ON THE INVESTIGATION OF A LINEAR EIGENVALUE PROBLEM FOR MATRICES

Galip Oturanç

Abstract. This paper investigates the properties of eigenvalues and eigenvectors of the problem Jy +λRy = (1/λ)Cy, where J is a “tridiagonal” real symmetric matrix and R and C are positive diagonal matrices. The results obtained are used to solve the corresponding system of differential equations with boundary and initial conditions.

1. Introduction. Let us consider the system of linear differential equations

a_{n−1} \frac{du_{n−1}(t)}{dt} + b_{n} \frac{du_{n}(t)}{dt} + r_{n} \frac{d^{2}u_{n}(t)}{dt^{2}} + a_{n} \frac{du_{n+1}(t)}{dt} = c_{n}u_{n}(t),  (1)

for n = 0, 1, 2, …, N − 1 with boundary conditions

u_{−1}(t) = 0, u_{N}(t) + hu_{N−1}(t) = 0, t ≥ 0  (2)

and initial conditions

u_{n}(0) = f_{n}, \frac{du_{n}(0)}{dt} = g_{n}, n = 0, 1, …, N − 1  (3)

where {u_{n}(t)}_{n=−1}^{N} is a desired solution; f_{n}, g_{n} (n = 0, 1, …, N − 1) are given complex numbers; the coefficients a_{n}, b_{n}, r_{n}, and c_{n} of the equation (1) and the number h in boundary condition (2) real besides

a_{n} ≠ 0, r_{n} > 0, c_{n} > 0.  (4)

We seek the solution of equation (1), which has the form

u_{n}(t) = e^{λt}y_{n}, n = −1, 0, 1, …, N  (5)

where λ is a complex constant, and the y_{n}’s are complex numbers depending only upon λ and not upon t. We desire {y_{n}}_{−1}^{N} to be nontrivial, that is not equal to 0, the zero vector. Substituting (5) into (1) and (2) we obtain λ ≠ 0 and

a_{n−1}y_{n−1} + (b_{n}y_{n} + r_{n}λ)y_{n} + a_{n}y_{n+1} = \frac{1}{λ}c_{n}y_{n}, n = 0, 1, 2, …, N − 1 (6)
y_{−1} = 0, y_{N} + hy_{N−1} = 0  (7)

\underline{Definition}. The complex number λ is said to be an eigenvalue of boundary problem (6), (7) if for this value λ there exists a nonzero vector {y_{n}}_{−1}^{N} satisfying equation (6) and boundary conditions (7). Further, the vector y = {y_{n}}_{0}^{N−1} is called an eigenvector of problem (6), (7) corresponding to eigenvalue λ.

Thus, the functions in (5) are a nontrivial solution of problem (1), (2) if and only if λ is an eigenvalue and y = {y_{n}}_{0}^{N−1} is the corresponding eigenvector of problem (6), (7).

Denote all the eigenvalues of problem (6), (7) by λ_{1}, …, λ_{m} and the corresponding eigenvectors by y^{(1)} = {y_{n}^{(1)}}_{0}^{N−1}, …, y^{(m)} = {y_{n}^{(m)}}_{0}^{N−1}. Then by the linearity of problem (1), (2) the functions

u_{n}(t) = ∑_{j=1}^{m} α_{j}e^{λ_{j}t}y_{n}^{(j)}, n = −1, 0, 1, …, N  (8)

will form a solution of the problem (1), (2) where α_{j} (j = 1, 2, …, m) are arbitrary constants (independent of t and n). Now we must try to choose the constants α_{j} (j = 1, 2, …, m) so that (8) will also satisfy the initial conditions

∑ α_{j}y_{n}^{(j)} = f_{n}, ∑ α_{j}λ_{j}y_{n}^{(j)} = g_{n}, n = 0, 1, … N − 1.  (9)

\underline{Definition}. If for the arbitrary vectors f = \{f_{n}\}_{0}^{N−1} and g−\{g_{n}\}_{0}^{N−1} belonging to ℂ^{N} the unique expansions (9) hold with the same coefficients α_{j} (j = 1, 2, …, m) in both expansions, then we will say that the eigenvectors y^{(1)}, …, y^{(m)} of problem (6), (7) form twofold basis in ℂ^{N}.

