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The proof of assertions 1–7 follows from the equalitiesWe prove assertion 9... Since from assertion 8 of Section 2it follows that the function μ is strictly increasing on [0,s γε] and st

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 satisfysome special conditions By virtue of nonlinearity of the equation the sets depend on thesolution gradient quantity We show double-ended estimates of the given sets and someproperties of these estimates

Copyright © 2006 V A Klyachin et al This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution,and reproduction in any medium, provided the original work is properly cited

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, Section2], [2, Section 15, Chapter IV]) Forγ =1±0 we assume

Hindawi Publishing Corporation

Journal of Inequalities and Applications

Volume 2006, Article ID 21693, Pages 1 29

DOI 10.1155/JIA/2006/21693

The case ofγ = −1 is known as the minimal surface equation (Chaplygin’s gas):

Forγ = −∞, (1.1) becomes the Laplace equation

In general, a solution of (1.1) with a function σ of variables (x1, ,x n) is called harmonic function Such functions were studied in many works (see., e.g., [3,4] and liter-ature quoted therein)

In general, the latter inequalities are valid only on subsets ofΩγ ×Ωγ withc1 andc2

depending on these subsets The purpose of the present paper is to describe that dence

depen-Introduce the sets

For everyγ ∈ R, define the functionsI γ −andI+

γ(| ξ |,| η |)≥ ε },V γ −(ε) = {(ξ,η) ∈Ωγ ×Ωγ:I γ −(| ξ |,| η |)≤ ε },V+

γ(ε) = {(ξ,η) ∈Ωγ ×Ωγ:I+

γ(| ξ |,| η |)≤ ε }.Also we will need the setsD γ = {(ξ,ξ) ∈Ωγ ×Ωγ },Q γ = {(ξ,η) ∈Ωγ ×Ωγ:ξσ( | ξ |)= ησ( | η |)}

The main result of our paper are the following theorems

Theorem 1.1 For every γ ∈ R ,

whereθ(t) = tσ(t) and ε is an arbitrary parameter It is easy to verify that for γ =1, (2.1)can be rewritten in the following form:

2

γ −1σ2− γ(t) − γ + 1

γ −1σ(t) + ε =0. (2.2)For arbitraryε ∈(0, 1) we set

Observe thatr γ(ε) ∈Σγfor everyγ ∈ Rand everyε ∈(0, 1)

The following assertions hold

(1) Letγ ∈ R Then the domain ofσ is the set Σ γ Moreover,σ(0) =1,σ(+ ∞)=0forγ ≤1 andσ(2/(γ −1))=0 forγ > 1.

(2) For eachγ ∈ Rwe have

0< σ(t) ≤1 ∀ t ∈Σγ (2.4)(3) Letγ ∈ R Thenσ (0)=0 and

Proof The proof of assertions (1)–(7) follows from the equalities

We prove assertion (9) Letx, y ∈Σγ,x2+y2> 0 If x = y, then

I −

γ(x, y) = σ(x) + xσ (x) < σ(x) = I+

γ(x, y) < 1. (2.16)Suppose thatx > y Since

The following assertions hold.

γ(ε) is linearly connected for every γ ∈ Randε ∈(0, 1)

(13) For everyγ ∈ Randε ∈(0, 1), we have



(ξ,η) ∈Ωγ ×Ωγ:| ξ | ≤ s γ(ε), | η | ≤ s γ(ε)⊂ W −

Heres γ(ε) is a unique positive solution of (2.1)

(14) For everyγ ∈ Randε ∈(0, 1), we have

W −

γ(ε) ⊂(ξ,η) ∈Ωγ ×Ωγ:| ξ | ≤ r γ(ε), | η | ≤ r γ(ε). (3.2)(15) Ifγ > −1, then

Proof of assertions (1)–(9) The proof follows from assertions (4) and (9) ofSection 2.

contains the segmentᏸ= {(ξt,ηt) : 0 ≤ t ≤1}with the endpoints 0 andζ.

Indeed, letζ ,ζ  ∈ W γ −(ε) be arbitrary Let ᏸ ,ᏸbe the segments with the endpoints

0,ζ and 0,ζ , respectively Denote byᏸ ∪ᏸthe double curve which consists of twosegmentsᏸandᏸ Then this double curve will join the pointsζ ,ζ and it will lie on

ε ≤ I −

γ(x, y) = θ (x) = σ(x) + xσ (x) ≤ σ(x) = σ(y), (3.10)and consequentlyx, y ∈[0,r γ(ε)] Thus if I γ(x, y) ≥ , thenx, y ∈[0,r γ(ε)].

