SEISMIC WAVES IN LINEAR ELASTIC MEDIUM

M.A.MALKOV RUSSIAN RESEARCH CENTER FOR ARTIFICIAL INTELLIGENCE

Abstract. Using Sobolev’s principle of wave superposition and Smirnov-Sobolev method of functionally invariant decisions of elastodynamics we have found the exact analytical solution for instantaneous point source of seismic waves in linear elastic half-space. We have proved that surface waves propagate with Rayleigh velocity and the front of the waves is perpendicular to direction of propagation since rotations are infinite in the front (but deformations are finite). It means the surface waves especially destructive at earthquakes since they creates the maximum slope of buildings. More that, the energy of surface waves decreases as the square of the distance from the epicenter but the energy of the other waves decreases as the cube of the distance due to reflection from earth surface.

Introduction

The solution of problem of seismic wave propagation in linear elastic medium allows to create a model of earthquakes [1]. The first exact analytical solution of the problem was given by S. L. Sobolev [2]. He used the new method [3] of the solution named now Smirnov-Sobolev’s. He investigated waves created by instantaneous point source of seismic waves. But he got the solution by using potentials of displacements [4]. This solution cannot be used to get numerical results due its complicity. In [5] we had found the solution for instantaneous point source on the earth surface by using functions of dilatations and rotations. The solution was more simple and allowed to get numerical results for any other surface sources by using the Green function. In the paper we have found the solution for body source again using functions of dilatations and rotations.

The paper consists of 4 sections. In first section we give basic equations of elastodynamics in 3-dimension space, formulate initial and boundary equations and find the initial body waves created by the instantaneous point source. In the second section we reduce the solutions of 3-dimension space to the solutions of 2-dimension by using the Sobolev principle of wave superposition. In the third section we represent the received solutions in complex form by using Smirnov-Sobolev’s metod of functionally invariant solutions. These solutions are for initial waves which are incident body waves. The solutions for reflected and surface waves are presented in the four section. We prove that surface waves are especially destructive since rotations are infinite in the waves.

1. Problem statement

1.1. Basic equations in 3-dimension space. Let an instantaneous source of dilatations or rotations arises in elastic half-space at the time t = 0 and at the point x = 0, y = 0, z = −z_{0}.

For t > 0 the wave propagation is described by the equations of elastodynamics,

\frac{∂^{2}u⃗}{∂t^{2}} = c_{l}^{2} div Δ − c_{t}^{2} rot ω⃗, Δ = div u⃗, ω⃗ = rot u⃗  (1)
\frac{∂^{2}Δ}{∂t^{2}} = c_{l}^{2}(\frac{∂^{2}Δ}{∂x^{2}} + \frac{∂^{2}Δ}{∂y^{2}} + \frac{∂^{2}Δ}{∂z^{2}}), \frac{∂^{2}ω⃗}{∂t^{2}} = c_{t}^{2}(\frac{∂^{2}ω⃗}{∂x^{2}} + \frac{∂^{2}ω⃗}{∂y^{2}} + \frac{∂^{2}ω⃗}{∂z^{2}} ) (2)

where c_{l} is the speed of longitudinal waves, c_{t} is the speed of transverse waves, u⃗ is a vector of displacements, Δ is the function of dilatations, ω⃗ is the vector of rotations.

1.2. Initial and boundary conditions. For initial conditions we use the spherical coordinate system ρ, φ, θ, with the pole x = 0, y = 0, z = −z_{0}.

The source of dilatation wave must satisfy the initial conditions at t = 0 (ω_{ρ} ≡ 0):

Δ = δ(ρ), ω_{φ} = ω_{θ} = 0

and two sources of rotation waves must satisfy initial conditions at t = 0:

ω_{φ} = δ(ρ), ω_{θ} = Δ = 0
ω_{θ} = δ(ρ), ω_{φ} = Δ = 0

For boundary conditions we use the Cartesian coordinates. At z = 0 all sources must satisfy the equations

Δ − 2γ^{2}(\frac{∂u_{x}}{∂x} + \frac{∂u_{y}}{∂y} ) = 0, ω_{x} + 2\frac{∂u_{z}}{∂y} = 0, ω_{y} − 2\frac{∂u_{z}}{∂x} = 0  (3)

where γ = \frac{c_{t}}{c_{l}}. The derivatives with respect to z are excluded in the equations since

they cannot be used at z = 0.

We can express the boundary conditions through the functions Δ and ω if we differentiate boundary conditions (3) twice with respect to time and use the basic equations (1) and (2) to remove the time derivatives.

1.3. Dilatation wave at t ⩽ z_{0}/c_{l} and rotation wave at t ⩽ z_{0}/c_{t}. We will use again the spherical coordinate system with the pole x = 0, y = 0, z = −z_{0}. Since Δ depend only on t and ρ then the elastodynamics equation becomes

\frac{∂^{2}Δ}{∂t^{2}} = c_{l}^{2}(\frac{∂^{2}Δ}{∂ρ^{2}} + \frac{2}{ρ} \frac{∂Δ}{∂ρ})  (4)

The solution of the equation is self-similar since there are only two dimensionless values: Δ and ρ/c_{l}t.

