For two body-force-free <Stokes flow>[Stokes flows] $(\mathbf u^{(1)},\boldsymbol\sigma^{(1)})$ and $(\mathbf u^{(2)},\boldsymbol\sigma^{(2)})$ in the same domain, the <Lorentz reciprocal theorem for Stokes flow> is
$$
\int_{\partial\mathcal D}
\mathbf u^{(1)}\mathbin\cdot\boldsymbol\sigma^{(2)}\mathbf n\,dS
=
\int_{\partial\mathcal D}
\mathbf u^{(2)}\mathbin\cdot\boldsymbol\sigma^{(1)}\mathbf n\,dS.
$$
Indeed, the difference of the two volume integrands is
$$
\nabla\mathbf u^{(1)}:\boldsymbol\sigma^{(2)}
-\nabla\mathbf u^{(2)}:\boldsymbol\sigma^{(1)}=0
$$
because both flows are incompressible and the <Newtonian fluid stress tensor> is symmetric. The <divergence theorem> proves the boundary identity.

On a rigid body, $\mathbf u^{(r)}=\mathbf U^{(r)}+\boldsymbol\Omega^{(r)}\times\mathbf x$. The reciprocal theorem becomes
$$
\mathbf U^{(1)}\mathbin\cdot\mathbf F^{(2)}
+\boldsymbol\Omega^{(1)}\mathbin\cdot\mathbf G^{(2)}
=
\mathbf U^{(2)}\mathbin\cdot\mathbf F^{(1)}
+\boldsymbol\Omega^{(2)}\mathbin\cdot\mathbf G^{(1)}.
$$
Writing $(\mathbf F,\mathbf G)^T=\mathsf R(\mathbf U,\boldsymbol\Omega)^T$ and choosing arbitrary pairs of rigid velocities shows that
$$
\boxed{\mathsf R=\mathsf R^T}.
$$
Thus the <hydrodynamic resistance matrix> is symmetric.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-329/2/a/solution