγ. We have
|η| (cid:3)
|ξ|
(cid:4) |η|2 − 2σ (cid:3) (cid:3) σ
|η|
(cid:3)
|ξ|
= σ
|ξ| (cid:3) |ξ|
Ψ(ϕ) = σ 2 (cid:3) Φ(ϕ) = σ
|ξ||η|cosϕ (cid:4) (cid:3) |ξ|2 cosϕ
|ξ|
(cid:3)
(cid:4) (cid:3) (cid:3) |ξ|2 + σ 2 |ξ| |η| (cid:4) (cid:4) (cid:3) |ξ|2 + σ |η|2 − |η| (cid:4) (cid:4) (cid:3) |ξ|2 + σ |ξ||η| − σ |ξ| (cid:4) (cid:3) (cid:4) (1 − cosϕ). |ξ| + |η| |ξ|
|ξ|
= σ
+ σ (cid:4) |ξ||η|cosϕ − σ
(4.22)
It is easy to see that cos ϕ (cid:7)= 1. Indeed, if cosϕ = 1, then ξσ(|ξ|) = ησ(|η|). Next, we find
(cid:4) (cid:3) |ξ|, |η| .
= I + γ
= 2|ξ|σ(|ξ|) |ξ| + |η|
(cid:4)
(4.23) Ψ(ϕ) Φ(ϕ)
≤ ε.
(4.24) Thus inequality (4.2) assumes the form (cid:3) |ξ|, |η| I + γ
Then
γ (ε) ∩ Pγ.
(4.25) (cid:3)γ(ε) ∩ Pγ = V +
γ . By assertion (3) of Section 2 we find that inequality (4.18) is valid.
(cid:4)
(cid:4)
(cid:4)
⊂
⊂
Let (ξ,η) ∈ U + Therefore inequalities (4.20) are true. Hence
(cid:3) V +
(cid:3) V −
γ (ε) ∩ U + γ
(cid:3) (cid:3)γ(ε) ∩ U + γ
γ (ε) ∩ U + γ
(4.26) .
γ . Observe that the set U −
γ is not empty. It is
(cid:4)
(cid:4)(cid:4)(cid:3)
(cid:4)
(cid:4)(cid:4)
(cid:3)
(cid:3)
Consider the remaining case (ξ,η) ∈ U − easy to see that
(cid:3) σ
|ξ|
− σ
|η|
|ξ|2σ 2
(cid:3) |ξ|
− |η|2σ 2
(cid:3) |η|
(4.27) > 0.
"
#
(cid:4) ≤
≤
(cid:4) ,
Hence for all (ξ,η) ∈ U − γ ,
(cid:3) |ξ| σ ! (cid:3) !σ
(cid:4) ξ − σ (cid:4) ξ − σ
(cid:3) |η| (cid:3) |η|
(cid:4) η,ξ − η ! (cid:4) !2 η
|ξ|
(4.28) 1 (cid:3) |ξ|, |η| 1 (cid:3) |ξ|, |η| I − γ I + γ
V. A. Klyachin et al. 15
(cid:4)
⊂
which implies that
(cid:4) .
(cid:3) γ (ε) ∩ U − V +
(cid:3) (cid:3)γ(ε) ∩ U − γ
γ
(4.29)
From this, by (4.25) and (4.26),
γ (ε) ⊂ (cid:3)γ(ε) ⊂ F − F+
γ (ε).
(4.30)
(cid:4)
(cid:4) ,
It is not hard to establish that
= Ωγ × Ωγ.
(cid:3) Pγ ∪ Qγ ∪ U + γ
(cid:3) Pγ ∪ Qγ ∪ U + γ
γ (0) ⊂
∪ U − γ
(4.31) W −
(cid:3)
(cid:4)
Then, using assertion (9) of Section 3, we find
⊂ F+
(cid:4) .
(cid:3) γ (ε) ∩ W − V +
γ (ε),
γ (0)
γ (ε) ∪ Qγ
(4.32) V − F − γ (ε) ⊂
(cid:2)
From this, by assertion (6) of Section 3 we obtain (1.15).
5. Properties of xγ(ε)
(cid:18)
For every γ ∈ R and ε ∈ (0,1) we set
(cid:17) x ∈ Σγ : ∃y ∈ Σγ,I +
γ (x, y) ≥ ε
, Xγ(ε) = (5.1) Xγ(ε). xγ(ε) = sup x
(cid:18) ,
(cid:17) (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≤ xγ(ε), |η| ≤ xγ(ε)
γ (ε) ⊂
If xγ(ε) ∈ Σγ, then the following relations are true:
(cid:18)
⊂ V +
γ (ε).