Further we will need the function

It is easy to see that for allx, y ∈[0,r γ(ε)], x = y,

I −

γ(x, y) = ε ⇐⇒ μ(x) = μ(y). (3.12)Define the monotonicity intervals ofμ Since

from assertion (8) of Section 2it follows that the function μ is strictly increasing on

[0,s γ(ε)] and strictly decreasing on [s γ(ε),r γ(ε)] Moreover,

Note that ifI −

γ(x, y) = ε and x = y, then x = y = s γ(ε) Consequently for each x ∈

[0,r γ(ε)] there is a unique number y ∈[0,r γ(ε)], satisfying (3.12) Therefore there existsthe functiong : [0,r γ(ε)] →[0,r γ(ε)] such that for all x, y ∈[0,r γ(ε)],

as well as

g(0) = r γ(ε), gs γ(ε)= s γ(ε), gr γ(ε)=0. (3.17)Note that the functionI γ −(x, y) is infinitely differentiable at each point of [0,r γ(ε)] ×

[0,r γ(ε)] Fix arbitrary x0,y0∈[0,r γ(ε)], x0= y0, satisfying (3.15) We have

We prove that the segmentᏸ lies in W −

W −

γ(ε) contains ᏸ.

The proof of assertion (11) is analogous

Now we prove assertion (12) We fixγ ∈ R,ε ∈(0, 1), and a nonzero pointζ =(ξ,η) ∈

Proof of assertions (13), (15), (17), and (19) Let

Therefore (ξ,η) ∈ W γ −(ε) The case | ξ | > | η |is analogous

The proof of assertion (15) is analogous Assertion (17) follows from assertion (13),

Proof of assertions (14) and (18) Let (ξ,η) ∈ W γ −(ε) Assume that | ξ | = | η | We have

σ| ξ |= σ| η |≥ ε (3.31)imply

Hence

(ξ,η) ∈(ξ,η) ∈Ωγ ×Ωγ:| ξ | ≤ r γ(ε), | η | ≤ r γ(ε). (3.33)Now we assume that| ξ | > | η | We have

(ξ,η) ∈(ξ,η) ∈Ωγ ×Ωγ:| ξ | ≤ r γ(ε), | η | ≤ r γ(ε). (3.35)

The case| ξ | < | η |is analogous.

Proof of assertions (16) and (20) Let

(ξ,η) ∈(ξ,η) ∈Ωγ ×Ωγ:| ξ | ≤ r γ(ε), | η | ≤ r γ(ε). (3.36)Then

σ| ξ |≥ ε, σ| η |≥ ε. (3.37)Suppose| ξ | = | η | Then

I+

γ

| ξ |,| η |= σ| ξ |≥ ε. (3.38)Hence (ξ,η) ∈ W+

γ(ε) The case | ξ | < | η |is analogous

4 Proofs of main theorems

Introduce the sets H γ = {(ξ,η) ∈Ωγ ×Ωγ:| ξ | = | η |,ξ = η },G γ = {(ξ,η) ∈Ωγ ×Ωγ:

We set

Υ(ϕ) = | ξ |2+| η |2−2| ξ || η |cosϕ, Φ(ϕ) = σ| ξ || ξ |2+σ| η || η |2−σ| ξ |+σ| η || ξ || η |cosϕ,

From this, by (4.7), and assertions (1), (2) ofSection 3we obtain (1.12) 

Proof of Theorem 1.2 (a) We fix γ ≤ −1 andε > 0 It is clear that inequality (4.2) holdsfor all (ξ,η) ∈ D γ

Let (ξ,η) ∈Ꮾγ(ε) ∩ H γ In this case inequality (4.2) becomes

σ| ξ |= σ| η |≤ ε. (4.13)Then

Thus for all (ξ,η) ∈ G γ,

From this, by (4.15) and assertion (6) ofSection 3, we obtain (1.13) and (1.14)

(b) We fixγ > −1 andε > 0 It is clear that the inequality (4.2) holds for all (ξ,η) ∈ Q γ

2| ξ | σ( | ξ |)

| ξ |+| η | = I+

γ

| ξ |,| η |. (4.23)Thus inequality (4.2) assumes the form

Consider the remaining case (ξ,η) ∈ U −

γ Observe that the setU −

γ is not empty It iseasy to see that



σ| ξ |− σ| η || ξ |2σ2 

| ξ |− | η |2σ2 

| η |> 0. (4.27)Hence for all (ξ,η) ∈ U −

γ,1

which implies that

the functionx γ(ε) is defined everywhere on (0,1) Moreover, for every γ ∈ R,

The functionx γ(ε) has the following properties.

for everyγ and ε, satisfying (5.8) or (5.9)

(5) (a) Ifγ ∈(−∞, 1], then the functionx γ(ε) ∈ C ∞(0, 1)

