The transverse force density is $f=-Ah_{xxxx}=\zeta h_t$. Its resultant is
$$
\int_0^Lf\,dx=-A[h_{xxx}]_0^L=0
$$
by the free-end shear conditions. To linear order, its torque about $x=0$ is
$$
\int_0^Lxf\,dx=-A[xh_{xxx}-h_{xx}]_0^L
=A[h_{xx}]_0^L=0,
$$
because both end curvatures equal the same imposed $\kappa_0$. Thus every decaying mode is force-free and torque-free, consistently leaving the translational and rotational zero modes unchanged.


Solution

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

 Solution

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