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By V.I. Fabrikani

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Example 3. 27). 25). The technique used in the previous example can be employed here for further integration. The final result depends on the value of n being even or odd. 5 External mixed boundary value problem for a half-space V (ρ,φ, z ) = 2√πσ0 Σ(2m 1 −    a 2m-1    B m+1 Σ mz + ( a 2 − l 21)1/2 − 1) z 2m-1 m=1 k-1  Am k-1 + Γ[( n − 1)/2]  2 B 1ln Q Γ( n /2) m m m m m [ Q 1 − Q 2 − ( Q 3 − Q 4 )]. 33), and 1 A B k-m = k-m+1 2 3 4 1 dm-1  ( t − z 2)k-1  ( m − 1)! d t m-1 (ρ2 + z 2 − t )k = for t =0; 1 dm-1  ( t 2 − z 2)k-1 , ( m − 1)!

3), we get, after integration, σ(ρ,φ) = Γ( n + 1/2) v nein φ π3/2Γ( n ) ρn(ρ2 − a 2)1/2 . 40) with respect to z for z =0. Example 3. 27). 25). The technique used in the previous example can be employed here for further integration. The final result depends on the value of n being even or odd. 5 External mixed boundary value problem for a half-space V (ρ,φ, z ) = 2√πσ0 Σ(2m 1 −    a 2m-1    B m+1 Σ mz + ( a 2 − l 21)1/2 − 1) z 2m-1 m=1 k-1  Am k-1 + Γ[( n − 1)/2]  2 B 1ln Q Γ( n /2) m m m m m [ Q 1 − Q 2 − ( Q 3 − Q 4 )].

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).

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### Applications of potential theory in mechanics. Selection of new results by V.I. Fabrikani

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