Existence and uniqueness of positive solitons for
a second order di¤erence equation
Xinfu Li, Guang Zhangyand Ying Wang
School of Science, Tianjin University of Commerce
Tianjin 300134, P. R. China
Abstract
In this paper, a discrete logistic steady-state equation with both positive and negative birth rate of population will be considered. By using suband upper-solution method, the existence of bounded positive solutions
and the existence and uniqueness of positive solitons will be established.
To this end, the Dirichlet eigenvalue problem with positive and negative
coe¢ cients is considered, and a general sub- and upper-solution theorem
is also obtained.
Keywords: Soliton, eigenvalue problem, logistic equation, sub- and
upper-solution.
MSC: 39A12, 39A70.
1
Introduction
In this paper, we consider the second order di¤erence equation
2
xi
1
=
pi xi
x1+
i
; i 2 Z;
(1)
where Z is a set of all integers, > 0 and > 0 are constants, xi = xi+1 xi ,
2
xi 1 = ( xi 1 ), and fpi gi2Z will be speci…ed later.
This equation is associated to the famous discrete logistic equation with
di¤usion:
1+
xt+1
xti = d 2 xti 1 + pi xti
xti
(2)
i
for i 2 Z and t 2 Z+ = f0; 1; 2; g, where the parameter d > 0 corresponds
to the rate at which the population di¤uses, and the unknown function x corresponds to the density of a population. The term x1+ in the equation corresponds to the fact that the population is self-limiting and the p corresponds
to the birth rate of population if the self-limiting is ignored. At points where
This work was supported by the National Natural Science Foundation of China (No.
11371277).
y Corresponding author. E-mail address: [email protected](G. Zhang)
1
pi > 0 (< 0) the population, ignoring self-limitation, has positive (negative)
birth rate. Hence, we assume throughout this paper that p takes on both positive and negative values on Z; we further assume that the sequence fpi gi2Z is
bounded and there exists an integer i0 2 Z such that pi0 > 0.
If we write
= 1=d, we see that the steady-state solutions of (2) must
satisfy equation (1). Since the sequence x represents a population density, it is
nonnegative. Such solutions correspond to possible steady-state distributions of
population. So our purpose in this paper is to establish the existence of positive
solutions or the existence and uniqueness of positive solitons for equation (1). A
discrete soliton is a spatially localized solution of (1) and decays to 0 at in…nity,
that is,
lim xi = 0:
jij!1
Recently, the existence of nontrivial discrete solitons for the general equation
2
xi
1
= f (i; xi ) ; i 2 Z;
(3)
has been extensively established by a number of authors [1-12]. Among the
methods used are the principle of anticontinuity [1-3], variational methods [410], center manifold reduction [11], and Nehari manifold approach [12], etc.
On the other hand, a soliton of (3) is also its homoclinic orbit or homoclinic
solution. By using the variational methods, the existence of homoclinic orbits or
homoclinic solutions has also been extensively discussed by a number of authors,
see [4-10] and [13-16]. However, as far as our knowledge, there is few papers
concerned with the existence of positive solitons.
Equation (2) is a discrete analogue of the well-known logistic equation of the
form
ut = Duxx + p (x) u u1+ ; x 2 R, t 2 R+ :
(4)
Indeed, by means of standard …nite di¤erence methods, we set up a grid in the
x; t plane with grid spacings x and t, and then replace the second derivative
uxx with a central di¤erence and ut with a forward di¤erence. By writing
xi = i x, tn = n t, pi = p (xi ), and uni
u (xi ; tn ), a …nite di¤erence scheme
for equation (4) is obtained:
un+1
i
uni
t
or
un+1
i
uni =
=
D
2
( x)
D t
uni+1
uni+1
2
( x)
2uni + uni
2uni + uni
1
1
+
+ pi uni
tpi uni
1+
(uni )
1+
t (uni )
;
which has the steady-state equation:
2
2
xi
1
=
( x)
D
pi xi
x1+
i
We note that the steady-state equation of (4) is
Du00 + p (x) u
2
u1+ = 0
; i 2 Z:
(5)
or
u00 =
where
= 1=D. When
p (x) u
u1+
;
(6)
= 1, equation (6) is reduced to
u00 =
(p (x) u
u) :
(7)
In [17], the authors considered the existence and uniqueness of positive solitons
of (7) by using sub- and upper-solution method. The present work is motivated
by [17].