W + (cid:17) (5.2) (ξ,η) ∈ Ωγ × Ωγ : |ξ| ≥ xγ(ε) or |η| ≥ xγ(ε)
(cid:4)
(cid:4)
For every γ ∈ R we will study the function xγ(ε) of variable ε. Let γ ∈ R and ε ∈ (0,1). Then
= σ
= ε,
(cid:3) 0,rγ(ε)
(cid:3) rγ(ε)
(5.3) I + γ
where rγ(ε) is defined in Section 2. From this, rγ(ε) ∈ Xγ(ε). Therefore for every γ ∈ R the function xγ(ε) is defined everywhere on (0,1). Moreover, for every γ ∈ R,
(5.4) rγ(ε) ≤ xγ(ε) ∀ε ∈ (0,1).
(cid:28)
(cid:27)(cid:15)
As above, let sγ(ε) be a unique positive solution of (2.1) for every γ ∈ R and ε ∈ (0,1). For γ > 1 we put
$εγ =
y∈[0,
2/(γ−1)]
max √ (5.5) . , y I + γ 2 γ − 1
16 Some elementary inequalities in gas dynamics equation
(cid:15)
(cid:30) ,
The function xγ(ε) has the following properties. (1) Let γ > 1. Then
∀ε ∈
(cid:3) 0, $εγ
(cid:15)
(5.6) xγ(ε) = 2 γ − 1
∀ε ∈
(cid:4) .
(cid:3) $εγ,1
(5.7) xγ(ε) < 2 γ − 1
(2) Let
(5.8) γ ∈ (−∞,1], ε ∈ (0,1),
or
(cid:4) .
(cid:3) $εγ,1
(5.9) γ ∈ (1,+∞), ε ∈
(cid:4)
Then xγ(ε) ∈ Σγ and
= ε.
(cid:3) xγ(ε),sγ(ε)
(5.10) I + γ
(cid:15)
(3) For every γ > 1 we have
(5.11) . xγ(ε) = lim ε→$εγ+0 2 γ − 1
(cid:4)
(4) The function xγ(ε) is strictly decreasing on (0,1) for γ ≤ 1 and strictly decreasing on ($εγ,1) for γ > 1. Moreover,
(5.12) < 0 x(cid:8) γ(ε) = xγ(ε) + sγ(ε) (cid:3) − ε θ(cid:8) xγ(ε)
for every γ and ε, satisfying (5.8) or (5.9). (5) (a) If γ ∈ (−∞,1], then the function xγ(ε) ∈ C∞(0,1). (b) If γ ∈ (1,2], then the function xγ(ε) ∈ C∞((0, $εγ) ∪ ($εγ,1)) and it is continuous at the point $εγ; (c) If γ ∈ (2,3], then the function xγ(ε) ∈ C∞((0, $εγ) ∪ ($εγ,1)) and it has the con- tinuous derivative at the point $εγ; (d) If γ ∈ (3,+∞], then the function xγ(ε) ∈ C∞((0, $εγ) ∪ ($εγ,1)) and it has the second continuous derivative at the point $εγ. (6) For every γ ∈ R we have
(5.13) xγ(ε) = 0. lim ε→1−0
(7) For every γ ≤ 1 we have
(5.14) xγ(ε) = +∞. lim ε→0+
V. A. Klyachin et al. 17
(8) (a) If γ ∈ (−∞, −1), then
(5.15) for every α < . xγ(ε)ε−α = 0 lim ε→0+ γ − 1 2
(b) If γ = −1, then
(5.16) xγ(ε)ε = 2. lim ε→0+
(cid:27)
(cid:28)(γ+1)/(2γ−2)
(c) If γ ∈ (−1,1), then
(5.17) . xγ(ε)ε = lim ε→0+ γ + 1 2
(cid:7)
(cid:8)
(d) If γ = 1, then
− 1 2
(5.18) . xγ(ε)ε = exp lim ε→0+
(9) For every γ ∈ R we have
= +∞ for every α >
(cid:28)
(5.19) . lim ε→1−0 1 2 xγ(ε) (1 − ε)α
=
Proof of property (1). Let γ > 1. We set (cid:27)(cid:15)
(cid:11)
(5.20) . , y α(y) = I + γ 2 γ − 1 y + θ(y) (cid:11) 2/(γ − 1) (cid:11) It is easy to see that the function α(y) is positive on (0, 2/(γ − 1)) and it is continuous
on [0, 2/(γ − 1)]. Therefore there exists
$εγ =
y∈[0,
2/(γ−1)]
max √ (5.21) α(y) > 0.
%
&
(cid:15)
(cid:11)
Next,
0, (5.22) . α(y) ≤ < 1 ∀y ∈ 2 γ − 1 y 2/(γ − 1) y +
Hence $εγ < 1. Therefore for every ε ∈ (0, $εγ] the equation
(cid:11)
(5.23) α(y) = ε
2/(γ − 1)). Otherwise the equation does not have has at the least one solution y0 ∈ (0, any solution.
