’iauliai Math. Semin.,
5 (13), 2010, 87100
ON THE EIGENVALUE PROBLEMS FOR
FINITE-DIFFERENCE OPERATORS WITH
COUPLED BOUNDARY CONDITIONS
Svaj
unas SAJAVIƒIUS
Vilnius University, Naugarduko str. 24, LT-03225, Vilnius, Lithuania;
e-mail: [email protected]
Mykolas Romeris University, Ateities str. 20, LT-08303, Vilnius, Lithuania;
e-mail: [email protected]
In the paper, the eigenvalue problems for one- and two-dimensional second order nite-dierence operators with nonlocal coupled boundary conditions are considered. Conditions for the existence of zero, positive, negative or complex eigenvalues are proposed and analytical expressions of eigenvalues are provided.
Abstract.
Key words and phrases: coupled boundary conditions, eigenvalue problem,
nite-dierence operator.
2000 Mathematics Subject Classication: 65L15, 65L12.
1.
Introduction
The present paper deals with the eigenvalue problems for one- and twodimensional second order nite-dierence operators with given nonlocal coupled boundary conditions.
Let us introduce the uniform discrete grid on the interval (0, 1),
{
}
1
ωh = xi : xi = ih, i = 1, 2, . . . , N − 1, h =
,
N
and the uniform discrete grid on the domain (0, 1) × (0, 1),
ωh×h = ωh × ωh = {(xi , yj ) : xi ∈ ωh , yj ∈ ωh }.
S. Sajavi£ius
88
First of all, we will consider the eigenvalue problem for one-dimensional nite-difference operator with given coupled boundary conditions,
Ui−1 − 2Ui + Ui+1
+ λUi = 0,
h2
U0 = γ0 UN ,
U1 − U0
UN − UN −1
= γ1
,
h
h
xi ∈ ωh ,
(1)
(2)
(3)
where γ0 , γ1 ∈ R, γ0 + γ1 ̸= 0. We will also briey discuss the similar
two-dimensional problem
Ui,j−1 − 2Uij + Ui,j+1
Ui−1,j − 2Uij + Ui+1,j
+
+ λUij = 0,
2
h
h2
(xi , yj ) ∈ ωh×h ,
Ui0 = UiN = 0,
xi ∈ ωh ,
(4)
(5)
U0j = γ0 UN j ,
UN j − UN −1,j
U1j − U0j
= γ1
,
h
h
(6)
y j ∈ ωh .
(7)
Such values of λ that the problem (1)(3) or (4)(7) has the non-trivial
solution are called eigenvalues, and the set of all eigenvalues is called the
spectrum of the problem.
Since conditions (2), (3) and (6), (7) are nonlocal, the corresponding
nite-dierence operators are non-self-adjoint. Therefore, the analysis of the
spectra of these problems leads to the problems on the existence of both real
and complex eigenvalues.
Let us introduce a parameter γ ,
γ=
1 + γ0 γ1
.
γ0 + γ1
The main aim of this paper is to investigate the dependence of the qualitative
structure of the spectra of the nite-dierence problems (1)(3) and (4)(7)
on the parameters γ0 , γ1 (to be precise, on the generalized parameter γ ), i.e.,
to formulate conditions for the existence of zero, positive, negative or complex
eigenvalues, and (when it is possible) to provide the analytical expressions for
eigenvalues. We use technique and arguments which are used, for example,
in the papers [10], [12], [15], in order to investigate similar problems with
other types of nonlocal conditions.
On the eigenvalue problems for nite-dierence operators...
89
If one of the parameters γ0 , γ1 in one of boundary condition is equal to
1 or −1, then the corresponding condition is called periodic or antiperiodic, respectively. In [17], [18], the existence of eigenvalues of periodic (antiperiodic) or coupled boundary value problems for self-adjoint second-order
dierence equations is proved, numbers of eigenvalues are calculated, and
their relationships are obtained. The similar problems with γ0 = 0 and γ1 =
β , |β| > 1, as auxiliary problems are investigated by A. V. Gulin et al. (see
[3] and references therein). In the paper [6], the nite-dierence eigenvalue
problem similar to the problem (1)(3) also appears as an auxiliary problem
in the investigation of the stability of nite-dierence scheme for a parabolic
equation subject to initial condition and coupled boundary conditions. But
due to the dierent way of approximation, instead of boundary conditions
(2) and (3), in [6], the dierent form conditions are considered.
2.
The one-dimensional problem
Let us consider the one-dimensional nite-dierence eigenvalue problem (1)
(3). We rewrite conditions (2) and (3) in the form of a system of two linear
algebraic equations with two unknowns U0 and UN
{
U0
−γ0 UN = 0,
(8)
− h1 U0 − γh1 UN = − h1 (U1 + γ1 UN −1 ).
Since γ0 + γ1 ̸= 0, the determinant of the latter system
1 −γ0 1
D= 1
γ1 = − (γ0 + γ1 ) ̸= 0.
−h − h
h
Therefore, we can write the solution of the system (8) in the form
U0 =
γ0 γ1
γ0
U1 +
UN −1 ,
γ0 + γ1
γ0 + γ1
Now we dene the square

