By V.I. Fabrikani

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**Example text**

26) 0 The next operator to apply is ρ tdt 1 d ⌠ L L( t ), 2 ρ dρ ⌡ (ρ − t 2)1/2 a and the final result takes the form 2 ⌠ σ(ρ,φ) = − 2 π(ρ − a 2)1/2 ⌡ a(a2 ρ2 − ρ20 0 = − 2π a 2 (a 1 π (ρ − a ) 2 2 2 1/2 − ρ20)1/2ρ0dρ0 ρ 0 L σ(ρ ,φ) 0 ρ − ρ20)1/2σ(ρ0,φ0) ρ0dρ0dφ0 ⌠⌠ ⌡ ⌡ ρ2 + ρ20 − 2ρρ0cos(φ−φ0) 0 . 27) defines the value of σ outside the circle ρ= a directly through its value inside. 3) allows us to express the potential function V directly through the prescribed value of σ. The first integration yields l1 dx V (ρ,φ, z ) = 4⌠ 2 2 1/2 ⌡ (ρ − x ) 0 ∞ a ρ0dρ0 x σ(ρ ,φ) ⌠ 2 2 1/2 L ρρ 0 0 ⌡ [ρ0 − g ( x )] 2 g(x) ρ0dρ0 ρρ dx 0 σ(ρ ,φ).

3) 0 0 x σ(ρ ,φ) = v (ρ,φ). 5) is now presented as a sequence of two Abel-type operators and one L-operator. 4 Internal mixed boundary value problem for a half-space a 2cos(π u /2) d ⌠ f( x) xdx . 5). 5) is t ζ d ⌠ ρdρ ρ L L . 3). We call the parameter ζ ’dummy’ because it was introduced for formal reasons only; it will disappear in the final result, and has no real bearing on the transformations to follow. 3). Such a validation is beyond scope of this book: the author is satisfied by the fact that the final result everywhere is proven to be correct.

D t m-1 t 2k( r 2 + t )k for t =√1+ρ2/ z 2; A k-m+1 Q0 = Q1 = Q3 = ( l 22 − ρ2)1/2 l2 , l 2[(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] Q = z [(ρ2 + z 2)1/2 + ( l 22 − a 2)1/2] l 21 z [(ρ2 + z 2)1/2 + z ] ρ2 , a [(ρ2 + z 2)1/2 + z ] , Q4 = , Q2 = z [(ρ2 + z 2)1/2 − ( l 22 − a 2)1/2] z [(ρ2 + z 2)1/2 − z ] ρ2 l 21 . 5 External mixed boundary value problem for a half-space where Cm = 1 dk-m ( t + z 2)k , k-m 2 2 k+1 ( k − m )! d t ( t + ρ + z ) for t =0; Dm = 1 dk+1-m ( t + z 2)k , ( k + 1 − m )!