On Global Solutions of Two-Dimensional Hyperbolic Equations with General-Kind Nonlocal Potentials

Abstract

In the case of one spatial independent variable, we study hyperbolic differential-difference equations with potentials represented as linear combinations of translations of the desired function along the spatial variable. The qualitative novelty of this investigation is that, unlike previous research, it is not assumed that the real part of the symbol of the differential-difference operator contained in the equation has a constant sign. Previously, it was possible to remove that substantial restriction (i.e., the specified sign constancy) only for the case where the nonlocal term (i.e., the translated potential) is unique. In the present paper, we consider the case of the general-kind one-variable nonlocal potential, i.e., equations with an arbitrary amount of translated terms. No commensurability assumptions are imposed on the translation lengths. The following results are presented: We find a condition relating the coefficients at the nonlocal terms of the investigated equation and the length of the translations, providing the global solvability of the investigated equation. Under this condition, we explicitly construct a three-parametric family of smooth global solutions of the investigated equation.

1. Introduction

This paper is devoted to hyperbolic equations with potentials undergoing spatial translations, i.e., hyperbolic differential-difference equations containing sums of differential operators and translation operators. For the motivation to study differential-difference equations, i.e., equations containing translation operators acting (in addition to differential ones) on the desired function, readers are referred to [1,2,3,4] (see also the references therein). In particular, among the applications of partial differential-difference equations are the theory of multi-layer plates and envelopes (see [1,5]); nonclassical diffusion, including biomathematical applications (see [6,7,8]); and nonlinear optics (see [9,10,11,12,13,14]).

In [15], global solutions of the equation

$$\frac{{\partial }^{2}u}{\partial t^2}-\frac{{\partial }^{2}u}{\partial x^2}=\sum _{k=1}^m{a}_{k}u(x+h_k,t),$$

(1)

where $a_1,\dots ,a_m$ and $h_1,\dots ,h_m$ are real constants, are obtained under the assumption that

$${\xi }^2-\sum _{k=1}^m{a}_{k}cos{h}_k\xi >0\mathrm{on}(-\infty ,+\infty ),$$

(2)

which implies, for instance, the negativity of ${\displaystyle \sum _{k=1}^m}{a}_k$. In [16,17], the investigation is extended to more general classes of hyperbolic equations with nonlocal potentials, but requirements such as those in (2), which involve the sign constancy of the real part of the symbol of the differential-difference operator with respect to the spatial variable, are still imposed. Actually, they mean the following restriction: we can add nonlocal terms to the classical differential hyperbolic operator, but the obtained differential-difference operator should remain hyperbolic. Here, hyperbolicity is treated in the following (quite natural) way: we say that an operator ${\displaystyle \frac{{\partial }^2}{\partial t^2}-L}$ is hyperbolic if the operator L is elliptic, while a differential-difference operator L is said to be elliptic if the real part of its symbol is nonpositive everywhere.

This explains the importance of the question of whether the above restriction can be removed: it is reasonable to find out whether properties specific to the hyperbolic type are preserved even if the investigated equation stops being hyperbolic. For the first time (to the best of the author’s knowledge), the said restriction is removed in [18], where the prototype case of Equation (1) with a unique nonlocal term is considered. We allow the coefficient at the said nonlocal term (translated potential) to be positive, but we are still able to construct smooth global solutions of the investigated equation. In the present paper, we pass to the case of general-kind potentials: the number of nonlocal terms in the equation is arbitrary, no commensurability of translations is required, and the sum ${\displaystyle \sum _{k=1}^m}{a}_k=:a_0$ is allowed to be non-negative. Thus, the real part of the symbol of the differential-difference operator might change its sign, but we still succeed in constructing a family of smooth global solutions.