0
2 − γ0γ+γ
 −1 1


0

−2 
A = h  ···

0


0

1
− γ0 +γ
1
UN =
1
γ1
U1 +
UN −1 .
γ0 + γ1
γ0 + γ1
matrix of order (N − 1),

γ1
−1 0 . . . 0
0
− γγ00+γ
1

2 −1 . . . 0
0
0


−1 2 . . . 0
0
0


··· ··· ··· ··· ···
··· .

0
0 . . . 2 −1
0


0
0 . . . −1 2
−1 
1
0
0 . . . 0 −1 2 − γ0γ+γ
1
S. Sajavi£ius
90
One can see that the nite-dierence eigenvalue problem (1)(3) is equivalent
to the eigenvalue problem
AU = λU
(9)
for the matrix A. Note that if γ0 γ1 = 1, then the matrix A is symmetric,
i.e., the considered nite-dierence operator is self-adjoint (see [5]).
We can rewrite equation (1) in the form
(
λh2 )
Ui−1 − 2 1 −
Ui + Ui+1 = 0.
(10)
2
Let us consider several cases of possible values of λ.
λh2
Case 1. λ < 0. In this case, 1 −
2 > 1. We introduce the notation
2
1 − λh2 = cosh (αh), α ̸= 0, and rewrite the nite-dierence equation (10) in
the form
Ui−1 − 2 cosh (αh)Ui + Ui+1 = 0.
By substituting the general solution of the latter equation
Ui = c1 cosh (αih) + c2 sinh (αih)
into nonlocal conditions (2) and (3), we get a system of two linear algebraic
equations with unknowns c1 and c2
{
(1 − γ0 cosh α)c1 −γ0 sinh α · c2 = 0,
ν1
+ νh2 c2 = 0,
h c1
where
(
)
ν1 = cosh (αh) − γ1 cosh α − cosh (α(1 − h)) − 1,
(
)
ν2 = sinh (αh) − γ1 sinh α − sinh (α(1 − h)) .
This system has a non-trivial solution if its determinant is equal to zero, i.e.,
(
))
1(
Dh,1 =
(1 + γ0 γ1 ) sinh (αh) − (γ0 + γ1 ) sinh α − sinh (α(1 − h)) = 0.
h
After simple transformations we get the equation
f1 (α; h) :=
sinh α − sinh (α(1 − h))
= γ,
sinh (αh)
α ̸= 0.
(11)
By using L'Hospital's rule, we prove that
lim f1 (α; h) = 1,
α→+0
lim f1 (α; h) = cosh α.
h→0
Since f1 (α; h) is an increasing and even function (see examples in Figure 1),
we can prove the following statement.
On the eigenvalue problems for nite-dierence operators...
91
1. The inequality γ > 1 is the necessary and sucient condition for the existence of one and only one negative eigenvalue of the problem
(9),
( αh )
4
λ−1 = − 2 sinh2
,
h
2
where α is the positive root of equation (11).
Proposition
Case 2.
λ = 0. The general solution of the equation (10) is
Ui = c1 + c2 ih.
By substituting this expression into conditions (2) and (3), we nd that
{
(1 − γ0 )c1
−γ0 c2 = 0,
(1 − γ1 )c2 = 0.
A non-trivial solution exists if
Dh,2 = 1 + γ0 γ1 − (γ0 + γ1 ) = 0,
i.e., if γ = 1.
2. The number λ0 = 0 is an eigenvalue of the problem (9) if
and only if γ = 1.
4
λh2 λh2
Case 3. 0 < λ < 2 . Now we have 1 −
