Gen. Math. Notes, Vol. 1, No. 2, December 2010, pp. 115-129 ISSN 2219-7184; Copyright © ICSRS Publication, 2010 www.i-csrs.org
Available free online at http://www.geman.in
About Uniform Limitation of Normalized Eigen Functions of T. Regge Problem in the Case of
Weight Functions, Satisfying to Lipschitz Condition
1Jwamer .K.H and 2Aigounv .G.A
1University of Sulaimani, College of Science, Department of Mathematics, Sulaimani, Iraq
2Daghestan State University, College of Mathematics, Department of Mathematical Analysis, South of Russian
Abstract
In this work, we estimate normalize eigenfunctions to the T.Regge problem whenever the weight functions satisfies Lipschitz condition.
KeyWords: Eigenfunctions, normalize, Lipschitz condition
1. INTRODUCTION
Let's consider spectral problem (q(x)∈C[ ]0;a ,ρ(x)∈Lip1and m≤ρ(x)≤M )
) ( ) ( )
( ) ( )
(x q x y x 2 x y x
y′′ + =λ ρ
− (0<x<a)
(1)
0 ) 0
( =
y , y′(a)−iλy(a)=0 (2)
1 )
( ) (
2 1
0
2 =
∫
aρ x y x dx , where λ - is spectral parameter.(3)
The problem (1) - (2) arises in different questions of mathematical physics. T.Regge [1], who studied it (in case of ρ(x)≡1) in connection with the theory of dispersion has shown, that if q(x) in left semi-neighborhood of point a satisfies to condition q(x)~ Cµ(a−x)µ x→a−0; µ≥0Cµ ≠0, the problem has discrete spectrum λn and system of eigenfunctions of problem (1) - (2) is full of century L2[ ]0 a, . In work [2] is studied asymptotes of proper values and received 2 multiple decomposition in uniform converging numbers on eigenfunctions from which 2 multiple completeness of eigenfunctions of century L2[ ]0 a, . In case of equation of n2 order and ρ(x)≡1 similar problem is considered in works [3] - [4]. And for ρ(x)≡/1 asymptotic of eigenvalues for more general problem is studied in works [5] - [6].
In work [7] is considered the case of weight functions close to Holder class where maximal growth rate of eigenfunctions of problem (1) - (3) is studied.
The purpose of the present work is reception of uniform estimations for normalized eigenfunctions of problem (1) - (3) in case of weight functions, satisfying to Lipschitz condition.
The following is true:
Lemma: For any ρ(x)∈Lip1 and ε >0 there is function 2[0, ]
)
(x ∈C a
ρε
such, that ρε(a)=ρ(a), ρε(0)=ρ(0), x dx x dx
a a
∫
∫
=0 0
) ( )
( ρ
ρε ,
N x x
x x a
a
x ( ) ( ) ,max ( ) 2
max[0,] − ≤ [0, ] ′ ≤
∈
∈ ρ ρε ε ρε and
ρε x εc
a
x ′′ ≤
∈ ( )
max[0, ] , where C is constant independent from ρ(x)and ε.
Proof:
Let’s divide interval
[ ]
0,a on m equal parts ( m -arbitrary) by points0=x0 <x1 <...<xm−1 <xm =a, and middle of intervals [xi−1,xi] we shall designate through x′i(such pointsm , namely x1′,x′2,...,xm′ ). We shall consider broken line, that connected points(x0,ρ(x0)),(x1,ρ(x1)),...,(xm,ρ(xm)).Obviously, this broken line is function graphρ0(x), satisfying to inequalities
m a x N x
a x
) . ( ) (
max 0
] , 0
[ − ≤
∈ ρ ρ and ρ0′(х) ≤N there, where ρ0′(x) exists.
Let's consider now other broken line connecting points(x0,ρ0(x0)),(x1′,ρ0(x1′)),(x2′,ρ0(x2′))...,(xm′ ,ρ0(xm′ )),(xm,ρ0(xm)).
Obviously, this broken line is the schedule of function ρ1(x), satisfying to parities ρ1′(х) ≤N there, where ρ1′(x) exists and
m a x N x
a x
) . ( ) (
max 1
] , 0
[ − ≤
∈ ρ ρ .