2. Results and Proofs

Let $a_0\ge 0$ and $h\in {\mathbb{R}}^m$. In $\left\{(x,t)|x\in (-\infty ,+\infty ),t\in (-\infty ,+\infty )\right\}=:{\mathbb{R}}^2$, consider Equation (1). Assuming that

$$\sum _{k=1}^m\left|a_k\right|h_k^2\le 2,$$

(3)

consider the one-variable function $f\left(\xi \right):={\xi }^2-{\displaystyle \sum _{k=1}^m}{a}_{k}cos{h}_k\xi$; due to its evenness, it suffices to consider it on the positive semi-axis. Its derivative is equal to

$$2\xi +\sum _{k=1}^m{a}_k{h}_{k}sin{h}_k\xi =2\xi \left(1+\sum _{k=1}^m\frac{a_k{h}_k^2}{2}\frac{sin{h}_k\xi }{h_k\xi }\right),$$

i.e., the continuous function f monotonously increases on $(0,+\infty ).$ Further, $f\left(0\right)=-a_0\le 0$ and $f\left(\xi \right)>0$ for each $\xi$ exceeding $\sqrt{{\displaystyle \sum _{k=1}^m}|a_k|}=:a_{*}$. Thus, the function f has one and only one positive zero, and it belongs to $[0,a_{*}].$ Denote this zero by ${\xi }_0$.

Denote ${\displaystyle \sum _{k=1}^m}{a}_{k}cos{h}_k\xi$ and ${\displaystyle \sum _{k=1}^m}{a}_{k}sin{h}_k\xi$ by $a\left(\xi \right)$ and $b\left(\xi \right)$, respectively, and introduce the function

$$\phi \left(\xi \right)=\frac{1}{2}arctan\frac{b\left(\xi \right)}{|{\xi }^2-a\left(\xi \right)|}.$$

(4)

It is well defined in $(-\infty ,-{\xi }_0)\cup (-{\xi }_0,{\xi }_0)\cup ({\xi }_0,\infty )$ provided that ${\xi }_0\ne 0$ and is well defined in $(-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty )={\mathbb{R}}^1\backslash \left\{0\right\}$ otherwise.

Also, introduce the function

$$\rho \left(\xi \right)={\left({[{\xi }^2-a\left(\xi \right)]}^2+b^2\left(\xi \right)\right)}^{\frac{1}{4}}$$

(5)

that is well defined in $(-\infty ,\infty )$.

The following assertions are valid.

Theorem 1. 

If inequality (3) holds, then each function

$$\alpha F(x,t;\xi )+\beta H(x,t;\xi ),$$

(6)

where

$$F(x,t;\xi )=e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right],$$

(7)

$$H(x,t;\xi )=e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right],$$

(8)

and

$$\begin{array}{c}{G}_1\left(\xi \right)=\rho \left(\xi \right)sin\phi \left(\xi \right),\\ G_2\left(\xi \right)=\rho \left(\xi \right)cos\phi \left(\xi \right),\end{array}$$

(9)

is an infinitely smooth solution of Equation (1) for each ξ from $(-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty ),$ each real $\alpha ,$ and each real β.

Proof. 

Assuming that $\xi \in (-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty )$, we have,

$${\displaystyle \phi \left(\xi \right)=\frac{1}{2}arctan\frac{b\left(\xi \right)}{{\xi }^2-a\left(\xi \right)},}$$

and then substitute function (7) into Equation (1):

$$\frac{\partial F}{\partial t}=G_1\left(\xi \right)e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+G_2\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right],$$

$$\begin{array}{c}\frac{{\partial }^{2}F}{\partial t^2}=G_1^2\left(\xi \right)e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+G_1\left(\xi \right)G_2\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ +G_1\left(\xi \right)G_2\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]-G_2^2\left(\xi \right)e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ =\left[G_1^2\left(\xi \right)-G_2^2\left(\xi \right)\right]e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+2G_1\left(\xi \right)G_2\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right],\end{array}$$

and

$$\frac{{\partial }^{2}F}{\partial x^2}=-{\xi }^2{e}^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right].$$

Further, $G_1^2\left(\xi \right)-G_2^2\left(\xi \right)={\rho }^2\left(\xi \right){sin}^2\phi \left(\xi \right)-{\rho }^2\left(\xi \right){cos}^2\phi \left(\xi \right)=-{\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)$ and $2G_1\left(\xi \right)G_2\left(\xi \right)={\rho }^2\left(\xi \right)sin2\phi \left(\xi \right)$.