Let r = ρ/c_{l}t. Then Eq. (4) becomes

r^{2} \frac{d^{2}Δ}{dr^{2}} + 2r \frac{dΔ}{dr} = \frac{d^{2}Δ}{dr^{2}} + \frac{2}{r} \frac{dΔ}{dr}

or

\frac{d^{2}Δ}{dr^{2}} + \frac{2}{r} \frac{dΔ}{dr} = 0

Using the initial conditions we get the solution of the equation for Δ and by analogy for ω⃗

Δ = \frac{1}{r^{3}} − 1, r = \frac{\sqrt{x^{2} + y^{2} + (z + z_{0})^{2}}}{c_{l}t}

(5)

ω_{φ} = ω_{θ} = \frac{1}{r^{3}} − 1, r = \frac{\sqrt{x^{2} + y^{2} + (z + z_{0})^{2}}}{c_{t}t}

2. Reduction 3-dimensional problems to 2-dimensional

2.1. Sobolev’s principle of wave superposition. Let Δ(x, y, z, t) and ω⃗(x, y, z, t) be waves in three dimension space. Then the waves become waves in two dimensional space, if x equals some x_{0}. This property is a base for Sobolev’s principe of wave superposition.

Let the plane wave Δ(x, y, t) be the solution of equation

\frac{∂^{2}Δ(x, y, t)}{∂t^{2}} = c_{l}^{2}(\frac{∂^{2}Δ(x, y, t)}{∂x^{2}} + \frac{∂^{2}Δ(x, y, t)}{∂y^{2}})

and let Δ_{1}(x_{1}, y_{1}, z_{1}, t) = Δ(x_{1} cos α + z_{1} sin α, y_{1}, t) for some α. Then Δ_{1}(x_{1}, y_{1}, z_{1}, t) is the solution of equation

\frac{∂^{2}Δ_{1}(x_{1}, y_{1}, z_{1}, t)}{∂t^{2}} = c_{l}^{2}(\frac{∂^{2}Δ_{1}(x_{1}, y_{1}, z_{1}, t)}{∂x_{1}^{2}} + \frac{∂^{2}Δ_{1}(x_{1}, y_{1}, z_{1}, t)}{∂y_{1}^{2}} + \frac{∂^{2}Δ_{1}(x_{1}, y_{1}, z_{1}, t)}{∂z_{1}^{2}})

Indeed, since x = x_{1} cos α + z_{1} sin α and y = y_{1} then

\frac{∂^{2}Δ_{1}}{∂x_{1}^{2}} = \frac{∂^{2}Δ}{∂x^{2}} cos^{2} α, \frac{∂^{2}Δ_{1}}{∂y_{1}^{2}} = \frac{∂^{2}Δ}{∂y^{2}}, \frac{∂^{2}Δ_{1}}{∂z_{1}^{2}} = \frac{∂^{2}Δ}{∂x^{2}} sin^{2} α, \frac{∂^{2}Δ_{1}}{∂t^{2}} = \frac{∂^{2}Δ}{∂t^{2}}

and

\frac{∂^{2}Δ_{1}}{∂x_{1}^{2}} + \frac{∂^{2}Δ_{1}}{∂y_{1}^{2}} + \frac{∂^{2}Δ_{1}}{∂z_{1}^{2}} = \frac{∂^{2}Δ}{∂x^{2}} + \frac{∂^{2}Δ}{∂y^{2}} = c_{l}^{2} \frac{∂^{2}Δ_{1}}{∂t^{2}}

where Δ_{1} = Δ_{1}(x_{1}, y_{1}, z_{1}, t), Δ = Δ(x, y, t).

Sobolev called

Δ_{1}(x, y, z, t) = \frac{1}{2π} ∫_{0}^{2π}Δ(x cos α + z sin α, y, t)dα

wave superposition. This wave is a solution of Eq. (2) and Δ_{1} = 1 for Δ = 1. By analogy

ω_{1_{φ}}(x, y, z, t) = ω_{1_{θ}}(x, y, z, t) = \frac{1}{2π} ∫_{0}^{2π}ω(x cos α + z sin α, y, t)dα

If we know a two-dimension solution then we can find the three-dimension solution by integrating. If we know a three-dimension solution then we shall solve the integral equation (see [6]).

Let Δ_{00}(x, y, t) and ω_{00}(x, y, t) be solutions of integral equations for the dilatation and rotation waves.

So our problem is reduced to the plane problem. Further we solve the plane problem.

