PARTIAL DIFFERENTIAL EQUATIONS
Global Classical Solutions of Some Two-Dimensional Hyperbolic Differential-Difference Equations
N. V. Zaitseva
Lomonosov Moscow State University, Moscow, 119991 Russia e-mail: [email protected], [email protected]
Received December 4, 2019; revised February 26, 2020; accepted February 27, 2020
Abstract— For a two-dimensional hyperbolic equation with a differential-difference operator acting with respect to the spatial variable, we construct a one-parameter family of global solu- tions. We prove that these solutions are classical for all values of the real parameter provided that the real part of the symbol of the difference operator occurring in the equation is positive.
We present a class of equations for which this condition is satisfied.
DOI: 10.1134/S0012266120060063
INTRODUCTION
The interest in equations with deviating arguments, in particular, differential-difference equa- tions, arises when studying problems from various applications for which the classical models of mathematical physics using only differential equations prove insufficient (see, e.g., [1–4] and the bib- liography therein). By now, problems in bounded domains for differential-difference equations have been studied rather comprehensively (see [1–4]). In unbounded domains, problems for parabolic (see [5]) and elliptic (see [6–10]) differential-difference equations have been studied. Hyperbolic differential-difference equations have been studied for the case in which the shift operators occur- ring in the equation act with respect to the variablet (see [11, 12] and the references therein).
The aim of the present papers is to cover hyperbolic differential-difference equations with shift operators acting with respect to the spatial variables.
In the half-plane (x, t) ∈R1× (0, +∞), consider the equation
∂2u(x, t)
∂t2 = a1∂2u(x − h1, t)
∂x2 + a2∂2u(x − h2, t)
∂x2 , (1)
whereaj and hj (j = 1, 2) are given real numbers, no commensurability conditions being imposed on the parametersh1 and h2.
To find solutions of Eq. (1), we use the classical Gelfand–Shilov operational scheme (see [13]).
Generally speaking, this scheme leads to solutions in the sense of generalized functions, but in our case one can prove that the resulting solutions are classical; i.e., they are functions all of whose derivatives occurring in the equation (these derivatives are understood in the classical sense as limits of finite difference ratios) exist at each point of the half-plane (x, t) ∈ R1× (0, +∞), and Eq. (1) is satisfied for these derivatives at each point of this half-plane.
Thus, along with Eq. (1), consider the equation
∂2E(x, t)
∂t2 − a1∂2E(x − h1, t)
∂x2 − a2∂2E(x − h2, t)
∂x2 = δ(x, t), (2)
whereδ(x, t) is the Dirac delta function.
Applying the Fourier transform Fx := F to relation (2), we pass to the dual variable ξ and, taking into account the fact that the translation operator, as well as differential operators, is a Fourier multiplier, namely,F [f(x − h)] = e−ihξF [f], we obtain (according to the cited operational scheme) the initial value problem
Z(t) + (a1e−ih1ξ+ a2e−ih2ξ)ξ2Z(t) = 0, Z(0) = 0, Z(0) = 1,
whose solution is given by the function
Z(t) = sin√
a1eih1ξ + a2eih2ξξt
√a1eih1ξ+ a2eih2ξξ . (3)
We introduce the functions ρ(ξ) :=
a21+ a22+ 2a1a2cos ((h1− h2)ξ)1/4
(4) and
θ(ξ) := 1
2arctan a1sin (h1ξ) + a2sin (h2ξ)
a1cos (h1ξ) + a2cos (h2ξ) (5) (note that the function (4) is well defined for all real values ofaj,hj(j = 1, 2), and ξ) and transform the radicand in the function (3) as follows:
a1eih1ξ+ a2eih2ξ = (a1cos (h1ξ) + a2cos (h2ξ)) − i(a1sin (h1ξ) + a2sin (h2ξ))
=
a21+ a22+ 2a1a2cos ((h1− h2)ξ) exp
−i arctan a1sin (h1ξ) + a2sin (h2ξ) a1cos (h1ξ) + a2cos (h2ξ)
= ρ2(ξ)e−2iθ(ξ). The function (3) thus acquires the form
Z(t) = sin
tξρ(ξ)e−iθ(ξ) ξρ(ξ)e−iθ(ξ) .
Then one can conclude that smooth solutions to Eq. (1) must be sought in the form G(x, t; ξ) := sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx)eρ(ξ)ξt sin θ(ξ)
+ sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx)e−ρ(ξ)ξt sin θ(ξ). (6) We prove below that the last formula gives a family of smooth solutions of Eq. (1).
1. ONE-PARAMETER FAMILY OF SMOOTH SOLUTIONS OF EQ. (1) The following assertion holds.
Theorem. If the inequality
a1cos (h1ξ) + a2cos (h2ξ) > 0, (7) holds for each real ξ , then the function G(x, t; ξ) defined by formula (6) satisfies Eq. (1) for each real value of the parameter ξ.
