= Uniform-load bending of a clamped filament
{title2=$h_s(x)$}
Let positive transverse displacement point along a uniform load per unit length $w$. The <filament bending modulus> $B$ gives energy $\int_0^L[B(h_{xx})^2/2-wh]dx$. The <variational derivative> is $Bh_{xxxx}-w$, so <resistive-force theory> gives $\zeta_\perp h_t=-Bh_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution is
$$
h_s(x)=\frac{w}{24B}x^2(x^2-4Lx+6L^2).
$$
Its tip displacement is $wL^4/(8B)$. For a submerged cylindrical filament under <Newtonian gravity>, <buoyancy> gives $w=(\rho_f-\rho)\pi a^2g$. This linear approximation requires small slope and negligible inertia.
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