On sites [x′i,x′i+1] where i=1,2,...,m−1 we shall construct curve ρε(x) as
polynomials parities 1 3 4 ( )2
4 ) 3 8 (
) ( )
( i i pi x xi
x p x
x
x −
− ∆
∆ − +
=ρ
ρε where
2m, x a xi − i′=
=
∆ 2
) ( )
( 0 1
0′ ′ − ′ ′+
= i i
i
x p p ρ x
. On sites [0,x′1] and [xm′ ,a]we shall get ρε(x)=ρ1(x) (in the same place ρ1(x)=ρ0(x)).
Let's put =2.[ ]+2 ε
m Na . Direct check shows, that all conditions of
lemma except for equality x dx x dx
a a
∫
∫
=0 0
) ( )
( ρ
ρε are executed. In addition,
inequality ρε′(x) ≤N takes place ( N and N2 in condition of lemma) now let’s
find number δ from condition x dx x dx x a
a a
∫
∫
+ =0 0
) ( )
sin 1 ( )
( δ π ρ
ρε ,
hence
∫
∫
−= a
a
dx x a x
dx x x
0 0
) ( sin
) ) ( )
( (
ε ε
π ρ ρ ρ
δ . Obviously at small ε number δ is also not
enough, and function ( ) ( )(1 sin x)2 x a
x ρ δ π
ρε = ε + satisfies to conditions of lemma.
Let's designate through Q[0,a] class of continuous on
[ ]
0,a functionsq(x),satisfying to inequality Q
a
a
C dx x q <
∫
10
)
( , where CQ =contand[a0,a1]⊆[0,a].
Let's consider countable subset
{
qi(x) i∈N}
≡Q[0,a] of class Q[0,a] satisfying to condition Limi→∞∫ ∫
x t qi s dsdt ≡ fo x0 0
) ( )
( , where fo(x) function satisfying to Lipschitz condition, and convergence is uniform on
[ ]
0,a .Let ρ ≠1,ρ>0,λ∈C− is complex, Im(λ)<const (that is ρ - is fixed and λ- is arbitrary of strip Im(λ)<const of complex plane).
Let's designate through y(x,λ,q) solution of Cauchy problem
. 1 ) 0 ( , 0 ) 0 (
) , 0 ( , ) ( )
( ) ( )
( 2
′ =
=
∈
= +
′′
−
y y
a x x y x
y x q x
y λ ρ
Then the following is true:
Theorem: There is constant C0 ≡C0(Q[0,a]) (uniform for all classQ[0,a]) such, that
0 2 1
0 ] 2 , 0 [
) ) , , ( (
) , ,
max ( C
dx q x y
q x y
a a
x <
∈
∫
λ ρ
λ for every value large enough by module λ.
From this theorem and previous lemma follows important consequence Consequence: Let q(x)- is continuous function, andρ(x)∈Lip1. Then solution of Cauchy problem
. 1 ) 0 ( , 0 ) 0 (
1 ) ( ), , 0 ( , ) ( )
( ) ( )
( 2
′ =
=
≠
∈
=
′′ +
−
y y
a a x x y x
y x q x
y λ ρ ρ
Satisfies to parity < <∞
∈
∫
constdx x y x
x y
a a x
2 1
0 ] 2 , 0 [
) ) ( ).
( (
) max (
ρ
For every value large enough λ from stripIm(λ)<const. Proof:
As solution of Cauchy problem continuously depends on weight function )
ρ(x and functional
∫
a x y x dx0
2 1
2 )
) , ( ) (
( ρ λ also continuously depends onρ(x).
Hence, functional
2 1
0
2 ] , 0 [
) ) , ( ).
( (
) , ( max
∫
∈ a
a x
dx x
y x
x y
λ ρ
λ
also continuously depends on weight
functionρ(x). Hence, there is number ε(R)such, that
2 1
0
2 ]
, 0 [ 2
1
0
2 ]
, 0 [
) ) , , ( ).
( (
) , , ( max 2.
1 ) ) , , ( ).
( (
) , , ( max
∫
∫
∈
∈ ≥ a x a
a a x
dx x
y x
x y dx
x y x
x y
ρ λ ρ
ρ λ ρ
λ ρ
ρ λ
, if λ ≤R and
) ( ) ( )
(x ρ x ε R
ρ − ≤
Where R>0is arbitrary constant . Let’s take some R and by ε(R) and lemma let’s plot function ρε(x) approaching ρ(x)(ρ(x)−ρ(x) ≤ε(R))
) ( ) (a ρ a
ρε = . Now let’s consider Cauchy problem with weight functionρε(x) instead of ρ(x). In this problem 2[0, ]
)
(x ∈C a
ρε and consequently we can make
double replacement , ( ) ( ). ( ( )) )
0 (
2 y x A x x
t A
x dt
ξ η
ξ =
∫
= ,where ( ) ( ). 4( )
1 4
1
a x x
A =ρ− ε ρ .
