On some classes of difference equations of infinite order
Vasilyev and Vasilyev Advances in Difference Equations
On some classes of difference equations of infinite order
Alexander V Vasilyev 1
Vladimir B Vasilyev 0
0 Chair of Pure Mathematics, Lipetsk State Technical University , Moskovskaya 30, Lipetsk, 398600 , Russia
1 Department of Mathematical Analysis, National Research Belgorod State University , Studencheskaya 14/1, Belgorod, 307008 , Russia
We consider a certain class of difference equations on an axis and a halfaxis, and we establish a correspondence between such equations and simpler kinds of operator equations. The last operator equations can be solved by a special method like the WienerHopf method.
difference equation; symbol; solvability

ak(x)u(x + k) = v(x), x ∈ M ⊂ R,
ak(x)u(x + βk) = v(x), x ∈ M ⊂ R,
where the functions ak(x), k = , . . . , n, v(x) are defined on M and given, and u(x) is an
unknown function. Since n ∈ N is an arbitrary number and all points x, x + , . . . , x + n,
∀x ∈ M, should be in the set M, this set M may be a ray from a certain point or the whole R.
A more general type of difference equation of finite order is the equation
where {βk}kn= ⊂ R.
Further, such equations can be equations with a continuous variable or a discrete one,
and this property separates such an equation on a class of properly difference equations
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and discrete equations. In this paper we will consider the case of a continuous variable x,
and a solution on the righthand side will be considered in the space L(R) for all equations.
1.1 Difference equation of a finite order with constant coefficients
This is an equation of the type
and it easily can be solved by the Fourier transform:
Indeed, applying the Fourier transform to () we obtain
has the following symbol:
+∞
–∞
+∞
–∞
+∞
–∞
Proof The proof of this assertion can be obtained immediately.
If we consider the operator () for x ∈ Z only
D : u(xd) −→
The function pn(ξ ) is called a symbol of a difference operator on the lefthand side ()
(cf. []). If pn(ξ ) = , ∀ξ ∈ R, then () can easily be solved,
1.2 Difference equation of infinite order with constant coefficients
The same arguments are applicable for the case of an unbounded sequence {βk}+–∞∞. Then
the difference operator with complex coefficients
D : u(x) −→
then its symbol can be defined by the discrete Fourier transform [, ]
+∞
–∞
1.3 Difference and discrete equations
Obviously there are some relations between difference and discrete equations.
Particularly, if {βk}+–∞∞ = Z, then the operator () is a discrete convolution operator. For studying
discrete operators in a halfspace the authors have developed a certain analytic technique
[–]. Below we will try to enlarge this technique for more general situations.
2 General difference equations
We consider the equation
(Du)(x) = v(x), x ∈ R+,
where R+ = {x ∈ R, x > }.
For studying this equation we will use methods of the theory of multidimensional
singular integral and pseudo differential equations [, , ] which are nonusual in the theory
of difference equations. Our next goal is to study multidimensional difference equations,
and this onedimensional variant is a model for considering other complicated situations.
This approach is based on the classical Riemann boundary value problem and the theory
of onedimensional singular integral equations [–].
2.1 Background
The first step is the following. We will use the theory of socalled paired equations [] of
the type
in the space L(R), where a, b are convolution operators with corresponding functions
a(x), b(x), x ∈ R, P± are projectors on the halfaxis R±. More precisely,
a(x – y)U(y) dy,
(bP–U)(x) =
b(x – y)U(y) dy.
–∞
Applying the Fourier transform to () we obtain [] the following onedimensional
singular integral equation [–]:
(aP+ + bP–)U = V
(aP+U)(x) =
+∞
(Hu)(x) = v.p. πi –∞+∞ xu–(y)y dy,
where P, Q are two projectors related to the Hilbert transform
Equation () is closely related to the Riemann boundary value problem [, ] for
upper and lower halfplanes. We now recall the statement of the problem: finding a pair
of functions ±(ξ ) which admit an analytic continuation on upper (C+) and lower (C–)
halfplanes in the complex plane C and of which their boundary values on R satisfy the
following linear relation:
d arg G(t) = .
where G(ξ ), g(ξ ) are given functions on R.
