ON THE EIGENVALUE OF INFINITE MATRICES WITH NONNEGATIVE OFF-DIAGONAL ELEMENTS

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1 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 Perron- Frobenius 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,u 0,u 1,...) with the norm and the operator L acting in E, u = sup u j (1.1) j (Lu) j = a j mu j m + + a j 0u j + + a j mu j+m, j = 0,±1,±2,..., (1.2) where m is a positive integer and a j k R, m k m are given coefficients. We assume that there exist the limits Consider the limiting operators L ±, a ± k = lim aj k, k = 0,±1,...,±m. (1.3) j ± ( L ± u ) j = a± mu j m + + a ± 0 u j + + a ± mu j+m, j = 0,±1,±2,... (1.4) Hindawi Publishing Corporation Proceedingsof the Conference on Differential & Difference Equationsand Applications,pp.81 89

2 82 On the eigenvalue of infinite matrices Let and suppose that the equations a ± m 0, a ± m 0, (1.5) 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 ImL 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) andβ(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 P ± λ (σ) = a± mσ 2m + + a ± 1 σ m+1 + ( a ± 0 λ) σ m + a ± 1σ m 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 P λ ± (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,whereq isasequencewithallelementsequal1. Proof. Suppose that the assertion of the corollary does not hold. Then P 0 ± (1) 0. On the other hand, for λ sufficiently large, P λ ± (1) < 0. Therefore for some λ, P± λ (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 )bethe dimension of kerl and let f ={f j } j= E be fixed. The below solvability conditions are established in [2].

3 Theorem 1.4. The equation Lu = f is solvable if and only if j= N. Apreutesei and V. Volpert 83 f j v l j = 0, l = 1,...,α ( L ), (1.10) where v l ={v l j} 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 a j k > 0, k =±1,±2,...,±m, j = 0,±1,±2,..., (1.11) and that there existsa positive solutionw 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 Auxiliary results Supposethatconditions(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) = ( u N m,...,u N 1 ), U+ (N) = ( u N+1,...,u N+m ). (2.1) Lemma 2.1. Let Lu = f,where f 0, u 0, u 0. Then u>0. Proof. Suppose that u j = 0forsomej. Sinceu 0, there exists i such that u i = 0, and either u i+1 0oru i 1 0. The equation (Lu) i = f i gives a contradiction in signs. The lemma is proved. Lemma 2.2. If the initial condition u 0 of the problem du dt = Lu, u(0) = u0 (2.2) is nonnegative, then the solution u(t) is also nonnegative for all t (0, ).

4 84 On the eigenvalue of infinite matrices Proof. Consider the approximate problem du i dt = (Lu) i, N i N, t 0, U ( N) = 0, U + (N) = 0, t 0, (2.3) u(0) = u 0, where the unknown function is u = (u N,u N+1,...u 0,...,u N 1,u N ). Since u 0 0andLu has nonnegative off-diagonal coefficients, it follows that the solution u N = (u N N,u N N+1,...u N 0,...,u N N 1,u N N) of the above problem is nonnegative. If we compare the solution u N at the interval [ N,N] and the solution u N+1 at the interval [ N 1,N +1],wefindu N+1 u N. Indeed, the difference u N+1 u N 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, u N+1 u N.So the sequence is monotonically increasing with respect to N. The sequence is also bounded with respect to N: u N (t) M,forallN and t [0,T], where T is any positive number, M>0dependsonu 0 and on the coefficients a i k of L, which are bounded. Being bounded and monotone, it follows that u N is convergent as N in C([0,T];E), say u N u.by the equations, we have also u N u in C 1 ([0,T];E). Then u verifies the problem (2.2)and u 0 (because u N 0), as claimed. Corollary 2.3 (comparison theorem). Let u 1 (t) and u 2 (t) be solutions of the equation du = Lu, (2.4) dt with the initial conditions u 1 (0) and u 2 (0),respectively.Ifu 1 (0) u 2 (0), then u 1 (t) u 2 (t) for t 0. Lemma 2.4. If the initial condition u 0 of the problem du dt = L+ u, u(0) = u 0 (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)convergestothetrivialsolutionu = 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,andn is sufficiently large, then u j 0 for j N. Proof. By virtue of Corollary 1.3, there exists a constant ɛ > 0suchthatL + q< ɛ.letus take N large enough such that (( L L +) q ) j ɛ, j N. (2.7) 2