We will show that the boundary problem (6), (7) has precisely 2N simple real eigenvalues λ_{1}, …, λ_{2N}, half are negative and half are positive. The corresponding eigenvectors make up a twofold basis in ℂ^{N}, and we will give formulas for the coefficients α_{j} in (9). In addition, we will show that the eigenvectors corresponding only to the negative (positive) eigenvalues form an ordinary basis for ℂ^{N}.

2. Eigenvalue Problem. We consider the boundary value problem (6), (7) under condition (4). If {y_{n}}_{−1}^{N} is a solution of problem (6), (7) then

(b_{0} + λr_{0})y_{0} + a_{0}y_{1} = (1/λ)c_{0}y_{0}
a_{n−1}y_{n−1} + (b_{n} + λr_{n})y_{n} + a_{n}y_{n} + a_{n}y_{n+1} = (1/λ)c_{n}y_{n}
a_{n−2}y_{n−2} + (b_{N−1} − ha_{N−1} + λr_{N−1})y_{N−1} = (1/λ)c_{N−1}y_{N−1}  (10)

for n = 1, 2, …, N. Consequently, finding a nontrivial solution {y_{n}}_{−1}^{N} of problem (6), (7) is equivalent to finding a nontrivial solution {y_{n}}_{0}^{N−1} of system (10).

Setting

y = ( \begin{matrix}y_{0} \\ y_{1} \\ ⋯ \\ y_{N−1} \end{matrix} ) , J = ( \begin{matrix}b_{0} & a_{0} & 0 & 0 & ⋯ & 0 & 0 & 0 \\ a_{0} & b_{1} & a_{1} & 0 & ⋯ & 0 & 0 & 0 \\ 0 & a_{1} & b_{2} & a_{2} & ⋯ & 0 & 0 & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 & 0 & ⋯ & a_{N−3} & b_{N−2} & a_{N−2} \\ 0 & 0 & 0 & 0 & ⋯ & 0 & a_{N−2} & b_{N−1} − ha_{N−1} \end{matrix} ) ,

(11)

R = ( \begin{matrix}r_{0} & 0 & ⋯ & 0 \\ 0 & r_{1} & ⋯ & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & ⋯ & r_{N−1} \end{matrix} ) , C = ( \begin{matrix}c_{0} & 0 & 0 & ⋯ & 0 \\ 0 & c_{1} & 0 & ⋯ & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ 0 & 0 & 0 & ⋯ & c_{N−1} \end{matrix} ) ,

we can write system (10) in the form

Jy + λRy = (1/λ)Cy.  (12)

We will investigate equation (12) in the space

ℂ^{N} = {y = {y_{n}}_{0}^{N−1} : y_{n} ∈ ℂ, n = 0, 1, …, N − 1}

with the inner product

(y, z) = ∑_{n=0}^{N−1} y_{n}\overline{z}_{n}  (13)

where the bar over a number denotes complex conjugation.

The matrices J, R, and C defined by (11) are selfadjoint, that is each of them satisfies the relation

(Ay, z) = (y, Az), for all y, z ∈ ℂ^{N}.  (14)

At first assuming the existence of eigenvalues and eigenvectors of problem (12) we establish some of their properties.

\underline{Lemma 1}. The eigenvalues of equation (12) are real.

\underline{Proof}. Let the complex number λ be an eigenvalue of equation (12) and

y = {y_{n}}_{0}^{N−1} ≠ 0

be a corresponding eigenvector. By forming the inner product of both sides of equation (12) by the vector y, we get

λ^{2}(Ry, y) + λ(Jy, y) − (Cy, y) = 0.