2.2. Basic equations in 2-dimension space. We use the coordinate system presented in Fig. 1. Then elastodynamics equations are

\frac{∂^{2}u_{x}}{∂t^{2}} = c_{l}^{2} \frac{∂Δ}{∂x} − c_{t}^{2} \frac{∂ω}{∂y} , \frac{∂^{2}u_{y}}{∂t^{2}} = c_{l}^{2} \frac{∂Δ}{∂y} + c_{t}^{2} \frac{∂ω}{∂x}  (6)
Δ = \frac{∂u_{x}}{∂x} + \frac{∂u_{y}}{∂y} , ω = \frac{∂u_{y}}{∂x} − \frac{∂u_{x}}{∂y}  (7)
\frac{∂^{2}Δ}{∂t^{2}} = c_{l}^{2}(\frac{∂^{2}Δ}{∂x^{2}} + \frac{∂^{2}Δ}{∂y^{2}}), \frac{∂^{2}ω}{∂t^{2}} = c_{t}^{2}(\frac{∂^{2}ω}{∂x^{2}} + \frac{∂^{2}ω}{∂y^{2}} )  (8)

I^{II}^{III}^{IV}  ^{IV}  x y

where ω = ω_{z}.

2.3. Boundary conditions. When t > y_{0}/c_{l}t for the dilatation wave and t > y_{0}/c_{t}t for the rotation wave then boundary conditions must be used at y = 0,

(1 − 2γ^{2})\frac{∂u_{x}}{∂x} + \frac{∂u_{y}}{∂y} = 0, \frac{∂u_{y}}{∂x} + \frac{∂u_{x}}{∂y} = 0  (9)

Since the derivatives with respect to y at y = 0 cannot be got then we will replace them by functions Δ and ω. Using (7) we have:

2γ^{2} \frac{∂u_{x}}{∂x} − Δ = 0, 2\frac{∂u_{y}}{∂x} − ω = 0  (10)

3. Complex solutions

3.1. Functionally invariant solutions. Further we will present the solutions as complex by using Sobolev-Smirnov method [2].

The solution of Laplace equation is Re f(z), where z = x + iy, f is an arbitrary analytic function, and z depends on x and y linearly.

Sobolev and Smirnov had found the analogous solutions of Eq. (8). They named the solutions functionally invariant.

So the solution of Eq. (8) is Re f(Ω), where Ω is a function of x, y, t and f is an arbitrary analytic function. Then Eq. (8) becomes

\frac{∂^{2}f(Ω)}{∂Ω^{2}}((\frac{∂Ω}{∂x})^{2}+ (\frac{∂Ω}{∂y})^{2}− \frac{1}{c^{2}} (\frac{∂Ω}{∂t})^{2}) + \frac{∂f(Ω)}{∂Ω}(\frac{∂^{2}Ω}{∂x^{2}} + \frac{∂^{2}Ω}{∂y^{2}} − \frac{1}{c^{2}} \frac{∂^{2}Ω}{∂t^{2}}) = 0

where c = c_{l} or c = c_{t} Since f is an arbitrary function then

(\frac{∂Ω}{∂x})^{2}+ (\frac{∂Ω}{∂y})^{2}− \frac{1}{c^{2}} (\frac{∂Ω}{∂t})^{2}= 0, \frac{∂^{2}Ω}{∂x^{2}} + \frac{∂^{2}Ω}{∂y^{2}} − \frac{1}{c^{2}} \frac{∂^{2}Ω}{∂t^{2}} = 0

The solution of these equations is

δ ≡ m(Ω)x + n(Ω)y + k(Ω) − l(Ω)ct = 0  (11)

where m, n, k are arbitrary analytic functions of Ω and

l^{2}(Ω) = m^{2}(Ω) + n^{2}(Ω)

Indeed,

\frac{∂Ω}{∂x} = −\frac{m(Ω)}{δ′}, \frac{∂Ω}{∂y} = −\frac{n(Ω)}{δ′}, \frac{1}{c} \frac{∂Ω}{∂t} = −\frac{l(Ω)}{δ′}

where ′ is the differentiation with respect to Ω. Then

(\frac{∂Ω}{∂x})^{2} + (\frac{∂Ω}{∂y})^{2} − \frac{1}{c}(\frac{∂Ω}{∂t})^{2} = 0

Since

\frac{∂^{2}Ω}{∂x^{2}} = \frac{∂}{∂Ω} (−\frac{m(Ω)}{δ′}) \frac{∂Ω}{∂x} + \frac{∂}{∂x} (−\frac{m(Ω)}{δ′}) =

= (\frac{m(Ω)}{δ′})′ \frac{m(Ω)}{δ′} + \frac{m′(Ω)m(Ω)}{δ′^{2}} = \frac{1}{δ′} (\frac{m^{2}(Ω)}{δ′})′

by analogy

\frac{∂^{2}Ω}{∂y^{2}} = \frac{1}{δ′} (\frac{n^{2}(Ω)}{δ′})′ , \frac{1}{c^{2}} \frac{∂^{2}Ω}{∂t^{2}} = \frac{1}{δ′} (\frac{l^{2}(Ω)}{δ′})′

then

\frac{∂^{2}Ω}{∂x^{2}} + \frac{∂^{2}Ω}{∂y^{2}} − \frac{1}{c^{2}} \frac{∂^{2}Ω}{∂t^{2}} = 0

