Journal of Mathematical Sciences, Vol. 117, No. 2, 2003
ON THE PROPAGATION OF LOVE WAVES IN ANISOTROPIC ELASTIC MEDIA Z. A. Yanson
UDC 550.344.55
The Love waves concentrated near the surface of an anisotropic elastic body are studied. A uniform asymptotics of the wave field is constructed with the use of the nonstationary caustic expansion (Yu. A. Kravtsov’s ansatz) in the form of a space-time ray series. Using three types of waves, which propagate along any direction in an elastic medium, as a vector basis, sufficient conditions for the existence of a nonzero asymptotic solution of the problem under study are obtained. The procedure for constructing asymptotic series is illustrated with the model of a transversely isotropic medium. Bibliography: 9 titles.
Dedicated to Vasilii Mikhailovich Babich on the occasion of his jubilee Introduction The statements of the problems on the mathematical description of Love waves in anisotropic and isotropic media are essentially the same. For this reason, the constructions of the asymptotics for waves of the whisperinggallery type, in particular, for surface Love modes, are similar in the above-mentioned media. Paper [1] is one of the early papers, where the asymptotics of Love waves propagating near the surface of an anisotropic elastic body was presented. In [1], as well as in [2, 3], Kravtsov’s ansatz [4] is used (in particular, its nonstationary version) as a tool for mathematically describing and constructing an asymptotics for Love waves. Note that precisely this ansatz allows one to obtain a uniform asymptotics (caustic expansion) for the wave field near a simple caustic. When applied to an isotropic elastic medium, Kravtsov’s ansatz yields two well-known types of surface waves (see [2, 3]) or two types of oscillations: (a) torsional (SH modes) and (b) spheroidal (SV modes). Similarly, in an anisotropic elastic medium, the boundary conditions also formally yield two types of possible solutions of the problem. The results of [1] correspond to Love waves of SV type1 both in isotropic and transversely isotropic elastic media, see [3, 5]. The other possible asymptotic solution is related to SH Love waves. Paper [6] is concerned with the construction of the asymptotics for this type of waves in a transversely isotropic medium. The aim of the present paper is to advance the results of [1] in proving the solvability of the problem on the construction of a uniform asymptotics for Love waves, which is represented in [1] as a space-time (ST ) ray solution concentrated in the vicinity of the boundary surface. Such a proof is necessary for the analysis of a linear system for the coefficients of ray series, which is generated by the boundary conditions. Necessary and sufficient conditions for the existence of a nontrivial solution of this linear system are derived in Sec. 2. In Sec. 3, expressions for some of the first coefficients of the asymptotic expansion are given, and the transport equation for the amplitude of the principal term of the asymptotics is derived. Finally, by way of illustration, an algorithm suggested for the development of the high-frequency asymptotics for Love waves is applied to a special type of anisotropy, namely, to a transversely isotropic elastic medium. In this case, we give expressions for the coefficients of asymptotic series, which determine the amplitude, phase, and dispersion of individual Love modes. 1. The general form of an asymptotic solution 1. In this section, we rewrite the asymptotic expansion of Love waves derived in [1], using the notation from [2, 3]. Taking Kravtsov’s ansatz from [4] as a test solution, we represent the Love wave in the form 2/3 (−p2/3 m) eip(x,t) , u(x, t) = Av(−p m) + ip−1/3 Bv x, t) = A(
∞ k A , (ip)k
k=0 1 In
x, t) = B(
∞ k B , (ip)k
(1.1) x = (x1 , x2, x3 ),
k=0
the scientific literature, this type of waves is known as surface Rayleigh waves.
Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 275, 2001, pp. 286–309. Original article submitted April 19, 2001. c 2003 Plenum Publishing Corporation 1072-3374/03/1172-4049 $25.00
4049
k, B k , m, and are functions of coordinates x1 , x2 , x3 , and time t; v and v are the Airy function where A (as defined by V. A. Fok) and its derivative and p is a large parameter (frequency). We naturally relate the coordinates x1 , x2 , and x3 with the surface Σ of an elastic body by setting x3 = n, where n = nn0 is the normal to Σ and x1 , x2 , n = 0 are the coordinates of the points of the surface Σ. For simplicity, assume that the coordinates x1 and x2 are Cartesian (one can see that the final asymptotic formulas remain valid if x1 and x2 are replaced by curvilinear coordinates q1 and q2 on Σ). The unit normal n0 is directed inward the elastic body. Expressions (1.1) must be a solution of the following boundary problem: ∂σij ∂ 2 Uj − ρ 2 = 0, ∂xi ∂t ∂U α σ3j = a3jαβ = 0, i, j, α, β = 1, 2, 3. Σ ∂xβ )j ≡ (LU
(1.2)
(1.3)
Σ
Moreover, by analogy with [2, 3], we assume that the following relation holds: m = γ > 0,
γ = const,
(1.4)
Σ
which ensures (near Σ) the phenomenon of interference of the waves forming a Love mode. In relations (1.2) and (1.3), σij is the stress tensor and aijαβ (x) is a positive symmetric tensor of rank 4 (stiffness tensor): aijαβ = aαβij = ajiαβ . Substituting (1.1) into boundary conditions (1.3), we obtain a linear combination of the functions v(−p2/3 γ) and v (−p2/3 γ) that vanishes on Σ. Formally, there are two choices: (a) v (−p2/3 γ) = 0 or (b) v(−p2/3 γ) = 0. We choose case (b), which, if the medium is homogeneous, yields asymptotic solutions for SV surface waves (see [3]), which are referred to as the class of spheroidal oscillations for an elastic model of the Earth. Therefore, γ = γν =
sν , ν = 1, 2, . . . , p2/3
(1.5)