This equality is a quadratic equation with respect to λ, the discriminant of which is

(Jy, y)^{2} + 4(Ry, y)(Cy, y) > 0

since the number (Jy, y) is real, in view of (14), and the numbers (Cy, y) and (Ry, y) are positive in view of (4). Consequently, the number λ will be real.

\underline{Lemma 2}. The eigenvectors y and z in equation (12) corresponding to distinct eigenvalues λ and μ, respectively satisfy the “orthogonality” relation

λμ(Ry, z) + (Cy, z) = 0.  (15)

\underline{Proof}. Multiplying in the sense of inner product the first of the equalties

Jy + λRy = (1/λ)Cy, Jz + μRz = (1/μ)Cz

from the right by z and the second one from the left by y, and remembering that λ and μ are real in view of Lemma 1, we get

\begin{matrix}(Jy, z) + λ(Ry, z) = \frac{1}{λ}(Cy, z) \\ (y, Jz) + μ(y, Rz) = \frac{1}{μ}(y, Cz). \end{matrix}

Subtracting the second result from the first one, and using property (14) of the matrices J, R and C, we have

(λ − μ)(Ry, z) = (1/λ − 1/μ)(Cy, z).

Hence, the Lemma follows from the condition λ ≠ μ.

Now we investigate the existence of other properties of eigenvalues and eigenvectors of equation (12), which is equivalent to problem (6), (7). For this purpose we define the solution {φ_{n}(λ)}_{−1}^{N} of equation (6) that satisfies the initial conditions

φ_{−1}(λ) = 0, φ_{0}(λ) = 1.  (16)

Using (16), we can recursively find φ_{n}(λ), n = 1, 2, 3, …, N and will have the form

φ_{n}(λ) = (1/λ^{n})P_{2n}(λ), n = 0, 1, …, N,  (17)

where P_{2n}(λ) is a polynomial in λ of degree 2n and

P_{2n}(λ) = \frac{r_{0}r_{1} ⋯ r_{n−1}}{a_{0}a_{1} ⋯ a_{n−1}} λ^{2n} + ⋯ + (−1)^{n} \frac{c_{0}c_{1} ⋯ c_{n−1}}{a_{0}a_{1} ⋯ a_{n−1}} .  (18)

It is easy to see that every solution {y_{n}(λ)}_{−1}^{N} of equation (10) satisfying the initial condition y_{−1} = 0 is equal to {φ_{n}(λ)}_{−1}^{N} up to a constant factor

φ_{n}(λ) = αφ_{n}(λ), n = −1, 0, 1, …, N.  (19)

Consequently, we have the following lemma.

\underline{Lemma 3}. To each eigenvalue λ_{0} of problem (6), (7) corresponds up to a constant factor a single eigenvector, which can be taken to be the vector {φ_{n}(λ_{0})}_{0}^{N−1}.

Set

χ(λ) = φ_{N}(λ) + hφ_{N−1}(λ).  (20)

The function χ(λ) is called the characteristic function of problem (6), (7).

\underline{Lemma 4}. The eigenvalues of problem (6), (7) coincide with the roots of the function χ(λ).

\underline{Proof}. Let λ_{0} be an eigenvalue of problem (6), (7) and \{y_{n}(λ_{0})\}_{−1}^{N} be a nontrivial solution of (6), (7) with λ = λ_{0}. Then (17) holds with α ≠ 0 and from boundary condition y_{N}(λ_{0}) + hy_{N−1}(λ_{0}) = 0 we get χ(λ_{0}) = 0.

Conversely, if χ(λ_{0}) = 0, then {φ_{n}(λ_{0})}_{−1}^{N} will be a nontrivial (remembering that by (16) we have φ_{0}(λ) ≠ 0) solution of boundary problem (6), (7).

\underline{Lemma 5}. The roots of the function χ(λ) are simple.