To obtain a positive subsolution of equation (1), we need a positive eigenvalue
and its corresponding positive eigenfunction of the eigenvalue problem
2
xa
1
xi 1 = pi xi ; i 2 fa; a + 1;
= 0 = xb+1 :
; bg ;
Note that p takes on both positive and negative values, thus, it is inde…nite
in a very strong sense. The corresponding problem for ordinary di¤erential
equation had been considered in [18] by using variational method. However,
such problem is new for the above discrete problem, see [19, 20, 21]. In Section
2, we will consider the above discrete eigenvalue problem by using the matrix
and vector method.
As far as our knowledge, for the discrete problem (1) there is no a general sub- and upper-solution theorem on the set Z. Thus, a general sub- and
upper-solution theorem will be …rstly obtained in Section 3 and such theorem
will be used for equation (1). On the other hand, our solutions are classical,
however, the reference [17] can not insure this fact because some points of their
subsolution are not derivative.
Our results will give some theory groundwork for the numerical calculation
2
2
of (7). Note that = ( x) =D 6= when ( x) 6= 1. This will induce to
di¤erent results between (1) and (7).
2
Preliminaries
For any a; b 2 Z with a < b, we consider the eigenvalue problem of the form
2
xi 1 = pi xi ; i 2 [a; b] ;
xa 1 = 0 = xb+1 ;
b
(8)
where [a; b] = fa; a + 1; ; bg, fpi gi=a is a real sequence and there exists i0 2
[a; b] such that pi0 > 0.
Denote
0
1
0
1
2
1 0
0
xa
B 1 2
1
0 C
B xa+1 C
B
C
B
C
B
C
x=B .
; A=B
(9)
C
C
@ ..
A
@ 0
2
1 A
xb
0
0
1 2
(b a+1) (b a+1)
3
and
0
pa
B 0
B
P =@
0
1
0
0 C
C
A
pb (b
0
pa+1
0
:
a+1) (b a+1)
Then problem (8) can be rewritten by matrix and vector in the form
Ax = P x:
(10)
b+1
Let H be a set of all real sequences fxi gi=a 1 with xa 1 = 0 = xb+1 . For
Pb
any u; v 2 H, the innerq
product is de…ned as (u; v) = i=a ui vi and the norm
Pb
2
k k is de…ned as kuk =
i=a ui .
For any v 2 H, we de…ne
b
X
Q (v) = v T Av
pi vi2 :
i=a
Consider the Rayleigh quotient
and let
1
For such de…ned
= inf
1,
(
v T Av
K (v) = Pb
2
i=a pi vi
K (v) : v 2 H;
b
X
pi vi2
(11)
)
>0 :
i=a
(12)
we have the following important result.
Lemma 1 1 is an positive eigenvalue of the problem (8) or (10). Moreover 1
is simple and the corresponding eigenfunction ' can be chosen such that 'i > 0
for all i 2 [a; b].
Proof. First, we show
1
Q
> 0. Clearly
(v) = v T Av
1
1
b
X
pi vi2
0
i=a
for all v 2 H. Moreover, by the spectral theorem, (Av; v)
1
= 4 sin2
2 (b
a + 2)
is the …rst eigenvalue of the eigenvalue problem
2
xi
xa
= xi ; i 2 [a; b] ;
1 = 0 = xb+1 :
1
4
1
(v; v), where
Hence, if v 2 H and
Pb
i=a
pi vi2 > 0, then
v T Av
K (v) = Pb
2
i=a pi vi
1
Pb
(v; v)
1
max fjpi j ; i = a; a + 1;
2
i=a pi vi
; bg
> 0:
Hence 1 > 0.
Second, consider the linear eigenvalue problem
2
ui
1 pi u i
1
ua
1
= ui ; i 2 [a; b] ;
= 0 = ub+1 ;
(13)
2
and de…ne (Su)i =
ui 1
1 pi ui . It is easy to see that 1 is an eigenvalue
for (8) with corresponding eigenfunction ' if and only if 0 is an eigenvalue of S
and so of (13) with corresponding eigenfunction '. The minimal eigenvalue 1
of S is given by
1
T
=
inffv Av
=
inffQ
1
b
X
i=a
1
pi vi2 : v 2 H; kvk = 1g
(v) : v 2 H; kvk = 1g.