As a result of such replacement we shall obtain problem:
) ( ) ( )
( ) ( )).
( ) ( ) ( ( )
(ξ 3 η ξ λ2ρ η ξ
η′′ + q x A x −A′′ x A x = a
− , )
) , (
0 (
0
∫
2∈ a
t A
ξ dt ,
, 0 ) 0
( =
η
) 0 ( ) 0 ( = A
η′ , As A(0)=ρ−41(0).ρ41(a)and
∫
∫
∫
a = a = a t dta t
a dt t
A dt
0 0
0
2 ( )
) ( 1 ))
( ) ( ( )
( ε
ε
ρ ρ ρ
ρ
const a a
dt t
a
=
=
=
∫
) (
) (
0
ρ ρ
(Independent from ε), designating
) ( )]
( ) ( ) ( )[
3(x q x A x A x qε ξ
A − ′′ ≡ we shall obtain problem:
4 2
) 0 (
) ) (
0 ( , 0 ) 0 (
, ) , 0 ( , ) ( ) ( )
( ) ( ) (
ρ η ρ
η
ξ ξ η ρ λ ξ η ξ ξ
η ε
a
a a
q
′ =
=
∈
=
′′ +
−
Estimated in theorem functional does not depend on value y′(0) (as all solutions of our equation, satisfying to condition y(0)=0, can be obtained from solution of problem with conditions y(0)=0,y′(0)=1, by multiplication to constant, which will be reduced in our functional) and consequently if to show, that
∫
tq d0
) (ξ ξ
ε in regular intervals on ε and t∈[0,a] is limited for all small
>0
ε under theorem there will be constant C0 >0such, that
, )
) ( ).
( (
)
max ( 0
2 1
0
2
C d
a
a ≤
∫
ρ η ξ ξξ η
From here and from parities
∫
=
= x
t A x dt x
x A x y
0 2( ) )
( )), ( ( ).
( )
( η ξ ξ obviously follows, that exists C0 >0such,
that 0
2 1
0 ] 2 , 0 [
) ) , , ( ).
( (
) , ,
max ( C
dx x
y x
x y
a a
x <
∈
∫
ρ λ ρ
ρ λ
For every λ ≤R, and by arbitrariness R , and for all considered λ(let’s remind, that Im(λ)<const).
Let's estimate
∫
tq d0
) (ξ ξ
ε . Passing to variable x in integral we shall get
′′ = +
′′ ≤
−
′′ =
−
′ =
− ′′
=
∫
∫
∫
∫
∫
∫
∫
−
−
s S
s S
s s
t
dx x A x A x dx
A x dx q
x A x A x dx
A x q
x A x dx A x A x A x q dx x x A x A x A x q d
q
0 0
2 0
0 2
0
2 3 0
3 0
) ( ) ) (
( ) ) (
( ) ) (
( ) (
) ) ( ( )]
( ) ( ) ( [ )
( ).
( )]
( ) ( ) ( [ )
(ξ ξ ξ
ε
].
, 0 [ ,
)]
( [ ) 0 ( ) 0 (
) ( ) ) (
( ) )] (
( [ )]
( ) ( ) [
( ) (
0
2
0 2 0
2 0
0 2
a s where dx
x A A
A
s A s A x dx
A x dx q
x A x
A x A x dx
A x q
s
S s
s S
′ ∈
′ +
′ + +
′ ≤
′ − +
∫
∫
∫
∫
− −From definition 4 ) (
) ) (
( x
x a A
ρρε
= follows, that
4 4
) 0 (
) ( )
0 (
) ) (
0
( ρ
ρ ρ
ρ
ε
a
A = a = ,
4 5 4 4
5 4
1
) ( 4
) ( . ) ) (
( ).
( ).