\underline{Proof}. Differentiating the equation

a_{k−1}φ_{k−1}(λ) + (b_{k} + λr_{k})φ_{k}(λ) + a_{k}φ_{k+1}(λ) = \frac{1}{λ}c_{k}φ_{k}(λ)

with respect to λ, we get

a_{k−1}φ̇_{k−1}(λ) + (b_{k} + λr_{k})φ̇_{k}(λ) + a_{k}φ̇_{k+1}(λ) − \frac{1}{λ}c_{k}φ_{k}(λ) = −r_{k}φ_{k}(λ) − \frac{1}{λ^{2}} c_{k}φ_{k}(λ)

where the dot over the function indicates the derivative with respect to λ. Multiplying the first equation by φ̇_{k}(λ) and the second one by φ_{k}(λ), and subtracting the left and right members of the resulting equations, we get

a_{k−1}[φ_{k−1}(λ)φ̇_{k}(λ) − φ̇_{k−1}(λ)φ_{k}(λ)] − a_{k}[φ_{k+1}(λ)φ̇_{k}(λ) − φ̇_{k+1}(λ)φ_{k}(λ)]

= (r_{k} + \frac{1}{λ^{2}} c_{k})φ_{k}^{2}.  (21)

Summing up the last equation for the values k = 0, 1, …, n (n ≤ N − 1) and using the initial conditions (16), we get

a_{n}[φ_{n+1}(λ)φ̇_{n}(λ) − φ̇_{n+1}(λ)φ_{n}(λ)] = ∑_{k=0}^{n}(r_{k} + \frac{1}{λ_{0}^{2}} c_{k})φ_{k}^{2}(λ_{0}).  (22)

Let χ(λ_{0}) = 0. In particular, setting (22), n = N − 1 and λ = λ_{0}, and using the equality φ_{N}(λ_{0}) = −hφ_{N−1}(λ_{0}) which follows from the boundary condition (7), we have

a_{N−1}χ̇(λ_{0})φ_{N−1}(λ_{0}) = ∑_{k=0}^{N−1}(r_{k} + \frac{1}{λ_{0}^{2}} c_{k})φ_{k}^{2}(λ_{0}).  (23)

The right–hand side of (23) is not zero since r_{k} > 0, c_{k} > 0, λ_{0} is in view of Lemma 1, φ_{k}(λ_{0}) (k = 0, 1, …, N − 1) are real and not all zero. Besides, in the left–hand side of (23), the value φ_{N−1}(λ_{0}) can not be equal to zero. Indeed, if φ_{N−1}(λ_{0}) = 0, and hence, by the uniqueness property of the solution of equality (6), we get φ_{n}(0) = 0, for all n, which is a contradiction. Thus, from formula (23) it follows that χ(λ_{0}) ≠ 0, that is the root λ_{0} of the function χ(λ) is simple.

\underline{Lemma 6}. The function χ(λ) has precisely 2N distinct roots.

\underline{Proof}. Since φ_{n}(λ) for each n is a polynomial in λ of degree 2n, the function χ(λ) = φ_{N}(λ) + hφ_{N−1}(λ) will be a polynomial in λ of degree 2N. Therefore, the function χ(λ) has 2N roots, which are distinct by virtue of Lemma 5.

\underline{Lemma 7}. Half of the roots of the function χ(λ) are negative and the other half are positive.

\underline{Proof}. The eigenvalue (12) is equivalent to the eigenvalue problem

(λ^{2}R + λJ − C)y = 0.  (24)

(Note that the value λ = 0 is not an eigenvalue of (24)). Therefore, the roots of the function χ(λ) coincide with the roots of the polynomial det(λ^{2}R + λJ − C).

Now we consider the auxiliary eigenvalue problem

(λ^{2}C − λϵR − C(ϵ))y = 0  (25)

depending on parameter ϵ ∈ [0, 1], where the matrix C(ϵ) is obtained from the matrix C by means of multiplying all its nondiagonal elements by ϵ. It is obvious that the analog of condition (4) is fulfilled for all ϵ ∈ (0, 1].