2 < 1. Set 1 − 2 = cos (αh),
h
0 < α < πh . Then the equation (10) becomes
Proposition
Ui−1 − 2 cos (αh)Ui + Ui+1 = 0,
and its general solution is
Ui = c1 cos (αih) + c2 sin (αih).
By substituting the expression of the general solution into nonlocal conditions
(2) and (3), we obtain
{
(1 − γ0 cos α)c1 −γ0 sin α · c2 = 0,
ν1
+ νh2 c2 = 0,
h c1
where
(
)
ν1 = cos (αh) − γ1 cos α − cos (α(1 − h)) − 1,
(
)
ν2 = sin (αh) − γ1 sin α − sin (α(1 − h)) .
S. Sajavi£ius
92
By equating the determinant of this system with zero, we get
(
))
1(
(1 + γ0 γ1 ) sin (αh) − (γ0 + γ1 ) sin α − sin (α(1 − h)) = 0,
Dh,3 =
h
i.e.,
( π)
sin α − sin (α(1 − h))
= γ, α ∈ 0,
.
(12)
sin (αh)
h
( 4)
Proposition 3. The positive eigenvalues λk ∈ 0, 2
of the problem (9), if
h
they exist at all, can be computed by the formula
f3 (α; h) :=
(
)
4
2 αk h
sin
,
h2
2
(
)
where αk are the roots from the interval 0, πh of the equation (12) .
λk =
Remark 1.
Note that
lim f3 (α; h) = 1, limπ f3 (α; h) = (−1)N −1 (2N − 1), lim f3 (α; h) = cos α.
α→0
h→0
α→ h
(
)
The number of eigenvalues λk ∈ 0, h42 depends on h and γ . In Figure 1, the
graphs of function γ(α) = f3 (α; h) with dierent values of h( are exhibited.
)
For every xed value γ = γ ∗ , the number of eigenvalues λk ∈ 0, h42 is equal
to the number of points where the straight line γ = γ ∗ intersects with curve
γ = f3 (α; h).
Case 4.
λ=
4
.
h2
In this case the equation (10) becomes
Ui−1 + 2Ui + Ui−1 = 0,
and the general solution can be expressed as
Ui = (−1)i (c1 + c2 ih).
Similarly as in previous cases, we get
{
(1 − (−1)N γ0 )c1
−(−1)N γ0 c2 = 0,
N
N
−2N (1 + (−1) γ1 )c1 −(1 + (−1) (2N − 1)γ1 )c2 = 0.
The condition for the existence of a non-trivial solution is
(
)
Dh,4 = − 1 + γ0 γ1 − (−1)N −1 (2N − 1)(γ0 + γ1 ) = 0.
On the eigenvalue problems for nite-dierence operators...
Proposition
and only if
λh2
2
4
h2
is the eigenvalue of the problem (9) if
f4 (α; h) := (−1)N −1 (2N − 1) = γ.
2
λ > h42 . Since 1 − λh2 < −1, in this case, after denoting
= − cosh (αh), α ̸= 0, we have
Case 5.
1−
4. The number λ =
93
Ui−1 + 2 cosh (αh)Ui + Ui+1 = 0,
and the general solution of this equation is
(
)
Ui = (−1)i c1 cosh (αih) + c2 sinh (αih) .
The corresponding system of two linear equations is
{
(1 − (−1)N γ0 cosh α)c1 −(−1)N γ0 sinh α · c2 = 0,
−N ν1 c1
−N ν2 c2 = 0,
where
(
)
ν1 = cosh (αN −1 ) + (−1)N γ1 cosh α + cosh (α(1 − N −1 )) + 1,
(
)
ν2 = sinh (αN −1 ) + (−1)N γ1 sinh α + sinh (α(1 − N −1 )) .
