Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 5D / i

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- Solution i

Solution

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} .

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) .

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