ON THE EIGENVALUE OF INFINITE MATRICES
WITH NONNEGATIVE OFF-DIAGONAL ELEMENTS
N. APREUTESEI AND V. VOLPERT
The paper is devoted to infinite-dimensional difference operators. Some spectral properties of such operators are studied. Under some assumptions on the essential spectrum,
it is shown that a real eigenvalue with a positive eigenvector is simple and that the real
parts of all other eigenvalues are less than for this one. It is a generalization of the PerronFrobenius theorem for infinite matrices.
Copyright © 2006 N. Apreutesei and V. Volpert. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Consider the Banach spaces E of infinite sequences u = (...,u−1 ,u0 ,u1 ,...) with the norm
u = sup u j (1.1)
j
and the operator L acting in E,
j
j
j
(Lu) j = a−m u j −m + · · · + a0 u j + · · · + am u j+m ,
j = 0, ±1, ±2,...,
(1.2)
j
where m is a positive integer and ak ∈ R, −m ≤ k ≤ m are given coefficients. We assume
that there exist the limits
j
a±k = lim ak ,
j →±∞
k = 0, ±1,..., ±m.
(1.3)
Consider the limiting operators L± ,
L± u j = a±−m u j −m + · · · + a±0 u j + · · · + a±m u j+m ,
j = 0, ±1, ±2,....
Hindawi Publishing Corporation
Proceedings of the Conference on Differential & Difference Equations and Applications, pp. 81–89
(1.4)
82
On the eigenvalue of infinite matrices
Let
= 0,
a±−m a±m = 0,
(1.5)
and suppose that the equations
L± u − λu = 0
(1.6)
do not have nonzero bounded solutions for any real λ ≥ 0. We will call it Condition NS(λ).
Recall that a linear operator L : E → E is normally solvable if its image Im L is closed. If L
is normally solvable with a finite-dimensional kernel and the codimension of its image is
also finite, then L is called Fredholm operator. Denoting by α(L) and β(L) the dimension
of kerL and the codimension of ImL, respectively, we can define the index κ(L) of the
operator L as κ(L) = α(L) − β(L).
In [1], the following result is proved (Theorem 4.10).
Theorem 1.1. If Condition NS(λ) is satisfied, then L is a Fredholm operator with the zero
index.
Consider the polynomials
Pλ± (σ) = a±m σ 2m + · · · + a±1 σ m+1 + a±0 − λ σ m + a±−1 σ m−1 + · · · + a±−m .
(1.7)
Lemma 2.1 in [2] for L − λI leads to the following conclusion.
Theorem 1.2. Condition NS(λ) is satisfied if and only if the polynomials Pλ± (σ) do not have
roots σ with |σ | = 1.
As a consequence, we can obtain the following corollary.
Corollary 1.3. If Condition NS(λ) is satisfied, then
a±−m + · · · + a±m < 0,
(1.8)
that is, L± q < 0, where q is a sequence with all elements equal 1.
Proof. Suppose that the assertion of the corollary does not hold. Then P0± (1) ≥ 0. On the
other hand, for λ sufficiently large, Pλ± (1) < 0. Therefore for some λ, Pλ± (1) = 0. We obtain
a contradiction with Theorem 1.2, so the conclusion is proved.
We recall that the formally adjoint operator L∗ is defined by the equality
(Lu,v) = u,L∗ v .
(1.9)
If we consider L as an infinite matrix, then L∗ is the adjoint matrix. Let α(L∗ ) be the
dimension of kerL∗ and let f = { f j }∞
j =−∞ ∈ E be fixed. The below solvability conditions
are established in [2].
N. Apreutesei and V. Volpert 83
Theorem 1.4. The equation Lu = f is solvable if and only if
∞
j =−∞
f j vlj = 0,
l = 1,...,α L∗ ,
(1.10)
∗
where vl = {vlj }∞
j =−∞ are linearly independent solutions of the equation L v = 0.
In what follows, we say that u is positive (nonnegative) if all elements of this sequence
are positive (nonnegative).
From now on, we suppose that
j
ak > 0,
k = ±1, ±2,..., ±m, j = 0, ±1, ±2,...,
(1.11)
and that there exists a positive solution w of the equation
Lu = 0.
