We consider the equation
$$ (\mathcal{D}u) (x)=v(x), \quad x\in{\mathbb{R}}_{+}, $$
(6)
where \({\mathbb{R}}_{+}=\{x\in{\mathbb{R}}, x>0\}\).
For studying this equation we will use methods of the theory of multi-dimensional singular integral and pseudo differential equations [3, 6, 12] which are non-usual in the theory of difference equations. Our next goal is to study multi-dimensional difference equations, and this one-dimensional variant is a model for considering other complicated situations. This approach is based on the classical Riemann boundary value problem and the theory of one-dimensional singular integral equations [13–15].
Background
The first step is the following. We will use the theory of so-called paired equations [15] of the type
$$ (aP_{+}+bP_{-})U=V $$
(7)
in the space \(L_{2}(\mathbb{R})\), where a, b are convolution operators with corresponding functions \(a(x)\), \(b(x)\), \(x\in{\mathbb{R}}\), \(P_{\pm }\) are projectors on the half-axis \({\mathbb{R}}_{\pm}\). More precisely,
$$(aP_{+}U) (x)=\int_{0}^{+\infty}a(x-y)U(y)\, dy, \qquad (bP_{-}U) (x)=\int_{-\infty}^{0}b(x-y)U(y)\, dy. $$
Applying the Fourier transform to (7) we obtain [12] the following one-dimensional singular integral equation [13–15]:
$$ \tilde{a}(\xi) (P\widetilde{U}) (\xi)+\tilde{b}(\xi) (Q\widetilde {U}) (\xi)=\widetilde{V}(\xi), $$
(8)
where P, Q are two projectors related to the Hilbert transform
$$\begin{aligned}& (Hu) (x)=\mathrm{v.p.}\, \frac{1}{\pi i}\int_{-\infty}^{+\infty} \frac {u(y)}{x-y}\, dy, \\& P=\frac{1}{2}(I+H), \qquad Q=\frac{1}{2}(I-H). \end{aligned}$$
Equation (8) is closely related to the Riemann boundary value problem [13, 14] for upper and lower half-planes. We now recall the statement of the problem: finding a pair of functions \(\Phi^{\pm}(\xi )\) which admit an analytic continuation on upper (\({\mathbb{C}}_{+}\)) and lower (\({\mathbb{C}}_{-}\)) half-planes in the complex plane \(\mathbb {C}\) and of which their boundary values on \(\mathbb{R}\) satisfy the following linear relation:
$$ \Phi^{+}(\xi)=G(\xi)\Phi^{-}(\xi)+g(\xi), $$
(9)
where \(G(\xi)\), \(g(\xi)\) are given functions on \(\mathbb{R}\).
There is a one-to-one correspondence between the Riemann boundary value problem (9) and the singular integral equation (8), and
$$G(\xi)=\tilde{a}^{-1}(\xi)\tilde{b}(\xi), \qquad \widetilde{V}(\xi )= \tilde{a}^{-1}(\xi)g(\xi). $$
Topological barrier
We suppose that the symbol \(G(\xi)\) is a continuous non-vanishing function on the compactification \(\dot{\mathbb{R}}\) (\(G(\xi)\neq0\), \(\forall\xi\in\dot{\mathbb{R}}\)) and
$$ \operatorname{Ind} G\equiv\frac{1}{2\pi}\int _{-\infty}^{+\infty }d\arg G(t)=0. $$
(10)
The last condition (10), is necessary and sufficient for the unique solvability of the problem (9) in the space \(L_{2}({\mathbb{R}})\) [13, 14]. Moreover, the unique solution of the problem (9) can be constructed with a help of the Cauchy type integral
$$\Phi^{+}(t)=G_{+}(t)P \bigl(G_{+}^{-1}(t)g(t) \bigr),\qquad \Phi ^{-}(t)=-G_{-}^{-1}(t)Q \bigl(G_{+}^{-1}(t)g(t) \bigr), $$
where \(G_{\pm}\) are factors of a factorization for the \(G(t)\) (see below),
$$G_{+}(t)=\exp \bigl(P \bigl(\ln G(t) \bigr) \bigr),\qquad G_{-}(t)=\exp \bigl(Q \bigl(\ln G(t) \bigr) \bigr). $$
Difference equations on a half-axis
Equation (6) can easily be transformed into (7) in the following way. Since the right-hand side in (6) is defined on \(\mathbb{R}_{+}\) only we will continue \(v(x)\) on the whole \({\mathbb {R}}\) so that this continuation \(\mathit{lf}\in L_{2}({\mathbb{R}})\). Further we will rename the unknown function \(u_{+}(x)\) and define the function
$$u_{-}(x)=(\mathit{lf}) (x)-(\mathcal{D}u_{+}) (x). $$
Thus, we have the following equation:
$$ (\mathcal{D}u_{+}) (x)+u_{-}(x)=(\mathit{lf}) (x),\quad x\in \mathbb{R}, $$
(11)
which holds for the whole space \({\mathbb{R}}\).
