5D (Methods)

Away from zero the Dirac delta distribution is absent. On either half-line, use

u = 1 + k ∣ z ∣

and the power-law ansatz

ψ = u^{α}

. The Euler-Cauchy equation becomes

α (α - 1) - \frac{15}{4} = 0, α = \frac{5}{2} or - \frac{3}{2} .

(1)

The growing term is excluded by boundedness at either end. Thus the independently bounded half-line solutions are

ψ (z) = C_{+} (1 + k z)^{- 3/2} (z > 0), ψ (z) = C_{-} (1 - k z)^{- 3/2} (z < 0) .

(2)

The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes

3 k ψ (0)

. The jump condition for an impulse is therefore

ψ^{'} (0^{+}) - ψ^{'} (0^{-}) = - 3 k ψ (0) .

(1)

Continuity at zero gives

C_{+} = C_{-} = C

. The two one-sided derivatives are

- 3 k C /2

and

+ 3 k C /2

, whose difference is exactly

- 3 k C

. Hence the matching condition introduces no further restriction on

C

, and the globally bounded solution is

ψ (z) = C (1 + k ∣ z ∣)^{- 3/2}, C \in R or C .

(2)

This continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.