(1.12)
This means that L has a zero eigenvalue. The goal of this paper is to show that it is
simple and all other eigenvalues lie in the left half-plane. Moreover, the adjoint operator
L∗ has a positive solution, which is unique up to a constant factor. It is a generalization
of the Perron-Frobenius theorem for infinite matrices. The method of the proof follows
the method developed for elliptic problems, in unbounded domains [3, 4]. Similarly to
elliptic problems, it is assumed that the essential spectrum lies to the left of the eigenvalue
with a positive eigenvector.
We note that the operator L can be considered as infinite-dimensional (2m + 1)-diagonal matrix with positive elements in all nonzero diagonals except for the main diagonal
where the signs of the elements are not prescribed.
In Section 2, we present some auxiliary results. The main result is proved in Section 3.
2. Auxiliary results
Suppose that conditions (1.3), (1.5), (1.11) are satisfied. In order to prove our main result,
we first present some auxiliary results. We begin with the positiveness of the solution of
equation Lu = f for f ≤ 0. We will use the notations
U− (N) = uN −m ,...,uN −1 ,
U+ (N) = uN+1 ,...,uN+m .
(2.1)
≡ 0. Then u > 0.
Lemma 2.1. Let Lu = f , where f ≤ 0, u ≥ 0, u ≡ 0, there exists i such that ui = 0, and
Proof. Suppose that u j = 0 for some j. Since u = 0 or ui−1 = 0. The equation (Lu)i = fi gives a contradiction in signs. The
either ui+1 lemma is proved.
Lemma 2.2. If the initial condition u0 of the problem
du
= Lu,
dt
u(0) = u0
is nonnegative, then the solution u(t) is also nonnegative for all t ∈ (0, ∞).
(2.2)
84
On the eigenvalue of infinite matrices
Proof. Consider the approximate problem
dui
= (Lu)i ,
dt
U− (−N) = 0,
−N ≤ i ≤ N, t ≥ 0,
U+ (N) = 0,
t ≥ 0,
(2.3)
u(0) = u0 ,
where the unknown function is u = (u−N ,u−N+1 ,...u0 ,...,uN −1 ,uN ).
Since u0 ≥ 0 and Lu has nonnegative off-diagonal coefficients, it follows that the solution uN = (uN−N ,uN−N+1 ,...uN0 ,...,uNN −1 ,uNN ) of the above problem is nonnegative.
If we compare the solution uN at the interval [−N,N] and the solution uN+1 at the
interval [−N − 1,N + 1], we find uN+1 ≥ uN . Indeed, the difference uN+1 − uN verifies a
problem similar to the above one, but with a nonnegative initial condition and with zero
boundary conditions. The solution of this problem is nonnegative, that is, uN+1 ≥ uN . So
the sequence is monotonically increasing with respect to N. The sequence is also bounded
with respect to N: ||uN (t)|| ≤ M, for all N and t ∈ [0,T], where T is any positive number,
M > 0 depends on u0 and on the coefficients aik of L, which are bounded. Being bounded
and monotone, it follows that uN is convergent as N → ∞ in C([0,T];E), say uN → u. By
the equations, we have also uN → u in C 1 ([0,T];E). Then u verifies the problem (2.2) and
u ≥ 0 (because uN ≥ 0), as claimed.
Corollary 2.3 (comparison theorem). Let u1 (t) and u2 (t) be solutions of the equation
du
= Lu,
dt
(2.4)
with the initial conditions u1 (0) and u2 (0), respectively. If u1 (0) ≤ u2 (0), then u1 (t) ≤ u2 (t)
for t ≥ 0.
Lemma 2.4. If the initial condition u0 of the problem
du
= L+ u,
dt
u(0) = u0
(2.5)
is constant (independent of j), then the solution u(t) is also constant. For any bounded initial
condition, the solution of problem (2.5) converges to the trivial solution u = 0.
The proof of this lemma follows from Corollaries 1.3 and 2.3.
Lemma 2.5. If u is a solution of the problem
Lu = f ,
j ≥ N, U− (N) ≥ 0,
(2.6)
where f ≤ 0, u j → 0 as j → ∞, and N is sufficiently large, then u j ≥ 0 for j ≥ N.