Using the new function θ = \frac{m(Ω)}{l(Ω)} in (11) and replacing \frac{k(Ω)}{l(Ω)} by the new function

k(θ) we get

xθ ± iy\sqrt{θ^{2} − 1} + k(θ) − ct = 0  (12)

The sign + before iy corresponds waves moving to the positive direction of axis y, the sign – corresponds waves moving to the opposite direction.

3.2. Incident waves. We call waves falling on the surface y = 0 incident. We shall represent the waves in the complex form.

The incident wave Δ_{00}(x, y, t) equals Re Δ_{00}(θ_{1}) and ω_{00}(x, y, t) equals Re ω_{00}(θ_{2}), where θ_{1} and θ_{2} are defined by the equation (12),

xθ_{1} − iy\sqrt{θ_{1}^{2} − 1} + k(θ_{1}) − c_{l}t = 0, \frac{x}{γ}θ_{2} − i\frac{y}{γ} \sqrt{θ_{2}^{2} − 1} + \frac{k(θ_{2})}{γ} − c_{l}t = 0

The functions k(θ_{1}) and k(θ_{2}) are different in the equations.

We will replace θ/γ in the last of the two equations by the new θ_{2} and replace k(θ_{2})/γ by the new k(θ_{2}) with the new θ_{2}. So the equations become

xθ_{1} − iy\sqrt{θ_{1}^{2} − 1} + k(θ_{1}) − c_{l}t = 0, xθ_{2} − iy\sqrt{θ_{2}^{2} − 1/γ^{2}} + k(θ_{2}) − c_{l}t = 0 (13)

Then k(θ_{1}) = iy_{0}\sqrt{θ_{1}^{2} − 1} and k(θ_{2}) = iy_{0}\sqrt{θ_{1}^{2} − 1/γ^{2}} due to self-similarity,

xθ_{1} + iy\sqrt{θ_{1}^{2} − 1} + iy_{0}\sqrt{θ_{1}^{2} − 1} − c_{l}t = 0

(14)

xθ_{2} + iy\sqrt{θ_{2}^{2} − 1/γ^{2}} + iy_{0}\sqrt{θ_{2}^{2} − 1/γ^{2}} − c_{l}t = 0

We call Δ_{00}(θ_{1}), ω_{00}(θ_{2}1) amplitudes of the waves and θ_{1}, θ_{2} phases of the waves. The phases do not equal since c_{l} ≠ c_{t}. It is important that by formality

Δ_{00}(θ_{1}) = ω_{00}(θ_{2}) if θ_{1} = θ_{2} since then c_{l} = c_{t}. But if c_{l} = c_{t} then accordingly (5) Δ = ω_{φ} = ω_{θ}. Therefore Δ_{00}(x, y, t) = ω_{00}(x, y, t) in this case. Further we will use the formal equation Δ_{00}(θ_{1}) = ω_{00}(θ_{2}) for θ_{1} = θ_{2}.

4. Reflected waves

4.1. Phases of reflected waves for incident dilatation wave. There are two reflected waves created by the incident dilatation wave. One of reflected waves is dilatation and the other is rotation. The dilatation wave will be named Δ_{10}(θ_{1}) and the rotation wave will be named ω_{10}(θ_{2}), where θ_{1} and θ_{2} are solutions of equations (13). Since phases of reflected waves must be equal to the phase of incident wave (14) at y = 0 then k(θ_{1}) = iy_{0}\sqrt{θ_{1}^{2} − 1} and k(θ_{2}) = iy_{0}\sqrt{θ_{2}^{2} − 1},

xθ_{1} − iy\sqrt{θ_{1}^{2} − 1} + iy_{0}\sqrt{θ_{1}^{2} − 1} − c_{l}t = 0

(15)

xθ_{2} − iy\sqrt{θ_{2}^{2} − 1/γ^{2}} + iy_{0}\sqrt{θ_{2}^{2} − 1} − c_{l}t = 0

4.2. Phases of reflected waves for incident rotation wave. There are two reflected waves created by the incident rotation wave. One of reflected waves is dilatation and the other is rotation. The dilatation wave is Δ_{01}(θ_{1}) and the rotation wave is ω_{01}(θ_{2}), where θ_{1} and θ_{2} are solutions of equations (13). Since phases of reflected waves must be equal to the phase of incident wave (14) at y = 0 then k(θ_{1}) = iy_{0}\sqrt{θ_{1}^{2} − 1/γ^{2}} and k(θ_{2}) = iy_{0}\sqrt{θ_{2}^{2} − 1/γ^{2}},

xθ_{1} − iy\sqrt{θ_{1}^{2} − 1} + iy_{0}\sqrt{θ_{1}^{2} − 1/γ^{2}} − c_{l}t = 0
xθ_{2} − iy\sqrt{θ_{2}^{2} − 1/γ^{2}} + iy_{0}\sqrt{θ_{2}^{2} − 1/γ^{2}} − c_{l}t = 0