A non-trivial solution to the latter system exists if
(
Dh,5 = −N (1 + γ0 γ1 ) sin (αN −1 )
(
))
− (−1)N −1 (γ0 + γ1 ) sinh α + sinh (α(1 − N −1 )) = 0,
i.e.,
(
)
(−1)N −1 sinh α + sinh (α(1 − N −1 ))
f5 (α; h) :=
= γ,
sinh (αN −1 )
α ̸= 0.
(13)
Since f5 (α; h) is a monotonous and even function (see examples in Figure 1),
we can prove the following statement.
5. The inequality (−1)N −1 γ > 2N − 1 is the necessary and
e > 42
sucient condition for the existence of one and only one eigenvalue λ
h
of the problem (9),
( )
e = 4 cosh2 αh ,
λ
h2
2
where α is the positive root of the equation (13).
Proposition
S. Sajavi£ius
94
g
g
30
10
20
0
2
4
6
8
10
12
a
K10
10
0
1
2
3
4
5
6
7
8
a
9
K20
(a)
(b)
g
g
30
30
20
10
20
0
10
K10
K20
K30
K40
10
0
2
4
6
8
10
12
14
16 a
20
30
a
K50
(c)
(d)
g
g
60
80
40
20
60
0
10
20
30
K20
40
40
50
60
a
K40
20
K60
0
10
20
30
40
(e)
Figure 1:
K80
(f)
γ(α) = f1 (α; h) (dashed line), γ(α) = f3 (α; h)
γ(α) = f4 (α; h) (dotted line), γ(α) = f5 (α; h) (dash-dot line), α ∈ (0, πh ):
1
1
1
1
1
1
(a) h = 3 ; (b) h = 4 ; (c) h = 5 ; (d) h = 10 ; (e) h = 15 ; (f) h = 20 .
(solid line),
The graphs of the functions
a
On the eigenvalue problems for nite-dierence operators...
95
2
λh
λ∈
√ C. Let us introduce notation 1 − 2 = cosh (qh), where
q = α ± iβ , i = −1, and α ̸= 0, β ̸= 0. We assume that α ̸= 0 and β ̸= 0.
If α = 0, β ̸= 0 or α ̸= 0, β = 0, then this case coincides with Case 1 or Case
3, respectively. However, when α = β = 0, a situation is the same as in Case
2.
Now the equation (10) becomes
Case 6.
Ui−1 − 2 cosh (qh)Ui + Ui+1 = 0,
and the general solution of this equation is
Ui = c1 eiqih + c2 e−iqih .
By substituting the general solution into nonlocal conditions (2) and (3), we
obtain the system of linear equations
{
(1 − γ0 eiq )c1 +(1 − γ0 e−iq )c2 = 0,
ν1
+ νh2 c2 = 0,
h c1
where
ν1 = eiqh − γ1 (eiq − eiq(1−h) ) − 1,
ν2 = e−iqh − γ1 (e−iq − e−iq(1−h) ) − 1.
This system has a non-trivial solution if
Dh,6 = −
)
2i (
(1 + γ0 γ1 ) sin (qh) − (γ0 + γ1 )(sin q − sin (q(1 − h))) = 0,
h
i.e.,
f6 (α; h) :=
sin q − sin(q(1 − h))
= γ,
sin (qh)
q = α ± iβ,
α ̸= 0,
β ̸= 0. (14)
6. The complex eigenvalues of the problem (9), if they exist at
all, can be computed by the formula
Proposition
λc,k =
(
)
4
2 qk h
sin
,
h2
2
where qk = αk ± iβk are the non-trivial complex roots of the equation (14).
S. Sajavi£ius
96
Remark 2. By separating the real and imaginary parts in the equation (14),
we get the equations

