46
5 Elastic Half-Space
τ xz = 0, τ yz = 0, σ z = −p(x, y).
(5.3)
Four conditions (5.2) and (5.3) allow finding the sought functions ϕ, ψ, f , and P .
The first condition (5.3) gives
∂ϕ
∂z
z=0
= −
∂(f + P )
∂x
.
(5.4)
Harmonic functions are in the last formula in the right and left parts, since
derivatives from harmonic functions are also harmonic functions. We should also
note that if this equation takes place at z = 0, then it is also true for any z, since the
harmonic function is totally defined by its value on the surface. By differentiating
formula (5.4) upon x, we obtain
∂ 2 ϕ
∂x∂z
= −
∂ 2
∂x 2 (f + P ).
A similar condition τ yz (x, u, 0) = 0 gives
∂ 2 ψ
∂y∂z
= −
∂ 2
∂y 2 (f + P ).
By summing the left and right parts of the two last formulas taking into account the
harmonic nature of the function (f + P )
∂ 2
∂x 2 +
∂ 2
∂y 2 +
∂ 2
∂z 2
(f + P ) = 0
we find
∂
∂z
∂ϕ
∂x
+
∂ψ
∂y
=
∂ 2
∂z 2 (f + P ).
(5.5)
By integrating Eq. (5.5) and taking into account the condition on the infinity, we
have
∂ϕ
∂x
+
∂ψ
∂y
=
∂
∂z
(f + P ).
Substituting this expression into the right part of formula (5.2) gives
∂P
∂z
= −
1
3 − 4ν
∂
∂z
(2f + P )