4.3. Boundary conditions through functions Δ and ω. The boundary conditions (9) contain derivatives of u_{x} and u_{y}. We must exclude the derivatives to replace them by derivatives of Δ and ω. For that we differentiate the boundary conditions with respect to t and use equations (6) to exclude derivatives with respect to t. Then

(1 − 2γ^{2})\frac{∂^{2}Δ}{∂x^{2}} + \frac{∂^{2}Δ}{∂y^{2}} + 2γ^{4} \frac{∂^{2}ω}{∂x∂y} = 0, 2\frac{∂^{2}Δ}{∂x∂y} + γ^{2} (\frac{∂^{2}ω}{∂x^{2}} − \frac{∂^{2}ω}{∂y^{2}} ) = 0 (16)

4.4. Derivatives of Δ and ω. Before using the boundary conditions we shall find derivatives of Δ and ω.

At first we denote the left sides of (15) by δ(t, x, y, θ_{1}),

δ(t, x, y, θ_{1}) ≡ xθ_{1} ± iy\sqrt{θ_{1}^{2} − 1} + iy_{0}\sqrt{θ_{1}^{2} − 1} − c_{l}t = 0

where ± means + for incident waves and – for reflecting waves. But below ∓ means – for incident waves and + for reflecting waves.

Then

\frac{∂Δ(θ_{1})}{∂x} = Δ′(θ_{1})\frac{∂θ_{1}}{∂x} = −Δ′(θ_{1})\frac{θ_{1}}{δ′}

\frac{∂Δ(θ_{1})}{∂y} = Δ′(θ_{1})\frac{∂θ_{1}}{∂y} = ∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}

where δ′ = δ′(x, y, t, θ) and ′ is the differentiation with respect to θ_{1}.

The last parts of the equalities are composite. So

\frac{∂^{2}Δ(θ_{1})}{∂x^{2}} = \frac{∂}{∂θ_{1}} (−Δ′(θ_{1})\frac{θ_{1}}{δ′} ) \frac{∂θ_{1}}{∂x} + \frac{∂}{∂x} (−Δ′(θ_{1})\frac{θ_{1}}{δ′} ) =

= (Δ′(θ_{1})\frac{θ_{1}}{δ′} )′ \frac{θ_{1}}{δ′} + Δ′(θ_{1})\frac{θ_{1}}{δ^{′2}} = \frac{1}{δ′} (Δ′(θ_{1})\frac{θ_{1}^{2}}{δ′} )′

\frac{∂^{2}Δ(θ_{1})}{∂y^{2}} = \frac{∂}{∂θ_{1}} (∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}) \frac{∂θ_{1}}{∂y} + \frac{∂}{∂y} (∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}) =

= − (Δ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′})′ \frac{\sqrt{θ_{1}^{2} − 1}}{δ′} − Δ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}\sqrt{θ_{1}^{2} − 1}′}{δ′} =

= −\frac{1}{δ′} (Δ′(θ_{1})\frac{θ_{1}^{2} − 1}{δ′})′

\frac{∂^{2}Δ(θ_{1})}{∂x∂y} = \frac{∂}{∂x} (∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}) =

= \frac{∂}{∂θ_{1}} (∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}) \frac{∂θ}{∂x} + \frac{∂}{∂x} (∓iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′}) =

= (±iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ′})′ \frac{θ}{δ′} ± iΔ′(θ_{1})\frac{\sqrt{θ_{1}^{2} − 1}}{δ^{′2}} = ±\frac{1}{δ′} (iΔ′(θ_{1})\frac{θ_{1}\sqrt{θ_{1}^{2} − 1}}{δ′})′

By analogy

\frac{∂^{2}ω(θ_{2})}{∂x^{2}} = \frac{1}{δ′} (ω′(θ_{2})\frac{θ_{2}^{2}}{δ′} )′ , \frac{∂^{2}ω(θ_{2})}{∂y^{2}} = −\frac{1}{δ′} (ω′(θ_{2})\frac{θ_{2}^{2} − 1/γ^{2}}{δ′} )′ ,

\frac{∂^{2}ω(θ_{2})}{∂x∂y} = ±\frac{1}{δ′} (iω′(θ_{2})\frac{θ_{2}\sqrt{θ_{2}^{2} − 1/γ^{2}}}{δ′} )′

4.5. Amplitudes of reflected waves for incident dilatation wave. Phases of all waves are equal on the surface y = 0: θ_{1} = θ_{2} = θ. Amplitudes are defined by boundary conditions (16),