The eigenvalues of equation (25) are nonzero for all ϵ ∈ [0, 1] and coincide with the roots of the polynomial

det(λ^{2}R + λϵJ − C(ϵ)).  (26)

For each ϵ ∈ (0, 1] the roots of the polynomial (26) are distinct by virtue of Lemma 5 being applicable to equation (25). Denote them by

λ_{1}(ϵ) < λ_{2}(ϵ) < ⋯ < λ_{2N}(ϵ).

Since λ_{j}(ϵ) (j = 1, 2, …, 2N) are the eigenvalues of a matrix of order 2N being continuous in ϵ ∈ [0, 1] (see section 3) they will be continuous functions of ϵ [3]. Note that at the point ϵ = 0 we do not state that λ_{1}(ϵ), λ_{2}(ϵ), …, λ_{2N}(ϵ) are distinct.

Now we show that for all values of ϵ ∈ (0, 1] half of the λ_{j}(ϵ) (j = 1, 2, …, 2N) are negative and the other half positive:

λ_{j}(ϵ) < 0 (j = 1, …, N), λ_{j}(ϵ) > 0 (j = N + 1, …, 2N).

Hence, in particular, for ϵ = 1, the statement of the lemma will follow. Assume the contrary. For some value of ϵ ∈ (0, 1] let

λ_{j}(ϵ) < 0 (j = 1, …, K), λ_{j}(ϵ) > 0 (j = K + 1, …, 2N)  (27)

where 0 ≤ K ≤ 2N and K ≠ N (for K = 0 all the eigenvalues λ_{j}(ϵ) are understood to be positive, and for K = 2N negative). Since λ_{j}(ϵ) (j = 1, …, 2N) are different from zero and are distinct and continuous functions for all values of ϵ ∈ (0, 1], taking inequalities (27) to the limit as ϵ → 0, we get

λ_{j}(0) ≤ 0 (j = 1, …, K), λ_{j}(0) ≥ 0 (j = K + 1, …, 2N).

But this is a contradiction, since for ϵ = 0 the roots of the polynomial (26) are the numbers

±\sqrt{\frac{c_{j}}{r_{j}}} (j = 0, 1, …, N − 2)

half of which are negative and the other half positive. Thus, the lemma is proved.

Now we can summarize the results obtained above in the following theorem.

\underline{Theorem 1}. The boundary value problem (6), (7) has precisely 2N real distinct eigenvalues λ_{j} (j = 1, 2, …, 2N). These eigenvalues are different from zero, half of them are negative and the other half positive. To each eigenvalue λ_{j} corresponds up to constant factor a single eigenvector which can be the vector φ^{(j)} = {φ_{n}(λ_{j})}_{n=0}^{N−1} where {φ_{n}(λ)}_{−1}^{N} is a solution of equation (6) satisfying the initial condition (16).

\underline{Theorem 2}. The eigenvector φ^{(j)} = \{φ_{n}(λ_{j})\}_{n=0}^{N−1}, j = 1, …, 2N of problem (6), (7) form a twofold basis in ℂ^{N}, that is for arbitrary vectors f = {f_{n}}_{0}^{N−1} and g = {g_{n}}_{0}^{N−1} belonging to ℂ^{N} the expansion formula hold

f_{n} = ∑_{j=1}^{2N} α_{j}φ_{n}(λ_{j}), g_{n} = ∑_{j=1}^{2N} α_{j}λ_{j}φ_{N}(λ_{j}), n = 0, 1, …, N − 1  (28)

where the coefficients α_{j} of expansion are defined by the formula

α_{j} = (1/ρ_{j}) ∑_{k=0}^{N−1}(c_{k}f_{k} + λ_{j}r_{k}g_{k})φ_{k}(λ_{j}), j = 1, 2, …, 2N  (29)

in which

ρ_{j} = ∑_{k=0}^{N−1}(c_{k} + λ_{j}^{2}r_{k})φ_{k}^{2}(λ_{j}), j = 1, 2, …, 2N.  (30)

\underline{Proof}. Consider the space ℂ^{N} × ℂ^{N} of vectors denoted by [y, z], where y, z ∈ ℂ^{N}. Define in this space the inner product by the formula

⟨[y, z], [u, v]⟩ = (Ry, u) + (Cz, v)

where (·, ·) in the right–hand side denotes the inner product in ℂ^{N} defined by the formula (13). In view of Lemma 2 the vectors

Φ_{j} = [φ^{(j)}, λφ^{(j)}], j = 1, 2, …, 2N

are orthogonal:

⟨Φ_{j}, Φ_{i}⟩ = 0, j ≠ i.

