On Hill's Equation with Piecewise Constant and Alternating Coefficient

1 (1), 2007, 157-166

©BEYKENT UNIVERSITY

On Hill's e q u a t i o n w i t h p i e c e w i s e c o n s t a n t a n d

a l t e r n a t i n g coefficient

In this paper we study the lengths of instability intervals of a Hill's equation having piecewise constant and alternating coefficient which tend to infinity.

MSC: 34L20; 47EU5

Keywords: Hill's equation, eigenvalue, stability.

1 Introduction

A well-known a Hill's equation is in the form

where A is a complex parameter and function p has the period w > 0, In most investigations on Hill's equation the Function p is assumed to be nonnegative (see, [!-3, 5-7]). The equation (1.1) having a positive piecewise constant function has been studied in [5,6]. Later Guseinov and Karaca [4] are concerned w i t h the equation (1,1) having a piecewise constant and alternating function

where a is a fixed point between 0 and u>. In this paper we consider the case in which the coefficient p of (1.1) is in the form

where a and 0 are positive constants satisfying \32 — a2| < a0 and a is a fixed point between 0 and u .

Recall that periodic

II kay Yaslan Karaca

Department of Mathematics, Ege University

35100 Bornova, Izmir, Turkey

ilkay.karacaOege.edu.tr

A b s t r a c t

y" = Ap(x)y (-oo < x < oo),

(1.1)

(1.2)

-y" =

Ap{x)y (0 <

x

< lj)

y(0) = y{u), y'( 0)

=

?

/(w)

and the semi-per iodic (or anti-periodic)

-y" = \p{x)y (0 < x < w),

y( 0) = -V(u), m =

(1.4)

boundary value problems associated with the equation (1.1).

The periodic and the semi-periodic boundary value problems associated with equa-tion (1.1) have a nontrivial soluequa-tion y(x, A) for values of complex parameter A. Those values A are said to be eigenvalues of the periodic and semi-periodic boundary value problems.

We shall denote by $(x, A) and <fi(x, A) the solutions of the equation (1.1) satisfying the initial conditions

Then the eigenvalues of the periodic and semi-periodic problems coincide with the roots of characteristic equation F(A) — 2 = 0 and F(A) + 2 = 0 respectively. It is clear that each of the problems (1.3) and (1.4) has an infinitely countable real eigenvalues with accumulation points at both infinity and negative infinity (see [9, Chapter VIII]). The eigenvalues fi^b a nd Vik+l (k £ Z) of the problem (1.3) and (1.4) respectively

occur in the order

••• < < fCi <t¿i<toi <¡4 <i¿¡ </4 < fa <t4 <••••

If A lies in any of the open intervals ( p . ( n e Z), then all non-trivial solu-tions of (1.1) are unbounded in (—00,00). These intervals are called the instability

intervals of the equation (1.1). If A lies in any of the open complementary intervals

{¡"i-itPQ) (n £ Z), then all solutions of (1.1) are bounded in (—00,00), and these

intervals are called the stability intervals of (1.1).

In the order to study the asymptotic behaviour of the lengths

of the instability intervals (fin, f i f ) as 11 —» ±00, we must investigate eigenvalues

of the problems (1.3) and (1.4).

2 Preliminaries

When we take A = s2 and assume that the p has the form (1.2) in the equation (1.1) the solutions 6(x, A) and tp(x, A) of equation (1.1) satisfying the initial condition (1.5) will be in the form

0(0, A) = 1, ff(0,A) = 0 ; ï?(0,A) = 0 , */(0,A) = 1. (1.5)

The Hill discriminant of the equation (1.1) is defined by the function

F(A)

= 0(w,A) + (pV,A).

