����������� ������ 45 (1993), 11–17

UDC 517.984.5

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ON THE EXPANSION THEOREM FOR A CERTAIN BOUNDARY VALUE PROBLEM FOR A FUNCTIONAL DIFFERENTIAL EQUATION

M. Dostanić

Abstract. The boundary value problem

−y′′ + q(x)y = λy + ∫_{0}^{π} y dσ(x), y(0) = y(π) = 0,

is concerned, where q ∈ C[0, π] and σ is a function of bounded variation. It is proved that the system of eigenfunctions of the given problem is complete and minimal in L^{2}(0, π), and also that functions of a certain class can be expanded into uniformly convergent series with respect to the mentioned system.

Introduction

In [1] and [2] problems concerning the asymptotics of spectra as well as determining regularized traces of the following two boundary problems were examined

−y′′ + q(x)y = λy + y (\frac{π}{2} ) , y(0) = y(π) = 0;  (1)
−y′′ + q(x)y = λy + ∑_{k=1}^{n−1} α_{k}y (\frac{kπ}{n} ) + α_{n} ∫_{0}^{π} y(t) dt, y(0) = y(π) = 0, (2)

where q is a sufficiently smooth function with complex values.

In [3] we see that, under certain conditions, the system of eigenfunctions of the boundary problem (1) is a Riesz base of L^{2}(0, π) and also that any function f ∈ C^{2}[0, π] for which f(0) = f(π) = 0 can be expanded into a uniformly convergent series with respect to a system of eigenfunctions of the boundary problem (1).

In this paper we examine the following boundary problem

−y′′ + q(x)y = λy + ∫_{0}^{π} y dσ(x), y(0) = y(π) = 0,  (3)

where q is a real continuous function and σ is a function of bounded variation on [0, π]. We prove a similar expansion theorem.

1. Preliminaries

Let L_{0} be the differential operator generated by the differential expression l_{0}(y) = −y′′ + q(x)y and boundary conditions y(0) = y(π) = 0. Note that L_{0} is a selfadjoint operator. Let L be the operator generated by the differentially-integral expression l(y) = −y′′ +q(x)y −∫_{0}^{π} y dσ(x) with the same boundary conditions. We say that λ = λ_{0} is an eigenvalue of the boundary problem (3) if there is a function y_{0} ∈ C^{2}[0, π] for which y_{0} ≢ 0 and

−y′′_{0} + q(x)y_{0} = λ_{0}y_{0} + ∫_{0}^{π} y_{0} dσ(x), y_{0}(0) = y_{0}(π) = 0.

An eigenvalue is simple if there is one and only one corresponding eigenfunction (up to a multiplicative constant).

The Green’s function of the operator L_{0} − λ is given by [4]

G(x, ξ, λ) = \frac{H(x, ξ, λ)}{Δ(λ)},  (4)

where Δ is the characteristic determinant of the boundary problem

−y′′ + q(x)y = λy, y(0) = y(π) = 0.  (5)

For any fixed x and ξ from [0, π], the function H(x, ξ, λ) is entire. The function Δ is also entire and its roots are eigenvalues of the problem (5).

Then, the function Δ_{1}(λ) = Δ(λ) − ∫_{0}^{π}∫_{0}^{π} H(x, ξ, λ) dσ(x) dξ is also entire. We suppose:

1^{∘} No eigenvalue of the problem (5) is an eigenvalue of the problem (3). 2^{∘} Roots λ_{1}, λ_{2}, … of the function Δ_{1} are simple and Δ(λ_{n}) ≠ 0. 3^{∘} Boundary problems −y′′ + q(x)y = 0, y(0) = y(π) = 0 and −y′′ + q(x)y =

∫_{0}^{π} y dσ(x), y(0) = y(π) = 0 have only trivial solutions.

2. Main results

Lemma. If the conditions 1^{∘} and 2^{∘} are satisfied then the eigenvalues of the boundary problem (3) are simple and they are roots of the function Δ_{1}.

Proof. Let λ_{0} be an eigenvalue of the boundary problem (3). It means that there is a function y_{0} ∈ C^{2}[0, π] for which

−y′′_{0} + q(x)y_{0} = λ_{0}y_{0} + ∫_{0}^{π} y_{0} dσ(x), y_{0}(0) = y_{0}(π) = 0  (6)

and y_{0} ≢ 0 on [0, π].

From 1^{∘} we see that Δ(λ_{0}) ≠ 0. Applying the Green’s function of the operator L_{0} − λ_{0} to (6) we get

y_{0}(x) = ∫_{0}^{π} G(x, ξ, λ_{0}) dξ ∫_{0}^{π} y_{0} dσ(x).  (7)

From (7) we have y_{0}(x) = K ∫_{0}^{π} G(x, ξ, λ_{0}) dξ (for some constant K). Also from (7) we see that ∫_{0}^{π} y_{0} dσ(x) ≢ 0 (otherwise we have y_{0} ≡ 0 on [0, π]).