(b) Ifγ ∈(1, 2], then the functionx γ(ε) ∈ C ∞((0,ε$γ)∪(ε$γ, 1)) and it is continuous

has at the least one solutiony0∈(0,

2/(γ −1)) Otherwise the equation does not haveany solution

Fixε ∈(0,ε$γ] andx ∈Σγ Lety0∈Σγbe a solution of (5.23) We have

From thisx ∈ X γ(ε) Hence X γ(ε) =Σγfor allε ∈(0,ε$γ] This proves (5.6)

Now we prove (5.7) Fixε ∈(ε$γ, 1) Suppose that

Lettingn → ∞in the inequality

Further we will need the following lemma

Lemma 5.1 If ( 5.8 ) or ( 5.9 ) holds, then x γ(ε) ∈Σγ and there exists a number y γ(ε) ∈Σγ

Assume that (5.8) holds The setZ γ(ε) is closed since the function I+

The right part of this inequality tends to zero asn → ∞ Thus we obtain a contradiction

to (5.8) For an unbounded subsequence of{ y n }we have

The right part of this inequality tends to zero asn → ∞ Again we obtain a contradiction

to (5.8) HenceZ γ(ε) is bounded Therefore Z γ(ε) is compact Because the mapping π is

continuous, the setX γ(ε) = π(Z γ(ε)) is compact too.

Assume that (5.9) holds By (5.7) it follows thatZ γ(ε) ⊂Σγ ×Σγ HereZ γ(ε) denotes

the closure ofZ γ(ε) Since the function I+

γ(x, y) is continuous, Z γ(ε) is compact Therefore

X γ(ε) is compact too.

Similarly we establish that the set

X γ(ε) =x ∈Σγ:∃ y ∈Σγ,I+

γ(x, y) = ε (5.38)

is compact for everyγ and satisfying (5.8) or (5.9)

We fixγ and satisfying (5.8) or (5.9) Prove that

maxx X γ(ε) =maxx X γ(ε). (5.39)

We set

a =maxx X γ(ε), b =maxx X γ(ε). (5.40)Obviously,a ≥ b Show that a ≤ b Since a ∈ X γ(ε), there exists a number y0∈Σγ suchthat

Since the functionI+

γ(x, y) is continuous, there exists a number x  > a such that

Then there exists a numberx  ∈(a,2/(γ −1)) satisfying (5.45) Hencex  ∈ X γ(ε)

There-forea < x  ≤ b and we arrive at a contradiction.

Thus we establish that

x γ(ε) =max

Proof of properties (2)–(5) Fix γ and ε0satisfying (5.8) or (5.9) ByLemma 5.1the berx γ(ε0)∈Σγand there exists a numbery γ(ε0)∈Σγsuch that

Observe that the functionF(x, y,ε) is C ∞-differentiable in some neighborhood U⊂ R3

of the pointp0=(x γ(ε0),y γ(ε0),ε0) andF(p0)=0 We have

By the well-known implicit function theorem, there exist a 3-dimensional intervalI =

I x × I y × I ε ⊂ U and a function f ∈ C ∞(I y × I ε) such that for all (x, y,ε) ∈ I x × I y × I ε,

F(x, y,ε) =0⇐⇒ x = f (y,ε). (5.52)Here

By the implicit function theorem, the functions γ(ε) is C ∞-differentiable at the point ε0.Therefore there is an interval

By assertion (8) ofSection 2we conclude that the functionμ(t) is strictly increasing on

(0,s γ(ε)) and strictly decreasing on (s γ(ε),+ ∞)∩Σγ Moreover,μ(0) = μ(r γ(ε)) =0 and

by property (2),μ(x γ(ε)) = − μ(s γ(ε)) Then it is not hard to check that x = x γ(ε) Thus

x γ(ε) = fs γ(ε),ε ∀ ε ∈ I  (5.67)Hence the functionx γ(ε) is C ∞-differentiable at the point ε0and

Letγ > 1 We show that

Lety0∈Σγbe a solution of the equation

holds with some constantm > 0 Note that x γ(ε) ∈[r γ(ε),(2/γ −1)) for everyε ∈(ε$γ, 1).

By assertion (8) of Section 2it follows that the function μ(t) is strictly decreasing on

Using L’Hospital rule and property (3), we find

Therefore the functionx γ(ε) is not doubly differentiable at the point ε$γforγ ∈(2, 3] and

it has second continuous derivative at the pointε$γ forγ ∈(3, +∞) Thus property (5) is

Proof of property (6) By assertion (8) ofSection 2,

0< s γ(ε) < r γ(ε) (5.87)for everyε and γ satisfying (5.8) or (5.9) Lettingε →1−0 we obtain

lim

Proof of property (7) Letting ε →0+ in the inequalityx γ(ε) ≥ r γ(ε), we obtain (5.7) 

Proof of property (8) (a) Let γ < −1 By assertion (8) ofSection 2,

0< s γ(ε) < r γ(ε) ∀ ε ∈(0, 1). (5.93)From this

Proof of property (9) Assume that γ > 1 Then

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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