D (r) = b + (b + r)^{2} = r + 2 b r + 2 b, Ψ (r) = - Φ (r) = \frac{GM}{D ( r )} .

The spherical Poisson equation gives

ρ (r) = \frac{1}{4 π G r ^{2}} \frac{d}{d r} (r^{2} \frac{d Φ}{d r}) = \frac{M [ 6 b ^{3/2} r + 10 b r + 3 b r ^{3/2} ]}{8 π r ^{2} D ^{3}} .

The numerator identity

6 b^{3/2} r + 10 b r + 3 b r^{3/2} = 3 b r D + 4 b r

ρ (r) = \frac{3 b}{8 π G ^{2} M} r^{- 3/2} Ψ^{2} + \frac{b}{2 π G ^{3} M ^{2}} r^{- 1} Ψ^{3} .

Thus

(γ_{1}, p_{1}, A_{1}) = (\frac{3}{2}, 2, \frac{3 b}{8 π G ^{2} M}),

(γ_{2}, p_{2}, A_{2}) = (1, 3, \frac{b}{2 π G ^{3} M ^{2}}) .

At small radius,

D \sim 2 b

and the first term dominates:

ρ (r) \sim \frac{3 M}{32 π b ^{3/2}} r^{- 3/2} .

The cusp has finite enclosed mass because

r^{2} ρ \sim r^{1/2}

. Moreover,

v_{c}^{2} = r \frac{d Φ}{d r} = \frac{GM r [ 1 + b / r ]}{D ^{2}} \sim \frac{GM}{4 b ^{3/2}} r,

so the circular speed tends to zero as

r^{1/4}

. At large radius,

D \sim r

and

ρ (r) \sim \frac{3 M b}{8 π} r^{- 7/2} .

Since

r^{2} ρ \sim r^{- 3/2}

is integrable at infinity and

Φ \sim - GM / r

, the total mass is finite and equals

M

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