= Solution
\b[Kernel conditions.] The <clamped-beam Green function> must obey $G(0,\xi)=G_x(0,\xi)=G(1,\xi)=G_x(1,\xi)=0$. Away from $x=\xi$, $G_{xxxx}=0$. At the load point, $G,G_x,G_{xx}$ are continuous and
$$
\boxed{G_{xxx}(\xi+,\xi)-G_{xxx}(\xi-,\xi)=1.}
$$
A jump in any lower <derivative> would create a <derivative> of the <Dirac delta>, rather than the single point load required by $D_x^4G=\delta(x-\xi)$.
\b[Construction.] A particularly simple representation is
$$
G(x,\xi)=ax^2+bx^3+\frac{(x-\xi)_+^3}{6}.
$$
It already enforces the left-end conditions and the correct source jump. The right-end conditions give $a+b=-(1-\xi)^3/6$ and $2a+3b=-(1-\xi)^2/2$, hence
$$
a=\frac{\xi(1-\xi)^2}{2},\qquad b=-\frac{(1-\xi)^2(1+2\xi)}6.
$$
For $x<\xi$ this is already factored; for $x>\xi$, expanding and refactoring gives
$$
\boxed{G(x,\xi)=\begin{cases}\dfrac{x^2(1-\xi)^2}{6}[3\xi-(1+2\xi)x],&x\leq\xi,\\\dfrac{\xi^2(1-x)^2}{6}[3x-(1+2x)\xi],&x\geq\xi.\end{cases}}
$$
The formulas agree at the junction. The forced displacement is obtained by integrating this kernel against $f$; for continuous loads it satisfies the equation and all four end conditions. Uniqueness follows because a cubic with zero value and slope at both ends is identically zero.
\b[Reciprocity.] In terms of $x_<=\min(x,\xi)$ and $x_>=\max(x,\xi)$, the same expression is
$$
G(x,\xi)=\frac{x_<^2(1-x_>)^2}{6}[3x_>-(1+2x_>)x_<].
$$
It is visibly symmetric under interchange of $x$ and $\xi$. Thus the <reciprocity of the clamped-beam Green function> gives
$$
\boxed{y_1(x_2)=G(x_2,x_1)=G(x_1,x_2)=y_2(x_1).}
$$
Equivalently, two <integrations by parts> show <self-adjointness> of $D_x^4$ under the clamped conditions, with bilinear form $\int u''v''dx$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-q3-clamped-beam-kernel.png]
{title=Point-load deflections and the symmetric Green function of a clamped beam}
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