where the −sν are the roots of the equation v(τ ) = 0, so that γν (where ν is the number of the Love wave) is a small parameter of the problem for limited values of ν. k and B k (obtained by the substitution of (1.1) into The system of recurrent equations for the coefficients A (1.2)) is of the same form as in an isotropic medium: k + 2mNm B k = F k−1, (N + mNm )A 1
(1.6) k + (N + mNm )B k = F k−1, 2Nm A 2
k + (Nm + mMm )B k + LA k−1, −F1k ≡ M A
(1.6a) −F
k 2
k + M B k + LB k−1 ; ≡ Mm A
−r = B −r = 0, A
r = 1, 2,
where Nξ , Nξξ˜, and Mξ are known ray operators, which for an anisotropic elastic medium are of the form
2 j ≡ aijαβ ξ i ξ α Dβ − ρ ∂ξ Dj , (Nξ D) ∂t
j ≡ aijαβ (ξ i ξ α + ξ α ξ i )Dβ − 2ρ ∂ξ ∂ ξ Dj ; 2(N D) ξξ ∂t ∂t ξi ≡ j≡ (Mξ D) 4050
∂ξ , ∂xi
(1.7)
∂ ξ ξ i ≡ , ∂xi
∂
∂Dβ ∂Dβ ∂ξ ∂Dj ∂2 ξ − ρ 2 Dj . aijαβ ξ α Dβ + aijαβ ξ i + aijαβ ξ α − 2ρ ∂xi ∂xα ∂xi ∂t ∂t ∂t
(1.8)
Now we recall the zeroth approximation of the asymptotics and set k = 0 in (1.6). For n = 0 (i.e., on Σ), in the approximation γ = 0 the first equation in system (1.6) yields
0 N A
j
≡ aijαβ i α (A0 )β − ρ2t (A0 )j = 0,
j = 1, 2, 3,
(1.9)
∂ ∂ 0 ≡ so that A 0 is an eigenfunction of the matrix aijαβ ∂x and ρ2t is its eigenvalue. The second equation i ∂xα 0 , is solvable only if the relation of system (1.6), which determines B
0, A 0 Nm A
n=0 γ=0
= 0 or
∂m a3αiβ (A0 )α (A0 )β i n=0 = 0 ∂n γ=0
(1.10)
is fulfilled. 0 and B 0 in the form As in [1], we represent A 0 n=0 = ϕ0 η, A γ=0
0 n=0 = ψ0 η + ϕ0 σ, B
(1.11)
γ=0
where the polarization vector η obeys Eq. (1.9) and is specified, for example, by the normalization condition 0 |η|2 = 1, η is real; σ is a solution of the nonhomogeneous equation (which follows from the equation for B mentioned above) Nσ = −2Nm η, (1.12) where (Nm η, η)n=0 = 0. The functions ϕ0 and ψ0 must be determined. γ=0
Substituting expansion (1.1) into boundary conditions (1.3), using (1.11), and setting v(−p2/3 γ) = 0, in the zeroth approximation of the asymptotics (for k = 0) on Σ, we find [ϕ0 (r1 + r2 ) + ψ0 r0 ]n=0 = 0, γ=0
(1.13)
where the components of the vectors r1 , r2 , and r0 take the form (r1 )j ≡ a3jαβ α σβ , (r2 )j ≡ a3jαβ mα η β , (r0 )j ≡ a3jαβ α η β , j = 1, 2, 3.
(1.13a)
Now we prove that Eq. (1.13) results in the values ϕ0 = ψ0 = 0. For convenience, we take the following linearly independent vectors as a basis: η, σ, ω = η × σ . (1.14) ≡ r1 + r2 . Since Assume that ϕ0 = 0 and ψ0 = 0, and let the vectors r1 and r2 be linearly dependent. Denote R (r0 , η) = 0 (with n = 0 and γ = 0), see (1.10), we have (R, η) = (r1 + r2 , η) = 0. From the fact that the vectors r1 and r2 are linearly dependent, it also follows that R = 0 (because η is one of the basis vectors (1.14) and, n=0 γ=0
in addition, (r2 , η)n=0 = 0, see below). This implies that for n = 0 and γ = 0, γ=0
a3jαβ α σβ = −m1 a3j3β η β ,
m1 =
∂m = 0. ∂n
(1.15)
The vector σ has already been found from the nonhomogeneous equation (1.12) in terms of η . Equation (1.15) holds only if either systems (1.12) and (1.15) are equivalent or (1.15) follows from (1.12). We note that one of the conditions under which system (1.15) can be reduced to (1.12) is a linear dependence of the vectors Nm η and r2 of the right-hand sides in relations (1.12) and (1.15). With respect to vector basis (1.14), by (1.10) we have Nm η = 0. At the same time, r2 = 0 (for n = 0 and γ = 0). Indeed, η
η
(r2 , η)n=0 = m1 (a3j3β η j η β )n=0 = 0, γ=0
γ=0
4051
because a3j3βη j η β > 0 in view of the positive definiteness of the tensor aijαβ . Consequently, the vectors Nm η and r2 are linearly independent. Next, if Rn=0 = 0 and ψ0 = 0, then (1.13) yields r0 n=0 = 0. Then, from (N η, n0 )n=0 = 0, see (1.9), γ=0 γ=0 γ=0 it follows that η0 = 0. If elastic media are isotropic or transversely isotropic, this relation contradicts the n n=0 = 0, statement of the boundary problem under study in [3, 5]. One can easily verify that in these media, R γ=0
see Sec. 3. = 0. Under this assumption, it is easy to see that the vectors r1 , r2 , and In what follows, we assume that R η) r0 must be linearly independent if (R, = 0, because (r1 , η) = −(r2 , η) = 0 and (r0 , η) = 0 (see Sec. 2).1 n=0 γ=0 0 n=0 = 0 and B 0 n=0 = 0. Then it follows from (1.13) that ϕ0 = ψ0 = 0, i.e., A γ=0
γ=0
In order to overcome this difficulty and to ensure that the problem on the construction of the asymptotics for surface Love waves is solvable, we consider the contribution made by two faster waves (in comparison with Love wave (1.1)), which decay with distance from the boundary. In [1] and [5], these waves are represented as ray series with complex eikonals and are in essence nonhomogeneous waves. 2. We now represent the wave field of a surface wave in the form u = u + u1 + u2 .
(1.16)
Here, u is expansion (1.1) and u1,2 are the ray series u1,2(x, t) = α
∞ k C1,2 (x, t) k=0
(ip)k
eipθ1,2 (x,t),
(1.17)
where the θ1,2 are complex eikonals such that θ1,2 = , Σ
Σ
∂θ1,2 Im > 0. ∂n
(1.18)
Σ
k of the ray series satisfy the recurrent equations of the ray method The coefficients C 1,2 k + Mθ C k−1 + L C k−2 = 0, k = 0, 1, 2, . . . ; Nθ1,2 C 1,2 1,2 1,2 1,2 −r = 0, C 1,2 and take the form
k = φ k κ k0 C 1,2 1,2 1,2 + U 1,2 ,
(1.19)
r = 1, 2, 00 = 0, k = 0, 1, 2, . . . , U 1,2
(1.19a)
where L, M , and N are operators of the form (1.2), (1.8), and (1.7); the κ 1,2 (the polarization vectors) are solutions of the homogeneous equations obtained from (1.19) with k = 0: r )j ≡ (Nθr κ
aijαβ θri
θrα κrβ
2 ∂θr −ρ (κr )j = 0, r = 1, 2. ∂t
(1.20)
Substituting (1.16) into boundary conditions (1.3) and taking (1.5) into account, i.e., v(−p2/3 γ) = 0, we have k k k k σ3j (u) = [σ3j (u) + σ3j (u1 ) + σ3j (u2 )]n=0 = 0, (1.21) n=0
k where σ3j (u) are the coefficients of the expansion of σ3j (u) in powers of 1/(ip)k . The terms in square n=0 n=0 brackets are of the form (see [1, 5]) k k−1 ) σ3j (u) ≡ ip−1/3 a3jαβ α (B k )β + a3jαβ mα (Ak )β + σ3j (B v (−p2/3 γ) exp[ip0 (x1 , x2 , t)], (1.22) n=0
1 If
4052
(R, η ) = 0, relation (1.13) immediately implies the relations ϕ0 = ψ0 = 0.