(1.6)

= ( n f Z )

sin sax

ıp( i . A) = _{sın saa si rıh s/3(x - a)} sa O < X < a,

- cash sf}[x - a) + cos sata ^ —, a < x < u>.

sa _s/3

Substituting these into (1.6) we have t!ie following formula for the Hill discriminant: /•'(A) = 2 cos saa cosh sfJ(<jj - a) - ( | ~ f ) s i n scxasinh$/.i(uJ - a), (2,1) From this formula we can define

$+(A) = F(A) - 2 = - 8 sin2 ^ s i n h2 i ^ t p l - 4 sin3 + 4sinh2 1

+ ( § — 5) sinsaasirıh S0{w — a) whore A = s2. Setting implies where ••i'-i , s ' (2.2) (2.3) (2.4)

$t(z) = - 8 sin2 jrzsiiih?P—^-vz - 4sin2 nz + 4sinh3 ^ ^ i r z

+ ( ^ - | ) s i n 2 ^ s i n h ^ l « . (2.5)

For any positive integer n we can define the rectangular contour

r

B

=

{z € C : |Re z\ = n + |Im z\ = + §)}.

Let's state the well-known theorem;

R o u t lie's T h e o r e m . I f f and y are analytic functions inside and on a closed contour r, and \g(z) \ < \f(z)\ on F, then f and f + g have the same number of zeros inside

r .

In order to apply Rouche theorem we put

T = r„, f ( z ) = -8sin2?rzsmh2 Si~1Trz,

g(z) = —4sin3 ttz + 4sinhs + ( g - | ) a i n 2 j r i s i n hM i i .

Thus = f ( z ) + g(z) and then we must show that the inequality \g(z)\ < |/(c)| is equivalent to the inequality

It is clear that for z — ± ( n + 1) + iy the formulas | sin 7T2p = cosh2 Try,

|sinh > j sinh + | cot ltz\ — \ tanh ny\,

and + Ê - ^ f f + D " - 2cos iry (2.7) (2.8) (2.9) (2.10)

hold. From (2.7), (2.8), (2.9), (2.10), we have the following inequality on the vertical sides of r „ ;

as n goes to infinity. Thus we have the inequality (2.6) on the vertical sides of r „ for a, 0 satisfying \02 — a2\ < a(3 and for all n > m , where raj is a sufficiently large positive integer.

Now consider the horizontal sides of r „ . It is obvious that for z — x ± ' + 5)

I cotirr|2 and ! s m ^ P > s i n h2^ ( 7 i + i K I sinh — eoshi + e-wSfrt»*-'»' + 2 cos 2KX ~~ + _ 2 cos 2irx

I coth fils^siwz\ = I tanh ^ ^ t t i I

1 aa 1 1 aa

( 2 . 1 1 )

(2.12)

(213)

(2.14) hold. By the equations (2.11), (2.12), (2.13), (2.14) we get following inequality on the horizontal sides of r„:

I J & I S J I 1 + 1 *z5!ll «.a

as ti —* 00. Hence the inequality (2,6) holds on the horizontal sides of T„ for all a, 0

satisfying \0* — a2 \ < a f ) and for all n > n2, whore n2 is a sufficiently large positive

integer.

Take n0 = m a i f n i , ^ } - Then the inequality (2.6) holds for all a, 0 satisfying

¡/J3 — a21 < ad and for all n > tiq. So we prove the following lemma.

L e m m a 2.1 There exist a positive integer no such that the inequality (2.6) holds for all a, 0 satisfying \02 - a2\ < afi and for all n>n0 and for all z £Tn.

From Lemma 2.1 and Rouche theorem we conclude that for n > «o the number of zeros of ^ f f z ) lying inside r „ is the same as that for sin2 itz sinh2 fcilf;, Zeros

of latter function inside T„ are

0 ±1 ±2

±n ±>—22— +2i +ni ""

where i denotes the imaginary unit. Note that zero is a root of multiplicity four and each nonzero root is double. Hence, must have 8n + 4 zeros inside r „ (n > Since is an even function, denote the zeros of $J"(z) lying inside I\, by

± Z L2 N, ± Z ± 2 „ , - . . , ± Z Z2,

±zt

2

,

±Zq,

± 4 , ±Z2,

±4,

-, ±s£

n

, ±z}„.

We note that since the eigenvalues of the periodic problem (1,3) are real (see Intro-duction), by (2.3) and (2,4) the zeros of (z) are either real or pure imaginary.