Now, we note that ${\displaystyle -\frac{\pi }{4}<\phi \left(\xi \right)<\frac{\pi }{4}}$ on ${\mathbb{R}}^1$ by definition. Then, ${\displaystyle 2\phi \left(\xi \right)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)}$ on ${\mathbb{R}}^1$; therefore, the function $cos2\phi \left(\xi \right)$ is positive everywhere. Then,

$$cos2\phi \left(\xi \right)=\frac{1}{\sqrt{1+{tan}^{2}2\phi \left(\xi \right)}}={\left[1+\frac{b^2\left(\xi \right)}{|{\xi }^2{-a\left(\xi \right)|}^2}\right]}^{-\frac{1}{2}}=\sqrt{\frac{{\left[{\xi }^2-a\left(\xi \right)\right]}^2}{{\left[{\xi }^2-a\left(\xi \right)\right]}^2+b^2\left(\xi \right)}}.$$

Under the assumptions of the theorem, ${\xi }^2-a\left(\xi \right)$ is positive because $\left|\xi \right|>{\xi }_0.$ Hence, the denominator of the last fraction is positive a fortiori. Thus, the last relation is meaningful under the assumptions of the theorem. Further, the denominator of the last fraction is equal to ${\rho }^4\left(\xi \right)$. Therefore,

$${\displaystyle cos2\phi \left(\xi \right)=\frac{\sqrt{{\left[{\xi }^2-a\left(\xi \right)\right]}^2}}{{\rho }^2\left(\xi \right)}.}$$

Since $\xi \in (-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty )$, it follows that the function ${\xi }^2-a\left(\xi \right)$ is positive. Hence,

$${\displaystyle {\displaystyle cos2\phi \left(\xi \right)=\frac{{\xi }^2-a\left(\xi \right)}{{\rho }^2\left(\xi \right)}}}$$

and, therefore,

$${\displaystyle sin2\phi \left(\xi \right)=tan2\phi \left(\xi \right)cos2\phi \left(\xi \right)=\frac{b\left(\xi \right)}{|{\xi }^2-a\left(\xi \right)|}\frac{{\xi }^2-a\left(\xi \right)}{{\rho }^2\left(\xi \right)}=\frac{b\left(\xi \right)}{{\xi }^2-a\left(\xi \right)}\frac{{\xi }^2-a\left(\xi \right)}{{\rho }^2\left(\xi \right)}=\frac{b\left(\xi \right)}{{\rho }^2\left(\xi \right)}.}$$

Thus,

$$\begin{array}{c}\frac{{\partial }^{2}F}{\partial t^2}-\frac{{\partial }^{2}F}{\partial x^2}=-{\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ +{\rho }^2\left(\xi \right)sin2\phi \left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+{\xi }^2{e}^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ =\left[a\left(\xi \right)-{\xi }^2\right]e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ +b\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+{\xi }^2{e}^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ =e^{t{G}_1\left(\xi \right)}\left(a\left(\xi \right)sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+b\left(\xi \right)cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\right)\\ =e^{t{G}_1\left(\xi \right)}\left({\displaystyle \sum _{k=1}^m}{a}_{k}cos{h}_k\xi sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+{\displaystyle \sum _{k=1}^m}{a}_{k}sin{h}_k\xi cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\right)\\ =e^{t{G}_1\left(\xi \right)}{\displaystyle \sum _{k=1}^m}{a}_k\left(cos{h}_k\xi sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+sin{h}_k\xi cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\right)\\ =e^{t{G}_1\left(\xi \right)}{\displaystyle \sum _{k=1}^m}{a}_{k}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi +h_k\xi \right]=e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+(x+h_k)\xi \right]={\displaystyle \sum _{k=1}^m}{a}_{k}F(x+h_k,t),\end{array}$$

(10)

i.e., function (7) satisfies Equation (1) for each $\xi$ from $(-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty )$ (in the classical sense).