5 N. Apreutesei and V. Volpert 85 Suppose that u j < 0forsomej>N.Bytheassumptionu j 0asj,wecanchoose τ>0suchthatv j = u j + τq j 0forallj N, and there exists i>nsuch that v i = 0. Since V (N) > 0andv j > 0forallj sufficiently large, there exists k>n such that v k = 0and either v k+1 0orv k 1 0 (i.e., v k+1 > 0orv k 1 > 0). 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< ɛ and f 0, the right-hand side of this equality is less than or equal to 0 for j N.AsintheproofofLemma 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 > 0andαu k + τl + q k + τ(l L + )q k <αu k ɛτ/2 = τ(α + ɛ/2) < 0, because v k = The main result In this section, we present the main result of this work and study some spectral properties of infinite-dimensionalmatrices with nonnegativeoff-diagonal elements. Theorem 3.1. Let (1.12) have a positive boundedsolution w. Then, the following hold. (i) The equation Lu = λu (3.1) does not have nonzero bounded solutions for Reλ 0, λ 0. (ii) Each solution of (1.12)hastheformu = kw,wherek is a constant. (iii) The equation has a positive solution unique up to a constant factor. L u = 0 (3.2) Proof. (1) In order to prove the first assertion, we analyze two cases. Case 1. We consider first the case where in (3.1) λ = α + iβ, α 0, β 0. Suppose by contradiction that there exists a bounded nonzero solution u = u 1 + iu 2 of this equation. Then Lu 1 = αu 1 βu 2 and Lu 2 = βu 1 + αu 2. Consider the equation Its solution is dv dt = Lv αv, v(0) = u1. (3.3) v(t) = u 1 cosβt u 2 sinβt. (3.4)

6 86 On the eigenvalue of infinite matrices For the sequence u ={u j }={u 1 j + iu 2 j }, we denote û ={ u j }. Let us take the value of N as in Lemma 2.5 and choose τ>0suchthat û j τw j, where at least for one j 0 with j 0 N, we have the equality For j N, consider the problem j N, (3.5) û j0 = τw j0. (3.6) dy = Ly αy, dt y N k (t) = û N k, k = 1,...,m, y (t) = 0, (3.7) and the corresponding stationary problem y(0) = û, (3.8) Lȳ αȳ = 0, ȳ N k = û N 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)ofproblem(3.7)-(3.8)convergestoȳ as t. For this, we consider the solution y (t) ofproblem(3.7) with the initial condition y (0) = ρq, where ρ is such that ρq j û j, j N. (3.10) By Corollary 1.3,wehaveL ± q<0. Since L + is close to L for j N,withN large enough, it follows that (Lq) j < 0, j N. Theny (t) monotonically decreases in t for each j N fixed. From the positiveness and the decreasing monotonicity of y,wededucethaty (t) converges as t to some x = lim t y (t) 0. It satisfies the equation Lx αx = 0. Taking the limit also in the boundary conditions, one obtains that x N+k = û N+k,fork = 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 lim t 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 t y (t) = lim t y (t) = ȳ. (3.11) By virtue of the comparison theorem applicable in this case (because 0 û j ρq j, j N), we have y (t) y(t) y (t), j N. (3.12)

