ρ δx · ÿ = δ (P · ∂y/∂x).

(38)

Neglecting the vertical acceleration we have P = gρx, whence

∂2y= g ( x ∂y).
∂t2 ∂x∂x

(39)

Assuming that y varies as eiσt we have

( x ∂y) + ky = 0.
∂x ∂x

(40)

provided k = σ2/g. The solution of (40) which is finite for x = 0 is readily obtained in the form of a series, thus

y = C ( 1 − kx+ k2x2− ... ) = CJ0(z),
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