Proof. By virtue of Corollary 1.3, there exists a constant > 0 such that L+ q < −. Let us
take N large enough such that
L − L+ q j ≤ ,
2
j ≥ N.
(2.7)
N. Apreutesei and V. Volpert 85
Suppose that u j < 0 for some j > N. By the assumption u j → 0 as j → ∞, we can choose
τ > 0 such that v j = u j + τq j ≥ 0 for all j ≥ N, and there exists i > N such that vi = 0. Since
V− (N) > 0 and v j > 0 for all j sufficiently large, there exists k > N such that vk = 0 and
= 0 or vk−1 = 0 (i.e., vk+1 > 0 or vk−1 > 0).
either vk+1 We have
Lv = Lu + τL+ q + τ L − L+ q = f + τL+ q + τ L − L+ q.
(2.8)
In view of (2.7), L+ q < − and f ≤ 0, the right-hand side of this equality is less than or
equal to 0 for j ≥ N. As in the proof of Lemma 2.1, we obtain a contradiction in signs in
the equation corresponding to k. The lemma is proved.
Remark 2.6. The assertion of the lemma remains true if we replace (2.6) by
Lu ≤ αu,
j ≥ N,
U− (N) ≥ 0,
(2.9)
for some positive α. Indeed, one obtains Lv ≤ αu + τL+ q + τ(L − L+ )q instead of (2.8),
where (Lv)k > 0 and αuk + τL+ qk + τ(L − L+ )qk < αuk − τ/2 = −τ(α + /2) < 0, because
vk = 0.
3. The main result
In this section, we present the main result of this work and study some spectral properties
of infinite-dimensional matrices with nonnegative off-diagonal elements.
Theorem 3.1. Let (1.12) have a positive bounded solution w. Then, the following hold.
(i) The equation
Lu = λu
(3.1)
does not have nonzero bounded solutions for Re λ ≥ 0, λ = 0.
(ii) Each solution of (1.12) has the form u = kw, where k is a constant.
(iii) The equation
L∗ u = 0
(3.2)
has a positive solution unique up to a constant factor.
Proof. (1) In order to prove the first assertion, we analyze two cases.
= 0. Suppose by
Case 1. We consider first the case where in (3.1) λ = α + iβ, α ≥ 0, β contradiction that there exists a bounded nonzero solution u = u1 + iu2 of this equation.
Then Lu1 = αu1 − βu2 and Lu2 = βu1 + αu2 . Consider the equation
dv
= Lv − αv,
dt
v(0) = u1 .
(3.3)
Its solution is
v(t) = u1 cosβt − u2 sinβt.
(3.4)
86
On the eigenvalue of infinite matrices
For the sequence u = {u j } = {u1j + iu2j }, we denote u = {|u j |}. Let us take the value of
N as in Lemma 2.5 and choose τ > 0 such that
j ≤ N,
u j ≤ τw j ,
(3.5)
where at least for one j0 with | j0 | ≤ N, we have the equality
u j0 = τw j0 .
(3.6)
For j ≥ N, consider the problem
dy
= Ly − αy,
dt
yN −k (t) = uN −k , k = 1,...,m,
(3.7)
y∞ (t) = 0,
y(0) = u,
(3.8)
and the corresponding stationary problem
L ȳ − α ȳ = 0,
ȳN −k = uN −k ,
k = 1,...,m, ȳ∞ = 0.
(3.9)
The operator corresponding to problem (3.9) satisfies the Fredholm property (see [2]).
The corresponding homogeneous problem has only the zero solution. (For L+ instead of
L, it follows from the explicit form of the solution, see [1]; for N big enough, L and L+
are close.) Therefore, problem (3.9) is uniquely solvable.
We show that the solution y(t) of problem (3.7)-(3.8) converges to ȳ as t → ∞. For this,
we consider the solution y ∗ (t) of problem (3.7) with the initial condition y ∗ (0) = ρq,
where ρ is such that
ρq j ≥ u j ,
j ≥ N.
(3.10)
By Corollary 1.3, we have L± q < 0. Since L+ is close to L for j ≥ N, with N large enough,
it follows that (Lq) j < 0, j ≥ N. Then y ∗ (t) monotonically decreases in t for each j ≥ N
fixed. From the positiveness and the decreasing monotonicity of y ∗ , we deduce that y ∗ (t)
converges as t → ∞ to some x = limt→∞ y ∗ (t) ≥ 0. It satisfies the equation Lx − αx = 0.
