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Drag and lift
2023, December 02
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:
\[-u \cdot \La u = \div ((D u) u) - |D u| ^2.\]