Now, substitute function (8) into Equation (1):

$$\frac{\partial H}{\partial t}=-G_1\left(\xi \right)e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]+G_2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right],$$

$$\begin{array}{c}\frac{{\partial }^{2}H}{\partial t^2}=G_1^2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-G_1\left(\xi \right)G_2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ -G_1\left(\xi \right)G_2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-G_2^2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ =\left[G_1^2\left(\xi \right)-G_2^2\left(\xi \right)\right]e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-2G_1\left(\xi \right)G_2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ =-{\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-{\rho }^2\left(\xi \right)sin2\phi \left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ =\left[a\left(\xi \right)-{\xi }^2\right]e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-b\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right],\end{array}$$

and

$$\frac{{\partial }^{2}H}{\partial x^2}=-{\xi }^2{e}^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right].$$

Hence,

$$\begin{array}{c}\frac{{\partial }^{2}H}{\partial t^2}-\frac{{\partial }^{2}H}{\partial x^2}=e^{-t{G}_1\left(\xi \right)}\left(a\left(\xi \right)sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-b\left(\xi \right)cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\right)\\ =e^{-t{G}_1\left(\xi \right)}\left({\displaystyle \sum _{k=1}^m}{a}_{k}cos{h}_k\xi sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-{\displaystyle \sum _{k=1}^m}{a}_{k}sin{h}_k\xi cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\right)\\ =e^{-t{G}_1\left(\xi \right)}{\displaystyle \sum _{k=1}^m}{a}_k\left(cos{h}_k\xi sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]-sin{h}_k\xi cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\right)\\ =e^{-t{G}_1\left(\xi \right)}{\displaystyle \sum _{k=1}^m}{a}_{k}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi -h_k\xi \right]=e^{-t{G}_1\left(\xi \right)}{\displaystyle \sum _{k=1}^m}{a}_{k}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-(x+h_k)\xi \right]\\ ={\displaystyle \sum _{k=1}^m}{a}_{k}H(x+h_k,t),\end{array}$$

(11)

i.e., function (8) satisfies Equation (1) for each $\xi \in (-\infty ,-{\xi }_0)\cup ({\xi }_0,\infty )$ (in the classical sense). □

Theorem 2. 

If ${\xi }_0\ne 0$ and Condition (3) is satisfied, then each function (6), where $F(x,t;\xi )$ and $H(x,t;\xi )$ are introduced by relations (7) and (8) respectively, with

$$\begin{array}{c}{G}_1\left(\xi \right)=\rho \left(\xi \right)cos\phi \left(\xi \right),\\ G_2\left(\xi \right)=\rho \left(\xi \right)sin\phi \left(\xi \right),\end{array}$$

(12)

is an infinitely smooth solution of Equation (1) for each real $\alpha ,$ each real $\beta ,$ and each ξ from $(-{\xi }_0,{\xi }_0)$.

Proof. 

Assuming that $\xi \in (-{\xi }_0,{\xi }_0)$, and therefore ${\displaystyle \phi \left(\xi \right)=\frac{1}{2}arctan\frac{b\left(\xi \right)}{a\left(\xi \right)-{\xi }^2}}$, we conclude that ${\displaystyle \phi \left(\xi \right)\in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)}$, i.e., ${\displaystyle 2\phi \left(\xi \right)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)}$ on ${\mathbb{R}}^1$. Consequently, the function $cos2\phi \left(\xi \right)$ is positive everywhere. Therefore,

$$\begin{array}{c}cos2\phi \left(\xi \right)=\frac{1}{\sqrt{1+{tan}^{2}2\phi \left(\xi \right)}}={\left(1+\frac{b^2\left(\xi \right)}{{[a\left(\xi \right)-{\xi }^2]}^2}\right)}^{-\frac{1}{2}}=\sqrt{\frac{{\left[a\left(\xi \right)-{\xi }^2\right]}^2}{{\left[a\left(\xi \right)-{\xi }^2\right]}^2+b^2\left(\xi \right)}}.\end{array}$$

Under the assumptions of the theorem, $a\left(\xi \right)-{\xi }^2$ is positive because $\left|\xi \right|<{\xi }_0.$ Hence, the denominator of the last fraction is positive a fortiori. Thus, the last relation is meaningful under the assumptions of the theorem. Further, the denominator of the last fraction is equal to ${\rho }^4\left(\xi \right)$. Therefore,

$${\displaystyle cos2\phi \left(\xi \right)=\frac{\sqrt{{\left[a\left(\xi \right)-{\xi }^2\right]}^2}}{{\rho }^2\left(\xi \right)}.}$$

