= Green kernel and endpoint compatibility for a singular Euler equation
{c}
{title2=$Lu=x^2u''+xu'-u$}
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 $x\leq s$ and $(x-x^{-1})/2$ for $x\geq s$. It is continuous at $x=s$ and its first derivative has jump $1/s^2$, so $L_xG=\delta(x-s)$.
For continuous forcing $f$ on $[0,1]$, its integral is
$$
u(x)=\frac{x-x^{-1}}2\int_0^xf(s)ds+\frac x2\int_x^1(1-s^{-2})f(s)ds.
$$
This is bounded at zero and vanishes at one, but $u(0^+)=-f(0)$. The first singular term uses the average $x^{-1}\int_0^xf\to f(0)$. In the other, the substitution $t=x/s$ gives $x\int_x^1f(s)s^{-2}ds=\int_x^1f(x/t)dt\to f(0)$ 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.
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