In this section, we consider the boundary value problem (2.1), (2.2) with . We count all eigenvalues with their proper multiplicities and develop a formula for the asymptotic distribution of the eigenvalues, which is used to obtain the corresponding formula for general g. Observe that for , the quasi-derivatives coincide with the standard derivatives . We take the canonical fundamental system , , of (2.1) with if for . It is well known that the functions are analytic on ℂ with respect to λ. Putting
the eigenvalues of the boundary value problem (2.1), (2.2) are the eigenvalues of the analytic matrix function M, where the corresponding geometric and algebraic multiplicities coincide, see [[15], Theorem 6.3.2].
Setting and
it is easy to see that
The second row of has exactly two non-zero entries (for ), and these non-zero entries are:
In Cases 1 and 2, and ;
In Cases 3 and 4, and .
Since the first row of has exactly one entry 1 and all other entries zero, an expansion of with respect to the second row shows that , where
with
In view of (2.7), (2.8) this gives
Each of the summands in ϕ is a product of a power in μ and a product of two sums of a trigonometric and a hyperbolic functions. The terms with the highest μ-powers in are non-zero constant multiples of
For the above four cases, we obtain:
We next give the asymptotic distributions of the zeros of with proper counting.
Lemma 3.1 Case 1: has a zero of multiplicity 8 at 0, simple zeros at
simple zeros at , and for , and no other zeros.
Case 2: has a zero of multiplicity 4 at 0, exactly one simple zero in each interval for positive integers k with asymptotics
simple zeros at , and for , and no other zeros.
Case 3: has a zero of multiplicity 6 at 0, simple zeros at
simple zeros at , and for , and no other zeros.
Case 4: has a zero of multiplicity 6 at 0, exactly one simple zero in each interval for positive integers k with asymptotics
simple zeros at , and for , and no other zeros.
Proof The result is obvious in Cases 1 and 3. Cases 2 and 4 only differ in the factor with the power of μ, and the multiplicity of the corresponding zero of at 0 is easy to verify. The choice of the indexing for the non-zero zeros of in each case will become apparent later.
It, therefore, remains to describe the behavior of the non-zero zeros of in Case 2. First, we are going to find the zeros of on the positive real axis. One can observe that for , implies and , whence the positive zeros of are those for which . Since and for all where the functions are defined, the function is increasing with a positive derivative on each interval , . On each of these intervals, the function moves from −∞ to ∞, thus we have exactly one simple zero of in each interval , where k is a positive integer, and no zero in . Since as , we have
The location of the zeros on the other three half-axes follows by repeated application of .
The proof will be complete if we show that all zeros of lie on the real or the imaginary axis. To this end, we observe that the product-to-sum formula for trigonometric functions gives
(3.1)
Putting , , it follows for that
Since has a positive derivative on , this function is strictly increasing, and therefore, implies by (3.2) that and thus . Then
is either real or pure imaginary. □
Proposition 3.2 For , there exists a positive integer such that the eigenvalues , counted with multiplicity, of the problem (2.1), (2.5)-(2.8), where in Cases 1 and 2 and in Cases 3 and 4, can be enumerated in such a way that the eigenvalues are pure imaginary for , and for . For , we can write , where the have the following asymptotic representation as :
In particular, the number of pure imaginary eigenvalues is even in Cases 1 and 2 and odd in Cases 3 and 4.
Proof Case 4: A straightforward calculation gives
(3.3)
Up to the constant factor , the second term equals . It follows that for μ outside the zeros of , and , we have
(3.4)
Fix and for let be the squares determined by the vertices , . These squares do not intersect due to . Since if and only if and , it follows from the periodicity of tan that the number
is positive and independent of ε. Since uniformly in the strip as , there is an integer such that
By periodicity, there are numbers and such that and for all and all k. Observing , it follows that there is such that for all μ on the squares with the estimate holds. Further, we may assume by Lemma 3.1 that is inside for and that no other zero of has this property. Hence, it follows by Rouché’s theorem that there is exactly one (simple) zero of ϕ in each for . Replacing μ with iμ only changes the sign of the second term in (3.3) and thus the sign of . Hence, the same estimates apply to corresponding squares along the other three half-axes, and we therefore have that ϕ has zeros , for with the same asymptotic behavior as the zeros , of as discussed in Lemma 3.1.
Next, we are going to estimate on the squares , , whose vertices are . For and ,
(3.5)
Therefore, we have for , where and , that
(3.6)
For , and , we have
uniformly in y as . Hence, there exists such that for all , and ,
(3.7)
It follows from (3.6) and (3.7) for , , and that
(3.8)
Furthermore, we are going to make use of the estimates
which hold for all with and all . Therefore, it follows from (3.6), (3.8)-(3.10) and the corresponding estimates with μ replaced by iμ that there is such that for all with . Again from the definition of in (3.4) and Rouché’s theorem, we conclude that the functions and ϕ have the same number of zeros in the square , for with .
Since has zeros inside and thus zeros inside , it follows that ϕ has no large zeros other than the zeros found above for sufficiently large, and that account for all eigenvalues of the problem (2.1)-(2.2) since each of these eigenvalues gives rise to two zeros of ϕ, counted with multiplicity. By Proposition 2.3, all eigenvalues with non-zero real part occur in pairs , , which shows that we can index all such eigenvalues as . Since there is an odd number of remaining indices, the number of pure imaginary eigenvalues must be odd.
Case 2: The function ϕ in this case is
Then all the estimates are as in Case 4, and the result in Case 2 immediately follows from that in Case 4 if we observe that each for k large enough contains two fewer zeros of ϕ than in Case 4.
Case 1: A straightforward calculation gives
Then
The result follows with reasonings and estimates as in the proof of Case 4, replacing μ by and , respectively.
Case 3: The function ϕ in this case is
and a reasoning as in Case 1 completes the proof. □