4 ( ) 1 (
x x x a
x a
x A
ε ε ε
ε ρ
ρ ρ ρ
ρ
ρ ′ =− ′
−
′ = − ,
4
4 5
4
) 0 (
) . ( ) 0 ( 4
) 0 ( )
0 ( 4
) 0 ( . ) ) (
0
( ρ
ρ ρ
ρ ρ
ρ
ρ ε
ε
ε a
A a ′ = − ′
−
′ = and consequently
∫
∫
∫
≤ + ′ + ′ + ′a a
t
s s a a
s s dx a
a x x d q
q
0
5 2 3
0
4 4
0 16. ( )
)]
( .[
) ( )
0 (
) . ( ) 0 ( . 4
) 0 ( )
( . 4
) ( . ) ) (
( ). (
) ) (
(
ε ε ε
ε ε ε
ε ρ
ρ ρ ρ
ρ ρ
ρ ρ
ρ ρ ρ
ξ ρ ξ
On lemma ρε′(x) ≤2.Nand ρ(x)−ε ≤ρε(x)≤ρ(x)+ε, hence
. ] ) ( [ . 4
. ) (
) 0 (
) . ( ) 0 ( ] 2
) min ( .[
2
. ) ) (
( ). (
) ) (
(
0
5 2
0 2
3
] , 0 [ 4
0 4
∫
∫
∫
+ −
+
− +
+
≤
∈
a a
a x t
dx x
N a
a N
x N dx a
a x x d q
q
ε ρ ρ
ρ ρ ε ρ
ρ ε ρ
ρ ρ ξ
ε ξ
As ε is not enough, consequence is proved.
Let's prove now theorem for what we shall evaluate ϕ0′(x,λ),ϕ(x,λ) and )
, ( ( ) ,
( λ ϕ λ
ϕ′ x x is solution of equation (1) satisfying to entry conditions ,
1 ) , 0 ( , 0 ) , 0
( λ = ϕ′ λ =
ϕ and ϕ0(x,λ) is solution of such Cauchy problem with constant coefficientρ(x)≡ρ). As it has been established [8, p. 22]
) sin . cosh cos
. sinh ( )
,
( 2 1 1 1 1
1 2 1
1 1
0 i x x i x x
x σ δ σ δ
δ σ
δ λ σ
ϕ − +
+
−
= − and
consequently ϕ0′(x,λ)=(coshσ1x.cosδ1x−isinhσ1x.sinδ1x)(simple
transformations are lowered). As at greater λ parities σ1≈ ρ.σ and δ1≈ ρ.δ take place, then obviously ϕ′0(x,λ) <const is regular on x∈[0,a] and λ from considered strip. Let’s consider number
1 0
1 0 0
0 1 0 2 0
1 1
1 0 1 0 1
0
....
) , ( ) , (
] ) ( [
...
) , ( . ] ) ( [ )
, ( ) , ( ].
) ( [ ) , ( ) , (
1
1
τ τ λ τ ϕ λ τ τ ϕ τ
λ τ ϕ τ
τ λ τ ϕ λ τ ϕ τ
λ ϕ λ ϕ
τ
τ
d d q
q
x q q
d x
q q
x x
i i i
i i
i x x
o
i − −
−
− +
−
− +
=
−
∞
=
∫
∑ ∫ ∫
∫
−
Let's enter designations fi(x)for i− member of series
∫
− −=
x
d x
q q
x f
0
1 1 0 1
1( ) [ ( ) ]. ( , ) ,...)
( τ ϕ τ λ τ , As result we shall get:
∑
∞=
=
−
1
0( , ) ( )
) , (
i i x f x
x λ ϕ λ
ϕ (4)
If i≥1 , then obviously
∫
− −+ =
x
i
i x q q x f d
f
0
1 1 1
0 1
1( ) [ (τ ) ]ϕ ( τ ,λ). (τ ) τ and consequently, integrating in parts, we shall obtain
{
1 0
0
1 1
0 1 1
0 0 0
1 1
0 1
] ) ( [
. )]
( ).
, ( ) ( ).
, ( [ ]
) ( [ ).
( ).
