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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 $y^{\prime \prime }=f$. Its transmission conditions for a delta coefficient areIndeed, the distributional second derivative across a jump contributes $[y^{\prime }{]}_1\delta (x-1)$, which must cancel the delta coefficient because $f$ has no point source.

Put $s=y^{\prime }(0)$ and $F(x)={\int }_0^x(x-t)f(t)dt$. Integrating the ordinary equation and inserting the required derivative jump giveswhere $(x-1{)}_{+}=\max (x-1,0)$. 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 equationConsequently a unique $s$, and hence a unique solution for every $f,c_1,c_2$, exists precisely whenAt the exceptional value, the nonzero homogeneous solution with zero endpoint values isIt has derivative jump $-\pi /(\pi -1)=-{\alpha }_{\ast }\varphi (1)$. Thus uniqueness fails. More precisely, there are infinitely many solutions ifand none otherwise. This follows either from the scalar equation or by integration by parts against $\varphi$. It is the compatibility obstruction at Dirichlet resonance for a point interaction.

For a point interaction in a second-order differential equation, the equation $y^{\prime \prime }+\alpha y(c){\delta }_c=f+J{\delta }_c$, with nonsingular continuous $f$, requiresThe distributional second derivative across a jump explains both conditions: there is no derivative-of-delta source to balance a value jump, and the coefficients of the delta source must agree. In particular, without a point source, $[y^{\prime }{]}_c=-\alpha y(c)$. With a unit point source at the same location, the condition is $[y^{\prime }{]}_c=1-\alpha y(c)$; one must combine the two contributions rather than impose two incompatible jump conditions.