Lemma 2.2 For n >n0 + 1 the function <l>f (2) has exactly two zeros in the region Dn = {z£C:n-l<Rez<n + i, |Im z\ < ^ ( n + i ) } ,

and exactly two zeros in the region

K„ = {z

e C : |Re z| < < Im 2

<

+

Therefore, for n > n0 + 1 we have

n-t<zi

n

<n

+ i and

zt

2n

=

»wi*, ^ ( n - j) < w*

ln

<

s

« g ( n

+

J ) .

(2,15)

Proof, From the proof of Lemma 2.1 the inequality (2.6) holds also on the bound boundaries of the

regions Dn and Kn for > no -I- 1. By the Rouche theorem

the function has in the D„ and Kn as many zeros as the function f ( z ) =

—8sin2jrzsinh2 rz, i.e. exactly two zeros in Dn and exactly two zeros in Kn.

This completes the proof. •

3 Results

Theorem 3.1 The eigenvalues of the periodic and the eigenvalues } of the semi-periodic boundary value problems have the form

± _ /B2±2r£V

_{) for n > »0 + 1.}

f<" « « - ( n o + 1),

w/iere 0 < 7-,i < f (|n| > tj0 + 1 ) and the equations (3.7). (3.8), (3.10), and (3.11) hold.

Proof.

for k > no + 1,

i - i l i r a

4

)

for k<-K

+

l).

Since the numbers are the zeros of the function (2,2) hik and kih are, for k > ri0 + l, the roots of the equation

2sin2 1 s i n h2S ^ ( f c 7 r + f) + s i n2i - s m h2f i ^ ( f c 7 r + t ) - J (fl _ J) sm2t sinh (far + t) = 0:

and for k < — (no + 1) the roots of the equation

2 sin2 f sinh2 ^ ( f c * - f) + sin21 - sinh2 kir - t) - }(« - f )

sin21 s i n h^ j ( t o r t ) = 0

-The equation (3,1) is equivalent to the union of equations sin( = ^ A + - a)sin2t tanh ^ — ^ ( k T r + /)

and

sini = - ^ 4 4- ±(£ - | ) sin 2ftanh ^ ^ ( f c r r + t) and the equation (3.2) is equivalent to the union of equations

sint = V' 0 + - | )s i n 2 f t a n h and sini = -JB + 1 ( | - | ) s i n 2 ( tanh ^( f a r - t) sinh 2 « ( * * + () where A — -,

B =

(3.1) (3,2) (3.3) (3,4) (3.5) (3.6) 1 + 2sinh2 ^ ( f c r r + t ) ' 1 + 2sinh2 ^ ( k n - t ) ' Define a function

V>(0 = sini - + | ) sin 2i tanh ^ ^ ( f o r + ()•

It is obvious that function lp(t) is continuous on an interval [0, f ] and besides is negative and positive at the points 0 and | respectively. Hence ip(t) has at least one zero in [0, f ) .

Analogously the equation (3.4) has at least one root in ( —O j .

In order that one of the equation (3.3) and (3.4) has more than one root it is

necessary that the function i ' f (f) has more than two zeros in a region D„. On the other hand, the latter is impossible in view of lemma 2.2 if k > + 1. Similarly one can prove that for k < — - 1 - 1 ) the equation (3.5) has exactly one root in [0, and the equation (3.6) has exactly one root in (—1,0],

Let r j t £ ¡0, f ) denote the root of equation (3.3) if k > % 4- 1, and the root of equation (3.5) if k < - ( n o + 1). Also, let - r ^ £ ( - f , 0 ] denote the root of (3.4) if

k > + 1, and the root of equation (3.6) if k < ~(n0 + 1). Then 0 < r*. < | for all |A;| > «o + 1> and = ±r^k if k > no + 1, and h%k = if k < -(<j0 + 1). Hence

the eigenvalues of the problem (1,3) are in the form

=

( 4 )

2

= k>n

0 +

l.

f4k = ( 4 )