Integrating (7) with respect to the function σ we have

∫_{0}^{π} y_{0} dσ(x) = ∫_{0}^{π}∫_{0}^{π} G(x, ξ, λ_{0}) dξ dσ(x) ∫_{0}^{π} y_{0} dσ(x).  (8)

From (8), knowing that ∫_{0}^{π} y_{0} dσ ≠ 0, we get

1 − ∫_{0}^{π}∫_{0}^{π} G(x, ξ, λ_{0}) dξ dσ(x) = 0

and (because of Δ(λ_{0}) ≠ 0 and (4)) Δ_{1}(λ_{0}) = 0.

Let now λ_{0} be a root of the function Δ_{1}. Then, from 2^{∘}, we have Δ(λ_{0}) ≠ 0. Let us examine the function

y_{0}(x) = ∫_{0}^{π} G(x, ξ, λ_{0}) dξ.  (9)

As Δ_{1}(λ_{0}) = 0 we have

1 − ∫_{0}^{π}∫_{0}^{π} G(x, ξ, λ_{0}) dξ dσ(x) = 0.  (10)

From (9) we get (using characteristics of the Green’s function)

−y′′_{0} + q(x)y_{0} = λ_{0}y_{0} + 1, y_{0}(0) = y_{0}(π) = 0.  (11)

Integrating (9) with respect to the function σ (and knowing (10)) we get

∫_{0}^{π} y_{0} dσ(x) = 1.  (12)

From (11) and (12) we conclude that λ_{0} is an eigenvalue and y_{0} an eigenfunction of the boundary problem (3).

Lemma is proved.

The operator L_{0} is selfadjoint, so from (L_{0} − λ)^{∗} = L_{0} − λ̄ we get (see [4])

\overline{G(x, ξ, λ̄)} = G(ξ, x, λ)  (13)

if λ ∈ ρ(L_{0}). Let us denote by G_{0}(x, ξ) the Green’s function of the operator L_{0}, i.e. G_{0}(x, ξ) = G(x, ξ, 0). From 3^{∘} we get Δ(0) ≠ 0 and Δ_{1}(0) ≠ 0.

Solving the equation

−y′′ + q(x)y − ∫_{0}^{π} y dσ(x) = f(x), y(0) = y(π) = 0

we get

y(x) = (L^{−1}f)(x) = ∫_{0}^{π} G_{0}(x, ξ)f(ξ) dξ + ∫_{0}^{π} G_{0}(x, ξ) dξ \frac{∫_{0}^{π}∫_{0}^{π} G_{0}(x,ξ)f(ξ) dξ dσ(x)}{1−∫_{0}^{π}∫_{0}^{π} G_{0}(x,ξ) dξ dσ(x)}.

Now, we put

c = 1 − ∫_{0}^{π}∫_{0}^{π} G_{0}(x, ξ) dξ dσ(x), φ(ξ) = \frac{1}{c̄} ∫_{0}^{π} G_{0}(ξ, x) d\overline{σ(x)}.

So (14) becomes

L^{−1}f(x) = ∫_{0}^{π} G_{0}(x, ξ)f(ξ) dξ + (f, φ) ∫_{0}^{π} G_{0}(x, ξ) dξ,  (15)

where (·, ·) is the scalar product in L^{2}(0, π). As L_{0}^{−1}f(x) = ∫_{0}^{π} G_{0}(x, ξ)f(ξ) dξ, from (15) we get

L^{−1}f = L_{0}^{−1}f + (f, φ)L_{0}^{−1}1.  (16)

Let us now define linear operators A and A_{0} by

\begin{aligned}Af(x) & = ∫_{0}^{π} G_{0}(x, ξ)f(ξ) dξ + (f, φ) ∫_{0}^{π} G_{0}(x, ξ) dξ & (A & = L^{−1}), \\ A_{0}f(x) & = ∫_{0}^{π} G_{0}(x, ξ)f(ξ) dξ & (A_{0} & = L_{0}^{−1}). \end{aligned}  (17)

From (15), (16) and (17) we get

A = A_{0} + (·, φ)A_{0}1.  (18)

We define an operator S: L^{2}(0, π) → L^{2}(0, π) by Sf(x) = (f, φ) · 1. Then we have (from (18))

A = A_{0}(I + S).  (19)

The operators A, A_{0}, S act on L^{2}(0, π). The operator A has the eigenvalues that are reciprocal to the eigenvalues of the operator L; therefore

σ(A) = \{ \frac{1}{λ_{n}} : Δ_{1}(λ_{n}) = 0 \} ∪ \{0\}

(because A ∈ 𝔖_{∞}, 𝔖_{∞} – the set of compact operators) and the eigenvectors are equal to the eigenfunctions of the operator L. Similarly, the operator A_{0} has the eigenvalues that are reciprocal to the eigenvalues of L_{0} and the eigenvectors are equal to the eigenfunctions of the operator L_{0}.

