= Dirichlet resonance for a point interaction
{c}
{title2=$\alpha_*={\ell}/{(c(\ell-c))}$}
The <boundary value problem> $y''+\alpha y(c)\delta_c=f$ on $(0,\ell)$ with <Dirichlet boundary conditions> has a unique solution for every continuous forcing and prescribed endpoint values exactly when $\alpha\ne\alpha_*:=\ell/(c(\ell-c))$. The left-normalized homogeneous solution is
$$
u_\alpha(x)=x-\alpha c(x-c)_+,\qquad u_\alpha(\ell)=\ell-\alpha c(\ell-c),
$$
where $(t)_+=\max(t,0)$. Continuity at $c$ and the <transmission conditions for a delta coefficient> give this formula. The final boundary condition fixes the initial slope unless the displayed endpoint value is zero. This is a one-dimensional <kernel> obstruction, analogous to the <Fredholm alternative>.
At resonance the normalized null function is
$$
\phi(x)=\begin{cases}x/c,&0\leq x\leq c,\\(\ell-x)/(\ell-c),&c\leq x\leq\ell.\end{cases}
$$
For $y(0)=a$ and $y(\ell)=b$, the compatibility condition is
$$
\int_0^\ell\phi(x)f(x)\,dx=\frac ac+\frac b{\ell-c}.
$$
Indeed, integrate $\phi L_\alpha y-yL_\alpha\phi$ on both sides of $c$. The internal terms cancel by the matching derivative jumps, leaving $[\phi y'-\phi'y]_0^\ell$. This equals the right-hand side. The initial-slope construction also proves sufficiency: when compatible there is a one-parameter family $y_0+t\phi$; otherwise no solution exists.
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