= Solution
The <functional derivative> is $\delta\mathcal E/\delta h=Ah^{(4)}$, so the <overdamped filament bending equation> is $\zeta h_t=-Ah^{(4)}$ with the free-end conditions from (a). Expand the initial shape in the complete <orthonormal basis>:
$$
c_j=\int_0^L W_j(x)H(x)\,dx.
$$
The <overdamped relaxation of a free filament> is
$$
\boxed{h(x,t)=c_{\rm tr}W_{\rm tr}(x)+c_{\rm tilt}W_{\rm tilt}(x)
+\sum_{n\geq1}c_nW_n(x)e^{-Ak_n^4t/\zeta}.}
$$
Each bending mode relaxes on time $\tau_n=\zeta/(Ak_n^4)$; the two zero modes remain constant. Consequently the long-time shape is the affine projection of $H$,
$$
h_\infty(x)=\frac1L\int_0^L H(s)ds
+\frac{12(x-L/2)}{L^3}\int_0^L(s-L/2)H(s)ds.
$$
The total displacement and its first moment are conserved: integrate $h_t$ and $(x-L/2)h_t$, using $h''=h^{(3)}=0$ at the ends. This also checks why the final affine part cannot generally be set to zero. The series defines the $L^2$ solution for arbitrary square-integrable initial data; a classical solution at time zero additionally requires compatible boundary values.
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