© Hindawi Publishing Corp. ON HILL’S EQUATION WITH A DISCONTINUOUS COEFFICIENT

We research the asymptotic formula for the lengths of the 
instability intervals of the Hill's equation with coefficients 
 q ( x ) and r ( x ) , where q ( x ) is piecewise continuous and 
 r ( x ) has a piecewise continuous second derivative in open 
intervals ( 0 , b ) and ( b , a ) ( 0 b a ) .


Introduction.
We consider the second-order equation where q(x) and r (x) are real valued and all have period a.Also q(x) is piecewise continuous, r (x) is piecewise continuous in (0,b) and (b, a) where 0 < b < a and r (x) ≥ r 0 (> 0).Those values of complex parameter λ, for which periodic or antiperiodic problem associated with (1.1) and the x-interval (0,a) has a nontrivial solution y(x,λ), are called eigenvalues.An equation giving the eigenvalues of the periodic or the antiperiodic problem can be found.For the purpose, after we use the Liouville transformation the periodic boundary value problem becomes the boundary value problem and the antiperiodic boundary value problem becomes the boundary value problem z + λ − Q(t) z(t) = 0, 0 ≤ t ≤ A, z(A) = −σ z(0), σ ż(A) + ρz(A) = −ż(0) − τz(0), (1.4) where , ρ = r −1/4 (0) r −1/4 (x) x=a , τ= r −1/4 (0) r −1/4 (x) x=0 . (1.5) Let θ(t, λ) and ϕ(t, λ) denote the solutions of problem (1.3), satisfying the initial conditions (1.6) Then, we can define the Hill discriminant (see [1]) of (1.1) by the function Thus, the eigenvalues of the periodic boundary value problem coincide with the roots λ of and also the eigenvalues of the antiperiodic boundary value problem coincide with the roots of Each of the periodic and antiperiodic boundary value problems has a countable infinity real eigenvalues with the points accumulate at +∞. Denote by the eigenvalues of the periodic boundary value problem, and by the eigenvalues of the antiperiodic boundary value problem (the equality holds in the case of double eigenvalue).These values occur in the order , then all nontrivial solutions of (1.1) are unbounded in (−∞, ∞).These kind of intervals are called the instability intervals of (1.1).Apart from (−∞,µ 0 ), some or all of the instability intervals vanish for the case of double eigenvalues.If λ lies in any of the complementary open intervals (µ , then all solutions of (1.1) are bounded in (−∞, ∞), and these intervals are called the stability intervals of (1.1).We are interested in the lengths I n of the instability intervals Eastham [2] has studied (1.1) where the second derivative of r (x) is piecewise continuous on interval (0,a).But we studied (1.1) where the second derivative of r (x) is piecewise continuous on intervals (0,b) and (b, a) (0 < b < a).On the other hands, our method is different from his method and is based on using Rouche's theorem about roots of analytic functions.

Proof. It is clear that θ(t, λ) and ϕ(t, λ) verify the integral equations for 0 ≤ t < B.
Let B < t ≤ A. Since θ(t, λ) and ϕ(t, λ) are solutions of problem (1.3), we have First, we multiply the equalities above by sin s(t − ξ)/s and take integral of the obtained equalities from B to t.Then, we get (2.5) When these equalities are written in (2.4), the proof is completed.

The asymptotic formulas for the lengths of the instability intervals.
First, we research the eigenvalues of the periodic boundary value problem, using Rouche's theorem which is stated as follows.
In order to apply Rouche's theorem to our case, we need the following lemma.

Lemma 3.2.
There is a positive number C such that where C does not depend on s and n. (3.4) On the vertical edge of square contour Γ 2n+1/2 , take When the value of u is written in (3.4), we have and hence On the other hand, we take that on the horizontal edge of square contour Γ 2n+1/2 .Since the function |f (s)| 2  has minimum values at the points u = pπ /A where p is even, we have (3.9) This completes the proof.
From Lemma 3.2, we have a positive number C such that and there exits a natural number n 0 such that, for all n ≥ n 0 , (3.15) where ρ is any number such that 0 < ρ < π/2A.Then functions f (s) and g(s) are analytic on these circles and region bounded by these circles.
Then, for all s on these circles, there is a positive number K such that f (s) ≥ Kρ 2 e A| Im s| , (3.16) where K does not depend on s and n ∈ N.

Proof. Let s = u + iv and consider |s −
(3.17) On the other hand, This function has a maximum value and a minimum value at the points respectively.So, for all u and v on the circles, we have and hence Therefore, there is a positive number 0 not depending on ρ and n such that (3.23) Thus, for all s on the circles, we get ).So, we can work on interval [0,ρ] because the function g is even for all v ∈ [−ρ, ρ].First, take the derivative of g on the variable v and we get So, g is increasing and then g(0 where Similarly, when we take (3.30) the inequality is verified.
From Lemma 3.3 and definition of g(s), we have , where C 2 = C 1 /K because there is a positive number m such that for all n ≥ m, ρ < π/2A.Moreover, for all s on these circles, we have By Rouche's theorem, the function Φ + (s) has one root inside the circle and one root inside the circle Denote these roots by s − 2n and s + 2n , respectively.Let )), we have Sup r ∓ 2n < ∞.Hence, for periodic boundary value problem, we have (3.37) Similarly, for antiperiodic boundary value problem, we have (3.38) Combining these two results, we get Proof.We know that eigenvalues of periodic and antiperiodic boundary value problems are real and go to infinity.So, it is sufficient to take positive values of parameters λ.By (2.7) and (2.8), we have (3.41) When the values θ(t, λ) and ϕ(t, λ) are written in (2.1) and in derivative of (2.2), we get (3.42)Moreover, we have (3.43) After putting the values θ(A, λ), ϕ(A, λ), and φ(A, λ) in equality (3.50)