image (193)

where

Ba=1νa2Eata1Γ2C (194)

image (194)

Proof: substitution of Eq. (193) into Eq. (189) yields

Ba(d2I1(Γr)dr2+1rdI1(Γr)drI1(Γr)r2)=1νa2Ea1taCI1(Γr) (195)

image (195)

But I1(Γr)image satisfies Eq. (180), i.e.,

d2I1(Γr)dr2+1rdI1(Γr)drI1(Γr)r2=Γ2I1(Γr) (196)

(196)

Substitution of Eq. (196) into Eq. (195) gives Eq. (194). QED.

Addition of Eqs. (192) ...

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