n=0
k σ3j (u1,2 )
α k β k−1 ) ≡ α a3jαβ θ1,2 (C1,2 ) +σ3j (C exp[ip0 (x1 , x2 , t)], 1,2 n=0 n=0 0 (x1 , x2 , t) ≡ (x, t) , j = 1, 2, 3.
(1.23)
n=0
We put α = ip−1/3 and replace the φ k1,2 in (1.19a) by the coefficients φk1,2 ≡ φ k1,2 v (−p2/3 γ). Let k = 0. Here, by (1.11), boundary conditions (1.21) yield linear equations that connect the four variables ϕ0 , ψ0 , φ1 , and φ2 on the surface Σ. Setting γ = 0, we write these equations as the system (1.24) (r0 )j ψ0 + (s1 )j φ01 + (s2 )j φ02 n=0 = −ϕ0 Rj n=0 , j = 1, 2, 3, γ=0
γ=0
= r1 + r2 and R = where R 0; r1 , r2, and r0 are defined by relations (1.13a), whereas the components of the vectors s1 and s2 are of the form (s1 )j = a3jαβ θ1α κ1β ,
(s2 )j = a3jαβ θ2α κ2β ,
j = 1, 2, 3.
(1.25)
The coefficients ψ0 and φ01,2 are uniquely found in terms of ϕ0 (x1 , x2 ), provided that the determinant of system (1.24) is different from zero. In turn, the function ϕ0 is obtained from the condition on the solvability of system (1.6) with k = 1 (see Sec. 3 and [5]). The coefficients and eikonals of asymptotic expansions (1.1) and (1.17) are functions of the parameter γν (see (1.5)) (we omit the number of the Love wave ν in the sequel). In view of the smallnes of γ, the above coefficients and eikonals l and θ1,2 can be approximated, by analogy with [2, 3], by the double power series ∞
k = A
ksj ns γ j , B k= A
s,j=0
=
∞
ksj ns γ j , C k1,2 = B
s,j=0
∞
ij ni γ j ,
m=
i,j=0
∞
mij ni γ j ,
∞
k1,2 C
s,j=0
θ1,2 =
i,j=0
∞
ns γ j ,
(1.26)
sj
(θ1,2 )ij ni γ j .
(1.27)
i,j=0
Note that, as opposed to [2, 6], Eqs. (1.26) contain no terms with negative powers of γ. The coefficients of the above expansions are found successively by a general recursion process, which includes in addition the construction of higher asymptotic approximations for k ≥ 1. 2. Proof of the existence of an asymptotic solution Expression (1.16) represents an individual mode (Love wave) as the sum of caustic expansion (1.1) and two correction terms (1.17) (which are ray solutions damped exponentially with depth). Wave (1.16) is an asymptotic solution of boundary problem (1.2) and (1.3) only if linear system (1.24) is solvable and has a unique solution. This means that the determinant of the system must be different from zero. The present section is concerned with the proof of this fact. We give some formulas and prove some auxiliary propositions, which will be used in the main proof. The following relations hold for the polarization vectors η and x1,2 with n = 0 and γ = 0: β a3jαβ κ j1,2 (α η β ) = −a3jαβ η j (θ α 1,2 κ 1,2 ),
j
β
j
β
a3jαβ κ 1 (θ2α κ2 ) = −a3jαβ κ2 (θ α 1 κ 1) β a3jαβ κ j2 (θ1α κ1β ) = −a3jαβ κ1j (θ α 2 κ 2 ),
j where θ1,2 ≡ j κ1,2
∂θ1,2 , ∂xj j κ 1,2
θ j1,2 ≡
∂θ 1,2 , ∂xj
and lj ≡
∂l , ∂xj
(2.1)
(2.2)
j = 1, 2, 3, are the components of the vectors ∇θ1,2 , ∇θ 1,2 , and
∇; and are the components of the vectors κ 1,2 and x1,2 , which are the solutions of homogeneous equations (1.20) ((θ1,2 )t = t for n = 0). The vector η (an eigenfunction of the matrix (aijαβ i α )) is a solution of the homogeneous equation (N η )j ≡ aijαβ i α η β − ρ2t ηj = 0,
n = 0, γ = 0.
(2.3) 4053
Relations (2.1) and (2.2), which are valid under the conditon ∂θ1 ∂θ2 ∂ = = , ∂n ∂n ∂n
n = 0,
γ = 0,
(2.4)
can easily be proved, see Appendix, Sec. 1. ∂θ1,2 Now we give a relation obtained by analogy with (2.1) and (2.2) under the condition Im ∂n n=0 = 0, namely, γ=0
β a3jαβκ r (θrα κrβ ) + a3jαβ κrj (θ α r κ r ) = 0,
r = 1, 2,
(2.2a)
whence it follows that for n = 0 and γ = 0 β Re a3jαβ κ jr (θrα κrβ ) = Re κrj (θ α r κ r ) = 0,
r = 1, 2.
In the sequel, we use the notation (θ1,2 )3 ≡
∂θ1,2 , ∂n
(θ1,2 )3 ≡
∂θ 1,2 , ∂n
3 ≡
∂ . ∂n
∂τ ∂τ has multiple eigenvalues; Inequalities (2.4) exclude the case where the matrix operator A = aijαβ ∂x i ∂xα here, τ represents the eikonals of waves related to three eigenfunctions of this operator. From the physical point of view, this implies that on the boundary Σ, the velocities of these waves are distinct. Using the notation from (1.25), we write (2.1) and (2.2) in the form (r0 , κ 1,2 ) = −(η , s1,2 ) = −(s1,2 , η), (s2 , κ 1 ) = −( κ2 , s1 ) = −(s1 , κ 2 ),
(2.5)
(sr , κ r ) = −( κr , sr ) = −(sr , κ r ),
r = 1, 2.