Consequently, the vectors ϕ_{1}, …, ϕ_{2N} are linearly independent in space ℂ^{N} × ℂ^{N}. Since the number of them is equal to 2N and dim(ℂ^{N} × ℂ^{N}) = 2N, they form a basis in this space. Therefore, for the arbitrary vector [f, g] belonging to ℂ^{N} × ℂ^{N} we have the unique expansion

[f, g] = ∑_{j=1}^{2N} α_{j}ϕ_{j},

and

\begin{matrix}α_{j} = (1/ρ_{j})⟨[f, g], ϕ_{j}⟩ = (1/ρ_{j})\{(C_{f}, φ^{(j)}) + λ_{j}(R_{g}, φ^{(j)})\} \\ ρ_{j} = ⟨ϕ_{j}, ϕ_{j}⟩ = (Cφ^{(j)}, φ^{(j)}) + λ_{j}^{2}(Rφ^{(j)}, φ^{(j)}). \end{matrix}

Hence, we have the following theorem.

\underline{Theorem 3}. All the eigenvectors of the problem (6), (7) form an ordinary basis in the space ℂ^{N}.

\underline{Proof}. We may assume that

λ_{1} < ⋯ < λ_{N} < 0 < λ_{N+1} < ⋯ < λ_{2N}.

Let x = {x_{n}}_{0}^{N−1} ∈ ℂ^{N} and

(x, φ^{(j)}) = 0, j = 1, …, N.  (31)

We must show that x = 0. Applying Theorem 2 to the vectors f = C^{−1}x and g = 0 we have

C^{−1}x = ∑_{j=1}^{2N} α_{j}φ^{(j)}, 0 = ∑_{j=1}^{2N} α_{j}λ_{j}φ^{(j)}  (32)

where

α_{j} = (1/ρ_{j})(x, φ^{(j)}), j = 1, 2, …, 2N.  (33)

From (33), in view of (31), we have α_{j} = 0, j = 1, …, N and therefore, (32) takes the form:

C^{−1}x = ∑_{j=N+1}^{2N} α_{j}φ^{(j)}, 0 = ∑_{j=N+1}^{2N} α_{j}λ_{j}φ^{(j)}.

Multiplying the last equalities by x in the sense of the inner product in ℂ^{N}, we get

(C^{−1}x, x) = ∑_{j=N+1}^{2N} α_{j}(φ^{(j)}, x) = ∑_{j=N+1}^{2N} ρ_{j}|α_{j}|^{2},  (34)

0 = ∑_{j=N+1}^{2N} α_{j}λ_{j}(φ^{(j)}, x) = ∑_{j=N+1}^{2N} λ_{j}ρ_{j}|α_{j}|^{2}.  (35)

Since λ_{j} > 0, ρ_{j} > 0, j = N + 1, …, 2N, from (31) it follows that α_{j} = 0, j = N + 1, …, 2N. Consequently, from (30) we have (R^{−1}x, x) = 0.

Hence, x = 0 since

(R^{−1}x, x) = ∑_{0}^{N−1}(1/r_{n})|x_{n}|^{2}.

3. Application. We now give an application of the results that were obtained above.

\underline{Theorem 4}. Problem (1), (2), (3) has a unique solution \{u_{n}(t)\}_{−1}^{N} which can be represented in the form

u_{n}(t) = ∑_{1}^{2N} α_{j}e^{λ_{j}t}φ_{n}(λ_{j}), n = −1, 0, 1, …, N  (36)

where λ_{1}, …, λ_{2N} are the eigenvalues of problem (6), (7), and φ^{(1)} = {φ_{n}(λ_{1})}_{0}^{N−1}, …, φ^{(2N)} = {φ_{N}(λ_{2N})}_{0}^{N−1} are the corresponding eigenvectors. Furthermore, the coefficients α_{1}, …, α_{2N} are defined by the formula (29), (30).