Proof. Let us substitute the function (6) directly into Eq. (1). First, we calculate its derivatives with respect to the time variable,
Gt(x, t; ξ) = ρ(ξ)ξ cos (ρ(ξ)ξt cos θ(ξ) + ξx)eρ(ξ)ξt sin θ(ξ)
+ ρ(ξ)ξ cos (ρ(ξ)ξt cos θ(ξ) − ξx)e−ρ(ξ)ξt sin θ(ξ), Gtt(x, t; ξ) = −ρ2(ξ)ξ2sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) + ξx)eρ(ξ)ξt sin θ(ξ)
− ρ2(ξ)ξ2sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) − ξx)e−ρ(ξ)ξt sin θ(ξ).
The derivatives of the function G(x, t; ξ) with respect to the spatial variable are expressed by the formulas
Gx(x, t; ξ) = ξ cos (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + xξ)eρ(ξ)ξt sin θ(ξ)
− ξ cos (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − xξ)e−ρ(ξ)ξt sin θ(ξ), Gxx(x, t; ξ) = −ξ2sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + xξ)eρ(ξ)ξt sin θ(ξ)
− ξ2sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − xξ)e−ρ(ξ)ξt sin θ(ξ).
It follows that
Gx(x − hj, t; ξ) = ξ cos (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − hjξ)eρ(ξ)ξt sin θ(ξ)
− ξ cos (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + hjξ)e−ρ(ξ)ξt sin θ(ξ), while
Gxx(x − hj, t; ξ) = −ξ2sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − hjξ)eρ(ξ)ξt sin θ(ξ)
− ξ2sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + hjξ)e−ρ(ξ)ξt sin θ(ξ).
Substituting the resulting expressions for the derivatives into Eq. (1), after obvious equivalent transformations, we conclude that we need to prove the relation
ρ2(ξ)[sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) + ξx)eρ(ξ)ξt sin θ(ξ)
+ sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) − ξx)e−ρ(ξ)ξt sin θ(ξ)]
= [a1sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − h1ξ)
+ a2sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − h2ξ)]eρ(ξ)ξt sin θ(ξ)
+ [a1sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + h1ξ)
+ a2sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + h2ξ)]e−ρ(ξ)ξt sin θ(ξ).
(8)
Obviously, it suffices to prove the relations
ρ2(ξ) sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) + ξx) = a1sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − h1ξ)
+ a2sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) + ξx − h2ξ) (9) and
ρ2(ξ) sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) − ξx) = a1sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + h1ξ)
+ a2sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) − ξx + h2ξ). (10) Let us prove the former. For convenience of the subsequent calculations, we introduce the notationρ(ξ)ξt cos θ(ξ)−θ(ξ)+ξx =: ϕ = ϕ(x, t, ξ) and transform the right-hand side of relation (9) as follows:
a1sin (ϕ + (2θ(ξ) − h1ξ)) + a2sin (ϕ + (2θ(ξ) − h2ξ))
= a1[sin ϕ cos (2θ(ξ) − h1ξ) + cos ϕ sin (2θ(ξ) − h1ξ)]
+ a2[sin ϕ cos (2θ(ξ) − h2ξ) + cos ϕ sin (2θ(ξ) − h2ξ)]
= a1[sin ϕ(cos (2θ(ξ)) cos (h1ξ) + sin (2θ(ξ)) sin (h1ξ)) + cos ϕ(sin (2θ(ξ)) cos (h1ξ) − cos (2θ(ξ)) sin (h1ξ))]
+ a2[sin ϕ(cos (2θ(ξ)) cos (h2ξ) + sin (2θ(ξ)) sin (h2ξ)) + cos ϕ(sin (2θ(ξ)) cos (h2ξ) − cos (2θ(ξ)) sin (h2ξ))]
= (a1cos (h1ξ) + a2cos (h2ξ)) sin ϕ cos (2θ(ξ)) + (a1sin (h1ξ) + a2sin (h2ξ)) sin ϕ sin (2θ(ξ)) + (a1cos (h1ξ) + a2cos (h2ξ)) cos ϕ sin (2θ(ξ))
− (a1sin (h1ξ) + a2sin (h2ξ)) cos ϕ cos (2θ(ξ)).