We also cite the relation (which follows from Eqs. (1.10) and (1.11)) (r0 , η)n=0 = 0, γ=0
(r0 )j ≡ a3jαβ α η β ,
j = 1, 2, 3.
(2.6)
2 are linearly independent. Lemma. The vector functions η , κ 1 , and κ We prove the lemma by contradiction. Let η = a κ1 + b κ2 , where either a = 0 and b = 0, or a = 0 and b = 0 (or else b = 0 and a = 0). Now we make use of Eqs. (2.1), multiplying them (with κ =κ 1 and κ =κ 2 ) by a and b, respectively. Upon summation of the results, we find 1 + |b|2Ψ 2 ) − (abΨ1 + abΨ2 )]n=0 = [(θ1 )3 |a|2 a3j3β κ j κ β + (θ 2 )3 |b|2 a3j3β κ j κ β ]n=0 , [(|a|2Ψ 1 1 2 2 γ=0
γ=0
where we denote r ≡ a3jαβ (κ j κ β + κ j κ β )α + 3 a3j3β κ j κ β , Ψ r r r r r r
r = 1, 2,
Ψ1 ≡ a3jαβ κ1j (θ2α κ2β ) + a3jαβ κ j2 κ1β α , Ψ2 ≡ a3jαβ κ2j (θ1α κ1β ) + a3jαβ κ j1 κ2β α . 1 and κ 2 , we obtain Then, multiplying (2.3) by κ Ψ1 = 0, 4054
Ψ2 = 0,
n = 0,
γ = 0,
α = 1, 2,
(2.7)
whence it is clear that the left-hand side of Eq. (2.7) is real. Therefore, the imaginary part on the right-hand side in Eq. (2.7) must vanish, which, by virtue of a3j3βκ rj κrβ > 0, implies Im
∂θ1,2 n=0 = 0, ∂n γ=0
which contradicts the statement of the problem, i.e., condition (1.18). In the cases κ1 or η = b κ2 ), from (2.7) it follows that where a = 0 and b = 0, or a = 0 and b = 0 (i.e., η = a ∂θ1 ∂θ2 either Im ∂n n=0 = 0 or Im ∂n n=0 = 0. γ=0
γ=0
The linear independence of the vectors κ 1 and κ 2 can be established by using (1.20) with κ 1,2 and κ 1,2 . Here we omit the proof for brevity. Thus, the vectors η , κ 1, and κ 2 are linearly independent, as required. Now we turn to linear system (1.24). Let A be the matrix of the system and T be the matrix composed of the linearly independent vectors η, κ 1 , and κ 2 ; namely,
(r0 )1
(s1 )1
(s2 )1
A = (r0 )2
(s1 )2
(s2 )2 ;
(r0 )3
(s1 )3
(s2 )3
η1
(κ1 )1
(κ2 )1
T = η2
(κ1 )2
(κ2 )2 ,
η3
(κ1 )3
(κ2 )3
(2.8)
where the vectors r0 , s1 , and s2 are the columns of the matrix A, whereas the vectors η , κ 1 , and κ 2 are the columns of the matrix T = (tik ). Then we use the matrix T to put system (1.24) in the form
(R, η0 ) ψ0 φ0 = −ϕ0 (R, κ A 1) , 1 0 κ2) φ2 (R,
n = 0,
γ = 0,
(2.9)
= T ∗ A and the matrix T ∗ = (tki) is the Hermite conjugate to T with det T ∗ = det T . where A in (2.9) has the form The matrix A
(r0 , η)
(s1 , η)
(s2 , η)
0
= A 1) (s1 , κ 1 ) (s2 , κ 1 ) = (r0 , κ (r0 , κ 1) (r0 , κ 2) (s1 κ 2 ) (s2 , κ 2)
(r0 , κ 2)
−(r0 , κ 1 ) −(r0 , κ 2) (s1 , κ 1) (s1 , κ 2)
−(s1 , κ 2)
(2.10)
(s2 , κ 2)
is transformed with the use of relations with n = 0 and γ = 0. In the second relation in (2.10), the matrix A = (2.6) and (2.5). Since the determinants of the matrices A, A, and T are connected by the relation det A ∗ ∗ is equal to (det T ) · (det A) and since det T = 0 (by virtue of the linear independence of η, κ 1 , and κ 2 ), det A zero if and only if det A is equal to zero. = 0, then det A = 0, and vice versa. Thus, if det A The vectors κ 1,2 and s1,2 are elements of the complex Euclidean space, where the scalar product of vectors y and z is defined by the formula (y, z) = yi z i,
i = 1, 2, 3,
(y, z) = (z, y),
(2.11)
and if λ is complex, then (λy, z) = λ(y, z) and (y, λz) = λ(y, z). These relations are used in (2.10) to transform and will be invoked again in the sequel. the matrix A Now we represent system (1.24) in vector form, which is used below: ψ0 r0 + φ01s1 + φ02s2 = −ϕ0 R,
R = 0.
(2.12)
4055
Proposition. If conditions (2.4) hold, then the determinant of the matrix A is different from zero if and only ∂ β if r0 = 0, where (r0 )j = a3jαβ ∂x η and j = 1, 2, 3.2 α n=0 γ=0
= 0.) We prove this proposition by contradiction. (Note that if r0 = 0, then det A = 0 and det A Thus, we assume that = 0, r0 = 0. det A = det A Since det A = 0, the columns of the matrix A must be linearly dependent, i.e., the following relation is valid: δr0 + αs1 + βs2 = 0,
(2.13)
where at least one of the coefficients α, β, and δ is different from zero. We split the proof into three parts. I. Let r0 = αs1 + βs2 , α = 0, β = 0, (r0 , κ 1 ) = 0,
(r0 , κ 2 ) = 0.
(2.14) (2.14a)
By the above lemma, the vectors η , κ 1 , and κ 2 are linearly independent at the points of the surface Σ. We choose them as a basis for the complex space and write 1 + b0 κ 2 + e0 η, r0 = a0 κ s1,2 = a1,2 κ 1 + b1,2 κ 2 + e1,2 η ,
(2.15)
where, by (2.14), the vectors r0 and s1 (as well as r0 and s2 ) are linearly independent. is determined by five independent entries As follows from the second relation in (2.10), the matrix A (r0 , κ 1), (r0 , κ 2 ), (s1 , κ 1 ), (s1 , κ 2 ), (s2 , κ 2 ).