Taking the limit also in the boundary conditions, one obtains that xN+k = uN+k , for k =
1,...,m and x∞ = 0, so x is a solution of problem (3.9). By the uniqueness, we get x = y,
that is, there exists the limit limt→∞ y ∗ (t) = ȳ.
On the other hand, let y∗ be the solution of (3.7) with the initial condition y∗ (0) = 0.
It can be shown that y∗ increases in time and it has an upper bound. As above, we can
deduce that y∗ converges to ȳ. Therefore,
lim y∗ (t) = lim y ∗ (t) = ȳ.
t →∞
t →∞
(3.11)
By virtue of the comparison theorem applicable in this case (because 0 ≤ u j ≤ ρq j , j ≥
N), we have
y∗ (t) ≤ y(t) ≤ y ∗ (t),
j ≥ N.
(3.12)
N. Apreutesei and V. Volpert 87
Hence
lim y j (t) = ȳ j ,
j ≥ N.
t →∞
(3.13)
One can easily verify that
v j (t) ≤ u j
∀ j ∈ Z.
(3.14)
Then it follows from the comparison theorem that
v j (t) ≤ y j (t),
j ≥ N, t ≥ 0.
(3.15)
From this, we have
v j (t) = v j t +
2πn
2πn
≤ yj t +
.
β
β
(3.16)
j ≥ N, t ≥ 0.
(3.17)
Passing to the limit as n → ∞, we obtain
v j (t) ≤ ȳ j ,
Observe that L(τw − y) ≤ α(τw − y), j ≥ N, and τwN − y N ≥ 0. We can apply
Remark 2.6 to τw − ȳ. Therefore,
ȳ j ≤ τw j ,
j ≥ N.
(3.18)
Hence,
v j (t) ≤ τw j
(3.19)
for j ≥ N, t ≥ 0. The similar estimate can be obtained for j ≤ −N. Together with (3.5),
these prove (3.19) for all j ∈ Z.
The sequence z(t) = τw − v(t) is a solution of the equation
dz
= Lz − αz + ατw.
dt
(3.20)
Since z(t) ≥ 0 (via (3.19) for all j ∈ Z), z is not identically zero, and is periodic in t, it
follows that z j (t) > 0 for all j and t ≥ 0. Indeed, suppose that for some t = t1 and j = j1 ,
z j1 (t1 ) = 0. Consider first the case where α > 0. Since (dz j1 /dt)(t1 ) ≤ 0 and w j1 > 0, we
obtain a contradiction in signs in the equation for z j1 . If α = 0, then the equation becomes
dz
= Lz.
dt
(3.21)
Assuming that z(t) is not strictly positive, we easily obtain that it is identically zero for
all j. We have (dz j1 /dt)(t1 ) ≤ 0 and (Lz) j1 (t1 ) ≥ 0. Then (Lz) j1 (t1 ) = 0, so all z j (t1 ) = 0.
Since z j1 verifies dz j1 /dt = (Lz) j1 , z j1 (t1 ) = 0, by the uniqueness we find z j1 (t) = 0, t ≥ t1 .
Combining this with z j (t1 ) = 0, (∀) j ∈ Z, we get z j (t) = 0, (∀) j ∈ Z, (∀)t ∈ (0, ∞).
88
On the eigenvalue of infinite matrices
Thus in both cases, z j (t) is positive for all j and t. We take t ≥ 0 such that
e−iβt = u j0
,
u j0 (3.22)
with j0 from (3.6), that is, cosβt = u1j0 / |u j0 | and sinβt = −u2j0 / |u j0 |. Then, v j0 (t) =
u1j0 cosβt − u2j0 sin βt = |u j0 |, hence with the aid of (3.6) we obtain the contradiction
z j0 (t) = τw j0 − u j0 = 0.
(3.23)
The first assertion of the theorem is proved for nonreal λ.
Case 2. Assume now that λ ≥ 0 is real and that u is a nonzero bounded solution of (3.1).