18 Some elementary inequalities in gas dynamics equation
(cid:4)
(cid:3)
(cid:4)
(cid:3) θ (cid:11)
=
(cid:3) ε = α
(cid:4) .
(cid:3) x, y0
= I + γ
≤ θ(x) + θ x + y0
Fix ε ∈ (0, $εγ] and x ∈ Σγ. Let y0 ∈ Σγ be a solution of (5.23). We have (cid:4) y0 (5.24) y0 y0 2/(γ − 1) y0 +
(cid:15)
From this x ∈ Xγ(ε). Hence Xγ(ε) = Σγ for all ε ∈ (0, $εγ]. This proves (5.6). Now we prove (5.7). Fix ε ∈ ($εγ,1). Suppose that
(5.25) . xγ(ε) = 2 γ − 1
(cid:15)
Then for arbitrary n ∈ N there exists a number xn ∈ Xγ(ε) such that
− 1 n
(5.26) < xn. 2 γ − 1
(cid:15)
Moreover,
n→∞xn = lim
(5.27) 2 γ − 1
(cid:4)
and for arbitrary n ∈ N there exists yn ∈ Σγ satisfying the inequality
≥ ε,
(cid:3) xn, yn
(5.28) I + γ
(cid:4)
(cid:3)
which implies
(cid:4) .
(cid:3) xn
− εxn ≥ εyn − θ
(5.29) θ yn
(cid:4)
=
Further, we have
(cid:3) α
≤ $εγ ∀n ∈ N.
(5.30) yn θ(yn) (cid:11) 2/(γ − 1) yn +
(cid:27)
(cid:28)
(cid:15)
(cid:3)
(cid:4)
Then
∀n ∈ N.
≤ $εγ
(cid:4)
(cid:3)
(cid:4)
(5.31) θ yn yn + 2 γ − 1
− εxn ≥ εyn − θ
(cid:28)
(cid:15)
(cid:15)
≥ εyn − $εγ
≥ −$εγ
θ Using (5.29), for all n ∈ N we deduce (cid:3) xn yn (cid:27) (5.32) . yn + 2 γ − 1 2 γ − 1
V. A. Klyachin et al. 19
(cid:15)
(cid:4)
Letting n → ∞ in the inequality
(cid:3) xn
− εxn ≥ −$εγ
, (5.33) θ 2 γ − 1
(cid:2)
we see that ε ≤ $εγ and we arrive at a contradiction.
Further we will need the following lemma.
(cid:4)
Lemma 5.1. If (5.8) or (5.9) holds, then xγ(ε) ∈ Σγ and there exists a number yγ(ε) ∈ Σγ such that
= ε.
(cid:3) xγ(ε), yγ(ε)
(5.34) I + γ
(cid:18)
Proof. Show that the set Xγ(ε) is compact for every γ and every (cid:2) satisfying (5.8) or (5.9). Introduce the set
(cid:17) (x, y) ∈ Σγ × Σγ : I +
γ (x, y) ≥ ε
(5.35) . Zγ(ε) =
γ (x, y) is contin- uous. The set Zγ(ε) is bounded. Indeed, we can find a sequence Zγ(ε) (cid:21) (xn, yn) → ∞. Assume that xn → ∞. Then for the bounded subsequence of {yn} we have
(cid:4)
(cid:3)
(cid:4)
(cid:4)
(cid:4)
Let π : R2 → R,π(x, y) = x be natural projection. It is clear that π(Zγ(ε)) = Xγ(ε). Assume that (5.8) holds. The set Zγ(ε) is closed since the function I +
≤ xnσ
= xnσ
(cid:3) xn, yn
(cid:3) + ynσ xn xn + yn
(cid:3) xn xn
yn + yn (5.36) . ε ≤ I + γ
(cid:4)
(cid:4)
≤ σ
(cid:4) .
(cid:3) xn, yn
(cid:3) xn
The right part of this inequality tends to zero as n → ∞. Thus we obtain a contradiction to (5.8). For an unbounded subsequence of {yn} we have (cid:3) (5.37) + σ yn ε ≤ I + γ
The right part of this inequality tends to zero as n → ∞. Again we obtain a contradiction to (5.8). Hence Zγ(ε) is bounded. Therefore Zγ(ε) is compact. Because the mapping π is continuous, the set Xγ(ε) = π(Zγ(ε)) is compact too.