(2.16)
Taking this into account, as well as the fact that the vectors r0 and s1 (and also r0 and s2 ) are linearly independent, we can find linear relations between the elements from (2.16). The derivation of these relations forms the basis for proving the above proposition. Projecting (2.14) onto η, κ 1 , and κ 2 , respectively, and taking (2.5) and (2.6) into account, we obtain (r0 , α κ1 + β κ2 ) = 0, (r0 , η) = 0, κ1 + β κ2 − η) = 0, (s1 , α (s2 , α κ1 + β κ2 − η) = 0.
(2.17) (2.18)
Now, from (2.15) we choose a pair of linearly independent vectors r0 and s2 and represent them in the form c1 (α κ1 + β κ2 ) + c1 κ1 + e0 η, r0 = (2.19) c2 (α κ1 + β κ2 ) + c2 κ 2 + e0 η, s2 = d2 (α κ1 + β κ2 ) + d 2 κ 2 + e2 η,
(2.19a)
where the coefficients c1,2 and c1,2 are connected by the relations (c1 − c2 )α = − c1 ,
(c1 − c2 )β = c2 ,
c1 / c2 = −α/β.
(2.20)
We note that c1 = 0 or c2 = 0 in (2.19) implies, by (2.17), that r0 = 0. Then we express the scalar product (s2 , r0 ) from (2.19) and (2.19a), apply (2.5), and obtain the relations (s2 , c1 κ 1 + (c1 + d 2 + e0 )η) = 0, c2 κ 2 + (c2 + d 2 + e0 )η) = 0, (s2 ,
(2.21)
2 For ( are different from zero. The second or the third column of 1,2 ) = 0 the second and the third columns of the matrix A r0 , κ 1 = 0 or κ 2 = 0, s2 = 0, then from (1.20) it necessarily follows that κ the matrix A is also different from zero. Indeed, if s1 = 0 or which is not allowed.
4056
Since c1 = 0, c2 = 0, and c1 = c2 , relations (2.21) are independent both of each other and of the relation (s1 , α κ1 + β κ2 − η) = 0 from (2.18). As a result, we derive a homogeneous linear system for s2 , which has a nonzero determinant and a unique solution s2 = 0. Indeed, for a pair of vectors r0 and s1 , we obtain relations that are similar to (2.21) with s2 and d˜2 replaced by s1 and d˜1 , and moreover c1 − c2 = d˜2 − d˜1 = 0. Employing these relations and (2.18), we deduce that (s1 , r0 ) = (s2 , r0 ) = 0. Multiplying (2.19) by s2 (or by s1 ), one can 1) = 0 if d˜2 = 0). Hence it follows that convince ourselves that (s2 , η) = 0 holds if d˜2 = 0 (or (s1 , η) = (r0 , κ (r0 , κ 2) = −(s2 , η) = 0 and, by (2.17), r0 = 0, which contradicts (2.14a). II. Asume that the second and third columns of the matrix A are linearly dependent, i.e., αs1 + βs2 = 0,
α = 0,
β = 0,
(2.22)
and relations (2.14a) are valid. It is easy to see that Eqs. (2.17) remain valid, and instead of (2.18) we now have (s1,2 , α κ1 + β κ2 ) = 0,
(2.23)
where the s1,2 are now linearly dependent. As above, we assume that the vectors r0 and s2 are linearly independent. Calculating (r0 , s2 ) from (2.19) and (2.19), respectively, we obtain (s2 , c1 κ 1 + (d 2 + e0 )η) = 0,
(2.24)
(s2 , c2 κ 2 + (d 2 + e0 )η) = 0.
It is obvious (see Eq. (2.20)) that, as opposed to case I, Eqs. (2.23) are now a consequence of Eqs. (2.24). Using relations of (2.5), (2.17), and (2.22), for five variables 1 ), (s1 , κ
(s1 , κ 2 ),
(r0 , κ 1),
(r0 , κ 2 ),
(s1 , η)
(2.25)
we obtain five independent relations, namely, the relations (r0 , α κ1 + β κ2 ) = 0,
(s1 , κ 2) = −(s2 , κ 1 ),
(s1 , η) = −(r0 , κ 1)
and two relations from (2.24). These relations form a linear homogeneous system for variables (2.25), which has a unique zero solution (note that the determinant of the system is different from zero if (d˜2 + e0 ) = 0, i.e., (sr , κ ˆ r ) = 0, r = 1, 2 (see Appendix, Sec. 2). We thus arrive at the relations (r0 , κ 1 ) = (r0 , κ 2 ) = 0, which contradict (2.14a). III. Assume that the first and second (or third) columns of the matrix A are linearly dependent, i.e., r0 = αs1 ,
(r0 , η) = 0,
α = 0,
(2.26)
and r0 = 0 as before. 1. From (2.26) and (2.5), we deduce that (r0 , κ 1 ) = −(s1 , η) = 0 and, consequently, (r0 , κ 2 ) = 0. For the determinant of the matrix A we find 0 0 −(r0 , κ 2 ) = det A 1 ) |r0 , κ 2 |2 = 0, 0 (s1 , κ 1 ) −(s1 , κ 2 ) = −(s1 , κ (r0 , κ 2 ) (s1 , κ 2 ) (s2 , κ 2) whence it follows that (s1 , κ 1 ) = 0. The reader is referred to Appendix, Sec. 2, where we show that, in general, the condition (s1 , κ 1) = 0 or (s2 , κ 2 ) = 0 yields Im(θ1,2 )3 = 0, which contradicts (1.18). One can easily verify that the inequalities (sr , κ r ) = 0, r = 1, 2, are valid, for example, in the case of a transversely isotropic medium (see [5]). 2. We give another proof for the case of (2.26) considered here. We write s2 and r0 in the form 1 + c0 κ 2 + e0 η, r0 = c0 κ
(2.27)
s2 = d2 κ 1 + d 2 κ 2 + e2 η, 4057
where s2 and r0 are linearly independent, and as above (r0 , κ 1) = 0. As in parts I and II, we calculate (r0 , s2 ) from these relations and obtain (s2 , c0 κ 1 + c0 κ 2 + (e0 + d 2 )η) = 0,
c˜0 = 0.