(1 − 2γ^{2})\frac{1}{δ′} (Δ′(θ)\frac{θ^{2}}{δ′} )′ + \frac{1}{δ′} (Δ′(θ)\frac{1 − θ^{2}}{δ′} )′ ± 2γ^{4} \frac{1}{δ′} (iω′(θ)\frac{θ\sqrt{θ^{2} − 1/γ^{2}}}{δ′} )′ = 0

±2\frac{1}{δ′} (iΔ′(θ)\frac{θ\sqrt{θ^{2} − 1}}{δ′})′ + γ^{2} (\frac{1}{δ′} (ω′(θ)\frac{θ^{2}}{δ′} )′ − \frac{1}{δ′} (ω′(θ)\frac{1/γ^{2} − θ^{2}}{δ′} )′) = 0

Integrating the equations we get

(1 − 2γ^{2})Δ′(θ)θ^{2} + Δ′(θ)(1 − θ^{2}) ± 2iγ^{4}ω′(θ)θ\sqrt{θ^{2} − 1/γ^{2}} = C_{1}δ′
±2iΔ′(θ)θ\sqrt{θ^{2} − 1} + γ^{2} (ω′(θ)θ^{2} − ω′(θ)(1/γ^{2} − θ^{2})) = C_{2}δ′

where C_{1} and C_{2} do not depend on θ. But the left parts of the equalities do not explicitly depend on x, y, t, and the right parts depend on. This means that C_{1} = C_{2} = 0. So we have

(1 − 2γ^{2}θ^{2})Δ′(θ) ± 2iγ^{4}ω′(θ)θ\sqrt{θ^{2} − 1/γ^{2}} = 0
±2iΔ′(θ)θ\sqrt{θ^{2} − 1} − (1 − 2γ^{2}θ^{2})ω′(θ) = 0

Since Δ = Δ_{00} + Δ_{10}, ω(θ) = ω_{10}(θ), and ±Δ(θ) = (Δ_{00}(θ) − Δ_{10}(θ)), ±ω = −ω_{10} then

(1 − 2γ^{2}θ^{2})Δ′_{10}(θ) − 2iγ^{4}θ\sqrt{θ^{2} − 1/γ^{2}}ω′_{10}(θ) = −(1 − 2γ^{2}θ^{2})Δ′_{00}(θ)
−2iθ\sqrt{θ^{2} − 1}Δ′_{10}(θ) − (1 − 2γ^{2}θ^{2})ω′_{10}(θ) = −2iθ\sqrt{θ^{2} − 1}Δ′_{00}(θ)

This is the system of equations. The solutions of the system are

Δ′_{10}(θ) = −\frac{(1 − 2γθ^{2})^{2} + 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}}{(1 − 2γθ^{2})^{2} − 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}} Δ′_{00}(θ)

(17)

ω′_{10}(θ) = \frac{−4iθ(1 − 2γ^{2}θ^{2})\sqrt{θ^{2} − 1}}{(1 − 2γθ^{2})^{2} − 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}} ω′_{00}(θ)

In the last equation we have used the formal equality Δ′_{00}(θ) = ω′_{00}(θ) to replace Δ′_{00}(θ) by ω′_{00}(θ).

4.6. Amplitudes of reflected waves for incident rotation wave. Amplitudes are defined by boundary conditions (16). Since Δ = Δ_{01}, ω = ω_{00} + ω_{01}, and ±Δ = −Δ_{01}, ±ω = ω_{00} − ω_{01} then

(1 − 2γ^{2}θ^{2})Δ′_{01}(θ) − 2iγ^{4}θ\sqrt{θ^{2} − 1/γ^{2}}ω′_{01}(θ) = −2iγ^{4}θ\sqrt{θ^{2} − 1/γ^{2}}ω′_{00}(θ)

−2iθ\sqrt{θ^{2} − 1}Δ′_{01}(θ) − (1 − 2γ^{2}θ^{2})ω′_{01}(θ)) = (1 − 2γ^{2}θ^{2})ω′_{00}(θ)

The solution of the system of equations is

Δ′_{01}(θ) = \frac{4iγ^{4}θ(1 − 2γ^{2}θ^{2})\sqrt{θ^{2} − 1/γ^{2}}}{(1 − 2γθ^{2})^{2} − 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}} Δ′_{00}(θ)

(18)

ω′_{01}(θ) = \frac{−4iθ(1 − 2γ^{2}θ^{2})\sqrt{θ^{2} − 1}}{(1 − 2γθ^{2})^{2} − 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}} ω′_{00}(θ)

In the first equation we have used the formal equality Δ′_{00}(θ) = ω′_{00}(θ) to replace ω′_{00}(θ) by Δ′_{00}(θ).

4.7. Rayleigh waves. The expressions (17) and (18) have the same denominator and zeros of the denominator are singular points. We will prove that the denominator has zeros which are points moving along surface y = 0 with the Rayleigh velocity in different directions.

