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I need to construct a divergence free, smooth, vector function on a unit circle such that $ \mathbf{u} = (u_1,u_2) = (0,0) $ on $\partial B$, and $\int\limits_B u_i \neq 0, \ i=1,2$. I was able to find/construct few such functions, but all of them fail to satisfy the last condition. Examples: 1) Hill's Spherical vortex $u = ( y(1-x^2-y^2), -x(1-x^2-y^2))$. 2) In 2d polar coordinates, $\nabla \cdot u = 0 \Rightarrow \frac{\partial (r v_r )}{\partial r} = - \frac{\partial v_\theta}{\partial \theta}$. So, e.g. let $ v_r = r^4(r-1)^3 \sin\theta \Rightarrow v_\theta = \left(5r^4(r-1)^3+ 3r^5(r-1)^2\right)\cos\theta $. From these, $v_1 = r^2(r-1)^2xy(4-7r), v_2 = r^2(r-1)^2\left( (r-1)y^2 + (8r-5)x^2 \right)$, where $r = \sqrt{x^2+y^2}$. But, $v_1, v_2$ have average value zero over the unit circle. So, is it always the case that average value is zero if we require smoothness? Anybody know any such function which does satisfy all conditions( maybe from Fluid Mechanics textbooks)?
Any help on these would be appreciated. Thanks!

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  • $\begingroup$ Perhaps this is a bit of pedantry, but when you say "circle," I picture the boundary of a disc, but you then talk about the boundary of a circle, which conflicts with this picture. $\endgroup$
    – Muphrid
    May 4, 2013 at 22:48

1 Answer 1

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A divergence-free vector field vanishing on the boundary can be thought of as the velocity field of an ideal fluid constrained to remain in $B$. The integral $\int_B \mathbf{u}$ (divided by the volume of $B$) is the velocity of the center of mass of the fluid. But the center of mass has nowhere to go: it's the center of $B$. Therefore $\int_B \mathbf{u}=0$.

If you prefer a "mathematical" proof, integrate by parts: $$\int_B \mathbf u\cdot \nabla f = -\int_B (\mathrm{div}\,\mathbf u)f = 0$$ with $f(x)=x_1$ or $f(x)=x_2$. Boundary term is zero because $\mathbf u=0$ on $\partial B$.

Another way, specific to two dimensions, is to use the fact that a divergence-free field, rotated by 90 degrees, becomes a curl-free (conservative) field, i.e., one of the form $\nabla f$. Since $\nabla f$ vanishes on the boundary, $f$ is constant on the boundary. Therefore, the integral of $f_{x_1}$ along every horizontal line is zero, and so is the integral of $f_{x_2}$ along every vertical line.


A somewhat similar problem (from Cambridge Mathematical Tripos) was quoted by Littlewood in his Miscellany:

An ellipsoid surrounded by frictionless homogeneous liquid begins to move in any direction with velocity $V$. Show that if the outer boundary of the liquid is a fixed confocal ellipsoid, the momentum set up in the liquid is $-MV$, where $M$ is the mass of the liquid displaced by the ellipsoid.

Littlewood remarks

(The result was later extended to other pairs of surfaces, e.g. two coaxial surfaces of revolution.)

and destroys the problem with one sentence

Whatever the two surfaces are we can imagine the inner one to be filled with the same liquid; then the centre of mass does not move.

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  • $\begingroup$ Thanks a lot for the in-depth explanation! $\endgroup$
    – Ter
    May 5, 2013 at 16:31

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