Drag and lift
Let K _t = K + t u _\infty be an object moving with constant horizontal velocity u _\infty. It experiences a thrust force and gravity F _{\text{thrust}} \parallel u _\infty and F _{\text{gravity}} \perp u _\infty. Fluid around it obeys the Navier-Stokes equation:
\begin{align} & \pt u + u \cdot \grad u = \div (-p \,\Id + \tau), \qquad \div u = 0 \\ & u \at{\partial K _t} = u _\infty, \qquad u \at{|x| = \infty} = 0. \end{align}Here \tau = 2 \nu D u is the deviatoric stress tensor, and -p\,\Id is the volumetric stress tensor. \div \tau = \nu \La u. Make a change of variable:
\begin{align*} v (t, x) &= u (t, x + t u _\infty) - u _\infty \\ \pi (t, x) &= p (t, x + t u _\infty) \\ R (t, x) &= \tau (t, x + t u _\infty) = 2 \nu D v. \end{align*}Then v, \pi solves
\begin{align*} & \pt v + v \cdot \grad v + \grad \pi = \div R, \qquad \div v = 0 \\ & v \at{\partial K} = 0, \qquad v \at{|x| = \infty} = -u _\infty. \end{align*}Total force exerted by the fluid to body: (n is inward of K _t)
F = \int _{\partial K _t} (p \,\Id - \tau) n \d S = \int _{\partial K} (\pi \,\Id - R) n \d S = F _{\text{form}} + F _{\text{skin}}.Form force is
F _{\text{form}} = \int _{\partial K} \pi n \d S = F _{\text{form drag}} + F _{\text{form lift}}.Skin force is
F _{\text{skin}} = -\int _{\partial K} R n \d S = -2 \nu \int _{\partial K} D v \d S = - \nu \int _{\partial K} \nabla v + \nabla v ^\top \d S = -\nu \int _{\partial K} \partial _n v \d S = F _{\text{skin drag}} + F _{\text{skin lift}}.If u has fast decay, then
F \cdot u _\infty = \int _{\partial K _t} u _\infty ^\top (p \,\Id - \tau) n \d S = \int _{\partial K _t} u ^\top (p \,\Id - \tau) n \d S = \int _{\RR3 \setminus K _t} \div ((p \,\Id - \tau) ^\top u) \d x.Here
\begin{align*} \div ((p \,\Id - \tau) ^\top u) &= \div (p \,\Id - \tau) \cdot u + (p \,\Id - \tau) : \grad u \\ &= (-\pt u - u \cdot \nabla u) \cdot u + p \div u - 2 \nu D u : \nabla u \end{align*}So
F \cdot u _\infty = -\int _{\RR3 \setminus K _t} (\pt + u \cdot \grad) \frac{|u| ^2}2 \dx - 2 \nu \int _{\RR3 \setminus K _t} |D u| ^2 \d x.If the total energy is conserved, then
\int _{\RR3 \setminus K _t} (\pt + u \cdot \grad) \frac{|u| ^2}2 \dx = \int _{\partial K _t} \hfsq u (u \cdot n) \dx = \int _{\partial K _t} \hfsq {u _\infty} (u _\infty \cdot n) \dx = 0.So we have
F \cdot u _\infty = F _{\text{drag}} \cdot u _\infty = -2 \nu \int _{\RR3 \setminus K _t} |D u| ^2 \dx = -2\nu \int _K |D v| ^2 \dx.There are two types of integration by part:
\begin{align} u \cdot \La u &= \div(\grad u ^\top u ) - |\grad u| ^2. \\ u \cdot \La u &= -u \cdot \curl \omega = \div (u \cross \omega) - |\omega| ^2 \\ \notag \implies 2 u \cdot \La u &= \div (-(S u) \omega) - |S u| ^2. \end{align}Here S u = \frac12 (\grad u - \grad u ^\top), \grad u = D u + S u. Note that |\grad u| ^2 = |S u| ^2 + |D u| ^2 and 2 u \cross \omega = (Su) u. So we have a third identity:
-u \cdot \La u = \div ((D u) u) - |D u| ^2.