 Green kernel and endpoint compatibility for a singular Euler equation

For $0<x<1$, the homogeneous Cauchy-Euler equation has solutions $x,x^{-1}$. The left-vanishing solution $x$ and right-vanishing solution $x-x^{-1}$ give the unweighted Green function $G(x,s)=x(1-s^{-2})/2$ for $x\le s$ and $(x-x^{-1})/2$ for $x\ge s$. It is continuous at $x=s$ and its first derivative has jump $1/s^2$, so $L_{x}G=\delta (x-s)$.

For continuous forcing $f$ on $[0,1]$, its integral isThis is bounded at zero and vanishes at one, but $u(0^{+})=-f(0)$. The first singular term uses the average $x^{-1}{\int }_0^{x}f\to f(0)$. In the other, the substitution $t=x/s$ gives $x{\int }_x^{1}f(s)s^{-2}ds={\int }_x^{1}f(x/t)dt\to f(0)$ by dominated convergence. Thus the two zero endpoint values require $f(0)=0$. When compatible, the solution is unique among bounded solutions because no nonzero combination of $x,x^{-1}$ meets both homogeneous endpoint conditions. This example shows why pointwise endpoint behavior of a Green function does not itself justify passing a limit through its integral.