Note that Q 1 (v) 0 for all v 2 H, thus, we have 1 0. Because of how we
P
(n)
de…ned 1 , there exists a sequence v (n) 2 H and
pi (vi )2 > 0 such that
T
which implies that
v (n) Av (n)
lim P
2 =
n!1
(n)
b
i=a pi vi
lim Q
1
n!1
1
v (n) = 0:
Thus, we have 1 0. In this case, we know that 1 = 0 is the minimal eigenvalue of (13). By Lemma 10 in Appendix, 1 is simple and the corresponding
eigenfunction can be chosen to be positive on H. Thus, the statement in this
lemma is right.
Let
>
1,
consider the eigenvalue problem of the form
2
ui
1
ua
pi ui = ui ; i 2 [a; b] ;
= 0 = ub+1 :
(14)
1
We have the following result.
Lemma 2 If > 1 , then the minimal eigenvalue 1 of (14) is negative and the
(1)
corresponding eigenfunction '(1) can be chosen so that 'i > 0 for all i 2 [a; b].
5
Proof. Let ' be the eigenvector obtained in Lemma 1 and k'k = 1 , then we
have
b
X
'T A'
pi '2i =
i=a
'T (A
1
1 P )'
1
'T A'
1
1
=
'T A' < 0:
1
Consequently,
1
T
= minf
A
b
X
i=a
pi
2
i
j k k = 1g < 0:
In view of Lemma 10 in Appendix, the eigenvector corresponding to
chosen to be positive. This completes the proof.
3
1
can be
Main results
First of all, we introduce the de…nitions of the subsolution and uppersolution
and give a general sub- and upper-solution theorem.
For any a; b 2 Z with a < b, consider the Dirichlet boundary value problem
of the form
2
xi 1 = f (i; xi ) for i 2 [a; b] ;
(15)
xa 1 = 0 = xb+1 :
We will denote by (15)n the problem (15) when a =
b+1
De…nition 3 A sequence x = fxi gi=a
2
xi
+
A sequence x = fxi gi=
1
is said to be a uppersolution of (15) if
xi 1 f (i; xi ) for i 2 [a; b] ;
0 for i = a 1 and b + 1:
is said to be a subsolution of (15) if
2
xi
1
n and b = n.
xi 1 f (i; xi ) for i 2 [a; b] ;
0 for i = a 1 and b + 1:
Theorem 4 Suppose that x; x : Z ! R are functions with x x (where x x
is de…ned by xi xi for i 2 Z) and xjn ; xjn are, respectively, a subsolution and
a uppersolution of (15)n for all large n. Assume that f (i; z) is continuous with
z 2 [ ; ] and there exists a constant k > 0 such that
f (i; s2 )
f (i; s1 )
k (s2
s1 )
for all i 2 Z and s2 > s1 with s1 ; s2 2 [ ; ](where
= inf i2Z xi and
supi2Z xi ). Then problem (3) has a solution x such that x x x.
6
(16)
=
Proof. Since x and x provide sub and uppersolutions for (15)n , there exists a
solution ' of (15)n such that xi 'i xi for all jij n.
n
In fact, for any sequence u = fui gi= n , clearly, the problem
2
wi
1
+ kwi = f (i; ui ) + kui for i 2 [ n; n];
w n 1 = 0 = wn+1 ;
(17)
has a unique solution w. This de…nes a mapping T : u ! w. We claim that
T is increasing on [x; x]. In fact, for any sequences f~
ui gni= n ; f~
vi gni= n 2 [x; x]
with v~ u
~, we have
(T u
~
T v~) = f (i; u
~i ) + k~
ui f (i; v~i ) k~
vi
k(~
ui v~i ) + k(~
ui v~i ) 0;
2
where is an operator de…ned as ( u)i =
ui 1 + kui for i 2 [ n; n]. Since
(T u
~) n 1 (T v~) n 1 0; (T u
~)n+1 (T v~)n+1 0, by the strongly monotonicity
of , we obtain T u
~ T v~, see [21].
Let u(m) = T u(m 1) ; u(0) = x; v (m) = T v (m 1) ; v (0) = x. In the following,
we claim that
x = u(0)
u(1)
u(2)
v (2)
v (1)
v (0) = x:
First of all, by using (17) and the de…nition of subsolution, we see
2
(0)
(0)
1
ui
=
2 (0)
ui 1
=
2
(1)
xi
(0)
1
(1)
1
ui
(0)
+ kui
f (i; xi )
(0)
(1)
+ k ui
ui
h
i
(0)
(0)
f i; ui
+ kui
0 for i 2 [ n; n];
(1)
and u n 1 u n 1
0; un+1 un+1
0. This implies that u(0) u(1)
0
(1)
(0)
by the strongly monotonicity of . A similar argument gives v
v . The
monotonicity of and T gives the rest. So there exist u and v such that
lim u(m) = u and lim v (m) = v:
m!1
m!1
By the de…nition of u and v, we see that u and v satisfy equation (15)n and
x u v x. Let x(n) denote such a solution corresponding to (15)n .
