Regular inverse of Y d2/dY2 plus nu d/dY

Core idea

For $\nu>0$ and smooth $F$ near $Y=0$ , the regular solution of

YG''+\nu G'=F,\qquad G(0)=0

(J_\nu F)(Y)=\int_0^Y\int_0^1s^{\nu-1}F(st)\,ds\,dt.

The integral is smooth through zero. For $Y>0$ , differentiating $Y^\nu G'$ verifies the equation. The homogeneous derivative is a constant times $Y^{-\nu}$ , so $C^1$ regularity and $G(0)=0$ give uniqueness on the nonnegative side.

Coefficients

For an analytic input $F(Y)=\sum_{\alpha\ge0}F_\alpha Y^\alpha$ , termwise integration gives

(J_\nu F)_0=0,\qquad (J_\nu F)_{\alpha+1}=\frac{F_\alpha}{(\alpha+1)(\alpha+\nu)}.

The operator raises the smallest radial degree by one and divides by two degree factors. Its boundedness in any chosen coefficient norm must be checked against that norm's weights.