= Solution
Use the same <first variation> on the half-plane, with no restrictions on the edge <traces> of $\eta$. Write $M=w_{xx}+\nu w_{yy}$. Two $x$ integrations of the $M\eta_{xx}$ term produce the boundary term $-M\eta_x+M_x\eta$ at $x=0$. The mixed term, after one integration in $x$ and one along the edge in $y$, contributes $2(1-\nu)w_{xyy}\eta$. All other boundary terms vanish at infinity. Hence
$$
\delta\mathcal U=D\iint_{x>0}\eta\nabla^4w\,dx\,dy
-D\int_{-\infty}^{\infty}
\{\eta_x(w_{xx}+\nu w_{yy})-\eta[w_{xxx}+(2-\nu)w_{xyy}]\}_{x=0}\,dy.
$$
The bulk equation makes its area term zero. Stationarity therefore gives exactly
$$
\boxed{\int_{-\infty}^{\infty}
\{\eta_x(w_{xx}+\nu w_{yy})-\eta[w_{xxx}+(2-\nu)w_{xyy}]\}_{x=0}\,dy=0.}
$$
The two edge <traces> can be prescribed independently: use $\eta(x,y)=\chi(x)[f(y)+xg(y)]$ with $\chi=1$ near zero and compactly supported smooth $f,g$. The fundamental lemma on the edge then gives
$$
\boxed{w_{xx}+\nu w_{yy}=0,\qquad
w_{xxx}+(2-\nu)w_{xyy}=0\quad(x=0).}
$$
They are zero <bending moment> and zero effective transverse shear at a free straight edge. The second includes the edge derivative of the twisting moment; omitting the mixed contribution would give the wrong coefficient.
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