So we shall find all decisions of the equation

(1 − 2γ^{2}θ^{2})^{2} = 4γ^{4}θ^{2}\sqrt{θ^{2} − 1}\sqrt{θ^{2} − 1/γ^{2}}

If we denote 1/γθ by ζ then the equation becomes

(ζ^{2} − 2)^{2} = 4\sqrt{1 − ζ^{2}γ^{2}}\sqrt{1 − ζ^{2}}  (19)

Raising both parts of the equality in square we get

ζ^{6} − 8ζ^{4} + 8ζ^{2}(3 − 2γ^{2}) − 16(1 − γ^{2}) = 0  (20)

The decision of the equation has roots as of (19) as of

(ζ^{2} − 2)^{2} = −4\sqrt{1 − ζ^{2}γ^{2}}\sqrt{1 − ζ^{2}}  (21)

The determinant of the equation (20) is \sqrt{q^{2} + p^{3}} where q = \frac{8}{27}(17 − 45γ^{3}),

p = \frac{8}{9}(1 − 6γ^{2}), and the determinant equals zero at γ^{2} = 0.321. This means that equation (20) has one real root and for γ > 0.321 has two complex roots additionally. But the complex roots belong only to the equation (21).

The real root is very known and it equals c_{R}/c_{t}, where c_{R} is the velocity of Rayleigh waves. Then θ = c_{l}/c_{R}. We denote this θ by θ_{R}. Therefore the denominator in (17) and (18) has only the real roots θ = ±θ_{R}.

We shall find the decisions of (17) and (18) near θ_{R}.

The denominator equals X_{0}(θ^{2} − θ_{R}^{2}) + O(θ^{2} − θ_{R}^{2})^{2}, where

X_{0} = −4γ^{2} \frac{1 − 2(3 − 2γ^{2})γ^{2}θ_{R}^{2} + 6(1 − β^{2})β^{4}θ_{R}^{4}}{(2γ^{2}θ_{R}^{2} − 1)^{2}}

Then

Δ_{10}(θ) = iX_{1} log |θ^{2} − θ_{R}^{2}| + O(θ^{2} − θ_{R}^{2})^{0}, ω_{10}(θ) = X_{2} log |θ^{2} − θ_{R}^{2}| + O(θ^{2} − θ_{R}^{2})^{0}

Δ_{01}(θ) = iX_{3} log |θ^{2} − θ_{R}^{2}| + O(θ^{2} − θ_{R}^{2})^{0}, ω_{01}(θ) = iX_{4} log |θ^{2} − θ_{R}^{2}| + O(θ^{2} − θ_{R}^{2})^{0}

where

X_{1} = −\frac{(2γ^{2}θ_{R}^{2} − 1)^{2}}{θ_{R}X_{0}} \frac{Δ′_{00}(θ_{R})}{i}, X_{2} = \frac{2(2γ^{2}θ_{R}^{2} − 1)\sqrt{θ_{R}^{2} − 1}}{X_{0}} \frac{ω′_{00}(θ_{R})}{i}

X_{3} = −\frac{2γ^{4}(2θ_{R}^{2} − 1)\sqrt{θ_{R}^{2} − 1/γ^{2}}}{X_{0}} \frac{Δ′_{00}(θ_{R})}{i}, X_{4} = −\frac{(2γ^{2}θ_{R}^{2} − 1)^{2}}{θ_{R}X_{0}} \frac{Δ′_{00}(θ_{R})}{i}

The value of Δ′_{00}(θ_{R})/i mast be real since Re Δ(θ) must be finite and since Re i log |θ^{2} − θ_{R}^{2}| is finite but Re log |θ^{2} − θ_{R}^{2}| is infinite. This means that Δ_{10}(θ), Δ_{01}(θ), ω_{01}(θ), are finite and ω_{10}(θ) is infinite at θ = θ_{R}. But all strains are finite at this point in spite of infinity of ω_{10}(θ).

4.7. Destructive waves. The Rayleigh waves are especially destructive at earthquakes since they are surface and their energy decreases as the square of the distance from the epicenter but the energy of the other waves decreases as the cube of the distance due to reflection from earth surface. More that, earth surface is perpendicular to the direction of propagation of seismic waves at Rayleigh point. So the Rayleigh wave creates the maximum slope of buildings.

We shall prove that displacement at Rayleigh point is perpendicular to direction of propagation.

Let the form of the surface be given by the curve Y (X), where Y = u_{y} and X = x + u_{x}.

At Rayleigh point

\frac{dY}{dX} = \frac{\frac{dY}{dx}}{\frac{dX}{dx}} = \frac{\frac{∂u_{y}}{∂x}}{1 + \frac{∂u_{x}}{∂x}}

Using the boundary condition (10)

2γ^{2} \frac{∂u_{x}}{∂x} − Δ = 0, 2\frac{∂u_{y}}{∂x} − ω = 0,

and using infinity of ω and finiteness of Δ we get \frac{dY}{dX} = ∞. This means that displacements are perpendicular to direction of wave propagation at Rayleigh point.

