= Solution
Let $\mathbf u$ be the <Stokes flow> and let $\mathbf v$ be any other admissible incompressible flow with the same <no-slip boundary condition> on the sphere and the same decay at infinity. Put
$$
\mathbf w=\mathbf v-\mathbf u,
\qquad
\mathbf e(\mathbf q)
=\frac12\left(\nabla\mathbf q+\nabla\mathbf q^T\right).
$$
Then $\nabla\mathbin{\cdot}\mathbf w=0$ and $\mathbf w=0$ on the sphere. The <viscous dissipation> satisfies
$$
\begin{aligned}
\mathcal D[\mathbf v]
&=2\mu\int_{\mathcal D}
\mathbf e(\mathbf u+\mathbf w):
\mathbf e(\mathbf u+\mathbf w)\,dV\\
&=\mathcal D[\mathbf u]
+4\mu\int_{\mathcal D}\mathbf e(\mathbf u):\mathbf e(\mathbf w)\,dV
+2\mu\int_{\mathcal D}\mathbf e(\mathbf w):\mathbf e(\mathbf w)\,dV.
\end{aligned}
$$
For the Stokes stress
$$
\boldsymbol\sigma=-p\mathbf I+2\mu\mathbf e(\mathbf u),
$$
incompressibility of $\mathbf w$ gives
$$
2\mu\mathbf e(\mathbf u):\mathbf e(\mathbf w)
=\boldsymbol\sigma:\nabla\mathbf w.
$$
The <divergence theorem>, the homogeneous boundary data, decay at infinity, and $\nabla\mathbin{\cdot}\boldsymbol\sigma=0$ give
$$
\int_{\mathcal D}\boldsymbol\sigma:\nabla\mathbf w\,dV
=\int_{\partial\mathcal D}\mathbf w\mathbin{\cdot}
\boldsymbol\sigma\mathbf n\,dS
-\int_{\mathcal D}\mathbf w\mathbin{\cdot}
(\nabla\mathbin{\cdot}\boldsymbol\sigma)\,dV
=0.
$$
Therefore
$$
\boxed{
\mathcal D[\mathbf v]-\mathcal D[\mathbf u]
=2\mu\int_{\mathcal D}
\mathbf e(\mathbf w):\mathbf e(\mathbf w)\,dV\geq0.}
$$
This proves the <Minimum-dissipation theorem for Stokes flow>.
Back to article page