(cid:18)
Assume that (5.9) holds. By (5.7) it follows that Zγ(ε) ⊂ Σγ × Σγ. Here Zγ(ε) denotes γ (x, y) is continuous, Zγ(ε) is compact. Therefore the closure of Zγ(ε). Since the function I + Xγ(ε) is compact too. Similarly we establish that the set
(cid:17) x ∈ Σγ : ∃y ∈ Σγ,I +
γ (x, y) = ε
(5.38) X γ(ε) =
is compact for every γ and (cid:2) satisfying (5.8) or (5.9). We fix γ and (cid:2) satisfying (5.8) or (5.9). Prove that
x
(5.39) Xγ(ε) = max X γ(ε). max x
20 Some elementary inequalities in gas dynamics equation
We set
x
x
(5.40) a = max b = max Xγ(ε), X γ(ε).
(cid:4)
Obviously, a ≥ b. Show that a ≤ b. Since a ∈ Xγ(ε), there exists a number y0 ∈ Σγ such that
≥ ε.
(cid:3) a, y0
(5.41) I + γ
(cid:4)
Assume that
= ε.
(cid:3) a, y0
(5.42) I + γ
(cid:4)
Then a ∈ X γ(ε) and hence a ≤ b. Now we assume
(cid:3) a, y0
(5.43) > ε. I + γ
(cid:4)
For γ ≤ 1 we have
= 0.
(cid:3) x, y0
γ
x→+∞ I + lim
(5.44)
γ (x, y) is continuous, there exists a number x(cid:8) > a such that
(cid:4)
Since the function I +
= ε.
(cid:3) x(cid:8), y0
(5.45) I + γ
(cid:28)
(cid:27)(cid:15)
Then x(cid:8) ∈ X γ(ε). Hence a < x(cid:8) ≤ b and we arrive at a contradiction. For γ > 1 we have
≤ $εγ < ε.
(cid:11)
(5.46) , y0 I + γ 2 γ − 1
2/(γ − 1)) satisfying (5.45). Hence x(cid:8) ∈ X γ(ε). There- Then there exists a number x(cid:8) ∈(a, fore a < x(cid:8) ≤ b and we arrive at a contradiction. Thus we establish that
x
(5.47) xγ(ε) = max X γ(ε)
(cid:2)
and arrive at the desired result.
(cid:4)(cid:4)
Proof of properties (2)–(5). Fix γ and ε0 satisfying (5.8) or (5.9). By Lemma 5.1 the num- ber xγ(ε0) ∈ Σγ and there exists a number yγ(ε0) ∈ Σγ such that
(cid:3) ε0
(cid:3) ε0
= ε0.
(cid:3) xγ
(cid:4) , yγ
(5.48) I + γ
We set
γ (x, y) − ε.
(5.49) F(x, y,ε) = I +
(cid:4)(cid:4)
(cid:4)(cid:4)
(cid:4)(cid:4)
(cid:3)
(cid:4)
V. A. Klyachin et al. 21
(cid:3) ε0
(cid:3) xγ
(cid:4) , yγ (cid:4)
(cid:4)
=
=
(cid:3) ε0 (cid:3) ε0
(cid:3) xγ (cid:3) ε0
− ε0 (cid:4) . (cid:3) ε0
(cid:3) − I + xγ γ (cid:4) (cid:3) + yγ ε0
(cid:3) ε0 + yγ
θ(cid:8) Observe that the function F(x, y,ε) is C∞-differentiable in some neighborhood U ⊂ R3 of the point p0 = (xγ(ε0), yγ(ε0),ε0) and F(p0) = 0. We have (cid:3) ε0 (5.50) p0 ∂F ∂x xγ θ(cid:8) xγ
(cid:3)
(cid:4)
By assertion (8) of Section 2, 0 < sγ(ε0) < rγ(ε0). Therefore the inequality rγ(ε0) ≤ xγ(ε0) yields
(5.51) < 0. p0 ∂F ∂x
By the well-known implicit function theorem, there exist a 3-dimensional interval I = Ix × Iy × Iε ⊂ U and a function f ∈ C∞(Iy × Iε) such that for all (x, y,ε) ∈ Ix × Iy × Iε,
(5.52) F(x, y,ε) = 0 ⇐⇒ x = f (y,ε).
(cid:17)
(cid:17)
(cid:3)
(cid:18)
(cid:18)
Here
(cid:4)! ! < b
(cid:3) ε0
! !x − xγ
! !y − yγ
(cid:4)! ! < a , (cid:17) ε ∈ R :
, x ∈ R : Ix = y ∈ R : (cid:18) (5.53) . Iy = ! ! ! < c !ε − ε0 ε0 Iε =
(cid:3)
(cid:4)(cid:4)
(cid:3)
(cid:4)
(cid:29)
(cid:3)
(cid:4)(cid:30)
(cid:30)−1(cid:29)
(cid:4)(cid:4)
= −
= −
Moreover,
(cid:3) ε0
(cid:4) ,ε0
(cid:3) ε0 (cid:3) ε0
− ε0 − ε0 (cid:4)
(cid:3)
(cid:4)
(cid:29)
(cid:3)
(cid:3)
(cid:4)(cid:30)
= −
=
(cid:4)(cid:30)−1[F (cid:8)
, p0 yγ F (cid:8) x(p0) F (cid:8) y ∂ f ∂y yγ (cid:3) xγ (cid:4) (5.54)
(cid:3) ε0
(cid:4) ,ε0
ε
(cid:3) ε0 − ε0
. p0 p0 yγ F (cid:8) x ∂ f ∂ε xγ θ(cid:8) + yγ (cid:4)(cid:4) (cid:3) ε0 θ(cid:8) θ(cid:8) (cid:3) ε0 (cid:3) xγ
(cid:3)
(cid:4)
It is easy to see that at the point yγ(ε0) the function x = f (y,ε0) reaches a maximum on Iy. Therefore
= 0.