(11)
Now, considering definitions (4) and (5) and applying the relations arctan x = arccos√ 1
1 + x2 = arcsin√ x 1 + x2,
we transform the functionscos (2θ(ξ)) and sin (2θ(ξ)) occurring in the expression (11) as follows:
cos (2θ(ξ)) = cos arctan a1sin (h1ξ) + a2sin (h2ξ) a1cos (h1ξ) + a2cos (h2ξ) =
1 +
a1sin (h1ξ) + a2sin (h2ξ) a1cos (h1ξ) + a2cos (h2ξ)
2−1/2
=
(a1cos (h1ξ) + a2cos (h2ξ))2
a21+ a22+ 2a1a2cos ((h1− h2)ξ) = |a1cos (h1ξ) + a2cos (h2ξ)|
ρ2(ξ) ,
sin (2θ(ξ)) = sin arctana1sin (h1ξ) + a2sin (h2ξ) a1cos (h1ξ) + a2cos (h2ξ)
= (a1sin (h1ξ) + a2sin (h2ξ))|a1cos (h1ξ) + a2cos (h2ξ)|
(a1cos (h1ξ) + a2cos (h2ξ))ρ2(ξ) .
According to the assumptions of the theorem, inequality (7) is satisfied for each realξ, and hence cos (2θ(ξ)) = a1cos (h1ξ) + a2cos (h2ξ)
ρ2(ξ) , sin (2θ(ξ)) = a1sin (h1ξ) + a2sin (h2ξ)
ρ2(ξ) .
We substitute the resulting expressions for cos (2θ(ξ)) and sin (2θ(ξ)) into the right-hand side of relation (11) to obtain
ρ−2(ξ)[(a1cos (h1ξ) + a2cos (h2ξ))2sin ϕ + (a1sin (h1ξ) + a2sin (h2ξ))2sin ϕ + (a1cos (h1ξ) + a2cos (h2ξ))(a1sin (h1ξ) + a2sin (h2ξ)) cos ϕ
− (a1cos (h1ξ) + a2cos (h2ξ))(a1sin (h1ξ) + a2sin (h2ξ)) cos ϕ]
= ρ−2(ξ)[(a1cos (h1ξ) + a2cos (h2ξ))2+ (a1sin (h1ξ) + a2sin (h2ξ))2] sin ϕ
= ρ−2(ξ)[a21+ a22+ 2a1a2cos ((h1− h2)ξ)] sin ϕ
= ρ−2(ξ)ρ4(ξ) sin ϕ = ρ2(ξ) sin (ρ(ξ)ξt cos θ(ξ) − θ(ξ) + ξx).
This proves relation (9).
To prove relation (10), we introduce the notation
ρ(ξ)ξt cos θ(ξ) + θ(ξ) − ξx =: ψ = ψ(x, t, ξ)
and make similar transformations of the right-hand side of this relation. As a result, we obtain the relation
a1sin (ψ − (2θ(ξ) − h1ξ)) + a2sin (ψ − (2θ(ξ) − h2ξ))
= (a1cos (h1ξ) + a2cos (h2ξ)) sin ψ cos (2θ(ξ)) + (a1sin (h1ξ)
+ a2sin (h2ξ)) sin ψ sin (2θ(ξ)) − (a1cos (h1ξ) + a2cos (h2ξ)) cos ψ sin (2θ(ξ)) + (a1sin (h1ξ) + a2sin (h2ξ)) cos ψ cos (2θ(ξ))
= ρ−2(ξ)[a21+ a22+ 2a1a2cos ((h1− h2)ξ)] sin ψ = ρ−2(ξ)ρ4(ξ) sin ψ
= ρ2(ξ) sin (ρ(ξ)ξt cos θ(ξ) + θ(ξ) − ξx),
which proves (10) and hence (8). Thus, if condition (7) holds for each realξ, then the function (6) satisfies Eq. (1) in the classical sense for each realξ.
If we relax the condition of the theorem by letting the inequality in the statement of the theorem to be violated on some subsets of the real line, then, reproducing the argument in the proof of the theorem word for word, we conclude that the following assertion holds.
Corollary. The function G(x, t; ξ) defined by formula (6) satisfies Eq. (1) for any real value of the parameter ξ for which inequality (7) holds.
Remark. If inequality (7) holds, then the denominator in the function (5) does not vanish, and hence every solution representable by formula (6) is indeed smooth under the assumptions of the Theorem and the Corollary.
2. MEANING OF CONDITION ON THE COEFFICIENTS AND EXAMPLES OF EQUATIONS SATISFYING THIS CONDITION
The differential-difference operator occurring on the right-hand side in Eq. (1) is the superposition of the differential operatorD2x and the difference operatorR that acts as follows:
Ru(x, t) = a1u(x − h1, t) + a2u(x − h2, t).
The symbol of the operator R is a1cos (h1ξ) + a2cos (h2ξ) − i(a1sin (h1ξ) + a2sin (h2ξ)); i.e., in- equality (7) implies that the real part of the symbol of the operatorR (or equivalently, the symbol of the operator R + R∗) is positive at each point ξ ∈ R. Thus, the assumption of the theorem is equivalent to saying that the real part of the symbol of the (only) difference operator occurring in Eq. (1) is positive on the entire real line (see [1, Sec. 8, 9] and [5, Sec. 1.6]).