Then standard a priori estimates and a diagonalization argument show that
there exists a subsequence of fx(n) g which converges to a solution x of (3) on
every bounded subset of Z. Moreover, since x
x(n)
x for all n, it follows
that x x x on Z. The proof is complete.
In the following, we use the sub- and upper-solution theorem to establish
the existence of positive solutions or the existence and uniqueness of positive
solitons for equation (1).
For problem (1), we have assumed that there exists i0 2 Z such that pi0 > 0.
For any n > ji0 j, consider the eigenvalue problem (8) when a = n and b = n
7
and denote the corresponding positive eigenvalue
Then
(1)
1
Theorem 5 For any
is de…ned in (18).
(1)
,
1
>
(1)
Proof. For any > 1
that i0 2 [ n1 ; n1 ] and
problem:
(n)
1
:= lim
n!1
de…ned in (12) by
1
0:
(n)
1 .
(18)
equation (1) has a positive solution, where
(1)
1
, clearly, there exists a positive integer n1 ji0 j such
(n )
< 1 1 < . In view of Lemma 2, the eigenvalue
(1)
1
2
xi
pi xi = xi ; i 2 [ n1 ; n1 ]
n1 1 = 0 = xn1 +1 ;
1
x
(n )
has a negative eigenvalue 1 1 and its corresponding eigenfunction '(n1 ) with
(n )
'(n1 ) = 1 can be chosen so that 'i 1 > 0 for all i 2 [ n1 ; n1 ]. For any " > 0,
we de…ne
(n )
"'i 1 ;
i 2 [ n1 ; n 1 ] ;
(19)
xi =
0;
i 2 Zn [ n1 ; n1 ] :
At this time, we have
=
(n1 )
1
2
"'i
(n1 )
1
"'i
(n1 )
(n1 )
pi "'i
(n1 )
+
xi
1
(n1 )
1+
"'i
1+
"'i
for i 2 [ n1 ; n1 ] and su¢ ciently small
2
+
0
. When i = n1 + 1,
=
xn1 +2 + 2xn1 +1
=
1)
"'(n
n1
xn1
<0
2
2
and
xi 1 0 for i > n1 + 1. Similarly, we can also prove that
xi 1 0
for i
n1 1. Thus, the sequence x de…ned in (19) is a subsolution of equation
(1).
Note that > 0 and the sequence fpi g is bounded. Thus, the su¢ ciently
large positive constant M x is a supersolution of (1).
For f (i; xi ) = pi xi x1+
, we have
i
f (i; s2 ) f (i; s1 )
=
pi s2 s1+
pi s1 s1+
2
1
=
(pi (1 + ) ) (s2 s1 ) ;
(20)
where 2 (s1 ; s2 ). Note that the sequence fpi g ; fxi g and M are bounded,
thus, equation (1) satis…es the condition (16). By Theorem 4, problem (1) has
a solution u such that 0
x
u
M . We claim that u > 0. Suppose
there exists i1 2 Z such that ui1 = 0, then we obtain ui1 1 + ui1 +1 = 0 since
2
u i1 1 =
pi1 ui1 u1+
. Thus we obtain ui1 1 = ui1 +1 = ui1 = 0. So
i1
u 0, which is contradict to u 6 0. The proof is complete.
8
Theorem 6 Assume that
n0 > ji0 j such that
pi
>
(1)
1
(
C
[i(i+1)] ;
C
i(i+1) ;
and there exists a constant C > 6= and
2 (0; 1];
>1
for jij > n0 , Then equation (1) has a positive soliton and there exists a constant
M > 0 and n2 n0 such that
M
for jij
i(i + 1)
xi
n2 :
Proof. Let x be the subsolution of (1) obtained in Theorem 5 and
=
i
M
; i 6= 0; 1; 2;
i (i + 1)
where M > 0 is a constant determined later. By simple calculation, we have
i 1
2
i 1
2
i 1
=
pi
=
M
i (i + 1)
2M
i (i + 1) (i + 2)
i
1+
i
+
=
M
=
i (i 1)
2M
=
i (i2 1)
M
i(i + 1)
(i
By the assumption on p, there exists n2
(i
2M
;
i (i2 1)
2M
i (i + 1) (i
M
i(i + 1)
6
+
1)(i + 2)
pi :
n0 and M1 > 0 such that
M
i(i + 1)
6
+
1)(i + 2)
3
;
1) (i + 2)
pi
0
for M M1 and jij n2 .