After the Fourier transform we have
$$\sigma(\xi)\tilde{u}_{+}(\xi)+\tilde{u}_{-}(\xi)=\widetilde {\mathit{lv}}(\xi), $$
where \(\sigma(\xi)\) is called a symbol of the operator \(\mathcal{D}\).
To describe a solving technique for (11) we recall the following (cf. [13, 14]).
Definition
A factorization for an elliptic symbol is called its representation if it is in the form
$$\sigma(\xi)=\sigma_{+}(\xi)\cdot\sigma_{-}(\xi), $$
where the factors \(\sigma_{+}\), \(\sigma_{-}\) admit an analytic continuation into the upper and lower complex half-planes \({\mathbb{C}}_{\pm}\), and \(\sigma^{\pm1}_{\pm}\in L_{\infty}({\mathbb{R}})\).
Example 1
Let us consider the Cauchy type integral
$$\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac {u(t)}{z-t}\, dt\equiv\Phi(z), \quad z=x\pm iy. $$
It is well known this construction plays a crucial role for a decomposition \(L_{2}({\mathbb{R}})\) on two orthogonal subspaces, namely
$$L_{2}({\mathbb{R}})=A_{+}({\mathbb{R}})\oplus A_{-}({\mathbb{R}}), $$
where \(A_{\pm}({\mathbb{R}})\) consists of functions admitting an analytic continuation onto \({\mathbb{C}}_{\pm}\).
The boundary values of the integral \(\Phi(z)\) satisfy the Plemelj-Sokhotskii formulas [13, 14], and thus the projectors P and Q are corresponding projectors on the spaces of analytic functions [15].
The simple example we need is
$$\exp(u)=\exp(Pu)\cdot\exp(Qu). $$
Theorem 2
Let
\(\sigma(\xi)\in C(\dot{\mathbb{R}})\), \(\operatorname{Ind}\sigma =0 \). Then (6) has unique solution in the space
\(L_{2}({\mathbb {R}}_{+})\)
for arbitrary right-hand side
\(v\in L_{2}({\mathbb{R}}_{+})\), and its Fourier transform is given by the formula
$$\tilde{u}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta. $$
Proof
We have
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi)= \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi). $$
Further, since \(\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)\in L_{2}({\mathbb{R}})\) we decompose it into two summands
$$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)=P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr)+Q \bigl( \sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi) \bigr) $$
and write
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde { \mathit{lv}}(\xi) \bigr)-\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi). $$
The left-hand side of the last quality belongs to the space \(A_{+}({\mathbb{R}})\), but the right-hand side belongs to \(A_{-}({\mathbb {R}})\), consequently these are zeros. Thus,
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=0 $$
and
$$ \tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr), $$
(12)
or in the complete form
$$\tilde{u}_{+}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta. $$
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Remark 1
This result does not depend on the continuation lv. Let us denote by \(M_{\pm}(x)\) the inverse Fourier images of the functions \(\sigma ^{-1}_{\pm}(\xi)\). Indeed, (12) leads to the following construction:
$$\begin{aligned} u_{+}(x)&=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t) (\mathit{lv}) (t)\, dt \biggr)\, dy \\ &=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t)v(t)\, dt \biggr)\, dy. \end{aligned}$$
Remark 2
The condition \(\sigma(\xi)\in C(\dot{\mathbb{R}})\) is not a strong restriction. Such symbols exist for example in the case that \(\sigma (\xi)\) is represented by a finite sum, and \(\beta_{k}\in{\mathbb {Q}}\). Then \(\sigma(\xi)\) is a continuous periodic function.