Analyzing this relation, together with (2.5) and (r0 , κ 2 ) = α(s1 , κ 2), we establish that (r0 , κ 2) = 0. In view of (r0 , η) = (r0 , κ 1 ) = 0, we conclude that r0 = 0, which contradicts the assumption of the theorem. Thus we proved that the columns of the matrix A are linearly independent, whence it follows that det A = 0. Remarks. 1. If r0 n=0 = 0, then the boundary problem has a unique zero solution. System (2.12) takes the form γ=0
= 0, φ1s1 + φ2s2 + ϕ0 R
= r1 + r2 , R
= 0, R
η) = 0, then ϕ0 = 0. In the case where, by (2.5), (s1,2 , η) = −(r0 , κ 1,2) = 0, provided that r0 = 0. Hence, if (R, η) = 0, we have the relation where (R, ϕ0r2 + ϕ0r1 + τ = 0,
τ ≡ φ1s1 + φ2s2 ,
(2.28)
where r1 and r2 are linearly independent vectors (with (r2 , η) = 0, see above). Then it follows from (2.28) that ϕ0 n=0 = 0. γ=0 2. If r0 n=0 = 0, then ηn = 0. This fact was mentioned above. γ=0
In this case, projecting (2.3) onto n, we have (N η, n) = a3iαβ i α η β − ρ2t ηn = 0,
where for n = 0 and γ = 0, a3iαβ i α η β = (r0 , ∇) = 0, because r0 = 0. Therefore, ηn n=0 = 0. Moreover, if γ=0
ηn = 0, then also r0 = 0. The proposition proved above can be stated as follows. If conditions (2.4) are fulfilled at all points of the boundary surface Σ (i.e., the three eigenvalues of the matrix operator A =
∂τ aijαβ ∂x i
∂τ ∂xα
are distinct) and (r0 , η) = 0, then an expansion of the form (1.16), (1.1), and (1.17) is a nontrivial asymptotic solution of boundary problem (1.2) and (1.3) only if ηn = 0 on Σ, where n is the normal to Σ and the vector η determines the polarization of the principal term of the asymptotics. If η n = 0 on Σ, the problem on the construction of a ray field on the surface Σ (with Eq. (1.5)) is not well defined (see Sec. 3). 3. The eikonal equation and the transport equation. The example of a transversely isotropic medium 1. In describing the propagation of high-frequency waves with the help of the ray method, the eikonal equation usually arises when the problem on the construction of the principal term of the asymptotics for a wave field is solved. In the case under study, we also use the zeroth asymptotic approximation, namely Eqs. (1.6) with k = 0, in order to describe the ray field (in essence, to solve the eikonal equation) on the boundary surface Σ and in 0 and B 0 can be put in the form the vicinity of it. The resulting homogeneous system for A 2 ∂τs 0 i α 0 β (Us0 )j = 0, s = 1, 2, (3.1) (Nτs U s )j ≡ aijαβ τs τs (Us ) − ρ ∂t τ1,2 = l ∓
2 3/2 m , 3
0 ∓ m1/4 B 0, 0 = m−1/4 A U 1,2
τsi ≡
∂τs , ∂xi
0 are the eikonals and amplitudes, respectively, of the waves incident on the surface Σ where the τ1,2 and U 1,2 and reflected from it. The quantities λ = ρ(τs )2t are eigenvalues of the matrix A = (aijαβ τ i τ α ) of Eq. (3.1) and are found from the characteristic equation of the matrix (see [1]), so that 1 ∂τ (−τ )t = 1/2 H xi, . (3.2) ρ ∂xi 4058
Relation (3.2) is the space-time (ST ) eikonal equation for the ray field near Σ. By the proposition in Sec. 2, an eigenfunction from (3.2) must have a nonzero component in the plane of a normal cross-section of the surface Σ (it is also the amplitude of the “quasi-transverse” surface wave under study (see [7])). Using (3.1), one can obtain well-known expressions for the group velocity g of propagation of waves whose components are of the form 1/4 0 i 1/4 0 β α ajiαβ m (Us ) m (Us ) τs , 0 |2 (−τs )t ρ|m1/4 U s (3.3) gj = ∂H 1 ∂τs j , τs = ∂xj , j = 1, 2, 3, ρ1/2 ∂τsj 0 /|U 0 | are assumed to be real. where the unit vectors ζ1,2 = U 1,2 1,2 In (3.3), the first relation is identified with the velocity of energy propagation (see, for example, [7]) and the second one is derived from the system of equations for the characteristics (ST rays), which corresponds to eikonal equation (3.2). Put j = 3 in (3.3), and let a point M tend to Σ, where mn=0 = γ. In the limit, for γ = 0 and n = 0, by (1.10), (1.11), and (2.6) we obtain the relation a3iαβ (A0 )i (A0 )β α (r0 , η) g n = g3 = = = 0, 0 |2 (−t ) ρ|η|2 (−t ) ρ|A
(3.4)
which means that the velocity of the surface wave is directed along Σ. The second equation in (3.3) yields (see [1]) ∂ ∂H = 0, 3 ≡ . (3.5) ∂3 n=0 ∂n γ=0
∂ , i = 1, 2, at all points of the surface Σ. Substituting the This relation defines the function 3 = f ∂x , x i i
1 ∂ value of l3 in (3.2) and setting n = 0 and γ = 0, we derive the eikonal equation (−l00 )t = 1/2 H xi , ∂x n=0 i ρ γ=0
for a ray field on the surface Σ. Thus, according to the proposition given at the end of Sec. 2, the condition ηn = 0 and the other conditions ensuring the construction of a nontrivial asymptotic solution of problem (1.2) and (1.3) (in the form of the sum of expansions (1.1), (1.16) and (1.17)) may hold. Now we proceed to the proof that η n = 0 is not only a sufficient but also a necessary condition for the existence of the above-mentioned asymptotic solution. Let η n = 0 with n = 0 and γ = 0. Projecting (2.3) to n, we find 0 , ξ) = 0, ξ = (a3iα1 i α , a3iα2i α , a3iα3i α ). (A (3.5a) Next, projecting (2.3) to ξ and taking (3.5a) into account, at each point on Σ we obtain a cone of order four ∂l ∂ ∂ ∂ ∂l in the space of directions ∂x , ∂x , ∂x . Here, 3 = ∂x proves to be a function of the variables ∂x and xi , 1 2 3 3 i
∂ i = 1, 2, and it is different from the function f ∂x , xi derived from (3.5). This proves the fact that the i eikonal equation cannot be constructed on the surface Σ. Now we give the results of computations of the coefficients m10 , 01 , and 20 of expansions (1.27), which are essential in the determination of the amplitude and dispersion of individual Love modes. We assume that the function 00 (x1 , x2, t), i.e., a solution of the eikonal equation on Σ, is known. Differentiating the first equation of system (1.6) with respect to γ (for k = 0), we obtain M1 ∂ −1 (m10 )2 a3j3β η j η β + 2(Nm σ, η) 01 = = dt, (3.6) ∂γ n=0 (−00 )t M0 2ρ|η|2 γ=0
where the integral is taken over an ST ray. To derive the above formula, we used the following expression for the group velocity on the surface Σ: dxj aj iαβ η i η β α = (g )j = , j = 1, 2. (3.7) dt ρ|η|2 (−t ) n=0 γ=0
4059
The coefficient m10 =
∂m ∂n n=0 , γ=0
occurring in the integrand of (3.6), determines, along with the small parameter
γ, the width of the boundary layer where the Love mode is concentrated, provided that m10 < 0. In [1], the coefficient m10 was obtained in the form 1/3 (−00 )t 1 m10 = − , R g33 n=0
where R−1 > 0 is the normal isotropic curvature in the pseudo-Finsler metrics: g33 = is the Lagrangian (a known function of the variables xi and 1 ∂2 expression for 20 = 2 ∂n2 n=0 is given below.