For any M M1 , de…ne
i
M
i(i+1) ;
=
jij n2 ;
jij < n2 :
M;
We will show that is a uppersolution of (1) by appropriate choice of M . In
fact,
2
(a). For jij > n2 , we have
pi i + 1+
0 by the choice of M ;
i 1
i
(b). For jij < n2 1, we have
2
i 1
pi
i
+
1+
i
=
pi M + M 1+
0
by choosing M > M1 large enough since > 0 and p is bounded;
(c). Similarly, for i = n2 ; n2 1; n2 ; n2 + 1, one can show that
pi i + 1+
0 for M > M1 large enough.
i
In view of (a)-(c), we construct a uppersolution of (1) with
the sub- and upper-solution theorem 4, we complete the proof.
9
2
i 1
x. Using
Theorem 7 Assume that there exists a positive integer m > ji0 j such that
pi 0 for jij m. Let x be a bounded positive solution of (1), then
lim xi = 0:
jij!1
Proof. At this time, we have
2
xi
1
=
x1+
i
pi xi
0 for jij
m:
Hence xi 1 is an increasing sequence for jij m and so there exists a positive
integer n1 such that xi is one sign for jij > n1 . Thus x is eventually a
monotone sequence for jij > n1 . Note that x is bounded. Thus, limn!+1 xn
and limn! 1 xn exist.
If limn!+1 xn 6= 0, we have
2
xi
1
x1+
i
for i
1
n
X
and
xn
xn0
m
x1+
;
i
(21)
i=n0
which implies that
lim xn = +1:
n!+1
This is impossible since x is bounded and so we must have limn!+1 xn = 0.
Similarly, we also have limn! 1 xn = 0. The proof is complete.
Theorem 8 For any 6= 0, there exists at most one positive solution of (1)
such that limjij!1 xi = 0.
Proof. Suppose that u and v are two distinct such solutions. As before we can
construct an arbitrarily small subsolution and so there must exists a solution
w of (1) such that w
u and w
v. Multiplying the u-equation by w, the
w-equation by u, then, we have
n
X
ui
u
w
2
wi
1
wi
2
ui
=
1
i= n
or
un wn
n
n
X
ui wi (wi
n 1
=
ui )
i= n
n 1
wn un + w
n
u
n
X
ui wi (wi
ui ) :
i= n
Let n ! +1, we have
1
X
ui wi (wi
ui ) = 0
i= 1
which implies that u = w. Similarly, we can also prove that v = w. The proof
is complete.
Corollary 9 Assume that all conditions of Theorem 8 hold, then equation (1)
has a unique positive soliton.
10
4
Conclusion
This paper studied a discrete logistic steady-state equation with both positive
and negative birth rate of population. By using sub- and upper-solution method
(Theorem 4), the existence of positive solution and positive soliton are obtained
(Theorem 5 and 6). Uniqueness of homoclinic type solution is also obtained
(Theorem 8).
5
Acknowledgments
The authors would like to thank the anonymous referees for their comments and
suggestions on the manuscript.
6
Con‡ict of Interests
The authors declare that there is no con‡ict of interests regarding the publication of this paper.
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case, Bull. Amer. Math. Sot. 21(1914), 6-9.
[19] Y. M. Shi and S. Z. Chen, Spectral theory of second-order vector di¤erence
equations, J. Math. Anal. Appl., 239(1999), 195-212.
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Sturm–Liouville problems, Linear Algebra and its Applications, 323(2001),
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12
7
Appendix
To obtain a positive eigenfunction of (8) or a positive eigenvector of (10), we
need the following lemma. We give the proof by the method of [22].
Lemma 10 For any (b a + 1)
0
qa
0
B 0 qa+1
Q=B
@
0
(b
a + 1) diagonal matrix
1
0
0 C
C ; qi 2 R; i 2 [a; b] ;
A
qb
0
let 1 be the minimal eigenvalue of A + Q, where A is de…ned in (9). Then
is simple and the corresponding eigenvector '(1) can be chosen to be positive.
