2 Excitation of the Waves with a Focused Source …
33
where η α =
D
c α
, λ
2
1 =
η 2
α − 1
, α = 1, 2.
Note that in the third equation of the system (2.48) there is only the potential of
shear waves. Let us integrate this expression with respect to y:
∂
2
ψ
∂ x∂ y
+
∂
2
ψ
∂ y 2 + const = 0.
(2.49)
Assuming the integration constant to be zero, expression (2.49) is substituted into
the second equation of system (2.48). We obtain the following system of equations:
η
2
2 − 2
∂
2
ϕ
∂ x 2 + 2
∂
2
ψ
∂ x∂ y
y=0
= −
P
μ
δ(x),
2
∂
2
ϕ
∂ x∂ y
+
η
2
2 + 2
∂
2
ψ
∂ x 2
y=0
= 0.
(2.50)
The solution to the differential equations (2.25), (2.27) will be integrals
ϕ(x, y) =
1
√
2π
∞
−∞
Ae
−ia(x−λ 1 y) da, ψ(x, y) =
1
√
2π
∞
−∞
Be
−ia(x−λ 2 y) da
(2.51)
found by applying the integral Fourier transform.
It is easy to verify that these functions satisfy the conditions of radiation at infinity.
Applying the Fourier transform to the boundary conditions (2.50) and given that
the Fourier transform for the function δ(x) has the form:
F[δ(x)] =
1
√
2π
∞
−∞
δ(x)Be
−iax dx =
1
√
2π
we get the following system of equations:
−a
η
2
2 − 2
A + 2a
2
λ 2 B = −
P
μ
√
2π
, 2a
2
λ 1 A − a
2
η
2
2 + 2
B = 0. (2.52)
We define the values of A and B. For this, from the second equation of system
(2.52), we express B through A:
B =
2λ 1
η
2
2 + 2
A,
then we find