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Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

In the dark, energy minimization gives $h_{xx}=-H_0$. Translation and rotation are zero-energy freedoms; one representative is $h_{-}(x)=-H_0{x}^2/2$. After illumination, choose the equivalent light-adapted equilibriumThen $u=h-h_{+}$ has homogeneous free-end conditions $u_{xx}=u_{xxx}=0$, and its initial valueis orthogonal to the two rigid zero modes $1$ and $x$.

Let $q_n>0$ solve the free--free biharmonic eigenvalue equationand chooseThese are orthogonal eigenfunctions of the biharmonic operator with ${\varphi }_n^{\prime \prime }={\varphi }_n^{\prime \prime \prime }=0$ at both ends. The shape iswhereThe omitted zero modes would only translate or rotate the whole filament.