(cid:3) ε0
(cid:4) ,ε0
(5.55) yγ ∂ f ∂y
(cid:3)
(cid:4)(cid:4)
From this
(cid:3) ε0
= ε0.
(5.56) θ(cid:8) yγ
Hence yγ(ε0) = sγ(ε0) and property (2) is proved. Further, we set
(5.57) G(y,ε) = θ(cid:8)(y) − ε.
(cid:3)
(cid:4)
(cid:4)(cid:4)
Observe that the function G(y,ε) is C∞-differentiable in some neighborhood V ⊂ R2 of the point q0 = (sγ(ε0),ε0) and G(q0) = 0. By assertions (6)–(8) of Section 2 we have
= θ(cid:8)(cid:8)
(cid:3) ε0
(cid:3) sγ
(5.58) < 0. q0 ∂G ∂y
22 Some elementary inequalities in gas dynamics equation
(cid:18)
=
By the implicit function theorem, the function sγ(ε) is C∞-differentiable at the point ε0. Therefore there is an interval
(cid:17) ε ∈ R :
! ! < c(cid:8)
! !ε − ε0
⊂ Iε
(5.59) I (cid:8) ε
such that
(cid:4)
(5.60) sγ(ε) ∈ Iy ∀ε ∈ I (cid:8) ε .
= 0 ⇐⇒ x = f
(cid:4) .
(cid:3) sγ(ε),ε
(5.61) F Hence for all (x,ε) ∈ Ix × I (cid:8) ε , (cid:3) x,sγ(ε),ε
ε . Next,
(cid:4)
We fix ε ∈ I (cid:8)
(cid:3) sγ(ε),ε
(5.62) x = f
and hence
(5.63) F(x,sγ(ε),ε) = 0.
(cid:4) ,
Rewrite the latter equality in the form
(cid:3) sγ(ε)
(5.64) μ(x) = −μ
where
(5.65) μ(t) = μ(t,ε) = θ(t) − tε.
We have
(5.66) μ(cid:8)(t) = θ(cid:8)(t) − ε.
(cid:4)
By assertion (8) of Section 2 we conclude that the function μ(t) is strictly increasing on (0,sγ(ε)) and strictly decreasing on (sγ(ε),+∞) ∩ Σγ. Moreover, μ(0) = μ(rγ(ε)) = 0 and by property (2), μ(xγ(ε)) = −μ(sγ(ε)). Then it is not hard to check that x = xγ(ε). Thus
(cid:3) sγ(ε),ε
∀ε ∈ I (cid:8) ε .
(5.67) xγ(ε) = f
(cid:4)
(cid:4)
(cid:4)
(cid:3)
(cid:4)
(cid:4)
(cid:4)
=
Hence the function xγ(ε) is C∞-differentiable at the point ε0 and
(cid:3) ε0
(cid:4) ,ε0
(cid:3) ε0
(cid:3) ε0
(cid:4) ,ε0
(cid:3) ε0
(cid:4) ,ε0
(cid:3) sγ
(cid:3) sγ
(cid:3) sγ
(cid:4) s(cid:8) γ
= ∂ f ∂y
= ∂ f ∂ε
(cid:3) ε0 − ε0
(cid:3) xγ ε0 (cid:3) θ(cid:8) xγ
+ < 0. ε0 x(cid:8) γ ∂ f ∂ε + sγ (cid:4)(cid:4) (cid:3) ε0 (5.68)
(cid:15)
This proves property (4). Let γ > 1. We show that
(5.69) . xγ(ε) = lim ε→$εγ+0 2 γ − 1
V. A. Klyachin et al. 23
Let y0 ∈ Σγ be a solution of the equation
(5.70) α(y) = $εγ.
(cid:28)
(cid:27)(cid:15)
Here, as above,
(5.71) . , y α(y) = I + γ 2 γ − 1
(cid:4)
(cid:3) θ (cid:11)
= $εγ,
Then
(cid:11)
y0 2/(γ − 1)
(cid:3)
(cid:3)
(cid:4)
(cid:3)
(cid:4)
=
( 2/(γ − 1) (cid:11)
= 0.