dxi dσ );
1 ∂2L 2 ∂ x˙ 23
and x˙ 3 =
dx3 dσ ;
L
dσ is a length increment, see also [8]. The
γ=0
2. Now we dwell upon the computation of the yet unknown function ϕ0 (x1 , x2 , t) of (1.11), which is the amplitude of the principal asymptotic term. To this end, we consider the first equation of system (1.6) with k = 1. This 0 n=0 = ϕ0 η, i.e., vector equation (for n = 0 and γ = 0) is solvable only if its right-hand side is orthogonal to A γ=0
0 0 0 (M A + Nm B , A )n=0 = 0,
(3.8)
γ=0
where the operator M is defined in (1.8). Make use of the relation 0, A 0, B 0 ) = m(2Nm B 0 ), (2Nm A
(3.9) which, as can easily be verified, follows from system (1.6) with k = 0. Using Eq. (1.7) and noting that mt n=0 = 0 ∂m and ∂x = 0, i = 1, 2, we find i n=0 ∂ ∂m10 0, A 0 ) 0 )2 (Nm A = ai jαβ (A0 )j (A0 )β α ) + ρ(−t )mt (A n=0 ∂n ∂x i γ=0 ∂ ∂ (a3jαβ (A0 )j (A0 )β α ) + 3 m10 a3j3β (A0 )j (A0 )β ∂n ∂n 0, B 0 )n=0 . +2m10 20 a3j3β(A0 )j (A0 )β n=0 = m10 (Nm B
+m10
γ=0
γ=0
(3.10)
0 n=0 in accordance with (1.8) and compare the resulting expression with (3.10) (omitting We compute M A γ=0
details here). Then we can write relation (3.8) in the form ∂ 0 2 1 dm10 0 2 ∂ (ai jαβ (A0 )j (A0 )β α ) + ρ(−t ) (A ) − ρ(−t ) (A ) ∂xα ∂t m10 dt
where A2 ≡
dm10 dt 0 )2 ρ (A |m10 |
= =
10 gj ∂m + ∂x j 2
(ϕ ) | η| ρ |m10 | , 0 2
∂m10 ∂t
0, B 0 ) + (Nm B 0, A 0 ) = 0, +(Nm B
is the derivative along an ST ray and gj is defined in (3.7). Introducing
by Eq. (3.7) we obtain the final expression for (3.10): ∂ 2 2 (−l00 )t divs g A + A + A2 E = 0, ∂t E≡−
≡ where div D
∂ ∂xj
(3.11)
(3.12)
1 0, B 0 ) + (Nm B 0, A 0 )], [(Nm B (ϕ0 )2 |η|2 ρ
Dj .
0 and B 0 are proportional to ϕ0 , (see (1.11) and (2.9)), Eq. (3.12) Since for n = 0 and γ = 0, the coefficients A 2 is a homogeneous differential equation for A . This is the transport equation for the amplitude of a surface wave on Σ. Integration of Eq. (3.12) along an ST ray (at any point on the surface Σ) yields a modified formula of the ray method. We give this formula below in the case of a transversely isotropic medium. 4060
3. Let an elastic medium be transversely isotropic. In this case, the tensor aijαβ is reduced to a symmetric matrix (crs ), r, s = 1, 2, . . . , 6, that is determined by five independent parameters. The following entries of (crs ) are different from zero: c11 = c22 , c12 = c21 , c13 = c31 = c23 = c32 , c33 , c44 = c55 ,
2c66 = c11 − c12 .
(3.13)
The conditions for the positiveness of all minors of the matrix (crs ) (which ensure the positiveness of the elastic energy of deformation) are of the form c11 > 0,
c33 > 0,
c44 = c55 > 0,
c11 > |c12|,
(c12 + c66 )c33 > c213 .
(3.14)
In accordance with the results obtained above, for the special case of anisotropy in question (for n = 0, i.e., on Σ, and for γ = 0) we find c13 + c44 η = ηn n0 , (σ∇s ) = −m10 ηn , σn = 0. (3.15) c11 − c44 In the basis of the vectors n0 , ∇s , and ζ = n0 ×∇s that are orthogonal on Σ (where ∇s is the surface gradient of the function ), we have η = (ηn , 0, 0), σ = (0, σ∇s , 0).
∂ ∂ Relations (1.10) and (2.6) take the form ∂n c33 ηn2 n=0 = 0, whence 3 = ∂n n=0 = 0. γ=0
γ=0
The x3 axis of a Cartesian coordinate system x1 , x2 , and x3 is chosen to be parallel with the only axis of symmetry of a transversely isotropic medium. In this case, the surface is assumed to be a plane and its equation is x3 = n = 0. in (1.7), which was derived under the notation from (3.13) in [5], the first Using the expression for Nξ D equation of system (1.6) (with k = 0) enables us to find both the surface eikonal equation and the coefficient m10 : 1/3 c44 2(−00 )2t 2 2 (−00 )t = (∇s 00 ) , m10 = − , (3.16) ρ c33/ρR e n=0
∂ where c33 = P 2 /(c11 − c44 ), P 2 ≡ (c11 − c44 )c33 − (c13 + c44 )2 , R2 e = ∂n ln cρ44 , and the inequality R e > 0 n=0
∂θ1,2 must be fulfilled. For the quantities ∂n n=0 of (1.20), we have (see [5]) γ=0
(θ1 )2n = −
P2 (∇s )2 , c33 c44
(θ2 )2n = −
m (∇s )2 , c44
m ≡ c66 − c44 .