\underline{Proof}. From the explanation given in the Introduction from Theorem 2 it follows that formula (36) gives a solution of problem (1), (2), (3). We now prove the uniqueness of this solution. To do this, write down problem (1), (2), (3) in the form

J(du(t)/dt) + r(d^{2}u(t)/dt^{2}) = Cu(t), 0 ≤ t < ∞  (37)
u(0) = f, du(0)/dt = g  (38)

where J, R, and C are matrices defined by (11), and u(t) = {u_{n}(t)}_{0}^{N−1}, f = {f_{n}}_{0}^{N−1}, g = {g_{n}}_{0}^{N−1}. Furthermore, we can rewrite problem (33), (34) in the following equivalent form

\begin{matrix}du(t)/dt = v(t), dv(t)/dt = R^{−1}Cu(t) − R^{−1}Jv(t) \\ u(0) = 0, v(0) = g \end{matrix}

or

\frac{d}{dt} ( \begin{matrix}u(t) \\ v(t) \end{matrix} ) = ( \begin{matrix}0 & I \\ R^{−1}C & −R^{−1}J \end{matrix} ) ( \begin{matrix}u(t) \\ v(t) \end{matrix} ) , ( \begin{matrix}u(0) \\ v(0) \end{matrix} ) = ( \begin{matrix}f \\ g \end{matrix} ) .

The uniqueness of the solution of this last problem is well known [2].

\underline{Theorem 5}. For arbitrary vector f = \{f_{n}\}_{0}^{N−1} ∈ ℂ^{N} the problem (1), (2) has a unique solution {u_{n}(t)}_{−1}^{N} that satisfies the conditions

u(n) = f_{n}, lim u_{n}(t) = 0 (t → ∞), n = 0, 1, …, N − 1.  (39)

\underline{Proof}. In view of Theorem 3 the functions

u_{n}(t) = ∑_{j=1}^{N} β_{j}e^{λ_{j}t}φ_{n}(λ_{j}), n = −1, 0, 1, …, N

form a solution of problem (1), (2), (39), where λ_{1}, …, λ_{N} are the negative eigenvalues of problem (6), (7), and β_{1}, …, β_{N} are defined by the help of the expansion

f = ∑_{j=1}^{N} β_{j}φ^{(j)}.

For the proof of uniqueness we note that in view of Theorem 4 the general solution of problem (1), (2) has the representation

u_{n}(t) = ∑_{j=1}^{2N} α_{j}e^{λ_{j}t}φ_{n}(λ_{j}), n = −1, 0, 1, …, N.

From the last equality and second condition of (39) it follows that α_{N+1} = ⋯ = α_{2N} = 0. Furthermore, setting t = 0 we get α_{j} = β_{j}, j = 1, …, N.

We remark that from the proof of Theorem 5, it follows that for the solution of problem (1), (2), (39) we have

u_{n}(t) = O(e^{−δt}), n = −1, 0, 1, …, N,

as t → ∞, where δ > 0 is the modulus of the greatest negative eigenvalue of boundary problem (6), (7).

\underline{References}

1. F. V. Atkinson, Discrete and Continuous Boundary Problems, Academic Press, New York, 1964.

2. R. E. Bellman, Stability Theory of Differential Equations, McGraw–Hill, New York, 1953.

3. T. Kato, Perturbation Theory for Linear Operators, Springer–Verlag, New York, 1966.

4. G. SH Guesinov and G. Oturanç, “On the Theory of a Certain Class of Quadratic Pencils of Matrices and its Applications,” Turkish Journal of Mathematics, 21 (1997), 409–422.

Galip Oturanç Department of Mathematics Selçuk University Kampüs – Konya TURKEY email: goturanc@selcuk.edu.tr