There is a onetoone correspondence between the Riemann boundary value problem
() and the singular integral equation (), and
2.2 Topological barrier
We suppose that the symbol G(ξ ) is a continuous nonvanishing function on the
compactification R˙ (G(ξ ) = , ∀ξ ∈ R˙) and
The last condition (), is necessary and sufficient for the unique solvability of the
problem () in the space L(R) [, ]. Moreover, the unique solution of the problem () can
be constructed with a help of the Cauchy type integral
+(t) = G+(t)P G+–(t)g(t) ,
–(t) = –G––(t)Q G+–(t)g(t) ,
where G± are factors of a factorization for the G(t) (see below),
G+(t) = exp P ln G(t) ,
G–(t) = exp Q ln G(t) .
2.3 Difference equations on a halfaxis
Equation () can easily be transformed into () in the following way. Since the righthand
side in () is defined on R+ only we will continue v(x) on the whole R so that this
continuation lf ∈ L(R). Further we will rename the unknown function u+(x) and define the
function
u–(x) = (lf )(x) – (Du+)(x).
Thus, we have the following equation:
R
(Du+)(x) + u–(x) = (lf )(x), x ∈ ,
which holds for the whole space R.
After the Fourier transform we have
To describe a solving technique for () we recall the following (cf. [, ]).
Definition A factorization for an elliptic symbol is called its representation if it is in the
form
where the factors σ+, σ– admit an analytic continuation into the upper and lower complex
halfplanes C±, and σ±± ∈ L∞(R).
Example Let us consider the Cauchy type integral
dt ≡
(z), z = x ± iy.
L(R) = A+(R) ⊕ A–(R),
It is well known this construction plays a crucial role for a decomposition L(R) on two
orthogonal subspaces, namely
where A±(R) consists of functions admitting an analytic continuation onto C±.
The boundary values of the integral (z) satisfy the PlemeljSokhotskii formulas [,
], and thus the projectors P and Q are corresponding projectors on the spaces of analytic
functions [].
The simple example we need is
exp(u) = exp(Pu) · exp(Qu).
Theorem Let σ (ξ ) ∈ C(R˙ ), Ind σ = . Then () has unique solution in the space L(R+)
for arbitrary righthand side v ∈ L(R+), and its Fourier transform is given by the formula
u˜ (ξ ) = σ –(ξ )lv(ξ ) + σ+–π(ξi ) v.p.
–∞
Proof We have
σ––(ξ )lv(ξ ) = P σ––(ξ )lv(ξ ) + Q σ––(ξ )lv(ξ )
σ+(ξ )u˜ +(ξ ) – P σ––(ξ )lv(ξ ) = Q σ––(ξ )lv(ξ ) – σ––(ξ )u˜ –(ξ ).
The lefthand side of the last quality belongs to the space A+(R), but the righthand side
belongs to A–(R), consequently these are zeros. Thus,
or in the complete form
u˜ +(ξ ) = σ –(ξ )lv(ξ ) + σ+–π(ξi ) v.p.
–∞
Remark This result does not depend on the continuation lv. Let us denote by M±(x)
the inverse Fourier images of the functions σ±–(ξ ). Indeed, () leads to the following
construction:
u+(x) =
+∞
–∞
+∞
+∞
M–(y – t)(lv)(t) dt dy
M–(y – t)v(t) dt dy.
Remark The condition σ (ξ ) ∈ C( R˙) is not a strong restriction. Such symbols exist for
example in the case that σ (ξ ) is represented by a finite sum, and βk ∈ Q. Then σ (ξ ) is a
continuous periodic function.
3 General solution
Since σ (ξ ) ∈ C(R˙ ), and Ind σ is an integer, we consider the case ae ≡ Ind σ ∈ N in this
section.