− θ (2
(5.72) y0 + (cid:4)’ θ(cid:8) y0 y0 y0 + ’ α(cid:8) y0 2/(γ − 1) y0 +
(cid:28)
(cid:15)
(cid:3)
(cid:4)
(cid:3)
(cid:27) (cid:4)
From this
= θ(cid:8)
(5.73) . θ y0 y0 y0 + 2 γ − 1
(cid:3)
(cid:4)
Using (5.72), we conclude that
= $εγ,
(5.74) θ(cid:8) y0
(cid:4)
that is, y0 = sγ($εγ). We rewrite the equality
= ε
(cid:3) xγ(ε),sγ(ε)
(5.75) I + γ
(cid:4)
(cid:3)
(cid:4)
in the form
(cid:3) θ
(cid:4) .
(cid:3) xγ(ε)
− xγ(ε)ε = −
− sγ(ε)ε
(5.76) θ sγ(ε)
(cid:15)
(cid:3)
(cid:4)
(cid:4)
(cid:3)
(cid:4)(cid:4)
(cid:4)
= −
Using (5.72), we obtain
(cid:3) θ
(cid:3) θ
− xγ(ε)ε
(cid:3) $εγ
(cid:4) $εγ
− sγ
= −$εγ
(5.77) . xγ(ε) sγ($εγ lim ε→$εγ+0 2 γ − 1
24 Some elementary inequalities in gas dynamics equation
(cid:15)
(cid:4)
Thus
(cid:3) xγ(ε),ε
= −$εγ
(5.78) μ . lim ε→$ε+0 2 γ − 1
(cid:15)
(cid:4)
≤
Suppose that (5.69) is not true. That is, for some sequence εi → $εγ + 0 of numbers the inequality
− m
(cid:3) εi
(cid:11)
(cid:11)
(5.79) xγ 2 γ − 1
(cid:28)
(cid:28)
(cid:27)(cid:15)
(cid:27)(cid:15)
(cid:4)
(2/γ − 1)). We have (2/γ − 1)) for every ε ∈ ($εγ,1). holds with some constant m > 0. Note that xγ(ε) ∈ [rγ(ε), By assertion (8) of Section 2 it follows that the function μ(t) is strictly decreasing on [rγ(ε),
≥ μ
− m
(cid:3) xγ
(cid:3) εi
(cid:4) ,εi
− m,εi
(5.80) > − μ εi. 2 γ − 1 2 γ − 1
(cid:4)
(cid:3) $εγ
(cid:4)
Letting εi → $εγ + 0, we obtain a contradiction to (5.78). Thus property (3) is proved. Hence the function xγ(ε) is continuous at the point $εγ for every γ > 1. For γ > 1 we have
= lim ε→$εγ+0
(5.81) . Aγ = lim ε→$εγ+0 x(cid:8) γ(ε) = lim ε→$εγ+0 xγ(ε) − xγ ε − $εγ xγ(ε) + sγ(ε) (cid:3) − ε θ(cid:8) xγ(ε)
(cid:11)
(cid:4)
(cid:3) $εγ
By assertion (5) of Section 2 and by property (3) we obtain
(cid:11)
(cid:4)
< 0 for γ ∈ (1,2), Aγ = − 2/(γ − 1) + sγ $εγ
(cid:3) $εγ
(5.82) < 0 for γ = 2, Aγ = − 2/(γ − 1) + sγ 2 + $εγ
for γ ∈ (2,+∞). Aγ = 0
(cid:4)
(cid:3) $εγ
(cid:4)
Hence xγ(ε) is not differentiable at the point $εγ for γ ∈ (1,2] and it has the continuous derivative at the point $εγ for γ ∈ (2,+∞). For γ > 2 we have
(cid:4) .