We suppose that at least one of the elements of the sequence {u j } is negative. Otherwise,
we could change the sign of u. We consider the sequence v = u + τw, where τ > 0 is chosen
such that v ≥ 0 for | j | ≤ N, but v j0 = 0 for some j0 , | j0 | ≤ N. We have
Lv = λv − λτw,
(3.24)
and therefore v j ≥ 0 for all j by virtue of Lemma 2.5. Indeed, for | j | ≤ N, the inequality
holds because of the way we have chosen τ. For j ≥ N, one applies Lemma 2.5 for (3.24)
written in the form (L − λI)v = −λτw, j ≥ N, with vN ≥ 0. If j ≤ −N, the reasoning is
similar.
If λ > 0, then the equation for v j0 leads to a contradiction in signs. Thus, (3.1) cannot
have different-from-zero solutions for real positive λ.
(2) If λ = 0, then we define v = u + τw as in Case 2 above. Here u is the solution of (3.1)
with λ = 0, that is, Lu = 0. Using the above reasoning for λ ≥ 0, we have v j ≥ 0, (∀) j ∈ Z,
but it is not strictly positive (at least v j0 = 0). In addition, v satisfies the equation Lv = 0.
It follows from Lemma 2.1 that v ≡ 0. This implies that u j = −τw j , (∀) j ∈ Z.
(3) The limiting operators L± are operators with constant coefficients. The corresponding matrices are (2m + 1)-diagonal matrices with constant elements along each
diagonal. The matrices associated to the limiting operators L+∗ , L−∗ of L∗ are the transposed matrices, which are composed by the same diagonals reflected symmetrically with
respect to the main diagonal. Therefore, the polynomials (Pλ∗ )± (σ) for the operator L∗
will be the same as for the operator L. By virtue of Theorem 1.2, the operator L∗ satisfies
the Fredholm property and it has the zero index.
We note first of all that (3.2) has a nonzero bounded solution v. Indeed, if such solution
does not exist, then by virtue of the solvability conditions, the equation
Lu = f
(3.25)
is solvable for any f . This implies that ImL = E, and hence codim(ImL) = 0. Since the
index of L is zero, it follows that dim(kerL) = 0. But by part two of the theorem, we
get dim(kerL) = 1. This contradiction shows that a nonzero bounded solution v of (3.2)
exists and it is exponentially decreasing at infinity (see [2, Theorem 3.2]).
N. Apreutesei and V. Volpert 89
We recall next (see Theorem 1.4) that (3.25) is solvable if and only if
( f ,v) = 0.
(3.26)
Case 1. If v ≥ 0, then from Lemma 2.1 for equation L∗ v = 0, it follows that v is strictly
positive, as claimed.
Case 2. If we assume that a nonnegative solution of (3.2) does not exist, then it has an
alternating sign. Then we can find a bounded sequence f < 0 such that (3.26) is satisfied.
Let u be the corresponding solution of (3.25). There exists a τ (not necessarily positive), such that u
= u + τw ≥ 0 for | j | ≤ N, but not strictly positive. Since Lu
= f and
f < 0, u
N ≥ 0, and u
j → 0 as j → ∞, by virtue of Lemma 2.5, one finds u
≥ 0 for all j. But
for those j where u
vanish, this leads to a contradiction in signs in the equation. Therefore
u
> 0. The theorem is proved.
References
[1] N. Apreutesei and V. Volpert, Some properties of infinite dimensional discrete operators, Topological Methods in Nonlinear Analysis 24 (2004), no. 1, 159–181.
, Solvability conditions for some difference operators, Advances in Difference Equations
[2]
2005 (2005), no. 1, 1–13.
[3] A. I. Volpert and V. Volpert, Spectrum of elliptic operators and stability of travelling waves, Asymptotic Analysis 23 (2000), no. 2, 111–134.
[4] A. I. Volpert, V. Volpert, and V. Volpert, Traveling Wave Solutions of Parabolic Systems, Translations of Mathematical Monographs, vol. 140, American Mathematical Society, Rhode Island,
1994.
N. Apreutesei: Department of Mathematics, “Gh. Asachi” Technical University of Iaşi,
700506 Iaşi, Romania
E-mail address: [email protected]
V. Volpert: Camille Jordan Institute of Mathematics, UMR 5208 CNRS, University Lyon 1,
69622 Villeurbanne, France
E-mail address: [email protected]