Since the function $a\left(\xi \right)-{\xi }^2$ is positive, it follows that

$$\begin{array}{c}cos2\phi \left(\xi \right)=\frac{a\left(\xi \right)-{\xi }^2}{{\rho }^2\left(\xi \right)},\\ {\displaystyle sin2\phi \left(\xi \right)=tan2\phi \left(\xi \right)cos2\phi \left(\xi \right)=\frac{b\left(\xi \right)}{a\left(\xi \right)-{\xi }^2}\frac{a\left(\xi \right)-{\xi }^2}{{\rho }^2\left(\xi \right)}=\frac{b\left(\xi \right)}{{\rho }^2\left(\xi \right)}.}\end{array}$$

(13)

Now, taking into account that, unlike the case of Theorem 1, $G_1^2\left(\xi \right)-G_2^2\left(\xi \right)={\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)$, and substituting function (7) into Equation (1), we have

$$\begin{array}{c}\frac{{\partial }^{2}F}{\partial t^2}-\frac{{\partial }^{2}F}{\partial x^2}={\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ +{\rho }^2\left(\xi \right)sin2\phi \left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+{\xi }^2{e}^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ =\left[a\left(\xi \right)-{\xi }^2\right]e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]+b\left(\xi \right)e^{t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right]\\ +{\xi }^2{e}^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right],\end{array}$$

which coincides with (10). Therefore, function (7) satisfies Equation (1) (in the classical sense).

Function (8) is substituted into Equation (1) in the same way. As above,

$$\begin{array}{c}\frac{{\partial }^{2}H}{\partial t^2}=\left[G_1^2\left(\xi \right)-G_2^2\left(\xi \right)\right]e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ -2G_1\left(\xi \right)G_2\left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right],\end{array}$$

but now $G_1^2\left(\xi \right)-G_2^2\left(\xi \right)=cos2\phi \left(\xi \right)$ (under the assumptions of Theorem 2). Therefore,

$$\begin{array}{c}\frac{{\partial }^{2}H}{\partial t^2}={\rho }^2\left(\xi \right)cos2\phi \left(\xi \right)e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ -{\rho }^2\left(\xi \right)sin2\phi \left(\xi \right)e^{-t{G}_1\left(\xi \right)}cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right].\end{array}$$

Now, taking into account relation (13), we conclude that

$$\begin{array}{c}\frac{{\partial }^{2}H}{\partial t^2}-\frac{{\partial }^{2}H}{\partial x^2}=e^{-t{G}_1\left(\xi \right)}\left(\left[a\left(\xi \right)-{\xi }^2\right]sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\right.\\ \left.-b\left(\xi \right)cos\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\right)+{\xi }^2{e}^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right]\\ ={\displaystyle \sum _{k=1}^m}{a}_{k}H(x+h_k,t)\end{array}$$

in the same way as in (11).

Thus, function (8) satisfies Equation (1) (in the classical sense) as well. □

Remark 1. 

If $b\left({\xi }_0\right)\ne 0,$ then function (4) can be defined (as a continuous function) at the point ${\xi }_0$: its left- and right-side limits at that point are equal to ${\displaystyle \frac{\pi }{4}\mathit{sgn}\left[b\left({\xi }_0\right)\right]}$. If $b\left({\xi }_0\right)=0,$ then ${\xi }_0$ might be a discontinuity point for function (4), and at most a finite jump is possible at that point. Thus, each global solution constructed above is continuous with respect to ξ, unless ${\displaystyle \sum _{k=1}^m}{a}_{k}sin{h}_k{\xi }_0=0;$ if the said continuity is violated, then only a denumerable amount of discontinuity points is possible, and they all have finite jumps.

3. Heuristic Considerations

As seen above, Theorems 1 and 2 are proved directly: we just take function (6) and substitute it into Equation (1). Such a proof is strict and clear, but it does not explain how to find the solution. In this section, we show how to apply the well-known Gel’fand–Shilov operational scheme (see, e.g., [19] (Sec. 10)) in the considered case.

