= Solution
Start with a <fundamental solution> of the second <derivative> operator:
$$
\widetilde g(x,\xi)=(x-\xi)H(x-\xi),\qquad \partial_x\widetilde g=H(x-\xi),\qquad \partial_x^2\widetilde g=\delta(x-\xi).
$$
Here $H$ is the <Heaviside step function> and the last equality holds in the sense of <distributions>. The factor $x-\xi$ removes a jump in the <function> itself, while its first <derivative> has a unit jump. Adding any <homogeneous solution> preserves the <Dirac delta> source.
Choose $y_1(x)=x-2$ and $y_2(x)=x-3$. They are independent <homogeneous solutions>, with $B_1y_1=0$, $B_2y_2=0$, $B_1y_2=-1$, and $B_2y_1=1$. Since $1<\xi<2$, the <fundamental solution> has $B_1\widetilde g=1$ and $B_2\widetilde g=2-\xi$. Imposing the two <boundary conditions> on the proposed combination therefore gives $-A_2+1=0$ and $A_1+2-\xi=0$. Consequently
$$
\boxed{A_1(\xi)=\xi-2,\qquad A_2(\xi)=1,\qquad
g(x,\xi)=(\xi-2)(x-2)+(x-3)+(x-\xi)H(x-\xi).}
$$
Equivalently,
$$
g(x,\xi)=\begin{cases}(\xi-1)x+1-2\xi,&1\leq x\leq\xi,\\ \xi x+1-3\xi,&\xi\leq x\leq2.\end{cases}
$$
At $x=\xi$ both expressions equal $\xi^2-3\xi+1$ and the right slope minus the left slope is $\xi-(\xi-1)=1$. At the endpoints, $g(1,\xi)=-\xi$ and $g_x(2,\xi)=\xi$, while $g_x(1,\xi)=\xi-1$ and $g(2,\xi)=1-\xi$. These explicitly verify both <coupled boundary conditions>. The homogeneous equation has only affine solutions; its two <boundary conditions> give $2a+b=3a+b=0$, hence $a=b=0$, proving <uniqueness> of the <Green function>.
The reciprocity defect is $g(x,\xi)-g(\xi,x)=2(x-\xi)$. It records how the <coupled boundary conditions> affect <self-adjointness>, even though the differential expression $D^2$ is formally <self-adjoint>. The construction is an instance of the <boundary-matrix construction of an ordinary differential Green function>.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2015-coupled-green-kernel.png]
{title=Unit slope jumps and the reciprocity defect of the coupled-endpoint Green function}
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