Theorem Let Ind σ ∈ N. Then a general solution of () in the Fourier image can be
written in the form
u˜ +(ξ ) = σ+–(ξ )ω–ae(ξ )P σ––(ξ )lv(ξ ) + (ξ – i)–aeσ+–(ξ )Pae–(ξ ),
and it depends on ae arbitrary constants.
Proof The function
has the index [–], thus the function
has the index , and we can factorize this function
Further, we write after ()
ωae(ξ )ω–ae(ξ )σ (ξ )u˜ +(ξ ) + u˜ –(ξ ) = lv(ξ ),
ωae(ξ )σ+(ξ )u˜ +(ξ ) + σ––(ξ )u˜ –(ξ ) = σ––(ξ )lv(ξ ).
Taking into account our notations we have
= ω–ae(ξ )P σ––(ξ )lv(ξ ) + ω–ae(ξ )Q σ––(ξ )lv(ξ ) ,
σ––(ξ )lv(ξ ) = P σ––(ξ )lv(ξ ) + Q σ––(ξ )lv(ξ )
= ω–ae(ξ )Q σ––(ξ )lv(ξ ) – ω–ae(ξ )σ––(ξ )u˜ –(ξ )
and we conclude from the last that the lefthand side and the righthand side also are a
polynomial Pae–(ξ ) of order ae – . It follows from the generalized Liouville theorem [,
] because the lefthand side has one pole of order ae in C in the point z = i. So, we have
u˜ +(ξ ) = σ+–(ξ )ω–ae(ξ )P σ––(ξ )lv(ξ ) + (ξ – i)–aeσ+–(ξ )Pae–(ξ ).
Remark This result does not depend on the choice of the continuation l.
Corollary Let v(x) ≡ , ae ∈ N. Then a general solution of the homogeneous equation ()
is given by the formula
Since we work with L(R) both the lefthand side and the righthand side are equal to
zero at infinity, hence these are zeros, and
4 Solvability conditions
Theorem Let – Ind σ ∈ N. Then () has a solution from L(R+) iff the following conditions
hold:
+∞
–∞
(ξ – i)k–σ––(ξ )lv(ξ ) dξ = , k = , , . . . , ae.
Proof We argue as above and use the equality (); we write it as
= (ξ + i)aeQ σ––(ξ )lv(ξ ) – (ξ + i)aeσ––(ξ )u˜ –(ξ ).
But there is some inaccuracy. Indeed, this solution belongs to the space A+(R), but more
exactly it belongs to its subspace Ak+(R). This subspace consists of functions analytic in C+
with zeros of the order –ae in the point z = i. To obtain a solution from L(R+) we need
some corrections in the last formula. Since the operator P is related to the Cauchy type
integral we will use certain decomposition formulas for this integral (see also [–]).
Let us denote σ––(ξ )lv(ξ ) ≡ g(ξ ) and consider the following integral:
–∞
Using a simple formula for a kernel
we obtain the following decomposition:
–∞
+∞
–∞
hold, then we obtain
So, we have a following property. If the conditions
(η – i)k–g(η) dη = , k = , , . . . , ae,
–∞
= (z – i)ae
–∞
+∞
–∞
–∞
Hence the boundary values on R for the lefthand side and the righthand one are equal,
and thus
P g(ξ ) = (ξ – i)aeP (ξ – i)aeg(ξ ) .
Substituting the last formula into the solution formula we write
u˜ +(ξ ) = σ+–(ξ )(ξ + i)aeP (ξ – i)aeσ––(ξ )lv(ξ ) .
5 Conclusion
It seems this approach to difference equations may be useful for studying the case that
the variable x is a discrete one. We have some experience in the theory of discrete
equations [–], and we hope that we can be successful in this situation also. Moreover, in
our opinion the developed methods might be applicable for multidimensional difference
equations.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Acknowledgements
The authors are very grateful to the anonymous referees for their valuable suggestions. This work is supported by Russian
Fund of Basic Research and government of Lipetsk region of Russia, project No. 144103595a.
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