, ( )
(
1
τ
τ λ τ ϕ τ λ τ ϕ τ
λ τ
ϕ τ
d ds q s q
f x
f x
ds q s q f
x x
f
x
x
i i
x
i i
−
′ −
′ −
−
−
− −
=
∫
∫
∫
+
Or
∫
∫
′ − − ′ − ′ −+ =
τ τ
τ λ τ ϕ τ λ τ ϕ
0 0
0 0
1(x) [ (x , )f ( ) (x , )f ( )]. [q(s) q]dsd f
x
i i
i ,
(5)
Absolutely similarly for f1(x)it is possible to get parity
∫
∫
′ − − ′ − ′ −= ϕ τ λ ϕ τ ϕ τ λ ϕ τ τ τ
0 0
0 0
0 0
1(x) [ (x , ) ( ) (x , ) ( )]. [q(s) q]dsd
f
x
(6)
Differentiating parities (5) and (6) on x we shall get parities for derivatives
, ] ) ( [ )].
( ) , (
) ( ) , ( ] [ [ ] ) ( [ ).
( ) (
0 0
0
0 0
2 1
∫
∫ ∫
′ −
′ −
−
−
− +
−
′ =
+
τ τ τ
τ λ τ ϕ
τ λ τ ϕ ρ λ τ
τ
dsd q q f
x
f x q
d q q x f x f
i
i
x x
i i
(7)
, ] ) ( [ )].
, ( ) , (
) , ( ) , ( ] [
[ ] ) ( [ ).
, ( ) (
0 0
0
0 0
0 0
2 0
1
∫
∫ ∫
′ −
′ −
−
−
− +
−
′ =
τ τ τ
λ τ ϕ λ τ ϕ
λ τ ϕ λ τ ϕ ρ λ τ
τ λ ϕ
dsd q q x
x q
d q q x
x f
x x
(8)
(Here, it is considered, that ϕ0′′(x,λ)≡(q−λ2ρ)ϕ0(x,λ).)
From choice of class Q[0,a] follows, that
∫
τ q s −q ds <Q=const<∞0
] ) ( [
and consequently from (5) and (8) follows, that
] )
. [(
) ( . ,
. ) 2
( 0 0 1 0 2 02 02
1 q C C
C Q x Q f C x C x Q
f ′ ′ ≤ + + + ′
≤ ρ λ
λ
λ , where C0
and C′0
Such constants for which inequalities
λ λ ϕ0( , ) C0
x ≤ and ϕ0′(x,λ) ≤C0′
are executed (as 2
1 2 1
1 2 1
2 0
sin ) sinh
,
( δ σ
δ λ σ
ϕ +
≤ x+ x
x then, obviously C0 and C′0
exist). Using recurrent parities (5) and (7) we shall obtain, that number (4) converges and moreover
λ λ ϕ λ
ϕ( , ) 0( , ) C0 x
x − ≤ (evaluations f2(x), f3(x),..., and f2′(x), f3′(x),... are made consistently for i=2,3, …).
Then ϕ x λ ϕ x λ qτ q ϕ x τ λ dτ
x
) , ( ] ) ( [ ) , ( ) ,
( 0
0
0′ + − ′ −
′ =
∫
, or, integrating inparts, we shall get
{
[ ( , ) ( , ) ( , ). ( , )] [ ( ) ] ,] ) ( [ ).
, ( ) , ( )
, ( ) , (
0 0
0 0
0 0 0
0
τ λ
τ ϕ λ τ ϕ λ τ ϕ λ τ ϕ
λ τ ϕ λ τ ϕ λ ϕ λ ϕ
τ τ
d ds q s q x
x
ds q s q x
x x
x
x
−
′′ −
′ −
′ −
−
−
′ − −
′ =
′ −
∫
∫
∫
As ϕ0′(0,λ)≡1 and ϕ0′′(x−τ,λ)≡(q−λ2ρ)ϕ0(x−τ,λ),
{
∫ ∫
∫ ∫ ∫
∫
∫ ∫
′ − −
−
− − −
+
−
=
−
−
−
′ −
′ −
−
−
′ =
′ −
x
x x a
a
x x
d ds q s q x
d ds q s q x
q ds
q s q x
d ds q s q x
q
x ds
q s q x x
x
0 0
0
0 0 0
0 2
0 0
2
0 0
0 0
, ] ) ( [ ) , ( ) , (
] ) ( [ ).
, ( ) , ( ) (
] ) ( [ ).
, (
} ] ) ( [ )].
, ( ) , ( ) (
) , ( ) , ( [ ]
) ( [ ).