= lim ε→$εγ+0
(cid:3) θ(cid:8)
γ(ε) − x(cid:8) x(cid:8) γ ε − $εγ
(5.83) Bγ = lim ε→$εγ+0 xγ(ε) + sγ(ε) (cid:4)(cid:3) (cid:3) − ε xγ(ε) ε − $εγ
V. A. Klyachin et al. 25
(cid:27)
(cid:28)(2−γ)/(γ−1)(cid:3)
(cid:4)
Using L’Hospital rule and property (3), we find
(cid:4)(γ−2)/(γ−1)
= lim ε→$εγ+0
(cid:3) 1 −
(cid:3) (γ − 1)/2
ε − $εγ x2 γ(ε) 1 − γ − 1 2 lim ε→$εγ+0
(cid:4)−1/(γ−1)
= lim ε→$εγ+0
−(γ − 2)xγ(ε)x(cid:8)
(cid:4) x2 γ(ε)
ε − $εγ (cid:4) x2 γ(ε)
(cid:4)
γ(ε) (cid:4)(cid:3)
− ε (cid:4)
= − 1
(cid:4)
(cid:4)
= − 1
(5.84) 1 (cid:3) (γ − 1)/2 (cid:4) σ lim ε→$εγ+0 γ − 2
(cid:3) 1 − (cid:3) (cid:3) θ(cid:8) xγ(ε) xγ(ε) (cid:3) xγ(ε) + sγ(ε) xγ(ε) (cid:4) (cid:3) (cid:3) θ(cid:8) xγ(ε) xγ(ε) (cid:3) xγ(ε) + sγ(ε)
(cid:4)(cid:3)
(cid:4)(3−γ)/(γ−1)
lim ε→$εγ+0 γ − 2
(cid:4) x2 γ(ε)
(cid:3) (γ − 1)/2 (cid:4)
= − 1
(cid:4) x2 γ(ε) xγ(ε)
σ xγ(ε) (cid:3) (cid:3) 1 − (γ + 1)/2 . lim ε→$εγ+0 γ − 2 1 − (cid:3) xγ(ε) + sγ(ε)
(cid:4)
Then
(cid:4)2
(cid:4)2(cid:3)
(cid:4)(3−γ)/(γ−1) .
= −(γ − 2) lim ε→$εγ+0
(cid:3) 1 −
(cid:3) (γ + 1)/2
(cid:3) xγ(ε) + sγ(ε) (cid:3) (γ − 1)/2)x2 1 −
Bγ = lim ε→$εγ+0 θ(cid:8) xγ(ε) + sγ(ε) (cid:4)(cid:3) (cid:3) ε − $εγ xγ(ε) (5.85)
γ(ε)
xγ(ε) (cid:4) x2 γ(ε)
By property (3) we find
(cid:4)(cid:4)2
Bγ = −∞ for γ ∈ (2,3),
(cid:3) 1 + sγ
(cid:3) $εγ
(5.86) < 0 for γ = 3, Bγ = −
for γ ∈ (3,+∞). Bγ = 0
Therefore the function xγ(ε) is not doubly differentiable at the point $εγ for γ ∈ (2,3] and it has second continuous derivative at the point $εγ for γ ∈ (3,+∞). Thus property (5) is (cid:2) proved.
Proof of property (6). By assertion (8) of Section 2,
(5.87) 0 < sγ(ε) < rγ(ε)
for every ε and γ satisfying (5.8) or (5.9). Letting ε → 1 − 0 we obtain
(5.88) sγ(ε) = 0. lim ε→1−0
26 Some elementary inequalities in gas dynamics equation
Show that
(5.89) xγ(ε) = 0. lim ε→1−0
Indeed, suppose that this is not true, that is, there will be a number ε0 ∈ (0,1) and a
(cid:4)
(cid:3)
(cid:3) εi
≤ xγ
(cid:29)
(cid:4)
(cid:4)(cid:30)
sequence εi → 1 (ε0 < εi < 1) such that the inequalities (cid:4) (5.90) ε0 m ≤ xγ
= a ∈
(cid:3) εi
(5.91) . hold with some constant m > 0. We can consider that (cid:3) ε0 m,xγ xγ lim εi→1
(cid:4)(cid:4)
Using property (2), we have
= I +
(cid:3) xγ
(cid:3) εi
(cid:4) ,sγ
(cid:3) εi
γ (a,0) = σ(a).
(5.92) I + γ 1 = lim εi→1 εi = lim εi→1
(cid:2)
Then a = 0 < m and we arrive at a contradiction. Proof of property (7). Letting ε → 0+ in the inequality xγ(ε) ≥ rγ(ε), we obtain (5.7). (cid:2) Proof of property (8). (a) Let γ < −1. By assertion (8) of Section 2,
(5.93) 0 < sγ(ε) < rγ(ε) ∀ε ∈ (0,1).
From this
(5.94) for every α < . sγ(ε)ε−α = 0 lim ε→0+ γ − 1 2
(cid:27)
(cid:28)(cid:27)
(cid:28)−1
We set
(5.95) . ϑ(t) = t2 t2 1 − γ + 1 2 1 − γ − 1 2
Obviously
t→+∞ ϑ(t) = γ + 1 γ − 1
lim (5.96) .
It is easy to see that the function ϑ(t) is strictly decreasing on [0,+∞). Therefore
∀t ≥ 0.