sin α · cosh β − sin (α(1 − h)) · cosh (β(1 − h))


= γ,

sin (αh) · cosh (βh)
(15)

cos α · sinh β − cos (α(1 − h)) · sinh (β(1 − h))


= 0.
cos (αh) · sinh (βh)
It is possible to construct the iterative method for the solution of the system
(15) with known value of γ .
If N is an odd number, then there exists such a value γ ∗ that all
eigenvalues of the problem (9) are complex numbers for all γ < γ ∗ . However,
for all values of γ there exists at least one real eigenvalue of the problem (9)
if N is an even number.
Remark 3.
3.
The two-dimensional problem
Now let us consider the two-dimensional nite-dierence eigenvalue problem
(4)(7). By separating variables, i.e., by representing the solution of the
problem (4)(7) in the form
Uij = Vi Wj ,
i, j = 0, 1, . . . , N,
we get two one-dimensional eigenvalue problems:
Vi−1 − 2Vi + Vi+1
+ ηVi = 0,
h2
VN − VN −1
V1 − V0
= γ1
,
h
h
V0 = γ0 VN ,
(16)
and
Wj−1 − 2Wj + Wj+1
+ µWj = 0,
h2
W0 = 0,
WN = 0,
(17)
where (xi , yj ) ∈ ωh×h and η + µ = λ. The problem (16) was considered in
Section 2, while the problem (17) is classic. It is well-known (see, e.g., [11])
that all eigenvalues of the problem (17) are real, positive, algebraically simple,
and can be computed by the formula
( πlh )
4
µl = 2 sin2
, l = 1, 2, . . . , N − 1.
(18)
h
2
Let us denote
λkl = ηk + µl .
(19)
On the eigenvalue problems for nite-dierence operators...
97
It is easy to see that the positivity of λkl is conditioned by the positivity of
ηk . Therefore, the following statements are valid.
( 8)
Proposition 7. 1) The positive eigenvalues λkl ∈ 0, 2
of the problem
h
(4)(7), if they exist at all, can be computed by the formula
(α h)
( πlh ))
4(
k
λkl = 2 sin2
+ sin2
, l = 1, 2, . . . , N − 1,
h
2
2
(
)
where αk are the roots of the equation (12), αk ∈ 0, πh .
2) Moreover, the numbers
( πlh ))
4(
λl = 2 1 + sin2
, l = 1, 2, . . . , N − 1,
h
2
are eigenvalues of the problem (4)(7) if and only if (−1)N −1 (2N − 1) = γ .
3) The inequality (−1)N −1 γ > 2N − 1 is the necessary and sucient
condition that the numbers
(
( )
(
))
el = 4 cosh2 αh + sin2 πlh , l = 1, 2, . . . , N − 1,
λ
h2
2
2
where α is the positive root of the equation (13), should be the eigenvalues of
the problem (4)(7).
Now let us investigate the existence of zero and negative eigenvalues of
the problem (4)(7). If λkl = 0, then the equations (18) and (19) imply that
( πlh )
4
ηk = − 2 sin2
.
h
2
However, a unique negative eigenvalue of the problem (16) has the form
( αh )
4
η−1 = − 2 sinh2
,
h
2
where α is the positive root of the equation (11). Since the numbers
( ( πlh ) √
( πlh )
)
2
∗
αl = log sin
+ sin2
+ 1 , l = 1, 2, . . . , N − 1,
h
2
2
are the positive roots of the equations
( αh )
( πlh )
sinh2
= sin2
,
2
2
the following statement is valid.
l = 1, 2, . . . , N − 1,
S. Sajavi£ius
98
8. If γ = γl = f1 (αl∗ ; h), l = 1, 2, . . . , N − 1, then, for each
l = 1, 2, . . . , N −1, the problem (4)(7) has an algebraically simple eigenvalue
λkl = 0.
If γ = f1 (αs∗ ; h), then problem (16) has a negative eigenvalue
Proposition
ηs = −
( ∗ )
4
2 αs h
sinh
.
h2
2
Hence, the problem (4)(7) has s − 1 negative eigenvalues
λsl = −
( ∗ )
(
))
4(
2 αs h
2 πlh
sinh
−
sin
,
h2
2
2
l = 1, 2, . . . , s − 1,
and an algebraically simple eigenvalue λss = 0. Finally, we note that
limh→0 α1∗ = π , and prove the following proposition.
Proposition 9. The number of negative eigenvalues of the problem (4)(7)
depends on γ . If
∗
f1 (αs∗ ; h) < γ < f1 (αs+1
; h),
where s = 1, 2, . . . , N − 1, then there exist exactly s negative eigenvalues
of the problem (4)(7). Consequently, when γ < f1 (α1∗ ; h) −−−→ cosh π ≈
h→0
11.59195 . . . , then all real eigenvalues of the problem (4)(7) are positive.
4.
Concluding remarks
The qualitative information about the spectral structure of a nite-dierence
operator is useful, for example, in order to analyze the stability of nitedierence schemes for parabolic equations [1][4], [6][10], [13], or to justify
the convergence of iterative methods for nite-dierence equations [12], [14],
[16].
As a rule, any nonlocal condition, implies that, depending on nonlocal condition parameters, both real numbers (positive or non-positive) and
complex numbers (with positive or non-positive real parts) can be the eigenvalues of the corresponding nite-dierence problem. Using quite a simple
technique allows us to investigate the qualitative structure of the spectra of
the nite-dierence problems (1)(3) and (4)(7).
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99
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Received
16 October 2009