Formally applying the Fourier transformation with respect to the variable x to Equation (1), which is a partial differential-difference equation, we obtain the following ordinary differential equation depending on the parameter $\xi$:

$$\frac{d^2\widehat{u}}{d{t}^2}+\left[{\left|\xi \right|}^2-a\left(\xi \right)-ib\left(\xi \right)\right]\widehat{u}=0.$$

(14)

Its general solution (up to arbitrary constants depending on the parameter $\xi$) is equal to

$$\frac{1}{\rho \left(\xi \right)}\left(e^{-t{G}_1\left(\xi \right)}{e}^{i[t{G}_2\left(\xi \right)-\phi \left(\xi \right)]}-e^{t{G}_1\left(\xi \right)}{e}^{-i[t{G}_2\left(\xi \right)+\phi \left(\xi \right)]}\right),$$

where the functions $\phi \left(\xi \right)$ and $\rho \left(\xi \right)$ are defined by relations (4) and (5), respectively. The functions $G_1\left(\xi \right)$ and $G_2\left(\xi \right)$ are defined by relation (12) in the interval $(-{\xi }_0,{\xi }_0)$ (if it exists) and by relation (9) outside it.

Now, it remains to (formally) apply the inverse Fourier transformation to eliminate terms with odd integrands and choose the arbitrary constants depending on the parameter $\xi$ such that purely imaginary terms are eliminated. Note that the remaining actions of the Gel’fand–Shilov procedure cannot be performed in our case because no convergence of the arising improper integral with respect to $\xi$ is guaranteed. However, if we truncate the Gel’fand–Shilov procedure before integrating with respect to the dual variable $\xi$ and treat that variable as a parameter, then the obtained function, which is represented by (6), satisfies Equation (1) for each value of that parameter apart from $\pm {\xi }_0$. To verify this, we substitute it into Equation (1), as shown in in Section 2.

4. Novel Insights

As noted above, the fundamental novelty of the presented results is as follows: previously, the real part of the symbol of the differential-difference operator acting with respect to spatial variables was always required to be of a constant sign. This is not merely a technical restriction. In the classical case of differential equations, the sign constancy of the symbol is the criterion of ellipticity. For differential-difference equations and operators, no canonical classification has been established yet, but it is rather conventional to treat an operator ${\displaystyle \frac{{\partial }^2}{\partial t^2}-L}$ as hyperbolic if the differential-difference operator L is elliptic.

The case of differential-difference operators is substantially more complicated than the classical case of differential ones. This is because of the polynomial nature of symbols of differential operators. Moreover, second-order homogeneous differential operators reduced to the canonical form are most frequently studied in the classical case; their symbols are just quadratic forms. In the case of an elliptic operator (i.e., if the operator has no real characteristics), the corresponding quadratic form maintains a constant sign. However, if the operator is hyperbolic, the sign of the corresponding quadratic form varies, and its sign-constancy sets are divided by the conical surface (pair of lines in the case of one-dimensional spatial variable), which is the only real characteristic of the operator. When studying differential-difference operators, their symbols stop being polynomials. This causes a qualitatively greater diversity, allowing us to omit the sign-constancy requirement for the symbol.

Assumptions on the symbol are still imposed in the investigated case, but their nature is fundamentally changed. Instead of requiring sign constancy, we now require the symbol to be monotonous. It should be noted that no restrictions of this kind arise in the classical theory of differential equations.

5. Examples

Example 1. 

Consider the equation

$$\frac{{\partial }^{2}u}{\partial t^2}-\frac{{\partial }^{2}u}{\partial x^2}=100u(x+0.03,t)-0.01u(x-10,t).$$

(15)