It is hence natural to ask whether the positiveness condition for the real part of the symbol of the operatorR is satisfied.
Assume that allaj = 0 and hj = 0, j = 1, 2, in Eq. (1). By denoting hj = hj/2 (j = 1, 2) and by using the double argument formula for the cosine, we conclude that inequality (7) is equivalent to the inequality
2a1cos2 h1ξ
+ 2a2cos2 h2ξ
> a1+ a2. (12)
The following four cases are possible for the coefficientsa1 and a2.
1. Let a1 > 0 and a2 > 0. We set ξ = π/(2 h2). Then cos2( h2ξ) = 0, and cos2( h1ξ) is at most 1. For the ξ selected, since a1 > 0, the left-hand side of inequality (12) is not greater than2a1, and therefore, the inequality2a1> a1+ a2must hold; i.e.,a1> a2. Given the choice of ξ = π/(2 h1), in a similar way we obtain the opposite inequality a2 > a1. Consequently, ifa1 > 0 and a2> 0, then inequality (12), and hence inequality (7) as well, cannot be satisfied for allξ ∈R.
2. Let a1 < 0 and a2 < 0. Let us take ξ = π/ h2. Then cos2( h2ξ) = 1 and cos2( h1ξ) is nonnegative. Therefore, for such aξ, since a1< 0, the left-hand side of inequality (12) is not greater than 2a2, and hence one must have the inequality 2a2 > a1+ a2, i.e., a2 > a1. By takingξ = π/ h1, in a similar way we obtain the opposite inequality a1 > a2. Consequently, fora1< 0 and a2< 0, the equivalent inequalities (7) and (12) cannot be satisfied for all ξ ∈R.
3. Now let a1 > 0 and a2 < 0. Suppose that ξ = π/(2 h1). Then cos2( h1ξ) = 0, and hence the left-hand side of inequality (12) is nonpositive, becausea2 < 0; therefore, a1+ a2< 0. If we take ξ = π/ h2, then cos2( h2ξ) = 0, and hence the left-hand side of inequality (12), is not greater than2a1+ 2a2 becausea1 > 0. Therefore, the inequality 2a1+ 2a2> a1+ a2, i.e., the inequalitya1+ a2> 0 opposite to the one established, must hold. Consequently, if a1> 0 and a2< 0, then inequality (12), and hence inequality (7) as well, cannot be satisfied for all ξ ∈R.
4. Leta1 < 0 and a2> 0. This case can be considered by analogy with case 3.
We have thus proved that if none of the parametersa1,a2 andh1,h2is zero, then condition (7) in the Theorem cannot be satisfied for allξ ∈R.
Now consider equations of the form (1) in which none of the shifts is zero while the modulus of the coefficient multiplying the (only) remaining term does not exceed the coefficient of the first term on the right-hand side, i.e., equations of the form
∂2u(x, t)
∂t2 = a1∂2u(x, t)
∂x2 + a2∂2u(x − h, t)
∂x2 ,
where |a2| < a1. For equations of this class, the symbol of the corresponding difference operator isa1+ a2cos hξ − a2i sin hξ, and hence the assumptions of the Theorem are satisfied for all real ξ.
Now assume that one of the coefficients on the right-hand side rather than one of the shifts vanishes in Eq. (1). This case is special, viz., for the function G(x, t; ξ) to exist for each real value of the parameter, it suffices that the (only) coefficient on the right-hand side of the equation stays positive. In this case, the equation takes the form
∂2u(x, t)
∂t2 = a∂2u(x − h, t)
∂x2 , (13)
while the one-parameter family of its solutions can be written as G(x, t; ξ) = sin √
at coshξ
2 + x + h 2
ξ
e√aξt sin(hξ/2)
+ sin √
at coshξ
2 − x − h 2
ξ
e−√aξt sin(hξ/2).
(14)
If a > 0, then, regardless of the real value of h, the function (14) is a smooth (classical) solution of Eq. (13) for each real value of ξ; this can be verified by a straightforward substitution of the function (14) into Eq. (13).
ACKNOWLEDGMENTS
The author expresses her gratitude to A.B. Muravnik for posing this problem and helpful remarks, to A.L. Skubachevskii for constant attention to this work, to I.S. Lomov for useful advice, and to the participants of the scientific seminar of the Mathematical Institute of PFUR on differential and functional–differential equations, headed by A.L. Skubachevskii, for constructive discussions of this work.
FUNDING
This work was supported by the Center for Fundamental and Applied Mathematics at Lomonosov Moscow State University.
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