1
Proof. Without loss of generality, we assume qi > 0 for i 2 [a; b]. If this is not
the case, choose a constant C > 0 large enough such that qi + C > 0. Consider
the matrix A + Q + CE instead, where E is a (b a + 1) (b a + 1) identity
~ = Q + CE, then q~i > 0 for any i 2 [a; b].
matrix. Denote Q
Since A + Q is a real (b a + 1) (b a + 1) symmetric positive de…nite
matrix, A+Q has b a+1 real eigenvalues. We repeat each eigenvalue according
to its multiplicity as follows:
0<
1
2
b a+1
and choose eigenvectors '(1) ; '(2) ;
; '(b
f'(1) ; '(2) ;
a+1)
; '(b
for A + Q such that
a+1)
g
is an orthonormal basis for Rb a+1 . In the following we give the proof of Lemma
10 in four steps.
Step 1. For any u; v 2 Rb a+1 , de…ne B[u; v] = v T (A + Q)u. It is obvious
that
0;
i 6= j;
B['(i) ; '(j) ] =
i; j 2 [a; b]
;
i
= j;
i
and
1
= minfB[u; u]ju 2 Rb
Step 2. We claim that if u 2 Rb
a+1
a+1
; kuk = 1g:
(22)
and kuk = 1, then u is a solution of
(A + Q)u =
1u
(23)
if and only if
B[u; u] =
1:
(24)
Obviously (23) implies (24). On the other hand, suppose (24) is valid. Then,
writing dk = (u; '(k) ), we have
u=
bX
a+1
(u; '(k) )'(k) =
k=1
bX
a+1
k=1
13
dk '(k)
and
bX
a+1
d2k
1
=
1
= B[u; u] =
k=1
Hence
bX
a+1
d2k
k:
k=1
bX
a+1
(
2
1 )dk
k
= 0:
k=1
Consequently dk = (u; '(k) ) = 0, if
m
X
u=
>
k
1.
It follows that
(u; '(k) )'(k)
k=1
for some m, where (A + Q)'(k) =
(A + Q)u =
m
X
1'
(k)
for k = 1; 2;
; m. Therefore
(u; '(k) )(A + Q)'(k) =
1 u:
k=1
This proves (23).
Step 3. We will show that if u 2 Rb a+1 satis…es (A + Q)u = 1 u; u 6= 0,
then either u > 0 or else u < 0.
To see this, let us assume without loss of generality that kuk = 1, and
note + = 1 for
= ku+ k2 ;
= ku k2 , where u+
i = max fui ; 0g and
ui = max f ui ; 0g. In this case, we have
1
= B[u; u] = B[u+ ; u+ ] + B[u ; u ]
1 ku
+ 2
k +
1 ku
k2 =
1:
Thus
B[u+ ; u+ ] =
1 ku
+ 2
k ;
B[u ; u ] =
1 ku
k2 :
By Step 2, u+ and u satisfy (A + Q)u+ = 1 u+ ; (A + Q)u
(A + Q)u+ 0; (A + Q)u
0. That is
8
+
u+
0;
2u+
>
a
a+1 + qa ua
>
>
+
+
+
+
>
< ua + 2ua+1 ua+2 + qa+1 ua+1 0;
+
+
+
ua+1 + 2ua+2 ua+3 + qa+2 u+
0;
a+2
>
>
>
>
:
+
+
u+
0:
b 1 + 2ub + qb ub
=
1u
. Thus
+
If there exists an i0 2 [a; b] such that u+
i0 = 0, then u = 0. Similar arguments
apply to u , and so either u > 0 or else u < 0. Thus, we can choose a positive
eigenvector '(1) corresponding to the minimal eigenvalue 1 .
Step 4. Finally assume that u and u
~ are two eigenvectors corresponding to
,
in
view
of
Step
3,
1
b
b
X
X
ui 6= 0 and
u
~i 6= 0
i=a
i=a
14
and so there exists a real constant k such that
b
X
ui = k
i=a
or
b
X
(ui
b
X
u
~i
i=a
k~
ui ) = 0:
(25)
i=a
But since u k~
u also satis…es (A+Q)(u k~
u) = 1 (u k~
u), by Step 3, u k~
u=0
or u k~
u > 0 or u k~
u < 0. In view of (25), we have u k~
u = 0. Hence the
eigenvalue 1 is simple. This completes the proof of Lemma 10.
15