Since q is a real, continuous function on [0, π], the eigenvalues of the operator L_{0} have the following asymptotics

μ_{n} = n^{2} + O(1).

From this we see that the asymptotics of eigenvalues of the operator A_{0} ( = A_{0}^{∗}) is

μ_{n}^{−1} = n^{−2}(1 + O(n^{−2}))

and so we conclude that A_{0} ∈ 𝔖_{1} (nuclear operator).

As y_{n}(x) = ∫_{0}^{π} G(x, ξ, λ_{n}) dξ are eigenfunctions of the operator L corresponding to the eigenvalues {λ_{n}}, they are the eigenvectors of the operator A corresponding to the eigenvalues {λ_{n}^{−1}}.

Theorem 1. If the conditions 1^{∘}, 2^{∘} and 3^{∘} are satisfied, then the system of eigenfunctions of the boundary problem (3) is complete and minimal in L^{2}(0, π).

Proof. Because of the previous results it is enough to prove that the system of eigenvectors of the operator A is complete in L^{2}(0, π).

Since the operator S is compact (its rank is one) and the operator A_{0} is selfadjoint and nuclear, by the Keldysh’s theorem (see [5]), for proving that our system is complete it is enough to prove that Ker A = {0}.

As Af = 0 then, from (19), we have

A_{0}(I + S)f = 0.  (20)

As the operator A_{0} is 1–1 (because the operator L_{0} is 1–1), we get (I + S)f = 0 i.e.

f + (f, φ)1 = 0.  (21)

From this we get

(f, φ) · (1 + (1, φ)) = 0.  (22)

It is easy to check that 1 + (1, φ) ≠ 0 (if 3^{∘} is satisfied). Then, from (22), we get (f, φ) = 0 and so, from (21), we conclude that f = 0.

That proves the completeness of the system {y_{n}(x)}_{1}^{∞}.

For the proof of minimality, it is enough to construct a system biorthogonal to the system {y_{n}}_{1}^{∞}.

Since Ay_{n} = λ_{n}^{−1}y_{n} and all eigenvalues are simple, if one denotes by z_{n}(x) the eigenvectors of the adjoint operator A^{∗} corresponding to the eigenvalues (\overline{λ_{n}})^{−1}, then A^{∗}z_{n} = (\overline{λ_{n}})^{−1}z_{n} (all eigenvalues (\overline{λ_{n}})^{−1} are simple). It is easy to check that

(z_{n}, y_{n}) ≠ 0, n ∈ N; (z_{n}, y_{m}) = 0, m ≠ n.

Because of this, we will suppose that the system {z_{n}}_{1}^{∞} is chosen in such a way that (y_{n}, z_{m}) = δ_{nm}. This system {z_{n}}_{1}^{∞} is biorthogonal to the system {y_{n}}_{1}^{∞}. That proves the minimality.

Since the operator A is compact, with eigenvalues that are simple, (I−λA)^{−1} is a meromorphic operator function with simple poles in points λ_{n} (see [5]). Moreover, the principal part of the Laurent expansion in the neighborhood of the point λ = λ_{n} is −λ_{n} \frac{(·,z_{n})y_{n}}{λ−λ_{n}}, i.e.

(I − λA)^{−1} = −λ_{n} \frac{(·, z_{n})y_{n}}{λ − λ_{n}} + G_{1}(λ),  (23)

where the function G_{1} is holomorphic in the neighborhood of the point λ = λ_{n}. As Ay_{n} = λ_{n}^{−1}y_{n}, from (23) we get

Res_{λ=λ_{n}} A(I − λA)^{−1} = −(·, z_{n})y_{n}.  (24)

As before, applying the Green’s function to the equation

−y′′ + q(x)y − λy − ∫_{0}^{π} y dσ(x) = f, y(0) = y(π) = 0

we get

(L−λ)^{−1}f = ∫_{0}^{π} G(x, ξ, λ)f(ξ) dξ+∫_{0}^{π} G(x, ξ, λ) dξ \frac{∫_{0}^{π}∫_{0}^{π} G(x,ξ,λ)f(ξ) dξ dσ(x)}{1−∫_{0}^{π}∫_{0}^{π} G(x,ξ,λ) dξ dσ(x)}. (25)

We also note that

(L − λ)^{−1}f = A(I − λA)^{−1}f (f ∈ C^{2}[0, π]).  (26)

Now, from (25) and (26), we have (for f ∈ C^{2}[0, π])

A(I − λA)^{−1}f = ∫_{0}^{π} G(x, ξ, λ)f(ξ) dξ + ∫_{0}^{π} G(x, ξ, λ) dξ \frac{∫_{0}^{π}∫_{0}^{π} G(x,ξ,λ)f(ξ) dξ dσ(x)}{1−∫_{0}^{π}∫_{0}^{π} G(x,ξ,λ) dξ dσ(x)}.

