In the dark, energy minimization gives $h_{xx}=-H_0$. Translation and rotation are zero-energy freedoms; one representative is $h_-(x)=-H_0x^2/2$. After illumination, choose the equivalent light-adapted equilibrium
$$
h_+(x)=\frac{H_0x^2}{2}-H_0Lx+\frac{H_0L^2}{6}.
$$
Then $u=h-h_+$ has homogeneous free-end conditions $u_{xx}=u_{xxx}=0$, and its initial value
$$
u(x,0)=-H_0\left(x^2-Lx+\frac{L^2}{6}\right)
$$
is orthogonal to the two rigid zero modes $1$ and $x$.

Let $q_n>0$ solve the <free--free biharmonic eigenvalue equation>
$$
\cos(q_nL)\cosh(q_nL)=1,
$$
and choose
$$
\phi_n(x)=\cosh(q_nx)+\cos(q_nx)
-\frac{\cosh(q_nL)-\cos(q_nL)}{\sinh(q_nL)+\sin(q_nL)}
[\sinh(q_nx)-\sin(q_nx)].
$$
These are orthogonal eigenfunctions of the <biharmonic operator> with $\phi_n''=\phi_n'''=0$ at both ends. The shape is
$$
\boxed{h(x,t)=h_+(x)+\sum_{n=1}^{\infty}a_n\phi_n(x)
e^{-Aq_n^4t/\zeta},}
$$
where
$$
a_n=\frac{\int_0^L u(x,0)\phi_n(x)dx}
{\int_0^L\phi_n(x)^2dx}.
$$
The omitted zero modes would only translate or rotate the whole filament.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-355/1/b/solution