Read Back Signals in Magnetic Recording - Research Group Fidler

Appendix A
In case we have the function as given in (A.2), we have to divide the volume V into a small
sphere with radius ε around the singularity r ′ denoted with Kε ( r ′ ) and the rest of V. Then
the theorem (A.1) can be written as
∫ grad f dV + ∫ grad f dV = �∫ f dA− �∫ f dA+ � ∫ f dA.
(A.7)
V / Kε( r′ ) Kε( r′ ) ∂V ∂Kε( r′ ) ∂Kε(
r′
)
For the volume V / Kε ( r ′ ) with the outer surface ∂ V and the inner surface ∂Kε ( r ) the
theorem holds, because the singularity is excluded. To prove the theorem, we only have to
show, that
∫ ∫
grad f dV = � f dA
. (A.8)
Kε( r′ ) ∂Kε(
r′
)
The equality is valid, if both sides vanish for ε → 0 . Without loss of generality, it is sufficient
to show the equality for the first vector component. Then the integrand of the left hand side is
( f ) = 3
grad x
x − x′
, (A.9)
r−r′ which is asymmetric around r ′ . Therefore the left hand side in (A.8) vanishes. For the right
hand side we have
1
⎛ ⎞
4πε
= ∫ dA = f dA ≥ ⎜ f d ⎟
Kε( ′ − ′ ∫ ⎜ ∫ A
r r
⎟
∂ r ) ∂Kε( r′ ) ⎝∂Kε( r′
) ⎠
� � � . (A.10)
So this side also vanishes, if ε→ 0 , q.e.d.
x
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