, ( ) , ( ) , (
τ λ
τ ϕ λ τ ϕ
τ λ
τ ϕ λ τ ϕ ρ λ λ
ϕ
τ λ
τ ϕ λ τ ϕ ρ λ
λ τ ϕ λ τ ϕ λ
ϕ λ ϕ λ ϕ
τ
Subtracting and adding in last integral ϕ0′(τ,λ)to ϕ′(τ,λ) and representing integral in the form of sum of two integrals we shall obtain
∫ ∫ ∫
− − −
− +
−
′ =
′ x − x x x q s qds q x x q s qds d
0 0 0
0 2
0( , ) ( , ). [ ( ) ] ( ) ( , ) ( , ). [ ( ) ]
) ,
( λ ϕ λ ϕ λ λ ρ ϕ τ λ ϕ τ λ τ
ϕ τ
τ λ
τ ϕ λ τ ϕ λ τ ϕ τ λ
τ ϕ λ τ
ϕ x τ q s qds d x τ q s qds d
x x
∫ ∫ ∫ ∫
′ − ′ − ′ −
−
′ − ′ −
−
0 0 0
0 0
0 0
0( , )[ ( , ). [ ( ) ] ( , )[ ( , ) ( , )]. [ ( ) ]
From this equality follows, that
∫ ∫
∫
∫
∫
∫
∫
− ′
− ′
′ −
+
′ −
′ − +
−
−
− +
−
′ ≤
′ −
x
x
x x
d ds
q s q x
d ds q s q x
d ds q s q
x q
ds q s q x
x x
0
0 0
0
0 0
0 0
0
0 0 2
0 0
) , ( ) , ( . ] ) ( [ . ) , (
] ) ( [ . ) , ( . ) , ( ]
) ( [
. ) , ( . ) , ( . ]
) ( [ . ) , ( ) , ( ) , (
τ λ τ ϕ λ τ ϕ λ
τ ϕ
τ λ
τ ϕ λ τ ϕ τ
λ τ ϕ λ τ ϕ ρ
λ λ
ϕ λ ϕ λ ϕ
τ
τ τ
And consequently, using estimations obtained before for ,
) ,
0( λ
ϕ x ϕ(x,λ) , ϕ0′(x,λ) and considering, that
, )
( )
( ]
) ( [
0 0
0 0
∞
<
<
+
= +
≤
−
∫ ∫ ∫
∫
q s q ds τ q s ds τ qds q τ q s ds constτ τ we shall obtain
τ λ τ ϕ λ τ ϕ τ λ
ϕ λ
ϕ′ x − ′ x <R+
∫
xB ′ − ′ d0
0
0( , ) ( ). ( , ) ( , )
) , (
, where R>0 and B(τ)>0 are limited.
Hence, by Gronwall’s lemma ∫ ≤ <∞
′ ≤
′ x − x Re const
x
d B
0
) (
0( , ) .
) , (
τ
λ τ
ϕ λ
ϕ .
As ϕ′0(x,λ) ≤const<∞, from last inequality follows, that
∞
<
′(x,λ) ≤const
ϕ is regular on x∈[ a0, ]and λ from considered strip.
Let now ϕ(x,λ) reaches maximum ϕm in point x0∈[0,a],
Maximum ϕ′(x,λ) is equal ϕm′ .Then function graph ϕ(x,λ) lies above triangle with top in point (x0,ϕm)and lateral faces with angular coefficients
m
m ϕ
ϕ′ ,− ′ accordingly.
And consequently
x y
0 a x
0
ϕm
•
. ] ) ( [ 3 ]
) [(
.
] ) ( ) ( . 2 [ .
] ) ( ) ( . 2 [ . )
( .
) ( )
, ( . )
, ( . )
, ( ) (
2 0 2
2 0 3 0 3 2 0
2
2 0 2 0 2
0
2 0 2 0 2 2
0
0 0
2 0 2
2
0 0
2
0 0
0
0
+ ′ − + − ′ + −
=
′ − +
′ −
−
+
′ − +
′ − +
=
′ −
− +
′ − +
≥
=
≥
∫
∫ ∫
∫ ∫
∫ ∫
x a x x
x a a
m
dx x x x
x m
dx x x x
x m
dx x x m
dx x x m
dx x
m dx x
m dx x
x
m m m
m m
a
x
m m
m m a
x
x
m m
m m m
m
a x
m m
a a
ϕ ϕ ϕ ϕ
ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ ϕ
ϕ
ϕ ϕ λ
ϕ λ
ϕ λ
ϕ ρ
From inequality we got follows, that
m m m
m a
a x
x a x x
x a a
dx m x x
x
ϕ ϕ ϕ
λ ϕ ϕ ρ
λ ϕ
− ′ +
′ − +
+ −
≤
∫
∈
] ) ( [ ) 3 .(
) . (
1 )
) , ( ) ( (
) , ( max
2 0 2
0 2 3
0 3 2 0
1
0
2 ] , 0 [
.