7 N. Apreutesei and V. Volpert 87 Hence lim y j(t) = ȳ j, j N. (3.13) t One can easily verify that Then it follows from the comparison theorem that v j (t) û j j Z. (3.14) v j (t) y j (t), j N, t 0. (3.15) From this, we have ( v j (t) = v j t + 2πn ) ( y j t + 2πn ). (3.16) β β Passing to the limit as n,weobtain v j (t) ȳ j, j N, t 0. (3.17) Observe that L(τw y) α(τw y), j N, andτw N y N 0. We can apply Remark 2.6 to τw ȳ. Therefore, Hence, ȳ j τw j, j N. (3.18) 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)forallj Z. The sequence z(t) = τw v(t) is a solution of the equation dz = Lz αz + ατw. (3.20) dt Since z(t) 0(via(3.19) forallj Z), z is not identically zero, and is periodic in t, it follows that z j (t) > 0forallj and t 0. Indeed, suppose that for some t = t 1 and j = j 1, z j1 (t 1 ) = 0. Consider first the case where α>0. Since (dz j1 /dt)(t 1 ) 0andw j1 > 0, we obtain a contradiction in signs in the equation for z j1.ifα = 0, then the equation becomes dz = Lz. (3.21) dt Assuming that z(t) is not strictly positive, we easily obtain that it is identically zero for all j. Wehave(dz j1 /dt)(t 1 ) 0and(Lz) j1 (t 1 ) 0. Then (Lz) j1 (t 1 ) = 0, so all z j (t 1 ) = 0. Since z j1 verifies dz j1 /dt = (Lz) j1, z j1 (t 1 ) = 0, by the uniqueness we find z j1 (t) = 0, t t 1. Combining this with z j (t 1 ) = 0, ( )j Z,wegetz j (t) = 0, ( )j Z,( )t (0, ).

8 88 On the eigenvalue of infinite matrices Thus in both cases, z j (t) is positive for all j and t.wetaket 0suchthat e iβt = u j 0 u j0, (3.22) with j 0 from (3.6), that is, cosβt = u 1 j 0 / u j0 and sinβt = u 2 j 0 / u j0.then,v j0 (t) = u 1 j 0 cosβt u 2 j 0 sin βt = u j0, hence with the aid of (3.6) we obtain the contradiction The first assertion of the theorem is proved for nonreal λ. z j0 (t) = τw j0 u j0 = 0. (3.23) 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.weconsiderthesequencev = u + τw,whereτ>0 is chosen such that v 0for j N,butv j0 = 0forsomej 0, j 0 N.Wehave Lv = λv λτw, (3.24) and therefore v j 0forallj by virtue of Lemma 2.5. Indeed, for j N, the inequality holds because of the way we have chosen τ.forj N,oneappliesLemma 2.5 for (3.24) writtenintheform(l λi)v = λτw, j N, withv N 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 (P λ )± (σ) fortheoperatorl will be the same as for the operator L.ByvirtueofTheorem 1.2,theoperatorL 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 contradictionshows thata nonzeroboundedsolutionv of (3.2) exists and it is exponentially decreasing at infinity (see [2, Theorem 3.2]).

9 We recall next (see Theorem 1.4)that(3.25) is solvable if and only if N. Apreutesei and V. Volpert 89 ( 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<0suchthat(3.26)issatisfied. Let u be the corresponding solution of (3.25). There exists a τ (not necessarily positive), such that ũ = u + τw 0for j N, but not strictly positive. Since Lũ = f and f<0, ũ N 0, and ũ j 0asj,byvirtueofLemma 2.5,onefindsũ 0forallj.But for those j where ũ vanish, this leads to a contradiction in signs in the equation. Therefore ũ>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, [2], Solvability conditions for some difference operators, Advances in Difference Equations 2005 (2005), no. 1, [3] A. I. Volpert and V. Volpert, Spectrum of elliptic operators and stability of travelling waves, Asymptotic Analysis 23 (2000), no. 2, [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, N. Apreutesei: Department of Mathematics, Gh. Asachi Technical University of Iaşi, Iaşi, Romania address: napreut@math.tuiasi.ro V. Volpert: Camille Jordan Institute of Mathematics, UMR 5208 CNRS, University Lyon 1, Villeurbanne, France address: volpert@math.univ-lyon1.fr