A planar rigid motion is controlled by its trace on a segment

The globally affine rigid motion $r$ is continuous on $\overline{U_D}$, so [F1] identifies $T_{D}r$ with its boundary restriction. Put $A:=a+\omega J{x}_0$ and decompose $A=A_{\parallel }J\tau +A_{\perp }$ with $A_{\perp }\cdot J\tau =0$. Since $J$ is a quarter-turn, $r(x(s))=A+\omega sJ\tau$. Integrating the two orthogonal components and completing the square in ${\int }_0^L(A_{\parallel }+\omega s{)}^{2}ds$ gives $\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}^2=L\mathrm{\mid }{A}_{\perp }{\mathrm{\mid }}^2+L(A_{\parallel }+\omega L\mathrm{/}2{)}^2+{\omega }^2{L}^3\mathrm{/}12$.

The last term of step 1.1 controls $\mathrm{\mid }\omega {\mathrm{\mid }}^2$, the first controls $\mathrm{\mid }{A}_{\perp }{\mathrm{\mid }}^2$, and the middle term together with $\mathrm{\mid }\omega \mathrm{\mid }$ controls $\mathrm{\mid }{A}_{\parallel }{\mathrm{\mid }}^2$; hence $\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }\omega \mathrm{\mid }\le C_L\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}$. Because $a=A-\omega J{x}_0$ and $\mathrm{\Omega }$ is bounded, direct integration of $r$ and its constant gradient gives $\mathrm{\parallel }r{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\le C_{\mathrm{\Omega }}(\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }\omega \mathrm{\mid })\le C(\mathrm{\Omega },{\mathrm{\Gamma }}_D)\mathrm{\parallel }{T}_{D}r{\mathrm{\parallel }}_{L^2({\mathrm{\Gamma }}_D)}$. If the trace vanishes, this also forces $a=0$ and $\omega =0$.