(27)

There is a sequence of circles Γ_{k} (whose centers are in the point λ = 0 and radii R_{k} → +∞ (k → ∞)) so that, on Γ_{k}, we have

|G(x, ξ, λ)| ⩽ \frac{M}{\sqrt{|λ_{k}|}} = \frac{M}{\sqrt{R_{k}}} ,  (28)

where the constant M does not depend on k and x, ξ ∈ [0, π] (see [4]). Let us examine the integral

\frac{1}{2πi} ∫_{Γ_{k}} \frac{A(I − λA)^{−1}f}{λ} dλ,

where f is a fixed, continuous function (on [0, π]). By the Cauchy’s residue theorem we get

\frac{1}{2πi} ∫_{Γ_{k}} \frac{A(I − λA)^{−1}f}{λ} dλ = Res_{λ=0} \frac{A(I − λA)^{−1}f}{λ} + ∑_{|λ_{n}|<R_{k}} Res_{λ=λ_{n}} \frac{A(I − λA)^{−1}f}{λ} .

(29)

As Res_{λ=0} \frac{A(I−λA)^{−1}f}{λ} = Af and (because of (24))

Res_{λ=λ_{n}} \frac{A(I − λA)^{−1}f}{λ} = −\frac{1}{λ_{n}} (f, z_{n})y_{n},

we have (from (29))

\frac{1}{2πi} ∫_{Γ_{k}} \frac{A(I − λA)^{−1}f}{λ} dλ = Af − ∑_{|λ_{n}|<R_{k}} \frac{1}{λ_{n}} (f, z_{n})y_{n}.  (30)

This can be written as

\frac{1}{2πi} ∫_{Γ_{k}} \frac{A(I − λA)^{−1}f}{λ} dλ = Af − ∑_{ν=0}^{k}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}} \frac{1}{λ_{n}} (f, z_{n})y_{n}),  (31)

where we take R_{0} = 0.

From (31) we get

|Af(x) − ∑_{ν=0}^{k}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}} \frac{1}{λ_{n}} (f, z_{n})y_{n}(x))| ⩽ max_{\begin{matrix}λ∈Γ_{k} \\ x∈[0,π] \end{matrix}} |A(I − λA)^{−1}f|. (32)

From (27) and (28) we get |A(I − λA)^{−1}f| ⩽ C/\sqrt{R_{k}}, where C is a constant that does not depend on k and x ∈ [0, π]. From this and (32) we have

Af − ∑_{ν=0}^{k}( ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}} \frac{1}{λ_{n}} (f, z_{n})y_{n}) ⇉ 0

when k → ∞, i.e.

Af(x) = ∑_{ν=0}^{∞}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}} \frac{1}{λ_{n}} (f, z_{n})y_{n}).  (33)

We will put Af = g. Since f ∈ C[0, π], for the function g we know that g ∈ C^{2}[0, π], g(0) = g(π) = 0 and

−g′′ + q(x)g − ∫_{0}^{π} g dσ(x) = f.

Now, from (33) we get

g(x) = ∑_{ν=0}^{∞}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}}(f, A^{∗}z_{n})y_{n}(x)) = ∑_{ν=0}^{∞}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}}(Af, z_{n})y_{n}(x))

or

g(x) = ∑_{ν=0}^{∞}(  ∑_{R_{ν}⩽|λ_{n}|<R_{ν+1}}(g, z_{n})y_{n}(x)).  (34)

All this can be formulated as a theorem:

Theorem 2. If the conditions 1^{∘}, 2^{∘} and 3^{∘} are satisfied and if a function g ∈ C^{2}[0, π] and g(0) = g(π) = 0, then it can be expanded into the uniformly convergent series (34) with respect to a system of eigenfunctions of the boundary value problem (3).

REFERENCES

[1] M. Martinović, On a boundary value problem for a functional differential equation, Diff. Equation, XVIII, 2 (1982), 239–245

[2] M. Martinović, Zeta function and trace formulae for a boundary value problem with a functional differential equation, Diff. Equation, XVIII, 3 (1982), 537–540

[3] M. Martinović, On some problems in the spectral theory of functional differential operators of second order, PhD thesis, Belgrade University, 1986

[4] M. A. Naimark, Linear Differential Operators, Moscow, 1969

[5] M. V. Keldysh, On eigenvalues and eigenfunctions for some classes of nonselfadjoint equations, Dokl. Akad. Nauk SSSR, 77, 1, (1951), 11–14

(received 11 12 1992)