(3.17)
If m > 0 (i.e., c44 < c66 ), then the inequality P 2 > 0 holds if (c12 + c66 )c33 > (c13 + c66 )2 .
(3.18)
Inequalities (3.14) also hold if c13 > 0.
∂θ1,2 Thus, if m > 0 and inequality (3.18) is satisfied, then n=0 is purely imaginary. The principal term of ∂n γ=0
0 = ϕ0 η, the asymptotics of a surface wave (individual mode) is then a slow transverse wave with amplitude A where η = ηnn0 . The function ψ0 in (1.11) is proportional to ϕ0 and is found from nonhomogeneous system (2.12) in the form ψ0 = (−i)
ϕ0 D2 m10 , |(θ1 )n |c33 c244 (c11 − c44 )
(3.19)
where |(θ1 )n | = 0 and D ≡ (c11 − c44 )c33 − c13 (c13 + c44 ) > 0, c11 − c44 = c12 + c66 + m > 0 if m > 0 and (3.18) holds. 0 = ψ0 η + ϕ0 σ in (3.12) and applying some transformations, for the function E from Eq. (3.12) Substituting B we find 2(−00 )2t D2 E=i . |(θ1 )n |c33 c244 (c11 − c44 )Re n=0 γ=0
4061
Integration of Eq. (3.12) along an ST ray yields ϕ0 = χ0 (α1 , α2 )
|m10 | ρJ
1/2
−i 2(−1
00 )t
e
M
(Im E)ds
M0
,
(3.20)
where ds = dt, and integration under the exponent proceeds along an ST ray. Expression (3.20) is a modified formula of the ST ray method and is similar to the formula that holds for the principal term of the asymptotics D(x1 ,x2 ) for surface waves of SV type in an isotropic medium (see [3]). In (3.20), J = D(α is the Jacobian of the 1 ,α2 ) 0 transformation of the coordinates x1 and x2 into ray coordinates α1 and α2 , χ (α1 , α2) = const along an ST ray. It is easy to verify that the exponent in (3.20) is identical to that in [3, 9] if the medium is isotropic. Appendix ∂θ2 1 1. We give a proof of formulas (2.2) under the condition that ∂θ ∂n = ∂n for n = 0 and γ = 0. Consider 2 . This yields Eq. (1.20), for example, for κ =κ 1 , and find the scalar product of it by κ i
ρ2t ( κ1 , κ 2 ) = aijαβ κ j2 (θ1i θ1α κ1β ) = ((θ1 )3 − (θ2 )3 )a3jαβ κ j2 (θ1α κ1β ) + aijαβ (κ j2 θ2 )θ1α κ1β j β = ((θ1 )3 − (θ 2 )3 ))a3jαβ κ j2 (θ1α κ1β ) + ((θ1 )3 − (θ 2 )3 )aij3β (κ j2 θ i2 )κ1β + aijαβ (θ i2 θ α 2 κ 2 )κ1 .
(A.1)
β 2 Since aij3β = a3βij and aijαβ θ i2 θ α 2 κ 2 = ρt (κ 2 )j (see (1.20)), it follows from (1), where, by (1.8), (θ1 )3 = (θ2 )3 , that the second relation in (2.2) is true and thus the first one is also true. The validity of relation (2.1) is proved in a similar way (see [6]). 2 ) implies, in particular, relation (2.2a). We also note that expression (A.1) (with κ 1 = κ
We multiply (1.20) by κ 2. Consider the diagonal elements (sr , κ r ), r = 1, 2, of the matrix A. r and put the resulting expression in the form ρ2t | κ|2 = aijαβ θi θα κ j κ β = (θ3 − θ 3 )(s, κ ) + aijαβ (θ i κ j )(θα κ β ), where by s, κ , and θ we mean sr , κ r and θr , r = 1 (or r = 2); (s, κ ) = sj κ ¯ j. Now assume that (r0 , κ 1 ) = 0, (r0 , κ 2 ) = 0,
(A.2)
(A.3)
1 ) = 0 (see part III in the proof of the theorem). and let (s1 , κ In general, it follows from (A.2) and (1.20) that θ = θ, i.e., Im θ = 0. Taking (1.18) into account, we conclude that (s1 , κ 1 ) = 0. Now we show that this holds also in the special case of anisotropy where relations (A.3) are valid. By way of example, we consider a transversely isotropic medium. On Σ, in the basis of the orthogonal vectors n0 , ∇s , and ζ (see Sec. 3), we have (see [5]) κ 2 = (κn , κ ∇ , 0), κ 1 = (0, 0, κζ). s
Using (3.13) and some trivial transformations, we conclude that (s1 , κ 1 ) = 0,
(s2 , κ 2 ) = 0,
where Re(sr , κ r ) = 0, r = 1, 2. The paper was supported by the Russian Foundation for Basic Research under grant No. 99-01-00107. Translated by Z. A. Yanson. REFERENCES 1. V. D. Azhotkin and V. M. Babich, “Love waves propagation along the surface of an anisotropic body of arbitrary shape,” Zap. Nauchn. Semin. LOMI, 165, 9–14 (1987). 4062
2. V. M. Babich and Z. A. Yanson, “On the propagation of Love waves along the surface of an elastic body of arbitrary shape,” Izv. Akad. Nauk SSSR, Fizika Zemli, No. 5, 17–27 (1985). 3. Z. A. Yanson, “Nonstationary waves of Rayleigh type near the surface of an inhomogeneous elastic body,” Zap. Nauchn. Semin. LOMI, 156, 168–183 (1986). 4. Yu. A. Kravtsov, “On a modification of the method of geometric optics,” Izv. Vuzov, Radiofizika, 7, No. 4, 664–673 (1964). 5. Z. A. Yanson, “On the propagation of Rayleigh waves of SV type in transversely isotropic elastic media,” Zap. Nauchn. Semin. POMI, 230, 278–292 (1995). 6. Z. A. Yanson, “High-order approximations of the asymptotics for Love waves of SH type in transversely isotropic elastic media,” Zap. Nauchn. Semin. POMI, 257, 323–345 (1999). 7. G. I. Petrashen, Wave Propagation In Anisotropic Elastic Media [in Russian], Nauka, Leningrad (1980). 8. V. D. Azhotkin, “On the concept of normal curvature in pseudo-Finsler geometry,” Zap. Nauchn. Semin. LOMI, 165, 3–8 (1987). 9. V. P. Krauklis and N. V. Tsepelev, “On the construction of high-frequency asymptotics of a wave field concentrated near the boundary of an elastic medium,” Zap. Nauchn. Semin. LOMI, 34, 72–92 (1973).
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