If to enter designations x0 =ε.a and z
m m′ = ϕ
ϕ we shall get
4].
) 1 3 12(
) 1 2 )( 1 2 ( [
] ) 1 ( 3 [
) 1 (
] ) ( [ ) 3 .(
) (
2 2
2 2
2 2 3 3 3 3 3
2 0 2
0 2 3
0 3 0
+
− +
−
−
=
=
− + + −
+ −
=
′ =
− +
′ − +
+ −
az az
az a
z a
a a z
a a
x a x x
x a a
m m m
m
ε
ε ε ε
ε
ϕ ϕ ϕ
ϕ
Where ε∈[0,1] and z∈(0,∞). Let’s enter now designations
4 ) 1 3 12(
) 1 2 )( 1 2 ( ) ,
( z =az az− − 2 + az− 2 +
f ε ε and estimate from below
) , ( z
f ε . It is obvious, that
4. ] 1 4 ) 3 2 [( 3 3 1
4 ) 1 9 6 12(
) 1 2 4(
) 1 , 1 ( ) , 0 (
2
2 2 2
2
≥ +
−
=
+ +
− +
−
=
=
az
az z
a az
z a z
f z f
Inside of interval 0≤ε ≤ 1 is unique critical point 2
= 1
ε , in this point we
have .
4 1 4 ) 1 3 12(
) 1 2,
(1 z = az− 2 + ≥
f . Hence, for any ε∈[0,1] and z∈(0,∞) estimation
4 ) 1 , ( z ≥
f ε is fair, therefore inequality is
a a m
dx m x
x
x
a
a x
. 2 4 .1 .
1 )
) , ( ) ( (
) , ( max
2 1
0
2 ] , 0
[ ≤ =
∫
∈
λ ϕ ρ
λ ϕ
So theorem is proved.
Thus we have proved, that normalized eigenfunctions of problem (1) - (3) in case of weight functions satisfying to Lipschitz condition are limited in regular intervals, the obtained result is proved by statement proved in [7] at α =1, as
0 )
(
lim 0
2 1 _____
] , 0
[ = >
∞ −
→ С
x y
n n C n
a
λ α .
References
[1] T.Regge. Analytical properties of matrix of dispersion. Mathematics (Col.
Translations) 7 (4) (1963), 83-89.
[2] Kravitsky A.O. About decomposition in number by eigenfunctions of one non self-interfaced regional problem // Daghestan Academy of science USSR, T. 170 (6) (1966), 1255-1258.
[3] Gehtman М.М. About some analytical properties of kernel resolvent of ordinary differential operator of even order on Riemann surface, Daghestan academy of science USSR, Moscow, 201(5) (1971), 1025 - 1028.
[4] Aigunov G.А. About one boundary value problem generated by non self- adjoin differential operator of n2 order on axle, Daghestan academy of science USSR, Moscow, 213(5) (1973), 1001-1004.
[5] Aigunov G.A. Spectral problem of T.Regge type for ordinary differential operator of n2 order // Col. Functional analysis, theory of functions and their
appendices, Issue 2. P.I. Makhachkala, 1975, 21-41.
[6] Aigunov G.A., Gadzhiev T.U. Studying of asymptotic of eigenvalues of one regular boundary value problem, generated by differential equation of n2 order on interval [ a // News of high schools of North Caucasus, Rostov-on-Don, 0, ]
№5, 2008,14-19.
[7] Aigunov G.A., Jwamer K.H, Dzhalaeva G.A. Asymptotic behavior of normalized eigenfunctions of problem of T.Regge type in case of weight functions close to functions from Holder classes. Journal of mathematics collection, Makhachkala (South of Russian), 2008, 7-10.
[8] Aigunov G.A., Gadzhiev T.U. Estimation of normalized eigenfunctions of problem of T.Regge type in case of smooth coefficients. Interhighschool Col.
«FDU and their appendices », Makhachkala, 2009, 18-26.