2

= k<-(n

o + 1),

where 0 < r ^ < § (|Jt| > n0 + 1) and

sin

r i

= p ±

-

J) sin 2 ^ tanh ^^{kn ± k > n

a

+

1, (3.7)

sin,-* = V/ D ± I ( f -1 )s i n 2 r ± tanh T k < - ( n o + I), (3.8)

where C =

,

j

=

1 + 2sinh ^ ( f o r ± 4 ) l + 2 s i n h2^ ( A 7 rT4 )

Similarly one can investigate the eigenvalues of the semi-per iodic boundary value problem, that is the roots of the function

F(A) + 2

= 8 cos2 i f cosh2 ^ p i - 4 cos2 s f - 4 cosh2 ^ f c i i + 4

— ( | — ¡j) sin saa sinh s0(u> — a) (3.9)

where A = s2. When we take saa — 2-kz, we can define

$ - ( A ) = $j"(s)

= 8 cos2 irz cosh2 ^ ^ t t z - 4 cos2 nz - 4 cosh2 ^^irz + 4

f { z ) = 8 cos3 7:z cosh2 ^^ırz,

g(z) = c o s2« - 4cosh2 + 4 - ( J - s i n 2 « s i n h ^ ^ t t i

and let

r„- = {

Z

£ C : [Re

z\

= n

+

1, |Im

z|

=

^ ^ ( r t +

1)},

From the Rouche theorem and cos2 nz cosh2 ^ ^ • a z have the same

num-ber of zeros inside r ; for all a ,f t satisfying \0* - u2| < a0 and for all n > n0,

where nD is a sufficiently large positive integer. On the other hand the function

cos2 Trzcosh3 ^ ttz has double zeros inside F~ at the points

+ k and * = +

where k = 0, ±1. ± 2 , . , . , ± n , -n-1. Thus the function (z) has 8n 4- 8 zeros inside which are denoted by

±z

_{- (211+1) >}

. • • • »

±z

-l> ğ

z

-l< =^1 1

±z

î> •••< İ^ün+l.

±Z

în+l-Let Sana = j ^ a U - n w he r e k e Z. Then following formulas can easily be obtained

for the eigenvalues = (s2it+i)2> k 7,

where 0 < < f (|fc| > no + 1 ) and for k > n0 + 1

c o s r ±+ 1 = | ) s i n 2 r ±+ 1 t a n h^ ^ ( k n + § ± r ±+ 1) ,

for

k

< -(«o + l)

c o s r ^ + i = | ) s i n 2 ^+ 1 tanh + f T r ±+ 1) , (3.10) (3.11)

where

E

=

l - c o s h

2

^ ( ^

+ f ± 4

This completes the proof.

Corollary 3.1 For the length I„ = — fi~ of the instability interval the formula

i . = + r~) + rf - r~3) forn>ntl + 1; (3.12)

'« = " rnhpir? - r f ) for n < -(ru> + 1); (3.13) hold, where r^ is the same as in Theorem 3.1.

Theorem 3.2 If a, ¡3 satisfy \/32 - < a0, then the length I„ an unbounded as

« —*

oo.

Proof, In view of (3,12), (3.13) it will be sufficient to show that the sequences 4 and 4 + i do not tend to zero as n —* oo. Let us assume the contrary: let 4 0 as k —* oo. Then the equation

sinh

2

S o f t e r + r

2+fe

) =

si"2 4 - K g - g ) s i n 2 4 tanh ^ ( f a r + 4 )

1 - 2sin2 r+ + - a )s i n 2 4 tanh + 4 )

arising from (3.7), gives

lim sinh ^ • — (far + r.tt) = 0.

aa

Thus, we arrive at a contradiction. Similarly, the case k —> —oo and the case ?i =

2k + 1 (k —» ±oo) can be considered. The theorem is proved. o

References

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[3] Easthain, M.S.P., Results and problems in the spectral theory of periodic differ-ential equations, Lecture Notes in Mathematics, Springer-Verlag, New York, 448 (1975), 126-135.

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