4.8. Final solutions. We shall extend the solutions (17) and (18) analytically out of θ_{1} = θ and θ_{2} = θ.

So we have finally for reflected waves generated by incident wave Δ_{00}(θ_{1})

Δ′_{10}(θ_{1}) = −\frac{(1 − 2γθ_{1}^{2})^{2} + 4γ^{4}θ_{1}^{2}\sqrt{θ_{1}^{2} − 1}\sqrt{θ_{1}^{2} − 1/γ^{2}}}{(1 − 2γθ_{1}^{2})^{2} − 4γ^{4}θ_{1}^{2}\sqrt{θ_{1}^{2} − 1}\sqrt{θ_{1}^{2} − 1/γ^{2}}} Δ′_{00}(θ_{1})

(22)

ω′_{10}(θ_{2}) = \frac{−4iθ_{2}(1 − 2γ^{2}θ_{2}^{2})\sqrt{θ_{2}^{2} − 1}}{(1 − 2γθ_{2}^{2})^{2} − 4γ^{4}θ_{2}^{2}\sqrt{θ_{2}^{2} − 1}\sqrt{θ_{2}^{2} − 1/γ^{2}}} ω′_{00}(θ_{2})

where

xθ_{1} − i(y − y_{0})\sqrt{θ_{1}^{2} − 1} − c_{l}t = 0, xθ_{2} − iy\sqrt{θ_{2}^{2} − 1/γ^{2}} + iy_{0}\sqrt{θ_{2}^{2} − 1} − c_{l}t = 0

By analogy we get for reflected waves generated by incident wave ω_{00}(θ_{2})

Δ′_{01}(θ_{1}) = \frac{4iγ^{4}θ_{1}(1 − 2γ^{2}θ_{1}^{2})\sqrt{θ_{1}^{2} − 1/γ^{2}}}{(1 − 2γθ_{1}^{2})^{2} − 4γ^{4}θ_{1}^{2}\sqrt{θ_{1}^{2} − 1}\sqrt{θ_{1}^{2} − 1/γ^{2}}} Δ′_{00}(θ_{1})

(23)

ω′_{01}(θ_{2}) = \frac{−4iθ_{2}(1 − 2γ^{2}θ_{2}^{2})\sqrt{θ_{2}^{2} − 1}}{(1 − 2γθ_{2}^{2})^{2} − 4γ^{4}θ_{2}^{2}\sqrt{θ_{2}^{2} − 1}\sqrt{θ_{2}^{2} − 1/γ^{2}}} ω′_{00}(θ_{2})

where

xθ_{1} − iy\sqrt{θ_{1}^{2} − 1} + iy_{0}\sqrt{θ_{1}^{2} − 1/γ^{2}} − c_{l}t = 0, xθ_{2} − i(y − y_{0})\sqrt{θ_{2}^{2} − 1/γ^{2}} − c_{l}t = 0

Integrating (22) and (23) we can get reflected waves Δ_{10}(θ_{1}), Δ_{01}(θ_{1}), ω_{10}(θ_{2}), ω_{01}(θ_{2}). Using the real parts of them we get Δ_{10}(x, y, t), Δ_{01}(x, y, t), ω_{10}(x, y, t), ω_{01}(x, y, t). Then using Sobolev’s principle of wave superposition we can finally have reflected waves Δ_{10}(x, y, z, t), Δ_{01}(x, y, z, t), ω_{10_{φ}}(x, y, z, t), ω_{01_{φ}}(x, y, z, t), ω_{10_{θ}}(x, y, z, t),ω_{01_{θ}}(x, y, z, t), by integrating.

References

1. Bolt Bruce, 2003, Earthquakes, Freeman, N.Y., 5th edition.

2. Sobolev S. L., 1930, ”On one limit problem of logarithmic potential and application of the problem to reflection of plane elastic waves,” Trudy Seismologic Institute, 11, pp. 1-18 (in Russian).

3. Smirnov V. I., and Sobolev, S. L., 1932, ”New Method of Solution for Plane problem of Elastic Vibration,” Trudy Seismologic Institute, 20, pp. 1-31 (in Russian).

4. Sobolev S.L., 1937, ”Some problems of vibration propagation theory,”, chapter 12 in Differential and Integral Equations of mechanics and physics by Frank Ph., Mises R., M.-L. (in Russian).

5. Malkov M.A., 1966, ”Asymptotic of source function of elastic perturbations in plane layer”, International Congress of Mathematicians, abstracts of brief scientific communications 12, p. 15.

6. Sobolev S. L., 1954, ”Some notes on numerical solution of integral equations”, Uspehi mat. nauk, 9(3), pp. 3-50 (in Russian).