Here, $m=2$, $a_1=100$, $a_2=-0.01$, $h_1=0.03$, and $h_2=-10$. To verify the fulfillment of Condition (3), one has to compute

$$|a_1|h_1^2+\left|a_2\right|h_2^2=100·0.0009+0.01·100=1.09.$$

This value does not exceed two, which guarantees the existence and uniqueness of a non-negative zero of the function $f\left(\xi \right)={\xi }^2-100cos0.03\xi +0.01cos10\xi$. This zero, denoted as ${\xi }_0$, does not exceed $\sqrt{100.01}$, and it divides the real axis into the following three (two if ${\xi }_0=0$) intervals of sign constancy of the function $f:$ $(-\infty ,-{\xi }_0),$ $(-{\xi }_0,{\xi }_0),$ and $({\xi }_0,+\infty )$. Thus, function (6), constructed under the assumption that $a\left(\xi \right)=100cos0.03\xi -0.01cos10\xi$ and $b\left(\xi \right)=100sin0.03\xi -0.01sin10\xi$, is a global solution of Equation (15) for each real α and β.

Example 2. 

Consider the equation

$$\frac{{\partial }^{2}u}{\partial t^2}-\frac{{\partial }^{2}u}{\partial x^2}=0.5u(x-0.5,t)+2u(x+1,t).$$

(16)

Here, $m=2$, $a_1=0.5$, $a_2=2$, $h_1=-0.5$, and $h_2=1$. Again, computing

$$|a_1|h_1^2+\left|a_2\right|h_2^2=0.5·0.25+2·1=2.125,$$

we conclude that this value exceeds two. Therefore, Condition (3) is not satisfied for Equation (16). Thus, the question about the sign-constancy intervals of the function f and therefore, about solutions of Equation (16) remains open.

6. Conclusions

In this paper, we continue the investigation of differential-difference hyperbolic equations with nonlocal potentials, removing restrictions on the signs of real (or imaginary) parts of symbols of operators contained in the investigated equations. We pass from the prototype case, investigated in [18], which involves an equation with a unique translated potential, to Equation (1), which involves scenarios with multiple potentials of varying translation lengths.

For Equation (1), we impose Condition (3), which means neither smallness of the coefficients at the nonlocal terms nor smallness of their translations are required. Under this assumption, we explicitly construct the following three-parameter family of smooth global solutions of Equation (1):

$$u(x,t)=\alpha F(x,t;\xi )+\beta H(x,t;\xi ),$$

where

$$F(x,t;\xi )=e^{t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)+\phi \left(\xi \right)+x\xi \right],$$

$$H(x,t;\xi )=e^{-t{G}_1\left(\xi \right)}sin\left[t{G}_2\left(\xi \right)-\phi \left(\xi \right)-x\xi \right],$$

$$G_1\left(\xi \right)=\left\{\begin{array}{c}\rho \left(\xi \right)sin\phi \left(\xi \right)\mathrm{for}\left|\xi \right|>{\xi }_0,\hfill \\ \rho \left(\xi \right)cos\phi \left(\xi \right)\mathrm{for}\left|\xi \right|<{\xi }_0,\hfill & \end{array}\right.$$

$$G_2\left(\xi \right)=\left\{\begin{array}{c}\rho \left(\xi \right)cos\phi \left(\xi \right)\mathrm{for}\left|\xi \right|>{\xi }_0,\hfill \\ \rho \left(\xi \right)sin\phi \left(\xi \right)\mathrm{for}\left|\xi \right|<{\xi }_0,\hfill & \end{array}\right.$$

$$\phi \left(\xi \right)=\frac{1}{2}\left\{\begin{array}{c}arctan\frac{b\left(\xi \right)}{{\xi }^2-a\left(\xi \right)}\mathrm{for}\left|\xi \right|>{\xi }_0,\hfill \\ arctan\frac{b\left(\xi \right)}{a\left(\xi \right)-{\xi }^2}\mathrm{for}\left|\xi \right|<{\xi }_0,\hfill & \end{array}\right.$$

$$\rho \left(\xi \right)={\left({\left[{\xi }^2-a\left(\xi \right)\right]}^2+b^2\left(\xi \right)\right)}^{\frac{1}{4}},$$

where ${\xi }_0$ is the only positive root of the equation ${\xi }^2=a\left(\xi \right),$ $\alpha \in {\mathbb{R}}^1,$ $\beta \in {\mathbb{R}}^1,$ and $\xi$ belongs to ${\mathbb{R}}^1\backslash \{-{\xi }_0,{\xi }_0\}$.