(5.97) ϑ(t) > γ + 1 γ − 1
(cid:27)
(cid:28)
(cid:28)(1/(γ−1))−1(cid:27)
(cid:4)
=
Next, for all ε ∈ (0,1),
(cid:3) sγ(ε)
(cid:3) sγ(ε)
(cid:4) γ + 1 γ − 1
(5.98) ε = θ(cid:8) > σ . s2 γ(ε) s2 γ(ε) 1 − γ − 1 2 1 − γ + 1 2
V. A. Klyachin et al. 27
(cid:4)
From this
(cid:3) sγ(ε) ε
σ (5.99) 1 < < γ − 1 γ + 1
γ (x, y) = ε can be written as (cid:28)
(cid:28)
(cid:27)
− 1
for all ε ∈ (0,1). We note that the equality I +
= y
(cid:27) 1 − σ(y) ε
(cid:27)
(cid:28)
(cid:27)
(cid:28)
(cid:4)
(cid:4)
(cid:3)
(5.100) . x σ(x) ε
− 1
= lim ε→0+
σ σ 1 − (5.101) sγ(ε)ε−α xγ(ε)ε−α 0 = lim ε→0+ By (5.94) and (5.99) we obtain (cid:3) sγ(ε) ε xγ(ε) ε
for each α < (γ − 1)/2. Assume that there exits α < (γ − 1)/2 such that
(5.102) xγ(ε)ε−α (cid:7)= 0. lim ε→0+
Then for some sequence εi → 0 of positive numbers the inequality
≥ m
(cid:3) εi
(cid:4) ε−α i
(5.103) xγ
(cid:3)
(cid:4)(cid:4)
holds with some constant m > 0. By (5.101) we find
(cid:3) εi
= 1.
σ (5.104) lim εi→0+ xγ εi
(cid:4)(cid:4)
(cid:3)
(cid:4)
By (5.103),
(cid:3) εi
= 0
≤ lim εi→0+
(cid:3) xγ εi
σ σ (5.105) lim εi→0+ mεα i εi
(cid:4)
(cid:3)
(cid:4)−3/2
=
and we arrive at a contradiction. (b) Let γ = −1. We have
(cid:3) sγ(ε)
γ(ε)
(5.106) 1 + s2 . ε = θ(cid:8)
(cid:26)
(cid:4)
Then
= ε1/3.
(cid:3) sγ(ε)
(5.107) ε−2/3 − 1, σ sγ(ε) =
(cid:3)
(cid:4)
By property (7),
= 1.
(5.108) θ xγ(ε) lim ε→0+
(cid:3)
(cid:4)
(cid:4)
We have
(cid:3) θ
(cid:4)(cid:4) .
− sγ(ε)ε
(cid:3) xγ(ε)ε − θ
(cid:3) xγ(ε)
= lim ε→0+
(5.109) sγ(ε) 1 = lim ε→0+
28 Some elementary inequalities in gas dynamics equation
From this
(5.110) xγ(ε)ε = 2. lim (cid:2)→0+
(cid:3)
(cid:4)
(c) Let γ ∈ (−1,1). By property (7),
= 0.
(5.111) θ xγ(ε) lim ε→0+
(cid:15)
Assertions (7) and (8) of Section 2 yield
(cid:27)
(cid:28)
(cid:27)(cid:15)
(5.112) . sγ(ε) = lim ε→0+ 2 γ + 1
(cid:3)
(cid:4)
(cid:4)
= θ
(cid:3) θ
Then, using (5.111), we obtain (cid:28)(γ+1)/(2γ−2)
− sγ(ε)ε
(cid:4)(cid:4)
(cid:3) xγ(ε)ε − θ
= lim ε→0+ (cid:3) xγ(ε)
sγ(ε) γ + 1 2 2 γ + 1 (5.113)
= lim (cid:2)→0+
= lim ε→0+
xγ(ε)ε.
(cid:2)
(cid:4)
(d) The proof is analogous.
(cid:3) 1 − εγ−1 γ − 1
Proof of property (9). Assume that γ > 1. Then (cid:23) (cid:24) (cid:24) (cid:25) 2 (5.114) . xγ(ε) ≥ rγ(ε) =
Using L’Hospital rule, we find
= +∞
= γ − 1 2α
(5.115) lim ε→1−0 lim ε→1−0 1 − εγ−1 (1 − ε)2α εγ−2 (1 − ε)2α−1
(cid:2)
for every α > 1/2. From this we obtain desired result. The case γ ≤ 1 is analogous.
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V. A. Klyachin: Department of Mathematics,Volgograd State University, Universitetsky Prospekt 100, 400062 Volgograd, Russia E-mail address: klchnv@mail.ru
A. V. Kochetov: Department of Mathematics, Volgograd State University, Universitetsky Prospekt 100, 400062 Volgograd, Russia E-mail address: kochetov.alexey@mail.ru
V. M. Miklyukov: Department of Mathematics, Volgograd State University, Universitetsky Prospekt 100, 400062 Volgograd, Russia E-mail address: miklyuk@mail.ru
