= Solution
Interpret the <Dirac delta> coefficient as a <point interaction in a second-order differential equation>. The solution is continuous at $x=1$ and is twice differentiable on either side, where $y''=f$. Its <transmission conditions for a delta coefficient> are
$$
[y]_1=0,\qquad [y']_1=-\alpha y(1).
$$
Indeed, the <distributional second derivative across a jump> contributes $[y']_1\delta(x-1)$, which must cancel the delta coefficient because $f$ has no point source.
Put $s=y'(0)$ and $F(x)=\int_0^x(x-t)f(t)\,dt$. Integrating the ordinary equation and inserting the required derivative jump gives
$$
y(x)=c_1+sx+F(x)-\alpha y(1)(x-1)_+,\qquad y(1)=c_1+s+F(1),
$$
where $(x-1)_+=\max(x-1,0)$. This formula is continuous, has the stated jump, satisfies the equation, and parametrizes every solution with the left endpoint value. Imposing the right endpoint value gives the scalar equation
$$
\bigl[\pi-\alpha(\pi-1)\bigr]s
=c_2-c_1-F(\pi)+\alpha(\pi-1)\bigl(c_1+F(1)\bigr).
$$
Consequently a unique $s$, and hence a unique solution for every $f,c_1,c_2$, exists precisely when
$$
\boxed{\alpha\ne\alpha_*:=\frac\pi{\pi-1}.}
$$
At the exceptional value, the nonzero homogeneous solution with zero endpoint values is
$$
\phi(x)=\begin{cases}x,&0\leq x\leq1,\\(\pi-x)/(\pi-1),&1\leq x\leq\pi.\end{cases}
$$
It has derivative jump $-\pi/(\pi-1)=-\alpha_*\phi(1)$. Thus uniqueness fails. More precisely, there are infinitely many solutions if
$$
\int_0^\pi\phi(x)f(x)\,dx=c_1+\frac{c_2}{\pi-1},
$$
and none otherwise. This follows either from the scalar equation or by <integration by parts> against $\phi$. It is the compatibility obstruction at <Dirichlet resonance for a point interaction>.
Back to article page
